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Je tiens tout d'abord à remercier Olivier qui a accepté d'encadrer mon mémoire quand je lui ai demandé durant mon master 2 à Paris. Je n'étais pas encore sûr de vouloir faire une thèse à ce moment-là, j'avais décidé de ne le faire que si je trouvais un sujet qui me motivait à fond. Je me souviens encore des réunions de plusieurs heures que nous avions. J'ai ensuite commencé ma thèse avec Ivan et lui. Il m'a aidé durant toute la thèse, que ce soit pour des problèmes mathématiques, pour rencontrer des gens à des conférences ou en relisant une bonne partie de ma thèse avec minutie, ce qui m'a aidé à la rendre bien plus lisible. Je le remercie de toute son aide et son soutien durant cette thèse et en dehors. Je m'étais engagé au début du contrat de la thèse à ramener du vin de Bordeaux, je vais enfin tenir ma promesse au pot de thèse.

Quand j'ai demandé à Olivier s'il pouvait me proposer un sujet de thèse, il m'a dit qu'il avait un projet avec un certain Ivan Marin. J'ai pu le rencontrer durant son cours de M2 spécialisé à Amiens. Son cours était fantastique et j'ai eu droit à un entretien privé au Lutétia où il avait l'air plutôt sympa. J'en ai alors profité pour me sortir de la ville de Paris et déménager à Amiens durant ma thèse. Nous avions une équipe de choc incroyable avec Georges et Eirini, à nous quatre, nous formions le groupe de travail d'Ivan Marin et ses thésards. Au fil des années, il a tenté de contenir mes envies de tout démontrer par des calculs interminables et à trouver des preuves plus lisibles. Il m'a accueilli chez lui à plusieurs reprises, je remercie Christine pour son accueil, j'ai été honoré de devenir responsable de production de grenouilles en papier de Claire. Ivan a réussi à coencadrer ma thèse à merveille et à devenir un chef, un patron, un guide et sourtout un ami.

Je remercie aussi tous les membres du LAMFA. Je veux remercier en particulier Serge, le spécialiste en de nombreux domaines des mathématiques et surtout en jeu de mots du labo, il est souvent le Bouc Emiserge du laboratoire quand il s'agit de trouver quelqu'un à qui poser des questions mathématiques ultra-spécifiques, il m'a par exemple aidé en me précisant où trouver la référence exacte pour le Lemme de Goursat, ce qui est un exploit assez remarquable.

Je veux remercier tous les chercheurs que j'ai pu rencontrer aux cours des diverses conférences. En particulier, Jérémie qui m'a expliqué les cellules de Kazdan-Lusztig. I want to thank Kay for all the kind help he gave me during the thesis. He helped at multiple times in conferences answering the many questions which I asked him daily. He helped make some unreadable proofs much more readable and also gave me some ideas which helped me begin to write some new unreadable proofs. I also want to thank Kay and Götz for accepting to review my thesis and Cédric, David, Donna and Jérémie for accepting to be on the jury.

Je remercie tous les membres de l'équipe de handball de Saint-Maurice qui m'ont aidé à me défouler chaque semaine, ce qui était vraiment nécessaire pour survivre à ces trois années. Pouvoir disputer la finale de la coupe de Somme trois semaines avant ma soutenance était un moment vraiment génial.

Je veux maintenant remercier tous les doctorants du LAMFA de ces trois dernières années. Merci à Vianney qui m'a aidé à prendre mes marques dès le début de ma thèse et à stopper ma première crise de panique qui est survenue après avoir appuyé sur la touche inser de mon clavier. Merci aussi à Maxime qui a été très amical dès mon arrivée. Merci à Pierrot, Malal et Emna pour leur compagnie dans le bureau BC010 qui était encore assez vide au début de ma thèse, merci à Pierrot et Emna de m'avoir présenté Youcef qui est trop mignon.

Merci à Sylvain et Anne-Sophie qui m'ont intégré à leur merveilleux duo en première année pour que je cesse de manger tout seul. Vous m'avez attendu cinq minutes avant de manger pour

Dans le chapitre 4, nous déterminons l'image du groupe A Dn dans son algèbre de Iwahori-Hecke associée. Ici, l'algèbre est définie sur le corps F p (α) où α correspond au paramètre de déformation de la relation d'ordre 2. Il n'y a qu'un seul paramètre car tous les générateurs sont conjugués. Cela rend les extensions de corps plus simples car il suffit de considérer F p (α) au-dessus de F p (α). Il faut utiliser la Proposition 2.2.4 pour montrer que la spécialisation de cette algèbre est semi-simple déployée dans le Théorème 4.1.1.

Les représentations sont encore indexées par des doubles-partitions de n. Il y a dans ce cas-là un isomorphisme naturel entre la représentation indexée par la double-partition (λ 1 , λ 2 et la représentation indexée par la représentation indexée par la double-partition (λ 2 , λ 1 ) car l'action sur le double-tableau (T 1 , T 2 ) correspond à l'action sur le double-tableau (T 2 , T 1 ). Les représentations indexées par des doubles-partitions de type (λ 1 , λ 1 ) ne sont alors plus irréductibles, elles se scindent en deux représentations irréductibles de même dimension que l'on note (λ 1 , λ 1 , +) et (λ 1 , λ 1 , -). La règle de branchement est ainsi plus complexe et est donnée ci-dessous.

Lemme 1. Soit n ≥ 5 et (λ, µ) n, λ > µ. On a alors :

Si n

Si

. Les arguments habituels permettent alors de démontrer le résultat suivant.

L'étude en type H 4 est nettement plus compliquée. De nombreuses représentations irréductibles sont auto-duales. De plus, il y a quatre représenations irréductibles de dimension 16. Les H 4 -graphes sont donnés dans la section 10.2 de l'appendice. Il est clair en étudiant la symétrie des graphes auto-duaux munis d'un 2-coloriage que les forme bilinéaires associées sont toutes symétriques. En particulier, les deux représentations auto-duales de dimension 8 font intervenir le groupe Ω + 8 (q). Cela complique considérablement l'étude de l'image de A H 4 au sein de ces représentations. On sépare donc l'étude de A H 4 en 4 sous-sections.

Dans la section 8.2, nous démontrons que l'on a bien les propriétés voulus pour la spécialisation du modèle aux corps finis. Nous montrons aussi certaines propriétés générales qui seront utiles pour l'étude de chaque représentation.

Dans la section 8.3, nous déterminons l'image de A H 4 dans les représentations de dimension inférieures à 8. Pour déterminer ρ 8r (A H 4 , nous montrons que ce groupes contient un groupe suffisament grand. Pour montrer cela, nous utilisons des preuves très calculatoires qui font intervenir le Théorème de Dickson (voir [27] Théorème 8.27) et le Lemme de Goursat (voir Lemme 3.3.1) à diverses reprises. Il faut alors montrer que des quantités sont non-nulles en effectuant des opérations qui s'apparentent à des divisions euclidiennes de polynômes en α. Certains de ces calculs ont été mis dans la section 10.3.

Dans la section 8.4, nous montrons que les représentations ρ 8r et ρ 8rr sont liées par la trialité. Cela fait apparaître une construction intéressante du groupe Spin + 8 (q). Enfin, dans la section 8.5, nous déterminons l'image de A H 4 dans les représentations de dimensions supérieures à 9. Nous concluons la section par l'image de A H 4 dans toute l'algèbre de Iwahori-Hecke. Un des cas pour la représentation de dimension 48 ne peut pas se traiter à l'aide des arguments habituels. Nous émettons alors la conjecture 8.5.1 qui donne ce à quoi devrait être isomorphe ρ 48rr (A H 4 ). Nous avons alors le résultat suivant qui conclut le chapitre 8.

Pour finir, dans la section 10.1 de l'appendice, nous donnons un erratum des articles [11] et [12]. Les résultats de ces articles sont utilisés à diverses reprises dans cette thèse, notamment dans le chapitre 3. Les résultats de ces articles sont correctes mais certaines preuves sont imprécises ou contiennent des erreurs qui sont listés dans cette section.

. If λ = (λ 1 , λ 2 ) is a double-partition with λ 1 a partition of r and λ 2 a partition of n -r, we let ν(λ) = ν(λ 1 )ν(λ 2 )(-1) r (n-r) .

.

Proposition 3.1.1. For all standard double-tableaux T, T, we have the following properties.

1. (S i .T|S i . T) = (-α)(T| T) and (T.T|T. T) = (-β)(T| T).

2. For all b ∈ A Bn , (b.T|b. T) = (T| T).

3. The restriction of (.|.) to V λ when λ = λ and to V λ ⊕ V λ when λ = λ is non-degenerate. Assume that λ = λ . Then (., .) is symmetric on V λ if ν(λ) = 1 and skew-symmetric otherwise. Moreover, its Witt index is positive.

 Prop 2.4.], we have (S i .T|S i . T) = (T| T) because in the same way, we have ω(T) = -ω(T i↔i+1 ) for any standard double-tableau T and m i (T) = m i (T τ T (i) ) when τ T (i) = τ T (i + 1).

We now assume that τ T (i) = τ T (i + 1). We have that S i .T = m i (T)T + (1 + m i (T))T i↔i+1 and S i . T = m i ( T) T + (1 + m i ( T)) Ti↔i+1 . It follows that (S i .T|S i . T) = m i (T)m i ( T)(T| T) + (1 + m i (T))m i ( T)(T i↔i+1 , T) + m i (T)(1 + m i ( T))(T| Ti↔i+1 ) + (1 + m i (T))(1 + m i ( T))(T i↔i+1 | Ti↔i+1 ).

This is non-zero only if T = T or T = T i↔i+1 . We now have two possible cases.

que je finisse le type B plus de fois que je ne peux le compter. Merci Anne-Sophie pour ta bonne humeur et la campagne terrible que j'ai eu l'honneur de mener face à toi pour l'organisation du séminaire doctorants que nous avons pu coorganiser à merveille pendant notre deuxième année. Merci Sylvain pour ta capacité à parler sans fin et ton aide quand j'envisageais encore de faire un peu de topologie dans ma thèse.

Merci Ruxi pour les cours de chinois que tu m'as prodigués, je me dois te demander : "Ni Xi Huan Tu Do Ma?". Merci Valérie pour toutes tes relectures, je ne noterai malheureusement jamais les références comme tu le souhaites, c'était un plaisir de travailler aux côtés d'une personne aussi sage qui a déjà vu passer une bonne partie de sa vie devant ses yeux. Merci Clément pour tes cours de ponctuation et ton unending combat contre les anglicismes. Merci Jérémy d'avoir coorganisé à 80% le séminaire doctorants la troisième année. Merci à toi et Sylvain d'avoir été à plusieurs reprises mes consultants en algorithmes avec des avis parfois quelque peu divergents mais qui m'ont bien aidés.

Merci Clara pour tes centaines de selfies qui me permettront de garder plein de souvenirs de la thèse. Merci Gauthier, le bout-en-train du laboratoire, tes blagues sont parfois très drôles, parfois moins mais le moment de gêne ou d'hilarité qui suit est toujours mémorable. Thank you Arrianne for your company in office BC010, I hope you will keep working on your French word lists while I'm gone, here's a new one "paillasson =mat". Thank you Afaf for smiling all the time which helped creating a great atmosphere in the office everytime you are there. Sebastian, gracias por tu sumario y lo siento por la calidad de mi español. Merci à Marouan pour les petits airs d'harmonica en fin de journée. Thank you Hatice for your support during this last year, I hope you manage to realize that life is a competition and that you manage to win at life.

Merci enfin Guillaume pour ta bonne humeur et tes histoires plus folles les unes que les autres. Je te remercie aussi pour ton niveau d'exigence très élevé en humour, j'essaierai d'atteindre les 8 ou 9 sur 10 bien plus souvent dans le futur. Merci aussi pour tes "à demain en super forme" qui ont rythmés nos journées et pour m'avoir fait connaître de grands artistes tels que Trent Reznor, François Asselineau ou Franck Syx, ça me fera bizarre de ne pas travailler avec toi au bureau l'année prochaine.

Bonne chance à tous ceux qui doivent encore finir leur thèse, merci d'avoir supporté mes nombreux mails pendant les deux dernières années, je vous laisse avec deux dernières questions mathématiques.

Question mathématique du jour : Combien ai-je de frères et soeurs et combien d'entre elles (accord de proximité) pouvez-vous reconnaître et nommer dans cet amphithéâtre?

Mathematical question of the day : Which mathematical word did I use the most during the presentation and how many times did I say it? Je vais remercier mes demi-frères et demi-soeurs de thèse de la moins jeune au plus jeune. Eirini, je te remercie pour ta gentillesse sans égale. Avoir pu te voir surmonter ta dernière année de thèse pendant ma première année m'a montré qu'il était possible de survivre à cette épreuve. Je me souviens des week-ends au laboratoire avec Georges où on travaillait dur pour tenir les rythmes infernaux imposés par le patron. Merci de m'avoir invité à Stuttgart et de m'avoir fait visité la ville alors que tu étais sur le point de démontrer la conjecture de trace pour un nouveau groupe.

Reda, tu as beau être bien plus agé, on a pu faire le master ensemble et on a pu explorer le monde de la recherche à de multiples conférences. Merci de m'avoir hébergé de nombreuses fois lors de mes séjours à Paris pour travailler avec Olivier. Merci aussi pour les discussions sur la recherche et le futur. Je te fais confiance pour assurer la descendance d'Olivier et honorer le badge de Michael Jackson que tu as récupéré à Birmingham. Merci à toi et à Edgar pour les trajets à 7h du matin gare du nord pour assister aux cours d'Ivan Marin. On a vécu l'aventure le premier jour en parcourant Amiens plusieurs fois pour mettre 1h45 à aller de la gare d'Amiens jusqu'au LAMFA.

Georges, tu m'as accueilli mon premier jour au LAMFA pour me présenter mon bureau et toute l'équipe des doctorants. Tu m'as ensuite fait assisté à des exposés uniques en leur genre au groupe de travail IMEST en parlant à un débit d'au moins 300 mots par minute et n'autorisant pas la moindre baisse d'attention. Tes histoires complotistes pendant la première année me permettaient de me concentrer à la perfection. Tu as pu te rapprocher d'I.M. à coups de Baklavas, je te félicite à nouveau pour l'obtention de son numéro. C'est un peu le comble pour un complotiste de trouver un emploi à "Bielefeld", je te souhaite une bonne continuation dans ta ville imaginaire. Merci à toi et Eirini d'avoir établi des standards très élevés pour la qualité de la thèse mais aussi et surtout pour la longueur des remerciements.

Henry, le petit dernier, ça a été un plaisir de te former cette année. Je pense que tu es déjà passé de l'état de larve à l'état de chrysalide. Tu seras à l'issue de ma soutenance le seul thésard du chef et il faudra que tu deviennes un papillon et que tu commences à voler de tes propres ailes.

Merci à Félix de m'avoir écouté lui expliquer que d était en fait forcément inférieur ou égal à 2 place de la République alors que ça n'avait a priori aucun intérêt pour lui qui a arrêté les mathématiques il y a bien longtemps. Merci à Planès de ne pas trop m'en vouloir de lui devoir encore 799 croissants 5 ans après le début de ma dette et d'avoir accepté de faire un peu de théorie des corps sur des serviettes à Burger King. Merci à Marc de nous avoir fait découvrir Saint-Crépin-Ibouvillers, son PMU et son kébab. Merci à Le Manach pour ses rappels sur l'ensemble vide qui m'ont remis les idées en place plus d'une fois. Merci à Sabrina pour son petit aller-retour à Amiens avant de quitter le pays qui m'a permis de dire "je me casse" à mes cobureaux un matin à 11h, merci à toi et Bastien pour votre super mariage où j'ai pu m'illustrer en faisant des maths immobile en costard et en plein soleil le lendemain matin pendant que les gens se baignaient dans la piscine. Merci aussi à Nicolas pour les billards qu'on a pu faire à défaut de babyfoot et pour le match de basket en 50 points serré jusqu'au bout qu'on a eu l'occasion de disputer. Thank you Loretta for all your support, coming to visit me in Amiens and initiating the email instead of facebook way of communication which will save the world. Thank you to both of you for hosting me in your home and letting me take care of Lucien a little bit, I hope he still remembers uncle Bop.

Je remercie mon oncle Alain et ma tante Maryse d'être venus à ma soutenance de thèse. J'espère que l'exposé ne sera pas trop pénible à suivre.

I want to thank Chuck, Soraya, Taisa, Jacob and Danny for putting up with me even when I kept excluding maximal subgroups of the special linear groups of dimensions 6 and 8 even during the Olympics. I cannot express how grateful I am to Chuck, Soraya and Taisa for coming all the way from Miami to attend my defense. I still remember a talk with Chuck at Thomas' graduation where I told him I wasn't sure about starting a doctoral thesis. He told me about his experience which did not go so well but also that he never regretted beginning it. It helped me make my decision and I'm very thankful for this.

I also want to thank T-dawg for hosting me in Seattle multiple times during this thesis. His craziness helped me relax in this stressful period. Ezell's fried chicken and hiking in Washington state were also very helpful and I hope to do this many more times with you, Alice and Guthrie in the future.

I thank Skipper for coming to my defense on his trip across Europe. He was always ready to meet me in a cool restaurant in San Francisco.

I want to thank my mom for all her support all my life and during this thesis. Even though it was a tough couple of years, she was there all the time and gave all her support. Guillaume can testify that the phone rang multiple times a week and my answer was usually "yah". She even listened to me talk about my thesis and group theory. I'm pretty sure she still doesn't know what a perfect group is but she always asks and remembers for a few days. I know you do your best to be a great mother and sometimes feel unappreciated. I rarely say it but I'll write it down this time so you remember it, I love you mom. Also, thanks for paying for the red wine.

Je remercie mon père de m'avoir tenu informé avec assiduité des scores du Biarritz Olympique durant mon exil dans le nord.

Sophia, ma petite soeur chérie, je sais que tu ne veux plus que je t'appelle comme ça car tu n'as plus huit ans. Malheureusement pour toi, tu auras toujours huit ans pour moi. Je te remercie d'être venu depuis le Québec pour ma soutenance. Tu ne sais pas ce qu'est le type B mais tu dois savoir que j'ai travaillé dessus toute la première année comme je te l'ai dit au téléphone à chaque fois que l'on se parlait. Tes cadeaux à base de patate et ton drapeau de la reine d'Angleterre rendent mon retour à l'appartement après le boulot plus sympatoches. Mes visites à Montréal m'ont permis de me changer les idées plus d'une fois même si tu as menti quand tu as dit qu'il allait faire en mars une température supportable pour un être humain. Merci aussi d'avoir accepté d'aller au bureau pour que je finisse vite fait le type B alors que tu venais de terminer ton baccalauréat. Heureusement, tu as pu profiter des hortillonnages qui ont dus être une expérience inoubliable pour toi. Tu es la meilleure petite soeur du monde et je te remercie d'accepter d'avoir huit ans jusqu'à la fin de ta vie.

Je continue mes remerciements des frères et soeurs de la plus jeune à la moins jeune avec mon frérot Thomas. Tu es en train de devenir le patriarche de la famille, merci de m'avoir aidé à organiser la soutenance et à réunir la famille. La prochaine grande occasion devrait être ton mariage. Je te remercie pour ton aide durant la thèse, notamment quand tu as été à l'initiative de l'achat de la PS4 qui m'a permis de mettre les bijections en pause plus d'une fois. Merci à Astrid pour sa bonne humeur à chaque fois que j'ai eu la chance de la voir, je ne pense pas connaître quelqu'un qui sourit autant, j'espère que Thomas a quelque chose à voir là-dedans.

Elsa, j'ai eu la chance de venir te voir à Annecy pendant la thèse plusieurs fois, c'était toujours avec grand plaisir. Tu m'as toujours impressionné par ta capacité de travail, j'espère avoir travaillé presque autant que toi pendant ces trois années. Tes compliments et encouragements m'ont aidé à avoir confiance en moi et à venir à bout de cette thèse. Tu m'as donné des précieux conseils pour l'enseignement, qui représente une partie très importante de ces dernières années et j'aurai encore plus besoin de tes conseils dans les années à venir.

Laure, je veux commencer par m'excuser pour ma tenue, tu n'es pas arrivée assez tôt pour qu'on puisse faire des courses de dernière minute cette fois-ci. Merci de m'avoir accueilli chez toi, j'ai découvert l'image des groupes Artin pour de nombreux types mais je ne pensais pas découvrir ce qu'est une figue pendant cette thèse.

Je peux remercier Cécile pour beaucoup de choses durant ces trois années. Je te remercie en particulier pour le voyage à Rome que nous avons fait avec Pablo pendant une des périodes les plus intenses de la thèse. Je pense que tu as apprécié le voyage aussi mais ça m'a permis de penser complétement à autre chose dans notre hôtel miteux au milieu d'une des plus belles villes du monde. J'ai hâte de pouvoir faire notre troisième voyage en Italie ensemble. Je remercie Pablo pour avoir été capable de sortir de belles phrases comme "Ces ruines, je m'en fiche comme de l'an 40" au Foro Romano à côté de ruines qui datent de l'an 40. Tu as été quand même assez agréable pour un jeune de 12 ans que l'on faisait marcher 20km par jour. Merci à Lio pour les histoires incroyables qu'elle m'a racontées au téléphone. Je remercie Gallien pour son soutien pendant la thèse, merci pour le match de rugby en loges à Bordeaux qu'il m'a offert et qui restera un très bon souvenir.

Je remercie enfin Anne pour son accueil pendant des phases obscures de ma thèse où mon esprit était parfois ailleurs. J'espère pouvoir être plus présent après ma thèse pour te soutenir à mon tour. Merci aussi à Zavier qui m'appelle déjà monsieur le professeur. Merci à mes nièces Afidy et Eding pour les petits exercices de mathématiques que vous m'envoyez de temps à autre et pour le jeu le logo des marques où j'ai eu l'occasion rêvée d'être "l'anim".

Pour finir, je veux remercier Mélyssa. Merci d'avoir changé ma vie un soir de juin où j'essayais de résoudre à la main un système à 2304 inconnues en me demandant, "ça vous dirait d'aller boire ou manger un truc un soir?". Ta rencontre est la meilleure chose qui me soit arrivée. Je t'ai rencontré le 9 Novembre 2016 après une nuit blanche où j'avais du mal à parler avant d'arriver, j'ai oublié mes soucis pendant 2 heures. Je ne savais pas encore à ce moment là à quel point tu es exceptionnelle et je n'imaginais pas une seconde que j'emménagerai avec toi quelques mois plus tard. Je ne pense pas que j'aurais pu réussir à venir à bout de ces derniers mois où nous nous levions à 7h47 tous les matins et où tu m'accompagnais à l'université que ce soit pendant la semaine ou le week-end ou les jours fériés. C'était beaucoup plus facile de mettre la tête dans la rédaction de ma thèse en me disant qu'à n'importe quel moment, la porte du bureaux des doctorants pouvait s'ouvrir avec toi derrière, un coca ou des croissants et chocolatines à la main pour me donner la force de continuer à travailler. Tes cadeaux surprises et tes petits mots que je trouve souvent dans les poches de mon jean ou mon portefeuille ou mes feuilles de calcul m'ont aidé à passer la plus belle année de ma vie alors que ça aurait probablement été une année d'épuisement physique et moral autrement. Merci aussi de m'avoir écouté plusieurs fois faire mon exposé de soutenance en notant à la seconde près le temps passé sur chaque diapositive et d'avoir noté mon étang. Cet exposé est le plus stressant que j'aurai à faire dans ma vie et il le sera beaucoup moins grâce à toi. Je t'aime et je te dédie cette thèse.

Résumé en français de la thèse

Nous déterminons dans cette thèse l'image des groupes de Artin associés à des groupes de Coxeter irréductibles dans leur algèbre de Iwahori-Hecke finie associée. Cela a été fait en type A dans [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] et [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. Dans le cas générique, la clôture de Zariski de l'image a été déterminée dans tous les cas [START_REF] Marin | Infinitesimal Hecke algebras II[END_REF]. L'approximation forte suggère que les résultats devraient être similaire dans le cas fini. Il est néanmoins impossible d'utiliser l'approximation forte sans utiliser de lourdes hypothèses et limiter l'étendue des résultats. Nous démontrons dans cette thèse que les résultats sont similaires mais que de nouveaux phénomènes interviennent de par la complexification des extensions de corps considérées. Les arguments principaux proviennent de la théorie des groupes finis. Nous utiliserons notamment un Théorème de Guralnick et Saxl [START_REF] Guralnick | Generation of finite almost simple groups by conjugates[END_REF] qui utilise la classification des groups finis simples pour les représentations de hautes dimensions. Ce théorème donne des conditions pour que des sous-groupes de groupes linéaires soient des groupes classiques dans une représentation naturelle. En petite dimension, nous utiliserons la classification des sous-groupes maximaux des groupes classiques de [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] pour les cas les plus compliqués.

Les résultats que nous démontrons peuvent avoir diverses applications. Par exemple, des groupes classiques finis ou des produits directs de groupes classiques finis apparaissent comme des quotients finis des groupes de Artin. Ces derniers sont des groupes fondamentaux de variétés algébriques donc cela définit des recouvrements finis de ces variétés qui peuvent être intéressant. Comme ces variétés sont définies sur le corps des rationnels Q, cela peut avoir des applications au problème de Galois inverse (voir par exemple [START_REF] Strambach | Generalized braid groups and rigidity[END_REF] pour un exemple en type A n ).

Ces résultats sont aussi intéressants du point de vue des groupes finis classiques. En effet, nous obtenons des générateurs explicites de ces groupes vérifiant les relations de tresses. Cela entraine des constructions intéressantes de ces groupes et de certains de leurs sous-groupes en s'intéressant à la restriction à des sous-groupes de Artin paraboliques. Nous obtenons par exemple une desciption intéressante du group Spin + 8 (q) à l'aide des deux représentations de dimension 8 en type H 4 (voir section 8.4).

Dans cette thèse, nous donnerons tout d'abord dans la section 2.1 une introduction aux groupes de Coxeter, aux groupes de Artin et aux algèbres de Iwahori-Hecke. Nous rappelerons la classification des groupes de Coxeter finis. Il y a quatre familles infinies A n , B n , D n et I 2 (n). Les groupes restant sont les groupes de Coxeter finis exceptionnels et sont noté E 6 , E 7 , E 8 , F 4 , H 3 et H 4 . Ils correspondent tous à des objets géométriques et ont été classifiés en utilisant des argument géométriques par Coxeter en 1932 [START_REF] Coxeter | The complete enumeration of finite groups of the form r 2 i = (r i r j ) k i,j = 1[END_REF]. Nous expliquerons ensuite comment définir les algèbres de Iwahori-Hecke dans un cadre général avant de donner des définitions plus précises sur les corps finis dans le chapitre correspondant à chaque type. Nous donnerons à la fin de cette section une idée des modèles pour les représentations irréductibles de ces algèbres.

Nous donnerons ensuite dans la section 2.2 des rappels sur les algèbres symétriques et les élements de Schur. Les éléments de Schur sont des outils qui permettent d'avoir un contrôle sur la semi-simplicité des algèbres symétriques. Les algèbres de Hecke sont des algèbres symétriques et leurs éléments de Schur dans le cas générique ont été déterminés dans tous les cas dans [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]. Nous démontrerons après ces rappels une version du Théorème de déformation de Tits. Nous utiliserons cette version du théorème de Tits dans les différents chapitres pour montrer que sous les bonnes conditions sur les paramètres et la caractéristique du corps fini, nous pouvons spécialiser les modèles matriciels issus du cas générique aux corps finis F q . Les algèbres de Iwahori-Hecke seront alors semi-simples déployées et serons donc isomorphes à ⊕ ρ irr M nρ (q).

Nous pourrons alors considérer le groupe multiplicatif A Wn =< S 1 , S 2 , . . . , S n > comme sousgroupe des éléments inversibles de l'algèbre < S 1 , S 2 , . . . , S n >. Ce groupe sera donc un sousgroupes du produit ρ irr GL nρ (q). Nous déterminerons à quoi ce groupe est isomorphe dans les différents types. Dans la section 2.3, nous rappelons le théorème de Aschbacher [4] sur les sous-groupes maximaux des groupes classiques finis. Il définit dans ce théorème 8 classes de sous-groupes géométriques noté C i . Nous rappelons dans cette section une description rapide de ces classes. Nous donnerons ensuite des théorèmes qui permettent d'identifier les groupes classiques finis sous certaines conditions. Nous utiliserons ces théorèmes afin de déterminer l'image du groupe A Wn dans les diverses représentations irréductibles des algèbres de Hecke sur les corps finis où A Wn est le groupe dérivé de A Wn .

Dans le chapitre 3, nous déterminons l'image du groupe A Bn dans son algèbre de Iwahori-Hecke associée. Nous définissons d'abord l'algèbre de Iwahori-Hecke finie sur le corps F p (α, β). Les représentations irréductibles en type B n sont indexées par des doubles-partitions de n. Les paramètres α et β correspondent aux déformations de la relation d'ordre 2 des groupes de Coxeter dans les algèbres de Iwahori-Hecke finies. Il y a deux paramètres en type B n car les générateurs ne sont pas tous conjugués. Nous devons alors considérer l'extension de corps F p (α, β) au-dessus de F p (α + α -1 , β + β -1 ). Les doubles-partitions avec une composante vide nous donne les mémes résultats qu'en type A n . Ces résultats dépendent de l'extension de corps F p (α) au-dessus de F p (α + α -1 ). Nous devrons alors distinguer les différents cas possibles pour ces extensions de corps. Cela donne 6 possibilités et les résultats pour l'image de A Bn sont alors différents. La preuve est dans tous les cas basée sur une récurrence où nous déterminons l'image pour n ≤ 5 de diverses manières en regardant les représentations irréductibles une à une. Nous déterminons aussi les différentes factorisations (Proposition 3.2.4) entre les représentations suivant les doubles-partitions qui les indexent. Ces factorisations dépendent des extensions de corps considérées et permettent de deviner quelle sera l'image de A Bn . Nous montrons ensuite par récurrence que le résultat annoncé est bien correct. Nous démontrons que les hypothèses du Théorème 2.3.2 sont vérifiées par ρ(A Bn ) pour chaque représentation irréductible ρ. Nous donnons ci-dessous le résultat pour ce type dans le cas le plus simple et un des cas les plus compliqués.

Notons

A 1,n = {(λ 1 , ∅), λ 1 n}, A 2,n = {(∅, λ 2 ), λ 2 n}, A n = A 1,n ∪ A 2,n . A n = {(λ 1 , ∅) ∈ A 1,n , λ 1 pas une équerre}, n = {λ n, λ / ∈ A n , λ pas une équerre}, F q = F p (α). Théorème 1. Si F q = F p (α, β) = F p (α + α -1 , β + β -1 ) et F p (α) = F p (α + α -1
), alors le morphisme :

A Bn → H × Bn,α,β λ n
GL(λ) se factorise à travers le morphisme surjectif

Φ 1,n : A Bn → SL n-1 (q) × (λ 1 ,∅)∈A n,λ1<λ 1 SL n λ (q) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SL n (q) 2 × λ∈ n,λ<λ SL n λ (q) × λ∈ n,λ=λ OSP (λ) . Théorème 2. Si F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ) et F p (α) = F p (α + α -1
), alors le morphisme A Bn → H × Bn,α,β λ n GL(λ) se factorise à travers le morphisme surjectif

Φ 6,n : A Bn → SL n-1 (q) × (λ 1 ,∅)∈A n,λ1<λ 1 SL n λ (q) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SL n (q) ×
λ∈ n,λ<λ ,λ<(λ 1 ,λ 2 ),λ =(λ 2 ,λ 1 )

SL n λ (q) × λ∈ n,λ<λ ,λ=(λ 2 ,λ 1 )

SL n λ (q 1
2 )× Cette règle de branchement permet de faire un raisonnement par récurrence comme dans le chapitre 3. Il faut alors traiter déterminer l'image de A Dn dans l'algèbre de Iwahori-Hecke associée. On utilise pour cela les résultats de [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] et les classifications des sous-groupes maximaux de certains groupes classiques sur des corps finis. La règle de branchement et le lemme de Goursat permettent d'obtenir le résultat suivant.

On écrit A 1,n = {(λ 1 , ∅), λ 1 n}, A 2,n = {(∅, λ 2 ), λ 2 n}, A n = A 1,n ∪ A 2,n et n = {λ n, λ not a hook} Théorème 3. Si F q = F p (α) = F p (α + α -1 ) et n est impair, alors le morphisme de A Dn dans H × Dn,α λ n λ 1 >λ 2

GL n λ (q) se factorise à travers le morphisme surjectif

Φ 1 ,n : A Dn → SL n-1 (q) × SL n (q) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SL n λ (q) × λ∈ n,λ=ϕ(λ) n λ >nµ
OSP (λ) .

Si F q = F p (α) = F p (α + α -1 ) et n ≡ 0 (mod 4), alors le morphisme de

A Dn dans H × Dn,α λ n λ 1 >λ 2 GL n λ (F q ) × λ=(λ 1 ,λ 1 ) n
GL n λ,+ (q) × GL n λ,-(q) se factorise à travers le morphisme surjectif

Φ 1 ,n : A Dn → SL n-1 (q) × SL n (q) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SL n λ (q) × λ∈ n,λ=ϕ(λ) λ 1 >λ 2 OSP (λ) × λ=(λ 1 ,λ 1 )∈ n λ>ϕ(λ) SL n λ 2 (q) 2 × λ=(λ 1 ,λ 1 )∈ n λ=ϕ(λ)
OSP (λ, +) 2 .

Si F q = F p (α) = F p (α + α -1 ) et n ≡ 2 (mod 4) alors le morphisme de A Dn dans H × Dn,α λ n λ 1 >λ 2 GL n λ (F q ) × λ=(λ 1 ,λ 1 ) n

GL n λ,+ (q) × GL n λ,-(q) se factorise à travers le morphisme surjectif

Φ 1 ,n : A Dn → SL n-1 (q) × SL n (q) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SL n λ (q) × λ∈ n,λ=ϕ(λ) λ 1 >λ 2 OSP (λ) × λ=(λ 1 ,λ 1 )∈ n λ>ϕ(λ) SL n λ 2 (q) 2 × λ=(λ 1 ,λ 1 )∈ n λ=ϕ(λ) SL n λ 2 (q).
Dans tout ce qui précéde, OSP (λ) désigne le groupe des isométries de la forme bilinéaire définie en Proposition 4.1.3.

Le résultat correspondant dans le cas F q = F p (α) = F p (α + α -1 ) est similaire. Les groupes spéciaux linéaires sont alors remplacés par des groupes unitaires et les groupes symplectiques et orthogonaux sont définis sur des corps plus petits. Le résultat est donné dans le théorème 4.2.3. Cela conclut l'étude pour les cas classiques.

La seule famille infinie restante et la famille I 2 (m), m ≥ 5. Dans le cas m impair, les deux générateurs du groupe de Coxeter sont conjugués. Dans le cas m pair, ils ne le sont et on a donc deux paramètres pour l'algèbre de Iwahori-Hecke. On sépare donc l'étude dans le chapitre 5 en deux sections suivant la parité de m. Dans les deux cas, les représentations irréductibles sont de dimension 1 ou 2. Nous déterminons dans les théorèmes 5.1.2 et 5.2.2 alors l'image de ρ(A I 2 (m) pour les représentations irréductibles de dimension 2 en utilisant le Théorème de Dickson (voir [START_REF] Huppert | I. Die Grundlehren der Mathematischen Wissenschaften[END_REF] Théorème 8.27) qui classifie les sous-groupes de SL 2 (q). La difficulté principale pour ces types provient des différentes factorisations possibles et de l'étude des extensions de corps lorsque m est pair. Les différentes extensions de corps sont décrites dans les Figures 5.1 à 5.7. Pour étudier les factorisations possibles, il faut introduire une relation d'équivalence sur les entiers. Elle est donnée dans les lemmes suivants qui dépendent de la parité de m. Lemme 2. Supposons m impair et ξ j = θ j + θ -j où θ est une racine primitive m-ième de l'unité dans F p . Soit j, l ∈ [[1, m- 1 2 ]] 2 . Il existe un automorphisme Ψ l,j de F q j = F p (α, ξ j + ξ -j ) qui vérifie Ψ l,j (α + α -1 ) = α + α -1 et Ψ l,j (ξ j + ξ -j ) = ξ l + ξ -l si et seulement si il existe r ∈ N tel que jp r ≡ l (mod m) ou jp r ≡ -l (mod m) et (α + α -1 ) p r = α + α -1 .

On dit que j ∼ l si une de ces conditions est vérifiée. Cela définit une relation d'équivalence et lorsque j ∼ l, on a ρ l|A I 2 (m) = Ψ l,j • ρ j|A I 2 (m) . Lemme 3. Supposons m pair et ξ j = θ j + θ -j où θ est une racine primitive m-ième de l'unité dans F p . On dit que j ∼ l si F p (α + α -1 , β + β -1 , ξ j ) F p (α + α -1 , β + β -1 , ξ j ) et il existe Φ j,l ∈ Aut(F q j ) tel que Φ j,l (α + α -1 ) = α + α -1 ), Φ j,l (β + β -1 ) = β + β -1 et Φ j,l (ξ j ) = ξ l . Cela définit une relation d'équivalence et si j ∼ l alors Φ j,l • ρ j|A I 2 (m) ρ l|A I 2 (m) .

On a alors les théorèmes suivants Théorème 4. Supposons m impair et que α vérifie les conditions données au début de la section 5.1. Pour j ∈ [[1, m- 1 2 ]], on pose G j = SL 2 (q j ) si F q j = F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j ) et G j SU 2 (q 1 2 j ) si F q j = F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j ).

On a alors que le morphisme de

A I 2 (m) dans H × I 2 (m),α GL 1 (q j ) 2 × j∈[[1, m-1 2 ]]
GL 2 (q j ) se factorise à travers le morphisme surjectif

Φ : A I 2 (m) → j∈[[1, m-1 2 ]]/∼ G j .
Théorème 5. Supposons m pair et que α et β vérifient les conditions données au début de la section 5.2. Pour j ∈ [[1, m-2 2 ]], on pose G j = ρ j ([< T t , T s >, < T t , T s >]) . On alors que le morphisme de A I 2 (m) dans H × I 2 (m),q

GL 1 (q j ) 2 × j∈[[1, m-1 2 ]]
GL 2 (q j ) se factorise par le morphisme surjectif Φ :

A I 2 (m) → j∈[[1, m-2 2 ]]/∼ G j .
Cela conclut l'étude pour les familles infinies de groupes de Coxeter finis irréductibles. Il reste ensuite à traiter les groupes de Coxeter exceptionnels. C'est à dire les algèbres de Hecke associées aux groupes de Coxeter de type E 6 , E 7 , E 8 , H 3 , H 4 et F 4 . Les représentations pour ces algèbres de Hecke sont données par des W -graphes. Nous décrivons ces objets dans le chapitre 6. La définition est la suivante. Soit W un Définition 1. Soit W un groupe de Coxeter, K son corps de définition, H son algèbre de Iwahori-Hecke de paramètres (α s ) s∈S et K = K ((α s ) s∈S ). Pour X un ensemble, on note D(X) = {(x, x), x ∈ X} sa diagonale. Un W -graphe Γ est la donnée d'un triplet (X, I, µ) tel que 1. X est un ensemble et I est une application de X dans P(S), 2. µ est une application de (X × X \ D(X) × S) dans K stable par l'involution du corps K qui envoie √ α s sur √ α s -1 .

Soit V le K ((α s ) s∈S )-espace vectoriel de base (e y ) y∈X . Pour tout s ∈ S, on définit ρ s : V → V par e y → -e y if s ∈ I(y), e y → α s e y + x∈X,s∈I(x) √ α s µ s x,y e x if s / ∈ I(y).

3. L'application T s → ρ s est une représenation de H.

On sait qu'un tel W -graphe existe pour n'importe quel représentation irréductible d'une algèbre de Iwahori-Hecke [START_REF] Gyoja | On the existence of a W -graph for an irreducible representation of a Coxeter group[END_REF]. Dans le chapitre 6, nous donnons des propriétés sur les représentations qui peuvent se déduire de ces modèles. La 2-colorabilité est une notion qui intervient dans plusieurs propositions de ce chapitre. Les W -graphes 2-coloriables dans le cas des paramètres égaux ont été classifié par Gyoja (Voir la remarque après le Théorème 6.1). Les notions de W -graphe dual de [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] et de représentation auto-duale sont données dans la Proposition 6.4 et la Définition 6.3. Le théorème principal de cette section est le suivant Théorème 6. Soit Γ = (X, I, µ) un W -graphe associé à une représentation irréductible de H tel que Γ soit 2-coloriable et tel que Γ soit isomorphe en tant que graphe orientée pondéré au graphe (X, Ĩ, -μ).

Soit ϕ : X → X l'automorphisme de graphe de Γ dans (X, Ĩ, -μ) et x 1 , x 2 , . . . , x n une numérotation de X telle que ϕ(x i ) = x n+1-i .

Soit ., . la forme bilinéaire définie par < e x i , e x j >= ω(e x i )δ i,n+1-j , où ω est un coloriage de Γ par 1 et -1.

On a alors ∀s ∈ S,

∀v 1 , v 2 ∈ V, ρ Γ (T s )v 1 , ρ Γ (T s )v 2 = -α v 1 , v 2 .
Cette forme bilinéaire est non-dégénérée et elle est symétrique si ω(x 1 )ω(x n ) = 1 et antisymétrique si ω(x 1 )ω(x n ) = -1.

La représentation associée est alors auto-duale.

On dit ensuite qu'un W -graphe auto-dual est un W -graphe vérifiant les propriétés du Théorème précédent. Cette propriété est vérifiée par certains W -graphes associés à des représentations irréductibles auto-duales des algèbres de Hecke mais pas par tous. Les W -graphes n'étant pas unique pour une représentation donnée, on établit la conjecture suivante. Conjecture 1. Soit W un groupe de Coxeter. Pour toute représentation irréductible autoduale, il existe un W -graphe auto-dual défini sur K assoicé cette représentation. Si il existe un W -graphe Γ défini sur K associé à la représentation alors il existe un W -graphe auto-dual Γ associé à la représentation et une matrice M ∈ GL |X| ( K) telle que pour tout h ∈ H K , M ρ Γ (h)M -1 = ρ Γ (h).

La deuxième partie de la conjecture est formulée uniquement dans l'optique de montrer la première partie de la conjecture. Afin d'utiliser cette deuxième partie, nous montrons deux conditions restrictives pour deux W -graphes soient associés à la même représentation irréductible dans les Propositions 6.2 et 6.3. Nous avons ensuite démontré par des calculs avec la plateforme de calcul HPC MatriCS [1] la conjecture pour les types E 6 , E 7 , E 8 , H 3 et H 4 . Les W -graphes auto-duaux obtenus peuvent être téléchargés sur [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF].

Les résultats du chapitre 6 permettent d'étudier les types exceptionnels. Dans le chapitre 7, nous déterminons l'image de A E 6 , A E 7 et A E 8 dans leur algèbre de Iwahori-Hecke associée. Le groupe de Artin A D 5 s'injecte naturellement dans le groupe de Artin A E 6 . Cela permet d'utiliser les résultats du chapitre 4 pour appliquer un raisonnement par récurrence.

Nous montrons dans un premier temps en utilisant la Proposition 2.2.4 et les éléments de Schur des algèbres de Iwahori-Hecke dans le cas générique que ces algèbres de Iwahori-Hecke sont semi-simples déployées après spécialisation sous les bonnes conditions sur les paramètres de l'algèbre.

Dans la section 7.1, nous utilisons le Théorème 2.3.2 et les Théorème 4.2.2 et 4.2.3 pour déterminer l'image de A E 6 dans chaque représentation irréductible de l'algèbre de Iwahori-Hecke H E 6 ,α . Les W -graphes auto-duaux obtenus et disponibles sur [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF] permettent de déterminer quel type de forme bilinéaire est préservée par les ρ(A E 6 ) pour les représentations irréductibles auto-duales. Les nouveaux E 6 -graphes auto-duaux de dimension 10 et 20 sont donnés dans la section 10.4. On voit sur ces figures que l'opération de symétrie sur les graphes inverse les couleurs, les formes bilinéaires associées sont donc anti-symétriques. C'est le cas pour toutes les représentations irréducibles auto-duales en type E 6 . On utilise enfin le Lemme de Goursat (Lemme 3.3.1) pour récupérer l'image totale dans l'algèbre de Iwahori-Hecke A E 6 ,α . Théorème 7. Si F q = F p (α) = F p (α + α -1 ), alors le morphisme de A E 6 dans H E 6 ,α ρ irr GL nρ (q) se factorise à travers le morphisme surjectif Φ : A E 6 → SL 6 (q) × SP 10 (q) × SL 15 (q) 2 × SL 20 (q) × SP 20 (q) × SL 24 (q) × SL 30 (q) ×SP 60 (q) × SL 60 (q) × SL 64 (q) × SP 80 (q) × SL 81 (q) × SP 90 (q). Si F q = F p (α) = F p (α + α -1 ), alors le morphisme de A E 6 dans H E 6 ,α ρ irr GL nρ (q) se factorise à travers le morphisme

Φ : A E 6 → SU 6 (q 1
2 ) × SP 10 (q

2 ) × SU 15 (q

2 ) 2 × SU 20 (q

2 ) × SP 20 (q

2 ) × SU 24 (q

2 ) × SU 30 (q

2 ) Théorème 8. Si F q = F p ( √ α) = F p (α + α -1 ), alors le morphisme de A E 7 dans H E 7 ,α ρ irr GL nρ (q) se factorise à travers le morphisme surjectif Φ : A E 7 → SL 7 (q) × SL 15 (q) × SL 21 (q) 2 × SL 27 (q) × SL 35 (q) 2 × SL 56 (q) × SL 70 (q) × SL 84 (q)

×SL 105 (q) 3 × SL 120 (q) × SL 168 (q) × SL 189 (q) 3 × SL 210 (q) 2 × SL 216 (q) × SL 280 (q) 2 ×SL 315 (q) × SL 336 (q) × SL 378 (q) × SL 405 (q) × SL 420 (q) × SL 512 (q).

Si F p ( √ α) = F q = F p (α) = F p (α + α -1 ) alors le morphisme de A E 7 dans H E 7 ,α ρ irr GL nρ (q) se factorise à travers le morphisme Φ : A E 7 → SL 7 (q) × SL 15 (q) × SL 21 (q) 2 × SL 27 (q) × SL 35 (q) 2 × SL 56 (q) × SL 70 (q) × SL 84 (q)

×SL 105 (q) 3 × SL 120 (q) × SL 168 (q) × SL 189 (q) 3 × SL 210 (q) 2 × SL 216 (q) × SL 280 (q) 2 ×SL 315 (q) × SL 336 (q) × SL 378 (q) × SL 405 (q) × SL 420 (q) × SU 512 (q).

Si F q = F p (α) = F p (α + α -1 ) alors le morphisme de A E 7 dans H E 7 ,α ρ irr GL nρ (q) se factorise à travers le morphisme

Φ : A E 7 → SU 7 (q 1
2 )×SU 15 (q

2 )×SU 21 (q

2 ) 2 ×SU 27 (q

1 2 )×SU 35 (q 1
2 ) 2 ×SU 56 (q

2 )×SU 70 (q

2 )×SU 84 (q

2 )

×SU 105 (q

1 2 ) 3 × SU 120 (q 1
2 ) × SU 168 (q

2 ) × SU 189 (q

1 2 ) 3 × SU 210 (q 1
2 ) 2 × SU 216 (q

2 ) × SU 280 (q

2 ) 2

×SU 315 (q

2 ) × SU 336 (q

2 ) × SU 378 (q

2 ) × SU 405 (q

2 ) × SU 420 (q

2 ) × SU 512 (q

2 ).

Dans la section 7.3, nous déterminons l'image de A E 8 dans son algèbre de Iwahori-Hecke associée. Les preuves se font par récurrence en utilisant les résultats de la section 7.2. La difficulté provient principalement des dimensions des représentations irréductibles auto-duales. La représentation irréductible auto-duale de plus haute dimension est de dimension 7168. En utilisant la conjecture, l'obtention de la forme bilinéaire associée nécessite plus d'une semaine de calcul. Nous avons utilisé la conjecture pour obtenir toutes les formes bilinéaires associées aux représentations disponibles dans le package CHEVIE [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF] de GAP. Une fois les formes bilinéaires obtenues, nous avons démontré la conjecture pour chaque représentation irréductible auto-duale. Les E 8 -graphes auto-duaux obtenus sont téléchargeables depuis [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF]. Cela permet d'obtenir les formes bilinéaires correspondantes en utilisant uniquement un 2-coloriage de la représentation. En utilisant ces résultats, nous avons démontré le résultat suivant qui conclut le chapitre 7. Si F p ( √ α) = F q = F p (α) = F p (α + α -1 ) alors le morphisme de A E 8 dans H E 8 ,α ρ irr GL nρ (q) se factorise à travers le morphisme surjectif

Φ : A E 8 → ρ∈A SL nρ (q) × SL 4096 (q 2 ) × ρ∈B Ω + nρ (q).
Si F q = F p ( √ α) = F p (α) = F p (α + α -1 ), alors le morphisme de A E 8 dans H E 8 ,α ρ irr GL nρ (q) se factorise à travers le morphisme surjectif

Φ : A E 8 → ρ∈A SL nρ (q) × SL 4096 (q) 2 × ρ∈B Ω + nρ (q).
Si F q = F p (α) = F p (α + α -1 ), alors le morphisme de A E 8 dans H E 8 ,α ρ irr GL nρ (q) se factorise à travers le morphisme surjectif

Φ : A E 8 → ρ∈A SU nρ (q 1
2 ) × SU 4096 (q

1 2 ) × ρ∈B Ω + nρ (q 1
2 ).

Dans le chapitre 8, nous considérons l'image des groupes A H 3 et A H 4 dans leur algèbre de Iwahori-Hecke finie associée. L'inclusion naturelle de A I 2 (5) dans A H 3 fait intervernir les résultats de la section 5.1. Le corps de définition de H 3 et de

H 4 est Q[ √ 5
], la relation d'équivalence définie dans la section 5.1 appliquée au cas m = 5 sera alors nécessaire pour bien distinguer les différents cas. Il y a peu de représentations irréductibles en type H 3 . En utilisant les résultats de la section 5.1 et les classifications de sous-groupes maximaux classiques en petite dimension [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF], nous obtenons le résultat suivant. Théorème 10. Supposons p / ∈ {2, 5} et que l'ordre de α ne divise ni 6 ni 20.

1. Supposons 1 ∼ 2.

(a) Si F q = F p ( √ α) = F p (α) = F p (α + α -1 ), alors le morphisme de A H 3 dans H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) se factorise à travers le morphisme surjectif Φ : A H 3 → SL 3 (q 2 ) × SL 4 (q) × SL 5 (q).

(b) Si

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et Φ 1,2 ( √ α) = √ α -1

alors le morphisme de

A H 3 dans H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) se factorise à travers le morphisme surjectif

Φ : A H 3 → SL 3 (q) × SU 4 (q 1
2 ) × SU 5 (q

2 ).

(c) Si F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et Φ 1,2 ( √ α) = - √ α -1
, alors le morphisme de A H 3 dans H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) se factorise à travers le morphisme surjectif

Φ : A H 3 → SL 3 (q) × SL 4 (q 1
2 ) × SU 5 (q

2 ).

(d) Si F q 2 = F p ( √ α) = F p (α) = F p (α + α -1 ), alors le morphisme de A H 3 dans H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) se factorise à travers le morphisme surjectif Φ : A H 3 → SL 3 (q 2 ) × SU 4 (q) × SL 5 (q). 2. Supposons 1 2. Lorsqu'il existe, on note l'automorphisme d'ordre 2 de F q .

(a) Si

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et ( √ α) = √ α -1 ,

alors le morphisme de

A H 3 dans H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) se factorise à travers le morphisme surjectif

Φ : A H 3 → SU 3 (q 1
2 ) 2 × SU 4 (q

2 ) × SU 5 (q

2 ).

(b) Si F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et ( √ α) = - √ α -1 ,

alors le morphisme de

A H 3 dans H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) se factorise à travers le morphisme surjectif

Φ : A H 3 → SU 3 (q 1
2 ) 2 × SL 4 (q

2 ) × SU 5 (q

2 ). (a) Si F q = F p ( √ α) = F p (α) = F p (α + α -1 ), alors le morphisme de A H 4 dans H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1 )) se factorise à travers le morphisme surjectif Φ :→ SL 4 (q 2 ) × Ω + 6 (q 2 ) × Spin + 8 (q) × SL 9 (q 2 ) × Ω + 10 (q) × SL 16 (q) 2 × Ω + 16 (q) × Ω + 18 (q) ×Ω + 24 (q) 2 × SL 25 (q) × Ω + 30 (q) × SL 36 (q) × Ω + 40 (q) × Ω + 48 (q

1 2 ). (b) Si F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et Φ 1,2 ( √ α) = √ α -1 ,

alors le morphisme de

A H 4 dans H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1
)) se factorise à travers le morphisme surjectif Φ :→ SL 4 (q)×Ω + 6 (q)×Spin + 8 (q

2 )×SL 9 (q 2 )×Ω + 10 (q

2 )×SU 16 (q

2 ) 2 ×Ω + 16 (q

2 )×Ω + 18 (q

2 )

×Ω + 24 (q 1
2 ) 2 × SU 25 (q

2 ) × Ω + 30 (q

2 ) × SU 36 (q

1 2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q

2 ).

(c) Si

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et Φ 1,2 ( √ α) = - √ α -1 ,
Φ :→ SL 4 (q) × Ω + 6 (q) × Spin + 8 (q 1
2 ) × SL 9 (q 2 ) × Ω + 10 (q

2 ) × SL 16 (q) × Ω + 16 (q

2 ) × Ω + 18 (q

2 )

×Ω + 24 (q 1
2 ) 2 × SU 25 (q

1 2 ) × Ω + 30 (q 1
2 ) × SU 36 (q

1 2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q

2 ).

(d) Si F q 2 = F p ( √ α) = F p (α) = F p (α + α -1
), alors le morphisme de A H 4 dans H H 4 ,α ρ irr

GL nρ (F p ( √ α, ξ + ξ -1
)) se factorise à travers le morphisme surjectif Φ :→ SL 4 (q 2 ) × Ω + 6 (q 2 ) × Spin + 8 (q) × SL 9 (q 2 ) × Ω + 10 (q) × SL 16 (q 2 ) × Ω + 16 (q) × Ω + 18 (q)

×Ω + 24 (q) 2 × SL 25 (q) × Ω + 30 (q) × SL 36 (q) × Ω + 40 (q) × Ω + 48 (q

2 ).

2. Supposons 1 2. Lorsqu'il existe, on note l'automorphisme d'ordre 2 de F q .

(a) Si

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et ( √ α) = √ α -1 ,
2 ) 2 × Ω + 6 (q

2 ) 2 × Spin + 8 (q

2 ) × SU 9 (q

2 ) 2 × Ω + 10 (q

2 ) × SU 16 (q

2 ) 2 × Ω + 16 (q

1 2 ) 2 ×Ω + 18 (q 1
2 ) × Ω + 24 (q

2 ) 4 × SU 25 (q

2 ) × Ω + 30 (q

2 ) 2 × SU 36 (q

1 2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q

2 ).

(b) Si

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) et ( √ α) = - √ α -1 ,

alors le morphisme de

A H 4 dans H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1
)) se factorise à travers le morphisme surjectif

Φ :→ SU 4 (q 1
2 ) 2 × Ω + 6 (q

1 2 ) 2 × Spin + 8 (q 1
2 ) × SU 9 (q

1 2 ) 2 × Ω + 10 (q 1
2 ) × SL 16 (q) × Ω + 16 (q

1 2 ) 2 ×Ω + 18 (q 1
2 ) × Ω + 24 (q

2 ) 4 × SU 25 (q

2 ) × Ω + 30 (q

2 ) 2 × SU 36 (q

1 2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q

2 ).

Dans le chapitre 9, nous déterminons l'image de A F 4 dans son algèbre de Iwahori-Hecke associée. Il y a deux injections naturelles de A B 3 dans A F 4 donc nous utiliserons les résultats du chapitre 3 pour démontrer les résultats. Il y a deux classes de conjugaison pour les générateurs du groupe de Coxeter de type F 4 . L'algèbre de Iwahori-Hecke associée dépend alors de deux paramètres α et β. Comme les représentations sont données par des F 4 -graphes, les modèles sont définis sur F p ( √ α, √ β). Il faut alors considérer l'extension de corps F p ( √ α, √ β) audessus de F p (α + α -1 , β + β -1 ). Une liste exhaustive des tours d'extensions possible permet de démontrer que cette extension est de degré au plus 2. Toutes les extensions sont décrites dans la section 10.5. Les F 4 -graphes de [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] sont aussi rappelés dans cette section, ils sont présentés de manière à faire apparaître les symétries pour les F 4 -graphes associés à des graphes auto-duaux. Nous n'avons pas démontré la conjecture dans ce cas car les propriétés d'unicité démontrés dans les propositions 6.2 et 6.3 ne sont démontrées que dans le cas des paramètres égaux. Les représentations étant toutes de dimension associée inférieure à 16, on peut déterminer par le calcul la forme bilinéaire associée sans émettre d'hypothèses supplémentaires. Le groupe A F 4 n'est pas parfait et cela complique considérablement les preuves dans certains cas. De plus, les restrictions aux sous-groupes paraboliques n'apportent pas autant d'informations que dans les chapitres précédents car la clôture normale des sous-groupes paraboliques isomorphes à A B 3 est différente de A F 4 . Cela nécessite de nouveaux arguments dans ce chapitre avec des preuves parfois plus calculatoires. Le théorème suivant donne l'image de A F 4 dans son algèbre de Iwahori-Hecke associée suivant les possibles tours d'extensions de corps considérées et conclut le chapitre 9.

Théorème 12. On note

F q = F p ( √ α, √ β), F rα = F p (α + α -1 ) et F r β = F p (β + β -1
). Dans les cas 1, 4, 5 et 10, le morphisme de A F 4 dans H F 4 ,α,β ρ irr GL nρ (q) se factorise à travers le morphisme surjectif

Φ : A F 4 → (SL 2 (r α ) • SL 2 (r β )) × SL 4 (q) 2 × Ω + 6 (
q) 2 ×SL 8 (q) 2 × SL 9 (q) 2 × Ω + 12 (q) × Ω + 16 (q). Dans les cas 11, 12, 13 et 16, le morphisme de A F 4 dans H F 4 ,α,β ρ irr GL nρ (q) se factorise à travers le morphisme surjectif

Φ : A F 4 → ×(SL 2 (r α ) • SL 2 (r β )) × SU 4 (q 1
2 ) 2 × ×Ω + 6 (q

1 2 ) 2 ×SU 8 (q 1
2 ) 2 × SU 9 (q

1 2 ) 2 × Ω + 12 (q 1
2 ) × Ω + 16 (q

2 ). Dans les cas 2, 6, 9 et 15, le morphisme de A F 4 dans H F 4 ,α,β ρ irr GL nρ (q) se factorise à travers le morphisme surjectif

Φ : A F 4 → (SL 2 (r α ) • SL 2 (r β )) × SL 4 (q) × ×Ω + 6 (q) ×SL 8 (q 1
2 ) × SU 8 (q

2 ) × SL 9 (q) × Ω + 12 (q

1 2 ) × Ω + 16 (q 1
2 ). Dans les cas 3, 7, 8 et 14, le morphisme de A F 4 dans H F 4 ,α,β ρ irr GL nρ (q) se factorise à travers le morphisme surjectif

Φ : A F 4 → (SL 2 (r α ) • SL 2 (r β )) × SL 4 (q) × Ω + 6 (q) ×SL 8 (q) × SU 8 (q 1
2 ) × SL 9 (q) × Ω + 12 (q

2 ) × Ω + 16 (q

2 ).

Chapter 1 Introduction 1.1 General Introduction

In this doctoral thesis, we will determine the image of Artin groups associated to all finite irreducible Coxeter groups inside their associated finite Iwahori-Hecke algebra. This was done in type A in [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] and [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. The Zariski closure of the image was determined in the generic case in [START_REF] Marin | Infinitesimal Hecke algebras II[END_REF]. It is suggested by strong approximation that the results should be similar in the finite case. However, the conditions required to use are much too strong and would only provide a portion of the results. We show in this thesis that they are but that new phenomena arise from the different field factorizations. The techniques used in the finite case are very different from the ones in the generic case. The main arguments come from finite group theory. In high dimension, we will use a theorem by Guralnick-Saxl [START_REF] Guralnick | Generation of finite almost simple groups by conjugates[END_REF] which uses the classification of finite simple groups to give a condition for subgroups of linear groups to be classical groups in a natural representation. In low dimension, we will mainly use the classification of maximal subgroups of classical groups in [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] for the complicated cases.

Our results about the image of Artin groups inside the finite Hecke algebras may have various applications. For instance, finite classical groups and direct products of finite classical groups appear in this way as finite quotients of the Artin groups. Since the latter are fundamental groups of algebraic varieties, this also defines étale coverings of these varieties. Since these varieties are defined over Q, this may have applications to the inverse Galois problem (see for example [START_REF] Strambach | Generalized braid groups and rigidity[END_REF] or [START_REF] Michael | The inverse Galois problem and rational points on moduli spaces[END_REF] for applications in type A n ).

They also provide new information on the finite classical groups because we get explicit generators verifying the braid relations for those groups. This can provide interesting constructions of these groups and some of their subgroups by looking at restrictions to parabolic subgroups of the Artin groups. In particular, we get a somewhat unexpected description of the 8-dimensional spin group from the two 8-dimensional representations in type H 4 (see section 8.4).

We find some new W -graphs in types H 4 , E 6 and E 8 which provide different information from the usual ones. They are all associated in a natural way to a bilinear form which is very complicated to obtain in the previous models. In this model, the bilinear form is obtained using only the two-colorability and its matrix in a well chosen basis is anti-diagonal. The uniqueness properties can probably be extended in a more general setting and understanding which setting this is may be worth considering.

Outline of the thesis

We will begin by giving an introduction of Coxeter groups, Artin groups and Iwahori-Hecke algebras. We then recall some properties of symmetric algebras and show that Iwahori-Hecke algebras are symmetric. In this section, we show a specialization result which we will use throughout the thesis to show that under certain conditions, we can specialize the models for Iwahori-Hecke algebras in the generic case to the finite case. We will then give Aschbacher's theorem [4] on maximal subgroups of classical groups and recall the different classes C i and S which are defined in this theorem.

We will then start by determining the image of Artin groups of classical types inside their associated finite Iwahori-Hecke algebras. The matrix models in those types are given by doublepartitions of an integer n. The general idea of the proof is then similar to the proof in type A. The image is first determined for small n. The branching rule is then used to give an inductive proof on n to determine the image in the general case. It is first necessary to determine which representations factor through each other via field automorphisms or the transposed inverse automorphism. The second parameter in type B will give rise to new factorizations which did not appear in the generic case. We will then have to separate the study in six different cases for the field extensions which will give different results. We get that in type B, if the fields extensions occuring are all trivial then the image of the representations associated to partitions of n are special linear groups defined over F q if λ = λ and symplectic or orthogonal groups if λ = λ . In the case when the field extensions are trivial, the only factorizations appearing are between the representations labeled by hook partitions and between the representation labeled by a partition and the representation labeled by its transposed partition. This is summarized in Theorem 3.2.1. When the field extensions a more complex, we get more factorizations and we get both unitary groups or special linear groups depending on the partition we are considering. The result for those cases are then given in Theorem 3.2.2 to Theorem 3.2.6.

In type D, the matrix models are similar but there is only one conjugacy class for the generators therefore the field extensions are less complicated. However, there are additional factorizations appearing which make the branching rule more complex. The result is then similar to the one in type B except there are more representations affording groups preserving non-degenerate bilinear forms. The results for this chapter are given in Theorem 4.2.2 and Theorem 4.2.3.

Those results will be useful for the groups of exceptional types because of the natural inclusions of Coxeter groups of classical types inside the exceptional types. Before treating the groups of type H, E and F , we will determine the image for dihedral groups and general results and W -graphs which will give us the matrix models for those exceptional types.

All the representations are 1-dimensional or 2-dimensional in dihedral type therefore determining the images inside each given representation will not be too difficult. There is a theorem by Dickson (see [START_REF] Huppert | I. Die Grundlehren der Mathematischen Wissenschaften[END_REF], Theorem 8.27) classifying the subgroups of P SL 2 (q) which we will use to determine the images in those cases and for 2-dimensional representations in all types. The difficulties will arise from the field extensions which depend on primitive roots of unity. This will lead us to define an equivalence relation between integers. The results for I 2 (5) will then be useful in type H since there is a natural inclusion from I 2 (5) to H 3 . Since SU 2 (q

1 2 ) is con- jugate to SL 2 (q 1
2 ) inside GL 2 (q), we will have that the image inside every given representation is conjugate to SL 2 (q 1 2 ) or SL 2 (q). The image inside the full Iwahori-Hecke is then given in Theorem 5.1.3 and Theorem 5.2.3.

The matrix models for exceptional types are given by W -graphs. We will recall some general properties for those graphs. We will then prove some uniqueness properties and establish a conjecture for W -graphs associated to self-dual representations (see Definition 6.3 in Conjecture 6.1. We proved the conjecture in types I, H and E by computation using the HPC platform MatriCS [1]. We assume the conjecture is true in order to find non-degenerate bilinear forms preserved by the image of derived subgroups of Artin groups under self-dual representations. It is clear that such a bilinear form exists for self-dual representations, however it is hard to determine the type of this bilinear form. The conjecture will allow us to determine explicitly those bilinear forms and to give new W -graphs, where a condition on the 2-colorability of the graph is sufficient to determine the type of the form. We draw them in type H 4 and some of them in type E 6 . The remaining ones can be downloaded from [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF].

We then determine the image in type H using the results in type I 2 (5). In type H 3 , we encounter some W -graphs which are not 2-colorable. It is then necessary to consider more complicated field extensions. By simple computations, we can treat those more complicated extensions and get the image inside the corresponding representations. As in the dihedral type, we need to use the equivalence relation on integers defined in Lemma 5.1.1. In this case, we only consider fifth roots of unity therefore the equivalence relation is only on 1 and 2. The factorizations between the different representations will depend on whether 1 ∼ 2 or 1 2. All the groups considered in type H 3 are linear or unitary depending on the field extensions. In type H 4 , the dimensions of the representations go up to 48. There are many self-dual representations for which we have verified the conjecture and found corresponding self-dual H 4 -graphs given in the Appendix. The study of this type is quite complicated so we separate the study in four parts. First, in section 8.2, we prove some general properties appearing in type H 4 as in the other types. Then, in section 8.3, we begin by considering the low-dimensional representations with the representations of dimension going up to 8. In these cases, we cannot using Theorem 2.3.2 therefore we need to use some more elementary theorems and some computational proofs. The image in the 8-dimensional representation is isomorphic to Ω + 8 (q) or Ω + 8 (q

2 ). It is very difficult to exclude the case 2 • Ω 7 (q). In order to do it, we provide a very computational proof using some elements in the normalizer of the image of the derived subgroup of < S 2 , S 3 >. Once this result is proven, we get a nice description of the Spin + 8 (q) groups appearing via the two 8dimensional representations. We explain in section 8.4 how the triality automorphism appears between the projections of the two 8-dimensional representations. We then prove that the image of the derived subgroup A H 4 inside the product of the two 8-dimensional representations is the universal central extension of P Ω + 8 (q) which proves it is isomorphic to Spin + 8 (q). In section 8.5, we determine the image of A H 4 inside the representations of dimension greater than 8 using Theorem 2.3.2. There are four 16-dimensional representations which are not 2-colorable which makes the study slightly more complicated but using the usual arguments, we get the image for all groups except for the image inside the 48-dimensional representation. We did not succeed in proving that the image is what we expect for the 48-dimensional representation but we conjecture it is. We then use Goursat's Lemma to recover the image inside the full Iwahori-Hecke algebra given in Theorem 8.5.1.

The image in type E is determined using the results in type D 5 . The main difficulty in this type was to determine the types of the bilinear forms. The high-dimensional representations required long computations using Matrics [1]. Those computations were done assuming the conjecture was true and we then proved the conjecture is true. Another difficulty arose in the proof that the specialization to finite fields still gave a split semisimple algebra. Some of the weights of the edges on the W -graphs considered vanish and it is then necessary to prove that the graphs remain connected once those edges vanish. Once we know if the forms are symplectic or orthogonal and the algebra is split semisimple, the usual arguments cover most of the proofs in those types. There are two representations in type E 7 which are not 2-colorable, we compute the traces of well-chosen elements in order to treat those cases. The main results in this chapter are given in Theorem 7.1.1, Theorem 7.2.1 and Theorem 7.3.1.

In type F 4 , there are two parameters and the representations are defined over F p ( √ α, √ β). This makes the field extensions much more complex. We also have that the normal closure of parabolic subgroups does verify the properties which we can observe and use in other types. The derived subgroup A F 4 is not perfect. This complicates the use of Goursat's lemma to determine the image inside the full Iwahori-Hecke algebra. The uniqueness properties for W -graphs are only proven for one parameter. It seems they do not hold in type F 4 and the 2-colorability of F 4graphs may not be stable by isomorphism of representations. We did not prove the conjecture in this type and it may not hold. Nevertheless, the representations are of dimension at most 16, therefore we can determine by hand the non-degenerate bilinear forms preserved by the groups. We still get the usual results depending on the field extensions except for the 4-dimensional representation 4 1 which is the central product of the 2-dimensional representations 2 1 and 2 3 . The main result in this section is given in Theorem 9.1.

Perspectives

Conjecture 6.1 does not need to be proved for classical types in order to conclude the doctoral thesis. After the Ph.D. is finished, one area to explore would be to prove Conjecture 6.1 for types A, B and D. A description of W -graphs affording the representations given by the combinatorial model associated to partitions in types A and B has been given by Naruse in [START_REF] Naruse | On an isomorphism between Specht module and left cell of S n[END_REF] and [START_REF] Naruse | Representation theory of Weyl group of type C n[END_REF]. Using those descriptions and trying to extend his work to type D, we hope to be able to prove the conjectures for types A, B and D.

The notion of Hecke algebras has been generalized to complex reflection groups by M.Broué, G. Malle and R.Rouquier [START_REF] Broué | Complex reflection groups, braid groups, Hecke algebras[END_REF]. It has been conjectured that there exists a symmetrizing trace for all those algebras. If that conjecture is true then it will be possible to use Schur elements to determine when those algebras are semisimple and specialize those algebras to finite fields. In type G(d, 1, n), the representations are known and are labeled by d-partitions of unity and we know the Schur elements. It is then possible to study the image of the associated braid group inside those representations and this work seems like a natural extension of the work done in the Ph.D. The specialization to finite fields could become complicated due to the high number of parameters involved.

It is still an open question whether Artin groups are linear or not. A linear group is a group for which there exists a monomorphism into some group of invertible matrices. It is also an open question whether the morphisms from the Artin groups onto the Iwahori-Hecke algebras are one-to-one in the generic case. If we could determine the kernels of those morphism in the finite case, it might be possible to find elements of the kernel in the generic case and show those morphisms are not monomorphisms. We have found some interesting elements in the kernels of some morphisms composed with projections upon irreducible representations in type H 4 but we have not yet had the opportunity to explore that question more.

In order to determine the image of the Artin group inside finite Iwahori-Hecke algebras, assumptions on the parameters were made for it to be semisimple. One could consider the cases where those parameters are not semisimple and still study the same question. The Hoefsmit models can only be specialized under some assumptions for the parameters which excludes most of the semisimple cases. On the other hand, W -graphs afford well-defined representations for nearly all possible representations.

The image of the Artin group inside each representation gives us a finite quotient of the Artin group. Using rigidity techniques which can be found in [START_REF] Malle | Inverse Galois theory[END_REF] and [START_REF] Völklein | Groups as Galois groups[END_REF], it should be possible to get some results in inverse Galois theory using methods as in [START_REF] Strambach | Generalized braid groups and rigidity[END_REF]. The key fact in this problem is that Artin groups can be seen as fundamental groups of certain algebraic varieties defined over Q. : finite field with q elements M n (q)

Notations

: algebra of matrices over the field F q GL n (q)

: group of invertible matrices over the field F q SL n (q) : group of invertible matrices of determinant 1 over F q SU n (q) : group of unitary matrices over F q 2 SP n (q) : group of symplectic matrices over F q Ω n (q)

: kernel of the spinor norm of the orthogonal group of type We here recall the definitions of Coxeter groups, Artin groups and Iwahori-Hecke algebras and give the classification of finite irreducible Coxeter groups. Those are the main objects we will use throughout the thesis. They appear in many different fields, they can be seen as groups generated by involutions verifying certain relations. They can also be considered as real reflection groups as a subclass of the complex reflection groups.

) ∈ S 1 × S 2 , we have s 1 s 2 = s 2 s 1 .
The irreducible finite Coxeter groups have been classified by Coxeter [START_REF] Coxeter | The complete enumeration of finite groups of the form r 2 i = (r i r j ) k i,j = 1[END_REF] and are given in Figure 2.1 Note that the Coxeter group of type A n is the symmetric group S n+1 generated by the transpositions s i = (i i + 1). The Coxeter group of type B n is the group of symmetries of the hypercube of dimension n. The Coxeter group of type I 2 (n) is the dihedral group with 2n elements.

There are natural inclusions between the groups W n and W n+1 and between A n-1 and B n , A n-1 and D n , I 2 (5) and H 3 , A 3 and H 4 , D 5 and E 6 , B 3 and F 4 . We will use those inclusions which remain in the Artin group context which we will define below. Definition 2.1.2. Let (W, S) be a finite Coxeter system. To the Coxeter group W , we associate the following Artin group A W , where the order 2 condition has been removed < S, ∀(s, t) ∈ S 2 , sts... The Artin group A An associated to the A n type Coxeter group admits the following presentation :

ms,t = tst... ms,t > s 1 s 2 s n A n t s 1 s n-1 B n s 1 s 2 s 3 s n D n s 1 s 2 n I 2 (n), n ≥ 5 s 1 s 2 s 3 5 H 3 s 1 s 2 s 3 s 4 5 H 4 s 1 s 3 s 4 s 2 s 5 s n E n , n ∈ {6, 7, 8} s 1 s 2 s 3 s 4 F 4
A An =< (s i ) i∈[[1,n]] , s i s i+1 s i = s i+1 s i s i+1 , |i -j| ≥ 2, s i s j = s j s i >.
The Artin group A Bn associated to the B n type Coxeter group admits the following presentation :

A Bn =< t, (s i ) i∈[[1,n-1]] , ts 1 ts 1 = s 1 ts 1 t, s i s i+1 s i = s i+1 s i s i+1 , |i -j| ≥ 2, s i s j = s j s i >.
Definition 2.1.3. Let A be an Artin group with generators {s i } i∈[ [1,n]] . Let a ∈ A, two expression a = s i k ..s i 1 and a = s j k ...s j 1 of a are said to be equivalent if one can be deduced from the other using the braid relations.

The following result is a fundamental result in the study of Iwahori-Hecke algebras, it can be found for example in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] We can now define the Iwahori-Hecke algebra associated to a given Coxeter group. The Matsumoto theorem will enable us to see it is a free algebra. We define it here in a general setting, we will give more precise definitions in the finite case in each type in the following chapters. Definition 2.1.4. Let (W, S) be a Coxeter system, W the associated Artin group, R a ring and (u s ) s∈S indeterminates such that α s = α t if s and t are conjugate in W .

The R[u ±1

s ]-Iwahori-Hecke Algebra H W,R,(us) s∈S associated to W is given by the following presentation

H W,R,(us) s∈S =< T 1 , ..., T n |T i T j T i ... ms i ,s j = T j T i T j ... ms i ,s j , (T i -α s i )(T i + 1) = 0 > By Theorem 2.
1.1, we can define the element T σ = T i 1 . . . T ir for σ = s i 1 . . . s ir in a reduced expression. The following proposition is then fairly easy to show Proposition 2.1.1. The Iwahori-Hecke algebra

H W,R,(us) s∈S is a free R[u ±1 s ]-module of rank |W |.
If we consider the case R = Z, we have that H W,Z,(us) s∈S is a split semi-simple algebra and models for its representations are then known. In types A n , B n and D n , the irreducible representations are labeled by partitions or double-partitions of n. A partition of n is a nonincreasing sequence (λ i ) i∈N such that +∞ i=1 λ i = n. To each partition of n, we can associate a Young diagram which is a diagram with λ i boxes in the ith row. We write [2 2 , 1] the partition of 5 with λ 1 = λ 2 = 2 and λ 3 = 1. There are 5 partitions of 4 given by the Young diagrams , , , , which we note respectively [1 4 ], [4], [3,1], [2, 1 2 ] and [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]. The irreducible representations of the Iwahori-Hecke algebra of type A n are labeled by partitions of n. There is a basis for each module associated to a partition of n given by standard tableaux associated to that partition.

A tableau associated to a partition of n is numbering with the integers from 1 to n. A standard tableau is a tableau such that the numbering is increasing towards the right and downwards. The representations labeled by the above partitions are therefore of respective dimensions 1, 1, 3, 3 and 2. A basis for the representation labeled by the partition [3,1] is given by the standard tableaux 1 2 3 4 ,

1 2 4 3 , 1 3 4 2 .
In type B n and D n , the irreducible representations are labeled by double-partitions of n. A double-partition of n is pair (λ, µ) with λ a partition of r and µ a partition of n -r. The double-diagram associated to the double-partition (λ, µ) is the pair given by the Young diagrams associated to λ and µ. A standard double-tableau associated to a double-partition is a numbering of the associated double-diagram with the integers from 1 to n increasingly towards the right and downwards within each component of the double-diagram. There are 5 double-partitions of 2 given by the double-diagrams ( , ∅), ( , ∅), ( , ), (∅, ), (∅, ) which we note respectively ([

1 2 ], ∅), ([2], ∅), ([1], [1]), (∅, [1 2 ]
) and (∅, [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]). The representations labeled by the above representations are of respective dimensions 1, 1, 2, 1 and 1. A basis for the representation labeled by the double-partition ([1], [1]) is given by the standard double-tableaux ( 1 , 2 ) ( 2 , 1 ). We will give the explicit models in the following chapters.

The representations for Coxeter groups of exceptional types are afforded by W -graphs, they will be explained in Chapter 6.

In the finite field setting, we will encounter some field automorphism through which the representations will factor. We give here a condition for a field automorphism to give a representation when composed with a representation of an Iwahori-Hecke algebra.

Proposition 2.1.2. Let (W, S) be a Coxeter system and H (αs) s∈S be a finite Iwahori-Hecke F p ((α s ) s∈S )-algebra. Let Φ ∈ Aut(F q ), where F q = F p ((α s ) s∈S ) and ρ be a finite dimensional representation of H (αs) s∈S . We then have that there exists a character η such that

(Φ • ρ) ⊗ η is a representation of H (αs) s∈S if and only if for all s ∈ S, Φ(α s ) ∈ {α s , α -1 s }. Moreover, if for all s ∈ S, Φ(α s ) ∈ {α s , α -1 s }, then η(T s ) = 1 if Φ(α s ) = α s and η(T s ) = -α s if Φ(α s ) = α -1 s .
Proof. Assume first that for all s ∈ S, we have Φ(α s ) ∈ {α s , α -1 s }. It is clear that the braid relations are verified so we only need to check that the deformations of the relations of order 2 are verified. Let s ∈ S, assume Φ(α s ) = α s . Let η(T s ) = 1, we then have

(Φ(ρ(T s ))η(T s ) -α s )(Φ(ρ(T s ))η(T s ) + 1) = Φ((ρ(T s ) -α s )(ρ(T s ) + 1)) = Φ(0) = 0. Assume Φ(α s ) = α -1 s , let η(T s ) = -α s , we then have (Φ • ρ)(T s ))η(T s ) -α s )(Φ • ρ(T s )η(T s ) + 1) = Φ((-α -1 s ρ(T s ) -α -1 s )(-α -1 s ρ(T s ) + 1)) = Φ(-α -2 s (ρ(T s ) + 1)(ρ(T s ) -α s )) = Φ(0) = 0. This proves that (Φ • ρ) ⊗ η is a representation of H (αs) s∈S .
Assume now that there exists a character η such that (Φ • ρ) ⊗ η is a representation of H (αs) s∈S . Let s ∈ S, the eigenvalues of Φ(ρ(T s ))η(T s ) are then α s and -1. We also have that the eigenvalues of Φ(ρ

(T s ))η(T s ) are η(T s )Φ(α s ) and -η(T s ). It follows that either -η(T s ) = -1 and η(T s )Φ(α s ) = α s or -η(T s ) = α s and η(T s )Φ(α s ) = -1. This implies that either Φ(α s ) = α s or Φ(α s ) = α -1
s . The proof is thus concluded.

Symmetric algebras and specializations

In this section, we recall the definition and some basic properties of symmetric algebras. We then give a version of Tits deformation theorem, which we will need to prove that the finite Iwahori-Hecke algebras are split semisimple under the right conditions. We then give a general result on representations of finite Iwahori-Hecke algebras, which we will use throughout the thesis.

Definition and first properties

In this subsection, we define symmetric algebreas and give some elementary properties of the corresponding trace. The results in this section are taken from Chapter 7 of [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]. Throughout this section A is a commutative ring and H is an A-algebra of finite rank.

H W,R,(us) s∈S is then a symmetric R[u ±1
s ] algebra with symmetrizing trace τ defined by τ

(T 0 ) = 1 and τ (T σ ) = 0 if σ = 1 W .
In what follows H is a symmetric A-algebra of finite rank and τ is a trace of H. For σ = s i 1 . . . s in a reduced expression of σ, we define u σ = u s i 1 . . . u s in . For the Iwahori-Hecke algebra H W,R,(us) s∈S , (T σ ) σ∈W is a basis. Its dual basis with regards to τ is (u σ -1 T σ -1 ).

We now consider a fixed basis B of H in what follows.

Definition 2.2.3. Let V and V be two right H-modules. For all ϕ ∈ Hom A (V, V ), we define

I(ϕ) from V to V by I(ϕ)(v) = b∈B ϕ(vb) b.
We consider A as a subset of H by considering that for a ∈ A, a = a.1 H . Proposition 2.2.2. Let V and V be two right H-modules and ϕ ∈ Hom A (V, V ). We then have that I(ϕ) is independant of the chosen basis and I(ϕ) ∈ Hom H (V, V ).

The previous proposition shows that I gives us a way to obtain H-linear map from A-linear maps.

Proposition 2.2.3. Let V, V and V be right H-modules and ϕ ∈ Hom H (V, V ), ψ ∈ Hom A (V , V ) and θ ∈ Hom A (V , V ). We have

I(ψ • ϕ) = I(ψ) • ϕ, I(ϕ • θ) = ϕ • I(θ).
Definition 2.2.4. We say an H-module V is projective if for every surjective H-module morphism from M to V with M an H-module, there exists i ∈ Hom H (V, M ) such that π • i = id V . Lemma 2.2.1 (Gaschütz-Ikeda). Let V be an H-module which is projective as an A-module. V is projective as an H-module if and only if there exists ϕ ∈ End A (V ) such that I(ϕ) = id V .

Schur elements

In this subsection, we recall the definition of Schur elements which will give us tools to understand the specializations to finite fields of the irreducible representations.

Let K be a field and H a symmetric K-algebra with symmetrizing trace τ . We will use the map I in order to define the Schur elements. We fix a K-basis B of H. Definition 2.2.5. An H-module V is sais to be split semi-simple if it is simple and we havedim K (End H (V )) = 1.

A K-algebra H is said to be split if all its simple modules are split.

Theorem 2.2.1. Let V be a split simple H-module. There exists a unique c V ∈ K such that for all ϕ ∈ End K (V ), we have

I(ϕ) = c V T r(ϕ)id V and c V only depends on the isomorphism class of V . c V is
= n V = dim(V ) and m = n V = dim(V ). Let ρ(resp ρ ) be a representation of H in M n (K)(resp M m (K)), we then have for all (i, l, k, j) ∈ [[1, n]] 2 × [[1, m]] 2 b∈B ρ(b) i,l ρ ( b) k,j = δ i,j δ k,l c V if V is isomorphic to V' ρ = ρ , = 0 if V is not isomorphic to V'.
We now recall Wedderburn's theorem on semi-simple algebras.

Theorem 2.2.2. (Wedderburn)

Let H be a finite dimensional split semi-simple K-algebra. We then have

H = V H(V )
, where the sum is over the isomorphism classes of simple H-modules. In the above,

H(V ) M n V (D V ),
where we have

D V = End H (V ).
We now give a semi-simplicity criteria with a condition on the Schur elements.

Theorem 2.2.3. A split semi-simple H-module V is projective if and only if c

V = 0.
If H is split then H split semi-simple if and only if all its Schur elements are non-zero and we then have

τ = V 1 c V χ V .
where the sum is over the isomorphism classes of H-modules.

Specialization theorem

We now give a proposition which we will use to show that the specializations of the Iwahori-Hecke algebras to finite fields remain split semisimple under the right conditions. This is a version from specialization theory of symmetric algebras and Schur elements which we have not found as such in the litterature. Let θ be a ring homomorphism from A to a field L such that L is the field of fractions of θ(A).

Let O be a valuation ring of F such that A ⊂ O and J (O) ∩ A = ker(θ), where J (O) is the unique maximal ideal of O. Let k = O/J (O) be the residue field of O. The restriction to A of the projection π from O to k has kernel equal to J (O) ∩ A = ker(θ). We can therefore see L as a subfield of k.

Let F be the field of fractions of A and B the subring of F formed of elements of the form a 1 a 2 , (a 1 , a 2 ) ∈ A 2 and θ(a 2 ) = 0. Assume F H is split. We can define an extension θ of θ to B. Assume there exists a representation of every simple module V such that for (i, We then have that LH is split semi-simple if and only if θ(c V ) = 0 for every simple F Hmodule V Proof. Assume θ(c V ) = 0 for every simple F H-module V . We then have that c V = 0 for all F H-module simple V .

j) ∈ [[1, n]] 2 , there exists b ∈ B such that θ(ρ V (b) j,i ) = 0.
Let B be a basis as in the theorem, by Proposition 7.2.7 of [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF],

e i,j,V = 1 c V b∈B ρ V (b) j,i b i,j,V is a basis of BH corresponding to the isomorphism BH V M n V (B).
We have that the θ(e i,j,V ) are in LH. Let us show that they give us a basis affording an explicit isomorphism from LH to V M n V (L). Since L is the field of fractions of θ(A), θ is a surjective morphism from B to L and the θ(e i,j,V ) give us a generating family of LH.

We have the relations θ(e i,j,V ) θ(e i ,j ,V ) = δ V,V δ j,i θ(e i,j ,V ).

Let (a

i,j,V ) ∈ L V n 2 V , assume V 1≤i,j≤n V a i,j,V θ(e i,j,V ) = 0. Let V 0 be a simple F H-module and 1 ≤ i 0 , i 1 , j 0 , j 1 ≤ n V 0 .
Multiplying by θ(e i 0 ,j 0 ,V 0 ) on the right and θ(e i 1 ,j 1 ,V 0 ) on the left, we have that 0

= V 1≤i,j≤n V a i,j,V θ(e i 1 ,j 1 ,V 0 ) θ(e i,j,V ) θ(e i 0 ,j 0 ,V 0 ), 0 = n V 0 i=1 a i,i 0 ,V 0 θ(e i 1 ,j 1 ,V 0 ) θ(e i,j 0 ,V 0 ) = a j 1 ,i 0 ,V 0 θ(e i 1 ,j 0 ,V 0 ).
Since each θ(e i,j,V ) is non-zero, these vectors are linearly independent and the relations they verify give us an isomorphism from LH to

V M n V (L). It follows that LH is split semi-simple.
Assume now that LH is split semi-simple. By Tits's deformation theorem (Theorem 7.4.6 of [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]), there is an isomorphism d k L between the Grothendieck groups of finite dimensional LHmodules and finite dimensional kH-modules sending each isomorphism class of simple modules to a unique isomorphism class of simple modules. If V is a simple LH-module, the unique

kH-module V such that [V ] = d k L ([V ]) verifies dim k (V ) = dim L (V ). We have dim k (kH) = dim L (LH) = V dim L (V ) 2 = V dim k (V ) 2 = dim k (kh/ rad(kH)).
kH is therefore semi-simple. Since LH is split, kH is split too. By Tits's deformation theorem applied to the specialization π from O to k, F H is semi-simple and there is an isomorphism d π sending isomorphism classes of F H-simple modules and isomorphism classes of simple kHmodules.

Let V be a F H-module and

ρ V a representation of F H in M n V (F ) such that for all h ∈ H, we have ρ V (h) ∈ O. By Corollary 7.2.2 of [20], we have that b∈B ρ V (b) 1,1 ρ V ( b) 1,1 = c V , b∈B π(ρ V (b) 1,1 )π(ρ V ( b) 1,1 ) = π(c V ) = θ(c V ).
This equality corresponds to the relation verified by the Schur element of a kH simple module V = k Ṽ . We then have c V = θ(c V ), therefore θ(c V ) = 0 because kH is split semisimple. This concludes the proof.

Maximal subgroups of classical groups

In this section, we recall Aschbacher's theorem on maximal subgroups of classical groups [4] and describe the different classes. We then give some important theorems about subgroups of classical groups which we will use throughout the thesis.

We first recall the following definitions from [5] Definition 2.

H = G 0 ⊂ G 1 ⊂ ... ⊂ G n = G and for all i ∈ [[1, n]], G i-1 G i A group G is quasisimple if it is perfect and G/Z(G) is simple.
The components of a group G are its subnormal quasisimple subgroups. Write Comp(X) for the set of components of X and set E(X) = Comp(X) .

The Fitting subgroup F (G) of G is the largest nilpotent normal subgroup of G.

The generalized Fitting group of G is the group

F (G) = F (G)E(G).
We begin by stating Aschbacher's theorem on maximal subgroups [4] Theorem 2.3.1. Let G be a finite group whose generalized Fitting group is a simple classical group G 0 over a finite field such that G 0 P Ω + 8 (q). Let H be a proper subgroup of G such that G = HG 0 . Then either H belongs to one of the geometric classes (C i ) i∈[ [1,[START_REF] Brauer | On the modular characters of groups[END_REF]] or H belongs to the class S.

We will not use this theorem directly but results that follow from this theorem. In particular, we will use many tables from [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] of the maximal subgroups of classical groups in low dimension, where all the conjugacy classes of maximal subgroups are determined. In table 2.1, we copy table 2.1 from [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] giving a rough description of the Aschbacher classes for classical groups over F q (the unitary groups SU n (q) are defined over F q 2 ).

C i

Rough description C 1 stabilizers of totally singular or non-singular subspaces

C 2 stabilizers of decompositions V = t ⊕ i=1 , dim(V i ) = n t C 3 stabilizers of extension fields of F q of prime index dividing n C 4 stabilizers of tensor product decompositions V = V 1 ⊗ V 2 C 5
stabilizers of subfields of F q of prime index C 6 normalizers of symplectic-type or extraspecial groups in absolutely irreducible representations

C 7 stabilizers of decompositions V = t ⊗ i=1 V i , dim(V i ) = a, n = a t
C 8 groups of similarities of non-degenerate classical forms can be included in a maximal group of class C 5 . We will use this type of argument to reduce the lists of possible maximal groups containing the groups we consider for groups of low dimension in the following sections.

