Commensurability in Artin groups of spherical type

TL;DR

This paper characterizes when two Artin groups of spherical type are commensurable, showing it depends on their irreducible components and ranks, and provides a classification for groups of fixed rank.

Contribution

It establishes a criterion for commensurability of Artin groups of spherical type based on their irreducible components and ranks, and classifies those of fixed rank.

Findings

Two Artin groups are commensurable iff their irreducible components are pairwise commensurable.

Commensurable Artin groups of spherical type have the same rank.

Complete classification of irreducible Artin groups of rank n commensurable with type A_n.

Abstract

Let $A$ and $A'$ be two Artin groups of spherical type, and let $A_1,\dots,A_p$ (resp. $A'_1,\dots,A'_q$) be the irreducible components of $A$ (resp. $A'$). We show that $A$ and $A'$ are commensurable if and only if $p=q$ and, up to permutation of the indices, $A_i$ and $A'_i$ are commensurable for every $i$. We prove that, if two Artin groups of spherical type are commensurable, then they have the same rank. For a fixed $n$, we give a complete classification of the irreducible Artin groups of rank $n$ that are commensurable with the group of type $A_n$. Note that it will remain 6 pairs of groups to compare to get the complete classification of Artin groups of spherical type up to commensurability.

Equations

$$A[\Gamma ]=⟨S\mid \Pi (s,t,m_{s,t})=\Pi (t,s,m_{s,t}),\text{ for }s,t\in S,s\ne t,m_{s,t}\ne \infty ⟩.$$

$$CA[\Gamma ]=\overline{\mathrm{CA}[{\Gamma }_1]}\times \cdots \times \overline{\mathrm{CA}[{\Gamma }_k]}\times Z(\mathrm{CA}[{\Gamma }_1])\times \cdots \times Z(\mathrm{CA}[{\Gamma }_p])$$

$$A[\Gamma {]}^{+}=⟨S\mid \Pi (s,t,m_{s,t})=\Pi (t,s,m_{s,t})\text{ for }s,t\in S,s\ne t,m_{s,t}\ne \infty {⟩}^{+}.$$

$$M=V_{\mathbb{C}}\setminus \left(\bigcup _{H\in \mathcal{H}}{H}_{\mathbb{C}}\right).$$

$$\mathrm{CA}[\Gamma ]=\overline{\mathrm{CA}[{\Gamma }_1]}\times \cdots \times \overline{\mathrm{CA}[{\Gamma }_k]}\times Z(\mathrm{CA}[{\Gamma }_1])\times \cdots \times Z(\mathrm{CA}[{\Gamma }_p]).$$

$$Z_{\mathrm{CA}[\Gamma ]}({\hat{K}}_i)=(Z_{\mathrm{CA}[\Gamma ]}({\hat{K}}_i)\cap \overline{\mathrm{CA}[{\Gamma }_1]})\times \cdots \times (Z_{\mathrm{CA}[\Gamma ]}({\hat{K}}_i)\cap \overline{\mathrm{CA}[{\Gamma }_k]})\times (Z_{\mathrm{CA}[\Gamma ]}({\hat{K}}_i)\cap \tilde{B}).$$

$$K_i\hookrightarrow L_i\stackrel{f_{\chi (i)}}{\to }{L}_i\cap H_{\chi (i)}$$

$$K_i\hookrightarrow L_i\stackrel{f_j}{\to }{L}_i\cap H_j=Z(H_j)$$

$$\nu (s_1)=\nu (s_2)=(1,2)(3,4)(5,6),\nu (s_3)=(2,3)(1,5)(4,6),\nu (s_4)=(1,3)(2,4)(5,6),$$

$$\nu (s_5)=(1,2)(3,5)(4,6),\nu (s_6)=(2,3)(1,4)(5,6).$$

$${\nu }^{\prime }(s_1^{\prime })=(1,2)(3,4)(5,6),{\nu }^{\prime }(s_2^{\prime })=(2,3)(1,5)(4,6),{\nu }^{\prime }(s_3^{\prime })=(1,3)(2,4)(5,6),$$

$${\nu }^{\prime }(s_4^{\prime })=(1,2)(3,5)(4,6),{\nu }^{\prime }(s_5^{\prime })=(2,3)(1,4)(5,6).$$

$$u_1=(1,2)(3,4)(5,6),u_2=(2,3)(1,5)(4,6),u_3=(1,3)(2,4)(5,6),$$

$$u_4=(1,2)(3,5)(4,6),u_5=(2,3)(1,4)(5,6).$$

$${\nu }^{\prime }(s_1^{\prime })=(1,2)(3,4)(5,6),{\nu }^{\prime }(s_2^{\prime })=(2,3)(1,5)(4,6),{\nu }^{\prime }(s_3^{\prime })=(1,3)(2,4)(5,6),$$

$${\nu }^{\prime }(s_4^{\prime })=(1,2)(3,5)(4,6),{\nu }^{\prime }(s_5^{\prime })=(2,3)(1,4)(5,6).$$

$$u_1=(1,2)(3,4)(5,6),u_2=(2,3)(1,5)(4,6),u_3=(1,3)(2,4)(5,6),$$

$$u_4=(1,2)(3,5)(4,6),u_5=(2,3)(1,4)(5,6).$$

$$\displaystyle E_{1}^{0}=\big{\{}\big{(}(1,2),(2,3),(4,5,6)\big{)},\big{(}(1,2,3,4,5,6),(1,6,3,2,5,4),(1,3,5)(2,4,6)\big{)},$$

$$\displaystyle\big{(}(1,2,3,4,5,6),(1,6,3,2,5,4),(1,5,3)(2,6,4)\big{)},$$

$$\displaystyle\big{(}(1,4)(2,5)(3,6),(1,2)(3,4)(5,6),(1,3,5)(2,4,6)\big{)},$$

$$\displaystyle\big{(}(2,3)(4,5,6),(1,2)(4,5,6),(4,5,6)\big{)},\big{(}(2,3)(4,5,6),(1,2)(4,5,6),(4,6,5)\big{)}\big{\}}\,.$$

$$\displaystyle E_{2}^{0}=\big{\{}\big{(}(1,2),(2,3),(5,6),(4,5)\big{)},$$

$$\displaystyle\big{(}(1,4)(2,5)(3,6),(1,2)(3,4)(5,6),(1,4)(2,3)(5,6),(1,6)(2,5)(3,4)\big{)}\big{\}}\,.$$

$$\zeta ({\bar{s}}_1)=(2,4)(3,5),\zeta ({\bar{s}}_2)=(1,2)(4,5),\zeta ({\bar{s}}_3)=(2,3)(4,5).$$

$${\zeta }_1({\bar{s}}_1)=(1,2,3,4,5),{\zeta }_1({\bar{s}}_2)=(1,4,2,3,5),{\zeta }_1({\bar{s}}_3)=(1,5,4,3,2),$$

$${\zeta }_2({\bar{s}}_1)=(2,4)(3,5),{\zeta }_2({\bar{s}}_2)=(1,2)(4,5),{\zeta }_2({\bar{s}}_3)=(2,3)(4,5).$$

Full text

2010 Mathematics Subject Classification: Primary 20F36, Secondary 57M07; 20B30.

\newaliascnt

lemmatheorem \aliascntresetthelemma

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propositiontheorem \aliascntresettheproposition

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corollarytheorem \aliascntresetthecorollary

1 Introduction

We start by recalling the definitions of Coxeter groups and Artin groups. Let $S$ be a finite set. A Coxeter matrix over $S$ is a square matrix $M=(m_{s,t})_{s,t\in S}$ indexed by the elements of $S$, having coefficients in $\mathbb{N}\cup\{\infty\}$, and satisfying $m_{s,s}=1$ for every $s\in S$, and $m_{s,t}=m_{t,s}\geq 2$ for every $s,t\in S$, $s\neq t$. This matrix is represented by a labeled graph $\Gamma$, called Coxeter graph and defined by the following data. The set of vertices of $\Gamma$ is $S$. Two vertices $s,t\in S$, $s\neq t$, are connected by an edge if $m_{s,t}\geq 3$, and this edge is labeled with $m_{s,t}$ if $m_{s,t}\geq 4$.

If $s,t\in S$ and $m$ is an integer $\geq 2$, we denote by $\Pi(s,t,m)$ the word $sts\cdots$ of length $m$. In other words, $\Pi(s,t,m)=(st)^{\frac{m}{2}}$ if $m$ is even and $\Pi(s,t,m)=(st)^{\frac{m-1}{2}}s$ if $m$ is odd. Let $\Gamma$ be the Coxeter graph associated to such a Coxeter matrix. The Artin group associated to $\Gamma$ is the group $A=A[\Gamma]$ defined by the following presentation.

[TABLE]

The Coxeter group $W=W[\Gamma]$ of $\Gamma$ is the quotient of $A[\Gamma]$ by the relations $s^{2}=1$, $s\in S$. We say that $\Gamma$ is of spherical type if $W[\Gamma]$ is finite.

Let $\Gamma_{1},\dots,\Gamma_{p}$ be the connected components of $\Gamma$ and, for $i\in\{1,\dots,p\}$, let $S_{i}$ be the set of vertices of $\Gamma_{i}$, $A_{i}$ be the subgroup of $A$ generated by $S_{i}$ and $W_{i}$ be the subgroup of $W$ generated by $S_{i}$. We can easily check that $A_{i}$ is the Artin group of $\Gamma_{i}$ and $W_{i}$ is the Coxeter group of $\Gamma_{i}$ for every $i$, and that $A=A_{1}\times\cdots\times A_{p}$ and $W=W_{1}\times\cdots\times W_{p}$. In particular, $\Gamma$ has spherical type if and only if $\Gamma_{i}$ has spherical type for every $i\in\{1,\dots,p\}$. The classification of Coxeter graphs of spherical type has been known for a long time and it is given in the following theorem:

Theorem 1** (Coxeter,, 1935).**

A Coxeter graph $\Gamma$ is connected and has spherical type if and only if it is isomorphic to one of the graphs $A_{n}$ ($n\geq 1$), $B_{n}$ ($n\geq 2$), $D_{n}$ ($n\geq 4$), $E_{n}$ ($n\in\{6,7,8\}$), $F_{4}$, $H_{3}$, $H_{4}$ and $I_{2}(p)$ ($p\geq 5$) represented in Figure 1.

Actually, this classification is also the classification of Artin groups of spherical type up to isomorphism because, by (Paris,, 2004, Theorem 1.1), two Artin groups of spherical type are isomorphic if and only if their associated Coxeter graphs are isomorphic. It is then natural to ask if such a result remains valid when changing the word “isomorphic” by “commensurable”. The answer has been known for a long time: it is no because the Artin groups associated to $A_{n}$ and $B_{n}$ are commensurable (see section 6) and they are not isomorphic by (Paris,, 2004, Theorem 1.1). However, the classification of Artin groups of spherical type up to commensurability was a very open question before this article. For instance, no example of two non-commensurable Artin groups of spherical type having the same rank was known before. This article almost gives the entire classification of Artin groups of spherical type up to commensurability, meaning that there are only 6 comparisons of groups that we do not treat. Two of them have been solved by Soroko, (2020) after the first version of the present paper.

We recall that two groups $G_{1}$ and $G_{2}$ are commensurable if there are two finite index subgroups $H_{1}$ of $G_{1}$ and $H_{2}$ of $G_{2}$ such that $H_{1}$ is isomorphic to $H_{2}$. The study of commensurability is useful when studying virtual properties of groups. There is also a strong relationship between commensurable groups and quasi-isometric groups. In particular, for a finitely generated group $G$ endowed with any word metric, the inclusion map of a finite index subgroup in $G$ is a quasi-isometry. This implies that, if two finitely generated groups are commensurable, then they are also quasi-isometric. The converse implication is true only under certain conditions.

The commensurator (also called abstract commensurator) of a group $G$ will be denoted by ${\rm Com}(G)$. We recall its definition. Let ${\widetilde{{\rm Com}}}(G)$ be the set of triples $(U,V,f)$ where $U$ and $V$ are finite index subgroups of $G$, and $f:U\to V$ is an isomorphism. Let $\sim$ be the equivalence relation on ${\widetilde{{\rm Com}}}(G)$ such that $(U,V,f)\sim(U^{\prime},V^{\prime},f^{\prime})$ if there is a finite index subgroup $W$ of $U\cap U^{\prime}$ such that $f(\alpha)=f^{\prime}(\alpha)$ for every $\alpha\in W$. Hence we define ${\rm Com}(G)$ as ${\widetilde{{\rm Com}}}(G)/\sim$ and the group operation is induced by the composition. We can easily show that, if $A$ and $B$ are two commensurable groups, then ${\rm Com}(A)$ and ${\rm Com}(B)$ are isomorphic. Commensurators are in general difficult to compute. Fortunately, the commensurator of the Artin group associated to $A_{n}$ (the braid group) is well understood (Charney & Crisp,, 2005, Leininger & Margalit,, 2006) and it is indeed used to prove the results in this paper.

So far, the results regarding commensurability for Artin groups in general are quite limited. In (Crisp,, 2005), the author studies commensurability for Artin groups of large type (each $m_{s,t}\geq 3$ for $s\neq t$) associated to triangle-free connected Coxeter graphs having at least three vertices. In the last years, the research on this topic has been focused on right-angled Artin groups (RAAGs). A RAAG is an Artin group whose only relations in its presentation are commutations. It is often represented by a commutation graph, $\Upsilon$, which is defined by the following data. The set of vertices of $\Upsilon$ is the set of standard generators of the group. Two vertices are connected by an edge if and only if the corresponding generators commute. Apart from the classifications made for free and free-abelian groups, commensurability studies are made for RAAGs with commutation graphs $\Upsilon$ in the following cases:

• •

$\Upsilon$ is connected, triangle-free and square-free without any vertices of degree one (Kim & Koberda,, 2014);

• •

$\Upsilon$ is star-rigid with no induced 4-cycles and the outer automorphism of the Artin group is finite (Huang,, 2018);

• •

$\Upsilon$ is a tree of diameter $\leq 4$ (Behrstock & Neumann,, 2008, Casals-Ruiz et al.,, 2019a);

• •

$\Upsilon$ is a path graph (Casals-Ruiz et al.,, 2019b). In this work they also compared these commensurability classes to the ones of RAAGs defined by trees of diameter 4.

Remark. The results of this paper, notably Part (3) of Theorem 4, are being used in a paper in preparation of Ursula Hamenstädt (Hamenstädt,, 2019) to refute a conjecture made by Kontsevich and Zorich (Kontsevich,, 1997). We fix a tuple of non-negative integers $d=(p_{1},p_{2},\dots,p_{k})$ and consider the vector space of holomorphic one-forms of a Riemann surface with genus $g$ bigger or equal to 2. We denote by $M_{d}$ the moduli space of these one-forms having zeros $x_{1},x_{2},\dots,x_{k}$ with multiplicity $p_{1},p_{2},\dots,p_{k}$, respectively. The conjecture says that each connected component of $M_{d}$ has homotopy type $K(G,1)$, where $G$ is a group commensurable to some mapping class group. Hamenstädt uses the results in (Looijenga & Mondello,, 2014), to show that there are components in genus 3 that are classifying spaces for the quotients of the Artin groups $A[E_{6}]$ and $A[E_{7}]$ by their centers. She proves that the only mapping class group which could be commensurable to $A[E_{6}]/Z(A[E_{6}])$ is the quotient of the braid group on 7 strands by its center, that is, $A[A_{6}]/Z(A[A_{6}])$. By section 3, the non-commensurability of $A[E_{6}]/Z(A[E_{6}])$ and $A[A_{6}]/Z(A[A_{6}])$ is equivalent to the non-commensurability of $A[E_{6}]$ and $A[A_{6}]$. These components provide a counterexample to the conjecture.

