John Guaschi

Carolina De
email: carolinapereiro@gmail.com

Miranda E Pereiro

LOWER CENTRAL AND DERIVED SERIES OF SEMI-DIRECT PRODUCTS, AND APPLICATIONS TO SURFACE BRAID GROUPS

Keywords: Mathematics Subject Classification. Primary, 20F36, 20F14; Secondary, 20E26

For an arbitrary semi-direct product, we give a general description of its lower central series and an estimation of its derived series. In the second part of the paper, we study these series for the full braid group B n (M ) and pure braid group P n (M ) of a compact surface M , orientable or non-orientable, the aim being to determine the values of n for which B n (M ) and P n (M ) are residually nilpotent or residually soluble. We first solve this problem in the case where M is the 2-torus. We then use the results of the first part of the paper to calculate explicitly the lower central series of P n (K), where K is the Klein bottle. Finally, if M is a non-orientable, compact surface without boundary, we determine the values of n for which B n (M ) is residually nilpotent or residually soluble in the cases that were not already known in the literature.

Theorem 1.1. Let G and H be groups, and let ϕ : G -→ Aut(H) be an action of G on H. We define recursively the following subgroups of H: L 1 = V 1 = H, and if n ≥ 2:

Introduction

Let G be a group. If g, g ∈ G then [g, g ] = gg g -1 g -1 denotes their commutator, and if H and K are subgroups of G, then the commutator subgroup of H and K, denoted by [H, K], is defined by [H, K] = [h, k] : h ∈ H and k ∈ K , the subgroup of G generated by the commutators of H and K. The lower central series {Γ i (G)} i≥1 of G is defined inductively by Γ 1 (G) = G, and for i ≥ 1, Γ i+1 (G) = [Γ i (G), G], and the derived series G (i) i≥0 of G is defined inductively by G (0) = G, and for i ≥ 0, G (i+1) = G (i) , G (i) . The quotient G/Γ 2 (G) is the Abelianisation of G that we denote by G Ab . Following P. Hall, for any group-theoretic property P, a group G is said to be residually P if for any (non-trivial) element x ∈ G, there exists a group H that possesses property P and a surjective homomorphism ϕ : G -→ H such that ϕ(x) = 1 (see also [START_REF] Mal'cev | On homomorphisms onto finite groups[END_REF]). It is well known that a group G is residually nilpotent (resp. residually soluble) if and only if i≥1 Γ i (G) = {1} (resp. i≥0 G (i) = {1}). zz If p is a prime number, the lower F p -linear central filtration {γ p i (G)} i≥1 of G is defined inductively by γ p 1 (G) = G, and for i ≥ 1, γ p i+1 (G) = [γ p i (G), G] , x p : x ∈ γ p i (G) [START_REF] Paris | Residual p properties of mapping class groups and surface groups[END_REF]. If the group G is finitely generated, then G is residually p-finite if and only if i≥1 γ p i (G) = {1} [30, Proposition 2.3 [START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF]]. For any group G, G (i) ⊂ Γ i+1 (G) ⊂ γ p i+1 (G), so if G is residually p-finite then it is residually nilpotent, which in turn implies that it is residually soluble. The combinatorial study of the lower central and derived series of a group is an interesting and important problem, see [START_REF] Falk | The lower central series of a fiber-type arrangement[END_REF][START_REF] Gaglione | Factor groups of the lower central series for special free products[END_REF][START_REF] Gruenberg | Residual properties of infinite soluble groups[END_REF][START_REF] Kohno | Série de Poincaré-Koszul associée aux groupes de tresses pures[END_REF][START_REF] Labute | On the descending central series of groups with a single defining relation[END_REF][START_REF] Magnus | Combinatorial group theory[END_REF] for example.

The first part of this paper is devoted to the analysis of the lower central and derived series of arbitrary semi-direct products. Our first main result describes the lower central series of such a group, and gives some information about its derived series.

Then ϕ induces an action, which we also denote by ϕ, of Γ n (G) on L n (resp. of G (n+1) on V n+2 ), and for all n ∈ N, we have:

(1) Γ n (H ϕ G) = L n ϕ Γ n (G).

(2) (H ϕ G) (n-1) ⊂ V n ϕ G (n-1) .

For the case of the commutator subgroup, namely n = 2, part [START_REF] Artin | Theorie der Zöpfe[END_REF] was obtained in [15, Proposition 3.3].

In the rest of this paper, we will be interested in computing the lower central and derived series of the full and pure braid groups of compact surfaces without boundary, and we will apply Theorem 1.1 to part of this calculation. We first recall some facts about these braid groups and their lower central and derived series. The braid groups of the disc, also called the Artin braid groups, were introduced by E. Artin [START_REF] Artin | Theorie der Zöpfe[END_REF]. If n ≥ 1, the n-string Artin braid group, denoted by B n , is generated by elements σ 1 , . . . , σ n-1 that are subject to the Artin relations:

σ i σ i+1 σ i = σ i+1 σ i σ i+1 for all 1 ≤ i ≤ n -2 σ j σ i = σ i σ j if |i -j| ≥ 2 and 1 ≤ i, j ≤ n -1.
The notion of braid group was generalised to surfaces by Fox and Neuwirth using configuration spaces as follows [START_REF] Fox | The braid groups[END_REF]. Let M be a compact, connected surface, and let n ∈ N. The n th configuration space of M , denoted by F n (M ), is defined by:

F n (M ) = {(x 1 , . . . , x n ) : x i ∈ M , and x i = x j if i = j, i, j = 1, . . . , n} .

The n-string pure braid group P n (M ) of M is defined by P n (M ) = π 1 (F n (M )). The symmetric group S n on n letters acts freely on F n (M ) by permuting coordinates, and the n-string braid group B n (M ) of M is defined by B n (M ) = π 1 (F n (M )/S n ). This gives rise to the following short exact sequence:

1 -→ P n (M ) -→ B n (M ) -→ S n -→ 1.

(1.1) If m ≥ 1, the projection p : F n+m (M ) -→ F n (M ) defined by p(x 1 , . . . , x n , . . . , x n+m ) = (x 1 , . . . , x n ) induces a homomorphism p * : P n+m (M ) -→ P n (M ). Geometrically, p * is the homomorphism that 'forgets' the last m strings. If M is without boundary, Fadell and Neuwirth showed that p is a locally-trivial fibration [9, Theorem 1], with fibre F m (M \ {x 1 , . . . , x n }) over the point (x 1 , . . . , x n ), which we consider to be a subspace of the total space via the map i : F m (M \ {x 1 , . . . , x n }) -→ F n+m (M ) defined by i((y 1 , . . . , y m )) = (x 1 , . . . , x n , y 1 , . . . , y m ). Applying the associated long exact sequence in homotopy to this fibration, we obtain the Fadell-Neuwirth short exact sequence of pure braid groups:

1 -→ P m (M \ {x 1 , . . . , x n })

i * -→ P n+m (M ) p * -→ P n (M ) -→ 1, (1.2
) where n ≥ 3 if M is the sphere S 2 [START_REF] Fadell | Homotopy groups of configuration spaces and the string problem of Dirac[END_REF][START_REF] Fadell | The braid groups of E 2 and S 2[END_REF], n ≥ 2 if M is the projective plane RP 2 [START_REF] Fadell | The braid groups of E 2 and S 2[END_REF], and n ≥ 1 otherwise [START_REF] Fadell | Configuration spaces[END_REF], and i * is the homomorphism induced by the map i. This sequence has been widely studied. If M is the torus T or the Klein bottle K, the existence of a non-vanishing vector field on M allows one to construct a section for p [9, Theorem 5]. This implies that the short exact sequence (1.2) splits for all n, m ∈ N, and that P n (M ) may be decomposed as an iterated semi-direct product (see Proposition 5.1 for an explicit section for p * in the case M = K).

We then use the above results to study the derived series of the braid groups of the torus and the lower central series and derived series of non-orientable surfaces. Theorem 1.1 will be used in the computation of the lower central series of P n (K), but we believe that it is of independent interest, and that it may be applicable to other groups. We first recall some facts about these series for surface braid groups. The lower central series of the Artin braid groups were analysed by Gorin and Lin who gave a presentation of the commutator subgroup Γ 2 (B n ) of B n for n ≥ 3, and who showed that (B n ) (1) = (B n ) (2) for all n ≥ 5, which implies that (B n ) (1) is perfect [START_REF] Gorin | Algebraic equations with continuous coefficients and some problems of the algebraic theory of braids[END_REF]. As a consequence, Γ 2 (B n ) = Γ 3 (B n ) for all n ≥ 3, so B n is not residually nilpotent. The lower central series of the pure braid group P n was studied by Falk and Randell [START_REF] Falk | The lower central series of a fiber-type arrangement[END_REF] and by Kohno [START_REF] Kohno | Série de Poincaré-Koszul associée aux groupes de tresses pures[END_REF], who proved independently that P n is residually nilpotent for all n ≥ 1.

The braid groups of orientable surfaces were studied by Bellingeri, Gervais and Guaschi [START_REF] Bellingeri | Lower central series of Artin-Tits and surface braid groups[END_REF]. If M g,m is a compact, connected, orientable surface of genus g ≥ 1 with m ≥ 0 boundary components, then B n (M g,m ) is not residually nilpotent if n ≥ 3, and B 2 (T) is residually nilpotent. In the case of the pure braid groups, P n (M g,m ) is residually torsion-free nilpotent for all n ≥ 1 if m ≥ 1, or if g = 1 and m = 0 (the torus). If m = 0 and g ≥ 1, Bardakov and Bellingeri proved that P n (M g,m ) is residually torsion-free nilpotent for all n ≥ 1, and the braid group B 2 (M g,m ) is residually 2-finite, in particular, it is residually nilpotent [START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF]. Gonçalves and Guaschi studied the lower central and derived series of the braid groups of the sphere S 2 and the projective plane RP 2 [START_REF] Gonçalves | The lower central and derived series of the braid groups of the sphere[END_REF][START_REF] Gonçalves | The lower central and derived series of the braid groups of the projective plane[END_REF]. For the sphere, B n (S 2 ) is residually nilpotent if and only if n ≤ 2, and residually soluble if and only if n ≤ 4. In the case of the projective plane, B n (RP 2 ) is residually nilpotent if and only if n ≤ 2, and if n = 4, B n (RP 2 ) is residually soluble if and only if n < 4. More recently, if M is a non-orientable surface different from RP 2 , Bellingeri and Gervais showed that P n (M ) is residually 2-finite, and so is residually nilpotent [START_REF] Bellingeri | On p-almost direct products and residual properties of pure braid groups of nonorientable surfaces[END_REF].

In the second part of this paper, we study the derived series of the torus and the lower central series and derived series of non-orientable surfaces. Our main results in this direction are as follows.

Theorem 1.2. The group B n (T) is residually soluble if and only if n ≤ 4.

For non-orientable surfaces, we first analyse the case of the Klein bottle. Using Theorem 1.1, we compute explicitly Γ n (P 2 (K)) and γ 2 n (P 2 (K)) in Theorems 5.4 and 5.13 respectively. From this it will follow that P 2 (K) is residually nilpotent and residually 2-finite. In Theorem 5.25, we show that P n (K) is residually nilpotent for all n ∈ N. This will allow us to determine the values of n for which B n (K) is residually nilpotent or residually soluble as follows.

Theorem 1.3. Let n ≥ 1. Then:

(1) P n (K) is residually nilpotent for all n ≥ 1.

(2) B n (K) is residually nilpotent if and only if n ≤ 2, and residually soluble if and only if n ≤ 4.

For a non-orientable surface M without boundary of higher genus, we may decide whether B n (M ) is residually nilpotent or residually soluble using results of [START_REF] Bellingeri | On p-almost direct products and residual properties of pure braid groups of nonorientable surfaces[END_REF][START_REF] Gonçalves | The lower central and derived series of the braid groups of the projective plane[END_REF]. Theorem 1.4. Let n, g ∈ N, and let M be a compact non-orientable surface of genus g without boundary. Then B n (M ) is residually nilpotent if and only if n ≤ 2, and is residually soluble if and only if n ≤ 4.

Although Theorem 1.4 contains Theorem 1.3(2) as a special case, we state the latter separately because the braid groups of the Klein bottle will be the focus of most of the second part of the paper.

The manuscript is organised as follows. In Section 2, we give presentations of the braid groups used in this paper, as well as the statement of Theorem 2.5 due to Gruenberg that will be required in the proofs of some of our results. Theorem 1.1 is proved in Section 3. In Section 4, we study the case of the torus and we prove Theorem 1.2. In Section 5, our focus is on the braid groups of the Klein bottle, and we use Theorem 1.1 in the proof of Theorem 1.3. Theorem 1.4 is proved in Section 6. If M is a compact surface different from K and the Möbius band, the centre Z(B n (M )) of B n (M ) is known [START_REF] Birman | On braid groups[END_REF][START_REF] Chow | On the algebraical braid group[END_REF][START_REF] Gillette | The word problem and its consequences for the braid groups and mapping class groups of the 2-sphere[END_REF][START_REF] Murasugi | Seifert fibre spaces and braid groups[END_REF][START_REF] Paris | Geometric subgroups of surface braid groups[END_REF][START_REF] Van Buskirk | Braid groups of compact 2-manifolds with elements of finite order[END_REF]. We determine Z(B n (K)) in Proposition 5.2, and for the sake of completeness, in Proposition A1 of the Appendix, we compute the centre of the braid groups of the Möbius band.

Acknowledgements. The authors would like to thank P. Bellingeri, S. Gervais, D. Gonçalves, L. Paris and D. Vendrúscolo for stimulating conversations. C. M. Pereiro was supported by project grant n o 2010/18930-6 and 2012/01740-5 from FAPESP. During the writing of this paper, J. Guaschi was partially supported by the CNRS/FAPESP PRC project n o 275209.

Generalities

In this section, we give the presentations of the braid and pure braid groups that will be used in this paper. If M = T or K, we will make use of the following presentations of P n (M ) and B n (M ).

Theorem 2.1 ([32]

). Let n ≥ 1, and let M be the torus T or the Klein bottle K. The following constitutes a presentation of the pure braid group P n (M ) of M : generators: {a i , b i , i = 1, . . . , n} ∪ {C i,j , 1 ≤ i < j ≤ n}. relations:

(1) [START_REF] Pereiro | Os grupos de tranças do toro e da garrafa de Klein[END_REF]). Let n ≥ 1, and let M be the torus T or the Klein bottle K. The following constitutes a presentation of the braid group B n (M ) of M : generators: a, b, σ 1 , . . . , σ n-1 . relations:

a i a j = a j a i , (1 ≤ i < j ≤ n) (2) a -1 i b j a i = b j a j C -1 i,j C i+1,j a -1 j , (1 ≤ i < j ≤ n) (3) a -1 i C j,k a i = C j,k , (1 ≤ i < j < k ≤ n) or (1 ≤ j < k < i ≤ n) a k C -1 i+1,k C i,k a -1 k C j,k C -1 i,k C i+1,k , (1 ≤ j ≤ i < k ≤ n) (4) C -1 i,l C j,k C i,l = C j,k , (1 ≤ i < l < j < k ≤ n) or (1 ≤ j ≤ i < l < k ≤ n) C i,k C -1 l+1,k C l,k C -1 i,k C j,k C -1 l,k C l+1,k , (1 ≤ i < j ≤ l < k ≤ n) (5) n j=i+1 C -1 i,j C i+1,j = a i b i C 1,i a -1 i b -1 i , (1 ≤ i ≤ n), if M = T n j=i+1 C i,j C -1 i+1,j = b i C 1,i a -1 i b -1 i a -1 i , (1 ≤ i ≤ n), if M = K (6) b j b i = b i b j , (1 ≤ i < j ≤ n), if M = T b j b i = b i b j C i,j C -1 i+1,j , (1 ≤ i < j ≤ n), if M = K (7) b -1 i a j b i = a j b j C i,j C -1 i+1,j b -1 j , (1 ≤ i < j ≤ n), if M = T b -1 i a j b i = a j b j (C i,j C -1 i+1,j ) -1 b -1 j , (1 ≤ i < j ≤ n), if M = K (8)
       b -1 i C j,k b i = C j,k , (1 ≤ i < j < k ≤ n) or (1 ≤ j < k < i ≤ n) C i+1,k C -1 i,k C j,k b k C i,k C -1 i+1,k b -1 k , (1 ≤ j ≤ i < k ≤ n) if M = T b -1 i C j,k b i = C j,k , (1 ≤ i < j < k ≤ n) or (1 ≤ j < k < i ≤ n) C i+1,k C -1 i,k C j,k b k (C i,k C -1 i+1,k ) -1 b -1 k , (1 ≤ j ≤ i < k ≤ n) if M = K. Theorem 2.2 ([
(1)

σ i σ i+1 σ i = σ i+1 σ i σ i+1 ; (2) σ j σ i = σ i σ j , if |i -j| ≥ 2; (3) aσ j = σ j a, if j ≥ 2; (4) bσ j = σ j b, if j ≥ 2; (5) b -1 σ 1 a = σ 1 aσ 1 b -1 σ 1 ; (6) a(σ 1 aσ 1 ) = (σ 1 aσ 1 )a; (7) b(σ -1 1 bσ -1 1 ) = (σ -1 1 bσ -1 1 )b, if M = T, b(σ -1 1 bσ 1 ) = (σ -1 1 bσ -1 1 )b, if M = K; (8) σ 1 σ 2 • • • σ n-2 σ 2 n-1 σ n-2 • • • σ 2 σ 1 = bab -1 a -1 if M = T, ba -1 b -1 a -1 if M = K.
We consider the torus and the Klein bottle to be a square whose edges are identified as indicated in Figure 1. Geometric representatives of the generators of P n (T) and P n (K) given in Theorem 2.1 are illustrated in Figure 2, and may be interpreted as follows. For 1 ≤ i ≤ n, the i th string is the only non-trivial string of the braid a i (resp. of b i ), and it passes through the edge α (resp. β). If 1 ≤ i < j ≤ n, the j th string is the only non-trivial string of the braid C i,j , and it encircles all of the basepoints between the i th and j th points. If i = j, it will be convenient to define C i,i to be the trivial braid. The figures represent the projection of the braids onto M , so the constant paths in each figure correspond to vertical strings of the braid. The generators of B n (T) and B n (K) given in Theorem 2.2 may be taken to be the standard Artin generators σ 1 , . . . , σ n-1 of B n as shown in Figure 3, and a = a 1 and b = b 1 . Various presentations of the braid and pure

M = T M = K α α α α β β β β Figure 1. Squares representing T and K i a i i b i i j C i,j i a i i b i i j C i,j M = T M = K α α α α α α β β β β β β α α α α α α β β β β β β Figure 2.
The generators of P n (T) and P n (K)

• • • • • • 1 i -1 i i + 1 i + 2 n σ i Figure 3
. The braid σ i braid groups of the torus and the Klein bottle may be found in the literature [START_REF] Bellingeri | On presentations of surface braid groups[END_REF][START_REF] Birman | On braid groups[END_REF][START_REF] González-Meneses | New presentations of surface braid groups[END_REF][START_REF] Scott | Braid groups and the group of the homeomorphisms of a surface[END_REF], but we choose to work with those of Theorems 2.1 and 2.2 because they highlight the similarities and differences between the braid groups of T and K. For example, the word C i,j C -1 i+1,j that appears in our presentation of P n (T) is often replaced by its inverse in P n (K). To prove Theorem 2.1 (resp. Theorem 2.2), one may use the Fadell-Neuwirth short exact sequence (1.2) (resp. the short exact sequence (1.1)), induction on n, and the following standard method for obtaining a presentation of a group extension [START_REF] Johnson | Presentations of groups[END_REF]Proposition 1,p. 139]. Given a short exact sequence 1

-→ A i -→ B p -→ C -→ 1 and presentations C = X | R and A = Y | S , then B = X, Y | S, R, T , where Y = { y = i(y) : y ∈ Y }, X = { x :
x ∈ X} is a transversal for Im(i) in B such that p( x) = x for all x ∈ X, S = { s : s ∈ S} is the set of words in Y obtained from S by replacing each letter y by y. For each r ∈ R, let r is the word in X obtained from r by replacing each letter x by x. Then r ∈ Ker(p), so it may be written as a word, v r say, in the elements of Y . Since Im(i) is normal in B, for all x ∈ X and y ∈ Y , x -1 y x ∈ Ker(p), so may be written as a word, w x,y say, in the

elements of Y . Then R = { rv -1 r : r ∈ R} and T = { x -1 y xw -1
x,y : x ∈ X, y ∈ Y }. The details of the proofs of Theorems 2.1 and 2.2 are left to the reader. Remark 2.3. Using Theorem 2.2, it is straightforward to check that:

B n (T) Ab ∼ = Z ⊕ Z ⊕ Z 2 = a, b, σ : [a, b] = [a, σ] = [b, σ] = σ 2 = 1 B n (K) Ab ∼ = Z ⊕ Z 2 ⊕ Z 2 = a, b, σ : [a, b] = [a, σ] = [b, σ] = σ 2 = a 2 = 1 ,
for all n ≥ 2, where a (resp. b, σ) represents the Γ 2 -coset of a (resp. of b, σ 1 ).

