# Subgroup of hyperbolic group generated by non-torsion elements

Let $G$ be a hyperbolic group. I know that it is an open problem whether $G$ has a torsion-free subgroup of finite index. But if we let $N$ be the subgroup of $G$ generated by its non-torsion elements, then is $G/N$ necessarily finite?

If not, $G/N$ would be a finitely generated infinite torsion group with finitely many conjugacy classes.

• If $x \in G$ is finite-order (i.e. elliptic) and $y \in G$ is infinite-order (i.e. hyperbolic), then for $n \gg 0$ it should be the case that $y^n x$ is also hyperbolic. Your subgroup would thus contain $y$ and $y^n x$, and thus $x$ as well. So $N = G$. Feb 13, 2018 at 18:37
• @AndyPutman if $y$ and $x$ generate an infinite dihedral group (or something mapping onto it with finite kernel), $y^nx$ is elliptic for all $n$.
– YCor
Feb 13, 2018 at 18:44
• @AndyPutman: What you wrote is almost correct: You need to assume that the fixed-point set of $y$ (at infinity) is not contained in the fixed-point set of $x$. A clean way to deal with this is to assume first that $G$ is non elementary and replace $G$ with its quotient by maximal normal finite subgroup. Feb 13, 2018 at 18:56
• @Misha: Good point! I figured that something like this had to work, but I only wrote it in a comment since I hadn't thought that hard about it and wasn't sure I hadn't missed some minor technical issues. Feb 13, 2018 at 20:35
• (and a similar comment to @YCor) Feb 13, 2018 at 20:36

Yes. The following answer is inspired by Andy Putman's comment. Let $N_\infty(G)$ be the subgroup generated by elements of infinite order, in a group $G$.

Every non-elementary hyperbolic group $G$ with trivial finite radical has a useful property, which bears the ridiculous name "Pnaive", namely that for every finite subset $F$ there exists a $x$ such that the natural homomorphism from $F\ast\mathbf{Z}$ to $G$ mapping $F$ identically to itself and the generator of $\mathbf{Z}$ to $x$, is injective. This is a result of Arzhantseva and Minasyan: G. Arzhantseva, A. Minasyan, Relatively hyperbolic groups are C*-simple. J. Funct. Anal. 243 (2007), no. 1, 345–351.

This applies to $F$ equal to an arbitrary singleton $\{y\}$, $y\in G$: then both $x$ and $yx$ have infinite order, hence $y\in N_\infty(G)$. So Pnaive for a group $G$ implies $G=N_\infty(G)$, that is, $G/N_\infty(G)=\{1\}$.

Moreover, the property $N_\infty(G)=G$ for nontrivial (!) groups, is also stable under taking extensions with torsion kernel. Hence it is satisfied by all non-elementary hyperbolic groups, since these have a maximal finite normal subgroup and we inherit Pnaive from the nontrivial quotient.

For completeness, the elementary cases: if $G$ is finite then $N_\infty(G)=\{1\}$ and the quotient $G/N_\infty(G)$ is $G$ itself. If $G$ is 2-ended, then, denoting by $W(G)$ the finite radical, $G/W(G)$ is infinite cyclic or infinite dihedral according to whether $G$ acts trivially or not on its 2-element boundary, and then the desired quotient $G/N_\infty(G)$ is either trivial, or cyclic of order 2.

Thus for an arbitrary hyperbolic group, $G/N_\infty(G)$ is finite.

(Note that we use much much less than the full result of Arzhantseva-Minasyan! we only need it for a singleton $\{y\}$, and we need much less than injectivity: only injectivity on the cyclic subgroup $\langle xy\rangle$.)

• Yes I meant "with trivial finite radical", which is what you formulate.
– YCor
Feb 14, 2018 at 9:23
• Misha's answer (with the use of his lemma 1) prove the Arzhantseva-Minasyan result as I'm quoting it, for non-elementary hyperbolic groups with trivial finite radical (arXiv link: arxiv.org/abs/math/0603305). I should mention that I specified their result, which is more generally stated (Theorem 1) in their paper for arbitrary non-elementary relatively hyperbolic groups with trivial finite radical.
– YCor
Feb 14, 2018 at 13:53
• Yves: Not that it matters in any way, but, actually, the same proof I gave (with trivial modifications) also works in relatively hyperbolic case, all I am using is some standard staff about convergence groups. I did not read their paper, but, most likely, they are using similar arguments. Feb 14, 2018 at 18:03