Acknowledgements.

The first author was financed by a postdoctoral fellowship of the University of Burgundy and supported by the research grants MTM2016-76453-C2-1-P (financed by the Spanish Ministry of Economy and FEDER) and US-1263032 (financed by the Andalusian Ministry of Economy and Knowledge and the Operational Program FEDER 2014–2020). The second author was supported by the French project “AlMaRe” (ANR-19-CE40-0001-01) of the ANR. Both authors also thank the two referees for their carreful reading of the manuscript and for their interesting comments and suggestions.

2 Statements

Recall that our aim is to partially classify the Artin groups of spherical type up to commensurability. Our starting point is the following result which can be easily proven. It allows to reduce the question to the case where both Coxeter graphs have the same number of vertices.

Proposition \theproposition.

Let $\Gamma$ and $\Omega$ be two Coxeter graphs of spherical type. If $A[\Gamma]$ and $A[\Omega]$ are commensurable, then $\Gamma$ and $\Omega$ have the same number of vertices.

Proof.

Suppose that $A[\Gamma]$ and $A[\Omega]$ are commensurable. Let $n$ be the number of vertices of $\Gamma$ and let $m$ be the number of vertices of $\Omega$. We know that the cohomological dimension of $A[\Gamma]$ is $n$ and the cohomological dimension of $A[\Omega]$ is $m$ (Paris,, 2004, Proposition 3.1). As every finite index subgroup of $A[\Gamma]$ has the same cohomological dimension as $A[\Gamma]$ and every finite index subgroup of $A[\Omega]$ has the same cohomological dimension as $A[\Omega]$, we have $n=m$. ∎

In Section 5 we will prove the following result, which allows to reduce our problem to the study of two connected Coxeter graphs having the same number of vertices.

Theorem 2**.**

Let $\Gamma$ and $\Omega$ be two Coxeter graphs of spherical type. Let $\Gamma_{1},\dots,\Gamma_{p}$ be the connected components of $\Gamma$ and $\Omega_{1},\dots,\Omega_{q}$ be the connected components of $\Omega$. Then $A[\Gamma]$ and $A[\Omega]$ are commensurable if and only if $p=q$ and $A[\Gamma_{i}]$ and $A[\Omega_{i}]$ are commensurable for every $i\in\{1,\dots,p\}$, up to permutation of the indices.

Let $G$ be a group. A subgroup $H$ of $G$ is a direct factor of $G$ is there is a subgroup $K$ of $G$ such that $G=H\times K$. We say that $G$ is indecomposable if $G$ does not have any non-trivial proper direct factor. We say that $G$ is strongly indecomposable if $G$ is infinite and every finite index subgroup $H$ of $G$ is indecomposable. A strong Remak decomposition of $G$ is a finite index subgroup $H$ of $G$ with a direct product decomposition $H=H_{1}\times\cdots\times H_{p}$ such that $H_{i}$ is strongly indecomposable for every $i\in\{1,\dots,p\}$. Two strong Remak decompositions of $G$, $H=H_{1}\times\cdots\times H_{p}$ and $H^{\prime}=H_{1}^{\prime}\times\cdots\times H_{q}^{\prime}$, are said to be equivalent if $p=q$ and $H_{i}$ and $H_{i}^{\prime}$ are commensurable for every $i\in\{1,\dots,p\}$, up to permutation of the indices.

The center of a group $G$ will be denoted by $Z(G)$. If $\Gamma$ is a connected Coxeter graph of spherical type then, thanks to (Brieskorn & Saito,, 1972) and (Deligne,, 1972), the center of $A[\Gamma]$ is a cyclic infinite group. The quotient $A[\Gamma]/Z(A[\Gamma])$ will be denoted by $\overline{A[\Gamma]}$ and it will play an important role in our study. Moreover, we denote by $\theta:A[\Gamma]\to W[\Gamma]$ the canonical projection and by ${\rm CA}[\Gamma]$ the kernel of $\theta$. As before, we let $\overline{{\rm CA}[\Gamma]}={\rm CA}[\Gamma]/Z({\rm CA}[\Gamma])$. In Section 3, we will prove that $Z({\rm CA}[\Gamma])\simeq\mathbb{Z}$ and ${\rm CA}[\Gamma]\simeq\overline{{\rm CA}[\Gamma]}\times Z({\rm CA}[\Gamma])$ (see section 3). If $\Gamma$ is reduced to a single vertex, then ${\rm CA}[\Gamma]=Z({\rm CA}[\Gamma])\simeq\mathbb{Z}$ and $\overline{{\rm CA}[\Gamma]}=\{1\}$. Otherwise $\overline{{\rm CA}[\Gamma]}\neq\{1\}$.

The proof of Theorem 2 is based on the following result which will be proven in Section 4.

Theorem 3**.**

• (1)

Let $\Gamma$ be a connected Coxeter graph of spherical type which is not reduced to a single vertex. Then $\overline{{\rm CA}[\Gamma]}$ is strongly indecomposable.

• (2)

Let $\Gamma$ be a Coxeter graph of spherical type and let $\Gamma_{1},\dots,\Gamma_{p}$ be its connected components. We suppose that each $\Gamma_{1},\dots,\Gamma_{k}$ has at least two vertices and each of $\Gamma_{k+1},\dots,\Gamma_{p}$ is reduced to a single vertex. Then

[TABLE]

is a strong Remak decomposition of $A[\Gamma]$, and it is unique up to equivalence.

A similar result for Coxeter groups is obtained in (Paris,, 2007). In order to finish the classification, we just need to compare the Artin groups associated to connected Coxeter graphs of spherical type with the same number of vertices. In Section 6 we prove the following result, which compares every group of this type with the corresponding Artin group of type $A_{n}$.

Theorem 4**.**

• (1)

Let $n\geq 2$. Then $A[A_{n}]$ and $A[B_{n}]$ are commensurable.

• (2)

Let $n\geq 4$. Then $A[A_{n}]$ and $A[D_{n}]$ are not commensurable.

• (3)

Let $n\in\{6,7,8\}$. Then $A[A_{n}]$ and $A[E_{n}]$ are not commensurable.

• (4)

$A[A_{4}]$* and $A[F_{4}]$ are not commensurable.*

• (5)

Let $n\in\{3,4\}$. Then $A[A_{n}]$ and $A[H_{n}]$ are not commensurable.

• (6)

Let $p\geq 5$. Then $A[A_{2}]$ and $A[I_{2}(p)]$ are commensurable.

The strategy of the proof of this theorem is the following. We use direct proofs to show Part (1) and Part (6). Using the fact that the abstract commensurator of $\overline{A[A_{n}]}$ is known to be a mapping class group of a punctured sphere (see Charney & Crisp,, 2005), we show that, if $A[\Gamma]$ is commensurable with $A[A_{n}]$, then there is a homomorphism $\varphi:\overline{A[\Gamma]}\to\mathfrak{S}_{n+2}\times\{\pm 1\}$ whose kernel has no generalized torsion. Then, in Parts (2) to (5), in order to prove that $A[\Gamma]$ and $A[A_{n}]$ are not commensurable, we check in each case that the kernel of every homomorphism $\varphi:\overline{A[\Gamma]}\to\mathfrak{S}_{n+2}\times\{\pm 1\}$ has generalized torsion.

The description of $\overline{A[D_{4}]}$ as the pure mapping class group of the three times punctured torus (Labruère & Paris,, 2001) as been recently used by Soroko, (2020) to apply the same techniques presented in this article to show that $A[D_{4}]$ is not commensurable with $A[F_{4}]$ and $A[H_{4}]$. For the remaining cases, we have no hint on how to describe the abstract commensurator of one of the two groups, and this is needed in our argument. So, the following cases remain open:

• •

For $n=6,7,8$, we do not know if $A[D_{n}]$ and $A[E_{n}]$ are commensurable.

• •

For $n=4$, we do not know if $A[F_{4}]$ and $A[H_{4}]$ are commensurable.

3 A technical and useful result

This section is devoted to some technical results (see section 3) that will be the key to prove the main theorems of the forthcoming sections. These results are also interesting by themselves.

Let $\Gamma$ be a Coxeter graph of spherical type. The Artin monoid associated to $\Gamma$ is the monoid $A[\Gamma]^{+}$ having the same presentation as $A[\Gamma]$, that is,

[TABLE]

By (Brieskorn & Saito,, 1972) (see also (Paris,, 2002)), $A[\Gamma]^{+}$ naturally injects in $A[\Gamma]$. We define a partial order $\leq_{L}$ on $A[\Gamma]$ by $\alpha\leq_{L}\beta$ if $\alpha^{-1}\beta\in A[\Gamma]^{+}$. Also by (Brieskorn & Saito,, 1972), the ordered set $(A[\Gamma],\leq_{L})$ is a lattice. We denote by $\wedge_{L}$ and $\vee_{L}$ the lattice operations in $(A[\Gamma],\leq_{L})$. In this case, the Garside element of $A[\Gamma]$ is defined as $\Delta=\vee_{L}S$. Again by (Brieskorn & Saito,, 1972) and (Deligne,, 1972) we know that, if $\Gamma$ is connected, then the center of $A[\Gamma]$ is infinite and cyclic, and it is generated by an element $\delta$ of the form $\delta=\Delta^{\kappa}$, where $\kappa\in\{1,2\}$. This element $\delta$ will be called the standard generator of $Z(A[\Gamma])$. We can also express $\delta$ as follows. Let $S=\{s_{1},\dots,s_{n}\}$. Then, by (Brieskorn & Saito,, 1972), $\delta=(s_{1}s_{2}\cdots s_{n})^{\frac{h}{2}}$ if $\kappa=1$ and $\delta=(s_{1}s_{2}\cdots s_{n})^{h}$ if $\kappa=2$, where $h$ is the Coxeter number of $\Gamma$, that is, the order of $s_{1}s_{2}\cdots s_{n}$ in the associated Coxeter group. These equalities do not depend on the choice when numbering the elements of $S$.

Let $\Gamma$ be a connected Coxeter graph of spherical type. Let $z:A[\Gamma]\to\mathbb{Z}$ be the homomorphism such that $z(s)=1$ for every $s\in S$. Hence, considering the later expression of $\delta$, we have that $z(\delta)>0$. The quotient $A[\Gamma]/Z(A[\Gamma])$ is denoted by $\overline{A[\Gamma]}$. Moreover, recall that $\theta:A[\Gamma]\to W[\Gamma]$ is the canonical projection, ${\rm CA}[\Gamma]$ is the kernel of $\theta$, and $\overline{{\rm CA}[\Gamma]}={\rm CA}[\Gamma]/Z({\rm CA}[\Gamma])$.

Proposition \theproposition.

Let $\Gamma,\Omega$ be two connected Coxeter graphs of spherical type.

• (1)

If $U$ is a finite index subgroup of $A[\Gamma]$, then $Z(U)=Z(A[\Gamma])\cap U$. In particular, $Z(U)$ is an infinite cyclic group.

• (2)

We have ${\rm CA}[\Gamma]\simeq\overline{{\rm CA}[\Gamma]}\times Z({\rm CA}[\Gamma])\simeq\overline{{\rm CA}[\Gamma]}\times\mathbb{Z}$.

• (3)

$A[\Gamma]$* and $A[\Omega]$ are commensurable if and only if $\overline{A[\Gamma]}$ and $\overline{A[\Omega]}$ are commensurable.*

• (4)

The group $\overline{A[\Gamma]}$ injects in its commensurator ${\rm Com}(\overline{A[\Gamma]})$.

Proof.

Proof of Part (1). Let $U$ be a finite index subgroup of $A[\Gamma]$. The inclusion $Z(A[\Gamma])\cap U\subset Z(U)$ is obvious. We need to show $Z(U)\subset Z(A[\Gamma])\cap U$. Let $\alpha\in Z(U)$ and $s\in S$. As $U$ is a finite index subgroup, there is $k\geq 1$ such that $s^{k}\in U$. Then $\alpha s^{k}\alpha^{-1}=s^{k}$ and, by (Paris,, 1997b, Corollary 5.3), $\alpha s\alpha^{-1}=s$. This proves that $\alpha$ belongs to $Z(A[\Gamma])$. To see that $Z(U)$ is infinite cyclic, notice that $Z(U)$ is a finite index subgroup of $Z(A[\Gamma])$, which is infinite cyclic because $\Gamma$ is connected.

Proof of Part (2). Let $V=\oplus_{s\in S}{\mathbb{R}}e_{s}$ be a real vector space with a basis in one-to-one correspondence with $S$. By (Bourbaki,, 1968), $W=W[\Gamma]$ has a faithful linear representation $\rho:W\to{\rm GL}(V)$ and $\rho(W)$ is generated by reflections. We denote by $\mathcal{H}$ the set of reflection hyperplanes of $W$. We let $V_{\mathbb{C}}=\mathbb{C}\otimes V$ and $H_{\mathbb{C}}=\mathbb{C}\otimes H$ for every $H\in\mathcal{H}$. Let also

[TABLE]

Notice that $M$ is a connected manifold of dimension $2\,|S|$. By (Brieskorn,, 1971), $\pi_{1}(M)={\rm CA}[\Gamma]$.

Let $h:V_{\mathbb{C}}\setminus\{0\}\to\mathbb{P}V_{\mathbb{C}}$ be the Hopf fibration. Let $\overline{M}=h(M)$ and denote by $h_{\mathcal{H}}:M\to\overline{M}$ the restriction of $h$ to $M$. Recall that the fiber of $h_{\mathcal{H}}$ is $\mathbb{C}^{*}$. As $\mathcal{H}$ is non-empty, we know that $h_{\mathcal{H}}$ is topologically a trivial fibration (Orlik & Terao,, 1992, Proposition 5.1). In other words, $M$ is homeomorphic to $\overline{M}\times\mathbb{C}^{*}$, hence ${\rm CA}[\Gamma]=\pi_{1}(M)\simeq\pi_{1}(\overline{M})\times\mathbb{Z}$. From this decomposition it follows that $Z({\rm CA}[\Gamma])\simeq Z(\pi_{1}(\overline{M}))\times\mathbb{Z}$. But, thanks to Part (1), $Z({\rm CA}[\Gamma])$ is isomorphic to $\mathbb{Z}$, which does not have any non-trivial direct product decomposition, hence $Z(\pi_{1}(\overline{M}))=1$, $\pi_{1}(\overline{M})\simeq{\rm CA}[\Gamma]/Z({\rm CA}[\Gamma])=\overline{{\rm CA}[\Gamma]}$, and ${\rm CA}[\Gamma]\simeq\overline{{\rm CA}[\Gamma]}\times\mathbb{Z}=\overline{{\rm CA}[\Gamma]}\times Z({\rm CA}[\Gamma])$.

