A certain free product of groups is virtually torsion-free

Question

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.

Answer 1

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).

Answer 2

There is a general theorem by Karass, Pietrowski and Solitar, in their work "Finite and infinite cyclic extensions of free groups". J. Austral. Math. Soc. 16 (1973), 458–466.

The theorem says, in the language of graphs of groups, that for any finite graph of finite groups, its Bass-Serre fundamental group has a free subgroup of finite index.

Your example can be expressed as a finite graph of finite groups in the following manner. Take a graph $\Gamma$ with one vertex $V_0$ of valence $m+n$, $n$ vertices $V_1,...,V_n$ of valence $1$, $m$ loop edges based at $V_0$, and $n$ edges connecting $V_0$ to $V_1,...,V_n$ respectively. Each edge is labelled by the trivial group, $V_0$ is labelled by the trivial group, and $V_1,...,V_n$ are labelled by the groups $G_1,...,G_n$ respectively.

You can think of the subgraph of loop edges as having fundamental group $F_m$, and the subgraph consisting of the edge connecting $V_i$ to $V_0$ as having fundamental group $G_i$. By identifying the $V_0$ points of all of those subgraphs one is taking the free product of all of those pieces.