Suppose that $G_1,\ldots,G_n$ are finite groups, and $m\geqslant 0$ is some integer. Set
$$G=G_1\ast\cdots\ast G_n*F_m,$$
(where $F_m$ is the free group on $m$ generators). Then, is $G$ virtually torsion-free (i.e. has a finite index torsion-free subgroup)?
It seems conceivable that the method given here just works in this context, but I am not familiar enough with geometric group theory.
Best Answer
The simplest proof I know is to consider the natural homomorphism $$ r: G\to G_1\times ... \times G_n $$ Its kernel is a finite index subgroup in $G$. Moreover, each nontrivial element conjugate to some $G_i$ maps nontrivially by $r$. Lastly, you use the fact that each torsion element of $G$ is conjugate to some $G_i$. (You can see this by using, for instance, normal forms or you can use the action of $G$ on its Bass-Serre tree.)
The fact that $G$ is linear is a bit trickier, one constructs a suitable discrete and faithful representation of $G$ in some $O(N,1)$ (the ping-pong argument).