For compact non-orientable surfaces of genus g ≥ 3 without boundary, we shall make use of the following presentation of their braid groups due to Bellingeri.

Theorem 2.4 ( [START_REF] Bellingeri | On presentations of surface braid groups[END_REF]). Let N g be a compact, connected non-orientable surface of genus g ≥ 3 without boundary. The braid group B n (N g ) admits the following presentation: generators: σ 1 , . . . , σ n-1 , a 1 , . . . , a g . relations:

(1)

σ i σ i+1 σ i = σ i+1 σ i σ i+1 . (2) σ j σ i = σ i σ j , if |i -j| ≥ 2. ( 3
) a r σ i = σ i a r (1 ≤ r ≤ g; i = 1). (4) σ -1 1 a r σ -1 1 a r = a r σ -1 1 a r σ 1 (1 ≤ r ≤ g). (5) σ -1 1 a s σ 1 a r = a r σ -1 1 a s σ 1 (1 ≤ s < r ≤ g). (6) a 2 1 • • • a 2 g = σ 1 σ 2 • • • σ 2 n-1 • • • σ 2 σ 1 .
To prove some of our results, we will also require the following theorem of Gruenberg.

Theorem 2.5 ([22]). Let P denote one of the following classes:

(1) the class of soluble groups.

(2) the class of finite groups.

(3) the class of p-finite groups for a given prime number p. Let K and H be groups, and suppose that K is P and that H is residually P. Then, for any group extension 1 -→ H -→ G -→ K -→ 1, the group G is residually P.

The lower central and derived series of semi-direct products

The main aim of this section is to establish the general decomposition of the lower central series and an estimate of the derived series of an arbitrary semi-direct product given in the statement of Theorem 1.1, which will be used in later computations of the lower central and derived series of P n (K). We first prove two lemmas that will be used in what follows. If x 1 , . . . , x n are elements of a group G, we set:

[x 1 , x 2 , . . . , x n-1 , x n ] = x 1 , x 2 , . . . , [x n-1 , x n ] ,
and if X is a subset of G then we denote the normal closure of X in G by X G .

Lemma 3.1. Let G be a group, and let x, y ∈ G. For all n ∈ N, we have:

[x 2 n , y] = [x, x, x 2 , . . . , x 2 n-1 , y].[x, x 2 , . . . , x 2 n-1 , y] 2 .[x 2 , . . . , x 2 n-1 , y] 2 • • • [x 2 n-1 , y] 2 . (3.1)
Proof. We prove the lemma by induction on n. Observe that:

[x 2 , y] = x.x.y.x -1 .x -1 .y -1 = x[x, y]yx -1 y -1 = x[x, y]x -1 [x, y] = [x, x, y].[x, y] 2 , (3.2)
which proves (3.1) in the case n = 1. Now let n ≥ 2, and suppose that the result holds for all 1 ≤ i ≤ n. Applying (3.2) to the elements x 2 n and [x 2 n , y], we have:

[x 2 n+1 , y] = [(x 2 n ) 2 , y] = [x 2 n , x 2 n , y][x 2 n , y] 2 ,
and applying (3.2) to the elements x 2 n and y, we obtain:

x 2 n , [x 2 n , y] = x, x, x 2 , . . . , x 2 n-1 , [x 2 n , y] x, x 2 , . . . , x 2 n-1 , [x 2 n , y] 2 • • • x 2 n-1 , [x 2 n , y] 2 .
Thus:

x 2 n+1 , y = x, x, x 2 , . . . , x 2 n-1 , [x 2 n , y] x, x 2 , . . . , x 2 n-1 , [x 2 n , y] 2 • • • x 2 n-1 , [x 2 n , y] 2 x 2 n ,
K n+1 ⊂ K n , H n+1 ⊂ H n , H n+1 ⊂ H n , L n+1 ⊂ L n and V n+1 ⊂ V n .
Proof. The proof is by induction on n. The proof in the case n = 2 was given in [START_REF] Gonçalves | The lower central and derived series of the braid groups of the sphere[END_REF]Proposition 3.3]. So suppose that n ≥ 2, and assume that L n (resp.

V n ) is a normal subgroup of H, let x ∈ L n+1 (resp. V n+1 ) and let h ∈ H. Since L n+1 = K n+1 , H n+1 , [H, L n ] (resp. V n+1 = H n+1 , [H, V n ] ), it suffices to show that hxh -1 ∈ L n+1 , where x is a generator of K n+1 , H n+1 or [H, L n ] (resp. of H n+1 or [H, V n ]
), in the sense of Remark 3.2.

• Suppose that x = ϕ(g)(y)y -1 ∈ K n+1 , where g ∈ Γ n (G) and y ∈ H. Then ϕ(g) ∈ Aut(H), and there exists h ∈ H such that ϕ(g)(h ) = h, so:

hxh -1 = h(ϕ(g)(y).y -1 )h -1 = (ϕ(g)(h y)y -1 h -1 )(ϕ(g)(h )h -1 ) -1 ∈ K n+1 ⊂ L n+1 .
This also implies that K n is a normal subgroup of H for all n ≥ 2. • Suppose that x = ϕ(g)(y)y -1 is an element of H n+1 (resp. of H n+1 ), where g ∈ G and y ∈ L n (resp. y ∈ V n ), and let h ∈ H be such that ϕ(g)(h ) = h. Then:

hxh -1 = h(ϕ(g)(y).y -1 )h -1 = (ϕ(g)(h yh -1 ).(h yh -1 ) -1 )[h , y][y, ϕ(g)(h )] ∈ L n+1 (resp. V n+1 ),
because h yh -1 ∈ L n (resp. V n ) by the normality of L n (resp. V n ) in H using the induction hypothesis.

• Suppose that x = [y, l] ∈ [H, L n ] (resp. [H, V n ]), where y ∈ H and l ∈ L n (resp. l ∈ V n ).
Then:

hxh -1 = [hyh -1 , hlh -1 ] ∈ [H, L n ] ⊂ L n+1 (resp. [H, V n ] ⊂ V n+1 ), because hlh -1 ∈ L n (resp. V n ) by the normality of L n (resp. V n ) in H. This proves that L n (resp. V n ) is a normal subgroup of H for all n ≥ 2.
To prove the second part of the statement, notice that the inclusion Γ n (G)

⊂ Γ n-1 (G) implies that K n+1 ⊂ K n for all n ≥ 2. It is straightforward to see that H 3 ⊂ H 2 (resp. H 3 ⊂ H 2 ) because L 2 ⊂ H (resp. V 2 ⊂ H). By induction, suppose that H n ⊂ H n-1 (resp. H n ⊂ H n-1 ) for some n ≥ 3. Since L n-1 (resp. V n-1 ) is normal in H, we have [H, L n-1 ] ⊂ L n-1 (resp. [H, V n-1 ] ⊂ V n-1
). Further, using the definitions and the induction hypothesis, we have the inclusions

K n ⊂ K n-1 ⊂ L n-1 and H n ⊂ H n-1 ⊂ L n-1 (resp. H n ⊂ H n-1 ⊂ V n-1 ). It follows that L n ⊂ L n-1 (resp. V n ⊂ V n-1 ), and then that H n+1 ⊂ H n (resp. H n+1 ⊂ H n ). Consequently, L n+1 ⊂ L n and V n+1 ⊂ V n for all n ≥ 2.
Proof of Theorem 1.1. The proof is by induction on n. The case n = 1 is trivial. If n = 2, part (1) was proved in [START_REF] Gonçalves | The lower central and derived series of the braid groups of the sphere[END_REF]Proposition 3.3], and part (2) follows from part (1) and the fact that L 2 = V 2 . Now suppose that parts (1) and (2) hold for some n ≥ 2, and let us prove the result for n + 1. Let ϕ : Γ n (G) -→ Aut(L n ) be the action (also denoted by ϕ) induced by ϕ such that

L n ϕ Γ n (G) = Γ n (H ϕ G).
We claim that ϕ also induces an action ϕ : Γ n+1 (G) -→ Aut(L n+1 ). To see this, let g ∈ Γ n+1 (G). To prove that ϕ(g)(L n+1 ) ⊂ L n+1 , it suffices to take x ∈ L n+1 to be of the form x = ϕ(g )(h).h -1 , where either g ∈ Γ n (G) and h ∈ H, or g ∈ G and h ∈ L n , or of the form x = [h, l] ∈ [H, L n ], where h ∈ H and l ∈ L n . The result will then follow for all elements of L n+1 because ϕ(g) is a homomorphism.

•

If x = ϕ(g )(h).h -1 ∈ K n+1 , where g ∈ Γ n (G) and h ∈ H, or x = ϕ(g )(h).h -1 ∈ H n+1 ,
where g ∈ G and h ∈ L n , then:

ϕ(g)(x) = ϕ(g) ϕ(g )(h).h -1 = ϕ(gg )(h).h -1 ϕ(g)(h).h -1 -1 . If g ∈ Γ n (G) and h ∈ H then ϕ(g)(x) ∈ K n+1 ⊂ L n+1 since g and gg belong to Γ n (G). If g ∈ G and h ∈ L n , then ϕ(g)(x) ∈ H n+1 ⊂ L n+1 because h ∈ L n+1 . • If x = [h, l] ∈ [H, L n ],
where h ∈ H and l ∈ L n then:

ϕ(g)(x) = [ϕ(g)(h), ϕ(g)(l)] ∈ [H, L n ], since g ∈ Γ n+1 (G) ⊂ Γ n (G), so ϕ(g)(l) ∈ L n . Since ϕ(g) : L n+1 -→ L n+1
is the restriction of an automorphism, it is injective, so to show that it is an automorphism, it suffices to prove surjectivity. We first consider the following two cases: (a) If x = ϕ(g )(h).h -1 , where either g ∈ Γ n (G) and h ∈ H, or g ∈ G and h ∈ L n , let:

y = ϕ(g -1 g )(h).h -1 h ϕ(g -1 )(h -1 ).h h -1 ∈ L n+1 ,
because K n+1 and L n+1 are normal in H, and one may check that ϕ(g

)(y) = x. (b) If x = [h, l] ∈ [H, L n ],
where h ∈ H and l ∈ L n , there exist l ∈ L n and h ∈ H such that ϕ(g)(l ) = l and ϕ(g)(h ) = h by the induction hypothesis. Taking

y = [h , l ] ∈ [H, L n ] ⊂ L n+1 ,
we see that ϕ(g)(y) = x. This shows that if x is a generator of K n+1 , H n+1 or [H, L n ], there exists y ∈ L n+1 such that ϕ(g)(y) = x. Given an arbitrary element x ∈ L n+1 , there exist x 1 , . . . , x s , each of which satisfies one of the conditions of cases (a) and (b) above, such that x = x 1 • • • x s . So for i = 1, . . . , s, there exists y i ∈ L n+1 such that ϕ(g)(y i ) = x i , and we have ϕ(g)(y 1 • • • y s ) = x, which proves the surjectivity of ϕ(g) : L n+1 -→ L n+1 . Therefore the semi-direct product L n+1 ϕ Γ n+1 (G) is well defined. Similar computations show that the same is true for the semi-direct product V n+1 ϕ (G) (n) .

To complete the proof of part (1) of Theorem 1.1, it remains to show that

L n+1 ϕ Γ n+1 (G) = Γ n+1 (H ϕ G). We first prove that L n+1 ϕ Γ n+1 (G) ⊂ Γ n+1 (H ϕ G). Let (x, g) ∈ L n+1 ϕ Γ n+1 (G), where x ∈ L n+1 and g ∈ Γ n+1 (G). Since (x, g) = (x, 1)(1, g), it suffices to show that (x, 1) and (1, g) belong to Γ n+1 (H ϕ G). Clearly, (1, g) ∈ Γ n+1 (H ϕ G). Further, (x, 1
) is a product of elements each of which is of one of the following forms:

• (ϕ(g)(h).h -1 , 1) = [(1, g), (h, 1)], where g ∈ Γ n (G), h ∈ H, and (1, g) ∈ Γ n (H ϕ G). Then (ϕ(g)(h).h -1 , 1) ∈ Γ n+1 (H ϕ G). • (ϕ(g)(h).h -1 , 1) = [(1, g), (h, 1)], where g ∈ G and h ∈ L n . Then (h, 1) ∈ L n ϕ Γ n (G) = Γ n (H ϕ G) by the induction hypothesis, and (ϕ(g)(h).h -1 , 1) ∈ Γ n+1 (H ϕ G). • ([h, l], 1) ∈ [H, L n ], where h ∈ H and l ∈ L n . Then ([h, l], 1) = [(h, 1), (l, 1)], and l ∈ L n ϕ Γ n (G) = Γ n (H ϕ G) by the induction hypothesis, so ([h, l], 1) ∈ Γ n+1 (H ϕ G). Since all of these elements belong to Γ n+1 (H ϕ G), it follows that (x, 1) ∈ Γ n+1 (H ϕ G), whence L n+1 ϕ Γ n+1 (G) ⊂ Γ n+1 (H ϕ G).
For the other inclusion, let [(h, g),

(x, y)] ∈ Γ n+1 (H ϕ G), where (h, g) ∈ H ϕ G and (x, y) ∈ Γ n (H ϕ G). By the induction hypothesis, Γ n (H ϕ G) = L n ϕ Γ n (G), so x ∈ L n and y ∈ Γ n (G), and thus: [(h, g), (x, y)] = (h.ϕ(g)(x).ϕ(gyg -1 )(h -1 ).ϕ([g, y])(x -1 ), [g, y]). (3.
3) The second factor [g, y] on the right-hand side of (3.3) belongs to Γ n+1 (H ϕ G), and the first factor, denoted by ρ, may be written in the following form:

ρ = [h, x].xhx -1 ϕ(g)(x).x -1 xh -1 x -1 .xh ϕ(gyg -1 )(h -1 ).h h -1 x -1 .x ϕ([g, y])(x -1 ).x x -1 .

Note that:

•

[h, x] ∈ [H, L n ] ⊆ L n+1 , since h ∈ H and x ∈ L n . • ϕ(g)(x).x -1 ∈ H n+1 ⊆ L n+1 , since x ∈ L n . • ϕ(gyg -1 )(h -1 ).h ∈ K n+1 ⊆ L n+1 , since y ∈ Γ n (G), so gyg -1 ∈ Γ n (G) because Γ n (G) is a normal subgroup of G. • ϕ([g, y])(x -1 ).x ∈ H n+1 ∩ K n+1 ⊆ L n+1 , since x ∈ L n and [g, y] ∈ Γ n (G).
By Lemma 3.3, the conjugates by elements of H of the elements [h, x], ϕ(g)(x).x -1 , ϕ(gyg -1 )(h -1 ).h and ϕ([g, y])(x -1 ).x also belong to L n+1 , therefore ρ ∈ L n+1 as required. This proves part (1) of the statement.

To prove part (2), suppose by induction that ( 1) . Then: 1) . 1) . Then:

H ϕ G) (n-1) ⊂ V n ϕ G (n-
(H ϕ G) (n) = [(H ϕ G) (n-1) , (H ϕ G) (n-1) ] ⊂ V n ϕ G (n-1) , V n ϕ G (n-
To show that V n ϕ G (n-1) , V n ϕ G (n-1) ⊂ V n+1 ϕ G (n) , let (h, g), (x, y) ∈ V n ϕ G (n-
• [h, x] ∈ [H, V n ] ⊆ V n+1 because h, x ∈ V n ⊂ H. • the three elements ϕ(g)(x).x -1 , ϕ(gyg -1 )(h -1
).h and ϕ([g, y])(x -1 ).x belong to H n+1 because h, x ∈ V n , so they belong to V n+1 . Arguing in a manner similar to that for part [START_REF] Artin | Theorie der Zöpfe[END_REF]

from (3.3) onwards, it follows that [(h, g), (x, y)] ∈ V n+1 ϕ G (n) as required.
The following lemma will help us simplify some of the calculations in the following sections. Lemma 3.4. With the notation of Theorem 1.1, let G be a subgroup of G, let H be a subgroup of H, and let X (resp. Y ) be a generating set of G (resp. H).

