Torsion subgroups of quasi-abelianized braid groups

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1 Torsion subgrous of quasi-abelianized braid grous Vincent Beck, Ivan Marin To cite this version: Vincent Beck, Ivan Marin. Torsion subgrous of quasi-abelianized braid grous <hal > HAL Id: hal htts://hal.archives-ouvertes.fr/hal Submitted on 5 Se 2017 HAL is a multi-discilinary oen access archive for the deosit and dissemination of scientific research documents, whether they are ublished or not. The documents may come from teaching and research institutions in France or abroad, or from ublic or rivate research centers. L archive ouverte luridiscilinaire HAL, est destinée au déôt et à la diffusion de documents scientifiques de niveau recherche, ubliés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires ublics ou rivés. Distributed under a Creative Commons Attribution - NonCommercial 4.0 International License

2 Torsion subgrous of quasi-abelianized braid grous Vincent Beck MAPMO (UMR CNRS 7349), Université d Orléans, F-45067, Orléans, France FDP - FR CNRS 2964 vincent.beck@univ-orleans.fr Ivan Marin LAMFA (UMR CNRS 7352), Université de Picardie Jules Verne, Amiens Cedex 1 France ivan.marin@u-icardie.fr Abstract This article extends the works of Gonçalves, Guaschi, Ocamo [GGO] and Marin [MAR2] on finite subgrous of the quotients of generalized braid grous by the derived subgrou of their ure braid grou. We get exlicit criteria for subgrous of the (comlex) reflection grou to lift to subgrous of this quotient. In the secific case of the classical braid grou, this enables us to describe all its finite subgrous : we show that every odd-order finite grou can be embedded in it, when the number of strands goes to infinity. We also determine a comlete list of the irreducible reflection grous for which this quotient is a Bieberbach grou. 1 Introduction Let V be a finite dimensional comlex vector sace and W GL(V ) be an irreducible comlex reflection grou. We define A to be the set of reflection hyerlanes for W and V reg = V \ H A H. Let x 0 V reg. We consider P = π 1 (V reg, x 0 ) the ure braid grou of W and B = π 1 (V reg /W, x 0 ) its braid grou where x 0 is the image of x 0 in V reg /W. We have the following defining exact sequence (see [BMR] for more on this subject) giving rise to the abelian extension of W 1 P B W 1 0 P ab i B/[P, P ] W 1 This last extension whose order has been studied in [B4] will lay a key role in the sequel. Let us say that B/[P, P ] is the relative abelianization of the braid grou of W. Let us also remind that as a W -module P ab is the ermutation module ZA (see [O-T, Section 6.1]). We get a very concrete criterion for a subgrou of W to embed in B/[P, P ] in terms of the stabilizers of the hyerlanes of W (Theorem 2); for the case of the symmetric grou S n (and more generally for a wide class of comlex reflection grous including the real ones) and the standard braid grou B n, we are able to describe all finite subgrous of B/[P, P ] (see Corollary 13 and Proosition 17). In articular we show that every odd-order subgrou of S n can be embedded in B n /[P n, P n ] and that every odd-order grou can be embedded in B /[P, P ] where B (res. P ) denotes the infinite (ure) braid grou. We are also able to give a comlete list of the irreducible reflection grous such that B/[P, P ] is a Bieberbach grou (Corollary 12). Let us start with the notations we will use in this article. The grou W acts on A : if H is the hyerlane of the reflection s W, then wh is the hyerlane of the reflection wsw 1. Moreover wh = H if and only if wsw 1 = s. For H A, we denote by N H = {w W, wh = H} = {w W, sw = ws} the stabilizer of H which is also the centralizer of any reflection of W with hyerlane H. For a subgrou G of W and H A, the stabilizer of H under the action of G is N H G. We denote it by N H,G. Since W is generated by reflections, we then deduce wh = H for every H A if and only w Z(W ) where Z(W ) is the centre of W. In articular, the grou W = W/Z(W ) acts faithfully on A. All along this article, we will consider subgrous G of W and subgrous G of W. For a subgrou G of W, we denote by G its direct image in W. Finally since W is finite, we may assume that, is an hermitian roduct invariant under W. For H A, we set C H = {w W, x H, w(x) = x} the arabolic subgrou of W associated to H. Let us end this introduction with a cohomological lemma that we will need later. Lemma 1 Let G be a finite grou, X a set on which G acts and ZX the corresonding ermutation module. Then H 1 (G,ZX) is trivial. Proof. Decomosing X into G-orbits X = A X/G A, we get H 1 (G,ZX) = A X/G H1 (G,ZA). Choosing for each A X/G a reresentative a and using Shairo s isomorhism ([BR, Proosition III.6.2]) we obtain that H 1 (G,ZX) = A X/G H1 (G a,z) where G a is the stabilizer of a. But G a acts trivially on Z and then H 1 (G a,z) = Hom gr. (G a,z) = {0} since G a G is finite.