Proof of Part (3). Suppose that $A[\Gamma]$ and $A[\Omega]$ are commensurable. There is a finite index subgroup $U$ of $A[\Gamma]$ and a finite index subgroup $V$ of $A[\Omega]$ such that $U$ is isomorphic to $V$. Let $\pi:A[\Gamma]\to\overline{A[\Gamma]}$ and $\pi^{\prime}:A[\Omega]\to\overline{A[\Omega]}$ be the corresponding canonical projections. Then $\pi(U)=U/(Z(A[\Gamma])\cap U)$ is a finite index subgroup of $\overline{A[\Gamma]}$, $\pi^{\prime}(V)=V/(Z(A[\Omega])\cap V)$ is a finite index subgroup of $\overline{A[\Omega]}$, and, by Part (1), we have $\pi(U)=U/Z(U)$ and $\pi^{\prime}(V)=V/Z(V)$. Hence $\pi(U)$ is isomorphic to $\pi^{\prime}(V)$. Therefore, $\overline{A[\Gamma]}$ and $\overline{A[\Omega]}$ are commensurable.

Suppose that $\overline{A[\Gamma]}$ and $\overline{A[\Omega]}$ are commensurable. By Part (1), $Z({\rm CA}[\Gamma])={\rm CA}[\Gamma]\cap Z(A[\Gamma])$, then $\pi({\rm CA}[\Gamma])=\overline{{\rm CA}[\Gamma]}$ and $\overline{{\rm CA}[\Gamma]}$ is a finite index subgroup of $\overline{A[\Gamma]}$. Likewise, $\overline{{\rm CA}[\Omega]}$ is a finite index subgroup of $\overline{A[\Omega]}$, then $\overline{{\rm CA}[\Gamma]}$ and $\overline{{\rm CA}[\Omega]}$ are commensurable. This means that there are finite index subgroups $\bar{U}$ of $\overline{{\rm CA}[\Gamma]}$ and $\bar{V}$ of $\overline{{\rm CA}[\Omega]}$ such that $\bar{U}$ and $\bar{V}$ are isomorphic. By Part (2), ${\rm CA}[\Gamma]=\overline{{\rm CA}[\Gamma]}\times\mathbb{Z}$ and ${\rm CA}[\Omega]=\overline{{\rm CA}[\Omega]}\times\mathbb{Z}$. Let $U=\bar{U}\times\mathbb{Z}\subset{\rm CA}[\Gamma]$ and $V=\bar{V}\times\mathbb{Z}\subset{\rm CA}[\Omega]$. Hence $U$ is a finite index subgroup of ${\rm CA}[\Gamma]$, $V$ is a finite index subgroup of ${\rm CA}[\Omega]$, and $U$ and $V$ are isomorphic. Thus, ${\rm CA}[\Gamma]$ and ${\rm CA}[\Omega]$ are commensurable, so $A[\Gamma]$ and $A[\Omega]$ are commensurable.

Proof of Part (4). If $G$ is a group and $\alpha\in G$ we denote by $c_{\alpha}:G\to G$, $\beta\mapsto\alpha\beta\alpha^{-1}$, the conjugation by $\alpha$. Then we have a homomorphism $\iota_{G}:G\to{\rm Com}(G)$ sending $\alpha$ to the class of $(G,G,c_{\alpha})$. Let $\iota=\iota_{\overline{A[\Gamma]}}:\overline{A[\Gamma]}\to{\rm Com}(\overline{A[\Gamma]})$, and let $\alpha\in A[\Gamma]$ be such that $\pi(\alpha)\in{\rm Ker}(\iota)$, where $\pi:A[\Gamma]\to\overline{A[\Gamma]}$ is the corresponding canonical projection. There is a finite index subgroup $\overline{U}$ of $\overline{A[\Gamma]}$ such that $\pi(\alpha)\,\pi(\beta)\,\pi(\alpha^{-1})=\pi(\beta)$ for every $\beta\in\pi^{-1}(\bar{U})$. Let $s\in S$. As $\bar{U}$ has finite index in $\overline{A[\Gamma]}$, there is $k\geq 1$ such that $\pi(s^{k})\in\bar{U}$. We have $\pi(\alpha)\,\pi(s^{k})\,\pi(\alpha^{-1})=\pi(s^{k})$, so $\pi(\alpha s^{k}\alpha^{-1}s^{-k})=1$ and then $\alpha s^{k}\alpha^{-1}s^{-k}\in{\rm Ker}(\pi)=Z(A[\Gamma])=\langle\delta\rangle$. Hence there is $\ell\in\mathbb{Z}$ such that $\alpha s^{k}\alpha^{-1}s^{-k}=\delta^{\ell}$. Recall that $z:A[\Gamma]\to\mathbb{Z}$ is the homomorphism sending every element of $S$ to $1$ and $z(\delta)>0$. Then $0=z(\alpha s^{k}\alpha^{-1}s^{-k})=z(\delta^{\ell})=\ell\,z(\delta)$ and $z(\delta)>0$, having that $\ell=0$ and $\alpha s^{k}\alpha^{-1}=s^{k}$. By (Paris,, 2004, Corollary 5.3) it follows that $\alpha s\alpha^{-1}=s$. This shows that $\alpha$ belongs to $Z(A[\Gamma])$, so $\pi(\alpha)=1$ and $\iota$ is injective. ∎

The proof of the following corollary is completely and explicitly included in the proof of the proposition above.

Corollary \thecorollary.

Let $\Gamma$ be a connected Coxeter graph of spherical type. Then $Z({\rm CA}[\Gamma])$ is an infinite cyclic group. On the other hand, $\overline{{\rm CA}[\Gamma]}$ can be viewed as a subgroup of $\overline{A[\Gamma]}$, it has finite index in $\overline{A[\Gamma]}$, and its center is trivial.

4 Strong Remak decomposition

In this section, we denote by $\Gamma$ a Coxeter graph of spherical type associated to a Coxeter matrix $M=(m_{s,t})_{s,t\in S}$. Recall that our aim is to show Theorem 3.

Let $G$ be a group and $E$ be a subset of $G$. Recall that the normalizer of $E$ in $G$ is $N_{G}(E)=\{\alpha\in G\mid\alpha E\alpha^{-1}=E\}$ and the centralizer of $E$ in $G$ is $Z_{G}(E)=\{\alpha\in G\mid\alpha e\alpha^{-1}=e\text{ for every }e\in E\}$. If $E=\{e\}$, we just write $Z_{G}(e)=Z_{G}(\{e\})$ to refer to the centralizer of $e$. We also recall that the center of $G$ is denoted by $Z(G)$.

Lemma \thelemma.

Suppose that $\Gamma$ is connected and different from a single vertex. Let $U$ be a finite index subgroup of $\overline{{\rm CA}[\Gamma]}$. Then $Z(U)=Z_{\overline{{\rm CA}[\Gamma]}}(U)=\{1\}$.

Proof.

We just need to show that $Z_{\overline{{\rm CA}[\Gamma]}}(U)=\{1\}$ because $Z(U)\subset Z_{\overline{{\rm CA}[\Gamma]}}(U)$. This follows directly from the forth statement of section 3. Indeed, as $U$ has finite index, any $\alpha$ element in $Z(U)$ is sent to the class $(G,G,1)$ via the homomorphism $\iota$ of the afore-mentioned proposition. Hence $\alpha$ sits in the kernel of $\iota$, which is trivial. ∎

Lemma \thelemma.

Let $G$ be a group, let $G_{1},G_{2}$ be two subgroups of $G$ such that $G=G_{1}\times G_{2}$, and let $H$ be a subgroup of $G$. Then $Z_{G}(H)=(Z_{G}(H)\cap G_{1})\times(Z_{G}(H)\cap G_{2})$.

Proof.

The inclusion $(Z_{G}(H)\cap G_{1})\times(Z_{G}(H)\cap G_{2})\subset Z_{G}(H)$ is obvious. Then we just need to show that $Z_{G}(H)\subset(Z_{G}(H)\cap G_{1})\times(Z_{G}(H)\cap G_{2})$. Let $\alpha\in Z_{G}(H)$ and $\gamma\in H$. We write $\alpha=(\alpha_{1},\alpha_{2})$ and $\gamma=(\gamma_{1},\gamma_{2})$ with $\alpha_{1},\gamma_{1}\in G_{1}$ and $\alpha_{2},\gamma_{2}\in G_{2}$. We have $1=\alpha\gamma\alpha^{-1}\gamma^{-1}=(\alpha_{1}\gamma_{1}\alpha_{1}^{-1}\gamma_{1}^{-1},\alpha_{2}\gamma_{2}\alpha_{2}^{-1}\gamma_{2}^{-1})$, hence $\alpha_{1}\gamma_{1}\alpha_{1}^{-1}\gamma_{1}^{-1}=1$. Moreover, $\alpha_{1}\gamma_{2}\alpha_{1}^{-1}\gamma_{2}^{-1}=1$, because $\alpha_{1}\in G_{1}$ and $\gamma_{2}\in G_{2}$, so $\alpha_{1}\gamma\alpha_{1}^{-1}\gamma^{-1}=1$. Thus, $\alpha_{1}\in(Z_{G}(H)\cap G_{1})$. Analogously, we can prove that $\alpha_{2}\in(Z_{G}(H)\cap G_{2})$. ∎

Proof of Theorem 3. We suppose that $\Gamma$ is connected and different from a single vertex. Let $U$ be a finite index subgroup of $\overline{{\rm CA}[\Gamma]}$. Let $U_{1},U_{2}$ be two subgroups of $U$ such that $U=U_{1}\times U_{2}$ and let $\tilde{U}=U\times Z({\rm CA}[\Gamma])$, which is included in $\overline{{\rm CA}[\Gamma]}\times Z({\rm CA}[\Gamma])={\rm CA}[\Gamma]$. Let $\tilde{U_{1}}=U_{1}$ and $\tilde{U_{2}}=U_{2}\times Z({\rm CA}[\Gamma])$, having $\tilde{U}=\tilde{U_{1}}\times\tilde{U_{2}}.$ As ${\rm CA}[\Gamma]$ has finite index in $A[\Gamma]$, $\tilde{U}$ has finite index in $A[\Gamma]$ and, by applying (Marin,, 2007, Theorem 5B), we know that either $\tilde{U_{1}}\subset Z(A[\Gamma])$ or $\tilde{U_{2}}\subset Z(A[\Gamma])$. Also by section 3, we have that $Z(A[\Gamma])\cap\tilde{U}=Z(\tilde{U})\subset Z(A[\Gamma])\cap{\rm CA}[\Gamma]=Z({\rm CA}[\Gamma])$. Then $\tilde{U}_{1}\subset Z({\rm CA}[\Gamma])$ or $\tilde{U}_{2}\subset Z({\rm CA}[\Gamma])$, so $U_{1}=\{1\}$ or $U_{2}=\{1\}$. This shows the first part of the theorem. We still have to prove the second part.

Let $\Gamma_{1},\dots,\Gamma_{p}$ be the connected components of $\Gamma$. We suppose that every $\Gamma_{1},\dots,\Gamma_{k}$ has at least two vertices and that each of $\Gamma_{k+1},\dots,\Gamma_{p}$ is reduced to a single vertex. We have that ${\rm CA}[\Gamma]={\rm CA}[\Gamma_{1}]\times\cdots\times{\rm CA}[\Gamma_{p}]$. By section 3, ${\rm CA}[\Gamma_{i}]=\overline{{\rm CA}[\Gamma_{i}]}\times Z({\rm CA}[\Gamma_{i}])$ for every $i\in\{1,\dots,k\}$ and ${\rm CA}[\Gamma_{i}]=Z({\rm CA}[\Gamma_{i}])$ for every $i\in\{k+1,\dots,p\}$, hence

[TABLE]

Then, $\overline{{\rm CA}[\Gamma_{i}]}$ is strongly indecomposable for every $i\in\{1,\dots,k\}$. Moreover, $Z({\rm CA}[\Gamma_{i}])$ is strongly indecomposable, because $Z({\rm CA}[\Gamma_{i}])\simeq\mathbb{Z}$, for every $i\in\{1,\dots,p\}$. Therefore, (1) is a strong Remak decomposition of $A[\Gamma]$. Now, we take a strong Remak decomposition of $A[\Gamma]$ of the form $H=H_{1}\times\cdots\times H_{m}$ and turn to prove that it is equivalent to (1).

Claim 1*.*

We can assume that $H\subset{\rm CA}[\Gamma]$.

Proof of Claim 1. Let $H_{i}^{\prime}=H_{i}\cap{\rm CA}[\Gamma]$ for every $i\in\{1,\dots,m\}$ and $H^{\prime}=H_{1}^{\prime}\times\cdots\times H_{m}^{\prime}$. Since ${\rm CA}[\Gamma]$ is a finite index subgroup of $A[\Gamma]$, $H_{i}^{\prime}$ has finite index in $H_{i}$ for every $i\in\{1,\dots,m\}$. This means that $H^{\prime}$ has finite index in $H$ and therefore $H^{\prime}$ has finite index in $A[\Gamma]$. As $H_{i}$ is strongly indecomposable and $H_{i}^{\prime}$ has finite index in $H_{i}$, $H_{i}^{\prime}$ is strongly indecomposable, for every $i\in\{1,\dots,m\}$. Then $H^{\prime}=H_{1}^{\prime}\times\cdots\times H_{m}^{\prime}$ is a strong Remak decomposition of $A[\Gamma]$. By construction, this decomposition is equivalent to $H=H_{1}\times\cdots\times H_{m}$ and $H^{\prime}$ is included in ${\rm CA}[\Gamma]$. This finishes the proof of Claim 1.

Let $\tilde{B}=Z({\rm CA}[\Gamma_{1}])\times\cdots\times Z({\rm CA}[\Gamma_{p}])\simeq\mathbb{Z}^{p}$ and $B=H\cap\tilde{B}$. Set $K_{i}=H\cap\overline{{\rm CA}[\Gamma_{i}]}$ for every $i\in\{1,\dots,k\}$. As $H$ has finite index in ${\rm CA}[\Gamma]$, $K_{i}$ has finite index in $\overline{{\rm CA}[\Gamma_{i}]}$ for every $i\in\{1,\dots,k\}$ and $B$ has finite index in $\tilde{B}$.

Claim 2*.*

We have that $Z(H)=B$.

Proof of Claim 2. Let $\alpha\in Z(H)\subset{\rm CA}[\Gamma]$. Then, by section 4, $\alpha$ can be expressed as $\alpha=\alpha_{1}\cdots\alpha_{k}\beta$, where $\alpha_{i}\in\overline{{\rm CA}[\Gamma_{i}]}\cap Z_{{{\rm CA}[\Gamma]}}(H)$ for every $i\in\{1,\dots,k\}$ and $\beta\in\tilde{B}$. Since $K_{i}\subset H$, $\alpha_{i}$ commutes with every element in $K_{i}$, so $\alpha_{i}\in Z_{\overline{{\rm CA}[\Gamma_{i}]}}(K_{i})$. By section 4, $Z_{\overline{{\rm CA}[\Gamma_{i}]}}(K_{i})=\{1\}$, hence $\alpha_{i}=1$. Therefore, $\alpha=\beta\in\tilde{B}\cap H=B$. This proves that $Z(H)\subset B$. To see $B\subset Z(H)$, just notice that $B=Z({\rm CA}[\Gamma])\cap H\subset Z(H)$, because $H\subset{\rm CA}[\Gamma]$ by Claim 1. This finishes the proof of Claim 2.

Let $\hat{K}_{i}=K_{1}\times\cdots\times K_{i-1}\times K_{i+1}\times\cdots\times K_{k}\times B$ and $L_{i}=(\overline{{\rm CA}[\Gamma_{i}]}\times\tilde{B})\cap H$, for every $i\in\{1,\dots,k\}$.

