A quotient of the Artin braid groups related to crystallographic groups

Journal of Algebra 474 (2017) 393–423

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Journal of Algebra www.elsevier.com/locate/jalgebra

A quotient of the Artin braid groups related to crystallographic groups Daciberg Lima Gonçalves a , John Guaschi b,∗ , Oscar Ocampo c a

Departamento de Matemática, IME-USP, Caixa Postal 66281, Ag. Cidade de São Paulo, CEP: 05314-970, São Paulo, SP, Brazil b Normandie Univ., UNICAEN, CNRS, Laboratoire de Mathématiques Nicolas Oresme UMR CNRS 6139, 14000 Caen, France c Departamento de Matemática, Instituto de Matemática, Universidade Federal da Bahia, CEP: 40170-110, Salvador, BA, Brazil

a r t i c l e

i n f o

Article history: Received 25 August 2016 Available online 11 November 2016 Communicated by Michel Broué Keywords: Braid groups Crystallographic groups

a b s t r a c t Let n ≥ 3. In this paper, we study the quotient group Bn /[Pn , Pn ] of the Artin braid group Bn by the commutator subgroup of its pure Artin braid group Pn . We show that Bn /[Pn , Pn ] is a crystallographic group, and in the case n = 3, we analyse explicitly some of its subgroups. We also prove that Bn /[Pn , Pn ] possesses torsion, and we show that there is a one-to-one correspondence between the conjugacy classes of the finite-order elements of Bn /[Pn , Pn ] with the conjugacy classes of the elements of odd order of the symmetric group Sn , and that the isomorphism class of any Abelian subgroup of odd order of Sn is realised by a subgroup of Bn /[Pn , Pn ]. Finally, we discuss the realisation of non-Abelian subgroups of Sn of odd order as subgroups of Bn /[Pn , Pn ], and we show that the Frobenius group of order 21, which is the smallest non-Abelian group of odd order, embeds in Bn /[Pn , Pn ] for all n ≥ 7. © 2016 Elsevier Inc. All rights reserved.

* Corresponding author. E-mail addresses: [email protected] (D.L. Gonçalves), [email protected] (J. Guaschi), [email protected] (O. Ocampo). http://dx.doi.org/10.1016/j.jalgebra.2016.11.003 0021-8693/© 2016 Elsevier Inc. All rights reserved.

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1. Introduction Let n ∈ N. Quotients of the Artin braid group Bn have been studied in various contexts, and may be used to study properties of Bn itself. It is well known that one such  Bn quotient is the symmetric group Sn , which may be expressed in the form Bn / σ12 , B where σ1 , . . . , σn−1 are the standard generators of Bn (see Section 3), and X n denotes the normal closure subgroup of a subset X of Bn . Similar quotients of the form B Bn / σ1m  n , where m ∈ N, were analysed by Coxeter in [5], who showed that this quotient is finite if and only if (n, m) ∈ {(3, 3), (3, 4), (3, 5), (4, 3), (5, 3)}, and computed the quotient groups in each case, and by Marin in [15] in the case (n, m) = (5, 3) with the aim of studying cubic Hecke algebras. The Brunnian braid groups Brunn have been studied in connection with homotopy groups of the 2-sphere S2 [1,13,19] by considering quotients of Bn . For example, for all n ≥ 3, there exists a subgroup Gn of Brunn that is normal in the Artin pure braid group Pn such that the centre of Pn /Gn is isomorphic to the direct product πn (S2 ) × Z (see [13, Theorem 1] and [19, Theorem 4.3.4]). In this paper, we study the quotient Bn /[Pn , Pn ] of Bn for n ≥ 3, where [Pn , Pn ] is the commutator subgroup of Pn . Our initial motivation emanates from the observation that B3 /[P3 , P3 ] is isomorphic to B3 / Brun3 (see [19, Corollary 2.1.4] as well as [19, Section 5.2] for other results about B3 / Brun3 , and [13, Proposition 3.9] and [19, Proposition 4.3.10(1)] for a presentation of B3 / Brun3 ). The quotient Bn /[Pn , Pn ] belongs to a family of groups known as enhanced symmetric groups (see [16, page 201]) and analysed in [21]. It also arises in the study of pseudo-symmetric braided categories by Panaite and Staic. They consider the quotient, denoted by P Sn , of Bn by the normal subgroup −1 generated by the relations σi σi+1 σi = σi+1 σi−1 σi+1 for i = 1, 2, . . . , n − 2, and they show that it is isomorphic to Bn /[Pn , Pn ] [20]. The results that we obtain in this paper for Bn /[Pn , Pn ] are different in nature to those of [20], with the exception of some basic properties. Crystallographic groups play an important rôle in the study of the groups of isometries of Euclidean spaces (see Section 2 for precise definitions, as well as [3,6,22] for more details). As we shall prove in Proposition 1, another reason for studying the quotient Bn /[Pn , Pn ] is the fact that it is a crystallographic group: Proposition 1. Let n ≥ 2. There is a short exact sequence: σ

1 −→ Zn(n−1)/2 −→ Bn /[Pn , Pn ] −→ Sn −→ 1, and the middle group Bn /[Pn , Pn ] is a crystallographic group. The aim of this paper is to analyse Bn /[Pn , Pn ] in more detail, notably its torsion, the conjugacy classes of its finite-order elements, and the realisation of abstract finite groups as subgroups of Bn /[Pn , Pn ]. Since B1 is trivial and B2 is isomorphic to Z, in what follows we shall suppose that n ≥ 3. In Section 2, we recall the basic definitions and some results about crystallographic and Bieberbach groups. In Section 3, we recall some

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standard information about Bn and Pn , and using the fact that the quotient Bn /Pn is isomorphic to Sn , we shall see that Bn /[Pn , Pn ] is an extension of the free Abelian group Pn /[Pn , Pn ] by Sn , and we shall compute the associated action, which will enable us to prove it is crystallographic. By analysing the action in more detail, we prove that the torsion of Bn /[Pn , Pn ] is odd. Theorem 2. If n ≥ 3 then the quotient group Bn /[Pn , Pn ] has no finite-order element of even order. By restricting the short exact sequence involving Bn /[Pn , Pn ], Pn /[Pn , Pn ] and Sn to 2-subgroups of the latter (see equation (8)), we are able to construct Bieberbach groups of dimension n(n − 1)/2 (which is the rank of Pn /[Pn , Pn ]), and show that there exist flat manifolds of the same dimension whose holonomy group is the given 2-subgroup (see Theorem 20). We also define the normal form of an element x of Bn /[Pn , Pn ], which will be the product of a (unique) element of Pn /[Pn , Pn ] with the positive permutation braid associated to x. Such a normal form will enable us to simplify a number of proofs that follow. In Section 4, we analyse the torsion of Bn /[Pn , Pn ] in more detail. In order to do so, we shall make use of the induced action of certain elements α0,r of Bn /[Pn , Pn ], where 2 ≤ r ≤ n, on the basis (Ai,j )1≤i
Bnt Bm Bn1 × ··· × → . [Pn1 , Pn1 ] [Pnt , Pnt ] [Pm , Pm ]

(b) Bm /[Pm , Pm ] possesses elements of order lcm(n1 , . . . , nt ). Further, there exists such an element whose cycle type is (n1 , . . . , nt ). As we shall see, Theorem 3(a) may be proved by studying the normal form of the B B nt image under ι of an element of [Pn n,P1n ] × · · · × [Pn ,P , and part (b) will follow from nt ] t 1 1 part (a) and Proposition 21. One consequence of Theorems 2 and 3 is the characterisation of the torsion of Bn /[Pn , Pn ] as that of the odd torsion of the symmetric group Sn : Corollary 4. Let n ≥ 3. The torsion of the quotient Bn /[Pn , Pn ] is equal to the odd torsion of the symmetric group Sn . Moreover, given an element θ ∈ Sn of odd order r,

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there exists β ∈ Bn /[Pn , Pn ] of order r such that σ(β) = θ. So given any cyclic subgroup  of order r of Bn /[Pn , Pn ] such that H of Sn of odd order r, there exists a subgroup H  σ(H) = H. In Section 5, we focus on the simplest non-trivial case, that of B3 /[P3 , P3 ], and we describe the structure of the preimages of the subgroups of S3 under the induced homomorphism B3 /[P3 , P3 ] −→ S3 . In the cases where these preimages are Bieberbach groups, we describe the corresponding flat 3-manifold. We also carry out this analysis for the group B3 /[P3 , P3 ] itself, and identify it in the international tables of crystallographic groups given in [2,10], as well as for the quotient of B3 /[P3 , P3 ] by the subgroup generated by the class of the full-twist braid. In Section 6, we study the conjugacy classes of the elements and the cyclic subgroups of Bn /[Pn , Pn ]. It is straightforward to see that if any two elements of Bn /[Pn , Pn ] are conjugate then their permutations have the same cycle type, and the use of a specific product of certain δr,k enables us to prove the converse: Theorem 5. Let n ≥ 3, and let k ≥ 3 be odd. Two elements of Bn /[Pn , Pn ] of order k are conjugate if and only if their permutations have the same cycle type. Thus two finite cyclic subgroups of Bn /[Pn , Pn ] of order k are conjugate if and only if their images under σ are conjugate in Sn . Consequently, given n ≥ 3, we may determine the number of conjugacy classes of elements of odd order k in Bn /[Pn , Pn ]. We shall see in Lemma 9 that the set of isomorphism classes of the finite subgroups of Bn /[Pn , Pn ] is contained in the corresponding set of finite subgroups of Sn of odd order. One may ask whether this inclusion is strict or not. By Corollary 4, any cyclic subgroup of Sn of odd order is realised as a subgroup of Bn /[Pn , Pn ]. Applying a result of [11] enables us to extend this result to the Abelian subgroups of Sn . Theorem 6. Let n ≥ 3. Then there is a one-to-one correspondence between the isomorphism classes of the finite Abelian subgroups of Bn /[Pn , Pn ] and the isomorphism classes of the Abelian subgroups of Sn of odd order. In Section 7, we turn our attention to what is probably a more difficult open problem, namely the realisation of finite non-Abelian groups of Sn as subgroups of the group Bn /[Pn , Pn ]. As a initial experiment, we consider the smallest value of n, n = 7, for which Sn possesses a non-Abelian subgroup of odd order. This subgroup is isomorphic to the Frobenius group of order 21 that we denote by F. We show that F is indeed realised as a subgroup of B7 /[P7 , P7 ]. Theorem 7. The quotient group B7 /[P7 , P7 ] possesses a subgroup isomorphic to the Frobenius group F.