The main theorem we will use in high dimension is a theorem by Guralnick and Saxl [START_REF] Guralnick | Generation of finite almost simple groups by conjugates[END_REF] giving conditions for subgroups of the special linear group to be classical groups. The proof of Theorem 2.3.2 uses the classification of finite simple groups. Theorem 2.3.2 (Gulralnick-Saxl). Let V be a finite-dimensional vector space of dimension d > 8, d = 10 or (d = 10 and p = 2) over an algebraically closed field F p of characteristic p > 0. Let G be a primitive tensor-indecomposable finite irreducible subgroup of GL(V ). We write v G (V ) the minimal dimension of [βg, V ] = (βg -1)V , for g ∈ G and β ∈ F p such that βg = 1. We then have either v

G (V ) > max(2, √ d
2 ) or one of the following assertions.

1. G is a classical group in a natural representation.

2. G is the alternating or the symmetric group of degree c and V is the permutation module of dimension c -

1 or c -2.
The other theorems we will use are on groups generated by long-root elements, we first recall the definition of long-root elements.

Definition 2.3.2. If G SL n (q), G SU n (q) or G SP n (q) then a long-root element of G is a transvection. If G Ω + n (q), n ≥ 4 then a long-root element x of G is an element of the form x(v) = v-v, a b+ v, b a for a, b
in a totally singular 2-space T and ., . the non-degenerate symmetric bilinear form associated to G.

For example, if n = 2m ≥ 4 is an integer then elements conjugate to elements of the form T a 0 0 t (T a ) -1 , where T a is a transvection of SL m (q) are long-root elements of Ω + 4 (q).

We can now state Kantor's Theorem on subgroups of orthogonal groups generated by long root elements [START_REF] Kantor | Subgroups of classical groups generated by long root elements[END_REF]. We only consider the irreducible subgroups of Ω + n (q) with q an odd prime because it will be the only case we will need to consider in this thesis.

Theorem 2.3.3. Let G be an irreducible subgroup of Ω + n (q), where n ≥ 4 and q = p s for some prime p and positive integer s generated by a conjugacy class of long root elements, such that

O p (G) ≤ [G, G] ∩ Z(G).
We then have that G belongs to the following list 1. Ω ± (q ) in a natural representation over F q , q |q,

2. Ω - 2m (q 1
2 ) ≤ Ω + 2m (q ), q |q, n = 2m in a natural representation over F q , 3. SU 2m (q ) ≤ Ω + 4m (q ), n = 4m in a natural representation over F q , q |q 4. SU 2m+1 (q ) ≤ Ω 4m+2 (q ), n = 4m + 2 in a natural representation over F q , q |q,

5. G/Z(G) P Ω(7, q ), |Z(G)| = (2, q -1), G ≤ Ω + 8 (q ) in a natural representation over F q , q |q, 6. 3 D 4 (q ) ≤ Ω + 8 (q 3
) in a natural representation over F q 3 , q 3 |q.

Finally we give a Theorem by Serezkin and Zaleskii for irreducible groups of the remaining classical groups generated by transvections. (First theorem of [START_REF] Serežkin | Linear groups generated by transvections[END_REF]). Theorem 2.3.4. If G is an irreducible subgroup of GL n (q) generated by transvections with q = p r , p > 3, n > 2, then G is conjugate inside GL n (q) to SL n (q), Sp n (q) or SU n (q 1 2 ) for some q dividing q.

Chapter 3

Type B

In this section, we will determine the image of the derived subgroup of the Artin group of type B n . This is a natural continuation of the work for type A n in [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] and [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. The general outline of the proof is based on inductive reasoning. We first give the factorizations through field automorphisms and the transposed inverse automorphisms between irreducible representations. The main result concerning those factorizations is given in Proposition 3.2.4. This allows us to determine what the image in the full finite Iwahori-Hecke algebra appears to be. The remainder of the section is the proof that the image appearing is indeed the image of the derived subgroup A Bn of the Artin group. The main results of this section are given in Theorems 3.2.1 to 3.2.6.

Let p be a prime, n ≥ 2 be an integer, α ∈ F p of order a greater than n and not in {1, 2, 3, 4, 5, 6, 8, 10} and

β ∈ F p \ {-α i , -(n -1) ≤ i ≤ n -1} different from 1. We set F q = F p (α, β).
The Artin group of type B is the group generated by the elements T = S 0 , S 1 , . . . , S n-1 verifying the relation

S 0 S 1 S 0 S 1 = S 1 S 0 S 1 S 0 , for i ∈ [[1, n -2]], S i S i+1 S i = S i+1 S i S i+1 and for (i, j) ∈ [[0, n -1]] such that |i -j| ≥ 2, S i S j = S j S i .
The associated Iwahori-Hecke F q -algebra H Bn,α,β is defined by the generators indexed in the same way as for the Artin group and verifying the previous relations and deformations of the relations of order 2 of the Coxeter groups : (T -β)(T +1) = 0 and for i ∈ [ [1, n-1]], (S i -α)(S i +1) = 0. In the sequel, we identify the Artin group with its image inside the Iwahori-Hecke algebra. We write 1 , 2 for the length functions on

A Bn = T, S i i∈[[1,n-1]] such that for all i ∈ [[1, n -1]], 1 (S i ) = 1, 1 (T ) = 0, 2 (S i ) = 0 and 2 (T ) = 1.
In Section 3.1 we give in Theorem 3.1.1 the irreducible representations described by the Hoefsmit model given in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] and [START_REF] Hoefsmit | Representations of Hecke algebras of finite groups with BN-pairs of classical type[END_REF] and define a weight on standard tableaux and doublepartitions of n which allows us to define in Proposition 3.1.1 a bilinear form verifying nice properties.

In Section 3.2, we determine all the isomorphisms between different irreducible representations and then state the main results for type B in Theorems 3.2.1 up to 3.2.6.

In Section 3.3, we prove the result in all possible cases depending on the properties of the field extensions F q of F p (α + α -1 , β + β -1 ) and F p (α) of F p (α + α -1 ).

Hoefsmit model and first properties

In this section, we define the matrix model we will be considering throughout this section and define a natural bilinear form appearing when we consider those models. This will establish the groundwork for the rest of this section. Theorem 3.1.1. Assume α is of order greater than n and β ∈ F p \{-α i , -(n-1) ≤ i ≤ n-1}. The following matrix model gives a list of the pairwise non-isomorphic absolutely irreducible modules V λ of the Iwahori-Hecke algebra H Bn,α,β labeled by double-partitions of n.

If T = (T 1 , T 2 ) is a standard double-tableau, then • if 1 ∈ T 1 , then T.T = βT, • if 1 ∈ T 2 , then T.T = -T, • S i .T = m i (T)T + (1 + m i (T)) T, where T = T i↔i+1 if T i↔i+1 is standard and 0 otherwise. Above, we have m i (T) = α-1 1- ct(T:i) ct(T:i+1)
, ct(T : j) = α c j (T)-r j (T) β if j ∈ T 1 and ct(T : j) = -α c j (T)-r j (T) otherwise and r j (T) (resp c j (T)) is the row (resp column) of j in the tableau of T containing j.

Proof. Let λ = (λ 0 , λ 1 ) n such that n λ = dim(V λ ) > 1.
We will show that ρ λ (A Bn ) contains diagonal matrices generating the algebra of diagonal matrices in GL n (q). By Proposition 5 of [START_REF] Marin | Irréductibilité générique des produits tensoriels de monodromies[END_REF], it is sufficient to show that for any standard double-tableau T associated to λ, there exist two diagonal invertible matrices D 1 and D 2 in the basis of standard double-tableaux such that

D 1 T = b 1 T and D 2 T = b 2 T and b 1 = b 2 .
By [START_REF] Marin | Irréductibilité générique des produits tensoriels de monodromies[END_REF], it will then be sufficient to show that for any couple of standard double-tableaux (T ρ , T γ ), there exists a matrix in ρ λ (A Bn ) such that a ρ,γ is non-zero.

In order to do that, we will use the Jucys-Murphy elements (a 1 , ..., a n ) whose expressions in this model are given in [3,Prop 3.16] and are a i .T ρ = u τ (i) α c i -r i +i-1 T ρ , where i is in box (r i , c i ) of the τ (i)-th tableau of T ρ and (u 1 , u 2 ) = (β, -1). (1) or (j ∈ T ρ,τ (1) and (r j , c j ) = (j, 1))} and we have ≥ 2.

Let T ρ = (T ρ 1 , T ρ 2 ) ∈ λ. Since dim(V λ ) > 1, we can define = min{j ∈ [[1, n]], j / ∈ T ρ,τ
We then have k ∈ T ρ,τ (1) and (r k , c k ) = (k, 1) for all k < . It follows that either ∈ T ρ,3-τ (1) and (r , c ) = (1, 1) or ∈ T ρ,τ (1) and (r , c ) = (1, 2). We have

a 1 .T ρ = u τ (1) T ρ and a .T ρ ∈ {u 3-τ (1) α -1 T ρ , u τ (1) α T ρ }, therefore a 1 .T ρ = a .T ρ since the order of α is strictly greater than n and β / ∈ {-α i , -(n -1) ≤ i ≤ n -1}.
Let now (T ρ , T γ ) be a pair of standard double-tableaux associated to λ. There exists a permutation of S n which affords T γ after permuting the numbers inside T ρ . We can decompose this permutation in a product of transpositions (i, i + 1) such that the path given by the successive standard double-tableaux is composed only of standard double-tableaux. It is thus sufficient to show that for any i ∈ [ [1, n]] and for any standard double-tableau T ρ such that T ρ,i↔i+1 is standard, we have 1 + m i (T ρ ) = 0.

We have

m i (T ρ ) = α-1 1- u τ (i) α c i -r i +r i+1 -c i+1 u τ (i+1) = -1 ⇔ u τ (i) u τ (i+1) = α 1+c i+1 -r i+1 +r i -c i and u τ (i) u τ (i+1) ∈ {1, -β, -β -1 }.
By the assumptions made on α and β, it now only remains to show that -n ≤ 1 + c i+1 -r i+1 + r i -c i ≤ n and 1 + c i+1 -r i+1 + r i -c i = 0 if i and i + 1 are in the same tableau and -n

+ 1 ≤ 1 + c i+1 -r i+1 + r i -c i ≤ n -1 if i and i + 1 are in different tableaux.
If i and i + 1 are in the same tableau, |c i+1 -r i+1 + r i -c i | + 1 represents the number of boxes to go through vertically and horizontally in order to go from i to i + 1, therefore we have the required bounds. Moreover, this quantity is non-zero by the conditions for a tableau to be standard and the fact that T ρ,i↔i+1 is standard.

If i and i+1 are in different tableaux of respective sizes j and n-j, we have 1-j

≤ c i -r i ≤ j -1 and 1 -(n -j) ≤ r i+1 -c i+1 ≤ n -j -1 since 1 ≤ c i , r i ≤ j and 1 ≤ c i+1 , r i+1 ≤ n -j. It follows that -n + 1 ≤ 3 -n ≤ 1 + c i+1 -r i+1 + r i -c i ≤ n -1.
The representations V λ are therefore irreducible. We have the same dimensions over C by taking α and β to be irreducible parameters and the same mode. It follows that we only need to show that those modules are pairwise non-isomorphic to conclude the proof.

Let us show that R λ R µ ⇔ λ = µ. Assume R λ R µ
, the eigenvalues of the Jucys-Murphy elements must be the same for both representations. We then have that for any

i ∈ [[1, n]], {u τ T (i) α c i,T -r i,T +i-1 , T ∈ λ} = {u τ T(i) α c i, T-r i, T+i-1
, T ∈ µ} and each of the elements of those sets appear with the same multiplicity. We will show that this implies that for any anda(i, λ, β, γ) be the above quantity.

i ∈ [[1, n]]], γ ∈ Z, we have card{T ∈ λ, u τ T (i) = β, c i,T -r i,T = γ} = card{ T ∈ µ, u τ T(i) = β, c i, T -r i, T = γ} Let λ = (λ 1 , λ 2 ), µ = (µ 1 , µ 2 ), λ 1 = (λ 1,l ) l∈N , µ 1 = (µ 1,l ) l∈N
We first show that r 1 = r 2 , where λ 1 r 1 and λ 2 r 2 . We have

a(1, λ, β, 0) = n-1 n-r 1 M λ 1 M λ 2 ,
where M λ i is the number of standard tableaux associated to the partition λ i . This is true because counting the number of double-tableaux with 1 in the left tableau is equivalent to choosing n -r -1 numbers in [ [2, n]] and counting the number of ways the numbers can be arranged in each tableau to get a standard double-tableau. In the same way, wa have a(1, λ, -1, 0) = n-1 r 1 M λ 1 M λ 2 . Since 1 is either in the first row and first column of the left tableau or in the first row and first column of the right tableau for any standard double-tableau, we have a(1, λ, β, 0) = a(1, µ, β, 0) and a(1, λ, -1, 0) = a(1, µ, -1, 0). It follows that

n -1 n -r 1 M λ 1 M λ 2 n -1 r 2 M µ 1 M µ 2 = n -1 r 1 M λ 1 M λ 2 n -1 n -r 2 M µ 1 M µ 2 (n -1)!(n -1)! (n -r 1 )!(r 1 -1)!(n -1 -r 2 )!r 2 ! = (n -1)!(n -1)! r 1 !(n -r 1 -1)!(n -r 2 )!(r 2 -1)! 1 (n -r 1 )r 2 = 1 r 1 (n -r 2 ) nr 2 -r 1 r 2 = nr 1 -r 1 r 2 r 1 = r 2 .
In order to show that the quantities are equal, it is sufficient to show that for any

i ∈ [[1, n]], T ∈ λ and T ∈ µ, we have u τ T (i) α c i,T -r i,T +i-1 = u τ T(i) α c i, T-r i, T+i-1 if and only if u τ T (i) = u τ T(i) and c i,T -r i,T = c i, T -r i, T. Let T ∈ λ, T ∈ µ such that for all i ∈ [[1, n]], u τ T (i) α c i,T -r i,T +i-1 = u τ T(i) α c i, T-r i, T+i-1 . Let i ∈ [[1, n]].
Assume by contradiction that u τ T (i) = u τ T(i) . We then have that

α c i, T-r i, T+r i,T -c i,T = u τ T (i) u τ T(i) ∈ {-β, -β -1 }.
Since i is in a tableau on a different side for T and T and r = r λ = r µ , we have that either

1 -r ≤ r i,T -c i,T ≤ r -1 and 1 -(n -r) ≤ c i, T -r i, T ≤ n -r -1 or 1 -(n -r) ≤ r i,T -c i,T ≤ n -r -1 and 1 -r ≤ c i, T -r i, T ≤ r -1. In both cases, we have 2 -n ≤ c i, T -r i, T + r i,T -c i,T ≤ n -2. By assumption, we have -β / ∈ {-α i , 1 -n ≤ i ≤ n -1}, therefore it follows by contradiction that u τ T (i) = u τ T(i) .
Let us show now that λ 1,1 = µ 1,1 and m λ = m µ . Assume by contradiction that µ 1,1 < λ 1,1 . We write i = µ 1,1 + 1. By the equality of the eigenvalues of the Jucys-Murphy elements, there exists T ∈ µ such that u τ T(i) α c i, T-r i, T = βα i-1 since there exists a standard double-tableau associated to λ with µ 1,1 +1 in box (1, µ 1,1 +1) of the left tableau. By the above reasonning, we get that u τ T(i) = β, therefore α µ 1,1 +r i, T-c i, T = 1. For any standard double-tableau associated to µ and for any number i in a box of its left tableau, we have

c i -r i ≤ µ 1,1 -1 < µ 1,1 , therefore 0 < µ 1,1 + r i, T -c i, T ≤ 2n < 2a
, where a is the order of α. It follows that µ 1,1 + r i, T -c i, T = a. We have in the same way that r i, T -c i, T ≤ m µ -1, where m µ is the number of boxes in the first column of µ 1 . Recall that by assumption, we have

a > n. It follows that n < a = µ 1,1 + r i, T -c i, T ≤ µ 1,1 + m µ -1.
We have that µ 1,1 + m µ -1 is equal to the number of boxes to go through horizontally and then vertically in order to get from box (1, µ 1,1 ) to box (m µ , 1), therefore µ 1,1 + m µ -1 ≤ n. This implies that n < n and we get by contradiction that µ 1,1 ≥ λ 1,1 . In the same way, we get

λ 1,1 ≥ µ 1,1 , therefore λ 1,1 = µ 1,1 .
Assume by contradiction that m µ < m λ . We write i = m µ + 1. There exists T ∈ µ such that u τ T(i) α c i, T-r i, T = βα 1-i since there exists a standard double-tableau associated to λ with m µ +1 in box (m µ +1, 1). It follows that α c i, T-r i, T+mµ = 1. We then get that 0

< c i, T -r i, T +m µ < 2n < 2a since c i, T -r i, T ≥ 1 -m µ > -m µ . It follows that n < a = c i, T -r i, T + m µ ≤ µ 1,1 + m µ -1 ≤ n.
This is a contradiction, therefore m λ ≤ m µ and in the same way m µ ≤ m λ , therefore m µ = m λ . We can now complete the proof of the equalities of the quantities.

Let T ∈ λ, T ∈ µ such that for all i ∈ [[1, n]], u τ T (i) α c i,T -r i,T +i-1 = u τ T(i) α c i, T-r i, T+i-1 . Let i ∈ [[1, n]], we have that u τ T (i) = u τ T(i) and α c i,T -r i,T +r i, T-c i, T = 1. We have that 1 -m λ ≤ c i,T -r i,T ≤ λ 1,1 -1 and 1 -m µ ≤ c i, T -r i, T ≤ µ 1,1 -1 so 1 -n ≤ 2 -m λ -λ 1,1 = 2 -m λ -µ 1,1 ≤ c i,T -r i,T + r i, T -c i, T ≤ λ 1,1 + m µ -2 = λ 1,1 + m λ -2 ≤ n -1.
It follows by the assumptions on the order of α that c i,T -r i,T = c i, T -r i, T.

We have proved for all i ∈ [ [1, n]] and γ ∈ Z that a(i, λ, β, γ) = a(i, µ, β, γ) and λ 1,1 = µ 1,1 . We will prove that this implies that

λ 1 = µ 1 . Let (λ 1,l i , l i ) 1≤i≤s λ (resp (µ 1,k i , k i ) 1≤i≤sµ ) be the extremal boxes of λ 1 (resp µ 1 ). For all i, j ∈ [[1, s λ ]] (resp [[1, s µ ]]), i < j, we have that λ 1,l i -l i > λ 1,l j -l j (resp µ 1,k i -k i > µ 1,k j -k j ). We have that for all i ∈ [[1, s λ ]], 0 = a(n, λ, β, λ 1,l i -l i ) = a(n, µ, β, λ 1,l i -l i ) and for all i ∈ [[1, s µ ]], 0 = a(n, µ, β, µ 1,k i -k i ) = a(n, λ, β, µ 1,k i -k i ) since
i ∈ [[1, s λ ]], λ 1,l i -l i = µ 1,k i -k i . Since λ 1,l 1 = µ 1,k 1 , we have that k 1 = l 1 .
Let us show by induction that for all i ∈ [[1, s λ ]], we have

l i = k i and λ 1,l i = µ 1,k i . Let j ≥ 2, assume that for all i ≤ j -1, (λ 1,l i , l i ) = (µ 1,k i , k i ).
Assume by contradiction that µ 1,k j < λ 1,l j . We then have k j < l j since µ 1,k j -k j = λ 1,l j -l j . We have that a(n -(k jl j-1 ), µ, β, µ 1,k j -l j-1 ) = 0 since there are k j -l j-1 boxes below (l j-1 , µ 1,k j ) and l j -l j-1 boxes to its right and no boxes both below and to its right. On the other hand, there exist standard double-tableaux associated to λ such that n -(k j -l j-1 ) is in box (l j -k j + l j-1 , λ 1,l j ) because l j > k j . It follows that a(n -(k j -l j-1 ), λ, β, λ 1,l j -(l j -k j + l j-1 )) = 0. We have that λ 1,l j -(l j -k j + l j-1 = µ 1,k j -k j + k j -l j-1 = µ 1,k j -l j-1 , therefore we get a contradiction. It follows that µ 1,k j ≥ λ 1,l j and in the same way, λ 1,l j ≥ µ 1,k j , therefore λ 1,l j = µ 1,k j . We then have that k j = l j because λ 1,l j -l j = µ 1,k j -k j .

We can then conclude by induction that λ 1 have µ 1 the same extremal boxes. This proves that λ 1 = µ 1 because a partition is completely determined by its extremal boxes. We get in the same way that λ 2 = µ 2 , therefore λ = µ.

We now generalize the work done in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] for type A to type B, that is we define a bilinear form which is fixed by the image of the derived subgroup of the Artin group of the Iwahori-Hecke algebra. We define a weight by ω(T) = ω 1 (T)ω 2 (T)ω 3 (T), where ω

1 (T) = i<j,i∈T 1 ,j∈T 2 (-1), ω 2 (T) = i<j,i,j∈T 2 ,r i (T)>r j (T)
(-1) and ω 3 (T) = i<j,i,j∈T 1 ,r i (T)>r j (T) (-1). We then define a bilinear form (.|.) by (T| T) = ω(T)δ T , T, where T = (T 2 , T 1 ) for T = (T 1 , T 2 ). In the same way λ = (λ 2 , λ 1 ) is the transpose of λ = (λ 1 , λ 2 ).

If µ is a partition of an integer m with diagonal size b(µ) = max{i,

µ i ≥ i}, we let ν(µ) = 1 if m-b(µ)
2 is even and -1 otherwise. Note that by Lemma 6 of [START_REF] Marin | L'algèbre de Lie des transpositions[END_REF]

, if µ = µ then m-b(µ) 2 is an integer and ν(µ) = (-1) m-b(µ) 2
The first case is T = T . We write a = c i -r i + r i+1 -c i+1 with (r i , c i ) and (r i+1 , c i+1 ) the boxes in T, we then have

(S i .T|S i . T) = m i (T)m i (T )ω(T) + (1 + m i (T))(1 + m i (T ))ω(T i↔i+1 ) = -ω(T)(1 + m i (T) + m i (T )) = -ω(T)   1 + α -1 1 + u τ (i) u τ (i+1) α a + α -1 1 + u τ (i+1) u τ (i) α -a   = -ω(T)1 + (α -1)   1 + u τ (i) u τ (i+1) α a + 1 + u τ (i+1) u τ (i) α -a 1 + u τ (i) u τ (i+1) α a + u τ (i+1) u τ (i) α -a + 1   = -α(T| T).
The second case is T = T i↔i+1 , we then have

(S i .T|S i . T) = (1 + m i (T))(m i (T i↔i+1 ))ω(T i↔i+1 ) + m i (T)(1 + m i (T i↔i+1 ))ω(T) = -ω(T)(m i (T i↔i+1 ) -m i (T)) = -ω(T)    α -1 1 + u τ (i) α r i+1 -c i+1 u τ (i+1) α r i -c i - α -1 1 + u τ (i) α c i -r i u τ (i+1) α c i+1 -r i+1    = 0 = -α(T| T).
This concludes the proof of 1. 2 follows from 1.

3. By definition, the bilinear form is non-degenerate. Assume λ = (λ 1 , λ 2 ) = λ . We consider T = (T 1 , T 2 ) ∈ λ. Since substituting i 1 < i 2 < ... < i l in T 1 by 1 < 2 < ... < l does not change the product and the weight ω on T ∈ µ for µ n satisfies ω(T)ω(T ) = ν(µ) by Lemme 6 of [START_REF] Marin | L'algèbre de Lie des transpositions[END_REF], we have that ω(T)ω(T ) = ν(λ 1 )ν(λ 2 ) i<j,i∈T 1 ,j∈T 2 or i∈T 2 ,j∈T 1 (-1). The cardinality of the set {i < j, i ∈ T 1 , j ∈ T 2 or i ∈ T 2 , j ∈ T 1 } is equal to the number of pairs (i, j) with i in T 1 and j in T 2 , which equals ( n 2 ) 2 . It follows that ω(T)ω(T ) = ν(λ) for any standard double-tableau T associated to λ. For any pair (T, T), we have that

( T|T) = ω( T)δ T, T = ν(λ)ω( T )δ T, T = ν(λ)ω(T)δ T,T = ν(λ)(T| T).
The Witt index is positive since the basis can be partitioned in pairs (T|T ).

We remark that we have proved that ω(T)ω(T ) = ν(λ) for any double-partition λ and any standard double-tableau T in V λ .

Factorization of the image of the Artin group in the Iwahori-Hecke algebra

In this section, we first see how the different representations are related when restricted to the derived subgroup. We see some factorisations appear between the representation associated to a given double-partition λ and the representation associated to its transposed double-partition in Proposition 3.2.1. We also see that when the field extension is non-trivial, we have factorizations through field automorphism. All those results are summarized in Proposition 3.2.4. Finally, we see that all the representations associated to hook diagrams factor through two representations as is shown in Proposition 3.2.5.

Isomorphisms between representations

Let L ∈ End(V ) be defined by T → ω(T)T . We give a generalization of Lemma 3.2. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF].

Proposition 3.2.1. Let λ be a double-partition of n such that λ = λ (resp λ = λ ). The map L induces an endomorphism of V λ ⊕ V λ (resp V λ ) which switches V λ and V λ (resp leaves V λ stable) such that the actions of S r and T satisfy

LS r L -1 (-α) -1 = t S -1 r , LT L -1 (-β) -1 = t T -1 .
Proof. This follows directly from Proposition 3.1.1 by writing the matrix of the bilinear form and the matrix of L.

We now suppose

F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ).
We then have an F q -automorphism of order 2 such that (α) = α -1 and (β) = β -1 . In [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], a function d was defined on any standard tableau T associated to a partition λ of n by d(T) = i,j,r i >r j α c j -r j -α c i -r i +1 α c j -r j +1 -α c i -r i , where for i ∈ [ [1, n]], r i denotes the row of i in T and c i denotes the column of i in T.

Let ., . be the hermitian form defined by T, T = d(T)δ T, T, where

d(T) = d(T 1 ) d(T 2 ) i∈T 1 ,j∈T 2 ,i<j 2 + βα a i,j -1 + β -1 α 1-a i,j α + α -1 + βα a i,j + β -1 α -a i,j
with a i,j = c i -r i + r j -c j and d induced by the d defined in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] applied to T 1 and T 2 by seeing them as standard tableaux using the ordered bijections onto [ [1, r]] and [ [1, n -r]].

We now check that d(T) is well-defined and non-zero for any standard double-tableau. We prove in what follows that the big product in the expression of d is indeed well-defined and non-zero for any double-tableau with no empty components.

Let λ n, T = (T 1 , T 2 ) ∈ λ and (i, j) a pair of integers such that i < j, i ∈ T 1 and j ∈ T 2 . We set r to be the number of boxes of T 1 , we have 1

-r ≤ c i -r i ≤ r -1 and 1 -(n -r) ≤ c j -r j ≤ n -r -1, therefore 2 -n ≤ a = a i,j ≤ n -2. We have α + α -1 + βα a + β -1 α -a = α(1 + βα a-1 ) + α -a β -1 (1 + βα a-1 ) = α(1 + βα a-1 )(1 + α -a-1 β -1 ). This product never cancels because β / ∈ {-α i , 1 -n ≤ i ≤ n -1}. In the same way 2 + βα a-1 + β -1 α 1-a = (1 + βα a-1 )(1 + β -1 α 1-a ) never cancels.
Now we have defined this hermitian form, we can generalize Proposition 3.6 from [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF].

Proposition 3.2.2. The group A Bn acts in a unitary way on V with respect to this hermitian form and this form is non-degenerate on V λ for any double-partition λ of n. In particular, for any double-partition λ of n, there exists a matrix P ∈ GL n λ (q) such that P R λ (T )P -1 = (R λ (T )) = R λ (T ) and P R λ (S r )P -1 = (R λ (S r )).

Proof. The action of T is indeed unitary with regards to this hermitian form because β (β) = (-1) (-1) = 1. Let T be a standard double-tableau and r ∈

[[1, n -1]]. If τ T (r) = τ T (r + 1)
then the result is a consequence of Proposition 3.6. in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF].

We now assume that τ T (r) = τ T (r + 1), up to switching T and T r↔r+1 we can assume that τ T (r) = 1 and τ T (r + 1) = 2. It remains to show that T, T = S r .T, S r .T , T r↔r+1 , T r↔r+1 = S r .T r↔r+1 , S r .T r↔r+1 and S r .T, S r .T r↔r+1 = T, T r↔r+1 . In the following computation, we write a = a i,i+1 and T = T r↔r+1 . We have

S r .T, S r .T = m r (T) (m r (T))d(T) + (1 + m r (T)) (1 + m r (T))d( T) = d(T)(m r (T) (m r (T)) + α + α -1 + βα a + β -1 α -a 2 + βα a-1 + β -1 α 1-a (1 + m r (T)) (1 + m r (T))) = d(T) α -1 1 + βα a α -1 -1 1 + β -1 α -a + α + α -1 + βα a + β -1 α -a 2 + βα a-1 + β -1 α 1-a α + βα a 1 + βα a α -1 + β -1 α -a 1 + β -1 α -a = d(T) 2 -α -α -1 2 + βα a + β -1 α -a + α + α -1 + βα a + β -1 α -a 2 + βα a + β -1 α -a = d(T) = T, T .
We also have

S r . T, S r . T = m r ( T) (m r ( T))d( T) + (1 + m r ( T)) (1 + m r ( T))d(T) = d( T) α -1 1 + β -1 α -a α -1 -1 1 + βα a + 2 + βα a-1 + β -1 α 1-a α + α -1 + βα a + β -1 α -a α + β -1 α -a 1 + β -1 α -a α -1 + βα a 1 + βα a = d( T) 2 -α -α -1 2 + βα a + β -1 α -a + 4 + 2βα a+1 + 2β -1 α -a-1 + 2βα a-1 + β 2 α 2a + α -2 + 2β -1 α 1-a + α 2 + β -2 α -2a (α + α -1 + βα a + β -1 α -a )(2 + βα a + β -1 α -a ) = d( T) 2 -α -α -1 2 + βα a + β -1 α -a + (α + α -1 + βα a + β -1 α -a ) 2 (α + α -1 + βα a + β -1 α -a )(2 + βα a + β -1 α -a ) = d( T) = T, T .
Finally, we have

S r .T, S r . T = m r (T) (1 + m r ( T))d(T) + (1 + m r (T)) (m r ( T))d( T) = d(T) α -1 1 + βα a α -1 + βα a 1 + βα a + α + α -1 + βα a + β -1 α -a 2 + βα a-1 + β -1 α 1-a α + βα a 1 + βα a α -1 -1 1 + βα a = d(T) α -1 (1 + βα a ) 2 (2 + βα a-1 + β -1 α 1-a ) (α -1 + βα a )(2 + βα a-1 + β -1 α 1-a )- α -1 (α + α -1 + βα a + β -1 α -a )(α + βα a ) = d(T) α -1 (1 + βα a ) 2 (2 + βα a-1 + β -1 α 1-a ) 2α -1 + βα a-2 + β -1 α -a + 2βα a + β 2 α 2a-1 + α -α -α -1 -βα a -β -1 α -a -βα a -βα a-2 -β 2 α 2a-1 -α -1 = 0 = T, T .
We recall that A Bn = [A Bn , A Bn ] is the derived subgroup of A Bn . When it exists, we write for the automorphism of order 2 of F q = F p (α, β). Lemma 3.2.1. If λ is a double-partition of n, then the restriction of R λ to A Bn is absolutely irreducible.

Proof. Assume first it is true for n = 2. Since A Bn is generated by A B n-1 and A Bn , we have the result for n ≥ 3 by the same method as in the Lemma 3.4(i) of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF].

We now show the result is true for n = 2. We only have to show it for ( [1], [1]) since the other representations are 1-dimensional. We will show in Section 3.3 (Lemmas 3.3.2 and 3.3.8) that R [1], [1] (A B 2 ) SL 2 (q ) for some q . The irreducibility then follows.

We now show a lemma computing the normal closure of A Bn . This is a generalization to type B of Lemma 2.1 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] Lemma 3.2.2. For n ≥ 4, the normal closure

A B n-1 A Bn of A B n-1 in A Bn is A Bn . Proof.
G → k such that R 2 = R 1 ⊗ η. Proposition 3.2.3. If λ 1 a partition of n then R (λ 1 ,∅)|A Bn R (∅,λ 1 )|A Bn .
Proof. The action of T is diagonal and the action of S i on (T 1 , ∅) is identical to the one on (∅, T 1 ), therefore the proof of the result is straightforward.

We now give a proposition stating the different possible factorizations between doublepartitions of n. In Proposition 2.7 of [START_REF] Marin | Infinitesimal Hecke algebras II[END_REF], a version of this proposition is given in the generic case. Note that the different possible field extensions in the finite case yield many additional factorizations. Proposition 3.2.4. Let λ and µ be double-partitions of n with no empty components. We then have the following properties.

1. If F q = F p (α, β) = F p (α + α -1 , β + β -1 ), then (a) R λ|A Bn R µ|A Bn ⇔ λ = µ, (b) R λ|A Bn R µ|A Bn ⇔ λ = µ . 2. If F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ), then (a) R λ|A Bn R µ|A Bn ⇔ λ = µ, (b) R λ|A Bn R µ|A Bn ⇔ λ = µ , (c) R λ|A Bn • R µ|A Bn ⇔ λ = µ , (d) R λ|A Bn • R µ|A Bn ⇔ λ = µ. 3. If F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ), then (a) R λ|A Bn R µ|A Bn ⇔ λ = µ, (b) R λ|A Bn R µ|A Bn ⇔ λ = µ , (c) R λ|A Bn • R µ|A Bn ⇔ (λ 1 , λ 2 ) = (µ 1 , µ 2 ), (d) R λ|A Bn • R µ|A Bn ⇔ (λ 1 , λ 2 ) = (µ 2 , µ 1 ). 4. If F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ), then (a) R λ|A Bn R µ|A Bn ⇔ λ = µ, (b) R λ|A Bn R µ|A Bn ⇔ λ = µ , (c) R λ|A Bn • R µ|A Bn ⇔ (λ 1 , λ 2 ) = (µ 2 , µ 1 ), (d) R λ|A Bn • R µ|A Bn ⇔ (λ 1 , λ 2 ) = (µ 1 , µ 2 ).
Proof. In all cases, (a) and (b) are the same and the proofs are identical.

(a) By Lemma 3.2.3 and Theorem 3.1.1, it is sufficient to show that if there exists η : (b) for some u, v ∈ F q . Taking the eigenvalues of S 1 and T on both sides of the equality we get that {α, -1} = {uα, -u} and {β, -1} = {vβ, -v}. Since α 2 = 1 = β 2 , we have u = v = 1 which implies that R λ and R µ are isomorphic representations. By theorem 3.1.1, this implies λ = µ.

A Bn → F q such that R λ R µ ⊗ η, then λ = µ. Assume such a character exists, since the abelianization of A Bn is < T , S 1 > Z 2 , up to conjugation we have R λ (b) = R µ (b)u 1 (b) v 2
(b) The implication λ = µ ⇒ R λ|A Bn R µ|A Bn follows from Proposition 3.2.1. Assume now R λ|A Bn R µ|A Bn , we then have R λ |A Bn R λ|A Bn (R µ|A Bn ) = R µ|A Bn . The result follow from (a).
In the same way, it is enough to show the converse implication in the remainder of the proof.

d)

This result follows directly from Proposition 3.2.2. 2. c) By 2. d) and 2. b), we have

• R λ |A Bn • ( • R λ |A Bn ) = R λ |A Bn R λ|A Bn .
3) In this case (α) = α -1 and (β) = β.

c)

For every standard double-tableau T = (T 1 , T 2 ), we define T by T = (T 1 , T 2 ). Let η : A Bn → F q be the character of A Bn defined by η(T ) = 1 and η(S r ) = -α for all r.

Let Q :

V (λ 1 ,λ 2 ) → V (λ 1 ,λ 2 ) , (T 1 , T 2 ) → (T 1 , T 2 ), U : V λ → V λ , T → ω(T)T.
Using the same notations as in Proposition 3.2.2, we will show that for all r ∈ [[1, n -1]] :

Q -1 (-α) (R λ 1 ,λ 2 (S r ))Q = U -1 P R λ (S r )P -1 U, Q -1 (R λ 1 ,λ 2 (T ))Q = U -1 P R λ (T )P -1 U.
Let T = (T 1 , T 2 ) be a standard double-tableau. The second equality follows from the equalities P R λ (T )P -1 = R λ (T ) and (β) = β. If T r↔r+1 is non-standard, the first equality is verified by S r . Assume T r↔r+1 is standard, write a = a r,r+1 . If τ T (r) = τ T (r + 1) then in the basis (T, T r↔r+1 ), we have :

R λ (S r ) = α-1 1-α a α-α -a 1-α -a α+α a 1-α a α-1 1-α -a , U -1 P R λ (S r )P -1 U = -α α -1 -1 1-α a α -1 -α -a 1-α -a α -1 -α a 1-α a α -1 -1 1-α -a , -αQ -1 (R λ 1 ,λ 2 (S r ))Q = -α ( α-1 1-α -a α-α a 1-α a α-α -a 1-α -a α-1 1-α a = U -1 P R λ (S r )P -1 U.
If τ T (r) = 1 and τ T (r + 1) = 2 then we have

R λ (S r ) =   α-1 1+βα a α+β -1 α -a 1+β -1 α -a α+βα a 1+βα a α-1 1+β -1 α -a   , U -1 P R λ (S r )P -1 U = -α   α -1 -1 1+βα a β -1 α -a +α -1 1+β -1 α -a βα a +α -1 1+βα a α -1 -1 1+β -1 α -a   , -αQ -1 (R λ 1 ,λ 2 )Q = -α (   α-1 1+βα -a α+β -1 α a 1+β -1 α a α+βα -a 1+βα -a α-1 1+β -1 α a   = U -1 P R λ (S r )P -1 U. It follows that R (λ 1 ,λ 2 )|A Bn • R (λ 1 ,λ 2 )|A Bn . 3. d)
This is a consequence of 3. c) and 3. b). 4. c) For each standard double tableau T = (T 1 , T 2 ), we set T = (T 2 , T 1 ). Let η : A Bn → F q be the character defined by η(T ) = -β and η(S r ) = 1 for all S r .

Let

Q : V λ 2 ,λ 1 → V λ 1 ,λ 2 be T → T. Let us show that R (λ 1 ,λ 2 ) = Q(( • R (λ 2 ,λ 1 ) ) η)Q -1
, where in this case (α) = α and (β) = β -1 . The proof of the result is straightforward for T . Let r ∈ [ [1, n -1]]. If r and r + 1 are in the same tableau then the result is clear. Assume r is in the right tableau, r + 1 is in the left tableau and set a = a r,r+1 . In the basis (T, T r↔r+1 ), we have

R λ (S r ) =   α-1 1+βα a α+β -1 α -a 1+β -1 α -a α+βα a 1+βα a α-1 1+β -1 α -a   Q -1 (R λ 2 ,λ 1 (S r ))Q = (   α-1 1+β -1 α a α+βα -a 1+βα -a α+β -1 α a 1+β -1 α a α-1 1+βα -a   = R λ (S r ).
This proves that we have The following proposition is a generalization of Proposition 3.5 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. The analogous proposition in the generic case is given by Proposition 2.13 of [START_REF] Marin | Infinitesimal Hecke algebras II[END_REF].

R (λ 1 ,λ 2 )|A Bn • R (λ 2 ,
Proposition 3.2.5. For r ∈ [[1, n -1]], there exists η 1,r , η 2,r : A Bn → F q such that R λ (r) (Λ r R λ (1) ) ⊗ η 1,r and R λ (r) (Λ r R λ (1) ) ⊗ η 2,r .
Proof. Every double-tableau associated with λ (r) or λ (r) can be mapped in a one-to-one way to a set {i 1 , i 2 , ..., i r } ⊂ [ [1, n]] such that i 1 < i 2 < ... < i r , where i k is the number in the k-th box of the right component. We write v I the corresponding double-tableau and

v i = v {i} . After computation, for k ∈ [[1, n -1]],
we get the following properties.

1. If 1 ∈ I then R λ (r) (T )v I = -v I . 2. If 1 / ∈ I then R λ (r) (T )v I = βv I . 3. If k, k + 1 / ∈ I then R λ (r) (S k )v I = -v I . 4. If k, k + 1 ∈ I then R λ (r) (S k )v I = αv I . 5. If k ∈ I, k + 1 / ∈ I then R λ (r) (S k )v I = α-1 1+β -1 α k-1 v I + α+β -1 α k-1 1+β -1 α k-1 v I∆{k,k+1} . 6. If k / ∈ I, k + 1 ∈ I then R λ (r) (S k )v I = α-1 1+βα 1-k v I + α+βα 1-k 1+βα 1-k v I∆{k,k+1} .
Above, ∆ is the symmetric difference : A∆B = (A ∪ B) \ (A ∩ B).

To each set I = {i 1 , i 2 , ..., i r } can be given in a one-to-one way an element u I of Λ r R λ (1) writing u I = v i 1 ∧ v i 2 ∧ ... ∧ v ir and these u I form a basis.

For k ∈ [[1, n -1]], we have the following properties.

1. If 1 ∈ I then Λ r R λ (1) (T )u I = -β r-1 u I . 2. If 1 / ∈ I then Λ r R λ (1) (T )u I = β r u I . 3. If k, k + 1 / ∈ I then Λ r R λ (1) (S k )u I = (-1) r u I . 4. If k, k + 1 ∈ I then Λ r R λ (1) (S k )u I = (-1) r-1 αu I . 5. If k ∈ I, k + 1 / ∈ I then Λ r R λ (1) (S k )u I = (-1) r-1 α -1 1 + β -1 α k-1 u I + (-1) -r-1 α + β -1 α k-1 1 + β -1 α k-1 u I∆{k,k+1} . 6. If k / ∈ I, k + 1 ∈ I then Λ r R λ (1) (S k )u I = (-1) r-1 α-1 1+βα 1-k u I + (-1) r-1 α+βα 1-k 1+βα 1-k u I∆{k,k+1} .
Looking at the basis change v I → u I and the character η 1,r (h) = (-1) (r-1) 1 (h) β (r-1) 2 (h) , we have the first part of the proposition.

In the same way, writing η 2,r (h) = (-1) (r-1) 1 (h) (-1) (r-1) 2 (h) , we have the second part of the proposition.

Factorization depending on the field

In this section, we use the work from the previous sections to state the main results in type B. As for the factorizations shown in Proposition 3.2.4, we have to distinguish cases depending on the field extensions appearing.

The result depends on the properties of the field extension

F q = F p (α, β) of F p (α + α -1 , β + β -1
) and the field extension F q = F p (α) of F p (α + α -1 ). Let us consider the Hasse diagram for the field extension

F p (α, β) of F p (α + α -1 , β + β -1 ). F p (α, β) F p (α, β + β -1 ) F p (α + α -1 , β) F p (α + α -1 , β + β -1 )
All the extensions represented by edges are of degree at most 2 since X 2 -(α+α -1 )X +1 and (α, β). This proves in particular that F p (α, β) cannot be an extension of degree 4 of F p (α+α -1 , β+β -1 ), otherwise we would have that F p (α, β + β -1 ) and F p (α + α -1 , β) are two subfields of degree 2 of F p (α, β) and would therefore be equal. This would then imply that they are equal to F p (α, β) and contradict the fact that the corresponding extensions are of degree 2. Note that F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ) implies that we have F p (α) = F p (α + α -1 ). This proves that we have the six following cases to consider

X 2 -(β + β -1 )X + 1 are the polynomials involved. Note that if F p (α, β + β -1 ) = F p (α + α -1 , β) then β ∈ F p (α, β + β -1 ). Therefore F p (α, β + β -1 ) = F p
1. F q = F p (α, β) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1 ). 2. F q = F p (α, β) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1 ). 3. F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ). 4. F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ). 5. F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1 ). 6. F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1 ).
We remark that in the third and fourth cases, we have F p (α) = F p (α + α -1 ). Before stating the main results for type B, we recall the two following lemmas, the first one is Lemma 2.4 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] and the proof of the second one is included in the proof of Proposition 4.1 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. Lemma 3.2.4. Let ρ be an absolutely irreducible representation of a group G in GL r (q), where F q is a finite field such that there exists an automorphism of order 2 of F q . If ρ

• ρ , then there exists S ∈ GL r (q) such that S -1 ρ(g)S ∈ GU r (q

2 ) for all g ∈ G.

Lemma 3.2.5. Let ρ and G be as in the previous lemma. If ρ

• ρ, then there exists S ∈ GL r (q) such that S -1 ρ(g)S ∈ GL r (q

2 ) for all g ∈ G.

In certain cases, (λ 1 , λ 2 ) factorizes through (λ 2 , λ 1 ) or (λ 1 , λ 2 ), therefore we need a good order on double-partitions of n. We first choose for r ≤ n an order on partitions of r such that if r has 2l partitions different from their transpose {a i , a i } i∈[ [1,l]] and s partitions {a l+i } i∈[ [1,s]] equal to their transpose then a 1 < a 1 < a 2 < a 2 < ... < a l < a l < a l+1 = a l+1 < ... < a l+s = a l+s . We also require a 1 = [r]. This gives us that λ < µ implies that λ < µ whenever λ = µ . If λ n 1 and µ n 2 then we say λ > µ if n 1 > n 2 or n 1 = n 2 and λ > µ. We then define the order < on double-partitions of n in the following way, where λ 1 is a partition of r λ : (λ 1 , λ 2 ) < (µ 1 , µ 2 ) if r λ < r µ or (r λ = r µ and λ 1 < µ 1 ) or (r λ = r µ , λ 1 = µ 1 and λ 2 < µ 2 ). Lemma 3.2.6. If λ = (λ 1 , λ 2 ) is a double-partition such that λ = λ , λ = (λ 2 , λ 1 ) and λ = (λ 1 , λ 2 ) then exactly one element of {λ, λ , (λ 1 , λ 2 ), (λ 2 , λ 1 )} verifies the property :

( * ) λ < λ and λ < (λ 1 , λ 2 ).
Proof. Let λ = (λ 1 , λ 2 ) be a double-partition verifying the conditions in the lemma. Assume λ > λ and λ < (λ 1 , λ 2 ). Since λ > λ , we have r λ ≥ r λ and since r λ + r λ = n, we get r λ ≥ n 2 . Let's show that either (λ < λ and λ < (λ 2 , λ 1 )) or ((λ 2 , λ 1 ) < (λ 1 , λ 2 ) and (λ 2 , λ 1 ) < (λ 2 , λ 1 )), i.e. either λ verifies ( * ) or (λ 2 , λ 1 ) verifies ( * ). Those two cases are indeed distinct because either λ < (λ 2 , λ 1 ) or (λ 2 , λ 1 ) < λ . If λ < (λ 2 , λ 1 ) then we are in the first case because we assumed λ > λ . Let's now assume λ > (λ 2 , λ 1 ), we must show (λ 2 , λ 1 ) < (λ 1 , λ 2 ). This is obvious if

r λ > n 2 . If r λ = n 2 , then (λ 1 , λ 2 ) = λ > λ = (λ 2 , λ 1 ) implies that λ 1 > λ 2 or (λ 1 = λ 2 and λ 2 > λ 1 )
, which is a contradiction. Therefore λ 1 > λ 2 and since λ 1 = λ 2 , this implies λ 1 > λ 2 by definition of our order on partitions of r λ . This shows that (λ 2 , λ 1 ) < (λ 1 , λ 2 ).

Assume λ > λ and λ > (λ 1 , λ 2 ). We then have that either λ verifies ( * ) or (λ 2 , λ 1 ) verifies ( * ) in exactly the same way as in the previous case.

Assume λ < λ and λ > (λ 1 , λ 2 ), let us show that (λ 1 , λ 2 ) < (λ 2 , λ 1 ) and (λ 1 , λ 2 ) < (λ 1 , λ 2 ), i.e. (λ 1 , λ 2 ) verifies ( * ). It is enough to show the second inequality since we have the first one by assumption. This is obvious if

r λ < n 2 . If r λ = n 2 then λ 1 < λ 2 because λ 1 = λ 2 , therefore λ 1 < λ 2 because λ 1 = λ 2 and (λ 1 , λ 2 ) < (λ 2 , λ 1 ).
Assume λ < λ and λ < (λ 1 , λ 2 ). To conclude the proof, it is enough to show that not one of λ , (λ 1 , λ 2 ) and (λ 2 , λ 1 ) verifies ( * ) in this case. It is obvious for λ and (λ

1 , λ 2 ). If r λ < n 2 , it is also obvious for (λ 2 , λ 1 ) since (λ 2 , λ 1 ) > (λ 1 , λ 2 ). If r λ = n 2 then since λ 1 < λ 2 and λ 1 = λ 2 , we have that λ 2 > λ 1 , therefore (λ 2 , λ 1 ) > (λ 1 , λ 2 ).
We are now able to state the main results for type B which are a generalization of Theorem 1.1 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. The end of the proof will be in the next section. The main difference arises from the additional factorizations in the last cases for the field extensions explicited in six different cases at the beginning of Section 3.2. The results are given in Theorems 3.2.1 up to 3.2.6. The analogous versions of these theorems in the generic case can be seen in Theorem 2.21 of [START_REF] Marin | Infinitesimal Hecke algebras II[END_REF]. Note here again that in type B, the complexity of the field extensions involved yields a wider variety of results.

We write

A 1,n = {(λ 1 , ∅), λ 1 n}, A 2,n = {(∅, λ 2 ), λ 2 n}, A n = A 1,n ∪ A 2,n . A n = {(λ 1 , ∅) ∈ A 1,n , λ 1 not a hook}, n = {λ n, λ / ∈ A n , λ not a hook}, F q = F p (α). Theorem 3.2.1. If F q = F p (α, β) = F p (α + α -1 , β + β -1
) and F p (α) = F p (α + α -1 ), then the morphism :

A Bn → H × Bn,α,β λ n GL(λ) factors through the epimorphism Φ 1,n : A Bn → SL n-1 (q) × (λ 1 ,∅)∈A n,λ1<λ 1 SL n λ (q) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SL n (q) 2 × λ∈ n,λ<λ SL n λ (q) × λ∈ n,λ=λ OSP (λ) ,
where OSP (λ) is the group of isometries of the bilinear form considered in Proposition 3.1.1.

Theorem 3.2.2. If F q = F p (α, β) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1 ), then the morphism : A Bn → H × Bn,α,β λ n GL(λ) factors through the epimorphism Φ 2,n : A Bn → SU n-1 (q 1 2 ) × (λ 1 ,∅)∈A n,λ1<λ 1 SU n λ (q 1 2 ) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SL n (q) 2 × λ∈ n,λ<λ SL n λ (q) × λ∈ n,λ=λ OSP (λ) ,
where OSP (λ) is the group of isometries of the bilinear form considered in Proposition 3.1.1 and Õ SP (λ) is the group of isometries of a form of the same type but defined over subfield of degree 2.

Theorem 3.2.3. If

F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ), then the morphism A Bn → H × Bn,α,β λ n GL(λ) factors through the epimorphism Φ 3,n : A Bn → SU n-1 (q 1 2 ) × (λ 1 ,∅)∈A n,λ1<λ 1 SU n λ (q 1 2 ) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SU n (q 1 2 ) 2 × λ∈ n,λ<λ SU n λ (q 1 2 ) × λ∈ n,λ=λ OSP (λ) ,
where OSP (λ) is the group of isometries of a bilinear form of the same type as the one considered in proposition 3.1.1 but defined over F

q 1 2 .
By Proposition 3.2.4 and Lemmas 3.2.4 and 3.2.5, we have the following theorem

Theorem 3.2.4. If F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ), then the morphism A Bn → H × Bn,α,β λ n GL(λ) factors through the epimorphism Φ 4,n : A Bn → SU n-1 (q 1 2 ) × (λ 1 ,∅)∈A n,λ1<λ 1 SU n λ (q 1 2 ) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SL n (q) × λ∈ n,λ<λ ,λ<(λ 1 ,λ 2 ),λ =(λ 2 ,λ 1 ) SL n λ (q) × λ∈ n,λ<λ ,λ=(λ 2 ,λ 1 ) SU n λ (q 1 2 )× λ∈ n,λ<λ ,λ=(λ 1 ,λ 2 ) SL n λ (q 1 2 ) × λ∈ n,λ=λ ,λ<(λ 1 ,λ 2 ) OSP (λ) × λ∈ n,λ=λ ,λ=(λ 1 ,λ 2 )
OSP (λ) .

Theorem 3.2.5.

If F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 )
and F p (α) = F p (α + α -1 ), then the morphism

A Bn → H × Bn,α,β λ n GL(λ) factors through the epimorphism Φ 5,n : A Bn → SU n-1 (q 1 2 ) × (λ 1 ,∅)∈A n,λ1<λ 1 SU n λ (q 1 2 ) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SL n (q) × λ∈ n,λ<λ ,λ<(λ 1 ,λ 2 ),λ =(λ 2 ,λ 1 ) SL n λ (q) × λ∈ n,λ<λ ,λ=(λ 2 ,λ 1 ) SL n λ (q 1 2 )× λ∈ n,λ<λ ,λ=(λ 1 ,λ 2 ) SU n λ (q 1 2 ) × λ∈ n,λ=λ ,λ<(λ 1 ,λ 2 ) OSP (λ) × λ∈ n,λ=λ ,λ=(λ 1 ,λ 2 ) OSP (λ) . Theorem 3.2.6. If F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1 ), then the morphism A Bn → H × Bn,α,β λ n GL(λ) factors through the epimorphism Φ 6,n : A Bn → SL n-1 (q) × (λ 1 ,∅)∈A n,λ1<λ 1 SL n λ (q) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ) × SL n (q) × λ∈ n,λ<λ ,λ<(λ 1 ,λ 2 ),λ =(λ 2 ,λ 1 ) SL n λ (q) × λ∈ n,λ<λ ,λ=(λ 2 ,λ 1 ) SL n λ (q 1 2 )× λ∈ n,λ<λ ,λ=(λ 1 ,λ 2 ) SU n λ (q 1 2 ) × λ∈ n,λ=λ ,λ<(λ 1 ,λ 2 ) OSP (λ) × λ∈ n,λ=λ ,λ=(λ 1 ,λ 2 )
OSP (λ) .

Surjectivity of the morphisms Φ i,n

In this section, we conclude the proof of the theorems in the previous section by showing that for all i ∈ [ [1,6]], the morphism Φ i,n is surjective. The core of the proof for all i will be in Section 2.3.1. However, for small n, the different factorizations which appear in cases 4, 5 and 6 of the field extensions described at the beginning of Section 3.2.2 change the proofs for small cases. They require to introduce new tools and consider the maximal subgroups of some classical groups in low dimension. Those were classified in [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] and we will use their tables in order to treat those cases. In all cases, the proof is based on inductive reasoning. We start by proving the result for small n and we then use Goursat's lemma and a Theorem by Guralnick and Saxl [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] which relies on the classification of finite simple groups and gives us a list of conditions for a group to be a classical group. Our induction assumptions will allow us to check that all the conditions are verified. It will be more tricky to verify those conditions in cases 4, 5 and 6 because of the additional factorizations.

First case :

F q = F p (α, β) = F p (α+α -1 , β +β -1 ), F p (α) = F p (α+α -1 )
In this subsection, we prove the surjectivity of the morphism in the easiest case and establish groundwork for the other cases. This will conclude the proof of Theorem 3.2.1. We first prove the result for n ≤ 4 and then use induction to get the result for all n.

We recall Goursat's Lemma also used in [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] and [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], it was originally proven in [START_REF] Goursat | Sur les substitutions orthogonales et les divisions régulières de l'espace. Annales Scientifiques de l'ENS 3ème série tome[END_REF] and an english version of the proof can be found in [6] (Lemma 2.3.5) or [START_REF] Ribet | Galois action on division points of Abelian varieties with real multiplications[END_REF] Lemma 5.2.1:

Lemma 3.3.1 (Goursat's Lemma). Let G 1 and G 2 be two groups, K ≤ G 1 × G 2 ,
and write π i : K -→ G i the projection. Let K i = π i (K) and K i = ker(π i ) for {i, i } = {1, 2}. There exists an isomorphism ϕ :

K 1 /K 1 → K 2 /K 2 such that K = {(k 1 , k 2 ) ∈ K 1 × K 2 , ϕ(k 1 K 1 ) = k 2 K 2 }.
We first prove that if for any λ n, the composition of R λ with the projection on its corresponding quasi-simple factor is surjective, then Φ 1,n is surjective. We will then prove by induction that each composition is indeed surjective. In order to get the images of the hook partitions it is enough to get the images inside the representations associated with the partitions ([1 n-1 ], [1]) and ( [1], [1 n-1 ]) . We recall Wagner's theorem which can be found for example in [START_REF] Malle | Inverse Galois theory[END_REF]II,Theorem 2.3]. Theorem 3.3.1. Let F r be a finite field, n ∈ N, n ≥ 3 and G ⊂ GL n (r) a primitive group generated by pseudo-reflections of order greater than or equal to 3. Then one of the following properties is true.

1. SL n (r) ⊂ G ⊂ GL n (r) for some r dividing r. 2. SU n (r 1 2 ) ⊂ G ⊂ GU n (r 1
2 ) for some r dividing r.

([1 n-1 ], [1]) (resp ([1], [1 n-1 ])). We have R 1 (A Bn ) R 2 (A Bn ) SL n (q).
Proof. Let n ≥ 3, we will use Theorem 3.3.1. The eigenvalues of R 1 (T ) are β with multiplicity n -1 and -1 with multiplicity 1. The eigenvalues of R 1 (S i ) are α with multiplicity 1, and -1 with multiplicity n -1. The group G = β -1 R 1 (T ), -R 1 (S 1 ), ..., -R 1 (S n-1 ) is generated by pseudo-reflections. To apply Wagner's Theorem (Theorem 3.3.1), we must show that the group is primitive. If G was imprimitive, we could write

F n q = V 1 ⊕ V 2 ⊕ ... ⊕ V r
, where for all i and for all g ∈ G, there exists a j such that g.

V i = V j . Since R 1 is irreducible, either β -1 R 1 (T ).V 1 = V 1 or there exists i ≤ n -1 such that -R 1 (S i ).V 1 = V 1 . Assume there exists i such that -R 1 (S i ).V 1 = V 1 . Up to reordering, we have V 2 = -R(S i ).V 1 . If dim(V 1 ) ≥ 2, then H -R 1 (S i ) (the hyperplane fixed by -R 1 (S i )) has a non-empty intersection with V 1 . It follows that V 1 ∩ V 2 = ∅,
which is a contradiction, which proves that dim(V 1 ) = 1. This reasoning is valid for any V i , therefore they are all one-dimensional. Let x ∈ V 1 be a non-zero vector, it can be written in a unique way as

x = x 1 + x 2 with x 1 ∈ ker(R 1 (S i ) + α) and x 2 ∈ H -R 1 (S i ) . We then have that -R 1 (S i )x = -αx 1 + x 2 and -R(S i )(-R(S i )x) = α 2 x 1 + x 2 = α(x 1 + x 2 ) + (1 -α)(-αx 1 + x 2 ) ∈ V 1 ⊕ V 2 . Since α /
∈ {0, 1} this contradicts the fact that there exists j such that -R(S i ).

V 2 = V j . If V 1 = β -1 R 1 (T )V 1 = V 2 , then if x = x 1 + x 2 = 0 with x ∈ V 1 , x 1 ∈ ker(R 1 (T ) + β)x 2 ∈ H β -1 R 1 (T ) , we have that β -1 R 1 (T )x = -β -1 x 1 + x 2 and β -1 R 1 (T )(β -1 R 1 (T )x) = β -2 x 1 + x 2 = β -1 (x 1 + x 2 ) + (1 -β -1 )(-β -1 x 1 + x 2 ) ∈ V 1 ⊕ V 2 .
This is absurd because β -1 / ∈ {0, 1}. This shows that G is primitive and in the same way, G = -R 2 (T ), -R 2 (S 1 ), ..., -R 2 (S n-1 ) is primitive and generated by pseudo-reflections of order greater than or equal to 3. By Theorem 3.3.1, we have

SL n (q) ⊂ G, G ⊂ GL n (q) or SU n (q 1 2 ) ⊂ G, G ⊂ GU n (q 1
2 ) for some q dividing q. If we were in the unitary case then there would exist an automorphism of order 2 of F q2 such that det(M ) = (det(M )) -1 for all M in G or G. We also have det(

β -1 R 1 (T )) = -β -1 , det(-R 1 (S 1 )) = det(-R 2 (S 1 )) = -α and det(-R 2 (T )) = -β. If G or G is unitary then (β) = β -1 and (α) = α -1
, therefore α+α -1 and β +β -1 are in F q. This contradicts the fact that q2 divides q and F q = F p (α+α -1 , β +β -1 ). This proves we have SL n (q) ⊂ G, G ⊂ GL n (q) for some q dividing q. Using the determinants again, we have α and β in F q, therefore q = q. We have

SL n (q) = [G, G] = [R 1 (A Bn ), R 1 (A Bn )] = R 1 (A Bn ) and SL n (q) = [ G, G] = [R 2 (A Bn ), R 2 (A Bn )] = R 2 (A Bn ) which concludes the proof.
By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A Bn is perfect for n ≥ 5 but not for n ≤ 4. Those cases must then be treated separately.

Lemma 3.3.2. If n ≤ 4 then Φ 1,n is surjective.
Proof. The representations labeled by double-partitions of n = 2 are all one-dimensional except for ( [1], [1]). We then only need to show that R [1], [1] (A B 2 ) = SL 2 (q). We write t = R [1], [1]

(T ) = β 0 0 -1 and s = R [1],[1] (S 1 ) = 1 β+1 α -1 α + β αβ + 1 αβ -β . First note that if P = 1 1 αβ+1 α+β -1 then P -1 tP = 1 α+1 β -1 α + β αβ + 1 βα -α and P -1 sP = α 0 0 -1.
. This proves that the roles of α and β are completely symmetrical in this case. It follows that up to conjugating by P , we can exchange the conditions on α and the conditions on β. We write G =< t, s >. We have det(t) = -β and det(s) = -α. Let (u, v) ∈ F p 2 such that u 2 = -β -1 and v 2 = -α -1 . We set

F q = F q (u, v).
We then have G = t, s = ut, vs ⊂ P SL 2 (q ). We write S n the permutation group of n elements and A n its derived subgroup. By Dickson's Theorem [27, Chapter II, HauptSatz 8.27], we have that G is either abelian by abelian or isomorphic to A 5 , S 4 , P SL 2 (q) or P GL 2 (q) for a given q greater than or equal to 4 and dividing q .

If ut r = 1 in P SL 2 (q ), then ((-u) r ) 2 = 1. Therefore β r = (-1) r and by the condition on the order of α, G cannot be isomorphic to A 4 or A 5 .

We now exclude the case G abelian by abelian. If G is abelian by abelian, then [G, G] is abelian, i.e. ab = ba for all a, b ∈ [G, G] or equivalently ab = ±ba for all a, b ∈ [G, G]. We have that (tst

-1 s -1 )(s -1 tst -1 ) -(s -1 tst -1 )(tst -1 s -1 ) =   -(β-1)(α-1) 2 (αβ+1)(α+β) βα 2 (β+1) -(α 2 β 2 +αβ 3 -αβ 2 -α 2 β+αβ+β 2 -α-β)(α-1)(αβ+1) βα 2 (β+1) (α 2 β 2 +αβ 3 -αβ 2 -α 2 β+αβ+β 2 -α-β)(α-1)(α+β) β 2 α 2 (β+1) (β-1)(α-1) 2 (αβ+1)(α+β) βα 2 (β+1)   .
This matrix is non-zero because the diagonal coefficients are non-zero by the conditions on β. This means that if [G, G] is abelian, then we have (tst -1 s -1 )(s -1 tst -1 ) + (s -1 tst -1 )(tst -1 s -1 ) = 0, but this matrix equals

  α 4 β+α 3 β 2 -2α 3 β-2α 2 β 2 +α 3 +4α 2 β+αβ 2 -2α 2 -2αβ+α+β α 2 β - (α 2 β+α β 2 -2 α β+α+β)(α-1)(α β+1) α 2 β -(α 2 β+α β 2 -2 α β+α+β)(α+β)(α-1) α 2 β 2 α 4 β+α 3 β 2 -2 α 3 β-2 α 2 β 2 +α 3 +4 α 2 β+α β 2 -2 α 2 -2 α β+α+β α 2 β   .
The non-diagonal coefficients are non-zero if

A = α 2 β + α β 2 -2 α β + α + β is non-zero. If A = 0 then, the bottom right coefficient of (tst -1 s -1 )(st -1 s -1 t) + (st -1 s -1 t)(tst -1 s -1 ) is equal to - 1 (β+1)α 2 β 2 multiplied by α 4 β 3 + α 3 β 4 -α 4 β 2 -3 α 3 β 3 -2 α 2 β 4 + 5 α 3 β 2 + 4 α 2 β 3 + α β 4 -3 α 3 β -8 α 2 β 2 -3 α β 3 + 4 α 2 β + 5 α β 2 + β 3 -2 α 2 -3 α β -β 2 = (α 2 β 2 -α 2 β + 2αβ -2α)A + β((β -1)A -2α 2 (β 3 + 1)) = -2βα 2 (β 3 + 1).
This is non-zero by the condition on β. The diagonal coefficients of the difference of these two commutators are identical to the ones of the difference of the previous commutators, therefore they are non-zero. This proves that G is not abelian by abelian and there exists q greater than or equal to 4 such that [G, G] P SL 2 (q). For H a group and A an H-module, we write

Z 2 (H, A) = {f : H × H → A, ∀x, y, z ∈ H, z.f (x, y)f (xy, z) = f (y, z)f (xy, z)} the group of cocyles and B 2 (H, A) = {f : H × H → A, ∃t : H → A, ∀x, y ∈ H, f (x, y) = t(y)t(xy) -1 y.t(x)} the group of coboundaries. We write M (H) = H 2 (H, C ) = H 2 (H, Z) its Schur multiplier. We have [G, G] = [G, G] ⊂ P SL 2 (q),
this inclusion gives a projective representation of SL 2 (q) with associated cocyle c ∈ Z 2 (SL 2 (F q), F q ). We will show that H 2 (SL 2 (F q), F q ) is trivial which is equivalent to Z 2 (SL 2 (F q), F q ) = B 2 (SL 2 (q), F q ) which implies that this cocycle is a coboundary. We write H = SL 2 (q). By the Universal Coefficients Theorem [25, Theorem 3.2], we have the following exact sequence

1 → Ext(H 1 (H, Z), F q ) → H 2 (H, F q ) → Hom(H 2 (H, Z), F q ) → 1. We have H 1 (H, Z) = H/[H, H] [29] and H 2 (H, Z) = M (H). Since q ≥ 4, we have that SL 2 (q)
is perfect and the exact sequence becomes

1 → 1 → H 2 (H, F q ) → Hom(M (H), F q ) → 1.
By [START_REF] Karpilovsky | The Schur multiplier, volume 2 of London Mathematical Society Monographs[END_REF]Theorem 7.1.1], if q / ∈ {4, 9} then the Schur multiplier M (H) is trivial, therefore this reduces to H 2 (H, F q ) {1}.

It remains to consider the cases q = 4 and q = 9. If q = 4, we have M (H) = Z/2Z and p = 2, therefore Hom(M (H), F q ) = 1. Indeed, every morphism ϕ from M (H) to F q satisfies 1 = ϕ(2x) = ϕ(x) 2 for all x ∈ M (H). It follows that 0 = ϕ(x) 2 -1 = (ϕ(x) -1) 2 for all x ∈ M (H). This proves that ϕ is trivial. If q = 9 then we have M (H) = Z/3Z and H 2 (H, F q ) is trivial by the same reasoning as for q = 4. In all cases, we can define a representation ρ of SL 2 (q) in SL 2 (q). By [START_REF] Brauer | On the modular characters of groups[END_REF], any representation σ of SL 2 (q) in GL 2 (q) is up to conjugation of the form σ(M ) = ψ(M ), where ψ(M ) is the matrix obtained from M by applying ψ ∈ Aut(F q ) to all its coefficients. We have F q = F q(w) for any w generating the cyclic group F q . There exists a homomorphism from F q to F q sending 1 to w and stabilizing F q. We define a representation ρ of SL 2 (q) in SL 2 (q) such that ρ(M ) = ρ(M ) for all M in SL 2 (q). We have

ρ(M ) = ρ(M ) = ψ(M ) for all M in SL 2 (q). We have [G, G] P SL 2 (q), therefore ψ([G, G]) P SL 2 (q) is conjugate to P SL 2 (q) in GL 2 (q ). We have ψ(tst -1 s -1 ) ∈ ψ([G, G]), therefore its trace 2 -(α + α -1 + β + β -1 ) belongs to F q. This shows that α + α -1 + β + β -1 ∈ F q.
We also have that the trace T 1 of s 2 t -1 s -2 t and the trace T 2 of st -2 s -1 t 2 are in F q. We have

T 1 = α 4 β+α 3 β 2 -2α 3 β-2α 2 β 2 +α 3 +4α 2 β+αβ 2 -2α 2 -2αβ+α+β α 2 β . We write B = α + α -1 + β + β -1 . We then have T 1 = (α + α -1 )B -2B + 2.
The quantity T 2 has the same expression as T 1 with α and β switched, therefore T 2 = (β+β -1 )B-2B+2. Since B, T 1 and T 2 are in F q, we have (α+α -1 )B and (β +β -1 )B are in F q. We have B = α+α -1 +β +β -1 . It follows that B = 0 implies α ∈ {-β, -β -1 }; which contradicts the assumptions on α and β. The quantity B is therefore non-zero, therefore α + α -1 and β + β -1 are in F q, and F q = F q . We conclude using Lemma 2.1 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF]. [1]). We want to show that the image of Φ 3 is equal to SL 2 (q) × SL 3 (q) × SL 3 (q). If we restrict ourselves to the image inside SL 3 (q) × SL 3 (q), we show that it is SL 3 (q) × SL 3 (q). By Proposition

The double-partitions of

n = 3 to consider are ([2, 1], ∅), ([1], [1 2 ]) and ([1 2 ],
3.3.1, SL 3 (q) = R [1 2 ],[1] (A B 3 ) and R [1],[1 2 ] (A B 3 ) = SL 3 (q)
. We now use Goursat's Lemma : we write as in Lemma 3.3.1,

K = R(A B 3 ), K 1 = R [1 2 ],[1] (A B 3 ), K 2 = R [1],[1 2 ] (A B 3 ) π 1 (resp π 2 ) the projection onto SL 3 (q)(corresponding to ([1 2 ], [1])) (resp SL 3 (q)) (corresponding to ([1], [1 2 ]))), K 1 = ker(π 2 ), K 2 = ker(π 1
) and ϕ the isomorphism given by Goursat's Lemma. We have

K = {(x, y) ∈ K 1 × K 2 , ϕ(xK 1 ) = yK 2 }.
By the same reasoning as the one in Proposition 3.1. of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], either SL 3 (q) × SL 3 (q) ⊂ K or K 1 /K 1 is non-abelian and ϕ is an isomorphism of P SL 3 (q) and using the same notations, up to conjugation

R 2 (b) = S φ (R 1 (b))z(b) for all b ∈ A B 3 (S = Id or S = M → t M -1 ).
Let us show that the second possibility is absurd by choosing the right elements in

A B 3 . For any element b of A B 3 , Tr(R [1],[1 2 ] (b)) = z(b) Tr(S φ (R [1 2 ],[1] (b)). We write U = S 1 S -1 2 , V = T S 1 T -1 S -1 2 , W = S 2 S 1 S -2 2 and X = S 2 T S 1 T -1 S -2 2
, they are all elements of A B 3 . By explicit computation, for both choices of S, we have :

Tr(R [1],[1 2 ] (U )) = Tr(R [1],[1 2 ] (V )) = Tr(R [1],[1 2 ] (W )) = Tr(R [1],[1 2 ] (X)) = - (α -1) 2 α , Tr(S(R [1 2 ],[1] (U ))) = Tr(S(R [1 2 ],[1] (V ))) = Tr(S(R [1 2 ],[1] (W ))) = Tr(S(R [1 2 ],[1] (X))) = - (α -1) 2 α .
This shows that z(U

) = z(V ) = z(W ) = z(X) and 1 = z(U )z(W ) -1 = z(U W -1 ) = -(α-1) 2 α φ(-(α-1) 2 α
) .

This proves φ(α + α -1 ) = α + α -1 . We also have

1 = z(U V -1 ) = 3 -α -α -1 -β -β -1 φ(3 -α -α -1 -β -β -1 )
.

Using φ(α + α -1 ) = α + α -1 , we have φ(β + β -1 ) = β + β -1 , therefore φ = 1. We deduce that 1 = z(U X -1 ) = (α -1)(αβ 2 -2αβ + 2β -1) αβ -αβ (α -1)(2αβ + β 2 -α -2β) = 1 + 2αβ -2β -αβ 2 β 2 + 2αβ -2β -α , 1 -αβ 2 = β 2 -α, (1 -β 2 )(1 + α) = 0.
Since β 2 = 1 and α 2 = 1, we get a contradiction. This shows that

SL 3 (q) × SL 3 (q) = R(A B 3 ). We now set G 1 = SL 3 (q) × SL 3 (q) and G 2 = SL 2 (q), the image of Φ 3 is a subgroup of G 1 × G 2 for
which the projections onto G 1 and G 2 are surjective. Using again Goursat's Lemma and the notation there, we have

K 1 /K 1 K 2 /K 2 . We have a surjective morphism ψ from K 1 = G 1 = SL 3 (q) × SL 3 (q) to K 2 /K 2 , where K 2 = SL 2 (q). If K 2 /K 2 was non-abelian then we would have K 2 /K 2 P SL 2 (q). If the restriction ψ 1 (resp ψ 2 ) of ψ to SL([1 2 ], [1]) (resp SL([1], [1 2 ]))
was not trivial then ψ 1 (resp ψ 2 ) would factor into an isomorphism from P SL 3 (q) onto P SL 2 (q) since the center of SL 3 (q) would again be in the center of ψ 1 and ψ 2 .This would lead to a contradiction, therefore their image is trivial and ψ is not surjective, therefore the quotients are abelian. This shows that

K 1 = [K 1 , K 1 ] ⊂ K 1 and K 2 = [K 2 , K 2 ] ⊂ K 2 then using Goursat's Lemma we conclude that the image of Φ 3 is equal to G 1 × G 2 . This shows that Φ 3 is surjective.
The double-partitions of 4 in our decomposition are ([

1 4 ], ∅), ([2 2 ], ∅), ([2, 1, 1], ∅), ([1], [1 3 ]), ([1 3 ], [1]), ([1 2 ], [1 2 ]
) and ([2, 1], [1]) of respective dimensions 1, 2, 3, 4, 4, 6 and 8 (we removed the hooks, ([3], [1]) and ( [1], [3])). We know the restriction to the first five is surjective by [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] and Proposition 3.3.1, therefore we only need to show that R

[1 2 ],[1 2 ] (A B 4 ) = SL 6 (q) and R ([2,1],[1]) (A B 4 ) = SL 8 (q).
Let us first consider the double-partition ([1 2 ], [1 2 ]). By the branching rule and the case n = 3 above, we have

SL 3 (q) × SL 3 (q) = R [1 2 ],[1] (A B 3 ) × R [1],[1 2 ] (A B 3 ) = R [1 2 ],[1 2 ] (A B 3 ) ⊂ R [1 2 ],[1 2 ] (A B 4 ) ⊂ SL 6 (q).
We can now use Theorem 3 from [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF]. Theorem 3.3.2. Let F r be a finite field and Γ < GL N (r) with N ≥ 5 and q > 3 such that

1. Γ is absolutely irreducible, 2. Γ contains SL a (r) in a natural representation with a ≥ N 2 . If N = 2a, then Γ contains SL N (r). Otherwise, either Γ contains SL N (r), or Γ is a subgroup of GLN 2 (r) S 2 .
We use this theorem on R [1 2 ],[1 2 ] (A B 4 ). To get the desired result, we only need to show that R [1 2 ],[1 2 ] (A B 4 ) cannot be a subgroup of GL 3 (r) S 2 . If it were true, then we would have

R [1 2 ],[1 2 ] (A B 4 ) ⊂ SL 3 (q) × SL 3 (q) which would contradict the irreducibility shown in Lemma 3.2.1. This shows that we have R [1 2 ],[1 2 ] (A B 4 ) = SL 6 (q).
We now consider the double-partition ([2, 1], [1]). Again by the branching rule and the case n = 3, we have that the restriction to A B 3 is SL 3 (q) × SL 3 (q) × SL 2 (q). We now use the fact that each of these groups is generated by transvections and the fact that A B 4 is normally generated by A B 3 . When the characteristic is different from 2, we can use Theorem 2.3.4

We write

G = R ([2,1],[1]) (A B 4 ), H = R ([2,1],[1]) (A B 3 ) = SL 3 (q) × SL 3 (q) × SL 2 (q) and we pick t 1 (resp t 2 (resp t 3 )) a transvection of SL 3 (q) × {I 5 } (resp {I 3 } × SL 3 (q) × {I 2 } (resp {I 6 } × SL 2 (q)). We then have H = ht i h -1 , h ∈ H, i ∈ {1, 2, 3} , therefore G = ghg -1 , h ∈ H, g ∈ G = gt i g -1 , g ∈ G, i ∈ {1, 2,
3} is generated by transvections and we can apply the theorem. We also recall the following lemma [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]Lemma 5.6].

Lemma 3.3.3. For any prime p and m ≥ 2, the field generated over F p by {Tr(g), g ∈ SL m (q)} is F q and for all m ≥ 3, the field generated over F p by {Tr(g), g ∈ SU m (q

1 2 )} is F q .
By Proposition 3.2.4, we know that R [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]1], [1] (A B 4 ) preserves no non-degenerate bilinear form.

It also shows that it can preserve no non-degenerate hermitian form. Indeed, if it were to preserve a hermitian form then we would have Tr(M ) = (Tr( t (M ) -1 )) for any M in G and we have diag

([α, α -1 , 1, 1, 1, 1, 1, 1]) and diag([β, β -1 , 1, 1, 1, 1, 1, 1]) in H ⊂ G. Therefore we would have (α+α -1 ) = α+α -1 and (β +β -1 ) = β +β -1 . Since F q = F p (α, β) = F p (α+α -1 , β +β -1
), the automorphism of order 2 would be trivial which is a contradiction. This proves that G is conjugate in GL 8 (q) to SL 8 (q) for some q dividing q. By Lemma 3.3.3, the field generated over F p by the traces of the elements of G is F q, therefore q = q because G contains SL 3 (q) in a natural representation. This proves that the field generated by its elements contains F q . This shows that when

p = 2, G = R ([2,1],[1]) (A B 4 ) = SL 8 (q).
Assume now that p = 2, we can use the following theorem [40, Theorem 1].

Theorem 3.3.3. Let V be a F q -vector space of dimension n ≥ 4 with q even. If G is an irreductible proper subgroup of SL(V ) = SL n (q) generated by a set D of transvections of G, then D is a conjugacy class of odd transpositions of G.

Assume that G = R ([2,1],[1]) (A B 4
) is different from SL 8 (q). We again have that G is generated by transvections and by applying the above theorem, those transvections are in a single conjugacy class of G.

Since O p (G) is normal in G and V = F 8
q is an irreducible F q G-module, we apply Clifford's Theorem [START_REF] Curtis | Methods of representation theory[END_REF]Theorem 11.1]

and get that Res

G Op(G) (V ) is semisimple. Since O p (G) is a p-group, the unique irreducible F q O p (G)-module is the trivial module, therefore O p (G) acts trivially on V . It follows that O p (G)
is trivial. We can thus apply Kantor's Theorem [28, Theorem II] : Theorem 3.3.4. Let p be a prime and q = p l for some l ∈ N. Assume G is an irreducible subgroup of SL N (q) generated by a conjugacy class of transvections, such that

O p (G) ≤ [G, G] ∩ Z(G). Then G is one of the following subgroups. 1. G = SL n (q ) or G Sp N (q ) in SL N (q ) or G SU N (q 1 2 ) in SL N (q ), with q |q. 2. G O ± N (q ) < SL n (q ), with q |q. 3. G S n < SL N (2), where N = n -d and d = Gcd(n, 2). 4. G S 2n in SL 2n-1 (2) or in SL 2n (2). 5. G SL 2 (5) < SL 2 (9). 6. G 3.P Ω -,π 6 < SL 6 (4). 7. G SU 4 (2) < SL 5 (4). 8. G A S N in SL N (2 i ),
where A is a subgroup of diagonal matrices.

Since α is of order greater than 4, we have q ≥ q = 2 r > 8. The group G contains H = SL 3 (q) × SL 3 (q) × SL 2 (q), therefore cases (3) to (7) are excluded. If we were in case [START_REF] Brauer | On the modular characters of groups[END_REF], then G would have at mose (q -1) 10×9 2 = 45(q -1) transvections (see proof of Theorem 1.3. page 661 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF]). SL 3 (q) has (q 3 -1)(q 2 -1) q-1 = (q -1)(q 2 + q + 1)(q + 1) transvections and (q 2 + q + 1)(q + 1) ≥ 847(q -1) > 45(q -1). For the same reasons as when p = 2, G is neither unitary nor symplectic nor orthogonal. The only remaining possibility is G = SL 8 (q) which is a contradiction since we assumed G = SL 8 (q). This proves that G = SL 8 (q). The restriction to each double-partition of 4 is thus surjective, it remains to show that Φ 1,4 is surjective using Goursat's Lemma (Lemma 3.3.1). This means that we have to show that the image is SL 2 (q) × SL 3 (q) × SL 4 (q) × SL 4 (q) × SL 8 (q). By Theorem 1.2. of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], the restriction to SL 2 (q) × SL 3 (q) is surjective. We write G 1 the image of this restriction and

G 2 = R ([1],[1 3 ]) (A B 4 ) = SL 4 (q). Let K be the image A B 4 in G 1 × G 2 , using the corresponding notations in Goursat's Lemma, we have K 1 = G 1 , K 2 = G 2 and K 1 /K 1 K 2 /K 2 .
If these quotients are abelian then the proof of K = G 1 × G 2 is straightforward using Goursat's Lemma. Since the only non-abelian decomposition factor of G 2 is P SL 4 (q) and the only non-abelian decomposition factors of G 1 are P SL 2 (q) and P SL 3 (q), we have a contradiction if these quotients are non-abelian.

Write now K = R ([1],[3]) (A B 4 ) = SL 4 (q)
and let us consider the image J of A B 4 inside K × K. Using again Goursat's Lemma, this time with

K 1 = K, K 2 = K, we have K 1 /K 1 K 2 /K 2 .
If the quotients are abelian then J = K × K. If the quotients are non-abelian then there is an isomorphism S φ from P SL 4 (q) to P SL 4 (q), where the first one corresponds to R ([1],[1 3 ]) (A B 4 ) and the second one to R ([1], [3]) (A B 4 ). This implies that there exists a character z from A B 4 to F q such that up to conjugation, for every h ∈ H 4 , we have

R [1],[1 3 ] (h) = S φ (R ([1],[3]) (h))z(h). The isomorphism S φ is of the form [46, Section 3.3.4] M → φ(M ) or M → φ( t (M -1 ))
, where φ is a field automorphism of F q . We would then have that for all h ∈ A B 4 ,

Tr(R ([1 3 ],[1]) (h)) = φ(Tr(S(R ([1],[3]) (h)))z(h). Writing U = S 1 S -1 2 , V = T S 1 T -1 S -1 2 , X = S 2 T S 1 T -1 S -2 2 , P = S 3 S 2 S -2 3 , Q = T S 1 T -1 S -1 3 , R 1 = R ([1 3 ],[1]) and R 2 = R ([1],[3]) , we have 3 -α -α -1 = Tr(R 1 (P )) = Tr(R 1 (P Q -1 )) = Tr(R 2 (P )) = Tr(R 2 (P Q -1 ) = Tr( t (R 2 (P Q -1 ) -1 )) = Tr( t (R 2 (P Q -1 ) -1 ).

It follows that z(P Q

-1 ) = z(P ) = 3-α-α -1 Φ(3-α-α -1 ) . This shows that z(Q) = z(P )z(P Q -1 ) -1 = 1. We also have Tr(R 1 (Q)) = Tr(R 2 (Q)) = Tr( t (R 2 (Q) -1 )) = 2 -α -α -1 , therefore 1 = z(Q) = 2-α-α -1 Φ(2-α-α -1 ) and Φ(α + α -1 ) = α + α -1 . We have Tr(R 1 (U )) = Tr(R 1 (V )) = Tr(R 1 (X)) = Tr(R 2 (U )) = Tr(R 2 (V )) = Tr(R 2 (X)) = Tr( t (R 2 (X) -1 )) = Tr( t (R 2 (V ) -1 )) = Tr( t (R 2 (U ) -1 ) = 3-α-α -1 .
This leads to z(U

) = z(V ) = z(X) = 1. We have Tr(R 1 (U V -1 )) = Tr(R 2 (U V -1 )) = Tr( t (R 2 (U V -1 ) -1 )) = 4 -α -α -1 -β -β -1 . It follows that z(U V -1 ) = z(U )z(V ) -1 = 1 = 4-(α+α -1 )-(β+β -1 ) Φ(4-(α+α -1 )-(β+β -1 )) , therefore Φ(β + β -1 ) = β + β -1 . Since F q = F p (α + α -1 , β + β -1 ), we have Φ = I d . Tr(R 1 (U X -1 )) = - 2α 2 β + αβ 2 -α 2 -5αβ -β 2 + α + 2β αβ , Tr(R 2 (U X -1 )) = Tr( t (R 2 (U X -1 ) -1 ))) = α 2 β 2 -2α 2 β -αβ 2 + 5αβ -α -2β + 1 αβ . Since z(U X -1 ) = z(U )z(X) -1 = 1 and Φ = I d , it follows that α 2 β 2 -2α 2 β -αβ 2 + 5αβ -α -2β + 1 = -2α 2 β -αβ 2 + α 2 + 5αβ + β 2 -α -2β.
This shows that

α 2 β 2 + 1 = α 2 + β 2 , therefore (α 2 -1)(β 2 -1) = 0.
This contradicts the conditions on α and β. This contradiction shows that J = K × K. We conclude using Goursat's Lemma with R ([

1 2 ],[1 2 ]) (A B 4 ) = SL 6 (q) and then with R [2,1],[1] (A B 4 ) = SL 8 (q).
We now show for n ≥ 5 that if the representation associated with each double-partition is surjective, then Φ 1,n is surjective.

Lemma 3.3.4. If n ≥ 5, F p (α, β) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1
) and the composition of Φ 1,n and the projection upon each quasi-simple group associated with each double-partition is surjective, then Φ 1,n is surjective.

Proof. Let n ≥ 5, we know by [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]Theorem 1.1] that the restriction to double-partitions with an empty component is surjective. We first show that we can add the hook partitions. We then show by induction on the double-partitions using the order we defined that Φ 1,n is surjective.

We write G 0,0 = SL n-1 (q)× (λ 1 ,∅)∈A n,λ1<λ 1 SL n λ (q)× (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ), where OSP (λ)
is the derived subgroup of the group of isometries of the F q-bilinear form defined in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], which identifies to the one defined in this article. We then have by Theorem 1.1. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] that the image of

A Bn inside G 0,0 is G 0,0 . We have G 0,1 = R [1],[n-1] (A B 4 ) = SL n (q) by Proposition 3.3.1. We use Goursat's Lemma to show that the image of A Bn inside G 0,0 × G 0,1 is equal to G 0,0 × G 0,1 . Using the notations in Goursat's Lemma, we have K 1 = G 0,0 , K 2 = G 0,1 and K 1 /K 1 K 2 /K 2 .
If the quotients are abelian then we are done since the groups we consider are perfect. We assume that they are non-abelian and show there is a contradiction. The only non-abelian decomposition factor of K 2 is P SL n (q). Since the finite classical simple groups are non-isomorphic as long as n ≥ 4 and q ≥ 4 [46, Section 1.2], there would exist a decomposition factor of K 1 corresponding to a double-partition λ of n with its right component empty such that

P G(λ) = R λ (A Bn ) P SL n (q) = R ([1],[n-1]) (A Bn ). Therefore, up to conjugation [46, Section 3.3.4], we have that R λ (h) = S Φ (R ([1],[n-1]) (h))z(h) for all h ∈ A Bn with z : A Bn → F q ,
Φ an automorphism of F q and S being either the identity or the transpose of the inverse. Since n ≥ 5, we have A Bn perfect [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF]Corollary], therefore z is trivial. By Lemma 3.2.3 and since the abelianization of A Bn is the group

{T , S 1 } Z 2 , we have R λ (h) = S Φ (R ([1],[n-1]) (h))u 1 (h) v 2 (h) .
Since the right component of λ is empty, the only eigenvalue of R λ (T ) is β. On the other hand, the eigenvalues of

S φ (R ([1],[n-1]) (T ))v are equal to {vΦ(β), -v} or {vΦ(β -1 ), -v}. Therefore we would have -v = vΦ(β) or -v = vΦ(β -1 ) which is not possible because we have β = -1. This contradiction shows that the image is equal to G 0,0 × G 0,1 . Assume now G 0,2 = G 0,0 ×G 0,1 and G 0,3 = R ([1],[1 n -1]) (A Bn ) = SL n (q) and consider the image of A Bn inside G 0 = G 0,2 × G 0,3 . We use Goursat's Lemma with K 1 = G 0,2 and K 2 = G 0,3 .
In the same way as before, it is sufficient to show that the quotients

K 1 /K 1 K 2 /K 2 are abelian. The sets of eigenvalues of S Φ (R ([1],[1 n-1 ]) (T ) are again {Φ(β), -1} or {Φ(β) -1 , -1}.
If the quotients were non-abelian, we would have

R ([1],[n-1]) (h) = S φ (R ([1],[1 n-1 ]) (h))z(h)
for all h ∈ A Bn with S, φ and z as before. We have z trivial since A Bn is perfect, therefore

R ([1],[n-1]) (h) = S φ (R ([1],[1 n-1 ]) (h))u 1 (h) v 2 (h) . Let us show that Φ is trivial. We have that R ([1],[n-1]) (T ) and R ([1],[1 n-1 ]) (T )
both have for eigenvalues -1 with multiplicity n-1 and β with multiplicity 1. This shows that either β = vΦ(β) and

-1 = -v or β = vΦ(β) -1 and -1 = -v. In both cases, v = 1 and Φ(β + β -1 ) = β + β -1 . The eigenvalues of R ([1],[n-1]) (S 1 ) are -1 with multiplicity 1 and -α with multiplicity n -1. The eigenvalues of R ([1],[1 n-1 ]) (S 1
) are -1 with multiplicity n -1 and α with multiplicity 1. Therefore we have either -1 = uΦ(α) and α = -u or -1 = uΦ(α -1 ) and α = -u. In both cases u = -α and Φ(α + α -1 ) = α + α -1 . We have Φ trivial, therefore

F q = F p (α+α -1 , β+β -1 ). This would imply R ([1],[n-1])|A Bn S(R ([1],[1 n-1 ])|A Bn ) but ([1], [1 n-1 ]) / ∈ {([1], [n -1]), ([1], [n -1]) } when n > 2. By Proposition 3.2.4, this is absurd. This shows the image of A Bn in G 0 is equal to G 0 .
For

λ 0 ∈ n = {λ n, λ / ∈ A n , λ not a hook}, we set G λ 0 = SL n-1 (q) × (λ 1 ,∅)∈A n,λ1<λ 1 SL n λ (q) × (λ 1 ,∅)∈A n,λ1=λ 1 OSP (λ)× SL n (q) 2 × λ∈ n,λ<min(λ ,λ 0 ) SL n λ (q) × λ∈ n,λ=λ <λ 0 OSP (λ).
where OSP (λ) is the group of isometries of the bilinear form defined before Proposition 3.1.1.

