This is a cross post from Math.StackExchange after 2 weeks without an answer and a bounty being placed on the question.

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 $k$?

I will accept answers which assume that $H$ and $K$ are finite index subgroups.


2 Answers 2


Yes, here's an example with an HNN over finite index subgroups as requested. It's based on constructing an amalgam of two f.g. virtually free groups, that has no proper finite index subgroups, using Burger-Mozes groups.

Fact (proved below): for every $n\ge 3$ there exists a non-torsion-free, virtually free group $G$ with a subgroup of finite index $H$, isomorphic to $F_n$, such that $G$ is normally generated by $H$.

By Burger-Mozes, there exists $n<m$ with two embeddings $u,v$ of $F_m$ as finite index subgroup of $F_n$, such that the resulting amalgam $A(u,v)$ of $F_n$ and $F_n$ over $F_m$ using the embeddings $u,v$ is simple. (D. Rataggi improved and made explicit the values, providing for instance $(n,m)=(9,81)$.)

Now use $G$ as in the fact, with $H$ free of rank $n$ as above. Identify $H$ to $F_n$ to deduce two embeddings $u',v'$ from $F_m\to H\subset G$. Consider the amalgam $A=A(u',v')$ (of $G$ and $G$ over $F_m$ using $u'$ and $v'$).

Claim: $A$ has no proper finite index subgroup. Proof: the finite residual contains the Burger-Mozes subgroup given as subamalgam of $F_n$ and $F_n$ over $F_m$, hence, since $F_n$ normally generates $G$, contains both amalgamated factors $G$, hence is all of $G$. (Since $A$ is not torsion-free, it follows that $A$ is not virtually torsion-free.)

Consider the HNN extension $B$ given by the pair of embeddings $u',v'$ of $F_m$ into $G$. Namely, this is $(\langle t\rangle\ast G)/R$, with $R$ normally generated by the $tu'(h)t^{-1}v'(h)^{-1}$ for $h\in F_m$. Then $B$ is also not virtually torsion-free (the quotient by its finite residual is $\mathbf{Z}$), since it contains the above amalgam.

Proof of the fact: it is enough to do it for $n=3$, yielding $H\le G$: in general just use the projection $G'=F_{n-3}\ast G\to G$ and define $H'$ as the inverse image of $H$.

Consider the virtually free group $G=\langle a,b:b^3=1\rangle$. Map it by $p$ onto the dihedral group $D_6$ mapping $a$ to an element of order 2 and $b$ to an element of order 3. Then the kernel $K$ of $p$ has index 6 and is free.

Now consider a subgroup $L$ of $G$ of index 3 containing $K$, this generating normally $G$. Since $L$ has a torsion-free subgroup of index 2, it has no element of order 3, and hence $L$ is free. Moreover $L$ has rank 3 (subfact below).$\square$

Subfact: the free group $L$ has rank 3. Proof: Note that $G$ also has a normal subgroup $N$ of index 3, free of rank $3=1+2$ (namely the kernel of the retraction from $G$ to $\langle b\rangle$ killing $a$, which is freely generated by $a$, $bab^{-1}$, $b^2ab^{-2}$). Since in a virtually free group all free subgroups of a given index have the same rank, we deduce that $L$ has rank 3.

  • $\begingroup$ Yves, if you'd like to post this on MSE for the bounty, you're more than welcome to! $\endgroup$ May 3, 2019 at 21:36
  • 1
    $\begingroup$ @RyleeLyman thanks for letting me know, I posted an answer there! $\endgroup$
    – YCor
    May 3, 2019 at 21:53
  • $\begingroup$ Nice answer, Yves! For the first fact you could just take any semidirect product $\mathbb Z/3\mathbb Z \rtimes F_n$, where some element of $F_n$ acts on the cyclic 3-group by a non-trivial automorphism. $\endgroup$ May 3, 2019 at 22:15
  • $\begingroup$ @AshotMinasyan indeed you're right, this is simpler (for some reason this lemma was the most time-consuming fact when I tried to fill in details writing the answer)! And of course it works directly with $C_3\rtimes F_n$ for any $n\ge 1$. $\endgroup$
    – YCor
    May 3, 2019 at 22:19
  • $\begingroup$ Thanks, this is a fantastic answer! $\endgroup$
    – Sam Hughes
    May 5, 2019 at 10:34

It is easy to construct a counter-example when $H$ and $K$ are not of finite index in $G$.

Let $A$ be a finitely presented torsion-free group which is not residually finite (e.g., the Baumslag-Solitar group $BS(2,3)$). Since $A$ is not residually finite, its finite residual $$Res(A):=\bigcap_{L\leqslant A, |A:L|<\infty} L$$ must be non-trivial, so take some $a \in Res(A)$, $a \neq 1$.

Define $G=A \times \langle b \rangle_2 \cong A \times \mathbb{Z}/2\mathbb{Z}$. Then $G$ is finitely presented and virtually torsion-free. We now set $$\Gamma=\langle G,t \mid tat^{-1}=ab\rangle.$$

I claim that $\Gamma$ is not virtually torsion-free. Indeed, it is easy to see that $Res(A) \subseteq Res(\Gamma)$ as $A\leqslant \Gamma$, so $a \in Res(\Gamma)$ and $tat^{-1}=ab \in Res(\Gamma)$, since the finite residual is a normal subgroup. Hence $b=a^{-1}(ab) \in Res(\Gamma)$, i.e., the element $b$, of order $2$, belongs to every finite index subgroup of $\Gamma$.

In the above example the associated subgroups $H$ and $K$ are infinite cyclic. I think that it should be possible to produce an example with $H$ and $K$ of finite index, but a different approach would have to be used.

On the positive side, the following two conditions together are sufficient to ensure that the HNN-extension $\Gamma$ is virtually torsion-free:

  • Suppose that the base group $G$ has finitely many conjugacy classes of finite order elements (this would yield the same property for $\Gamma$);
  • Suppose that $\Gamma$ is residually finite (then it has a finite index normal subgroup $N$ which avoids the finitely many conjugacy class representatives of torsion elements in $\Gamma$, hence $N$ must be torsion-free).
  • $\begingroup$ Remark: the first condition alone is not sufficient as your own example shows. Whether the second condition is sufficient is mysterious to me (but in practice requiring that the HNN is residually finite is a strong condition, not obvious to check). $\endgroup$
    – YCor
    May 3, 2019 at 22:20

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