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It then follows from Theorem 3 that F is realised as a subgroup of Bn /[Pn , Pn ] for all n ≥ 7. In Proposition 32, we prove that B7 /[P7 , P7 ] admits a single conjugacy class of subgroups isomorphic to F. We remark that we do not currently know of an example of a subgroup of odd order of Sn whose isomorphism class is not represented by a subgroup of Bn /[Pn , Pn ]. Note added in proof: using different techniques, I. Marin has since generalised the results of this paper to generalised braid groups associated to arbitrary complex reflection groups [17]. 2. Crystallographic and Bieberbach groups In this section, we recall briefly the definition of crystallographic and Bieberbach groups, and we characterise crystallographic groups in terms of a representation that arises from certain group extensions whose kernel is a free Abelian group of finite rank and whose quotient is finite. We also review some results concerning Bieberbach groups and the fundamental groups of flat Riemannian manifolds. For more details, see [3, Section I.1.1], [6, Section 2.1] or [22, Chapter 3]. From now on, we identify Aut(Zn ) with GL(n, Z). Let G be a Hausdorff topological group. A subgroup H of G is said to be a discrete subgroup if it is a discrete subset. If H is a closed subgroup of G then the coset space G/H has the quotient topology for the projection π : G → G/H, and H is said to be uniform if G/H is compact. Definition. A discrete and uniform subgroup Π of Rn  O(n, R) ⊆ Aff(Rn ) is said to be a crystallographic group of dimension n. If in addition Π is torsion free then Π is called a Bieberbach group of dimension n. Definition. Let Φ be a group. An integral representation of rank n of Φ is defined to be a homomorphism Θ : Φ → Aut(Zn ). Two such representations are said to be equivalent if their images are conjugate in Aut(Zn ). We say that Θ is a faithful representation if it is injective. The following characterisation of crystallographic groups seems to be well known to the experts in the field. Since we did not find a suitable reference, we give a short proof. Lemma 8. Let Π be a group. Then Π is a crystallographic group if and only if there exist an integer n ∈ N and a short exact sequence 0 such that: (a) Φ is finite, and

Zn

Π

ζ

Φ

1

(1)

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(b) the integral representation Θ : Φ → Aut(Zn ), induced by conjugation on Zn and defined by Θ(ϕ)(x) = πxπ −1 , where x ∈ Zn , ϕ ∈ Φ and π ∈ Π is such that ζ(π) = ϕ, is faithful. Definition. If Π is a crystallographic group, the integer n that appears in the statement of Lemma 8 is called the dimension of Π, the finite group Φ is called the holonomy group of Π, and the integral representation Θ : Φ → Aut(Zn ) is called the holonomy representation of Π. Proof of Lemma 8. Let Φ and Π be groups, and suppose that there exist n ∈ N and a short exact sequence of the form (1) such that conditions (a) and (b) hold. Assume on the contrary that Π is not crystallographic. The characterisation of [6, Theorem 2.1.4] implies that Zn is not a maximal Abelian subgroup of Π, in other words, there exists an Abelian group A for which Zn  A ⊆ Π. Let a ∈ A \ Zn . Then ζ(a) = 1 and Θ(ζ(a))(x) = axa−1 = x for all x ∈ Zn . Hence Θ(ζ(a)) = IdZn , which contradicts the hypothesis that Θ is injective. We conclude that Π is a crystallographic group of dimension n with holonomy Φ. The converse follows from the paragraph preceding [3, Definition I.6.2], since the short exact sequence (1) gives rise to an integral representation Θ : Φ → Aut(Zn ) that is faithful by [3, Proposition I.6.1]. 2 The following lemma will be very useful in what follows. Lemma 9. Let G, G be groups, and let f : G → G be a homomorphism whose kernel is torsion free. If K is a finite subgroup of G then the restriction f |K : K → f (K) of f to K is an isomorphism. In particular, with the notation of the statement of Lemma 8, if Π is a crystallographic group then the restriction ζ |K : K → ζ(K) of ζ to any finite subgroup K of Π is an isomorphism. Proof. Since Ker (f ) is torsion free, the restriction of f to the finite subgroup K is injective, which yields the first part, and the second part then follows directly. 2 Corollary 10. Let Π be a crystallographic group of dimension n and holonomy group Φ, and let H be a subgroup of Φ. Then there exists a crystallographic subgroup of Π of dimension n with holonomy group H. Proof. The result follows by considering the short exact sequence (1), and by applying Lemma 8 to the subgroup ζ −1 (H) of Π. 2 Definition. A Riemannian manifold M is called flat if it has zero curvature at every point. As a consequence of the first Bieberbach Theorem, there is a correspondence between Bieberbach groups and fundamental groups of closed flat Riemannian manifolds (see [6, Theorem 2.1.1] and the paragraph that follows it). We recall that the flat manifold

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determined by a Bieberbach group Π is orientable if and only if the integral representation Θ : Φ → GL(n, Z) satisfies Im (Θ) ⊆ SO(n, Z). This being the case, we say that Π is an orientable Bieberbach group. By [22, Corollary 3.4.6], the holonomy group of a flat manifold M is isomorphic to the group Φ. It is a natural problem to classify the finite groups that are the holonomy group of a flat manifold. The answer was given by L. Auslander and M. Kuranishi in 1957. Theorem 11 (Auslander and Kuranishi [22, Theorem 3.4.8], [3, Theorem III.5.2]). Any finite group is the holonomy group of some flat manifold. 3. Artin braid groups and crystallographic groups In this section we prove Proposition 1 and Theorem 2, namely that if n ≥ 3 then the quotient group Bn /[Pn , Pn ] of the Artin braid group Bn by the commutator subgroup [Pn , Pn ] of its pure braid subgroup Pn is crystallographic and does not have 2-torsion. As we shall see in Section 4, Bn /[Pn , Pn ] possesses (odd) torsion. We first recall some facts about the Artin braid group Bn on n strings. We refer the reader to [9] for more details. It is well known that Bn possesses a presentation with generators σ1 , . . . , σn−1 that are subject to the following relations: σi σj = σj σi for all 1 ≤ i < j ≤ n − 1 such that |i − j| ≥ 2

(2)

σi+1 σi σi+1 = σi σi+1 σi for all 1 ≤ i ≤ n − 2.

(3)

Let σ : Bn → Sn be the homomorphism defined on the given generators of Bn by σ(σi ) = (i, i + 1) for all 1 ≤ i ≤ n − 1. Just as for braids, we read permutations from left to right so that if α, β ∈ Sn then their product is defined by α · β(i) = β(α(i)) for i = 1, 2, . . . , n. The pure braid group Pn on n strings is defined to be the kernel of σ, from which we obtain the following short exact sequence: σ

1 −→ Pn −→ Bn −→ Sn −→ 1.

(4)

A generating set of Pn is given by {Ai,j }1≤i
(5)

For 1 ≤ i < j ≤ n, we also set Aj,i = Ai,j , and if Ai,j appears with exponent mi,j ∈ Z, then we let mj,i = mi,j . It follows from the presentation of Pn given in [9] that Pn /[Pn , Pn ] is isomorphic to Zn(n−1)/2 , and that a basis is given by the Ai,j , where 1 ≤ i < j ≤ n, and where by abuse of notation, the [Pn , Pn ]-coset of Ai,j will also be denoted by Ai,j . Using equation (4), we obtain the following short exact sequence: σ

1 −→ Pn /[Pn , Pn ] −→ Bn /[Pn , Pn ] −→ Sn −→ 1,

(6)

where σ : Bn /[Pn , Pn ] → Sn is the homomorphism induced by σ. This short exact sequence may also be found in [20, Proposition 3.2].

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Since B1 is the trivial group and B2 /[P2 , P2 ] ∼ = Z, we shall suppose in most of this paper that n ≥ 3. We shall be interested in the action by conjugation of Bn on Pn and on Pn /[Pn , Pn ]. Recall from [13, Lemma 3.1] (see also [18, Proposition 3.7, Chapter 3]) that for all 1 ≤ k ≤ n − 1 and for all 1 ≤ i < j ≤ n,

σk Ai,j σk−1 =

⎧ ⎪ Ai,j ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Ai,j+1 ⎪ ⎪ ⎪ ⎨A−1 A i,j

if k = i − 1, i, j − 1, j if j = k

i,j−1 Ai,j

⎪ Ai,j ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A ⎪ ⎪ i+1,j ⎪ ⎩A−1 A i,j i−1,j Ai,j

if j = k + 1 and i < k

(7)

if j = k + 1 and i = k if i = k < j − 1 if i = k + 1.

The study of this action now allows us to prove Proposition 1. Proposition 12. Let α ∈ Bn /[Pn , Pn ], and let π be the permutation induced by α−1 . Then αAi,j α−1 = Aπ(i),π(j) in Pn /[Pn , Pn ]. Proof. Let α be an element of Bn /[Pn , Pn ], and let π = σ(α−1 ) be the permutation induced by α−1 . Let t denote the length of α with respect to the generating set {σ1 , . . . , σn−1 }. We prove the proposition by induction on t. If t = 1, the result follows immediately from equation (7). So suppose that the result is valid for any braid of length t for some t ≥ 1, and let β ∈ Bn /[Pn , Pn ] be of length t + 1. Thus there exist α ∈ Bn /[Pn , Pn ] of length t, 1 ≤ k ≤ n − 1 and ε ∈ {1, −1} such that β = ασkε . If τ = σ(β −1 ), then for all 1 ≤ i < j ≤ n: βAi,j β −1 = ασkε Ai,j σk−ε α−1 = αAσ(σ−ε )(i),σ(σ−ε )(j) α−1 = Aπ(σ(σ−ε )(i)),π(σ(σ−ε )(j)) k

k

k

k

= Aσ(σ−ε α−1 )(i),σ(σ−ε α−1 )(j) = Aσ(β −1 )(i),σ(β −1 )(j) = Aτ (i),τ (j), k

k

as required. 2 Proof of Proposition 1. Suppose first that n = 2. Since B2 = Z and [P2 , P2 ] = 1, we obtain B2 /[P2 , P2 ] = B2 ∼ = Z, and thus the group B2 /[P2 , P2 ] is crystallographic. So assume that n ≥ 3, and consider the short exact sequence (6) and the induced action ϕ : Sn → Aut(Zn(n−1)/2 ). From Proposition 12, ϕ(θ) is the identity automorphism if and only if θ is the trivial permutation. It follows that ϕ : Sn → Aut(Zn(n−1)/2 ) is injective, hence the group Bn /[Pn , Pn ] is crystallographic. 2 Using Proposition 1 and Corollary 10, we may produce other crystallographic groups as follows. Let H be a subgroup of Sn , and consider the following short exact sequence: 1 −→

Pn σ  n −→ −→ H H −→ 1 [Pn , Pn ]

(8)

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 n is defined by: induced by that of equation (6), where H −1  n = σ (H) . H [Pn , Pn ]

(9)