(1) The subgroup

ϕ(g)(h).h -1 : g ∈ G, h ∈ H is contained in the normal closure Z H of Z = ϕ(g)(h).h -1 : g ∈ X, h ∈ Y in H. In particular, if this subgroup is a normal subgroup of G, it is equal to Z H . Consequently, if X (resp. Y ) is a generating set of Γ n-1 (G) (resp. of H) then to calculate the subgroup K n , it suffices to compute the elements ϕ(g)(h).h -1 , where g ∈ X and h ∈ Y . (2) Let W be a subset of H such that L n = W H (resp. V n = W H ) is the normal closure of W in H. Let X (resp. Y ) be a generating set of G (resp. of H). Then H n+1 is contained in {ϕ(g)(w).w -1 : g ∈ X, w ∈ W } H ∪ [H, L n ] (resp. H n+1 is contained in {ϕ(g)(w).w -1 : g ∈ X, w ∈ W } H ∪ [H, V n ]
). Therefore:

L n+1 = K n+1 , ϕ(g)(w).w -1 , [h, w] : g ∈ X, h ∈ Y, w ∈ W H V n+1 = ϕ(g)(w).w -1 , [h, w] : g ∈ X, h ∈ Y, w ∈ W H . Remark 3.5.
With the notation of Lemma 3.4(1), we will say that the elements of Z are generators of the subgroup Z H . It follows from part (2) that to determine L n+1 and V n+1 , we need only compute K n+1 in the case of L n+1 , and calculate the elements of the set {ϕ(g)(w).w

-1 , [h, w] : g ∈ X, w ∈ W }.
Proof of Lemma 3.4.

(1) To prove the first part of the statement, note that it suffices to prove the result for elements of the subgroup of the form ϕ(g)(h).h -1 , where g ∈ G and h ∈ H. If g ∈ G, there exist g 1 , . . . , g p ∈ G and 1 , . . . , p , ∈ {1, -1} such that g i i ∈ X for all i = 1, . . . , p and g = g 1 1 • • • g p p . Now:

ϕ(g)(h).h -1 = p i=1 ϕ(g i i ) ϕ g i+1 i+1 • • • g p p (h) . ϕ g i+1 i+1 • • • g p p (h) -1 = p i=1 ϕ(g i i )(h i ).h -1 i , (3.4)
where for all i = 1, . . . , p, h i = ϕ g i+1 i+1 • • • g p p (h). Further, for all h ∈ H, there exist h 1 , . . . , h q ∈ H and δ 1 , . . . , δ q ∈ {1, -1} such that h

δ j j ∈ Y for all j = 1, . . . , q and h = h δ 1 1 • • • h δq q . Since ϕ(g )(h ).h -1 = q j=1 h δ 1 1 • • • h δ j-1 j-1 ϕ(g )(h δ j j ).h -δ j j h -δ j-1 j-1 • • • h -δ 1 1 (3.5)
for all g ∈ G, the first part of the statement follows by combining (3.4) and (3.5). The second and third parts are consequences of the first part.

(2) Let ϕ(g)(h).h -1 ∈ H n+1 (resp. H n+1 ), where g ∈ G and h ∈ L n (resp. V n ). As in (1)
above, (3.4) holds. For all h ∈ L n (resp. V n ), there exist x 1 , . . . , x q ∈ W , δ 1 , . . . , δ q ∈ {1, -1} and α 1 , . . . , α q ∈ H, such that x

δ j j ∈ W and h = α 1 x δ 1 1 α -1 1 • • • α q x δq q α -1 q .
Then we obtain an equation similar to (3.5), where for all j = 1, . . . , q, h

δ j j is replaced by α j x δ j j α -1 j .
Further, for all j = 1, . . . , q, ϕ(g

)(α j x δ j j α -1 j ).(α j x δ j j α -1 j ) -1 is equal to: ϕ(g )(α j ) ϕ(g )(x δ j j ).x -δ j j ϕ(g )(α -1 j ) • α j α -1 j ϕ(g )(α j ), x δ j j ∈[H,Ln] α -1 j . (3.6)
Part ( 1) then follows from (3.4), (3.5) and (3.6).

The case of the torus

In this section, we study the derived series of B n (T), the aim being to prove Theorem 1.2. We shall consider two cases, n ≤ 4 and n ≥ 5. To study the case n ≥ 5, we start by exhibiting a presentation of (B n (T)) (1) . Proposition 4.2. A presentation of (B n (T)) (1) is given by: generators: for k, m ∈ Z and i = 1, . . . , n -1:

• b k,m = b k a m ba -m b -k-1 • d k,m = b k a m σ 1 bσ -1 1 a -m b -1-k • a k,m = b k a m (σ 1 aσ -1 1 a -1 )a -m b -k • θ i,k,m = b k a m σ i σ -1 1 a -m b -k • ρ i,k,m = b k a m σ 1 σ i a -m b -k relations: (1) θ i,k,m ρ i+1,k,m θ i,k,m = θ i+1,k,m ρ i,k,m θ i+1,k,m ρ i,k,m θ i+1,k,m ρ i,k,m = ρ i+1,k,m θ i,k,m ρ i+1,k,m (2)
θ i,k,m ρ j,k,m = θ j,k,m ρ i,k,m ρ i,k,m θ j,k,m = ρ j,k,m θ i,k,m if |i -j| ≥ 2.
(3)

a k,m = θ -1 j,k,m θ j,k,m+1 a k,m = ρ j,k,m ρ -1 j,k,m+1 for j ≥ 2. (4) b k,m θ j,k+1,m = θ j,k,m d k,m d k,m ρ j,k+1,m = ρ j,k,m b k,m for j ≥ 2. (5) b -1 k-1,m a k-1,m b k-1,m+1 ρ -1 1,k,m+1 a -1 k,m = 1 d -1 k-1,m ρ 1,k-1,m ρ -1 1,k-1,m+1 d k-1,m+1 ρ -1 1,k,m = 1 (6) a k,m+1 ρ 1,k,m+2 = a k,m ρ 1,k,m+1 ρ 1,k,m a k,m+1 = a k,m ρ 1,k,m+1 (7) b k,m ρ -1 1,k+1,m d k+1,m = ρ -1 1,k,m d k,m b k+1,m b k,m ρ -1 1,k+1,m d k+1,m = d k,m b k+1,m ρ -1 1,k+2,m (8) if n is odd: θ 1,k,m ρ 2,k,m θ 3,k,m • • • ρ n-1,k,m θ n-1,k,m • • • ρ 3,k,m θ 2,k,m ρ 1,k,m = b k,m b -1 k,m+1 ρ 1,k,m θ 2,k,m ρ 3,k,m • • • θ n-1,k,m ρ n-1,k,m • • • θ 3,k,m ρ 2,k,m θ 1,k,m = d k,m a k+1,m d -1 k,m+1 a -1 k,m (9)
if n is even:

θ 1,k,m ρ 2,k,m θ 3,k,m • • • θ n-1,k,m ρ n-1,k,m • • • ρ 3,k,m θ 2,k,m ρ 1,k,m = b k,m b -1 k,m+1 ρ 1,k,m θ 2,k,m ρ 3,k,m • • • ρ n-1,k,m θ n-1,k,m • • • θ 3,k,m ρ 2,k,m θ 1,k,m = d k,m a k+1,m d -1 k,m+1 a -1 k,m . Proof.
One applies the Reidemeister-Schreier rewriting process [START_REF] Hansen | Braids and Coverings: selected topics[END_REF]Appendix 1] to the short exact sequence:

1

-→ (B n (T)) (1) -→ B n (T) -→ B n (T) Ab Z⊕Z⊕Z 2 -→ 1,
using the presentation of the group B n (T) given in Theorem 2.2, and taking the Schreier transversal to be b k a m ; b k a m σ 1 : k, m ∈ Z . The details are left to the reader.

Proposition 4.3. If n ≥ 5, then B n (T)
is not residually soluble. Moreover, (B n (T)) (2) = (B n (T)) (3) .

Theorem 1.2 then follows directly from Propositions 4.1 and 4.3.

Proof of Proposition 4.3. The first step is a standard procedure that may be found in [15, Theorem 1.4, p. 3389], and uses just the Artin relations and some properties of the derived series. For future reference, we note that it may also be applied to the braid groups of non-orientable surfaces.

If M is a compact surface, consider the following short exact sequence:

1 -→ (B n (M )) (1) (B n (M )) (2) i -→ B n (M ) (B n (M )) (2) p -→ B n (M ) Ab -→ 1,
where p is the canonical projection. By using the above-mentioned procedure, for i = 1, . . . , n -1, the (B n (M )) (2) -cosets of the σ i coincide in B n (M )/(B n (M )) (2) , and are equal to an element that we denote by σ.

Now suppose that M = T. Using relations (3) and (4) of Theorem 2.2, the (B n (T)) (2) -cosets of a and b commute with σ in B n (T)/(B n (T)) (2) . Using this fact and relations ( 5) and (8) of Theorem 2.2, it follows that σ -2 = bab -1 a -1 and σ 2n = 1, and so σ has order at most 2n. To show that the order of σ in B n (T)/(B n (T)) (2) is exactly 2n, using Proposition 4.2, we note that:

(B n (T)) (1) /(B n (T)) (2) Θ = ρ 1,0,0 ∼ = Z n , (4.1)
where Θ is the normal closure in (B n (T)) (1) /(B n (T)) (2) of the (B n (T)) (2) -cosets of the elements of the set {θ i,k,m , k, m ∈ Z, i = 1, . . . , n -1}. Let q be the canonical projection of (B n (T)) (1) /(B n (T)) (2) onto ((B n (T)) (1) /(B n (T)) (2) ) Θ . The order of σ in B n (T)/(B n (T)) (2) is even because p(σ) is the generator of Z 2 . Suppose that the order of σ is 2r, where r < n. Then i(ρ 1,0,0 ) = σ 2 , and i(ρ r 1,0,0 ) = σ 2r = 1. Since i is injective, ρ r 1,0,0 = 1, and it follows that 1 = q(ρ r 1,0,0 ) = ρ r 1,0,0 in ((B n (T)) (1) /(B n (T)) (2) ) Θ . Thus ρ 1,0,0 is of order r < n, which contradicts (4.1). Hence:

B n (T)/(B n (T)) (2) = σ, a, b : [a, σ] = [b, σ] = σ 2n = 1, [b, a] = σ -2 .
To complete the proof, consider the short exact sequence:

1 -→ (B n (T)) (2) (B n (T)) (3) -→ B n (T) (B n (T)) (3) p -→ B n (T) (B n (T)) (2) -→ 1,
where p is the canonical projection. Then p(σ i ) = σ for all i = 1, . . . , n -1, and as above, we see that for i = 1, . . . , n -1, the (B n (T)) (3) -cosets of the σ i coincide in B n (T)/(B n (T)) (3) , and are equal to an element that we denote by σ. Using the same relations of B n (T) as before, it follows that the (B n (T)) (3) -cosets of a and b commute with σ in B n (T)/(B n (T)) (3) , [b, a] = σ -2 and σ 2n = 1, so p is an isomorphism, and hence (B n (T)) (2) = (B n (T)) (3) . We conclude that B n (T) is not residually soluble if n ≥ 5, because (B n (T)) (2) is non trivial.

The case of the Klein bottle

In this section, we study the lower central and derived series of the (pure) braid groups of the Klein bottle, and we prove Theorem 1.3. In Section 5.1, we exhibit an algebraic section of the Fadell-Neuwirth short exact sequence (1.2) for the Klein bottle, and we determine the centre of B n (K). These results will be used in the rest of the section, and we believe that they are interesting in their own right. In Section 5.2, we focus on the case n = 2, and in Theorems 5.4 and 5.13, we describe the lower central and the lower F 2 -linear central filtration respectively of P 2 (K) in an explicit manner. In Section 5.4, we study the lower central series of P n (K), if n ≥ 3, and in Section 5.5, we complete the proof of Theorem 1.3 by extending our results to B n (K).

5.1.

A section in the case of the Klein bottle. As we mentioned previously, the Fadell-Neuwirth short exact sequence (1.2) admits a section for geometric reasons. We start by giving an explicit section.

Proposition 5.1. In terms of the presentation of P n (K) given by Theorem 2.1, the map s : P n (K) -→ P n+1 (K) defined on the generators of P n (K) by:

                   a i -→ a i , i = 1, . . . , n -1 b i -→ b i , i = 1, . . . , n -1 C i,j -→ C i,j , 1 ≤ i < j ≤ n -1 C i,n -→ C i,n C i,n+1 C -1 n,n+1 , 1 ≤ i ≤ n -1 a n -→ a n a n+1 b n -→ b n+1 b n
extends to an algebraic section for the Fadell-Neuwirth short exact sequence arising from equation (1.2):

1 -→ π 1 (K \ {x 1 , . . . , x n }) -→ P n+1 (K) p * -→ P n (K) -→ 1. (5.1)
The geometric idea behind this section is to make use of the non-vanishing vector field on K to duplicate the last string of the elements of P n (K).

Proof of Proposition 5.1. In what follows, the numbers (1)-( 8) refer to the relations of P n (K) given by Theorem 2.1. To prove the proposition, it suffices to check that the images under s of these relations remain valid in P n+1 (K). We do this for relations (2) and [START_REF] Birman | On braid groups[END_REF]. For relation [START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF], which is

a -1 i b n a i = b n a n C -1 i,n C i+1,n a -1 n , we consider two cases. • If i + 1 = n then: s(a -1 i b n a i ) = a -1 i b n+1 b n (6)
a i = a -1 i (b n b n+1 C n,n+1 )a i (3) = a -1 i b n a i (2) a -1 i b n+1 (a i (2) C n,n+1 ) = (b n a n C -1 i,n a -1 n )(b n+1 (2)
a n+1 C -1 i,n+1 C n,n+1 a -1 n+1 )C n,n+1 = b n a n C -1 i,n (b n+1 a n+1
(3) and ( 8)

C -1 n,n+1 a -1 n+1 a -1 n )a n+1 (1)
C -1 i,n+1 C n,n+1 a -1 n+1 C n,n+1 = b n a n (b n+1 ( * * ) a n+1 C -1 i,n )C -1 n,n+1 (a -1 n )C -1 i,n+1 C n,n+1 ( * ) a -1 n+1 C n,n+1 = b n (b n+1 C n,n+1 a n )a n+1 C -1 i,n C -1 n,n+1 (C n,n+1 C -1 i,n+1 a -1 n )a -1 n+1 C n,n+1 (3)
= b n b n+1 C n,n+1 .a n a n+1 . C -1 i,n C -1 i,n+1 (C n,n+1 (4)
a

-1 n a -1 n+1 ) = b n b n+1 C n,n+1 .a n a n+1 .(C n,n+1 C -1 i,n C -1 i,n+1 ).a -1 n a -1 n+1 = s(b n a n C -1 i,n a -1 n ),
where ( * ) (resp. ( * * )) is a consequence of relation (3) (resp. relations ( 2) and (3)).

• If i + 1 < n then: s(a -1 i b n a i ) = a -1 i b n+1 b n (6)
a i = a -1 i (b n b n+1 C n,n+1 )a i (3) = a -1 i b n a i (2) a -1 i b n+1 (a i (2) C n,n+1 ) = (b n a n C -1 i,n C -1 i+1,n a -1 n )(b n+1 (2)
a n+1 C -1 i,n+1 C i+1,n+1 a -1 n+1 )C n,n+1 = b n a n C -1 i,n C i+1,n (b n+1 a n+1
(3) and ( 8)

C -1 n,n+1 a -1 n+1 a -1 n )a n+1 (1)
C -1 i,n+1 C i+1,n+1 a -1 n+1 C n,n+1 = b n a n (b n+1 ( * * ) a n+1 C -1 i,n C i+1,n )C -1 n,n+1 (a -1 n )C -1 i,n+1 C i+1,n+1 ( * ) a -1 n+1 C n,n+1 = b n (b n+1 C n,n+1 a n )a n+1 C -1 i,n C i+1,n C -1 n,n+1 (C n,n+1 C -1 i,n+1 C i+1,n+1 C -1 n,n+1 a -1 n )a -1 n+1 C n,n+1 (3)
= b n b n+1 C n,n+1 a n a n+1 C -1 i,n C i+1,n C -1 i,n+1 (4)
C i+1,n+1 C -1 n,n+1 (C n,n+1 a -1 n a -1 n+1 ) = b n b n+1 C n,n+1 .a n a n+1 .C -1 i,n (C -1 i,n+1 C i+1,n )C i+1,n+1 .a -1 n a -1 n+1 = s(b n a n C -1 i,n C i+1,n a -1 n ),
where ( * ) is a consequence of relation (3), similar to that in the case i + 1 = n, and ( * * ) is the same relation as in the case i + 1 = n.

Thus s respects relation [START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF]. For relation [START_REF] Birman | On braid groups[END_REF], which is

b n b i = b i b n C i,n C -1 i+1
,n , we also consider two cases.

• if i + 1 = n then: s(b n b i ) = b n+1 b n (6) b i = (b n b n+1 C n,n+1 )b i (8) = b n b n+1 (b i (6) C n,n+1 ) = b n (b i (6) b n+1 C i,n+1 C -1 n,n+1 )C n,n+1 = (b i b n C i,n )b n+1 (8)
C i,n+1 = b i b n (b n+1 C i,n )C i,n+1 = b i b n b n+1 (C n,n+1 C -1 n,n+1 )C i,n C i,n+1 (4)
= b i b n b n+1 C n,n+1 (C i,n C i,n+1 C -1 n,n+1 ) = s(b i b n C i,n ). • if i + 1 < n: s(b n b i ) = b n+1 b n (6) b i = (b n b n+1 C n,n+1 )b i (8) = b n b n+1 (b i (6) C n,n+1 ) = b n (b i (6) b n+1 C i,n+1 C -1 i+1,n+1 )C n,n+1 = (b i b n C i,n C -1 i+1,n )b n+1 (8) C i,n+1 C -1 i+1,n+1 C n,n+1 = b i b n (b n+1 C i,n C -1 i+1,n )C i,n+1 (4)
C -1 i+1,n+1 C n,n+1 = b i b n b n+1 (C n,n+1 C -1 n,n+1 )C i,n (C i,n+1 (4)
C -1 i+1,n )C -1 i+1,n+1 C n,n+1 (4)
= b i b n b n+1 C n,n+1 (C i,n C i,n+1 C -1 n,n+1 )(C n,n+1 C -1 i+1,n+1 C -1 i+1,n ) = s(b i b n C i,n C -1 i+1,n
). Thus s respects relation [START_REF] Birman | On braid groups[END_REF]. The computations for the other relations are similar, and are left to the reader.

As we mentioned at the end of the introduction, for any compact surface M and for all n ∈ N, the centre Z(B n (M )) of B n (M ) is known, with the exception of the Klein bottle and the Möbius band. The section given by Proposition 5.1 allows us to determine Z(B n (M )) if M = K. For the sake of completeness, in Proposition A1 of the Appendix of this paper, we also compute the centre of the braid groups of the Möbius band.