3 2 Finite subgrous of B/[P, P] 2.1 A criterion for finite subgrou In this subsection, we give a criterion for a subgrou of W to lift in B/[P, P ]. The criterion (Theorem 2) has many consequences such that the descrition of the finite subgrous of B n /[P n, P n ] and B /[B, B ] which are exlored in the following subsections. Theorem 2 Let G be a subgrou of W then the extension 1 P ab i 1 (G) G 1 slits if and only if for every H A we have N H G C H. When these conditions are fulfilled, G identifies to a finite subgrou of B/[P, P ] and G is isomorhic to G. Moreover 1 (G) is a crystallograhic grou. Let G be a torsion subgrou of B/[P, P ] then G is finite and is isomorhic to its image in W and the extension 1 P ab i 1 (( G)) ( G) 1 slits. Moreover every subgrou of B/[P, P ] isomorhic through to ( G) is conjugated to G by an element of P ab and the intersection of the normalizer of G in B/[P, P ] with P ab is isomorhic to ZA /G. Proof. Let us consider the extension 1 P ab i 1 (G) This is the image of the extension 1 P ab i B/[P, P ] G 1 as an element in H 2 (G, P ab ). Res: H 2 (W, P ab ) H 2 (G, P ab ). Following the roof of Proosition 5 of [B4], we get the following commutative diagram H 2 (W, P ab ) [ H A H2 (N H,Z) ] W W 1 through the restriction ma [ H A Hom gr(n H,C ) ] W Res H 2 (G, P ab ) Res [ H A H2 (N H G,Z) ] G Res [ H A Hom gr(n H G,C ) ] G Hence the extension 1 P ab i 1 (G) G 1 can be identified with the family of grou homomorhisms r H : N H G C given by the restriction to the line H whose kernel is by definition C H. In this case, G ZW = {1} since for every non trivial element w of the center of W and every hyerlane H A, we have r H (w) 1. So G and G are isomorhic. Finally, 1 (G) is a crystallograhic grou since G acts faithfully on A since G ZW = {1}. Let us now consider G a torsion subgrou of B/[P, P ]. For x G, let us first show that the order of x is the same that the order of (x) W. Let m be the order of x and assume that (x) n = 1 for some n dividing m and n m. Then x n P but x n / [P, P ]. But P ab is a torsion-free grou, so it is imossible. The ma : G W is then injective and G is finite and : G ( G) is an isomorhism. The P ab -conjugacy classes of subgrous of B/[P, P ] isomorhic to ( G) through are in bijection with H 1 (( G), P ab ) (see [BR, Proosition 2.3 Chater IV]). But H 1 (( G), P ab ) is trivial (Lemma 1) and we get the result. Finally, let x P ab such that x Gx 1 = G. Then for every g G, we have xgx 1 g 1 G P ab. But G is finite and P ab is torsion-free, hence xgx 1 g 1 = 1. We then deduce gxg 1 = x for all g G and so ) x = c H. C A /( G) α C ( H C Since w W has a finite order lifting in B/[P, P ] if and only if the extension 1 P ab i 1 ( w ) w 1 is slit, we get the following corollary. Corollary 3 H A. An element w W has a lifting in B/[P, P ] of finite order if and only if w N H C H for every We can then easily generalize Theorem 2.5 of [MAR2]. Corollary 4 Let w ZW \ {1}; the order of every lifting of w in B/[P, P ] is infinite. Let w W such that w has two eigenvalues one of those is 1 then the order of every lifting of w in B/[P, P ] is infinite. In articular, if w W has order 2 or is a reflection then the order of every lifting of w in B/[P, P ] is infinite.