The following corollary is then a consequence of Corollary 10 and Proposition 1.  n defined by Corollary 13. Let n ≥ 3, and let H be a subgroup of Sn . Then the group H equation (9) is a crystallographic group of dimension n(n −1)/2 with holonomy group H. We now recall the definition of the crossing number of strings of a braid in Bn from [18, p. 10], but taking into account the fact that we use the opposite sign convention for crossings to that given in [18]. Let β ∈ Bn , and for 1 ≤ k ≤ n, let dk denote its kth string. For 1 ≤ i, j ≤ n, i = j, let p(di , dj ) (resp. n(di , dj )) denote the number of times di passes over dj starting from the left (resp. from the right). Then the number p(di , dj ) − n(di , dj ) is called crossing number of β from di to dj and shall be denoted by cr(β | i, j). The notion of crossing number extends in an obvious way to the quotient Bn /[Pn , Pn ]. Remarks 14. We highlight the following immediate properties of the crossing number. Let 1 ≤ i, j ≤ n, i = j. (a) Let 1 ≤ k ≤ n −1. Then cr(σk | k, k +1) = 1, and cr(σk | i, j) = 0 if (i, j) = (k, k +1). (b) Let 1 ≤ s < t ≤ n. Then cr(Ai,j | s, t) = 1 if {s, t} = {i, j}, and cr(Ai,j | s, t) = 0 if {s, t} = {i, j}. (c) Let α, β ∈ Bn , and let τ = σ(α). Then: cr(αβ | i, j) = cr(α | i, j) + cr(β | τ (i), τ (j)). In particular, if α, β ∈ Pn then: cr(αβ | i, j) = cr(α | i, j) + cr(β | i, j). (d) Let α ∈ [Pn , Pn ]. Then by property (c), we have: cr(α | i, j) = 0.

mi,j Proposition 15. Let x ∈ Pn , and let σ(x) = ∈ Pn /[Pn , Pn ], where 1≤i
m sth and tth strings of 1≤i
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cr

m Ai,ji,j s, t

 =

1≤i




cr



m Ai,ji,j s, t

 2

= ms,t .

1≤i
Remark 16. Let α ∈ Bn /[Pn , Pn ], let π be the permutation of α−1 , and let

mi,j ∈ Pn /[Pn , Pn ]. Then by Proposition 12, we have: 1≤i
⎞



m Ai,ji,j ⎠ α−1



=

1≤i
1≤i


=



m

i,j Aπ(i),π(j) =

m

Ai,jπ

−1 (i),π −1 (j)

1≤i
(10)

1≤i
Following [7], we recall the definition of positive braids and positive permutation braids. A positive braid is an element of Bn that may be written as a product of positive powers of the generators {σ1 , . . . , σn−1 } (without using the inverse elements  −1  −1 σ1 , . . . , σn−1 ). A positive braid is called a positive permutation braid, or simply a permutation braid, if it may be represented by a geometric braid in which every pair of strings cross at most once. Let Sn+ denote the set of permutation braids. By [7, Lemma 2.3], there is a bijection between Sn+ and the set Sn of permutations. Further, the set Sn+ is a set of Pn -coset representatives in Bn . For every θ ∈ Sn , let δθ ∈ Sn+ be the permutation braid for which σ(δθ ) = θ. By abuse of notation, we will also denote the projection of δθ in Bn /[Pn , Pn ] by δθ . Thus every element x ∈ Bn /[Pn , Pn ] may be written uniquely in the following form: ⎛ x=⎝



⎞ Ai,ji,j ⎠ δσ(x) . m

(11)

1≤i


 mi,j We shall refer to (11) as the normal form of x. If x = A δσ(x) and y = 1≤i
 ni,j δσ(y) belong to Bn /[Pn , Pn ] then using Proposition 12, the product of 1≤i


1≤i
⎞⎛ Ai,ji,j ⎠ ⎝



m

1≤i
⎞⎛ i,j ⎠⎝ A(σ(x)) −1 (i),(σ(x))−1 (j)



n

⎞ i,j ⎠ Ai,j δσ(x)σ(y) ,

t

1≤i
where the ti,j are obtained using Proposition 15 as the crossing numbers of the strings −1 of the pure braid δσ(x) δσ(y) δσ(x)σ(y) . We now prove Theorem 2, which says that the quotient groups Bn /[Pn , Pn ] do not have 2-torsion. Proof of Theorem 2. Let n ≥ 3. Suppose on the contrary that there exists β ∈ Bn whose [Pn , Pn ]-coset, which we also denote by β, is of even order in Bn /[Pn , Pn ]. By taking

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a power of β if necessary, we may suppose that β is of order 2 in Bn /[Pn , Pn ]. Since Pn /[Pn , Pn ] is torsion free, it follows that β ∈ Bn \ Pn , and hence σ(β) is conjugate to (1, 2)(3, 4) · · · (k, k + 1) in Sn , where 1 ≤ k ≤ n − 1 and k is odd. Conjugating β by an element of Bn if necessary, we may suppose that σ(β) = (1, 2)(3, 4) · · · (k, k + 1), and thus β 2 ∈ [Pn , Pn ]. Let  θ = σ(β). Then δθ = σ1 σ3 · · · σk−2 σk , and by Proposi

ni,j tion 15, β = A δθ in Bn /[Pn , Pn ]. Thus δθ δθ = A1,2 A3,4 · · · Ak,k+1 , and 1≤i
1≤i
⎞⎛ Ai,ji,j ⎠ ⎝



n

⎞ n Ai,jθ(i),θ(j) ⎠ (A1,2 A3,4

· · · Ak,k+1 ).

1≤i
By Proposition 15, the coefficient of A1,2 in this expression is equal to cr(β 2 | 1, 2) = 2n1,2 + 1, but this contradicts the fact that β 2 is trivial in Pn /[Pn , Pn ]. The result then follows. 2 Remarks 17. Let n ≥ 3. (a) Theorem 2 generalises [20, Proposition 3.6], where it is shown that there is no element Bn /[Pn , Pn ] of order two whose image under σ is the transposition (1, 2). (b) Theorem 2 implies that any finite-order subgroup of Bn /[Pn , Pn ] is of odd order. (c) Applying Proposition 1, Theorem 2 and Lemma 9 to the short exact sequence (6), the restriction σ |K : K → σ(K) of σ to any finite subgroup K of Bn /[Pn , Pn ] is an isomorphism. In particular, the set of isomorphism classes of the finite subgroups of Bn /[Pn , Pn ] is contained in the set of isomorphism classes of the odd-order subgroups of Sn . As we shall now see, by choosing H appropriately, we may use Corollary 13 to construct Bieberbach groups of dimension n(n − 1)/2 in Bn /[Pn , Pn ]. In Theorem 20, we will give a statement for Bn /[Pn , Pn ] analogous to that of Theorem 11 in the case that the holonomy group is a finite 2-group.  n given by Lemma 18. Let n ≥ 3, and let H be a 2-subgroup of Sn . Then the group H equation (9) is a Bieberbach group of dimension n(n − 1)/2. Proof. Let n ≥ 3, and let H be a 2-subgroup of Sn . Consider the short exact sequences (6)  n is a crystallographic group of dimension n(n − 1)/2 with and (8). By Corollary 13, H holonomy group H. Since the kernel is torsion free, σ respects the order of the torsion  n [8, Lemma 13]. In particular, the fact that H is a 2-group implies that elements of H  n is a positive power of 2. On the other the order of any non-trivial torsion element of H hand, by Theorem 2, the group Bn /[Pn , Pn ] has no such torsion elements, so the same is  n . It follows that H  n is torsion free, hence it is a Bieberbach group of dimension true for H n(n − 1)/2 because Pn /[Pn , Pn ] ∼ = Zn(n−1)/2 . 2

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Remark 19. It is not clear to us whether the family of groups that satisfy the conclusions of Corollary 13 (resp. of Lemma 18) contains all of the isomorphism classes of crystallographic (resp. Bieberbach) subgroups of Bn /[Pn , Pn ] of dimension n(n − 1)/2. Lemma 18 enables us to give an alternative proof of Theorem 11 in the case that the finite group in question is a 2-group, and to estimate the dimension of the resulting flat manifold. Theorem 20. Let H be a finite 2-group. Then H is the holonomy group of some flat manifold M . Further, the dimension of M may be chosen to be n(n − 1)/2, where n is an integer for which H embeds in the symmetric group Sn , and the fundamental group of M is isomorphic to a subgroup of Bn /[Pn , Pn ]. Proof. Let H be a finite 2-group. Cayley’s Theorem implies that there exists an integer  n is a Bieberbach n ≥ 3 such that H is isomorphic to a subgroup of Sn . From Lemma 18, H group of dimension n(n −1)/2 with holonomy group H and is a subgroup of Bn /[Pn , Pn ]. By the first Bieberbach Theorem, there exists a flat manifold M of dimension n(n − 1)/2  n (see [6, Theorem 2.1.1] and the paragraph with holonomy group H such that π1 (M ) = H that follows it). 2 For a given finite group H, it is natural to ask what is the minimal dimension of a flat manifold whose holonomy group is H. Theorem 20 provides an upper bound for this minimal dimension when H is a 2-group. This upper bound is not sharp in general, for example if H = Z2 . 4. The torsion of the group Bn /[Pn , Pn ] Let n ≥ 3. In this section, we study the torsion elements of the group Bn /[Pn , Pn ]. The main aim is to show that if θ ∈ Sn is of odd order r then there exists β ∈ Bn whose [Pn , Pn ]-coset projects to θ in Sn and is of order r in Bn /[Pn , Pn ] (see Corollary 4). The first result of this section shows that if n ≥ 3 is odd, then there are infinitely many elements of order n in Bn /[Pn , Pn ]. Before proceeding to the statement and the proof, we introduce some notation that will be used in the rest of the paper. If θ is a permutation in Sn , it acts on the set { {i, j} | 1 ≤ i < j ≤ n} of unordered pairs of distinct elements of {1, . . . , n}. If 1 ≤ i < j ≤ n, let Oθ (i, j) be the orbit of the pair {i, j} under this action, and let Tθ be a transversal for this action (so Tθ contains exactly one representative of each orbit). Proposition 21. If n ≥ 3 is odd then Bn /[Pn , Pn ] possesses infinitely many elements of order n. Proof. Let n ≥ 3 be odd, let α0,n = σ1 · · · σn−1 , and let θ = (1, 2, . . . , n) denote −1 the permutation associated to α0,n . The full-twist braid Δ2n of Bn may be written

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n  j−1 as (α0,n )n = (σ1 · · · σn−1 )n , or alternatively in the form j=2 A i,j . So in i=1



a 2 n Bn /[Pn , Pn ], Δn is equal to (α0,n ) = 1≤i




(Aα0,n )n =

a Ai,ji,j

1≤i
⎛

···⎝





=

···



a Ai,ji,j





=⎝



a 1−n 1−n Ai,jθ (i),θ (j)

n−1

Ai,jt=0

1≤i
Ai,j



 Ai,j

1≤i
⎞⎛ ai+t,j+t



a −1 −1 Ai,jθ (i),θ (j)

1≤i




1≤i
1≤i
⎛

⎞

ai,j ⎠ Aθn−1 (i),θ n−1 (j)

1≤i


ai,j Aθ(i),θ(j)

1≤i
1≤i






⎠⎝



⎞

Ai,j ⎠ ,

1≤i
where the indices i + t and j + t are taken modulo n (with values in {1, . . . , n}), and where we write aq,p = ap,q if p > q. If {i, j} ∈ Tθ , then: Oθ (i, j) = { {i + t, j + t} | t ∈ {0, 1, . . . , n − 1}} . Thus (A · α0,n )n is equal to the trivial element of Pn /[Pn , Pn ] if and only if: ⎛ ⎝



⎞ ap,q ⎠ + 1 = 0 for all {i, j} ∈ Tθ .

