TITLE: Are free products of finite groups virtually free?
QUESTION [18 upvotes]: Is the free product $A*B$ of two nontrivial finite groups always virtually free? 
If yes, is it easy to show?

REPLY [2 votes]: In fact, in his book (property 11 p. 120), Serre shows that the fundamental group of a finite graph of groups whose vertex groups are finite is virtually free. (The converse is also true, giving a characterization of virtually free groups.) I describe below the argument for the free product of two finite groups.
First, $A \ast B$ acts on a bipartite tree $T$ such that the edge stabilizers are trivial and the vertex stabilizers are conjugated to $A$ or $B$.
Let $|A|=n$, $|B|=m$ and $p=\max(n,m)$. Notice that $A$ (resp. $B$) acts freely on $A$ (resp. $B$) by multiplication on the left, hence a monomorphism $\varphi : A \hookrightarrow S_n \subset S_p$ (resp. $\phi : B \hookrightarrow S_m \subset S_p$). Now define $$ \psi = \varphi \ast \phi : A \ast B \to S_p,$$ and let $H= \mathrm{ker}(\psi)$. Clearly, $H$ is a finite-index subgroup of $A \ast B$. By construction, $H$ meets trivially $A$ and $B$; moreover, because $H$ is a normal subgroup, it meets trivially any conjugate of $A$ and $B$. Therefore, the action $H \curvearrowright T$, induced by $A \ast B \curvearrowright T$, is free. 
We deduce that $H$ is a finite-index free subgroup of $A \ast B$; so $A \ast B$ is virtually free.