Claim 3*.*

Let $i\in\{1,\dots,k\}$. Then $Z_{H}(\hat{K}_{i})=L_{i}$ and $L_{i}=(L_{i}\cap H_{1})\times\cdots\times(L_{i}\cap H_{m})$.

Proof of Claim 3. By section 4, we have that

[TABLE]

Let $j\in\{1,\dots,k\}$ be such that $j\neq i$. Then, by section 4, $(Z_{{\rm CA}[\Gamma]}(\hat{K}_{i})\cap\overline{{\rm CA}[\Gamma_{j}]})\subset Z_{\overline{{\rm CA}[\Gamma_{j}]}}(K_{j})=\{1\}$. Moreover, $(Z_{{\rm CA}[\Gamma]}(\hat{K}_{i})\cap\overline{{\rm CA}[\Gamma_{i}]})=\overline{{\rm CA}[\Gamma_{i}]}$ and $(Z_{{\rm CA}[\Gamma]}(\hat{K}_{i})\cap\tilde{B})=\tilde{B}$. Therefore $Z_{{\rm CA}[\Gamma]}(\hat{K}_{i})=\overline{{\rm CA}[\Gamma_{i}]}\times\tilde{B}$ and $Z_{H}(\hat{K}_{i})=Z_{{\rm CA}[\Gamma]}(\hat{K}_{i})\cap H=L_{i}$. Finally, by section 4, $L_{i}=Z_{H}(\hat{K}_{i})=(L_{i}\cap H_{1})\times\cdots\times(L_{i}\cap H_{m})$. This finishes the proof of Claim 3.

Claim 4*.*

Let $i\in\{1,\dots,k\}$. Then $Z(L_{i})=B$ and $L_{i}/B$ is strongly indecomposable. Also, there is $\chi(i)\in\{1,\dots,m\}$ such that $L_{i}/B=(L_{i}\cap H_{\chi(i)})/Z(H_{\chi(i)})$ and $L_{i}\cap H_{j}=Z(H_{j})$ if $j\neq\chi(i)$.

Proof of Claim 4. Since $B\subset Z({\rm CA}[\Gamma])$ and $B\subset L_{i}\subset H$, we have $B\subset Z(L_{i})$. Now, take $\alpha\in Z(L_{i})$. As $L_{i}$ is a subgroup of $\overline{{\rm CA}[\Gamma_{i}]}\times\tilde{B}$, by section 4 we can write $\alpha$ in the form $\alpha=\alpha_{i}\beta$, where $\alpha_{i}\in\overline{{\rm CA}[\Gamma_{i}]}\cap Z_{CA[\Gamma]}(L_{i})$ and $\beta\in\tilde{B}$. Since $K_{i}\subset L_{i}$, $\alpha_{i}$ commutes with every element in $K_{i}$, so $\alpha_{i}\in Z_{\overline{{\rm CA}[\Gamma_{i}]}}(K_{i})$. By section 4, $Z_{\overline{{\rm CA}[\Gamma_{i}]}}(K_{i})=\{1\}$, hence $\alpha_{i}=1$. Therefore, $\alpha=\beta\in\tilde{B}\cap H=B$ and then $Z(L_{i})\subset B$.

Let $\pi:\overline{{\rm CA}[\Gamma_{i}]}\times\tilde{B}\rightarrow\overline{{\rm CA}[\Gamma_{i}]}$ be the projection homomorphism and let $\pi^{\prime}$ be the restriction of $\pi$ to $L_{i}$. Then ${\rm Ker}(\pi^{\prime})={\rm Ker}(\pi)\cap L_{i}=\tilde{B}\cap L_{i}\subset Z(L_{i})=B$. On the other hand, $B\subset\tilde{B}\cap L_{i}$, hence ${\rm Ker}(\pi^{\prime})=B$. Using the first isomorphism theorem, we have that $L_{i}/B\simeq\pi(L_{i})$. As $L_{i}$ has finite index in $\overline{{\rm CA}[\Gamma_{i}]}\times\tilde{B}$, $\pi(L_{i})$ has finite index in $\overline{{\rm CA}[\Gamma_{i}]}$, which is strongly indecomposable. This implies that $L_{i}/B$ is strongly indecomposable.

By Claim 3, we have that $L_{i}=(L_{i}\cap H_{1})\times\cdots\times(L_{i}\cap H_{m})$. If we quotient this equality by $B=Z(H)=Z(H_{1})\times\cdots\times Z(H_{m})$, we get $L_{i}/B=(L_{i}\cap H_{1})/Z(H_{1})\times\cdots\times(L_{i}\cap H_{m})/Z(H_{m})$. We already know that $L_{i}/B$ is strongly indecomposable, so there is $\chi(i)\in\{1,\dots,m\}$ such that $L_{i}/B=(L_{i}\cap H_{\chi(i)})/Z(H_{\chi(i)})$ and $(L_{i}\cap H_{j})/Z(H_{j})=\{1\}$ if $j\neq\chi(i)$. This implies that $L_{i}\cap H_{j}\subset Z(H_{j})$ if $j\neq\chi(i)$. Finally, notice that $Z(H_{i})\subset B\subset L_{i}$. This proves Claim 4.

For $j\in\{1,\dots,m\}$ we denote by $f_{j}:H\to H_{j}$ the projection of $H$ on $H_{j}$. Let $K=K_{1}\times\cdots\times K_{k}\times B$. For $i\in\{1,\dots,k\}$ we denote by $g_{i}:K\to K_{i}$ the projection of $K$ on $K_{i}$, and we denote by $h:K\to B$ the projection of $K$ on $B$. Notice that, since $K_{i}\times B$ is a subgroup of $L_{i}$, $K_{i}$ is a subgroup of $L_{i}/B$. Then, $K_{i}$ injects into $L_{i}$ and into $L_{i}/B$, which by Claim 4 is isomorphic to $(L_{i}\cap H_{\chi(i)})/Z(H_{\chi(i)})$. This means that the composition

[TABLE]

has to be injective. In other words, the restriction $f_{\chi(i)}|_{K_{i}}:K_{i}\to H_{\chi(i)}$ is injective. Also by Claim 4, if $j\neq\chi(i)$,

[TABLE]

Hence, we have $f_{j}(K_{i})\subset Z(H_{j})$.

Let $\psi_{i}:K_{i}\to B$ be the map defined by $\psi_{i}(\alpha)=\prod_{j\neq\chi(i)}f_{j}(\alpha)^{-1}$. As $B$ is an abelian group, $\psi_{i}$ is a well-defined homomorphism. Let $\psi:K\to B$ be the map defined by $\psi(\alpha)=\prod_{i=1}^{k}(\psi_{i}\circ g_{i})(\alpha)$. Then, again, $\psi$ is a well-defined homomorphism because $B$ is abelian. Also, notice that $\psi(\beta)=1$ for every $\beta\in B$. If $\varphi:K\to K$ is the map defined by $\varphi(\alpha)=\alpha\,\psi(\alpha)$, it is clear that $\varphi$ is a homomorphism. In addition, as $\psi(\beta)=1$ for every $\beta\in B$, $\varphi$ is invertible and $\varphi^{-1}$ is defined by $\varphi^{-1}(\alpha)=\alpha\,\psi(\alpha)^{-1}$.

Claim 5*.*

For every $i\in\{1,\dots,k\}$ we have $\varphi(K_{i})\subset H_{\chi(i)}$.

Proof of Claim 5. Let $i\in\{1,\dots,k\}$ and $\alpha\in K_{i}$. For $\ell\in\{1,\dots,k\}$, $\ell\neq i$, we have that $g_{\ell}(\alpha)=1$, then $(\psi_{\ell}\circ g_{\ell})(\alpha)=1$ and $\psi(\alpha)=\psi_{i}(\alpha)$. Moreover, $\alpha=\prod_{j=1}^{m}f_{j}(\alpha)$, having $\varphi(\alpha)=f_{\chi(i)}(\alpha)\in H_{\chi(i)}$. This finishes the proof of Claim 5.

Up to applying $\varphi$ we can assume that $K_{i}\subset H_{\chi(i)}$ for every $i\in\{1,\dots,k\}$.

Claim 6*.*

• (1)

For every $i\in\{1,\dots,k\}$ we have $f_{\chi(i)}(K)=K_{i}$ and $f_{\chi(i)}(\hat{K}_{i})=\{1\}$. Moreover, $K_{i}$ is a finite index subgroup of $H_{\chi(i)}$.

• (2)

For every $i,j\in\{1,\dots,k\}$, $i\neq j$, we have $\chi(i)\neq\chi(j)$.

Proof of Claim 6. As $K$ has finite index in $H$, $f_{\chi(i)}(K)$ has finite index in $H_{\chi(i)}$. Notice also that $f_{\chi(i)}(K_{i})=K_{i}$ because $K_{i}\subset H_{\chi(i)}$. We have that $K=K_{i}\times\hat{K}_{i}$, so $f_{\chi(i)}(K)=f_{\chi(i)}(K_{i})\,f_{\chi(i)}(\hat{K}_{i})=K_{i}\,f_{\chi(i)}(\hat{K}_{i})$ and $[K_{i},f_{\chi(i)}(\hat{K}_{i})]=\{1\}$. Moreover, by section 4, $K_{i}\cap f_{\chi(i)}(\hat{K}_{i})\subset Z(K_{i})=\{1\}$. Hence, $f_{\chi(i)}(K)\simeq K_{i}\times f_{\chi(i)}(\hat{K}_{i})$. As $H_{\chi(i)}$ is strongly indecomposable, we have that $f_{\chi(i)}(\hat{K}_{i})=\{1\}$ and $f_{\chi(i)}(K)=K_{i}$. On the other hand, let $j\in\{1,\dots,k\}$ be such that $j\neq i$. As $K_{j}$ is a subgroup of $\hat{K}_{i}$, we have that $f_{\chi(i)}(K_{j})=\{1\}\neq K_{j}$, and then $\chi(i)\neq\chi(j)$. This finishes the proof of Claim 6.

By the results from above, $m\geq k$ and we can suppose that $\chi(i)=i$ for every $i\in\{1,\dots,k\}$, up to renumbering the $H_{i}$’s. Recall that $B=Z(H)=Z(H_{1})\times\cdots\times Z(H_{m})$ and $Z(H_{i})=f_{i}(B)$ for every $i\in\{1,\dots,m\}$. If $i\in\{1,\dots,k\}$, then $B\subset\hat{K}_{i}$, so by Claim 6, $\{1\}=f_{i}(B)=Z(H_{i})$. Hence, $B=Z(H_{k+1})\times\cdots\times Z(H_{m})$. We also have that $K=K_{1}\times\cdots\times K_{k}\times B$ is a subgroup of $K_{1}\times\cdots\times K_{k}\times H_{k+1}\times\cdots\times H_{m}$, and that $K$ is a finite index subgroup of $H=H_{1}\times\cdots\times H_{k}\times H_{k+1}\times\cdots\times H_{m}$. Therefore $B$ has finite index in $H_{k+1}\times\cdots\times H_{m}$.

For $j\in\{k+1,\dots,m\}$, we let $B_{j}=B\cap H_{j}$. As $B$ is a finite index subgroup of $H_{k+1}\times\cdots\times H_{m}$, $B_{j}$ has finite index in $H_{j}$. In addition, as $H_{j}$ is strongly indecomposable, $B_{j}$ is indecomposable. The group $B_{j}$ is a subgroup of $B\simeq\mathbb{Z}^{p}$ and it is indecomposable, so $B_{j}\simeq\mathbb{Z}$. Let $B^{\prime}=B_{k+1}\times\cdots\times B_{m}$. Then $B^{\prime}$ is a finite index subgroup of $H_{k+1}\times\cdots\times H_{m}$ and $B^{\prime}$ has finite index in $B$. As $B^{\prime}\simeq\mathbb{Z}^{m-k}$ and $B\simeq\mathbb{Z}^{p}$, it follows that $m-k=p$.

For $i\in\{1,\dots,k\}$, $K_{i}$ is a finite index subgroup of both $H_{i}$ and $\overline{{\rm CA}[\Gamma_{i}]}$, so $H_{i}$ and $\overline{{\rm CA}[\Gamma_{i}]}$ are commensurable. Moreover, for every $i\in\{k+1,\dots,m\}$, $Z({\rm CA}(\Gamma_{i-k}))\simeq\mathbb{Z}\simeq B_{i}$, and $B_{i}$ is a finite index subgroup of $H_{i}$. Therefore $Z({\rm CA}[\Gamma_{i-k}])$ and $H_{i}$ are commensurable. ∎

5 Reduction to the connected case

Theorem 2 is a consequence of Theorem 3 and the following proposition.

Proposition \theproposition.

Let $G$ and $G^{\prime}$ be two infinite groups. We suppose that $G$ (resp. $G^{\prime}$) has a unique strong Remak decomposition up to equivalence, $H=H_{1}\times\cdots\times H_{p}$ (resp. $H^{\prime}=H_{1}^{\prime}\times\cdots\times H_{q}^{\prime}$). Then $G$ is commensurable with $G^{\prime}$ if and only if $p=q$ and, up to permutation of the factors, $H_{i}$ is commensurable with $H_{i}^{\prime}$ for every $i\in\{1,\dots,p\}$.

Proof.

Suppose that $G$ and $G^{\prime}$ are commensurable. There is a finite index subgroup $K$ of $G$ and a finite index subgroup $K^{\prime}$ of $G^{\prime}$ such that $K\simeq K^{\prime}$. Let $\varphi:K\to K^{\prime}$ be an isomorphism between $K$ and $K^{\prime}$. For every $i\in\{1,\dots,p\}$ we take $K_{i}=K\cap H_{i}$ and $U=K_{1}\times\cdots\times K_{p}$. As $K$ has finite index in $G$, $K_{i}$ has finite index in $H_{i}$. It follows that $U$ is a finite index subgroup of $H$ (and of $G$) and $U=K_{1}\times\cdots\times K_{p}$ is a strong Remak decomposition of $G$. The group $U$ is a finite index subgroup of $K$, so $\varphi(U)=\varphi(K_{1})\times\cdots\times\varphi(K_{p})$ is a finite index subgroup of $\varphi(K)=K^{\prime}$ and then also a finite index subgroup of $G^{\prime}$. The subgroups $\varphi(K_{i})$ ($i\in\{1,\dots,p\})$ are strongly indecomposable, hence $\varphi(U)=\varphi(K_{1})\times\cdots\times\varphi(K_{p})$ is a strong Remak decomposition of $G^{\prime}$. As $G^{\prime}$ has only one decomposition of that form (up to equivalence), we have that $p=q$ and $\varphi(K_{i})$ is commensurable with $H_{i}^{\prime}$ for every $i\in\{1,\dots,p\}$, up to permutation of the factors. Also, as $K_{i}\simeq\varphi(K_{i})$ is a finite index subgroup of $H_{i}$, it follows that $H_{i}$ and $H_{i}^{\prime}$ are commensurable for every $i\in\{1,\dots,p\}$.