Proposition 5.2. For all n ∈ N, the centre of B n (K) is equal to (b n • • • b 1 ) 2 , and is isomorphic to Z.
Proof. The idea of the proof is similar to that of [START_REF] Paris | Geometric subgroups of surface braid groups[END_REF]Proposition 4.2]. Let

β n = b n • • • b 1 , and let Z n = β 2
n . We will show by induction on n that Z n = Z(B n (K)). Arguing as in [31, Proposition 4.2, step 4], we see that Z(B n (K)) ⊂ P n (K), so Z(B n (K)) ⊂ Z(P n (K)). Thus it suffices to show that Z(P n (K)) = Z n and Z n ⊂ Z(B n (K)). We prove that Z(P n (K)) = Z n by induction on n. If n = 1, by Theorem 2.1, we have:

π 1 (K) = a 1 , b 1 : a 1 b 1 = b 1 a -1 1 , (5.
2) and it is well known that Z(π 1 (K)) = Z 1 . Now suppose by induction that

Z(P n-1 (K)) = Z n-1 for some n ≥ 2.
We first prove that Z(P n (K)) ⊂ Z n in a manner similar to that of [31, Proposition 4.2, step 3]. Let g ∈ Z(P n (K)), and consider the Fadell-Neuwirth short exact sequence arising from (1.2):

1 -→ π 1 (K \ {x 1 , . . . , x n-1 }) -→ P n (K) p * -→ P n-1 (K) -→ 1.
Since p * is surjective, p * (g) ∈ Z(P n-1 (K)) = Z n-1 , and since p * (Z n ) = Z n-1 , there exists h ∈ Z n such that p * (h) = p * (g). If g = gh -1 then g belongs to Z(P n (K)) and to the free group π 1 (K \ {x 1 , . . . , x n-1 }) by exactness. Hence g ∈ Z(π 1 (K \ {x 1 , . . . , x n-1 })) = {1}, so g = 1, and thus g = h ∈ Z n , which shows that Z(P n (K)) ⊂ Z n .

Still under the above induction hypothesis, we now prove that Z n ⊂ Z(P n (K)). Using the section s : P n-1 (K) -→ P n (K) given by Proposition 5.1, we have s(β 2 n-1 ) = β 2 n . By the induction hypothesis,

β 2 n-1 a i = a i β 2 n-1 and β 2 n-1 b i = b i β 2 n-1 for all 1 ≤ i ≤ n -1, and β 2 n-1 C i,j = C i,j β 2 n-1
for all 1 ≤ i < j ≤ n -1 in P n-1 (K). Taking the image of both sides of these equations by s, we obtain the following relations:

β 2 n a i = a i β 2 n and β 2 n b i = b i β 2 n if 1 ≤ i < n -1 β 2 n . a n-1 a n = a n-1 a n . β 2 n and β 2 n . b n b n-1 = b n b n-1 . β 2 n if i = n -1, (5.3)
and

β 2 n C i,j = C i,j β 2 n if 1 ≤ i < j < n -1 β 2 n . C i,n-1 C i,n C -1 n-1,n = C i,n-1 C i,n C -1 n-1,n . β 2 n if 1 ≤ i < j = n -1. (5.4)
Let us prove that a n and b n commute with β 2 n , from which it will follow from (5.3) that β 2 n commutes with a i and b i for all 1 ≤ i ≤ n. Using relations ( 6)-(8) of Theorem 2.1, we have:

C 1,n a -1 n b i = C i+1,n b i C -1 i,n C 1,n a -1 n (5.5) b n C i+1 b i = b i b n C i,n (5.6)
for all i = 1, . . . , n -1. To prove relation (5.5), notice that by relations ( 7) and ( 8) we have:

b -1 i a n b i = a n b n (C i,n C -1 i+1,n ) -1 b -1 n (8) = a n (C -1 i+1,n b -1 i C i,n b i ),
and taking the inverse of both sides, it follows that a -1

n b i = C -1 i,n b i C i+1,n a -1 n .
We thus obtain:

C 1,n a -1 n b i = C 1,n (C -1 i,n b i C i+1,n a -1 n ) = (b i b -1 i )C 1,n (b i (8) b -1 i )C -1 i,n b i (8) C i+1,n a -1 n = b i (C i+1 (8)
C -1 i,n C 1,n b n C i+1,n C -1 i,n b -1 n )(b n C i,n C -1 i+1,n b -1 n C -1 i+1,n )C i+1,n a -1 n = (C i+1,n b i )C -1 i,n C 1,n a -1 n .
To prove relation (5.6), one may use relation ( 6) and the fact that b i commutes with C i+1,n by relation [START_REF] Fadell | Homotopy groups of configuration spaces and the string problem of Dirac[END_REF]. We now claim that:

a n β n = b n • • • b i+1 C -1 i+1,n C 1,n a -1 n b i b i-1 • • • b 1 and b n β n = b n • • • b i+1 b n C i+1,n b i • • • b 1 for all i = 0, . . . , n -1.
We shall prove the claim by reverse induction on i. First, we have

a n β n = a n (b n b n-1 • • • b 1 ) (5) = (b n C 1,n a -1 n )b n-1 • • • b 1 and b n β n = b n b n • • • b 1 , so the claim is valid if i = n -1.
Suppose that it holds for some 1 ≤ i ≤ n -1. Then:

a n β n = b n • • • b i+1 C -1 i+1,n C 1,n a -1 n b i b i-1 • • • b 1 (5.5) = b n • • • b i+1 C -1 i+1,n C i+1,n b i C -1 i,n C 1,n a -1 n b i-1 • • • b 1 = b n • • • b i+1 b i C -1 i,n C 1,n a -1 n b i-1 • • • b 1 , and
b n β n = b n • • • b i+1 b n C i+1,n b i • • • b 1 (5.6) = b n • • • b i+1 b i b n C i,n b i-1 • • • b 1
so the claim holds. Taking i = 0, we obtain

a n β n = β n a -1 n and b n β n = β n b n C 1,n . Hence a n β 2 n = β n a -1 n β n = β 2
n a n , and applying ( 5) with i = n twice, we obtain:

b n β 2 n = β n b n C 1,n β n = β n a n b n a n β n = β 2 n a -1 n b n C 1,n a -1 n = β 2 n b n . Thus β 2
n commutes with a n and b n , and so with a i and b i for all i ∈ {1, . . . , n}. Finally, by relation [START_REF] Bellingeri | Lower central series of Artin-Tits and surface braid groups[END_REF],

C 1,n = b -1 n a n b n a n and C i+1,n = C i,n a -1 n b -1 n a -1 i b n a i a n by relation (2). Since β 2
n commutes with a i and b i for all 1 ≤ i ≤ n, it follows that β 2 n commutes with C 1,n by relation [START_REF] Bellingeri | Lower central series of Artin-Tits and surface braid groups[END_REF], and by induction that β 2 n commutes with C i,n for all 1 ≤ i < n. So by (5.4), β 2 n commutes with C i,j for all 1 ≤ i < j ≤ n, and therefore commutes with all of the elements of a generating set of P n (K), hence Z n = Z(P n (K)).

To complete the proof of the proposition, it remains to show that Z n ⊂ Z(B n (K)). To do so, from above, it suffices to prove that σ i commutes with β 2 n for all i = 1, . . . , n -1. One may see by Figure 2 that σ 2 i = C i,i+1 . Further:

σ -1 i b j σ i =      b i+1 σ 2 i if j = i σ -2 i b i if j = i + 1 b j otherwise.
(5.7)

The case j = i is illustrated in Figure 4.

If j = i + 1 then σ -1 i b i+1 σ i = σ -1 i . σ -1 i b i σ i . σ -2 i . σ i = σ -2
i b i using the case j = i. For i = 1, . . . , n -1, using (5.7) and relation (6) of Theorem 2.1, we have:

σ -1 i b i σ i b i+1 σ 2 i Figure 4. The relation σ -1 i b i σ i = b i+1 σ 2 i . σ -1 i (b n • • • b 1 )σ i = b n • • • b i+2 • C -1 i,i+1 b i • b i+1 C i,i+1 •b i-1 • • • b 1 = C -1 i,i+1 b n • • • b 1 = σ -2 i b n • • • b 1 , from which it follows that (b n • • • b 1 ) 2 σ i = σ i (b n • • • b 1 ) 2 as required.
Remark 5.3. For n = 2, we modify slightly the presentation of P 2 (K) given by Theorem 2.1 by removing the generator C 1,2 using relation ( 5), so C 1,2 = b -1 2 a 2 b 2 a 2 . Hence P 2 (K) is generated by a 1 , a 2 , b 1 and b 2 , subject to the relations:

(1) a -1 1 a 2 a 1 = a 2 . (2) a -1 1 b 2 a 1 = a -1 2 b 2 a -1 2 . (3) b -1 1 a 2 b 1 = a 2 b 2 a -1 2 b -1 2 a -1 2 . (4) b -1 1 b 2 b 1 = a 2 b 2 a 2 . (5) b -1 2 a 2 b 2 a 2 = b 1 a -1 1 b -1 1 a -1
1 . Using Proposition 5.1 and the Fadell-Neuwirth short exact sequence (5.1), P n+1 (K) may be written as a semi-direct product of the free group π 1 (K \ {x 1 , . . . , x n }) by s(P n (K)) for all n ∈ N. In particular, if n = 1 then:

P 2 (K) ∼ = π 1 (K \ {x 1 }) ϕ s(P 1 (K)), (5.8)
where

π 1 (K \ {x 1 }) = a 2 , b 2 is a free group of rank 2, s(P 1 (K)) = a 1 a 2 , b 2 b 1 : (b 2 b 1 )(a 1 a 2 ) = (a 1 a 2 ) -1 (b 2 b 1 )
by (5.2), and where the action ϕ : s(P 1 (K)) -→ Aut(π 1 (K \ {x 1 })) is given by:

           ϕ(a 1 a 2 ) : a 2 -→ a 2 b 2 -→ a -2 2 b 2 ϕ(b 2 b 1 ) : a 2 -→ a -1 2 b 2 -→ a 2 b 2 a 2 , (5.9)
using Remark 5.3.

5.2.

The lower central series of P 2 (K). In this section, we use Theorem 1.1 to calculate explicitly the lower central series of P 2 (K). This will enable us to prove that P 2 (K) is residually nilpotent.

Theorem 5.4. Let n ≥ 2. With the notation of Theorem 2.1, we have:

Γ n (P 2 (K)) = a 2 n-1 2 , x 2 n-i : x ∈ Γ i (π 1 (K \ {x 1 })), 2 ≤ i ≤ n π 1 (K\{x 1 }) ϕ (a 1 a 2 ) 2 n-1 ,
where ϕ is as defined in equation (5.9). Consequently, P 2 (K) is residually nilpotent.

The idea of the proof is to apply Theorem 1.1(1) to the semi-direct product structure of P 2 (K) given by (5.8). From now on, we shall make use of the notation of that theorem, taking H = π 1 (K \ {x 1 }), G = s(P 1 (K)) and P 2 (K) = H ϕ G. In the lemmas that follow, we first compute the subgroups K n , H n and L n for these choices of G and H, and in Proposition 5.12, we calculate the subgroup L n , which will enable us to prove Theorem 5.4. We first need to know the lower central series of P 1 (K). Applying Theorem 1.1 to the semi-direct product structure a 1 b 1 of π 1 (K) given by (5.2), it is straightforward to see that

K 2 = H 2 = L 2 = a 2 1 , and if n ≥ 3, K n = {1} and H n = L n = a 2 n-1 1
, therefore the lower central series of s(P 1 (K)) is given by: Γ n (s(P 1 (K))) = (a 1 a 2 ) 2 n-1

(5.10) for all n > 1. We now turn to the case of P 2 (K). We first determine K n .

Lemma 5.5. With the notation of Theorem 1.1, and taking

H = π 1 (K \ {x 1 }), G = s(P 1 (K)) and P 2 (K) = H ϕ G, the subgroup K n is equal to a 2 2 , Γ 2 (H) H if n = 2, and to a 2 n-1 2 H if n ≥ 3.
Proof. First suppose that n = 2. By (5.9), we have:

     ϕ(a 1 a 2 )(a 2 ).a -1 2 = 1 ϕ(a 1 a 2 )(b 2 ).b -1 2 = ϕ(b 2 b 1 )(a 2 ).a -1 2 = a -2 2 ϕ(b 2 b 1 )(b 2 ).b -1 2 = a 2 b 2 a 2 b -1 2 = a 2 2 [a -1 2 , b 2 ],
(5.11) and so a

2 2 , [a -1 2 , b 2 ] ⊂ K 2 . Since K 2 is normal in H by Lemma 3.3, a 2 2 , [a -1 2 , b 2 ] H is a subgroup of K 2 , and therefore a 2 2 , Γ 2 (H) H ⊂ K 2 because Γ 2 (H) = [a -1 2 , b 2 ] H .
For the other inclusion, (5.11) implies that ϕ(g)(h).h -1 ∈ a 2 2 , Γ 2 (H) H for all h (resp. all g) belonging to the generating set {a 2 , b 2 } (resp.

{a 1 a 2 , b 2 b 1 }) of H (resp. of G). The inclusion K 2 ⊂ a 2
2 , Γ 2 (H) H then follows from Lemma 3.4(1). This proves the result for n = 2. Now assume that n ≥ 3. Then Γ n-1 (G) = (a 1 a 2 ) 2 n-2 by (5.10). Using (5.9), we have ϕ((a 1 a 2 ) 2 n-2 )(a 2 ).a -1 2 = 1 and:

ϕ((a 1 a 2 ) 2 )(b 2 ) = ϕ(a 1 a 2 )(a -2 2 b 2 ) = a -2 2 (a -2 2 b 2 ) = a -4 2 b 2 . Suppose by induction that ϕ((a 1 a 2 ) j )(b 2 ) = a -2j
2 b 2 for some j ≥ 2. Then:

ϕ((a 1 a 2 ) j+1 )(b 2 ) = ϕ(a 1 a 2 )ϕ((a 1 a 2 ) j )(b 2 ) = ϕ(a 1 a 2 )(a -2j 2 b 2 ) = a -2j 2 (a -2 2 b 2 ) = a -2(j+1) 2 b 2 .
In particular, if j = 2 n-2 , we have ϕ((a

1 a 2 ) 2 n-2 )(b 2 ).b -1 2 = (a -2(2 n-2 ) 2 b 2 ).b -1 2 = a -2 (n-1)

2

, and hence [START_REF] Bellingeri | On p-almost direct products and residual properties of pure braid groups of nonorientable surfaces[END_REF](1) and using (5.10), we obtain K n ⊂ Z H , where:

a 2 n-1 2 H ⊂ K n . Conversely, taking G = {(a 1 a 2 ) 2 n-2 } and H = {a 2 , b 2 } in Lemma 3.
Z = ϕ(g)(h).h -1 : g ∈ (a 1 a 2 ) 2 n-2 , h ∈ {a 2 , b 2 } = a -2 (n-1) 2 , whence the inclusion K n ⊂ a 2 n-1
L 2 = a 2 2 , Γ 2 (H) = a 2 2 , [a 2 , b 2 ] H .
(5.12)

Let W 2 = W 2 = L 2 , and for n ≥ 3, define:

W n = Γ n (H), x 2 : x ∈ W n-1 (5.13) W n = a 2 n-1 2 , x 2 n-i : x ∈ Γ i (H), 2 ≤ i ≤ n H . (5.14)
Note that W n is normal in H for all n ≥ 2. This follows from the fact that Γ n (H) is normal in H for all n ≥ 2 and arguing by induction on n as follows. If n = 2 then W 2 = L 2 by Lemma 3.3, so suppose that n ≥ 3, and that W n-1 is normal in H. Then hxh -1 ∈ W n-1 for all x ∈ W n-1 and h ∈ H, thus hx 2 h -1 = (hxh -1 ) 2 ∈ W n , and W n is normal in H as claimed. If x ∈ Γ i (H) and 2 ≤ i ≤ n, we will refer to the elements a 2 n-1 2 and x 2 n-i as generators of W n . In order to prove Theorem 5.4, we will show in Proposition 5.12 that L n = W n = W n for all n ≥ 2. Supposing this to be the case, the following lemma implies that to determine the lower central series of P 2 (K) using Theorem 1.1, it is not necessary to calculate the subgroups H n . Lemma 5.6. Suppose that

L i = W i = W i for all 2 ≤ i ≤ n. Then H n+1 ⊂ K n+1 , [H, L n ] . In particular, L n+1 = K n+1 , [H, L n ] .
Proof. We prove the given inclusion by induction on n. If n = 2, using (5.9) notice that:

• ϕ(a 1 a 2 )(a 2 2 )a -2 2 = 1. • ϕ(b 2 b 1 )(a 2 2 )a -2 2 = a -4 2 ∈ K 3 by Lemma 5.5. • ϕ(a 1 a 2 )([a 2 , b 2 ])[b 2 , a 2 ] = [a -2 2 , a 2 b 2 a -1 2 b -1 2 a -1 2 ] ∈ [L 2
, H] by (5.12).

• ϕ(b 2 b 1 )([a 2 , b 2 ])[b 2 , a 2 ] = [b 2 , a 2 2 ] ∈[H,L 2 ] [a 2 , [a 2 , b 2 ]] ∈Γ 3 (H) ∈ [L 2
, H] by (5.12).

Using these calculations and the description of L 2 given in (5.12), it follows from the first part of Lemma 3.4(2) and Lemma 5.5 that:

H 3 = [H, L 2 ], ϕ(g)(w).w -1 : g ∈ {a 1 a 2 , b 2 b 1 } , w ∈ a 2 2 , [a 2 , b 2 ] H ⊂ K 3 , [H, L 2 ] , which proves the result if n = 2.
Now suppose that the given inclusion holds for n -1 for some n ≥ 3, and assume that L i = W i = W i for all 2 ≤ i ≤ n. Let ϕ(g)(x).x -1 be an element of H n+1 , where g ∈ G and x ∈ L n = K n , [H, L n-1 ] by the induction hypothesis. We wish to show that ϕ(g)(x).x -1 ∈ K n+1 , [H, L n ] . By Lemma 3.4(1), we only need to check the following two possibilities: Lemma 5.5 and (5.11).

