2.1. Braid groups and examples

Let S be a surface as above. We introduce a particular element of $\mathrm {Mod}(S)$ that is used throughout the paper. Let A be an annulus. The homeomorphism depicted in Figure 1 is called a twist map.

Now, let $c \subset S$ be a simple closed curve. The regular neighbourhood $\mathcal {N}(c)$ of c is homeomorphic to an annulus A. Consider the homeomorphism $f_c$ that acts as a twist map on $\mathcal {N}(c)$ and as the identity on $S \setminus \mathcal {N}(c)$ . The homotopy class of $f_c$ is called a Dehn twist about c, denoted by $T_c$ [Reference Farb and MargalitFM12, Section 3.1].

Braid groups can be defined in several equivalent ways, long known to be equivalent; see for instance [Reference Birman and BrendleBB05, Reference Kassel and TuraevKT08]. In this work, it is convenient to define them in terms of mapping class groups. Let $D_n$ be a disc with $n \in \mathbb {N}$ marked points in its interior. The braid group $B_n$ is $\mathrm {Mod}(D_n)$ . For a geometric insight into twists in the context of braid groups, let $D_n$ lie on the $xy$ -plane with its centre on the x-axis. Denote the punctures from left to right by $p_1, p_2, \ldots , p_n$ : the arc connecting $p_i$ and $p_{i+1}$ is denoted by $a_i$ (see Figure 4). Consider $a_i$ to be the diameter of a circle c such that the points $p_i$ and $p_{i+1}$ lie on c. Interchanging the points $p_i$ and $p_{i+1}$ by rotating them half way along c in the clockwise direction gives a homeomorphism of $D_n$ , and its homotopy class in $\mathrm {Mod}(D_n)$ is called a half-twist, denoted by . Note that all conjugates of are called half-twists. In terms of presented groups, half-twists correspond to the Artin generators from Artin’s presentation for $B_{n}$ [Reference ArtinArt25]:

Let $c\in D_n$ be a curve surrounding the points $p_i, p_{i+1}$ . This curve is homotopic to the circle described above. We note that if is a half-twist, then $\sigma _{\hspace {-0.3ex}i}^{2}$ is a Dehn twist about the curve c. This Dehn twist is generalised, for $1\leq i<j\leq n$ , as

$$ \begin{align*} A_{i,j} = (\sigma_{j-1} \sigma_{j-2} \ldots \sigma_{i+1}) \sigma_{\hspace{-0.3ex}i}^{2} (\sigma_{j-1} \sigma_{j-2} \ldots \sigma_{i+1})^{-1}. \end{align*} $$

We recall that a generating set of $P_{n}$ is given by $\{A_{i,j}\}_{1\leq i<j\leq n}$ . Geometrically, the element $A_{i,j}$ can be represented as a Dehn twist about a curve surrounding punctures $p_i$ and $p_j$ . For instance, in Figure 2, we describe $A_{2,5}$ as the Dehn twist about the curve that surrounds punctures $p_2$ and $p_5$ .

Figure 1 Twist map acts on an annulus.

Figure 2 Dehn twist along the curve that surrounds the punctures $p_2,p_5$ is $A_{2,5}$ .

We are interested in the action by conjugation of $B_n$ on $P_n$ . Recall from [Reference Murasugi and KurpitaMK99, Proposition 3.7, Ch. 3] that for all $1\leq k\leq n-1$ and for all $1\leq i<j\leq n$ ,

$$ \begin{align*} \sigma_kA_{i,j}\sigma_k^{-1}= \begin{cases} A_{i,j} & \text{if}\ k\neq i-1, i, j-1, j,\\ A_{i,j+1} & \text{if}\ j=k,\\ A_{i,j}^{-1}A_{i,j-1}A_{i,j} & \text{if}\ j=k+1\ \text{and}\ i<k,\\ A_{i,j} & \text{if}\ j=k+1\ \text{and}\ i=k,\\ A_{i+1,j} & \text{if}\ i=k<j-1,\\ A_{i,j}^{-1}A_{i-1,j}A_{i,j} & \text{if}\ i=k+1. \end{cases} \end{align*} $$

This action induces an action of on ; see [Reference Gonçalves, Guaschi and OcampoGGO17, Proposition 12]: let and let $\pi $ be the permutation induced by $\alpha ^{-1}$ , then $\alpha A_{i,j}\alpha ^{-1}=A_{\pi (i), \pi (j)}$ in .