Suppose that $p=q$ and that $H_{i}$ is commensurable with $H_{i}^{\prime}$ for every $i\in\{1,\dots,p\}$. There is a finite index subgroup $K_{i}$ of $H_{i}$ and a finite index subgroup $K_{i}^{\prime}$ of $H_{i}^{\prime}$ such that $K_{i}\simeq K_{i}^{\prime}$. Take $U=K_{1}\times\cdots\times K_{p}$ and $U^{\prime}=K_{1}^{\prime}\times\cdots\times K_{p}^{\prime}$. As $K_{i}$ has finite index in $H_{i}$ for every $i$, the subgroup $U$ has finite index in $H$ and also has finite index in $G$. Analogously, $U^{\prime}$ has finite index in $G^{\prime}$. It is obvious that $U$ and $U^{\prime}$ are isomorphic. Therefore, $G$ and $G^{\prime}$ are commensurable. ∎

Proof of Theorem 2. Let $\Gamma$ and $\Omega$ be two Coxeter graphs of spherical type. Let $\Gamma_{1},\dots,\Gamma_{p}$ be the connected components of $\Gamma$ and $\Omega_{1},\dots,\Omega_{q}$ be the connected components of $\Omega$. If $p=q$ and $A[\Gamma_{i}]$ and $A[\Omega_{i}]$ are commensurable for every $i\in\{1,\dots,p\}$, then it is clear that $A[\Gamma]$ and $A[\Omega]$ are commensurable. Then suppose that $A[\Gamma]$ and $A[\Omega]$ are commensurable. We need to show that $p=q$ and that $A[\Gamma_{i}]$ and $A[\Omega_{i}]$ are commensurable for every $i\in\{1,\dots,p\}$ up to permutation of the indices.

Suppose that every $\Gamma_{1},\dots,\Gamma_{k}$ has a least two vertices and that each of $\Gamma_{k+1},\dots,\Gamma_{p}$ is reduced to a single vertex. Analogously, suppose that every $\Omega_{1},\dots,\Omega_{\ell}$ has a least two vertices and that each of $\Omega_{\ell+1},\dots,\Omega_{q}$ is reduced to a single vertex. By Theorem 3, ${\rm CA}[\Gamma]=\overline{{\rm CA}[\Gamma_{1}]}\times\cdots\times\overline{{\rm CA}[\Gamma_{k}]}\times Z({\rm CA}[\Gamma_{1}])\times\cdots\times Z({\rm CA}[\Gamma_{p}])$ and ${\rm CA}[\Omega]=\overline{{\rm CA}[\Omega_{1}]}\times\cdots\times\overline{{\rm CA}[\Omega_{\ell}]}\times Z({\rm CA}[\Omega_{1}])\times\cdots\times Z({\rm CA}[\Omega_{q}])$ are strong Remak decompositions of $A[\Gamma]$ and $A[\Omega]$, respectively, and they are unique up to equivalence. Let $i\in\{1,\dots,k\}$ and $j\in\{1,\dots,q\}$. Let $U$ be a finite index subgroup of $\overline{{\rm CA}[\Gamma_{i}]}$ and let $V$ be a finite index subgroup of $Z({\rm CA}[\Omega_{j}])$. By section 4 we have that $Z(U)=\{1\}$. Moreover, $V$ is a finite index subgroup of $Z({\rm CA}[\Omega_{j}])\simeq\mathbb{Z}$, hence $V\simeq\mathbb{Z}$. Then, $U$ and $V$ are not isomorphic. This shows $\overline{{\rm CA}[\Gamma_{i}]}$ and $Z({\rm CA}[\Omega_{j}])$ are not commensurable.

By applying section 5, we know that $k=\ell$, $p=q$ and that $\overline{{\rm CA}[\Gamma_{i}]}$ and $\overline{{\rm CA}[\Omega_{i}]}$ are commensurable for every $i\in\{1,\dots,k\}$, up to permutation of the indices. Let $i\in\{1,\dots,k\}$. As $\overline{{\rm CA}[\Gamma_{i}]}$ and $\overline{{\rm CA}[\Omega_{i}]}$ are commensurable, by section 3, $\overline{A[\Gamma_{i}]}$ and $\overline{A[\Omega_{i}]}$ are commensurable. Then, by section 3, $A[\Gamma_{i}]$ and $A[\Omega_{i}]$ are commensurable. Let $i\in\{k+1,\dots,p\}$. Thus, $A[\Gamma_{i}]\simeq\mathbb{Z}\simeq A[\Omega_{i}]$, having that $A[\Gamma_{i}]$ and $A[\Omega_{i}]$ are commensurable. ∎

Remark 5*.*

An alternative proof of Theorem 2, based on (Kapovich et al.,, 1998, Theorem B), has been communicated to us by one of the referees. The idea is as follows. Consider the decomposition (1) given in the proof of Theorem 3. We can show using (Calvez & Wiest,, 2017), (Sisto,, 2016) and (Kapovich et al.,, 1998, Proposition 4.2) that each $\overline{{\rm CA}[\Gamma_{i}]}$ is of coarse type I. It follows by (Kapovich et al.,, 1998, Theorem B) that the decomposition (1) is unique up to quasi-isometry. To pass from quasi-isometry to commensurability we have to apply again (Kapovich et al.,, 1998, Theorem B) in the following manner. Let $\Gamma$ and $\Omega$ be two Coxeter graphs of spherical type. We assume that ${\rm CA}[\Gamma]$ and ${\rm CA}[\Omega]$ are commensurable and we consider the same decompositions of ${\rm CA}[\Gamma]$ and ${\rm CA}[\Omega]$ as in the above proof. Then, there exist finite index subgroups $U$ of ${\rm CA}[\Gamma]$ and $V$ of ${\rm CA}[\Omega]$ and an isomorphism $f\colon U\rightarrow V$. We extend $f$ to a quasi-isometry $f\colon{\rm CA}[\Gamma]\rightarrow{\rm CA}[\Omega]$. By applying (Kapovich et al.,, 1998, Theorem B), we obtain that $p=q$, $k=\ell$ and, for each $i\in\{i,\dots,k\}$ there exists $j\in\{1,\dots,k\}$ such that the composition $\overline{{\rm CA}[\Gamma_{i}]}\rightarrow{\rm CA}[\Gamma]\rightarrow{\rm CA}[\Omega]\rightarrow\overline{{\rm CA}[\Omega_{j}]}$ is a quasi-isometry, where the first map is the inclusion, the second map is $f$, and the third map is the projection. This map restricted to $U\cap\overline{{\rm CA}[\Gamma_{i}]}$ is an injective homomorphism whose image must be of finite index in ${\rm CA}[\Omega_{j}]$.

6 Comparison with the Artin group of type $A_{n}$

Let $n\in\mathbb{N}$, $n\geq 2$, and let $\Gamma$ be a connected Coxeter graph of spherical type with $n$ vertices. Recall that the aim of this section is to determine whether $A[\Gamma]$ and $A[A_{n}]$ are commensurable or not. We start with the cases where $A[\Gamma]$ and $A[A_{n}]$ are commensurable.

Lemma \thelemma.

Let $n\geq 2$. Then $A[A_{n}]$ and $A[B_{n}]$ are commensurable.

Proof.

Let $\theta:A[A_{n}]\to W[A_{n}]$ be the quotient homomorphism and let $H$ be the subgroup of $W[A_{n}]$ generated by $\{s_{2},\dots,s_{n}\}$. By (Lambropoulou,, 1994), $\theta^{-1}(H)$ is isomorphic to $A[B_{n}]$. It has finite index in $A[A_{n}]$ because $W[A_{n}]$ is finite, hence $A[A_{n}]$ and $A[B_{n}]$ are commensurable. ∎

Lemma \thelemma.

Let $p\geq 5$. Then $A[A_{2}]$ and $A[I_{2}(p)]$ are commensurable.

Proof.

Let $\Gamma=I_{2}(p)$. Then $A[\Gamma]=\langle s,t\mid\Pi(s,t,p)=\Pi(t,s,p)\rangle$. We consider the construction of the proof of section 3 (2). Let $V={\mathbb{R}}e_{s}\oplus{\mathbb{R}}e_{t}$. By (Bourbaki,, 1968), $W=W[\Gamma]$ has a faithful linear representation $\rho:W\to{\rm GL}(V)$, and $\rho(W)$ is generated by reflections. In our case, $W$ is the dihedral group of order $2p$ and $\rho:W\to{\rm GL}(V)$ is the standard representation of $W$. Let $\mathcal{H}$ be the set of reflection lines of $W$. Take $V_{\mathbb{C}}=\mathbb{C}\otimes V$, $H_{\mathbb{C}}=\mathbb{C}\otimes H$ for every $H\in\mathcal{H}$, and $M=V_{\mathbb{C}}\setminus\left(\cup_{H\in\mathcal{H}}H_{\mathbb{C}}\right)$. Let $h:V_{\mathbb{C}}\setminus\{0\}\to\mathbb{P}V_{\mathbb{C}}$ be the Hopf fibration and $\overline{M}=h(M)$. Thanks to the proof of section 3 (2), we know that $\pi_{1}(\overline{M})=\overline{{\rm CA}[\Gamma]}$.

In this case, $\mathbb{P}V_{\mathbb{C}}$ is the complex projective line and $\overline{M}$ is the complement of $|\mathcal{H}|=p$ points in $\mathbb{P}V_{\mathbb{C}}$, hence $\overline{{\rm CA}[\Gamma]}=\pi_{1}(\overline{M})$ is isomorphic to the free group $F_{p-1}$ of rank $p-1$. Analogously, $\overline{{\rm CA}[A_{2}]}$ is isomorphic to $F_{2}$. As $F_{p-1}$ is isomorphic to a finite index subgroup of $F_{2}$, it follows that $\overline{{\rm CA}[\Gamma]}$ and $\overline{{\rm CA}[A_{2}]}$ are commensurable. By section 3, we have that $\overline{A[A_{2}]}$ and $\overline{A[\Gamma]}$ are commensurable. Therefore, by section 3, $A[A_{2}]$ and $A[\Gamma]$ are commensurable. ∎

Let $\Sigma=\Sigma_{g,b}$ be the orientable surface of genus $g$ with $b$ boundary components. Let $\mathcal{P}_{n}$ be a collection of $n$ different points in the interior of $\Sigma$. Recall that the mapping class group of the pair $(\Sigma,\mathcal{P}_{n})$, denoted by $\mathcal{M}(\Sigma,\mathcal{P}_{n})$, is the group of isotopy classes of homeomorphisms $h:\Sigma\to\Sigma$ that preserve the orientation, fix the boundary of $\Sigma$ pointwise and preserve $\mathcal{P}_{n}$ setwise. The extended mapping class group of the pair $(\Sigma,\mathcal{P}_{n})$, denoted by $\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})$, is the group of isotopy classes of homeomorphisms $h:\Sigma\to\Sigma$ that fix the boundary of $\Sigma$ pointwise and preserve $\mathcal{P}_{n}$ setwise. Notice that, if the surface $\Sigma$ has non-empty boundary, the homeomorphisms fixing this boundary pointwise cannot change the orientation of $\Sigma$ and we have $\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})=\mathcal{M}(\Sigma,\mathcal{P}_{n})$. Otherwise, $\mathcal{M}(\Sigma,\mathcal{P}_{n})$ has index $2$ in $\mathcal{M}^{*}(\Sigma_{n},\mathcal{P}_{n})$.

Denote by $\mathfrak{S}_{n}$ the permutation group of $\{1,\dots,n\}$. The action of $\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})$ on $\mathcal{P}_{n}$ induces a homomorphism $\theta^{\prime}:\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})\to\mathfrak{S}_{n}$, whose kernel is denoted by $\mathcal{P}\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})$. On the other hand, we can define another homomorphism $\omega:\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})\to\{\pm 1\}$ sending an element $h\in\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})$ to $1$ if it preserves the orientation and to $-1$ otherwise. Notice that the kernel of $\omega$ is $\mathcal{M}(\Sigma,\mathcal{P}_{n})$. These two homomorphisms lead to the construction of the homomorphism $\hat{\theta}:\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})\to\mathfrak{S}_{n}\times\{\pm 1\}$ defined by $h\mapsto(\theta^{\prime}(h),\omega(h))$. The kernel of $\hat{\theta}$ is called the pure mapping class group of the pair $(\Sigma,\mathcal{P}_{n})$ and it is denoted by $\mathcal{P}\mathcal{M}(\Sigma,\mathcal{P}_{n})$.

These mapping class groups and the problem that we are studying are related by the following theorem.

Theorem 6** (Charney & Crisp,, 2005).**

Let $\Sigma=\Sigma_{0,0}$ and let $\mathcal{P}_{n+2}$ be a family of $n+2$ points in $\Sigma$. Then ${\rm Com}(\overline{A[A_{n}]})\simeq\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n+2})$.

Lemma \thelemma.

Let $\Sigma=\Sigma_{0,0}$ and let $\mathcal{P}_{n+2}$ be a family of $n+2$ points in $\Sigma$. Then ${\rm Ker}(\hat{\theta})=\mathcal{P}\mathcal{M}(\Sigma,\mathcal{P}_{n+2})\simeq\overline{{\rm CA}[A_{n}]}$.

Proof.

Let $\mathcal{B}_{n+1}$ be the braid group on $n+1$ strands. By (Artin,, 1947b), $\mathcal{M}(\Sigma_{0,1},\mathcal{P}_{n+1})=\mathcal{B}_{n+1}=A[A_{n}]$ and $\mathcal{P}\mathcal{M}(\Sigma_{0,1},\mathcal{P}_{n+1})={\rm CA}[A_{n}]$. Let $\delta$ be the standard generator of $Z(A[A_{n}])$. It is well-known that $\delta\in{\rm CA}[A_{n}]$ and $Z(A[A_{n}])=Z({\rm CA}[A_{n}])=\langle\delta\rangle$. Notice that $\delta$, seen as an element of $\mathcal{P}\mathcal{M}(\Sigma_{0,1},\mathcal{P}_{n+1})$, is the Dehn twist about the boundary component of $\Sigma_{0,1}$. Then $\overline{{\rm CA}[A_{n}]}=\mathcal{P}\mathcal{M}(\Sigma_{0,1},\mathcal{P}_{n+1})/\langle\delta\rangle=\mathcal{P}\mathcal{M}(\Sigma_{0,0},\mathcal{P}_{n+2})$; see, for example (Paris & Rolfsen,, 2000). ∎

Let $G$ be a group. We say that an element $\alpha\in G$ is a generalized torsion element if there are $p\geq 1$ and $\beta_{1},\dots,\beta_{p}\in G$ such that $(\beta_{1}\alpha\beta_{1}^{-1})(\beta_{2}\alpha\beta_{2}^{-1})\cdots(\beta_{p}\alpha\beta_{p}^{-1})=1$. We say that $G$ has generalized torsion if it contains a non-trivial generalized torsion element. For most of our cases, the criterion we will use to show that $A[\Gamma]$ and $A[A_{n}]$ are not commensurable is given by the following two results.

Lemma \thelemma.

Let $\Gamma$ be a connected Coxeter graph of spherical type with $n$ vertices. Let $\Phi:\overline{A[\Gamma]}\to\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{n+2})$ be a homomorphism and set $\varphi=\hat{\theta}\circ\Phi:\overline{A[\Gamma]}\to\mathfrak{S}_{n+2}\times\{\pm 1\}$. If ${\rm Ker}(\varphi)$ has generalized torsion, then $\Phi$ is not injective.

Proof.