For the minimal element λ 0 of n , we just showed the composition Φ 1,n with the projection onto G λ 0 = G 0 is surjective. Let us show by induction (numbering the double-partitions of n with the order defined previously) that for all λ 0 , the composition of Φ 1,n with the projection onto G λ 0 is surjective.

Let λ 0 ∈ n . Assume that the composition is surjective onto G λ 0 and let us show that the composition onto

G λ 0 +1 = G λ 0 × G(λ 0 ) is surjective, where G(λ 0 ) = SL N (q) if λ 0 = λ 0 and G(λ 0 ) OSP (λ 0 ) ∈ {SP N (q), Ω + N (q)} if λ 0 = λ 0 . We use Goursat's Lemma with K 1 = G λ 0 and K 2 = G(λ 0 ) on the image Φ 1,n in K 1 × K 2 .
As before, it is sufficient to show that the quotients K 1 /K 1 K 2 /K 2 are abelian. Assume they are non-abelian. The only non-abelian Jordan-Hölder factor of G(λ 0 ) is P G(λ 0 ), therefore there exists λ less than λ 0 such that up to conjugation (see [START_REF] Wilson | The finite simple groups[END_REF] 3.3.4., 3.5.5. and 3.7

.5) R λ (h) = S Φ (R λ 0 (h)z(h) for all h ∈ A Bn (there is no triality involved since if n ≥ 5, λ = λ and λ ∈ n then dim(V λ ) > dim(V ([2,1],[2,1]) ) = 80 > 8)
. By the same arguments as in the induction initialization, we have that λ has no empty components. Since n ≥ 5, A Bn is perfect and z is trivial. We then have R λ 0 |A Bn S Φ (R λ|A Bn ). Let us show that Φ is trivial. By Lemma 3.2.3, there exists u, v ∈ F q such that up to conjugation, for all h ∈ A Bn , we have

R λ 0 (h) = S Φ (R λ (h))u 1 (h) v 2 (h) . Comparing eigenvalues of T , we get either {β, -1} = {vΦ(β), -v} or {β, -1} = {vΦ(β -1 ), -v}. In the first case, either v = 1 and Φ(β) = β or v = -β and -1 = vΦ(β), therefore Φ(β + β -1 ) = β + β -1 . In the second case either v = 1 and Φ(β -1 ) = β or v = -β and vΦ(β -1 ) = -1, therefore Φ(β + β -1 ) = β + β -1 .
In the same way using S 1 , we show Φ(α + α -1 ) = α + α -1 . This shows that Φ is trivial because

F q = F p (α + α -1 , β + β -1 ). We then have R λ 0 |A Bn S(R λ|A Bn ) which contradicts Proposition 3.2.4 since λ < λ 0 ≤ λ 0 .
To get that Φ 1,n is surjective, it now only remains to show that what we assumed in Lemma 3.3.4 is true. Theorem 3.3.5. If n ≥ 5 then for all λ n double-partitions in our decomposition, we have R λ (A Bn ) = G(λ), where G(λ) is the corresponding group in the following list.

1. SL n-1 (q) if λ = ([n -1, 1], ∅). 2. SL N (q) if λ = (λ 1 , ∅), λ 1 < λ 1 . 3. SP N (q) if λ = (λ 1 , ∅), λ 1 = λ 1 and ( p = 2 or (p ≥ 3 and ν(λ 1 ) = -1)). 4. Ω + N (q) if λ = (λ 1 , ∅), λ 1 = λ 1 , p ≥ 3 and ν(λ 1 ) = 1. 5. SL n (q) if λ ∈ {([1], [n -1]), ([1], [1 n-1 ])}. 6. SL N (q) if λ ∈ n , λ < λ . 7. SP N (q) if λ = λ and (p = 2 or ν(λ) = -1). 8. Ω + N (q) if λ = λ , p ≥ 3 and ν(λ) = 1.
Proof. Let n ≥ 5. By [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] (Theorem 1.1.), it is sufficient to show it for λ ∈ n . The result is true for n = 4, therefore we can use induction and assume Φ n-1 is surjective. The first thing to do is to take care of the double-partitions such that n λ > 8 and n λ = 10. For n = 5, the double partitions to consider are ([

1 3 ], [1 2 ]), ([1 2 ], [1 3 ]), ([1], [2, 2]), ([1], [2, 1 2 ]), ([1], [3, 1]), ([2], [2, 1]) and ([1 2 ], [2, 1]
) of respective dimensions 10, 10, 10, 15, 15, 20 and 20. For n = 6, they are ([ 15, 15, 20, 24, 24, 30, 30, 36, 30, 30, 45, 45, 45, 45, 40, 40 and 80. We can now note that if n = 6, we have n λ ≥ 15, therefore by the branching rule, if n ≥ 6 and λ is a doublepartition of n then n λ ≥ 15. The only double-partitions λ such that n λ ≤ 8 or n λ = 10 are ([ [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]) which are of dimension 10. By Lemma 3.3.2 and the branching rule, we have that R

1 2 ], [1 4 ]), ([1 4 ], [1 2 ]), ([1 3 ], [1 3 ]), ([1], [4, 1]), ([1], [2, 1 3 ]), ([1], [3, 2]), ([1], [2 2 , 1]), ([1], [3, 1, 1]), ([2], [2 2 ]), ([1 2 ], [2 2 ]), ([2], [3, 1]), ([1 2 ], [3, 1]), ([2], [2, 1 2 ]), ([1 2 ], [2, 1 2 ]), ([3], [2, 1]), ([1 3 ], [2, 1]) and ([2, 1], [2, 1]) of respective dimensions
1 3 ], [1 2 ]), ([1 2 ], [1 3 ]) and ([1],
[1],[2,2] (A B 4 ) = SL 8 (q)×SL 2 (q), R [1 3 ],[1 2 ] (A B 4 ) = SL 4 (q)×SL 6 (q) and R [1 3 ],[1 2 ] (A B 4 ) = SL 4 (q) × SL 6 (q). By Theorem 3.3.2, we have that R ([1],[2,2]) (A B 5 ) = R [1 3 ],[1 2 ] (A B 5 ) = R [1 2 ],[1 3 ] (A B 5 ) SL 10 (q).
We now show in the same way as in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] (Part 5), that the other assumptions of Theorem 2.3.2 are verified. In order to do this, we use the following results shown in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF].

Lemma 3.3.5. If d ≥ 6 and G ≤ GL d (q) contains an element conjugate to an element of the form diag(ξ, ξ -1 , 1, 1, ...) with ξ 2 = 1, then G is tensor-indecomposable. Lemma 3.3.6. If d ≥ 16 and G ≤ GL d (q) contains an element of order prime to p conjugate to an element of the form diag(ξ, ξ, ξ -1 , ξ -1 , 1, 1, .., 1) with ξ 2 = 1, then G is tensor-indecomposable except possibly if G ≤ G 1 ⊗ G 2 with G 1 ≤ GL 2 (q).
For a block diagonal matrix with blocks B 1 , . . . , B r , we write diag(B 1 , . . . , B r ).

Lemma 3.3.7. If G contains a natural SL 2 (q) and q ≥ 8 or G contains a twisted diagonal embedding of SL 2 (q) (G ⊃ {diag(M, t (M -1 ), I N -4 ), M ∈ SL 2 (q)}), then case (2) of Theorem 2.3.2 is excluded.
By the proof of the imprimitivity of G in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], it is sufficient to show that A Bn is normally generated by A B n-1 and that G contains either a transvection or an element of Jordan form diag(

I 2 + E 1,2 , I 2 + E 1,2 , I N -4 ) to get that G is imprimitive.
In order to show that we are in case (1) of Theorem 2.3.2, we must show that for n ≥ 5, that we have q ≥ 8 and that for any double-partition λ of n, G = R λ (A Bn ) contains either a natural SL 2 (q) and n λ > 6 or contains a twisted diagonal embedding of SL 2 (q) and n λ > 16. We must also prove that A Bn is normally generated A B n-1 and the exceptional case of Lemma 3.3.6 is impossible when n λ > 16, G contains a twisted diagonal embedding of SL 2 (q) but no natural SL 2 (q) in an obvious way.

Let n ≥ 5, assume the lemma is true for all m ≤ n -1. By Lemmas 3.3.4 and 3.3.2, we have Φ m surjective for all m ≤ n -1. By assumption, α is of order strictly greater than 5 and not belonging to {1, 2, 3, 4, 5, 6, 8, 10}. This implies that α is of order at least 7 and that q ≥ 8. If λ has at most two columns then since λ ∈ n , λ contains a natural SL 2 (q). Assume now λ ∈ n has at least three rows or three columns.

Assume that for all µ ⊂ λ containing ([2, 1], [1]) or ([1], [2, 1]), we have µ ⊂ λ. We then have that λ = λ and n is even. Since n is even, we have µ = µ for any double-partition µ ⊂ λ. Since λ ∈ n and contains strictly more than two rows and two columns, there exists

µ ⊂ λ containing ([1], [2, 1]) or ([2, 1], [1]). Since Φ m-1 is surjective and µ ⊂ λ, we have a twisted diagonal embedding of SL nµ (q) in G = R λ (A Bn ) and since n µ ≥ 8 ≥ 2, we have a twisted diagonal embedding of SL 2 (q). Otherwise there exists µ ⊂ λ containing ([2, 1], [1]) or ([1], [2, 1]) such that µ ⊂ λ. Since Φ m-1 is surjective, we get that λ contains a natural SL nµ (q)
and, therefore contains a natural SL 2 (q). For double-partitions which are not of dimension strictly greater than 16, i.e. ([1], [2, 1 2 ]) and ( [1], [3,1]), we are in the second case.

We now show that A Bn is normally generated by A B n-1 for n ≥ 5. By [12, Lemma 2.1], we have that A An is normally generated by A A n-1 for n ≥ 4. Since T commutes with S i for all i ≥ 2, we have the same result for A Bn for n ≥ 4.

It now only remains to show that the exception of Lemma 3.3.6 is impossible when there is no obvious natural SL 2 (q) in G. In order to do this, we show a proposition analogous to Proposition 2.4. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF].

Proposition 3.3.2. Let K be a field. If n ≥ 7 and ϕ : A Bn → P SL 2 (K) is a group morphism then ϕ = 1.
Proof. Let K be a field, n ≥ 7 and ϕ such a morphism. The restriction of ϕ to A An ≤ A Bn is trivial by Proposition 2.4. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. By Theorem 3.9. of [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A Bn is generated by

p 0 = S n-2 S -1 n-1 , p 1 = S n-1 S n-2 S -2 n-1 , q 3 = S n-3 S -1 n-1 , b = S n-2 S -1 n-1 S n-3 S -1 n-2 , r l = T l S 1 T -l S n-1 , q i = S n-i S -1 n-1 , l ∈ Z, 4 ≤ i ≤ n - 2
and the following relations.

1. For 4 ≤ j ≤ n -2, p 0 q j = q j p 1 and p 1 q j = q j p -1 0 p 1 .

2. For l ∈ Z, p 0 r l = r l p 1 and p 1 r l = r l p -1 0 p 1 .

For 3

≤ i < j ≤ n -2, |i -j| ≥ 2, q i q j = q j q i . 4. For 3 ≤ i ≤ n -3, q i r l = r l q i . 5. p 0 q 3 p -1 0 = b, p 0 bp -1 0 = b 2 q -1 3 b 6. p 1 q 3 p -1 1 = q -1 3 b, p 1 bp -1 1 = (q -1 3 b) 3 q -2 3 b. 7. For 3 ≤ i ≤ n -3, q i q i+1 q i = q i+1 q i q i+1 .
8. For l ∈ Z, q n-2 r l q n-2 = r l q n-2 r l . 9. For l ∈ Z, r l r l+1 = r l+1 r l+2 .

By [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]Proposition 2.4], the images of all the generators except for (r l ) l∈Z are trivial. By the eighth relation, we get that the images of the r l are also trivial and the desired result follows.

This shows that if n ≥ 7 and

G ≤ G 1 ⊗ G 2 with G 1 ≤ GL 2 (q), then G ⊂ SL N 2 (q) × SL N 2 (q).
This contradicts the irreducibility. Since we need n ≥ 7 to apply this reasoning, we must consider separately the cases where n ∈ {5, 6} and G does not contain a natural SL 2 (q). Looking at all the cases enumerated previously, the only one to consider is λ = ([2, 1], [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]1]). Up to conjugation, we have

H = R [2,1],[2,1] (A B 5 ) {diag(M, t (M -1 ), N, t (N -1 )), M, N ∈ SL 20 (q)} SL 20 (q) × SL 20 (q). Assume that G = R ([2,1],[2,1]) (A B 6 ) ⊂ G 1 ⊗ G 2 and that G 1 ⊂ GL 2 (q).
We then have a morphism θ from G to SL 2 (q) since A Bn is perfect for n ≥ 5. If we consider the restriction of θ to H, its kernel is a subgroup of H and its image is a subgroup of SL 2 (q). Since P SL 20 (q) is the only non-abelian composition factor of H, we have that if the image is non-abelian then there exists a subgroup of SL 2 (q) isomorphic to P SL 20 (q). This is absurd, therefore the image is abelian and the kernel contains the derived subgroup of H which is equal to H since H is perfect. In the same way, for all g ∈ G, the restriction of θ to gHg -1 is trivial. Since H normally generates G, θ is trivial which contradicts the irreducibility of G in the same way as in the proof of the previous proposition.

We have thus shown that we are in the first case of Theorem 2.3.2. By the same reasoning as in [12, page 16], we have in all cases that q = q. If λ = λ , we have G ⊂ G(λ) by Proposition 3.2.1, therefore G = G(λ). If λ = λ , G preserves no bilinear form since R λ is not isomorphic to R λ . If G preserves a hermitian form then there exists an automorphism Φ of order 2 of F q such that M is conjugate to Φ( t (M ) -1 ) for all M ∈ G. Since G contains a natural SL 2 (q), we then have Tr(diag(α,

α -1 , 1, 1, ..., 1)) = Φ(Tr(diag(α -1 , α, 1, ..., 1))) and Tr(diag(β, β -1 , 1, 1, ..., 1)) = Φ(Tr(diag(β -1 , β, 1, ..., 1))), therefore Φ(α + α -1 ) = α + α -1 and Φ(β + β -1 ) = β + β -1 . This implies that Φ = I d because F q = F p (α + α -1 , β + β -1
). This is absurd and we conclude that G = SL n λ (q). By Theorem 3.3.5, Lemma 3.3.2 and Lemma 3.3.4, we have that for all n, Φ 1,n is surjective.

Cases (2) and (3)

We have shown the surjectivity of Φ 1,n , this corresponds to the first of the six possible field extension configurations described at the beginning of subsection 3.2.2. The proof in cases ( 2) and (3) only requires small changes to the one in the first case, but the new factorizations appearing in cases (4) to (6) require more work, especially for the low dimensional representations. We treat in this section cases (2) and (3) emphasizing on the differences with the first case. This will conclude the proof of Theorem 3.2.3 and the corresponding statement in the second of the six cases listed at the beginning of subsection 3.2.2.

In case (2), i.e.,

F q = F p (α, β) = F p (α + α -1 , β + β -1 ) and F p (α) = F p (α + α -1
), the same arguments as the ones in case 1 work at every step of the proof. Indeed, SU 2 (q 1 2 ) is also generated by a conjugacy class of transvections. Since q is a square and α is order not diving 8 by assumption, we have that q ≥ 16 and q 1 2 > 3. We also still have that SL 8 (q) × SU 2 (q

2 ) contains SL 8 (q) × {I 2 }, therefore all the arguments work in the same way. This shows that in case (2), Φ 2,n is surjective for all n.

In case (3), i.e.,

F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1
), all representations are unitary. The main differences occur in the proof that when n = 4, the direct product of two SU 4 (q

1 2
) is in the image, and in the conclusion of the proof of this version of Theorem 3.3.5.

Theorem 3.3.6. If n ≥ 5, then for all λ n in our decomposition, R λ (A Bn ) = G(λ)
, where G(λ) is the corresponding group in the following list.

1. SU n-1 (q 1 2 ) if λ = ([n -1, 1], ∅). 2. SU N (q 1 2 ) if λ = (λ 1 , ∅), λ 1 < λ 1 . 3. SP N (q 1 2 ) if λ = (λ 1 , ∅), λ 1 = λ 1 and ( p = 2 or (p ≥ 3 and ν(λ 1 ) = -1)). 4. Ω + N (q 1 2 ) if λ = (λ 1 , ∅), λ 1 = λ 1 , p ≥ 3 and ν(λ 1 ) = 1. 5. SU n (q 1 2 ) if λ ∈ {([1], [n -1]), ([1], [1 n-1 ])}. 6. SU N (q 1 2 ) if λ ∈ n , λ < λ . 7. SP N (q 1
2 ) if λ = λ and (p = 2 or (p ≥ 3 and ν(λ) = -1.

8. Ω + N (q 1
2 ) if λ = λ , p ≥ 3 and ν(λ) = 1. Proof. We recall Proposition 4.1. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. Proposition 3.3.3. Let q = u 2 , ϕ be a non-degenerate bilinear form over F N q , ψ a nondegenerate hermitian form over

F N q . If G ⊂ OSP N (ϕ) ∩ U N (ψ)
is absolutely irreducible, then there exists x ∈ GL N (q) and a non-degenerate bilinear form ϕ over F N u such that x G ⊂ OSP (ϕ ) and ϕ is of the same type as ϕ.

When n = 4, the proof that Φ 4 is surjective is the same up to the point, where we prove Φ is trivial using Φ(α + α -1 ) = α + α -1 and Φ(β + β -1 ) = β + β -1 . In case 3, Φ could also be equal to the automorphism of order 2 of F q . It is thus necessary to show that the following is absurd :

α 2 β 2 -2α 2 β -αβ 2 + 5αβ -α -2β + 1 αβ = ( -2α 2 β -αβ 2 + α 2 + 5αβ + β 2 -α -2β αβ ).
This would imply

α 2 β 2 -2α 2 β -αβ 2 + 5αβ -α -2β + 1 = -2α -2 β -1 -α -1 β -2 + α -2 + 5α -1 β -1 + β -2 -α -1 -2β -1 α -2 β -2 = -2β -α + β 2 + 5αβ + α 2 -αβ 2 -2α 2 β.
This is absurd because it is the same equality we proved to be impossible in the first case. We now adapt the end of the proof of the corresponding version of Theorem 3.3.5. By [12, page 18], we are in case (1) of Theorem 2.3.

2. If λ = λ , G contains a natural SU 3 (q 1 2 ), therefore q = q by Lemma 3.3.3. Since G ⊂ SU n λ (q 1
2 ) and G preserves no bilinear form by Proposition 3.2.4, we have G SU n λ (q

1 2 ). If λ = λ , we use Proposition 3.3.3 to get that G ⊂ OSP (q 1
2 ). By Lemma 3.3.3, we have that F q contains {x + (x), x ∈ F q }. This implies that q = q 1 2 because F q contains α + α -1 and β + β -1 and q divides q 1 2 . We conclude that G OSP (q 1 2 ).

Cases (4), (5) and (6)

In this subsection, we finish the proof for type B by considering the last three cases for the field extensions listed at the beginning of subsection 3.2.2. This will conclude the proofs of Theorems 3.2.4, 3.2.5 and 3.2.6. In these cases more factorizations appear and this complicates greatly the proof for small n. We will use the tables of maximal subgroups of finite classical groups in low dimension from [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF]. This gives interesting techniques to determine if a certain subgroup G of a classical group is the group itself, when given information on the subgroups of G.

In these cases, we can still use various arguments from the first case, but except for Proposition 3.3.1 which remains true in all these cases, all the low-dimensional cases must be done again. It is not necessary to use new arguments for Lemma 3.3.4. We start by studying the case n ≤ 4. Lemma 3.3.8. For i ∈ {4, 5, 6} and n ≤ 4, we have Φ i,n surjective.

Proof. For n = 2, using the same arguments as in the first case and Lemma 3.2.5, we have that Im(Φ 2 ) SL 2 (q 1 2 ). For n = 3, we have by the factorizations in Proposition 3.2.4 that in all cases Φ 3 is surjective. The only case left to consider is n = 4 and the double-partitions ([

1 2 ], [1 2 ]) and ([1], [2, 1]) of respective dimensions 6 and 8. We have to prove that R ([1 2 ],[1 2 ]) (A B 4 ) SL 6 (q 1 2 ) or SU 6 (q 1 2 ) and R ([2,1],[1]) (A B 4 ) SL 8 (q 1 2 ) or SU 8 (q 1
2 ) depending on the case. We start by 5) or (6), where

G = R ([1 2 ],[1 2 ]) (H 4 ) in case (
F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ). We then have H = R ([1 2 ],[1 2 ]) (A B 3 ) SL 3 (q). Since ([1 2 ], [1 2 ]) = ([1 2 ], [1 2 ]
), by Proposition 3.2.4 (4.c) and Lemma 3.2.5, up to conjugation, we have G ⊂ SL 6 (q 1 2 ). We use the classification of maximal subgroups SL 6 (q 1 2 ) [9, Tables 8.24 and 8.25] . Using the fact that H is a subgroup of G, we exclude the possibility that G is included in all but two of these groups, using the divisibility of the orders that would ensue. We start by considering the sporadic maximal subgroups in Table 8.25 and get the orders of these groups using the atlas [START_REF] Conway | Maximal subgroups and ordinary characters for simple groups, With computational assistance from[END_REF]. We list below those groups and their order or a quantity their order divides

1. 2 × 3 . A 6 .2 3 , 4320, 2. 2 × 3 . A 6 , 2160, 3. 6 . A 6 , 2160, 4. (q 1 2 -1, 6) • 2 . P SL 2 (11), 12 × 660 = 7920, 5. 6 . A 7 , 15120, 6. 6 . P SL 3 (4) . 2 - 1 , 6 × 2 × 20160 = 241920, 7. 6 . P SL 3 (4), 6 × 20160 = 120960, 8. 2 . M 12 , 2 × 95040 = 190080, 9. 6 . 1 P SU 4 (3) . 2 - 2 , 6 × 2 × 3265920 = 39191040, 10. 6 . 1 P SU 4 (3), 6 × 3265920 = 19595520, 11. (q 1 2 -1, 6) • SL 3 (q 1 2 ), 6 × q 3 2 (q -1)(q 3 2 -1).
Since q is a square and α is of order greater than 4, we have q ≥ 9. This implies that |SL 3 (q)| = q 3 (q 2 -1)(q 3 -1) ≥ 9 3 (9 2 -1)(9 3 -1) = 42456960, which is greater than all the orders in the list (the last one is of order 6q 1 2 (q -1)(q 3 -1) and 6 < q (q + 1)(q 3 2 + 1)). We now look at the list in table 8.24 of the 18 geometric maximal subgroups of SL 6 (q 1 2 ), which we provide below with their order or a quantity which is divisible by their order

1. E 5 q 1 2 : GL 5 (q 1 2 ), q 15 2 (q 5 2 -1)(q 2 -1)(q 3 2 -1)(q -1)(q 1 2 -1), 2. E 8 q 1 2 : (SL 4 (q 1 2 ) × SL 2 (q 1 2 )) : (q 1 2 -1), q 15 2 (q 2 -1)(q 3 2 -1)(q -1) 2 (q -1), 3. E 9 q 1 2 : (SL 3 (q 1 2 ) × SL 3 (q 1 2 )) : (q 1 2 -1), q 15 2 (q 3 2 -1) 2 (q -1) 2 (q 1 2 -1), 4. E 1+8 q 1 2 : (GL 4 (q 1 2 ) × (q 1 2 -1)), q 15 2 (q 2 -1)(q 3 2 -1)(q -1)(q 1 2 -1) 2 , 5. E 4+8 q 1 2 : SL 2 (q 1 2 ) 3 : (q 1 2 -1) 2 , q 15 2 (q -1) 3 (q 1 2 -1), 6. GL 5 (q 1 2 ), q 5 (q 5 2 -1)(q 2 -1)(q 3 2 -1)(q -1)(q 1 2 -1), 7. (SL 4 (q 1 2 ) × SL 2 (q 1 2 )) : (q 1 2 -1), q 7 2 (q 2 -1)(q 3 2 -1)(q -1) 2 (q 1 2 -1), 8. (q 1 2 -1) 5 × S 6 , 6!(q 1 2 -1) 5 , 9. SL 2 (q 1 2 ) 3 : (q 1 2 -1) 2 .S 3 , 6q 3 2 (q 1 2 -1) 2 (q -1) 3 , 10. SL 3 (q 1 2 ) 2 : (q 1 2 -1).S 2 , 2q 3 2 (q 3 2 -1)(q -1)(q 1 2 -1), 11. SL 3 (q).(q 1 2 + 1).2, 2(q 1 2 + 1)|SL 3 (q)|, 12. SL 2 (q 3 2 ).(q + q 1 2 + 1).3, 3q 3 2 (q 3 -1)(q + q 1 2 + 1), 13. SL 2 (q 1 2 ) × SL 3 (q 1 2 ), q 2 (q 3 2 -1)(q -1) 2 , 14. SL 6 (q 0 ).[( q 1 2 -1 q 0 -1 , 6)],
where q 1 2 = q r 0 and r prime, 6|SL 6 (q 0 )|, 15. (q

1 2 -1, 3) × SO + 6 (q 1
2 ).2, q odd, 6q 3 (q 2 -1)(q

3 2 -1)(q -1), 16. (q 1 2 -1, 3) × SO - 6 (q 1
2 ).2, q odd, 6q 3 (q 2 -1)(q

3 2 + 1)(q -1), 17. (q 1 2 -1, 3) × SP 6 (q 1
2 ), 3q 4 ).(q

9 2 (q -1)(q 2 -1)(q 3 -1) = 3q 3 2 (q -1)|SL 3 (q)|, 18. SU 6 (q 1
1 4 -1, 6), 6q 15
4 (q 3 2 -1)(q 5 4 + 1)(q -1)(q 3 4 + 1)(q 1 2 -1).
In cases 1, 2, 3, 4, 5, 6, 7 and 13, the order of the maximal subgroup divides q 15 2 (q

5 2 -1)(q 2 - 1)(q 3 2 -1) 2 (q -1) 2 (q 1 2 -1)
. This implies that it is sufficient to show that |SL 3 (q)| = q 3 (q 3 -1)(q 2 -1) does not divide this quantity to exclude these cases. It can be true only if q 3 -1 divides (q 5 2 -1)(q 3 2 -1) 2 (q -1) 2 (q 1 2 -1). The Euclidean remainder of those two quantities seen as polynomials in q 1 2 is 4q

5 2 + 2q 2 -2q 3 2 -4q -2q 1 2 + 2.
Therefore, if q 3 -1 divides the first quantity then it divides the remainder, which is positive. Therefore it is less than or equal to it. We have (α) = α -1 = α q 1 2 , therefore α q 1 2 +1 = 1. Since α is of order strictly greater than 6 by assumption, we have that q 1 2 ≥ 6 and 4q

5 2 + 2q 2 -q -2q 3 2 -4q -2q 1 2 + 2 ≤ 4q 5 2 + 2q 2 + 2 ≤ 4q 5 2 + 3q 2 ≤ 5q 5 2 < q 3 -1.
This gives us the desired contradiction. Cases 8, 9, 10 and 12 are excluded because q 3 is coprime to q r -1 and (q + q 1 2 + 1) for every integer r and q 3 does not divide 6! or 6q

3 2 since q = (q 1 2 ) 2 ≥ 36.
In case 14, the order of the maximal subgroup M divides the quantity

6q 15 2r (q 3 r -1)(q 5 2r -1)(q 2 r -1)(q 3 2r -1)(q 1 r -1),
where q 1 2r = q 0 and r is a prime. If |SL 3 (q)| divides this quantity, then q 3 divides 6q

15 2r . If r ≥ 3, then 6q 15 2r ≤ 6q 15 6 < q 3 because q 1 2 ≥ 8 > 6 when r ≥ 3.
It only remains to consider the case r = 2. We then have that (q

3 2 -1)(q 3 2 + 1)(q -1) divides 6(q 3 2 -1)(q 5 4 -1)(q -1)(q 3 4 -1)(q 1 2 -1), therefore q 3 2 + 1 divides 6(q 5 4 -1)(q 3 4 -1)(q 1 2 -1)
. The Euclidean remainder of the division of those two polynomials in q 1 4 is -6q

4 4 +6q 3 4 +12q 2 4 +6q 1 4 -6.
This implies that q 3 2 + 1 divides the above quantity. We have q 1 2 ≥ 6 and it is a square when r = 2. Therefore q 1 2 ≥ 9 and q

1 4 ≥ 3. It follows that -6q 4 4 +6q 3 4 +12q 2 4 +6q 1 4 -6 ≤ -6q 4 4 +11q 3 4 ≤ -4q 4 4 < 0. We then have that q 3 2 + 1 divides 6q 4 4 -6q 3 4 -12q 2 4 -6q 1 4 + 6 < 6q 4 4 ≤ 2q 5 4 < q 3 2 + 1,
which is absurd since both quantities are positive. Case 14 is therefore excluded.

The four last cases are of class C 8 , therefore they preserve a non-degenerate bilinear form. We cannot have P GP -1 included in one of those groups because

R [1 2 ],[1 2 ]|A B 4 R [1 2 ],[1 2 ]|A B 4 by Proposition 3.2.4.
The only remaining case is now cases 11 which is SL 3 (q).(q

1 2 + 1).2. We know that H = R ([1 2 ],[1 2 ]) (A B 3 ) SL 3 (q) normally generates G = R ([1 2 ],[1 2 ]) (A B 4 ) ⊂ P -1 SL 6 (q 1
2 )P for a certain matrix P in GL 6 (q). Assume P GP -1 is a subgroup of M = SL 3 (q).(q 1 2 + 1).2. Since SL 3 (q) is perfect, P HP -1 is perfect and the image of P HP -1 in the quotient Z/2Z of M is trivial. The group H is thus included in SL 3 (q).(q 1 2 + 1). Using the same argument, we have that P HP -1 is included in the SL 3 (q) appearing in the expression of M , therefore P HP -1 is equal to that SL 3 (q). For all g ∈ P GP -1 , we can apply the same reasoning to gP HP -1 g -1 = SL 3 (q) = P HP -1 . It follows that P GP -1 = SL 3 (q) = P HP -1 because H normally generates G, therefore H = G. This leads to a contradiction because G is irreducible and H is not.

This concludes the study of double-partition ([1 2 ], [1 2 ]) in the field cases 5 and 6 and we have R

[1 2 ],[1 2 ] (A B 4 ) SL 6 (q 1
2 ) in those cases.

Assume now we are in case 4, i.e.,

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1
). Then by Lemma 3.2.4 and Proposition 3.2.4, there exists a matrix P such that P GP -1 ⊂ SU 6 (q 1 2 ) and H SL 3 (q), writing again

G = R ([1 2 ],[1 2 ]) (A B 4 ) and H = R ([1 2 ],[1 2 ]) (A B 3 ).
The goal this time is to show that P GP -1 SU 6 (q 1 2 ). We first consider the maximal subgroups of class S of SU 6 (q 1 2 ) given in Table 8.27 of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] and give their order or a quantity their order divides. [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], 12 × 660 = 7920, 5. 6 . A 7 , 15120, 6. 6 . P SL 3 (4), 6 × 20160 = 120960, 7.

1. 2 × 3 . A 6 , 2160, 2. 2 × 3 . A 6 .2 3 , 4320, 3. 6 . A 6 , 2160, 4. (q 1 2 + 1, 6) • 2 . L 2
6 . P SL 3 (4) . 2 - 1 , 241920, 8. 3 . M 22 , 3 * 443520 = 1330560, 9. 3 . 1 U 4 (3) : 2 2 , 6 × 3265920 = 19595520, 10. 6 . 1 U 4 (3), 19595520, 11. 6 . 1 U 4 (3) . 2 - 2 , 39191040, 12. (q 1 2 + 1, 6) • SU 3 (q 1
2 ), 6(q

3 2 (q -1)(q 3 2 + 1).
As before, we have q ≥ 9. Therefore |SL 3 (q)| ≥ 42456960 and the last case is excluded since q 3 does not divide 6q Consider now the maximal subgroups of geometric type. We here omit the groups of class C 1 because we know that P GP -1 is irreducible. The remaining maximal subgroups obtained from Table 8.26. of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] of SU 6 (q 1 2 ) are the following 1. (q 1 2 + 1) 5 .S 6 , 720(q

1 2 + 1) 5 , 2. SU 2 (q 1 2 ) 3 : (q 1 2 + 1) 2 .S 3 , 6q 3 2 (q 1 2 + 1) 2 (q -1) 3 , 3. SU 3 (q 1 2 ) 2 : (q 1 2 + 1).S 2 , 2q 3 (q 1 2 + 1)(q -1) 2 (q 3 2 + 1) 2 , 4. SL 3 (q).(q 1 2 -1).2, 2(q 1 2 -1)|SL 3 (F q )|, 5. SU 2 (q 3 2 ).(q -q 1 2 + 1).3, 3q 3 2 (q 3 -1)(q -q 1 2 + 1), 6. SU 2 (q 1 2 ) × SU 3 (q 1 2 ), q 2 (q -1) 2 (q 3 2 + 1), 7. SU 6 (q 0 ).[( q+1 q 0 +1 , 6)], q 0 = q 1 2r
, r odd prime, 6q 15 0 (q 2 0 -1)(q 3 0 -1)(q 4 0 -1)(q 5 0 -1)(q 6 0 -1), 8. (q

1 2 + 1, 3) × SP 6 (q 1
2 ), 3q

3 2 (q -1)|SL 3 (q)|, 9. (q 1 2 + 1, 3) × SO + 6 (q 1
2 ).2, 6q 3 (q 2 -1)(q 3 2 -1)(q -1), 10. (q

1 2 + 1, 3) × SO - 6 (q 1
2 ).2, 6q 3 (q 2 -1)(q 3 2 + 1)(q -1).

In case 6, we have that the order of the maximal subgroup is q 2 (q -1) 2 (q 3 2 + 1). If |SL 3 (q)| divides this quantity, then q 3 -1 divides (q -1)(q

3 2 + 1) = q 5 2 -q 3 2 + q -1 < q 5 2 -1 < q 3 -1.
This is absurd, therefore this case is excluded.

In case 1, we have that q 3 divides 720 which is absurd since q ≥ 36. In the same way, cases 2 and 5 are excluded because q 3 does not divide 6q 3 2 . In case 3, we have that q 3 -1 divides 2(q 1 2 + 1)(q -1) 2 (q 3 2 + 1). The Euclidean remainder of the division of those two polynomials in q 1 2 is -4q

5 2 + 8q 2 -4q 3 2 -4q + 8q 1 2 -4n which is negative. We have 4q 5 2 -8q 4 2 + 4q 3 2 + 4q -8q 1 2 + 4 ≤ 4q 5 2 + 6q 3 2 ≤ 5q 5 2 < q 3 -1 when q 1 2 ≤ 3.
This case is therefore also excluded.

In case 7, we have that q 3 divides 6q 15 0 = 6q 15 2r ≤ 6q 15 6 < q 3 since q 1 2 = q r 0 ≥ 8. In cases 8, 9 and 10, we have P GP -1 included in a subgroup preserving a non-degenerate bilinear form. This would imply that

•R [1 2 ],[1 2 ]|A B 4 R [1 2 ],[1 2 ]|A B 4 which is absurd by Proposition 3.2.4.
The last remaining case is case 4. In the same way as in case 11 when

F p (α, β) = F p (α + α -1 , β), we would have that H = G because R ([1 2 ],[1 2 ]) (S 2 S -1 3 ) ∈ G \ H.
Since G is normally generated by H and gHg -1 is perfect for all g ∈ G, we would have that G ⊂ SL 3 (q) and, therefore G = H, which is absurd.

We have shown that P GP -1 cannot be included in any maximal subgroup of SU 6 (q

1 2 ). It follows that P GP -1 SU 6 (q 1 2 ).
The only double-partition remaining for n ≤ 4 now is λ = ([2, 1], [1]). It affords a representation of dimension 8 and satisfies λ = (λ 1 , λ 2 ).

We start by case 4, i.e.,

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1
) and, therefore F q = F p (α + α -1 ) = F p (α). We then have by Goursat's Lemma and the result for n = 3 that

H = R ([2,1],[1]) (A B 3 ) SL 3 (q) × SU 2 (q 1 2 ) ⊂ G = R ([2,1],[1]) (A B 4 )
. By Proposition 3.2.4, we know that there exists P ∈ GL 8 (q) such that for all h ∈ H 4 , P R ([2,1], [1]) (h)P -1 = (R ([2,1], [1]) )(h). By Lemma 3.2.5, this implies that there exists S ∈ GL 8 (q) such that

S -1 R ([2,1],[1]) (A B 4 )S ⊂ GL 8 (q 1
2 ) with γ -1 P = (S)S -1 and (P )P = (γ)γ. We can use the arguments used previously to see that our group is primitive, irreducible, tensor-indecomposable, preserves no symmetric, skew-symmetric or hermitian form over F 1 2 q and cannot be included in GL 8 (q ) for q < q 1 2 . We then get that G is included in no maximal subgroup of class C 1 , C 2 , C 4 , C 5 and C 8 . It contains a transvection, therefore it cannot be included in a maximal subgroup of class C 3 . We list below the maximal subgroups remaining obtained from Tables 8.44 and 8.45 of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF]. We give the order of those groups or a quantity their order divides. We have that α is of order greater than or equal to 7 and (α) = α -1 , where is the unique automorphism of order 2 of F q . It follows that α q 1 2 +1 = 1, therefore q 1 2 + 1 ≥ 7 and q 1 2 ≥ 6. We have F q = F p (α) = F p (α + α -1 ), therefore ˜ (α) = α -1 , where ˜ is the unique automorphism of order 2 of F q. This implies that q 1 2 ≥ 6. It follows that |H| ≥ 591963268176000. This excludes all the maximal subgroups in the list. It follows that S -1 GS is included in no maximal subgroup of SL 8 (q 1 2 ), therefore G SL 8 (q 1 2 ). We now consider cases 5 and 6, where our representation is now unitary by Proposition 3.2.4. In both cases, there exists a matrix P such that P GP -1 ⊂ SU 8 (q

((8,

q 1 2 + 1) • 2 1+6 ) . (SP 6 (2)),
1 2 ) with G = R ([2,1],[1]) (A B 4 ) and we have H = R ([2,1],[1]) (A B 3 ) SL 3 (q)×SU 2 (q 1
2 ) (resp SL 2 (q)) in case 5 (resp case 6). This proves that G contains either a natural SL 2 (q) or a natural SU 2 (q). We then have by Lemma 3.3.5 that G is tensor-indecomposable, therefore it is not included in any maximal subgroup of class C 4 of SU 8 (q 1 2 ). It contains a transvection, therefore it cannot be included in any group of class C 3 . We also have that G is a primitive irreducible group preserving no symmetric or skew-symmetric form over

F q 1 2 si G is included in no maximal subgroup of class C 1 or C 2 or C 5 for q 0 = q 1 2
. Consider now the maximal subgroups of SU 8 (q

1 2 ) which are not of class C 1 , C 2 , C 3 , C 4 or C 5 with q 0 = q 1 2
. They are given in Tables 8. [START_REF] Wilson | The finite simple groups[END_REF] and 8.47 of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF] and we list them below with their order or a quantity their order divides 1. SU 8 (q 0 ), q 28 0 8 i=2 We have that α of order greater than or equal to 7 and (α) = α, where α is the automorphism of order 2 of F q . It follows that α q 1 2 -1 = 1 and, therefore q 1 2 -1 ≥ 7 and q 1 2 ≥ 8. In case 5, we have F q = F p (α) = F p (α + α -1 ), therefore ˜ (α) = α -1 , where ˜ is the automorphism of order 2 of F q and, therefore q 1 2 ≥ 6. In case 6, we have that q ≥ 7 because α is of order greater than or eaqual to 7. This proves that in both cases we have |H| ≥ 6 × (6 2 -1) × 64 3 × (64 2 -1)(64 3 -1) = 59095088588390400. This excludes the last three cases.

(q i 0 -(-1) i ) q 1 2 = q r 0 , r odd prime 2. ((8, q 1 2 + 1) • 2 1+6 ) . (SP 6 ( 2
Assume that G is included in the first maximal subgroup. Let r be the prime such that q r 0 = q 1 2 . We have that q 3 divides q 28 0 = q 28 2r = q 14 r . This implies that 3 ≤ 14 r and, therefore r ≤ 14 3 < 5. Since r is an odd prime, we have that r = 3. We then have that (q 3 -1) divides

8 i=2 (q i 6 -(-1) i ).
The Euclidean remainder of those two polynomials in q 1 6 is then 2q 17 6 -2q 16 6 + 2q 15 6 -q 14 6 + q 13 6 -q 12 6 -q 11 6 + q 10 6 -q 9 6 -q

8 6 + q 7 6 -q 6 6 -q 5 6 + q 4 6 -q 3 6 + 2q 2 6 -2q 1 6 + 2.
The latter quantity is positive, therefore we have that q 3 -1 ≤ 2q 17 6 -q 16 6 < q 3 -1. This contradiction shows that the first cases is also excluded, therefore G is included in no maximal subgroup of SU 8 (q 1 2 ) and, therefore G SU 8 (q 1 2 ). This concludes the proof of the lemma.

We must now show that we can use Theorem 2.3.2. The factorizations of λ = (λ 1 , λ 2 ) by (λ 1 , λ 2 ) and by (λ 2 , λ 1 ) change the arguments for the natural SL 2 (q) and twisted diagonal embeddings of SL 3 (q). Let λ = (λ 1 , λ 2 ) be a double-partition of n ≥ 5.

We then have five different cases.

1. λ = λ , λ = (λ 2 , λ 1 ) and λ = (λ 1 , λ 2 ). Let us show R λ (A Bn ) contains a natural SL 3 (q). It is sufficient to show that there exists µ ⊂ λ such that µ ⊂ λ, (µ 2 , µ 1 ) ⊂ λ and (µ 1 , µ 2 ) ⊂ λ.

We write λ 1 partition of n 1 and λ 2 partition of n 2 with n = n 1 + n 2 ≥ 5. We only consider double-partitions with no empty component. This implies that n 1 and n 2 are greater than or equal to 1. Since the roles of λ 1 and λ 2 are symmetrical for this, we can assume without loss of generality n 1 ≥ n 2 .

(a) n 2 = 1, we then have that λ 2 = λ 2 , therefore λ 1 = λ 1 . There exists

µ 1 ⊂ λ 1 such that µ 1 ⊂ λ 1 . We then have that µ = (µ 1 , λ 2 ) ⊂ λ, but µ ⊂ λ and (λ 2 , µ 1 ) ⊂ λ because n 1 -1 ≥ 4 > 1 and (µ 1 , λ 2 ) ⊂ λ, because µ 1 ⊂ λ 1 . (b) n 1 > n 2 = 2 and λ 1 = λ 1 . We set µ = (λ 1 , [1]), we have µ and ([1], λ 1 ) ⊂ λ because n 1 > n 2 and (λ 1 , [1]) ⊂ λ because λ 1 = λ 1 . (c) n 1 > n 2 = 2 and λ 1 = λ 1 . If for all µ 1 ⊂ λ 1 , µ 1 ⊂ λ 2 or µ 1 ⊂ λ 2 , then n 1 = 3 and λ 1 = [2, 1]. This implies that either ([2], [1 2 ]) ⊂ λ or ([1 2 ], [2]) ⊂ λ. By Proposition 3.2.5, R λ (A Bn ) contains a natural SL 3 (q). (d) n 1 > n 2 ≥ 3 and λ 2 = λ 2 . There exists µ 2 ⊂ λ 2 such that µ 2 ⊂ λ 2 . We then set µ = (λ 1 , µ 2 ). We have that (µ 2 , λ 1 ) ⊂ λ, (µ 2 , λ 1 ) ⊂ λ because n 1 > n 2 and (λ 1 , µ 2 ) ⊂ λ because µ 2 ⊂ λ 2 .
(e) n 1 > n 2 ≥ 3 and λ 2 = λ 2 , therefore λ 1 = λ 1 . We know that there exists

µ 1 ⊂ λ 1 such that µ 1 ⊂ λ 1 . If (λ 2 , µ 1 ) ⊂ λ or (λ 2 , µ 1 ) ⊂ λ, then µ 1 = λ 2 or µ 1 = λ 2 . We have that λ 2 = λ 2 , therefore this contradicts µ 1 ⊂ λ 1 . This shows that µ 1 = µ 1 . We have that (µ 1 , λ 2 ) ⊂ λ because µ 1 ⊂ λ 1 . (f) n 1 = n 2 ≥ 3. We then have that λ 1 = λ 1 or λ 2 = λ 2 . If λ 1 = λ 1 , we pick µ 1 ⊂ λ 1 such that µ 1 ⊂ λ 1 and set µ = (µ 1 , λ 2 )
, by the assumption on µ 1 , (µ 1 , λ 2 ) ⊂ λ, (λ 2 , µ 1 ) ⊂ λ and (λ 2 , µ 1 ) ⊂ λ because λ 2 = λ 1 and λ 2 = λ 1 . If λ 2 = λ 2 , we pick µ 2 ⊂ λ 2 such that µ 2 ⊂ λ 2 and µ = (λ 1 , µ 2 ) verifies the required property.

2. λ = (λ 1 , λ 2 ), λ = (λ 2 , λ 1 ) and λ = λ . We then have that µ 1 ⊂ λ for all µ 1 ⊂ λ 1 and that for all µ 2 ⊂ λ 2 , µ 2 ⊂ λ 2 . We also have that

n 1 + n 2 ≥ 5. (a) n 1 ≥ n 2 = 1. Let µ 1 ⊂ λ 1 , we set µ = (µ 1 , λ 2 ). We have that (λ 2 , µ 1 ) ⊂ λ and µ ⊂ λ because n 1 -1 ≥ 3 > 1 and (µ 1 , λ 2 ) ⊂ λ. (b) n 1 > n 2 ≥ 2. We pick µ 2 ⊂ λ 2 and set µ = (λ 1 , µ 2 ). We have that (µ 2 , λ 1 ) ⊂ λ and µ ⊂ λ because n 1 > n 2 and (λ 1 , µ 2 ) ⊂ λ. (c) n 1 = n 2 ≥ 2. We pick µ 1 ⊂ λ 1 and set µ = (µ 1 , λ 2 ). We have that (λ 2 , µ 1 ) ⊂ λ because λ 2 = λ 1 and µ ⊂ λ because λ 2 = λ 1 and (µ 1 , µ 2 ) ⊂ λ.
In case 4 for the fields, i.e.,

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β), if µ = (µ 1 , µ 2 ) then R λ (A Bn ) contains up to conjugation {diag(M, (M ), I n λ -6
), M ∈ SL 3 (q)}, and a natural SL 3 (q

2 ) if µ = (µ 1 , µ 2 ) (it is possible for this to be the case for all µ ⊂ λ if we have square partitions). In cases 5 and 6 for the fields, i.e.,

F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ), if µ = (µ 1 , µ 2 ) then R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ), I n λ -6 ), M ∈ SL 3 (q)}, and a natural SU 3 (q 1 2 ) if µ = (µ 1 , µ 2 ). 3. λ = (λ 2 , λ 1 ) = λ . We then have n 1 = n 2 ≥ 3 and λ 1 = λ 2 = λ 1 because λ = λ . We can then pick µ 1 ⊂ λ 1 such that µ 1 ⊂ λ 1 and µ = (µ 1 , λ 2 ). We have (µ 1 , λ 2 ) ⊂ λ and µ ⊂ λ because λ 2 = λ 1 = λ 2 and (λ 2 , µ 1 ) ⊂ λ but µ = (λ 2 , µ 1 ) because n 1 -1 < n 1 = n 2 .
In case 4 for the fields, R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ), I n λ -6 ), M ∈ SL 3 (q)}.

In cases 5 and 6 for the fields, R λ (A Bn ) contains up to conjugation {diag(M, (M ),

I n λ -6 ), M ∈ SL 3 (q)}. 4. λ = λ = (λ 1 , λ 2 ) = (λ 2 , λ 1 ), we have n 1 = n 2 ≥ 3 and there exists µ 1 ⊂ λ 1 such that µ 1 ⊂ λ 2 because λ 1 = λ 2 . We have µ ⊂ λ because µ 1 ⊂ λ 1 = λ 2 , (λ 2 , µ 1 ) ⊂ λ since λ 2 = λ 1 and (µ 1 , λ 2 ) ⊂ λ because λ 2 = λ 2 = λ 1 . We have µ = µ because λ 2 = µ 1 . R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ), I n λ -6 ), M ∈ SL 3 (q)}. 5. λ = λ = (λ 2 , λ 1 ) = (λ 1 , λ 2 ). We then have n 1 = n 2 ≥ 3. If λ 1 and λ 2 are square partitions, then for all µ ⊂ λ, we have that µ = (µ 1 , µ 2 ) = µ = (µ 2 , µ 1 ), because n 1 = n 2 > n 1 -1 = n 2 -1.
In case 4 for the fields, R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ),

I n λ -6 ), M ∈ SU 3 (q 1
2 )}. In cases 5 and 6 for the fields, R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ),

I n λ -6 ), M ∈ SL 3 (q 1 2 )}.
If λ 1 or λ 2 is a square partition, then there exists µ ⊂ λ such that µ = µ , µ = (µ 2 , µ 1 ) and µ = (µ 1 , µ 2 ). This implies that R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ), (M ), ( t (M -1 )), I n λ -12 ), M ∈ SL 3 (q)}.

We now use the notations of Theorem 2.3.2. In all of the above cases except for the last one, there exists

g in R λ (A Bn ) such that [g, V ] ≤ 2. This implies that v G (V ) ≤ 2, therefore v G (V ) ≤ max(2, √ d
2 ). In the last case, we have in the same way an element g such that [g, V ] = 4. We also have in that case that λ = λ = (λ 1 , λ 2 ) and n ≥ 6,.This implies that λ contains ([2, 1], [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]1]), which is of dimension 6 3 × 2 × 2 = 80. It follows that d ≥ 80 and

√ d 2 ≥ √ 80 2 > 4. This shows that we still have v G (V ) ≤ max(2, √ d
2 ).

It remains to check that all the assumptions of the theorem are again verified and the classical group we get is the one we want.

The first step is to take care separately of double-partitions λ such that n λ ≤ 10. If n λ = 10, then by the conditions of Theorem 2.3.2, we can assume p = 2. The second step is to verify that the remaining double-partitions are tensor-indecomposable. The third step is to verify that they are imprimitive in the monomial case. The fourth step is to verify that they are imprimitive in the non-monomial case. The fifth step is to check that we are not in case 2. of Theorem 2.3.2. The sixth and last step is to verify that we have the desired classical groups in each of the above cases.

First step. For n = 5, it is enough to consider ([2, 2], [1]), ([2, 1, 1], [1]), ([2, 1], [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]) and ([1 3 ], [1 2 ]), for which the respective n λ is 10, 15, 20 and 10. We must show that R ([

1 3 ],[1 2 ]) (A B 5 ) = SL 10 (q) and R ([2,2],[1]) (A B 5 ) SL 10 (q 1
2 ) in case 4 for the fields and R ([2,2], [1]) (A B 5 )

SU 10 (q 1
2 ) in cases 5 and 6 for the fields. The other doublepartitions are of dimensions greater than 10. We know that G

= R ([1 3 ],[1 2 ]) (A B 5 ) contains R ([1 3 ],[1 2 ]) (A B 4 ) SL 4 (q) × SL 6 (q 1
2 ) and it is normally generated by this group, which is generated by transvections. Since p = 2, Theorem 2.3.4 implies that G is conjugate in GL 10 (q) to SL 10 (q ), SP 10 (q ) or SU 10 (q 1 2 ) for some q dividing q. Lemma 3.3.3 implies that q = q. The groups SP 10 (q) and SU 10 (q 1 2 ) are excluded by Proposition 3.2.4, because R ([1 3 ],[1 2 ]) is not isomorphic to its dual representation or its dual representation composed with the automorphism of order 2 of F q . This shows that G = SL 10 (q). In case 4, we know that

G = R ([2,2],[1]) (A B 5 ) is conjugate to a subgroup of SL 10 (q 1
2 ) by Proposition 3.2.4 and Lemma 3.2.5 and that

G contains R ([2,2],[1]) (A B 4 ) SL 8 (q 1 2 ) × SL 2 (q). It follows that it contains a natural SL 8 (q 1
2 ) and we can apply Theorem 3.3.2 to get that G SL 10 (q 1 2 ). In cases 5 and 6,

G = R ([2,2],[1]) (A B 5 ) is conjugate to a subgroup of SU 10 (q 1
2 ) by Proposition 3.2.4 and Lemma 3.2.4 and contains R ([2,2], [1]) (A B 4 ), which contains a natural SU 8 (q 1 2 ) in both cases. By Theorem 1.4 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], we have indeed G SU 10 (q 1 2 ).

Second step. We now show that those representations are tensor-indecomposable. Since ([2, 1, 1], [1]) contains a natural SL 3 (q), doubles-partitions with at most two rows or at most two columns are tensor-indecomposable by Lemmas 3.3.5 and 3.3.6. By the enumeration of the different cases, those lemmas cover all double-partitions of

n except if λ = (λ 1 , λ 2 ) = λ = (λ 2 , λ 1 ) = (λ 1 , λ 2 ) and neither λ 1 nor λ 2 contains a sub-partition µ such that µ = µ . In such a case, λ contains ([2, 1], [2, 1]
) which is of dimension 80 and we can use the following lemma. Lemma 3.3.9. If d ≥ 80 and G ⊂ GL d (q) contains an element of order coprime to p and conjugate in GL d (q) to the diagonal matrix diag(ξ, ξ, ξ, ξ, ξ

-1 , ξ -1 , ξ -1 , ξ -1 , 1, ..., 1) with ξ 2 = 1, then G is tensor-indecomposable, except possibly if G ⊂ G 1 ⊗ G 2 with G 1 ⊂ GL a (q), a ∈ {2, 4}.
Proof. Let g = P diag(ξI 4 , ξ -1 I 4 , I d-8 )P -1 . Assume that g = g 1 ⊗ g 2 with g 1 ∈ GL a (F q ), g 2 ∈ GL b (F q ) with 3 ≤ a ≤ b and ab = d. We have that b ≥ √ d, therefore b ≥ 9 because d ≥ 80. We write λ 1 , ..., λ a the eigenvalues of g 1 and µ 1 , ..., µ b the eigenvalues of g 2 . We then have that ∀i ∈

[[1, a]], ∀j ∈ [[1, b]], λ i µ j ∈ {1, ξ, ξ -1 }.
The numbers ξ and ξ -1 only appear 4 times each. This implies the number of couples (λ 1 µ i , λ 2 µ i ) ∈ {(1, ξ), (ξ, 1), (ξ, ξ -1 )} is less than or equal to 4 as is the number of couples (λ

1 µ i , λ 2 µ i ) ∈ {(1, ξ -1 ), (ξ -1 , 1), (ξ -1 , ξ)}. For any i ∈ [[1, a]], the inequal- ity λ 1 µ i = λ 2 µ i implies that (λ 1 µ i , λ 2 µ i ) ∈ {(1, ξ), (ξ, 1), (1, ξ -1 ), (ξ -1 , 1), (ξ, ξ -1 ), (ξ -1 , ξ)}. It follows that there are at most 8 couples (λ 1 µ i , λ 2 µ i ) such that λ 1 µ i = λ 2 µ i . Since b ≥ 9, there exists i ∈ [[1, a]] such that λ 1 µ i = λ 2 µ i . It follows that λ 1 = λ 2 .
In the same way, we have that λ 1 = λ j for all j ∈ [ [1, a]]. Up to reordering, we can assume λ 1 µ 1 = ξ. We then have λ 2 µ 1 = λ 3 µ 1 = ξ. Since there are exactly 4 ways ξ appears as a λ i µ j , we have that a = 4.

By the assumptions on λ,

H = R λ (A B n-1 ) is a direct product of groups isomorphic to some SL m (q) with m ≥ n ([2,1],[2]) = 20. If G = R λ (A Bn ) is not tensor-indecomposable, then G ⊂ SL 2 (q) ⊗ SL d 2 (q) or G ⊂ SL 4 (q) ⊗ SL d 4
(q). We then have a morphism from G into SL 2 (q) or SL 4 (q). If we consider the restriction of this morphism to H, its kernel is a normal subgroup of H. The only non-abelian decomposition factors of H are P SL m (q) with m ≥ 20. If the image is non-abelian, then there exists a subgroup of SL 2 (q) or a subgroup of SL 4 (q) isomorphic to some P SL m (q). This leads to a contradiction because m ≥ 20. It follows that the image is abelian and since H is perfect, the kernel is equal to H. Since H normally generates G, the morphism is trivial on G which contradicts the irreducibility of G.

Third step. In the monomial case, the only additional case to consider is the same one as in the second step. Looking at the corresponding proof in [12, page 14], we get that (p-1)r ≤ 4 with q = p r . We know that q is a square, n ≥ 6, α is of order greater than n and (α) ∈ {α, α -1 }.

Therefore α q 1 2 -1 = 1 or α q 1 2 +1 = 1. In both cases q 1 2 + 1 > 6, and, therefore q 1 2 ≥ 6 and q ≥ 36. The condition (p -1)r ≤ 4 implies that q ≤ max(5 1 , 4 1 , 3 2 , 2 4 ) = 16, therefore we have a contradiction.

Fourth step. We know that there exists a matrix t of order p such as the one in [12, page 14] or with Jordan form diag(

I 2 + E 1,2 , I 2 + E 1,2 , I 2 + E 1,2 , I 2 + E 1,2 , I n λ -8 ).
If p = 2, we can use the same arguments as in page 15 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] because we still have (t-1)

2 = 0. Assume now that p = 2. Assume that G ⊂ H S m = (H 1 × H 2 × • • • × H m ) S m with H 1 , . . . H m the m-copies of GL N/m (q) permuted by S m , that V = U 1 ⊕ U 2 ⊕ • • • ⊕ U m is the direct sum corresponding to the wreath product and that t / ∈ H 1 × • • • × H m . Assume t / ∈ H 1 × • • • × H m .
Up to reordering, we can assume tU 1 = U 2 . If dim(U i ) ≥ 5 then we can consider linearly independent vectors v 1 , v 2 , v 3 , v 4 , v 5 in U 1 and by completing the family of vectors (v 1 , tv 1 , v 2 , tv 2 , v 3 , tv 3 , v 4 , tv 4 , v 5 , tv 5 ) which are linearly independent because tU 1 = U 2 = U 1 , we get a basis upon which t acts as a matrix of the form

M 2 ⊕ M 2 ⊕ M 2 ⊕ M 2 ⊕ M 2 ⊕ X for a certain X with M 2 = 0 1 1 0
. This implies that the rank of t -1 is greater than or equal to 5, which is a contradiction. We can thus assume that dim

(U i ) ≤ 4. Note that G = R λ (A Bn ) and A Bn is perfect for n ≥ 5 [36], therefore G is perfect. If G ⊂ (H 1 × H 2 × • • • × H m ) S m , we get G ⊂ (H 1 × H 2 × • • • × H m ) A m because [S m , S m ] ⊂ A m .
If t is a transvection then by the same reasoning as above on the dimensions of U i , we are in the monomial case which was done in the third step.

If t is of rank 2, then either we are in the monomial case or dim(U i ) = 2. The monomial case is done, therefore it is sufficient to prove that dim(U i ) = 2 leads to a contradiction. We take t 1 and t 2 two such elements of rank 2. Assume dim(U i ) = 2, since we have t(U 1 ) = U 2 and

t 1 (U 2 ) = t 2 1 (U 1 ) = U 1 . If (u a , u b ) are linearly independent then (t 1 u a -u a , t 1 u b -u b ) is a basis of Im(t 1 -1), which is of dimension 2 and included in U 1 ⊕ U 2 for all i / ∈ {1, 2}, t i (U i ) = U i .
It follows that the projection of t 1 upon S m from the semi-direct product is a transposition. This is a contradiction because the projection of G upon S m is included in A m .

If t is of rank 4 and R λ (A B n-1 ) does not contain in an obvious way any transvections or elements t of rank 2, then G contains up to conjugation {diag(M, t (M -1 ), (M ), t (M -1 ),

I n λ -8 ), M ∈ SL 2 (q)}. We consider two elements t 1 and t 2 of rank 4. If dim(U 1 ) = 4, then if (u 1 , u 2 , u 3 , u 4 ) is a basis of U 1 , (u 1 -t 1 u 1 , u 2 -t 1 u 2 , u 3 -t 1 u 3 , u 4 - t 1 u 4 ) is a basis of Im(t 1 -1), which is of dimension 4. It follows that the projection of t 1 upon S m is a transposition, which is absurd. If dim(U 1 ) = 3, then if (u 1 , u 2 , u 3 ) is a basis of U 1 , we have that Vect{t 1 u 1 -u 1 , t 1 u 2 - u 2 , t 1 u 3 -u 3 } ⊂ Im(t 1 -1). If there exists i / ∈ {1, 2} such that t i (U i ) = U i
then in the same way as before, there would exist a subspace of dimension 6 of Im(t 1 -1), which is of dimension 4. This shows that the projection of t 1 upon S m is a transposition, which is absurd.

If dim(U i ) = 2, then we can take 4 distinct non-zero elements a 1 , a 2 , a 3 , a 4 of F q . This is possible because q 1 2 ≥ 6. We know that G contains up to conjugation the elements t j for j ∈ {1, 2, 3, 4} with

t j = diag(I 2 +a j E 1,2 , I 2 +a j E 1,2 , I 2 + (a j )E 1,2 , I 2 + (a j )E 1,2 , I n λ -8
). We have that Im(t j -1) is independent of j. We also have that t

1 (U 1 ) = U 2 and t 1 (U 2 ) = t 2 1 (U 1 ) = U 1 .
Since Im(t 1 -1) ∩ U 1 ⊕ U 2 is then of dimension 2 and the projection of t 1 upon S m is not a transposition, there exists i / ∈ {1, 2} such that t 1 (U i ) = U i . Up to reordering, we can assume

t 1 (U 3 ) = U 4 and t 1 (U 4 ) = t 2 1 (U 3 ) = U 3 . This shows that for all j ∈ {1, 2, 3, 4}, Im(t j -1) = Im(t 1 -1) ⊂ U 1 ⊕ U 2 ⊕ U 3 ⊕ U 4 .
Since each t j is of order 2, it follows writing π the projection of G upon S m that we have {π(t 1 ), π(t 2 ), π(t 3 ), π(t 4 )} ⊂ {I d , (12)(34), (13)( 24), ( 14)(23)}.

Let us show that π(t j ) = I d for all j. They are all conjugate in G. Since

H 1 × H 2 × • • • × H m is a normal subgroup of (H 1 × H 2 × • • • × H m ) S m ,
it is sufficient to show it for one of them. Assume it is false for all of them. We then have {π(t 1 ), π(t 2 ), π(t 3 ), π(t 4 )} ⊂ {(12)(34), (13)( 24), ( 14)(23)}. Therefore, there exists a pair (i, j), i = j such that π(t i ) = π(t j ) and, therefore π(t i t j ) = I d . But the matrix of t i t j in the basis we chose is diag(

I 2 + (a i + a j )E 1,2 , I 2 + (a i + a j )E 1,2 , I 2 + (a i + a j )E 1,2 , I 2 + (a i + a j )E 1,2 , I n λ -8 ).
We have a i + a j = 0 because p = 2 and the elements a l are pairwise distinct. It follows that t i t j is conjugate to each t l , therefore we have a contradiction. This shows that for all j ∈ {1, 4},

π(t j ) = I d . It follows that t j ∈ H 1 × H 2 × • • • × H m , which is normal in (H 1 × H 2 × • • • × H m ) S m . Since G is normally generated by R λ (A B n-1 )
, which is normally generated by elements of the form t j , we have that

G ⊂ H 1 × H 2 × • • • × H m .
This contradicts the irreducibility of G. This is absurd and it follows that G is a primitive group.

Fifth step. If G contains a natural SL 2 (q 1 2 ) or a natural SU 2 (q 1
2 ) then we can apply the same arguments as in [12, page 13]. If G contains a twisted diagonal embedding or a twisted diagonal embedding composed with the automorphism of order 2 of F q of SL 3 (q), then we can apply the arguments of [12, page 14]. If we are not in any of the above cases, then λ = λ = (λ 2 , λ 1 ), therefore n ≥ 6 and we are in one of the following cases.

1. R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ),

I n λ -6 ), M ∈ SU 3 (q 1 2 )}. 2. R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ), I n λ -6 ), M ∈ SL 3 (q 1 2 )}. 3. R λ (A Bn ) contains up to conjugation {diag(M, t (M -1 ), (M ), t (M -1 ), I n λ -12 ), M ∈ SL 3 (q)}.
In the first two cases, we have an element g conjugate to diag(ξ, ξ, ξ -1 , ξ -1 , 1, . . . , 1) with ξ of order q 1 2 -1 but the order of α is less than or equal to q 1 2 +1 in both cases. If g is an element of S n λ such that [g, V ] = 4, then we have (g = σ 1 σ 2 σ 3 σ 4 is the product of 4 disjoint transpositions and g is of order 2) or (g is the product of 2 disjoint 3-cycles and g is of order 3) or (g is a 5-cycle and g is of order 5) or (g is the disjoint product of 2 transpositions and a 3-cycle and g is of order 6). Since n λ ≥ 6 and the order of α is greater than n, q

1 2 + 1 > 7, therefore q 1 2 -1 > 5
which contradicts all the cases except for the last one. In the last case, we have that n λ ≥ 7 by the decomposition of g. Since λ = λ , n λ is even and q

1 2 -1 = q 1 2 + 1 -2 > n λ -2 > 6,
which contradicts the last case.

In the third case, we have an element g conjugate to diag(ξ, ξ, ξ, ξ, ξ -1 , ξ -1 , ξ -1 , ξ -1 , 1, . . . , 1) which is of order o(g) = q -1. However q 1 2 + 1 > 7, therefore q > 36. Since q is an even power of a prime number, it follows that q > 49 and q -1 ≥ 49. We have [g, V ] = 8. By considering the decomposition into disjoint cycles of g and using the fact that the rank of σ -1 of a cycle σ is equal to the length of the cycle minus 1, we get o(g) ∈ {lcm({n i + 1} i∈I ),

i∈I n i = 8, n i ∈ N }. It follows that o(g) ≤ 30 < 49 ≤ q -1 = o(g)
which is a contradiction.

Sixth step.

We have shown that G = R λ (A Bn ) is a classical group in a natural representation. The last step is to show that we have the following theorem. Theorem 3.3.7. If n ≥ 5, then for all double-partition λ n in our decomposition, R λ (A Bn ) = G(λ), where G(λ) is given by the following list.

1. When F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β), and F q = F p (α) = F p (α + α -1 ) (a) SU n-1 (q 1 2 ) if λ = ([n -1, 1], ∅). (b) SU n λ (q 1 2 ) if λ = (λ 1 , ∅), λ 1 < λ 1 . (c) SP n λ (q 1 2 ) if λ = (λ 1 , ∅), λ 1 = λ 1 and ( p = 2 or (p ≥ 3 and ν(λ 1 ) = -1)). (d) Ω + N (q 1 2 ) if λ = (λ 1 , ∅), λ 1 = λ 1 , p ≥ 3 and ν(λ 1 ) = 1. (e) SL n (q) if λ = ([1], [n -1]). (f) SL n λ (q) if λ = λ , λ = (λ 1 , λ 2 ) and λ = (λ 2 , λ 1 ). (g) SU n λ (q 1 2 ), if λ = (λ 2 , λ 1 ) = λ . (h) SL n λ (q 1 2 ),if λ = (λ 1 , λ 2 ) = λ . (i) SP n λ (q), if λ = λ = (λ 2 , λ 1 ) and (p = 2 or ν(λ) = -1). (j) Ω + n λ (q),if λ = λ = (λ 2 , λ 1 ), p = 2 and ν(λ) = 1. (k) SP n λ (q 1 2 ), if λ = λ = (λ 2 , λ 1 ) and (p = 2 or ν(λ) = -1). (l) Ω + n λ (q 1 2 ) if λ = λ = (λ 2 , λ 1 ), p = 2 and ν(λ) = 1.

When

F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ), (a) when F q = F p (α) = F p (α + α -1 ), i. SL n-1 (q) if λ = ([n -1, 1], ∅). ii. SL n λ (q) if λ = (λ 1 , ∅), λ 1 < λ 1 . iii. SP n λ (q) if λ = (λ 1 , ∅), λ 1 = λ 1 and (p = 2 or (p ≥ 3 and ν(λ 1 ) = -1)). iv. Ω + N (q) if λ = (λ 1 , ∅), λ 1 = λ 1 , p ≥ 3 and ν(λ 1 ) = 1. (b) when F q = F p (α) = F p (α + α -1 ), i. SU n-1 (q 1 2 ) if λ = ([n -1, 1], ∅). ii. SU n λ (q 1 2 ) if λ = (λ 1 , ∅), λ 1 < λ 1 . iii. SP n λ (q 1 2 ) if λ = (λ 1 , ∅), λ 1 = λ 1 and ( p = 2 or (p ≥ 3 and ν(λ 1 ) = -1)). iv. Ω + N (q 1 2 ) if λ = (λ 1 , ∅), λ 1 = λ 1 , p ≥ 3 and ν(λ 1 ) = 1. (c) SL n (q) if λ = ([1], [n -1]), (d) SL n λ (q) if λ = λ , λ = (λ 1 , λ 2 ) and λ = (λ 2 , λ 1 ), (e) SL n λ (F q 1 2 ) if λ = (λ 2 , λ 1 ) = λ , (f) SU n λ (q 1 2 ) if λ = (λ 1 , λ 2 ) = λ , (g) SP n λ (q) if λ = λ = (λ 2 , λ 1 ) and (p = 2 or ν(λ) = -1), (h) Ω + n λ (q) if λ = λ = (λ 2 , λ 1 ), p = 2 and ν(λ) = 1, (i) SP n λ (q 1 2 ) if λ = λ = (λ 2 , λ 1 ) and (p = 2 or ν(λ) = -1), (j) Ω + n λ (q 1 2 ) if λ = λ = (λ 2 , λ 1 ), p = 2 and ν(λ) = 1.
Proof. It is sufficient to prove the result for double-partitions with no empty components which are not hooks. We know by Theorem 2.3.2 and the previous steps that G(λ) is a classical group in a natural representation. The proof uses Proposition 3.2.4 and the separation of the cases made before the enumeration of the six steps. We write F q the field over which our classical group is defined. In all cases G(λ) ⊂ SL n (q), therefore q divides q.

Assume

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β). 1. If λ = λ , λ = (λ 1 , λ 2 ) and λ = (λ 2 , λ 1 ), then G(λ) contains a natural SL 2 (q)
. By Lemma 3.3.3, we have that q = q. By Proposition 3.2.4, G(λ) preserves no hermitian or bilinear form, therefore

G(λ) = SL n λ (q).
2. If λ = (λ 2 , λ 1 ) = λ , then by Proposition 3.2.4 and Lemma 3.2.4, we have up to conjugation G(λ) ⊂ SU n λ (q

1 2 ). Up to conjugation, G(λ) contains {diag(M, t (M -1 ), I n λ -6 ), M ∈ SL 3 (q)}, therefore G(λ) contains {diag(M, M, I n λ -6 ), M ∈ SU 3 (q 1 2 )}.
Let ϕ be the natural representation of SU 3 (q 1 2 ) in GL 3 (F p ) and ρ the diagonal representation of SU 3 (q 1 2 ) in GL n λ (F p ), given by the above subgroup of G(λ). We have ρ ϕ ⊕ ϕ ⊕ 1 n λ -6 with 1 the trivial representation. Let σ be a generator of Gal(F q /F q ). Since G(λ) is a classical group over F q , we have that ρ ρ σ , therefore ϕ ϕ σ . It follows that for every M ∈ SU 3 (q 1 2 ), we have σ(Tr(M )) = Tr(M ). By Lemma 3.3.3, we have that F q = F Gal(Fq/F q ) q and, therefore q = q. By Proposition 3.2.4, G(λ) preserves no bilinear form, therefore G(λ) = SU n λ (q 1 2 ).

3. If λ = (λ 1 , λ 2 ) = λ , then by Proposition 3.2.4 and Lemma 3.2.5, up to conjugation, we have that

G(λ) ⊂ SL n λ (q 1
2 ). The group G(λ) contains either a natural SL 3 (q

2 ) or a group of the form {diag(M, (M ),

I n λ -6 ), M ∈ SL 3 (q)}. If G(λ) contains a natural SL 3 (q 1
2 ) then by Lemma 3.3.3, we have q = q 1 2 . We know by Proposition 3.2.4 that G(λ) preserves no symmetric or skew-symmetric bilinear form. If we had G(λ) ⊂ SU n λ (q 1 4 ), then the natural SL 3 (q

1 2 ) in G(λ) would inject itself in some SU 3 (q 1 4
). This is absurd because of their orders, therefore we have G(λ) SL n λ (q

1 2 ). If G contains up to conjugation a group of the form {diag(M, (M ), I n λ -6 ), M ∈ SL 3 (q)} then it contains {diag(M, M, I n λ -6 ), M ∈ SL 3 (q 1 2 )}. Let ϕ be the natural representation of SL 3 (q 1 2 ) in GL 3 (F p ) and ρ the diagonal representation of SL 3 (q 1
2 ) in GL n λ (F p ) given by the above subgroup of G(λ). We then have ρ ϕ ⊕ ϕ ⊕ 1 n λ -6 . Let σ be a generator of Gal(F q 1 2 /F q ). We have ρ ρ σ , therefore ϕ ϕ σ . By Lemma 3.3.3, we have that

F q 1 2 = F Gal(F q 1 2 /F q ) q 1 2
, therefore q = q 1 2 . We cannot have G(λ) SU n λ (q

1 4 ) because SL 3 (q) would inject itself in SU 6 (q 1
4 ) and we know that |SU 6 (q

1 4 )| q 15 4 < |SL 3 (q)| q 3
. By Proposition 3.2.4, G(λ) cannot preserve any symmetric or skew-symmetric bilinear form, therefore

G(λ) SL n λ (q 1
2 ). . This shows that q ≤ q 1 2 and it is enough to show that q = q 1 2 to conclude the proof. If λ 1 and λ 2 are square partitions, then G(λ) contains up to conjugation the group {diag(M, t (M -1 ),

I n λ -6 ), M ∈ SU 3 (q 1 2 )}.
Let ϕ be the natural representation of SU 3 (q 1 2 ) in GL 3 (F p ), and ρ the twisted diagonal representation of SU 3 (q 1 2 ) in GL n λ (F p ) given by the above subgroup of G(λ). We have ρ ϕ ⊕ ϕ ⊕ 1 n λ -6 . Let σ be a generator of Gal(F

q 1 2 /F q ). Since G(λ) is a classical group over F q , we have ρ ρ σ . It follows that ϕ ϕ σ or ϕ (ϕ ) σ . The first possibility implies that F q 1 2 = F Gal(F q 1 2 /F q ) q 1 2
, therefore q = q 1 2 . The second possibility implies that ϕ ϕ σ 2 . Therefore q = q 1 2 or σ is of order 2 and SU 3 (q

2 ) injects into SU 3 (q

1 4
), which is a contradiction. In both cases, we have q = q 1 2 and the desired result follows. If λ 1 or λ 2 is not a square partition, then G(λ) contains up to conjugation the group {diag(M, t (M -1 ), (M ), t (M -1 ), I n λ -12 ), M ∈ SL 3 (q)}, and, therefore contains its subgroup {diag(M, t (M -1 ), M, t (M -1 ),

I n λ -12 ), M ∈ SL 3 (q 1 2 )}. Let ϕ be the natural repre- sentation of SL 3 (q 1 2 ) in GL 3 (F p ) and ρ be the representation of SL 3 (q 1 2 ) in GL n λ (F p )
given by the above subgroup G(λ). We have ρ ϕ ⊕ ϕ ⊕ ϕ ⊕ ϕ ⊕ 1 n λ -6 . Let σ be a generator of Gal(F q 1 2 /F q ). Since G(λ) is a classical group defined over F q , we have that ρ ρ σ . It follows that ϕ ϕ σ or ϕ (ϕ ) σ . By the same arguments as before, we have q = q 1 2 or SL 3 (q 1 2 ) injects itself in SU 3 (q 1 4 ), which is not possible. This proves q = q 1 2 and concludes the case

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β). If F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1
), then all the arguments are the same up to permutation of the different cases.

We have determined the image of the derived subgroup of the Artin group in all cases in type B. This is close to determining the image of the Artin group itself since A Bn /A Bn Z 2 . We give in the following subsection an example of how to recover the group G = R(A Bn ) from H = R(A Bn ) with the representation of H B 2 ,α,β labeled by the double-partition ([1], [1]).

Image of the full Artin group for the 2-dimensional representation of H B 2 ,α,β

We first recall the results for the 2-dimensional representation depending on the fields.

Proposition 3.4.1. Assume the order of β does not belong to {1, 2, 3, 4, 5, 6, 10} or the order of α does not belong to {1, 2, 3, 4, 5, 6, 10}

1. If F q = F p (α, β) = F p (α + α -1 , β + β -1 ) then R [1],[1] (A B 2 ) = SL 2 (q). 2. If F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ) then R [1 2 ],[1 2 ] (A B 2 ) SU 2 (q 1
2 ).

If

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ) then R [1 2 ],[1 2 ] (A B 2 ) SL 2 (q 1
2 ).

If

F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ) then R [1 2 ],[1 2 ] (A B 2 ) SL 2 (q 1
2 ).

Corollary 3.4.1. Under the same assumptions as in the previous proposition, we have that if

G = R [1],[1] (A B 2 )
, a is the order of -α and b is the order of -β then

1. If F q = F p (α, β) = F p (α + α -1 , β + β -1 ) then G SL 2 (q) Z/ lcm(a, b)Z. 2. F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ) then G SU 2 (q 1 2 ) Z/ lcm(a, b)Z. Proof. We write s = R [1],[1] (S 1 ) = α-1 β+1 αβ+1 β+1 α+β β+1 αβ-β β+1 and t = R [1],[1] (T ) = β 0 0 -1 . Assume first that F q = F p (α, β) = F p (α + α -1 , β + β -1
). By Proposition 3.4.1 and the fact that Det(s) = -α and Det(t) = -β, we have the following exact sequence

1 -→ SL 2 (q) -→ G det -→< -α, -β >-→ 1.
In order to prove the result, we must show that the exact sequence is split and that < -α, -β > is isomorphic to Z/ lcm(a, b)Z. We first show that the exact sequence is split. This is equivalent to finding a subgroup N of G isomorphic to < -α, -β > and such that

Det(N ) =< -α, -β >. We have SL 2 (q) ≤ G, therefore U = 0 1 -1 0 ∈ G. It follows that V = stsU t -1 U = -α 0 0 -α ∈ G. We also have sts = -α 0 0 αβ ∈ G and -t = -β 0 0 1 ∈ G.
We now distinguish five possibilities depending on the orders of α and the order of β. We write c = 2 c (resp d = 2 k d ) the order of α (resp β) with c (resp d odd). First case : k = 1 or = 1. By symmetry of the roles of α and β, it is sufficient to show the exact sequence is split when l = 1. We then have (-α)

c 2 = (-1) c α c = 1, therefore -α is of order m for some m dividing c 2 . We also have α 2m = (-α) 2m = 1, therefore c divides 2m. It follows that m = c 2 .
The order of α being even, we have that the order of α 2 is also c 2 . The subgroup generated by V and -t then verifies the desired conditions.

Second case : k = = 0. We then have that the order of -α is equal to the order of -α 2 . We have (sts

) cd = (-1) cd (α c ) d 0 0 (α c ) d (β d ) c = -1 0 0 1 , therefore M = (sts) cd V = α 0 0 -α ∈ G.
The subgroup < M, -t > then verifies the desired conditions.

Third case : k > 1 > or > 1 > k. It is sufficient to consider the case k > 1 > l.
We then have that -α and -α 2 have the same order since c is odd. We have

t d 2 = β d 2 0 0 (-1) d 2 = -1 0 0 1 , therefore M = t d 2 V = α 0 0 -α ∈ G0
The subgroup < M, -t > then verifies the desired conditions.

Fourth case : k > > 1 or > k > 1.
It is sufficient to consider the case k > > 1. We then have that the order of α is equal to the order of -α. We have

t d 2 l = β 2 k-l d 0 0 (-1) 2 k-l d = β 2 k-l d 0 0 1 . We set γ = β 2 k-l d
, γ is then of order 2 . We have that γα 2 is of order r, where r is an integer dividing c. We also have that (γα 2 )

c 2 = γ 2 -1 c α c = -1, therefore r does not divide c 2 .
This implies that there exists an odd integer c such that r = 2 c and c divides c . We have 1 = (γα 2 ) r = α 2r and c divides 2r. Therefore c divides c and γα 2 is of order a. We then

set M = t d 2 l V = -γα 0 0 -α ∈ G.
The matrix M is of order c = a and its determinant is of order a because it is equal to γα 2 . This proves that we can take < M, -t > as our subgroup.

Fifth case : k = l ≥ 2. We then have that -α is of order c. We have

t d = β d 0 0 (-1) d = β d 0 0 -1
. We set γ = β d , we have that the order of γ is 2 k . The element -γα 2 is then of order c by the same reasonning as in the fourth case. We set

M = t d V = -γα 0 0 α ∈ G.
The matrix M is of order c, therefore the subgroup < M, -t > verifies the desired conditions. It follows that < -α, -β > (Z/aZ × Z/bZ)/Z/dZ. Its order is therefore equal to ab d = lcm(a, b). Since it is a subgroup of Z/nZ, it is cyclic and therefore isomorphic Z/ lcm(a, b)Z. This concludes the proof for

This shows that G G < -α, -β >. It now only remains to show that < -α, -β > Z/ lcm(a, b)Z. Let n = q -1, we have < -α, -β > < n a , n b > when it is seen as a subgroup of Z/nZ. Let ϕ : Z/aZ × Z/bZ →< n a , n b > be the map that maps (u, v) to u n a + v n b .
F p (α, β) = F p (α + α -1 , β + β -1 ). Assume now that F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ). Recall that (α) = α -1 and (β) = β -1 . Let P = α+β αβ+1 0 0 1 , we then have P sP -1 = ( t s -1 )
and

P tP -1 = ( t t -1 ). If we set (X, Y ) = t (X)P Y , then for all M ∈ G, X, Y ∈ F 2 q , we have (M X, M Y ) = (X, Y ). This proves by Proposition 3.2.5 that there exists Q ∈ GL 2 (q) such that G = QGQ -1 is a subgroup of GU 2 (q 1
2 ). By Proposition 3.4.1, we have the following exact sequence 1 -→ SU 2 (q

1 2 ) -→ G det -→< -α, -β >-→ 1.
As in the previous case, it is sufficient to show that there is a splitting of

< -α, -β > in G or in G since G Q -1 GQ. We have that [ G, G] SU 2 (q 1
2 ) and for M ∈ GL 2 (F q ), M ∈ GU 2 (q) if and only if QM Q -1 stabilizes the sesquilinear form defined by (X, Y ) = t (X)P Y . We have that G contains all the matrices of determinant 1 preserving the above sesquilinear form. Let M = β 0 0 β -1 . We have ( t (M -1 ))P = M P = P M , therefore (M X, M Y ) = (X, Y ). Since det(M ) = 1, we have that M ∈ G. In the same way, we have -I 2 ∈ G. We then have that -α 0 0 -α = M stst -1 ∈ G and -t ∈ G. This proves that we can have all the matrices appearing in the previous case in G and, therefore that there exists a splitting of < -α, -β > in G. This concludes the proof in this case.

If

F q = F p (α, β) = F p (α, β +β -1 ) = F p (α+α -1 , β) = F p (α+α -1 , β +β -1 ) or F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1
) then the exact sequence we consider is not always split, therefore the situation is slightly more complex. The Gcd of the order of -α and the order of -β then divides 2 as we prove in the following lemma.

Lemma 3.4.1. If F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ) or F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ) then Gcd(a, b) ≤ 2,
where a is the order of -α and b is the order of -β.

Proof. Since the roles of α and β are symmetric, we can assume that

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ). Let a = a d and b = b d .
The group F q is isomorphic to Z/nZ, where n = q -1. Since -α is of order a, its image in Z/nZ is of the form u n a with u coprime to a. The group generated by β is mapped to the group generated by n b . The image of (-α) a is then equal to

ua n a = u a d n a = u b d n b ∈< n b >. This proves that (-α) a ∈< -β >, therefore (-α) a ∈ F q 1 2 . It follows that the polynomial R(X) = X a -(-α) a has its coefficients in F q 1 2 . Since F p (α + α -1 , β) = F p (α, β), the polynomial X 2 + (α + α -1 )X + 1 is irreducible over F q 1 2
and the unique automorphism of order 2 of

F q verifies (-α) = -α -1 . Since -α is a root of R ∈ F q 1 2 [X], (-α) = -α -1 is also a root of R. If k ∈ [[0, a -1]], we have that ((-α) 1+kd ) a = (-α) a is a root of R and since kd ∈ [[0, a -d]],
those roots are distinct. This proves that those are all the roots of R since its degree is a . It follows that there exists k ∈ [[0, a -1]] such that (-α) -1 = (-α) 1+kd and, therefore (-α) 2+kd = 1. It follows that a divides 2 + kd. We have 2 ≤ 2 + kd ≤ 2 + (a -1)d = 2 + a -d and a > 1, therefore a = 2 + kd.

If d ≥ 3 then we have 2 + kd ≤ 2 + a -d ≤ a -1 do we cannot have a = 2 + kd. This proves by contradiction that d ≤ 2. Note that we can have d = 2 or d = 1 since 2 + 2(a -1) = 2a and 2 + (a -2) = a . Proposition 3.4.2. Assume F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ) or F q = F p (α, β) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ). Let d = Gcd(a, b) ≤ 2,
where a is the order of -α and b is the order of -β.

If d = 1 then we have G (SL 2 (q 1 2 ) Z/abZ. If d = 2, then we have G J Z/ lcm(a, b)Z, where J = {M ∈ SL 2 (q), M (M ) -1 ∈ {I 2 , -I 2 }}
and is the unique automorphism of order 2 of F q .

Proof. Again, we can assume

F q = F p (α, β) = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ). Let P = -(α+β)(α-1) α(β+1) 0 0 1 . We will first show that P [G, G]P -1 ⊂ SL 2 (q 1
2 ).

Recall that we have t =

β 0 0 -1 and s =   α-1 1+β α+β -1 1+β -1 α+β 1+β α-1 1+β -1   . Note that β ∈ F q 1 2 , α+α -1 ∈ F q 1 2
and for x ∈ F q , x ∈ F q 1 2 if and only if (x) = x with the unique automorphism of order 2 of F q . We have (β) = β and (α) = α -1 . Let Q = (P -1 )P , we have

Q = α -1 (β+1) (α -1 +β)(α -1 -1) (α+β)(α-1) α(β+1) 0 0 1 = (β+1)(α+β)(α-1) (αβ+1)(1-α)(β+1) 0 0 1 = -α+β αβ+1 0 0 1 .
We then have QtQ -1 = t = (t) and

QsQ -1 =   α-1 β+1 -α+β β+1 -αβ+1 β+1 (α-1)β β+1   =   -α(1-α -1 ) 1+β --α(1+βα -1 ) β+1 -α(β+α -1 ) β+1 -α(1-α -1 ) 1+β -1   = -α (s).
It follows that for all g ∈ [G, G], QgQ -1 = (g) and, therefore (P -1 )P gP -1 (P ) = (g) and (P gP -1 ) = P gP -1 . This proves that we have

P [G, G]P -1 ⊂ SL 2 (q 1
2 ). We write in the following G = P GP -1 , s the image of P sP -1 in G/[ G, G] and t the image of

P tP -1 in G/[ G, G].
We have by Proposition 3.4.1 that [ G, G] SL 2 (q 1 2 ). We now show that a is the order of s and b is the order of t. Let r be the order of s. We have (P sP -1 ) r ∈ SL 2 (q 1 2 ), therefore (-α) r = det(s) r = 1 and a divides r. We have (-α) a = 1, therefore α a = (-1) a . The eigenvalues of P sP -1 are α -1 and -1, therefore (P sP -1 ) a is conjugate to the diagonal matrix diag(α -a , (-1) a ) = (-1) a I 2 and, therefore (P sP -1 ) a = (-1) a I 2 ∈ SL 2 (q 1 2 ). It follows that sa = 1 and, therefore r divides a. This proves that r = a. In the same way, we have b is the order of t.

We now determine G in terms of G ∩ SL 2 (q). The determinant gives us the following exact sequence

1 -→ G ∩ SL 2 (q) -→ G det -→< -α, -β >-→ 1.
The matrix P commutes with all diagonal matrices, therefore sts = -α 0 0 αβ ∈ G and t ∈ G.

Moreover, since SL 2 (q

1 2 ) ⊂ G, we have u = 0 1 -1 0 ∈ G and, therefore ut -1 u = 1 0 0 -β -1 ∈ G. It follows that M = stsut -1 u = -α 0 0 -α ∈ G, t ∈ G and -t ∈ G.
It follows that we can use the same arguments as for

F q = F p (α, β) = F p (α + α -1 , β + β -1 ) to get that there is a splitting of < -α, -β > in G. This proves that G ( G ∩ SL 2 (q) Z/ lcm(a, b)Z. It now only remains to determine G ∩ SL 2 (q) depending on d. First case : d = 1. Since d = 1, we have lcm(a, b) = ab. It is thus sufficient to show that SL 2 (q 1 2 ) [ G, G] = G ∩ SL 2 (q). The group [ G, G] is a normal subgroup of G ∩ SL 2 (q), therefore we can consider the quotient of those two groups. Let M ∈ ( G ∩ SL 2 (F q ))/[ G, G].
Since a is the order of s and b is the order of t, we can write M as sk t with 0 ≤ k ≤ a -1 and 0 ≤ ≤ b -1. The matrix M is of determinant 1, therefore (-α) k (-β) l = 1, which is only possible if l = k = 0 since a and b are coprime. This proves that the quotient is trivial and, therefore G ∩ SL 2 (q) SL 2 (q

1 2 ). It follows that G SL 2 (q 1 2 ) Z/abZ. Second case : d = 2. Since [ G, G] SL 2 (q 1 2 ), we have that SL 2 (q 1 2 ) G ∩ SL 2 (q). Let M ∈ G ∩ SL 2 (q) and A ∈ SL 2 (q 1 2 ), we then have (M )A (M ) -1 = (M AM -1 ) = M AM -1 since stabilizes SL 2 (q 1 2 ). It follows that (M ) -1 M belongs to the centralizer of SL 2 (q 1 2 ) in SL 2 (q) which is equal to {I 2 , -I 2 }. It follows that G ∩ SL 2 (q) ⊂ {M ∈ SL 2 (q), (M ) -1 M ∈ {I 2 , -I 2 }}.
Let us show that this inclusion is in fact an equality.

