13
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $N = G$. $\endgroup$ – Andy Putman Feb 13 '18 at 18:37
  • $\begingroup$ @AndyPutman if $y$ and $x$ generate an infinite dihedral group (or something mapping onto it with finite kernel), $y^nx$ is elliptic for all $n$. $\endgroup$ – YCor Feb 13 '18 at 18:44
  • $\begingroup$ @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. $\endgroup$ – Misha Feb 13 '18 at 18:56
  • $\begingroup$ @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. $\endgroup$ – Andy Putman Feb 13 '18 at 20:35
  • $\begingroup$ (and a similar comment to @YCor) $\endgroup$ – Andy Putman Feb 13 '18 at 20:36
18
$\begingroup$

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 $\langle xy\rangle$.)

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes I meant "with trivial finite radical", which is what you formulate. $\endgroup$ – YCor Feb 14 '18 at 9:23
  • $\begingroup$ 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. $\endgroup$ – YCor Feb 14 '18 at 13:53
  • $\begingroup$ 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. $\endgroup$ – Misha Feb 14 '18 at 18:03
9
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.