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?


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.

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    $\begingroup$ I can't exclude that some amalgam of two quasi-finite groups might provide an example. $\endgroup$
    – YCor
    Jun 24, 2015 at 20:11
  • $\begingroup$ @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. $\endgroup$ Jun 24, 2015 at 20:58
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Jun 24, 2015 at 22:04
  • $\begingroup$ Thanks, I only knew the vulgarised vague definition of a Tarski monster. $\endgroup$ Jun 24, 2015 at 22:11

1 Answer 1


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.

For more information on acylindrically hyperbolic groups, see Osin's article.

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.

  • $\begingroup$ 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) $\endgroup$
    – YCor
    Jun 25, 2015 at 5:31
  • $\begingroup$ 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. $\endgroup$ Jun 25, 2015 at 7:26
  • $\begingroup$ 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. $\endgroup$
    – Seirios
    Jun 25, 2015 at 7:33
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    $\begingroup$ 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/…) $\endgroup$ Jun 25, 2015 at 8:26
  • $\begingroup$ 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. $\endgroup$
    – Seirios
    Jun 25, 2015 at 9:51

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