Another important element of $B_n$ that plays a crucial role in the paper is the Dehn twist (or a full twist) along a curve surrounding all marked points of $D_n$ . We denote this element by $\Delta _n^2$ . In fact, $\Delta _n^2$ generates the centre of $B_n$ [Reference ChowCho48]; in terms of half-twists,

2.2. Burau representation and symplectic structures

Braid groups naturally act on the homology of topological spaces obtained from the punctured disk. A construction arising in such a way is the Burau representation [Reference BurauBur35]. One of the most famous representations of the braid group, originally introduced in terms of matrices assigned to the generators in Artin’s presentation of $B_{n}$ , the Burau representation is fundamental in low-dimensional topology. While this representation has been extensively studied, it still retains some mystery: a long standing candidate for proving the linearity of the braid group (later established independently in [Reference BigelowBig01, Reference KrammerKra02]), the question of its faithfulness has remained open for quite some time. The Burau representation, faithful for $n\leq 3$ [Reference Magnus and PelusoMP69], eventually proved to be unfaithful for $n \geq 5$ (Moody [Reference MoodyMoo91] proved unfaithfulness for $n \geq 9$ , Long and Paton [Reference Long and PatonLP93] for $n \geq 6$ , and Bigelow [Reference BigelowBig99] for $n = 5$ ). However, the case $n=4$ remains open, with advances towards closing the problem being published recently [Reference Bharathram and BirmanBB21, Reference Beridze and TraczykBT18, Reference DattaDat22].

In this work, we are going to take the viewpoint of the Burau representation as a homological representation. Let $\pi =\pi _1(D_n,q)$ denote the fundamental group of $D_n$ , where $q \in \partial D_n$ . The function $\pi \to \mathbb {Z}\cong \langle t \rangle $ defines a covering space . Let Q be a set of all lifts of q. The action of t on induces a $\mathbb {Z}[t]$ -module of dimension n. Every mapping class in $\mathrm {Mod}(D_n)$ lifts to a unique mapping class in . Hence, the (reducible) Burau representation is given by a map

This representation splits into a direct sum of an $(n-1)$ - and a one-dimensional representation.

Fixing $t=-1$ , the covering space becomes a two-fold branch cover $\Sigma \to D_n$ , where $\Sigma $ is homeomorphic to a surface of genus $g=(n-1)/2$ and one boundary component if n is odd, and $g=n/2 - 1$ and two boundary components if n is even [Reference Perron and VannierPV96]. As mentioned above, every mapping class in $\mathrm {Mod}(D_n)$ lifts to a unique mapping class in $\mathrm {Mod}(\Sigma )$ leading to an injection $\mathrm {Mod}(D_n) \to \mathrm {Mod}(\Sigma )$ . Let $q \in \partial D_n$ be a point and Q be a set of all lifts of q. The reducible Burau representation at $t=-1$ [Reference Brendle and MargalitBM18, Section 2] (see also [Reference Bloomquist, Patzt and ScherichBPS22]) is

$$ \begin{align*} \mathrm{Mod}(D_n) \to \mathrm{Mod}(\Sigma) \to \mathrm{Aut}(\mathrm{H}_1(\Sigma, Q;\mathbb{Z})). \end{align*} $$

For n odd, the module $\mathrm {H}_1(\Sigma , Q;\mathbb {Z})$ splits as $\mathrm {H}_1(\Sigma ;\mathbb {Z}) \times \mathbb {Z}$ and the induced action of $\mathrm {Mod}(D_n)$ preserves a symplectic form on $\mathrm {H}_1(\Sigma ;\mathbb {Z})$ . Hence, the image of the latter representation is conjugate to $\mathrm {Sp}_{n-1}(\mathbb {Z})$ [Reference Gambaudo and GhysGG16, Proposition 2.1]. When n is even, the module $\mathrm {H}_1(\Sigma ,Q;\mathbb {Z})$ carries a symplectic structure. More precisely, if g is the genus of $\Sigma $ , then let $\Sigma '$ be a surface obtained by gluing a pair of pants in the boundary of $\Sigma $ . Then $\Sigma '$ is a surface genus $g+1$ with one boundary component. We consider $\mathrm {H}_1(\Sigma ,Q;\mathbb {Z})$ as a submodule of $\mathrm {H}_1(\Sigma ';\mathbb {Z})$ . In Figure 3, we give a basis for each of the latter modules.