(12)

(p,q)∈Oθ (i,j)

This system of equations admits infinitely many solutions in Z. For each such solution, Aα0,n is of finite order, and its order divides n. On the other hand, since σ(Aα0,n ) = (1, n, n − 1, . . . , 2), the order of Aα0,n is at least n. We thus conclude that for any

a A = 1≤i
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B n1 [Pn1 ,Pn1 ]

× ··· ×

B nt [Pnt ,Pnt ] ,

ι induces a map ι : G →

Bm [Pm ,Pm ] defined as follows. Let B xk ∈ [Pn n,Pkn ] may be written in k k

x = (x1 , . . . , xt ) ∈ G. Then for all 1 ≤ k ≤ t, ⎛ ⎞ (k) m (k) normal form, xk = ⎝ Ai,ji,j ⎠ δσ(xk ) , where mi,j ∈ Z. We then define: 1≤i
ι(x) =

t k=1

βk δσ(xk ) , where βk =

1≤i
(k)

m

k−1  Ai+i,j , n ,j+ k−1 n l=1

l

l=1

(13)

l

and where σ(xk ) should be interpreted as the image of σ(xk ) ∈ Snk under the inclusion in the kth factor of the subgroup Sn1 × · · · × Snt of Sm . In parB ticular, if σi ∈ [Pn n,Pkn ] , where 1 ≤ k ≤ t and 1 ≤ i ≤ nk − 1 then k k ι(1, . . . , 1, σi , 1, . . . , 1) = σi+k−1 nl . By construction, ι is well defined. Further, if l=1 1 ≤ k, l ≤ t and k = l, then βk and δσ(xl ) commute (their strings lie in disjoint

t

t

t blocks), and thus ι(x) = ( k=1 βk )( k=1 δσ(xk ) ). But k=1 δσ(xk ) is a permutation braid because the only crossings in this braid are those that occur in each

t individual δσ(xk ) , and so any two strings of k=1 δσ(xk ) cross at most once (posi t

t tively). Hence ι(x) = ( k=1 βk )( k=1 δσ(xk ) ) is in normal form. If ι(x) = 1, then (k) mi,j = 0 and σ(xk ) = Id for all 1 ≤ k ≤ t and 1 ≤ i < j ≤ nk − 1, and thus xk = 1 for all 1 ≤ k ≤ t from which we deduce that ι is injective. Finally, if

t y = (y1 , . . . , yt ) ∈ G and ι(y) = k=1 γk δσ(yk ) is the analogue of (13) for ι(y),

t

t

t then ι(x). ι(y) = ( k=1 βk δσ(xk ) )( k=1 γk δσ(yk ) ) = k=1 βk δσ(xk ) γk δσ(yk ) = ι(xy) because γl δσ(yl ) ) and βk δσ(xk ) commute if k = l, and hence ι is a homomorphism. (b) If 1 ≤ k ≤ t, Proposition 21 implies that Bnk /[Pnk , Pnk ] possesses an element γk of order nk , and so γ = (γ1 , . . . , γk ) is an element of G of order lcm(n1 , . . . , nt ). By m part (a), we conclude that ι is an element of [PmB,P of the same order. The second m] part follows from the construction given in the proof of Proposition 21. 2 As a consequence, we are able to prove Corollary 4, which says that the torsion of Bn /[Pn , Pn ] is equal to the odd torsion of the symmetric group Sn , and that the map induced by σ from the set of finite cyclic subgroups of Bn /[Pn , Pn ] to the set of cyclic subgroups of Sn of odd order is surjective. Proof of Corollary 4. Let β ∈ Bn /[Pn , Pn ] be a non-trivial element of finite order r. By Theorem 2, r is odd. Lemma 9 implies that σ(β) is also of order r, so the torsion of Bn /[Pn , Pn ] is contained in the odd torsion of the symmetric group Sn . Conversely, suppose that θ is an element of Sn of odd order r ≥ 3, and let θ = θ1 θ2 · · · θt be a product of disjoint non-trivial cycles, where θi is an ni -cycle for all i = 1, . . . , t. Then t r = lcm(n1 , . . . , nt ), the ni are odd and greater than or equal to 3, and i=1 ni ≤ n since the θi are disjoint. By Theorem 3(b), Bn /[Pn , Pn ] possesses an element γ of order r whose permutation has cycle type (n1 , . . . , nt ). So σ(γ) is conjugate to θ, and thus a

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suitable conjugate of γ is an element of order r whose permutation is equal to θ. The last part of the statement follows in a straightforward manner. 2 Remark 22. In order to study the conjugacy classes of finite-order elements of the group Bn /[Pn , Pn ], we will describe some of these elements in more detail in Section 6. 5. A study of some crystallographic subgroups of dimension 3 of B3 /[P3 , P3 ] As we saw in Section 3, the group B3 /[P3 , P3 ] is crystallographic and has no 2-torsion. In this section, we further analyse this quotient and we study some of the crystallographic subgroups of B3 /[P3 , P3 ] of dimension 3, of the form σ −1 (H), where H is a subgroup of S3 . In order to study these subgroups, it suffices to consider a representative of each conjugacy class of subgroups of S3 . We shall also comment on some other subgroups of B3 /[P3 , P3 ].  3 be given by equation (9). Proposition 23. Let H be a subgroup of S3 , and let H  3 admits a presentation whose generators (a) Let H = {1}. The crystallographic group H are A1,2 , A1,3 , A2,3 , with defining relations [A1,2 , A1,3 ] = 1, [A1,2 , A2,3 ] = 1 and [A1,3 , A2,3 ] = 1.  3 is normal in B3 /[P3 , P3 ] and (b) Let H = (1, 3, 2). The crystallographic group H admits a presentation given by: • generators: A1,2 , A2,3 , A1,3 , α0,3 , where α0,3 = σ1 σ2 ∈ B3 /[P3 , P3 ]. • relations: (i) [A1,2 , A1,3 ] = 1, [A1,2 , A2,3 ] = 1, [A1,3 , A2,3 ] = 1. 3 (ii) α0,3 = Δ23 = A1,2 A1,3 A2,3 (Δ23 is the class of the full-twist braid in P3 /[P3 , P3 ]). −1 −1 −1 (iii) α0,3 A1,2 α0,3 = A , α0,3 A1,3 α0,3 = A1,2 , α0,3 A2,3 α0,3 = A1,3 .  2,3  3 is given by: 3 The Abelianisation H of H 

3 H

 Ab

Ab

  3 = A1,2 , α0,3 [A1,2 , α0,3 ] = 1, α0,3 = A31,2 ,

−1 and is isomorphic to Z ⊕ Z3 , where the factors are generated by A1,2 and A1,2 α0,3 .  3 admits a presentation given by: (c) Let H = (1, 2). The crystallographic group H • generators: A1,2 , A2,3 , A1,3 , σ1 . • relations: (i) [A1,2 , A1,3 ] = 1, [A1,2 , A2,3 ] = 1, [A1,3 , A2,3 ] = 1. (ii) σ12 = A1,2 . (iii) σ1 A1,2 σ1−1 = A1,2 , σ1 A1,3 σ1−1 = A2,3 , σ1 A2,3 σ1−1 = A1,3 . We have:



3 H

 Ab

  = A1,2 , A1,3 , σ1 [A1,2 , σ1 ] = 1, [A1,3 , σ1 ] = 1, σ12 = A1,2 ∼ = Z ⊕ Z.

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 3 = B3 /[P3 , P3 ] admits a presenta(d) Let H3 = S3 . The crystallographic group H tion whose generators are  σ1 , σ2 , with defining relations σ1 σ2 σ1 = σ2 σ1 σ2 and 3 (σ1−1 σ2 )3 = 1. We have H = σ1  ∼ = Z. Ab

 3 = P3 /[P3 , P3 ] if H is trivial. The presentaProof. Part (a) follows from the fact that H tions given in parts (b) and (c) may be obtained by applying the method of presentations  3 follows of group extensions given in [12, Section 10.2]. In part (b), the normality of H from that of Z3 in S3 . By [19, Lemma 4.3.9], the commutator subgroup [P3 , P3 ] is equal to the normal closure of [A1,2 , A2,3 ] in B3 . Since [A1,2 , A2,3 ] = (σ1−1 σ2 )3 and B3 = σ1 , σ2 | σ1 σ2 σ1 = σ2 σ1 σ2 , 3 we thus obtain the presentation given in part (d). In each case, H is obtained in  3. a straightforward manner from the presentation of H

2

Ab

Remark 24. The presentation of B3 /[P3 , P3 ] of Proposition 23(d) also appeared in [19, Proposition 4.3.10] and in [13, Proposition 3.9].  3 be given by equation (9). Theorem 25. Let H be a subgroup of S3 , and let H  3 is isomorphic to the quotient P3 /[P3 , P3 ], which is isomorphic (a) Let H = {1}. Then H 3 to Z . The corresponding flat manifold is the 3-torus.  3 is a Bieberbach group of dimension 3 with holon(b) Let H = (1, 2). Then H omy group Z2 . The corresponding flat Riemannian manifold is diffeomorphic to the non-orientable manifold B2 that appears in the classification of flat Riemannian 3-manifolds given in [22, Corollary 3.5.10].  3 is isomorphic to the semi-direct product Z3  Z3 , where (c) Let H = (1, 3, 2). Then H   the action is given by the matrix

0 0 1 1 0 0 0 1 0

with respect to the basis (A1,2 , A2,3 , A1,3 )

of P3 /[P3 , P3 ], this quotient being identified with Z3 . Proof. Part (a) is clear, so let us prove part (b). Theorem 2 implies that B3 /[P3 , P3 ]  3 is a Bieberbach group of dimension 3 has no 2-torsion, and so the subgroup H 3, by Lemma 18. Let X be the flat Riemannian manifold uniquely determined by H   so that π1 (X) = H3 . The holonomy representation of H3 is a homomorphism of the form Z2 −→ Aut(Z3 ), where we identify P3 /[P3 , P3 ] with Z3 . Relative to the basis (A1,2 , A1,3 , A2,3 ) of P3 /[P3 , P3 ], by Proposition 12,  theimage of the generator of Z2 by this homomorphism is given by the matrix

100 001 010

whose determinant is equal

to −1. Thus X is a non-orientable flat Riemannian 3-manifold with holonomy group Z2 . Up to affine diffeomorphism, X is one of the two manifolds B1 or B2 described  3 given in Proposition 23(b) we have in [22, Theorem 3.5.9]. Using the presentation of H 2 ∼ H1 (X; Z) = Z , and from the table in [22, Corollary 3.5.10], we conclude that X = B2 . Finally we prove part (c). The following short exact sequence:

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 3 −→ H −→ 1 1 −→ P3 /[P3 , P3 ] −→ H admits a section given by sending the generator (1, 3, 2) of H to the element σ1−1 σ2 of  3 , and so H  3 is isomorphic to the semi-direct product of the form Z3  Z3 . Relative to H the  basis  (A1,2 , A2,3 , A1,3 ) of P3 /[P3 , P3 ], the matrix of the associated action is equal to 001 100 010

.