• x = a 2 n-1 2 . Then x ∈ K n by Lemma 5.5, ϕ(a 1 a 2 )(x).x -1 = 1 and ϕ(b 2 b 1 )(x)x -1 = a -2 n 2 ∈ K n+1 by
• x = [h, l] ∈ [H, L n-1 ]
, where h ∈ H and l ∈ L n-1 . Then ϕ(g)([h, l])[h, l] -1 may be written in the following form:

ϕ(g)(h).l [(ϕ(g)(l -1 ).l) -1 , ϕ(g)(h -1 )] ∈[Ln,H] l -1 ϕ(g)(h -1 ) h (ϕ(g)(h -1 ).h) -1 , l ∈[L 2 ,L n-1 ] h -1 .
(5.15)

To complete the proof, it suffices to show that the subgroup

[L 2 , L n-1 ] is contained in [H, L n ].
To do so, first note that [L 2 , L n-1 ] is normal in H because L j is normal in H for all j ≥ 2 by Lemma 3.3. Using the fact that L 2 = a 2 2 , Γ 2 (H) by (5.12), it suffices to show that the following elements belong to [H, L n ]:

•

[[y, z], l] ∈ [Γ 2 (H), L n-1 ]
, where y, z ∈ H and l ∈ L n-1 . Then:

[[y, z], l] = y z, [y -1 , l] ∈[H,Ln] y -1 .lyz [z -1 , l -1 ], y -1 ∈[H,Ln] z -1 y -1 l -1 , because [H, L n-1 ] ⊂ L n . Thus [[y, z], l] ∈ [H, L n ] because [H, L n ] is a normal subgroup of H, and therefore [Γ 2 (H), L n-1 ] ⊂ [H, L n ]. • [a 2 2 , l] ∈ [ a 2 2 , L n-1 ]
, where l ∈ L n-1 . Then using (3.2), we have:

[a 2 2 , l] = [a 2 , [a 2 , l]] ∈[H,Ln] [a 2 , l] 2 , (5.16) since [H, L n-1 ] ⊂ L n . Further [a 2 , l] 2 ∈ [H, L n ] because l ∈ L n-1 = W n-1 by hypothesis, so l 2 ∈ W n = L n , and [l 2 , a 2 ] ∈[Ln,H] = [l, [l, a 2 ]] ∈[H,Ln] [l, a 2 ] 2 by (3.2). So [l, a 2 ] 2 ∈ [H, L n ], and thus [a 2 2 , l] ∈ [H, L n ] by (5.16). This shows that [L 2 , L n-1 ] ⊂ [H, L n ], hence H n+1 ⊂ K n+1 , [H, L n ]
as desired, which concludes the proof of the first part of the statement. The second part follows from the first part and the definition of L n+1 .

In order to prove Proposition 5.12, we shall require a couple of intermediate results. Let A be either the empty set or a normal subgroup of H, and for m ≥ 1, let:

B m = [x 1 , . . . , x i ] 2 m-i-k ∈ Γ i (H) : ∃1 ≤ j 1 < • • • < j k ≤ i : x j 1 , . . . , x j k ∈ A, 0 ≤ k ≤ m -i and i = 1, . . . , m .
For 1 ≤ l ≤ m, let:

E l,m = [x 1 , . . . , x i ] 2 m-i-k ∈ Γ i (H) : ∃1 ≤ j 1 < • • • < j k ≤ i : x j 1 , . . . , x j k ∈ A, 0 ≤ k ≤ m -i and l ≤ i ≤ m .
(5.17)

In the case that A = ∅, we will denote the subgroup E l,m by E l,m . This corresponds to taking k = 0 in (5.17), so:

E l,m = [x 1 , . . . , x i ] 2 m-i ∈ Γ i (H) : l ≤ i ≤ m . (5.18) If x = [x 1 , . . . , x i ] ∈ Γ i (H), the elements x 2 m-i-k (resp. x 2 m-i ) of E l,m (resp. of E l,m
) given in (5.17) (resp. in (5.18)) will be termed [START_REF] Magnus | Combinatorial group theory[END_REF]Problem 3,Section 5.3,p. 297] for all l = 1, . . . , m -1, and that for all 1 ≤ l ≤ m, E l,m is normal in H, since if h ∈ H and x 2 m-i-k is a generator of E l,m , where

generators of E l,m (resp. of E l,m ). Note that Γ m (H) = E m,m ⊂ E l+1,m ⊂ E l,m ⊂ E 1,m = B m by
x = [x 1 , . . . , x i ] ∈ Γ i (H), then h[x 1 , . . . , x i ] 2 m-i-k h -1 = [hx 1 h -1 , . . . , hx i h -1 ] 2 m-i-k ∈ E l,m because
A is normal in H or is empty. In particular, taking A = ∅, we have:

Γ m (H) = E m,m ⊂ E l+1,m ⊂ E l,m ⊂ E 1,m = B m , (5.19)
and that E l,m is normal in H for all 1 ≤ l ≤ m.

Lemma 5.7. Let y ∈ H, let m ≥ 1, and let

1 ≤ i ≤ m. If x = [x 1 , . . . , x i ] is an element of Γ i (H) for which x 2 m-i-k is a generator of E i,m , where 0 ≤ k ≤ m -i (resp. x 2 m-i is a generator of E i,m
), then: Proof. It suffices to prove (5.20), since then the congruence (5.21) follows by taking A = ∅. We will do so by induction on m. If m = 1 then i = 1 and k = 0, and the congruence is in fact an equality. If m = 2, we consider two cases:

[x 2 m-i-k , y] ≡ [x, y] 2 m-i-k mod E i+1,m+1 and (5.20) [x 2 m-i , y] ≡ [x, y] 2 m-i mod E i+1,m+1 (
• if i = 2, or if i = k = 1, then x ∈ Γ 2 (H)
, and the two sides of (5.20) are equal.

• if i = 1 and k = 0 then x ∈ Γ 1 (H) = H. Thus [x 2 , y] = [x, x, y][x, y] 2 by (3.
2), and [x, x, y] ∈ Γ 3 (H), so [x, x, y] ∈ E 2,3 , and we obtain (5.20).

This proves the result if m = 2. We now consider the general case.

Induction hypothesis 1: suppose that the congruence (5.20) holds for some m ≥ 1. Let us show by induction that the result holds for m

+ 1 i.e. if 1 ≤ i ≤ m + 1, x ∈ Γ i (H), x 2 m+1-i-k is a generator of E i,m+1 , 0 ≤ k ≤ m + 1 -i and y ∈ H, then: [x 2 m+1-i-k , y] ≡ [x, y] 2 m+1-i-k mod E i+1,m+2 . (5.22)
This will be achieved by making a second induction hypothesis as follows.

Induction hypothesis 2: let 1 ≤ i ≤ m + 1 be such that:

[x 2 m+1-j-k , y] ≡ [x, y] 2 m+1-j-k mod E j+1,m+2 (5.23)
for all i ≤ j ≤ m + 1, where x ∈ Γ j (H), x 2 m+1-j-k is a generator of E j,m+1 , 0 ≤ k ≤ m + 1 -j and y ∈ H. If i = m + 1 then (5.23) holds trivially. So suppose that (5.23) is valid for some 2 ≤ i ≤ m + 1, and let us prove by reverse induction on i that it also holds for i -1. By induction hypothesis 2, it suffices to prove (5.23) for j = i-1. Let y ∈ H, and let

x = [x 1 , . . . , x i-1 ] ∈ Γ i-1 (H), where x 2 m+1-(i-1)-k is a generator of E i-1,m+1 and 0 ≤ k ≤ m + 1 -(i -1). Then [x 2 (m+1)-(i-1)-k , y] = [x 2 m+2-i-k , y]. If k = m + 2 -i then (5.23) is an equality. So assume that 0 ≤ k ≤ m + 1 -i. By Proposition 3.1, [x 2 (m+1)-(i-1)-k , y] is equal to: [x, x, x 2 , . . . , x 2 m+1-i-k , y][x, x 2 , . . . , x 2 m+1-i-k , y] 2 • • • [x 2 m-i-k , x 2 m+1-i-k , y] 2 ( * ) [x 2 m+1-i-k , y] 2 ( * * )
. (5.24)

Using induction hypotheses 1 and 2, we will first show that the expression ( * * ) is congruent to [x, y] 2 m+2-i-k modulo E i,m+2 , and then that the expression ( * ) belongs to E i,m+2 , from which we will conclude by induction that [

x 2 n+2-i-k , y] ≡ [x, y] 2 n+2-i-k modulo E i,m+2 for all i = 1, . . . , m + 1. To show that [x 2 m+1-i-k , y] 2 ≡ [x, y] 2 m+2-i-k modulo E i,m+2 , since x = [x 1 , . . . , x i-1 ] ∈ Γ i-1 (H) and x 2 m+1-(i-1)-k is a generator of E i-1,m+1 , there exist 1 ≤ j 1 < • • • < j k ≤ i -1 such that x j 1 , . . . , x j k ∈ A, and therefore [x 1 , . . . , x i-1 ] 2 m-(i-1)-k is a generator of E i-1,m . Applying induction hypothesis 1, we have: [x 2 m+1-i-k , y] ≡ [x, y] 2 m+1-i-k mod E i,m+1 .
It follows that there exist generators α 1 , . . . , α t of E i,m+1 and δ 1 , . . . , δ t ∈ {1, -1} for which [

x 2 m+1-i-k , y] = [x, y] 2 m+1-i-k α δ 1 1 • • • α δt t , and so: [x 2 m+1-i-k , y] 2 =([x, y] 2 m+1-i-k α δ 1 1 • • • α δt t ) 2 = [x, y] 2 m+2-i-k [x, y] -2 n+1-i-k , α δ 1 1 • • • α δt t • α 2δ 1 1 [α -δ 1 1 , α δ 2 2 • • • α δt t ]α 2δ 2 2 • • • [α -δ t-1 t-1 , α δt t ]α 2δt t .
(5.25)

We claim that:

(1)

α 2δ 1 1 , . . . , α 2δt t ∈ E i,m+2 . (2) [α s , θ], [α -1 s , θ] ∈ E i,m+2
for all s = 1, . . . , t and all θ ∈ H.

(3) [x, y] -2 m+1-i-k , θ ∈ E i,m+2 for all θ ∈ H.
Claim (1) follows from (5.17). To prove (2), let s ∈ {1, . . . , t}. Since α s is a generator of E i,m+1 ,

α s = β 2 m+1-l-k , where β = [β 1 , . . . , β l ] ∈ Γ l (H), β j 1 , . . . , β j k ∈ A, i ≤ l ≤ m+1 and 0 ≤ k ≤ m+1-l. By induction hypothesis 2, [α s , θ] = [β 2 m+1-l-k , θ] ≡ [β, θ] 2 m+1-l-k mod E l+1,m+2 .
(5.26)

Moreover, [θ, β] = [θ, β 1 , . . . , β l ] ∈ Γ l+1 (H), where β j 1 , . . . , β j k ∈ A, so [θ, β] 2 (m+2)-(l+1)-k ∈ E l+1,m+2 since i ≤ l + 1 ≤ m + 2 and 0 ≤ k ≤ (m + 2) -(l + 1). Hence: [β, θ] 2 m+1-l-k = [θ, β] 2 m+1-l-k -1 ∈ E l+1,m+2 .
The fact that E l+1,m+2 ⊂ E i,m+2 implies that the congruence (5.26) is also valid modulo E i,m+2 , from which it follows using (5.26

) that [α s , θ] ∈ E l+1,m+2 ⊂ E i,m+2 . Further, [α -1 s , θ] = α -2 s [α s , θ](θα 2 s θ -1
) ∈ E i,m+2 , using also (1) and the fact that E i,m+2 is normal in H, which proves (2). To prove (3), since

x 2 m+1-(i-1)-k ∈ E i-1,m+1 , where x = [x 1 , . . . , x i-1 ] ∈ Γ i-1 (H) and 0 ≤ k ≤ m + 1 -i, and there exist 1 ≤ j 1 < • • • < j k ≤ i -1 such that x j 1 , . . . , x j k ∈ A, it follows that [y, x] = [y, x 1 , . . . , x i-1 ] ∈ Γ i (H) and [y, x] 2 (m+1)-i-k ∈ E i,m+1
. So applying induction hypothesis 2, we have:

[x, y] -2 m+1-i-k , θ = [y, x] 2 m+1-i-k , θ ≡ [y, x], θ 2 m+1-i-k mod E i+1,m+2 . (5.27) Now [[y, x], θ] = [θ, y, x] -1 = [[y, x], θ] = [θ, y, x 1 , . . . , x i-1 ] -1 ,
and since [θ, y, x 1 , . . . ,

x i-1 ] 2 (m+2)-(i+1)-k ∈ E i+1,m+2 , we conclude that [x, y] -2 m+1-i-k , θ ∈ E i+1,m+2 ⊂ E i,m+2
, and this proves claim [START_REF] Bellingeri | On presentations of surface braid groups[END_REF]. Thus it follows from (5.25) and claims (1), ( 2) and (3) that:

[x 2 m+1-i-k , y] 2 ≡ [x, y] 2 m+2-i-k mod E i,m+2 , which proves that the expression ( * * ) is congruent to [x, y] 2 m+2-i-k modulo E i,m+2 .
To see that the expression ( * ) belongs to E i,m+2 , notice that each of its terms is a commutator, so can be written as a product of conjugates of the element

x 2 m-i-k , [x 2 m+1-i-k , y] or its inverse. Since E i,m+2 is normal in H, it thus suffices to show that θ, [x 2 m+1-i-k , y] -1 = [x 2 m+1-i-k , y], θ ∈ E i,m+2
for all y, θ ∈ H and for all

x = [x 1 , . . . , x i-1 ] ∈ Γ i-1 (H) for which x 2 m+1-(i-1)-k ∈ E i-1,m+1 , where i ≥ 3, 0 ≤ k ≤ m + 1 -i, x j 1 , . . . , x j k ∈ A and 1 ≤ j i < • • • < j k ≤ i -1. To do so, note that x 2 m+1-i-k = x 2 m-(i-1)-k ∈ E i-1,m , so [x 2 m+1-i-k , y] ≡ [x, y] 2 m+1-i-k mod E i,m+1
by induction hypothesis 1. Thus there exist generators α 1 , . . . , α t of E i,m+1 , and

δ 1 , . . . , δ t ∈ {1, -1} such that [x 2 m+1-i-k , y] = [x, y] 2 m+1-i-k α δ 1 1 • • • α δt t , and hence: [x 2 m+1-i-k , y], θ = [x, y] 2 m+1-i-k α δ 1 1 • • • α δt t , θ = [x, y] 2 m+1-i-k [α δ 1 1 • • • α δt t , θ][x, y] -2 m+1-i-k [x, y] 2 m+1-i-k , θ . (5.28)
Further, [α δ 1 1 • • • α δt t , θ] belongs to the normal closure of α δ 1 1 , θ , . . . , α δt t , θ in H. This may be seen by applying reverse induction on 1 ≤ s ≤ t, and by noting that for s ≥ 2, [α

δ s-1 s-1 α δs s • • • α δt t , θ] = (α δ s-1 s-1 [α δs s • • • α δt t , θ]α -δ s-1 s-1 )[α δ s-1 s-1 , θ]. (5.29) Then [x 2 n+1-i-k , y], θ ∈ [x, y] 2 n+1-i-k , θ , α δ 1 1 , θ , .
. . , α δt t , θ H by (5.28) and (5.29). Now by claims ( 2) and ( 3), the elements [α δ 1 1 , θ], . . . , [α δt t , θ] and [x, y] 2 m+1-i-k , θ belong to E i,m+2 , and since E i,m+2 is normal in H, we conclude that the expression ( * ) belongs to E i,m+2 . This completes the proof of (5.23) for i -1, and so by induction, (5.23) holds for all 1 ≤ i ≤ m + 1, which is exactly (5.22). By induction, we conclude that (5.20) holds, and this completes the proof of the lemma.

Lemma 5.7 has the following consequences.

Corollary 5.9. For all

1 ≤ l ≤ m, [E l,m , H] ⊂ E l+1,m+1 and [ E l,m , H] ⊂ E l+1,m+1 . Proof. Let l ≤ i ≤ m. First assume that x = [x 1 , . . . , x i ] ∈ Γ i (H) is such that x 2 m-i-k is a generator of E l,m
, and let y ∈ H. Then by (5.20)

, [x 2 m-i-k , y] ≡ [x, y] 2 m-i-k mod E l+1,m+1 . Since [y, x] = [y, x 1 , . . . , x i ] ∈ Γ i+1 (H), it follows that [y, x] 2 (m+1)-(i+1)-k ∈ E l+1,m+1 , so [x 2 m-i-k , y] ∈ E l+1,m+1 . Now suppose that x = α δ 1 1 • • • α δt t ,
where for all i = 1, . . . , t, α i is a generator of E l,m , and δ i ∈ {1, -1}. Then [x, y] belongs to the normal closure of [α δ i , y], i = 1, . . . , t in H by (5.29), and so [x, y] ∈ E l+1,m+1 for all x ∈ E l,m , y ∈ H by the first paragraph of the proof and the fact that E l+1,m+1 is normal in H. Once more, the result for E l,m is obtained from that for E l,m by taking

A = ∅. Corollary 5.10. If m ≥ 2, then W m = a 2 m-1 2 , E 2,m H . Proof. It suffices to prove that E 2,m = x 2 m-i : x ∈ Γ i (H), 2 ≤ i ≤ m (5.30)
for all m ≥ 2. If m = 2, (5.30) follows from (5.12) and (5.18). Suppose by induction on m that (5.30) holds for some m ≥ 2. It is clear from (5.18

) that E 2,m+1 ⊂ x 2 m-i : x ∈ Γ i (H), 2 ≤ i ≤ m .
To prove the converse, let y = x 2 m+1-i , where x ∈ Γ i (H) and 2

≤ i ≤ m + 1. If i = m + 1 then y ∈ Γ m+1 (H) ⊂ E 2,m+1 by (5.19). So suppose that 2 ≤ i ≤ m. Then x 2 m-i ∈ E 2,m by the induction hypothesis, so x 2 m-i = α δ 1 1 • • • α δt t ,
where for all i = 1, . . . , t, α i is a generator of E 2,m and δ i ∈ {1, -1}. Hence:

y = (x 2 m-i ) 2 = (α δ 1 1 • • • α δt t ) 2 = α 2δ 1 1 [α -δ 1 1 , α δ 2 2 • • • α δt t ]α 2δ 2 2 [α -δ 2 2 , α δ 3 3 • • • α δt t ] • • • [α -δ t-1 t-1 , α δt t ]α 2δt t . So y ∈ E 2,m+1 because for all i = 1, . . . , t, α 2δ i i ∈ E 2,m+1 by (5.18), and [α -δ i i , θ] ∈ [ E 2,m , H] ⊂ E 3,m+1 ⊂ E 2,
m+1 by Corollary 5.9 and (5.19). The inclusion

x 2 m-i : x ∈ Γ i (H), 2 ≤ i ≤ m ⊂ E 2,m+1 then follows. Corollary 5.11. For all m ≥ 2, [H, W m ] ⊂ W m+1 .
Proof. Since [ E 2,m , H] ⊂ E 2,m+1 using Corollary 5.9, by Corollary 5.10, it suffices to prove that [h, a 2 m-1 2 ] ∈ W m+1 for all h ∈ H. To see this, observe that [h, a 2 m-1

2

] ≡ [h, a 2 ] 2 m-1 mod W m+1 by taking i = 1 and x = a 2 in (5.21) and using Remark 5.8. So there exists w (5.14), and we conclude that [h, a 2 m-1 2 ] ∈ W m+1 as required.