Figure 3 Generators for $\mathrm {H}_1(\Sigma ';\mathbb {Z})$ on the left, and $\mathrm {H}_1(\Sigma ,Q;\mathbb {Z})$ on the right.

The representation obtained by the construction above is

(2-1) $$ \begin{align} \rho \colon B_{n} \rightarrow \begin{cases} \mathrm{Sp}_{n-1}(\mathbb{Z}) &\mbox{for } n \mbox{ odd}, \\ (\mathrm{Sp}_{n}(\mathbb{Z}))_u &\mbox{for } n \mbox{ even}, \\ \end{cases} \end{align} $$

where, without loss of generality, we can choose $u = y_{2g+1}$ . For the detailed construction, see [Reference Brendle and MargalitBM18, Section 2.1].

An analogue of the principal congruence subgroups for the braid groups $B_{n}$ can be defined starting from the integral Burau representation. The level m congruence subgroup $B_{n}[m]$ is the kernel of the mod m reduction of the integral Burau representation

$$ \begin{align*} \rho_m \colon B_{n} \rightarrow \begin{cases} \mathrm{Sp}_{n-1}(\mathbb{Z}/m \mathbb{Z}) &\mbox{for } n \mbox{ odd}, \\ (\mathrm{Sp}_{n}(\mathbb{Z}/m \mathbb{Z}))_u &\mbox{for } n \mbox{ even}, \\ \end{cases} \end{align*} $$

for $m> 1$ .

In [Reference Arnol’dArn68], Arnol’d proved that the pure braid group $P_{n}$ is isomorphic to the level $2$ congruence subgroup $B_{n}[2]$ of the braid group $B_{n}$ ; see also [Reference Brendle and MargalitBM18, Section 2] for a sketch of the original argument. In [Reference Brendle and MargalitBM18], Brendle and Margalit go on to prove that $B_{n}[4]$ is isomorphic to the subgroup $P_{n}^2$ , where $P_{n}^2$ is the subgroup of $P_{n}$ generated by the squares of all elements.

A well-known family of elements in $B_{n}[m]$ are braid Torelli elements. Consider the symplectic representation (2-1). The kernel of this representation is denoted by $\mathcal {BI}_n$ and it is called braid Torelli group. Since the representation (1-1) is a $\bmod\ m$ reduction of $\rho $ , every element of $\mathcal {BI}_n$ is actually an element of $B_{n}[m]$ . In particular, $\mathcal {BI}_n$ is generated by squares of Dehn twists about curves surrounding an odd number of marked points in $D_n$ [Reference Brendle, Margalit and PutmanBMP15]. In terms of half-twists, these elements are of the form

where $k < n$ is even. This family of elements can be extended. If, for example, we denote by c a curve surrounding an odd number of marked points, then $T_{c}^2 \in \mathcal {BI}_n$ . Other families of elements in $B_{n}[m]$ , such as mod p involutions and centre maps, are described in [Reference StylianakisSty18, Section 4].

2.3. Actions of half-twists on symplectic groups

Recall that $B_n \cong \mathrm {Mod}(D_n)$ and $\Sigma \to D_n$ is a two-fold branched cover. The image of the monomorphism $\mathrm {Mod}(D_n) \to \mathrm {Mod}(\Sigma )$ is called the hyperelliptic mapping class group denoted by $\mathrm {SMod}(\Sigma )$ . Below, we explain how to lift elements of $\mathrm {Mod}(D_n)$ into $\mathrm {SMod}(\Sigma )$ . Then we use these lifts to explain their action on $\mathrm {H}_1(\Sigma ,Q;\mathbb {Z})$ .

Figure 4 An example of a 2-fold cover of a marked disc. The simple closed curve $c_i$ in the genus 3 surface becomes the arc $a_i$ in the disc.

