# On the number of ends of a countable simple group

At the beginning I thought that the following statement could be an easy exercise after Stallings' theorem, but I found myself incapable of proving it:

Any countable f.g. simple group has one end.

It is obvious that a f.g. simple group cannot have two ends, as $\mathbb Z$ has many quotients. If it has infinitely many ends, 1) it can't be an HNN extension over a finite group, since it is a semi-direct product with a surjection on $\mathbb Z$, also 2) it can't be a free product $A*B$ since it surjects onto $A\times B$. So I'm left with the case of an amalgamated product over a non-trivial finite group.

Maybe I'm wrong, maybe there are example of f.g. groups with infinitely many ends... do you know some example?

EDIT

This is something that I can easily say studying the amalgamated product of two simple groups.

Let us consider the group $G=M_1*_{C}M_2$ be the amalgamated product of two simple groups. Let $\phi:G\to H$ be a nontrivial morphism which is not injective. Then the restriction $\phi_i=\phi\vert_{M_i}:M_i\to H$ is either trivial or injective, for $M_i$ is a simple group. It is not possible that $\phi_1$ and $\phi_2$ are both trivial, since $M_1$ and $M_2$ generate $G$. Also, if $\phi_1$ is trivial, then the copy of $C$ in $M_2$ is in the kernel of $\phi_2$, so $\phi_2$ must be trivial because $M_2$ is simple. As a consequence, both $\phi_1$ and $\phi_2$ are injective.

• I can't exclude that some amalgam of two quasi-finite groups might provide an example.
– YCor
Commented Jun 24, 2015 at 20:11
• @YCor: Actually, your suggestion is related to what I've been discussing with some people: one candidate example proposed by Carderi is to take something like, or in the spirit of, the amalgamated product of two identical copies of a Tarski monster over the obvious finite subgroup. I have to confess that I have no idea of what such a group can look like. Commented Jun 24, 2015 at 20:58
• You maybe have in mind a 2-generated non-abelian groups with all proper nontrivial subgroup of order $p$ for some fixed prime $p$ (I'm not sure of a unique definition of Tarski monster). I'm not sure what is "the obvious subgroup", because there are many and do not necessarily look the same (in the sense they may a priori fail to be conjugate under automorphisms). Certainly it does not work if you consider a double $G*_HG$ where $H$ is (nontrivially) embedded in both copies of $G$ in the same way, since $G$ is naturally a quotient of this amalgam.
– YCor
Commented Jun 24, 2015 at 22:04
• Thanks, I only knew the vulgarised vague definition of a Tarski monster. Commented Jun 24, 2015 at 22:11

Let $G$ be a group with infinitely many ends. According to Stallings' theorem, $G$ splits non trivially over a finite subgroup. Now, consider the action of $G$ on the associated Bass-Serre tree $T$. Because the edge stabilizers are finite, it is clear that the action $G \curvearrowright T$ is acylindrical. Furthermore, since there at most two orbits of vertices and that $T$ has necessarily infinitely many ends, we deduce that the limit set $\Lambda(G) \subset \partial T$ has infinitely many points, ie. the action $G \curvearrowright T$ is non elementary. Since $G$ has a non elementary acylindrical action on a hyperbolic space, $G$ is acylindrically hyperbolic (as defined by Osin), and is in particular SQ-universal (according to a result of Dahmani-Guirardel-Osin). This implies that $G$ has uncountably many normal subgroups, so that $G$ is far from being simple.

EDIT 1: As suggested by Yves Cornulier, these groups are in fact relatively hyperbolic (a stronger property than acylindrical hyperbolicity if the group is not virtually cyclic). It is essentially a consequence of the following criterion due to Bowditch:

Theorem: A group $G$ is hyperbolic relative to a collection $\mathcal{G}$ of infinite subgroups if it acts on a connected graph $\Gamma$ such that

• $\Gamma$ is hyperbolic and fine (ie., for every $n \geq 1$, an edge belongs to finitely many simple cycles of length $n$),
• there are finitely many orbits of edges, whose stabilizers are finite,
• the elements of $\mathcal{G}$ are precisely the infinite vertex stabilizers of $\Gamma$,
• every element of $\mathcal{G}$ is finitely generated.

Notice that, if the vertex stabilizers are finite, then the action $G \curvearrowright \Gamma$ is properly discontinuous and cocompact, so that $G$ turns out to be hyperbolic.

Thus, because a finitely generated group $G$ with infinitely many ends splits non trivially over a finite subgroup, we deduce from the action on the associated Bass-Serre tree that $G$ is hyperbolic relatively to the factors of this splitting.

EDIT 2: In his article SQ-universality of free products with amalgamated finite subgoups, Lossov proves that the amalgamated product $A \underset{C}{\ast} B$ is SQ-universal provided that $[A:C] \geq 2$ and $[B:C] \geq 3$ (it corresponds to the case where the group has infinitely many ends). However, I did not find a similar reference for HNN extensions.

• Great! I'd guess that it was known before acylindricity that such $G$ would actually be relatively hyperbolic with respect to vertex groups? (relatively hyperbolic groups were known to be SQ-universal, Arzhantseva-Minasyan-Osin 2006)
– YCor
Commented Jun 25, 2015 at 5:31
• Well, thank you very much! Do you think that this fact was unknown before the work of Osin & co.? I would say that this sounds pretty strange, but on the other hand it is also true that the interest for relatively hyperbolic groups is only relatively recent. Actually I started wondering about this after having read a theorem claiming that Thompson's T and V (and related other simple groups) have one end. Luckily, the theorem was claiming something more (about the relative number of ends), otherwise it wasn't surprising me. Commented Jun 25, 2015 at 7:26
• Indeed, relative hyperbolicity follows from the work of Bowditch, so it was probably known soon. However, I am pretty sure that the SQ-universality was known even before. I will search a reference. Commented Jun 25, 2015 at 7:33
• I think this is the good reference: Lossov, SQ-universality of free products with amalgamated finite subgroups (see [link.springer.com/article/10.1007%2FBF00970007] and [ams.org/mathscinet/search/…) Commented Jun 25, 2015 at 8:26
• But it deals with amalgamated products only. The case of HNN extensions is still missing, and I did not find a reference. It is curious, it is however a natural question on SQ-universality. I suspect a direct proof should exist. Commented Jun 25, 2015 at 9:51