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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.

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    $\begingroup$ Yes it is. Such groups are even linear. $\endgroup$ Jan 23 at 14:04
  • $\begingroup$ @MoisheKohan Thanks very much! Do you happen to have a simple reference? $\endgroup$ Jan 23 at 14:05
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    $\begingroup$ In fact, they are virtually free. Linearity then follows as free groups are linear. $\endgroup$
    – user1729
    Jan 24 at 18:06

2 Answers 2

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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).

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  • $\begingroup$ Dear Moishe. Thanks for your answer! I am slightly confused though--what about the $F_m$? Perhaps I've made things too complicated, but I knew how to show that the free product of finite groups is VTF--it was actually the free group part I was worried about. Sorry if this wasn't clear/is very easy! $\endgroup$ Jan 23 at 14:30
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    $\begingroup$ $F_m$ is in the kernel of the homomorphism $r$. $\endgroup$
    – Derek Holt
    Jan 23 at 16:49
  • $\begingroup$ @AlexYoucis: You can map it any way to like, but the most natural choice is to make it trivially. $\endgroup$ Jan 23 at 16:53
  • $\begingroup$ @MoisheKohan I see, thanks! $\endgroup$ Jan 24 at 7:02
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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.

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  • $\begingroup$ Dear Lee Mosher. Thanks very much for your answer! I also saw this result stated in some places, and wondered if it could be applied here, but was worried that I was misinterpretting things as I thought I have seen a counter example that shows that my statement is false if one also allows finite groups as indices in an amalgamated product. But, maybe one cannot apply this idea there (or maybe for amalgamated products with groups of the form I said it's fine). $\endgroup$ Jan 23 at 14:35
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    $\begingroup$ The theory of graphs of groups is designed to incorporate both the "amalgated free product" construction and the "HNN amalgamation" construction. You might like to read about the general theory. There are two excellent and complementary resources for that purpose: Serre's book "Trees"; and the paper by Scott and Wall entitled "Topological methods in group theory" (which is not even mentioned on the wikipedia page... hrumph... I think I'll add it). $\endgroup$
    – Lee Mosher
    Jan 23 at 14:42
  • $\begingroup$ Thanks for the info! I will definitely check them out. How precisely then does the following fit into your comment: mathoverflow.net/a/330658/38867 ? $\endgroup$ Jan 23 at 14:43
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    $\begingroup$ I'm not sure what connection you are trying to draw. The main question in that link is not about a finite graph of finite groups, it is instead about an HNN amalgamation of a group along two finite index subgroups (as said in my previous comment, the finite graph of groups construction is a generalization of both the HNN construction and the amalgamated free product construction). $\endgroup$
    – Lee Mosher
    Jan 23 at 14:56
  • $\begingroup$ I see. Sorry for the silly quesiton--this question has come up in a sort of tangential way in my research and, unfortunately, I am very unfamiliar with this sort of group theory. $\endgroup$ Jan 23 at 15:00

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