∈ W m+1 such that [h, a 2 m-1 2 ] = [h, a 2 ] 2 m-1 . w. Now [h, a 2 ] ∈ Γ 2 (H), hence [h, a 2 ] 2 m-1 ∈ W m+1 by
The following result will enable us to obtain the explicit characterisation of Γ n (P 2 (K)) given in Theorem 5.4. Proposition 5.12. For all n ≥ 2, L n = W n = W n .

Proof. If n = 2, the statement is true by definition. So suppose by induction that L n = W n = W n for some n ≥ 2. Then we have the following inclusions:

• W n+1 ⊂ W n+1 . To see this, let y ∈ W n+1 . If y is a generator of W n+1 , then by (5.14), y either belongs to Γ n+1 (H), and so belongs to W n+1 by (5.13), or is of the form x 2 n+1-i , where 1 ≤ i ≤ n, x ∈ Γ i (H), and x = a 2 if i = 1. Hence x 2 n-i ∈ W n = W n by induction, and thus y = x 2 n+1-i = (x 2 n-i ) 2 ∈ W n+1 by (5.13). If y is an arbitrary element of W n+1 , it may be written as a product of conjugates of generators and their inverses, so it belongs to Lemma 5.5 and (5.14), and [H,

W n+1 because W n+1 is normal in H. • L n+1 ⊂ W n+1 , since K n+1 ⊂ W n+1 by
L n ] = [H, W n ] ⊂ W n+1
by induction and Corollary 5.11, so

L n+1 = K n+1 , [H, L n ] ⊂ W n+1 using Lemma 5.6. • W n+1 ⊂ L n+1 . First, Γ n+1 (H) = [H, Γ n (H)] ⊂ [H, W n ] = [H, L n ] ⊂ L n+1 by (5.13) and induction. Secondly, let x 2 ∈ W n+1 , where x ∈ W n = L n is a generator. If n = 2, by (5.12), x = a 2 2 or x = [b 2 , a 2 ], then x 2 = a 4 2 ∈ K 3 ⊂ L 3 by Lemma 5.5, or x 2 = [b 2 , a 2 2 ∈L 2 ][a 2 , [a 2 , b 2 ] ∈L 2 ] ∈ [H, L 2 ] ⊂ L 3 .
So assume that n ≥ 3. By Lemma 5.6, there are two possibilities for x:

-if x ∈ K n , then x = a 2 n-1
2 by Lemma 5.5, and

x 2 = a 2 n 2 ∈ K n+1 ⊂ L n+1 . -if x = [h, l] ∈ [H, L n-1 ], where h ∈ H and l ∈ L n-1 = W n-1 , then x 2 = [h, l] 2 = [h, l 2 ][l, [l, h]]. Now l 2 ∈ W n = L n by induction and [l, h] ∈ L n , so [h, l 2 ] and [l, [l, h]] belong to [H, L n ], which is contained in L n+1 . Finally, let x 2 ∈ W n+1
, where x is an arbitrary element of W n . Then there exists q ∈ N such that x = x 1 • • • x q , where for all i = 1, . . . , q, x i is a generator of W n = L n . Then as in (5.25), we have:

x 2 = x 2 1 [x -1 1 , x 2 • • • x q ]x 2 2 [x -1 2 , x 3 • • • x q ] • • • [x -1 q-1 , x q ]x 2 q .
(5.31)

From the second case above, for all i = 1, . . . , q, x 2 i ∈ L n+1 . Further, for all i = 1, . . . , q -1,

[x -1 i , x i+1 • • • x q ] ∈ [L n , H] ⊂ L n+1 . It then follows from (5.31) that x 2 ∈ L n+1 . It follows from these three inclusions that L n+1 = W n+1 = W n+1 .
Proof of Theorem 5.4. The result is a consequence of Theorem 1.1(1), Proposition 5.12, and equations (5.8), (5.10) and (5.14). To see that P 2 (K) is residually nilpotent, note first that W n ⊂ γ 2 n (H) for all n ≥ 2 by (5.13). Since H is a free group of finite rank, it is residually 2-finite, and it follows using Proposition 5.12 that n≥1 L n = n≥2 W n = {1}. The residual nilpotence of P 2 (K) is then a consequence of the first part of the statement, equation (5.14) and Proposition 5.12.

5.3.

The lower F 2 -linear central filtration of P 2 (K). Now that we have a good description of Γ n (P 2 (K)), we may obtain the following decomposition of γ 2 n (P 2 (K)). Theorem 5.13. Let n ≥ 2. With the notation of Theorem 2.1, γ 2 n (P 2 (K)) is equal to:

a 2 n-1 2 , b 2 n-1 2 , x 2 n-i : x ∈ Γ i (π 1 (K \ {x 1 })), 2 ≤ i ≤ n π 1 (K\{x 1 }) (a 1 a 2 ) 2 n-1 , (b 2 b 1 ) 2 n-1 .
Consequently, P 2 (K) is residually 2-finite.

Remark 5.14. Using the action given by (5.9) and the description of Γ n (P 2 (K)) given by Theorem 5.4, it is straightforward to see that:

a 2 n-1 2 , b 2 n-1 2 , x 2 n-i : x ∈ Γ i (π 1 (K \ {x 1 })), 2 ≤ i ≤ n π 1 (K\{x 1 }) (a 1 a 2 ) 2 n-1 , (b 2 b 1 ) 2 n-1
is equal to U n , where:

U n = Γ n (P 2 (K)), b 2 n-1 2 , (b 2 b 1 ) 2 n-1 P 2 (K) . ( 5
(K))) ⊂ Γ 2 (P 2 (K)). Further, if x = x 1 • • • x q , where for i = 1, . . . , q, x i ∈ a ±1
2 , (a 2 a 1 ) ±1 , b ±1 2 , (b 2 b 1 ) ±1 , then using the decomposition given in (5.31), we have

x 2 ∈ U 2 because x 2 i ∈ U 2 for all 1 ≤ i ≤ q and [x -1 i , x i+1 • • • x q ] ∈ Γ 2 (P 2 (K)) ⊂ U 2 for all 1 ≤ i ≤ q -1. Thus γ 2 2 (P 2 (K)) ⊂ U 2 ,
∈ {b 2 , b 2 b 1 } then x 2 n = (x 2 n-1 ) 2 , where x 2 n-1 ∈ U n = γ 2
n (P 2 (K)) by the induction hypothesis, so x 2 n ∈ γ 2 n+1 (P 2 (K)). To prove that γ 2 n+1 (P 2 (K)) ⊂ U n+1 , using the induction hypothesis and the fact that γ 2 n+1 (P 2 (K)) is generated by [P 2 (K), γ 2 n (P 2 (K))] ∪ {x 2 : x ∈ γ 2 n (P 2 (K))}, it suffices to show that [P 2 (K), U n ] ⊂ U n+1 , and that x 2 ∈ U n+1 for all x ∈ U n . We first show that [P 2 (K), U n ] ⊂ U n+1 . Let x ∈ P 2 (K), and let u ∈ U n .

( , x] ∈ L n+1 for all x ∈ {a 2 , (a 2 a 1 ) -1 , b 2 , (b 2 b 1 ) -1 }. The result is clear if x = b 2 , so we consider the three other cases. We proceed by induction on n. Suppose first that n = 2. If x = a 2 then by (3.2), (5.13) and Proposition 5.12, we have:

) If u ∈ Γ n (P 2 (K)) then [x, u] ∈ U n+1 . ( 1
[u, x] = [b 2 2 , a 2 ] = [b 2 , b 2 , a 2 ] ∈W 3 =L 3 [b 2 , a 2 ] 2 ∈W 3 =L 3 ∈ L 3 .
In the remaining two cases, by (5.9), (5.12), (5.14) and Proposition 5.12, we have:

[(a 2 a 1 ) -1 , u] = [(a 2 a 1 ) -1 , b 2 2 ] = (a -2 2 b 2 )(a -2 2 b 2 )b -2 2 = a -4 2 ∈ W 3 =L 3 [a 2 2 , b 2 ] ∈[L 2 ,H]⊂L 3 ∈ L 3 [(b 2 b 1 ) -1 , u] = [(b 2 b 1 ) -1 , b 2 2 ] = (a 2 b 2 a 2 )(a 2 b 2 a 2 )b -2 2 = [a 2 b 2 , a 2 2 ] ∈[H,L 2 ]⊂L 3 a 4 2 ∈ W 3 =L 3 [a -1 2 , b 2 ] 2 ∈ W 3 =L 3 [b 2 , a -1 2 ], b 2 ∈Γ 3 (H)⊂L 3 ∈ L 3 ,
which proves the claim in the case n = 2. Now suppose that [b 2 j-1

2

, x] ∈ L j+1 for all 2 ≤ j ≤ n and x ∈ {a 2 , (a 2 a 1 ) -1 , b 2 , (b 2 b 1 ) -1 }. Then by Proposition 5.12 and (5.13), [b 2 n-1

2

, x] ∈ L n+1 = W n+1 , and hence [b 2 n-1

2

, x] 2 ∈ W n+2 = L n+2 . So by (3.2), we have:

[b 2 n 2 , x] = b 2 n-1 2 , [b 2 n-1 2 , x] ∈L n+1 ∈[H,L n+1 ]⊂L n+2 [b 2 n-1 2 , x] 2 ∈L n+2 ∈ L n+2 ,
which proves the claim for all n ≥ 2. Now let x be an arbitrary element of P 2 (K). Since the set {a 2 , (a 2 a 1 ) -1 , b 2 , (b 2 b 1 ) -1 } generates P 2 (K), for some t ≥ 0, there exist x 1 , . . . , x t ∈ {a 2 , (a 2 a 1 ) -1 , b 2 , (b 2 b 1 ) -1 } and ε 1 , . . . , ε t ∈ {1, -1} for which

x = x ε 1 1 x ε 2 2 • • • x εt t .
As in (5.29), we have the following relation:

[u, x] = [u, x ε 1 1 ] x ε 1 1 [u, x ε 2 2 ]x -ε 1 1 • • • x ε 1 1 x ε 2 2 • • • x ε t-1 t-1 [u, x εt t ]x -ε t-1 t-1 • • • x -ε 2 2 x -ε 1 1 .
(5.33)

Using the fact that L n+1 is normal in H by Lemma 3.3, it follows from (5.33) that [b 2 n-1

2

, x] ∈ L n+1 , and since L n+1 ⊂ Γ n+1 (P 2 (K)) by Theorem 1.1, we deduce that [b

2 n-1 2 , x] ∈ U n+1 .
This concludes the proof of the inclusion [P 2 (K),

U n ] ⊂ U n+1 . It remains to prove that {x 2 : x ∈ U n } ⊂ U n+1 . If x = b 2 n-1 2 or x = (b 2 b 1 ) 2 n-1 then clearly x 2 ∈ U n+1 . Using Theorems 1.1(1) and 5.4, if x ∈ Γ n (P 2 (K)) = L n (a 1 a 2 ) 2 n-1
, then in terms of this semi-direct product, x = (x 1 , x 2 ), where x 1 ∈ L n and x 2 ∈ (a 1 a 2 ) 2 n-1 , and

x 2 = (x 1 , x 2 )(x 1 , x 2 ) = (x 1 .ϕ(x 2 )(x 1 ), x 2
2 ). Now x 2 2 ∈ (a 1 a 2 ) 2 n , and since x 1 ∈ L n = W n , we have x 2 1 ∈ W n+1 = L n+1 by Proposition 5.12, and therefore

x 1 .ϕ(x 2 )(x 1 ) = x 2 1 (ϕ(x 2 )(x -1 1 ).x 1 ) -1 ∈ L n+1 and x 2 ∈ L n+1 (a 1 a 2 ) 2 n = Γ n+1 (P 2 (K)) ⊂ U n+1 . If
x is a product of conjugates of generators of U n then x 2 ∈ U n+1 using (5.31). This shows that

{x 2 : x ∈ U n } ⊂ U n+1 . It follows that γ 2 n+1 (P 2 (K)) ⊂ U n+1 , so γ 2 n+1 (P 2 (K)) = U n+1 .
Using the explicit characterisation of Γ n (P 2 (K)) given by Theorem 5.4 and Remark 5.14, the first part of the statement follows. The second part is a consequence of the fact that n≥2 U n = {1} using the first part.

5.4. The lower central series of P n (K). For n ≥ 3, the situation is more complicated due to the complexity of the lower central series of P 2 (K). The idea is to use the Fadell-Neuwirth short exact sequence (5.1) and to calculate recursively Γ m (P n (K)) for m ≥ 1 and n ≥ 3. Thus will allows us to prove that P n (K) is residually nilpotent for all n ≥ 1 in Theorem 5. [START_REF] Kohno | Série de Poincaré-Koszul associée aux groupes de tresses pures[END_REF].

With the notation of Theorem 1.1 and equation (5.1), we may write P n+1 (K) = H ϕ G, where H = π 1 (K \ {x 1 , . . . , x n }), G = s(P n (K)), and s is the section for p * given by Proposition 5.1. In what follows, we will take G to be equipped with the generating set:

X = s(z) : z ∈ {a i , b i , C j,k : 1 ≤ i ≤ n and 1 ≤ j < k ≤ n} = a i , b i , C i,k , a n a n+1 , b n b n+1 C n,n+1 , C i,n C i,n+1 C -1 n,n+1 : 1 ≤ i ≤ k ≤ n -1 .
(5.34)

In P n+1 (K), for 1 ≤ j ≤ n, let D j = C -1 j,n+1 C j+1,n+1 . Since C j,n+1 = D -1 n • • • D -1 j+1 D -1 j
for all j = 1, . . . , n, the set:

Y = {a n+1 , b n+1 , D j : j = 1, . . . , n} (5.35)
generates H, and using Theorem 2.1, the action ϕ : s(P n (K)) -→ Aut(π 1 (K \ {x 1 , . . . , x n })) is given by: (5.36) where:

                                                                                           ϕ(a i )(z) =      a n+1 if z = a n+1 b n+1 a n+1 D i a -1 n+1 if z = b n+1 α i,j D j α -1 i,j if z = D j ϕ(b i )(z) =          a n+1 b n+1 C i,n+1 D i C -1 i,n+1 b -1 n+1 if z = a n+1 b n+1 C i,n+1 D -1 i C -1 i,n+1 if z = b n+1 β i,i D -1 i β -1 i,i if z = D i β i,j D j β -1 i,j if z = D j , j = i ϕ(C i,k )(z) =      a n+1 if z = a n+1 b n+1 if z = b n+1 δ i,j,k D j δ -1 i,j,k if z = D j ϕ(a n a n+1 )(z) =      a n+1 if z = a n+1 a -1 n+1 b n+1 a n+1 D n if z = b n+1 α j D j α -1 j if z = D j ϕ(b n b n+1 C n,n+1 )(z) =          D n b -1 n+1 a n+1 b n+1 if z = a n+1 b n+1 D -1 n if z = b n+1 D -1 n if z = D n b -1 n+1 D j b n+1 if z = D j , j = n ϕ(C i,n C i,n+1 C -1 n,n+1 )(z) =      C n,n+1 C -1 i,n+1 a n+1 C i,n+1 C -1 n,n+1 if z = a n+1 C n,n+1 C -1 i,n+1 b n+1 C i,n+1 C -1 n,n+1 if z = b n+1 δ i,j D j δ -1 i,j if z = D j ,
α i,j =      1 if i < j C -1 j+1,n+1 a n+1 if i = j C -1 i+1,n+1 C i,n+1 if i > j β i,j =      1 if i < j b n+1 C i,n+1 if i = j b n+1 C i,n+1 C -1 i+1,n+1 b -1 n+1 if i > j δ i,j,k =      1 if k < j or i > j C -1 j+1,n+1 C i,n+1 if k = j C -1 k+1,n+1 C k,n+1 if k > j ≥ i
for all 1 ≤ i ≤ k ≤ n -1 and 1 ≤ j ≤ n, and where α j = a -1 n+1 α n,j and δ i,j = C n,n+1 C -1 i,n+1 δ i,j,n . Our aim is to determine the subgroups L m that were defined in the statement of Theorem 1.1 for all m ≥ 2. For i ≥ 0, let:

A 2 i n = D 2 i j : j = 1, . . . , n H . (5.37) If i = 0, we write A n = A 1 n . To compute K 2 , recall that K 2 = H 2 ,
(a i )(b n+1 )b -1 n+1 = b n+1 a n+1 D i a -1 n+1 b -1 n+1 for all 1 ≤ i ≤ n-1, and ϕ(b n b n+1 C n,n+1 )(b n+1 )b -1 n+1 = b n+1 D -1 n b -1 n+1 .
So by normality of K 2 , D j belongs to K 2 for all 1 ≤ j ≤ n, and therefore A n ⊂ K 2 by (5.37). Moreover,

ϕ(b n b n+1 C n,n+1 )(a n+1 )a -1 n+1 = D n b -1 n+1 a n+1 b n+1 a -1 n+1 , so [b -1 n+1 , a n+1
] belongs to K 2 also. Applying Lemma 3.4(1) with X and Y as defined in (5.34) and (5.35), and using (5.36), we see that K 2 = A n , [a n+1 , b n+1 ] H , and therefore:

L 2 = Γ 2 (H), A n .
(5.38)

Let Y 1 = L 1 = H, and for m ≥ 2, let:

Y m = A 2 m-2 n , [Y i , Y k ] : 1 ≤ i ≤ k < m, i + k = m . (5.39)
In what follows, we will refer to the elements of the set for all j = 1, . . . , n by (5.36), so ϕ(g)(A

D 2 m-2 j , [y i , y k ] : j = 1, . . . , n, y i ∈ Y i , y k ∈ Y k and i + k = m as generators of Y m . Since Γ m (P n+1 (K)) = L m Γ m (P n (K))
2 i n ) ⊂ A 2 i n for all i ≥ 1. Also, if y k ∈ Y k and y l ∈ Y l then ϕ(g)([y k , y l ]) = [ϕ(g)(y k ), ϕ(g)(y l )]
, and the claim follows by induction on i.

Lemma 5.17. [G, Y m ] ⊂ Y m+1 for all m ≥ 1. Proof. If m = 1, then [G, Y 1 ] = [G, H] = H 2 ⊂ L 2 = Y 2 . Now, by induction on m, suppose that [G, Y i ] ⊂ Y i+1 , for all 1 ≤ i ≤ m. Let us prove that [G, Y m+1 ] ⊂ Y m+2 . To do so, let [g, h] = ϕ(g)(h).h -1 ∈ [G, Y m+1 ],
where g ∈ G and h ∈ Y m+1 . By Lemma 3.4(2), we need only analyse the following cases where g ∈ X and h is a generator of Y m+1 :

(

1) h = D 2 m-1 j
, where j ∈ {1, . . . , n}. By (5.36), if j ∈ {1, . . . , n -1} then:

ϕ(b j )(D 2 m-1 j ).D -2 m-1 j = [β j,j , D -2 m-1 j ]D -2 m j ∈ Y m+2 , because D 2 m-1 j ∈ Y m+1 , and if j = n then: ϕ(b n b n+1 C n,n+1 )(D 2 m-1 n ).D -2 m-1 n = D -2 m n ∈ Y m+2 .
Similarly, by (5.36), one may check that if g is any other element of X then ϕ(g)(D 2 m-1 j ).D -2 m-1 j is a commutator of D 2 m-1 j with an element of H, and so belongs to [H, Y m+1 ], which is contained in Y m+2 by (5.39).

