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It is well known that in a given $\delta$-hyperbolic group there are only finitely many conjugacy classes of finite subgroups. This is clearly false for relatively hyperbolic groups since we have no control over the parabolic subgroups. Fix a relatively hyperbolic group $\Gamma$ with parabolic subgroups $P_i$. Are there only finitely many conjugacy classes of finite subgroups $F<\Gamma$ with the property that $F$ is contained in a two-ended subgroup $H<\Gamma$ which is not conjugate into any $P_i$? Perhaps weaker: are there only finitely many isomorphism types of two-ended subgroups which are not conjugate into any parabolic? Weaker still: Is there an upper bound to the orders of elements in such subgroups $F$?

(Motivation: I am trying to find a law obeyed by non-parabolic two-ended subgroups of relatively hyperbolic groups, or at least a partition of them into finitely many families, each of which obeys a law.)

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If you find that they obey finitely many laws, can't you just take an iterated commutator to obtain a single law they all obey? – HJRW Apr 21 '12 at 6:42
Good point. In any case, now there is a law and I'm very happy. – nolte Apr 21 '12 at 19:01
up vote 5 down vote accepted

Here's an idea for a proof in hyperbolic groups which generalizes to relatively hyperbolic groups. Actually, this will prove something slightly weaker but which easily strengthens to what you want: namely, that there are only finitely many conjugacy classes of finite subgroups $F < \Gamma$ such that $F$ is contained in a 2-ended subgroup $H$, $H$ is not parabolic, and $F$ fixes each end of $H$.

Take $C$ to be the Cayley graph of a hyperbolic group. For every 2-ended group $H$ with fixed points $\xi,\eta$ at infinity, take its ``axis bundle'' consisting of all geodesics in $C$ connecting $\xi$ to $\eta$. This axis bundle has a uniform width, a number $W$ independent of $H$ such that for any point $p$ in the axis bundle, its $W$ neighborhood separates each geodesic in the axis bundle. Then take $N(p)$ to be, say, the $101W$ neighborhood of $p$, and so $N(p)$ separates each axis by removing a subsegment of length $100W$. It follows that any element of $H$ that fixes $\xi$ and $\eta$ and that moves $N(p)$ off itself has infinite order. So, every element of the subgroup $F$ moves $N(p)$ so that the image has nontrivial intersection with $N(p)$, which places $F$ amongst finitely many conjugacy classes of finite subgroups.

If $C$ is the Cayley graph of a relatively hyperbolic group, you can still define the axis bundle of a nonparabolic 2-ended group, but it does not have uniform width. The width can bloat out where the axis bundle runs through a parabolic coset. Nonetheless, there is still a uniform number $W$ such that there exists a point $p$ in the axis bundle whose $W$ neighborhood separates every geodesic in the axis bundle. The point $p$ will be located in the region where the axis bundle is passing from one parabolic coset to another. The proof above then goes through, using that choice of $p$.

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Thanks, Lee. I was trying to do something along these lines, but with the action on a fine hyperbolic graph, which is of course totally wrongheaded since I was never going to get $N(p)$ finite. – nolte Apr 21 '12 at 18:58
I struck through the wrong part after Agol's comment. – Lee Mosher Apr 21 '12 at 22:01
If you want a reference for the geometric facts about (quasi-)geodesics in Cayley graphs needed for Lee's argument, I think Corollary 8.16 of Hruska arXiv:0801.4596 does the job. Or I'm sure it's somewhere in the oeuvre of Drutu--Sapir (I think the phrase they use is 'saturation'). – HJRW Apr 22 '12 at 20:41

You should read Dahmani's thesis: He explains how to construct Rips complexes $R_d$ in the relatively hyperbolic (RH) case. In particular, he verifies that Rips complexes $R_d$ ($d$ sufficiently large) of a RH group $G$ satisfy the following: The group $G$ acts cocompactly (but not properly) on $R_d$, but it does act properly away from the vertices; $R_d$ is finite-dimensional, contractible and all links are finite except at vertices (stabilized by some $P_i$'s). Vertex stabilizers are trivial or conjugate to parabolic subgroups $P_i$ of $G$. With this, the same proof as in the absolute hyperbolic case goes through: There exists large $D$ so that every finite subgroup $F$ of $G$ stabilizes a simplex in $R_D$. Since $G$ acts on $R_D$ cocompactly and stabilizer of every simplex is finite or one of $P_i$'s, there are finitely many finite subgroups $F_1,...,F_m$ so that every finite subgroup $F< G$ is conjugate to a subgroup of one of $P_i$'s or one of $F_j$'s.

This proof was probably written down by somebody (most likely, Dahmani himself in one of his many papers). You can email him and ask.

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If you take two groups $P_1, P_2$ that have infinitely many conjugacy classes of involutions, then $P_1\ast P_2$ will be hyperbolic relative to $\{P_1, P_2\}$ (maybe one also needs to assume $P_1$ and $P_2$ are finitely generated depending on your notion of relative hyperbolicity), but will contain infinitely many dihedral subgroups generated by pairs of involutions in $P_1$ and $P_2$. Thus, I think the strict answer to your question is no, but this does not contradict Lee Mosher's answer, since he is assuming that the finite subgroups preserve the ends of the group.

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Very nice example – Lee Mosher Apr 21 '12 at 21:53

This question is stated as Corollary 9.6 in Strong accessibility for finitely presented groups but much of the actual proof is in Lemma 9.5.

The proof doesn't quite follow Lee's sketch (which I'm sure works.) I know this is an old question, but I'm providing this answer for completeness and for the cheap thrill of gratuitous self-promotion.

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The link doesn't work: here it is: Louder-Touikan: – YCor Feb 5 at 19:27
Oops. It's fixed. – NWMT Feb 6 at 21:13

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