4 Proof. Let w ZW \ {1} then w N H for every H A and w / C H. So Theorem 2 gives the result. This result follows also directly from the fact that, when W is irreducible 1 (ZW) is a free abelian grou of rank A generated by P ab and the class z of the ath t ex(2iπt/ Z(W ) )x 0 (see [MAR2]). Let us now consider w W with w having two eigenvalues 1 and λ 1. Steinberg s theorem imlies that there exists a reflection in W such that its hyerlane H contains the eigensace of w associated to 1. Then H is contained in ker(w λid) = ker(w id). Thus the restriction of w on H is λid id. If w W has order 2 then either w = id ZW \ {1} or w has two eigenvalues 1 and 1. If w W is a reflection, its eigenvalues are 1 and λ for some λ 1. Remark 5 In fact, Theorem 2.5 of [MAR2] and its roof can easily be extended to the case of w W such that w has two eigenvalues one of those is 1. So the revious corollary was already known but we give here a different roof. Lemma 6 If w W is such that a ower of w has only infinite lifting in B/[P, P ] then every lifting of w in B/[P, P ] has infinite order. This is in articular the case when w is of even order or if a ower of w is a non trivial element of ZW. Proof. If x is a lifting of w then x n is a lifting of w n. We may interret Theorem 2 as a kind of local roerty in the following sense: a subgrou G of W identifies to a subgrou of B/[P, P ] if and only if every element of G has a finite order lifting in B/[P, P ]. More recisely, we get the following corollary. Corollary 7 Let G be a subgrou of W. The extension 1 P ab i 1 (G) and only if every element of G has a finite order lifting in B/[P, P ]. G 1 is slit if Proof. Both conditions are equivalent to the following: for every H A and every w G N H, w C H. 2.2 Examles and Alications We start this subsection by describing the elements of the grous in the infinite series that can be lifted in their restricted braid grou. We then study the grous W such that B/[P, P ] is a Bieberbach grou and finally we give a list of grous W (containing the Coxeter grous) such that every odd-order element of W has a finite order lifting in B/[P, P ]. For an integer n, we denote by U n the set on n th root of unity in C. Let us consider (e 1,..., e r ) the canonical basis of C r. For w G(de, e, r), we denote by σ w S the image of w through the natural homomorhism G(de, e, r) S r. Thus, there exists a family (a i ) 1 i r U de r such that w(e i ) = a i e σw(i) for all i {1,..., r}. Let us write σ w = σ 1 σ s as a roduct of disjoint cycles. For such a cycle σ = (i 1,..., i α ), we denote by w,σ = a i1 a iα U de. Corollary 8 Finite order lifting for the infinite series. Let w G(de, e, r). If w has a finite order lifting in B/[P, P ] then either σ w id and for every cycle σ aearing in σ w, we have w,σ = 1, or σ w = id and for every i j {1,..., r} the order of a i a j 1 is a multile of the order of a i. If w is of odd order and for every cycle σ aearing in σ w, we have w,σ = 1 then w has a finite order lifting in B/[P, P ]. When d 2, then w has a finite order lifting in B/[P, P ] if and only if w is of odd order and for every cycle σ aearing in σ w, we have w,σ = 1. When d = 1, w has a finite order lifting in B/[P, P ] if and only if w is of odd order and for w for every cycle σ of σ w, we have w,σ = 1 or σ w = id and for every i j {1,..., r} the order of a i a j 1 is a multile of the order of a i. Proof. The result is clear for r = 1. Let us assume that r 2. Let us consider w W with a finite order lifting in B/[P, P ] and σ a cycle of σ w of length greater than 1. Set i j in the suort of σ and ζ U de and consider the hyerlane H i,j,ζ = {(z 1,..., z r ) C r, z i = ζz j }. For l Z, if w l N Hi,j,ζ then {σ l w (i), σ l w (j)} = {i, j}. But σ w is of odd order (Corollary 4), hence σ l w (i) = i and σ l w (j) = j. We then deduce that l is a multile of the length α of σ: we can write l = αm. Then the restriction of w l to the sace San(e i1,..., e iα ) (where σ = (i 1,..., i α )) is m w,σ id. But from Corollary 3, we get that w l C Hi,j,ζ, hence m w,σ = 1. Since we can choose m = 1, we get the result. If i is a fixed oint of σ w. We have w(e i ) = a i e i. We want to show that a i = 1 when there exists a non trivial orbit σ of σ w, we consider j in the suort of σ. If w l in N Hi,j,1 then San(e i, e j ) is stable by w l. Hence l is a multile of the length of σ. Hence w l (e j ) = e j by the receeding oint. But w l (e j e i ) = e j e i since w l C Hi,j,1 (Corollary 3). Hence w l (e i ) = e i. Let us now consider the case where σ w = id then we can write w(e i ) = a i e i for all i. For i j, we have w l N Hi,j,1 if and only if a l i = a l j. If w l N Hi,j,1 then w l (e i e j ) = (e i e j ) (since w l C Hi,j,l ). Hence a l i = a l j imlies a l i = a l j = 1. So for every i j the order of a i a 1 j is a multile of the order of a i and a j.