(2) h = [y i , y j ], where 1 ≤ i, j ≤ m, i+j = m+1, y i ∈ Y i and y j ∈ Y j . By Remark 5.16, we have that ϕ(g)(Y l ) ⊂ Y l for all l ≥ 1. By the induction hypothesis (ϕ(g)(y -1 j ).y (5.39), and writing [g, [y i , y j ]] = ϕ(g)([y i , y j ]).[y j , y i ] in the form of (5.15) and Remark 5.16, where we replace l by y j and h by y i , and using the fact that Y m+2 is normal in H, it follows that ϕ(g)([y i , y j ]).[y j , y i ] ∈ Y m+2 .

j ) -1 ∈ [G, Y j ] ⊂ Y j+1 and (ϕ(g)(y -1 i ).y i ) -1 ∈ [G, Y i ] ⊂ Y i+1 . Therefore (ϕ(g)(y -1 j ).y j ) -1 , ϕ(g)(y -1 i ) ∈ [Y j+1 , Y i ] ⊂ Y m+2 and [(ϕ(g)(y -1 i ).y i ) -1 , y j ] ∈ [Y i+1 , Y j ] ⊂ Y m+2 by
∈ G, g k ∈ Γ k (G) ⊂ G and h ∈ Y m of the commutator [Γ k+1 (G), Y m ] = [[Γ k (G), G], Y m ], we obtain: [g k , g], h g [g, h], g h k [h, g k ], g g k = 1. ( 5
], g g k ∈ [[Y m , Γ k (G)], G] ⊂ [Y m+k , G] ⊂ Y m+k+1
by the induction hypothesis and Lemma 5.17. Thus [g k , g], h g ∈ Y m+k+1 by (5.42). By (5.40), we see that:

[g k , g], h g = [g k , g], [g, h]h = [g k , g], [g, h] [g, h], [g k , g], h [g k , g], h . (5.43) Now [g k , g], [g, h] ∈ [Γ k (G), Y m+1 ] ⊂ Y m+k+1
by the induction hypothesis, and by Remark 5.16, we see that

[g, h], [g k , g], h ∈ [Y m+1 , [Γ k (G), Y m ]] ⊂ [Y m+1 , Y k+m ] ⊂ Y 2m+k+1 ⊂ Y m+k+1 .
We conclude from (5.43) that [g k , g], h ∈ Y m+k+1 , and the result follows. Proof. We prove the lemma by induction on m ≥ 1. For m = 1, 2, the result follows from (5.38) and (5.39). Suppose that m ≥ 2 is such that L i = Y i for all 1 ≤ i ≤ m.

We first show that Y m+1 ⊂ L m+1 . For all i, j ≥ 0, it follows from [27, Theorem 5.3] and Theorem 1.1(1) that:

[L i ϕ Γ i (G), L j ϕ Γ j (G)] = [Γ i (H ϕ G), Γ j (H ϕ G)] ⊂ Γ i+j (H ϕ G) = L i+j ϕ Γ i+j (G).
(5.44) Let x i ∈ L i and x j ∈ L j . Then (x i , 1) ∈ L i ϕ Γ i (G) and (x j , 1) ∈ L j ϕ Γ j (G), and using (3.3) and (5.44), we obtain: To prove that Z m+1 ⊂ Z m+1 , let us show that A 2 m-1 n and X m+1 are contained in Z m+1 . For all j = 1, . . . , n, we have D 2 m+1-2 j = (D 2 m-2 j ) 2 and D 2 m-2 j ∈ Z m = Z m by induction, so A 2 m-1 n ⊂ Z m+1 by (5.45). Now suppose that [x 1 , . . . , x i ] 2 m+1-i-k ∈ X m+1 , so k elements of {x 1 , . . . , x i } belong to A n , where 0 ≤ k ≤ m + 1 -i. If m + 1 -i > k then [x 1 , . . . , x i ] 2 m-i-k belongs to Z m = Z m by induction, so [x 1 , . . . , x i ] 2 m+1-i-k = ([x 1 , . . . , x i ] 2 m-i-k ) 2 belongs to Z m+1 . If m + 1 -i = k then [x 1 , . . . , x i ] 2 m+1-i-k = [x 1 , . . . , x i ] ∈ Γ i (H), where k = (m + 1) -i elements of this commutator belong to A n , so [x 1 , . . . , x i ] ∈ Z m+1 . Hence X m+1 ⊂ Z m+1 , and thus Z m+1 ⊂ Z m+1 .

([x i , x j ], 1) = [(x i , 1), (x j , 1)] ∈ L i+j ϕ Γ i+j (G). Hence [x i , x j ] ∈ L i+j , then [L i , L j ] ⊂ L i+j . So if 1 ≤ i, j ≤ m, where i + j = m + 1, we see that [Y i , Y j ] = [L i , L j ] ⊂ L m+1 using the induction hypothesis. To prove that A 2 m-1 n ⊂ L m+1 , by induction, we have D 2 m-2 j ∈ Y m = L m for all j = 1, . . . , n. Therefore ϕ(b j )(D 2 m-2 j ).D -2 m-2 j and ϕ(b n b n+1 C n,n+1 )(D 2 m-2 n ).D -2 m-2 n belong to H m+1 ⊂ L m+1 for all j = 1, . . . , n -1. Consequently [β j,j , D -2 m-2 j ]D -2 m-1 j and D -2 m-1 n belong to L m+1 for all j = 1, . . . , n -1. Now [β j,j , D -2 m-2 j ] ∈ [H, L m ] ⊂ L
We now show that Z m+1 ⊂ Z m+1 . First note that X m+1 ⊂ X m+1 , for if the commutator [x 1 , . . . , x i ] belongs to X m+1 then m + 1 -i of its elements belong to A n , and so [x 1 , . . . , x i ] = [x 1 , . . . , x i ] 2 m+1-i-k ∈ X m+1 , where k = m + 1 -i. Now let y ∈ Z m+1 be of the form y = x 2 , where x ∈ Z m = Z m by induction. If x is a generator of Z m then x 2 ∈ Z m+1 from the definition of Z m+1 . If x = x 1 • • • x l , where x i is a generator of Z m for all 1 ≤ i ≤ l, then making use of a decomposition of x analogous to that of (5.31), the previous sentence and the fact that [H, Z m ] ⊂ Z m+1 by Remark 5.23, it follows that y ∈ Z m+1 . We conclude that Z m+1 ⊂ Z m+1 , and hence Z m+1 = Z m+1 .

To show that Z m+1 ⊂ Y m+1 , let y ∈ Z m+1 . We first consider the following two possibilities: (i) y = x 2 , where x ∈ Z m = Y m by induction. If x = D 2 m-2 j for some j = 1, . . . , n, it follows from the definition of Y m+1 that y = x 2 = D 2 m-1 j ∈ Y m+1 . If x = [x i , x j ], where 1 ≤ i ≤ j < m, i + j = m, x i ∈ Y i and x j ∈ Y j , then:

y = [x i , x j ] 2 = x i , [x j , x -1 i ] ∈[Y i ,Y j ]⊂Ym ∈[H,Ym]⊂Y m+1 x j x -1 i x -1 j , x 2 i ∈Z i+1 =Y i+1 ∈[Y j ,Y i+1 ]⊂Y m+1
x i x -1 j .

Note that to obtain x 2 i ∈ Y i+1 , we have i < m, so Y i = Z i and Y i+1 = Z i+1 by the induction hypothesis. Therefore y ∈ Y m+1 since Y m+1 is normal in H by Remark 5.16. (ii) y = [x 1 , . . . , x i ] ∈ Γ i (H), where 2 ≤ i ≤ m + 1, x 1 , . . . , x i ∈ H, and there exist 1 ≤ j 1 < . . . < j m+1-i ≤ i such that x j 1 , . . . , x j m+1-i ∈ A n . If j 1 = 1, i.e. For the general case, if y is a product of conjugates of the two types of elements described in (i) and (ii) above, then y ∈ Y m+1 because Y m+1 is normal in H by Remark 5.16, and we conclude that Z m+1 ⊂ Y m+1 .

To complete the proof, it remains to see that Y m+1 ⊂ Z m+1 . From the definition of Z m+1 given in (5.45), it is clear that A 2 m-1 n ⊂ Z m+1 . Applying the induction hypothesis, we have [Y i , Y j ] = [ Z i , Z j ] for all 1 ≤ i, j ≤ m for which i + j = m + 1, so it suffices to show that [ Z i , Z j ] ⊂ Z i+j . We shall prove by induction on i that [ Z i , Z j ] ⊂ Z i+j for all 1 ≤ i ≤ m and j ≥ 1. The inclusion that we require then follows as a special case. So let j ≥ 1. If i = 1 then [ Z 1 , Z j ] = [H, Z j ] ⊂ Z j+1 by Remark 5.23. So assume that 1 < i ≤ m, and suppose by induction that:

[ Z k , Z j ] ⊂ Z k+j for all j ≥ 1 and 1 ≤ k < i ≤ m.

(5.47)

The aim is to prove that [ Z i , Z j ] ⊂ Z i+j . Let x ∈ Z i and y ∈ Z j . Then x ∈ Y i by the induction hypothesis (5.46) and the fact that 1 < i ≤ m. Assume first that x is a generator of Y i , so that one of the following conditions holds:

5.5. The case of the braid group B n (K). In this section, we prove Theorem 1.3 [START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF]. We start by giving two propositions in the cases where B n (K) is not residually nilpotent or is not residually soluble.

Proposition 5.26. If n ≥ 3 then B n (K) is not residually nilpotent. Further, for all i ≥ 3, Γ 2 (B n (K)) = Γ i (B n (K)) = σ -1 2 σ 1 Bn(K) .

Proof. In this proof, the relation numbers are those of Theorem 2.2. Let n ≥ 3, and consider the following short exact sequence: It remains to show that Γ 2 (B n (K)) = σ -1 2 σ 1 Bn(K) . From relations (1) and ( 2), for all i = 1, . . . , n-2, we have [σ i , σ i+1 ] = σ i σ i+1 σ -1 i σ -1 i+1 = σ -1 i+1 σ i , in particular, σ -1 2 σ 1 Bn(K) ⊂ Γ 2 (B n (K)). To prove the other inclusion, for all i = 2, . . . , n -2, note that:

1 -→ Γ 2 (B n (K)) Γ 3 (B n (K)) -→ B n (K) Γ 3 (B n (K))
[σ i , σ i+1 ] = σ -1 i+1 σ i = σ -1 i+1 σ i (σ i-1 σ i (1)
σ -1 i σ -1 i-1 ) = σ -1 i+1 (σ i-1

(2)

σ i σ i-1 )σ -1 i σ -1 i-1 = (σ i-1 σ -1 i+1 )σ i (1)
σ i-1 σ -1 i σ -1 i-1 = σ i-1 (σ i σ i+1 σ -1 i σ -1 i+1 )σ i-1 (2)
σ -1 i σ -1 i-1

= σ i-1 σ i σ i+1 (σ -1 i σ i-1 )σ -1 i+1 σ -1 i σ -1 i-1 . It follows by induction on i that [σ i , σ i+1 ] ∈ σ -1 2 σ 1 Bn(K) for all i = 1, . . . , n -2. Further, [a, σ 1 ] = aσ 1 (σ -1 2 σ 2 )a -1

(3)

σ -1 1 = a(σ 2 σ -1 1 ) -1 a -1 . (σ 2 σ -1 1 ) ∈ σ -1 2 σ 1 Bn(K) , (5.51)
and similarly, [b, σ 1 ] ∈ σ -1 2 σ 1 Bn(K) using relation (4). To see that [b, a] ∈ σ -1 2 σ 1 Bn(K) , first note that:

σ 2 1 = σ 1 b -1 σ 1 b -1 σ -1 1 bσ -1 1 b = (σ 1 b -1 [σ 1 , b -1 ]bσ -1 1 )[σ 1 , b -1 ] ∈ σ -1 2 σ 1 Bn(K)
(5.52) using relation [START_REF] Chow | On the algebraical braid group[END_REF], and:

bab -1 a -1 = ba(σ -1 1 σ 1 )b -1 (σ 1

(5)

σ -1 1 )a -1 = baσ -1 1 (a -1 σ -1 1 b -1 σ 1 a)σ -1 1 a -1 = b[a, σ -1 1 ]σ -2 1 b -1 [σ 1
, a] ∈ σ -1 2 σ 1 Bn(K) , by (5.51) and (5.52). Since the result is valid for the generators of B n (K), the result follows for an arbitrary element of Γ 2 (B n (K)) using the formula given by (5.29) and by the normality of σ -1 2 σ 1 Bn(K) . We conclude that Γ 2 (B n (K)) ⊂ σ -1 2 σ 1 Bn(K) , and hence that Γ 2 (B n (K)) = σ -1 2 σ 1 Bn(K) .

Proposition 4 . 1 .

 41 If n ≤ 4 then B n (T) is residually soluble. Proof. If n ≤ 4, the result follows by using the short exact sequence (1.1), Theorem 2.5, the solubility of S n if n ≤ 4, and the fact that P n (T) is residually soluble for all n ≥ 1 [5, Theorem 4].

5 . 8 .

 58 Let m ≥ 1, and let 1 ≤ i ≤ m. Since E i+1,m+1 ⊂ B m+1 (resp. E i+1,m+1 ⊂ E 2,m+1 ⊂ W m+1 by (5.14)), the congruence (5.20) (resp. (5.21)) is also valid modulo B m+1 (resp. modulo W m+1 ).

2 ,

 2 ) If u = (b 2 b 1 ) 2 n-1 then [x, u] = 1 because (b 2 b 1 ) 2 n-1 ∈ Z(P 2 (K)) by Proposition 5.2. (3) If u = b 2 n-1 we claim that [b 2 n-1 2

Remark 5 . 15 .

 515 Notice that H m = [G, L m-1 ] and K m = [Γ m-1 (G), H],for all m ≥ 2, since the action ϕ of G on H is defined by conjugation.

Remark 5 . 18 .

 518 For all m ≥ 1, Y m is a normal subgroup of H ϕ G. To see this, recall that Y m is normal in H by Remark 5.16, and if y ∈ Y m and g ∈ G, [g, y] ∈ Y m+1 ⊂ Y m by Lemma 5.17, and therefore gyg -1 = [g, y]y ∈ Y m . To prove the results that follow, we will make use of two commutator identities [27, Theorem 5.1]: [a, bc] = [a, b] b, [a, c] [a, c] (5.40) [a, b], c b [b, c], a c [c, a], b a = 1 (5.41) where c b = bcb -1

Lemma 5 . 21 .

 521 For all m ≥ 1, L m = Y m .

  m+1 for all j = 1, . . . , n -1, and thus D 2 m-1 j ∈ L m+1 . Using the fact that L m+1 is normal in H by Remark 5.16, Lemma 3.3 and (5.39), we conclude that Y m+1 ⊂ L m+1 . To prove that L m+1 ⊂ Y m+1 , the induction hypothesis implies that [H, L m ] = [H, Y m ] ⊂ Y m+1 . By Remark 5.15 and the induction hypothesis, we have H m+1 = [G, L m ] = [G, Y m ] and K m+1 = [Γ m (G), H]. So by Lemma 5.20 H m+1 and K m+1 are contained in Y m+1 . Since L m+1 = [H, L m ], H m+1 , K m+1 , it follows that L m+1 ⊂ Y m+1 , and hence L m+1 = Y m+1 .

x 1 ∈

 1 A n , then x 1 ∈ Y 2 , [x 2 , . . . , x i ] ∈ Γ i-1 (H), and m -i = (m -1) -(i -1) elements of {x 2 , . . . , x i } belong to A n , so [x 2 , . . . , x i ] ∈ Z m-1 = Y m-1 by induction. Hence: y = [x 1 , [x 2 , . . . , x i ]] ∈ [Y 2 , Y m-1 ] ⊂ Y m+1 . If j 1 > 1 then [x 2 , . . . , x i ] ∈ Γ i-1 (H) and m + 1 -i = m -(i -1) elements of {x 2 , . . . , x i } belong to A n . Therefore [x 2 , . . . , x i ] ∈ Z m = Y mby induction, and thus:y = [x 1 , [x 2 , . . . , x i ]] ∈ [H, Y m ] ⊂ Y m+1 .

  p * is the canonical projection. Using relations (1) and[START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF], and arguing as in[16, p. 680] or[START_REF] Bellingeri | Lower central series of Artin-Tits and surface braid groups[END_REF] Proposition 3], it follows that the Γ 3 (B n (K))-cosets of σ 1 , . . . , σ n-1 in B n (K)/Γ 3 (B n (K)) are all identified to a single element, which we denote by σ. Since n ≥ 3, it follows from relations (3) and (4) that the Γ 3 (B n (K))-cosets of a and b commute with σ. By relation (7) and (5.50), σ is of order 2, by relation (5), the Γ 3 (B n (K))-cosets of a and b commute, and from relation[START_REF] Fadell | Homotopy groups of configuration spaces and the string problem of Dirac[END_REF], the Γ 3 (B n (K))-coset of a is of order 2 in B n (K)/Γ 3 (B n (K)). By Remark 2.3, σ and the Γ 2 (B n (K))-cosets of a and b are non trivial in B n (K)/Γ 2 (B n (K)), therefore σ and the Γ 3 (B n (K))-cosets of a and b are also non trivial in B n (K)/Γ 2 (B n (K)) and satisfy the same relations in B n (K)/Γ 3 (B n (K)) as their images in B n (K)/Γ 2 (B n (K)) under p * . Hence p * is a isomorphism, and Γ2 (B n (K)) = Γ 3 (B n (K)), so Γ 2 (B n (K)) = Γ i (B n (K)) for all i ≥ 3. Since Γ 2 (B n (K)) is non trivial, we see that B n (K) is not residually nilpotent.

  Remark 3.2. With the notation of Theorem 1.1, In what follows, for the groups K n , H n or H n , we will use the word generator to mean a word of the form ϕ(g)(h).h -1 , where g ∈ Γ n-1 (G) and h ∈ H, g ∈ G and h ∈ V n-1 , or g ∈ Γ n-1 (G) and h ∈ H respectively. Similarly, a generator of the group L n (resp. V n ) will mean either a generator of K n or H n , or an element of the form [h, l], where h ∈ H and l ∈ L n-1 (resp. either a generator of H n , or an element of the form [h, v], where h ∈ H and v ∈ V n-1 ). Let n ≥ 2. With the notation of Theorem 1.1, the subgroups K n , L n and V n are normal in H for all n ≥ 2, and we have the inclusions

	y	2 ,
	which completes the proof by induction.
	Lemma 3.3.