Assume that $\Phi$ is injective and that ${\rm Ker}(\varphi)$ has generalized torsion. As $\overline{{\rm CA}[A_{n}]}={\rm Ker}(\hat{\theta})$, the homomorphism $\Phi$ induces an injective homomorphism $\Phi^{\prime}:{\rm Ker}(\varphi)\to\overline{{\rm CA}[A_{n}]}$. Recall that a group is called biorderable if it admits a total ordering invariant under left-multiplication and right-multiplication. We know by (Rolfsen & Zhu,, 1998) that ${\rm CA}[A_{n}]$ is biorderable. By section 3, $\overline{{\rm CA}[A_{n}]}$ is a subgroup of ${\rm CA}[A_{n}]$, hence $\overline{{\rm CA}[A_{n}]}$ is also biorderable, having that ${\rm Ker}(\varphi)$ is biorderable. However, a non-trivial biorderable group has no generalized torsion (Rolfsen & Zhu,, 1998). This is a contradiction. ∎

Corollary \thecorollary.

Let $\Gamma$ be a connected Coxeter graph of spherical type with $n$ vertices. If the kernel of every homomorphism $\varphi:\overline{A[\Gamma]}\to\mathfrak{S}_{n+2}\times\{\pm 1\}$ has generalized torsion, then $A[\Gamma]$ and $A[A_{n}]$ are not commensurable.

Proof.

Assume that $A[\Gamma]$ and $A[A_{n}]$ are commensurable. By section 3, $\overline{A[\Gamma]}$ injects in ${\rm Com}(\overline{A[\Gamma]})$. Again by section 3, $\overline{A[\Gamma]}$ and $\overline{A[A_{n}]}$ are commensurable, so ${\rm Com}(\overline{A[\Gamma]})\simeq{\rm Com}(\overline{A[A_{n}]})$. Moreover, by Theorem 6, ${\rm Com}(\overline{A[A_{n}]})=\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{n+2})$. Then we have an injective homomorphism $\Phi:\overline{A[\Gamma]}\to\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{n+2})$. Let $\varphi=\hat{\theta}\circ\Phi:\overline{A[\Gamma]}\to\mathfrak{S}_{n+2}\times\{\pm 1\}$. Therefore, by section 6, ${\rm Ker}(\varphi)$ does not have generalized torsion, having a contradiction. ∎

From here, in order to finish our proof, for each considered Coxeter graph $\Gamma$ and each homomorphism $\varphi\colon\overline{A[\Gamma]}\rightarrow\mathfrak{S}_{n+2}\times\{\pm 1\}$, we show an element in ${\rm Ker}(\varphi)$ having generalized torsion. To find such elements we apply the following strategy, well-known to some experts. Let $G$ be a group. An element $\beta\in G$ is quasi-central if there exists $n\geq 1$ such that $\beta^{n}$ lies in the center of $G$. Assume that $\beta$ is a quasi-central element and that $\alpha$ is an element of $G$ which does not commute with $\beta$. Then $\alpha\beta\alpha^{-1}\beta^{-1}$ is a generalized torsion non-trivial element of $G$. The quasi-central elements of the Artin groups of spherical type are well-understood, and we look for quasi-central elements in standard parabolic subgroups that belong to the kernel of $\varphi$ to find our generalized torsion elements.

We will use the following notations and definitions. For a group $G$ and $\alpha\in G$ we denote by $c_{\alpha}:G\to G$, $\beta\mapsto\alpha\beta\alpha^{-1}$, the conjugation by $\alpha$. We say that two homomorphisms $\varphi_{1},\varphi_{2}:G\to H$ are conjugate if there is $\alpha\in H$ such that $\varphi_{2}=c_{\alpha}\circ\varphi_{1}$. Moreover, a homomorphism $\varphi:G\to H$ is said to be cyclic if the image of $\varphi$ is a cyclic subgroup of $H$.

Lemma \thelemma.

The groups $A[D_{4}]$ and $A[A_{4}]$ are not commensurable.

Proof.

Let $\varphi:\overline{A[D_{4}]}\to\mathfrak{S}_{6}\times\{\pm 1\}$ be a homomorphism written in the form $\varphi=\varphi_{1}\times\varphi_{2}$, where $\varphi_{1}:\overline{A[D_{4}]}\to\mathfrak{S}_{6}$ and $\varphi_{2}:\overline{A[D_{4}]}\to\{\pm 1\}$ are two homomorphisms. By section 6, we just need to show that ${\rm Ker}(\varphi)$ has generalized torsion. We denote by $s_{1},s_{2},s_{3},s_{4}$ the standard generators of $A[D_{4}]$ numbered as in Figure 1. We also denote by $\pi:A[D_{4}]\to\overline{A[D_{4}]}$ the quotient homomorphism and $\bar{s}_{i}=\pi(s_{i})$ for every $i\in\{1,2,3,4\}$. Notice that $\varphi_{2}$ is always cyclic since its image is contained in $\{\pm 1\}$, which is a cyclic group. Then, the relations $s_{j}s_{3}s_{j}=s_{3}s_{j}s_{3}$, for $j\in\{1,2,4\}$, imply that there is $\epsilon\in\{\pm 1\}$ such that $\varphi_{2}(\bar{s}_{i})=\epsilon$ for every $i\in\{1,2,3,4\}$.

Firstly, suppose that $\varphi_{1}$ is cyclic. Let $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$ and $\beta=\bar{s}_{3}\bar{s}_{2}\bar{s}_{1}\bar{s}_{3}\bar{s}_{1}^{-4}$. Then $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$, and $\alpha\beta\alpha\beta^{-1}=1$, having that ${\rm Ker}(\varphi)$ has generalized torsion. Now, suppose that $\varphi_{1}$ is not cyclic. A direct computation using the software SageMath (see code in Cumplido & Paris,, 2020) shows that there are $14400$ non-cyclic homomorphisms from $\overline{A[D_{4}]}$ to $\mathfrak{S}_{6}$ divided into $40$ conjugacy classes. By using the same software, we check that in every case we have either $\varphi_{1}(\bar{s}_{1})=\varphi_{1}(\bar{s}_{2})$ or $\varphi_{1}(\bar{s}_{1})=\varphi_{1}(\bar{s}_{4})$ or $\varphi_{1}(\bar{s}_{2})=\varphi_{1}(\bar{s}_{4})$. Then we can assume without loss of generality that $\varphi_{1}(\bar{s}_{1})=\varphi_{1}(\bar{s}_{2})$. In this case we have $8640$ homomorphisms satisfying our conditions that are divided into $24$ conjugacy classes. Let $\beta=\bar{s}_{1}\bar{s}_{3}\bar{s}_{2}\bar{s}_{1}\bar{s}_{3}\bar{s}_{1}$ and $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$. Note that they both belong to ${\rm Ker}(\varphi_{2})$. We check that $\varphi_{1}(\beta)=1$ in every case. Moreover, as $\varphi_{1}(\bar{s}_{1})=\varphi_{1}(\bar{s}_{2})$, we also have that $\varphi_{1}(\alpha)=1$. Therefore, $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$ and $\alpha\beta\alpha\beta^{-1}=1$ and ${\rm Ker}(\varphi)$ has generalized torsion. ∎

Lemma \thelemma.

Let $n\geq 5$. Then $A[D_{n}]$ and $A[A_{n}]$ are not commensurable.

Proof.

We denote by $s_{1},\dots,s_{n}$ the standard generators of $A[D_{n}]$ numbered as in Figure 1. We also let $t_{i}=(i,i+1)\in\mathfrak{S}_{n+2}$ for every $i\in\{1,\dots,n+1\}$. Let $\zeta:A[D_{n}]\to\mathfrak{S}_{n+2}$ be the homomorphism defined by $\zeta(s_{1})=\zeta(s_{2})=t_{1}$ and $\zeta(s_{i})=t_{i-1}$ for every $i\in\{3,\dots,n\}$. Moreover, for $n=6$, let $\nu:A[D_{6}]\to\mathfrak{S}_{8}$ be the homomorphism defined by

[TABLE]

Claim. Let $\psi:A[D_{n}]\to\mathfrak{S}_{n+2}$ be a homomorphism. Then, we have one of the following situations, up to conjugation.

• (1)

$\psi$ is cyclic,

• (2)

$\psi=\zeta$,

• (3)

$n=6$ and $\psi=\nu$.

Proof of the claim. Let $s_{1}^{\prime},\dots,s_{n-1}^{\prime}$ be the standard generators of $A[A_{n-1}]$. Let $\zeta^{\prime}:A[A_{n-1}]\to\mathfrak{S}_{n+2}$ be the homomorphism defined by $\zeta^{\prime}(s_{i}^{\prime})=t_{i}$ for every $i\in\{1,\dots,n-1\}$. For $n=6$, let $\nu^{\prime}:A[A_{5}]\to\mathfrak{S}_{8}$ be the homomorphism defined by

[TABLE]

Let $\iota:A[A_{n-1}]\to A[D_{n}]$ be the homomorphism sending $s_{i}^{\prime}$ to $s_{i+1}$ for every $i\in\{1,\dots,n-1\}$, and $\psi^{\prime}=\psi\circ\iota:A[A_{n-1}]\to\mathfrak{S}_{n+2}$. By (Artin,, 1947a, Theorem 1) and (Lin,, 2004, Theorem A, Theorem E), we have one of the following possibilities, up to conjugation.

• (1)

$\psi^{\prime}$ is cyclic,

• (2)

$\psi^{\prime}=\zeta^{\prime}$,

• (3)

$n=6$ and $\psi^{\prime}=\nu^{\prime}$.

First assume that $\psi^{\prime}$ is cyclic. Then there is $w\in\mathfrak{S}_{n+2}$ such that $\psi^{\prime}(s_{i}^{\prime})=\psi(s_{i+1})=w$ for every $i\in\{1,\dots,n-1\}$. Let $\gamma=s_{1}s_{3}s_{2}s_{1}s_{3}s_{1}$. We have $\gamma s_{2}\gamma^{-1}=s_{1}$ and $\gamma s_{3}\gamma^{-1}=s_{3}$, hence $w=\psi(s_{3})=\psi(\gamma s_{3}\gamma^{-1})=\psi(\gamma s_{2}\gamma^{-1})=\psi(s_{1})$. Thus, $\psi$ is cyclic.

Now suppose that $\psi^{\prime}=\zeta^{\prime}$. We have $\psi(s_{i+1})=\psi^{\prime}(s_{i}^{\prime})=t_{i}$ for every $i\in\{1,\dots,n-1\}$. Let $u=\psi(s_{1})$. As $u$ commutes with $\psi(s_{i})=t_{i-1}$ for every $i\geq 4$, it follows that $u(k)=k$ for every $k\in\{3,4,\dots,n\}$. Moreover, $u$ commutes with $t_{1}=\psi(s_{2})$, so $u\in E=\{1,t_{1},t_{n+1},t_{1}t_{n+1}\}$. The only element $u$ of $E$ satisfying $ut_{2}u=t_{2}ut_{2}$ is $u=t_{1}$, hence $u=t_{1}$ and $\psi=\zeta$.

Assume that $n=6$ and $\psi^{\prime}=\nu^{\prime}$. Let

[TABLE]

A direct computation with the software SageMath (see code in Cumplido & Paris,, 2020) shows that the only element $v\in\mathfrak{S}_{8}$ satisfying $vu_{1}=u_{1}v$, $vu_{2}v=u_{2}vu_{2}$, $vu_{3}=u_{3}v$, $vu_{4}=u_{4}v$, $vu_{5}=u_{5}v$ is $v=u_{1}$, hence $\psi=\nu$. This finishes the proof of the claim.

Let $\Delta$ be the Garside element of $A[D_{n}]$ and let $\delta$ be the standard generator of $Z(A[D_{n}])$. By (Paris,, 1997a, Lemma 5.1), $\Delta=(s_{n}\cdots s_{3}s_{2}s_{1}s_{3}\cdots s_{n})\cdots(s_{3}s_{2}s_{1}s_{3})(s_{2}s_{1})$. Moreover, $\delta=\Delta$ if $n$ is even, and $\delta=\Delta^{2}$ si $n$ is odd. Notice that $\zeta(\Delta)=1$, so $\zeta(\delta)=1$. It follows that $\zeta$ induces a homomorphism $\bar{\zeta}:\overline{A[D_{n}]}\to\mathfrak{S}_{n+2}$. Similarly, if $n=6$, $\nu(\Delta)=1$ and $\nu(\delta)=1$, then $\nu$ induces a homomorphism $\bar{\nu}:\overline{A[D_{6}]}\to\mathfrak{S}_{8}$. Let $\varphi_{1}:\overline{A[D_{n}]}\to\mathfrak{S}_{n+2}$ be a homomorphism. Then, by the results from above and the claim, we have one of the following three possibilities, up to conjugation.

• (1)

$\varphi_{1}$ is cyclic,

• (2)

$\varphi_{1}=\bar{\zeta}$,

• (3)

$n=6$ and $\varphi_{1}=\bar{\nu}$.

Let $\varphi:\overline{A[D_{n}]}\to\mathfrak{S}_{n+2}\times\{\pm 1\}$ be a homomorphism written in the form $\varphi=\varphi_{1}\times\varphi_{2}$, where ${\varphi_{1}:\overline{A[D_{n}]}\to\mathfrak{S}_{n+2}}$ and $\varphi_{2}:\overline{A[D_{n}]}\to\{\pm 1\}$ are homomorphisms. By section 6, we just need to show that ${\rm Ker}(\varphi)$ has generalized torsion. Denote by $\pi:A[D_{n}]\to\overline{A[D_{n}]}$ the quotient homomorphism and $\bar{s}_{i}=\pi(s_{i})$ for every $i\in\{1,\dots,n\}$. Here again, $\varphi_{2}$ is always cyclic since its image is contained in $\{\pm 1\}$.

Suppose that $\varphi_{1}$ is cyclic. Let $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$ and $\beta=\bar{s}_{3}\bar{s}_{2}\bar{s}_{1}\bar{s}_{3}\bar{s}_{1}^{-4}$. In this case $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$ and $\alpha\beta\alpha\beta^{-1}=1$, and then ${\rm Ker}(\varphi)$ has generalized torsion. Assume either $\varphi=\bar{\zeta}$ or $n=6$ and $\varphi=\bar{\nu}$. Let $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$ and $\beta=\bar{s}_{1}\bar{s}_{3}\bar{s}_{2}\bar{s}_{1}\bar{s}_{3}\bar{s}_{1}$. In both cases $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$ and $\alpha\beta\alpha\beta^{-1}=1$, hence ${\rm Ker}(\varphi)$ has generalized torsion. ∎

Lemma \thelemma.

Let $n\in\{6,7,8\}$. Then $A[E_{n}]$ and $A[A_{n}]$ are not commensurable.

Proof.

We denote by $s_{1},\dots,s_{n}$ the standard generators of $A[E_{n}]$ numbered as in Figure 1. We also let $t_{i}=(i,i+1)\in\mathfrak{S}_{n+2}$ for every $i\in\{1,\dots,n+1\}$.

Claim. Every homomorphism $\psi:A[E_{n}]\to\mathfrak{S}_{n+2}$ is cyclic.