5 Conversely, assume that w,σ = 1 for every cycle σ of σ w and w is of odd order and let us aly the criterion of Theorem 2 to show that w has a finite order lifting in B/[P, P ]. Let 1 i r and H i = {(z 1,..., z r ) C r, z i = 0}. If w l N Hi then w l (e i ) C e i. Hence l is a multile of the length of the cycle y = (i, i 1,..., i α ) of σ w whose suort contains i. So that the restriction of w l to San(e i, e i1,..., e iα ) is the identity since w,σ = 1. So w l C Hi. For i j {1,..., r} and ζ U de, if w l N Hi,j,ζ then {σ l w (i), σ l w (j)} = {i, j}. If i and j belongs to different orbits σ and σ under σ w then necessarily σ l w (i) = i and σ l w (j) = j. Hence l is a multile of the lengths of σ and σ. But w,σ = w,σ = 1 so that w l (e i ) = e i and w l (e j ) = e j and w l C Hi,j,ζ. If i and j are in the same orbit σ under σ w then we also have σ l w (i) = i and σ l w (j) = j because if σ l w (i) = j and σ l w (j) = i then σ w would be of even order and w too, which is not. Hence l is a multile of the length of σ and since w,σ = 1, we get that w l C i,j,ζ. Let us now consider the case d 2. From the first two arts, the only thing to show is that if w verifies σ w = id and w has finite order lifting in B/[P, P ] then w = id. Let i {1,..., r}. We have w(e i ) = a i e i. Hence w N Hi where H i = {(z 1,..., z r ) C r, z i = 0}. We then deduce that w C Hi that is a i = 1. Let us consider the case d = 1. From the first two arts, the only thing to show is that if σ w = id and for i j {1,..., r}, the order of a i a 1 j is a multile of the order of a i then w has a finite order lifting in B/[P, P ]. For i j {1,..., r} and ζ U de, if w l N Hi,j,ζ then w l stabilizes three lines in the lane San(e i, e j ); namely Ce i,ce j and the line H i,j,ζ. So a i l = a j l and then a i a j 1 l = 1. Hence a i l = a j l = 1 and then w l C Hi,j,ζ. Corollary 3 gives then the result. Examle 9 Let j = 1/2 + i 3/2. The receding corollary ensures us that w = diag (j, j 2, 1..., 1) G(3, 3, r) verifies σ w = id and has a finite order lifting in B/[P, P ]. In articular, for r = 2, diag (j, j 2 ) gives an examle of an element of G(3, 3, 2) of order 3 that has a finite order lifting in B/[P, P ] but such that no element of order 3 of S 2 (since there are no such element) lifts in B/[P, P ]. More generally, if de = 2 m q with q odd, let ζ of q th root of unity. Then w = diag (ζ, ζ 1, 1,..., 1) G(de, de, r) has a finite order lifting in B/[P, P ]. Remark 10 Let G be a subgrou of G(de, e, r) where d 2 which lifts in B/[P, P ]. Then Corollary 8 shows that G D = {1} where D is the subgrou of diagonal matrices of G(de, e, r). In articular G is isomorhic to its image in S r. Moreover the criterion of Corollary 8 shows also that every odd-order subgrou of S r lifts into B/[P, P ]. So that the isomorhism classes of finite subgrous of B/[P, P ] are the same that the isomorhism classes of odd-order subgrous of S r. The receding examle shows that this is not the case when d = 1. In [MAR2], Marin shows that B/[P, P ] is a Bieberbach grou for G 4 and G 6. We are able to give the list of the irreducible grou such that B/[P, P ] is a Bieberbach grou (Corollary 12). But we start with two small grous that we can study by hands. Corollary 11 For W = G 5, the grou B/[P, P ] is a Bieberbach grou of holonomy grou A 4 and dimension 8. For W = G 7, the grou B/[P, P ] is a Bierberbach grou of holonomy grou A 4 and dimension 14. Proof. In G 5, there are elements of order 1, 2, 3, 4, 6, 12. Corollary 6 ensures us that we have only to study the case of the elements of order 3. There are 26 of them shared in 8 conjugacy classes. There are two elements in the center of G 5. There are 16 reflections. Both