Let $\Sigma $ be a genus g surface as in Figure 4. The surface $\Sigma $ is the 2-fold cover of the disc $D_n$ . Each simple closed curve $c_i$ is a lift of the arc $a_i$ . Recall that is a half-twist along $a_i$ . Then lifts to the Dehn twist $T_{c_i}$ . This association describes the homomorphism $B_{n} \to \mathrm {SMod}(\Sigma )$ by .

Suppose that $\Sigma $ is a genus $g \geq 1$ surface with one boundary component (similarly for two boundary components). Let $T_c$ be a Dehn twist about a simple closed curve c and let $[c]$ be its homology class in $\mathrm {H}_1(\Sigma ;\mathbb {Z})$ . Denote by $t_{[c]}$ a transvection induced by $T_c$ . The action of the transvection $t_{[c]}$ on a homology class u is defined by $t_{[c]}(u) = u + i(u,[c])[c]$ , where $i(,)$ is a symplectic form. Therefore, the homomorphism $\rho _m \colon B_n \to \mathrm {Sp}_{n-1}(\mathbb {Z}/m \mathbb {Z})$ is defined by $\sigma _i \mapsto t_{[c_i]}$ (similarly for two boundary components). The next two lemmas describe the images of particular elements of $B_{n}$ in the symplectic group over $\mathbb {Z}/m \mathbb {Z}$ .

Proof. Since is mapped to the transvection $t_{[c_i]}$ , we only need to compute the matrix form of $t_{[c_i]}$ . It is easy to calculate the action of $t_{[c_i]}$ based on Figure 3. The result is conjugate to the following matrix:

$$ \begin{align*} \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix} \oplus I, \end{align*} $$

where I is the identity matrix of dimension $n-2$ . The result follows by calculating the m th power of the latter matrix over $\mathbb {Z}/ m \mathbb {Z}$ .

Lemma 2.1 leads to the question of whether $B_{n}[m]$ coincides with the group normally generated by $\sigma _i^m$ . This is generally not the case (see [Reference Bellingeri, Damiani, Ocampo and StylianakisBDOS24] for further details): in fact, $B_{n}[m]$ is of finite index in $B_n$ , while the group normally generated by $\sigma _i^m$ is not (except pairs $(n,m)\in \{ (3,3), (3,4), (3,5), (4,3), (5,3) \}$ ; see [Reference CoxeterCox59]).

Recall that $\Delta _n^2$ denotes the element $(\sigma _1 \sigma _2 \cdots \sigma _{n-1})^{n}$ in $B_{n}$ , generating the centre of $B_{n}$ .

Proof. Suppose that n is odd. The lift of $(\sigma _1 \sigma _2 \cdots \sigma _{n-1})^{n}$ to $\Sigma $ is the product of Dehn twists $(T_{c_1} T_{c_2} \cdots T_{c_{n-1}})^{n}$ . Consider the basis $\{x_i, y_i \}$ depicted in Figure 3. Then the action of the product $(t_{[c_1]} t_{[c_2]} \cdots t_{[c_{n-1}]})^{n}$ reverses the orientation of $x_i, y_i$ [Reference StylianakisSty18]. Thus, it has order 2.

Suppose that n is even. The lift of $(\sigma _1 \sigma _2 \cdots \sigma _{n-1})^n$ to $\Sigma $ is the product $(T_{c_1} T_{c_2} \cdots T_{c_{n-1}})^{n}$ . By the chain relation, the latter product is $T_{q_1} T_{q_2}$ , where the curves $q_1,q_2$ are parallel to the boundary components of $\Sigma $ [Reference Farb and MargalitFM12, Proposition 4.12]. Since $[q_1]=[q_2] = y_{n-1}$ , we have that $T_{q_1} T_{q_2}$ is mapped into the square transvection $t_{y_{n-1}}^2$ . The transvection $t_{y_{n-1}}$ fixes all basis elements $\{ x_i, y_i\}$ except $x_{n-1}$ . Hence,

$$ \begin{align*} t_{y_{n-1}}^2(x_{n-1}) = x_{n-1} + 2 y_{n-1}.\\[-36pt] \end{align*} $$

3.1. A general result on crystallographic groups

In this subsection, we prove two results that are general and that are applied to the study of crystallographic structures on quotients of the braid group by commutator subgroups of congruence subgroups.