Proof of the claim. Denote by $s_{1}^{\prime},\dots,s_{n-1}^{\prime}$ the standard generators of $A[A_{n-1}]$. Let $\zeta^{\prime}:A[A_{n-1}]\to\mathfrak{S}_{n+2}$ be the homomorphism defined by $\zeta^{\prime}(s_{i}^{\prime})=t_{i}$ for every $i\in\{1,\dots,n-1\}$. For $n=6$, let $\nu^{\prime}:A[A_{5}]\to\mathfrak{S}_{8}$ be the homomorphism defined by

[TABLE]

Let $\iota:A[A_{n-1}]\to A[E_{n}]$ be the homomorphism sending $s_{i}^{\prime}$ to $s_{i}$ for every $i\in\{1,\dots,n-1\}$, and $\psi^{\prime}=\psi\circ\iota:A[A_{n-1}]\to\mathfrak{S}_{n+2}$. By (Artin,, 1947a, Theorem 1) and (Lin,, 2004, Theorem A, Theorem E), we have one of the following possibilities, up to conjugation.

• (1)

$\psi^{\prime}$ is cyclic,

• (2)

$\psi^{\prime}=\zeta^{\prime}$,

• (3)

$n=6$ and $\psi^{\prime}=\nu^{\prime}$.

First suppose that $\psi^{\prime}$ is cyclic. Then there is $w\in\mathfrak{S}_{n+2}$ such that $\psi^{\prime}(s_{i}^{\prime})=\psi(s_{i})=w$ for every $i\in\{1,\dots,n-1\}$. Let $\gamma=s_{2}s_{3}s_{n}s_{2}s_{3}s_{2}$. We have $\gamma s_{2}\gamma^{-1}=s_{n}$ and $\gamma s_{3}\gamma^{-1}=s_{3}$, hence $w=\psi(s_{3})=\psi(\gamma s_{3}\gamma^{-1})=\psi(\gamma s_{2}\gamma^{-1})=\psi(s_{n})$. Then $\psi$ is cyclic.

Now assume that $\psi^{\prime}=\zeta^{\prime}$. In this case we have $\psi(s_{i})=\psi^{\prime}(s_{i}^{\prime})=t_{i}$ for every $i\in\{1,\dots,n-1\}$. Let $u=\psi(s_{n})$. As $u$ commutes with $\psi(s_{i})=t_{i}$ for every $i\in\{1,2,4,\dots,n-1\}$, it follows that $u(k)=k$ for every $k\in\{1,2,3,4,5,\dots,n\}$, so $u\in E=\{1,t_{n+1}\}$. But there is no element $u$ of $E$ satisfying $ut_{3}u=t_{3}ut_{3}$, so we cannot have $\psi^{\prime}=\zeta^{\prime}$.

Finally, assume $n=6$ and $\psi^{\prime}=\nu^{\prime}$. Let

[TABLE]

A direct computation with the software SageMath (see code in Cumplido & Paris,, 2020) shows that there is no element $v\in\mathfrak{S}_{8}$ satisfying $vu_{1}=u_{1}v$, $vu_{2}=u_{2}v$, $vu_{3}v=u_{3}vu_{3}$, $vu_{4}=u_{4}v$ and $vu_{5}=u_{5}v$, hence we cannot have $n=6$ and $\psi^{\prime}=\nu^{\prime}$. This finishes the proof of the claim.

Denote by $\pi:A[E_{n}]\to\overline{A[E_{n}]}$ the quotient homomorphism and $\bar{s}_{i}=\pi(s_{i})$ for every $i\in\{1,\dots,n\}$. Let $\varphi:\overline{A[E_{n}]}\to\mathfrak{S}_{n+2}\times\{\pm 1\}$ be a homomorphism written in the form $\varphi=\varphi_{1}\times\varphi_{2}$, where $\varphi_{1}:\overline{A[E_{n}]}\to\mathfrak{S}_{n+2}$ and $\varphi_{2}:\overline{A[E_{n}]}\to\{\pm 1\}$ are homomorphisms. By the claim, $\varphi_{1}\circ\pi:A[E_{n}]\to\mathfrak{S}_{n+2}$ is cyclic, hence $\varphi_{1}$ is also cyclic. On the other hand, $\varphi_{2}$ is cyclic since the image of $\varphi_{2}$ is contained in $\{\pm 1\}$. Let $\alpha=\bar{s}_{2}\bar{s}_{n}^{-1}$ and $\beta=\bar{s}_{3}\bar{s}_{n}\bar{s}_{2}\bar{s}_{3}\bar{s}_{2}^{-4}$. We have that $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$ and $\alpha\beta\alpha\beta^{-1}=1$, and then ${\rm Ker}(\varphi)$ has generalized torsion. By section 6, it follows that $A[A_{n}]$ and $A[E_{n}]$ are not commensurable. ∎

Lemma \thelemma.

The groups $A[F_{4}]$ and $A[A_{4}]$ are not commensurable.

Proof.

Let $\varphi:\overline{A[F_{4}]}\to\mathfrak{S}_{6}\times\{\pm 1\}$ be a homomorphism written in the form $\varphi=\varphi_{1}\times\varphi_{2}$, where $\varphi_{1}:\overline{A[F_{4}]}\to\mathfrak{S}_{6}$ and $\varphi_{2}:\overline{A[F_{4}]}\to\{\pm 1\}$ are two homomorphisms. By section 6, we just need to show that ${\rm Ker}(\varphi)$ has generalized torsion. We denote by $s_{1},s_{2},s_{3},s_{4}$ the standard generators of $A[F_{4}]$ numbered as in Figure 1. We also denote by $\pi:A[F_{4}]\to\overline{A[F_{4}]}$ the quotient homomorphism and $\bar{s}_{i}=\pi(s_{i})$ for every $i\in\{1,2,3,4\}$. Notice that the relation $s_{1}s_{2}s_{1}=s_{2}s_{1}s_{2}$ implies $\varphi_{2}(\bar{s}_{1})=\varphi_{2}(\bar{s}_{2})$. Analogously, $\varphi_{2}(\bar{s}_{3})=\varphi_{2}(\bar{s}_{4})$.

If $g$ is an element of a group, we denote by ${\rm ord}(g)$ the order of $g$. Let $i\in\{1,2,3,4\}$. As $\varphi_{1}(\bar{s}_{i})\in\mathfrak{S}_{6}$, we have that ${\rm ord}(\varphi_{1}(\bar{s}_{i}))\in\{1,2,3,4,5,6\}$. It follows that ${\rm ord}(\varphi(\bar{s}_{i}))\in\{1,2,3,4,5,6,10\}$. Suppose that $\varphi(\bar{s}_{1})=\varphi(\bar{s}_{2})$ and ${\rm ord}(\varphi(\bar{s}_{1}))\in\{1,2,4\}$. Let $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$ and $\beta=(\bar{s}_{1}\bar{s}_{2})^{4}$. In this case $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$ and $\alpha(\beta\alpha\beta^{-1})(\beta^{2}\alpha\beta^{-2})=1$, and then ${\rm Ker}(\varphi)$ has generalized torsion. Now assume that $\varphi(\bar{s}_{1})=\varphi(\bar{s}_{2})$ and ${\rm ord}(\varphi(\bar{s}_{1}))=3$. If we let $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$, $\beta=\bar{s}_{1}\bar{s}_{2}\bar{s}_{1}$, then $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$, $\alpha(\beta\alpha\beta^{-1})=1$, and ${\rm Ker}(\varphi)$ has generalized torsion. Now suppose that $\varphi(\bar{s}_{1})=\varphi(\bar{s}_{2})$ and ${\rm ord}(\varphi(\bar{s}_{1}))\in\{5,10\}$. We let $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$ and $\beta=(\bar{s}_{1}\bar{s}_{2})^{10}$. Then $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha(\beta\alpha\beta^{-1})(\beta^{2}\alpha\beta^{-2})=1$, and ${\rm Ker}(\varphi)$ has generalized torsion.

By the reasoning above we can assume that, if $\varphi(\bar{s}_{1})=\varphi(\bar{s}_{2})$, then ${\rm ord}(\varphi(\bar{s}_{1}))=6$. We can also suppose that, if $\varphi(\bar{s}_{3})=\varphi(\bar{s}_{4})$, then ${\rm ord}(\varphi(\bar{s}_{3}))=6$.

Suppose that $\varphi(\bar{s}_{1})=\varphi(\bar{s}_{2})$ and $\varphi(\bar{s}_{3})=\varphi(\bar{s}_{4})$. Then we also have ${\rm ord}(\varphi(\bar{s}_{1}))={\rm ord}(\varphi(\bar{s}_{3}))=6$. If $\varphi_{1}(\bar{s}_{1})=\varphi_{1}(\bar{s}_{2})$ and $\varphi_{1}(\bar{s}_{3})=\varphi_{1}(\bar{s}_{4})$ are both of order $3$, then $\varphi_{2}(\bar{s}_{1})=\varphi_{2}(\bar{s}_{2})=\varphi_{2}(\bar{s}_{3})=\varphi_{2}(\bar{s}_{4})=-1$. In this case, we let $\alpha=\bar{s}_{1}\bar{s}_{2}^{-1}$ and $\beta=\bar{s}_{1}\bar{s}_{2}\bar{s}_{1}\bar{s}_{4}^{3}$, having $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$ and $\alpha(\beta\alpha\beta^{-1})=1$. Hence ${\rm Ker}(\varphi)$ has generalized torsion. We can then assume that $\varphi_{1}(\bar{s}_{1})$ or $\varphi_{1}(\bar{s}_{3})$ is of order $6$, say that $\varphi_{1}(\bar{s}_{1})$ has order $6$. Then $\varphi_{1}(\bar{s}_{1})$ is conjugate to $(1,2,3,4,5,6)$ or to $(1,2,3)(4,5)$ in $\mathfrak{S}_{6}$. In both cases it follows that the centralizer of $\varphi_{1}(\bar{s}_{1})$ in $\mathfrak{S}_{6}$ is a cyclic group of order $6$ generated by $\varphi_{1}(\bar{s}_{1})$. As $\varphi_{1}(\bar{s}_{3})$ belongs to this centralizer and it has order $3$ or $6$, there is $k\in\{1,2,-1,-2\}$ such that $\varphi_{1}(\bar{s}_{3})=\varphi_{1}(\bar{s}_{4})=\varphi_{1}(\bar{s}_{1})^{k}$. We let $\alpha=\bar{s}_{3}\bar{s}_{4}^{-1}$ and $\beta=\bar{s}_{3}\bar{s}_{4}\bar{s}_{1}^{-2k}$. Then, $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$, and $\alpha(\beta\alpha\beta^{-1})(\beta^{2}\alpha\beta^{-2})=1$, having generalized torsion in ${\rm Ker}(\varphi)$.

By (Brieskorn & Saito,, 1972), the standard generator of the center of $A[F_{4}]$ coincides with its Garside element and equals $(s_{1}s_{2}s_{3}s_{4})^{\frac{h}{2}}$ where $h$ is the Coxeter number of $F_{4}$. As $h=12$ (Humphreys,, 1990, Page 80), we have $\delta=\Delta=(s_{1}s_{2}s_{3}s_{4})^{6}$. Let $\hat{\alpha}_{0}=(s_{1}s_{2}s_{3}s_{4})^{3}$. Recall that $z:A[F_{4}]\to\mathbb{Z}$ is the homomorphism sending $s_{i}$ to $1$ for every $i\in\{1,2,3,4\}$. As $z(\delta)=24$, we have $z(Z(A[F_{4}]))=24\mathbb{Z}$, so $\hat{\alpha}_{0}\not\in Z(A[F_{4}])$ because $z(\hat{\alpha}_{0})=12$. On the other hand, $\hat{\alpha}_{0}^{2}=\delta$, so $\hat{\alpha}_{0}^{2}\in Z(A[F_{4}])$. Let $\alpha_{0}=\pi(\hat{\alpha}_{0})$. Then $\alpha_{0}\neq 1$ and $\alpha_{0}^{2}=1$. In the remaining cases, we will show that $\alpha_{0}\in{\rm Ker}(\varphi)$, which will immediately imply that ${\rm Ker}(\varphi)$ has (generalized) torsion.

Suppose that $\varphi(\bar{s}_{1})\neq\varphi(\bar{s}_{2})$ and $\varphi(\bar{s}_{3})=\varphi(\bar{s}_{4})$ (hence ${\rm ord}(\varphi(\bar{s}_{3}))=6$). Let $E_{1}$ be the set of triples $(u_{1},u_{2},u_{3})$ of elements of $\mathfrak{S}_{6}$ such that $u_{1}u_{2}u_{1}=u_{2}u_{1}u_{2}$, $u_{1}u_{3}=u_{3}u_{1}$, $u_{2}u_{3}=u_{3}u_{2}$, $u_{1}\neq u_{2}$ and ${\rm ord}(u_{3})\in\{3,6\}$. Another direct computation with SageMath (see code in Cumplido & Paris,, 2020) shows that $E_{1}$ has $1440$ elements divided into $6$ conjugacy classes. Again with SageMath, we compute a set $E_{1}^{0}$ of representatives of the conjugacy classes in $E_{1}$ and we get

[TABLE]

We check with a direct computation that $(u_{1}u_{2}u_{3}^{2})^{3}=1$ for every $(u_{1},u_{2},u_{3})\in E_{1}^{0}$. Up to conjugation, we can suppose that $(\varphi_{1}(\bar{s}_{1}),\varphi_{1}(\bar{s}_{2}),\varphi_{1}(\bar{s}_{3}))=(u_{1},u_{2},u_{3})\in E_{1}^{0}$. Then, as $(u_{1}u_{2}u_{3}^{2})^{3}=1$, we have $\varphi_{1}(\alpha_{0})=1$. It is obvious that $\varphi_{2}(\alpha_{0})=1$. So, $\varphi(\alpha_{0})=1$ and ${\rm Ker}(\varphi)$ has (generalized) torsion.

Suppose that $\varphi(\bar{s}_{1})\neq\varphi(\bar{s}_{2})$ and $\varphi(\bar{s}_{3})\neq\varphi(\bar{s}_{4})$. Let $E_{2}$ be the set of quadruples $(u_{1},u_{2},u_{3},u_{4})$ of elements of $\mathfrak{S}_{6}$ such that $u_{1}u_{2}u_{1}=u_{2}u_{1}u_{2}$, $u_{1}u_{3}=u_{3}u_{1}$, $u_{1}u_{4}=u_{4}u_{1}$, $u_{2}u_{3}u_{2}u_{3}=u_{3}u_{2}u_{3}u_{2}$, $u_{2}u_{4}=u_{4}u_{2}$, $u_{3}u_{4}u_{3}=u_{4}u_{3}u_{4}$, $u_{1}\neq u_{2}$ and $u_{3}\neq u_{4}$. A direct computation with SageMath (see code in Cumplido & Paris,, 2020) shows that $E_{2}$ has $1440$ elements divided into $2$ conjugacy classes. Again with SageMath, we compute a set $E_{2}^{0}$ of representatives of the conjugacy classes in $E_{2}$ and we get

[TABLE]

We check by a direct computation that $(u_{1}u_{2}u_{3}u_{4})^{3}=1$ for every $(u_{1},u_{2},u_{3},u_{4})\in E_{2}^{0}$. Up to conjugation, we can suppose that $(\varphi_{1}(\bar{s}_{1}),\varphi_{1}(\bar{s}_{2}),\varphi_{1}(\bar{s}_{3}),\varphi_{1}(\bar{s}_{4}))=(u_{1},u_{2},u_{3},u_{4})\in E_{2}^{0}$. Then, as $(u_{1}u_{2}u_{3}u_{4})^{3}=1$, we have $\varphi_{1}(\alpha_{0})=1$. It is clear that $\varphi_{2}(\alpha_{0})=1$. Then $\varphi(\alpha_{0})=1$ and ${\rm Ker}(\varphi)$ has (generalized) torsion. ∎

Lemma \thelemma.

