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.



On the number of ends of a countable simple group
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](http://arxiv.org/abs/1304.1246).


**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](http://homepages.warwick.ac.uk/~masgak/papers/bhb-relhyp.pdf):


**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.