cases are treated by revious results. It remains to consider two conjugacy classes which are roducts of an element of order 3 of ZG 5 with a reflection and are not reflections: they have two eigenvalues which are j and j 2 (the third non trivial roots of unity). Such an element is contained in the stabilizer of a hyerlane but not in the corresonding C H. So Theorem 2 allows us to conclude. The situation is exactly the same for G 7 since the elements of G 7 have order 1, 2, 3, 4, 6, 12 and every element of order 3 of G 7 is contained in the G 5 subgrou of G 7. Corollary 12 Bieberbach grous. For W = G(de, e, r), then B/[P, P ] is a Bieberbach grou if and only if r = 1 or r = 2 and d 2 or r = 2, d = 1 and e = 2 m for some m. Among the excetionnal grous, the grou B/[P, P ] is a Bieberbach grou for W = G 4, G 5, G 6, G 7, G 10, G 11, G 14, G 15, G 18, G 19, G 25, G 26. Proof. Assume that for G(de, e, r), B/[P, P ] is a Bieberbach grou. Corollary 8 ensures us that (1, 2, 3) S r has a finite order lifting in B/[P, P ]. Hence r 2. If r = 2 and d = 1, then Examle 9 shows that e has to be a ower of 2. Conversely, if r = 1, then B = B/[P, P ] = Z. If r = 2 and d 2 then a diagonal matrix D G(de, e, 2) has a finite order lifting only if D = id (since for every i {1, 2} we have a i = D,{i} = 1). And a non diagonal matrix is of even order so it does not have a finite order lifting in B/[P, P ]. For r = 2, d = 1, e = 2 m, let D = diag (ζ, ζ 1 ) be a diagonal matrix of G(e, e, 2) with ζ U e. If ζ 1 then the order of ζ 2 is half the order of ζ (since e is a ower of 2) and by Corollary 8, D does not have a finite order lifting in B/[P, P ]. A non diagonal matrix is of even order so it does not have a finite order lifting in B/[P, P ]. For the excetionnal grous, we use the ackage [CHEVIE] of [GAP] to comute stabilizers of hyerlanes and arabolic subgrous associated to the orthogonal line.

6 We now give a corollary generalizing Corollary 2.7 of [MAR2] and Theorem 16 of [GGO]. Corollary 13 Real reflection grous. Let W be a real reflection grou. Then every odd-order element has a finite order lifting in B/[P, P ]. More generally, if W is a reflection grou for which the roots of unity inside its field of definition all have for order of ower of 2 (The irreducible grous verifying this roerty are the following G 8, G 9, G 12, G 13, G 22, G 23, G 24, G 28, G 29, G 30, G 31, G 35, G 36, G 37 for the excetional ones, and G(de, e, r) with d and e owers of 2 and r 2, S r, G(d, 1, 1) with d ower of 2 and G(e, e, 2) with e 3 for the infinite series), then every odd-order element has a finite order lifting in B/[P, P ]. For those grous, let G W then 1 (G) the inverse image of G in B/[P, P ] is a Bieberbach grou if and only if G is a 2-subgrou of W. Again, for those grous, let G W with G odd then B/[P, P ] has a finite grou isomorhic to G and 1 (G) is a crystallograhic grou. Proof. Let w be an odd-order element of W and H A. If w N H then w stabilizes H and the restriction of w to H is still of odd order. In articular the eigenvalue of w along H is of odd order. But by hyothesis, this eigenvalue is a root of unity belonging to the field of definition of W. Its order is then a ower of 2. Hence w acts trivially on H which means that w C H. If G is a 2-subgrou of W then every element in 1 (G) has infinite order thanks to Corollary 4. Let P 0 = 1 (Z(W )) gr. Z A (see [MAR2]), the extension 0 P 0 1 (G) G 1 gives the result (let us remind that G is the direct image of G in W/ZW ). With Theorem 2, we get the converse. If G is not a 2-grou, then it contains an element of odd order. The first art ensures us that this element may be lifted in 1 (G) as a finite order element. Let us now