  .32)Proof of Theorem 5.13. Let n ≥ 2. By Remark 5.14, it suffices to prove by induction on n that U n = γ 2 n (P 2 (K)), where U n is defined by(5.32).If n = 2, U 2 ⊂ γ 2 2 (P 2 (K)) because b 2 2 and (b 2 b 1 ) 2 belong to {x 2 : x ∈ P 2 (K)}, which is contained in γ 2 2 (P 2 (K)), Γ 2 (P 2 (K)) is contained in γ 2 2 (P 2 (K)),and γ 2 2 (P 2 (K)) is normal in P 2 (K). For the converse inclusion, we know that Γ 2 (P 2 (K)) ⊂ U 2 , that P 2 (K) is generated by {a 2 , a 2 a 1 , b 2 , b 2 b 1 } by Remark 5.3, and that the square of each element of this set belongs to U 2 , since by Theorem 5.4, a 2 2 ∈ Γ 2 (P 2 (K)) and (a 2 a 1 ) 2 ∈ Γ 2 (s(π 1

  and therefore γ 2 2 (P 2 (K)) = U 2 . Now assume that n ≥ 2, and suppose by induction thatU i = γ 2 i (P 2 (K)) for all 2 ≤ i ≤ n. Then U n+1 ⊂ γ 2 n+1 (P 2 (K)) since Γ n+1 (P 2 (K)) ⊂ γ 2 n+1 (P 2 (K)), Γ n+1 (P 2 (K)) and γ 2 n+1 (P 2 (K)) are normal in P 2 (K), and if x

  and that this subgroup is normal in H by Lemma 3.3. Using (5.36), observe that ϕ

  by Theorem 1.1[START_REF] Artin | Theorie der Zöpfe[END_REF], to prove that P n+1 (K) is residually nilpotent by induction on n, it will suffice to show that L m ⊂ Y m for all m ≥ 2 (we will show in fact that L m = Y m ), and then that m≥1 Y m = {1}. Remark 5.16. For all m ≥ 1, Y m is a normal subgroup of H, and Y m+1 ⊂ Y m by induction on m. Further, we claim that ϕ(g)(Y i ) ⊂ Y i for all i ≥ 1 and g ∈ G. To see this, observe that ϕ(g)(D j ) is a conjugate of D j or D -1

	j

  Then y w = zy, so by(5.40), [x, y w ] = [x, zy] = [x, z] z, [x, y] [x, y]. Since z ∈ Y q by hypothesis, we have that [x, z] ∈ [Y p , Y q ] ⊂ Y p+q by Remark 5.16. Also, [x, y] ∈ Y p because Y p is normal in H ϕ Gby Remark 5.18, and so z, [x, y] ∈ [Y q , Y p ] ⊂ Y p+q by Remark 5.16. The result then follows. Lemma 5.20. For all k, m ≥ 1, [Γ k (G), Y m ] ⊂ Y k+m . Proof. If k = 1, the result is a consequence of Lemma 5.17. Now suppose by induction that [Γ k (G), Y m ] ⊂ Y m+k for some k ≥ 1 and all m ≥ 1, and let us prove that [Γ k+1 (G), Y m ] ⊂ Y k+m+1 for all m ≥ 1. Applying (5.41) to elements g

. The following two lemmas are inspired by those of [11, Section 3]. Lemma 5.19. Let x ∈ Y p , and y, w ∈ H ϕ G. Suppose that [w, y] ∈ Y q . Then [x, y] ∈ Y p+q if and only if [x, y w ] ∈ Y p+q . Proof. Set z = [w, y].

  .42) By Lemma 5.17, [g, h] ∈ Y m+1 , and[g, h], g k ∈ [Y m+1 , Γ k (G)] ⊂ Y m+k+1 using also the induction hypothesis. Further, [h, g k ] ∈ [Y m , Γ k (G)] ⊂ Y m+k ⊂ Y k bythe induction hypothesis and Remark 5.16, and consequently [g, h], g h k ∈ Y m+k+1 by Lemma 5.19. Also, [h, g k

To prove that P n+1 (K) is residually nilpotent, it remains to show that m≥1 Y m = {1}. To do so, we define two families (Z m ) m≥1 and ( Z m ) m≥1 of subgroups of H as follows. Let Z 1 = Z 1 = H, Z 2 = V 2 = Y 2 , and if m ≥ 3, let:

n ∪ X m-1 H , where:

(5.45) X m = [x 1 , . . . , x i ] ∈ Γ i (H) : ∃ 1 ≤ j 1 < . . . < j m-i ≤ i, x j 1 , . . . , x j m-i ∈ A n for all i = 2, . . . , m for all m ≥ 3 X m = [x 1 , . . . , x i ] 2 m-i-k ∈ Γ i (H) : ∃ 1 ≤ j 1 < . . . < j k ≤ i, x j 1 , . . . , x j k ∈ A n for all 0 ≤ k ≤ m -i, and i = 2, . . . , m for all m ≥ 2.

Note that if i = m (resp. k = 0), the elements x 1 , . . . , x i of H that appear in the definition of X m (resp. of X m ) are arbitrary. If m ≥ 3 (resp. m ≥ 2), we will refer to the elements of X m ∪ {x 2 : x ∈ Z m-1 } (resp. of X m ∪ D 2 m-2 j : j = 1, . . . , n ) as generators of Z m (resp. of Z m ).

Proposition 5.22. Let m ≥ 1. Then Z m ⊂ γ 2 m/2 (H), where x denotes the least integer greater than or equal to x. In particular, m≥1 Z m = {1}.

Proof. If m ∈ {1, 2} then m/2 = 1, γ 2 1 (H) = H and thus Z m ⊂ γ 2 1 (H). So suppose by induction on m that Z i ⊂ γ 2 i/2 (H) for some m ≥ 2 and all 1 ≤ i ≤ m. Since γ 2 (m+1)/2 (H) is normal in H, by (5.37) and (5.45), it suffices to show that {D 2 m-1 j

m/2 (H) using the induction hypothesis, and thus

(m+1)/2 (H) as required, and this completes the proof of the inclusion {D 2 m-1 j : j = 1, . . . , n} ∪ X m+1 ⊂ γ 2 (m+1)/2 (H). Finally, since H is a free group of finite rank, it is residually 2-finite and m≥1 γ 2 m (H) = {1}, so m≥1 Z m = {1}. The aim now is to prove that Y m = Z m = Z m for all m ≥ 2, from which we will conclude that m≥1 Y m = {1} and that P n+1 (K) is residually nilpotent. Remark 5.23. Let m ≥ 2. Taking A = A n in (5.17), we have X m = E 2,m . Further, since D j ∈ A n for all j = 1, . . . , n, we see that

m , and Corollary 5.9 then implies that

Proof. If m = 1, the given equality holds by definition. If m = 2 then X 2 = Γ 2 (H), and (5.38) and the fact that V 2 is normal in H by Lemma 3.3. So suppose by induction that:

(5.46)

and let us prove that [x, y] ∈ Z i+j . In case [START_REF] Artin | Theorie der Zöpfe[END_REF], [START_REF] Magnus | Combinatorial group theory[END_REF]Theorem 5.2] implies that:

(5.48)

By the induction hypothesis (5.47) and the fact that k, l < m, we have:

and thus [x, y] ∈ Z i+j for all y ∈ Z j using (5.48) and (5.49). In case (2

, where k ∈ {1, . . . , n}. We consider the following cases:

Suppose first that j = 1, let x ∈ A n , and let

, where for all k = 1, . . . , s, y k is a generator of Z j , α k ∈ H and δ k ∈ {1, -1}, then applying (5.29) and induction on s, we see that [x, y] may be written as a product of conjugates of commutators of the form [x, y δ k k ]. Then [x, y] ∈ Z j+2 using the above computations, the normality of Z j+2 in H, and the fact Z l = Z l for all l ≥ 1 from the first part of the proof. (5.47), and we deduce from Remark 5.23 that [D

y] ∈ Z i+j using the above computations, (5.29), and the normality of Z i+j . This shows that [ Z i , Z j ] = [Y i , Z j ] ⊂ Z i+j as claimed, and so Y m+1 ⊂ Z m+1 as required.

Theorem 5.25. For all n, m ≥ 1, we have

Proof. If n = 1, 2 then P n (K) is residually nilpotent by Theorem 5.4 and (5.10). Suppose by induction that P n (K) is residually nilpotent for some n ≥ 2. Using the Fadell-Neuwirth split short exact sequence (5.1), the result follows by induction, and by applying Theorem 1.1, Lemmas 5.21 and 5.24.

Proof of Theorem 1.3 [START_REF] Artin | Theorie der Zöpfe[END_REF]. If n = 1 (resp. n = 2), the result is a consequence of (5.10) (resp. Theorem 5.4). If n ≥ 3, the result follows by induction on n, Proposition 5.22 and Theorem 5.25.

Proposition 5.27. If n ≥ 5 then B n (K) is not residually soluble. Further, (B n (K)) (1) = (B n (K)) (i) for all i ≥ 2.

Proof. Let n ≥ 5. Once more, the relation numbers will refer to those of Theorem 2.2. As in the case of the torus (Proposition 4.3), first consider the following short exact sequence:

where p * is the canonical projection. Using relations (1) and ( 2), for all i = 1, . . . , n -1, the σ i belong to the same (B n (K)) (2) -coset, denoted by σ, in B n (K)/(B n (K)) (2) (the hypothesis that n ≥ 5 is used here, see [15, Theorem 1.4, p. 3389]). By Remark 2.3, σ and the (B n (K) (2) -cosets of a and b are non trivial. From relations (3) and (4), the (B n (K)) (2) -cosets of a and b commute with σ. Relation [START_REF] Chow | On the algebraical braid group[END_REF] implies that σ is of order 2, and so from relation [START_REF] Bellingeri | Lower central series of Artin-Tits and surface braid groups[END_REF], the (B n (K)) (2) -coset of a commutes with that of b. By relation [START_REF] Fadell | Homotopy groups of configuration spaces and the string problem of Dirac[END_REF], the (B n (K)) (2) -coset of a is of order 2. So p * is a isomorphism, and (B n (K)) (1) = (B n (K)) (2) . The second part then follows.

Proof of Theorem 1.3 [START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF]. By Proposition 5.26 (resp. Proposition 5.27), if n ≥ 3 (resp. n ≥ 5), B n (K) is not residually nilpotent (resp. not residually soluble). Conversely, using Theorem 2.5, the short exact sequence (1.1), and the fact that P 2 (K) is residually 2-finite by Theorem 5.13, B 2 (K) is residually 2-finite. In particular, B 2 (K) is residually nilpotent. By (5.10), B 1 (K) = π 1 (K) is residually nilpotent. The fact that P n (K) is residually soluble for all n ≥ 1 by Theorem 5.25 implies that B n (K) is residually soluble for all n ≤ 4 using Theorem 2.5.

The case of non-orientable surfaces of higher genus

In this short section, we prove Theorem 1.4, by generalising Propositions 5.26 and 5.27 to nonorientable surfaces of higher genus. Theorem 6.1. Let M be a compact, connected non-orientable surface without boundary and of genus g ≥ 3. Then B n (M ) is not residually nilpotent if n ≥ 3, and is not residually soluble if n ≥ 5.

Proof. The relation numbers will refer to those of Theorem 2.4. Using Theorem 2.4 it is easy to see that n , where σ denotes the Γ 2 (B n (M ))-coset of σ i for all i = 1, . . . , n-1. As in the proof of Proposition 5.26 (resp. Proposition 5.27), using relations (1) and ( 2), one may show that for all i = 1, . . . , n -1, the σ i belong to the same Γ 3 (B n (M ))-coset (resp. (B n (M )) (2) -coset), which we also denote by σ, in B n (M )/Γ 3 (B n (M )) (resp. in B n (M )/(B n (M )) (2) ). If n ≥ 3 (resp. n ≥ 5), by relation (3), the Γ 3 (B n (M ))-coset (resp. (B n (M )) (2) -coset) of a r commutes with σ for all 1 ≤ r ≤ g. By relation [START_REF] Bellingeri | Lower central series of Artin-Tits and surface braid groups[END_REF], the Γ 3 (B n (M ))-coset (resp. (B n (M )) (2) -coset) of a r commutes with that of a s for all 1 ≤ r, s ≤ g. By (6), 1) , and relation ( 4) does not give any new information. Thus

Proof of Theorem 1.4. If M = K, the result follows from Theorem 1.3 [START_REF] Bardakov | On residual properties of pure braid groups of closed surfaces[END_REF], and if M is a compact surface without boundary of genus g ≥ 3, the conclusion follows from [START_REF] Bellingeri | On p-almost direct products and residual properties of pure braid groups of nonorientable surfaces[END_REF] and from Theorem 6.1. If M = RP 2 , by [START_REF] Gonçalves | The lower central and derived series of the braid groups of the projective plane[END_REF], B n (RP 2 ) is residually nilpotent if n ≤ 2, and if n = 4, B n (RP 2 ) is residually soluble if n < 4. The result in the case n = 4 may be obtained by using Theorem 2.5 and the following Fadell-Neuwirth short exact sequence:

where n ≥ 2. Note that if n = 2, P 2 (RP 2 ) is the quaternion group of order 8 [START_REF] Van Buskirk | Braid groups of compact 2-manifolds with elements of finite order[END_REF], which is 2-finite, and P m (RP 2 \ {x 1 , . . . , x n }) is residually 2-finite by [START_REF] Bellingeri | On p-almost direct products and residual properties of pure braid groups of nonorientable surfaces[END_REF]. Therefore P m+2 (RP 2 ) is residually 2-finite for all m ≥ 1, in particular P 4 (RP 2 ) is residually soluble. Applying Theorem 2.5 to the short exact sequence (1.1), we see that B 4 (RP 2 ) is residually soluble.

Appendix

Let M be the Möbius band, and let n ≥ 1. The braid groups of M are those of RP 2 with a single point removed [17, proof of Theorem 2(a)], and so P n (M ) is the group Γ n,1 (RP 2 ) of [START_REF] Gonçalves | Inclusion of configuration spaces in Cartesian products, and the virtual cohomological dimension of the braid groups of S 2 and RP 2[END_REF]Proposition 11]. We use the notation and results of that proposition in what follows. In particular, P n (M ) is generated by the set {A i,j , ρ j | 1 ≤ i < j and 2 ≤ j ≤ n + 1}. may be written in the following form:

Since ρ n+1 commutes with A i,j for all 1 ≤ i < j < n + 1 [19, Proposition 11, relation (II)], it follows that ρ n+1 commutes with ∆ 2 n+1 . Now the relation

of [34, p. 83] for B n+1 (RP 2 ) also holds in B n (M ) for all i = 2, . . . , n, so

from which we conclude that ρ i commutes with ∆ 2 n+1 . Thus ∆ 2 n+1 commutes with all of the elements of a generating set of B n (M ), hence ∆ 2 n+1 ⊂ Z(B n (M )), and ∆ 2 n+1 ⊂ Z(P n (M )) since ∆ 2 n+1 ∈ P n (M ). To prove the converse, we consider the following two cases: (a) n = 2. Let p : P 2 (M ) -→ P 1 (M ) denote the surjective homomorphism given geometrically by forgetting the second string. The kernel of p is a free group of rank 2 for which (A 2,3 , ρ 3 ) is a basis, and P 1 (M ) = ρ 2 is infinite cyclic. Let Q = ρ 2 2 be the index 2 subgroup of P 1 (M ), and let G = p -1 (Q) be the index 2 subgroup of P 2 (M ). Then we have the following commutative diagram of short exact sequences:

where q : P 1 (M ) -→ Z 2 is defined by q(ρ 2 ) = 1, and q : P 2 (M ) -→ Z 2 is given by q = q • p. From (A1), we have ∆ 2 3 = A 1,2 ρ -2 3 , and since A 1,2 = ρ -1 2 A 2,3 ρ -1 2 by [19, Proposition 11, relation (V)], we see that:

using exactness of (A2). So the restriction p G : G -→ Q admits a section given by sending ρ 2 2 to ∆ -2 3 , and from this and the fact that ∆ 2 3 ∈ Z(P 2 (M )), the upper row of (A2) splits as a direct product. In particular, G is the internal direct product of Ker(p) and ∆ 2 3 , and Z(G) = ∆ 2 3 . Now P 2 (M ) (resp. G) is an index 2 subgroup of B 2 (M ) (resp. of P 2 (M )), and a transversal is given by {1, σ 2 } (resp. by {1, ρ 2 }). So every element z of B 2 (M ) may be written as z = aρ i 2 σ j 2 , where a ∈ G and i, j ∈ {0, 1} are unique. We shall refer to this expression as the normal form of z. Let z ∈ Z(B 2 (M )), and assume first that z / ∈ P 2 (M ), so j = 1. Consider the basis (u, v) of Ker(p), where u = A 2,3 ρ 3 and v = ρ 3 . By [START_REF] Gonçalves | Inclusion of configuration spaces in Cartesian products, and the virtual cohomological dimension of the braid groups of S 2 and RP 2[END_REF]Proposition 11, relations (III) and (IV)], we have:

) Hence the action by conjugation of ρ 2 on Ker(p) is given by composing the involution that exchanges u and v with conjugation by u -1 . The relation

), z and v commute, so:

If i = 1 then the left-and right-hand sides of (A5) are in normal form, and they clearly differ. If i = 0 then using the fact that ρ 2 2 = u -1 v -1 ∆ -2 3 ∈ G by (A3) and (A4), equation (A5) may be written as:

2 . ρ 2 u -1 vρ -1 2 . ρ 2 σ 2 = va∆ 2 3 u 2 ρ 2 σ 2 . Again the left-and right-hand sides are in normal form, and they differ also. In both cases, this contradicts the fact that z ∈ Z(B 2 (M )), and so we conclude that j = 0. Hence z ∈ P 2 (M ), and ∆ 2 3 ⊂ Z(B 2 (M )) ⊂ Z(P 2 (M )). It remains to show that z ∈ ∆ 2 3 . Suppose that i = 1. Since z ∈ Z(P 2 (M )), z commutes with u, so:

2 . ρ 2 = uau -1 v -1 uρ 2 by (A4). Both sides are in normal form, and thus a = uau -1 v -1 u in Ker(p), which gives rise to a contradiction under Abelianisation in this free group. Hence i = 0, and thus z ∈ G. So z ∈ Z(G), and therefore z ∈ ∆ 2 3 . We conclude that ∆ n+1 is analogous to that of the inclusion Z(P n (K)) ⊂ Z n given in the second paragraph of Proposition 5.2, where the kernel of the Fadell-Neuwirth short exact sequence involving the pure braid groups of M is a free group with trivial centre.