2

Remarks 26. (a) The subgroup of B3 /[P3 , P3 ] generated by the class of the full-twist A1,2 A1,3 A2,3 , given by (1, 1, 1) in terms of the basis (A1,2 , A1,3 , A2,3 ) of P3 /[P3 , P3 ], is a normal subgroup of B3 /[P3 , P3 ]. The associated quotient group admits the following presentation that is obtained from a presentation of B3 /[P3 , P3 ]:

 σ1 , σ2 σ2 σ1 σ2 = σ1 σ2 σ1 , (σ1−1 σ2 )3 = 1, A1,2 A1,3 A2,3 = 1 .



The group G13 given in the first theorem of [14, page 73] is generated by the set {α, β, σ, ρ}, with relations: [α, β] = [ρ, α] = σ 3 = ρ2 = (σρ)2 = 1, σασ −1 = α−1 β, σβσ −1 = α−1 , ρβρ−1 = αβ −1 . −1 A routine calculation shows that the map that sends A1,2 (resp. A−1 1,3 , A1,3 σ1 , σ1 σ2 ) to α (resp. β, ρ, σ) extends to an isomorphism of the two groups. (b) The group B3 /[P3 , P3 ] is the three-dimensional crystallographic group that appears as 5/4/1:SPGR:02 of [2, page 71], and that corresponds to IT 161; OBT 1 in the international table [10]. (c) Let L be a crystallographic subgroup of B3 /[P3 , P3 ] of dimension 3, and consider the subgroup σ(L) of S3 . If σ(L) = {Id} then clearly L is isomorphic to Z3 . If σ(L) = (1, 2) then L is a Bieberbach group. If σ(L) = (1, 3, 2) then the group L may be Bieberbach or not, with holonomy Z3 . For example, if L is the subgroup   generated by σ1−1 σ2 , A21,2 , A21,3 , A22,3 then L is a proper crystallographic subgroup of σ −1 ((1, 3, 2)) of dimension 3 with holonomy Z3 and that admits torsion elements, σ1−1 σ2 for example. On the other hand, if L is the subgroup generated   by A1,2 σ1−1 σ2 , A31,2 , A31,3 , A32,3 then L is a proper subgroup of σ −1 ((1, 3, 2)), and is a Bieberbach group of dimension 3 with holonomy Z3 . To see this, let L1 = L ∩ Ker (σ) = L ∩ A1,2 , A1,3 , A2,3 . Clearly L1 is a free Abelian group, so is torsion free. Using Proposition 12, we see that (A1,2 σ1−1 σ2 )3 = A1,2 A1,3 A2,3 ,   and since (A1,2 σ1−1 σ2 )j j∈{0,1,2} is a set of coset representatives of L1 in L, it   follows that L1 is generated by A1,2 A1,3 A2,3 , A31,2 , A31,3 , A32,3 . Note then that   A1,2 A1,3 A2,3 , A31,3 , A32,3 is a basis of L1 . Suppose that w is a non-trivial torsion

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element of L. By Lemma 9, w must be of order 3. Now w ∈ / L1 , so there exist θ ∈ L1 −1 j and j ∈ {1, 2} such that w = θ(A1,2 σ1 σ2 ) . Since θ ∈ L1 , there exist λ1 , λ2 , λ3 ∈ Z 3λ3 2 such that θ = (A1,2 A1,3 A2,3 )λ1 A3λ 1,3 A2,3 , and hence: 1 = w3 = θ(A1,2 σ1−1 σ2 )j θ(A1,2 σ1−1 σ2 )−j . (A1,2 σ1−1 σ2 )2j θ(A1,2 σ1−1 σ2 )−2j . (A1,2 σ1−1 σ2 )3j . Using once more Proposition 12, the relation (A1,2 σ1−1 σ2 )3 = A1,2 A1,3 A2,3 , and comparing the coefficients of A1,2 , A1,3 and A2,3 , we obtain the equality 3(λ1 + λ2 + λ3 ) + j = 0, which has no solution in Z. It follows that L is torsion free, and so is a Bieberbach group of dimension 3 with holonomy Z3 . (d) There is no Bieberbach subgroup of B3 /[P3 , P3 ] of dimension 3 that projects to S3 , since none of the ten flat Riemannian 3-manifolds have fundamental group with holonomy S3 (see [22, Theorems 3.5.5 and 3.5.9]). 6. Conjugacy classes of finite-order elements of Bn /[Pn , Pn ] In this section, we study the conjugacy classes of finite-order elements of Bn /[Pn , Pn ]. The aim is to prove Theorem 5, which states that there is a bijection between the conjugacy classes of cyclic subgroups of odd order of Bn /[Pn , Pn ] and the set of conjugacy classes of cyclic subgroups of odd order of the symmetric group Sn . We begin with an elementary fact about conjugacy classes that will help to simplify the study of our problem. Lemma 27. Let α, β ∈ Bn /[Pn , Pn ] be two conjugate elements of finite order. Then σ(α) and σ(β) are permutations of odd order and have the same cycle type. Proof. Since α, β are of finite order, their common order is odd by Theorem 2. The fact that α and β are conjugate in Bn /[Pn , Pn ] implies that the permutations σ(α) and σ(β) are conjugate in Sn . The result then follows since two permutations are conjugate in Sn if and only if they have the same cycle type. 2 In order to analyse the conjugacy classes of elements of finite order, Lemma 27 implies that it suffices to choose a single representative permutation for each conjugacy class of Sn of odd order and to study the conjugacy classes of elements of Bn/[Pn , Pn ] of finite order that project to the chosen permutation. Let us consider the action by conjugation of certain elements of Bn/[Pn , Pn ] on the group Pn /[Pn , Pn ]. If k, n ≥ 3 and r ≥ 0 are integers such that k is odd and r + k ≤ n, define δr,k , αr,k ∈ Bn /[Pn , Pn ] by: −1 −1 δr,k = σr+k−1 · · · σr+ k+1 σr+ k−1 · · · σr+1 and αr,k = σr+1 · · · σr+k−1 . 2

2

(14)

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Lemma 28. Let n, k ≥ 3 and r ≥ 0 be integers such that k is odd and r + k ≤ n. Then δr,k is of order k in Bn /[Pn , Pn ], and satisfies:   −1 δr,k = Ar+ k+1 ,r+k Ar+ k+3 ,r+k · · · Ar+k−1,r+k αr,k . 2

(15)

2

Proof. By (14), we have: −1 −1 δr,k αr,k = (σr+k−1 · · · σr+ k+1 σr+ k−1 · · · σr+1 )(σr+1 · · · σr+ k−1 σr+ k+1 · · · σr+k−1 ) 2

2

2

2

= σr+k−1 · · · σr+ k+1 σr+ k+1 · · · σr+k−1 2

2

= Ar+ k+1 ,r+k Ar+ k+3 ,r+k · · · Ar+k−1,r+k , 2

2

which yields (15). It remains to show that δr,k ∈ Bn /[Pn , Pn ] is of order k. Usj −j ing the Artin relations (2) and (3), one may show that α0,n σi α0,n = σi+j for all i = 1, . . . , n − 2 and j = 1, . . . , n − 1 − i. It follows from this and equation (15) −r r that δ0,k = α0,n δr,k α0,n . So without loss of generality, we may suppose that r = 0. Further, the homomorphism ι : Bk /[Pk , Pk ] → Bn /[Pn , Pn ] of Theorem 3 is injective because k ≤ n, and since ι(δ0,k ) = δ0,k , we may suppose that n = k. Setting −1 −1 A = (A k+1 ,k A k+3 ,k · · · Ak−1,k )−1 ∈ Pk /[Pk , Pk ], we have α0,k δ0,k α0,k = Aα0,k by (14), 2 2 and so δ0,k and Aα0,k are of the same order. On the other hand, A satisfies the system of  k+1   k+3   equations (12), since Tθ = 2 ,k , 2 , k , · · · , {k − 1, k} is a transversal for the −1 action of θ = (1, 2, . . . , k) that is the permutation associated to α0,k . It follows from the proof of Proposition 21 that Aα0,k is of order k in Bk /[Pk , Pk ], and so δr,k is of order k in Bn /[Pn , Pn ]. 2 We now prove Theorem 6, which states that there is a one-to-one correspondence between the finite Abelian subgroups of Bn /[Pn , Pn ] and the Abelian subgroups of Sn of odd order. Proof of Theorem 6. First, it follows from Remarks 17(c) that the isomorphism class of a finite Abelian subgroup of Bn /[Pn , Pn ] is realised by a subgroup of Sn (of odd order). Conversely, let H be an Abelian subgroup of Sn of odd order. Then H is isomorphic to a direct product of the form Zk1 × · · · × Zkr , where for i = 1, . . . , r, ki is a power of an r odd prime number. By [11], i=1 ki ≤ n. Let k0 = 0. Then for l = 1, . . . , r, the element δl−1 kj ,kl belongs to Bn /[Pn , Pn ] and is of order kl by Lemma 28. By construction, the j=0   δl−1 kj ,kl commute pairwise. The subgroup δ0,k1 , . . . , δr−1 kj ,kr is then isomorphic j=1 j=1    to H since σ δ l−1 kj ,kl is a kl -cycle in Sn , and the supports of such cycles are pairwise j=1 disjoint. 2 Let n ≥ 3, let k0 = 0, let 3 ≤ k1 ≤ k2 ≤ . . . ≤ ks be odd, and suppose that s j=1 kj ≤ n. We define:

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δ = δ(k0 , . . . , ks ) = δk0 ,k1 δk1 ,k2 δk1 +k2 ,k3 · · · δs−1 j=1

kj ,ks ,

(16)

where for all 0 ≤ l ≤ s − 1, the element δl kj ,kl+1 is given by equation (14). Since j=1 the δl kj ,kl+1 commute pairwise and δl kj ,kl+1 is of order kl+1 by Lemma 28, it j=1 j=1 follows that δ is of order lcm(k1 , . . . , ks ) in Bn /[Pn , Pn ] and: θ = σ(δ) = θ1 · · · θs ,

(17)

where for i = 1, . . . , s, θi is the ki -cycle defined by: ⎛ ⎞ i−1 i−1 i    θi = ⎝ kj + 1, kj + 2, . . . , kj ⎠ . j=1

j=1

(18)

j=1

The order of the permutation θ is also equal to lcm(k1 , k2 , . . . , ks ). The following proposition allows us to decide whether Bn /[Pn , Pn ] possesses elements of order lcm(k1 , . . . , ks ). Proposition 29. Let n ≥ 3, s ≥ 1 and k0 = 0, let 3 ≤ k1 ≤ k2 ≤ . . . ≤ ks be odd such that s defined by equation (16), and let θ = σ(δ) ∈ Sn be j=1 kj ≤ n, let δ ∈ Bn /[Pn , Pn ] be

m as defined in equation (17). If A = 1≤i
mp,q = 0,

(19)

(p,q)∈Oθ (i,j)

for all {i, j} ∈ Tθ . In particular, Bn /[Pn , Pn ] possesses infinitely many elements of order lcm(k1 , . . . , ks ). Proof. Let  = lcm(k1 , . . . , ks ). First, Lemma 28 implies that δ is of order . If A =

mi,j ∈ Pn /[Pn , Pn ] then: 1≤i
(20)

where At = δ t Aδ −t for all 0 ≤ t ≤  − 1. Let 1 ≤ i < j ≤ n. Then cr(At | i, j) = mθ−t (i),θ−t (j) by equation (10) and Remarks 14(c), and it follows from (20) that: cr((Aδ) | i, j) =

−1 

mθ−t (i),θ−t (j) .

