Let $G$ be a finitely generated group splitting non-trivially over a infinite virtually cyclic subgroup $V$, namely an amalgamated free product

$$ A*_V B,\; V \neq A,B. $$ Question: can the group $G$ be simple?

MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.

Sign up to join this community
Anybody can ask a question

Anybody can answer

The best answers are voted up and rise to the top

$\begingroup$
$\endgroup$

6
Let $G$ be a finitely generated group splitting non-trivially over a infinite virtually cyclic subgroup $V$, namely an amalgamated free product

$$ A*_V B,\; V \neq A,B. $$ Question: can the group $G$ be simple?

$\begingroup$
$\endgroup$

The answer is no: it's not simple.

It's an easy adaptation of the remarks done in Button's paper.

The background: (a) Minasyan-Osin: let $G$ be an amalgam $G=A\ast_C B$ of two groups over a proper subgroup $C$, such that $C$ is weakly malnormal in $B$. Then $A\ast_C B$ is acylindrically hyperbolic. [In case of index (2,2), there's a homomorphism onto the infinite dihedral group.] (b) (Osin) Every finitely generated acylindrically hyperbolic group is SQ-universal (and hence has continuum many normal subgroups).

Given this it is not hard to conclude. Consider an amalgam $A\ast_C B$ with $C$ virtually cyclic. It remains to consider the case when $C$ is weakly malnormal in neither $A$ nor $B$. Since $C$ is infinite virtually cyclic, it follows that it is a commensurated subgroup, in the sense that it is commensurate to all its conjugates, in both $A$ and $B$ (two subgroups of a group are commensurate if their intersection has finite index in both).

Given a group $H$, its abstract commensurator $\mathrm{Comm}(H)$ is the set of isomorphisms $f$ between two finite index subgroups $H_1$, $H_2$, modulo coincidence on a smaller finite index subgroup. If $L$ has finite index in $H$, then the inclusion $L\to H$ induces an isomorphism $\mathrm{Comm}(L)\simeq\mathrm{Comm}(H)$.

Hence for every amalgam $A\ast_C B$ such that $C$ is commensurated in both $A$ and $B$, we have a homomorphism $A\ast_C B\to\mathrm{Comm}(C)$. Here, $C$ being virtually (infinite cyclic), we have $\mathrm{Comm}(C)\simeq\mathbf{Q}^*$.

Hence assuming the amalgam has a finite abelianization, this homomorphism is trivial, which means that deep enough cyclic subgroup $N$ of finite index in $C$ are normal in both $A$ and $B$. Hence $G$ is not simple, and even better, admits the quotient $(A/N)\ast_{C/N}(B/N)$. In particular, unless $C$ has index 2 in both $A$ and $B$, $G$ is SQ-universal. (If the index is twice 2, we deduce that $G$ is virtually $\mathbf{Z}^2$.)

[Actually the latter conclusion (SQ-university except in index 2-2 case) holds if both homomorphisms $A\to\mathrm{Comm}(C)\leftarrow B$ have a finite image. If both have a nontrivial image and at least one has an infinite image, then since both vanish on $C$, we get a free product quotient and get SQ-universality. It remains the case where one has infinite image and the the other is trivial. In this case Button gets SQ-universality with a more complicated argument in the case $C$ is cyclic. I haven't tried to see if it adapts to $C$ virtually cyclic, but in all cases this is not asked by the OP.]

$\begingroup$
$\endgroup$

The answer is no. As Yves de Cornulier pointed out in a comment, this was proved by Jack Button when $V\cong \mathbb{Z}$. We follow the approach of his proof for the general case.

A subgroup $C$ of a group
$G$ is *weakly malnormal* if there exists $g \in G$ such that $|C^g ∩ C| < ∞$, where $C^g=gCg^{-1}$.

Suppose that $G = A\ast_V B$ is simple, where $V$ is virtually cyclic, and $V \neq A, B$. By Corollary 2.2 of Minasyan-Osin, if $V$ is weakly malnormal in $G$, then $G$ is either acylindrically hyperbolic or virtually cyclic. In either case, $G$ would not be simple. Acylindrically hyperbolic groups are SQ-universal, hence have many normal subgroups.

Thus, we conclude that $V$ is not weakly malnormal in $G$.

Thus, for every $g\in G$, $|V^g\cap V|=\infty$. But $V$ is virtually cyclic, so $[V:V^g\cap V] < \infty$. Hence we get a homomorphism $\varphi: G \to Comm(V) \cong Comm(\mathbb{Z}) \cong \mathbb{Q}^{\times}$, the abstract commensurator. Since $G$ is simple, $\varphi$ is the trivial homomorphism. Let $Z < V$ be a finite-index copy of $\mathbb{Z}\cong Z$. Then $|Z^g\cap Z|=\infty$, so the homomorphism $g : Z\cap Z^{g^{-1}} \to Z^g\cap Z$, $ a \mapsto gag^{-1}$ must be trivial (since it is trivial as an representative of $Comm(\mathbb{Z})$), which implies that $Z\cap Z^{g^{-1}} = Z^g \cap Z$. Since $G$ is finitely generated, say by $g_1, \ldots, g_n$, there exists a finite-index subgroup $Z^{g_1}\cap \cdots \cap Z^{g_n}\cap Z \leq Z$ which is normalized (in fact, centralized) by each $g_i$, and hence by $G$, a contradiction.

clearlybetter, for reasons of non-associativity of terminology? IANAIGT, but I genuinely spent a few seconds thinking “Eh? Isn’t ‘virtually infinite’ the same as ‘infinite’?” before it clicked. $\endgroup$ – Jeremy Rickard Mar 2 '18 at 14:001more comment