Proof. Since $[K,K]$ is characteristic in K and K is normal in G, then $[K,K]$ is normal in G. Hence, we may consider the action by conjugacy of on . This induces a representation . Let $z\in Z(G)$ be a nontrivial element in the centre of G such that $z\notin K$ . We note that $\overline {z}$ does not belong to . Furthermore, since $z\in Z(G)$ ,

(3-1)

Let $\overline {\phi }(\overline {z})=t$ , where . Notice that t is a nontrivial element in F. So, we conclude that $\eta $ is not injective since $\eta (t)$ is the identity homomorphism (see (3-1)).

In the following result, we consider the case where the holonomy representation defined in Lemma 3.2 is not injective and give conditions for the middle group to be a crystallographic group.

Proof. First, we note that $p^{-1}(\mathrm {Ker}(\varphi ))$ is a Bieberbach group, since it is finitely generated, torsion free and virtually abelian; see [Reference DekimpeDek96, Theorem 3.1.3(4)].

Now, we prove that $p^{-1}(\mathrm {Ker}(\varphi ))$ is free abelian. Since $p^{-1}(\mathrm {Ker}(\varphi ))$ is a Bieberbach group, it fits in a short exact sequence $ 1 \to A \to p^{-1}(\mathrm {Ker} (\varphi ) ) \to F \to 1 $ where F is a finite group and A is a free abelian group containing K as a normal subgroup of finite index. Suppose now that F is not the trivial group. Let $x \in p^{-1}(\mathrm {Ker} (\varphi ))$ be an element that is mapped onto a nontrivial element in F. We know that the induced map $F \to \mathrm {Aut}(A)$ is injective, so conjugation by x induces a nontrivial automorphism of A. However, since K is of finite index in the free abelian group A, this implies that conjugation by x also induces a nontrivial automorphism of K. But this is not possible since $x \in p^{-1}(\mathrm {Ker} (\varphi ) )$ .

Hence, $p^{-1}(\mathrm {Ker}(\varphi ))$ is free abelian and we obtain the sequence

such that the middle group is a crystallographic group.

3.2. Crystallographic structures and congruence subgroups of braid groups

In this subsection, we study a quotient of $B_{n}$ , namely, . Since is crystallographic [Reference Gonçalves, Guaschi and OcampoGGO17, Proposition 1], being isomorphic to the crystallographic braid group , it is reasonable to ask whether is crystallographic for any positive integer m. Here we give conditions for this statement to hold.

The following short exact sequence

induces a short exact sequence on the quotients

(3-2)

The action by conjugacy of on induces a homomorphism

(3-3)

As a consequence of Theorem 3.3, we have the following result.

Since the representation $\Theta _m$ is not injective for $m\geq 3$ , we cannot apply Lemma 3.2 in this case. However, we may give general conditions such that the group is crystallographic. We have the following result about crystallographic structures and quotients of braid groups by commutators of congruence subgroups.

Proof. From Theorem 3.4, if $\overline {\rho _m}^{-1}(\mathrm {Ker}(\Theta _m))$ is torsion free, where $\rho _m$ and $\Theta _m$ are the homomorphisms defined in (3-2) and (3-3), respectively, then the group is crystallographic.

We note that the $\mathrm {Ker}(\Theta _m)$ is isomorphic to $Z(\rho _m(B_n))$ the centre of $\rho _m(B_n)$ , since the full twist $\Delta _n^2$ generates the centre of $B_n$ and $\rho _m(\Delta _n^2)$ belongs to the normal subgroup $\mathrm {Ker}(\Theta _m)$ of the symplectic group $\rho _m(B_n)$ . Recall that, under the assumptions of the statement, $Z(\rho _m(B_n))$ is isomorphic to $\mathbb {Z}/2\mathbb {Z}$ . We consider now the following short exact sequence:

such that the kernel is torsion free (by hypothesis), the class of the element $\Delta _n^2\in B_n$ is a nontrivial element of $\overline {\rho _m}^{-1}(\mathrm {Ker}(\Theta _m))$ and .