Let ϕ : G∩SL 2 (q) → {I 2 , -I 2 } be the map M → (M ) -1 M . The above inclusion proves that this map is a group morphism. We have ker(ϕ

) = {M ∈ G ∩ SL 2 (q), (M ) = M } ⊂ SL 2 (q 1 2 ) and SL 2 (q 1 2 ) ⊂ G ∩ SL 2 (q), therefore ker(ϕ) SL 2 (q 1
2 ). To conclude the proof, we only need to show that there exists M ∈ G ∩ SL 2 (q) such that (M ) -1 M = -I 2 .

We have that (-α)

a 2 = (-β) b 2 = -1 and (α) = α -1 . Let M = P s a 2 t b 2 P -1 , we have M =    (-1) a 2 β b 2 +1 α+β b 2 α a 2 +1 +β b 2 +1 α a 2 +β b 2 (-1) a 2 ((α+1)(β+1) -(α+β)(α-1)(αβ+1)((-1) b 2 α a 2 -(-1) b 2 + a 2 ) α(β+1)(αβ+α+β+1) αβ b 2 ((-1) a 2 -α a 2 ) (α 2 -1) (-1) b 2 α a 2 +1 β+(-1) b 2 + a 2 ) α+(-1) b 2 + a 2 ) β+(-1) b 2 α a 2 αβ+α+β+1    =    (-1) a 2 + b 2 (-βα+α+β-1) αβ+α+β+1 (-1) b 2 + a 2 (2(α+β)(α-1)(αβ+1)) α(β+1)(αβ+α+β+1) -(-1) b 2 + a 2 2α α 2 -1 (-1) b 2 + a 2 (-αβ+α+β-1) αβ+α+β+1    = (-1) a+b 2   -βα+α+β+-1 αβ+α+β+1
2(α+β)(α-1)(αβ+1) α(β+1)(αβ+α+β+1) -2α α 2 -1 -αβ+α+β-1 αβ+α+β+1   (M ) = (-1) a+b 2   -βα -1 +α -1 +β-1 α -1 β+α -1 +β+1 2(α -1 +β)(α -1 -1)(α -1 β+1) α -1 (β+1)(α -1 β+α -1 +β+1) -2α -1 α -2 -1 -α -1 β+α -1 +β-1 α -1 β+α -1 +β+1   = (-1) a+b 2   -β+1+βα-α β+1+αβ+α
2(1+αβ)(1-α)(β+α) α(β+1)(β+1+αβ+α) -2α 1-α 2 -β+1+αβ-α β+1+αβ+α   = -M.
It follows that (M ) -1 M = -I 2 which concludes the proof.

Chapter 4 Type D

In this section, we determine the image of the Artin group of type D inside its associated finite Iwahori-Hecke algebra. The structure of this section is similar to the one in chapter 3 for type B. We first define the model and prove some basic properties on the irreducible representations.

We then determine the different factorizations appearing, they will be similar to the last cases we had to consider in type B. We then state the main results for type D in Theorem 4.2.2 and Theorem 4.2.3. The last two sections prove those theorems using induction. The main differences which will arise come from the fact that we only have to consider one parameter. The branching rule (Lemma 4.1.1) is quite a bit more complicated and we need to prove some results on orthogonal groups defined over finite fields in a less general setting as we did in type B.

Definition of the model

Let n ≥ 4, p a prime different from 2, α ∈ F p of order greater than 2n. We set in this section F q = F p (α). As in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF], we take for the Iwahori-Hecke algebra of type D, H Dn,α the sub-algebra of H Bn,α,1 generated by U = T S 1 T, S 1 , . . . S n-1 . More precisely, we have the following definition.

Definition 4.1.1. The Iwahori-Hecke algebra of H Dn,α of type D is the subalgebra of H Bn,α,1 generated by U = T S 1 T, S 1 , . . . S n-1 . We then have (see [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] 10.4) that H Dn,α is the algebra generated by the above generators and that they verify the following relations

1. U 2 = (α -1)U + α, 2. ∀i ∈ [[1, n -1]], S 2 i = (α -1)S i + α, 3. ∀i ∈ [[1, n -1]] \ {2}, U S i = S i U , 4. U S 2 U = S 2 U S 2 , 5. ∀i ∈ [[1, n -2]], S i S i+1 S i = S i+1 S i S i+1 , 6. ∀i, j ∈ [[1, n -1]], |i -j| > 1, S i S j = S j S i .
We will now give a decomposition into irreducible modules of this algebra. In order to do this, we give the action of H Dn,α on modules generated by standard double-tableaux associated to double-partitions of n.

Proposition 4.1.1. Let λ = (λ 1 , λ 2 ) n and T ∈ λ. For i ∈ [[1, n -1]], we write m i (T) = α-1 1-(-1) δ i α c i -r i +r i+1 -c i+1 , where i (resp i + 1) is in box (r i , c i ) (resp (r i+1 , c i+1 )) of a component of T, δ i = 0 if i and i + 1 are in the same component and δ i = 1 otherwise.
The action of the generators on the standard double-tableau T is then the following

1. U.T = m 1 (T)T -(1 + m 1 (T)) T, with T = 0 if T 1↔2 not standard T = T 1↔2 otherwise. 2. ∀i ∈ [[1, n -1]], S i .T = m i (T)T + (1 + m i (T)) T with T = 0 if T i↔i+1 is not standard and T = T i↔i+1 otherwise.
We write V λ for the module generated by the standard double-tableaux associated to λ. By [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] (prop 10.4.5), we have that the action on the generators of H Dn,α commutes with the action of σ defined by σ((T 1 , T 2 )) = (T 2 , T 1 ). We then have that for any λ

= (λ 1 , λ 2 ) n, V (λ 1 ,λ 2 ) is isomorphic to V (λ 2 ,λ 1 ) as H Dn,α -module.
If λ = (λ 1 , λ 1 ), then we can consider a basis (T 1 , T 2 , . . . , T r , σ(T 1 ), σ(T 2 ), . . . , σ(T r )) of V λ . We then have that

V λ,+ =< T i + σ(T i ) > i∈[[1,r]] and V λ,-=< T i -σ(T i ) > i∈[[1,r]] are submodules of V λ .
We know by [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] (10.4) that the irreducible modules of the Iwahori-Hecke algebra of type D in the generic case are the modules V λ labeled by double-partitions (λ 1 , λ 2 ) with λ 1 < λ 2 and the modules V (λ 1 ,λ 1 ),+ and V (λ 1 ,λ 1 ),-for λ 1 n 2 . We will use Proposition 2.2.4 in order to prove that this is also the case in the finite field setting. The Schur elements are quite complicated to write, therefore we need to introduce some new objects before giving the Schur elements. If r ≤ m, then we set

X λ,µ = {λ i + m-i} i∈[[1,r]] ∪ [[0, m-r -1]] and Y λ,µ = {µ i + m -i} i∈[[1,m]] . If m > r, then we set X λ,µ = {λ i + r -i} i∈[[1,r]] and Y λ,µ = {µ i + r -i} i∈[[1,m]] ∪ [[0, r -m]].
For X ⊂ N and u a given parameter, we set ∆(X, u)

= (k,l)∈X 2 ,k>l (u k -u l ).
By [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]

Dn,Z[u,u -1 ] -module V λ,µ is c λ,µ = 2 b-1 u b(b-1)(4b-5) 6 k∈X λ,µ k h=1 (u 2h -1) l∈Y λ,µ l h=1 (u 2h -1) (u -1) n ∆(X λ,µ , u)∆(Y λ,µ , u) (k,l)∈X λ,µ ×Y λ,µ (u k + u l ) where b = |X λ,µ | = |Y λ,µ |.
If (λ, λ) n, then the Schur elements associated to V (λ,λ),+ and V (λ,λ),-are equal and are given by

c λ,λ = 2 b u b(b-1)(4b-5) 6 k∈X λ,λ k h=1 (u 2h -1) l∈Y λ,λ l h=1 (u 2h -1) (u -1) n ∆(X λ,λ , u)∆(Y λ,λ , u) (k,l)∈X λ,λ ×Y λ,λ (u k + u l ) .
We can now state the theorem giving the semi-simple decomposition of H Dn,α .

Theorem 4.1.1. Assume the order of α is greater than 2n. We then have that H Dn,α is split semi-simple and its pairwise non-isomorphic irreducibles modules are V λ,µ (with λ > µ), V λ,λ,+ and V λ,λ,-.

In the following, we write V for the direct sum of these irreducible modules.

Proof. Let A = Z[u ±1 ], θ : Z[u ±1 ] → F q defined by θ(k) = k for k ∈ Z and θ(u) = α. By Proposition 2.2.4
, it is sufficient to show that c λ,µ ∈ B with B as in Proposition 2.2.4, that θ(c λ,µ ) = 0 and (1 + m i (T))m i (T) = 0 and that they are well-defined for (λ, µ) n, for i ∈ [[1, n -1]] and for T ∈ (λ, µ) such that T i↔i+1 is also standard. The condition on the m i implies that for any pair (i, j), there exists a matrix M in the representation associated to V λ,µ such that M i,j = 0 since there exists a path between any pair of standard double-tableaux such that any standard double-tableau in the path is obtained by transposing a pair (r, r + 1).

Let (λ, µ) n as in Definition 4.1.

2, i ∈ [[1, n -1]] and T ∈ (λ, µ). We first show that c λ,µ ∈ B, i.e. θ (u -1) n ∆(X λ,µ )∆(Y λ,µ ) (k, )∈X λ,µ ×Y λ,µ (u k + u ) = 0.
We have that θ((u -

1) n ) = (α -1) n = 0. We have ∆(X λ,µ , u) = (k, )∈X 2 λ,µ ,k> (α k -α ). Let (k, ) ∈ X 2 λ,µ such that k > . We have 0 ≤ < k ≤ λ 1 + max(m, r) -1 and, therefore k -≤ λ 1 + max(m, r) -1. Since λ 1 + r -1 is equal to the number of boxes in the first row of λ, we have that λ 1 + r -1 ≤ n λ .
Since m is the number of boxes in the first column of µ, we have that m ≤ n µ , therefore

λ 1 + max(m, r) -1 = λ 1 + max(r -1 + m -r, r -1) ≤ n λ + n µ ≤ n < 2n. The equality α k -α = 0 implies that α k-= 1.
This quantity is therefore non-zero because the order of α is greater than 2n.

In the same way, we have ∆(Y λ,µ , u) = 0. Let (k, ) ∈ X λ,µ × Y λ,µ . Assume by contradiction that θ(u k + u ) = 0. We then have α k + α = 0, therefore α k-= -1 and α 2(k-) = 1. We have 0

≤ k ≤ λ 1 + max(r, m) -1 and 0 ≤ ≤ µ 1 + max(r, m) -1. Therefore -2n ≤ 1 -max(m, r) -µ 1 ≤ 2(k -) ≤ 2(λ 1 + max(m, r) -1) ≤ 2n and we have α k-= 1 if k = since the order of α is greater than to 2n. If k = , then θ(u k + u ) = 2α k = 0 because p = 2.
It follows that c λ,µ ∈ B.

We now show that θ(c λ,µ ) = 0, i.e.

θ   2 b u b(b-1)(4b-5) 6 k∈X λ,µ k h=1 (u 2h -1) l∈Y λ,µ l h=1 (u 2h -1)   = 0. We have θ(2 b u b(b-1)(4b-5) 6 ) = 2 b α b(b-1)(4b-5) 6 = 0 because p = 2. Let k ∈ N such that k ∈ X λ,µ or (k ∈ Y λ,µ and 0 ≤ h ≤ k).
We have shown that k ≤ n, therefore θ(u 2h -1)) = α 2h -1 = 0 because α is of order greater than 2n. It follows that θ(c λ,µ ) = 0. We now show that m i (T) is well-defined and non-zero. We have that m i (T) = α-1

1-(-1) δ i α c i -r i +r i+1 -c i+1 = 0. It is sufficient to show that 1 -(-1) δ i α c i -r i +r i+1 -c i+1 = 0 or α c i -r i +r i+1 -c i+1 = (-1) δ i .
If i and i + 1 are in the same component of T, then δ i = 0 and |c i -

r i + r i+1 -c i+1 | = |c i -c i+1 +r i+1 -r i |
is less than the minimal number of boxes on a path within the Young diagram from the box where i is and the box where i + 1 is. It follows that α c i -r i +r i+1 -c i+1 = 1 because α is of order greater than 2n > max(n λ , n µ )

> |c i -r i + r i+1 -c i+1 | and c i -r i = c i+1 -r i+1 because T is standard.
If i and i + 1 are in distinct components then we have that δ i = 1. It is then sufficient to show that -2n ≤ c i -r i + r i+1 -c i+1 ≤ 2n because p = 2. If i is in the left tableau and i + 1 is in the right tableau then we have 1

-n ≤ 1 -r ≤ c i -r i ≤ λ 1 -1 ≤ n -1 and 1-n ≤ 1-m ≤ c i+1 -r i+1 ≤ µ 1 -1 ≤ n-1. It follows that 2-2n ≤ c i -r i +r i+1 -c i+1 ≤ 2n-2
which proves that m i (T) is well-defined since α is of order greater than 2n. The same reasoning shows that m i (T) is well-defined if i is in the right tableau and i + 1 is in the left tableau.

Finally, we show that 1 + m i (T) = 0. We have

1 + m i (T) = 1 + α -1 1 -(-1) δ i α c i -r i +r i+1 -c i+1 = α(1 -(-1) δ i α c i -r i +r i+1 -c i+1 -1 ) 1 -(-1) δ i α c i -r i +r i+1 -c i+1 .
We have shown in the previous step that -2n

+ 2 ≤ c i -r i + r i+1 -c i+1 ≤ 2n -2, therefore -2n + 1 ≤ c i -r i + r i+1 -c i+1 ≤ 2n -3. Since T i↔i+1 is standard if i and i + 1 are in the same component we have in that case that |c i -r i + r i+1 -c i+1 | -1 ≥ 2.
This shows that 1 + m i (T) = 0 and concludes the proof.

The branching rule for type D is more complicated, therefore we recall it in the following proposition (a proof in a more general setting can be found in [START_REF] Marin | Branching properties for the groups G(de, e, r)[END_REF]). Lemma 4.1.1. Let n ≥ 5 and (λ, µ) n, λ > µ. We then have:

1. If n λ > n µ + 1, then V λ,µ|H D n-1 ,α = ( λ,μ)⊂(λ,µ) V λ,μ . 2. If n λ = n µ + 1 and µ ⊂ λ, then V λ,µ|H D n-1 ,α = ( μ⊂µ V λ,μ ) ⊕ ( λ⊂λ λ>µ V λ,µ ) ⊕ ( λ⊂λ λ<µ V µ, λ).

If n

λ = n µ + 1 and µ ⊂ λ, then V λ,µ|H D n-1 ,α = ( μ⊂µ V λ,μ ) ⊕ ( λ⊂λ λ>µ V λ,µ ) ⊕ ( λ⊂λ λ<µ V µ, λ) ⊕ V µ,µ,+ ⊕ V µ,µ,-. 4. If n λ = n µ and λ > µ, then V λ,µ|H D n-1 ,α = ( μ⊂µ V λ,μ ) ⊕ ( λ⊂λ V µ, λ). 5. If λ = µ, then V λ,λ,+|H D n-1 ,α = V λ,λ,-|H D n-1 ,α = μ⊂µ V λ,μ .
Proof. Assume first that n λ > n µ + 1. We then have V λ,µ = T∈(λ,µ) F q T. Let r be the number of extremal boxes in the Young double-diagram associated to (λ, µ) (a box is said to be extremal if there exists a standard double-tableau containing n in that box). We write (r , c , δ ) l∈[ [1,r]] the extremal box, where r is the row and c is the column of the box in the component the box belongs to and δ indicates which component the box belongs to. We then have V λ,µ = r l=1 j∈I l T j,l for T j,l , j ∈ I l standard tableaux associated to (λ, µ) such that n is in box (r l , c l , δ l ).

We can then define a bijection from the basis of V λ,µ to the standard double-tableaux basis of ( λ,μ)⊂(λ,µ) V λ,μ by mapping a standard double-tableau T j,l to the standard double-tableau T j,l \ {(r l , c l , δ l )} and we have λ > μ for any ( λ, μ) ⊂ (λ, µ) since n λ > n μ. By construction, this bijection commutes with the action of H D n-1 because for any i ∈ [[1, n -2]], the position of i and i + 1 is unchanged.

Assume now that n λ = n µ + 1 and µ ⊂ λ. We can then apply a similar reasoning except that we can have λ < µ for some λ ⊂ λ. We then map the standard tableau T j,l to σ(T j,l \ {(r l , c l , δ l )}). The action of H D n-1 ,α then still commutes with the bijection since it commutes with action of σ.

Assume now that n λ = n µ + 1 et µ ⊂ λ. We keep the same bijection as in the previous case except for the extremal box (r 0 , c 0 , δ 0 ), which, when removed from the Yound doublediagram (λ, µ), affords the Young double diagram associated to (µ, µ). We then map the tableau T j, 0 = (T j, 0 ,1 , T j, 0 ,2 ) to T j, 0 \ {(r 0 , c 0 , δ 0 )} + σ(T j, 0 \ {(r l 0 , c l 0 , δ l 0 )}) if τ T j, 0 (1) = 1 and to T j, 0 \ {(r 0 , c 0 , δ 0 )} -σ(T j, 0 \ {(r 0 , c 0 , δ 0 )}) otherwise. The action of A Dn,α then again commutes with the bijection because it commutes with σ.

Assume now that n λ = n µ and λ > µ. The bijection is defined in the same way as before except when λ ⊂ λ, where we apply σ.

Assume now that λ = µ. We can then number the standard double-tableaux associated to (λ, λ) by (T 1 , T 2 , . . . , T m , σ(T 1 ), σ(T 2 ), . . . , σ(T m )) such that n is in left component of the m first ones and in the right component of the last m ones. We then have V λ,λ,+ = r j=1 F q (T j + σ(T j )) and V λ,λ,-= r j=1 F q (T j -σ(T j )).

For V λ,λ,+ , we map T j + σ(T j ) to the standard double-tableau obtained by removing the box of T j containing n. We have to check that the action of H D n-1 ,α on T j + σ(T j ) is the same as the one on this tableau. We have m i (T j ) = m i (σ(T j )), σ(T j ) i↔i+1 = σ(T j ) i↔i+1 and T j,i↔i+1 / ∈ {T j , σ(T j )}, therefore the action is indeed identical. For V λ,λ,-, we map in the same way T j -σ(T j ) to the standard double-tableau obtained by removing the box of T j containing n. In order to check that the action is the same, we can use the same arguments as for V λ,λ,+ and check that T j,i↔i+1 ∈ {T k } k∈[ [1,m]] . This is true since we chose a numbering such that n is in the left component only for the m first tableaux and n stays in the same box after permutation of i and i + 1 for any i ≤ n -2.

Remark : The two submodules of V λ,λ are not isomorphic, by the fact that n goes from the left component to the right one or from the right component to the left one after applying S n-1 .

We keep the same weight on double-tableaux as for type B. Let λ = (λ 1 , λ 2 ) n and T = (T 1 , T 2 ) ∈ λ. We define ϕ(T) to be T if µ > λ and σ(T ) otherwise. We define a new ν(λ) to be ν(λ 1 )ν(λ 2 )(-1) n λ 1 (n-n λ 1 ) if λ 2 ≥ λ 1 and ν(λ) = ν(λ 1 )ν(λ 2 ) otherwise. We define the bilinear form (T| T) = ω(T)δ ϕ(T), T.

Proposition 4.1.3. For any pair of standard double-tableaux (T, T), we have the following properties.

1. (S i .T|S i . T) = (-α)(T| T) and (U.T|U. T) = (-α)(T| T).

2. For all d ∈ A Dn , we have that (d.T|d. T) = (T| T).

Those relations stay true if we substitute one or two of the standard double-tableaux by the elements σ(T) -T and σ(T) + T, which form bases for V λ,+ and V λ,-for double-partitions λ of the form λ = (λ 1 , λ 1 ).

3. The restriction of (., .) to

V λ if λ = ϕ(λ) = (λ 2 , λ 1 ) and to V λ ⊕V ϕ(λ) if λ / ∈ {ϕ(λ), (λ 2 , λ 1 )} is non-degenerate with ϕ(λ) = λ if n λ = n µ and µ > λ , and ϕ(λ) = (λ 1 , λ 2 ) otherwise. If λ = ϕ(λ) = (λ 2 , λ 1 ) then (., .) is symmetric on V λ if ν(λ) = 1 and skew-symmetric otherwise.
Moreover, its Witt index is positive.

4. If n ≡ 0 (mod 4) and λ = (λ 1 , λ 1 ), then the restriction of (., .) to V λ,+ and V λ,-if λ = λ and to

V λ,+ ⊕ V λ,-if λ = λ , is non-degenerate.
If λ = λ then (., .) is symmetric on V λ,+ and V λ,-if μ(λ) = 1 and skew-symmetric otherwise. Moreover, its Witt index is positive.

5. If n ≡ 2 (mod 4) and λ = (λ 1 , λ 1 ) then the restriction of (., .

) to V λ,+ ⊕ V λ ,-is non- degenerate.
Proof. For 1. and 2., we have m i (σ(T)) = m i (T), therefore the same proof as for Proposition 3.1.1 applies. The extension to elements of the bases of V λ,+ and V λ,-follows from the bilinearity of (., .). For 3., the same proof also applies because ν(λ) = ω(T)ω(ϕ(T)). This is true because when ϕ(T) = T , ν(λ) does not change from the one in type B and when ϕ(T) = σ(T ), ν(λ) is multiplied by (-1) n λ 1 (n-n λ 1 ) = ω(T)ω(σ(T)).

4. We assume n ≡ 0 (mod 4). If λ = (λ 1 , λ 1 ) n and T ∈ λ then

ω(σ(T)) = (-1) n λ 1 (n-n λ 1 ) ) ω(T) = (-1) ( n 2 ) 2 ω(T) = ω(T).
For any standard double-tableaux T, T, we have that (T| T) = ω(T)δ T,ϕ(T) . Since λ = (λ 1 , λ 1 ), we have ϕ(λ) = λ and for all T ∈ λ and all T ∈ λ , we have ϕ(T) = T .

Let λ = (λ 1 , λ 1 ) and λ = ( λ1 , λ1 ) be double-partitions of n. If T ∈ λ and T ∈ λ, then we have

(T + σ(T)| T + σ( T)) = (T| T) + (T|σ( T)) + (σ(T)| T)) + (σ(T)|σ( T)) = ω(T)(δ T, T + δ T,σ( T) ) + ω(σ(T))(δ σ(T), T + δ σ(T),σ( T) ) = (δ T, T + δ T,σ( T) )(ω(T) + ω(σ(T)) = δ T+σ(T), T +σ( T ) (ω(T) + ω(σ(T))) = 2ω(T)δ T +σ(T) , T+σ( T) .
In the same way, we have that

(T + σ(T)| T -σ( T)) = (T -σ(T)| T + σ( T)) = 0 and (T -σ(T)| T -σ( T)) = 2ω(T)δ T -σ(T) , T-σ( T)
. The result follows.

5.

Assume n ≡ 2 (mod 4).

If λ = (λ 1 , λ 1 ) n and T ∈ λ, then ω(σ(T)) = (-1) n λ 1 (n-n λ 1 ) ) ω(T) = (-1) ( n 2 ) 2 ω(T) = -ω(T). It follows that if λ = ( λ1 , λ1 ) n and T ∈ λ, then (T + σ(T)| T + σ( T)) = (T - σ(T)| T -σ( T)) = 0, (T + σ(T)| T -σ( T)) = 2ω(T)δ T -σ(T) , T-σ( T) and (T -σ(T)| T + σ( T)) = 2ω(T)δ T +σ(T) , T+σ( T)
. The result follows.

Factorization of the image of the Artin group inside the finite Hecke algebra

In this section, we find the different factorizations between the irreducible representations of A Dn . Most of the factorization results are summarized in Proposition 4.2.4. We then state the main results for type D in Theorems 4.2.2 and 4.2.3. We define the linear map L from V to V which sends T to L(T) = ω(T)ϕ(T).

Proposition 4.2.1. Let r ∈ [[1, n -1]
] and T a standard double-tableau, we then have

LS r L -1 (T) = (-α) t (S -1 r )(T), LU L -1 = (-α) t (U -1 )(T). Let λ = (λ 1 , λ 2 ) n.
We have the following propositions.

1. If λ / ∈ {ϕ(λ), (λ 2 , λ 1 )}, then L stabilizes V λ ⊕ V ϕ(λ) and switches V λ and V ϕ(λ) . 2. If λ = ϕ(λ) = (λ 2 , λ 1 ), then L stabilizes V λ . 3. If n ≡ 0 (mod 4) and λ = (λ 1 , λ 1 ) = (λ 1 , λ 1 ), then L stabilizes V λ,+ ⊕ V λ ,+ (resp V λ,-⊕ V λ ,-
) and switches V λ,+ and V λ ,+ (resp V λ,-and V λ ,-).

4. If n ≡ 0 (mod 4) and λ = (λ 1 , λ 1 ) = (λ 1 , λ 1 ), then L stabilizes V λ,+ and V λ,-.

5. If n ≡ 2 (mod 4) and λ = (λ 1 , λ 1 ), then L stabilizes V λ,+ ⊕ V λ ,-and switches V λ,+ and V λ ,-.

Proof. This follows directly from Proposition 4.1.3 by writing the matrix of the bilinear form and the matrix of L. We then have by Proposition 3.2.

5 that R λr (Λ r R λ 1 ) ⊗ η 1,r for all r ∈ [[1, n -1]]. Assume F q = F p (α) = F p (α + α -1
). We write the unique automorphism of F q of order 2. We have that (α) = α -1 . We then define for every standard double-tableau in the same way as for type B,

d(T) = d(T 1 ) d(T 2 ) i∈T 1 ,j∈T 2 i<j 2 + α a i,j -1 + α 1-a i,j α + α -1 + α a i,j + α -a i,j
and the associated hermitian form ., . defined by T, T = d(T)δ T, T. We write Λ for the set of all double-partitions λ = (λ 1 , λ 2 ) of n such that λ 1 ≥ λ 2 . Proof. The proof of the first statement follows from Proposition 3.2.2 and the second follows from the expression of the bases of V λ,± and the Z-bilinearity of the hermitian form.

We now prove two lemmas which will allow us to restrict ourselves to the derived subgroup A Dn of A Dn . Proof. Assume first it is true for n = 2. Since A Dn is generated by A D n-1 and A Dn , we have the result for n ≥ 3 by the same method as in the Lemma 3.4(i) of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF].

We now show the result is true for n = 2. We only have to show it for ( [1], [1]) since the other representations are 1-dimensional. We will show in Section 3.3 (Lemmas 3.3.2 and 3.3.8) that R [1], [1] (A D 2 ) SL 2 (q ) for some q . The irreducibility then follows.

We now show a lemma computing the normal closure of A Dn . This is a generalization to type D of Lemma 3.2.2 Lemma 4.2.2. For n ≥ 4, the normal closure

A D n-1 A Dn of A D n-1 in A Dn is A Dn .
Proof. Let n ≥ 4. By Lemma 2.1 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], we have that A An = A A n-1 A An , where A An =< S 1 , S 2 , . . . , S n-1 >≤ A Dn . We have that A Dn is generated by A An and A D n-1 therefore the result follows.

The following proposition summarizes the results in this section : Proposition 4.2.4. Let λ, µ, γ and δ be doubles-partitions of n such that dim(V λ ) > 1,

λ 1 > λ 2 , µ 1 > µ 2 , γ 1 = γ 2 and δ 1 = δ 2 .
We have the following properties.

(a) R γ,±|A Dn R δ,±|A Dn . (b) R γ,±|A Dn R δ,∓|A Dn ⇔ γ = ϕ(δ). 10. If F q = F p (α) = F p (α + α -1 ), then (a) R λ|A Dn • R µ|A Dn ⇔ λ = µ. (b) R λ|A Dn • R γ,± . (c) R γ,±|A Dn • R δ,±|A Dn ⇔ γ = δ. (d) R γ,±|A Dn • R δ,∓|A Dn .
Proof. (d) . We have R λ (S 1 ) = uR µ (S 1 ). By considering the eigenvalues, we have that {α, -1} = {uα, -u}. Therefore u = 1 or α 2 = 1. By the conditions on α, u = 1 and R λ R µ , therefore λ = µ. Since the set of eigenvalues is of R γ,± (S 1 ) is also {α, -1}, the rest of the proof follows.

A Dn → F q such that R λ R µ ⊗ η. Since A Dn /A Dn =< S 1 >, there exists u ∈ F q such that for all d ∈ A Dn , η(d) = u
We now give a theorem for double-partitions with an empty component and then results for hook partitions. Theorem 4.2.1. Let λ = (λ 1 , ∅) n with λ 1 not a hook and G = R λ (A Dn ). We then have the following properties

1. If F q = F p (α) = F p (α + α -1 ), then (a) if λ 1 = λ 1 , then G = SL n λ (q), (b) if λ 1 = λ 1 and ν(λ) = -1, then G SP n λ (q), (c) if λ 1 = λ 1 and ν(λ) = 1, then G Ω + n λ (q). 2. If F q = F p (α) = F p (α + α -1 ), then (a) if λ 1 = λ 1 , then G SU n λ (q 1
2 ),

(b) if λ 1 = λ 1 and ν(λ) = -1, then G SP n λ (q 1
2 ),

(c) if λ 1 = λ 1 and ν(λ) = 1, then G Ω + n λ (q 1
2 ).

Proof. The restriction of R λ to A Dn is the same as the representation R λ 1 in type A. Since ν(λ) = ν(λ 1 ), the result follows directly from [12, Theorem 1.1] after noting that R λ (A An ) ⊂ R λ (A Dn ) and that we have the corresponding inclusions by Proposition 4.1.3.

Proposition 4.2.5.

If F q = F p (α) = F p (α + α -1 ), then R ([1 n-1 ],[ 1
]) (A Dn ) = SL n (q) and if

F q = F p (α) = F p (α + α -1 ) then R ([1 n-1 ],[1]) (A Dn ) SU n (q 1
2 ).

Proof. The proof is the same one as the proof of Proposition 3.3.1.

We write again

A 1,n = {(λ 1 , ∅), λ 1 n}, A 2,n = {(∅, λ 2 ), λ 2 n}, A n = A 1,n ∪ A 2,n and n = {λ n, λ not a hook} Theorem 4.2.2. If F q = F p (α) = F p (α + α -1
) and n is odd, then the morphism from A Dn to H × Dn,α λ n λ 1 >λ 2

GL n λ (q) factorizes through the epimorphism

Φ 1 ,n : A Dn → SL n-1 (q) × SL n (q) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SL n λ (q) × λ∈ n,λ=ϕ(λ) n λ >nµ OSP (λ) . If F q = F p (α) = F p (α + α -1
) and n ≡ 0 (mod 4), then the morphism from

A Dn to H × Dn,α λ n λ 1 >λ 2 GL n λ (F q ) × λ=(λ 1 ,λ 1 ) n GL n λ,+ (q) × GL n λ,-(q) factorizes through the epimorphism Φ 1 ,n : A Dn → SL n-1 (q) × SL n (q) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SL n λ (q) × λ∈ n,λ=ϕ(λ) λ 1 >λ 2 OSP (λ) × λ=(λ 1 ,λ 1 )∈ n λ>ϕ(λ) SL n λ 2 (q) 2 × λ=(λ 1 ,λ 1 )∈ n λ=ϕ(λ) OSP (λ, +) 2 . If F q = F p (α) = F p (α + α -1
) and n ≡ 2 (mod 4) then the morphism from

A Dn to H × Dn,α λ n λ 1 >λ 2 GL n λ (F q ) × λ=(λ 1 ,λ 1 ) n GL n λ,+ (q) × GL n λ,-(q) factorizes through the epimorphism Φ 1 ,n : A Dn → SL n-1 (q) × SL n (q) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SL n λ (q) × λ∈ n,λ=ϕ(λ) λ 1 >λ 2 OSP (λ) × λ=(λ 1 ,λ 1 )∈ n λ>ϕ(λ) SL n λ 2 (q) 2 × λ=(λ 1 ,λ 1 )∈ n λ=ϕ(λ) SL n λ 2 (q).
In all of the above, OSP (λ) is the group of isometries of the bilinear form defined in Proposition 4.1.3.

In the unitary case, we have an analogous result.

Theorem 4.2.3.

If F q = F p (α) = F p (α + α -1
) and n is odd, then the morphism from

A Dn to H × Dn,α λ n λ 1 >λ 2 GL n λ (q) factorizes through the morphism Φ 2 ,n : A Dn → SU n-1 (q 1 2 ) × SU n (q 1 2 ) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SU n λ (q 1 2 ) × λ∈ n,λ=ϕ(λ) n λ >nµ OSP (λ) . If F q = F p (α) = F p (α + α -1
) and n ≡ 0 (mod 4), then the morphism from

A Dn to H × Dn,α λ n λ 1 >λ 2 GL n λ (F q ) × λ=(λ 1 ,λ 1 ) n GL n λ,+ (q) × GL n λ,-(q) factorizes through the morphism Φ 2 ,n : A Dn → SU n-1 (q 1 2 ) × SU n (q 1 2 ) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SU n λ (q 1 2 ) × λ∈ n,λ=ϕ(λ) λ 1 >λ 2 OSP (λ) × λ=(λ 1 ,λ 1 )∈ n λ>ϕ(λ) SU n λ 2 (q 1 2 ) 2 × λ=(λ 1 ,λ 1 )∈ n λ=ϕ(λ) OSP (λ, +) 2 . If F q = F p (α) = F p (α + α -1
) and n ≡ 2 (mod 4), then the morphism from

A Dn to H × Dn,α λ n λ 1 >λ 2 GL n λ (F q ) × λ=(λ 1 ,λ 1 ) n GL n λ,+ (q) × GL n λ,-(q) factorizes through the morphism Φ 2 ,n : A Dn → SU n-1 (q 1 2 ) × SU n (q 1 2 ) × λ∈ n,λ>ϕ(λ) λ 1 >λ 2 SU n λ (q 1 2 ) × λ∈ n,λ=ϕ(λ) λ 1 >λ 2 OSP (λ) × λ=(λ 1 ,λ 1 )∈ n λ>ϕ(λ) SU n λ 2 (q 1 2 ) 2 × λ=(λ 1 ,λ 1 )∈ n λ=ϕ(λ) SU n λ 2 (q 1
2 ).

In all of the above, OSP (λ) is the group of isometries associated with the bilinear form over

F q 1 2 obtained

The case n = 4

In this section, we prove the result for n = 4.

The double-partitions to consider for n = 4 are ( [4], ∅),

([3, 1], ∅), ([2, 2], ∅), ([2, 1, 1], ∅), ([1 4 ], ∅), ([3], [1]), ([2, 1], [1]), ([1 3 ], [1]), ([2], [2]), ([2], [1 2 ]) and ([1 2 ], [1 2 ]).
By Proposition 4.2.1, if we know the image for λ, we know the image for ϕ(λ). By Proposition 4.2.1, we know the image for doubles-partitions with an empty component. By Proposition 4.2.5, we know the image for ([1 3 ], [1]) and by Proposition 4.2.2, we know the image for ( [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF], [1 2 ]) using the image of ([1 3 ], [1]). The only double-partitions left to consider are ([1 2 ], [1 2 ]) and ([2, 1], [1]).

Lemma 4.3.1. If F q = F p (α) = F p (α + α -1 ), then R [2,1],[1] (A D 4 ) SP 8 (q). If F q = F p (α) = F p (α + α -1 ), then R [2,1],[1] (A D 4 ) SP 8 (q 1
2 ).

Proof. Assume first that F q = F p (α) = F p (α + α -1 ). Using Proposition 4.1.3, there exists

P ∈ GL 8 (q) such that G = P R [2,1],[1] (A D 4 )P -1 ⊂ SP 8 (q). Using Lemma 4.1.1, we have that R [2,1],[1] (A D 3 ) = R [2],[1] × R [1 2 ],[1] × R [2,1],∅ (A D 3 ) SL 3 (q) × SL 2 (q)
, where SL 3 (q) is in a twisted diagonal embedding and SL 2 (q) is in a natural representation using Goursat's Lemma and the previous arguments. Using the same arguments as before and Lemma 3.3.3 with the natural representation of SL 2 (q), we know G is primitive, tensor-indecomposable, irreducible, perfect and cannot be realized in a natural representation over a proper subfield of F q . This implies that G cannot be included in a maximal subgroup of class C 1 , C 2 , C 4 or C 5 . Since it contains a transvection of SL 2 (q), we have that it cannot be contained in a maximal group of class C 3 . Assume that G is included in a maximal subgroup of SP 8 (q). By the Tables 8.48. and 8.49. in [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF], the only possible maximal subgroups and their order or a quantity their order divides are given below 1. 2 1+6.

-SO - 6 (2), 51840

2. 2 1+6.

-Ω - 6 (2), 25920

3. (SP 2 (q) • SP 2 (q) • SP 2 (q)).2 2 .S 3 , 24q 3 (q 2 -1) 3 9. 2 . P SL 2 (q), q(q 2 -1) 10. 2 . P SL 2 (q 3 ).3, 3q 3 (q 6 -1)

The order of SL 3 (q) × SL 2 (q) is q 4 (q 2 -1) 2 (q 3 -1), therefore cases 3, 9 and 10 are excluded. We have that α is of order greater than 16 and p = 2, therefore we have that q > 17 and, therefore q ≥ 19 and |G| ≥ 19 4 (19 2 -1)(19 3 -1) = 115828887772800. This excludes all the remaining cases. It follows that G can be included in no maximal subgroup of SP 8 (q), therefore G = SP 8 (q). Assume now that F q = F p (α) = F p (α + α -1 ). There exists P ∈ GL 8 (q) such that

G = P R [2,1],[1] (A D 4 )P -1 ⊂ SP 8 (q 1
2 ) and G contains H SU 3 (q

1 2 ) × SU 2 (q 1
2 ), where SU 3 (q

2 ) is in a twisted diagonal embedding and SU 2 (q

1 2
) is in a natural representation. We can no longer use Lemma 3.3.3 in this case, but since (α) = α -1 , we have up to conjugation that diag(I 6 , α 0 0 α -1 ) ∈ G. It follows that α + α -1 belongs to the field generated by the traces of the elements of G. This shows that any field over which G is realized in a natural representation contains F q 1 2 . By the above, in this case, we have that G is primitive, tensor-indecomposable, irreducible, perfect and cannot be realized in a natural representation over a proper subfield of F q 1 2 . This implies that G cannot be included in a maximal subgroup of SP 8 (q

1 2 ) of class C 1 , C 2 , C 4 or C 5 . It contains a transvection of SU 2 (q 1
2 ), therefore it cannot be included in a maximal subgroup of class C 3 . Assume G is included in a maximal subgroup of SP 8 (q 1 2 ). We list below the possible maximal subgroups and their order or a quantity their order divides 1. 2 1+6.

-SO - 6 (2), 51840

2. 2 1+6.

-Ω - 6 (2), 25920

3. (SP 2 (q 1 2 ) • SP 2 (q 1 2 ) • SP 2 (q 1 2 )).2 2 .S 3 , 24q 3
2 (q -1) 3

2 ), q 1 2 (q -1) 10. 2 . P SL 2 (q 3 ).3, 3q

2 (q 3 -1)

We have |H| = q 2 (q -1) 2 (q 3 2 + 1). This excludes cases 3, 9 and 10. We have α q 1 2 = (α) = α -1 . It follows that q 1 2 + 1 > 16 and, therefore q 1 2 ≥ 17 because p = 2. This implies that |H| ≥ 34042058459136. This proves that G = SP 8 (q 1 2 ) and concludes the proof of the lemma.

Lemma 4.3.2. If F q = F p (α) = F p (α + α -1 ), we have R ([1 2 ],[1 2 ]),+ (A D 4 ) = R ([1 2 ],[1 2 ]),-(A D 4 ) = SL 3 (q). If F q = F p (α) = F p (α + α -1 ), we have R ([1 2 ],[1 2 ]),+ (A D 4 ) = R ([1 2 ],[1 2 ]),-(A D 4 ) SU 3 (q 1
2 ).

Proof. The result follows from Lemma 4.1.1 and the fact that R ([

1 2 ],[1]) (A D 3
) is equal to the group we want in both cases.

Surjectivity of Φ 1 ,n for n ≥ 5

In this section, we use results of the previous sections to prove by induction on n the main results for type D. We will here conclude the proof of Theorem 4.2.2. Assume first that F q = F p (α) = F p (α + α -1 ). Using Proposition 4.2.4, by the same kind of arguments as for type B, we can use Goursat's Lemma to show the morphism is surjective upon each component. This means it is sufficient to show the following theorem. n not a hook, such that λ 1 ≥ λ 2 . We write

G(λ) = R λ (A Dn ) if λ 1 > λ 2 , G(λ, +) = R λ,+ (A Dn ) and G(λ, -) = R λ,-(A Dn ) otherwise.
We then have the following possibilities. i. If n ≡ 0 (mod 4) then

1. If λ = ([2, 1 n-2 ], ∅), then G(λ) = SL n-1 (q). 2. If λ = ([1 n-1 ], [1]), then G(λ) = SL n (q), 3. If λ ∈ n , λ 1 > λ 2 and λ > ϕ(λ), then G(λ) = SL n λ (q), 4. If λ ∈ n , λ 1 > λ 2 and λ = ϕ(λ), then we have the following possibilities. (a) If ν(λ) = -1, then G(λ) SP n λ (q). (b) If ν(λ) = 1, then G(λ) Ω + n λ (q).
A. if ν(λ) = -1 then G(λ, +) = G(λ, -) SP n λ (q), B. if ν(λ) = 1 then G(λ, +) = G(λ, -) Ω + n λ (q). ii. If n ≡ 2 (mod 4) then G(λ, +) = G(λ, -) = SL n λ 2 (q).
Proof. For n = 4, we have the result by the previous section. Theorem 4.2.1 gives us the result for double-partitions with an empty component and Proposition 4.2.5 gives us the result for double-partitions with two rows or two columns and one of the components of size one. For n ≥ 5, we proceed by induction but we must first treat the following cases separately :

([2, 2], [1]), ([1 3 ], [1 2 ]) and (([1 3 ], [1 3 ]), ±).
By Lemma 4.3.1, Theorem 4.2.1, Lemma 4.1.1 and Goursat's Lemma, we have that

R ([2,2],[1]) (A D 4 ) SP 8 (q) × SP 2 (q)
. By Theorem 2.3.4 and Lemma 5.6. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], we have that

R ([2,2],[1]) (A D 5 ) ∈ {SL 10 (q), SU 10 (q 1 2 ), SP 10 (q)}. We have that ([2, 2], [1]) = ϕ(([2, 2], [1])) and ν([2, 2], [1]) = (-1) 4-2 2 (-1) 1-1 2
= -1. This implies by Proposition 4.1.3 that up to conjugation in GL 10 (q), we have that

R ([2,2],[1]) (A D 5 ) ⊂ SP 10 (q), therefore we have that R ([2,2],[1]) (A D 5 ) SP 10 (q).
In the same way, we have that R ([

1 3 ],[1 2 ]) (A D 4 ) SL 4 (q) × SL 3 (q) × SL 3 (q), therefore G(([1 3 ], [1 2 ])) = R ([1 3 ],[1 2 ]) (A D 5 ) is in {SL 10 (q), SU 10 (q 1
2 ), SP 10 (q)}. By Proposition 4.2.4, we know that G(([1 3 ], [1 2 ])) preserves no bilinear form, therefore we only have to exclude the unitary case. Assume that G(([

1 3 ], [1 2 ])) is included up to conjugation in SU 10 (q 1
2 ). There then exists an automorphism of order 2 of F q such that each M in G(λ) is conjugate to t ((M -1 )).

In particular G(([

1 3 ], [1 2 ])) contains a natural SL 2 (q). This implies that diag(I 8 , α 0 0 α -1 )
is conjugate to diag(I 8 , t (( α 0 0 α -1 ) -1 )). Taking the traces of those matrices implies that (α + α -1 + 8) = α + α -1 + 8. We have that F q = F p (α) = F p (α + α -1 ), therefore this shows that is trivial which is a contradiction. It follows that G(([1 3 ], [1 2 ])) = SL 10 (q). By Lemma 4.1.1 and the fact that R ([

1 3 ],[1 2 ]) (A D 5 ) = SL 10 (q), we have that SL 10 (q) ⊂ R ([1 3 ],[1 3 ]),± (A D 6 ) ⊂ SL 10 (q). It follows that R ([1 3 ],[1 3 ]),± (A D 6 ) = SL 10 (q).
We now proceed to the induction on n using Theorem 2.3.2.

Let n ≥ 5 and λ n. Suppose the theorem is true for n -1. We use Lemma 4.1.1 for different possibilities to show that G(λ) or G(λ, ±) contains a subgroup verifying the same properties as in type B.

1. If λ = (λ 1 , λ 2 ) and λ 1 > λ 2 and λ = ϕ(λ) then ϕ(λ) = (λ 1 , λ 2 ) because the order we defined for partitions of n 2 verifies that, if

λ 1 = λ 2 and λ 1 > λ 2 , then λ 1 > λ 2 . We then have λ 1 = λ 1 or λ 2 = λ 2 . (a) If λ 2 = λ 2 , then there exists µ 2 ⊂ λ 2 such that µ 2 ⊂ λ 2 . We have that (λ 1 , µ 2 ) ⊂ (λ 1 , λ 2 ) because µ 2 ⊂ λ 2 and (µ 2 , λ 1 ) ⊂ (λ 1 , λ 2 ) because otherwise λ 1 = λ 2 and, therefore λ 2 = λ 1 . This shows that G(λ) contains a natural SL 3 (q). (b) If λ 2 = λ 2 and λ 1 = λ 1 , then there exists µ 1 ⊂ λ 1 such that µ 1 ⊂ λ 1 . We then have (µ 1 , λ 2 ) ⊂ (λ 1 , λ 2 ) because µ 1 ⊂ λ 1 and (λ 2 , µ 1 ) ⊂ (λ 1 , λ 2 ) because λ 2 = λ 1 .
This shows that G(λ) also contains a natural SL 3 (q) in this case.

2. If λ = (λ 1 , λ 2 ) = ϕ(λ) and λ 1 > λ 2 , then (a) If ϕ(λ) = (λ 1 , λ 2 ), then i. If λ 1 and λ 2 are square partitions, then R λ (A D n-1 ) = G(µ 1 , λ 2 ) × G(λ 1 , µ 2 ) and since ν(µ 1 , λ 2 ) = ν(λ 1 , µ 2 ) = ν(λ), we have that : A. If ν(λ) = 1, then Ω + n (µ 1 ,λ 2 ) (q) × Ω + n (λ 1 ,µ 2 ) (q) ⊂ G(λ) ⊂ Ω + n (λ 1 ,λ 2 ) (q). B. If ν(λ) = -1 then SP n (µ,λ 2 ) (q) × SP n (λ 1 ,µ 2 ) (q) ⊂ G(λ) ⊂ SP n λ (q). It follows that G(λ)
is an irreducible group generated by transvections because it is normally generated by the group on the left of our inclusions, therefore by Theorem 2.3.4, we have that G(λ) is equal to the group on the right and the theorem is proved in this case. ii. If λ 1 or λ 2 is not a square partition then there exists µ ⊂ λ such that ϕ(µ) = µ.

It follows that ϕ(µ) ⊂ λ or σ(ϕ(µ)) ⊂ λ, therefore G(λ) contains a twisted diagonal SL 3 (q). (b) If ϕ(λ) = (λ 2 , λ 1 ), then if µ ⊂ λ 2 , we have that (λ 1 , µ) ⊂ (λ 1 , λ 2 ), ϕ((λ 1 , µ)) = (λ 1 , µ ) ⊂ (λ 1 , λ 2 ) because λ 1 = λ 1 . We have that (µ , λ 1 ) ⊂ (λ 1 , λ 2 ), therefore G(λ) contains a twisted diagonal SL 3 (q). 3. If λ = (λ 1 , λ 1 ) = (λ 1 , λ 1 ), then there exists µ 1 ⊂ λ 1 such that µ 1 ⊂ λ 1 . It follows that (λ 1 , µ 1 ) ⊂ (λ 1 , λ 1 ) and (µ 1 , λ 1 ) ⊂ (λ 1 , λ 1 )
. This shows that G(λ, ±) contains a natural

SL 3 (q). 4. If λ = (λ 1 , λ 1 ) = (λ 1 , λ 1
) and λ 1 is not a square partition, then there exists

µ 1 ⊂ λ 1 such that µ 1 = µ 1 , therefore (λ 1 , µ 1 ) = ϕ((λ 1 , µ 1 )) = (λ 1 , µ 1 ). We have that (µ 1 , λ 1 ) ⊂ (λ 1 , λ 1 ), therefore G(λ, ±) contains a twisted diagonal SL 3 (q). 5. If λ = (λ 1 , λ 1 ) = (λ 1 , λ 1
) and λ 1 is a square partition, then we have the two following possibilities.

(a) If n ≡ 0 (mod 4), then for all µ ⊂ λ, we have that ν(λ) = ν(λ 1 ) 2 (-1) ( n 2 ) 2 = 1 = ν(µ). This is because if λ 1 is a square, then the only sub-partition µ 1 of λ 1 verifies ν(µ 1 ) = ν(λ 1 ). By the branching rule, we have that Ω

+ n λ 2 (q) ⊂ G(λ, ±) ⊂ Ω + n λ 2 (q). It follows that G(λ) Ω + n λ 2
(q) and the theorem is proved in this case. (b) If n ≡ 2 (mod 4), then ν(µ) = ν(λ 1 ) 2 = 1 for all µ ⊂ λ. The branching rule shows that Ω

+ n λ 2 (q) ⊂ G(λ, ±) ⊂ SL n λ 2 (q). By Proposition 4.2.4, G(λ, ±) preserves no bilinear form, therefore G(λ, ±) = SL n λ 2 (q).
In all the cases where G(λ) or G(λ, ±) contains a natural SL 3 (q) or a twisted diagonal SL 3 (q), we can use exactly the same arguments as in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] because if the morphism A An to A Dn defined by S i → S i is trivial then A Dn is trivial.

The only case we need to treat separately is (( 40 (q). This contradicts its irreducibility.

[2, 1], [2, 1]), ±) because n = 6. We need a separate argument to show that G(([2, 1], [2, 1]), ±) is tensor-indecomposable. In this case R ([2,1],[2,1]) (A D 5 ) = G([2, 1], [1 2 ]) × G([2, 1], [2]) = SL 20 (q) × SL 20 (q). If G(([2, 1], [2, 1]), ±) ⊂ SL 40 (q) ⊗ SL 2 (q), then the morphism from R ([2,1],[2,1]) (A D 5 ) to SL 2 (q) is trivial. Since R ([2,1],[2,1]) (A D 5 ) normally generates G(([2, 1], [2, 1]), ±), G(([2, 1], [2, 1]), ±) is included in SL 40 (q) × SL
This shows that it is sufficient to consider case 2.a.i.A. Assume we are in case 2.a.i.A. We then have that G(λ) ⊂ Ω + n λ (q) is generated by a conjugacy class of long root elements and Op(G(λ)) (V ) is semisimple and since O p (G(λ)) is a p-group, its only irreducible representation over F q is the trivial one. This shows that Res

G(λ) is irreducible. Since p = 2, if we check that O p (G(λ)) ⊂ [G, G] ∩ Z(G),
G(λ) Op(G(λ)) (V ) is trivial, therefore O p (G(λ)) = 1
and all the assumptions of Theorem I of Kantor are verified (the minimal dimension in this case is greater than or equal to the dimension of ( [3,3,3], [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]) and the dimension of ( [4,4,4,4], [1]) which are 42 × 2 × 13 4 ≥ 5 and 17 × 24024 ≥ 5). This shows that we are in one of the following cases :

1. G(λ) Ω + n λ (q ), and q |q,

2. G(λ) Ω - n λ (q ) ⊂ Ω + n λ (q 2
), q 2 |q and n λ is even,

3. G(λ) SU n λ 2 (q ) ⊂ Ω + n λ (q )
, n λ ≡ 0 (mod 4) and q |q, 4. G(λ) ⊂ Ω + 8 (q ) and q |q, 5. G(λ) [G 2 (q ), G 2 (q )] ⊂ Ω 7 (q ) and q |q, 6. G 3 D 4 (q ) ⊂ Ω + 8 (q 3 ) and q 3 |q.

Since n ≥ 13, α q-1 = 1 and α is of order greater than 2n, we have q ≥ 29 and n λ ≥ min(84 13 4 , 17 × 24024). This proves that Cases 4, 5 and 6 are excluded by cardinality arguments.

Let us show that 3. is also excluded by cardinality arguments. We write |G| p the order of a Sylow p-subgroup of a group G, therefore that

|SU n λ 2 (q )| p = q n λ 2 ( n λ 2 -1) 2
. We know that G(λ) contains Ω + n 1 (q) × Ω + n 2 (q). It follows that if λ 1 is the square partition of r and λ 2 is the square partition of n -r < r, writing a l for the number of standard tableaux associated with a square partition of l ∈ N , we have that n λ = n r a r a n-r , n 1 = n-1 r-1 a r a n-r and n 2 = n-1 r a r a n-r . Note that a r is even because r > 1 and using the branching rule twice, we get that a r is equal to twice the dimension of the two partitions we get by removing first the only extremal node and then one of the two extremal nodes of the resulting partition. It follows that

|Ω + n 1 (q) × Ω + n 2 (q)| p = q n 1 2 ( n 1 2 -1)+ n 2 2 ( n 2
2 -1) . To exclude 3, it is sufficient to show that this quantity is strictly greater than q

n λ 2 ( n λ 2 -1) 2
. If we write A the q-logarithm of the quotient of those two quantities, we have that :

A = n 1 2 n 1 2 -1 + n 2 2 n 2 2 -1 - n λ 2 ( n λ 2 -1) 2 = n 1 2 2 + n 2 2 2 - n λ 2 2 2 - n λ 4 = n λ 2 2 2 -2 n 1 2 n λ -n 1 2 - n λ 4 = ( n λ 2 -n 1 ) 2 2 - n λ 4 = ( n r )aran-r 2 -n-1 r-1 a r a n-r 2 2 - n r a r a n-r 4 = a r a n-r   ( n r ( 1 2 -r n )) 2 2 a r a n-r - n r 4   = a r a n-r 4 n r 2a r a n-r n r 2r -n 2n 2 -1 .
This shows that A > 0 if and only if 2a r a n-r n r

(2r-n) 2 4n 2 > 1.
Using the branching rule and the hook formula, we get : 2 , the last inequality being true because k ≥ 6. It follows that for all k ≥ 4, we have that a k 2 > 81 × (k 2 ) 2 . In our case, we have that r ≥ 16 or r = 9 and n -r = 4. If r ≥ 16 and n -r ≥ 2 then we have a r a nr ≥ r 2 (n -r) 2

a 1 = 1, a 4 = 2, a 9 = 42, a 16 = 16! 7×6 2 ×5 3 ×4 4 ×3 3 ×2 2 = 24024 > 81 × 16 2 , a 25 = 701149020 > 81 × 25 2 and a 36 > 81 × 36 2 . Let k ≥ 6, assume a k 2 > 81(k 2 ) 2 . The branching rule shows that a (k+1) 2 > 2a k 2 > 81(2k 4 ) > 81(k 4 + 4k 3 + 6k 2 + 4k + 1) = 81((k + 1) 2 )
≥ 4r 2 ≥ 2r 2 + 2r(n -r) ≥ (r + n -r) 2 ≥ n 2 . It follows that 2a r a n-r n r (2r-n) 2 4n 2 ≥ 2n 2 n r 1 4n 2 = ( n r ) 2 > 1. If r ≥ 16 and n -r = 1, then 2a r a n-r n r (2r-n) 2 4n 2 = 8na n-1 4n 2 = 2a n-1 n > 162 > 1. If r = 9 and n -r = 4, then 2a r a n-r n r (2r-n) 2 4n 2 = 2 × 42 × 2 13 9 (18-13) 2 4×13 2 > 1.
This shows that independently of r and n -r, we have that A > 0. This proves that 3. is excluded.

We have that |Ω + n λ (q

1 2 )| p = q n λ 2 ( n λ 2 -1) 2
. The previous arguments show that 2. is also impossible.

The only remaining possibility is 1 and using again the same arguments, we get q > q 1 2 , therefore q = q and this concludes the proof of Theorem 4.4.1.

In the unitary case, i.e. F q = F p (α) = F p (α + α -1 ), all the arguments are analogous.

Chapter 5

Type I 2 (m)

m odd

The main difficulty in finding the image of Artin groups of dihedral type inside their finite Hecke algebras arises from the various field extensions which intervene. When m is even they can be quite complex. In this section, we only consider m odd. The outline of the proof is to first determine the image inside each 2-dimensional using Dickson's Theorem and then recover the image inside the full Iwahori-Hecke algebra using Goursat's Lemma. The main difficulty will be in the use of Goursat's Lemma, we will need to introduce the equivalence relation from Lemma 5.1.1 and the the proof will be computational. The image in type I 2 (5) will be useful for inductive arguments in type H 3 .

Let m ≥ 5 be an odd integer and p a prime number such that there exists α ∈ (F p ) × of order not dividing m and not in {1, 2, 3, 4, 5, 6, 10} and θ ∈ F p a primitive m-th root of unity. Moreover, we assume α + α -1 = θ j + θ -j for all j from 1 to m-1 2 . For j ∈ N, we write F q j = F p (α, θ j + θ -j ). We write F q for the smallest field containing all F q j , for j between 1 and m-1 2 .

Note that we have F q = F q 1 , this can be seen using Chebyshev polynomials for example. Note also that when F q j = F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j ), the extension is always of degree 2 since X 2 -(α + α -1 )X + 1 is an irreducible polynomial of degree 2 such that F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j )/(X 2 -(α + α -1 )X + 1). This implies that q j is a prime and q 1 2 j is well-defined. Definition 5.1.1. The Iwahori-Hecke algebra of dihedral type I 2 (m) which we write H I 2 (m),q is the F q -algebra with the following presentation : Generators : T t , T s . Relations :

(T s -α)(T s + 1) = 0, (T t -α)(T t + 1) = 0, T s T t T s . . . m = T t T s T t . . . m .
We can then use the Kilomoyer-Solomon model [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] (Theorem 8.3.1).

Theorem 5.1.1. The following matrix model gives a decomposition into pairwise non-isomorphic irreducible modules of H I 2 (m),q :

1. Ind : T s → α, T t → α.

2. : T s → -1, T t → -1.

For

j ∈ [[1, m-1 2 ]], T s → ρ j (T s ) = -1 0 1 α , T t → ρ j (T t ) = α α(2 + θ j + θ -j ) 0 -1 .
Proof. We already know these models give us representations. It is thus sufficient to show that they are still irreducible and pairwise non-isomorphic in the finite field case.

The two 1-dimensional representations are non-isomorphic because α = -1 by the condition on its order.

Let us show that the 2-dimensional representations are indeed irreducible. Let j ∈ [[1, m-1 2 ]] and V the associated H I 2 (m),q -module.

Let W be a H I 2 (m),q -submodule of V associated to the representation ρ j and x = (x 1 , x 2 ) ∈ W \ {(0, 0)}.

We have ρ j (T s ).x + x = (0, x 1 + (α + 1)x 2 ) ∈ W , therefore it follows that (0, 1) ∈ W or

x 1 + (α + 1)x 2 = 0.
Assume first that (0, 1) ∈ W .

We have x = x 1 (1, 0) + x 2 (0, 1) ∈ W , therefore x 1 (1, 0) ∈ W , which implies that

x 1 = 0 or (1, 0) ∈ W . If (1, 0) ∈ W then since (0, 1) ∈ W , we have V = W . If x 1 = 0, then (1, 0) ∈ W or x 2 α(2 + θ j + θ -j ) = 0 since ρ j (T t ).x = (αx 1 + α(2 + θ j + θ -j )x 2 , -x 2 ) ∈ V . If (1, 0) ∈ W , then V = W by the same reasoning as above. We have 2 + θ j + θ -j = (1 + θ j )(1 + θ -j ) and θ 2j = 1 because m is odd, j ∈ [[1, m-1 2 ]
] and θ is a primitive m-th root of unity. It follows that x 2 α(2+θ j +θ -j ) = 0 implies x 2 = 0 and, therefore x = (0, 0), which is absurd because we chose x ∈ W \ {(0, 0)}.

Assume now x 1 + (α + 1)x 2 = 0. We have ρ j (T t ).x + x = ((α + 1)x 1 + α(2 + θ j + θ -j )x 2 , 0) ∈ W , therefore (α + 1)x 1 + α(2 + θ j + θ -j )x 2 = 0 or (1, 0) ∈ W . If (α +1)x 1 +α(2+θ j +θ -j )x 2 = 0 then by substituting x 1 by -(α +1)x 2 , we get x 2 (-α 2 -2α -1 + 2α + α(θ j + θ -j )) = 0, therefore x 2 = 0 or α 2 -α(θ j + θ -j ) + 1 = (α -θ j )(α -θ -j ) = 0. We cannot have x 2 = 0 because otherwise x 1 = -(α +1)x 2 = 0 and x = (0, 0). We also cannot have α = θ ±j because otherwise the order of α would divide m which contradicts the assumption on α.

If (1, 0) ∈ W , then x 2 = 0 or (0, 1) ∈ W since x ∈ W . We therefore have

x 2 = x 1 = 0 or W = V .
This shows that in all cases W = V . This proves that V is indeed irreducible. It now remains to show that these representations are non-isomorphic. Let j and l be two integers such that 1 ≤ j ≤ ≤ m-1 2 . We have Tr(ρ j (T s )ρ j (T t )) = α(θ j + θ -j ). This implies that if ρ j is isomorphic to ρ then α(θ j + θ -j ) = α(θ + θ -), therefore θ j + θ -j = θ + θ -and, therefore (θ j -θ l )(1 -θ --j ) = 0. This implies that j = l because 0 ≤ -j < m and 2

≤ j + ≤ m -1 < m. Theorem 5.1.2. Let j ∈ [[1, m-1 2 ]]. If G = A I 2 (m) , then 1. If F q j = F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j
) then we have ρ j (G) = SL 2 (q j ).

2. If F q j = F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j ) then we have up to conjugation in GL 2 (q j ) that ρ j (G) SU 2 (q 1 2 j ).

Proof. The proof of this result is done in the same way as Lemma 3.5. of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF]. The proof uses Dickson's theorem (see [START_REF] Huppert | I. Die Grundlehren der Mathematischen Wissenschaften[END_REF], Theorem 8.27. chapter 2) and shows that if we prove that ρ j (G) is not abelian by abelian, ρ j (A I 2 (m ) / ∈ {S 4 , A 5 } up to isomorphism and that α + α -1 and θ j + θ -j belong to the field generated by the traces of the elements of ρ j (G), then we have the result stated in the proposition. We also need to show that in the second case ρ j (G) ⊂ GU 2 (q 1 2 j ). We need to show that the groups considered are not abelian by abelian. In order to do this, we will show that in both cases, we have

A = [ρ j (T s ), ρ j (T t )][ρ j (T t ) -1 , ρ j (T s )] -[ρ j (T t ) -1 , ρ j (T s )][ρ j (T s ), ρ j (T t )] = 0 and B = [ρ j (T s ), ρ j (T t )][ρ j (T t ) -1 , ρ j (T s )] + [ρ j (T t ) -1 , ρ j (T s )][ρ j (T s ), ρ j (T t )] = 0, where [x, y] = xyx -1 y -1 .

This will prove that [ρ

j (T s ), ρ j (T t )][ρ j (T t ) -1 , ρ j (T s )] = [ρ j (T t ) -1 , ρ j (T s )][ρ j (T s ), ρ j (T t )]. If ρ j (G)
is abelian by abelian then there exists an abelian normal subgroup H of ρ j (G) such that ρ j (G)/H is abelian. This implies that the derived subgroup of ρ j (G) is included in H and is therefore abelian which contradicts the above inequality. We now prove that A = 0 and B = 0. We have that the entry

A 1,2 of A verifies A 1,2 = -(2+θ j +θ -j ) 2 (α-1) 2 (α-θ j )(θ -j -α) α 3
, therefore A is non-zero because we assumed that the order of α does not divide m and that θ is an m-th root of unity. Assume by contradiction that B = 0. We write γ j = θ j +θ -j +2. We have γ j = (θ j +1)(θ -j +1) = 0. We then have B 1,2 = -

γ j (α 2 -1)(-α 2 γ j +αγ 2 j -2αγ j +2α-γ j ) α 3 = 0, therefore -α 2 γ j + αγ 2 j -2αγ j + 2α - γ j = 0.
If p = 2 then γ j (-α 2 + αγ j -1) = 0, therefore α 2 + α(θ j + θ -j + 1) = 0 and (α + θ j )(α + θ -j ) = 0 This is absurd because α m = 1. Assume now p = 2, we set a = α 3 A 1,2 γ j (α 2 -1) , b = α 2 A 1,1 and c = -α 2 A 2,2 . We then have 1 4 (b + 2(α 2 -1)a -c -αa + a) = α(α 2 -α + 1). However a = b = c = 0, therefore α 2 -α + 1 = 0. It follows that α 6 -1 = (α 3 -1)(α + 1)(α 2 -α + 1) = 0, which is impossible by the assumption on the order of α.

We need to prove that ρ j (G) contains elements of order different from 1, 2, 3 and 5. This will show that ρ j (G) is not isomorphic to A 5 . We already have that it is not isomorphic to S 4 since S 4 is abelian by abelian. The eigenvalues of ρ j (T s ) are -1 and α, therefore if ρ j (T s ) r = I 2 , we have (α r , (-1) r ) ∈ {(1, 1), (-1, -1)}, therefore α 2r = 1 which implies that r / ∈ {1, 2, 3, 5} by the conditions on α.

We now show that α + α -1 and β + β -1 belong to the field generated by the traces of elements of ρ j (G). We have Tr(ρ j (T t T -1 s )) = γ j -α -α -1 and Tr([ρ j (T t ), ρ j (T s )]) -2 = γ j (γ j -(α + α -1 ) -2). We have γ j -(α + α -1 ) -2 = (θ j -α)(1 -α -1 θ -j ) = 0, therefore γ j and α + α -1 are indeed in the field generated by traces of the elements of ρ j (G).

It remains to show that if F q j = F p (α + α -1 , θ j + θ -j ) then we have up to conjugation in GL 2 (q j ) that ρ j (G) ⊂ GU 2 (q 1 2 j ). By Lemma 2.4. of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], it is sufficient to show that there exists a matrix P ∈ GL 2 (q j ) such that P ρ j (T s )P -1 = ( t ρ j (T s ) -1 ) and P ρ j (T t )P -1 = ( t ρ j (T t ) -1 ), where is the unique automorphism of order 2 of F q j . Since γ j ∈ F q 1 2 j , we have (γ j ) = γ j . We also have (X

-α)(X -α -1 ) = X 2 -(α + α -1 )X + 1 ∈ F q 1 2 j and α / ∈ F q 1 2 j
, therefore we have (α) = α -1 .

Set P = α+1 γ j 1 α α + 1 . We have det(P ) = (α-θ j )(α-θ -j ) γ j = 0. The matrix P verifies the desired property and the result follows.

We now provide a field-theoretic lemma to see when the representations are linked by composition with a field automorphism. This is necessary to determine the image of A I 2 (m) inside the full Iwahori-Hecke algebra.

Lemma 5.1.1. Let j, l ∈ [[1, m-1 2 ]] 2 .
There exists an automorphism Ψ l,j of F q j = F p (α, ξ j + ξ -j ) verifying Ψ l,j (α + α -1 ) = α + α -1 and Ψ l,j (ξ j + ξ -j ) = ξ l + ξ -l if and only if there exists r ∈ N such that jp r ≡ l (mod m) or jp r ≡ -l (mod m) and (α + α -1 ) p r = α + α -1 .

We say that j ∼ l if one of those conditions is verified. This is an equivalence relation and when j ∼ l, we have

ρ l|A I 2 (m) = Ψ l,j • ρ j|A I 2 (m) .
Proof. Assume there exists r ∈ N such that jp r ≡ l (mod m) or jp r ≡ -l (mod m) and (α + α -1 ) p r = α + α -1 . Let Ψ be the automorphism of F q j defined by Ψ(x) = x p r for all x ∈ F q j . We then have Ψ(α + α -1 ) = α + α -1 by assumption and Ψ(ξ j + ξ -j ) = (ξ j + ξ -j ) p r = ξ jp r + ξ -jp r = ξ l + ξ -l .

Assume now that there exists an automorphism Ψ of

F q j = F p (α, ξ j + ξ -j ) verifying Ψ(α + α -1 ) = α + α -1 and Ψ(ξ j + ξ -j ) = ξ l + ξ -l .
There exists r ∈ [[0, log p (q) -1]] such that Ψ(x) = x p r for all r ∈ N, therefore (α + α -1 ) p r = Ψ(α + α -1 ) = α + α -1 . The map Ψ can be extended to an automorphism Ψ of F p by defining Ψ to be the automorphism sending x to x p r for all x ∈ F p . We then have ξ l + ξ -l = Ψ(ξ j + ξ -j ) = Ψ(ξ j + ξ -j ) = Ψ(ξ j ) + Ψ(ξ -j ). It follows that ( Ψ(ξ j )ξ l -1)(ξ -l -Ψ(ξ -j )) = 0, therefore Ψ(ξ j ) ∈ {ξ l , ξ -l }. This proves that ξ jp r ∈ {ξ l , ξ -l }, therefore jp r ≡ l (mod m) or jp r ≡ -l (mod m).

The fact that ∼ is an equivalence relation follows from the fact that for all r ∈ N, Gcd(m, p r ) = 1.

Theorem 5.1.3. Assume m odd and α satisfies the conditions given at the beginning of this section. For

j ∈ [[1, m-1 2 ]], we set G j = SL 2 (q j ) if F q j = F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j ) and G j = SU 2 (q 1 2 j ) if F q j = F p (α, θ j + θ -j ) = F p (α + α -1 , θ j + θ -j ).
We then have that the morphism from

A I 2 (m) to H × I 2 (m),α GL 1 (q j ) 2 × j∈[[1, m-1 2 ]] GL 2 (q j )
factorizes through the surjective morphism :

Φ : A I 2 (m) → j∈[[1, m-1 2 ]]/∼ G j .
Proof. We know by Theorem 5.1.2 that the composition of the morphism from

A I 2 (m) to H × I 2 (m),α
with the projection upon each representation is surjective. We also know by lemma 5.1.1 that the morphism to H × I 2 (m),α factorizes through the morphism Φ. We will now use Goursat's lemma and induction on j ∈ [[1, m- 1 2 ]] in order to conclude the proof of this theorem. For

j 0 ∈ [[1, m-1 2 ]
]. We define Φ j 0 (A I 2 (m) ) to be the image of

A I 2 (m) inside j∈[[1,j 0 ]]/∼ GL 2 (q j ).
We know that Φ

1 (A I 2 (m) ) = G 1 . Let j 0 ∈ [[1, m-3 2 ]], assume that Φ j 0 (A I 2 (m) ) = j∈[[1,j 0 ]]/∼ G j . Consider Φ j 0 +l (A I 2 (m) ) ⊂ j∈[[1,j 0 ]]
G j × G j l for the smallest positive integer such that j j for all j ∈ [[1, j 0 ]]. We know that the projection upon each factor is surjective.

Let K 1 = j∈[[1,j 0 ]] G j
and K 2 = G j 0 + as in Goursat's Lemma. We then have

K 1 /K 1 K 2 /K 2 .
If the quotients are abelian then we are done since both groups are perfect. Assume that those quotients are non-abelian. There is only one non-abelian decomposition factor of K 2 and it is isomorphic to P SL 2 (q j 0 + ) or P SU 2 (q 1 2 j 0 + ) depending on the field F q j 0 + . We write that decomposition factor P G j 0 + . The isomorphism then implies that there exists

j 1 ∈ [[1, j 0 ]] such that ρ j 1 (A I 2 (m ) P G j 1 ρ j 0 + (A I 2 (m ).
We then have that there exists z : A I 2 (m) → F p × and Ψ ∈ Aut(F q j 0 + ) such that up to conjugation, for all h ∈ A I 2 (m) , ρ j 0 + (h) = Ψ(ρ j 1 (h))z(h). We will prove this is absurd by considering traces of some elements in A I 2 (m) under these representations. We may first note that for all M ∈ SL 2 (F q ), we have 1 = det(z(h)M ) = z(h) 2 det(M ) = z(h) 2 , therefore for all h ∈ A I 2 (m) , z(h) ∈ {-1, 1}. We write as before in the sequel γ j 0 + = θ j 0 + + θ -(j 0 + ) + 2 and

γ j 1 = θ j 1 + θ -j 1 + 2. Assume first that z(T t T -1 s ) = z([T t , T s ]) = 1.
We then consider the traces of those two elements inside each representation, Tr(ρ

j 0 + (T t T -1 s ) == Ψ(Tr(ρ j 1 (T t T -1 s ))) and Tr(ρ j 0 + ([T t , T s ])) = Ψ(Tr(ρ j 1 ([T t , T s ]))). This implies that -(α + α -1 ) + γ j 0 + = -Ψ(α + α -1 ) + Ψ(γ j 1 ) γ j 0 + (-(α + α -1 ) + γ j 0 +l -2) + 2 = Ψ(γ j 1 (-(α + α -1 ) + γ j 1 -2) + 2).
Since α + α -1 + γ j 0 + = 0, it follows that Ψ(θ j 1 + θ -j 1 ) = γ j 0 + and then Ψ(α + α -1 ) = α + α -1 . This implies j 1 ∼ j 0 + , which is absurd by assumption.

Assume now z(T t T -1

s ) = -z([T t , T s ]) = 1. By considering the traces of the same elements, we get

A = -(α + α -1 ) + γ j 0 + = -Ψ(α + α -1 ) + Ψ(γ j 1 ) = Ψ(A 1 ) B = γ j 0 + (-(α + α -1 ) + γ j 0 + -2) + 2 = -Ψ(γ j 1 (-(α + α -1 ) + γ j 1 -2) + 2) = -Ψ(B 1 ).

We have z(T

t T -1 s [T t , T s ]) = -1, therefore C = Tr(ρ j 0 + (T t T -1 1 [T t , T s ])) = -Ψ(Tr(ρ j 1 (T t T -1 s [T t , T s ]))) -Ψ(C 1 ) but C = AB -A and C 1 = A 1 B 1 -A 1 , therefore we have AB -A = C = -Ψ(C 1 ) = -Ψ(A 1 B 1 -A 1 )
= AB + A and it follows that A = 0, which is absurd as was proven before.

In the same way as above if z(

T t T -1 s ) = z([T t , T s ]) = 1, we have A = -Ψ(A 1 ), B = -Ψ(B 1 ) and AB -A = Ψ(A 1 B 1 -A 1 ) = A 1 B 1 + A 1 , therefore A = A 1 = 0, which is absurd. The last case remaining is -z(T t T -1 s ) = z([T t , T s ]) = 1.
We then have

-(α + α -1 ) + γ j 0 + = Ψ(α + α -1 ) -Ψ(γ j 1 ) γ j 0 + (-(α + α -1 ) + γ j 0 + -2) + 2 = Ψ(γ j 1 (-(α + α -1 ) + γ j 1 -2) + 2). so γ j 0 + (-(α + α -1 ) + γ j 0 + -2) = Ψ(γ j 1 )(α + α -1 -γ j 0 +l -2).
We consider a third element in this case and distinguish two possibilities.

Assume first that z([T 2 t , T

s ]) = 1. We then have that z([T t , T s ][T 2 t , T s ]T t T -1 s ) = -1, therefore E = Tr(ρ j 0 + ([T t , T s ][T 2 t , T s ]T t T -1 s )) = -Ψ(Tr(ρ j 1 ([T t , T s ][T 2 t , T s ]T t T -1 s ))) = -Ψ(E 1 ) and D = Tr(ρ j 0 + ([T 2 t , T s ])) = Ψ(Tr(ρ j 1 ([T 2 t , T s ]))) = Ψ(D 1 ). We also have ABD -AD -AB + D + A -2 = E = -Ψ(A 1 B 1 D 1 ) + Ψ(A 1 D 1 ) + Ψ(A 1 B 1 ) + Ψ(D 1 ) -Ψ(A 1 ) -2 and -Ψ(E 1 ) = -Ψ(A 1 B 1 D 1 -A 1 D 1 -A 1 B 1 + D 1 + A 1 -2). It follows that 2(Ψ(D 1 ) -2) = 0, therefore Ψ(D 1 -2) = 0. We then have 0 = D 1 -2 = γ j 1 (α-1) 2 (α 2 -αγ j 1 +2α+1 α 2
, therefore α + α -1 = γ j 1 -2 = θ j 1 + θ -j 1 . This would imply that α ∈ {θ j 1 , θ -j 1 } and, therefore α m = 1 which contradicts our assumptions on α.

Assume now that z([

T 2 t , T s ]) = -1. We then have z([T 2 t , T s ][T t , T s ]) = -1, therefore F = Tr(ρ j 0 + ([T 2 t , T s ][T t , T s ])) = -Ψ(Tr(ρ j 1 ([T 2 t , T s ][T t , T s ]))) = -Ψ(F 1 ). We also have F = BD - B = -Ψ(B 1 D 1 ) -Ψ(B 1 ) and -Ψ(F 1 ) = -Ψ(B 1 D 1 -B 1 ) = -Ψ(B 1 D 1 ) + Ψ(B 1 ). It follows that Ψ(B 1 ) = 0, therefore B 1 = 0 and B = Ψ(B 1 ) = 0. We also have z([T t , T s ][T 2 t , T s ](T t T -1 s )) = 1 and, therefore E = Ψ(E 1 ). We have E = ABD - AD -AB + D + A -2 = -AD + D + A -2 = -Ψ(A 1 D 1 ) -Ψ(D 1 ) -Ψ(A 1 ) -2 and Ψ(E 1 ) = Ψ(A 1 B 1 D 1 -A 1 D 1 -A 1 B 1 + D 1 + A 1 -2) = -Ψ(A 1 D 1 ) + Ψ(D 1 ) + Ψ(A 1 ) -2. It follows that Ψ(A 1 + D 1 ) = 0, therefore A 1 + D 1 = 0. We then have 0 = α -1 (α -1) 2 B 1 + (A 1 + D 1 ) = α -2 + α -1 + γ j 1 = 0, therefore α + α -1 = 2 -γ j 1 . We then have 0 = B 2 = -(α + α -1 )γ j 1 + γ 2 j 1 - 2γ j 1 + 2 = -(2 -γ j 1 )γ j 1 + γ 2 j 1 -2γ j 1 + 2 = 2γ 2 j 1 -4γ 2 j 1 + 2 = 2(γ j 1 -1) 2 . It follows that γ j 1 = 1, therefore α + α -1 = 2 -1 = 1 and A 1 = -α -α -1 + γ j 1 = 0. We then have A = -Ψ(A 1 ) = 0, therefore -α -α -1 + γ j 0 + = 0 and γ j 0 + = α + α -1 = γ j 1 .
This implies that j 0 + = j 1 , which is absurd. This concludes the proof.

m even

Let m ≥ 5 even, p a prime, α, β ∈ F p of orders not belonging to {1, 2, 3, 4, 5, 6, 10} and θ ∈ F p a primitive m-th root of unity. Note that p = 2 because there exists a primitive m -th root of unity with m even. We assume that j ∈ [[1, m- 2 2 ]], we have (α + √ αβ(θ j + θ -j ) + β)(αβ -√ αβ(θ j + θ -j ) + 1) = 0. For j ∈ N, we write F q j = F p (α, β, √ αβ(θ j + θ -j )). We write F q the smallest field containing F q j for j in [[1, m- 22 ]]. Note that we have F q = F q 1 as in section 5.1. Definition 5.2.1. The Iwahori-Hecke algebra of dihedral type I 2 (m) which we write H I 2 (m),q is the F q -algebra with the following presentation Generators : T t , T s . Relations :

(T s -α)(T s + 1) = 0, (T t -β)(T t + 1) = 0, T s T t T s • • • m = T t T s T t • • • m .
We then give the Kilmoyer-Solomon matrix model given in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] (Theorem 8.3.1) in the finite field case.

Theorem 5.2.1. Under the assumptions made on α, β and θ, the following matrix model gives us a decomposition into pairwise non-isomorphic irreducible H

I 2 (m),q -modules 1. Ind : T s → α, T t → β.

: T

s → -1, T t → -1. 3. For j ∈ [[1, m-2 2 ]], T s → ρ j (T s ) = -1 0 1 α , T t → ρ j (T t ) = β α + √ αβ(θ j + θ -j ) + β 0 -1 .
Proof. We know that these models give us representations of H I 2 (m),q . It is thus sufficient to show that they are irreducible and pairwise non-isomorphic. The two 1-dimensional representations are non isomorphic by the conditions on α and β. Let us show that the 2-dimensional representations are irreducible.

Let j ∈ [[1, m-1 2 ]].
Let W be a non-trivial H I 2 (m),q -submodule of the module V associated to the representation ρ j . Let x = (x 1 , x 2 ) ∈ W \ {(0, 0)}. We have that ρ j (T s ) • x + x = (0, x 1 + (α + 1)x 2 ) ∈ W , therefore (0, 1) ∈ W or x 1 + (α + 1)x 2 = 0.

Assume first that (0, 1) ∈ W . Since x ∈ W , we have that

x 1 = 0 or (1, 0) ∈ W . If (1, 0) ∈ W then W = V . Assume now x 1 = 0. We then have that ρ j (T t ) • x + x = ((β + 1)x 1 + (α + √ αβ(θ j + θ -j ) + β)x 2 , 0) ∈ W . It follows that (1, 0) ∈ W or (α + √ αβ(θ j + θ -j ) + β)x 2 = 0.
By the assumptions on α, β and θ, we have that α + √ αβ(θ j + θ -j ) + β = 0, therefore the latter possibility would imply x = (0, 0), which contradicts our assumptions. It follows that (1, 0) ∈ W and W = V . Assume now that x 1 + (α + 1)x 2 = 0. We consider again ρ j (T t ) • x + x and we get that (1, 0) ∈ W or (β + 1)x 1 + (α + √ αβ(θ j + θ -j ) + β)x 2 = 0. Assume (1, 0) ∈ W . Since x ∈ W , we then have (0, 1) ∈ W or x 2 = 0. The latter would imply that x 2 = x 1 = 0 since x 1 + (α + 1)x 2 = 0. It follows that (0, 1) ∈ W and, therefore V = W .

Assume now by contradiction that (β + 1)

x 1 + (α + √ αβ(θ j + θ -j ) + β)x 2 = 0. We have x 1 = -(α + 1)x 2 , it follows that (-αβ + √ αβ(θ j + θ -j ) -1)
x 2 = 0, therefore x 2 = 0 and x 1 = x 2 = 0. This is absurd, therefore we are in the first case and V = W . This proves that in all cases, we have that W = V and ρ j is therefore irreducible.

It remains to show that those representations are pairwise non-isomorphic. Let j and be two integers such that 1 ≤ j ≤ ≤ m-2 2 . We have Tr(ρ j (T t )ρ j (T t )) = √ αβ(θ j + θ -j ) and Tr(ρ (T t )ρ (T t )) = √ αβ(θ + θ -). Assume now that those representations are isomorphic. We then have that √ αβ(θ j + θ -j ) = √ αβ(θ + θ -). It follows that θ j + θ -j = θ + θ - and, therefore (θ j -θ )(1 -θ --j ) = 0. This implies that j = since 0 ≤ -j ≤ m and 2 ≤ j + ≤ m -2 < m.

The main difference between m even and m odd arises in the field extensions we have to consider. We describe the different cases we encounter in what follows. First note that for j ∈

[[1, m-2 2 ]], if we set P j = 0 α + √ αβ(θ j + θ -j ) 1 0
then by the assumptions made on α, β and θ, the matrix P j is invertible. Moreover, we have that P j ρ j (T s )P -1

j = α α + √ αβ(θ j + θ -j ) + β 0 -1 and P j ρ j (T t )P -1 j = -1 0 1 β
. This shows that the roles of α and β are perfectly symmetric.

We now fix a j ∈ [[1, m-2 2 ]] and we write γ = √ αβ, u j = θ j + θ -j and ξ = γu j (αβ-α-β+1)

αβ . We set L = F p (α, β, γu j ), L 1 = F p (α + α -1 , β, γu j ), L 2 = F p (α, β, ξ), L 3 = F p (α, β + β -1 , ξ), K 1 = F p (α + α -1 , β, ξ), K 2 = F p (α + α -1 , β + β -1 , γu j ), K 3 = F p (α, β + β -1 , θ) and K = F p (α + α -1 , β + β -1 , ξ). Note that L = L 2 since αβ -α -β + 1 = (α -1)(β -1) = 0 and αβξ αβ-α-β+1 = γu j . If we consider the polynomials X 2 -(α + α -1 )X + 1 and X 2 -(β + β -1 )X + 1, we get that L L 1 L 2 L 3 K 1 K 2 K 3 K Figure 5.1: Field extensions in type I 2 (m), m even in case 1 ([L : L 1 ], [L : L 3 ], [L 2 , K 1 ], [L 2 , K 3 ], [K 1 : K], [K 3 : K]) ∈ {1, 2} 6 .
We then have the following Hasse diagram

L L 1 L 2 L 3 K 1 K 2 K 3 K 1 1, 2 1, 2 1, 2 1, 2 1, 2 1, 2
We then see that

[L : K] = [L : L 2 ][L 2 : K 1 ][K 1 : K] ∈ {1, 2, 4}. Assume that [L : K] = 4, we then have that [L 2 : K 1 ] = [L 2 : K 3 ] = 2
. By unicity of the subfields of degree 2, we then have K 1 = K 3 and, therefore α ∈ K 1 . It follows that K 1 = L 2 and, therefore [K 1 : K 1 ] = 2 which is a contradiction. This proves that [L : K] ∈ {1, 2}. By uniqueness of the subfields of a given degree we cannot have [L :

L 1 ] = [L : L 3 ] = 2 or [L 1 : K 1 ] = [L 1 : K 2 ] = 2 or [L 2 : K 1 ] = [L 2 : K 3 ] = 2 or [L 3 : K 3 ] = [L 3 : K 2 ] = 2.
It follows that the possible ways L is an extension of K corresponds to Hasse diagrams described in Figures 5.1 to 5.7. We write in dashed red lines the extensions of degree 2, in red the subfields of degree 2 of L, in dotted black lines the extensions of degree 1 and in black the fields equal to L.

Theorem 5.2.2. For

j ∈ [[1, m-2 2 ]], let G = [< T t , T s >, < T t , T s >].
We then have 1. In case 1, ρ j (G) = SL 2 (q j ).

2. In case 2, ρ j (G) SU 2 (q 1 2 j ).

3. In case 3, ρ j (G) SU 2 (q 1 2 j ).

4.

In cases 4 and 5, ρ j (G) SL 2 (q

1 2 j ). L L 1 L 2 L 3 K 1 K 2 K 3 K Figure 5.2: Field extensions in type I 2 (m), m even in case L L 1 L 2 L 3 K 1 K 2 K 3 K Figure 5.3: Field extensions in type I 2 (m), m even in case L L 1 L 2 L 3 K 1 K 2 K 3 K Figure 5.4: Field extensions in type I 2 (m), m even in case L L 1 L 2 L 3 K 1 K 2 K 3 K Figure 5.5: Field extensions in type I 2 (m), m even in case L L 1 L 2 L 3 K 1 K 2 K 3 K Figure 5.6: Field extensions in type I 2 (m), m even in case L L 1 L 2 L 3 K 1 K 2 K 3 K Figure 5
.7: Field extensions in type I 2 (m), m even in case 5. In cases 6 and 7, ρ j (G) SU 2 (q 1 2 j ).

Proof. We write the unique automorphism of order 2 of F q j when it exists. By symmetry of the roles of α and β, it is sufficient to consider cases 1, 2, 3, 4 and 6. As for the proof of Theorem 5.1.2, it is sufficient to show that

1. A = [ρ j (T s ), ρ j (T t )][ρ j (T t ) -1 , ρ j (T s )] -[ρ j (T t ) -1 , ρ j (T s )][ρ j (T s ), ρ j (T t )] = 0, 2. B = [ρ j (T s ), ρ j (T t )][ρ j (T t ) -1 , ρ j (T s )] + [ρ j (T t ) -1 , ρ j (T s )][ρ j (T s ), ρ j (T t )] = 0,
3. ρ j (G) contains elements of order different from 1, 2, 3 and 5.

4. ξ, α + α -1 and β + β -1 are in the field L G generated by the traces of the elements of

ρ j ([G, G]),
5. In case 2 and 3, ρ j

• t ρ -1 j , 6. In case 4, ρ j|[G,G] • ρ j|[G,G] , 7. In case 6, ρ j|[G,G] • t ρ -1 j|[G,G] .
We now prove these assertions.

1. We have that A 1,2 = (β-1)(α-1)(αβ-γu j +1)(α+γu j +β) 2 α 2 β 2 = 0 by the assumptions on α, β and θ.

2. We have that B 1,2 = -(β-1)(α+1)(α+γu j +β)(α 2 β+αβ 2 +α+β-2αβ+(αβ-α-β+1-γu j )γu j )

α 2 β 2 . It follows that B 1,2 = 0 implies t = α 2 β + αβ 2 + α + β -2αβ + (αβ -α -β + 1 -γu j )γu j = 0. If p = 2 then we have 0 = 1 2α 1 β ( α 2 β 2 (B 1,1 +B 2,2 ) 2α -t) + (2 -β)t = Φ 6 (β)
.This is absurd by the conditions on β.

If p = 2 then we have 0 = t = α 2 β + αβ 2 + α + β + (αβ + α + β + 1 + γu j )γu j = (α + γu j + β)(αβ + γu j + 1) which contradicts our assumptions.

It follows that B = 0.

We have (2 -tr([ρ

j (T s ), ρ j (T t )])(α + α -1 -2) = tr([ρ j (T s ) 2 , ρ j (T t )]) -2 ∈ L G . We also have C = 2 -tr([rho j (T s ), ρ j (T t )]) ∈ L G . We have C = (αβ-γu j +1)(α+γu j +β) αβ = 0. It follows that α + α -1 ∈ L G . We have that C(β+β -1 -2) = tr([ρ j (T t ) 2 , ρ j (T s )])-2, therefore we also have β+β -1 ∈ L G . We have ξ = C -(α + α -1 + β + β -1 ) + 2-tr([ρ j (Ts)ρ j (Tt)ρ j (Ts),ρ j (Tt)] C ∈ L G . It follows that K ⊂ L G as required.
4. We can use the same arguments as for the proof of Theorem 5.1.2 to get that ρ j (G) contains elements of order different from 1, 2, 3 and 5.