(21)

t=0

Further, θ is also of order , so Aδ cannot be of order less than , and the length of the orbit Oθ (i, j) of the pair {i, j} under the induced action of θ is a divisor  of . In particular:

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−1 

mθ−t (i),θ−t (j) =

t=0

−1  t=0

mθt (i),θt (j)

⎛  ⎝ =  



413

⎞ mp,q ⎠ .

(22)

(p,q)∈Oθ (i,j)

Now Aδ is of order  if and only if cr((Aδ) | i, j) = 0 for all 1 ≤ i < j ≤ n by Proposition 15, and by (21) and (22), this is the case if and only if 

mp,q = 0 for all pairs {i, j} ∈ Tθ .

(23)

(p,q)∈Oθ (i,j)

This proves the first part of the statement. Since the orbits are disjoint, each mi,j appears exactly once in the system of equations (23), and the fact that s ≥ 1 implies that the system (23) possesses infinitely many solutions. 2 We now prove Theorem 5 that concerns the conjugacy classes of finite-order elements of Bn /[Pn , Pn ], and which is the main result of this section. Proof of Theorem 5. Let θ ∈ Sn be of order k. Conjugating θ if necessary, we may s suppose that there exist odd integers 3 ≤ k1 ≤ . . . ≤ ks such that i=1 ki ≤ n and  = lcm(k1 , . . . , ks ) for which θ is of the form given by equation (17), and where the elements θi of that equation are defined by equation (18). Let δ ∈ Bn /[Pn , Pn ] be defined by equation (16), which we know to be of order  using Lemma 28. Now let β ∈ Bn /[Pn , Pn ] be an element of finite order such that σ(β) = θ. By Lemma 9, β is of order . To prove Theorem 5, it suffices to show that β and δ are conjugate. Since they have the same permutation, there exists A ∈ Pn /[Pn , Pn ] such that β = Aδ, and we may write

m A = 1≤i
(24)

x

Ai,ji,j , where xi,j ∈ Z for all 1 ≤ i < j ≤ n, and applying

−x Proposition 12 and equation (10), we obtain δX −1 δ −1 = 1≤i
xi,j +mi,j −xθ(i),θ(j) XAδX −1 δ −1 = 1≤i
1≤i
xi,j + mi,j − xθ(i),θ(j) = 0, for all 1 ≤ i < j ≤ n.

(25)

If {i, j} ∈ Tθ , let ri,j denote the length of its orbit under the action of θ. The system (25) may be rewritten as the union of the following subsystems: xθk+1 (i),θk+1 (j) − xθk (i),θk (j) = mθk (i),θk (j) , where 0 ≤ k ≤ ri,j − 1,

(26)

and where {i, j} runs through the elements of Tθ . These subsystems are pairwise independent since each one corresponds to the orbit of an element of Tθ under θ. Choosing

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xi,j ∈ Z arbitrarily, the solution of the subsystem of (26) corresponding to the values k−1 0 ≤ k ≤ ri,j − 2 is given by xθk (i),θk (j) = xi,j + l=0 mθl (i),θl (j) . The remaining equation of (26) corresponding to k = ri,j − 1 is automatically satisfied, since if we take the sum of the equations corresponding to 0 ≤ k ≤ ri,j − 2, we obtain: ri,j −2

xi,j − xθri,j −1 (i),θri,j −1 (j) = −



mθl (i),θl (j) ,

l=0

which is equal to mθri,j −1 (i),θri,j −1 (j) by equation (19). Hence the system (25) possesses solutions for all {i, j} ∈ Tθ , so equation (24) also admits solutions, from which it follows that Aδ is conjugate to δ by an element of Pn /[Pn , Pn ]. This proves the first part of the statement. The second part is then a direct consequence. 2 Remarks 30. (a) The number of conjugacy classes of permutations of order k in Sn is equal to the r number of partitions (n1 , . . . , nr ) of n, where ni ∈ N, n1 ≤ n2 ≤ . . . ≤ nr , i=1 ni = n and lcm(n1 , . . . , nr ) = k. (b) It follows from Corollary 4 and Theorem 5 that if k is odd, σ induces a bijection between the set of conjugacy classes of elements of order k in Bn /[Pn , Pn ] and the set of conjugacy classes of elements of order k in Sn . The same result also holds for finite cyclic subgroups. (c) Given an Abelian subgroup H of finite odd order of Sn , we saw in Theorem 6 that Bn /[Pn , Pn ] contains a subgroup G isomorphic to H. An open and more difficult question is whether Bn /[Pn , Pn ] contains a subgroup G such that σ(G) = H. 7. Finite non-Abelian subgroups of Bn /[Pn , Pn ] As we saw in Theorem 2 and Lemma 9, any finite subgroup of Bn /[Pn , Pn ] is of odd order, and embeds in Sn . Following the discussion of the previous sections, it is natural to try to characterise the isomorphism classes of the finite subgroups of Bn /[Pn , Pn ] as well as their conjugacy classes. For the question of isomorphism classes, this was achieved for finite Abelian subgroups in Theorem 6, and for that of conjugacy classes, was solved in Theorem 5 and Corollary 4 for cyclic groups. Going a step further, we may also ask whether Bn /[Pn , Pn ] possesses finite non-Abelian subgroups. Since any group of order 9 or 15 is Abelian, the smallest non-Abelian group of odd order is the Frobenius group of order 21, which we denote by F. It admits the following presentation:

  F = s, t s3 = t7 = 1, sts−1 = t2 .

(27)

The group F is thus a semi-direct product of the form Z7  Z3 , and it possesses six (resp. fourteen) elements of order 7 (resp. of order 3). As we shall see in Lemma 31,

D.L. Gonçalves et al. / Journal of Algebra 474 (2017) 393–423

415

F embeds in S7 , and as a first step in deciding whether Bn /[Pn , Pn ] possesses finite non-Abelian subgroups, one may ask whether F embeds in B7 /[P7 , P7 ]. The main result of this section, Theorem 7, shows that the answer is positive. Theorem 3(a) then implies that F embeds in Bn /[Pn , Pn ] for all n ≥ 7. In Theorem 35, we show that in B7 /[P7 , P7 ], there is a single conjugacy class of subgroups isomorphic to F. The general questions regarding the embedding in Bn /[Pn , Pn ] of an arbitrary finite non-Abelian group of odd order (for large enough n) and the number of its conjugacy classes remain open. We first exhibit a subgroup F0 of S7 that is isomorphic to F. We shall see later in Proposition 32 that any subgroup of S7 that is isomorphic to F is conjugate to F0 . In what follows, we consider the following elements of S7 : α = (1, 3, 4, 2, 5, 6, 7) and β = (1, 2, 3)(4, 5, 6).

(28)

Let F0 denote the subgroup of S7 generated by {α, β}. As noted previously, we read our permutations from left to right, to coincide with our convention for the composition of braids. Lemma 31. The subgroup F0 of S7 is isomorphic to F. Further, if G is a subgroup of S7 that is isomorphic to F then G is generated by two elements α and β  , where α is a 7-cycle, the cycle type of β  is (3, 3, 1), and β  α β  −1 = α 2 . Proof. The first part of the statement is obtained from a straightforward computation using equations (27) and (28). For the second part, if G is a subgroup of S7 that is isomorphic to F then it possesses a generating set {α , β  }, where α is a 7-cycle, β  is of order 3, and β  α β  −1 = α 2 . The cycle type of β  is either (3, 3, 1) or (3, 1, 1, 1, 1). Suppose that we are in the second case. Then β  = (n1 , n2 , n3 ), where n1 , n2 and n3 are distinct elements of {1, . . . , 7}. Hence the remaining four elements m1 , m2 , m3 and m4 of {1, 2, 3, 4, 5, 6, 7} \ {n1 , n2 , n3 } are fixed by β  . So there are two consecutive elements of the 7-cycle α , denoted by mj , mk , that belong to {m1 , m2 , m3 , m4 }. Since β  α β  −1 = α 2 , we have βαβ −1 (mj ) = αβ −1 (mj ) = β −1 (mk ) = mk , but this is different from α2 (mj ) = α(mk ). This yields a contradiction, and shows that the cycle type of β  is (3, 3, 1). 2 The rest of this section is devoted to proving that F embeds in B7 /[P7 , P7 ] and to showing that in B7 /[P7 , P7 ], there is a single conjugacy class of subgroups isomorphic to F. In this quotient, we define: x = σ2 σ1−1 σ5 σ4−1 and y = σ2 σ3 σ6 σ5 σ4 σ3−1 σ2−1 σ1−1 σ3−1 σ2−1 .

(29)

Then σ(y) = α and using the notation of equation (14), y = σ2 σ3 δ0,7 σ3−1 σ2−1 . So y is of order 7 by Lemma 28. Similarly, σ(x) = β, x = δ0,3 δ3,3 , and x is of order 3 (see the discussion on page 412 just after equation (16)). We now prove Theorem 7 that asserts the existence of a subgroup of B7 /[P7 , P7 ] isomorphic to the Frobenius group F.

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Proof of Theorem 7. Consider the subgroup H of B7 /[P7 , P7 ] generated by {x, y}. By the above comments, we know that σ(y) = α and σ(x) = β, therefore σ(H) = F0 . Using Remarks 14(c) and Proposition 15, we may rewrite xyx−1 y −2 ∈ P7 /[P7 , P7 ] in terms of the Ai,j in a straightforward manner as follows: −1 −1 −1 xyx−1 y −2 = A4,7 A1,7 A1,6 A−1 2,7 A2,6 A2,4 A1,2 A4,6 .