Applying a standard method to give presentations for group extensions [Reference JohnsonJoh90, Ch. 10] and using the fact that the full twist generates the centre of $B_n$ , we conclude that the middle group $\overline {\rho _m}^{-1}(\mathrm {Ker}(\Theta _m))$ is free abelian and its rank corresponds to the rank of the free abelian group .

3.3. Symmetric quotients of congruence subgroups of braid groups

From the definition of congruence subgroups, we get an inclusion $\iota \colon B_n[m]\to B_n$ that induces a homomorphism . In general, $\overline {\iota }$ is not an isomorphism. In the following result, we study it in more detail.

Theorem 3.8. Let m be an odd positive integer and let $n \geq 3$ . The homomorphism induced from the inclusion $\iota \colon B_n[m]\to B_n$ ,

is injective. Furthermore, the group is a normal proper subgroup of such that the quotient is isomorphic to $(\mathbb {Z}/m \mathbb {Z})^{n(n-1)/2}$ .

Before delving into the proof, we state two technical lemmas that are needed.

Lemma 3.10. Let $N, H, G$ groups be such that $H\leq G$ and N is a normal subgroup of G. Then the inclusion homomorphism $\iota \colon H\hookrightarrow G$ induces an injective homomorphism

Lemma 3.11. Consider the following commutative diagrams of (vertical and horizontal) short exact sequences of groups in which every square is commutative:

1. (1)

   1. (a) If $\beta _i$ is an inclusion, for $i=1,2$ , and $\zeta $ is an isomorphism, then $\mu $ is an isomorphism.

   2. (b) If $\alpha _i$ is an inclusion, for $i=1,2$ , and $\mu $ is an isomorphism, then $\zeta $ is an isomorphism.

2. (2) Suppose that, for $i=1,2$ , the homomorphisms $\iota _i$ and $\psi _i$ are inclusions. Then $\eta $ is an isomorphism if and only if $\phi $ is.

Lemma 3.10 is a simple consequence of the standard isomorphism theorem between $H/N \cap H$ and the subgroup $NH/N$ of $G/N$ , while Lemma~3.11 can be easily proven using diagram chasing.

Proof of Theorem 3.8.

From [Reference Appel, Bloomquist, Gravel and HoldenABGH20, Theorem 3.1 and its proof], we have the following commutative diagram:

where $\tau $ is the natural surjective homomorphism that sends each braid generator $\sigma _i$ to the transposition $(i,\, i+1)$ , $\tau _m$ is the restriction of $\tau $ to the subgroup $B_n[m]$ , $\iota $ is the natural inclusion from the definition of congruence subgroups and $\psi $ is the restriction of $\iota $ to the subgroup $B_n[2m]$ .

Now, we consider the following diagram induced from the commutative square on the left, where the vertical arrows on this square are inclusion homomorphisms and $\psi |$ is the restriction of $\psi $ :

From Lemma 3.10, the third arrow $\overline {\psi }$ on the right is also injective. Since is a free abelian group of rank $n(n-1)/2$ , then is a free abelian group of finite rank, at most $n(n-1)/2$ . From [Reference Appel, Bloomquist, Gravel and HoldenABGH20, Corollary 2.4], the element $A_{i,j}^m$ belongs to $B_n[2m]$ for all $1\leq i<j\leq n$ , where $\{ A_{i,j} \mid 1\leq i<j\leq n \}$ is the set of Artin generators of $P_n$ . Since is generated by the set of cosets $\{ \overline {A_{i,j}} \mid 1\leq i<j\leq n \}$ , it follows that $\{ \overline {A_{i,j}^m} \mid 1\leq i<j\leq n \}$ is a basis of , so it has rank $n(n-1)/2$ . Furthermore, from the above,

(3-4)

Considering [Reference Bloomquist, Patzt and ScherichBPS22, Proposition 3.1], Arnol’d’s result $B_n[2]=P_n$ and with some set theoretical equivalences, we can see that

$$ \begin{align*} [P_{n}, P_{n}] \cap B_{n}[m] &= ([P_{n}, P_{n}] \cap P_{n} ) \cap B_{n}[m] \\ &= [P_{n}, P_{n}] \cap ( P_{n} \cap B_{n}[m] ) \\ &= [P_{n}, P_{n}] \cap B_{n}[{2m}]. \end{align*} $$

The following diagram is induced from the commutative square on the left, where the vertical arrows on this square are inclusion homomorphisms:

From Lemma 3.10, the third arrow $\overline {\iota }$ on the right is also injective. With this information and (3-4), we construct the following commutative diagram:

From Lemma 3.11 item (1), the homomorphism $\mu $ is an isomorphism and we get the result.