In case 2, we have

L = L 1 = L 2 = L 3 = K 1 = K 2 = K 3 = K, therefore (α) = α -1 , (β) = β -1 and (ξ) = ξ. It follows that (γu j ) α -1 β -1 -α -1 -β -1 +1 α -1 β -1 = γu j αβ-α-β+1 αβ
, therefore (γu j ) = α -1 β -1 γu j . We then set P = α(β+1) α+γu j +β α 1 α + 1 . We have that det(P ) = α(αβ-γu j +1) αγu j +β = 0 and

P ρ j (T s )P -1 = -1 α 0 α = ( -1 α -1 0 α -1 ) = ( t ρ j (T s ) -1 )
and

P ρ j (T t )P -1 = β 0 α+γu j +β α -1 = ( β -1 0 β+γu j +α β -1 ) = ( t ρ j (T t ) -1 ).
It follows that ρ j • t ρ -1 j . In case 3, we have

L = L 1 = L 2 = L 3 = K 1 = K 3 = K = K 2 , therefore (α) = α -1 , (β) = β -1 , (γu j ) = γu j et (ξ) = ξ. It follows that γu j = αβγu j , therefore u j = 0 or αβ = 1.
Assume first that u j = 0. We then have θ j + θ -j = 0, therefore θ 2j = -1 and θ 4j = 1. This implies that j = m 4 . Let P = α(β+1) α+β α 1 α + 1 . We have det(P ) = α(αβ+1) α+β and (P ρ j (T s )P -1 , P ρ j (T t )P -1 ) = ( ( t ρ j (T s ) -1 ), ( t (ρ j (T t ) -1 )).

Assume now that αβ = 1. We then have γ = 1, it follows that

α 2 + αu j + 1 = α(α + u j + α -1 ) = α(α + γu j + β) = 0. Moreover, we have that u j -2 = θ j + θ -j -2 = (θ j -1)(1 -θ -j ) = 0. Let P = α(α+1) α 2 +αu j +1 α 1 α + 1
, we then have det(P ) = -α 2 (u j -2) α 2 +αu j +1 = 0 and (P ρ j (T s )P -1 , P ρ j (T t )P -1 ) = ( ( t ρ j (T s ) -1 ), ( t ρ j (T t ) -1 )).

We then have that in both cases ρ j

• t ρ -1 j . 6. In case 4, we have

L = L 2 = L 3 = K 3 = L 1 = K 1 = K 2 = K, therefore (α) = α -1 ,
(β) = β, (γu j ) = γu j and (ξ) = ξ. We have that γu j

α -1 β-α -1 -β+1 α -1 β = γu j αβ-α-β+1 αβ . It
follows that γu j (α -1) 2 (1 -β) = 0, therefore u j = 0 and j = m 4 . Let P = 1 α + 1 0 -α , we have det(P ) = -α = 0. We have P ρ j (T s )P -1 = α 0 -α -1 = -α (ρ j (T s )) and

P ρ j (T t )P -1 = β β + α -1 0 -1 = (ρ j (T t )). We then have that ρ j|[G,G] • ρ j|[G,G] .
7. In case 6, we have that

L = L 1 = L 2 = L 3 = K 1 = K 2 = K = K 3 , therefore (α) = α, (β = β -1 ) and (ξ) = ξ. It follows that (γu j ) = αβ -1 (αβ-α-β+1) αβ(αβ -1 -α-β -1 +1) γu j = -β -1 γu j . Let P = β + 1 α + γu j + β αβ -γu j + 1 0
, we have that det(P ) = -(αβ-γu j +1)(α+γu j +β) = 0. We also have P ρ j (T s )P -1 = -α ( t (ρ j (T s ) -1 ) and P ρ j (T t )P

-1 = ( t (ρ j (T t ) -1 ). It follows that ρ j|[G,G] • t ρ -1 j|[G,G]
and the proof is concluded.

Lemma 5.2.1. We say that j ∼ l if F p (α + α -1 , β + β -1 , ξ j ) F p (α + α -1 , β + β -1 , ξ j ) and there exists Φ j,l ∈ Aut(F q j ) such that Φ j,l (α + α -1 ) = α + α -1 ), Φ j,l (β + β -1 ) = β + β -1 and Φ j,l (ξ j ) = ξ l . This defines an equivalence relation and if j ∼ l then Φ j,l • ρ j|A I 2 (m) ρ l|A I 2 (m) .

Proof. Let us show this is an equivalence relation.

Let j, l, k ∈ [[1, m-2 2 ]
]. We have j ∼ j. If j ∼ l and l ∼ k then it it clear that j ∼ k. This relation is symmetric because Φ l,j = Φ -1 j,l verifies the desired conditions if j ∼ l.

By Theorem 5.2.2, the second part of the statement is also true because SU 2 (q

1 2 ) SL 2 (q 1
2 ).

We now give the image of the derived subgroup of the Artin group in the full Iwahori-Hecke algebra.

Theorem 5.2.3. Assume m even and α and β verify the conditions given at the beginning of this section.

For j ∈ [[1, m-2 2 ]], we set G j = ρ j ([< T t , T s >, < T t , T s >]
) . We then have that the morphism from

A I 2 (m) to H × I 2 (m),q GL 1 (q j ) 2 × j∈[[1, m-1 2 ]] GL 2 (q j )
factorizes through the surjective morphism

Φ : A I 2 (m) → j∈[[1, m-2 2 ]]/∼ G j .
Proof. We know by Theorem 5.2.2 that the composition of the morphism from

A I 2 (m) to H × I 2 (m),q
with the projection upon each representation is surjective. We know by Lemma 5.2.1 that it factorizes through the morphism. We will now use Goursat's lemma and induction on j ∈

[[1, m-2 2 ]
] in order to conclude the proof of this theorem. For

j 0 ∈ [[1, m-2 2 ]], we define Φ j 0 (A I 2 (m) ) to be the image of A I 2 (m) inside j∈[[1,j 0 ]]/∼ GL 2 (q j ). We know that Φ 1 (A I 2 (m) ) = G 1 . Let j 0 ∈ [[1, m-4 2 ]], assume Φ j 0 (A I 2 (m) ) = j∈[[1,j 0 ]]/∼ G j . Consider Φ j 0 + (A I 2 (m) ) ⊂ j∈[[1,j 0 ]]
G j × G j l for the smallest positive integer such that j j for all j ∈ [[1, j 0 ]]. We know that the projection upon each factor is surjective.

Let K 1 = j∈[[1,j 0 ]] G j
and K 2 = G j 0 + as in Goursat's Lemma. We then have

K 1 /K 1 K 2 /K 2 .
If the quotients are abelian then we are done since both groups are perfect. Assume that those quotients are non-abelian. There is only one decomposition factor of K 2 and it is equal to P SL 2 (q j 0 + ), P SU 2 (q 1 2 j 0 + ) or P SL 2 (q 1 2 j 0 + ) depending on the field F q j 0 + . We write that decomposition factor P G j 0 + . The isomorphism then implies that there exists

j 1 ∈ [[1, j 0 ]] such that F p (α + α -1 , β + β -1 , ξ j 1 ) F p (α + α -1 , β + β -1 , ξ j 0 + ) and ρ j 1 (A I 2 (m ) P G j 1 ρ j 0 + (A I 2 (m ).
We then have that there exists z : A I 2 (m) → F p × and Ψ ∈ Aut(F q j 0 + ) such that up to conjugation, for all

h ∈ A I 2 (m) , ρ j 0 + (h) = Ψ(ρ j 1 (h))z(h).
We will prove this is absurd by considering traces of some elements in A I 2 (m) under these representations. We may first note that for all M ∈ SL 2 (F q ), we have

1 = det(z(h)M ) = z(h) 2 det(M ) = z(h) 2 , therefore for all h ∈ A I 2 (m) , z(h) ∈ {-1, 1}. We write as before in what follows γ = √ αβ, u j 0 + = θ j 0 + + θ -(j 0 + ) , u j 1 = θ j 1 + θ -j 1 , ξ j 1 = γu j 1 (αβ-α-β+1) αβ and ξ j 0 + = γu j 0 + (αβ-α-β+1) αβ .
1. Assume first that z([T t , T s ]) = 1 and Tr(ρ j 1 ([T t , T s ])) = 0. We then have that

A 1 = Tr(ρ j 0 + ([T t , T s ])) = Φ(Tr(ρ j 1 ([T t , T s ]))) = Φ(B 1 ).
We have Tr(ρ

j 0 + ([T -1 t , T s ])) = Tr(ρ j 0 + ([T t , T s ])) and Tr(ρ j 1 ([T -1 t , T s ])) = Tr(ρ j 1 ([T t , T s ])), therefore z([T -1 t , T s ]) = 1 because Tr(ρ j 1 ([T t , T s ])) = 0. It follows that z([T t , T s ][T -1 t , T s ]) = 1. We then have A 2 = Tr(ρ j 0 + ([T t , T s ][T -1 t , T s ])) = Φ(Tr(ρ j 0 + ([T t , T s ][T -1 t , T s ]))) = Φ(B 2 ).
We have

A 2 = A 2 1 -(β + β -1 -2)(A 1 -2) -2 and B 2 = B 2 1 -(β + β -1 -2)(B 1 -2) -2. It follows that A 2 1 -(β+β -1 -2)(A 1 -2)-2 = Φ(B 2 1 -(β+β -1 -2)(B 1 -2)-2) = A 2 1 -Φ(β+β -1 -2)(A 1 -2)-2. ((β + β -1 ) -Φ(β + β -1 ))(A 1 -2) = 0.
We have A 1 -2 = -(αβ-γu j +1)(α+γu j +β) αβ = 0 by assumption, therefore Φ(β+β -1 ) = β+β -1 . Note that we also have

A 1 = Tr(ρ j 0 + ([T s , T t ])) = Tr(ρ j 0 + ([T -1 s , T t ])) and B 1 = Tr(ρ j 1 ([T s , T t ])) = Tr(ρ j 1 ([T -1 s , T t ])). It follows that z([T s , T t ][T -1 s , T t ]) = 1 and
A 3 = Tr(ρ j 0 + ([T s , T t ][T -1 s , T t ])) = Φ(Tr(ρ j 1 ([T s , T t ][T -1 s , T t ]))) = Φ(B 3 ).
We have

A 3 = A 2 1 -(α + α -1 -2)(A 1 -2) -2 and B 3 = B 2 1 -(α + α -1 -2)(B 1 -2) -2.
It follows that by the same reasoning as before α + α -1 = Φ(α + α -1 ).

Assume now that z([T t T s T t , T

s ]) = 1. We then have A 4 = Tr(ρ j 0 + ([T t T s T t , T s ])) = Φ(tr(ρ j 1 ([T t T s T t , T s ]))) = Φ(B 4 ). We also have ξ j 0 + = 2-A 4 2-A 1 + 2 -A 1 -(α + α -1 + β + β -1 ) and ξ j 1 = 2-B 4 2-B 1 + 2 -B 1 -(α + α -1 + β + β -1
). It follows that ξ j 0 + = Φ(ξ j 1 ) and, therefore j 0 + j 1 , which contradicts our assumptions.

Assume now that z([T

t T s T t , T s ]) = -1. We then have A 4 = Tr(ρ j 0 + ([T t T s T t , T s ])) = Φ(tr(ρ j 1 ([T t T s T t , T s ]))) = -Φ(B 4 ). We have z([T t T s T t , T s ][T t , T s ]) = -1, therefore A 5 = Tr(ρ j 0 + ([T t T s T t , T s ][T t , T s ])) = -Φ(tr(ρ j 1 ([T t T s T t , T s ][T t , T s ]))) = -Φ(B 5 ). We have A 5 = A 4 A 1 -A 1 and B 5 = B 4 B 1 -B 1 , therefore A 4 A 1 -A 1 = -Φ(B 4 )Φ(B 1 ) + Φ(B 1 ) = A 4 A 1 + A 1 .
It follows that 2A 1 = 0 which contradicts our assumption since p = 2.

Assume now that z([T t , T s ]) = -1 and Tr

(ρ j 1 ([T t , T s ])) = 0. We have that A 1 = Tr(ρ j 1 ([T t , T s ])) = -Φ(Tr(ρ j 1 ([T t , T s ]))) = -Φ(B 1 ).
We then have that z

([T t , T s ]) = z([T -1 t , T s ]) = z([T -1 s , T t ]). It follows that z([T t , T s ][T -1 t , T s ]) = z([T s , T t ][T -1 s , T t ]) = 1.
We then have

A 2 = Tr(ρ j 0 + ([T t , T s ][T -1 t , T s ])) = Φ(Tr(ρ j 0 + ([T t , T s ][T -1 t , T s ]))) = Φ(B 2 ), A 3 = Tr(ρ j 0 + ([T s , T t ][T -1 s , T t ])) = Φ(Tr(ρ j 1 ([T s , T t ][T -1 s , T t ]))) = Φ(B 3 ). We have A 2 = A 2 1 -(β + β -1 -2)(A 1 -2) -2, B 2 = B 2 1 -(β + β -1 -2)(B 1 -2) -2, A 3 = A 2 1 -(α + α -1 -2)(A 1 -2) -2 and B 3 = B 2 1 -(α + α -1 -2)(B 1 -2) -2. It follows that A 2 1 -(β + β -1 -2)(A 1 -2) -2 = Φ(B 2 1 ) -Φ(β + β -1 -2)Φ(B 1 -2) -Φ(2) = A 2 1 -Φ(β + β -1 -2)(-A 1 -2) -2 and A 2 1 -(α + α -1 -2)(A 1 -2) -2 = Φ(B 2 1 ) -Φ(α + α -1 -2)Φ(B 1 -2) -Φ(2) = A 2 1 -Φ(α + α -1 -2)(-A 1 -2) -2
It follows that

(β + β -1 -2) = Φ(β + β -1 -2) -A 1 -2 A 1 -2 and (α + α -1 -2) = Φ(α + α -1 -2) -A 1 -2 A 1 -2 2.1. Assume now that z([T 2 t , T 2 s ]) = 1. We then have that A 8 = Tr(ρ j 0 + ([T 2 t , T 2 s ])) = Φ(Tr(ρ j 1 ([T 2 t , T 2 s ]))) = Φ(B 8 ). We have A 8 = (β + β -1 -2)(α + α -1 -2)(A 1 -2) + 2 and B 8 = (β + β -1 -2)(α + α -1 -2)(B 1 -2) + 2. It follows that (β + β -1 -2)(α + α -1 -2)(A 1 -2) + 2 = Φ(β + β -1 -2)Φ(α + α -1 -2)(Φ(B 1 ) -2) + 2 (β + β -1 -2)(α + α -1 -2)(A 1 -2) = (A 1 -2) 2 (-A 1 -2) (-A 1 -2) 5 (β + β -1 -2)(α + α -1 -2) (A 1 -2)(-A 1 -2) = (A 1 -2) 2 -A 2 1 + 4 = A 2 1 -4A 1 + 4 2(A 2 1 -2A 1 ) = 0 2A 1 (A 1 -2) = 0.
This is a contradiction since p = 2, A 1 = 2 and, by assumption, A 1 = 0.

Assume now that z([T 2

t , T 2 s ]) = -1. We then have that

A 8 = Tr(ρ j 0 + ([T 2 t , T 2 s ])) = -Φ(Tr(ρ j 1 ([T 2 t , T 2 s ]))) = -Φ(B 8 ).
We have

A 8 = (β+β -1 -2)(α+α -1 -2)(A 1 -2)+2 and B 8 = (β+β -1 -2)(α+α -1 -2)(B 1 -2)+2.
It follows that

(β + β -1 -2)(α + α -1 -2)(A 1 -2) + 2 = -Φ(β + β -1 -2)Φ(α + α -1 -2)(Φ(B 1 ) -2) -2 (β + β -1 -2)(α + α -1 -2)(A 1 -2) + 2 = (A 1 + 2)(A 1 -2) 2 (A 1 + 2) 2 (β + β -1 -2)(α + α -1 -2) -2 (β + β -1 -2)(α + α -1 -2)(A 2 1 -4 -A 2 1 + 4A 1 -4) = -4(A 1 + 2) (β + β -1 -2)(α + α -1 -2)(A 1 -2) = -A 1 -2 A 8 -2 = -A 1 -2 A 8 = -A 1 .

Assume now that z([T t T s T t , T

s ]) = -1. We then have z([T t T s T t , T s ][T t , T s ]) = 1, A 4 = Tr(ρ j 0 + ([T t T s T t , T s ])) = Φ(tr(ρ j 1 ([T t T s T t , T s ]))) = -Φ(B 4 ), A 5 = Tr(ρ j 0 + ([T t T s T t , T s ][T t , T s ])) = Φ(tr(ρ j 1 ([T t T s T t , T s ][T t , T s ]))) = Φ(B 5 ). We have A 5 = A 4 A 1 -A 1 and B 5 = B 4 B 1 -B 1 , therefore A 4 A 1 -A 1 = Φ(B 4 )Φ(B 1 ) -Φ(B 1 ) = A 4 A 1 + A 1 . It follows that A 4 = A 1 which contradicts our assumptions.

Assume now that z([T

t T s T t , T s ]) = 1. We then have A 4 = Tr(ρ j 0 + ([T t T s T t , T s ])) = Φ(tr(ρ j 1 ([T t T s T t , T s ]))) = Φ(B 4 ).

2.2.2.1.

Assume that z([T 2 t , T s ]) = 1. We then have

A 6 = Tr(ρ j 0 + ([T 2 t , T s ])) = Φ(tr(ρ j 1 ([T 2 t , T s ]))) = Φ(B 6 ). We then have z([T t , T s ][T 2 t , T s ][T t T s T t , T s ]) = -1, therefore A 7 = Tr(ρ j 0 + ([T t , T s ][T 2 t , T s ][T t T s T t , T s ])) = -Φ(Tr(ρ j 1 ([T t , T s ][T 2 t , T s ][T t T s T t , T s ]))) = -Φ(B 7 ).
We have

A 7 = A 1 A 4 A 6 -A 1 A 4 -A 1 A 6 -A 1 -ξ j 0 + (A 1 -2) and B 7 = B 1 B 4 B 6 -B 1 B 4 -B 1 B 6 - B 1 -ξ j 1 (B 1 -2). It follows that ξ j 0 + = Φ(ξ j 1 ) A 1 +2 A 1 -2 . We have z([T 2 t , T 2 s ][T t , T s ]) = 1, therefore A 9 = Tr(ρ j 0 + ([T 2 t , T 2 s ][T t , T s ])) = Φ(Tr(ρ j 1 ([T 2 t , T 2 s ][T t , T s ]))) = Φ(B 9 ).
We also have

A 9 = A 8 A 1 -A 8 + (ξ j 0 + -1)(A 1 -2) and B 9 = B 8 B 1 -B 8 + (ξ j 1 -1)(B 1 -2).
It follows that

A 8 A 1 -A 8 + (ξ j 0 + -1)(A 1 -2) = Φ(B 8 B 1 -B 8 + (ξ j 1 -1)(B 1 -2)) = A 8 A 1 + A 8 + (Φ(ξ j 1 ) -1)(-A 1 -2) = A 8 A 1 -A 1 + (ξ j 0 + A 1 -2 A 1 + 2 -1)(-A 1 -2) A 1 + (ξ j 0 + -1)(A 1 -2) = -A 1 -(ξ j 0 + (A 1 -2) -A 1 -2) A 1 + ξ j 0 + (A 1 -2) -A 1 + 2 = -A 1 -ξ j 0 + (A 1 -2) + A 1 + 2 2ξ j 0 + (A 1 -2) = 0.
This implies that ξ j 0 + = 0 = ξ j 1 . It then follows that u j 0 + = 0 = u j 1 . It follows that θ j 0 + + θ -(j 0 + ) = θ j 1 + θ -j 1 and, therefore (θ j 0 + +j 1 -1)(θ -j 1 -θ -(j 0 +l) ) = 0. Since we have 1 ≤ j 0 + , j 1 ≤ m-2 2 , this implies that j 0 + = j 1 which contradicts our assumptions.

Assume now that z([T 2

t , T s ]) = -1. We then have

A 6 = Tr(ρ j 0 + ([T 2 t , T s ])) = -Φ(tr(ρ j 1 ([T 2 t , T s ]))) = -Φ(B 6 ).
We then have z([T 2 t , T s ][T t , T s ]) = 1, therefore

A 10 = Tr(ρ j 0 + ([T 2 t , T s ][T t , T s ])) = Φ(Tr(ρ j 1 ([T 2 t , T s ][T t , T s ]))) = Φ(B 10 ).
We have A 10 = A 1 A 6 -A 1 and B 10 = B 1 B 6 -B 1 . We then have A 1 A 6 -A 1 = A 10 = Φ(B 10 ) = A 1 A 6 + A 1 , therefore 2A 1 = 0 and A 1 = 0. This contradicts our assumptions.

Assume now that

A 1 = Tr(ρ j 0 + ([T t , T s ])) = 0. We then have B 1 = Tr(ρ j 1 ([T t , T s ])) = 0. 3.1. Assume z([T 2 t , T s ]) = 1. We then have A 6 = Tr(ρ j 0 + ([T 2 t , T s ])) = Φ(Tr(ρ j 1 ([T 2 t , T s ]))) = Φ(B 6 ). We also have A 6 = 2(β + β -1 -1) -A 1 (β + β -1 -2) = 2(β + β -1 -1) and B 6 = 2(β + β -1 -1) -B 1 (β + β -1 -2) = 2(β + β -1 -1). It follows that Φ(β + β -1 ) = β + β -1 . 3.1.1. Assume z([T 2 t , T 2 s ]) = 1.
We then have

A 8 = Tr(ρ j 0 + ([T 2 t , T 2 s ])) = Φ(Tr(ρ j 1 ([T 2 t , T 2 s ]))) = Φ(B 8 ).
We have

A 8 = (β + β -1 -2)(α + α -1 -2)(A 1 -2) + 2 = 2 -2(β + β -1 -2)(α + α -1 -2), B 8 = (β + β -1 -2)(α + α -1 -2)(B 1 -2) + 2 = 2 -2(β + β -1 -2)(α + α -1 -2).
It follows that α + α -1 = Φ(α + α -1 ).

Assume z([T t T s T t , T s ]) = 1. We then have

A 4 = Tr(ρ j 0 + ([T t T s T t , T s ])) = Φ(tr(ρ j 1 ([T t T s T t , T s ]))) = Φ(B 4 ).
We have

ξ j 0 + = 2-A 4 2-A 1 +2-A 1 -(α+α -1 +β +β -1 ) and ξ j 1 = 2-B 4 2-B 1 +2-B 1 -(α+α -1 +β +β -1
). It follows that ξ j 0 + = Φ(ξ j 1 ) and, therefore, j 0 + ∼ j 1 . This contradicts our assumptions.

Assume now z([T t T s T t , T s ]) = -1.

We then have A 4 = -Φ(B 4 ). By the same computations as in 3.1.1.1, we get that

ξ j 0 + = 2 -A 4 2 +2-(α+α -1 +β +β -1 ) = Φ( 2 + B 4 2 +2-(α+α -1 +β +β -1 )) = Φ(ξ j 1 )+Φ(B 4 ).
We also have z([

T 2 t , T s ][T t T s T t , T s ]) = -1, therefore A 11 = Tr(ρ j 0 + ([T 2 t , T s ][T t T s T t , T s ])) = -Φ(Tr(ρ j 1 ([T 2 t , T s ][T t T s T t , T s ]))) = -Φ(B 11
). We also have

A 11 = A 4 A 6 -A 4 -A 6 + ξ j 0 + (A 1 - 2) + 2 and B 11 = B 4 B 6 -B 4 -B 6 + ξ j 1 (B 1 -2) + 2. It follows that ξ j 0 + = -A 6 -Φ(ξ j 1 ) + 2.
We then have Φ(ξ j 1 ) = -A 6 -ξ j 0 + + 2 = ξ j 0 + + A 4 and, therefore 2ξ j 0 + = -A 6 -A 4 -2. We have

-A 6 -A 4 + 2 = -A 2 1 -ξ j 0 + A 1 -(α + α -1 -2)A 1 + 2ξ j 0 + + 2(α + α -1 -1)
. It follows that α + α -1 -1 = 0 and, therefore, 0 = α 2 -α + 1 = Φ 6 (α), which contradicts our assumptions.

Assume now z([T 2 t , T 2

s ]) = -1. We then have

A 8 = Tr(ρ j 0 + ([T 2 t , T 2 s ])) = -Φ(Tr(ρ j 1 ([T 2 t , T 2 s ]))) = -Φ(B 8 ). We have z([T 2 t , T s ][T 2 t , T 2 s ]) = -1, therefore A 12 = Tr(ρ j 0 + ([T 2 t , T 2 s ][T 2 t , T s ])) = -Φ(Tr(ρ j 1 ([T 2 t , T 2 s ][T 2 t , T s ]))) = -Φ(B 12
). We also have

A 12 = A 8 A 6 -A 6 and B 12 = B 8 B 6 -B 6 , therefore A 8 A 6 -A 6 = A 8 A 6 + A 6 and A 6 = 0. We then have 0 = A 6 = -(β + β -1 -2)A 1 + 2(β + β -1 -1) = 2 β Φ 6 (β).
This contradicts our assumptions.

Assume now z([T 2 t , T s ]) = -1. We then have

A 6 = Tr(ρ j 0 + ([T 2 t , T s ])) = -Φ(Tr(ρ j 1 ([T 2 t , T s ])) = -Φ(B 6 ). We have A 6 = 2(β +β -1 -1) = B 6 , therefore Φ(β +β -1 -1) = -(β +β -1 -1). We have A 6 = 0, Tr(ρ j 0 + (T -2 t , T s ])) = A 6 and Tr(ρ j 1 (T -2 t , T s ])) = B 6 . It follows that z([T -2 t , T s ]) = z([T 2 t , T s ]) = -1 and z([T 2 t , T s ][T -2 t , T s ]) = 1. We then have A 12 = Tr(ρ j 0 + ([T 2 t , T s ][T -2 t , T s ])) = Φ(B 12
). We also have

A 12 = A 2 6 -(β + β -1 ) 2 (A 6 -2) -2 = B 12 . It follows that ( A 6 2 + 1) 2 (A 6 -2) = Φ ( A 6 2 + 1) 2 (A 6 -2) ( A 2 6 4 + A 6 + 1)(A 6 -2) = ( A 2 6 4 -A 6 + 1)(-A 6 -2) A 3 6 4 + A 2 6 + A 6 - A 2 6 2 -2A 6 -2 = - A 3 6 4 + A 2 6 -A 6 - A 2 6 2 + 2A 6 -2 A 3 6 2 -2A 6 = 0 A 6 2 (A 2 6 -4) = 0 A 6 2 (A 6 -2)(A 6 + 2) = 0 4(β + β -1 -1)(β + β -1 -2)(β + β -1 ) = 0 4β -3 Φ 6 (β)Φ 1 (β) 2 Φ 4 (β) = 0.
This is a contradiction by the assumptions on the order of β. This concludes the proof.

Chapter 6 W-graphs

Before extending our study to types H 3 and H 4 , we need to introduce the notion of W -graphs and give some properties which they verify. We also prove some new properties and propose a new conjecture. In this section, (W, S) is a Coxeter system with W a finite Coxeter group, K is the splitting field of W , K = K (( √ α s ) s∈S ), K = K ((α s ) s∈S ), C is the ring of integers of K and C = C (( √ α s ) s∈S ). We consider the Iwahori-Hecke algebra given by the presentation

H = H W,(αs) s∈S =< T 1 , ..., T n |T i T j T i ... ms i ,s j = T j T i T j ... ms i ,s j , (T i -α s i )(T i + 1) = 0 >, where α s i = α s j if
s i and s j are in the same conjugacy class of W . W-graphs were introduced in 1979 by Kazhdan-Lusztig [START_REF] Kazhdan | Representations of Coxeter groups and Hecke algebras[END_REF] for one-parameter families and the definition was extended to all Coxeter groups in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]. We here give the definition from [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]. We will prove some uniqueness properties (Proposition 6.2 and Proposition 6.3) in the one-parameter case and establish a conjecture (Conjecture 6.1) for certain W -graphs.

We here give the definition of W -graphs which can be found in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] Definition 6.1. For X a set, we write D(X) = {(x, x), x ∈ X} its diagonal. A W -graph Γ is given by a triple (X, I, µ) such that 1. X is a set and I is a map from X to P(S), √ α s µ s x,y e x if s / ∈ I(y).

2. µ is a map from (X × X \ D(X) × S) to K stable by the field involution sending √ α s to √ α s -1 . Let V be a K-

The map T s → ρ s affords a representation of H.

For Γ a W -graph, we write ρ Γ its associated representation and V Γ the corresponding H K,(αs) s∈S -module. Definition 6.2. A W -graph (X, I, µ) is said to be 2-colorable whenever there exists a map ω : X → {-1, 1} such that for any (s, x, y) ∈ S ×X 2 verifying µ s (x, y) = 0, we have ω(x) = -ω(y).

The data given by the triple (X, I, µ) can be represented by a weighted oriented graph for which the set X represents the vertices and the map I represents the weight of the vertices. The non-zero values of the map µ represent the oriented weighted edges of the graph. After they were introduced, it was shown by Alvis and Lusztig [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF] that there exist W -graphs affording all irreducible representations of Coxeter groups of types H 3 and H 4 . Using those results, Gyoja [START_REF] Gyoja | On the existence of a W -graph for an irreducible representation of a Coxeter group[END_REF] showed that any irreducible representation of an Iwahori-Hecke algebra associated to a finite non-crystalographic Coxeter group in the equal parameters case could be afforded by a W -graph. Moreover, he showed the following result Theorem 6.1. If all the parameters of the Hecke algebra are equal, then the following statements hold 1. Every irreducible H K,α -module is afforded by V Γ for a W -graph Γ over C.

2. An irreducible H q,K (W )-module is afforded by V Γ for a 2-colorable W -graph Γ over K( √ α) if and only it admits a form over K( √ α). The representation is then said to be nonexceptional.

Remark : He also classified all the exceptional representations for Iwahori-Hecke algebras, that is, the ones which are not non-exceptional, and obtained The link between 2-colorable representations and non-exceptionality can be seen through the following proposition.

Proposition 6.1. If (X, I, µ) is a 2-colorable W -graph then ρ (X,I,µ) ρ (X,I,-µ) . If σ : K(( √ α s ) s∈S ) → K(( √ α s ) s∈S ) leaves K stable and maps √ α s to - √ α s for all s ∈ S then ρ (X,I,-µ) = σ • ρ (X,I,µ) .
Proof. Let (X, I, µ) be a 2-colorable W -graph and ω : X → {-1, 1} be an associated coloring. Let L : V → V be the linear map defined by L(e x ) = ω(x)e x for all x ∈ X. We have L -1 = L. The second part of the proposition is proved in the same way. Proposition 6.1 shows that two different W -graphs can give isomorphic representations. We now provide some uniqueness conditions with the following propositions. Proposition 6.2. If Γ 1 = (X, I 1 , µ 1 ) and Γ 2 = (Y, I 2 , µ 2 ) are W -graphs such that ρ Γ 1 ρ Γ 2 and ρ Γ 1 is irreducible, then there exists a bijection ϕ : X → Y such that for all x ∈ X we have

I 2 (ϕ(x)) = I 1 (x).
Proof. Let Γ 1 and Γ 2 be as above. If n = 1, then the result is straightforward. Let us assume n ≥ 2. Let us show that for all S ⊂ S, we have

dim( ∩ s∈S ker(ρ Γ 1 (T s ) + 1)) = |{x ∈ X, S ⊂ I(x)}|.
If there exists x 0 ∈ X such that I(x 0 ) = ∅, then by definition of a W -graph, we would have that Ṽ = x∈X\{x 0 } Ke x is a stable subvector-space of dimension n -1 which contradicts the irreducibility of ρ Γ 1 .

Let S = {s i l } l∈[ [1,k]] ⊂ S for some k ∈ N . We label the vertices {e

x i } i∈[[1,n Γ 1 ]
] in a such a way that there exist r S ∈ N verifying S ⊂ I(x h ) for all h ∈ [[1, r S ]] and

S ⊂ I(x h ) for all h ∈ [[r S + 1, n Γ 1 ]]. Let x ∈ ∩ s∈S ker(ρ Γ 1 (T s ) + 1). There exists a unique family (λ i ) i∈[[1,n Γ 1 ]] ∈ Kn Γ 1 such that x = i∈[[1,n Γ 1 ]] λ i e x i . Let h 0 ∈ [[r S + 1, n Γ 1 ]], we will show that λ h 0 = 0. There exists j ∈ [[1, k]] such that s i j / ∈ I(x h 0 ). We then have ρ Γ 1 (T s i j )(x) = α s i j λ h 0 e x h 0 + h =h 0 a h e x h for some a h ∈ K. Since x ∈ ker(ρ Γ 1 (T s i j ) + 1), we also have ρ Γ 1 (T s i j )(x) = -x = - i∈[[1,n Γ 1 ]] λ i e x i . This implies that α s i j λ h 0 = -λ h 0 , therefore λ h 0 = 0. We then conclude that x ∈ Vect K ((e x i ) i∈[[1,r S ]] ), therefore ∩ s∈S ker(ρ Γ 1 (T s ) + 1) ⊂ Vect K ((e x i ) i∈[[1,r S ]] ).
The reverse inclusion follows from the definition of r S . This proves that dim(

∩ s∈S ker(ρ Γ 1 (T s ) + 1)) = |{x ∈ X, S ⊂ I(x)}|.
We now have that for all S ⊂ S, |{x ∈ X, S ⊂ I 1 (x)}| = |{x ∈ X, S ⊂ I 2 (x)}| because the representations are isomorphic. It follows by induction on |S| -|S | that for all S ⊂ S, |{x ∈ X, S = I 1 (x)}| = |{x ∈ X, S = I 2 (x)}| which concludes the proof.

The next proposition gives a fairly good uniqueness property which will be used for the computations on W -graphs. Proposition 6.3. Let Γ 1 = (X, I, µ) and Γ 2 = (X , I , µ ) be W -graphs as above and ϕ : X → X be a bijection such that for all x ∈ X, I (ϕ(x)) = I(x). We assume that X and X are labeled in such a way that if I(x i ) I(x j ) then i < j and ϕ(x i ) = x i . We also assume that the images of µ and µ are included in K and that µ and µ are independent of S. We then write µ x,y instead of µ s x,y . If there exists M ∈ GL n ( K) such that for all

T ∈ H K , M ρ Γ 1 (T )M -1 = ρ Γ 2 (T ) then M is block diagonal with blocks of length |{x ∈ X, I 1 (x) = I 1 (x i )}|.
Proof. Note first that by Proposition 6.2, we can choose a labeling as required.

We can choose numberings verifying desired by Proposition 6.2. We write (e i ) i∈[ [1,m]] the basis corresponding to Γ 1 and (e i ) i∈ [1,m]] the basis corresponding to Γ 2 . Assume there exists a matrix M ∈ GL n ( K) such that for all T ∈ H K , M ρ Γ 1 (T )M -1 = ρ Γ 2 (T ). We will first show that the matrix M is block lower-triangular and then that it is diagonal in the basis indexed by the numbering we chose on the vertices. We write ρ 1 = ρ Γ 1 , ρ 2 = ρ Γ 2 and m the dimension of both representations for the remainder of the proof.

Let x j ∈ X. Let s ∈ I(x j ). We then have M ρ 1 (T s )(e j ) = -M e j = - m i=1 m i,j e i . On the other hand, we have ρ 2 (T s )M e j = m i=1 m i,j ρ 2 (T s )e i = - i∈[[1,m]],s∈I(x i ) a i e i + i∈[[1,m]],s / ∈I(x i ) α s m i,j e i . Since
those two quantities are assumed to be equal, we have

m i,j = 0 for all i ∈ [[1, m]] such that s / ∈ I(x i ). Let now (i, j) ∈ [[1, m]
] 2 be such that I(x i ) = I(x j ) and i < j. By the assumption on the numbering, we have that there exists s ∈ I(x j ) such that s / ∈ I(x i ). The above computation implies that m i,j = 0, therefore M is block lower-triangular.

We have proven that for all i ∈ [ [1, m]], M e i = j,I(x i )⊂I(x j ) m i,j e j . Let i ∈ [ [1, m]] and j 0 ∈ [ [1, m]] such that I(x i ) I(x j 0 ) and s ∈ I(x j 0 ) \ I(x i ). We have

M ρ 1 (T s )(e i ) = M (α s e i + √ α s s∈I(x ) µ ,i e i ) = α s j I(x i )⊂I(x j ) m i,j e j + √ α s s∈I(x ) µ ,i k I(x )⊂I(x k ) m ,k e k = α s m i,j 0 e j 0 + α s j =j 0 I(x i )⊂I(x j ) m i,j e j + √ α s s∈I(x ) k I(x k )⊂I(x ) µ ,i m ,k e k =      α s m i,j 0 + √ α s s∈I(x )⊂I(x j 0 ) µ ,i m ,j 0      e j 0 + l =j 0 b e .
where the coefficients b are elements of K which can be deduced from the above equalities.

We also have

ρ 2 (s)M (e i ) = - j I(x i )⊂I(x j ) s∈I(x j ) m i,j e j + j,I(x i )⊂I(x j ),s / ∈I(x j ) m i,j   α s e j + √ α s ,s∈I(x ) µ ,j e   =      -m i,j 0 + √ α s j s / ∈I(x j ) m i,j µ j 0 ,j      e j 0 + =j 0 c e .
where the coefficients c are elements of K which can be deduced from the above equalities.

Since we have assumed those two quantities to be equal, we get that

α s m i,j 0 + √ α s l s∈I(x l )⊂I(x j 0 ) µ l,i m l,j 0 = -m i,j 0 + √ α s j s / ∈I(x j )
m i,j µ j 0 ,j .

Since 1 + α s and √ α s are K-linearly independant, we get that m i,j 0 = 0. This is true for all (i, j) such that I(x i ) I(x j 0 ), therefore we have that the matrix M is block-diagonal.

Remark : To prove that the matrix is block lower-triangular, we don't need to assume anything on the images of µ or µ or on the independence with regards to S and we can take M ∈ GL m (K). If we don't assume that the image of µ is in K then the result does not hold. The two following W -graphs give us a counter-example in the more general setting

µ s 2 4,1 = -1 and µ s 1 4,1 = µ s 3 4,1 = 0. Then the matrix M =      1 0 0 0 0 1 0 0 1 0 1 0 0 1 0 1      verifies for all s ∈ S, M ρ Γ (T s )M -1 = ρ Γ (T s ).
We now give a proposition which can be found in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] (11.1.7) providing a construction of a W -graph associated to the dual representation of a W -graph. Proposition 6.4. Let Γ = (X, I, µ) be a W -graph. We define its dual W -graph Γ = (X, Ĩ, μ) by

1. ∀x ∈ X, Ĩ(x) = S \ I(x), 2. ∀(x, y, s) ∈ X × X \ D(X) × S, μs
x,y = -µ s y,x . We then have that ρ Γ ρ Γ , where ρ (T s ) = ρ(-α s T -1 s ). Definition 6.3. We say a representation ρ of H K is self-dual if ρ ρ . Proposition 6.5. If ρ : H K → GL nρ (K) is an irreducible self-dual representation of a Iwahori-Hecke algebra then there is either a symmetric or skew-symmetric non-degenerate bilinear form ., . associated to ρ in the following way

∀s ∈ S, ∀(u, v) ∈ V 2 , ρ(T s )u, ρ(T s )v = u, v .
Proof. Let ρ : H K → GL nρ (K) be a self-dual representation. Then there exists P ∈ GL n (k) such that for all s ∈ S, P ρ(T s )P

-1 = -α s t ρ(T s ) -1 .
This implies that P ρ(T s )P -1 = -α s t (-α s P -1 t ρ(T s ) -1 P ) -1 = t P ρ(T s ) t P -1 . It follows that for all s ∈ S, t P -1 P ρ(T s )( t P -1 P ) -1 = ρ(T s ). Hence, for all h ∈ H K , t P -1 P ρ(h)( t P -1 P ) -1 = ρ(h). By Schur's lemma, there exists λ ∈ F q such that t P -1 P = λ. Thus P = λ t P = λ t (λ t P ) = λ 2 P . It follows that λ ∈ {-1, 1}. This bilinear form .|. associated to P is non-degenerate. It is symmetric when λ = 1 and skew-symmetric when λ = -1. For all s ∈ S,

(u, v) ∈ V 2 , we have ρ(T s )u, ρ(T s )v = t (ρ(T s )u)P ρ(T s )v = t u( t ρ(T s )P ρ(T s ))v = t u(-α s P )v = -α s u, v . This concludes the proof.
Although such a bilinear form always exists for any given self-dual representation, it is difficult to obtain the bilinear form explicitely and to determine whether it is symmetric or anti-symmetric. Since there is a combinatorial way to define a dual W -graph, it is natural to expect W -graphs affording self-dual representations to be isomorphic to their dual W -graph. This seems to never be the case. Nevertheless, we have (X, I, µ) (X, Ĩ, -μ) in many cases. We know by Proposition 6.1 that if (X, Ĩ, -μ) is 2-colorable, then we have ρ (X, Ĩ,-μ) ρ (X, Ĩ,μ) . When this is the case, we can define a bilinear form using only the 2-coloring of the graph. The construction is given in the following theorem. Theorem 6.2. Let Γ = (X, I, µ) be a W -graph affording an irreducible representation of H such that Γ is 2-colorable and Γ is isomorphic as an oriented weighted graph to (X, Ĩ, -μ).

Let ϕ : X → X be the graph automorphism from Γ to (X, Ĩ, -μ) and x 1 , x 2 , . . . , x n be a numbering of X such that ϕ(x i ) = x n+1-i .

Let ., . be the bilinear form defined by e x i , e x j = ω(e x i )δ i,n+1-j , where ω corresponds to a coloring of Γ with 1 and -1.

We then have

∀s ∈ S, ∀v 1 , v 2 ∈ V, ρ Γ (T s )v 1 , ρ Γ (T s )v 2 = -α v 1 , v 2 .

This bilinear form is non-degenerate and it is symmetric if

ω(x 1 )ω(x n ) = 1 and skew- symmetric if ω(x 1 )ω(x n ) = -1.
The associated representation is then self-dual.

Proof. Let Γ = (X, I, µ) be a W -graph as above and ω : X → {-1, 1} be the corresponding 2coloring. Let s ∈ S and x, y ∈ V . First note that since I(ϕ(x)) = S \ I(x), we have e x , e z = 0 for all z ∈ X such that I(z) = S \ I(x). We now have four different cases to consider.

+ √ α s µ ϕ(x),y ω(x) = √ α s (µ ϕ(y),ϕ(ϕ(x)) ω(ϕ(y)) + µ ϕ(x),y ω(x)) = √ α s (µ ϕ(x),y ω(ϕ(y)) + µ ϕ(x),y ω(x) = √ α s µ ϕ(x),y (ω(ϕ(y)) + ω(x)).
If µ ϕ(x),y = 0 then the above quantity is equal to zero. If µ ϕ(x),y = 0 then µ ϕ(y),x = 0, therefore ω(ϕ(y)) = -ω(x) and the above quantity is again zero. This proves that in all cases, ρ Γ (T s )e x , ρ Γ (T s )e y = -α s e x , e y . It only remains to show that this bilinear form is symmetric if ω(x 1 )ω(x n ) = 1 and skew-symmetric if ω(x 1 )ω(x n ) = -1.

To show this, we first prove that for all x ∈ X, we have ω(x)ω(ϕ(x)) = ω(x 1 )ω(x n ). If Γ was a disconnected graph, then ρ γ would be reducible, therefore Γ is connected. This means we only need to show that for a given x, we have ω(y)ω(ϕ(y)) = ω(x)ω(ϕ(x)) for all y ∈ X such that µ x,y = 0 or µ y,x = 0, . Let x ∈ X and y ∈ X such that µ x,y = 0 or µ y,x = 0. Then µ ϕ(y),ϕ(x) = 0 or µ ϕ(x),ϕ(y) = 0. This implies that ω(y) = -ω(x) and ω(ϕ(y)) = -ω(ϕ(x)), therefore ω(y)ω(ϕ(y)) = (-ω(x))(-ω(ϕ(x)) = ω(x)ω(ϕ(x)). This shows that for all x ∈ X, we have ω(x)ω(ϕ(x)) = ω(x 1 )ω(x n ). Let now (x, y) ∈ X 2 . We have e x , e y = ω(x)δ y,ϕ(x) = ω(x 1 )ω(x n )ω(ϕ(x))δ ϕ(y),x = ω(x 1 )ω(x n )ω(y)δ x,ϕ(y) = ω(x 1 )ω(x n ) e y , e x as required. Definition 6.4. A W -graph satisfying the conditions in Theorem 6.2 is said to be self-dual. Not all W -graphs affording self-dual representations are self-dual. For example, in type D 4 , there is an irreducible self-dual eight-dimensional representation which corresponds to the double Young diagram λ = ([2, 1], [1]). By Proposition 4.1.3, we know that the associated bilinear form must be skew-symmetric since ν

(λ) = ν([2, 1])ν([1]) = (-1) 3-1 2 (-1) 1-1 2
= -1. We can associate to this representation the following W -graph which we determined using the restrictions to the three different A A 3 and A 3

A 1 3 1, 2 1, 3 1, 4 1, 2, 4 3, 4 2, 4 2, 3 2 2 2
Here the W -graphs are presented as in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]. If I(x) = {s 3 }, then we write 3 inside the vertex x. We don't write the edges when µ x,y = 0. If µ x,y = µ y,x , then we write non-oriented edges. If µ x,y = 1, then we do not write the weight on the edge and we write µ x,y as a weight on the edge otherwise. The W -graph does not satisfy the desired properties because the edge between nodes 3 and 3, 4 is of weight 1 whereas the edge between notes 1, 2 and 1, 2, 4 is of weight 2.

However, since the properties appear to be natural and are verified by some W -graphs, we propose the following conjecture. Conjecture 6.1. Let W be a Coxeter group. For any irreducible self-dual representation, there exists a self-dual W -graph defined over K. If there exists a W -graph Γ = (X, I, µ) defined over K then there exist a self-dual W -graph Γ = (X , I , µ ) defined over K and a matrix

M ∈ GL |X| ( K) such that for all h ∈ H K , M ρ Γ (h)M -1 = ρ Γ (h).
In order to prove the conjecture for exceptional groups, we effectively find the W -graphs verifying the right properties by assuming they exist and finding the bilinear form preserving the W -graph we are working with. This can be seen in the following proposition. Proposition 6.6. Assume the conjecture holds and that α s = α s for all s, s ∈ S. We write α for the unique parameter of the Iwahori-Hecke algebra. Then any bilinear form associated to a W -graph affording an irreducible self-dual representation defined over K is represented by a block anti-diagonal matrix.

Proof. Assume the conjecture is true and that Γ = (X, I, µ) is a self-dual W -graph defined over K . There exists Γ = (X , I , µ ) such that µ (X × X \ D(X) × S) ⊂ K and ρ Γ ρ Γ . We order the vertices of X in order to have I(x i ) = I(x i ) for all i, and if I(x i ) I(x j ) then i < j. We consider the matrices with respect to the bases corresponding to those orders.

There exists M ∈ GL |X| ( K) such that for all h ∈ H K , M ρ Γ (h)M -1 = ρ Γ (h). By Propositions 6.2 and 6.3, M is block-diagonal. Since Γ is self-dual, there exists an anti-diagonal matrix P corresponding to its 2-coloring such that for all s ∈ S, P ρ Γ (T s ) P -1 = -α t ρ Γ (T s ) -1 . We also have by Proposition 6.5 that there exists P ∈ GL |X| (K) such that for all s ∈ S, P ρ Γ (T s )P

-1 = -α t ρ Γ (T s ) -1 . By substituting M ρ Γ (T s )M -1 = ρ Γ (T s ) in the first expression, we get P M ρ Γ (T s )M -1 P -1 = -α t (M ρ Γ (T s )M -1 ) -1 = -α t M -1 t ρ Γ (T s ) -1 t M . It follows that t M P M ρ Γ (T s )( t M P M ) -1 = -α t ρ Γ (T s ) -1 = P ρ Γ (T s )P -1 .
Hence, by Schur's lemma, there exists λ ∈ F p such that P = λ t M P M . Since M is block-diagonal and P is anti-diagonal, we get that P is block antidiagonal.

Remark : This proposition can be useful to to find the matrix P after assuming the conjecture is true. For the W -graph associated to ([2, 1], [1]), all the blocks are of size one, therefore we only need to look for an anti-diagonal matrix. This means we have only 8 unknowns, assuming the 2-coloring is preserved up to permutation of vertices having the same image under I, we get that P must be anti-symmetric, therefore we only have four unknowns.

After solving the equations afforded by each generator, we get

P =                0
0 0 0 0 0 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 -2 0 0 0 0 0 0 2 0 0 0 0 0 0 -2 0 0 0 0 0 0 2 0 0 0 0 0 0 -2 0 0 0 0 0 0 -1 0 0 0 0 0 0 0

              
We then look for a matrix M such that t M P M = P .

Using Gaussian reduction for quadratic forms, we get

M =               
1 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1

              
By computing M ρ(T s )M -1 for all s, we obtain the following W -graph. 2

We obtain in the same way self-dual W -graphs for all self-dual representations in types I 2 (m), H 3 , H 4 , E 6 , E 7 and E 8 . We give the new W -graphs in type H 4 in the Appendix. All the new W -graphs can be downloaded from [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF].

Chapter 7 Type E

In this section we determine the images of the Artin groups of type E inside their associated Iwahori-Hecke algebras. Their representations are given by W -graphs, therefore we will use the results from Chapter 6. We have defined in that chapter self-dual representations in Definition 6.3 and self-dual W -graphs in Definition 6.4. We have established the Conjecture 6.1 which states that there exists a self-dual W -graph affording any self-dual representation of a W -graph. There are no self-dual representations in type E 7 . In types E 6 and E 8 , we have proved the conjecture and obtained new W -graphs which are self-dual for each of the self-dual representations. We only provide in this Appendix the graphs for the 10-dimensional self-dual representation of E 6 and its 20-dimensional self-dual representation. All the remaining ones are of dimension greater than 60 and can be downloaded from [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF].

Type E 6

Let p / ∈ {2, 3} be a prime and α ∈ F p be of order not dividing 5, 8, 9 or 12. We write F q = F p (α). There are 25 irreducible representations of the Iwahori-Hecke algebra H E 6 ,α which we define below. They are all of dimension less than or equal to 90 and there are 5 self-dual representations, they associated to the E 6 -graphs 10 s , 20 s , 60 s , 80 s and 90 s . We have found self-dual E 6 -graphs [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF] for each of these representations. Using the 2-coloring, the bilinear form defined in Theorem 6.2 is skew-symmetric for each of those self-dual representations. Definition 7.1.1. The Iwahori-Hecke algebra H E 6 ,α of type E 6 is the F q -algebra generated by S 1 , S 2 , S 3 , S 4 , S 5 , S 6 and the following relations For σ in the Coxeter group E 6 , we set

1. ∀i ∈ {1, 2, 3, 4, 5, 6}, (S i -α)(S i + 1) = 0. 2. S 1 S 3 S 1 = S 3 S 1 S 3 . 3. ∀i ∈ {2, 4, 5, 6}, S 1 S i = S i S 1 . 4. S 2 S 4 S 2 = S 4 S 2 S 4 . 5. ∀i ∈ {3, 5, 6}, S 2 S i = S i S 2 . 6. ∀i ∈ {3, 4, 5}, S i S i+1 S i = S i+1 S i S i+1 .
T σ = S i 1 . . . S i k whenever σ = s i 1 . . . s i k is a reduced expression.
This means we consider E 6 as in the CHEVIE package of GAP3 [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF] Proof. We will apply Proposition 2.2.4.

Let A = Z[ √ u ±1 ] and F = Q( √ u).
We have a symetrizing trace defined by τ (T 0 ) = 1 and τ (T σ ) = 0 for all σ ∈ E 6 \ {1 E 6 }. By [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF], H E 6 ,u is then a free symmetric F -algebra of rank 51840. By [7] V.3. Corollary 1, A is integrally closed. Let θ be the ring homomorphism from A to L = F q defined by θ(u) = α and θ(k) = k. We know F H is split. The basis formed by the elements T σ , σ ∈ E 6 verifies the conditions of the Proposition 2.2.4. The E 6 -graphs are still connected after specialization since all the weights are in {-6, -4, -3, -2, -3/2, -1, -1/2, -1/3, -1/6, 1/3, 1/2, 1, 3/2, 2, 3}.

It now only remains to check that the Schur elements associated to the specialized representations are in B and do not vanish under θ with B as in Proposition 2.2.4. The Schur elements are given in Table 7.1. For a pair (ρ, ρ ) of representations, we only give the Schur element of one of the representations since the other is obtained by applying the involution √ u → √ u -1 . The conditions on α and p imply that the Schur elements verify the right conditions and the proof is concluded.

Using the CHEVIE package of GAP3 [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF], we give in Table 7.2 the restriction table from H E 6 ,α to its subalgebra H D 5 ,α generated by S 1 , S 2 , S 3 , S 4 and S 5 which is naturally isomorphic to the Iwahori-Hecke of type D 5 with parameter α. They correspond in the generic case to the induction/restriction tables of the corresponding Coxeter groups.

Proposition 7.1.2. The restrictions to A E 6 of the representations afforded by those E 6 -graphs are absolutely irreducible and the representations of dimension greater than one are pairwise non-isomorphic.

Proof. As in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] Lemma 3.4, we only need to prove that A E 6 is generated by A E 5 and A E6 . This true because s 6 = s 6 s -1

1 s 1 , s 6 s -1 1 ∈ A E 6 and s 1 ∈ A D 5 .
Let ρ 1 and ρ 2 be two irreducible representations of H E 6 ,α such that ρ 1|A E 6

ρ 2|A E 6 . We have A E 6 /A E 6 < S 1 > Z. It follows by Lemma 3.2.3 that there exists x ∈ F q such that for all i ∈ [ [1,6]], we have ρ 1 (S i ) conjugate to xρ 2 (S i ). Since the representations are irreducible we have that the set of eigenvalues of ρ 1 (S i ) is {-1, α} and the set of eigenvalues of ρ 2 (S i ) is {-x, xα}. This implies that x = 1 or (x = -α and α 2 = 1). The latter contradicts our assumptions on α therefore x = 1 and ρ 1 ρ 2 .

Before determining the image of the Artin groups inside this Iwahori-Hecke algebra, we need as in the other cases a Lemma on Artin groups which will allow us to use the restriction from

H E 6 ,α to H D 5 ,α . Lemma 7.1.1. The normal closure A D 5 A E 6 of A D 5 inside A E 6 is equal to A E 6
, where we identify A D 5 as a subgroup of A E 6 using the natural isomorphism from A D 5 to < S i , i ∈ [ [1,5]] >.

Proof. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], we have

A D 5 =< S 3 S -1 1 , S 1 S 3 S -1 1 , S 4 S -1 1 , S 2 S -1 1 S 5 S -1 1 , S 3 S -1 1 , S 4 S -1 3 > and A E 6 =< S 3 S -1 1 , S 1 S 3 S -1 1 , S 4 S -1 1 , S 2 S -1 1 S 5 S -1 1 , S 3 S -1 1 , S 4 S -1 3 , S 6 S -1 1
>. This proves we only need to show that S 6 S -1

1 ∈ A D 5 A E 6 .
We have S 6 S 5 S 6 = S 5 S 6 S 5 , therefore S 6 = S 5 S 6 S 5 (S 5 S 6 ) -1 and

S 6 S -1 1 = S 5 S 6 S 5 S -1 1 (S 5 S 6 ) -1 = ((S 5 S -1 1 )(S 6 S -1 1 ))(S 5 S -1 1 )((S 5 S -1 1 )(S 6 S -1 1 )) -1 .
This concludes the proof.

We will now use the above information and the usual techniques to determine the image of the Artin group inside H E 6 ,α .

Proposition 7.1.3. If F q = F p (α) = F p (α + α -1 ) then we have 1. ρ 6p (A E 6 ) SL 6 (q), 2. ρ 10s (A E 6 ) SP 10 (q), 3. ρ 15p (A E 6 ) SL 15 (q), 4. ρ 15q (A E 6 ) SL 15 (q), 5. ρ 20p (A E 6 ) SL 20 (q), 6. ρ 20s (A E 6 ) SP 20 (q), 7. ρ 24p (A E 6 ) SL 24 (q), 8. ρ 30p (A E 6 ) SL 30 (q), 9. ρ 60s (A E 6 ) SP 60 (q), 10. ρ 60p (A E 6 ) SL 60 (q), 11. ρ 64p (A E 6 ) SL 64 (q), 12. ρ 80s (A E 6 ) SP 80 (q), 13. ρ 81p (A E 6 ) SL 81 (q), 14. ρ 90s (A E 6 ) SP 90 (q). If F q = F p (α) = F p (α + α -1 ) then we have 1. ρ 6p (A E 6 ) SU 6 (q 1 2 ), 2. ρ 10s (A E 6 ) SP 10 (q 1 2 ), 3. ρ 15p (A E 6 ) SU 15 (q 1 2 ), 4. ρ 15q (A E 6 ) SU 15 (q 1 2 ), 5. ρ 20p (A E 6 ) SU 20 (q 1 2 ), 6. ρ 20s (A E 6 ) SP 20 (q 1 2 ), 7. ρ 24p (A E 6 ) SU 24 (q 1 2 ), 8. ρ 30p (A E 6 ) SU 30 (q 1 2 ), 9. ρ 60s (A E 6 ) SP 60 (q 1 2 ), 10. ρ 60p (A E 6 ) SU 60 (q 1 2 ), 11. ρ 64p (A E 6 ) SU 64 (q 1 2 ), 12. ρ 80s (A E 6 ) SP 80 (q 1 2 ), 13. ρ 81p (A E 6 ) SU 81 (q 1 2 ), 14. ρ 90s (A E 6 ) SP 90 (q 1 2 ). Proof. Assume first F q = F p (α) = F p (α + α -1 ). The representations considered are all defined over F p ( √ α). If F p ( √ α) = F p (α) then X 2 -α is an irreducible polynomial over F p (α)
. The unique automorphism σ of order 2 of F p ( √ α) fixes F p pointwise and verifies σ( √ α) = -√ α. Hence, Proposition 6.1 and Lemma 3.2.5 imply that the representations can be considered to be defined over F q .

We have found non-degenerate skew-symmetric bilinear forms defined over F p associated to the representation ρ 10s , ρ 20s , ρ 60s , ρ 80s and ρ 90s , therefore we have all the corresponding inclusions for the images of A E 6 .

It now only remains to prove that all the inclusions are isomorphisms. We prove it separately for 6 p and 10 s and we will use Theorem 2.3.2 for the remaining ones. By Table 7.2 and Theorem 4.4.1, we have that ρ 6p (A D 5 ) SL 5 (q)×{1}. It follows by Lemma 7.1.1 that ρ 6p (A E 6 ) is generated by transvections. Theorem 2.3.4 then shows that there exists q such that up to conjugation in GL 6 (q), we have ρ 6p (A E 6 ) ∈ {SL 6 (q ), SP 6 (q ), SU 6 (q 1 2 )}. Since it contains a natural SL 5 (q), we have that q = q and ρ 6p (A E 6 ) SL 6 (q). By Table 7.2, we have that ρ 10s (A D 5 ) ρ [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF], [1] (A D 5 ). We have ϕ( [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF], [1]) = ([2, 2], [1]) and ν( [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF], [1]

) = ν([2, 2])ν([1]) since [2, 2] > [1] . It follows that ν([2, 2], [1]) = (-1) 4-2 2 (-1) 1-1 2
= -1 and by Theorem 4.4.1, ρ [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF], [1] (A D 5 ) SP 10 (q). It follows that ρ 10s (A E 6 ) SP 10 (q). By Table 7.2, the non self-dual representations contain a natural SL 2 (q) and the self-dual ones contain a twisted diagonal SL 3 (q). We can therefore apply Theorem 2.3.2 and Lemmas 3.3.5, 3.3.6, 8.5.1 to conclude the proof.

Assume now F q = F p (α) = F p (α + α -1 ). The representations can again be considered to be defined over F q . We have that X 2 -(α + α -1 )X + 1 is an irreducible F p (α + α -1 )-polynomial over F p , therefore we have an automorphism of order 2 of F q mapping α to α -1 . To conclude the proof, we only need to use previous arguments if we prove that all the representations considered verify ρ

• ρ . First consider the non self-dual representations of dimension different from 15, 20 and 60. Let ρ be such a representation, we know that • ρ is an irreducible representation, therefore we have • ρ ρ or • ρ ρ since those are the only irreducible representations of the same dimension. Assume • ρ ρ . Then ρ • ρ and Lemma 3.2.4 implies that up to conjugation ρ ≤ SL nρ (q 12 ). By Table 7.2, those representations all contain a natural SU a (q 1 2 ) with multiplicity 1 for some a ≥ 5. This implies that SU a (q

1 2 ) is a subgroup of SL a (q 1
2 ) which is absurd by simple cardinality arguments since a ≥ 3. It follows by contradiction that ρ

• ρ . The result is obvious for the self-dual representations of dimension different from 15, 20 and 60 because there is only one possibility when ρ ρ .

Consider now the 60-dimensional representations, we have ρ 60s ρ 60s and ρ 60p ρ 60p . We know that • ρ 60s ρ 60s or • ρ 60s ρ 60p or • ρ 60s ρ 60p . We have • ρ 60s

• ρ 60s ( • ρ 60s ) . This proves that the only possibility is the first one, therefore • ρ 60s ρ 60s . We have

• ρ 60p ρ 60p or • ρ 60p ρ 60p or • ρ 60p ρ 60s .
The second possibility is excluded by the same reasonning as for the representations of dimension different from 15 or 20. The third possibility would imply that • ρ 60s ρ 60p . By the above, this would imply ρ 60s ρ 60p which is absurd. It follows that • ρ 60p ρ 60p . The arguments are identical for the 20-dimensional representations.

It only remains to consider the 15-dimensional representations. We have ρ 15p ρ 15p , ρ 15q ρ 15q . The are therefore four 15-dimensional representations. We have

• ρ 15p ρ 15p or • ρ 15p ρ 15p or • ρ 15p ρ 15q or • ρ 15p ρ 15q .
Using the same arguments as before we have [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF] . We know by Proposition 4.2.3 that • ρ ρ for every representation ρ of H D 5 . This implies that under the assumption • ρ 15p ρ 15q , we would have ρ [4], [1] ⊕ ρ [3], [1 2 ] ρ [3,[START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF],∅ ⊕ ρ [3], [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF] which is absurd. We exclude in the same way • ρ 15p ρ 15q . It follows that • ρ 15p ρ 15p . In the same way • ρ 15q ρ 15q . This concludes the proof.

• ρ 15p ρ 15p . Assume now • ρ 15p ρ 15q . We have that ρ 15p|H D 5 ,α ρ [4],[1] ⊕ ρ [3],[1 2 ] and ρ 15q|H D 5 ,α ρ [3,2],∅ ⊕ ρ [3],
We now state the main theorem for type E 6 . Theorem 7.1.1. Assume F q = F p (α) = F p (α + α -1 ), we then have that the morphism from A E 6 to H E 6 ,α ρ irr GL nρ (q) factorizes through the surjective morphism

Φ : A E 6 → SL 6 (q) × SP 10 (q) × SL 15 (q) 2 × SL 20 (q) × SP 20 (q) × SL 24 (q) × SL 30 (q)
×SP 60 (q) × SL 60 (q) × SL 64 (q) × SP 80 (q) × SL 81 (q) × SP 90 (q).

Assume F q = F p (α) = F p (α + α -1 ), we then have that the morphism from A E 6 to H E 6 ,α ρ irr GL nρ (q) factorizes through the surjective morphism

Φ : A E 6 → SU 6 (q 1
2 ) × SP 10 (q

1 2 ) × SU 15 (q 1 2 ) 2 × SU 20 (q 1 2 ) × SP 20 (q 1 2 ) × SU 24 (q 1
2 ) × SU 30 (q

1 2 ) ×SP 60 (q 1
2 ) × SU 60 (q

1 2 ) × SU 64 (q 1
2 ) × SP 80 (q

1 2 ) × SU 81 (q 1
2 ) × SP 90 (q

2 ).

Proof. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A E 6 is perfect. Furthermore, Lemma 3.3.1 gives that the morphism is surjective unless there exists two different representations ρ 1 and ρ 2 in the decomposition such that Ψ • ρ 1|A H 4 ρ 2|A H 4 for some field automorphism Ψ. By Proposition 2.1.2, we have that Ψ(α + α -1 ) = α + α -1 . This shows that Ψ must be trivial over F p (α + α -1 ). It follows by the previous propositions that there are no such representations in the decompositions and the proof is concluded.

Type E 7

Let p be a prime different from 2 and 3 and α ∈ F p of order not dividing 8, 10, 12, 14 and 18. We write F q = F p (α). There are 60 irreducible representations of H E 7 ,α , none of them are self-dual. The highest dimensional representation is of dimension 512. They are all 2-colorable except for the two 512-dimensional representations. Definition 7.2.1. The Iwahori-Hecke algebra H E 7 ,α of type E 7 is the F q -algebra generated by S 1 , S 2 , S 3 , S 4 , S 5 , S 6 , S 7 and the following relations 1. ∀i ∈ {1, 2, 3, 4, 5, 6, 7}, (S i -α)(S i + 1) = 0. Proof. We will apply Proposition 2.2.4.

S

Let A = Z[ √ u ±1 ] and F = Q( √ u).
We have a symetrizing trace defined by τ (T 0 ) = 1 and τ (T σ ) = 0 for all σ ∈ E 7 \ {1 E 7 }. By [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF], H E 7 ,u is then a free symmetric F -algebra of rank 2903040. By [7] V.3. Corollary 1, A is integrally closed. Let θ be the ring homomorphism from A to L = F q defined by θ(u) = α and θ(k) = k. We know F H is split. The basis formed by the elements T σ , σ ∈ E 7 verifies the conditions of the Proposition 2.2.4. The E 7 -graphs remains connected since all the weights lie in {-3, -2, -1, 1, 2, 3}.

It now only remains to check that the Schur elements associated to the specialized representations are in B and do not vanish under θ with B as in Proposition 2.2.4. The Schur elements are given in Table 7.3. They were obtained using Proposition 9.3.6 and Table E.6 of [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]. For a pair (ρ, ρ ) of representations, we only give the Schur element of one of the representations since the other is obtained by applying the involution

√ u → √ u -1
. The conditions on α and p imply that the Schur elements verify the right conditions and the proof is concluded.

The restriction table from H E 7 ,α to its subalgebra H E 6 ,α generated by S 1 , S 2 , S 3 , S 4 , S 5 and S 6 which is naturally isomorphic to the Iwahori-Hecke of type E 6 with parameter α is then given by Table 7.4. It is obtained using the CHEVIE package of GAP3 [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF]. They correspond in the generic case to the induction/restriction tables of the corresponding Coxeter groups.

Proposition 7.2.2. The restrictions to A E 7 of the representations afforded by those E 7 -graphs are absolutely irreducible and the representations of dimension greater than 1 are pairwise nonisomorphic.

Proof. As in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] Lemma 3.4, we only need to prove that A E 7 is generated by A E 6 and A E 7 . This true because s 7 = s 7 s -1

1 s 1 , s 7 s -1 1 ∈ A E 6 and s 1 ∈ A E 6 . We now prove the second part of the statement. Let ρ 1 and ρ 2 be two irreducible representations of H E 7 ,α such that ρ 1|A E 7 ρ 1|A E 7 . By Lemma 3.2.3, there exists a character ξ : A E 7 → F q such that ρ 1 ρ 2 ⊗ ξ. This means there exists x ∈ F q such that for all i ∈ [ [1,7]], ρ 1 (S i ) is conjugate to xρ 2 (S i ). We know for any representation ρ of dimension greater than 1, the set of eigenvalues of ρ is equal to {α, -1}. This implies that {α, -1} = {xα, -x}. We then have x = 1 or (x = -α and -α 2 = -1). It follows that x = 1 and ρ 1 ρ 2 .

We now prove the usual lemma computing the normal closure of A E 6 inside A E 7 .

Lemma 7.2.1. The normal closure

A E 6 A E 7 of A E 6 inside A E 7 is equal to A E 7
, where we identify A E 6 as a subgroup of A E 7 using the natural isomorphism from A E 6 to < S i , i ∈ [ [1,6]] >. Proof. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], we have

1 a 1 1 a 1 a 1 1 b 1 1 a 1 1 1 a 1 1 b 1 1 a 1 1 1 a 1 1 a 1 1 a 1 1 1 1 b 1 1 1 c 1 1 a 1 1 1 1 a 1 1 1 1 a 1 1 1 1 b 1 1 1 1 1 c 1 1 1 1 a 1 1 1 1 1 b 1 1 1 1 a 1 1 1 1 a 1 1 1 1 1 b 1 1 1 1 1 a 1 1 1 1 1 a 1 1 1 1 1 a 1 1 1 1 1 1 a 1 1 1 1 1 1 a 1 1 1 1 1 1 a 1 1 1 1 1 1 1
A E 7 =< S 1 S 3 S -1 1 , S 4 S -1 3 , S i S -1 1 , i ∈ [[1, 7]] > and A E 6 =< S 1 S 3 S -1 1 , S 4 S -1 3 , S i S -1 1 , i ∈ [[1, 6]] > .
This proves we only need to show that S 7 S -1

1 ∈ A E 6 A E 7 .
We have S 7 S 6 S 7 = S 6 S 7 S 6 , therefore S 7 = S 6 S 7 S 6 (S 6 S 7 ) -1 and

S 7 S -1 1 = S 6 S 7 S 6 S -1 1 (S 6 S 7 ) -1 = ((S 6 S -1 1 )(S 7 S -1 1 ))(S 6 S -1 1 )((S 6 S -1 1 )(S 7 S -1 1 )) -1 .
This concludes the proof.

Note now that there are no self-dual representations in type E 7 , they are all 2-colorable except for the two representations of dimension 512. We then have the following proposition.

Proposition 7.2.3. If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) then for any irreducible represen- tation ρ of H E 7 ,α , ρ(A E 7 ) SL nρ (q), where n ρ = dim(ρ). If F p ( √ α) = F q = F p (α) = F p (α + α -1
) then for any irreducible representation ρ of H E 7 ,α such that n ρ = 512, we have ρ(A E 7 ) SL nρ (q). We have ρ 512a (A E 7 ) SU 512 (q).

If

F q = F p (α) = F p (α + α -1 ) then for any irreducible representation ρ of H E 7 ,α , we have ρ(A E 7 ) SU nρ (q 1 2 ).
Proof. Assume first that F q = F p (α) = F p (α + α -1 ). Let ρ be an irreducible representation of H E 7 ,α of dimension different from 7 and 512. The associated E 7 -graph is then 2-colorable, therefore by Proposition 6.1, the image of A E 7 under ρ is inluded up to conjugation in GL nρ (q) even when F p ( √ α) = F p (α). By Table 7.2 and Proposition 7.1.3, ρ(A E 7 ) contains a natural SL r (q) for some r ≥ 6 whenever n ρ > 1. We can therefore apply Theorem 2.3.2. We get that ρ(A E 7 ) is a classical group over F q in a natural representation for some q dividing q. Since it contains a natural SL r (q), we get q = q and the representation is not unitary. By Corollary 7.2.2, it cannot preserve a bilinear form, therefore we get ρ(A E 7 ) SL nρ (q). We now have to consider the 7-dimensional representations ρ 7 a and ρ 7a = ρ 7 a . By Lemma 7.2.1 and Proposition 7.1.3, ρ 7a (A E 7 ) is normally generated by a natural SL 6 (q). It follows by Corollary 7.2.2 that it is an irreducible subgroup of SL 7 (q) generated by transvections. We can then apply Theorem 2.3.4 and the same arguments as above give ρ 7 a (q) SL 7 (q). This concludes the proof for the representations of dimension different from 512.

Consider now the representation

ρ 512 a . If F p ( √ α) = F p (α) then ρ(A E 7
) is included in SL 512 (q). By Proposition 7.1.3, it contains a natural SP 60 (q), therefore we can apply the above reasoning to get that ρ 512 a (A E 7 ) SL 512 (q). Assume now

F p ( √ α) = F p (α). We then have that X 2 -α is an irreducible polynomial of degree 2 of F p (α) such that F p ( √ α) = F p (α)/(X 2 -α). It follows that F p ( √ α
) is an extension of degree 2 of F p (α) and there exists a unique automorphism Ψ of F p ( √ α) which fixes F q pointwise and such that Ψ(

√ α) = - √ α.
Note that the restriction of ρ 512 a and ρ 512a to A E 6 are identical, therefore we cannot get any information from this restriction. Using CHEVIE [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF], we have

Tr(ρ 512 a (S 1 S -1 7 S 6 S -1 2 S 3 S 5 S -2 4 )) = 228 + 5α -4 -34α -3 + 104α -2 -189α -1 + √ α -1 - √ α -189α + 104α 2 -34α 3 + 5α 4 . It follows that ρ 512 a Ψ • ρ 512 a then √ α - √ α -1 = Ψ( √ α -1 - √ α) = √ α -1 - √ α. Hence 2 √ α = 2 √ α -1
and α = 1 which contradicts our assumptions on α. This proves that Ψ • ρ 512 a ρ 512a = ρ 512 a . It follows that ρ 512 a (A E 7 ) is included in SU 512 (q) up to conjugation in GL 512 (q 2 ). Since ρ 512a contains a natural SP 60 (q), we can apply Theorem 2.3.2 and ρ 512a is a classical group over F q for some q dividing q 2 . However q divides q because ρ 512 a (A E 7 ) contains a natural SP 60 (q), therefore q ∈ {q, q 2 }. We know ρ 512a does not preserve any non-degenerate bilinear form because ρ 512a|A E 7 ρ 512 a |A E 7 . It follows that up conjugation in GL 512 (q 2 ), we have ρ 512a (A E 7 ) ∈ {SL 512 (q), SU 512 (q

2 ), SL 512 (q 2 ), SU 512 (q)}. Assume we are in one the first two cases, we would then have using the trace of the same element as above that √ α -1 -√ α ∈ F q . This would imply that 1-α √ α ∈ F q , therefore √ α ∈ F q since α = 1 and α ∈ F q . The third possibility is excluded since SL 512 (q 2 ) cannot be injected inside SU 512 (q). We can then conclude that ρ 512a (A E 7 ) SU 512 (q).

Assume now that

F q = F p (α) = F p (α + α -1 ). Then X 2 -(α + α -1 )X + 1 is a F p (α + α -1 )- irreducible polynomial of degree 2, andF q = F p (α + α -1 )/(X 2 -(α + α -1 )X + 1
) is an extension of degree 2. There is a unique automorphism of degree 2 of F q . It fixes F p (α + α -1 ) pointwise and (α) = α -1 . We can then consider the extension

F p ( √ α) of F p (α). We have ( √ α) 2 = (α) = (α) = α -1 , therefore ( √ α) ∈ {- √ α -1 , √ α -1 }. It follows that 2 ( √ α) = √ α which implies that √ α ∈ F q , therefore F p (α) = F p ( √ α).
Let ρ be an irreducible representation of H E 7 ,α of degree greater than 1. We have that • ρ is an irreducible representation. Assume that the only representations of dimension n ρ are ρ and ρ . We then have • ρ ρ or • ρ ρ . The set of eigenvalues of ρ S 1 is {α, -1}. In the second case we would have ( t ρ(S i ) -1 ) conjugate to t ρ(S i ) -1 , therefore { (-1, (α -1 )} = {-1, α} = {-1, α -1 }, therefore α 2 = 1 which contradicts our assumptions. It follows that ρ

• ρ , and Lemma 3.2.5 gives

ρ(A E 7 ) ⊂ SU nρ (q 1
2 ). We can then apply the same reasoning as bove to conclude that ρ(A E 7 ) SU nρ (q 1 2 ). It only remains to consider the representations of dimension 21, 35, 105, 189, 210 and 280. By Proposition 7.1.3, we only need to check that the restrictions to H E 6 ,α are different for the other representations of the same dimension. This is true by Table 7.4.

Theorem 7.2.1. Assume

F q = F p ( √ α) = F p (α + α -1 ). Then the morphism from A E 7 to H E 7 ,α ρ irr
GL nρ (q) factorizes through the surjective morphism

Φ : A E 7 → SL 7 (q) × SL 15 (q) × SL 21 (q) 2 × SL 27 (q) × SL 35 (q) 2 × SL 56 (q) × SL 70 (q) × SL 84 (q) ×SL 105 (q) 3 × SL 120 (q) × SL 168 (q) × SL 189 (q) 3 × SL 210 (q) 2 × SL 216 (q) × SL 280 (q) 2
×SL 315 (q) × SL 336 (q) × SL 378 (q) × SL 405 (q) × SL 420 (q) × SL 512 (q).

Assume F p ( √ α) = F q = F p (α) = F p (α + α -1 ). Then the morphism from A E 7 to H E 7 ,α ρ irr
GL nρ (q) factorizes through the surjective morphism

Φ : A E 7 → SL 7 (q) × SL 15 (q) × SL 21 (q) 2 × SL 27 (q) × SL 35 (q) 2 × SL 56 (q) × SL 70 (q) × SL 84 (q)
×SL 105 (q) 3 × SL 120 (q) × SL 168 (q) × SL 189 (q) 3 × SL 210 (q) 2 × SL 216 (q) × SL 280 (q) 2 ×SL 315 (q) × SL 336 (q) × SL 378 (q) × SL 405 (q) × SL 420 (q) × SU 512 (q).

Assume F q = F p (α) = F p (α + α -1 ). Then the morphism from A E 7 to H E 7 ,α ρ irr GL nρ (q) factorizes through the surjective morphism

Φ : A E 7 → SU 7 (q 1
2 )×SU 15 (q

1 2 )×SU 21 (q 1 2 ) 2 ×SU 27 (q 1 2 )×SU 35 (q 1 2 ) 2 ×SU 56 (q 1
2 )×SU 70 (q

2 )×SU 84 (q

2 )

×SU 105 (q

1 2 ) 3 × SU 120 (q 1 2 ) × SU 168 (q 1
2 ) × SU 189 (q

1 2 ) 3 × SU 210 (q 1 2 ) 2 × SU 216 (q 1 2 ) × SU 280 (q 1 2 ) 2 ×SU 315 (q 1
2 ) × SU 336 (q

2 ) × SU 378 (q

1 2 ) × SU 405 (q 1
2 ) × SU 420 (q

1 2 ) × SU 512 (q 1
2 ).

Proof. The proof of this theorem is very similar to the proof of Theorem 7.1.1. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A E 7 is perfect. We have by Goursat's Lemma 3.3.1 that the morphism is surjective unless there exists two different representations ρ 1 and ρ 2 in the decomposition such that there exists a field automorphism Ψ verifying Ψ • ρ 1|A E 6 ρ 2|A E 6 . By Proposition 2.1.2, we have that Ψ(α + α -1 ) = α + α -1 . This shows that Ψ must be trivial over F p (α + α -1 ). It follows by the previous propositions that there are no such representations in the decompositions and the proof is concluded.

Type E 8

Let p be a prime different from 2, 3 and 5 and α ∈ F p of order not dividing 14, 18, 20, 24 and 30. We write F q = F p (α). There are 112 irreducible representations of H E 8 ,α , 18 of them are self-dual. For each self-dual representation, we have found a self dual-W-graph [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF]. Using the 2-coloring of those graphs, we have that all of the associated bilinear forms are symmetric. The highest dimensional representation is of dimension 7168. Definition 7.3.1. The Iwahori-Hecke algebra H E 8 ,α of type E 8 is the F q -algebra generated by S 1 , S 2 , S 3 , S 4 , S 5 , S 6 , S 7 , S 8 and the following relations 1. ∀i ∈ {1, 2, 3, 4, 5, 6, 7, 8}, (S i -α)(S i + 1) = 0.

S

1 S 3 S 1 = S 3 S 1 S 3 . 3. ∀i ∈ {2, 4, 5, 6, 7, 8}, S 1 S i = S i S 1 . 4. S 2 S 4 S 2 = S 4 S 2 S 4 . 5. ∀i ∈ {3, 5, 6, 7, 8}, S 2 S i = S i S 2 . 6. ∀i ∈ {3, 4, 5, 6, 7}, S i S i+1 S i = S i+1 S i S i+1 . 7. ∀i ∈ {5, 6, 7, 8}, S 3 S i = S i S 3 . 8. ∀i ∈ {6, 7, 8}, S 4 S i = S i S 4 . For σ in the Coxeter group E 8 , if σ = s i 1 . . . s i k is a reduced expression we set T σ = S i 1 . . . S i k .
This means we consider E 8 as in the CHEVIE package of GAP3 [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF] Proof. We will apply Proposition 2.2.4.

Let A = Z[ √ u ±1 ] and F = Q( √ u).
We have a symetrizing trace defined by τ (T 0 ) = 1 and τ (T σ ) = 0 for all σ ∈ E 8 \ {1 E 8 }. By [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF], H E 8 ,u is then a free symmetric F -algebra of rank 696729600. By [7] V.3. Corollary 1, A is integrally closed. Let θ be the ring homomorphism from A to L = F q defined by θ(u) = α and θ(k) = k. We know F H is split. The basis formed by the elements T σ , σ ∈ E 8 satisfies the conditions of Proposition 2.2.4. We work with the E 8 -graphs implemented in CHEVIE [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF] for the non-self dual E 8 -graphs. For the self-dual ones, we use the new ones we found [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF] which we denote by ρ if ρ is the initial representation, they verify the properties of Theorem 6.2 and they all verify ω(e x 1 )ω(e xn ) = -1 for any 2-coloring ω. For the 4536dimensional representation, we need to define two different E 8 -graphs over Q( √ u) because the bilinear form is different for p = 11, therefore we work with the one in CHEVIE. Going through the weights of the E 8 -graphs in CHEVIE [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF] and [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF], we see that most of the E 8graphs considered remain connected since the weights of the E 8 -graphs which do not belong to {2240 x , 2240 x , 4200 y , 4200 y , 4480 y , 5670 y , 7168 w } belong to the set composed of -24, -16, -10, -8, -20/3, -6, -5, -4, -10/3, -3, -8/3, -5/2, -2, -5/3, -3/2, -4/3, -5/4, -1, -8/9, -5/6, -3/4, -2/3, -5/8, -3/5, -1/2, -5/12, -2/5, -3/8, -1/3, -1/4, -1/8, 1/9, 1/4, 1/3, 3/8, 2/5, 1/2, 5/8, 2/3, 3/4, 5/6, 8/9, 1, 4/3, 3/2, 5/3, 2, 5/2, 8/3, 3, 10/3, 4, 5, 6, 8.

We now consider the remaining E 8 -graphs separately. Consider first 2240 x . We have four weighted edges which vanish when p = 7 and none otherwise. Using CHEVIE [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF], when p = 7, the edges 1572 → Consider now 4480 y . When p / ∈ {7, 11}, none of the weights vanish, therefore all the edges remain and the specialization is still connected. We give the proof of the connectedness for p ∈ {7, 11} in subsection 10.6.1 of the Appendix. The information is obtained using [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF].

Consider now 5670 y . When p = 7, none of the weights vanish, therefore all the edges remain. We give the proof of the connectedness for p = 7 in subsection 10.6.2 of the Appendix.

Consider now 7168 w . When p = 7, none of the weights vanish, therefore the E 8 -graph remains connected. We give the proof of the connectedness for p = 7 in subsection 10.6.3 of the Appendix.

It now only remains to check that the Schur elements associated to the specialized representations are in B and do not vanish under θ with B as in Proposition 2.2.4. The Schur elements are given in Tables 7.5, 7.6 and 7.7. They were obtained using Proposition 9.3.6 and Table E.7 of [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]. For a pair (ρ, ρ ) of representations, we only give the Schur element of one of the representations since the other is obtained by applying the involution

√ u → √ u -1
. The conditions on α and p imply that the Schur elements verify the right conditions and the proof is concluded.

The restriction table from H E 8 ,α to its subalgebra H E 7 ,α generated by S 1 , S 2 , S 3 , S 4 , S 5 , S 6 and S 7 which is naturally isomorphic to the Iwahori-Hecke of type E 7 with parameter α is then given by Tables 7.8 and 7.9. It is obtained using the CHEVIE package of GAP3 [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF]. They correspond in the generic case to the induction/restriction tables of the corresponding Coxeter groups.

Proposition 7.3.2. The restrictions to A E 8 of the representations afforded by those E 8 -graphs are absolutely irreducible and the representations of dimension greater than 1 are pairwise nonisomorphic.

Proof. As in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] Lemma 3.4, we only need to prove that A E 8 is generated by A E 7 and A E 8 . This true because s 8 = s 8 s -1

1 s 1 , s 8 s -1 1 ∈ A E 7 and s 1 ∈ A E 7 . We now prove the second part of the statement. Let ρ 1 and ρ 2 be two irreducible representations of H E 8 ,α such that ρ 1|A E 8 ρ 1|A E 8 . By Lemma 3.2.3,there exists a character ξ : A E 8 → F q such that ρ 1 ρ 2 ⊗ ξ. This means there exists x ∈ F q such that for all i ∈ [ [1,[START_REF] Brauer | On the modular characters of groups[END_REF]], ρ 1 (S i ) is conjugate to xρ 2 (S i ). We know for any representation ρ of dimension greater than 1, the set of eigenvalues of ρ is equal to {α, -1}. This implies that {α, -1} = {xα, -x}. We then have x = 1 or (x = -α and -α 2 = -1). It follows that x = 1 and ρ 1 ρ 2 .

We now prove a lemma computing the normal closure of A E 7 inside A E 8 as we did in the other types.

Lemma 7.3.1. The normal closure

A E 7 A E 8 of A E 7 inside A E 8 is equal to A E 8
, where we identify A E 7 as a subgroup of A E 8 using the natural isomorphism from

A E 7 to < S i , i ∈ [[1, 7]] >. Proof. By [36], we have A E 8 =< S 1 S 3 S -1 1 , S 4 S -1 3 , S i S -1 1 , i ∈ [[2, 8]] > and A E 7 =< S 1 S 3 S -1 1 , S 4 S -1 3 , S i S -1 1 , i ∈ [[2, 7]] > .
This proves we only need to show that S 8 S -1

1 ∈ A E 7 A E 8 .
We have S 8 S 7 S 8 = S 7 S 8 S 7 , therefore S 8 = S 7 S 8 S 7 (S 7 S 8 ) -1 and

S 8 S -1 1 = S 7 S 8 S 7 S -1 1 (S 7 S 8 ) -1 = ((S 7 S -1 1 )(S 8 S -1 1 ))(S 7 S -1 1 )((S 7 S -1 1 )(S 8 S -1 1 )) -1 .
This concludes the proof. We now determine the image of A E 8 inside each given representation. Note that the new self-dual W -graphs we found all verify ω(e 1 )ω(e n ) = 1 for any 2-coloring ω, therefore the bilinear forms appearing are self-dual. For the self-dual representation of dimension 4536, since we do not work with the new W -graph, we use the bilinear form which is availiable at [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF] and see that it is symmetric. The E 8 -graphs are all 2-colorable except for 4096 x , 4096 x , 4096 z and 4096 z .

Proposition 7.3.3. Assume F p ( √ α) = F q = F p (α) = F p (α + α -1 ). Let ρ be a representation of dimension n ρ associated to a W -graph. If ρ is not self-dual then ρ(A E 8 ) SL nρ (q) if n ρ = 4096 and ρ(A E 8 ) = SL nρ (q 2 ) if n ρ = 4096. If ρ is self-dual then ρ(A E 8 ) Ω + nρ (q). Assume F q = F p ( √ α) = F p (α) = F p (α + α -1 ). Let ρ be a representation of dimension n ρ associated to a W -graph. If ρ is not self-dual then ρ(A E 8 ) SL nρ (q). If ρ is self-dual then ρ(A E 8 ) Ω + nρ (q). Assume F q = F p (α) = F p (α+α -1 ). Let ρ be a representation of dimension n ρ associated to a W -graph. If ρ is not self-dual then ρ(A E 8 ) SU nρ (q 1 2 ). If ρ is self-dual then ρ(A E 8 ) Ω + nρ (q 1
2 ).

Proof. Assume first

F q = F p (α) = F p (α + α -1
). Let ρ be a non-self-dual representation of dimension n ρ associated to a W -graph. Assume first that the corresponding E 8 -graph is 2colorable. We can then consider the representation is defined over F q . If n ρ ≥ 28 then we can apply Theorem 2.3.2 since by Tables 7.8 and 7.9 and Proposition 7.2.3, ρ(A E 7 ) contains a natural SL 7 (q). It follows that ρ(A E 8 ) is a classical group over F q . We know that no non-degenerate bilinear form is preserved by this group because the representation is not self-dual. It cannot be unitary because it contains a natural SL 7 (q). We can then conclude that ρ(A E 8 ) = SL nρ (q). We know ρ 8z (A E 8 ) is an irreducible group normally generated by ρ 8z (A E 7 ). Since ρ 8z (A E 7 ) is a natural SL 7 (q), we have that ρ 8z (A E 8 ) is an irreducible subgroup of GL 8 (q) generated by transvections, therefore by Theorem 2.3.4, ρ 8z (A E 8 ) is isormorphic to SL 8 (q ), SU 8 (q

2 ) or SP 8 (q ) for some q dividing q. It contains a natural SL 7 (q), therefore q = q and ρ 8z (A E 8 ) = SL 8 (q). Assume now that the corresponding W -graph is not 2-colorable. The representations we have to consider are then the representations of dimension 4096. If F q = F p ( √ α) = F p (α) then we can apply the previous reasoning.

If F p ( √ α) = F q = F p (α) = F p (α + α -1 ) then X 2 -α is an irreducible polynomial over F q . We then have F q 2 = F p ( √ α) = F q /(X 2 -α)
and there is a unique field automorphism Ψ of degree 2 of F q 2 , ot fixes F q pointiwise and Ψ( √ α) = -√ α. By Proposition 7.2.3, for any representation ϕ of dimension different from 512 of A E 7 , we have Ψ • ϕ ϕ. We also have Ψ • ρ 512a ρ 512 a . By Table 7.9, we have that Ψ • ρ 4096x is not isomorphic to ρ 4096x or ρ 4096 z because otherwise we would have Ψ • ρ 512a ρ 512a , therefore ρ 512a ρ 512 a . We also have that Ψ • ρ 4096x is not isomorphic ρ 4096 x because otherwise, we would have Ψ • ρ 420a • ρ 420 a , therefore ρ 420a ρ 420 a . It follows that Ψ • ρ 4096x ρ 4096z . We know by the above reasonning that ρ 4096x (A E 8 ) is a classical group over F q for some q dividing q 2 . Furthermore, q divides q since it contains a natural SL 420 (q). It does not preserve any non-degenerate bilinear form since ρ is not self-dual. We cannot have ρ 4096x (A E 8 ) SL 4096 (q) because Ψ • ρ 4096x is not isomorphic to ρ 4096x . We also cannot have ρ 4096x (A E 8 ) SU 4096 (q) because ρ 4096x is not isomorphic to Ψ • ρ 4096 x . It follows that ρ 4096x (A E 8 ) SL 4096 (q 2 ). We get the same result for the remaining representations of dimension 4096 using ρ 4096 x ρ 4096x and ρ 4096z Ψ • ρ 4096x .