(30)

For example, the coefficient of A1,2 in the expression of xyx−1 y −2 is given by: cr(xyx−1 y −2 | 1, 2) = cr(x | 1, 2) + cr(y | 2, 3) + cr(x−1 | 5, 4) + cr(y −2 | 4, 6) = 0 + 0 + 1 + 0 = 1. The other coefficients may be computed in a similar manner. One may also draw the braid xyx−1 y −2 and determine the crossing numbers geometrically. Note also that using the Artin relations (2) and (3), one may show that the equality given in equation (30) holds in P7 . Equation (30) shows that xyx−1 y −2 is non-trivial in the free Abelian group P7 /[P7 , P7 ], which implies that H is not isomorphic to F. We now look for an element N ∈ P7 /[P7 , P7 ] such that if v = N y then the subgroup x, v is isomorphic to F, where v is of order 7, σ(v) = α and xvx−1 = v 2 . This last equality gives rise to the following equivalences: xvx−1 = v 2 ⇐⇒ xN yx−1 = N yN y ⇐⇒ xyx−1 y −2 = xN −1 x−1 . N. yN y −1 .

(31)

We seek solutions N ∈ P7 /[P7 , P7 ] of equation (31) taking into account equation (30) and the fact that N y is of order 7. Let N=



p

Ai,ji,j ∈ P7 /[P7 , P7 ].

(32)

1≤i
Arguing in a manner similar to that of the proof of Proposition 21, and using the fact that y is of order 7, we obtain: 0 = (N y)7 =



7

Ai,jt=1

pαt (i), αt (j)

,

1≤i
from which it follows that the sum of the coefficients corresponding to the elements of each of the orbits Oα (i, j) is zero: ⎧ ⎪ ⎪ p1,2 + p4,7 + p3,6 + p1,5 + p2,7 + p4,6 + p3,5 = 0 ⎨ p1,3 + p1,7 + p6,7 + p5,6 + p2,5 + p2,4 + p3,4 = 0 ⎪ ⎪ ⎩ p1,4 + p3,7 + p1,6 + p5,7 + p2,6 + p4,5 + p2,3 = 0. By equation (10), we also have:

(33)

D.L. Gonçalves et al. / Journal of Algebra 474 (2017) 393–423

⎛ xN −1 x−1 . N. yN y −1 = ⎝

⎞⎛



1≤i
=





−p Ai,j β(i),β(j) ⎠ ⎝

⎞⎛ Ai,ji,j ⎠ ⎝



p

1≤i
417

⎞ p Ai,jα(i),α(j) ⎠

1≤i
−p +pi,j +pα(i),α(j) Ai,j β(i),β(j) .

1≤i
So xN −1 x−1 . N. yN y −1 = xyx−1 y −2 if and only if the following system of equations admit a solution: ⎧ A1,2 : ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A4,7 : ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A3,6 : ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A1,5 : ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A2,7 : ⎨ A4,6 : ⎪ ⎪ ⎪ ⎪ ⎪ A3,5 : ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A1,4 : ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A3,7 : ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ A1,6 : ⎪ ⎪ ⎪ ⎪ ⎩ A5,7 :

−p2,3 + p1,2 + p3,5 = 1

A1,3 : −p1,2 + p1,3 + p3,4 = 0

−p5,7 + p4,7 + p1,2 = 1

A1,7 : −p2,7 + p1,7 + p1,3 = 1

−p1,4 + p3,6 + p4,7 = 0

A6,7 : −p4,7 + p6,7 + p1,7 = 0

−p2,6 + p1,5 + p3,6 = 0

A5,6 : −p4,6 + p5,6 + p6,7 = 0

−p3,7 + p2,7 + p1,5 = −1

A2,5 : −p3,6 + p2,5 + p5,6 = 0

−p4,5 + p4,6 + p2,7 = −1

A2,4 : −p3,5 + p2,4 + p2,5 = −1

−p1,6 + p3,5 + p4,6 = 0

A3,4 : −p1,5 + p3,4 + p2,4 = 0

−p2,5 + p1,4 + p2,3 = 0

A2,6 : −p3,4 + p2,6 + p5,7 = −1

−p1,7 + p3,7 + p1,4 = 0

A4,5 : −p5,6 + p4,5 + p2,6 = 0

−p2,4 + p1,6 + p3,7 = 1

A2,3 : −p1,3 + p2,3 + p4,5 = 0

(34)

−p6,7 + p5,7 + p1,6 = 0.

One may check that the systems (33) and (34) together admit a solution, taking for example all of the coefficients to be zero, with the exception of: p2,7 = p5,7 = −1 and p3,5 = p1,6 = 1. −1 For these values of pi,j , we have N = A3,5 A1,6 A−1 2,7 A5,7 , and it follows from above that the subgroup x, v of B7 /[P7 , P7 ] is isomorphic to F, which completes the proof of the theorem. 2

We now analyse the conjugacy classes of subgroups isomorphic to F in B7 /[P7 , P7 ]. We first show that S7 possesses a single such conjugacy class. Proposition 32. Any two subgroups of S7 isomorphic to F are conjugate. Proof. Let G be a subgroup of S7 isomorphic to F. It suffices to show that G is conjugate to F0 . By Lemma 31, G is generated by two elements α and β  , where α is a 7-cycle, the cycle type of β  is (3, 3, 1), and β  α(β  )−1 = (α )2 . Conjugating G if necessary, we may suppose that α = α. Now β  α(β  )−1 = α2 in G and βαβ −1 = α2 in F0 , from which

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it follows that β −1 β  belongs to the centraliser of α. But since α is a complete cycle in S7 , its centraliser is equal to α. So there exists k ∈ {0, 1, . . . , 6} such that β  = βαk , and hence G = α, β   = α, β = F0 as required. 2 Remark 33. For the purposes of the proof of Proposition 34, we shall study the elements of the form εαε−1 , where ε belongs to the centraliser of β in S7 . This centraliser may be seen to be of order 18, and consists of the elements of the form τ i(1, 2, 3)j , where τ = (1, 4, 2, 5, 3, 6), 0 ≤ i ≤ 5 and 0 ≤ j ≤ 2. Let:  α1 = (1, 3, 2)α(1, 3, 2)−1 = (1, 4, 3, 5, 6, 7, 2), −1

α2 = (4, 6, 5)α(4, 6, 5)

= (1, 3, 5, 2, 6, 4, 7).

and

(35)

A straightforward computation shows that:

−j

j

(1, 2, 3) α(1, 2, 3)

=

⎧ ⎪ ⎪ ⎨α

if j = 0

α2 ⎪ 2

if j = 1

⎪ ⎩α 1

if j = 2,

and τ ατ −1 = α2−2 , τ α2 τ −1 = α−1 and τ α1 τ −1 = α13 . It then follows that for all 0 ≤ i ≤ 5 and 0 ≤ j ≤ 2, there exists z ∈ {α, α1 , α2 } such that τ i (1, 2, 3)j α(1, 2, 3)−j τ −i is a generator of z. Proposition 34. Suppose that H is a subgroup of B7 /[P7 , P7 ] isomorphic to F. Then H is conjugate to a subgroup of the form x, v, where x is given by equation (29) and σ(v) = α. Proof. Let H be a subgroup of B7 /[P7 , P7 ] isomorphic to F. Since σ(H) is a subgroup of S7 isomorphic to F by Lemma 9, it follows from Proposition 32 that there exists ρ ∈ S7 such that F0 = ρσ(H)ρ−1 . So if ρ ∈ B7 /[P7 , P7 ] is such that σ( ρ) = ρ then H1 = ρH ρ −1 satisfies σ(H1 ) = F0 . Let x , y ∈ H1 be such that σ( x) = β and σ( y ) = α, where α and β are given by equation (28). Now β = σ(x), and since x and x  are of order 3 and have the same permutation, Theorem 5 implies that they are conjugate. So there exists λ1 ∈ B7 /[P7 , P7 ] such that λ1 x λ−1 = x. Hence σ(λ1 )σ( x)σ(λ1 )−1 = σ(x), and 1 since σ( x) = σ(x) = β, we conclude that σ(λ1 ) belongs to the centraliser of β in S7 . By Remark 33, this centraliser is equal to τ, (1, 2, 3), and the fact that σ( y ) = α implies −1 that there exists z ∈ {α, α1 , α2 } such that σ(λ1 yλ1 ) is a generator of z. Let: ⎧ ⎪ ⎪ ⎨e λ2 =

σ1 σ2−1 ⎪ ⎪ ⎩σ σ −1 4 5

if z = α if z = α1 if z = α2 .

D.L. Gonçalves et al. / Journal of Algebra 474 (2017) 393–423

419

Note that λ2 commutes with x, and by equation (35), σ(λ−1 λ−1 2 λ1 y 1 λ2 ) is a generator −1 −1 of α. Taking v to be the element of λ2 λ1  y  λ1 λ2 for which σ(v) = α, the subgroup −1 −1 λ−1 λ ρ  H(λ λ ρ  ) is then seen to be equal to x, v, which proves the proposition. 2 1 1 2 2 Theorem 35. The group B7 /[P7 , P7 ] possesses a unique conjugacy class of subgroups isomorphic to F. Proof. From the proof of Theorem 7, B7 /[P7 , P7 ] possesses a subgroup H0 = x, v0  −1 isomorphic to F, where v0 = N0 y, and N0 = A1,6 A3,5 A−1 2,7 A5,7 . Let H be a subgroup of B7 /[P7 , P7 ] isomorphic to F. By Proposition 34, up to conjugacy, we may suppose that H = x, v, where σ(v) = α = σ(y) = σ(v0 ). Thus v = N y, where N ∈ P7 /[P7 , P7 ]. Again from the proof of Theorem 7, the coefficients pi,j of N given by equation (32) satisfy the systems of equations (33) and (34), and one may check that the general solution of these two systems is of rank 6, and is given by: ⎧ p1,2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p4,7 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p3,6 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p1,5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p2,7 ⎪ ⎪ ⎨ p4,6 ⎪ ⎪ ⎪ ⎪ ⎪ p3,5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p1,4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p ⎪ ⎪ 3,7 ⎪ ⎪ ⎪ ⎪ p1,6 ⎪ ⎪ ⎪ ⎪ ⎩ p5,7

= −r6 + r4 + r3 − r2 + 1

p1,3 = −r6 − r2

= r6 − r 3 + r 2

p1,7 = r6 − r5 − r4 − r3 + r2

= r3 − r 2

p6,7 = r5 + r4

= r2

p5,6 = −r6 + r3 − r2 − r1

= −r5 − r4 − r3 − 1

p2,5 = r6 + r1

= −r6 + r5 + r4 + r3 − r2 − r1

p2,4 = −r4 − r3 + r2 − 1

= r6 − r 4 − r 3 + r 2 + r 1

p3,4 = r4 + r3 + 1

= r6

p2,6 = r3

= −r5 − r4 − r3 + r2

p4,5 = −r6 − r2 − r1

= r5

p2,3 = r1

(36)