consider G W whose order is odd. Then, for every H A and every w G N H, we have w C H by the above argument since the order of w is odd. Hence Theorem 2 gives the result. Remark 14 The irreducible comlex reflection grous W such that every element of odd order has a finite order lifting in B/[P, P ] are the same as the list of Corollary 13 that is to say G 8, G 9, G 12, G 13, G 22, G 23, G 24, G 28, G 29, G 30, G 31, G 35, G 36, G 37 for the excetional ones, and G(d, 1, 1) with d a ower of 2, G(de, e, r) with d and e owers of 2 and r 2, S r and G(e, e, 2) with e 3. The receding corollary shows that all these grous verify the roerty. For the excetional ones, a [GAP] comutation using [CHEVIE] gives that there are no more excetional grou such that every element of odd order has a finite order lifting in B/[P, P ]. Let us now study the infinite series and consider G(de, e, r) such that every element of odd order has a finite order lifting in B/[P, P ]. If r = 1, since B/[P, P ] = Z then G(d, 1, 1) should have only one odd order element, so that d is a ower of 2. Let us assume that r 2 and d 2. For every ζ U de \ {1}, the element diag (ζ, ζ 1, 1,..., 1) does not have a finite order lifting in B/[P, P ] (see Corollary 8). It imlies that de has to be a ower of 2. Let us consider the case where r 3 and d = 1. For ζ U de \ {1}, the element diag (ζ, ζ, ζ 2, 1,..., 1) does not have a finite order lifting in B/[P, P ] (see Corollary 8). It imlies that de has to be a ower of 2. When r = 2 et d = 1, we get G(e, e, 2) which verify the roerty. Remark 15 Following Cayley s argument, every odd-order grou may be identified to a subgrou of S n for n large enough and thus to a subgrou of B n /[P n, P n ] thanks to Corollary 13. In the course of writing, Gonçalves, Guaschi and Ocamo have communicated to us that they indeendently got this same result. 2.3 Infinite Braid Grou Let B the infinite braid grou. This is the direct limit of the family (B n ) n N where B n is the standard braid grou on n strands and the ma B n B n+1 consists to add one strand on the right. The ure braid grou P is the direct limit of the family (P n ) n N where P n is the ure braid grou on n strands. Since [P n 1, P n 1 ] = [P n, P n ] B n, B n 1 /[P n 1, P n 1 ] may be identified to a subgrou of B n /[P n, P n ] and then also to a subgrou of the direct limit of the family (B n /[P n, P n ]) n N which is nothing else than B /[P, P ]. Moreover, with the same kind of arguments P /[P, P ] is the direct limit of the grous P n /[P n, P n ] and is a free abelian grou on the set of 2-sets of the infinite set N. Finally, the grou S may be defined as the direct limit of the family (S n ) n N where an element of S n 1 is seen as an element of S n fixing n. But S may also be viewed as the grou of ermutation of N whose suort is finite. Proosition 16 For every odd-order grou G, there exists a subgrou of B /[P, P ] isomorhic to G and B /[P, P ] contains no even order element. Proof. The grou B /[P, P ] is the direct limit of the family (B n /[P n, P n ]) n N. Moreover Proosition 2.4 of [MAR2] ensures us that the mas B n 1 /[P n 1, P n 1 ] B n /[P n, P n ] are injective for all ositive n. Then the ma B n /[P n, P n ] B /[P, P ] is injective for all n. Let us now consider G a grou of odd order. Following Cayley s argument, we let G acts on itself by left translation, so that G identifies to an odd-order grou of S G the symmetric grou on G letters. But Corollary 13 ensures us that G identifies to a subgrou of B G /[P G, P G ] which itself is a subgrou of B /[P, P ]. Let x be an even order element of B /[P, P ], then it belongs to B n /[P n, P n ] for some n. But B n /[P n, P n ] has no even order element.