Assume now that F q = F p (α) = F p (α + α -1 ). Then the same argument as in the proof of Proposition 7.2.3 gives F q = F p ( √ α). Using the same reasoning as above, we only need to prove that for any representation ρ, we have • ρ ρ, where is the unique field automorphism of order 2 of F q . The polynomial X 2 -(α + α -1 )X + 1 is an irreducible F p (α + α -1 )-polynomial of degree 2, therefore we have (α) = α -1 . We know by Proposition 7.2.3 that • ϕ ϕ for any representation ϕ of A E 7 . Finally, by Table 7.8 and Table 7.9, no pair of distinct representations have the same restriction to A E 7 . The result follows.

We can now state the main theorem for type E 8 . Theorem 7.3.1. We let A be a set of representatives of the irreducible 2-colorable non selfdual representations for the equivalence relation ρ ≈ ϕ if ρ = ϕ and B be the set of irreducible self-dual representations.

Assume

F p ( √ α) = F q = F p (α) = F p (α + α -1 ). Then the morphism from A E 8 to H E 8 ,α ρ irr
GL nρ (q) factorizes through the surjective morphism

Φ : A E 8 → ρ∈A SL nρ (q) × SL 4096 (q 2 ) × ρ∈B Ω + nρ (q). Assume F q = F p ( √ α) = F p (α) = F p (α + α -1 ). Then the morphism from A E 8 to H E 8 ,α ρ irr
GL nρ (q) factorizes through the surjective morphism

Φ : A E 8 → ρ∈A SL nρ (q) × SL 4096 (q) 2 × ρ∈B Ω + nρ (q). Assume F q = F p (α) = F p (α + α -1 ). Then the morphism from A E 8 to H E 8 ,α ρ irr GL nρ (q)
factorizes through the surjective morphism

Φ : A E 8 → ρ∈A SU nρ (q 1
2 ) × SU 4096 (q

1 2 ) × ρ∈B Ω + nρ (q 1
2 ).

Proof. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A E 8 is perfect. We can then apply exctly the same arguments as in the proof of of Proposition 7.1.1.

Chapter 8

Type H

In this section, we determine the images of the Artin groups in types H 3 and H 4 . The main difference with the other types is that in the generic case, some of the irreducible representations are defined over

Q[ √ 5]( √ u) but not Q( √ u).
We therefore need arguments as in the dihedral cases to understand the field extensions. Lemma 8.1.2 summarizes the information we need about the field extensions. The proof is then similar to the proof in the other types. We first determine the image inside each irreducible representation and then recover the full image using Goursat's Lemma. In order to determine the image inside each representation, we will use inductive arguments using the image in type I 2 (5) for type H 3 and the image in type H 3 for type H 4 . The main results in this section are in Theorems 8.1.1 and 8.5.1. The 8dimensional irreducible representations in type H 4 use additional assumptions on the order of α which might not be necessaray. The proof is highly computational, therefore it might be complicated to ommit those assumptions on the order of α. The image in the product of those representations gives rise to a nice description of the Spin + 8 (q) group as can be seen in Proposition 8.4.3. This section uses many results from chapter 6. We proved Conjecture 6.1 in type H 4 and the self-dual H 4 -graphs we obtained are available in section 10.2 of the Appendix and at [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF].

Type H 3

Let p be a prime number, p / ∈ {2, 5} and α ∈ (F p ) × such that the order of α does not divide 20 and does not belong to {1, 2, 3, 4, 5, 6, 10}. Let ξ ∈ F p be a primitive fifth-root of unity. We set F q = F p (α) and F r = F p (α, ξ + ξ -1 ). Definition 8.1.1. The Iwahori-Hecke H H 3 ,α is the F q -algebra generated by the generators S 1 , S 2 , S 3 and the following relations : Under the conditions we assumed on α, the Iwahori-Hecke algebra is split semi-simple and the models given by specialization of the W -graphs are irreducible.

1. ∀i ∈ {1, 2, 3}, (S i -α)(S i + 1) = 0. 2. S 1 S 2 S 1 S 2 S 1 = S 2 S 1 S 2 S 1 S 2 , 3. S 1 S 3 = S 3 S 1 , 4. S 2 S 3 S 2 = S 3 S 2 S 3 . For σ in the Coxeter group H 3 , if σ = s i 1 . . . s i k is a reduced expression we set T σ = S i 1 . . . S i k . 1 r : (Φ 3 2 Φ 3 Φ 5 Φ 6 Φ 10 )(u). 3 s : 5 + √ 5 2 Φ 3 2 (u)Φ 5 (u)Φ 10 (u) uΦ 5,b (u)Φ 10,b (u) . 3 s : 5 - √ 5 2 Φ 3 2 (u)Φ 5 (u)Φ 10 (u) uΦ 5,b (u)Φ 10,b (u) . 5 r : 1 u 2 Φ 3 2 (u)Φ 3 (u)Φ 6 (u).
Proof. We want to apply Proposition 2.2.4.

Let A = Z[ 1+ √ 5 2 ][ √ u ±1 ], B = Frac(A) and F = Q[ √ 5]( √ u).
We have a symetrizing trace defined by τ (T 0 ) = 1 and τ

(T σ ) = 0 for all σ ∈ H 3 \ {1 H 3 }. H H 3 ,u is then a free F -algebra of rank 120. A is an integrally closed integral domain because Z[ 1+ √ 5
2 ] is integrally closed (see [7] V.3. Corollary 1 and [39] Theorem 9.20.). Let θ be the ring homomorphism from A to L = F q defined by θ( 1+

√ 5 2 ) = ξ + ξ -1 + 1, θ(u) = α and θ(k) = k.
We know F H is split. The basis formed by the elements T σ , σ ∈ H 3 verifies the conditions of the Proposition 2.2.4.

All the W -graphs are connected and remain connected after we specialize the weights because none of them vanishes under θ. Indeed, if they were not connected then they would afford reducible representations over Q( √ 5)( √ u). We now only need to check that the Schur elements associated to these irreducible representations are in B and don't vanish under θ. For n ∈ N, Φ n is the n-th cyclotomic polynomial and Φ 5,a (u

) = u 2 + 1+ √ 5 2 u + 1, Φ 5,b (u) = u 2 + 1- √ 5 2 u + 1, Φ 10,a (u) = u 2 -1+ √ 5 2 u + 1 and Φ 10,b (u) = u 2 + √ 5-1 2 u + 1.
If χ is an irreducible character then the character χ associated to the dual representation of χ has a Schur element c χ = a(c χ ), where a is the involution of

Q( √ 5)( √ u) sending √ u to √ u -1 . We define the field automorphism of Q( √ 5), written x → x by √ 5-1 2 = -1- √ 5 2
and q = q for all q ∈ Q. The Schur elements are given in Table 8.1 (obtained using Table E.2. of the Appendix and corollary 9.3.6 of [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF])

Since α is of order not dividing 20 and different from 6 and p = 2, we only need to check that θ( 5+ √ 5

2 ) = 0, θ( 5-

√ 5 2 ) = 0, α 2 ± (ξ + ξ -1 )α + 1 = 0, α 2 ± (ξ + ξ -1 + 1)α + 1 = 0, α 2 ∓ (ξ 2 + ξ -2 )α + 1 = 0.
We have θ( 5+

√ 5 2 ) = ξ + ξ -1 + 3 = -ξ 2 -ξ -2 + 2 = -(ξ -ξ -1 ) 2 = 0 and θ( 5- √ 5 2 ) = -ξ -ξ -1 + 2 = -(ξ 2 -ξ -2 ) 2 = 0.
We also have α 2 + (ξ + ξ -1 )α + 1 = (α + ξ)(α + ξ -1 ) = 0 because α is of order not dividing 20. In the same way, α 2 -(ξ + ξ -1 )α + 1 = (α -ξ)(α -ξ -1 ) = 0. The last two inequalities are shown in exactly the same way because the order of ξ is the same as the order of ξ 2 .

We now show a lemma on the Artin groups A H 3 and A I 2 (5) which will be useful later on. Lemma 8.1.1. We write A H 3 =< t, s 1 , s 2 , ts 1 ts 1 t = s 1 ts 1 ts 1 , s 1 s 2 s 1 = s 2 s 1 s 2 , ts 2 = s 2 t > and A I 2 (5) its subgroup generated by t and s 1 . It is naturally isomorphic to the Artin group of type I 2 (5).

F p (α, ξ + ξ -1 ) F p (α + α -1 , ξ + ξ -1 )
F p (α)

F p (α + α -1 )
Figure 8.2: Field extensions in types H 3 and H 4 .1

Proof. The Hasse diagram representing the inclusions between fields in this case can be found in Figure 8.2. Note that when 1 ∼ 2, we have

F p (α + α -1 , ξ + ξ -1 ) = F p (α + α -1 ).
Assume by contradiction that 1 ∼ 2 and

L 1 = F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1 ) = L 2 . Since 1 ∼ 2, we have L 2 = F p (α + α -1 , ξ + ξ -1 ) = F p (α + α -1 ) = L 4 , because otherwise, Φ 1,2 would stabilize F p (α + α -1 , ξ + ξ -1 ). It follows that [L 1 : L 2 ] = [L 2 : L 4 ] = 2 and [L 1 : L 4 ] = 4. Let L 3 = F p (α). We have [L 1 : L 3 ] ≤ 2, [L 3 : L 4 ] ≤ 2 and 4 = [L 1 : L 4 ] = [L 1 : L 3 ][L 3 : L 4 ]. This implies that [L 1 : L 3 ] = [L 3 : L 4 ] = 2. By uniqueness of the subfield of degree 2 of L 1 , we have L 2 = L 3 . It follows that α ∈ L 2 = F p (α + α -1 , ξ + ξ -1 ), therefore we have that L 2 = L 1 which is a contradiction. Assume now that F p (α + α -1 , ξ + ξ -1 ) = F p (α + α -1
). We then have that there exists an automorphism Φ of order 2 of F q permuting the roots of X 2 + X -1. We have (ξ

2 + ξ -2 ) 2 + ξ 2 + ξ -2 -1 = ξ + ξ -1 + 2 + ξ 2 + ξ -2 -1 = 0 so Φ(ξ 2 + ξ -2 ) = ξ + ξ -1
. This proves by definition of ∼ that 1 ∼ 2 Assume now 1 ∼ 2 and F p (α) = F p (α + α -1 ). We have that Φ 1,2 stabilizes F p (α + α -1 ) and α ∈ F p (α + α -1 , therefore we have that Φ 1,2 (α) = α.

Assume now 1 ∼ 2 and F p (α) = F p (α + α -1 ). Using the notations of the first part of the proof, we here have that [L 2 : L 4 ] = [L 3 : L 4 ] = 2. This implies that the unique automorphism of order 2 of

L 3 is Φ 1,2 and X 2 -(α + α -1 )X + 1 is an irreducible polynomial of L 4 [X]. It follows that Φ 1,2 (α) = α -1 . Assume 1 2. We then have F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1 ). This implies that α / ∈ F p (α + α -1 , ξ + ξ -1 ). It follows that F p (α) = F p (α, ξ + ξ -1
). We then have by the Hasse diagram in Figure 8.2 that F p (α) = F p (α + α -1 ).

We now determine the image in each of those representations before determining the image in the full Iwahori-Hecke algebra.

Proposition 8.1.3. If F r = F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1 ), then we have ρ 3 s (A H 3 ) SL 3 (r) and ρ 3 s (A H 3 ) = SL 3 (r). If F r = F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1 ), then we have ρ 3 s (A H 3 ) SU 3 (r 1
2 ) and

ρ 3 s (A H 3 ) SU 3 (r 1 2 ). If 1 ∼ 2 and F p (α) = F p (α + α -1 ), then Φ 1,2 • ρ 3 s |A H 3 ρ 3 s |A H 3 . If 1 ∼ 2 and F p (α) = F p (α + α -1 ), then Φ 1,2 • ρ 3 s |A H 3 ρ 3s|A H 3 . If F p (α) = F p (α + α -1 ), then r = q 2 . If F p (α) = F p (α + α -1 ), then r = q.
Proof. First note that by Proposition 6.1 above and Proposition 4.1. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], we can see this representation as a representation over

F q even if F p ( √ α, ξ + ξ -1 ) = F r = F p (α, ξ + ξ -1 ). Let G =< α -1 ρ 3 s (S 1 ), α -1 ρ 3 s (S 2 ), α -1 ρ 3 s (S 3 ) >.
Let us show that it is primitive. If G was imprimitive, we could write

F n r = V 1 ⊕ V 2 ⊕ ... ⊕ V r
, where for all i and for all g ∈ G, there exists a j such that g.

V i = V j . Since R 1 is irreducible, there exists i ∈ [[1, 3]] such that -R 1 (S i ).V 1 = V 1 . Assume there exists i such that -R 1 (S i ).V 1 = V 1 . Up to reordering, we have V 2 = -R(S i ).V 1 . If dim(V 1 ) ≥ 2 then H -R 1 (S i ) (the hyperplane fixed by -R 1 (S i )) has a non- empty intersection with V 1 , therefore V 1 ∩V 2 = ∅ which is a contradiction, therefore dim(V 1 ) = 1.
This reasoning is valid for any V i , therefore they are all one-dimensional. Let x ∈ V 1 be a nonzero vector, it can be written in a unique way as

x = x 1 + x 2 with x 1 ∈ ker(R 1 (S i ) + α) and x 2 ∈ H -R 1 (S i ) . We then have that -R 1 (S i )x = -αx 1 + x 2 and -R(S i )(-R(S i )x) = α 2 x 1 + x 2 = α(x 1 + x 2 ) + (1 -α)(-αx 1 + x 2 ) ∈ V 1 ⊕ V 2 . Since α /
∈ {0, 1} this contradicts the fact that there exists j such that -R(S i ).V 2 = V j . This shows that G is primitive. By Wagner's theorem on groups generated by reflections ([45] and Theorem 2.3. of [START_REF] Malle | Inverse Galois theory[END_REF]), since G is primitive and is generated by pseudo-reflections, there exists r dividing r such that

SL 3 (r ) ≤ G ≤ GL 3 (r ) or SU 3 (r 1 2 ) ≤ G ≤ GU 3 (r 1
2 ). We now show that r must be equal to r. We have det(α

-1 ρ 3 s (S 1 )) = -α -1 , therefore α ∈ F r . We also have Tr((α -1 ρ 3 s (S 1 )α -1 ρ 3 s (S 2 )) = 1 -2α -1 + α -1 ((ξ + ξ -1 + 1) 2 ) ∈ F r , therefore (ξ + ξ -1 + 1) 2 ∈ F r . We have (ξ + ξ -1 + 1) 2 = ξ 2 + ξ -2 + 1 + 2ξ + 2ξ -1 + 2 = ξ + ξ -1 + 2, therefore it follows that ξ + ξ -1 ∈ F r . Since F r = F p (α, ξ + ξ -1 ), we have r = r. Assume now F r = F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1 ). If G ≤ GU 3 (r 1
2 ), we let be the automorphism of order 2 of F r . We then have (det(α -1 ρ 3 s (S 1 )) = det( t (α -1 ρ 3 s (S 1 )) -1 ), therefore (-α -1 ) = -α and (α + α -1 ) = α + α -1 .

We also have (Tr((α

-1 ρ 3 s (S 1 )α -1 ρ 3 s (S 2 )))) = Tr( t (α -1 ρ 3 s (S 1 )α -1 ρ 3 s (S 2 )) -1 ), therefore (1- 2α -1 + α -1 (ξ + ξ -1 + 1) 2 ) = 1 -2α + α(ξ + ξ -1 + 1) 2 . This implies that ((ξ + ξ -1 + 1) 2 ) = (ξ + ξ -1 + 1) 2 , therefore (ξ + ξ -1 ) = ξ + ξ -1 . This would imply that ξ + ξ -1 ∈ F r 1 2 , therefore F r = F r 1 2 which is absurd. It follows that SL 3 (r) ≤ G ≤ GL 3 (r), therefore ρ 3 s (A H 3 ) = [ρ 3 s (A H 3 ), ρ 3 s (A H 3 )] = [ G, G] = SL 3 (r). Assume now F r = F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1
). The unique automorphism of order 2 of F r stabilizes F p and verifies (α) = α -1 and (ξ + ξ -1 ) = ξ + ξ -1 . By Lemma 2.4. of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], we only need to show that ρ 3 s

• ρ 3s in order to prove that

ρ 3 s (A H 3 ) ≤ GU 3 (r 1
2 ). Once this is shown, we will have that there exists P ∈ GL 3 (r) such that for all i ∈ {1, 2, 3}, P ρ 3 s (S i )P -1 = ( t ρ 3 s (S i ) -1 ). We then have that for all i ∈ {1, 2, 3}, P (α

-1 ρ 3 s (T s ))P -1 = α -1 ( t ρ 3 s (S i ) -1 ) = (α -1 ) -1 ( t ρ 3 s (S i ) -1 ) = ( t (α -1 ρ 3 s (S i )) -1 ), therefore G ≤ GU 3 (r 1 2 ). We have (α) = α -1 , therefore ( √ α 2 ) = ( √ α) 2 = √ α -2 and ( ( √ α) - √ α -1 )( ( √ α) + √ α -1 ) = 0. It follows that ( √ α) ∈ {± √ α -1 }.
We will show that ρ 3 s • ρ 3 s in both cases and the proof will be completed.

Assume (

√ α) = √ α -1 . Let P =     -α+1 √ α λ 0 λ -α+1 √ α 1 0 1 -α+1 √ α   
 , we have for all i ∈ {1, 2, 3},

P ρ 3 s (S i )P -1 = ( t ρ 3 s (S i ) -1
). Note that the matrix is invertible in F r because det(P ) =

-(α+1)(-αλ 2 +α 2 +α+1) α √ α = 0 would imply -ξ 2 -ξ -2 +1 = λ 2 = α+α -1 +1 and then α ∈ {-ξ 2 , -ξ -2 } which is absurd. Assume ( √ α) = - √ α -1 . Let P =     1 -λ √ α α+1 0 λ √ α α+1 -1 √ α α+1 0 - √ α α+1 1     .
It verifies the same conditions as in the previous case and the determinant is again non-zero, therefore the proof is completed for the W -graph 3 s .

The proof is identical for 3 s . Assume now 1 ∼ 2.

Assume first that F p (α) = F p (α + α -1 ). By Lemma 8.1.2, we have that Φ

1,2 (α) = α and Φ 1,2 (λ) = Φ 1,2 (ξ + ξ -1 + 1) = ξ 2 + ξ -2 + 1 = -ξ -ξ -1 = 1 -λ. It follows by Proposition 2.1.2 that there exists a character η : A H 3 → F × r such that Φ 1,2 •ρ 3 s is an irreducible representation of H H 3 ,α . We have ρ 3 s |A I 2 (5) ρ 3s|A I 2 (5) ρ 1 × 1, ρ 3 s |A I 2 (5) ρ 3s|A I 2 (5) ρ 2 × 1, where 1 represents the trivial representation and Φ 1,2 • ρ 2 ρ 1 . It follows that either Φ 1,2 • ρ 3 s |A H 3 ρ 3 s |A H 3 or Φ 1,2 • ρ 3 s |A H 3 ρ 3s|A H 3 . We have Tr(ρ 3 s (S 1 S 2 S -2 3 )) = α 2 -α + 1 -λα -1 . It follows that Φ 1,2 (Tr(ρ 3 s (S 1 S 2 S -2 3 ))) = α 2 -α + 1 + (λ -1)α -1 .
We have Tr(ρ

3 s (S 1 S 2 S -2 3 )) = α 2 -α + 1 + (λ -1)α -1 and Tr(ρ 3s (S 1 S 2 S -2 3 )) = (λ -1)α + 1 - α -1 + α -2 . Assume by contradiction that Φ 1,2 • ρ 3 s |A H 3 ρ 3s|A H 3 .
We then have

α 2 -α + 1 + (λ -1)α -1 = (λ -1)α + 1 -α -1 + α -2 α 2 -α -2 = λα -λα -1 (α -α -1 )(α + α -1 ) = λ(α -α -1 ) α + α -1 = -ξ 2 -ξ -2 (αξ 2 + 1)(ξ -2 + α -1 ) = 0.
It then follows that α 10 = 1 which contradicts our assumptions on the order of α. This proves that Φ

1,2 • ρ 3 s |A H 3 ρ 3 s |A H 3 .
Assume now that F p (α) = F p (α + α -1 ). By Lemma 8.1.2, we have that Φ 1,2 (α) = α -1 and Φ 1,2 (λ) = 1 -λ. By the same arguments as in the previous case, we have that either

Φ 1,2 • ρ 3 s |A H 3 ρ 3 s |A H 3 or Φ 1,2 • ρ 3 s |A H 3 ρ 3s|A H 3 . Assume by contradiction that Φ 1,2 • ρ 3 s |A H 3 ρ 3 s |A H 3 . We then have α -2 -α -1 + 1 + (λ -1)α = Φ 1,2 (Tr(ρ 3 s (S 1 S 2 S -2 3 ))) = Tr(ρ 3 s (S 1 S 2 S -2 3 )) = α 2 -α + 1 + (λ -1)α -1 .
The same computation as in the previous case shows this is a contradiction and the proof is thus completed.

We now show that r = q 2 if F p (α) = F p (α + α -1 ) and r = q if F p (α) = F p (α + α -1 ). If 1 ∼ 2, then the result follows by Lemma 8.1.2. Assume now that 1 2. Note that

F p (α) = F p (α+α -1 , ξ+ξ -1 ) then F p (α) = F p (α, ξ+ξ -1
) therefore the two former fields cannot be simultaneously subfields of degree 2 of the latter. This implies that [F p (α, ξ + ξ -1 ) :

F p (α)] ≤ 2. By Lemma 8.1.2, we have F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1
). The result then follows from the fact that [F p (α, ξ + ξ -1 ) :

F p (α)] = 2 [F p (α) : F p (α + α -1 )] . Proposition 8.1.4. If F q = F p ( √ α) = F p (α) = F p (α + α -1 ), then we have ρ 4 r (A H 3 ) = SL 4 (q). Assume that F q = F p ( √ α) = F p (α) = F p (α + α -1
) and is the unique automorphism of order 2 of F q .

If (

√ α) = √ α -1 , then we have ρ 4 r (A H 3 ) SU 4 (q 1 2 ). If ( √ α) = - √ α -1 , then we have ρ 4 r (A H 3 ) SL 4 (q 1
2 ).

If F q = F p ( √ α) = F q = F p (α) = F p (α + α -1 ), then we have ρ 4 r (A H 3 ) = SU 4 (q). Proof. Let β = ξ + ξ -1 and M =        1 - √ α(3+2β) (α+1)(β+1) β+2 β+1 - √ α(2+β) (α+1)(β+1) -1 α+1 1 √ α -2+β (α+1)(1+β) 1 √ α(β+1) 1 - √ αβ 2 α+1 -β √ αβ (α+1) -1 α+1 1 √ α β α+1 -1 √ αβ        , we have det(M ) = 5(β + 1)Φ 5 (α) α(α + 1) 4 (1 + β) 2 β .
Let ρ 1 and ρ 2 be the two 2-dimensional irreducible representations of

H I 2 (5) =< S 1 , S 2 > defined by ρ 1 (S 1 ) = -1 0 1 α , ρ 1 (S 2 ) = -1 0 1 α , ρ 2 (S 1 ) = α α(2 + β) 0 -1 and ρ 2 (S 2 ) = α αβ 2 0 -1 .
We then have

M (ρ 4 r (S 1 ))M -1 = ρ 1 (S 1 ) 0 0 ρ 2 (S 1 ) , M (ρ 4 r (S 2 ))M -1 = ρ 1 (S 2 ) 0 0 ρ 2 (S 2
) .

We set

H = (ρ 1 × ρ 2 )(A I 2 (5) ) and G = ρ 4 r (A H 3 ). Assume first that F q = F p ( √ α) = F p (α) = F p (α + α -1
) and 1 ∼ 2. By Theorem 5.1.3, we have ρ(A I 2 (5) ) SL 2 (q 2 ). More precisely, we have that

ρ 4 r (A I 2 (5) ) = { N 0 0 Φ 1,2 (N ) , N ∈ SL 2 (q 2 )} ⊂ M GM -1
. Consider now the maximal subgroups of SL 4 (q). They are given in Tables 8.8 and 8.9 of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF]. Since our group is irreducible, we can remove the groups of class C 1 . We can also remove the groups of class C 5 because the field generated by the traces of the elements of our group if F q . Note that ξ + ξ -1 / ∈ F q so q is an odd power of p and the representation is not self-dual. It follows that M GM -1 is included in no maximal subgroup of C 8 . The only remaining maximal subgroups are listed below with their order or a quantity their order divides.

1. (q -1) 3 .S 4 , 24(q -1) 3 2. SL 2 (q) 2 : (q -1).2, 2q 2 (q 2 -1) 2 (q -1) 3. SL 2 (q 2 ).(q + 1).2, 2(q + 1)q 2 (q 4 -1)

4. (4 • 2 1+4 ) . S 6 , 92160 5. (4 • 2 1+4 ) . A 6 , 46080 6. (q -1, 4) • 2 . P SL 2 (7), 1344 7. (q -1, 4) • 2 . A 7 , 20160 8. (q -1, 4) • 2 . P SU 4 (2), 103680.
By the conditions on the order of α, we have q ≥ 8 and q = 11. Since p = 2 and q is an odd power of p, we have q ≥ 13. It follows that |H| = q 2 (q 4 -1) ≥ 4826640. This implies that cases 4, 5, 6, 7 and 8 are excluded. If we were in case 1, we would have that q 2 divides 24 which is absurd since p = 2. In case 2, we would have that q 2 + 1 divides 2(q 2 -1)(q -1) = 2(q 2 + 1)(q -1) -4(q -1), therefore we would have that q 2 + 1 divides 4(q -1) < q 2 + 1 since q ≥ 13.

The only remaining case is case 3. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A H 3 is perfect, it would then follow that G SL 2 (q 2 ). This would imply that G = ρ 4 r (A I 2 (5) ). The coefficient on the first row and third column of

M ρ 4 r (S 1 S -1 3 )M -1 is equal to 1 5 (-α -3 √ α -1)β -(3α -4 √ α -3). If this is equal to zero and -α -3 √ α -1 = 0, then β = -3α+4 √ α+3 α+3 √ α+1 ∈ F p which is false. If the coefficient is equal to zero and -α -3 √ α -1 = 0 then α + 1 = -3 √ α, therefore the coefficient is equal to 5 √ α = 0 which is absurd. This proves that M ρ 4 r (S 1 S -1 3 )M -1 / ∈ ρ 4 r (A I 2 (5) ). This proves that if F q = F p ( √ α) = F p (α) = F p (α+α -1 ) and 1 ∼ 2 then G = SL 4 (q). Assume now that F q = F p ( √ α) = F p (α) = F p (α + α -1
) and 1 2. By Theorem 5.1.3, we have ρ(A I 2 (5) ) SL 2 (q ) × SL 2 (q ), where

F q = F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1 )
. By Lemma 8.1.2, we have that F q = F q . By Lemma 8.1.1, G is generated by transvections. We can then apply Theorem 2.3.4 to get that G ∈ {SL 4 (q), SU 4 (q 1 2 ), SP 4 (q)} for some q dividing q. We have q = q because G contains a natural SL 2 (q). We have that G cannot be preserved by any non-degenerate bilinear form because ρ 4 r is not self-dual. Assume by contradiction that G SU 4 (q

2 ). There exists an automorphism of order 2 of F q such that • ρ 4 r ρ 4 r . By Lemma 3.2.3, we can apply Proposition 2.1.2 to . It follows that we have that (α) ∈ {α, α -1 }, therefore (α + α -1 ) = α + α -1 . This implies that is of order 1 since F q = F p (α + α -1 ). This is a contradiction, therefore G = SL 4 (q).

Assume now

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and 1 ∼ 2. We have Φ 1,2 (α) = α -1 , therefore Φ 1,2 ( √ α) ∈ {± √ α -1 }. By Theorem 5.1.3, we have H SL 2 (q). Assume first Φ 1,2 ( √ α) = √ α -1 . Let R =         α- √ α+1 √ α 1 1 √ α α+1 1 α+1 √ α α+ √ α+1 α+1 1 1 α+ √ α+1 α+1 α+1 √ α 1 √ α α+1 1 1 α- √ α+1 √ α        
. We have

det(R) = Φ 5 ( √ α)Φ 10 ( √ α) 3 α 2 (α+1) 4 = 0 and (R -1 ρ 4 r (S 1 )R, R -1 ρ 4 r (S 2 )R, R -1 ρ 4 r (S 3 )R) = ( ( t ρ 4 r (S 1 ) -1 ), ( t ρ 4 r (S 2 ) -1 ), ( t ρ 4 r (S 3 ) -1 )).
It follows by Lemma 3.2.4 that up to conjugation in GL 4 (q), we have G ≤ SU 4 (q 1 2 ). We know G is irreducible so it is included in no subgroup of class C 1 . We list below the remaining maximal subgroups of SU 4 (q 1 2 ) with their order or a quantity their order divides. The tables are obtained using Table 8.10 and 8.11 of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF].

Assume now Φ 1,2 ( √ α) = - √ α -1 . Let R =         √ α α+1 1 1 α- √ α+1 √ α 1 α+ √ α+1 α+1 α+1 √ α 1 1 α+1 √ α α+ √ α+1 α+1 1 α- √ α+1 √ α 1 1 √ α α+1         . We have det(R ) = det(R) = 0 and (R -1 ρ 4 r (T )R , R -1 ρ 4 r (S 1 )R , R -1 ρ 4 r (S 2 )R ) = (-α (ρ 4 r (T )), -α (ρ 4 r (S 1 )), -α (ρ 4 r (S 2 ))).
It follows by Lemma 3.2.4 that up to conjugation in GL 4 (q), we have G ≤ SL 4 (q 1 2 ). We have here q ≥ 49 so |H| ≥ 117600. We can then apply the same reasoning as for

F q = F p ( √ α) = F p (α) = F p (α + α -1
) and 1 ∼ 2 to prove that G SL 4 (q

2 ).

Assume

F q = F p ( √ α) = F p (α) = F p (α + α -1
) and 1 2. Let be the unique automorphism of order 2 of F q . By the same arguments as in the previous case, if

( √ α) = √ α -1 then we have that up to conjugation, G ≤ SU 4 (q 1 2 ). If ( √ α) = - √ α -1 , then we have that, up to conjugation, G ≤ SL 4 (q 1
2 ). We have

H SU 2 (q 1 2 ) × SU 2 (q 1
2 ). By Lemma 8.1.1, we have that G is normally generated by H. It follows that G is an irreducible subgroup of GL 4 (q) generated by transvections.

Assume (

√ α) = √ α -1 . By Theorem 2.3.4, we have that G is conjugate in GL 4 (q) to SU 4 (q 1
2 ), SP 4 (q ) or SL 4 (q ) for some q dividing q. We know that G contains a natural SU 2 (q 1 2 ). By Lemma 3.3.3, we have that q 1 2 divides q. ρ 4 r is not self-dual so the symplectic case is excluded. The groups SL 4 (q) and SL 4 (q 1 2 ) are not included in SU 4 (q 1 2 ), therefore we have G SU 4 (q

1 4 ) or G SU 4 (q 1
2 ). The natural SU 2 (q 1 2 ) cannot be included in SU 4 (q 1 4 ), therefore we have that G SU 4 (q

1 2 ). Assume now that ( √ α) = - √ α -1
. By Theorem 2.3.4 and the facts that G contains a natural SU 2 (q

2 ) and ρ 4 r not self-dual, we have that G is conjugate in GL 4 (q) to SU 4 (q

2 ), SL 4 (q 1 2 ) or SL 4 (q). G is conjugate to a subgroup of SL 4 (q

2 ) therefore we have that G SL 4 (q

1 2 ). Assume now that F q 2 = F p ( √ α) = F p (α) = F p (α+α -1
) and 1 ∼ 2. By Lemma 8.1.2, we have that the unique automorphism of order 2 of

F q 2 is Φ 1,2 and Φ 1,2 (α) = α. The polynomial X 2 -α is irreducible in F q [X], therefore we have that Φ 1,2 ( √ α) = - √ α. Let Q = E 1,4 +E 2,3 +E 3,2 +E 4,1 , we have that (Qρ 4 r (S 1 )Q -1 , Qρ 4 r (S 2 )Q -1 , Qρ 4 r (S 3 )Q -1 ) = (-α ( t (ρ 4 r (S 1 ) -1 )), -α ( t (ρ 4 r (S 2 ) -1 )), -α ( t (ρ 4 r (S 3 ) -1 )).
It follows by Lemma 3.2.4 that up to conjugation in GL 4 (q 2 ), we have that G ≤ SU 4 (q). We have H SL 2 (q 2 ). We have here Φ 1,2 (α) = α, therefore α q-1 = 1. This implies that q ≥ 8 and q = 11. We also have that q is not a square since ξ + ξ -1 / ∈ F q . It follows that q ≥ 13 and we can apply the same reasonning as when

F q = F p ( √ α) = F p (α) = F p (α + α -1 ), 1 ∼ 2 and Φ 1,2 ( √ α) = √ α -1 to conclude that G SU 4 (q). Assume now that F q 2 = F p ( √ α) = F p (α) = F p (α + α -1
) and 1 2. The unique automorphism of order 2 of F q still verifies ( √ α) = -√ α, therefore we have that G ≤ SU 4 (q) up to conjugation in GL 4 (q 2 ). By Lemma 8.1.2 and Theorem 5.1.3, we have that H SU 2 (q) × SU 2 (q). By Lemma 8.2.1, we have that A H 3 is normally generated by A I 2 (5) . It follows that ρ 4 r (A H 3 ) is generated by transvections. The group G contains a natural SU 2 (q), therefore by Lemma 3.3.3, we have that the field generated by the traces of the elements of G contains F 1 2 q . We also have that G preserves no non-degenerate bilinear form because ρ 4 r is not self-dual. By Theorem 2.3.4, we have that G is conjugate in GL 4 (q 2 ) to SL 4 (q), SU 4 (q) or SL 4 (q 2 ). We can then conclude that G SU 4 (q) conclude from the fact that G ≤ SU 4 (q). Proposition 8.1.5. If F q = F p (α) = F p (α + α -1 ), then we have ρ 5r (A H 3 ) SL 5 (q). If

F q = F p (α) = F p (α + α -1 ), then we have ρ 5r (A H 3 ) SU 5 (q 1 2 )
Proof. Let G = ρ 5r (A H 3 ) and H = ρ 5r (A I 2 (5) ). First note that by Proposition 6.1, we can assume G ≤ SL 5 (q). Let us now consider the restriction to A I 2 (5) . Let M be the following matrix

        √ α(α + 1)(β + 1) -α(β + 2)(β + 1) √ α(α + 1)(β + 2) -α(β + 2) 0 - √ α(β + 1) (α + 1)(β + 1) - √ α(β + 2) α + 1 - √ α β(α + 1) - √ αβ 3 β(β 2 -1)(α + 1) - √ αβ(β 2 -1) 0 -β β α+1 √ α -β(β 2 -1) (β 2 -1)(α+1) √ αβ -β 2 -1 β 0 0 0 0 1        
.

We then have det(M ) = 5Φ 5 (α) = 0 and for all h ∈ H I 2 (5),α ,

M ρ 5s (h)M -1 =    ρ 1 (h) 0 0 0 ρ 2 (h) 0 0 0 Ind(h)    .
The representations above are given in Theorem 5.1.1.

Assume now F q = F p (α) = F p (α + α -1 ) and 1 ∼ 2. We then have q is an odd power of p. By Lemma 8.1.2 and Theorem 5.1.3, we have that H SL 2 (q 2 ). We now consider the maximal subgroups of SL 5 (q) given in Tables 8.18 and 8.19 of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF]. We know that G is irreducible, therefore it cannot be contained in any group of class C 1 . Since G contains H, we have that the field generated by the elements of G contains F q , therefore we have that G cannot be included in any group of class C 5 . The group G contains a SL 2 (q 2 ) twisted by the field automorphism of degree 2 of F q 2 , therefore it cannot be a subgroup of SU 5 (q 1 2 ). We know that ρ 5r is not selfdual, therefore G cannot preserve any non-degenerate bilinear form. It follows that G cannot be included in any group of class C 8 . We list below the remaining maximal subgroups remaining and their order or a quantity their order divides.

1. (q -1) 4 : S 5 , 120(q -1) 4 2. q 5 -1 q-1 : 5, 5(q 4 + q 3 + q 2 + q + 1)

5 1+2

+ : Sp 2 (5), 15000 4. (q -1, 5) × P SL 2 (11), 3300 5. M 11 , 7920 6. (q -1, 5) × P SU 4 (2), 129600

We have α q-1 = 1 so q -1 ≥ 7. It follows that q ≥ 9 and |H| ≥ 531360. This excludes cases 3 to 6.

Assume we are in case 1 or 2. We have |H| = q 2 (q 4 -1). It follows that q 2 divides 120. This is absurd because p = 2.

It follows that G SL 5 (q).

Assume now F q = F p (α) = F p (α + α -1 ) and 1 2. We then have that ρ 5s (A I 2 (5) ) SU 2 (q) × SU 2 (q). The group G is then irreducible and generated by transvections. Therefore, by Theorem 2.3.4, we have that G ∈ {SL 5 (q )Sp 5 (q ), SU 5 (q 1 2 ), q |q}. Since G contains a natural SL 2 (q), we have that q = q. We have G = SU 5 (q 1 2 ) because G contains a natural SU 2 (q) and G = Sp 5 (q) because ρ 5s ρ 5s . This proves that G SL 5 (q). Assume now that F q = F p (α) = F p (α + α -1 ) and 1 ∼ 2. By Lemma 8.1.2 and Theorem 5.1.3, we have that H SL 2 (q) and Φ 1,2 is the unique automorphism of order 2 of F q . We have Φ 1,2 (α) = α -1 , therefore by Proposition 2.1.2, we have that Φ

1,2 • ρ 5r|A H 3 ρ 5r|A H 3 or Φ 1,2 • ρ 5r|A H 3 ρ 5 r |A H 3 . We have Tr(ρ 5r (S 1 S 2 S -2 3 )) = α 2 -2α + 2 -α -1 Φ 1,2 (Tr(ρ 5r (S 1 S 2 S -2 3 ))) = α -2 -2α -1 + 2 -α Tr(ρ 5 r (S 1 S 2 S -2 3 )) = α -2 -2α -1 + 2 -α If α 2 -2α + 2 -α -1 = α -2 -2α -1 + 2 -α, then 0 = α 2 -α -2 -(α -α -1 ) = (α -α -1 )(α + α -1 -1) = α -2 (α 2 -1)Φ 6 (α).
This proves that Φ 1,2 • ρ 5r|A H 3 ρ 5 r |A H 3 , therefore we have that Φ 1,2 • ρ 5r|A H 3 ρ 5r|A H 3 . By Lemma 3.2.4, we have that G is conjugate in GL 5 (q) to a subgroup of SU 5 (q

2 ). We now consider the maximal subgroups of SU 5 (q 1 2 ) given in Tables 8.20 and 8.21 of [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF]. G is irreducible so it cannot be included in a maximal subgroup of class C 1 . It contains a diagonal SL 2 (q) twisted by , therefore it cannot be included in a maximal subgroup of class C 5 . We list below the remaining maximal subgroups with their order or a quantity their order divides.

1. (q 1 2 + 1) 4 : S 5 , 120(q 1 2 + 1) 4 2. q 5 +1 q+1 : 5, 5(q 4 -q 3 + q 2 -q + 1)

3. 5 1+2 + : SP 2 (5), 15000 4. (q 1 2 + 1, 5) × P SL 2 (11), 3300 5. (q 1 2 + 1, 5) × P SU 4 (2), 129600
We have |H| = q(q 2 -1). We have α q 1 2 +1 = 1, therefore q 1 2 ≥ 6. This implies that q 1 2 ≥ 7 and q ≥ 49. This implies that q = 49 or q ≥ 81 since q is as square and p = 2. We then have that |H| = 117600 or |H| ≥ 531360. Cases 3, 4 and 5 are therefore excluded. We have that q is a square and p = 2, therefore q does not divide 120 and cases 1 and 2 are excluded.

It follows that G SU 5 (q

2 ).

Assume now that F q = F p (α) = F p (α + α -1 ) and 1

2. We then have by the same arguments as in the previous case that G is conjugate in GL 5 (q) to a subgroup of SU 5 (q 1 2 ). We have H SU 2 (q 1 2 ) × SU 2 (q 1 2 ). By Lemma 8.1.1, we have that G is normally generated by H. This implies that G is an irreducible group generated by transvections. We also have that G contains a natural SU 2 (q 1 2 ) and ρ 5r is not self-dual. It follows by Theorem 2.3.4 that G is conjugate in GL 5 (q) to SU 5 (q 1 2 ), SL 5 (q 1 2 ) or SL 5 (q). G is conjugate to a subgroup of SU 5 (q 1 2 ) so G SU 5 (q 1 2 ) and the proof is concluded. Theorem 8.1.1. Under the assumptions on α and p, we have the following results.

1. Assume 1 ∼ 2. (a) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ), then the morphism from A H 3 to H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) factorizes through the surjective morphism Φ : A H 3 → SL 3 (q 2 ) × SL 4 (q) × SL 5 (q). (b) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and Φ 1,2 ( √ α) = √ α -1 , then the morphism from A H 3 to H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) factorizes through the surjective morphism Φ : A H 3 → SL 3 (q) × SU 4 (q 1
2 ) × SU 5 (q

1 2 ). (c) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and Φ 1,2 ( √ α) = - √ α -1
, then the morphism from A H 3 to H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) factorizes through the surjective morphism

Φ : A H 3 → SL 3 (q) × SL 4 (q 1
2 ) × SU 5 (q

2 ).

(d) If F q 2 = F p ( √ α) = F p (α) = F p (α + α -1 ), then the morphism from A H 3 to H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) factorizes through the surjective morphism Φ : A H 3 → SL 3 (q 2 ) × SU 4 (q) × SL 5 (q).

Assume 1 When it exists, we write the automorphism of order

2 of F q . (a) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and ( √ α) = √ α -1 , then the morphism from A H 3 to H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) factorizes through the surjective morphism Φ : A H 3 → SU 3 (q 1 2 ) 2 × SU 4 (q 1
2 ) × SU 5 (q

1 2 ). (b) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and ( √ α) = - √ α -1 , then the morphism from A H 3 to H H 3 ,α GL 1 (q) 2 × GL 3 (q) 2 × GL 4 (q) 2 × GL 5 (q) factorizes through the surjective morphism Φ : A H 3 → SU 3 (q 1 2 ) 2 × SL 4 (q 1
2 ) × SU 5 (q

2 ).

Proof. Assume first that 1 ∼ 2. We then have the result by Lemmas 3.

(α) ∈ {α, α -1 }. It follows that ϕ(α + α -1 ) = α + α -1 . If F p (α) = F p (α + α -1 ), this proves that ϕ is trivial. If F p (α) = F p (α + α -1 ), then ϕ = and ϕ • ρ 3s|A H 3 ρ 3 s |A H 3
. The result thus follows from Proposition 8.1.2.

Type H 4 , general aspects

Let p be a prime number, p / ∈ {2, 3, 5} and α ∈ F p such that the order of does not divide 20, 30 or 48. Let ξ ∈ F p be a primitive fifth-root of unity. We set F q = F p (α). The irreducible representations are given by the H 4 -graphs in subsection 10.2 of the Appendix. The highest dimensional representation is of dimension 48 and our results for this representation are still conjectural in some cases. Definition 8.2.1. The Iwahori-Hecke H H 4 ,α is the F q -algebra generated by the generators S 1 , S 2 , S 3 , S 4 and the following relations :

1. ∀i ∈ {1, 2, 3, 4}, (S i -α)(S i + 1) = 0. We then use the W -graphs given in the Appendix which are the W -graphs given in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] together with the new ones verifying the conditions in Theorem 6.2. Proposition 8.2.1. If the order of α does not divide 12, 20 or 30 then the Iwahori-Hecke algebra is split semi-simple and the models given by specialization of the W -graphs are irreducible and pairwise non-isomorphic. The restriction rules from the generic case then apply to the specialized case.

S

Proof. As in type H 3 , we let

A = Z[ 1+ √ 5 2 ][ √ u ±1 ] and F = Q[ √ 5]( √ u).
We have a symetrizing trace defined by τ (T 0 ) = 1 and τ (T σ ) = 0 for all σ ∈ H 4 \ {1 H 4 }. The algebra H H 4 ,u is then a free F -algebra of rank 14400. We have that A is an integrally closed integral domain because

Z[ 1+ √ 5
2 ] is integrally closed (see [7] V.3. Corollary 1 and [START_REF] Niven | An introduction to the theory of numbers[END_REF] Theorem 9.20.).

Let θ be the ring homomorphism from A to L = F q defined by θ( 1+ √ 5

2 ) = ξ + ξ -1 + 1, θ(u) = α and θ(k) = k. We know F H is split. The basis formed by the elements T σ , σ ∈ H 4 verifies the conditions of the Proposition 2.2.4.

We now only need to check that the Schur elements associated to these irreducible representations are in B and don't vanish when specialized under θ. For n ∈ N, Φ n is the n-th cyclotomic polynomial and Φ 5,a (u

) = u 2 + 1+ √ 5 2 u + 1, Φ 5,b (u) = u 2 + 1- √ 5 2 u + 1, Φ 10,a (u) = u 2 -1+ √ 5 2 u + 1, Φ 10,b (u) = u 2 + √ 5-1 2 u + 1, Φ 15,a (u) = u 4 -1+ √ 5 2 u 3 + 1+ √ 5 2 u 2 -1+ √ 5 2 u + 1, Φ 15,b (u) = u 4 + √ 5-1 2 u 3 + 1- √ 5 2 u 2 + √ 5-1 2 u+1, Φ 20,a (u) = u 4 -1+ √ 5 2 u 2 +1, Φ 20,b (u) = u 4 + √ 5-1 2 u 2 +1, Φ 30,a (u) = u 4 + 1+ √ 5 2 u 3 + 1+ √ 5 2 u 2 + 1+ √ 5 2 u + 1 and Φ 30,b (u) = u 4 + 1- √ 5 2 u 3 + 1- √ 5 2 u 2 + 1- √ 5
2 u + 1. If χ is an irreducible character then the character χ associated to the dual representation of χ has a Schur element c χ = a(c χ ), where a is the involution of

Q( √ 5)( √ u) sending √ u to √ u -1 .
We define the field automorphism of Q( √ 5), written x → x by

√ 5-1 2 = -1- √ 5 2
and k = k for all k ∈ Q. They are given Table 8.2 (obtained using Table E.2. of the Appendix and Corollary 9.3.6 of [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF]).

It then only remains to check that the H 4 -graphs are still connected after specialization. We must then check which weights vanish under θ since they are connected in the generic case. The 48-dimensional H 4 -graph we found verifying the conditions of Theorem 6.2 is not defined for p = 29. We therefore only use the bilinear form obtained using its existence. We consider the one found in [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF] which is also available in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] and the CHEVIE Package of GAP3 [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF]. Recall that β = ξ + ξ -1 + 1 and p / ∈ {2, 3, 5}. We first prove that for p = 19, we have

(1 -β)(β + 1)(2β + 3)(β + 2)(3β + 4)(2β + 1)(β + 3)(β + 5) = 0.
We have 1 -

β = 1 -ξ -ξ -1 = -ξ -1 (ξ 2 -ξ + 1) = -ξ -1 (Φ 6 (ξ)) = 0. λ + 1 = ξ -1 Φ 3 (ξ) = 0, λ + 2 = ξ -1 Φ 2 (ξ) = 0 and λ = ξ -1 Φ 4 (ξ) = 0.
Assume by contradiction that 2β + 3 = 0. We then have β = -3 2 . We have β 2 + β -1 = 0. It follows that 0 = 9 4 -3 2 -1 = 9-6-4 4 = -1 4 , which is absurd. Assume by contradiction that 3β + 4 = 0. We then have β = - 4 3 . It follows that 0 = 16 9 -4 3 -1 = -5 9 , which is absurd. Assume by contradiction that 2β + 1 = 0. We then have β = -1 2 . It follows that 0 = 1 4 -1 2 -1 = -5 4 , which is absurd. Assume by contradiction that β + 3 = 0. We then have ξ + ξ -1 = -3, therefore 0 = 9 -3 -1 = 5, which is absurd.

Assume by contradiction that β + 5 = 0. We then have ξ + ξ -1 = -5 therefore 0 = 25 -5 -1 = 19. This implies that p = 19, which contradicts our assumption on p.

For the 40-dimensional representation, the only weight which can vanish is 7 3 . We have that none of the weights vanish for p / ∈ {7, 19}. For p = 7, only the brown edges in the figure in the Appendix vanish for the 40-dimensional representation. There are only 2 such edges in the H 4 -graph 40 r . Since the graph is symmetric, we only need to check that the path represented by one of the brown edges can be replaced. One of the brown edges connects the right vertex x with I(x) = {s 1 , s 2 , s 4 } to a vertex y with I(y) = {s 2 , s 4 }. It can be replaced by the path going through yellow then green then black edges with vertices x 0 = x, x 1 , x 2 and x 3 = y such that I(x 1 ) = {s 2 , s 4 } and I(x 2 ) = {s 3 }.

For p = 19, we only need to consider the H 4 -graph 3 0 s . Only the blue edges vanish and it is clear from the figure in the Appendix that the graph remains connected without those edges. Proposition 8.3.3. If F q = F p (α) = F p (α + α -1 ), then we have ρ 8r (A H 4 ) Ω + 8 (q). If F q = F p (α) = F p (α + α -1 ), then we have ρ 8r (A H 4 ) Ω + 8 (q

2 ).

Proof. Assume first that F p (α) = F p (α + α -1 ). By Proposition 6.2, we have that G = ρ 8r (A H 4 ) ≤ Ω + 8 (q) up to conjugation in GL 8 (q). We will use the same theorem as for the 6-dimensional representation but one of the cases will be much more technical to exclude. We have by Table 8.3 and Theorem 8.1.1 that ρ 8r (A H 3 ) is conjugate in GL 8 (q) to a twisted diagonal SL 4 (q). It follows that G is irreducible and generated by long root elements. By Theorem 2.3.3, we have that G belongs to the following list

1. Ω + 8 (q) 2. Ω - 8 ( √ q) 3. SU 4 (q) 4. G/Z(G) = P Ω 7 (q), Z(G) = 2 5. 3 D 4 ( 3 √ q)
We first exclude cases 2, 3 and 5 because they cannot occur by simple cardinality arguments. We will then exclude case 4 showing that G contains a group which cannot be contained in a group of the same order as in case 4. We have |SL 4 (q)| = q 6 (q 2 -1)(q 3 -1)(q 4 -1).

We have |Ω - 8 ( √ q)| = √ q 12 ( √ q 4 + 1)( √ q 2 -1)( √ q 4 -1)( √ q 6 -1) = q 6 (q 2 + 1)(q -1)(q 2 -1)(q 3 -1) = |SL 4 (q)| (q 2 +1)(q-1) q 4 -1

< |SL 4 (q)|, therefore the second case is excluded.

We have |SU 4 (q)| = q 6 (q 2 -1)(q 3 + 1)(q 4 -1) and q 3 -1 cannot divide q 3 + 1 because it is greater than 2, therefore the third case is excluded.

We have

| 3 D 4 ( 3 √ q)| = 3 √ q 12 ( 3 √ q 8 + 3 √ q 4 + 1)( 3 √ q 6 -1)( 3 √ q 2 -1)
, therefore it cannot contain a group isomorphic to SL 4 (q) because q 6 does not divide q 4 . We now want to show that the fourth case is also excluded but it can contain a twisted diagonal SL 4 (q). We therefore have to construct a different subgroup of our group G.

We order the vertices with the graded lexicographic order (I(x 1 ) = {s 1 }, I(x 2 ) = {s 2 }, I(x 3 ) = {s 1 , s 3 }, I(x 4 ) = {s 1 , s 4 }, I(x 5 ) = {s 2 , s 3 }, I(x 6 ) = {s 2 , s 4 }, I(x 7 ) = {s 1 , s 3 , s 4 } and I(x 8 ) = {s 2 , s 3 , s 4 }) and consider the matrices with respect to the associated basis ordered the same way.

Let P =               
1 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 -1 0 0 -1 0 0 0 1 0 0 1 0 0 0 -1 0 0 -1 0 0 1 0 0 0 0

               and X =                 1 0 -v 2 -v+1 v 0 0 0 0 0 1 -v 2 -v+1 v 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 -v 2 +1 v -v 2 +1 v v 4 +v 3 +2v 2 +v+1 v 2 0 0 0 0 0 0 0 0 0 0 1 v v 2 +v+1 0 0 0 0 0 1 0 0 v v 2 +v+1 0 0 0 0 1 v v 2 -v+1 v v 2 -v+1 v 2 (v 2 +1)(v 2 -v+1) 0 0 0 0 0 0 0 1                 .
We then have for i ∈ {1, 2, 3}, P ρ 8r (S i )P -1 = ρ 4 r (S i ) 0 0 -α t ρ 4 r (S i ) -1 . We also have for i ∈ {2, 3}, (XP )ρ 8r (S i )(XP )

-1 =              ρ 2r (S i ) 0 0 ρ 1r (S i ) 0 0 -α t ρ 1r (S i ) -1    0 0    -α t ρ 2r (S i ) -1 0 0 -α t ρ 1r (S i ) -1 0 0 ρ 1r (S i )             
where 2 r and 1 r are given by the following W-graphs and the bases are ordered in the antilexicographic way for 2 r (I(e x 1 ) = {s 3 } and I(e

x 2 ) = {s 2 }) ∅ 1 r 2 3 2 r
We have ρ 4 r (A H 4 ) = SL 4 (q) and by Lemma 3.5 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], ρ 2r (A A 2 ) = SL 2 (q), where A A 2 =< S 2 , S 3 >. We note H = (XP )ρ 8r (A A 2 )(XP ) -1 . We will consider a large subgroup of the normalizer of H and show it cannot be contained in a group corresponding to the fourth case of our list. Let n = (XP )ρ 8r ((S 1 S 3 S 2 ) 5 (S 2 S 3 S 2 ) -5 )(XP ) -1 , m = (XP )ρ 8r ((S 3 S 2 S 4 ) 4 (S 2 S 3 S 2 ) -4 )(XP ) -1 .

Let then u = [n, m]. We have for i ∈ {2, 3},

(XP ) -1 u(XP )ρ 8r (S i ) = ρ 8r (S i )(XP ) -1 u(XP ), (XP ) -1 m(XP )ρ 8r (S i ) = ρ 8r (S i )(XP ) -1 m(XP ).
This shows that N 1 =< u, m >≤ C G (H). We will now determine what this group

N 1 is. If we let R =               
1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 -1 0 0 1 -1 0 0 0 0 0 0 0 0 0 0 0 0 1 -1 0 0 1 -1 0 0 0 0

               , we have R -1 N 1 R ⊂ {      M 1 0 0 0 0 M 2 0 0 0 0 M 3 0 0 0 0 M 4      , M 1 , M 2 , M 3 , M 4 ∈ GL 2 (q)}.
Let π 1 , π 2 , π 3 and π 4 be the corresponding projections and u = R -1 uR, m = R -1 mR and N 1 =< u , m >. We will first show that π 1 (N 1 ) = SL 2 (q) then determine the images under the other projections and then determine fully N 1 using Goursat's Lemma.

Let u 1 = π 1 (u ) and n 1 = π 1 (m ). We have det(u 1 ) = det(n 1 ) = 1. As in the proof of Theorem 5.1.2, if we prove that G 1 =< u 1 , m 1 > contains elements whose traces generate F q , that G 1 / ∈ {A 5 , S 4 } and that G 1 is not abelian by abelian, then we have

G 1 = SL 2 (q).
Let us first show that G 1 is not abelian by abelian. Assume by contradiction that G 1 is abelian by abelian, we would then have [u

1 , n 1 ][u 1 , n -1 1 ] ∈ {±[u 1 , n -1 1 ][u 1 , n 1 ]}. Let A 1 = [u 1 , n 1 ][u 1 , n -1 1 ] + [u 1 , n -1 1 ][u 1 , n 1 ] and A 2 = [u 1 , n 1 ][u 1 , n -1 1 ] -[u 1 , n -1 1 ][u 1 , n 1 ]. Assume A 1 = 0, we then have a 1 = 4A 1 [1,1]α 15 (α-1) 5 (α 2 +1) 5 (α 4 +1)Φ 3 ( √ α)Φ 6 ( √ α) = 0 and a 2 = - 4A 1 [1,2]α 15 (α-1) 6 (α 2 +1) 5 (α 4 +1)Φ 3 ( √ α)Φ 6 ( √ α) = 0, where A 1 [i, j] is the coefficient in row i and column j of A 1 . If we let v = √ α, we have a 1 = v 18 + v 17 -v 16 + 2v 14 -2v 12 + 2v 11 + 2v 10 + 2v 9 -2v 8 + 2v 7 + 2v 6 -2v 4 + v 2 + v -1, a 2 = v 16 -v 15 + 2v 14 -v 13 + 2v 12 -v 11 + 2v 10 + v 9 + 2v 8 -v 7 + 2v 6 + v 5 + 2v 4 + v 3 + 2v 2 + v + 1.
We will show that for any prime p, those two polynomials in F p [v] are coprime. We write Rem the Euclidean remainder in F p [v]. We have

a 3 = 1 4v 3 Rem(a 1 , a 2 ) = -v 12 + v 11 -v 10 -v 8 -v 4 -v 2 -v -1, a 4 = Rem(a 2 , a 3 ) = -v 11 + v 10 + v 9 + v 8 -v 7 + v 2 + v + 1, a 5 = Rem(a 3 , a 4 ) = -2v 10 -v 9 -v 4 -v 3 -2v 2 -2v -1, a 6 = 4 Rem(a 4 , a 5 ) = v 9 + 4v 8 -4v 7 + 2v 5 -v 4 + v 3 + 2v 2 + 1, a 7 = Rem(a 5 , a 6 ) = -36v 8 + 28v 7 + 4v 6 -16v 5 + 8v 4 -4v 3 -16v 2 -8, a 8 = 81 × Rem(a 6 , a 7 ) = -14v 7 + 7v 6 + 8v 5 -4v 4 + 2v 3 -10v 2 -18v -5, a 9 = 7 81 Rem(a 7 , a 8 ) = -v 6 + 2v 3 + 2v 2 -1, a 10 = 1 4 Rem(a 8 , a 9 ) = 2v 5 -8v 4 -3v 3 + v 2 -v -3, a 11 = 2 7 Rem(a 9 , a 10 ) = -5v 4 -v 3 + v 2 -v -2, a 12 = 25 × Rem(a 10 , a 11 ) = -23v 3 -27v 2 -3v + 9, a 13 = 23 2 50 Rem(a 11 , a 12 ) = -43v 2 -38v -1, a 14 = 43 2
4×23 2 Rem(a 12 , a 13 ) = 3v + 8, a 15 = - 9 43 2 Rem(a 13 , a 14 ) = 1.

These computations are only correct if p / ∈ {2, 3, 5, 7, 23, 43} because the computations are made in Q[v] and they can only be specialized if the polynomials considered are of same degree in F p and in Q. We have p / ∈ {2, 3, 5} by assumption. We check the remaining cases using the Gcd function in GAP4 with a 1 and a 2 polynomials in the indeterminate v:=Indeterminate(GF(p)). This shows that A 1 cannot be equal to zero.

Assume now A 2 = 0, we then have b

1 = 4α 15 A 2 [1,1] α 2 -α+1 = 0, b 2 = - 4α 15 A 2 [1,2] (α 2 -α+1)(α-1) 3 (α 2 +1) 3 = 0 and b 3 = 4α 15 A 2 [1,2] (α 2 -α+1)(α-1) 3 (α 2 +1) 3 (α+1) 2 = 0. We then have that b 4 = Rem( Rem(b 2 ,b 3 ) 8 , 9 Rem(b 3 , Rem(b 2 ,b 3 ) 8 )) × 1 -9v(v 2 +v+1)(v 4 +1) = 0.
We then show using the same techniques as for A 1 that b 1 and b 4 are coprime for p / ∈ {2, 3, 5, 7, 11, 13, 17, 47, 167, 233, 293, 449, 5303, 13649, 15797, 25913, 245071}. Again, by assumption, we have p / ∈ {2, 3, 5} and using GAP4, we get that b 1 and b 4 are coprime except for p = 11. For p = 11, we have Gcd(b 1 , b 4 ) = v 4 + 6v 2 + 1 = (v 2 -6v -1)(v 2 + 6v -1). Again, using GAP4, we see that this polynomial divides v 24 -1 in F 11 [v], therefore if √ α is a root of this polynomial then α 12 = 1 which contradicts our assumptions. This concludes the proof that G 1 is not abelian by abelian.

Assume now that G 1 ∈ {A 5 , S 4 }. We then have that its elements are of order less than or equal to 5. The eigenvalues of n 1 are α 2 and α -2 , therefore if n 1 r = I 2 then α 2r ∈ {-1, 1}, therefore α 4r = 1 and r ≥ 7 by the assumptions on α. This is absurd since the elements in G 1 are of order less than or equal to 5. It also implies that G 1 S 4 because A 5 contains no element of order 4. We have shown above that the commutator subgroup of G 1 was not abelian. This leads to a contradiction because the commutator subgroup of S 4 is the Klein group of order 4, which is abelian.

It only remains to show that F q is generated by traces of elements in G 1 . We have T r (n 1 ) = α 2 + α -2 and Tr(u This shows that α + α -1 belongs to the field generated by traces of the elements of G 1 , therefore F q is generated by traces of the elements of G 1 . It follows that G 1 = SL 2 (q). If u 3 = π 3 (u ), n 3 = π 3 (m ) and C 13 = 0 1 1 0 , then we have that C 13 u 1 C -1 13 = u 3 and C 13 n 1 C -1 13 = n 3 . We therefore get that G 3 =< u 3 , n 3 >= SL 2 (q) and the representation π 3 of N 1 factors through π 1 .

1 ) = α 4 + α -4 -(α 3 + α -3 ) + 2(α 2 + α -2 ) -3(α 2 + α -2 ) + 4.
We now show that if u 2 = π 2 (u ) and

n 2 = π 2 (m ) then G 2 =< u 2 , n 2 > SL 2 (q). < α 4 >,
where SL 2 (q). < α 4 > denotes a group having SL 2 (q) as a normal subgroup and the subgroup of F q generated by α 4 as its quotient group by SL 2 (q). Again as in the proof of Theorem 5.1.2, to prove that G 2 contains SL 2 (q) as a normal subgroup, we only need to show that G 2 is not abelian by abelian, that G 2 contains elements of order greater than 6 and that the field generated by the traces of elements of [G 2 , G 2 ] is F q .

We first show that G 2 is not abelian by abelian. Assume G 2 is abelian by abelian. We then have

B 1 = [u 2 , n 2 ][u 2 , n -1 2 ] -[u 2 , n -1 2 ][u 2 , n 2 ] = 0 or B 2 = [u 2 , n 2 ][u 2 , n -1 2 ] + [u 2 , n -1 2 ][u 2 , n 2 ] = 0. Assume B 1 = 0, we then have 0 = B 1 [1, 2] = (α 3 -1)(α 5 -1)(α-1) 3 (α 4 +1) 2 √ α 15
, which is absurd.

Assume B 2 = 0. We then have 0

= b 2 = √ α 15 B 2 [1,2] (α 2 -α+1)(α-1) 3 (α 4 +1) and 0 = b 3 = √ α 23 B 2 [2,1] (α 2 -α+1)(α 8 -1)(α-1) 2 . It follows that 0 = b 2 = (α 4 -α 3 -α 2 -α + 1)(α 6 + 2α 5 + 3α 4 + 2α 3 + 3α 2 + 2α + 1). 0 = b 3 = (α 4 + α 2 + α + 1)(α 4 -α 3 -α 2 -α + 1).
We then have 0 [1,1] α 2 -α+1 = 0, therefore 0 = α 17 -2α 16 -α 14 + α 13 -α 12 + 2α 11 + α 10 -2α 9 + 3α 8 -3α 7 + α 6 -α 5 + 5α 8 -2α 3 -α 2 -α + 1.

= b 4 = 1 (α+1)(α- √ α+1)(α+ √ α+1) Rem(b 2 , b 3 ) = α 4 -α 3 -α 2 -α + 1. We also have b 1 = α 9 B 2
We then have

b 5 = 1 2 Rem(b 1 , b 4 ) = 83α 3 + 60α 2 + 20α -48 b 6 = 83 2 Rem(b 4 , b 5 ) = 31α 2 -45α + 25 b 7 = 31 2 Rem(b 5 , b 6 ) = 2 × 3 × 5 × 83 2 α -186003 b 8 = 2 2 ×5 2 31 2 Rem(b 6 , b 7 ) = 1.
Those computations show that 1 = 0 for p / ∈ {31, 83}. For p ∈ {31, 83}, we check using GAP4 that b 1 and b 4 are coprime in F p [v]. It follows that B 2 = 0 and G 2 is not abelian by abelian.

We must now show that G 2 / ∈ {A 5 , S 4 }. Assume G 2 ∈ {A 5 , S 4 }. We have that the order of n 2 must belong to {1, 2, 3, 5}. The eigenvalues of n 2 are 1 and α -4 . Therefore, if n 2 r = I 2 , then we have α 4r = 1 r ∈ {-1, 1}. It follows that α 4r = 1. By the assumptions on the order of α, we cannot have r ∈ {1, 2, 3, 4, 5}, therefore we have a contradiction.

We now show that the traces of the elements of [G 2 , G 2 ] generate F q . Let F be the field generated by those traces. We have C 1 = -(Tr([u 2 , n 2 ]) -4) = α 5 + α -5 -2(α 4 + α -4 ) + (α 3 + α -3 ) -(α 2 + α -2 ) + 2(α + α -1 ) ∈ F. We then have C 1 -2 = (α 3 -1)(α 5 -1)(α-1) 2 α 5 = 0, therefore

C 2 = Tr([[u 2 , n 2 ], n 2 ]) -C 2 1 + 6C 1 -10 C 1 -2 = α 4 + α -4 ∈ F.
It follows by induction that α 4r + α -4r ∈ F for all r ∈ N.

We then have

C 3 = C 1 + 2C 2 = α 5 + α -5 + α 3 + α -3 -(α 2 + α -2 ) + 2(α + α -1 ) ∈ F. We also have C 4 = Tr([n 2 , u 2 ][u -1 2 , n 2 ]+(C 1 -2)(α 8 +α -8 )-2C 2 1 +11C 1 -(α 4 +α -4 )-16 ∈ F, therefore C 4 = α 7 + α -7 -2(α 6 + α -6 ) + 2(α 3 + α -3 ) -3(α 2 + α -2 ) + α + α -1 ∈ F.
We then have

C 5 = 1 2 4 × 3 (Tr([[[u 2 , n 2 ], n 2 ], n 2 ])-(α 20 +α -20 )-52(α 16 +α -16 )+(6C 4 +34C 3 -427)(α 12 +α -12 ))+ (60C 4 -5C 2 3 + 255C 3 -1601)(α 8 + α -8 ) + (210C 4 -13C 2 3 + 726C 3 -16 × 211)(α 4 + α -4 )+ 336C 4 + 1034C 3 -16 × 269) ∈ F.
Since C 5 = α 2 + α -2 , we get that α 2 + α -2 ∈ F. We then have C 3 +C 5 C 2

5

= α + α -1 ∈ F. It follows that F = F q and we have G 2 = SL 2 (q). Let u 4 = π 4 (u ) and n 4 = π 4 (m ).

Let C 24 = v 4 + v 3 + 3v 2 + v + 1 v 4 + v 3 + v 2 + v + 1 -v 4 -v 3 -v 2 -v -1 -v 4 -v 3 -3v 2 -v -1
, we have det(C 24 ) = -4α(α + 1)(v 2 + v + 1). We also have

C 24 u 2 C -1 24 = t u -1 4 and C 24 n 2 C -1 24 = t n -1 4 .
By what has been done above, we have that N 1 =< u , m > π 1 × π 2 (N 1 ) ≤ SL 2 (q) × GL 2 (q), π 1 (N 1 ) = SL 2 (q) and SL 2 (q) ⊂ π 2 (N 1 ). We will show that [N 1 , N 1 ] SL 2 (q)×SL 2 (q). By Goursat's lemma, we only need to show that we cannot have π 1 Φ(π 2 ), which is equivalent to π 1 Φ(π 2 ) ⊗ χ, where χ : N 1 → F × q is a character of N 1 and Φ ∈ Aut(F q ). Assume such a character exists, the following proof will be quite computational. We have that there exists M ∈ GL 2 (q) such that for all h ∈ [N 1 , N 1 ], Tr(π 1 (h)) = Φ(Tr(π 2 (h)))χ(h) and χ(h) ∈ {-1, 1}.

Let

D 1 = Tr([u 1 , n 1 ]) and E 1 = Tr([u 2 , n 2 ]
). We have

- v 14 Φ 6 (α) D 1 = v 24 -2v 22 + 4v 20 -8v 18 + 9v 16 -12v 14 + 14v 12 -12v 10 + 9v 8 -8v 6 + 4v 4 -2v 2 + 1, - v 10 Φ 6 (α)Φ 8 (α) E 1 = v 8 -v 6 -v 4 -v 2 + 1.
We have

D 1 = Φ(E 1 ) or D 1 = -Φ(E 1
). If one of them vanishes, then they both vanish because Φ is an automorphism. This proves it is sufficient to show that they cannot vanish at the same time to prove that neither of them vanishes. Assume D 1 = E 1 = 0. We then have that the following quantities vanish

v 1 = Rem(-v 14 Φ 6 (α) D 1 , -v 10 Φ 6 (α)Φ 8 (α) E 1 ) = 20v 6 + 10v 4 + 5v 2 -10 v 2 = 2 3 Rem(-v 10 Φ 6 (α)Φ 8 (α) E 1 , v 1 ) = -4v 4 -v 2 + 2 v 3 = 2 2 5 Rem(v 1 , v 2 ) = 11v 2 -6 v 4 = 11 2 2 5 Rem(v 2 , v 3 ) = 1.
For any p = 2, this proves that 0 = 1 which is absurd. It follows that neither of them vanishes, therefore χ([u, m ]) = Φ(E 1 ) D 1 . We have

D 1 = Tr([u -1 1 , n 1 ]) and E 1 = Tr([u -1 2 , n 2 ]), there- fore χ([u 1 , n 1 ]) = χ([u -1 1 , n 1 ]) and χ([u 1 , n 1 ][u -1 1 n 1 ]) = 1. We let D 2 = Tr(χ([u 1 , n 1 ][u -1 1 n 1 ]) and E 2 = Tr(χ([u 2 , n 2 ][u -1 2 n 2 ]). We have D 2 = Φ(E 2 ). Let D 3 = Tr([u 1 , n 2 1 ]
) and E 3 = [u 2 , n 2 2 ]. We have D 3 = Φ(E 3 ) or D 3 = -Φ(E 3 ), therefore if we show that they cannot vanish simultaneously, then we have that neither of them vanishes. Assume by contradiction that D 3 = E 3 = 0. We then use the Euclidean algorithm to prove that 1 = 0 modifying it slightly using the conditions on α to dividing by non-zero quantities. In order to complete the algorithm, we need to invert the primes 7, 13, 17, 19, 23, 47, 53, 193, 599, 881, 1471, 2503, 3559, 13967, 44101, 180811, 382843 and 981391. See subsection 10.3.1 for the detailed computation. We check using GAP4 that those polynomials remain coprime when p is a prime in the previous list. This is true except for p = 599.

For p = 599, we have Gcd(-

v 22 Φ 6 (α) D 3 , -v 18 Φ 6 (α) E 3 ) = v 4 + 394v 2 + 1 = 0. Let v 59901 = v 4 +394v 2 +1 v 2
= 0. We then have

D 1 = D 1 -(-(v 12 + v -12 ) + 397(v 10 + v -10 ) -156424(v 8 + v -8 ) + 61630673(v 6 + v -6 ) -24282328759(v 4 + v -4 ) +9567175900402(v 2 + v -2 ) -3769443022429664)v 59901 -2479367974099310 × 599) = 160 ∈ ±Φ(E 1 ) ∈ ±Φ(E 1 -(-v 8 -v -8 + 396(v 6 + v -6 ) -156024(v 4 + v -4 ) + 61473061(v 2 + v -2 ) -24220230012)v 59901 -15930964405 × 599) ∈ ±Φ (15)
∈ ±15.

We then have 145 = 0 or 175 = 0. This concludes the proof that D 3 and E 3 are non-zero.

Case

1 : D 1 = Φ(E 1 ). Case 1.1 : D 3 = Tr([u 1 , n 2 1 ]) = Φ(Tr([u 2 , n 2 2 ])) = Φ(E 3 ). We then have D 4 = D 3 1 + D 1 D 2 -4D 1 D 3 + D 2 3 -2D 2 = Φ(E 4 )
, where

E 4 = E 3 1 + E 1 E 2 - 4E 1 E 3 + E 2 3 -2E 2 .
We have

0 = v 46 (E 5 1 -4E 3 1 E 3 + 36E 3 1 + 7E 2 1 E 3 + 6E 1 E 2 3 -58E 2 1 + 18E 1 E 2 -116E 1 E 3 -6E 1 E 4 + 5E 2 E 3 ) (v 2 -1) 8 (v 4 + 1) 12 (v 6 -1) 4 Φ 6 (v 2 ) + v 46 (E 2 E 4 -4E 3 E 4 + 76E 1 -42E 2 + 50E 3 -10E 4 -36) (v 2 -1) 8 (v 4 + 1) 12 (v 6 -1) 4 Φ 6 (v 2 ) .
It follows that the same expression, where the E is replaced by D, is equal to zero but this is equal to (v 12 -2v 10 +4v 8 -2v 6 +4v 4 -2v 2 +1)(v 16 -v 14 +2v 12 -3v 10 +6v 8 -3v 6 +2v 4 -v 2 +1)(v 16 -4v 14 +8v 12 -8v 10 +7v 8 -8v 6 +8v 4 -4v 2 +1) v 22 This means that one of those three factors must vanish, we therefore treat those three cases separately.

Cases 1.1.1 : D 5 = v 12 -2v 10 +4v 8 -2v 6 +4v 4 -2v 2 +1 v 6 = 0. We then define the following elements in order to get another polynomial in v which will vanish and we will then prove that the two polynomials obtained this way cannot simultaneously vanish. Let us assume first that p = 7. We then let F 1 = v 6 D 5 and F 2 = -v 26 (46013E 1 + 7539E 3 -616E 2 -73360). Those two polynomials in v vanish. Let us prove that this is absurd using the Euclidean algorithm. This leads to the contradiction 1 = 0 if p is different from 7,[START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF][START_REF] Naruse | Representation theory of Weyl group of type C n[END_REF][START_REF] Ribet | Galois action on division points of Abelian varieties with real multiplications[END_REF]47,53,101,139,181,349,431,51071,16660499,262400449,1023778455893 or 20498163516987363083353. We are under the assumption that p = 7, therefore we first check the other primes. Using GAP4, we see that F 1 and F 2 are coprime unless p ∈ {139, 262400449}.

D 401 = - 1 7 (D 1 -(-(v 8 + v -8 ) + (v 6 + v -6 ) -(v 4 + v -4 ) + 6(v 2 + v -2 ) + 1)D 5 ) = v 8 + v 6 + 2v 4 + v 2 + 1 v 4 = - 1 7 Φ(E 1 ) D 402 = 1 2 (D 2 -(v 24 + v -24 -2(v 22 + v -22 + 4(v 20 + v -20 ) -12(v 18 + v -18 ) + 14(v 16 + v -16 ) -22(v 14 + v -14 ) +52(v 12 + v -12 ) -30(v 10 + v -10 ) + 55(v 8 + v -8 ) -158(v 6 + v -6 ) -56(v 4 + v -4 ) -118(v 2 + v -2 ) + 508)D 5 ) = 479v 8 -598v 6 + 311v 4 -598v 2 + 479 v 4 = 1 2 Φ(E 2 ) D 403 = D 3 -( -(v 32 + 1) + (v 30 + v 2 ) -(v 28 + v 4 ) + 6(v 26 + v 6 ) -(v 24 + v 8 ) + v 22 + v 10 -24(v 20 + v 12 ) -17(v 18 + v 14 ) + v 16 v 16 D 5 ) = 121v 8 -55v 6 + 146v 4 -55v 2 + 121 v 4 = Φ(E 3 ) D 201 = 479D 401 -D 402 = 1077v 4 + 647v 2 + 1077 v 2 = Φ(- 479 7 E 1 - 1 2 E 2 ) D 202 = 1 2 4 (121D 401 -D 403 ) = 11v 4 + 6v 2 + 11 v 2 = Φ(- 121 112 E 1 - 1 16 E 3 ) D 101 = 7 1 2 4 v 26 (1077D 202 -11D 201 + 655) = 0 = -v 26 (46013E 1 + 7539E 3 -616E 2 -73360) = 616(v 52 +
F 3 = 1 2 4 Rem(F 1 , F 2 ) = -1560220v 10 -984236v 8 -1419140v 6 -109343v 4 -108563v 2 -272023 F 4 = 2
For p = 139, we have Gcd(F 1 , F 2 ) = v 4 + 82v 2 + 1. Let then F 13901 = v 2 + v -2 + 82 = 0. We have

D 1 = D 1 -(-(v 12 + v -12 ) + 85(v 10 + v -10 ) -6976(v 8 + v -8 ) + 571961(v 6 + v -6 ) -46893847(v 4 + v -4 ) +3844723522(v 2 + v -2 ) -315220434992)F 13901 -185902059153 × 139 = 71 = Φ(E 1 ) = Φ(E 1 -(-(v 8 + v -8 ) + 84(v 6 + v -6 ) -6888(v 4 + v -4 ) + 564733(v 2 + v -2 ) -46301220)F 13901 -27306263 × 139) = Φ(21) = 21.
This implies 71 = 21, therefore 50 = 0 which is a contradiction since p = 139.

For p = 262400449, we have Gcd(F 1 , F 2 ) = v 4 + 144711873v 2 + 1. Let then F 26240044901 = v 2 + v -2 + 144711873 = 0. We have

D 1 = D 1 -(-(v 12 + v -12 + 144711876(v 10 + v -10 ) -20941526621303754(v 8 + v -8 ) + 3030487540848227798559380(v 6 + v -6 ) -438547528139311032518667960215007(v 4 + v -4 ) + 63463034196559901414852807139575347118760(v 2 + v -2 ) -9183854544847233051891171601164688387512952822508)F 26240044901 -5064826670873590871335015289738121630701687311347 × 262400449 = 19405199 = Φ(E 1 ) = Φ(E 1 -(-(v 8 + v -8 ) + 144711875(v 6 + v -6 ) -20941526476591875(v 4 + v -4 ) + 3030487519906700743120001(v 2 + v -2 ) -438547525108823428845860731880000)F 26240044901 -241855659926909504463506824998245 × 262400449) = Φ(212787997) = 212787997.
This implies 212787997 = 19405199, therefore 193382798 = 0 which is a contradiction since p = 262400449.

The only remaining prime is p = 7. We have

D 1 = D 1 -(-(v 8 + v -8 ) + (v 6 + v -6 ) -(v 4 + v -4 ) + 6(v 2 + v -2 ) + 1)D 5 = - 7(v 8 + v 6 + 2v 4 + v 2 + 1) v 4 = 0.
This leads to a contradiction because we have proven that D 1 = 0. This concludes the proof of case 1.1.1.

Case 1.1.2 : D 6 = v 16 -v 14 +2v 12 -3v 10 +6v 8 -3v 6 +2v 4 -v 2 +1 v 8 = 0.

We then have

D 601 = D 1 -(-(v 6 + v -6 ) + 2(v 4 + v -4 ) -3(v 2 + v -2 ) + 4)D 6
= 2(2v 12 -4v 10 + 6v 8 -7v 6 + 6v 4 -4v 2 + 2)

v 6 = Φ(E 1 ) D 602 = D 2 -(v 22 + v -22 -3(v 20 + v -20 ) + 7(v 18 + v -18 ) -14(v 16 + v -16 ) + 19(v 14 + v -14 ) -27(v 12 + v -12 ) + 37(v 10 + v -10 ) -48(v 8 + v -8 ) + 74(v 6 + v -6 ) -82(v 4 + v -4 ) + 70(v 2 + v -2 ) -52)D 6 = - 2(24v 12 -16v 10 -24v 8 + 63v 6 -24v 4 -16v 2 + 24) v 6 = Φ(E 2 ) D 603 = D 3 -(-(v 14 + v -14 ) + 2(v 12 + v -12 ) -3(v 10 + v -10 ) + 4(v 8 + v -8 ) -(v 6 + v -6 ) + 2(v 4 + v -4 ) -7(v 2 + v -2 ) + 12)D 6 = - 2(6v 12 -4v 10 -6v 8 + 15v 6 -6v 4 -4v 2 + 6) v 6 = Φ(E 3 ) D 404 = -12D 601 -D 602 = 2(32v 8 -96v 6 + 147v 4 -96v 2 + 32) v 4 = Φ(-12E 1 -E 2 ) D 405 = -3D 601 -D 603 = 8(2v 8 -6v 6 + 9v 4 -6v 2 + 2) v 4 = Φ(-3E 1 -E 3 ) D 102 = D 404 -4D 405 -6 = 0 = Φ(-E 2 + 4E 3 -6) = -Φ( (v 6 -1)(v 10 -1)(v 8 + 1) 2 (v 2 -1) 2 (v 8 -v 5 + v 3 + 1)(v 8 + v 5 -v 3 + 1) v 26 )
Since Φ is an automorphism, we have that G 2 = (v 8 -v 5 + v 3 + 1)(v 8 + v 5 -v 3 + 1) = 0. We also have by assumption 2 . This leads to a contradiction by the assumptions on α since α = v 2 .

G 1 = v 6 D 6 = 0. The Euclidean remainder of the division of G 1 by G 2 as polynomials in v is equal to -v 2 (v 6 -1)Φ 6 (v 2 )(v 2 -1)
Case 1.1.3 : D 7 = v 16 -4v 14 +8v 12 -8v 10 +7v 8 -8v 6 +8v 4 -4v 2 +1 v 8 = 0.

Assume first p / ∈ {11, 13}. We now define new quantities which will permit us to find polynomials which vanish in v. It follows that one of the two factors inside the last expression vanishes. We separate the two possibilities in order to make the computations easier.

D = D 1 -(-v 6 -v -6 -v 4 -v -4 -3(v 2 + v -2 ) + 2)D 7 = - 13v 12 -23v 10 + 23v 8 -16v 6 + 23v 4 -23v 2 + 13 v 6 = Φ(E 1 ) D = D 2 -(v 22 + v -22 + 4(v 18 + v -18 ) -6(v 16 + v -16 ) -v 14 -v -14 -32(v 12 + v -12 ) -7(v 10 + v -10 ) -10(v 8 + v -8 ) +137(v 6 + v -6 ) + 232(v 4 + v -4 ) + 331(v 2 + v -2 ) -158)D -7 = - 476v 12 -751v 10 + 838v 8 -645v 6 + 838v 4 -751v 2 + 476 v 6 = Φ(E 2 ) D = D 3 -(-v 14 -v -14 -v 12 -v -12 -3(v 10 + v -10 ) + 2(v 8 + v -8 ) + 8(v 6 + v -6 ) + 26(v 4 + v -4 ) + 28(v 2 + v -2 ) -2)D 7 = - 149v 12 -271v 10 + 283v 8 -208v 6 + 283v 4 -271v 2 + 149 v 6 = Φ(E 3 ) D = D 2 1 -(v 20 + v -20 -2(v 18 + v -18 ) + 7(v 16 + v -16 ) -18(v 14 + v -14 ) + 24(v 12 + v -12 ) -62(v 10 + v -10 ) + 77(v 8 + v -8 ) -82(v 6 + v -6 ) + 224(v 4 + v -4 ) -30(v 2 + v -2 ) + 207)D 7 = - 476v 12 -751v 10 + 838v 8 -645v 6 + 838v 4 -751v 2 + 476 v 6 = Φ(E 2 1 ) D = 1697D 604 + 13D 605 = 3245v 8 -4025v 6 + 3736v 4 -4025v 2 + 3245 v 4 = Φ(1697E 1 + 13E 2 ) D = 1 4 (149D 604 + 13D 606 ) = 24v 8 -63v 6 + 80v 4 -63v 2 + 24 v 4 = 1 4 Φ(149E 1 + 13E 3 ) D = -13D 605 -476D 604 = 1185v 8 -54v 6 -769v 4 -54v 2 + 1185 v 4 = Φ(-13E 2 1 -476E 1 ) D = 1 13 (24D 406 -3245D 407 ) = 8295v 4 -13072v 2 + 8295 v 2 = Φ(- 24661 4 E 1 + 24E 2 - 3245 4 E 3 ) D = 1 3 2 11 1 13 1 (-1185D 407 + 24D 408 ) = 57v 4 -88v 2 + 57 v 2 = Φ(- 395
2 2 3 1 11 1 E 3 - 5699 2 2 3 1 11 1 E 1 - 8
Case 1.1.3.1 : H 2 = v 16 + 4v 14 + 8v 12 + 8v 10 + 7v 8 + 8v 6 + 8v 4 + 4v 2 + 1 = 0. We set H 1 = v D 7 = 0. We then use the Euclidean algorithm to obtain a contradiction. ) -471434v 18 = 0. We set again I 1 = v 6 D 7 . We then prove that 0 = 1 for most primes using the Euclidean algorithm

H 3 = -1 8 Rem(H 1 , H 2 ) = v 14 + 2v
I 3 = Rem(I 2 , I 1 ) = 214002v
) = v 8 + v 6 + 4v 4 + v 2 + 1 = (v 2 + v -1) 2 (v 2 -v -1) 2 . If v 2 = 1 -v, then v 4 = 1 + v 2 -2v = 2 -3v, v 8 =
4 + 9v 2 -12v = 13 -21v = 13 = -1 and v 16 = 1, which contradicts the fact that α is of order not dividing 8.

If

v 2 = v + 1, then v 4 = v 2 + 2v + 1 = 2 + 3v, v 8 = 4 + 9v 2 + 12v = 13 + 21v = 13 = -1, v 16 = 1, which is a contradiction. We then have 0 = D 1 D 3 -D 8 -D 1 = Φ(-E 1 E 3 + E 8 -E 1 ) = 2Φ ( E 1
). It follows that E 1 = 0, which is absurd by what was proven before case 1.

Case 2 : D 1 = -Φ(E 1 ). Case 2.1 : D 3 = Φ(E 3 ). This is the worst case in terms of computations, we will not write all the polynomials appearing because their degree can be very high. We will give all the necessary elements to consider and the final results of the computations. We have 2 -E 1 = (v 10 -1)(v 6 -1)(v 2 -1) 2 v 10 = 0, therefore D 1 + 2 = Φ -1 (2 -E 1 ) = 0. The elements defined in subsection 10.3.2 of the Appendix and the corresponding computations then lead us to the contradiction 1 = 0 except for the primes which we needed to invert in order to do the computations. We check using GAP4 that v 86 D 13 and v 70 D 12 are coprime for the primes p we inverted in the computations. It is true except for p ∈ {7, 17, 43, 79, 1013, 1747}.

For p = 7, we have Gcd(

v 86 D 13 , v 70 D 12 ) = (v 2 + v -1) 3 (v 2 -v -1) 3 (v 8 + v 6 -v 4 + v 2 + 1), therefore v 8 + v 6 -v 4 + v 2 + 1 = 0, because (v 2 + v -1)(v 2 -v -1) = 0 would imply v 16 = 1 as we have seen in case 1.1.3.2. Let now K 701 = v 8 +v 6 -v 4 +v 2 +1 v 4
. We then have

D 1 = D 1 -(-(v 10 + v -10 + 4(v 8 + v -8 ) -12(v 6 + v -6 ) + 31(v 4 + v -4 ) -67(v 2 + v -2 ) + 135)K 701 = - 7(27v 4 -35v 2 + 27) v 2 = 0.
This is absurd because we proved that we cannot have D 1 = 0. For p = 17, we have Gcd(v 86 D 13 , v 70 D 12 ) = Φ 6 (α) = 0, which is a contradiction. For p = 43, we have Gcd(v 86 D 13 , v 70 D 12 ) = v 4 + 10v 2 + 1. We then set

K 4301 = v 4 +10v 2 +1 v 2
. We have

D 1 = D 1 -(-(v 12 + v -12 + 13(v 10 + v -10 ) -136(v 8 + v -8 ) + 1361(v 6 + v -6 ) -13495(v 4 + v -4 ) +133618(v 2 + v -2 ) -1322720)K 4301 -301395 × 43 = 17 = -Φ(E 1 ) = -Φ(E 1 -(-(v 8 + v -8 ) + 12(v 6 + v -6 ) -120(v 4 + v -4 ) + 1189(v 2 + v -2 ) -11772)K 4301 -2682 × 43) = -Φ(20) = -20.
This implies that 37 = 0 which is a contradiction because p = 43.

For p = 79, we have Gcd(

v 86 D 13 , v 70 D 12 ) = v 4 + 4v 2 + 1. Let K 7901 = v 4 +4v 2 +1 v 2
, we then have a contradiction because

D 1 = D 1 -(-(v 12 + v -12 ) + 7(v 10 + v -10 ) -34(v 8 + v -8 ) + 143(v 6 + v -6 ) -559(v 4 + v -4 ) + 2122(v 2 + v -2 ) -7964)K 7901 -350 × 79 = 0.
For p = 1013, we have Gcd(

v 86 D 13 , v 70 D 12 ) = v 4 + 179v 2 + 1. Let K 101301 = v 4 +179v 2 +1 v 2
, we have

D 1 = D 1 -(-(v 12 + v -12 ) + 182(v 10 + v -10 ) -32584(v 8 + v -8 ) + 5832368(v 6 + v -6 ) -1043961309(v 4 + v -4 ) +186863241972(v 2 + v -2 ) -33447476351714)K 101301 -5909895893852 × 1013 = 824 = -Φ(E 1 ) = -Φ(E 1 -(-(v 8 + v -8 ) + 181(v 6 + v -6 ) -32399(v 4 + v -4 ) + 5799241(v 2 + v -2 ) -1038031742)K 101301 -183411730 × 1013) = -Φ(850) = 163.
This would imply that 661 = 0 which is absurd because p = 1013. For p = 1747, we have Gcd(v 86 D 13 , v

70 D 12 ) = v 4 + 482v 2 + 1. Let K 174701 = v 4 +482v 2 +1 v 2
, we have

D 1 = D 1 -(-v 12 -v -12 + 485(v 10 + v -10 ) -233776(v 8 + v -8 ) + 112679561(v 6 + v -6 ) -54311314647(v 4 + v -4 ) +26177940980322(v 2 + v -2 ) -12617713241200592)K 174701 -3481216614983814 × 1747 = 1680 = -Φ(E 1 ) = -Φ(E 1 -(-(v 8 + v -8 ) + 484(v 6 + v -6 ) -233288(v 4 + v -4 ) + 112444333(v 2 + v -2 ) -54197935220)K 174701 -14953165361 × 1747) = -Φ(1711) = 36
This would imply 1644 = 0 which is absurd because p = 1747. This shows that Case 2.1 is absurd.

