# If a group acts properly and coboundedly on a hyperbolic space, each finite subgroup has a (uniformly) bounded orbit.

I am trying to solve the following problem:

"Let $G$ be a group acting properly, coboundedly and by isometries on a hyperbolic space $X$. Show that there is a constant $C$ such that any finite subgroup $H$ of $G$ has an orbit of diameter at most $C$. Deduce that any hyperbolic group has finitely many conjugacy classes of finite subgroups."

I have understood the proof of Theorem III$.\Gamma.3.2$ on Bridson and Haefliger's "Metric spaces of non-positive curvature", pages 459-460, but it deals directly with the second part of my question, using only the usual action of $G$ on its Cayley graph, and not on a general hyperbolic space. Since the first part of the question is itself interesting, I am looking for a hint on how to show:

"Let $G$ be a group acting properly, coboundedly and by isometries on a hyperbolic space $X$. Show that there is a constant $C$ such that any finite subgroup $H$ of $G$ has an orbit of diameter at most $C$."

My idea is to use the fact that in Cay$(G)$ the statement holds, in order to deduce it for a general $X$. From the Milnor-Svarc lemma we know that for any $x_0 \in X$ the map $$f \colon g \mapsto gx_0$$ extends to a quasi-isometry between Cay$(G)$ and $X$, and $f$ is also $G$-equivariant and therefore maps each $H$-orbit $A \subseteq$ Cay$(G)$ into an $H$-orbit $B \subseteq X$. But there could be points in $B$ that are far from the image $f(A)$, and I do not know whether coarse surjectivity of $f$ can be useful to show boundedness of $B$.

## 1 Answer

If a group $G$ acts on a $\delta$-hyperbolic space with a bounded orbit, say $O$, then Lemma III.$\Gamma.3.3$ produces a subset $C \subset O$ (namely, the set of the quasi-centers) of diameter at most $5 \delta$, and which is defined isometrically so that, because $O$ is $G$-invariant, $C$ must be $G$-invariant as well. Thus, taking some $x \in C$, the orbit $G \cdot x$ has diameter at most $5 \delta$.

I think this argument is due to Brady, Finite subgroups of hyperbolic groups (also available on his webpage). Alternatively, use your other question.

• Sorry, probably I was not clear. My question is how to show that a bounded orbit exists (for each finite subgroup $H \le G$) for each hyperbolic space $X$ on which $G$ acts nicely. The second part of my post was an attempt at using what we know about Cay$(G)$ to deduce the same about $X$. – 57Jimmy Jul 27 '16 at 14:03
• I am not sure to understand your problem: any orbit of a finite group is necessarily finite, so bounded. – Seirios Jul 27 '16 at 14:20
• Yes, sorry, I am quite ashamed of that. Then your answer definitely is what I was looking for :) – 57Jimmy Jul 27 '16 at 15:31