7 We can be even more recise. Proosition 17 Let S be the ermutation grou of the infinite set N = {1, 2,...} with finite suort. We have the following extension 1 ab P B /[P, P ] S 1 and P ab identifies to the free abelian grou over the set P 2 of 2-subsets of N. A finite subgrou of B /[P, P ] is of odd order and mas isomorhically to S through. For every odd-order subgrou G of S, there exists a grou homomorhism s : G B /[P, P ] such that s = id G. Moreover two such grou homomorhisms are conjugated by an element of P ab and the intersection of the normalizer of a finite subgrou G of B /[P, P ] with P ab is isomorhic to ZP 2 /( G). Proof. A finite subgrou of B /[P, P ] is contained in some B n /[P n, P n ] and hence of odd order (Corollary 13). Moreover since P /[P, P ] is a free abelian grou, the surjective ma reserves the order of the finite order elements (we have already seen this in the roof of theorem 2). So a finite subgrou of B /[P, P ] mas isomorhically onto a subgrou of S through. Let G be a finite subgrou of S. Then it is a subgrou of S n for some n and hence embeds in B n /[P n, P n ] and so in B /[P, P ]. The second art follows from the fact that H 1 (G, P /[P, P ]) = {0} since P /[P, P ] is a ermutation module (Lemma 1). Finally, let x P ab. Then, as in the roof of Theorem 2, x verifies x Gx 1 = G if and only if gxg 1 = x for every g G. 2.4 Some More Examles We have seen that every odd-order subgrou ofs n can be lifted in B n /[P n, P n ]. This has been alied to show that every odd-order grou G can be embedded in B /[P, P ]. Here we describe some odd-order subgrous of finite symmetric grous verifying a stronger roerty: they do not meet any stabilizer of a hyerlane. The study of such subgrous is quite natural because they are the subgrous of W that acts freely on A. Moreover, when G is a subgrou of W such that for every H A, G N H = {1} then the criterion of Theorem 2 is clearly verified for G. But we have in fact a stronger roerty: H 2 (G,ZA ) is trivial. In order to rove this, it suffices to consider the orbits of A under G and to aly Shairo s isomorhism The case of the symmetric grou To study the case of the symmetric grou, let us introduce the following notation. For an integer n, we denote by F n the set F n = {σ S n, σ is of cycle tye k [n/k] or 1 [1] k [(n 1)/k] with k odd}. Proosition 18 Let G be a subgrou of S n. For i j, we denote by C(i, j) the centralizer of the transosition (i, j): this is also the stabilizer N Hi,j. Then, G C(i, j) = {id} for all i j if and only if G F n. Proof. We have C i,j = S n 2 (i, j). In articular, w C i,j if and only if {w(i), w(j)} = {i, j}. Let now G such that G C(i, j) = {id} for all i j. If G has an element w of order n = 2k then w k is a non trivial roduct of transositions and then belongs to C(i, j) for some (i, j). So every element of G is of odd order and every cycle of an element of G is of odd length (to see this, we could also have alied the fact the G lifts into B/[P, P ] thanks to Theorem 2). Moreover every non trivial element w of G has at most one fixed oint: if there are two or more, then let {i, j} be two such fixed oints, then w C(i, j). We then deduce that the length of every non trivial cycle of w G is the same. Indeed if c 1 and c 2 are two non trivial cycles of length k 1 < k 2 then w k1 1 and has more than one fixed oint. This gives the wanted cycle decomosition. Conversely, assume that G F n. For g F n, every ower of g is still in F n since every cycle of g will decomoses in cycles of length a divisor of an odd number. So for every i j, the set {i, j} is not stable by any non trivial element of G. Examle 19 Grou of odd order. Let G be a grou of odd order. When G acts on itself by left multilication, then the cycle decomosition of an element g G S G is a roduct of [G : g ] cycles of length the order of g which is odd. And so G F G which a bit more recise that what was needed in Proosition 17. Let us know study a bit Frobenius grous who will rovide some more examles. Let G be a Frobenius grou with kernel N and H a Frobenius comlement. The grou G is a disjoint union G = N (ghg 1 \ 1). Lemma 20 g G/H The grou G acts on G/H with the following roerties. (i) An element n N \ {1} has no fixed oint on G/H. (ii) For g G, an element of ghg 1 \ 1 has only one fixed oint on G/H namely gh.