Case 2.2 :

D 3 = -Φ(E 3 ).
We then have

D 8 = Tr([u 1 , n 1 ][u 1 , n 2 1 ]) = Φ(E 8 ) = Φ(Tr([u 2 , n 2 ][u 2 , n 2 2 ]
). We have as in case

1.2, D 1 D 3 -D 8 -D 1 = 0 = Φ(E 1 E 3 -E 8 + E 1 ) and E 1 E 3 -E 8 -E 1 = 0, therefore Φ(2E 1 ) = 0 and E 1 = 0 which is absurd. We have proven (π 1 × π 2 )([N 1 , N 1 ]) = SL 2 (q) × SL 2 (q). We now want to determine (π 1 × π 2 )(N 1 ). Since det(u 1 ) = det(u 2 ) = det(n 1 ) = 1 and det(n 2 ) = 1 v 8 , we have that det((π 1 × π 2 )(N 1 )) < v 8 > Z/ o(α)
Gcd(o(α),4) Z, where o(α) is the order of α. Since (π 1 × π 2 )(N 1 ) ≤ SL 2 (q)×GL 2 (q), we have that the kernel of the determinant is contained in SL 2 (q)×SL 2 (q) and since (π

1 × π 2 )([N 1 , N 1 ]) = SL 2 (q) × SL 2 (q),
we get that the kernel is equal to SL 2 (q) × SL 2 (q). This shows that it is a normal subgroup of (π 1 ×π 2 )(N 1 ), the resulting quotient is a cyclic group of order o(α) Gcd(o(α),4) . It follows that N 1 (SL 2 (q) × SL 2 (q)).Z/ o(α) Gcd(o(α),4) Z, where this denotes an extension which may be split.

Recall now that H = (XP )ρ 8r (A A 2 )(XP ) -1 SL 2 (q), N 1 N 1 and N 1 ≤ C G (H). It follows that the order

|HN 1 | is equal to |H||N 1 | |H∩N 1 |
. The intersection of H and N 1 is at most

Z/2Z. By noting that R -1      -I 2 0 0 0 0 I 2 0 0 0 0 -I 2 0 0 0 0 I 2      R =      -I 2 0 0 0 0 I 2 0 0 0 0 -I 2 0 0 0 0 I 2     
∈ N 1 , we get that the intersection is exactly Z/2Z. It follows that the order of HN 1 is equal to

1 2 |SL 2 (q) 3 | o(α) Gcd(o(α),4) = 1
2 Gcd(o(α),4) q 3 (q 2 -1) 3 o(α). Assume now that we are in the fourth case of Theorem 2.3.3, we would have G 2 • Ω 7 (q). It would follow that 1 2 Gcd(o(α),4) q 3 (q 2 -1) 3 o(α) divides 2q 9 (q 2 -1)(q 4 -1)(q 6 -1)

Gcd(2,q-1) = q 9 (q 2 -1) 3 (q 2 + 1)(q 4 + q 2 + 1). We would then have that o(α) divides 2 Gcd(o(α), 4)(q 2 + 1)(q 4 + q 2 + 1) which divides 8(q 2 + 1)(q 4 + q 2 + 1). Since o(α) divides q 2 -1, q 2 + 1 = q 2 -1 + 2 and q 4 + q 2 + 1 = q 4 -1 + q 2 -1 + 3, we would have o(α) divides 48 which contradicts our assumptions on α.

Assume now that

F q 2 = F p ( √ α) = F p (α) = F p (α + α -1
). We have, by Table 8.3 and Proposition 8.1.4, that ρ 8r (A H 3 ) SU 4 (q). It follows that G is again an irreducible group generated by long root elements. By Propositions 6.1 and 6.2, we have that G ≤ Ω + 8 (q), up to conjugation in GL 8 (q 2 ). It follows by Theorem 2.3.3 that G is isomorphic to one of the following groups

1. Ω + 8 (q) 2. Ω - 8 ( √ q) 3. SU 4 (q) 4. G/Z(G) = P Ω 7 (q), Z(G) = 2 5. 3 D 4 ( 3 √ q)
By Lemma 3.5 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], we again have ρ 8r (A A 2 ,1 ) SL 2 (q), where A A 2 ,1 =< S 2 , S 3 >. This proves that we again have a subgroup of G of order 1 2 Gcd(o(α),4) q 3 (q 2 -1) 3 o(α). This excludes the fourth case by the same arguments as before. We also have that G contains a subgroup of order equal to q 6 (q 2 -1)(q 3 + 1)(q 4 -1).

Assume now by contradiction that G Ω - 8 ( √ q). We then have |G| = q 6 (q 2 + 1)(q -1)(q 2 -1)(q 3 -1). It then follows that (q 2 -1) divides 8(q 2 + 1)(q 2 -q + 1). We then have that q 2 -1 divides 8(q 2 + 1)(q 2 -q + 1) -8(q 2 -1)(q 2 -q + 1) = 16(q 2 -q + 1), therefore q 2 -1 divides -16(q 2 -q + 1) + 16(q 2 -1) = 16(q -2) = 16q -32. We have 16q -32 < q 2 -1 if q ≥ 16, therefore q ≤ 15. By the conditions on α and p, this is absurd.

Assume now by contradiction that G 3 D 4 ( 3 √ q). We then have

|G| = 3 √ q 12 ( 3 √ q 8 + 3 √ q 4 + 1)( 3 √ q 6 -1)( 3 √ q 2 -1)
. This is absurd because q 6 does not divide q 4 . Assume now by contradiction that G SU 4 (q). We then have |G| = q 6 (q 2 -1)(q 3 + 1)(q 4 -1). It follows that (q 2 -1) divides 8(q 3 + 1)(q 2 + 1). This implies that (q 2 -1) divides 8(q 3 + 1)(q 2 + 1) -8(q 3 + 1)(q 2 -1) = 16(q 3 + 1). We then have that q 2 -1 divides 16(q 3 + 1) -16(q 3 -q) = 16(q + 1). It follows that q -1 divides 16 which is absurd by the conditions on α and p.

This proves that we also have G Ω + 8 (q) when F q 2 = F p ( √ α) = F p (α) = F p (α + α -1 ).

Assume now that F q = F p ( √ α) = F p (α) = F p (α + α -1 ). Let be the unique automorphism of order 2 of F q . We have (α) = α -1 . It follows by Proposition 2.1.

2 that • ρ 8r|A H 4 ρ 8r|A H 4 or • ρ 8r|A H 4 ρ 8rr|A H 4 . We have Tr(ρ 8r (S 1 S 3 S -1 2 S -1 4 )) = α 2 + α -2 -5(α + α -1 ) + 8 (8.1) (Tr(ρ 8r (S 1 S 3 S -1 2 S -1 4 ))) = α 2 + α -2 -5(α + α -1 ) + 8 (8.2) Tr(ρ 8rr (S 1 S 3 S -1 2 S -1 4 )) = α 2 + α -2 -5(α + α -1 ) + 7 (8.3) It follows that • ρ 8r|A H 4 ρ 8r|A H 4 .
We then have by Proposition 4.1 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] that up to conjugation in GL 8 (q), we have G ≤ Ω + 8 (q

2 ). By Proposition 8.1.4, we have that ρ 8r (A H 3 ) SU 4 (q

1 2 ) or ρ 8r (A H 3 ) SL 4 (q 1
2 ). The computations in the first case show here that G contains a subgroup of order divisible by 1 2 |SL 2 (q 4) . By the same arguments as for the case F p (α) = F p (α + α -1 ), we have that G Ω + 8 (q

1 2 ) 3 | o(α) Gcd(o(α),
2 ).

Triality automorphism and the two 8-dimensional representations

The two 8-dimensional representations being linked through triality, we first recall a few facts about triality and we will then determine the image of the Artin group inside ρ 8r × ρ 8rr . We use the following notations, definitions and results of [START_REF] Carter | Simple groups of Lie type[END_REF]. This phenomenon was also observed in the generic case in Proposition 6.7 of [START_REF] Marin | Infinitesimal Hecke algebras II[END_REF]. The 24 roots can then be expressed in terms of r 1 , r 2 , r 3 and r 4 . We list here the twelve positive roots :

= h ⊕ r∈Φ L r , where Φ = {± i ± j , 1 ≤ i < j ≤ 4}, for 1 ≤ i < j ≤ 4, 1. e i -j = E i,j -E 4+j,4+i , 2. e j -i = E 4+i,4+j + E j,i , 3. e i + j = E i,4+j -E j,4+i , 4. e -i -j = -E 4+i,j + E 4+j,i . L r = Ce r and h =                                                  λ 1 0 0 0 0 λ 2 0 0 0 0 λ 3 0 0 0 0 λ 4      0 0      -λ 1 0 0 0 0 -λ 2 0 0 0 0 -λ 3 0 0 0 0 -λ 4                     , (λ 1 , λ 2 , λ 3 , λ 4 ) ∈ C 4                              . We let r 1 = -1 -2 , r 2 = 1 -2 ,
r 1 , r 2 , r 3 , r 4 , r 1 + r 3 , r 2 + r 3 , r 3 + r 4 , r 1 + r 2 + r 3 , r 1 + r 3 + r 4 , r 2 + r 3 + r 4 , r 1 + r 2 + r 3 + r 4 , r 1 + r 2 + 2r 3 + r 4 .
We have then fixed the constant structures {N r,s , r, s ∈ Φ, r + s ∈ Φ}, which are defined by the relation [e r , e s ] = N r,s e r+s . They have the following values :

N r 1 ,r 3 = N r 2 ,r 3 = N r 3 ,r 4 = N r 1 +r 3 ,r 4 = N r 1 +r 3 ,r 2 = N r 2 +r 3 ,r 4 = N r 1 ,r 3 +r 4 = N r 2 +r 3 ,r 1 = N r 2 ,r 3 +r 4 = N r 2 +r 3 +r 4 ,r 1 = N r 1 +r 3 +r 4 ,r 2 = N r 1 +r 2 +r 3 ,r 4 = N r 1 +r 2 +r 3 +r 4 ,r 3 = 1.
We get the remaining ones using the facts that N s,r = -N r,s and N -r,-s = -N r,s . We write L Z = Vect Z (h r , r ∈ π, e s , s ∈ Φ}. We then call (h r , r ∈ π, e s , s ∈ Φ) the Chevalley basis of L.

For an element x ∈ L, we write ad(x) : L → L, ad(x)(y) = [x, y] = xy -yx. For each r ∈ Φ, since ad(e r ) is a nilpotent endomorphism, we can define an element

x r (t) of End Q[[t]] (L Z ⊗ Z Q[[t]]
) by x r (t) = exp(t ad(e r )). We check on the Chevalley basis that x r (t) is actually an element of GL Z

[t] (L Z ⊗ Z [t]).
We write L Fq = F q ⊗ L Z . For an element u ∈ F q , let θ u be the morphism from Z[t] to F q by θ u (k) = k and θ u (t) = u. We can extend θ u to a morphism θu from GL Z[t] (L Z ⊗ Z [t]) to Aut(L Fq ). We then define for r ∈ Φ and u ∈ F q the elements x r (u) = θu (x r (t)) ∈ Aut(L Fq ).

The adjoint Chevalley group L(F q ) is then defined to be < x r (u), r ∈ Φ, u ∈ F q >.

We now determine an isomorphism between the Chevalley group L(F q ) and P Ω + 8 (q), it can be found in [START_REF] Carter | Simple groups of Lie type[END_REF]. We have the following classical lemma Lemma 8.4.1. Let A be a matrix algebra over F q with Lie bracket defined the usual way. We then have that for any nilpotent matrix y and any matrix x, exp(ad(y))(x) = exp(y)x exp(y) -1 . Proposition 8.4.1. We have P Ω + 8 (q) L(F q ).

Proof. The group Ω + 8 (q) is generated by long root elements which are elements of the form I 8 + ue r with r ∈ Φ and u ∈ F q (See for example [START_REF] Wilson | The finite simple groups[END_REF] 3.7.3). Since e 2 r = 0, we have exp(ue r ) = I 8 + ue r . We define the morphism Ψ from Ω + 8 (q) to L(F q ) on those generators by Ψ(I 8 + ue r ) = (x → (I 8 + ue r )x(I 8 + ue r ) -1 = exp(ue r )x exp(ue r ) -1 = exp(ad(ue r ))(x), where the last equality follows from Lemma 8.4.1.

We then have Ψ(I 8 + ue r ) = x r (u), therefore the morphism is surjective. Let y ∈ ker(Ψ), we have yxy -1 = x for all x ∈ L Fq , therefore y ∈ F q I 8 , where F q is the group of invertible elements of F q . It follows that ker(Ψ) = Z(Ω + 8 (q)) and we get the desired isomorphism.

We now recall Proposition 12.2.3. of [START_REF] Carter | Simple groups of Lie type[END_REF].

Proposition 8.4.2. Suppose all the roots of L have the same length and let r → τ (r) be a map of Φ into itself arising from a symmetry of the Dynkin diagram of L. Then there exists numbers γ r = ±1 such that the map x r (u) → x τ (r)(γ r u) can be extended to an automorphism of L(K). The γ r can be chosen, therefore that γ r = 1 for all r ∈ π ∪ -π. For r = r 1 + r 2 , we have γ r = γr 1 γr 2 N τ (r 1 ),τ (r 2 ) Nr 1 ,r 2

.

We now apply this proposition with τ the following triality automorphism of the Dynkin diagram of type D 4 τ therefore |G| = |Spin + 8 (q)|. By [5] (33.8), there exists an onto morphism from Spin + 8 (q) to G. This proves they are isomorphic since their order is equal.

All the arguments are identical for F p (α) = F p (α + α -1 ).

Remark : Note that the restrictions to A H 3 of ρ 8r and ρ 8rr are identical (this can be seen on the W -graphs by removing 4 from the vertices and deleting the edges from x to y if I(x) = I(y)), this proves that the projective twisted diagonal P SL 4 (q) inside P Ω + 8 (q) is stabilized by the triality automorphism.

We also have that τ does not extend to a morphism from Ω + 8 (q) into itself. Assume it does, there exists Ψ from K 1 Ω + 8 (q) to K 2 Ω + 8 (q) such that τ (x) = Ψ(x) for all x ∈ K 1 . We then have that ρ 8rr (u

i ) = τ (ρ 8r (u i )) = Ψ(ρ 8r (u i ) for all i ∈ [[1, 6]], therefore there exists i ∈ {-1, 1} such that ρ 8rr (u i ) = i Ψ(ρ 8r (u i ))
. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], we have that A H 4 is perfect, therefore i = 1 for all i ∈ [ [1,6]] and for all h ∈ A H 4 , we have Ψ(ρ 8r (h)) = ρ 8rr (h). This is absurd because

ρ 8r ((T s 1 T s 3 T s 2 T s 4 ) 15 (T s 1 T s 3 T s 2 ) -20 ) = I 8 = -ρ 8rr ((T s 1 T s 3 T s 2 T s 4 ) 15 (T s 1 T s 3 T s 2 ) -20 ).

Type H 4 , high-dimensional representations

We now give two propositions, where we determine the image of A H 4 in the remaining irreducible representations. We could use the theorem by Guralnick and Saxl [START_REF] Guralnick | Generation of finite almost simple groups by conjugates[END_REF] for all of them but we use arguments not requiring the classification of finite simple groups to determine those images when possible. The first of the two following propositions does not use the classification and the second one does. 1. If F q = F p (α) = F p (α + α -1 ), then ρ 9s (A H 4 ) SL 9 (q 2 ) and ρ 9s (A H 4 ) SL 9 (q 2 ). We also have that Φ

1,2 • ρ 9s|A H 4 ρ 9s|A H 4 . 2. If F q = F p (α) = F p (α + α -1
), then ρ 9s (A H 4 ) SL 9 (q 2 ) and ρ 9s (A H 4 ) SL 9 (q 2 ). We also have that Φ

1,2 • ρ 9s|A H 4 ρ 9 s |A H 4 .
If 1 2 then we have ρ 9s (A H 4 ) SU 9 (q

2 ) and ρ 9s (A H 4 ) SU 9 (q

2 ). If F q = F p (α) = F p (α + α -1 ) then ρ 10r (A H 4 ) Ω + 10 (q). If F q = F p (α) = F p (α + α -1 ) then ρ 10r (A H 4 ) Ω + 10 (q 1
2 ).

Proof. Assume 1 ∼ 2 and F p (α) = F p (α + α -1 ). By Table 8.3 and Theorem 8.1.1, we have ρ 9s (A H 3 ) SL 3 (q 2 ) × SL 5 (q) in a natural representation. By Lemma 8.2.1, we then have that ρ 9s (A H 4 ) is an irreducible group generated by transvections. We also have that ρ 9s is not self-dual, therefore ρ 9s (A H 4 ) is included in no symplectic group. We also have that the field generated by the traces of the elements of ρ 9s (A H 4 ) contains F q 2 . It follows by Theorem 2.3.4 that ρ 9s (A H 4 ) SL 9 (q 2 ). Assume 1 ∼ 2 and F p (α) = F p (α+α -1 ). We then have by Lemma 8.1.2 that F p (α, ξ +ξ -1 ) = F p (α). We have by Theorem 8.1.1 that ρ 9s (A H 3 ) SL 3 (q) × SU 5 (q 1 2 ). It follows that ρ 9s (A H 4 ) is neither unitary nor symplectic and is an irreducible group generated by transvections. We also have that the field generated by the traces of its elements contains F q . We also have by Proof. Let θ be a morphism from SL 3 ( ) to SL 2 ( 4 ). We have that ker(θ) is a normal subgroup of SL 3 ( ).

Assume this kernel is different from SL 3 ( ). The image of θ is then non-abelian. We get an isomorphism from P SL 3 ( ) to a subgroup of SL 2 ( 4 ). We have

|P SL 3 ( )| = 1 Gcd(3, -1) 3 ( 2 - 1)( 3 -1) and |SL 2 ( 4 )| = 4 ( 8 -1)
. It follows that 3 -1 divides 3( 8 -1). This implies that 3 -1 divides 3( 8 -1) -3 2 ( 6 -1) = 3( 2 -1). We have > 3 since p / ∈ {2, 3, 5}, therefore 3 2 -3 < 3 -1 which is a contradiction. This proves that there is no non-trivial morphsim from SL 3 ( ) to SL 2 ( 4 ). This implies in the same way that there exists no non-trivial morphism from SL r ( ) to SL 2 ( 4 ) if r ≥ 3 and ∈ { , 2 , 4 }.

In the same way, if there was a non trivial morphism from SU r ( ) to SL 2 ( 4 ) for r ≥ 3 and q ∈ { , 2 } then we would have that 3 + 1 divides 3( 2 -1) which is also absurd. Proposition 8.5.2. We write the unique automorphism of order 2 of F q when it exists. We then have the following results.

1. If F q = F p ( √ α) = F p (α) = F p (α + α -1 ), then ρ 16r (A H 4 ) = SL 16 (q) and ρ 16rr (A H 4 ) = SL 16 (q). 2. If F q 2 = F p ( √ α) = F p (α) = F p (α + α -1 ), then ρ 16r (A H 4 ) SL 16 (q 2 ) and ρ 16rr|A H 4 ϕ • ρ 16r|A H 4 , where ϕ is the unique automorphism of order 2 of F q 2 . 3. If F q = F p (α) = F p ( √ α) = F p (α + α -1 ) and ( √ α) = √ α -1 , then ρ 16r (A H 4 ) SU 16 (q 1
2 ) and ρ 16rr (A H 4 ) SU 16 (q 1 2 ).

If

F q = F p (α) = F p ( √ α) = F p (α + α -1 ) and ( √ α) = - √ α -1 , then ρ 16r (A H 4 ) SL 16 (q) and ρ 16 rr |A H 4 • ρ 16r|A H 4 .
Assume 1 ∼ 2. We then have the following results

1. We have Φ 1,2 • ρ 16t|A H 4 ρ 16t|A H 4 , Φ 1,2 • ρ 24s|A H 4 ρ 24s|A H 4 , Φ 1,2 • ρ 24t|A H 4 ρ 24t|A H 4 and Φ 1,2 • ρ 30s|A H 4 ρ 30s|A H 4 . 2. If F q = F p (α) = F p (α + α -1 ), then ρ 16t (A H 4 ) Ω + 16 (q), ρ 24s (A H 4 ) Ω + 24 (q), ρ 24t (A H 4 ) Ω +
24 (q) and ρ 30s (A H 4 ) Ω + 30 (q).

If

F q = F p (α) = F p (α+α -1 ), then ρ 16t (A H 4 ) Ω + 16 (q 1 2 ), ρ 24s (A H 4 ) Ω + 24 (q 1 2 ), ρ 24t (A H 4 ) Ω + 24 (q 1
2 ) and ρ 30s (A H 4 ) Ω + 30 (q

2 ).

Assume 1 2. We then have ρ 16t (A H 4 ) Ω + 16 (q

2 ), ρ 24s (A H 4 ) Ω + 24 (q 1 2 ), ρ 24t (A H 4 ) Ω + 24 (q 1
2 ) and ρ 30s (A H 4 ) Ω + 30 (q

2 ).

If F q = F p (α) = F p (α + α -1 ), then ρ 18r (A H 4 ) Ω + 18 (q), ρ 25r (A H 4 ) SL 25 (q), ρ 36rr (A H 4 ) SL 36 (q) and ρ 40r (A H 4 ) Ω + 40 (q). If F q = F p (α) = F p (α+α -1 ), then ρ 18r (A H 4 ) Ω + 18 (q 1
2 ), ρ 25r (A H 4 ) SU 25 (q

2 ), ρ 36rr (A H 4 ) SU 36 (q 1 2 ) and ρ 40r (A H 4 ) Ω + 40 (q

2 ).

Proof. We first check that in all cases, the assumptions of Theorem 2.3.2 are verified. We have as in the proof of primitivity in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] and in type B that all the groups considered are primitive finite irreducible subgroup of GL(V ) by considering Table 8.3, because they contain either a natural SL 2 (q ) or a twisted diagonal SL 3 (q ) or SU 3 (q ) for q ∈ {q 1 2 , q, q 2 }. Using again Table 8.3 and Lemmas 3.3.5, 3.3.6 and 8.5.1, all the groups considered are tensor-indecomposable. We also have v

G (V ) ≤ 2 ≤ max(2, √ d
2 ). As in page 13 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], the restriction also shows us that we are not in case 2 of Theorem 2.3.2. This shows that they are classical groups in a natural representation.

Consider first the H 4 -graphs 16 r and 16 rr . They are not 2-colorable, therefore we cannot apply Proposition 6.1.

Assume

F q = F p ( √ α) = F p (α) = F p (α + α -1
). We then have by Theorem 8.1.1 that ρ 16r (A H 4 ) contains a natural SL 5 (q), therefore the field generated by the traces of its elements contains F q . It follows that it is a classical group in a natural representation over F q . The natural SL 5 (q) also shows that it is preserves no non-degenerate bilinear or hermitian form over F q . It follows that ρ 16r (A H 4 ) = SL 16 (q). The same arguments show that ρ 16rr (A H 4 ) = SL 16 (q). Assume now . We also have that ρ 16r is not self-dual therefore ρ 16r (A H 4 ) = SL 16 (q 2 ). The result for 16 rr follows from the factorization.

F q 2 = F p ( √ α) = F p (α) = F p (α + α -1
Assume now

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and ( √ α) = √ α -1 .
2 ), up to conjugation in GL 16 (q). It contains a natural SU 5 (q

2 ), therefore it is a classical group in a natural representation over F q . It follows that ρ 16r (A H 4 ) SU 16 (q 1 2 ). The same arguments show that ρ 16rr (A H 4 ) SU 16 (q

1 2 ). Assume F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and ( √ α) = - √ α -1 .
We apply here the same reasoning as before and we get that Φ

2 ) therefore it is a classical group over F q . It follows that ρ 16r (A H 4 ) = SL 16 (q). The result for 16 rr follows from the factorization.

All the remaining H 4 -graphs are 2-colorable therefore we can assume they are defined over F p (α, ξ +ξ -1 ). We know they are classical groups in a natural representation. We will show that they are the groups given in the proposition. We first consider the H 4 -graphs which contain weights in F p (ξ + ξ -1 ) i.e. 16 t , 24 s , 24 t and 30 s . Note first that all of those representations are self-dual, and by Proposition 6.2 and the H 4 -graphs given in the Appendix, we have that they preserve a non-degenerate symmetric bilinear form.

Assume first 1 ∼ 2 and F q = F p (α) = F p (α + α -1 ). We then have by Lemma 8.1.2 that F q 2 = F p (α, ξ + ξ -1 ) = F p (α + α -1 , ξ + ξ -1 ), Φ 1,2 is the unique automorphism of order 2 of F q 2 and Φ 1,2 (α) = α.

Since ρ 16t is the only 16-dimensional self-dual representation, we have by Proposition 2.1.2 that Φ 1,2 • ρ 16t|A H 4 ρ 16t|A H 4 . The representation ρ 30s is the unique 30-dimensional irreducible representation therefore we have Φ 1,2 • ρ 30s|A H 4 ρ 30s|A H 4 . It follows by Proposition 4.1. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] that up to conjugation, we have that ρ 16t (A H 4 ) ≤ Ω + 16 (q) and ρ 30s (A H 4 ) ≤ Ω + 30 (q). They both contain a twisted diagonal SL 5 (q) therefore ρ 16t (A H 4 ) contains an element conjugate to diag(diag(α, α -1 , 1, 1, 1), diag(α, α -1 , 1, 1, 1), I 6 ) and ρ 30s (A H 4 ) contains an element conjugate to diag(diag(α, α -1 , 1, 1, 1), diag(α, α -1 , 1, 1, 1), I 20 ). It follows that the field generated by the traces of their elements contains 2(α + α -1 ). This implies that they are classical groups defined in a natural representation over F q . It follows that ρ 16t (A F 4 ) Ω + 16 (q) and ρ 30s (A F 4 ) Ω + 30 (q). By Proposition 8.1.3, we have Φ

1,2 • ρ 3s ρ 3s and Φ 1,2 • ρ 3 s ρ 3 s . It follows that Φ 1,2 • ρ 24s|A H 4 ρ 24s|A H 4 or Φ 1,2 • ρ 24s|A H 4 ρ 24t|A H 4 . We have Tr(ρ 24s (S 1 S 3 S -1 2 S -1 4 )) = 4(α 2 + α -2 ) + (ξ + ξ -1 -15)(α + α -1 ) + 22 -3(ξ + ξ -1 ) Φ 1,2 (Tr(ρ 24s (S 1 S 3 S -1 2 S -1 4 ))) = 4(α 2 + α -2 ) + (ξ 2 + ξ -2 -15)(α + α -1 ) + 22 -3(ξ 2 + ξ -2 ) Tr(ρ 24t (S 1 S 3 S -1 2 S -1 4 )) = 4(α 2 + α -2 ) + (ξ 2 + ξ -2 -15)(α + α -1 ) + 21 -2(ξ 2 + ξ -2 ). It follows that Φ 1,2 • ρ 24s|A H 4 ρ 24t|A H 4 would imply that 0 = 1 -ξ -ξ -1 = -ξ -1 Φ 6 (ξ), which is absurd. This proves that Φ 1,2 • ρ 24s|A H 4 ρ 24s|A H 4
. It follows by the same arguments as before that ρ 24s (A H 4 ) Ω + 24 (q) and ρ 24t (A H 4 ) Ω + 24 (q). Assume now 1 ∼ 2 and F q = F p (α) = F p (α + α -1 ). We then have by Proposition 8.1.2 that F p (α, ξ + ξ -1 ) = F q and Φ 1,2 is the unique automorphism of order 2 of F q . We then have using the same arguments as before that each of the groups considered preserves a nondegenerate bilinear form over F q 1 2 . They each contain a twised diagonal SU 5 (q 1 2 ), therefore they are classical groups in a natural representation over F q 1 2 . The result then follows. Assume now 1 2. By Lemma 8.1.2, we have F p (α, ξ + ξ -1 ) = F p (α) = F p (α + α -1 ). Let be the unique automorphism of order 2 of F q . We have (α) = α -1

(ρ 24t (S 1 S 3 S -1 2 S -1 4 )) = 4(α 2 + α -2 ) + (ξ + ξ -1 -15)(α + α -1 ) + 21 -2(ξ + ξ -1 ). Since ξ + ξ -1 ∈ F q 1 2 , we have (ξ + ξ -1 ) = ξ + ξ -1 .
It follows by the same computation as in the previous case that • ρ 24s|A H 4 ρ 24s|A H 4 . The result then follows from the same arguments as before.

The only remaining H 4 -graphs to consider are 18 r , 25 r , 36 rr and 40 r . By Proposition 6.1, we can assume they are defined over F q = F p (α). By Proposition 6.2, we have up to conjugation in GL 18 (q) that ρ 18r (A H 4 ) ≤ Ω + 18 (q) and up to conjugation in GL 40 (q) that ρ 40r (A H 4 ) ≤ Ω + 40 (q). Assume F p (α) = F p (α + α -1 ). By Table 8.3 and Theorem 8.1.1, we have that each of the groups associated to those representations contains a twisted diagonal SL 4 (q). It follows by the same arguments as before that the field generated by the traces of their elements contains F p (α + α -1 ) = F q . By Proposition 2.1.2, if there is a field automorphism ϕ and a character η such that (ϕ • ρ) ⊗ η is an irreducible representation of H H 4 ,α then ϕ(α + α -1 ) = α + α -1 . This proves that such an automorphism must be trivial. The result then follows.

Assume now F q = F p (α) = F p (α + α -1 ). Let be the unique automorphism of order 2 of F q . By Proposition 8.1.5, we have that • ρ 5r|A H 3 ρ 5 r |A H 3 . It follows by Table 8.3 and Proposition 2.1.2 that all the representations considered are unitary. We have by the same arguments as before that the field generated by the traces of its elements contains F

q 1 2 . It follows that ρ 18r (A H 4 ) Ω + 18 (q 1
2 ) and ρ 40r (A H 4 ) Ω + 40 (q

2 ). We also have that . This implies that they are classical groups in a natural representation over F q , therefore ρ 25r (A H 4 ) SU 25 (q 1 2 ) and ρ 36rr (A H 4 ) SU 36 (q 1 2 ) and the proof is concluded.

There is now only one case remaining. We cannot apply the previous theorem to it because the restriction to A H 3 only shows that v G (v) ≤ 4 and max(2, √

2 ).

Proof. By Proposition 6.1, we can assume that the representation is defined over

F p (α). Let G = ρ 48r (A H 4 ).
Assume now 1 2. By Lemma 8.1.2, we have

F p (α, ξ + ξ -1 ) = F q = F p (α) = F p (α + α -1
). Let be the unique automorphism of F q . We have (α) = α -1 . It follows by Proposition 2.1.2 that • ρ 48r|A H 4 ρ 48r|A H 4 . Since we have found a symmetric non-degenerate bilinear form preserved by G, we have by Proposition 4.1. of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] that up to conjugation in GL 48 (q), we have that G ≤ Ω + 48 (q

2 ). By Theorem 8.1.1, we have that G contains a twisted diagonal SU 3 (q

2 ) and ρ 48r (A H 3 ) SU 3 (q

1 2 ) × SU 4 (q 1 2 ) × SU 5 (q 1 2 ) or ρ 48r (A H 3 ) SU 3 (q 1 2 ) × SL 4 (q 1 2 ) × SU 5 (q 1
2 ). It follows by Lemma 3.3.6 and 8.5.1 that G is tensor-indecomposable. We have with the same arguments as in the previous proposition that G is a classical group in a natural representation over F

q 1 2 therefore G Ω + 48 (q 1
2 ).

Conjecture 8.5.1. If 1 ∼ 2 then we have ρ 48rr (A H 4 ) Ω + 48 (q

2 ).

Assuming the previous proposition is correct and has been proved. We show the following theorem which gives us the image of A H 4 inside the full Iwahori-Hecke algebra.

a) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ), then the morphism from A H 4 to H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1
)) factorizes through the surjective morphism Φ :→ SL 4 (q 2 ) × Ω + 6 (q 2 ) × Spin + 8 (q) × SL 9 (q 2 ) × Ω + 10 (q) × SL 16 (q) 2 × Ω + 16 (q) × Ω + 18 (q) ×Ω + 24 (q) 2 × SL 25 (q) × Ω + 30 (q) × SL 36 (q) × Ω + 40 (q) × Ω + 48 (q

2 ). (b) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and Φ 1,2 ( √ α) = √ α -1 , then the morphism from A H 4 to H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1 1
)) factorizes through the surjective morphism Φ :→ SL 4 (q)×Ω + 6 (q)×Spin + 8 (q

2 )×SL 9 (q 2 )×Ω + 10 (q

2 )×SU 16 (q

2 ) 2 ×Ω + 16 (q

2 )×Ω + 18 (q

2 )

×Ω + 24 (q 1
2 ) 2 × SU 25 (q

2 ) × Ω + 30 (q

2 ) × SU 36 (q

2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q

2 ).

(c) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and Φ 1,2 ( √ α) = - √ α -1 , then the morphism from A H 4 to H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1 )) factorizes through the surjective morphism Φ :→ SL 4 (q) × Ω + 6 (q) × Spin + 8 (q 1
2 ) × SL 9 (q 2 ) × Ω + 10 (q

2 ) × SL 16 (q) × Ω + 16 (q

2 ) × Ω + 18 (q

2 )

×Ω + 24 (q 1
2 ) 2 × SU 25 (q

2 ) × Ω + 30 (q

2 ) × SU 36 (q

2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q 1
2 ).

(d) If F q 2 = F p ( √ α) = F p (α) = F p (α + α -1 ), then the morphism from A H 4 to H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1
)) factorizes through the surjective morphism

Φ :→ SL 4 (q 2 ) × Ω + 6 (q 2 ) × Spin + 8 (q) × SL 9 (q 2 ) × Ω + 10 (q) × SL 16 (q 2 ) × Ω + 16 (q) × Ω + 18 (q) ×Ω + 24 (q) 2 × SL 25 (q) × Ω + 30 (q) × SL 36 (q) × Ω + 40 (q) × Ω + 48 (q 1
2 ).

2. Assume 1 2. We write the unique automorphism of order 2 of F q .

(a) If

F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and ( √ α) = √ α -1 , then the morphism from A H 4 to H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1 )) factorizes through the surjective morphism Φ :→ SU 4 (q 1 2 ) 2 × Ω + 6 (q 1
2 ) 2 × Spin + 8 (q

2 ) × SU 9 (q

2 ) 2 × Ω + 10 (q

2 ) × SU 16 (q

2 ) 2 × Ω + 16 (q 1
2 ) 2 ×Ω + 18 (q 1
2 ) × Ω + 24 (q

2 ) 4 × SU 25 (q

2 ) × Ω + 30 (q 1
2 ) 2 × SU 36 (q

2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q 1
2 ). (b) If F q = F p ( √ α) = F p (α) = F p (α + α -1 ) and ( √ α) = - √ α -1 , then the morphism from A H 4 to H H 4 ,α ρ irr GL nρ (F p ( √ α, ξ + ξ -1 )) factorizes through the surjective morphism Φ :→ SU 4 (q 1
2 ) 2 × Ω + 6 (q 1
2 ) 2 × Spin + 8 (q 1
2 ) × SU 9 (q

2 ) 2 × Ω + 10 (q 1
2 ) × SL 16 (q) × Ω + 16 (q 1
2 ) 2 ×Ω + 18 (q 1
2 ) × Ω + 24 (q 1
2 ) 4 × SU 25 (q 1
2 ) × Ω + 30 (q 1
2 ) 2 × SU 36 (q 1
2 ) × Ω + 40 (q 1
2 ) × Ω + 48 (q 1
2 ).

Proof. We have the factorizations by the previous propositions. We thus only have to show that the considered morphisms are surjective. Since A H 4 is perfect, we have by Goursat's Lemma 3.3.1 that the morphism is surjective unless there exists two different representations ρ 1 and ρ 2 in the decomposition such that there exists a field automorphism Ψ verifying Ψ•ρ 1|A H 4 ρ 2|A H 4 . By Proposition 2.1.2, we have that Ψ(α + α -1 ) = α + α -1 . This shows that Ψ must be trivial over F p (α + α -1 ). It follows by the previous propositions that there are no such representations in the decompositions and the proof is concluded.

Chapter 9

Type F 4

In this section, we determine the image of the Artin group of type F 4 . This group contains in a natural way two groups isomorphic to A B 3 . We will therefore use some results from the type B section. The F 4 -graphs are graphs with two parameters since there are two conjugacy classes in type F 4 . This makes the uniqueness conditions proven in the section on W -graphs fail. The irreducible representations are all in low dimension. We can therefore compute easily the bilinear forms for the self-dual representations and it is not necessary to use the conjecture on W -graphs associated to self-dual representations. It is unclear whether the conjecture is true in type F 4 because of the lack of rigidity when there are two parameters. The field extensions involved in type F 4 are quite complicated, we have to distinguish 15 cases which are described in the Appendix. Determining the image inside each representation uses arguments similar to the other cases. The main difference comes from the fact that A F 4 is not the normal closure of A B 3 in A F 4 , we use the fact that there are two copies of A B 3 inside A F 4 for the inductive arguments.

The image inside the full Iwahori-Hecke algebra is also slightly more complicated because A F 4 is not perfect. This requires some additional computations for the proof of Theorem 9.1.

Let p be a prime different from 2 and 3. Let α, β ∈ F p such that α 4 = 1, α 6 = 1, α 10 = 1,

β 4 = 1, β 6 = 1, β 10 = 1, ( α β ) 6 = 1, ( α β ) 4 = 1, (αβ) 6 = 1, (αβ) 4 = 1, α / ∈ {-β 2 , -β -2 } and β / ∈ {-α 2 , -α -2 }. Write F q = F p (α, β).
There are 25 irreducible representations of H F 4 ,α,β . The highest dimensional one is of dimension 16. Five of them are self-dual and they are represented by the F 4 -graphs 4 1 , 6 1 , 6 2 , 12 and 16 given in section 10.5 of the Appendix. Definition 9.1. The Iwahori-Hecke algebra H F 4 ,α,β of type F 4 is the F q -algebra generated by the generators S 1 , S 2 , S 3 , S 4 and the following relations

1. (S 1 -α)(S 1 + 1) = (S 2 -α)(S 2 + 1) = 0. 2. (S 3 -β)(S 3 + 1) = (S 4 -β)(S 4 + 1) = 0. 3. S 1 S 2 S 1 = S 2 S 1 S 2 . 4. S 1 S 3 = S 3 S 1 . 5. S 1 S 4 = S 4 S 1 . 6. S 2 S 3 S 2 S 3 = S 3 S 2 S 3 S 2 . 7. S 2 S 4 = S 4 S 2 . 8. S 3 S 4 S 3 = S 4 S 3 S 4 . For σ in the Coxeter group F 4 , if σ = s i 1 . . . s i k is a reduced expression we set T σ = S i 1 . . . S i k .
Note that we have two parameters here, therefore most of the results from the Chapter 6 do not hold. Moreover, we have to consider the different ways F p ( √ α, √ β) can be a field extension of F p (α + α -1 , β + β -1 ). The Hasse diagram representing the setup of the field extensions involved is given Figure 10.4.

Note that all the extensions represented by edges are of degree at most 2 because they involve of the following polynomials :

X 2 -( √ β + √ β -1 )X +1, X 2 -α, X 2 -β, X 2 -( √ α+ √ α -1 )X +1, X 2 -(β + β -1 + 2), X 2 -(β + β -1 )X + 1, X 2 -(α + α -1 )X + 1, X 2 -(α + α -1 + 2)
. The roles of α and β are perfectly symmetric in the graph, therefore we only have to consider the cases up to permutation of α and β. We now try to establish what all the possibilities are. Set

F q = F p ( √ α, √ β). Assume first that F p ( √ α, √ β) = F p ( √ α, √ β + √ β -1 ). We write q = q 1 2 . Then F q2 = F p ( √ α, √ β) = F p ( √ α, √ β + √ β -1 )/(X 2 -( √ β + √ β -1 )X + 1
). The field F q2 has a unique subfield of degree 2 and it is equal to F q. We have that √ β does not belong to F q, therefore

F q = F p ( √ α + √ α -1 , √ β) and F q = F p (α, √ β).
Note that in this case, we also have that β / ∈ F q because otherwise, we would have

√ β = √ β 1+β -1 1+β -1 = √ β+ √ β -1 1+β -1 ∈ F q. It follows that F q = F p ( √ α, β), and that F q2 = F p ( √ α, √ β) = F p √ α, β) = F p (α, √ β) = F p ( √ α + √ α -1 , √ β). We have in the same way F q2 = F p ( √ α + √ α -1 , β) = F p (α, β) = F p (α + α -1 , √ β) = F p (α + α -1 , β). This implies that F p ( √ α, √ β
) is an extension of degree at most 2 of F p (α + α -1 , β + β -1 ).

F p (α + α -1 , β + β -1 ) is included in F p ( √ α, √ β + √ β -1 ) = F q, therefore it is equal to F q.
We can now complete all the edges in the Hasse diagram where we put dotted edges for extensions of degree 1, full edges for extensions of degree 2, fields equal to F q2 in blue and fields equal to F q in red. This can be seen in Figure 10.5. We get by symmetry that if We can now assume that F

F p ( √ α, √ β) = F p ( √ α + √ α -1 , √ β
p ( √ α, √ β) = F p ( √ α, √ β + √ β -1 ) = F p ( √ α + √ α -1 , √ β). Assume F p ( √ α, √ β) = F p ( √ α, β). We then write F q = F p ( √ α, β) and F q2 = F p ( √ α, √ β). We then have that √ β / ∈ F q, therefore F q2 = F p (α, √ β) = F p (α + α -1 , √ β). Note that √ β = √ β+ √ β -1 1+β -1 , therefore √ β + √ β -1 / ∈ F q. It follows that F q2 = F p (α, √ β + √ β + √ β -1 ) = F p (α + α -1 , √ β + √ β -1
). This proves also that F q = F p (α + α -1 , β + β -1 ). The only Hasse diagram possible is then given in Figure 10.7.

We get by symmetry that if

F p ( √ α, √ β) = F p (α, √

We can now assume that

F p ( √ α, √ β) = F p ( √ α, √ β+ √ β+ √ β -1 ) = F p ( √ α, β) = F p (α, √ β) = F p ( √ α + √ α -1 , √ β). Assume F q = F p ( √ α, β + β -1 ) = F q . We then get that β / ∈ F q and √ β + √ β -1 / ∈ F q.
The only Hasse diagram possible is then given in Figure 10.9.

We get by symmetry that if F p (α + α -1 , √ β) = F q then we get the Hasse diagram given in Figure 10.10.

Assume now that F

q = F p ( √ α + √ α -1 , β) = F q . We then have that √ α / ∈ F q and √ β / ∈ F q. It follows that α / ∈ F q since √ α = √ α+ √ α -1
1+α -1 . We also have that

√ β + √ β -1 / ∈ F q since √ β = √ β+ √ β -1 1+β -1
. The only Hasse diagram possible is then given in Figure 10.11.

By symmetry, if F p (α, √ β + √ β -1 ) = F q then we get the Hasse diagram given in Figure 10.12. Assume now F q = F p (α, β) = F q . We then get that

√ α + √ α -1 / ∈ F q and √ β + √ β -1 / ∈ F q.
We then get the Hasse diagram given in Figure 10. [START_REF] Carter | Simple groups of Lie type[END_REF].

Assume now

F q = F p ( √ α + √ α -1 , √ β + √ β -1 ) = F q .
We have α / ∈ F q and β / ∈ F q. We then get the Hasse diagram given in Figure 10.14.

We can now assume that

F q = F p ( √ α, β+β-1) = F p ( √ α+ √ α -1 , β) = F p ( √ α+ √ α -1 , √ β+ √ β -1 ) = F p (α, β) = F p (α, √ β + √ β -1 ) = F p (α+α -1 ). Assume F q = F p ( √ α+ √ α -1 , β +β -1 ) = F q . We then have α / ∈ F q, β / ∈ F q and √ β + √ β -1 / ∈ F q.
This gives us the Hasse diagram in Figure 10. [START_REF] Coxeter | The complete enumeration of finite groups of the form r 2 i = (r i r j ) k i,j = 1[END_REF].

By symmetry, if

F p (α + α -1 , √ β + √ β -1 ) = F q then we get in Hasse diagram given in Figure 10.16. Assume now F q = F p (α+α -1 , β) = F q . We have α / ∈ F q, √ α+ √ α -1 / ∈ F q and √ β + √ β -1 / ∈ F q.
We then get the Hasse diagram given in Figure 10.17.

By symmetry, if F p (α, β + β -1 ) = F q we get the Hasse diagram given in Figure 10.18.

We can now assume

F q = F p ( √ α + √ α -1 , β + β -1 ) = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , √ β + √ β -1
). We then either have case 1 which is The F 4 -graphs we will be considering are given in Figures 10.1, 10.2 and 10.3. They are taken from [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] (11.3.2). They are slightly different because we consider left-actions instead of right actions. We here have two parameters, therefore the rules to read the F 4 -graphs are more complicated. When there is an edge between x and y in the graph then as in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF], we use the following conventions 1. µ s

F p ( √ α, √ β) = F p (α,
x,y = 0 if one of the following is satisfied (a) I(x) consists of three elements, I(y) consists of one element and s = s 1 .

(b) s = s 2 , {s 2 , s 4 } ⊂ I(x) and I(y) contains s 3 but not s 4 .

2. If the above conditions are not verified then µ s x,y = 1 except if (a) The is an integer labeling the edge between x and y, then µ s x,y is equal to that integer. (b) If s ∈ {s 3 , s 4 } and there is a label on the edge between x and y which is not an integer, then µ s x,y is equal to that label, where

a = √ u √ v -1 + √ v √ u -1 , b = u √ v -1 + √ vu -1 f = √ u + √ u -1 , g = u + u -1 , h = √ u 3 + √ u -3
.

We now prove that the algebra is split semisimple as we did with the other types using the Schur elements.

= Z[ √ u ±1 , √ v ±1 ] and F = Q( √ u, √ v).
We have a symetrizing trace defined by τ (T 0 ) = 1 and τ (T σ ) = 0 for all σ ∈ F 4 \ {1 F 4 }. H F 4 ,u,v is then a free symmetric F -algebra of rank 1152. A is integrally closed. Let θ be the ring homomorphism from A to L = F q defined by θ(

√ u) = √ α, θ( √ v) = √ β and θ(k) = k.
We know F H is split. The basis formed by the elements T σ , σ ∈ F 4 verifies the conditions of the Proposition 2.2.4. The F 4 -graphs considered remain connected since the weights don't vanish after specialization. In order to verify this, we only need to check that θ

(a) = 0, θ(b) = 0, θ(f ) = 0, θ(g + 1) = 0, θ(2 -g) = 0 and θ(h) = 0. We have θ(a) = √ α √ β -1 + √ β √ α -1 = √ α -1 √ βΦ 2 ( α β ) = 0, θ(b) = α √ β -1 + √ βα -1 = α √ β -1 (1 + βα -2 ) = 0, θ(f ) = √ α + √ α -1 = √ α -1 Φ 2 (α) = 0, θ(g + 1) = α + α -1 + 1 = α -1 Φ 3 (α) = 0, θ(2 -g) = 2 -α -α -1 = (1 -α)(1 -α -1 ) and θ( √ u 3 + √ u -3 ) = √ α 3 + √ α -3 = √ α -3 (α 3 + 1) = 0.
We now only need to check that the Schur elements can be specialized and do not vanish under the specialization. This is clear from Table 9 . We know by Lemma 3.2.3 that there exists a character χ; A F 4 → F q such that ρ ρ ⊗ χ. Since F 4 /A F 4 < s 1 , s 3 > Z 2 , there exists (x, y) ∈ F 2 q such that ρ(S 1 ) = xρ (S 1 ) and ρ(S 3 ) = yρ (S 3 ). The eigenvalues of ρ(S 1 ) are -1 and α, therefore {-1, α} = {-x, xα}. It follows that x = 1 since α 2 = 1. In the same way y = 1 since β 2 = 1. It follows that ρ ρ .

Before determining the image of the Artin groups inside this Iwahori-Hecke algebra, we need as in the other cases a Lemma on Artin groups which will allow us to use the restriction from E 6 to D 5 .

Lemma 9.1. A F 4 is generated by

A B 3 ,1 and A B 3 ,2 , where we identify A B 3 ,1 (resp A B 3 ,2 ) as a subgroup of A F 4 using the natural isomorphism from A B 3 to < s 1 , s 2 , s 3 > (resp < s 2 , s 3 , s 4 >).
Proof. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], we have that A F 4 is generated by s 2 s -1 1 , s 1 s 2 s -2 1 , s 3 s -1 4 and s 4 s 3 s -2 4 . We have that s 2 s -1

1 and s 1 s 2 s -2 1 belong to A B 3 ,1 and s 3 s -1 4 and s 4 s 3 s -2 4 are in A B 3 ,2 , therefore the proof is complete.

We now give a proposition where we determine the image of A F 4 with respect to the representations ρ 2 3 , ρ 4 1 and ρ 4 2 . The special phenomenon appearing for the representation ρ 4 1 was also observed in the generic case, see Lemma 2.22 of [START_REF] Marin | Infinitesimal Hecke algebras II[END_REF] for an analogous result.

(∅, [1 3 ]) ([3], ∅) ([1 3 ], ∅) (∅, [3]) ([2, 1], ∅) (∅, [2, 1]) ([2], [1]) ([1], [1 2 ]) ([1], [2]) ([1 2 ], [1]) 8 2 1 1 1 1 8 2 1 1 1 1 8 3 1 1 1 8 3 1 1 1
2 ) or SP 8 (q ). The symplectic case is exluded because ρ 8 2 ρ 8 2 . In cases 4, 5 and 10, we have q = q 2 and ρ 8

2 (A B 3 ,1 ) SL 3 (q) 2 × SL 2 (r) if F r = F p (α) = F p (α + α -1 ) and ρ 8 2 (A B 3 ,1 ) SL 3 (q) 2 × SU 2 (r 1 2 ) if F r = F p (α) = F p (α + α -1
). It follows that q divides q since G 8 2 contains a natural SL 3 (q). Since q = q 2 , there exists a unique automorphism Φ of order 2 of F q 2 . For any representation ϕ of H B 3 ,1 , we have Φ • ϕ ϕ, therefore by Table 9.2, we have that Φ • ρ 8 2 ρ 8 2 . It follows by Lemma 3.2.5 that up to conjugation in GL 8 (q 2 ), G 8 2 is a subgroup of SL 8 (q). Furthermore, G 8 2 SL 8 (q) because SU 8 (q 1 2 ) is not conjugate in GL 8 (q 2 ) to a subgroup of SL 8 (q).

In cases 1, 11, 12, 13 and 16, we have q = q and G 8 2 contains either a natural SL 3 (q) or a natural SU 3 (q 1 2 ). We then have by Lemma 3.3.3 that q = q. Hence G 8 2 = SL 8 (q) in case 1 and G 8 2 SU 8 (q

2 ) in cases 11, 12, 13 and 16. Assume now

F p ( √ α, √ β) = F p (α, β) = F q = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ). Then ρ 8 2 (A B 3 ,1 ) SL 3 (q) × SU 2 (r 1
2 ) if F r = F p (α) = F p (α + α -1 ) and ρ 8 2 (A B 3 ,1 ) SL 3 (q) × SL 2 (r) if F r = F p (α) = F p (α + α -1
). There exists a unique automorphism of order 2 of F q and if we consider the representations appearing in the restriction of ρ

8 2 to H B 3 ,1 , then • ρ [2],[1] ρ [1],[2] and • ρ [1 2 ],[1] ρ [1],[1
(q), G 8 2 is a subgroup of SU 8 (q 1
2 ). By the exact same arguments as in the proof of Lemma 3.3.8, we conclude that G 8 2 SU 8 (q

2 ). This corresponds to cases 2, 6, 9 and 15.

Assume now

F p ( √ α, √ β) = F p (α, β) = F q = F p (α, β + β -1 ) = F p (α + α -1 , β) = F p (α + α -1 , β + β -1 ). Then ρ 8 2 (A B 3 ,1 ) SL 3 (q) × SU 2 (r 1
2 ), where F r = F p (α) = F p (α + α -1 ). There exists a unique automorphism of order 2 of F q and if we consider the representations appearing in the restriction of ρ [1] by Proposition 3.2.4. It follows by Table 9.2 that • ρ 8 2 ρ 8 2 . We then have by Lemma 3.2.5 that up to conjugation in GL 8 (q), G 8 2 is a subgroup of SL 8 (q 1 2 ). We can then again use the arguments of the proof of Lemma 3.3.8 to get that G 8 2 SL 8 (q

8 2 to H B 3 ,1 , • ρ [2,1] ρ [1 2 ],[1] and • ρ [1 2 ],[1] ρ [2],
2 ). This corresponds to cases 3, 7, 8 and 14. The results for the representation ρ 8 3 are symmetric with regards to α and β to the results for ρ 8 2 since Tables 9.2 and 9.3 are identical after permutation of 8 2 and 8 3 . Proposition 9.5. In cases 11, 12, 13 and 16, we have ρ 9

1 (A F 4 ) SU 9 (q 1
2 ) and ρ 9 2 (A F 4 ) SU 9 (q 1 2 ). In all the remaining case, we have ρ 9 1 (A F 4 ) SL 9 (q) and ρ 9 2 (A F 4 ) SL 9 (q). In cases 2, 6, 9 and 15, if is the unique automorphism of order 2 of F q then • ρ 9

1 |A F 4 ρ 9 2 |A F 4 .
In cases 3, 7, 8 and 14, if is the unique automorphism of order 2 of F q then • ρ 9 ) is generated by transvections. We get by Theorem 2.3.4 that there exists q dividing q such that G 9 1 is conjugate in GL 9 (q) to SL 9 (q ), SU 9 (q 1 2 ) or SP 9 (q ). The symplectic case is excluded since we have ρ 9 1 ρ 9 1 . In case 1, we have q = q and F p (α, β) = F p (α + α -1 , β + β -1 ). By Theorem 3.2.1, G 9 1 contains a natural SL 3 (q), therefore q = q and G 9 1 = SL 9 (q). In cases 4, 5 and 10, we have q = q 2 and F q = F p (α, β) = F p (α + α -1 , β + β -1 ). We then have again by Theorem 3.2.1 that G 9 1 contains a natural SL 3 (q), therefore q ∈ {q, q 2 }. There exists a unique automorphism Φ of order 2 of F q 2 and if we consider the restrictions of ρ

1 |A F 4 ρ 9 2 |A F 4 . Proof. Let F q = F p ( √ α, √ β) and G 9 1 = ρ 9 1 (A F 4
9 1 to H B 3 ,1 , then Φ • ρ [2],[1]|A F 4 ρ [2],[1]|A F 4 and Φ • ρ [1],[2]|A F 4 ρ [1],[2]|A F 4 . By Table 9.4, we have either Φ • ρ 9 1 |A F 4 ρ 9 1 |A F 4 or Φ • ρ 9 1 |A F 4 ρ 9 2 |A F 4 . Assume now by contradiction that Φ • ρ 9 1 |A F 4 ρ 9 2 |A F 4 . There exists a character χ from A F 4 to F q 2 such that Φ • ρ 9 1 ρ 9 2 ⊗ χ by Lemma 3.2.3. Since A F 4 /A F 4 < S 2 , S 3 >, it
follows that there exists y ∈ F q 2 such that Φ(ρ 9 1 (S 3 )) is conjugate in GL 9 (q) to yρ 9 2 (S 3 ). The eigenvalues of ρ 9 1 (S 3 ) are β with multiplicity 3 and -1 with multiplicity 6, therefore the eigenvalues of Φ(ρ 9 1 (S 3 )) are β with multiplicity 6 and -1 with multiplicity 3. On the other hand, the eigenvalues of yρ 9 2 (S 3 ) are yβ with multiplicity 3 and -y with multiplicity 6. This implies that yβ = -1 and -y = β. It follows that y = -β = -β -1 , therefore β 2 = 1 which is absurd. This proves that Φ • ρ 9 1 ρ 9 1 . Hence, up to conjugation in GL 9 (q), G 9 1 is a subgroup of SL 9 (q) by Lemma 3.2.4. It follows that G 9 1 SL 9 (q) in cases 4, 5 and 10.

In cases 11, 12, 13 and 16 we have q = q and G 9 1 contains a natural SU 3 (q

2 ). By Lemma 3.3.3, we have q = q. There exists a unique automorphism of order 2 of F q and for any representation ϕ appearing in the decomposition of the restriction of ρ

9 1 from H F 4 to H B 3 ,1 , we have • ϕ ϕ. It follows that • ρ 9 1 ρ 9 1 or • ρ 9 1 ρ 9 2 .
The eigenvalues of (ρ 9 1 (S 3 )) are β -1 with multiplicity 3 and -1 with multiplicity 6, whereas the eigenvalues of vρ 9 2 (S 3 ) are vβ with multiplicity 3 and -v with multiplicity 6. We get in the same way as in cases 4, 5 and 10 that • ρ 9 1 ρ 9 1 . By Lemma 3.2.5, we get that G 9 2 is conjugate to a subgroup of SU 9 (q

1 2 ), therefore G 9 2 SU 9 (q 1 2 ). Assume now F p ( √ α, √ β) = F p (α, β) = F q = F p (α + α -1 , β) = F p (α, β + β -1 ) = F p (α + α -1 , β + β -1 ). We then have ρ 9 1 (A B 3 ,1 ) SL 3 (q) × SU 2 (r 1 2 ) if F r = F p (α) = F p (α + α -1 ) and ρ 9 1 (A B 3 ,1 ) SL 3 (q)×SL 2 (r) if F r = F p (α) = F p (α+α -1
). There exists a unique automorphism of order 2 of F q and if we consider the representations appearing in the restriction of ρ 9 1 to H B 3 ,1 , we have by Proposition 3.2.4 that

• ρ [2],[1] ρ [1],[2] and • ρ [1],[2] ρ [2],[1] . It follows that •ρ 9 1 |A F 4 ρ 9 1 |A F 4 or •ρ 9 1 |A F 4 ρ 9 2 |A F 4 .
We have here (α) = α and (β) = β -1 . Assume now by contradiction that • ρ 9 1 |A F 4 ρ 9 1 |A F 4 . There exists v ∈ F q such that (ρ 9 1 )(S 3 ) conjugate to vρ 9 1 (S 3 ). The eigenvalues of (ρ 9 1 (S 3 )) are β -1 with multiplicity 6 and -1 with multiplicity ) to a subgroup of Ω + 6 (q). Since it contains up to conjuugation a twisted diagonal SL 3 (q), it contains up to conjugation diag(α, α -1 , 1, α -1 , α, 1) and diag(β, β -1 , 1, β -1 , β, 1). This implies the field generated by the traces of its elements contains 2(α + α -1 ) + 2 and 2(β + β -1 ) + 2. It follows that q = q.

If G 6 1 SU 3 (q) then we would have that SL 3 (q) is isomorphic to a subgroup of SU 3 (q), therefore q 3 (q 2 -1)(q 3 -1) divides q 3 (q 2 -1)(q 3 + 1). This would imply that q 3 -1 divides q 3 + 1, therefore q 3 -1 divides q 3 + 1 -(q 3 -1) = 2. This is absurd, therefore G 6 1 SU 3 (q). Assume now by contradiction that G 6 1 Ω - 6 (q

2 ). We have |Ω - 6 (q

2 )| = 1 2 q 3 (q 3 2 + 1)(q -1)(q 2 -1) and |SL 3 (q)| = q 3 (q 2 -1)(q 3 -1). Since Gcd(q 3 -1, q 3 ) = 1, q 3 -1 divides (q

3 2 + 1)(q -1) = q 5 2 -q 3 2 + q -1 < q 5 2 -1 < q 3 -1. This is absurd, therefore G 6 1 Ω - 6 (q 1
2 ). It follows that in cases 4, 5 and 10 , we have G 6 1 Ω + 6 (q). We can now assume q = q, therefore q divides q. By the same reasoning as above using Lemma 3.3.3, we have that F q contains F p (α + α -1 , β + β -1 ) in all the remaining cases. In case 1, we can apply the same reasoning as above to get that G 6 1 Ω + 6 (q). In cases 11, 12, 13 and 16, there exists a unique automorphism of order 2 of F q and for any representation ϕ of H B 3 ,1 , we have • ϕ ϕ. It then follows by Table 9.

6 that • ρ 6 1 |A F 4 ρ 6 1 |A F 4 .
Then, by [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] (Proposition 4.1.), G 6 1 is conjugate in GL 6 (q) to a subgroup of Ω + 6 (q

2 ). It follows that q divides q 1 2 , therefore q = q

1 2 since F q contains F p (α + α -1 , β + β -1 ) = F q 1 2 .
This proves that G 6 1 is isomorphic to one of the following groups 1. Ω + 6 (q

2 ) in a natural representation, 2. Ω - 6 (q

4 ) as a subgroup of Ω + 6 (q

2 ), where Ω + 6 (q

2 ) is in a natural representation, 3. SU 3 (q

2 ) as a subgroup of Ω + 6 (q

2 ), where Ω + 6 (q

2 ) in a natural representation.

Assume now by contradiction that G 6 1 SU 3 (q

2 ). We know that ρ 6 1 (A B 3 ,1 ) SU 3 (q

2 ). We only need to show that ρ 6

1 (A B 3 ,1 ) = G 6 1 . Let X =                 0 1 0 0 0 α 2 +β α √ β √ α(β+1) αβ+1 -α+β αβ+1 0 0 (β+1)(α 2 +β) √ α √ β(αβ+1) -(α+β)(α 2 +β) α √ β(αβ+1) -α 2 +β √ α(αβ+1) 0 (α 2 +β)(α+β) α √ β(αβ+1) 0 -(α 4 +β) 2 α √ α √ β(αβ+1) 0 1 -α+β √ α(β+1) 0 α 2 +β α(β+1) 0 0 0 αβ+1 √ α(β+1) 0 -(αβ+1)(α 2 +β) α(α+β)(β+1) 0 0 0 0 0 αβ+1 α+β 0 0                
.

We have with respect to the previous basis that for all g ∈ A B 3 ,1 ,

Xρ 6 1 (g)X -1 = ρ [1],[2] (g) 0 0 ρ [1 2 ],[1] (g) .
Note that here, the matrix ρ 6 1 (q) with respect to the previous basis is again denoted by the same symbol in order to simplify the notations.

By Tables 9.6 and 9.7, we only need to check that ρ 6 2 (A F 4 ) = ρ 6 2 (A B 3 ,1 ) to get the same results for the representation ρ We consider the matrices corresponding to the representation ρ 6 2 with respect to the basis (e x 1 , e x 2 , e x 3 , e x 4 , e x 5 , e x 6 ) with I(x 1 ) = {s 1 , s 2 }, I(x 2 ) = {s 1 , s 3 }, I(x 3 ) = {s 1 , s 4 }, I(x 4 ) = {s 2 , s 3 }, I(x 5 ) = {s 2 , s 4 }, I(x 6 ) = {s 3 , s 4 }. We then have that for all g ∈ H B 3 ,1 ,

Y ρ 6 2 (g)Y -1 = ρ [1],[2] (g) 0 0 ρ [2],[ 1 2
] (g) .

We have that (Y ρ 6 2 (S 3 S -1 4 )Y -1 ) 1,5 = -α+β √ β = 0, therefore G 6 2 = ρ 6 2 (A B 3 ,1 ) and the proof is concluded.

Proposition 9.7. In cases 1, 4, 5 and 10, we have ρ 12 (A F 4 ) Ω + 12 (q) and ρ 16 (A F 4 ) Ω + 16 (q). In all the remaining cases, we have ρ 12 (A F 4 ) Ω + 12 (q

2 ) and ρ 16 (A F 4 ) Ω + 16 (q).

Proof. Let F q = F p ( √ α, √ β), G 12 = ρ 12 (A F 4 ) and G 16 = ρ 16 (A F 4 ). In cases 1, 4, 5, 10, 11, 12, 13 and 16, G 12 contains a twisted diagonal SL 3 (q) or a twisted diagonal SU 3 (q 1 2 ) by Table 9.8 ρ 12|A F 4 . This implies by Lemma 3.2.5 that G 12 is conjugate to a subgroup of Ω + 12 (q

2 ). We know by Theorem 3.2.5 and Tables 9.8 and 9.9 that ρ 12 (A B 3 ,1 ) SL 3 (q) and ρ 12 (A B 3 ,2 ) SL 3 (q). We have that G 12 is primitive, irreducible and that the field generated by its traces contains F q 1 2 . Let us show that G 12 is also tensor-indecomposable. If it was tensor-decomposable then we would have G 12 ≤ GL 2 (q 1 2 ) ⊗ GL 6 (q 1 2 ) or G 12 ≤ GL 3 (q 1 2 ) ⊗ GL 4 (q 1 2 ). By the same argument as in the above cases, we have that G 12 ≤ GL 2 (q 1 2 ) ⊗ GL 6 (q 1 2 ) is impossible. Assume now by contradiction that G 12 ≤ GL 3 (q 1 2 ) ⊗ GL 4 (q 1 2 ). We would then have a morphism from G 12 to GL 3 (q 1 2 ). The restriction of this morphism from G 12 to ρ 12 (A B 3 ,1 ) or to ρ 12 (A B 3 ,2 ) gives us a morphism from SL 3 (q) to GL 3 (q 1 2 ). If this morphism is non-trivial, we get an isomorphism from P SL 3 (q) to a subgroup of GL 3 (q 1 2 ). This is absurd since |P SL 3 (q)| = 1 (3,q-1) q 3 (q 2 -1)(q 3 -1), |GL 3 (q 1 2 )| = q 3 2 (q 1 2 -1)(q -1)(q 3 2 -1) and q 3 does not divide q 3 2 . This proves that the restriction of this morphism to ρ 12 (A B 3 ,1 ) and ρ 12 (A B 3 ,2 ) is trivial. It follows by Lemma 9.1 that this morphism is trivial and it contradicts the irreducibility of G 12 . This proves that G 12 is tensor-indecomposable. We will now explicitely find a matrix g ∈ G 12 such that dim(g -I 12 ) = 2. We consider our matrices in the same basis as before.

Let X = (X 1 X 2 X 3 ) and Y = (Y 1 Y 2 Y 3 Y 4 ), where X 1 , X 2 , X 3 and Y 1 , Y 2 , Y 3 and Y 4 are given in section 10.7 of the Appendix.

We then have that We then have L 3 = [L 2 , L 1 ] ∈ XG 12 X -1 . We have

Xρ 12|A B 3 ,1 X -1 =       ρ [2],[1] 0 0 0 0 ρ [2],[1] 0 0 0 0 • ρ [2],[1] 0 0 0 0 • ρ [2], [1]
L 3 = I 12 + 2 Φ 6 (α)Φ 2 (αβ) (α 2 β + 1)(β + 1)(α + 1) 2 E 6,9 -2 √ αΦ 6 (α)Φ 2 (αβ) 2 Φ 2 (α) 3 (α + β)Φ 2 (α 2 β)Φ 2 (β) E 12,3 .
Hence, dim((L 3 -I 12 )V ) = 2. We can then apply Theorem 2.3.2 to get that G 12 is a classical group in a natural representation over F q for some q dividing q. We have that q divides q 1 2 since G 12 is conjugate to a subgroup of Ω + 12 (q

2 ). We have diag(α, α -1 , 1, α -1 , α, 1, α, α -1 , 1, α -1 , α, 1) ∈ XG 12 X -1 , diag(β, β -1 , 1, β -1 , β, 1, β -1 , β, 1, β, β -1 , 1) ∈ XG 12 X -1 therefore the field F q generated by the traces of the elements of G 12 contains 4(α + α -1 + 1) and 4(β + β -1 + 1). It follows that F q contains F p (α + α -1 , β + β -1 ) = F q 1 2 , therefore q = q 1 2 . We can then conclude that G 12 Ω + 12 (q 1 2 ). By complete symmetry of α and β for the 12-dimensional representation, we get that in cases 3, 7, 8 and 14, we have G 12 Ω + 12 (q

2 ).

Consider now the 16-dimensional representation. We consider the matrices corresponding to the representation ρ 16 with respect to the basis (e x i ) i∈[ [1,[START_REF] Curtis | Methods of representation theory[END_REF]] , where (I(x i )) i∈[ [1,[START_REF] Curtis | Methods of representation theory[END_REF]] is ordered lexicographically i.e. I(x 1 ) = {s 2 }, I(x 2 ) = {s 3 }, I(x 3 ) = {s 1 , s 2 }, I(x 4 ) = I(x 5 ) = I(x 6 ) = {s 1 , s 3 }, I(x 7 ) = I(x 8 ) = {s 2 , s 4 } and I(x i ) = {s 1 , s 2 , s 3 , s 4 } \ I(x 17-i ) for all i ∈ [ [START_REF] Bray | The maximal subgroups of the lowdimensional finite classical groups[END_REF][START_REF] Curtis | Methods of representation theory[END_REF]] and such that µ s 1 x 4 ,x 2 = 1, µ s 3 x 5 ,x 3 = -1, µ s 2 x 10 ,x 2 = 1, µ s 2 x 15 ,x 7 = 1, µ s 3 x 14 ,x 11 = 2 and µ s 1 x 15 ,x 13 = 1. Let P 16 be the symmetric antidiagonal matrix with coefficients with coefficients in the first 8 rows respectively equal to 1, 1, -1, -1, 1, -2, -1 and 2. We then have for all g ∈ G 16 , P gP -1 = t g -1 . Since P is symmetric, we get that G 16 is up to conjugation a subgroup of Ω + 16 (q). Note that by Table 9 (q). In all the remaining cases, we have q = q. By Lemma 3.3.6, we have G 16 tensor-indecomposable or G 16 ≤ GL 2 (q) ⊗ GL 8 (q). Depending on the cases, we have that G 16 contains up to conjugation either or

H 3 =                        M 0 0 0 0 0 t M -1 0 0 0 0 0 (M ) 0 0 0 0 0 ( t M -1 ) 0 0 0 0 0 I 4         , M ∈ SL 3 (q)               
where is the unique automorphism of F q (H 2 and H 3 only appear when F p (α, β) = F p (α + α -1 , β + β -1 )). Assume by contradiction that G 16 ≤ GL 2 (q) ⊗ GL 8 (q). The restriction of the morphism from G 16 to GL 2 (q) to H 1 , H 2 or H 3 is then trivial since |P SL 3 (q)| does not divide |GL 2 (q)|. This implies that the eigenvalues of semisimple elements of H 1 , H 2 or H 3 all have multiplicity divisible by 8. This is absurd because there exists ξ ∈ F q such that ξ 2 = 1, H 1 and H 2 contain elements conjugate to diag(ξ, ξ, ξ -1 , ξ -1 , 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) or H 3 contains an element conjugate to diag(ξ, ξ, ξ, ξ, ξ -1 , ξ -1 , ξ -1 , ξ -1 , 1, 1, 1, 1, 1, 1, 1, 1). It follows that G 16 is tensor-indecomposable in all cases. We also have that in all cases G 16 contains up to conjugation diag(α, α, α, α, α -1 , α -1 , α -1 , α -1 , 1, 1, 1, 1, 1, 1, 1, 1) and diag(β, β, β, β, β -1 , β -1 , β -1 , β -1 , 1, 1, 1, 1, 1, 1, 1, 1), therefore the field F q generated by the traces of the elements of G 16 contains F p (α + α -1 , β + β -1 ). Hence, by Theorem 2.3.2, G 16 is a classical group in a natural representation over F q .

Assume now ρ 6 1 |A F 4 (Ψ • ρ 6 2 |A F 4 ) ⊗ z. We have )) = -2α -1 (α -1) 2 = 0.

Tr(ρ 6 1 (q 0 )) = Tr(ρ 6 1 (q 1 )) = Tr(ρ 6 1 (q 0 q -1 1 )) = Tr(ρ 6 2 (q 0 )) = Tr(ρ 6 2 (q 1 )) = Tr(ρ 6 2 (q 0 q -1 1 )) = -2β -1 (β -1) 2 = 0.

It follows that z(p 0 ) = z(p 1 ) = z(q 0 ) = z(q 1 ) = 1. We then have by Proposition 2.1.2 that Ψ is trivial therefore ρ It follows that z(p 0 ) = z(p 1 ) = z(p 0 p -1 1 ) = 1. We also have Tr(ρ 8 2 (q 0 )) = Tr(ρ 8 2 (q 1 )) = Tr(ρ 8 2 (q 0 q -1 1 )) = 6 -2(β + β -1 ), Tr(ρ 8 3 (q 0 )) = Tr(ρ 8 3 (q 1 )) = Tr(ρ 8 3 (q 0 q -1 1 )) = 5 -3(β + β -1 ). It follows that z(q 0 ) = z(q 1 ) = 1. We then have that z is trivial and Ψ is trivial therefore ρ )) = -3α -1 (α-1) 2 = 0.

Tr(ρ 91 (q 0 )) = Tr(ρ 91 (q 1 )) = Tr(ρ 91 (q 0 q -1 1 )) = Tr(ρ 92 (q 0 )) = Tr(ρ 92 (q 1 )) = Tr(ρ 92 (q 0 q -1 1 )) = -3β -1 (β-1) 2 = 0.

It follows that z(p 0 ) = z(p 1 ) = z(q 0 ) = z(q 1 ) = 1. We then have by Proposition 2.1.2 that Ψ is trivial therefore ρ 9 1 |A F 4 ρ 9 2 |A F 4 . All the computations give the same results if we substitute ρ 9 3 by ρ 9 3 . This concludes the proof in cases 1, 4, 5 and 10.

In cases 11, 12, 13 and 16, we have the result using the same computations since Ψ is either the trivial automorphism or the unique automorphism of order 2 of F q if z is trivial.

In [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], after Lemma 5.3, Proposition 2.4. is used in order to show that a morphism from R(A An ) to P SL 2 (q) must be trivial. It is used for n ≥ 6 eventhough Proposition 2.4 is only valid for n ≥ 7. The proof is therefore not correct for n = 6. The only partition to consider is [3,[START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]1] which is of dimension 16. By the induction assumption, we have R [3,2,1] (A A 5 ) SL 5 (q)×SL 6 (q). It follows by Lemma 8.5.1 of this Ph.D. thesis that the morphism to SL 2 (q) is indeed trivial and concludes the proof.

H 4 -graphs

Below are the W -graphs we use for type H 4 , they are taken from [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] except for the new ones we found verifying the properties of Theorem 6.2 ( 16 t , 18 r , 24 s , 24 t , 30 s , 40 r and 48 rr . We rearranged the vertices in the ones taken from [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] in order to male the connected components of the restrictions to A H 3 appear. In what follows β = ξ + ξ -1 and if the graph is 2-colorable then we give a possible coloring. Tr 4 (-5v 16 -18v 14 + 30v 12 -34v 10 + 31v 8 -21v 6 + 10v 4 -3v 2 + 1)

Computations on triality

(v 4 -v 2 + 1) 2 v 7 (v 2 -1) )T -r 4 (v 3 (v 6 -v 4 + v 2 -1))
Tr 4 ( v 8 -2v 6 + 2v 4 -3v 2 + 1 v 7 (v 2 -1) )T r 3 +r 4 (-v 12 -2v 10 + 2v 8 -4v 6 + 3v 4 -2v 2 + 1 T -r 1 -r 3 (-v 16 -2v 14 + 3v 12 -2v 10 + 4v 8 -3v 6 + 2v 4 -2v 2 + 1

v 4 (v 4 -v 2 + 1 )Tr 3 (- (v 6 -v 4 -1)v 3 v 4 -v 2 + 1 ) T -r 1 -r 3 -r 4 (- (v 6 -v 4 -1)v 3 v 4 -v 2 + 1 )T -r 1 -r 3 ( v 12 -2v 10 + v 8 -3v 6 + 2v 4 -2v 2 + 1 v 4 (v 4 -v 2 + 1) )T -r 1 ( v 3 (v 2 + 1) v 4 -v 2 + 1 ) τ (U 2 ) = T -r 3 -r 4 ( v 6 -v 4 -1 v 2 (v 4 -v 2
v 4 -v 2 + 1 ) = Ũ2 . U 3 = T -r 1 -r 3 (- v 8 -v 6 + v 4 + 1 v 2 (v 6 -v 4 + v 2 -
v 4 (v 6 -v 4 + v 2 -1)
)T -r 1 ( v 5 (v 2 + 1)

v 6 -v 4 + v 2 -1 ) τ (U 3 ) = T -r 3 -r 4 ( v 8 -v 6 + v 4 + 1 v 2 (v 6 -v 4 + v 2 -1)
) Tr 4 ( 2v 12 -4v 10 + 5v 8 -5v 6 + 4v 4 -2v 2 + 1 v 7

) T r 3 +r 4 (-v 6 + v 4 -v 2 + 1)

T r 1 +r 3 +r 4 ( (v 4 -v 2 + 1)(v 6 -v 4 + v 2 -1)

v 5
) T r 1 +r 3 (-2v 8 -2v 6 + 2v 4 -v 2 + 1

v 2 (v 6 -v 4 + v 2 -1) )
T -r 3 ( 5v 20 -15v 18 + 28v 16 -39v 14 + 44v 12 -40v 10 + 30v 8 -18v 6 + 9v 4 -3v 2 + 1

v 9 (v 6 -v 4 + v 2 -1) )T -r 1 -r 3 (v 6 -v 4 + v 2 -1)
Tr 1 (-5v 28 -20v 26 + 48v 24 -85v 22 + 125v 20 -155v 18 + 167v 16 -155v 14 + 126v 12 -89v 10 + 54v 8 -27v 6 + 12v 4 -3v 2 + 1

v 9 (v 8 -v 6 + v 4 -v 2 + 1)(v 6 -v 4 + v 2 -1) 2 )
T -r 1 (v 13 -v 11 + v 9 -v 7 + v 5 )Tr 1 ( v 16 -2v 14 + 3v 12 -4v 10 + 6v 8 -5v 6 + 3v 4 -3v 2 + 1 v 9 (v 8 -v 6 + v 4 -v 2 + 1) )

T r 1 +r 3 (-v 16 -2v 14 + 3v 12 -3v 10 + 5v 8 -4v 6 + 3v 4 -2v 2 + 1

v 4 (v 6 -v 4 + v 2 -1) ) Tr 3 ( (v 8 -v 6 + v 4 + 1)v 5 v 6 -v 4 + v 2 -1 ) T -r 1 -r 3 -r 4 ( (v 8 -v 6 + v 4 + 1)v 5 v 6 -v 4 + v 2 -1 )
T -r 3 -r 4 ( v 16 -2v 14 + 3v 12 -2v 10 + 4v 8 -3v 6 + 2v 4 -2v 2 + 1

v 4 (v 6 -v 4 + v 2 -1)
) T -r 4 ( v 5 (v 2 + 1)

v 6 -v 4 + v 2 -1 ) = Ũ3 .

E 6 -graphs

We give in this section the E 6 -graphs 10 s and 20 s which verify the properties of Theorem 6.2. We do not give the other new E 6 -graphs and E 8 -graphs which were obtained because of the large number of vertices and edges. They can be downloaded at [START_REF] Esterle | Self-dual W-graphs and bilinear form[END_REF]. )

Fp( √ α, β) Fp(α, β) Fp( √ α + √ α -1 , β) Fp( √ α, β + β -1 ) Fp( √ α + √ α -1 , β) Fp( √ α + √ α -1 , β + β -1
) Fp(α, β) Fp(α, β + β )

Fp( √ α, β) Fp(α, β) Fp( √ α + √ α -1 , β) Fp( √ α, β + β -1 ) Fp( √ α + √ α -1 , β) Fp( √ α + √ α -1 , β + β -1
) Fp(α, β) Fp(α, β + β ) Fp(α + α -1 , β)

Fp( √ α + √ α -1 , β + β -1 ) Fp(α + α -1 , β) Fp(α, β + β -1 ) Fp(α + α -1 , β + β -1
)

Fp(α + α -1 , β + β -1 )
Figure 10.12: Field extensions in type F 4 in case 9 )

Fp(α + α -1 , β + β -1 )
Figure 10.17: Field extensions in type F 4 in case 14

Fp(

√ α, β + β -1 ) Fp( √ α + √ α -1 , β) Fp( √ α + √ α -1 , β + β -1
) Fp(α, β) Fp(α, β + β Fp(

√ α + √ α -1 , β + β -1 ) Fp(α + α -1 , β) Fp(α, β + β -1 ) Fp(α + α -1 , β + β -1
)

Fp(α + α -1 , β + β -

Théorème 9 .

 9 Soit A un ensemble de représentants de représentations irréductibles 2-coloriables non auto-duales pour la relation d'équivalence ρ ≈ ϕ si ρ = ϕ et B l'ensemble des représentations irréductibles auto-duales.

Théorème 11 . 1 .

 111 Supposons p / ∈ {2, 3, 5} et α d'ordre ne divisant ni 20, ni 30, ni 48. Supposons 1 ∼ 2 et que la conjecture 8.5.1 est vraie.

Figure 2 . 1 :

 21 Figure 2.1: Classification of finite irreducible Coxeter groups

Definition 2 . 2 . 2 .

 222 Let B be a basis for H, its dual basis with regards to τ is defined to be ( b) b∈B , where for all b, b in B × B, we have τ ( bb ) = δ b,b .

  Assume there exists a basis B of BH such that for all b ∈ B, b ∈ BH and θ( b) = 0, where for b ∈ B, b is the unique element in B such that τ (b b) = 1.

λ 1 )

 1 |A Bn . 4. d) This is a consequence of 4.c) and 4.b). For r ∈ [[1, n -1]], we define the double-partitions λ (r) = ([1 n-r ], [r]) and λ (r) = ([r], [1 n-r ]).

3. n ≤ 4

 4 and the pseudo-reflections are of order 3 and G GU n (2).

Proposition 3 . 3 . 1 .

 331 Let n ≥ 3 and R 1 (resp R 2 ) be the representation associated with the double-partition

1 2 + 1 )

 11 • 4 1 ) . P SL 3 (4), 1935360 4. (8 • 4 1 ) . P SL 3 (4).2 3 , 3870720

4 .

 4 Case 4 is analogous to Case 3. 5. If λ = λ = (λ 2 , λ 1 ), then by Proposition 3.2.4 and Proposition 3.3.3, G(λ) preserves a bilinear form of the type given by Proposition 3.1.1 defined over F q 1 2

  The map ϕ is a group epimorphism and its Kernel is ker(ϕ) = {(u, v) ∈ Z/aZ×Z/bZ, u n a +v n b = 0}. Let d = Gcd(a, b). We have a group isomorphism ψ from Z/dZ to ker(ϕ) defined by ψ(k) = (ka/d, -kb/d). The map ψ is well-defined and is clearly one-to-one. We must show that it is onto. Let u ∈ Z, v ∈ Z such that u n a + v n b = n for some integer . We have that unb = ( b -v)na and, therefore, ub = ( b -v)na and u b d = ( b -v) a d . It follows that a d divides u b d and a d divides u. This implies that there exists k ∈ Z such that u = k a d and the projection of u in Z/aZ only depends on the projection of k in Z/dZ. It follows that ψ is indeed onto.

Definition 4 . 1 . 2 .

 412 Let (λ, µ) n with λ = (λ 1 , . . . , λ r , 0, . . . ) n λ and µ = (µ 1 , . . . , µ m , 0, . . . ) n µ .

Proposition 4 . 2 . 2 .

 422 For r ∈ [[1, n -1]], we write λ r the double-partition of n defined by λ r = ([r], [1 n-r ]) if r ≥ n 2 and λ r = ([1 n-r ], r) if r < n 2 .Taking the same notations as in Proposition 3.2.5 of type B, for all d ∈ A Dn , we have η 1,r (d) = η 2,r (d) because the length in T of such an element is even and we have β = 1. We define the character η r of A Dn by η r (d) = η 1,r (d) = η 2,r (d).

Proposition 4 . 2 . 3 .

 423 For all d ∈ A Dn , λ ∈ Λ and T ∈ λ, we have d.T, d. T = T, T . This shows that A Dn acts in a unitary way on those irreducible modules.

Lemma 4 . 2 . 1 .

 421 If λ is a double-partition of n then the restriction of R λ to A Dn is absolutely irreducible.

Theorem 4 . 4 . 1 .

 441 Let λ = (λ 1 , λ 2 )

5 .

 5 If λ = (λ 1 , λ 1 ) ∈ n , then we have the following possibilities. (a) If ϕ(λ) > λ, then G(λ, +) = G(λ, -) SL n λ 2 (q). (b) If ϕ(λ) = λ, then we have the following possibilities.

1. 2

 2 irreducible representations of dimension 512 of E 7 , 2. 4 irreducible representations of dimension 4096 of E 8 , 3. 2 irreducible representations of dimension 4 of H 3 , 4. 4 irreducible representations of dimension 16 of H 4 ,

  Let s ∈ S and y ∈ X. If s ∈ I(y), then L(ρ (X,I,µ) (L(e y ))) = -ω(y) 2 e y = -e y = ρ (X,I,-µ) (e y ). If s / ∈ I(y) then L(ρ (X,I,µ)(Ts) (L(e y ))) = L(ω(y)α s e y + x∈X,s∈I(x) √ α s µ s x,y ω(y)e x ) = ω(y) 2 α s e y + x∈X,s∈I(x) √ α s µ s x,y ω(x)ω(y)e x = α s e y + x∈X,s∈I(x) √ α s (-µ s x,y )e x = ρ (X,I,-µ) (T s )(e y ).

- 1 .

 1 The blue dashed edge indicates that

  If s ∈ I(x) ∩ I(y) then ρ Γ (T s )e x , ρ Γ (T s )e y = -e x , -e y = e x , e y = 0 = -α s e x , e y . If s ∈ I(x), s / ∈ I(y) then ρ Γ (T s )e x , ρ Γ (T s )e y = -e x , α s e y + √ α s z∈X,s∈I(z) µ z,y e z = -α s e x , e y -√ α s z∈X,s∈I(z) µ z,y e x , e z = -α s e x , e y . If s / ∈ I(x), s ∈ I(y) then ρ Γ (T s )e x , ρ Γ (T s )e y = α s e x + √ α s z∈X,s∈I(z) µ z,x e z , -e y = -α s e x , e y -√ α s z∈X,s∈I(z) e z , e y = -α s e x , e y . If s / ∈ I(x) ∪ I(y) then we have e x , e y = 0 and ρ Γ (T s )e x , ρ Γ (T s )e y = α s e x + √ α s z∈X,s∈I(z) µ z,x e z , α s e y + √ α s z ∈X,s∈I(z ) µ z ,x e z = √ α s z∈X,s∈I(z) µ z,x e z , e y + √ α s z ∈X,s∈I(z ) µ z ,y e x , e z = √ α s µ ϕ(y),x ω(ϕ(y))

7 .

 7 S 3 S 5 = S 5 S 3 . 8. S 3 S 6 = S 6 S 3 . 9. S 4 S 6 = S 6 S 4 .

  12, 2183 → 1, 2229 → 1 and 2229 → 121 all vanish, therefore we need to prove the E 8 -graph remains connected when those edges vanish. The edge 1572 → 12 can be replaced by the path 1572 → 24 → 100 → 12. The edge 2183 → 1 can be replaced by the path 2183 → 7 → 627 → 1. The edge 2229 → 1 can be replaced by 2229 → 22 → 59 → 1. The edge 2229 → 121 can be replaced by the path 2229 → 2230 → 2234 → 121. This proves 2240 x remains connected, this is true as well for 2240 x because it is its dual E 8 -graph. Consider now 4200 y . When p = 7, none of the weights vanish, therefore all the edges remain. For p = 7, the edges 2700 → 20, 3465 → 1, 3465 → 2, 4075 → 894, 4172 → 1, 4172 → 399, 4190 → 2, 4190 → 12 and 4190 → 399 are the only ones disappearing. The edge 2700 → 20 can be replaced by the path 2700 → 563 → 630 → 20. The edge 3465 → 1 can be replaced by the path 3465 → 103 → 119 → 1. The edge 3465 → 2 can be replaced by the path 3465 → 217 → 287 → 2. The edge 4075 → 894 can be replaced by the path 4075 → 908 → 2900 → 894. The edge 4172 → 1 can be replaced by the path 4172 → 141 → 178 → 1. The edge 4172 → 399 can be replaced by the path 4172 → 432 → 2150 → 399. The edge 4190 → 2 can be replaced by the path 4190 → 68 → 197 → 2. The edge 4190 → 12 can be replaced by the path 4190 → 31 → 136 → 12. The edge 4190 → 399 can be replaced by the path 4190 → 385 → 20 → 399. This proves 4200 y and 4200 y remain connected.

4 r : 2 u 3 Φ 3

 433 (u)Φ 5 (u).

  We then have Tr(n 1 ) + 2 = (α + α -1 ) 2 = 0 and Tr n 1 2 + 2 Tr(n 1 ) + 2 -Tr(u 1 ) Tr(n 1 ) + 2 = α + α -1 .

r 3 = 2 - 3 1 r 2 r 3 r 4

 323124 and r 4 = 3 -4 be the positive simple roots of Φ and π = {r 1 , r 2 , r 3 , r 4 }. They correspond to the following Dynkin diagram r

Proposition 8 . 5 . 1 .

 851 Assume 1 ∼ 2.

Theorem 8 . 5 . 1 .

 851 Under our assumptions on α and p, we have the following results. 1. Assume 1 ∼ 2 and Conjecture 8.5.1 is true.

(

  ) then we have the Hasse diagram of Figure 10.6.

  β) then we have the Hasse diagram in Figure 10.8.

  β) or the Hasse diagram given in Figure 10.19.

  (α 2 β+1)(αβ+1) α(β+1)(α+β) (α 2 β+1)(αβ+1) -√ αα(β+1) 2 (αβ+1)(α 2 β+1)

L 1 = 1 and L 2 =

 112 (XY )(I 12 + E 2,3 -E 6,5 + E 8,9 -E 12,11 )(XY ) -1 ∈ XG 12 X -I 12 + E 2,3 -E 6,5 + E 8,9 -E 12,11 ∈ XG 12 X -1 .

Tr(ρ 6 1 1 1 1 1

 111 (p 0 )) = Tr(ρ 6 1 (p 1 )) = Tr(ρ 6 1 (p 0 p -)) = Tr(ρ 6 2 (p 0 )) = Tr(ρ 6 2 (p 1 )) = Tr(ρ 6 2 (p 0 p -

6 1

 6 |A F 4 ρ 6 2 |A F 4 . Assume now ρ 8 2 |A F 4 (Ψ • ρ 8 3 |A F 4 ) ⊗ z. We have Tr(ρ 8 2 (p 0 )) = Tr(ρ 8 2 (p 1 )) = Tr(ρ 8 2 (p 0 p -1 1 )) = 6 -2(α + α -1 ), Tr(ρ 8 3 (p 0 )) = Tr(ρ 8 3 (p 1 )) = Tr(ρ 8 3 (p 0 p -1 1 )) = 5 -3(α + α -1). Note that those quantities are non-zero because we have Tr(ρ 8 2 (p 0 )) = z(p 0 )Ψ(Tr(ρ 8 3 (p 0 ))), 3 Tr(ρ 8 2 (p 0 )) -2 Tr(ρ 8 2 (p 0 )) = 18 -10 = 8 = 0.