Here is how I would prove it using ping-pong. The relevant form of ping-pong is the following proposition:

Proposition. Let $X$ be a set and $G_1, G_2$ be groups acting on $X$ such that there exist two subsets $X_1, X_2\subset X$ with disjoint complements and nonempty intersection, such that $$\forall g_i\in G_i -\{1\}, g_i(X_i)\cap X_i=\emptyset.$$ Then the group $\langle G_1, G_2\rangle$ of bijections of $X$ generated by $G_1, G_2$ is naturally isomorphic to $G_1\star G_2$.

Now, back to our hyperbolic group $G$. I will assume that $G$ is a non elementary hyperbolic group whose finite radical $W(G)$ is trivial. (Otherwise, see Yves' answer.) Given $g\in G$, let $Fix(g)$ denote the fixed-point set of $g$ in $\partial G$ (the Gromov boundary of $G$).

The following lemma answers Yves' question:

Lemma 1. If $g\in G-\{1\}$ has finite order then $Fix(g)$ has empty interior.

Proof. Suppose not. Then, since $W(G)=1$, $Fix(g)\ne \partial G$.

Recall:

Theorem. The set of pairs $(p_+, p_-)$ of attractive/repulsive fixed points of infinite order elements of $G$ is dense in $\partial G\times \partial G$.

Hence, one can find such fixed pair for an element $h\in G$ such that $p_+\notin Fix(g), p_-\in Fix(g)$. Then the sequence of conjugates $$g_n=h^n g h^{-n}, n\in {\mathbb N},$$ contains infinitely many distinct elements (since the sequence of complements to $Fix(g_n)$ in $\partial G$ converges to the singleton $p_+$). At the same time, the sequence $(h^{-n})_{n\in {\mathbb N}}$ is uniformly close to the quasiconvex hull of $Fix(g)$. Since $g$ acts on this quasiconvex hull with bounded displacement, we obtain that the sequence of word norms $|g_n|$ is bounded. A contradiction. qed

Lemma 2. Let $g\in G$ be nontrivial, generating a finite cyclic group $C< G$. Let $h\in G$ be an infinite order element such that $$g^k(Fix(h))\cap Fix(h)=\emptyset$$
for all $g^k\in C-\{1\}$. Then there exists $n$ such that the subgroup $\langle g, h^n\rangle$ of $G$ generated by $g$ and $h^n$ is naturally isomorphic to $C \star \langle h^n \rangle$.

Proof. Let $U$ be a small neighborhood of $Fix(h)$ in $\partial G$ such that $$g^k(U)\cap U=\emptyset$$
for all $g^k\in C-\{1\}$ (such a neighborhood existents since $C$ is finite). Let $V= \partial G -U$. Then there exists $n\in {\mathbb N}$ such that for all nontrivial elements $f\in \langle h^n\rangle$ we have $$f(V)\cap V=\emptyset.$$ It now follows from Tits' ping-pong that the subgroup $\langle g, h^n\rangle$ is naturally isomorphic to $C \star \langle h^n\rangle$. qed

Lemma 3. Given a nontrivial finite order element $g\in G$, there exists an infinite order element $h\in G$ such that $$g^k(Fix(h))\cap Fix(h)=\emptyset$$
for all $g^k\in \langle g \rangle -\{1\}$.

Proof. Let $U$ be a nonempty open subset of $\partial G -U$. Since $C=\langle g\rangle$ is finite, we can choose $U$ such that for all nontrivial elements $f\in C$ we have $$f(U)\cap U=\emptyset.$$ (For instance, one can use Lemma 1 here.)

On the other hand, by minimality of the action of $G$ on $\partial G$, there exists an infinite order element $h\in G$ such that $Fix(h)\subset U$. (This also follows from the Theorem stated above.) qed

Now, Lemmas 2 and 3 imply:

Corollary. Let $g\in G$ be a nontrivial finite order element. Then there exists an infinite order element $h\in G$ such that the subgroup $\langle g, h\rangle$ of $G$ is naturally isomorphic to $\langle g\rangle \star \langle h\rangle$. In particular, the product $gh$ also has infinite order.