8 In articular, G acts faithfully on G/H. Proof. (i) If ngh = gh then g 1 ng H and n ghg 1 which is absurd thanks to the set artition of G. (ii) If ghg 1 g H = g H (with h H \ {1}) then g 1 ghg 1 g H and h H g 1 g Hg 1 g. Hence, the set artition of G ensures us that g 1 g H. Corollary 21 For every g G, the cycle decomosition of g G seen as a ermutation of G/H is the following (i) if g N, then g is a roduct of [G : H]/k cycles of length k where k is the order of g. (ii) if g g Hg 1 \ {1} then g is a roduct of ([G : H] 1)/k cycles of length k where k is the order of g. Proof. It is trivial when g = 1. Let us consider g 1. (i) For g H G/H and 1 l < k, we have g l g H g H since g l N \ {1}. Hence the orbit of g H under g has k elements. (ii) Let 1 l < k and g H G/H with g H g H. We have g l g H g H since g l g Hg 1 \ {1} and g H g H. Hence the orbit of g H under g has k elements. Corollary 22 Let K be a subgrou of a Frobenius grou of odd order, then K is contained in F (G:H). This last corollary associated to Corollary 13 generalizes Corollary 3.10 and 3.11 of [MAR2] and Theorem 7 of [GGO] The infinite series Let us introduce the following subset of G(de, e, r) F(de, e, r) = {w G(de, e, r), the cycle tye of σ w is [k] r/k with k odd and w,σ = 1 for every σ cycle of σ w } where σ w is ermutation associated to w. Proosition 23 Let us consider G a subgrou of G(de, e, r) with d 2 and A be the set of hyerlanes of G(de, e, r). Then G N H = {1} for every H A if and only if G F(de, e, r). Proof. Assume that G N H = {1} for every H A then from Theorem 2 we deduce that every element of G is of odd order and so for every w G and every σ cycle of σ w is of odd order and from Corollary 8 that w,σ = 1 for every σ cycle of σ w and every w G. Moreover consider w G such that σ w has at least two cycles of different lengths. Suose that these lengths are l 1 < l 2. Then w l1 G \ {1} and stabilizes every hyerlane of the form {z i = 0} where i belongs to the orbit of length l 1. Conversely, assume that G F(de, e, r). If w G stabilizes the hyerlane {z i = 0} then σ w (i) = i and hence σ w = id since every cycle of σ w has the same length. Since w,σ = 1 for every σ cycle of σ w = 1, we get that w = 1. If w G stabilizes H i,j,ζ for i j and ζ U de then {σ w (i), σ w (j)} = {i, j}. Since every cycle of σ w is of odd length then σ w (i) = i and σ w (j) = j. Hence w stabilizes {z i = 0} so w = 1. References [B4] [BMR] V. Beck. Abelianization of subgrous of reflection grous and their braid grous; an alication to cohomology. Manuscrita Math., 136: , M. Broue, G. Malle, et R. Rouquier. Comlex reflection grous, braid grous, Hecke algebras. J. Reine und Angew. Math., 500: , [BR] K. S. Brown. Cohomology of Grous. Sringer, [GGO] D.L. Gonçalves, J. Guaschi, et O. Ocamo. Quotients of the Artin braid grous and crystallograhic grous. J. Algebra, 474: , [CHEVIE] M. Geck, G. Hiss, F. Lübeck, G. Malle, et G. Pfeiffer. CHEVIE A system for comuting and rocessing generic character tables for finite grous of Lie tye, Weyl grous and Hecke algebras. Al. Algebra Engrg. Comm. Comut., 7: , [MAR2] I. Marin. Crystallograhic grous and flat manifolds from comlex reflection grous. Geometriae Dedicata, 182: , [O-T] [GAP] P. Orlik et H. Terao. Arrangements of Hyerlanes, volume 300 of Grundlehren der mathematichen wissenshaften. Sringer-Verlag, Martin Schönert et others. GAP Grous, Algorithms, and Programming version 3 release 4 atchlevel 4. Lehrstuhl D für Mathematik, Rheinisch Westfälische Technische Hochschule, Aachen, Germany, 1997.