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.