The groups $A[H_{4}]$ and $A[A_{4}]$ are not commensurable.

Proof.

We denote by $s_{1},s_{2},s_{3},s_{4}$ the standard generators of $A[H_{4}]$ numbered as in Figure 1. We also consider the quotient homomorphism $\pi:A[H_{4}]\to\overline{A[H_{4}]}$ and $\bar{s}_{i}=\pi(s_{i})$ for every $i\in\{1,2,3,4\}$. Let $\varphi:\overline{A[H_{4}]}\to\mathfrak{S}_{6}\times\{\pm 1\}$ be a homomorphism written in the form $\varphi=\varphi_{1}\times\varphi_{2}$ where $\varphi_{1}:\overline{A[H_{4}]}\to\mathfrak{S}_{6}$ and $\varphi_{2}:\overline{A[H_{4}]}\to\{\pm 1\}$ are two homomorphisms. Notice that the relations $s_{3}s_{4}s_{3}=s_{4}s_{3}s_{4}$, $s_{2}s_{3}s_{2}=s_{3}s_{2}s_{3}$, $s_{1}s_{2}s_{1}s_{2}s_{1}=s_{2}s_{1}s_{2}s_{1}s_{2}$ imply $\varphi_{2}(\bar{s}_{1})=\varphi_{2}(\bar{s}_{2})=\varphi_{2}(\bar{s}_{3})=\varphi_{2}(\bar{s}_{4})$. Then $\varphi_{2}$ is always cyclic. For $\varphi_{1}$, a direct computation with SageMath (see code in Cumplido & Paris,, 2020) shows that there are $720$ homomorphisms from $\overline{A[H_{4}]}$ to $\mathfrak{S}_{6}$, all of them cyclic. We let $\alpha=\bar{s}_{3}^{-1}\bar{s}_{4}$ and $\beta=\bar{s}_{3}\bar{s}_{4}\bar{s}_{3}\bar{s}_{1}^{-3}$. They both belong to ${\rm Ker}(\varphi)$ and they satisfy $\alpha\beta\alpha\beta^{-1}=1$. Therefore, ${\rm Ker}(\varphi)$ has generalized torsion and, by section 6, $A[A_{4}]$ and $A[H_{4}]$ are not commensurable. ∎

Our last issue is to compare $A[H_{3}]$ and $A[A_{3}]$. In this case we cannot apply section 6, as we have done with the previous cases. Indeed, we can find homomorphisms sending $\overline{A[H_{3}]}$ to $\mathfrak{S}_{5}$ whose kernel does not have generalized torsion:

Lemma \thelemma.

Let $s_{1},s_{2},s_{3}$ be the standard generators of $A[H_{3}]$ numbered as in Figure 1, let $\pi\colon A[H_{3}]\to\overline{A[H_{3}]}$ be the quotient homomorphism, and, for each $i\in\{1,2,3\}$, let $\bar{s}_{i}=\pi(s_{i})$. Let $\zeta:\overline{A[H_{3}]}\to\mathfrak{S}_{5}$ be the homomorphism defined by

[TABLE]

Then ${\rm Ker}(\zeta)$ does not have generalized torsion.

Proof.

By (Brieskorn & Saito,, 1972), $Z(A[H_{3}])$ is an infinite cyclic group generated by $\delta=(s_{1}s_{2}s_{3})^{5}$. Let $u_{1}=(2,4)(3,5)$, $u_{2}=(1,2)(4,5)$ and $u_{3}=(2,3)(4,5)$. A direct computation shows that we have the relations $u_{1}u_{2}u_{1}u_{2}u_{1}=u_{2}u_{1}u_{2}u_{1}u_{2}$, $u_{1}u_{3}=u_{3}u_{1}$, $u_{2}u_{3}u_{2}=u_{3}u_{2}u_{3}$ and $(u_{1}u_{2}u_{3})^{5}=1$, hence $\zeta$ is well-defined. We are going to prove that ${\rm Ker}(\zeta)=\overline{{\rm CA}[H_{3}]}$. As $\overline{{\rm CA}[H_{3}]}$ projects into ${\rm CA}[H_{3}]$ by section 3 and ${\rm CA}[H_{3}]$ has no generalized torsion by (Marin,, 2007, Theorem 3), it will follow that ${\rm Ker}(\zeta)$ has no generalized torsion.

Let $H$ be the subgroup of $\mathfrak{S}_{5}$ generated by $\{u_{1},u_{2},u_{3}\}$. A direct computation with SageMath (see code in Cumplido & Paris,, 2020) shows that $|H|=60$. As $u_{1}^{2}=u_{2}^{2}=u_{3}^{2}=1$ and $\overline{{\rm CA}[H_{3}]}$ is the normal subgroup of $\overline{A[H_{3}]}$ generated by $\{\bar{s}_{1}^{2},\bar{s}_{2}^{2},\bar{s}_{3}^{2}\}$, we have $\overline{{\rm CA}[H_{3}]}\subset{\rm Ker}(\zeta)$. Then, to show that ${\rm Ker}(\zeta)=\overline{{\rm CA}[H_{3}]}$, we just need to prove that $|\overline{A[H_{3}]}/\overline{{\rm CA}[H_{3}]}|=60$. It is well known that $|A[H_{3}]/{\rm CA}[H_{3}]|=|W[H_{3}]|=120$ (Humphreys,, 1990, Page 46). The projection $\pi:A[H_{3}]\to\overline{A[H_{3}]}$ induces a surjective homomorphism $\bar{\pi}:A[H_{3}]/{\rm CA}[H_{3}]\to\overline{A[H_{3}]}/\overline{{\rm CA}[H_{3}]}$ whose kernel is the cyclic group generated by the class $[\delta]$ of $\delta$. We have that $\delta=\Delta$ is the Garside element of $A[H_{3}]$, so $\delta\not\in{\rm CA}[H_{3}]$. However, $\delta^{2}=\Delta^{2}\in{\rm CA}[H_{3}]$, hence ${\rm Ker}(\bar{\pi})$ is a cyclic group $\langle[\delta]\rangle$ of order $2$, having $|\overline{A[H_{3}]}/\overline{{\rm CA}[H_{3}]}|=|A[H_{3}]/{\rm CA}[H_{3}]|/2=60$. ∎

Let $\Sigma$ be a closed surface and $\mathcal{P}_{n}$ be a collection of $n$ different points in $\Sigma$. With such a pair $(\Sigma,\mathcal{P}_{n})$ we can associate a simplicial complex called the curve complex of $(\Sigma,\mathcal{P}_{n})$, denoted by $\mathcal{C}(\Sigma,\mathcal{P}_{n})$. The vertices of $\mathcal{C}(\Sigma,\mathcal{P}_{n})$ are the isotopy classes of simple closed curves on $\Sigma\setminus\mathcal{P}_{n}$ that are non-degenerate. Non-degenerate means that the curve does not bound a disk embedded in $\Sigma$ containing 0 or 1 point of $\mathcal{P}_{n}$. Every $n$-simplex is formed by $n+1$ classes having representatives that are pairwise disjoint. We say that a mapping class $f\in\mathcal{M}^{*}(\Sigma,\mathcal{P}_{n})$ is pseudo-Anosov if $f^{n}(\alpha)\neq\alpha$ for every $\alpha\in\mathcal{C}(\Sigma,\mathcal{P}_{n})$ and every $n\in\mathbb{Z}\setminus\{0\}$. We say that $f$ is periodic if it has finite order. The following lemma finishes the proof of Theorem 4.

Lemma \thelemma.

The groups $A[H_{3}]$ and $A[A_{3}]$ are not commensurable.

Proof.

Recall that, by section 3, we need to prove that $\overline{A[H_{3}]}$ and $\overline{A[A_{3}]}$ are not commensurable, and, to do this, it is enough to prove that ${\rm Com}(\overline{A[H_{3}]})$ and ${\rm Com}(\overline{A[A_{3}]})$ are not isomorphic. Also by section 3, $\overline{A[H_{3}]}$ injects in ${\rm Com}(\overline{A[H_{3}]})$ and recall that ${\rm Com}(\overline{A[A_{3}]})$ and $\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{5})$ are isomorphic by Theorem 6. Then, to prove our lemma it suffices to prove that there is no injective homomorphism from $\overline{A[H_{3}]}$ to $\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{5})$.

Let $\Phi:\overline{A[H_{3}]}\to\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{5})$ be a homomorphism. Recall $\hat{\theta}$ and $\theta^{\prime}$ defined right before Theorem 6 and consider $\varphi=\hat{\theta}\circ\Phi:\overline{A[H_{3}]}\to\mathfrak{S}_{5}\times\{\pm 1\}$ being of the form $\varphi=\varphi_{1}\times\varphi_{2}$, where $\varphi_{1}=\theta^{\prime}\circ\Phi:\overline{A[H_{3}]}\to\mathfrak{S}_{5}$ and $\varphi_{2}=\omega\circ\Phi:\overline{A[H_{3}]}\to\{\pm 1\}$. We denote by $s_{1},s_{2},s_{3}$ the standard generators of $A[H_{3}]$ numbered as in Figure 1. Moreover, we let $\pi:A[H_{3}]\to\overline{A[H_{3}]}$ be the quotient homomorphism and $\bar{s}_{i}=\pi(s_{i})$ for every $i\in\{1,2,3\}$.

Notice that the relations $s_{2}s_{3}s_{2}=s_{3}s_{2}s_{3}$ and $s_{1}s_{2}s_{1}s_{2}s_{1}=s_{2}s_{1}s_{2}s_{1}s_{2}$ imply $\varphi_{2}(\bar{s}_{2})=\varphi_{2}(\bar{s}_{3})$ and $\varphi_{2}(\bar{s}_{1})=\varphi_{2}(\bar{s}_{2})$. Notice also that the standard generator of the center of $A[H_{3}]$ is $\delta=(s_{1}s_{2}s_{3})^{5}$, hence $(\bar{s}_{1}\bar{s}_{2}\bar{s}_{3})^{5}=1$. Let $\epsilon=\varphi_{2}(\bar{s}_{1})=\varphi_{2}(\bar{s}_{2})=\varphi_{2}(\bar{s}_{3})\in\{\pm 1\}$. Then $1=\varphi_{2}(1)=\varphi_{2}((\bar{s}_{1}\bar{s}_{2}\bar{s}_{3})^{5})=\epsilon^{15}$, having that $\epsilon=1$.

Suppose that $\varphi_{1}$ is cyclic, that is, there is $w\in\mathfrak{S}_{5}$ such that $\varphi_{1}(\bar{s}_{1})=\varphi_{1}(\bar{s}_{2})=\varphi_{1}(\bar{s}_{3})=w$. We denote by ${\rm ord}(w)$ the order of $w$. As $w\in\mathfrak{S}_{5}$, we have ${\rm ord}(w)\in\{1,2,3,4,5,6\}$. On the other hand, as $(\bar{s}_{1}\bar{s}_{2}\bar{s}_{3})^{5}=1$, we have $w^{15}=1$, hence ${\rm ord}(w)$ divides $15$. Thus, ${\rm ord}(w)\in\{1,3,5\}$. Now, we let $\alpha=\bar{s}_{2}\bar{s}_{3}^{-1}$ and $\beta=(\bar{s}_{2}\bar{s}_{3}\bar{s}_{2})^{5}$. Then, $\alpha,\beta\in{\rm Ker}(\varphi)$, $\alpha\neq 1$, $\alpha\beta\alpha\beta^{-1}=1$, and ${\rm Ker}(\varphi)$ has generalized torsion. By section 6, $\Phi$ is not injective.

Suppose that $\varphi_{1}$ is not cyclic. Consider the two homomorphisms $\zeta_{1},\zeta_{2}:\overline{A[H_{3}]}\to\mathfrak{S}_{5}$ defined by

[TABLE]

A direct computation with SageMath (see code in Cumplido & Paris,, 2020) shows that every non-cyclic homomorphism from $\overline{A[H_{3}]}$ to $\mathfrak{S}_{5}$ is conjugate to either $\zeta_{1}$ or $\zeta_{2}$. We can then suppose that $\varphi_{1}\in\{\zeta_{1},\zeta_{2}\}$.

If $\varphi_{1}=\zeta_{1}$, by (Boyland,, 1994, Proposition 9.4) and (Bonatti & Paris,, 2009, Lemma 5.9) it follows that $\Phi(\bar{s}_{1})$ is periodic or pseudo-Anosov. If $\Phi(\bar{s}_{1})$ is periodic, then there is an integer $k\geq 1$ such that $\Phi(\bar{s}_{1})^{k}={\rm id}$, hence $\bar{s}_{1}^{k}$ is a non-trivial element of ${\rm Ker}(\Phi)$ and $\Phi$ is not injective. Suppose that $\Phi(\bar{s}_{1})$ is pseudo-Anosov. As $\Phi((\bar{s}_{1}\bar{s}_{2})^{5})$ is in the centralizer of $\Phi(\bar{s}_{1})$ in $\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{5})$ and the centralizer of a pseudo-Anosov element is virtually cyclic (Ivanov,, 1992, Lemma 8.13), there are integers $k,\ell\in\mathbb{Z}$, $\ell\neq 0$, such that $\Phi(\bar{s}_{1})^{k}=\Phi((\bar{s}_{1}\bar{s}_{2})^{5})^{\ell}$. Let $\alpha=(\bar{s}_{1}\bar{s}_{2})^{5\ell}\bar{s}_{1}^{-k}$. Then $\alpha$ is a non-trivial element of ${\rm Ker}(\Phi)$ and $\Phi$ is not injective.

Suppose that $\varphi_{1}=\zeta_{2}$. Then $\varphi_{1}(\bar{s}_{1}\bar{s}_{2})=(1,4,3,5,2)$, and again by (Boyland,, 1994, Proposition 9.4) and (Bonatti & Paris,, 2009, Lemma 5.9), $\Phi(\bar{s}_{1}\bar{s}_{2})$ is periodic or pseudo-Anosov. If $\Phi(\bar{s}_{1}\bar{s}_{2})$ is periodic, there is an integer $k\geq 1$ such that $\Phi(\bar{s}_{1}\bar{s}_{2})^{k}={\rm id}$ and $\alpha=(\bar{s}_{1}\bar{s}_{2})^{k}$ is a non-trivial element belonging to the kernel of $\Phi$. This means that $\Phi$ is not injective. If $\Phi(\bar{s}_{1}\bar{s}_{2})$ is pseudo-Anosov, then $\Phi((\bar{s}_{1}\bar{s}_{2})^{5})=\Phi(\bar{s}_{1}\bar{s}_{2})^{5}$ is also pseudo-Anosov and $\Phi(\bar{s}_{1})$ is in the centralizer of $\Phi((\bar{s}_{1}\bar{s}_{2})^{5})$ in $\mathcal{M}^{*}(\Sigma_{0,0},\mathcal{P}_{5})$, which is virtually cyclic. Hence there are integers $k,\ell\in\mathbb{Z}$, $\ell\neq 0$, such that $\Phi(\bar{s}_{1})^{\ell}=\Phi((\bar{s}_{1}\bar{s}_{2})^{5})^{k}$. Let $\alpha=(\bar{s}_{1}\bar{s}_{2})^{5k}\bar{s}_{1}^{-\ell}$. Therefore, $\alpha$ is a non-trivial element of ${\rm Ker}(\Phi)$ and $\Phi$ is not injective. ∎