Let $G=\langle X\ |\ R\rangle$ be a (finitely presented) virtually torsion-free group. Let $H,K<G$ be isomorphic (finite index) subgroups of $G$ and let $\varphi:H\rightarrow K$ be an isomorphism.

Define the HNN extension $\Gamma$ of $G$ and $\varphi$ in the usual way, i.e. $\Gamma=\langle G,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in H\rangle$.

Is $\Gamma$ virtually torsion-free?

My thought is that if $T$ is a finite index torsion-free subgroup of $G$ and if $H$ and $K$ are finite index we should be able to look at the intersection of each of them with $T$. So the group $\langle T,t\ |\ tht^{-1}=\varphi(h)\ \forall h\in T\cap H\rangle$ would be a finite index torsion-free subgroup of $\Gamma$.

Also, can we say anything about the smallest index of a torsion-free subgroup? For example if $G$ contains a torsion-free subgroup of index $k$, does $\Gamma$ contain a torsion-free subgroup of index $k$? Or is the index bounded by some function of $k$?


1 Answer 1


No, even assuming that $H,K$ have finite index in $G$: see https://mathoverflow.net/a/330658/14094, where I provided an example (where $G$ is finitely generated, virtually free, and $H,K$ are free subgroups of the same finite index). It relies on Burger-Mozes groups.

If one does not assume that $H,K$ have finite index, it's even simpler (see for instance https://mathoverflow.net/a/330655/14094).


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