= r4 ,

where r1 , . . . , r6 ∈ Z are arbitrary. So choose the values of the rl so that v = N y. We claim that there exists Θ ∈ P7 /[P7 , P7 ] such that: ΘxΘ−1 = x, Θv0 Θ−1 = v.

and

(37) (38)

This being the case, we have H = x, v = Θ x, v0  Θ−1 = ΘH0 Θ−1 , in particular H and H0 are conjugate in B7 /[P7 , P7 ], which proves the statement of the theorem. To θ prove the claim, let Θ = Ai,ji,j . We must determine the coefficients θi,j of Θ that 1≤i
D.L. Gonçalves et al. / Journal of Algebra 474 (2017) 393–423

420

satisfy equations (37) and (38). By Proposition 12, the orbits of {Ai,j | 1 ≤ i < j ≤ 7} under the action of conjugation by x are of the form: ⎧ ⎪  → A1,3 −  → A2,3 −  → A1,2 A1,2 − ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ A2,7 −  → A1,7 −  → A3,7 −  → A2,7 ⎪ ⎪ A3,6 −  → A2,5 −  → A1,4 −  → A3,6 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩A −  → A1,6 −  → A3,5 . 3,5  → A2,4 −

A4,6 −→ A5,6 −→ A4,5 −→ A4,6 A4,7 −→ A6,7 −→ A5,7 −→ A4,7 A1,5 −→ A3,4 −→ A2,6 −→ A1,5

(39)

By equation (39), equation (37) holds if and only if there exist s1 , . . . , s7 ∈ Z such that: ⎧ ⎪ s1 = θ1,2 = θ1,3 = θ2,3 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ s3 = θ3,6 = θ2,5 = θ1,4 ⎪ ⎪ s5 = θ4,6 = θ5,6 = θ4,5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩s = θ = θ = θ . 7 1,5 3,4 2,6

s2 = θ2,7 = θ1,7 = θ3,7 s4 = θ3,5 = θ2,4 = θ1,6 (40) s6 = θ4,7 = θ6,7 = θ5,7

Equation (38) may be written in the form Θ. N0 . yΘ−1 y −1 = N . Using equation (39), we obtain the following system of equations: ⎧ ⎪ p1,2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p4,7 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p3,6 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p1,5 ⎪ ⎪ ⎪ ⎪ ⎪ p2,7 ⎪ ⎪ ⎪ ⎨ p4,6 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p3,5 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p1,4 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ p ⎪ ⎪ 3,7 ⎪ ⎪ ⎪ ⎪ ⎪ p1,6 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩p 5,7

= θ1,2 − θ3,5 = s1 − s4

p1,3 = θ1,3 − θ2,3 = s1 − s7

= θ4,7 − θ1,2 = s6 − s1

p1,7 = θ1,7 − θ1,3 = s2 − s1

= θ3,6 − θ4,7 = s3 − s6

p6,7 = θ6,7 − θ1,7 = s6 − s2

= θ1,5 − θ3,6 = s7 − s3

p5,6 = θ5,6 − θ6,7 = s5 − s6

= θ2,7 − θ1,5 − 1 = s2 − s7 − 1

p2,5 = θ2,5 − θ5,6 = s3 − s5

= θ4,6 − θ2,7 = s5 − s2

p2,4 = θ2,4 − θ2,5 = s4 − s3

= θ3,5 − θ4,6 + 1 = s4 − s5 + 1

p3,4 = θ3,4 − θ2,4 = s7 − s4

= θ1,4 − θ3,5 = s3 − s1

p2,6 = θ2,6 − θ5,7 = s7 − s6

= θ3,7 − θ1,4 = s2 − s3

p4,5 = θ4,5 − θ2,6 = s5 − s7

= θ1,6 − θ3,7 + 1 = s4 − s2 + 1

p2,3 = θ2,3 − θ4,5 = s1 − s5

(41)

= θ5,7 − θ1,6 − 1 = s6 − s4 − 1.

It remains to show that by choosing the sk appropriately, we obtain a system of coefficients that satisfy the equations of system (41). Consider the system:

D.L. Gonçalves et al. / Journal of Algebra 474 (2017) 393–423

⎧ s1 − s4 = −r6 + r4 + r3 − r2 + 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ s6 − s1 = r6 − r3 + r2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ s3 − s6 = r3 − r2 ⎪ ⎪ s7 − s3 = r2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ s2 − s7 = −r5 − r4 − r3 ⎪ ⎪ ⎪ ⎪ ⎩ s5 − s2 = −r6 + r5 + r4 + r3 − r2 − r1 .

421

(42)

This system clearly possesses solutions in the sk in terms of the rl , obtained for example by taking s4 to be an arbitrary integer, and by rewriting the other sk in terms of s4 and the rl . For such a solution, the first six equations of the first column of (41) are satisfied using equation (36). Using just (36) and (42), we now verify the remaining equations of (41). For example: s4 − s5 + 1 = − ((s1 − s4 ) + (s6 − s1 ) + (s3 − s6 ) + (s7 − s3 )+ (s2 − s7 ) + (s5 − s2 )) + 1 = − ((−r6 + r4 + r3 − r2 + 1) + (r6 − r3 + r2 ) + (r3 − r2 ) + r2 + (−r5 − r4 − r3 ) + (−r6 + r5 + r4 + r3 − r2 − r1 )) + 1 =r6 − r4 − r3 + r2 + r1 = p3,5 . In a similar manner, one may check that the right-hand side of each of the equations of the system (41) is equal to the left-hand side, using first (42) to express the sk in terms of the rl , and then using (36) to obtain the corresponding pi,j . The straightforward computations are left to the reader. So with this choice of sk , we obtain values of the θi,j using equation (40) for which equations (37) and (38) are satisfied. Conversely, given arbitrary r1 , . . . , r6 ∈ Z and s1 , . . . , s6 satisfying equation (42), we see that if the pi,j are given by equation (41) and the θi,j are given by equation (40) then equations (37) and (38) are satisfied, and this completes the proof of the theorem. 2 Remark 36. We saw in Theorem 7 that the Frobenius group F embeds in B7 /[P7 , P7 ]. It is the only finite non-Abelian subgroup of S7 of odd order. To see this, besides 3 × 7, which is the order of F, the possible orders of non-Abelian subgroups of odd order of S7 are 32 × 5, 32 × 7, 3 × 5 × 7 and 32 × 5 × 7. Further, if H is a subgroup of Sn of odd order then it is necessarily a subgroup of An . Indeed, any element h ∈ H may be decomposed as a product of disjoint cycles each of which is of odd length, and so it follows that h ∈ An . From the table of maximal subgroups of A7 given in [4, page 10], we see that S7 has no subgroup of order 32 × 7, 3 × 5 × 7 or 32 × 5 × 7. It also has no subgroup of order 32 × 5. Suppose on the contrary that it had such a subgroup K. Then by the same table, K would be a subgroup of A6 , but this contradicts the information given in the corresponding table for A6 (see [4, page 4]).

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Acknowledgments The initial ideas for this paper occurred during the stay of the third author at the Laboratoire de Mathématiques Nicolas Oresme from the 5th January to the 5th June 2012, and were developed during the period 2012–2014 when the third author was at the Departamento de Matemática do IME – Universidade de São Paulo and was partially supported by a project grant no. 2008/58122-6 from FAPESP, and also by a project grant no. 151161/2013-5 from CNPq. Further work took place during the visit of the first author to the Departamento de Matemática, Universidade Federal de Bahia during the period 16th–20th May 2014 and also during the visit of the second author to the Departamento de Matemática do IME – Universidade de São Paulo during the periods 10th July–2nd August 2014, and to the Departamento de Matemática do IME – Universidade de São Paulo and the Departamento de Matemática, Universidade Federal de Bahia during the period 1st–17th November 2014, and was partially supported by the international CNRS/FAPESP programme no. 226555. We would like to thank an anonymous referee who generously made a number of suggestions to shorten and simplify many of the proofs of this paper. References [1] A.J. Berrick, F.R. Cohen, Y.-L. Wong, J. Wu, Configurations, braids and homotopy groups, J. Amer. Math. Soc. 19 (2006) 265–326. [2] H. Brown, R. Bülow, J. Neubüser, H. Wondratschek, H. Zassenhaus, Crystallographic Groups of Four-Dimensional Space, Wiley Monographs in Crystallography, Wiley-Interscience, 1978. [3] L. Charlap, Bieberbach Groups and Flat Manifolds, Springer-Verlag, New York, 1986. [4] J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker, R.A. Wilson, Atlas of Finite Groups, Oxford University Press, Eynsham, 1985. [5] H.S.M. Coxeter, Factor groups of the braid groups, in: Proc. Fourth Canad. Math. Congress, 1957, pp. 95–122. [6] K. Dekimpe, Almost-Bieberbach Groups: Affine and Polynomial Structures, Lecture Notes in Mathematics, vol. 1639, Springer, Berlin, 1996. [7] E.A. Elrifai, H.R. Morton, Algorithms for positive braids, Q. J. Math. 45 (1994) 479–497. [8] D.L. Gonçalves, J. Guaschi, The braid groups of the projective plane, Algebr. Geom. Topol. 4 (2004) 757–780. [9] V.L. Hansen, Braids and Coverings: Selected Topics, London Math. Soc. Student Text, vol. 18, Cambridge University Press, 1989. [10] N.F.M. Henry, K. Lonsdale (Eds.), International Tables for X-Ray Crystallography, vol. 1 (Symmetry Groups), 3rd edition, International Union of Crystallography, Kynoch Press, 1969. [11] M. Hoffman, An invariant of finite Abelian groups, Amer. Math. Monthly 94 (1987) 664–666. [12] D.L. Johnson, Presentation of Groups, London Math. Soc. Lecture Notes, vol. 22, Cambridge University Press, 1976. [13] J.Y. Li, J. Wu, Artin braid groups and homotopy groups, Proc. Lond. Math. Soc. 99 (2009) 521–556. [14] R. Lyndon, Groups and Geometry, London Math. Soc. Lecture Notes, vol. 101, Cambridge University Press, 1985. [15] I. Marin, The cubic Hecke algebra on at most 5 strands, J. Pure Appl. Algebra 216 (2012) 2754–2782. [16] I. Marin, On the representation theory of braid groups, Ann. Math. Blaise Pascal 20 (2013) 193–260. [17] I. Marin, Crystallographic groups and flat manifolds from complex reflection groups, Geom. Dedicata 182 (2016) 233–247. [18] K. Murasugi, B.I. Kurpita, A Study of Braids, Mathematics and Its Applications, vol. 484, Kluwer Academic Publishers, Dordrecht, 1999.

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