8 2

 8 |A F 4 ρ 8 3 |A F 4 which contradicts Proposition 8.2.2. All the computations above give the same results if we substitute ρ 8 3 by ρ 8 3 | therefore ρ 8 2 |A F 4 (Ψ • ρ 8 3 |A F 4 ) ⊗ z. Assume now ρ 9 1 |A F 4 (Ψ • ρ 9 2 |A F 4 ) ⊗ z. We have Tr(ρ 91 (p 0 )) = Tr(ρ 91 (p 1 )) = Tr(ρ 91 (p 0 p -1 1 )) = Tr(ρ 92 (p 0 )) = Tr(ρ 92 (p 1 )) = Tr(ρ 92 (p 0 p -1 1

10 - 6 ( 4 - 1 + ( 4 -E 1 10 - 6 ( 4 + 1 2 1 + 1 - 1 E 10 + E 3 9 + 4E 1 E 11 -180E 2 1 + 18E 1 E 9 -2E 2 9- 9 - 1 - 9 -

 1064141106411111031112192919 E 1 ) + E 3 -E 2 D 1 ) + D 3 -D 2 1 + (4 + D 1 ) 2 2 + D 1 = Φ(E 9 ) E = Tr([n 2 , u 2 ][u -, n 2 ]) -E 1 E E 1 E 10 -14E 1 E 9 -4E 10 E 9 -D 1 D 10 + 14D 1 D 9 -4D 10 D 9 -18E 10 -9E 11 + (856)E 1 -40E 9 -208 = 0 D = (D 1 + 2)v 2 (-D 1 D 10 + D 3 4D 1 D 11 -180D 2 18D 1 D 9 -2D 2 18D 10 -9D 11 -(856)D 1 -40D 9 -

U 0 = 3 )T r 3 +r 4 (v - 2 ) 5 ) 4 ) 4 -v 2 + 1 v 3 ) 1 v 5 ) 1 v 4 (v 5 - 1 v 5 (v 4 - 1 v 4 ) 5 ) T -r 4 ( 1 -r 3 (- v 6 -v 4 - 1 v 2 3 )T r 3 +r 4 ( 2v 6 - 1 v 2 (v 4 - 1 v 7 (v 4 -

 03254431514515414541612346124174 Tr 1 (-v -v -1 )T r 1 +r 3 (-1 -v -2 )T -r 1 (v)T r 3 +r 4 (1)T -r 3 -r 4 (1)T -r 3 (-v)Tr 4 (-v -v -1 )T -r 4 (v)Tr 4 (-2v -Tr 1 (-v -3 )τ (U 0 ) = Tr 4 (-v -v -1 ) T r 3 +r 4 (1 + v -2 ) T -r 4 (v) T r 1 +r 3 (-1) T -r 1 -r 3 (-1) T -r 3 (-v) Tr 1 (-v -v -1 ) T -r 1 (v) Tr 1 (-2v -3 ) T r 1 +r 3 (-v -2 ) Tr 4 (-v -3 ) = Ũ0 . U 1 = T -r 1 -r 3 (v -2 -1)T r 1 +r 3 (-v 2 + 1)Tr 1 (-v 4 -v 2 + 1 v 3 )T r 1 +r 3 +r 4 (v -1 -v -1)T -r 1 -r 3 -r 4 (2)T r 1 +r 3 +r 4 (-1) T r 3 +r 4 (v -2 -2)T -r 3 -r 4 (-v 2 )T r 3 +r 4 (2)T -r 3 -r 4 (-1)T -r 3 ( 3v 6 -4v 4 + 3v 2 -1 v Tr 4 (-6v 10 -11v 8 + 15v 6 -11v 4 + 4v 2 -1 v 4 (v 5 -v 3 + v) ) T -r 4 (-v 5 + v 3 -v)Tr 4 ( v 8 -2v 6 + 5v 4 -3v 2 + 1 v 5 (v 4 -v 2 + 1) )T r 3 +r 4 (-2v 8 -3v 6 + 6v 4 -3v 2 + 1 v Tr 3 (-2v 5 + v 3 -v) T -r 1 -r 3 -r 4 (-2v 4 -v 3 + 4v 2 + v -2)T -r 1 -r 3 ( 2v 8 + v 7 -6v 6 -3v 5 + 10v 4 + 3v 3 -6v 2 -v + 2 v 5 )T -r 1 (v -2) τ (U 1 ) = T -r 3 -r 4 (1 -v -2 ) T r 3 +r 4 (v 2 -1) Tr 4 (-v T r 1 +r 3 +r 4 (v -1 -v -1) T -r 1 -r 3 -r 4 (2) T r 1 +r 3 +r 4 (-1) T r 1 +r 3 (2 -v -2 ) T -r 1 -r 3 (v 2 ) T r 1 +r 3 (-2) T -r 1 -r 3 (1) T -r 3 ( 3v 6 -4v 4 + 3v 2 -Tr 1 (-6v 10 -11v 8 + 15v 6 -11v 4 + 4v 2v 3 + v) ) T -r 1 (-v 5 + v 3 -v) Tr 1 ( v 8 -2v 6 + 5v 4 -3v 2 + v 2 + 1) ) T r 1 +r 3 ( 2v 8 -3v 6 + 6v 4 -3v 2 + Tr 3 (-2v 5 + v 3 -v) T -r 1 -r 3 -r 4 (-2v 4 -v 3 + 4v 2 + v -2) T -r 3 -r 4 (-2v 8 + v 7 -6v 6 -3v 5 + 10v 4 + 3v 3 -6v 2 -v + 2 v (v 4 -v 2 + 1) )T r 1 +r 3 (-v 4 + v 2 -1)Tr 1 (-2v 8 -3v 6 + 3v 4 -2v 2 + 1 v 5 )T r 1 +r 3 +r 4 (-(v 2 -1)(v 4 -v 2 + 1) v 2v 4 + v 2v 2 + 1) )T -r 3 -r 4 (v 4 -v 2 +1)T -r 3 (-(5v 14 -13v 12 + 19v 10 -20v 8 + 15v 6 -8v 4 + 3v 2v 2 + 1) )

+ 1 ) ) T r 3 +r 4 (v 4 -v 2 + 1 ) 1 v 5 ) 3 )T r 1 +r 3 (- 2v 6 -2v 4 + v 2 - 1 v 2 (v 4 - 1 v 7 (v 4 -

 1442115364124174 Tr 4 (-2v 8 -3v 6 + 3v 4 -2v 2 + T r 1 +r 3 +r 4 (-(v 2 -1)(v 4 -v 2 + 1) v v 2 + 1) ) T -r 1 -r 3 (-v 4 + v 2 -1)T -r 3 (-(5v 14 -13v 12 + 19v 10 -20v 8 + 15v 6 -8v 4 + 3v 2v 2 + 1) ) Tr 1 (-5v 16 -18v 14 + 30v 12 -34v 10 + 31v 8 -21v 6 + 10v 4 -3v 2 + 1) (v 4 -v 2 + 1) 2 v 7 (v 2 -1) ) T -r 1 (v 3 (v 6 -v 4 + v 2 -1)) Tr 1 ( v 8 -2v 6 + 2v 4 -3v 2 + 1 v 7 (v 2 -1) ) T r 1 +r 3 ( v 12 -2v 10 + 2v 8 -4v 6 + 3v 4 -2v 2 + 1 v 4 (v 4 -v 2 + 1) ) Tr 3 (-(v 6 -v 4 -1)v 3 v 4 -v 2 + 1 ) T -r 1 -r 3 -r 4 (-(v 6 -v 4 -1)v 3 v 4 -v 2 + 1 ) T -r 3 -r 4 (-v 12 -2v 10 + v 8 -3v 6 + 2v 4 -2v 2 + 1 v 4 (v 4 -v 2 + 1) ) T -r 4 ( v 3 (v 2 + 1)

v 6 -v 4 + v 2 - 1 )T -r 1 -r 3 -r 4 ( (v 8 -v 6 + v 4 + 1 )v 5 v 6 -

 641134864156 v 4 + v 2 -1 )

U 4 = 1 ) 1 ) 2 ) 2 )

 41122 T r 1 +r 3 +r 4 (v -2 )T -r 1 -r 3 -r 4 (-v 2 -1)T -r 2 (v)T r 1 +r 3 +r 4 (1)T -r 4 (v -1 )T r 1 +r 2 +2r 3 +r 4(v)T -r 1 -r 2 -r 3 -r 4 (v + v -Tr 3 (v 2 + 1)T -r 1 (v -1 )T r 2 +r 3 (-v -v -1 )T -r 1 -r 3 -r 4 (-1 -v -2 ) τ (U 4 ) = T r 1 +r 3 +r 4 (v -2 ) T -r 1 -r 3 -r 4 (-v 2 -1) T -r 2 (v) T r 1 +r 3 +r 4 (1) T -r 1 (v -1 ) T r 1 +r 2 +2r 3 +r 4 (v) T -r 1 -r 2 -r 3 -r 4 (v + v -Tr 3 (v 2 + 1) T -r 4 (v -1 ) T r 2 +r 3 (-v -v -1 ) T -r 1 -r 3 -r 4 (-1 -v -2 ) = Ũ4 . U 5 = T -r 1 -r 2 -r 3 -r 4 (-1)T r 1 +r 2 +r 3 +r 4 (v 2 + 1)T r 1 +r 2 +r 3 (v 3 + v)T -r 4 (-v)T r 1 +r 2 +2r 3 +r 4 (v(v 2 + 2))T -r 1 -r 2 -r 3 -r 4 (-v -Tr 3 (-v -v -1 )T -r 1 (v -1 )T r 2 +r 3 +r 4 (v 3 )Tr 2 (v)T r 1 +r 2 +r 3 +r 4 (v 4 + v 2 ) τ (U 5 ) = T -r 1 -r 2 -r 3 -r 4 (-1) T r 1 +r 2 +r 3 +r 4 (v 2 + 1) T r 1 +r 2 +r 3 (v 3 + v) T -r 1 (-v) T r 1 +r 2 +2r 3 +r 4 (v(v 2 + 2)) T -r 1 -r 2 -r 3 -r 4 (-v -Tr 3 (-v -v -1) T -r 4 (v -1 ) T r 1 +r 2 +r 3 (v 3 ) Tr 2 (v) T r 1 +r 2 +r 3 +r 4 (v 4 + v 2 ) = Ũ5 .

2 Figure 10 . 1 : F 4 16 Figure 10 . 3 : F 4

 21014161034 Figure 10.1: F 4 -graphs

- 1 1 )Fp(α + α - 1 , β + β - 1 )Figure 10 . 4 : 4 Fp

 11111044 Figure 10.4: Field extensions in type F 4

- 1 1 )Fp(α + α - 1 , β + β - 1 )Figure 10 . 5 :

 1111105 Figure 10.5: Field extensions in type F 4 in case 2

- 1 1 )Figure 10 . 18 :

 111018 Figure 10.18: Field extensions in type F 4 in case 15

10. 7

 7 Matrices for the 12-dimensional representation of H F 4 ,α,β X 1 , X 2 and X 3 are respectively

-α 3 2 α 2 α 3 √ α 5 β 2 √ β 2 α 2 √ β 2 α 2 √ β 2 α 2 α 3 β+α 2 β 2 +4 α 3 +9 α 2 β+5 αβ 2 +5 α 2 +9 αβ+4 β 2 3 α 3 +2 α 2 2 - 3 β 2 2 - 3 β 2 +3 α √ α α 3 +2 α 2 β+αβ 2 +α 2 +2 αβ+β 2 β αβ+β 2 α α 3 +2 α 2 β+αβ 2 +α 2 +2 αβ+β 2 (α α 3 +2 α 2 β+αβ 2 +α 2 +2 αβ+β 2 ( 0 - 3 √ ββ α 3 +2 α 2 β+αβ 2 +α 2 +2 αβ+β 2 (

 32322222322233223223232232320332 +2 α 2 β+αβ 2 +α 2 +2 αβ+β 2 √ α α 2 β 2 +α 2 β+αβ+α (α+1) β+2 αβ 2 -α 2 -2 αβ-β 2 +2 α+2 β β+α 2 β 2 +4 α 3 +9 α 2 β+5 αβ 2 +5 α 2 +9 αβ+4 β 2 +α+β √ α β+2 αβ 2 -α 2 -2 αβ-β 2 +2 α+2 β √ α β+2 αβ 2 -α 2 -2 αβ-β 2 +2 α+2 β √ α β+2 αβ 2 -α 2 -2 αβ-β 2 +2 α+2 β √ α α 3 +2 α 2 β+αβ 2 +α 2 +2 αβ+β 2 √ α αβ 2 +1 α 2 β 2 +α 2 β+αβ+α (α+1) β+αβ 2 +α 2 +2 αβ+β 2 √ αβ (α+1) αβ 2 +α +3 α √ α α 3 +2 α 2β+αβ 2 +α 2 +2 αβ+β 2 β αβ+β 2 +α+β (α+1) αβ 2 +α β+1)(α+1) αβ 2 +α 2

2 Preliminaries 2.1 Definitions and first properties of Coxeter groups, Artin groups and Iwahori-Hecke algebras

	A Wn	: Artin group of type W n
	A Wn	: derived subgroup of the Artin group A Wn
	Z(G)	: center of a group G
	[G,G]	: derived subgroup of a group G
	(a, b)	: gcd of a pair of integers (a,b)
	I n	: identity matrix of size n x n
	E i,j	: elementary matrix with non-zero coefficient in position (i,j)
	λ	: transposed partition (or double-partition) of λ
	A : B	: split normal extension of A by B
	A • B	: normal extension of A by B which is not split
	A.B	: normal extension of A by B which may be split
	diag(a

1 , a 2 , . . . , a n ) : diagonal matrix with coefficients a 1 , a 2 , . . . , a n Φ n : n-th cyclotomic polynomial Chapter

  (Theorem 1.2.2). S) be a Coxeter system with W a finite irreducible Coxeter group. We let A + W be its braid monoid(the monoid of positive words in the generators of A W ). A reduced expressionσ = s i k ...s i 1 of an element σ in the braid monoid A + W isan expression where k is the minimal number of generators necessary to write the element σ in the braid monoid. s i k ...s i 1 and s j k ...s j 1 are reduced expressions of the same element in an Artin monoid A + W if and only if they are equivalent.

	Theorem 2.1.1. (Matsumoto's Theorem)
	Let (W,

Table 2 .

 2 1: Description of the geometric Aschbacher classes By definition of the classes, no irreducible group can be included in a group of class C 1 , no primitive group can be included in a group of class C 2 , no group containing a transvection can be included in a group of class C 3 , no tensor-indecomposable group can be included in a group of class C 4 and no group whose trace generate the field F q

  for any extremal box, there exists a tableau with n in that box. It follows that s λ = s µ and for all

Lemma 3.2.3. Let

  Let n ≥ 4. By Lemma 2.1 of[START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], we have that A An = A A n-1 A An , where A An =< S 1 , S 2 , . . . , S n-1 >≤ A Bn . We have that A Bn is generated by A An and A B n-1 therefore the result follows. G be a group, k a field and R 1 , R 2 two representations of G in GL N (k) such that the restrictions to the derived subgroup of G are equal and the restriction of at least one of them is absolutely irreducible. There exists a character η :

	We now recall Lemma 2.2 of [12].

  1. is shown in the same way as Lemma 3.4. of [12], because A Dn is generated by A Dn ans A D n-1 . Using Propositions 4.2.3 and 4.2.1, it is sufficient to show 2,3,4 and 5 to conclude the proof. In the same way as for type B, we need to use Lemma 3.2.3. If R λ|A Dn R µ|A Dn then there exists a character η :

  from the one in Proposition 4.1.3 using Proposition 3.3.3. Those two theorems (except for the surjectivity) follow from Propositions 4.1.3, 4.2.1, 4.2.2, 4.2.3, 4.2.5, Theorem 4.2.1 and Proposition 4.2.4. It now remains to check that Φ 1 ,n and Φ 2 ,n are surjective in all cases.

  . A 6 .2 2 , 1440 8. 2 . P SL 2 (17), 4896 9. 2 . P SL 2 (q

	4. 2 . P SL 2 (7), 336
	5. 2

. P SL 2 (7) . 2, 672 6. 2 . A 6 , 720 7. 2

  then we can apply Theorem 2.3.3. Applying Clifford's Theorem (Theorem 11.1 of[START_REF] Curtis | Methods of representation theory[END_REF]), we have that Res

G(λ)

  1 S 3 S 1 = S 3 S 1 S 3 .3. ∀i ∈ {2, 4, 5, 6, 7},S 1 S i = S i S 1 . 4. S 2 S 4 S 2 = S 4 S 2 S 4 . 5. ∀i ∈ {3, 5, 6, 7}, S 2 S i = S i S 2 . 6. ∀i ∈ {3, 4, 5, 6}, S i S i+1 S i = S i+1 S i S i+1 .7. S 3 S 5 = S 5 S 3 . 8. S 3 S 6 = S 6 S 3 . 9. S 3 S 7 = S 7 S 3 . 10. S 4 S 6 = S 6 S 4 .11. S 4 S 7 = S 7 S 4 . For σ in the Coxeter group E 7 , we set T σ = S i 1 . . . S i Under our assumptions on p and α, H E 7 ,α is split semisimple, the representations afforded by the W -graphs are irreducible and pairwise non-isomorphic over F q . The restrictions of the irreducible representations of H E 7 ,α to H E 6 ,α are the same as in the generic case.

			S 2
	S 1	S 3	S 4	S 5	S 6	S 7
	Proposition 7.2.1.
	Dynkin diagram

k for any reduced expression σ = s i 1 . . . s i k is a reduced expression. This means we consider E 7 as in the CHEVIE package of GAP3

[START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF]

with the following

  1 p 1 p 6 p 6 p 10 s 15 p 15 p 15 q 15 q 20 p 20 p 20 s 24 p 24 p 30 p 30 p 60 s 60 p 60 p 64 p 64 p 80 s 81 p 81 p 90 s a

Table 7 .

 7 4: Restriction table from H E 7 ,α to H E 6 ,α .

  Under our assumptions on p and α, H E 8 ,α is split semisimple, the representations afforded by the W -graphs are irreducible and pairwise non-isomorphic over F q . The restrictions of the irreducible representations of H E 8 ,α to H E 7 ,α are the same as in the generic case.

	Proposition 7.3.1.
							with the following
	Dynkin diagram
			S 2
	S 1	S 3	S 4	S 5	S 6	S 7	S 8

Table 7 .

 7 + 21 b + 27 a + 35 b 112 z = a + 7 a + 21 b + 27 a + 56 a 160 z = a + 21 a + 27 a + 105 a 168 y = a + 84 a 175 x = a + 105 b 210 x = a + 27 a + 56 a + 120 a 300 x = a + 105 a + 168 a 350 x = a + 35 a + 105 a + 189 a 400 z = a + 35 b + 56 a + 105 b + 189 b 420 y = b + 210 b 448 w = a + 35 a + 189 a + 189 a 448 z = b + 70 a + 168 a + 189 b 525 x = b + 105 c + 189 c + 210 a 560 z = a + 21 a + 27 a + 35 b + 56 a + 105 a + 120 a + 189 b 567 x = a + 21 a + 21 b + 27 a + 56 a + 105 a + 120 a + 210 a 700 x = a + 35 b + 56 a + 105 b + 120 a + 168 a + 189 b 700 xx = a + 84 a + 105 c + 216 a + 280 b 840 x = a + 168 a + 210 b + 378 a 840 z = a + 168 a + 189 a + 378 a 972 x = b + 84 a + 168 a + 189 c + 216 a + 280 b 1008 z = a + 27 a + 56 a + 105 a + 120 a + 189 c + 210 a + 280 a 1050 x = a + 35 b + 105 b + 189 b + 210 a + 216 a + 280 b 1134 y = a + 189 a + 378 a + 378 a 1296 z = a + 35 a + 105 a + 120 a + 189 a + 210 a + 280 a + 336 a 1344 x = b + 27 a + 35 b + 105 a + 120 a + 168 a + 189 b + 189 c + 210 a + 280 b 1344 w = a + 84 a + 210 b + 210 b + 378 a + 378 a 1400 x = a + 105 b + 120 a + 189 b + 210 a + 315 a + 405 a 1400 y = a + 280 a + 420 a + 420 a 1400 z = a + 27 a + 56 a + 105 a + 120 a + 168 a + 189 b + 189 c + 210 a + 315 a 1400 zz = a + 70 a + 105 b + 189 b + 210 b + 216 a + 280 b + 315 a 1575 x = a + 56 a + 105 a + 120 a + 189 a + 189 b + 210 a + 280 a + 405 a 1680 y = a + 35 a + 189 a + 189 a + 280 a + 280 a + 336 a + 336 a 2016 w = a + 70 a + 210 b + 210 b + 216 a + 216 a + 512 a + 512 a 2100 x = a + 105 a + 189 a + 189 c + 210 a + 280 a + 336 a + 336 a + 420 a 2100 y = c + 105 c + 189 c + 189 c + 336 a + 336 a + 420 a + 420 a 8: Restriction table from H E 8 ,α to H E 7 ,α . = 70 a + 105 b + 120 a + 168 a + 189 b + 210 b + 280 b + 315 a + 378 a + 405 a x = 56 a + 105 a + 120 a + 168 a + 189 c + 210 a + 280 a + 315 a + 405 a + 420 a z = 35 a + 189 a + 189 a + 210 a + 280 a + 336 a + 336 a + 405 a + 420 a y = 216 a + 216 a + 280 b + 280 b + 336 a + 336 a + 512 a + 512 a z = 105 c + 120 a + 189 c + 210 a + 280 a + 315 a + 336 a + 405 a + 420 a + 420 a x = 70 a + 105 b + 210 b + 210 b + 216 a + 280 b + 315 a + 405 a + 512 a + 512 a y = 70 a + 70 a + 210 b + 210 b + 378 a + 378 a + 405 a + 405 a + 512 a + 512 a x = 84 a + 105 c + 168 a + 189 c + 216 a + 280 b + 336 a + 378 a + 420 a + 512 a + 512 a z = 35 b + 56 a + 105 a + 105 b + 120 a + 168 a + 2 × 189 b + 189 c + 210 a +216 a + 280 a + 280 b + 315 a + 378 a + 405 a z = 105 b + 189 b + 210 a + 216 a + 280 a + 280 b + 315 a + 336 a + 405 a + 512 a + 512 a x = 105 a + 120 a + 168 a + 189 a + 189 b + 189 c + 210 a +280 a + 280 b + 315 a + 336 a + 378 a + 405 a + 420 a + 512 a z = 105 a + 120 a + 168 a + 189 a + 189 b + 189 c + 210 a +280 a + 280 b + 315 a + 336 a + 378 a + 405 a + 420 a + 512 a x = 105 b + 168 a + 189 b + 210 a + 210 b + 216 a + 280 a + 280 b + 315 a + 378 a + 405 a +420 a + 512 a + 512 a y = 84 a + 84 a + 210 b + 210 b + 216 a + 216 a + 280 b + 280 b + 378 a + 378 a +420 a + 420 a + 512 a + 512 a z = 84 a + 105 b + 105 c + 210 a + 216 a + 280 a + 280 b + 315 a + 378 a + 378 a +405 a + 420 a + 512 a + 512 a y = 210 b + 210 b + 315 a + 315 a + 378 a + 378 a + 405 a + 405 a + 420 a + 420 a + 512 a + 512 a y = 280 a + 280 a + 315 a + 315 a + 336 a + 336 a + 405 a + 405 a + 420 a + 420 a + 512 a + 512 a z = 70 a + 84 a + 168 a + 189 b + 189 c + 210 b + 210 b + 216 a + 280 b + 315 a +2 × 378 a + 405 a + 420 a + 512 a + 512 a w = 189 a + 189 a + 280 a + 280 a + 280 b + 280 b + 336 a + 336 a + 378 a + 378 a +405 a + 405 a + 420 a + 420 a + 512 a + 512 a z = 105 c + 168 a + 189 a + 189 c + 210 a + 280 a + 315 a + 336 a + 336 a + 378 a + 405 a + 405 a +2 × 420 a + 420 a + 512 a + 512 a y = 189 a + 189 a + 280 a + 280 a + 315 a + 315 a + 336 a + 336 a + 378 a + 378 a +405 a + 405 a + 420 a + 420 a + 512 a + 512 a x = 105 c + 189 a + 189 b + 189 c + 210 a + 216 a + 280 a + 280 b + 315 a + 336 a + 336 a +378 a + 378 a + 2 × 405 a + 420 a + 420 a + 512 a + 512 a w = 210 b + 210 b + 216 a + 216 a + 280 b + 280 b + 315 a + 315 a + 336 a + 336 a + 378 a + 378 a +405 a + 405 a + 420 a + 420 a + 2 × 512 a + 2 × 512 a

	1 x	= a
	8 z	= a + 7 a
	28 x	= a + 21 a
	35 x	= a + 7 a + 27 a
	50 x	= a + 35 b
	56 z	= a + 35 a
	70 y	= a + 35 a
	84 x	= a

x

Table 7 .

 7 9: Restriction table from H E 8 ,α to H E 7 ,α

Table 8 .

 8 1: Schur elements in type H 3

	Proposition 8.1.1.

  3.1 and 8.1.2 and Propositions 8.1.3, 8.1.4 and 8.1.5. Assume now that 1 2. By [36], A H 3 is perfect, it follows that by Lemmas 3.3.1 and 8.1.2 and Propositions 8.1.3, 8.1.4 and 8.1.5, we only need to prove that in all cases there exists no non-trivial field automorphism ϕ of F p (α) such that ϕ•ρ 3s|A H 3 ρ 3s|A H 3 or ϕ•ρ 3s|A H 3 ρ 3 s |A H 3 to conclude the proof. Assume there exists such an automorphism ϕ. By Proposition 2.1.2 and Lemma 3.2.3, we have that ϕ

  1 S 2 S 1 S 2 S 1 = S 2 S 1 S 2 S 1 S 2 , 3. S 1 S 3 = S 3 S 1 , 4. S 1 S 4 = S 4 S 1 , 5. S 2 S 3 S 2 = S 3 S 2 S4 , 6. S 2 S 4 = S 4 S 2 , 7. S 3 S 4 S 3 = S 4 S 3 S 4 . For σ in the Coxeter group H 4 , if σ = s i 1 . . . s i k is a reduced expression we set T σ = S i 1 . . . S i k .

  10 + 2v 6 + v 2 Rem(H 6 , H 7 , v) = 161v4 + 63H 9 = 23 Rem(H 7 , H 8 , v) = 420v 2 H 10 = 1 3 2 7 1 Rem(H 8 , H 9 , v) = 1.For p = 23, we get H 8 = 63 = 17 = 0, therefore for any prime different from 7, we get a contradiction. Using GAP4, we get for p = 7 that Gcd(H 1 , H 2 ) = v 4 + 1, which is also a contradiction.Case 1.1.3.2 : I 2 = 7524(v 36 + 1) -45144(v 34 + v 2 ) + 127908(v 32 + v 4 ) -214660(v 30 + v 6 ) + 262569(v 28 + v 8 ) -329886(v 26 + v 10 ) + 445989(v 24 + v 12 ) -513465(v 22 + v 14 ) + 494686(v 20 + v 16

	H 4	= Rem(H 2 , H 3 )	= 6v 12 + 5v 8 + 7v 4 + 1
	H 5	= 6 Rem(H 3 , H 4 )	= 7v 10 + 5v 6 + 5v 2
	H 6	= 7 Rem(H 4 , H 5 )	= 5v 8 + 19v 4 + 7
	H 7	= 5 Rem(H 5 , H 6 )	= -108v 6 -24v 2
	H 8 = 9

  . The simple Lie algebra L of type D 4 over C is the matrix algebra {T ∈ M 8 (C), t T A + AT = 0}. We write (E i,j ) 1≤i,j≤4 the basis of elementary matrices of M 8 (C).Its Cartan decomposition is given by L

	Definition 8.4.1. Let A =	0 I 4 I 4 0

  ). By Theorem 8.1.1, we have that ρ 16r (A H 4 ) contains a natural SU 4 (q). It follows by Lemma 3.3.3 that it is a classical group in a natural representation over F q 2 . Let then ϕ be the unique automorphism of order 2 of F q 2 . We have ϕ( It follows by Proposition 2.1.2 that ϕ • ρ 16r|A H 4 is isomorphic to the restriction of an irreducible representation of H H 4 ,α . By Table 8.3 and Propositions 8.1.3 and 8.1.4, we have that ϕ • ρ 16r|A H 4 ρ 16rr|A H 4 . This proves by Proposition 8.2.2 that ϕ • ρ 16r|A H 4 ρ 16r|A H 4

	√ α) = -	√ α and
	ϕ(α) = α.

  By Lemma 8.1.2, we have 1 ∼ 2, = Φ 1,2 and Φ 1,2 (α) = α -1. It follows by Proposition 2.1.2 that Φ 1,2 • ρ 16r|A H 4 is isomorphic to the restriction of an irreducible representation of H H 4 ,α . By Table 8.3, we have ρ 16r|A H 3 ρ 3 s × ρ 3 s × ρ 4r × ρ 5r . It follows by Propositions 8.1.3, 8.1.4 and 8.1.5 that Φ 1,2 • ρ 16r|A H 3 ρ 3s × ρ 3s × ρ 4 r × ρ 5 r . It follows by Table 8.3 that Φ 1,2 • ρ 16r|A H 3 ρ 16 r . By Lemma 3.2.4, we have that ρ 16r (A H 4 ) ≤ SU 16 (q

  1,2 • ρ 16r|A H 3 ρ 3s × ρ 3s × ρ 4r × ρ 5 r . It follows by Table 8.3 that Φ 1,2 • ρ 16r|A H 4 ρ 16 rr |A H 4 . It follows that ρ 16r (A H 4 ) preserves no non-degenerate bilinear or hermitian form over F q . It contains a natural SU 5 (q

  . It follows by Lemma 2.1.2 that • ρ 16t|A H 4 ρ 16t|A H 4 and • ρ 30s|A H 4 ρ 30s|A H 4 . By Proposition 8.1.3 and Table 8.3, we have that • ρ 24s|A H 4 ρ 24s|A H 4 or • ρ 24s|A H 4 ρ 24t|A H 4 . We have Tr

  • ρ 25r|A H 4 ρ 25r|A H 4 and • ρ 36rr|A H 4 ρ 36rr|A H 4 . It follows that they cannot be classical groups in a natural representation over F

	q	1 2

  Proposition 9.1. Under our assumptions on p, α and β, H F 4 ,α,β is split semisimple over F p ( the representations afforded by the W -graphs are irreducible and pairwise nonisomorphic over F q . The restrictions of the irreducible representations of H F 4 ,α,β to H B 3 ,α,β are the same as in the generic case.Proof. We will apply Proposition 2.2.4. Let A

	√	α,	√ β),

  .1. (This table is taken from Table 11.1. [20]) The restrictions to A F 4 of the representations afforded by those W -graphs are absolutely irreducible and the representations of dimension greater than one are pairwise non-isomorphic. Proof. As in [12] Lemma 3.4, we only need to prove that A F 4 =< s 1 , s 2 , s 3 , s 4 > is generated by A B 3 =< s 1 , s 2 , s 3 > and A F 4 . This true because s 4 = s 4 s -1 3 s 3 , s 4 s -1 3 ∈ A F 4 and s 3 ∈ A B 3 . Let now ρ and ρ be two representations of H F 4 ,α,β such that ρ |A F 4 ρ |A F 4

	Proposition 9.2.

Table 9 .

 9 3: Restriction of the 8-dimensional representations to H B 3 ,2 H B 3 ,β,α .

	SU 8 (q

  2 ] by Proposition 3.2.4. Hence Table 9.2 gives • ρ 8 2 ρ 8 2 ρ 8 2 , and it follows that • ρ 8 2 ρ 8 2 . By Lemma 3.2.4, we have that up to conjugation in GL 8

Table 9 . 5 :

 95 ). By Tables 9.4 and 9.5, in cases 1, 4, 5, 10, 11, 12, 13 and 16, ρ 9 1 (A B 3 ,1 ) and ρ 9 2 (A B 3 ,2 ) are generated by transvections. It follows that Restriction of the 9-dimensional representations to H B 3 ,2 H B 3 ,β,α . ρ 9 1 (A F 4

Φ

  • ρ 6 1 |A F 4 ρ 6 1 |A F 4 or Φ • ρ 6 1 |A F 4 ρ 6 2 |A F 4 . By Table 9.6, we cannot have Φ • ρ 6 1 |A F 4 ρ 6 2 |A F 4 . It follows that Φ • ρ 6 1 |A F 4 ρ 6 1 |A F 4 . Hence, by Lemma 3.2.4, G 6 1 is conjugate in GL 6 (q 2

  .8 and Theorem 4.2.1, G 16 contains up to conjugation a diagonal SL 2 (r) or a diagonal SL 2 (r 1 2 ), where F r = F p (α) in all cases. It follows that G 16 is irreducible, primitive, v G 16 (V ) ≤ 2 and we are not in the second case of Theorem 2.3.2. It only remains to show that G 16 is tensor-indecomposable in order to apply Theorem 2.3.2. Note that in cases 4, 5 and 10, we have Φ • ρ 16|A F 4 ρ 16|A F 4 , therefore G 16 is conjugate to a subgroup of Ω + 16

  (D 1 + 2) 2 (-D 2 1 D 9 + 34D 2 1 -12D 1 D 10 -4D 1 D 9 + D 2 10 + 64D 1 -12D 10 -4D 9 + 28) Φ 6 (α)1 7 1 419 2 5598301 2 11550103 2 2551712030421623 2 3631 2 3041 2 (1126366500715763371124174330938368977431333(1126366500715763371124174330938368977431333D 28 107 2 576391299578767 2 274076093 2 290942059 2 13931 2 28657 2 204137 2 23 2 1493 2 2830547 2 544917713 2 31598314303 2 24535887822332807 2 201101 2 17 2 29 2 31 1 4379 1 1013 1 36537661 2 368487522735331 2 19813 2 1747 2

	D	D	=	=	-7 2 1311258103 2 93383157195708795083578685538920598221647 2 1 3 1 5 2 677 2 224699 2 10708193576771347 2 24833257 2 (66321637240913770895312913339793214 1
					(66321637240913770895312913339793214D 24 -(202259863244780558381860453499585(v (189712945006349928111821235829016019978456271676888(189712945006349928111821235829016019978456271676888D 35 2 + v -2 ))D 25 )
					-14465919591689307128573983437824887265672029343238691854081861559749D 25 ) -(857145950896171400671963795556989894042194131289487(v 2 + v -2 ))D 36 )
	D	D	=	=	1 -3888871381671794196458802315710490482078226712051900740103091728501684497594902265929767805231194888739D 36 ) 2 2 223 2 597482859539 2 248882825907331565531 2 (2019335401386293786207166724075859341(2019335401386293786207166724075859341D 25 -1 -2 1 261 2 128478995994071 2 3025833381100035119025676342890881 2
					(66321637240913770895312913339793214(v (24773656778414901634803870348969176884565517861269127(24773656778414901634803870348969176884565517861269127D 36 2 + v -2 ))D 26 )
					+297580655462944983601072219405031933105913989851705016725521711922510649D 26 ) -(189712945006349928111821235829016019978456271676888(v 2 + v -2 ))D 37 )
	D	D	=	=	1 +116014108817226025636250906637610633082682989655401422927845421603685431302028558553419600166912131824443D 37 ) 569 2 1439 2 2466240348741368415391921411051 2 (4412343134447284474117530515743496149(61772803882261982637645427220408946086D 26 -(2019335401386293786207166724075859341(v 2 + v -2 ))D 27 ) 1 3 2 (1462774104358166447320328081145651681708102851750470(1462774104358166447320328081145651681708102851750470D 37
					+13337596486249616633416128365715003750495014302489045160099715949334689029D 27 ) -(24773656778414901634803870348969176884565517861269127(v 2 + v -2 ))D 38 )
	D	D	=	=	1 -107114048131428196602036042282893363776335322672552405610021367908555000955590515740746720015779761857821D 38 ) 31 2 218069171 2 28721196055172805209 2 22725361 2 (763361856645910935772663964451373952681(763361856645910935772663964451373952681D 27 1 2 2 5 2 472233067 2 335188641021288661 2 924126824062838315412481 2
					-(61772803882261982637645427220408946086(v (447126463542130736420808986261322507372482025744507(447126463542130736420808986261322507372482025744507D 38 2 + v -2 ))D 28 )
					-149724800808298139234895716448044923226445865165412568988595381125439091482374D 28 ) -(1462774104358166447320328081145651681708102851750470(v 2 + v -2 ))D 39 )
	D	D	=	=	1 +1362236736392092735711595816371027107045738556864046551337719758325370857704119596126985689847376416801D 39 ) -1 3 2 8278065226744470551 2 4566184955538414377682528233 2 3943 2 2 -((v 2 + v -2 )763361856645910935772663964451373952681)D 29 ) (7693200087216050876804839304036908158990939786094(7693200087216050876804839304036908158990939786094D 39
					+156243595081359809872606121440348819964297847254205462786652267503931502270227271D 29 ) -(447126463542130736420808986261322507372482025744507(v 2 + v -2 ))D 40 )
	D	D	=	=	1 +12380421501975427887574839221558916249368633135750618849551513659803018520966646313776000175920433209D 40 ) 2 2 19 2 326983565485534051144425519451185707 2 181301 2 (88867285707427504552815003873633980867964653(88867285707427504552815003873633980867964653D 29 1 2 4 23 2 28775297939 2 45354556404323956764088141663033 2 128147 2
					-(1126366500715763371124174330938368977431333(v (444262371078411275036530891280549760044894125887(444262371078411275036530891280549760044894125887D 40 2 + v -2 ))D 30 )
					-123184813670701842459356619811841515374511032937050012802961098943874095248072007393959D 30 ) -(7693200087216050876804839304036908158990939786094(v 2 + v -2 ))D 41 )
	D	D	=	=	1 +1359689975955312021939066413608950103421830908431132145007295475564434519858469029413502565524083D 41 ) 2 2 29 2 449 2 1091 2 18414023 2 339722369745704256525298189901 2 (1713951505154451849278412459322972408517919889(1713951505154451849278412459322972408517919889D 30 1 2 2 3 2 53 2 49609439671346607175679 2 56322002047749796817567 2
					-(88867285707427504552815003873633980867964653(v (2598201402519308463837536678307826538637637361(12991007012596542319187683391539132693188186805D 41 2 + v -2 ))D 31 )
					2 10 -64E 1 -12E 10 -4E 9 + 28 1 213353666340007623638526192809579151292119580681379269874979045556633724290006464855951269)D 31 ) -(444262371078411275036530891280549760044894125887(v +(2 2 + v -2 ))D 42 )
	= = = = v 2 D 0 D D -1423853873843032305071517991227836398944647254463545764863856507677353245106560774424962103775D 42 ) 1 -43 2 97273570621 2 26396954918953349 2 130471771 2 4157 2 28621 2 (29676624191545618144648111624944621294859696166(29676624191545618144648111624944621294859696166D 31 D 1 719 2 61 2 185233449385388602061 2 12102357082912717651939 2 = -1 2 (D 13 -(2(v 16 + v -16 ) -v 12 -v -12 -4(v 10 + v -10 ) -26(v 8 + v -8 ) -((v 2 + v (5724569388018125395560846055677725147411014(5724569388018125395560846055677725147411014D 42 -2 )1713951505154451849278412459322972408517919889)D 32 ) +8239748870740490166734825221497528333002908462542913649313150812445359919137914031034615093D 32 ) -(12991007012596542319187683391539132693188186805(v 2 + v -2 ))D 43 )
	D D D D	= = = =	+4(v 76((219)D 12 -(v 6 + v -6 ) + 13(v 2 + v 4 + v -2 )D 14 ) + 787D 14 -4 ) + 152(v 2 + v -1 2 3 19 2 (453385(453385D 14 -(76(v 2 + v 1 -(7 1 16858333738933659887870586331090479589286797477594035643445468458518447355054012844780395)D 43 ) -2 ) + 184)D 12 ) 2 2 53 2 107073391 2 3080296628471719923564479600753 2 848857 2 (1202360930403560931030222837534182401486698364611(1202360930403560931030222837534182401486698364611D 32 1 5 1 2 2 53 2 54005371585076654675102321279978539126519 2 -2 ))D 15 ) -250464003D 15 ) -(29676624191545618144648111624944621294859696166(v 2 + v -2 ))D 33 ) (173537434709794154672446152614045596469(173537434709794154672446152614045596469D 43
	D D D D D	= = = = =	1 +101467727823064281002362989326473766812185962223175981805992928804837837463110204469450768695497D 33 ) (805808162(805808162D 15 -(453385(v 2 + v -(5724569388018125395560846055677725147411014(v 2 + v -2 ))D 44 ) -2 ))D 16 ) -245248536209874D 16 ) 2 1 5 2 90677 2 -1 402904081 2 (109342364049/2(437369456196D 16 -(805808162(v 2 + v -2 ))D 17 ) -57979709384117131616D 17 ) -1 +(51729710475616727676629256114810220610438295967049693913920779869327786123824237459)D 44 ) 3 2 (50592899583888726481372641670551635013829487634578(50592899583888726481372641670551635013829487634578D 33 1 7 1 11836998751 2 14660594155687780275955370219 2 1 (1835659768232(1835659768232D 17 -(109342364049(v 2 + v -2 ))D 18 ) -(1202360930403560931030222837534182401486698364611(v 2 + v -2 ))D 34 ) (472466105843823212817581554083188(58585797124634078389380112706315312D 44 3 8 1349905729 2 -470719001514611845893347D 18 ) -18716563042659124012459968685711353673244482249337370871121262458352499400472107136941540852623D 34 ) -(173537434709794154672446152614045596469(v 2 + v -2 ))D 45 )
	D D D	= = =	1 2 5 233 2 984796013 2 +246084500040920924910400141D 19 ) (26716821821819(26716821821819D 18 -(1835659768232(v 1 -486773091634646981124420181550725259305095623574103826313324090433645617D 45 ) 2 + v -2 ))D 19 ) 2 4 29 2 1187 2 522300221197 2 7694404573578921876239719 2 182858701 2 (857145950896171400671963795556989894042194131289487(857145950896171400671963795556989894042194131289487D 34 1 2 3 118116526460955803204395388520797 2
	D	=	1 1291 2 20694672209 2 -(50592899583888726481372641670551635013829487634578(v (7431837830116907(37159189150584535D 19 -(106867287287276(v 2 + v -2 ))D 35 ) (7735916375013449867448090407(7735916375013449867448090407D 45 +4411923185656392383365159760236306142723974489779682676837077861812027157604126881190104182451178043D 35 ), 2 + v -2 ))D 20 ) -(14646449281158519597345028176578828(v 2 + v -2 ))D 46 )
					-161710748575151990511450485922D 20 ) -297284847843573291454908916261083417941832630990370255741557001D 46 ),
	D	=	1 2 1 64271 2 362953 2 318589 2	(207561548775276615471(207561548775276615471D 20 -(37159189150584535(v	2 + v	-2 ))D 21 )
					-(212489945002283355770270866923659197255)D 21 )
	D D	= = D 48	1 (2 1 3 2 5 1 229 2 302127436354114433 2 -(207561548775276615471(v 2 + v -2 ))D 22 ) + (212206056324453264983259133859103049535987217108)D 22 ) (32987388636176124480235619(32987388636176124480235619D 21 1 (185954429283234131709345979553(557863287849702395128037938659D 22 1 = 7 4 (2976484557725606944157223(2976484557725606944157223D 46 -(30943665500053799469792361628(v 2 + v -2 ))D 47 ) 2 4 7 2 1523 2 31068281 2 99593906868959 2 -(32987388636176124480235619(v 2 + v +99497340316827556722995396776286275093114733864673142D 47 ) -2 ))D 23 ) -2739412859329752149612296806686633985128432534267734101D 23 ) = 1.
	D	=	-	1 73 2 25284900439 2 21330668861813 2 4723 2	(202259863244780558381860453499585(202259863244780558381860453499585D 23
					-(557863287849702395128037938659(v	2 + v	-2 ))D 24 )
					-92984479634762534856210727781070979304376139845640812904182336D 24 ),

2 1 + 12E 1 E 10 + 4E 1 E 9 + E

  1))Tr 1 ( 2v12 -4v 10 + 5v 8 -5v 6 + 4v 4 -2v 2 + 1v 7 )T r 1 +r 3 (v 6 -v 4 + v 2 -1) T r 1 +r 3 +r 4 ( (v 4 -v 2 + 1)(v 6 -v 4 + v 2 -1) v 5 )T r 3 +r 4 ( 2v 8 -2v 6 + 2v 4 -v 2 + 1 v 2 (v 6 -v 4 + v 2 -1) ) T -r 3 ( 5v 20 -15v 18 + 28v 16 -39v 14 + 44v 12 -40v 10 + 30v 8 -18v 6 + 9v 4 -3v 2 + 1 v 9 (v 6 -v 4 + v 2 -1) )T -r 3 -r 4 (-v 6 + v 4 -v 2 + 1)Tr 4 (-5v 28 -20v 26 + 48v 24 -85v 22 + 125v 20 -155v 18 + 167v 16 -155v 14 + 126v 12 -89v 10 + 54v 8 -27v 6 + 12v 4 -3v 2 + 1v 9 (v 8 -v 6 + v 4 -v 2 + 1)(v 6 -v 4 + v 2 -1) 2) Tr 4 ( v16 -2v 14 + 3v 12 -4v 10 + 6v 8 -5v 6 + 3v 4 -3v 2 + 1v 9 (v 8 -v 6 + v 4 -v 2 + 1))T r 3 +r 4 ( v16 -2v 14 + 3v 12 -3v 10 + 5v 8 -4v 6 + 3v 4 -2v 2 + 1v 4 (v 6 -v 4 + v 2 -1))Tr 3 ( (v 8 -v 6 + v 4 + 1)v 5

	T -r 4 (v	13 -v	11 + v	9 -v	7 + v	5 )

  Figure 10.6: Field extensions in type F 4 in case 3 Figure 10.7: Field extensions in type F 4 in case 4 Figure 10.8: Field extensions in type F 4 in case 5 Figure 10.10: Field extensions in type F 4 in case 7

	Fp( √ α, Fp( √ α,	β + β +	β β	-1 -1	) )		Fp( √ Fp( √	α, β) α, β)					Fp(α, Fp(α,	β) β)		Fp( √ Fp( √	α + α +	√ √	α -1 , α -1 ,	β) β)
	Fp( √ α, β + β -1 ) Fp( √ α, β + β -1 )		Fp( √ Fp( √	α + α +	√ √	α -1 , β) α -1 , β)	Fp( √ Fp( √	α + α +	√ α -1 , √ α -1 ,	β + β +	β β	-1 -1	) )	Fp(α, β) Fp(α, β)	Fp(α, Fp(α,	β + β +	β β	-1 -1	) )	Fp(α + α -1 , Fp(α + α -1 ,	β) β)
													Fp( Fp( Fp(	√ α, √ α, √ α,		β) β) β)
	Fp( √ Fp( √	α + α +	√ α -1 , β + β -1 ) √ α -1 , β + β -1 )	Fp(α + α -1 , β) Fp(α + α -1 , β)			Fp(α, β + β -1 ) Fp(α, β + β -1 )	Fp(α + α -1 , Fp(α + α -1 ,	β + β +	β β	-1 -1
	Fp( √ α, Fp( √ α, Fp( √ α,	β + β + β +	β β β	-1 -1 -1	) ) )		Fp( √ Fp( √ Fp( √	α, β) α, β) α, β)					Fp(α, Fp(α, Fp(α,	β) β) β)		Fp( √ Fp( √ Fp( √	α + α + α +	√ √ √	α -1 , α -1 , α -1 ,	β) β) β)
	Fp( √ α, β + β -1 ) Fp( √ α, β + β -1 ) Fp( √ α, β + β -1 )		Fp( √ Fp( √ Fp( √	α + α + α +	√ √ √	α -1 , β) α -1 , β) α -1 , β)	Fp( √ Fp( √ Fp( √	α + α + α +	√ α -1 , √ α -1 , √ α -1 ,	β + β + β +	β β β	-1 -1 -1	) ) )	Fp(α, β) Fp(α, β) Fp(α, β)	Fp(α, Fp(α, Fp(α,	β + β + β +	β β β	-1 -1 -1	) ) )	Fp(α + α -1 , Fp(α + α -1 , Fp(α + α -1 ,	β) β) β)
	Fp( √ α, Fp( √ α,	β + β +	β β	-1 -1	) )		Fp( √ Fp( √	α, β) α, β)					Fp(α, Fp(α,	β) β)		Fp( √ Fp( √	α + α +	√ √	α -1 , α -1 ,	β) β)
	Fp( √ Fp( √ Fp( √	α + α + α +	√ α -1 , β + β -1 ) √ α -1 , β + β -1 ) √ α -1 , β + β -1 )	Fp(α + α -1 , β) Fp(α + α -1 , β) Fp(α + α -1 , β)			Fp(α, β + β -1 ) Fp(α, β + β -1 ) Fp(α, β + β -1 )	Fp(α + α -1 , Fp(α + α -1 , Fp(α + α -1 ,	β + β + β +	β β β	-1 -1 -1
	Fp( √ α, β + β -1 ) Fp( √ α, β + β -1 )		Fp( √ Fp( √	α + α +	√ √	α -1 , β) α -1 , β)	Fp( √ Fp( √	α + α +	√ α -1 , √ α -1 ,	β + β +	β β	-1 -1	) )	Fp(α, β) Fp(α, β)	Fp(α, Fp(α,	β + β +	β β	-1 -1	)	Fp(α + α -1 ,	β)
	Fp( √	α +	√ α -1 , β + β -1 )	Fp(α + α -1 , β)			Fp(α, β + β -1 )	Fp(α + α -1 ,	β +	β	-1	)
												Fp(α + α -1 , β + β -1 )

)

Fp(α + α -1 , β + β -1 ) ) Fp(α + α -1 , β + β -1 ) ) Fp(α + α -1 , β + β -1 ) ) Fp(α + α -1 , β + β -1 )

Figure

10

.9: Field extensions in type F 4 in case

6

)

Fp(α + α -1 , β + β -1 )

Figure 10.11: Field extensions in type F 4 in case 8

  Fp(α + α -1 , β + β -1 )Figure 10.16: Field extensions in type F 4 in case 13

	Fp( √ α, α, β + β -1 ) √ Fp( Fp( √ α + Fp( √ α, β + β -1 ) Fp( √ α + √ α, β + β -1 ) Fp( √ α + Fp( √ α, β + β -1 ) α Fp( β + Fp( √ β α + -1 √ α -1 , β + β -1 ) ) √ α -1 , β) Fp( √ α + √ α -1 , β) √ Fp( √ α + √ α -1 , β) √ α -1 , β + β -1 ) Fp( √ α + √ α -1 , β) Fp( √ α + √ α -1 , β + β -1 )	Fp( √ √ Fp( α + √ α, β) α -1 , Fp(α + α -1 , β) β + Fp( √ α + √ α -1 , β + Fp( √ α + √ α -1 , β + Fp(α + α -1 , β) Fp( √ α + √ α -1 , β + Fp(α + α -1 , β)	β β β β	-1 -1 -1 -1	) ) ) )	Fp(α, Fp(α, β) β) Fp(α, β + β -1 ) Fp(α, β) Fp(α, β) Fp(α, β + β -1 ) Fp(α, β) Fp(α, β + β -1 )	Fp( √ β + Fp(α + α -1 , α + Fp(α, β -1 √ ) β + α -1 , Fp(α + α -1 , β) β -1 Fp(α, β + β -1 ) Fp(α + α -1 , Fp(α, β + β -1 ) Fp(α + α -1 , Fp(α + α -1 , β + β -1 ) Fp(α, β + β -1 ) Fp(α + α -1 , β + β Fp(α + α -1 , -1	β) β) β) β)
	Fp( √ α,	β +	β	-1	)		Fp( √	α, β)					Fp(α,	β)		Fp( √	α +	√	α -1 ,	β)
	Fp( √ α, β + β -1 )		Fp( √	α +	√	α -1 , β)	Fp( √	α +	√ α -1 ,	β +	β	-1	)	Fp(α, β)	Fp(α,	β +	β	-1	)	Fp(α + α -1 ,	β)
	Fp( √	α +	√ α -1 , β + β -1 )	Fp(α + α -1 , β)				Fp(α, β + β -1 )	Fp(α + α -1 ,	β +	β	-1

)

Fp(α + α -1 , β + β -1 )

Figure

10

.13: Field extensions in type F 4 in case 10

) Fp(α + α -1 , β + β -1 ) Figure 10.14: Field extensions in type F 4 in case 11 -1 , β + β -1 ) Fp(α + α -1 , β) Fp(α, β + β -1 ) Fp(α + α -1 , β + β -1

)

Fp(α + α -1 , β + β -1 )

Figure

10

.15: Field extensions in type F 4 in case 12

1 )

 1 Figure 10.19: Field extensions in type F 4 in case 16 5264 → 3772 → 1414, 7012 → 6766 → 7083 → 5796 → 2176 → 2374, 7071 → 30 → 106 → 4, 7083 → 30 → 106 → 4, 7083 → 2166 → 510 → 181, 7083 → 5796 → 2176 → 2374, 7165 → 2009 → 344 → 86, 7165 → 81 → 91 → 98, 7165 → 418 → 1148 → 157, 7165 → 2522 → 2577 → 300, 7165 → 2022 → 2667 → 1231, 7165 → 2022 → 854 → 2176 and 7165 → 5406 → 5623 → 3174. This proves this E 8 -graph remains connected after specialization when p = 7.

√ α(β+1)(αβ+1) (α+β) 2

Remerciements

1 : ( [5], ∅), 4 : ([4, 1], ∅), 5 1 : ( [3,[START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF], ∅), 5 2 : ( [4], [1]), 6 : ([3, 1 2 ]), 10 1 : ([3], [1 2 ]), 10 2 : ([3], [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]), 10 3 : ( [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF][START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF], [1]), 15 : ( [3,1], [1]) and 20 : [2, 1], [START_REF] Alvis | The representations and generic degrees of the Hecke algebra of type H 4[END_REF]). The derived subgroup

Proof. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], we know that A I 2 (5) is generated by s 1 t -1 , ts 1 t -2 and t 2 s 1 t -3 and that A H 3 is generated by

It is thus sufficient to show that s 2 t -1 can be written as a product of conjugates of the generators of A I 2 (5) or of their inverses. We have

Proposition 8.1.2. The restrictions to A H 3 of the representations of dimension greater than 1 afforded by those W -graphs are absolutely irreducible and pairwise non-isomorphic.

Proof. As in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] Lemma 3.4, we only need to prove that A H 3 is generated by A I 2 (5) and A H 3 in order to prove the absolute irreducibility. This is true because s 3 = s 3 s -1 1 s 1 and s 3 s -1 1 ∈ A H 3 and s 1 ∈ A I 2 (5) .

Let ρ 1 and ρ 2 be irreducible representations afforded by H 3 -graphs. Assume ρ 1|A H 3 ρ 2|A H 3 . By Lemma 3.2.3, we have that there exists a character η such that ρ 1 ρ 2 ⊗ η. We have

It follows that there exists u ∈ F p such that ρ 1 (S 1 ) is conjugate to uρ 2 (S 1 ). The eigenvalues of ρ 1 (S 1 ) and ρ 2 (S 1 ) are -1 and α. It follows that {-1, α} = {-u, uα}. We then have either u = 1 or (u = -α and α 2 = 1). The latter contradicts our assumptions on the order of α therefore u = 1 and η is the trivial morphism. It follows that ρ 1 ρ 2 .

The four W -graphs provided in [START_REF] Geck | Characters of finite Coxeter groups and Iwahori-Hecke algebras[END_REF] (see Figure 8.1, where λ = ξ + ξ -1 + 1) then determine all the irreducible representations of the Iwahori-Hecke algebra over F q up to taking the dual W -graph or the algebraic conjugate by the involution x → x.

Before determining the image of the Artin groups inside each representation, we show that we cannot always have 1 ∼ 2 as in Lemma 5.1.1. Recall that 1 ∼ 2 if there exists a field automorphism Φ of

1. (q 1 2 + 1) 3 .S 4 , 24(q

2 ) 2 : (q 1 2 + 1).2, 2q(q

4. SU 4 (q 0 ), q 1 2 = q r 0 , r odd prime, q 6 0 (q 2 0 -1)(q 3 0 + 1)(q 4 0 -1)

2 ).[(q

2 ).[(q 12. (q

We have |H| = q(q 2 -1). We have Φ 1,2 (α) = α -1 so α q 1 2 +1 = 1. This implies that q 1 2 + 1 ≥ 7, therefore we have that q ≥ 49. It follows that |H| ≥ 117600. This excludes cases 7 to 11. We have that q is a square and p is odd, therefore q ≥ 49 implies that q = 49 or q ≥ 81. If q = 49 then |H| = 117600 does not divide 207360, therefore case 12 is excluded. If q ≥ 81 then |H| ≥ 531360. Therefore, case 12 is excluded for any q.

Assume by contradiction that we are in case 1. We then have that q divides 24 which is a contradiction. Case 1 is therefore excluded.

Assume by contradiction that we are in case 2. We then have that q+1 divides (q 1 2 +1)(q-1). It follows that q + 1 divides (q 1 2 + 1)(q + 1) -(q 1 2 + 1)(q -1) = 2q 1 2 < q + 1. This contradiction proves that case 2 is excluded.

Assume by contradiction that we are in case 4. We then have that q divides q 6 0 = q 3 r . This implies that r = 3, therefore (q 2 -1) divides (q 2 3 -1)(q + 1)(q 4 3 -1). It follows that q -1 divides (q

This implies that q -1 divides q 2 3 + q 1 3 -2 < q -1. This contradiction proves that case 4 is also excluded. Assume now by contradiction that we are in case 3. Since G is perfect, we have that G ≤ SL 2 (q). It follows that G = H but this is absurd by the computations made when

Assume by contradiction that we are in case 5 or 6. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], we have that A H 3 is perfect. It follows that we would have G ≤ SO ± 4 (q

2 ). This contradicts the fact that ρ 4 r is not self-dual. It follows that G is included in no maximal subgroup of SU 4 (q 1 2 ). This implies that G SU 4 (q The roles of ξ+ξ -1 and ξ 2 +ξ -2 are symmetric, therefore it remains to check that 3+ξ+ξ -1 = 0, 5 -8(ξ + ξ -1 ) = 0, 2 -3(ξ + ξ -1 ) = 0 and 1 -ξ -ξ -1 = 0.

. Therefore ( 5 8 ) 2 + 5 8 -1 = 0 and 25 + 40 -8 = 0, which implies that 57 = 0. We then have 19 = 0 because p = 3. In F 19 , we have 4 2 + 4 -1 = 0 and (-5) 2 + (-5) -1 = 0, therefore ξ + ξ -1 ∈ {4, 14}. It follows that 5 -8(ξ + ξ -1 ) ∈ {5 -8 × 4, 5 -8 × (-5)} = {-27, 45}. Therefore 27 = 0 or 45 = 0 in F 19 , which is absurd.

If 2 -3(ξ + ξ -1 ) = 0, then ξ + ξ -1 = 2 3 . Therefore ( 23 ) 2 + 2 3 -1 = 0 and 4 + 6 -3 = 0. This would imply 7 = 0, therefore p = 7. Note that X 2 + X -1 has no roots in F 7 and ξ + ξ -1 cancels X 2 + X -1, therefore ξ + ξ -1 / ∈ F 7 and ξ + ξ -1 = 2 3 = 10 = 3. This implies 3 / ∈ F 7 which is absurd.

If 1 -ξ -ξ -1 = 0, then 2 + ξ 2 + ξ -2 = 0. Therefore (1 + ξ)(1 + ξ -1 ) = 0, which is absurd. This concludes the proof.

The restrictions of the representations to H H 3 available in the CHEVIE package of GAP3 [START_REF] Geck | CHEVIE -A system for computing and processing generic character tables for finite groups of Lie type, Weyl groups and Hecke algebras[END_REF] are given in Table 8.3 (we don't write the restriction of a dual representation or a representation obtained after applying the field automorphism sending ξ + ξ -1 to ξ 2 + ξ -2 if the restriction of the representation is already given), where for a representation R, the dual representation is denoted by R . In the generic case, they are the same as the induction/restriction tables of the corresponding finite Coxeter groups, therefore they can be computed easily.

Lemma 8.2.1. The normal closure

Proof. We note s 1 , s

. This is true because

Type H 4 , low dimensional representations

We now determine the image of the Artin group of type H 4 inside the low-dimensional representations of the Iwahori-Hecke algebra.

Proposition 8.3.1.

2 ).

Proof. The proof of the first part of the statement is identical to the proof of Proposition

2. Assume 1 2. We then have ρ 6s (A H 4 ) Ω + 6 (q

2 ).

Proof. Let F r = F p (α, ξ + ξ -1 ). By Proposition 6.2, we have ρ 6s (A H 4 ) ≤ Ω + 6 (q ), where

. By Proposition 6.1, we also have ρ 6s σ • ρ 6s , where σ = I dF r if F q = F r and σ is the automorphism of order 2 of F q otherwise. We then have by Proposition 4.1 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] that up to conjugation, we have ρ 6s (A H 4 ) ≤ Ω + 6 (r). Assume first that 1 ∼ 2. By Proposition 8.1.2, we have that

We have

We cannot have ρ 6s (A H 4 ) Ω - 6 (r) because otherwise we would have Ω - 6 (r) ≤ Ω + 6 (r), therefore r 6 (r 3 + 1)(r 4 -1)(r 2 -1) would divide r 6 (r 3 -1)(r 4 -1)(r 2 -1). This would imply that r 3 + 1 divides r 3 -1 which is absurd.

This implies that ρ

. The Table 8.3 and Proposition 8.1.3 imply that the latter possibility is the only one possible.

Assume now 1 2. By Proposition 8.1.2, we have that F p (α, ξ +ξ -1 ) = F p (α+α -1 , ξ +ξ -1 ). There exists then a unique automorphism of order 2 of F r . We have (α

. By Table 8

2 ). We can again apply Theorem 2.3.3, and we have that ρ 6s (A H 4 ) is conjugate in GL 6 (r) to one of the following groups

Assume by contradiction that we are in the second case. We then have that (r

). The proof is then concluded by Proposition 8.1.3.

For p = 103, we have Gcd(

. We then have

This would imply 17 = 0 which is absurd since p = 103.

For p = 90023, we have Gcd(

. We then have

This would imply that 16295 = 0 which is absurd since p = 90023. For p = 159073, we have Gcd(

. We then have

This would imply that 13377 = 0, which is absurd since p = 159073. It now only remains to consider p ∈ {11, 13} to conclude Case 1.1.3. Assume first p = 13. We have, using the same notations as before, that 24D 406 -3245D

We then have using GAP4 that Gcd(H 1301 , H 1302 ) = 1 which leads to a contradiction.

Assume now p = 11. We have, using the same notations as before, that -1185D 407 + 24D 408 ) = 0 = Φ(-1185E 2 + 24E 3 ). Therefore H 1101 = v 26 (-1185E 2 + 24E 3 ) = 0. We let H 1102 = v 8 D 7 = 0. We then have using GAP4 that Gcd(H 1101 , H 1102 ) = 1 which leads to a contradiction.

Case 1.2 :

We choose as in the proposition γ r = 1 for all r ∈ π ∪ -π. We can then compute the remaining γ r using the induction relation in the proposition. We get -γ r 1 +r 3 = γ r 2 +r 3 = -γ r 3 +r 4 = 1, γ r 1 +r 2 +r 3 = γ r 2 +r 3 +r 4 = γ r 1 +r 3 +r 4 = γ r 1 +r 2 +r 3 +r 4 = γ r 1 +r 2 +2r 3 +r 4 = 1. For the negative roots, we use the fact that γ -r = γ r for all r ∈ Φ.

We write in the following T r (u) = I 8 + ue r . By Proposition 8.4.1, we get an isomorphism of P Ω + 8 (q) defined on the generators T r (u) by τ (T τ (r) (γ r u)). 5]] by explicit computations given in subsection 10.3.3 of the Appendix.

This proves that τ

2 ) since it is generated by long root elements and irreducible.

Assume now that F q = F p (α) = F p (α + α -1 ). We now want to use Goursat's Lemma (see Lemma 3.3.1). We consider G as a subgroup of Ω + 8 (q)×Ω + 8 (q). We write π 1 the projection upon the first factor and π 2 the projection upon the second factor. We write

. By Goursat's Lemma, there exists an isomorphism

Let x ∈ ker(π 1 ), we have π 1 (x) = I 8 . We know that projectively, we have τ

This proves that

. This implies that K 1 /K 1 P Ω + 8 (q) and K 1 = (-I 8 ) × I 8 . For every x ∈ K 1 , we have exactly two elements of K 2 such that ϕ(x) = y. This implies that |G| = 2|Ω + 8 (q)|. We also know that for all x ∈ Ω + 8 (q), there exists y x ∈ Ω + 8 (q) such that τ (x) = y x and (x, y x ) ∈ G. We know there exists h ∈ A H 4 such that

This proves that G contains {(x, y) ∈ K 1 × K 2 , τ (x) = y}. It follows using the cardinality of those two sets that

We also have that the center Z(G) of G is equal to {±I 8 }×{±I 8 } since this group is included in G and Z(Ω + 8 (q)) = {±I 8 }. It follows that G/Z(G) = {(x, τ (x)), x ∈ Ω + 8 (q)} P Ω + 8 (q). We also have that G has two normal subgroups Z 1 and Z 2 of order 2,

. This proves that G is a double cover of Ω + 8 (q). We recall here the definitions of a central extension and of a universal central extension from [5]. A central extension of a group Γ is a pair (H, π), where H is a group and Π :

there exists a unique morphism α : ( Γ, π) → (H, σ) of central extensions. By [5] (33.1), there exists at most one universal central extension of a group Γ. By [5] (33.4), any perfect group posseses a universal central extension. The universal central extension Γ is then called the universal covering group of Γ and ker(π) is the Schur multiplier of Γ.

We now show that G is the universal covering group of P Ω + 8 (q). We have by [START_REF] Daniel Gorenstein | The classification of the finite simple groups. Number 3. Part I. Chapter A[END_REF] (Theorem 6.1.4 and Table 6.1.2) that the Schur multiplier of P Ω + 8 (q) is (Z/2Z) 2 . By Theorem 1.10.7 of [START_REF] Daniel Gorenstein | The classification of the finite simple groups. Number 3. Part I. Chapter A[END_REF], we have that the universal cover of P Ω + 8 (q) is Spin + 8 (q). By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A H 4 is perfect, it follows that G is perfect. We have shown above that G/Z(G) P Ω + 8 (q), this proves that G is a perfect central extension of G. We have Z(G) (Z/2Z)

2 ) in a natural representation. Let be the unique automorphism of order 2 of F q . We have (α) = α -

2 ). We have, by Lemma 3.3.3, that the field generated by the traces of the elements of ρ 9s (A H 4 ) contains

2 ). The same arguments show that ρ 9s (A H 4 ) SU 9 (q

Consider now the 10-dimensional representation ρ 10r . Let G = ρ 10r (A H 4 ). By Proposition 6.1, we can assume that it is defined over F q .

Assume first that F q = F p (α) = F p (α + α -1 ). By Table 8.3 and Theorem 8.1.1, we have that ρ 10r (A H 3 ) SL 5 (q) in a twisted diagonal representation. We have by Proposition 6.2 that up to conjugation in GL 10 (q), we have G ≤ Ω + 10 (q). It follows by Lemma 8.2.1 that G is an irreducible subgroup of Ω + 10 (q) generated by long root elements. It follows by Theorem 2.3.3 that G is isomorphic to one of the following groups

2 )

If ρ 10r (A H 4 ) was conjugate to Ω - 10 (q

2 ), then we would have that |SL 5 (q)| = q 10 (q 2 -1)(q 3 -1)(q 4 -1)(q 5 -1) divides |Ω - 10 (q

2 )| = q 10 (q 5 2 + 1)(q 4 -1)(q 3 -1)(q 2 -1)(q -1). This would imply that q 5 -1 divides (q 5 2 + 1)(q -1) = q 7 2 -q 5 2 + q -1 < q 7 2 -1 = q 5 -1, which is absurd. This proves that ρ 10r (A H 4 ) = Ω + 10 (q). Assume now that F q = F p (α) = F p (α + α -1 ). Let be the unique automorphism of order 2 of F q , we have (α) = α -1 . It follows by Proposition 2.1.2 that • ρ 10r ρ 10r . We then have by Proposition 4.1 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] that up to conjugation in GL 10 (q), we have G ≤ Ω + 10 (q

2 ). We have

2 ) in a twisted diagonal representation. By the same arguments as above, we have G Ω + 10 (q

2 ) or G Ω - 10 (q 4 ). Assume by contradiction that G Ω - 10 (q 4 ). We then have that |SU 5 (q

2 )| = q 5 (q -1)(q

4 )| = q 5 (q

2 -1). This implies that q 5 2 + 1 divides (q

2 ).

The remaining cases will be proved as in [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] and types B and D using Theorem 2.3.2. We first give a Lemma which will help us prove that the groups considered are tensor-indecomposable. Lemma 8.5.1. Let = p k with p / ∈ {2, 3, 5} a prime and k ≥ 1. If r ≥ 3 then there exists no non-trivial morphism from SL r ( )

2 ). In cases 11, 12, 13 and 16, we have ρ 4 2 (A F 4 ) SU 4 (q 1 2 ) and ρ 4 4 (A F 4 ) SU 4 (q 1 2 ). In all the remaining cases, we have ρ 4 2 (A F 4 ) SL 4 (q) and ρ 4 4 (A F 4 ) SL 4 (q). Proof. We have that ρ

). The result then follows from Lemma 3.5 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF].

We now consider the representation ρ 4 1 . We order the basis indexed by the vertices in the lexicographic way, i.e. I(

. Note now that for any matrices M, N ∈ GL 2 (F p ), we

We have by [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF] that A F 4 is generated by

4 and s 4 s 3 s -2 4 . The result then follows from what was proven above for 2 1 and 2 3 .

Consider now the representation ρ 4 2 . We know, using Table 9.1 and Table 9

By Lemmas 3.3.2 and 3.3.8, ρ 4 2 (A B 3 ,1 ) ∈ {SL 3 (q), SU 3 (q

2 )}. Take now a transvection t of ρ 4 2 (A B 2 ). We know that for i ∈ {1, 2}, ρ 4 2 (A B 3 ,i ) is normally generated by that transvection in ρ 4 2 (A B 3 ,i . Since A F 4 is generated by A B 3 ,1 and A B 3 ,2 , we have that the normal closure in ρ 4 2 (A F 4 ) of t is equal to ρ 4 2 (A F 4 ). This proves that ρ(A F 4 ) is an irreducible subgroup of SL 4 (F p ( √ α, √ β)) generated by transvections. We can now use Theorem 2.3.4. Let F q 0 = F p ( √ α, √ β). There exists q dividing q 0 such that ρ 4 2 (A F 4 ) is conjugate in GL 4 (q 0 ) to SL 4 (q ), SU 4 (q 1 2 ) or SP 4 (q ). We also have that ρ 4 2 ρ 4 2 , therefore the symplectic case is excluded. Note that ρ 4 2 (A F 4 ) contains either a

2 ), therefore we have that q divides q and q ∈ {q, q 0 }. It now only remains to show that we have the correct groups. Assume

), therefore ρ 4 2 (A F 4 ) contains a natural SL 3 (q). There exists then a unique automorphism of order 2 of F q 2 . Since F q is fixed by , we have that • ρ 4 2 ρ 4 2 by considering the restriction to A B 3 ,1 . It follows by Lemma 3.2.5 that up to conjugation, we have ρ 4 2 (A F 4 ) ≤ SL 4 (q). This proves that q = q and ρ 4 2 (A F 4 ) SL (q) in cases 4, 5 and 10. We can now assume F p ( √ α, √ β) = F p (α, β), this implies that q 0 = q, therefore q = q. This implies that ρ 4 2 (A F 4 ) is conjugate in GL 4 (q) to SL 4 (q) or SU 4 (q

) then we have that there exists a unique automorphism of order 2 of F q and that for all representation ϕ of H B 3 ,α,β , we have

SL 4 (q). The only remaining possibility is

). The arguments are identical to the previous case.

To conclude, we have ρ 4 2 (A F 4 ) SU 4 (q 1 2 ) in cases 11, 12, 13 and 16 and ρ 4 2 (A F 4 ) SL (q) otherwise. By the restriction table, we can apply the same reasoning to ρ 4 4 and we get the same result. Proposition 9.4. In cases 1, 4, 5 and 10, we have ρ 8 2 (A F 4 ) SL 8 (q) and ρ 8 3 (A F 4 ) SL (q).

In cases 11, 12, 13 and 16, we have

2 ) and ρ 8 3 (A F 4 ) SU 8 (q

2 ). In cases 2, 6, 9 and 15, we have

2 ). In cases 3, 7, 8 and 14, we have

. Consider now the representation ρ 8 2 . We have in cases 1, 4, 5, 10, 11, 12, 13 and 16 by Tables 9.2 and 9.3 and Lemmas 3.3.2 and 3.3.8 that ρ 8 2 (A B 3 ,1 ) and ρ 8 2 (A B 3 ,2 ) are generated by transvections. Since A F 4 is generated by A B 3 ,1 and A B 3 ,2 , we have that G 8 2 is generated by transvections. We can then apply Theorem 2.3.4. We get that there exists q dividing q such that ρ 8 2 (A F 4 ) is conjugate in GL 8 (q) to SL 8 (q ),

Table 9.4: Restriction of the 9-dimensional representations to

3 and the eigenvalues of vρ 9 1 (S 1 ) are vβ with multiplicity 6 and -v with multiplicity 3. This implies that v = 1 = β -2 , therefore β 2 = 1 which is absurd. This implies that • ρ 9 1 |A F 4 ρ 9 2 |A F 4 . We will now use Theorem 2.3.2. We have that G 9 1 contains a natural SL 2 (r) or a natural

2 ) and that we are not in the second case of Theorem 2.3.2. It also implies by Lemma 3.3.5 that G 9 1 is tensor-indecomposable. However, A F 4 is not normally generated by A B 3 , therefore we cannot use the same arguments as before to show that G 9 1 is primitive in the non-monomial case. Nevertheless, we can show in the same way that if G 9 1 ⊂ (GL 3 (q)×GL 3 (q)×GL 3 (q)) S 3 then ρ 9 1 (A B 3 ,1 ) ⊂ GL 3 (q)×GL 3 (q)×GL 3 (q) and ρ 9 1 (A B 3 ,2 ) ⊂ GL 3 (q) × GL 3 (q) × GL 3 (q). Since G 9 1 is generated by ρ 9 1 (A B 3 ,1 ) and ρ 9 1 (A B 3 ,2 ), this contradicts the irreducibility, and G 9 1 is therefore primitive. It follows that G 9 1 is a classical group over F q for some q dividing q in a natural representation. We have that ρ

, therefore G 9 1 preserves no non-degenerate bilinear form. We have that

which is equal up to conjugation to

, where

) and r = r

. Thus, the field generated by the traces of the elements of G 9 1 contains 5 + 2(α + α -1 ) and 5 + 2(β + β -1 ). Since p = 2, this field contains α + α -1 and β + β -1 , therefore q 1 2 divides q . This implies that G 9 1 is conjugate to SU 9 (q 1 4 ), SL 9 (q 1 2 ), SU 9 (q 1 2 ) or SL 9 (q). We have

It follows that G 9 1 SL 9 (q). This corresponds to cases 2, 6, 9 and 15.

Assume now

). We then have ρ 9 1 (A B 3 ,1 ) SL 3 (q) × SU 2 (r 1 2 ), where F r = F p (α) = F p (α + α -1 ). There exists a unique automorphism of order 2 of F q and if we consider the representations appearing in the restriction of ρ

We have here (α) = α -1 and (β) = β. We have

We then get in the same ways as in cases 2, 6, 9 and 15 that we can apply Theorem 2.3.2 to get that G 9 1 is a classical group over F q in a natural representation and that q 1 2 divides q . This implies that G 9 1 is conjugate to SU 9 (q 1 4 ), SL 9 (q 1 2 ), SU 9 (q 1 2 ) or SL 9 (q). We have

Table 9.6: Restriction of the 6-dimensional representations to By what was proven in cases 2, 3, 6, 7, 8, 9, 14 and 15 and Table 9.5, we have the second part of the proposition. Proposition 9.6. In cases 11, 12, 13 and 16, we have

2 ) and ρ

2 ). In all the remaining cases, we have ρ 6 1 (A F 4 ) Ω + 6 (q) and ρ 6 2 (A F 4 ) Ω + 6 (q).

. Let P 1 be the anti-diagonal matrix with coefficients (1, -1, 1, 1, -1, 1). We write the matrices of ρ 6 1 with respect to the basis (e x 1 , e x 2 , e x 3 , e x 4 , e x 5 , e x 6 ), where I(x 1 ) = {s 1 , s 3 , s 4 }, I(x 2 ) = {s 2 , s 4 }, I(x 3 ) = {s 3 } and I(x j ) = {s 1 , s 2 , s 3 , s 4 } \ I(x 6-j ) for j ∈ {4, 5, 6}. We then have for all g ∈ G 6 1 , P 1 gP -1 1 = t g -1 . This proves that up to conjugation in GL 6 (q), G 6 1 ≤ Ω + 6 (q). By Proposition 6.2, we have that up to conjugation, G 6 2 ≤ Ω + 6 (q). By Tables 9.6 and 9.7 and Theorems 3.2.1 to 3.2.6, we have that (ρ 6 1 (A B 3 ,1 ) SL 3 (q) or ρ 6 1 (A B 3 ,1 ) SU 3 (q

2 )) and (ρ 6

2 ). Note that those isomorphisms are given by twisted diagonal embeddings, therefore a transvection in SL 3 (q) is mapped to a long root element of Ω + 6 (q) by the isomorphism ρ 6 1 (A B 3 ,1 ) SL 3 (q). This proves that ρ 6 1 (A B 3 ,1 ) and ρ 6 1 (A B 3 ,2 ) are generated by long root elements. If t is a long root element of ρ 6 1 (A B 2 ) then its normal closure in ρ 6 1 (A B 3 ,1 ) is equal to ρ 6 1 (A B 3 ,1 ) and its normal closure in ρ 6 1 (A B 3 ,2 ) is equal to ρ 6 1 (A B 3 ,2 ). It follows that the normal closure of t in G 6 1 is equal to G 6 1 since by Lemma 9.1, A B 3 ,1 and A B 3 ,2 generate A F 4 . This proves that G 6 1 is an irreducible subgroup of Ω + 6 (q) generated by a conjugacy class of long root elements. Since O p (G 6 1 ) is normal in G 6 1 and V = F 6 q is an irreducible F q G 6 1 -module, we apply Clifford's Theorem [START_REF] Curtis | Methods of representation theory[END_REF]Theorem 11.1] and get that Res

is isomorphic to one of the following groups for some q dividing q 1. Ω + 6 (q ) in a natural representation,

2 ) as a subgroup of Ω + 6 (q ), where Ω + 6 (q ) is in a natural representation, 3. SU 3 (q ) as a subgroup of Ω + 6 (q ), where Ω + 6 (q ) is in a natural representation.

Assume now that we are in cases 4, 5 or 10. We have q = q 2 and F q = F p (α, β) = F p (α + α -1 , β + β -1 ). There exists a unique automorphism Φ of order 2 of F q . We have We have (Xρ

2 ). Assume by contradiction that G 6 1 Ω - 6 (q 1 4 ). Then |Ω - 6 (q

2 )| = q 3 2 (q -1)(q 3 2 + 1)|, and q 3 2 + 1 divides (q 3 4 + 1)(q -1) 1 2 . Therefore, q 3 2 + 1 divides (q 3 4 +1)(q -1) 1 2 -(q 3 4 -1)(q -1) 1 2 = q -1 < q 3 2 +1. This is absurd. Thus, we have

). There exists then a unique automorphism of order 2 of F q . We have by Proposition 3.2.4 that

|A F 4 by Table 9.6. We know that q ∈ {q 1 2 , q}. We have by Proposition 9.2 that • ρ 6 1 |A F 4 ρ 6 1 |A F 4 , therefore we cannot have q = q 1 2 . It follows that q = q. The proof then uses the same arguments as in case 1, therefore we get that G 6 1 Ω + 6 (q) in cases 2, 6, 9 and 15. Finally, assume

). There exists then a unique automorphism of order 2 of F q . We have by Proposition 3.2.4 that

|A F 4 by Table 9.6. We know that q ∈ {q 1 2 , q}. We have by Proposition 9.2 that • ρ 6 1 |A F 4 ρ 6 1 |A F 4 , therefore we cannot have q = q 1 2 . It follows that q = q. The proof then uses the same arguments as in case 1, therefore we get that G 6 1 Ω + 6 (q) in cases 3, 7, 8 and 14.

Table 9.8: Restriction of the high dimensional representations to

where Φ is the unique automorphism of order 2 of F q. It follows by Lemma 3.2.5 that in cases 4, 5 and 10, G 12 is conjugate in GL 12 (q) to a subgroup of GL 12 (q). We then have that

2 ), G 12 is primitive and we are not in case (2) of Theorem 2.3.2. In order to apply the theorem we have to prove that G 12 is tensor-indecomposable but we cannot use Lemma 3.3.6 because 12 ≤ 16. The arguments in the proof of Lemma 3.3.6 [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] only require d ≥ 16 in order to have a ≥ 3 and b ≥ 4 if a, b ≥ 3. This is still true for d ≤ 16 unless d = 9, therefore we can still apply Lemma 3.3.6. We still have to prove that we do not have G 12 ≤ GL 2 (q)⊗GL 6 (q). Assume by contradiction that G 12 ≤ GL 2 (q)⊗GL 6 (q). We then have a morphim from G 12 to GL 2 (q). Consider the restriction of this morphism to ρ 12 (A B 3 ,1 ). Table 9.8 and Propositions 3.3.2 and 3.3.8 give us a morphism from SL 3 (q)×SL 3 (q) or SU 3 (q 1 2 )×SU 3 (q 1 2 ) to GL 2 (q). We can consider the restriction to each factor and we get a morphism from SL 3 (q) to GL 2 (q) or a morphism from SU 3 (q 1 2 ) to GL 2 (q). If this morphism is non-trivial then we get an isomorphism from P SL 3 (q) to a subgroup of GL 2 (q) or an isomorphism form P SU 3 (q 1 2 ) to a subgroup of GL 2 (q). This would imply by considering the orders of those groups that 1 (3,q-1) q 3 (q 2 -1)(q 3 -1) or 1 (3,q-1) q 3 (q 2 -1)(q 3 + 1) divides q 2 (q 2 -1)(q -1) which is absurd since q 3 cannot divide q 2 . This proves that the restriction of this morphism to each factor is trivial, therefore the restriction of this morphism to ρ 12 (A B 3 ,1 ) is trivial. The restriction to ρ 12 (A B 3 ,2 ) is also trivial by Table 9.9 and the same arguments as above. This would imply by Lemma 9.1 that this morphism is trivial, which contradicts the irreducibility of G 12 . Therefore G 12 is tensor-indecomposable. So, we can apply Theorem 2.3.2 and we get that G 12 is a classical group in a natural representation. Consider now this representation with respect to the basis (e x i ) i∈[ [1,[START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]] , where I(x 1 ) = {s 2 }, I(x 2 ) = {s 3 }, I(x 3 ) = {s 1 , s 2 }, I(x 4 ) = {s 1 , s 3 }, I(x 5 ) = {s 1 , s 3 }, I(x 6 ) = {s 1 , s 4 } and for j ∈ [ [7,[START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]], I(x j ) = {s 1 , s 2 , s 3 , s 4 } \ {I(x 13-j )}. Let P be the anti-diagonal matrix with coefficients (3, 3, -1, 1, -3, -1, -1, -3, 1, -1, 3, 3). We then have for all i ∈ {1, 2} that P ρ 12 (S i )P -1 = -α t ρ 12 (S i ) -1 and for all j ∈ {3, 4}, P ρ 12 (S j )P -1 = -β t ρ 12 (S j ). This proves that G 12 is conjugate in GL 12 (q) to a subgroup of Ω + 12 (q). Since it contains a twisted diagonal SL 3 (q) or a twisted diagonal SU 3 (q 1 2 ), the field generated by its traces contains F p (α + α -1 , β + β -1 ). In case 1, 4, 5 and 10 this proves that G 12 is a classical group over F q and since G 12 is conjugate to a subgroup of Ω + 12 (q), we get G 12 Ω + 12 (q). Assume now that we are in case 11, 12, 13 or 16. Then there exists an automorphism of order 2 of F q , and • ρ 12|A F 4 ρ 12|A F 4 . It follows by Lemma 3.2.5 that G 12 is conjugate to a subgroup of Ω + 12 (q

2 ). Hence, G 12 is a classical group in a natural representation over

). There exists then a unique automorphism of order 2 of F q . We have •ρ 12|A F 4 Assume now we are in case 1, 4, 5 or 10. We have F q = F p (α + α -1 , β + β -1 ), therefore q = q and G 16 is a classical group in a natural representation over F q . Since G 16 is conjugate to a subgroup of Ω + 16 (q), we get that G 16 Ω + 16 (q). In all the remaining cases, there exists a unique automorphism of order 2 of F q and • ρ 16|A F 4 ρ 16|A F 4 . It follows that G 16 is conjugate to a subgroup of Ω + 16 (q

2 ). It then follows that q = q 1 2 , therefore G 16 Ω + 16 (q

2 ).

Theorem 9.1. We write

) and F r β = F p (β + β -1 ). In cases 1, 4, 5 and 10, the morphism from A F 4 to H F 4 ,α,β ρ irr GL nρ (q) factorizes through the surjective morphism

×SL 8 (q) 2 × SL 9 (q) 2 × Ω + 12 (q) × Ω + 16 (q). In cases 11, 12, 13 and 16, the morphism from A F 4 to H F 4 ,α,β ρ irr GL nρ (q) factorizes through the surjective morphism

2 ) 2 × ×Ω + 6 (q

2 ).

In cases 2, 6, 9 and 15, the morphism from A F 4 to H F 4 ,α,β ρ irr GL nρ (q) factorizes through the surjective morphism

2 ) × SU 8 (q

2 ) × SL 9 (q) × Ω + 12 (q

2 ).

In cases 3, 7, 8 and 14, the morphism from A F 4 to H F 4 ,α,β ρ irr GL nρ (q) factorizes through the surjective morphism

2 ) × SL 9 (q) × Ω + 12 (q

2 ).

Proof. Note the by [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF], A F 4 is not perfect, this makes the proof of this theorem more complex than for the previous types. The result follows easily from Goursat's Lemma in cases 2, 3, 6, 7, 8, 9, 14 and 15.

Assume now we are in case 1, 4 5 or 10. We must show that there exists no field automorphism Ψ and no character z :

Assume by contradiction that there exists such a field automorphism Ψ and such a character z. By [START_REF] Mulholland | Local indicability and commutator subgroups of Artin groups[END_REF],

1.1. Assume 3 -(β + β -1 ) = 0. We have Tr(ρ 4 2 (q 1 )) = Tr(ρ 4 4 (q 1 )) = Tr(ρ 4 2 (q 0 q -1 1 )) = Tr(ρ 4 4 (q 0 q -1 1 )) = 3 -β -β -1 = 0.

It follows that z(q 0 ) = z(q 1 ) = z(q 0 q -1 1 ) = 3-β-β -1 Ψ(3-β-β -1 ) therefore z(q 1 ) -1 = 1 and z(q 0 ) = z(q 1 ) = 1. This proves that z is the trivial character. It follows that Ψ(α + α -1 ) = α + α -1 and Ψ(β + β -1 ) = β + β -1 . This implies that Ψ is the trivial automorphism. This would imply that ρ 4 2 |A F 4 ρ 4 4 |A F 4 which is absurd by Proposition 8.2.2.

Assume

It follows that z(q 2 0 ) = z(q 2 1 ) = 4 Ψ(4) = 1. This implies that z(q 0 ) ∈ {±1}. We have

It follows that z(q 0 q 2 1 ) = 1 therefore z(q 0 ) = 1. We also have

therefore z(q 1 ) = 1. We get as before that z is trivial. This implies that ρ

It follows that z(p 0 p 2 1 ) = 1 therefore z(p 0 ) = 1. We also have

By the same computations as in 1.1. we have that z(q 0 ) = z(q 1 ) = 1. It follows that z is trivial. This implies that ρ 2.2. Assume 3 -β -β -1 = 0. We then have α + α -1 = 3 = β + β -1 . This implies that α ∈ {β, β -1 } and contradicts our assumptions on α and β.

Note that all the computations above were made in A A 2 ,1 or A A 2 ,2 . Since the restrictions to those subgroups are stable by the transposed inverse operation, we have that ρ

Chapter 10 Appendix 10.1 Erratum of the papers [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] and [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF] in type A This Ph.D. thesis is mainly based on [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] and [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF]. It uses many results from those papers. All the results are correct, however some proofs are incomplete. We give here a few corrections of those papers.

In the proof of Proposition 3.1 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], B n-1 should be replaced by B n . In section 3.2. of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], R(B) should be replaced by -R(B n ). The determinant of a matrix M ∈ GU n (q 1 2 ) does not verify (det(M )) = det(M ) but (det(M )) = det(M ) -1 . We have det(-R(σ i )) = -α therefore (-α) = -α -1 . It follows that (α + α -1 ) = α + α -1 and the conclusion remains true since F q = F p (α + α -1 ).

In the proof of Lemma 3.5 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF], in order to prove that [G, G] is not abelian by abelian, it is necessary to check that (s

2 ) = 0. In the paper, it was only checked that (s

2 ) = 0 and the matrix given is wrong. We have

This matrix is non-zero because it is assumed in the Lemma that the order of α is not in {1, 2, 3, 4, 5, 6, 10}. We have

This matrix is also non-zero by the conditions on the order of α.

In the proof of Proposition 3.7, it is not proved that • ρ is a representation of the Hecke algebra. However, it is true by Proposition 2.1.2 of this Ph.D. thesis since (α) = α -1 .

In the proof of section 4, it should be k = N 2 instead of k = N 2 in order to prove that G r A S N has less transvection than the natural SL a q with a ≥ N 2 . The proof for N = 5 then becomes a subcase of the proof for N ≥ 6 because we have that k = 3 when N = 5.

In Lemma 3.2 of [START_REF] Brunat | Image of the braid groups inside the finite Iwahori-Hecke algebras[END_REF], the action of L satisfies LsrL -1 -α = t s -1 r but not LsrL -1 -αν(λ) = t s -1 r . After Theorem 5.1 and Lemma 5.5 of [START_REF] Brunat | Image of the braid groups inside the finite Temperley-Lieb algebras[END_REF] it should say that we need to prove that G is primitive not imprimitive.

In the 40-dimensional H 4 -graph, the blue edges are of weight 2, the red edges are of weight 4 3 , the cyan edges are of weight 6 5 , the orange edges are of weight 5 6 , the green edges are of weight 2 3 , the dark yellow edges are of weight 3 2 , the purple edges are of weight -2 3 , the dark gray edges are of weight -1 2 , the yellow edges are of werght -5 9 , the dark green edges are of weight 5 3 , the teal edges are of weight 3, the lime edges are of weight 4 9 , the pink edges are of weight - 4 5 , the brown edges are of weight 7 3 and the olive edges are of weight -2.

In the 48-dimensional H 4 -graph, we have omitted the weights on the edges for clarity. One can still observe the symmetry appearing in this self-dual H 4 -graph.