Let $n\geq 3$ . Recall from [Reference Appel, Bloomquist, Gravel and HoldenABGH20, Lemma 2.3] that the element $\sigma _i^m$ belongs to $B_n[m]$ for all $1\leq i\leq n-1$ , where $\{ \sigma _i \mid 1\leq i\leq n-1 \}$ is the set of Artin generators of $B_n$ . Although the set map $\xi \colon B_n\to B_n[m]$ defined by $\xi (\sigma _i)= \sigma _i^m$ , for all $1\leq i\leq n-1$ , is not a homomorphism, when m is odd, it induces an isomorphism on the quotient groups , as we show in the next result.

Theorem 3.12. Let m be a positive integer and let $n \geq 3$ . Consider the map

defined by $\overline {\xi }(\sigma _i)= \sigma _i^m$ for all $1\leq i\leq n-1$ . If m is odd, then $\overline {\xi }$ is an isomorphism. As a consequence, for $n\geq 3$ and m odd, is a crystallographic group of dimension $n(n-1)/2$ and holonomy group $S_n$ .

Proof. Suppose that $n\geq 3$ and m is an odd positive integer and consider the map

defined by $\overline {\xi }(\sigma _i)= \sigma _i^m$ for all $1\leq i\leq n-1$ . To show that $\overline {\xi }$ is a homomorphism, it is enough to verify that Artin’s relations are preserved by $\overline {\xi }$ .

Let $1\leq i,j\leq n$ such that $|i-j|\geq 2$ . From Artin’s relation $\sigma _i\sigma _j=\sigma _j\sigma _i$ , we obtain $\sigma _i^m\sigma _j^m=\sigma _j^m\sigma _i^m$ in $B_n[m]$ , which is then preserved by $\overline {\xi }$ .

Let $1\leq i\leq n-2$ . The equality $\sigma _i^m\sigma _{i+1}^m\sigma _i^m\sigma _{i+1}^{-m}\sigma _i^{-m}\sigma _{i+1}^{-m}=1$ is valid in . In fact, suppose that $m=2k+1$ , then from the action of conjugation in described in Section 2.1,

From Theorem 3.8, the homomorphism

is injective, and then $\sigma _i^m\sigma _{i+1}^m\sigma _i^m\sigma _{i+1}^{-m}\sigma _i^{-m}\sigma _{i+1}^{-m}=1$ in .

Now, consider the following commutative diagram of short exact sequences:

As seen in the proof of Theorem 3.8, the free abelian groups and of rank $n(n{\kern-1pt}-{\kern-1pt}1)/2$ have bases $\{ \overline {A_{i,j}^m} \mid 1{\kern-1pt}\leq{\kern-1pt} i{\kern-1pt}<{\kern-1pt}j{\kern-1pt}\leq{\kern-1pt} n \}$ and $\{ \overline {A_{i,j}} \mid 1{\kern-1pt}\leq{\kern-1pt} i{\kern-1pt}<{\kern-1pt}j{\kern-1pt}\leq{\kern-1pt} n \}$ , respectively. Since

is a homomorphism such that $\overline {\xi }|(A_{i,j})= A_{i,j}^m$ , for all $1\leq i<j\leq n$ , it is an isomorphism. Therefore, from the five lemma, $\overline {\xi }$ is an isomorphism.

The last part follows from the corresponding result on the crystallographic braid group ; see [Reference Gonçalves, Guaschi and OcampoGGO17, Proposition 1].

A group G is called co-Hopfian if it is not isomorphic to any of its proper subgroups, or equivalently, if every injective homomorphism $\phi \colon G\to G$ is surjective. It is known that the braid group $B_n$ is not co-Hopfian. However, for $n\geq 4$ , the quotient by its centre is co-Hopfian; see [Reference Bell and MargalitBM06].