Let G be a wordhyperbolic group with torsion and let ∂G be its boundary. Do there exist criteria that imply that all nontrivial finite order elements of G act fixedpoint freely on ∂G?

1$\begingroup$ Yes, this is true if $G$ is torsionfree. $\endgroup$ – Igor Rivin Aug 9 '13 at 17:34

$\begingroup$ Note: crossposted as math.stackexchange.com/q/463688 $\endgroup$ – MvG Aug 9 '13 at 18:16
Let $G$ be a hyperbolic group with the Cayley graph $X$. Let $F<G$ be a finite subgroup and $L\subset \partial G$ be the fixedpoint set of $F$. I will assume that $F$ is the maximal finite subgroup with the fixedpoint set $L$.
Lemma. The normalizer $H$ of $F$ in $G$ has the property that:
$H$ is quasiconvex in $G$.
The Gromov boundary of $H$ maps to $L$ under the natural embedding $\partial H\to \partial G$,
Proof. I will first give a proof assuming that $L$ contains at least 2 points and then explain what to do in the 1point case. Let $C\subset X$ denote the quasiconvex hull of $L$, i.e., the union of all complete geodesics which are asymptotic (in both directions) to points of $L$. Then there exists $r<\infty$ so that $$ d(x, gx)\le r, \forall g\in F, \forall x\in C. $$ Let $H<G$ denote the stabilizer of $C$ in $G$. Then (by maximality of $F$) $H$ normalizes $F$. Conversely, if $g\in G$ belongs to the normalizer of $F$ then $g(L)=L$ and, hence, $g(C)=C$.
Suppose now that $H$ does not act cocompactly on $C$. Then there exists a sequence of vertices $x_n\in C$ such that $$ x_i\ne H x_j, \forall i\ne j. $$ Let $g_n\in G$ be elements representing the vertices $x_n$. Consider the subgroups $$ F_i= g_i^{1} F g_i<G. $$ These subgroups have the property that $$ \max_{g\in F_i} g\le r, $$ where $g$ is the word metric on $G$ and $r$ is the constant defined above. The number of elements of $G$ whose norms are bounded above is finite. Hence, after passing to a subsequence in $(x_n)$ , we can assume that $F_i=F_j$ for all $i, j$. In particular, $$ (g_i g_j^{1}) F (g_ig_j^{1})^{1} = F. $$ Therefore, $h_{ij}=g_i^{1} g_j$ preserves $C$. By construction, $$ h_{ij}(x_j)=x_i, $$ which is a contradiction. Thus, $H$ acts cocompactly on $C$. Since $C$ is quasiconvex, so is $H$. The limit set of $C$ in $\partial G$ is $H$equivariantly homeomorphic to the ideal boundary of (the hyperbolic group) $H$, since $H$ acts on $C$ cocompactly and $C$ the inclusion map $C\to X$ is a q.i. embedding.
This finishes the proof except for the possibility that $L$ is a singleton. I claim that this cannot happen. The argument is similar to the main proof. Suppose that $L$ consists of a single point $z$. Consider a geodesic ray $\rho$ in $X$ asymptotic to $z$. Let $v_i$ denote the vertices on $\rho$ and let $g_i\in G$ be the corresponding group elements. Consider the sequence of rays $$ \rho_i= g_i^{1}(\rho). $$ These rays contain the vertex $e\in G\subset X$ and, hence, subconverge to a complete geodesic $\gamma$ in $X$. The conjugate groups $F_i=g_i^{1} F g_i$ move $e$ by uniformly bounded amount, hence, after passing to another subsequence, we can assume they are all equal to a subgroup $F'<G$. Hence, $F'$ is conjugate to $F$ and quasipreserves the limit geodesic $\gamma$. It is now clear that $\gamma$ is asymptotic to two fixed points of $F'$. Hence, $F$ also has at least 2 ideal fixed points. Contradiction. qed
Thus, you can give purely algebraic conditions for the free action of $F$ on the Gromov boundary of $G$. In particular, we have a short exact sequence $$ 1\to F\to H\to Q \to 1. $$ By passing to a finite index subgroup $Q'<Q$, we obtain another (infinite) quasiconvex subgroup $H'<H$ containing $F$ so that $H'$ centralizes $F$. Therefore, in order to exclude (nontrivial) finite subgroups of $G$ with nonempty fixedpoint sets in $\partial G$, it suffices to assume that every infinite quasiconvex subgroup in $G$ has trivial centralizer.

$\begingroup$ Thanks. Could you please give a reference for the existence of the quasiconvex subgroup $H$? $\endgroup$ – user68316 Aug 9 '13 at 18:31

$\begingroup$ @user68316: I do not have a reference (although I am sure it is written somewhere), I will write a proof when I have time. $\endgroup$ – Misha Aug 9 '13 at 22:44

$\begingroup$ Misha: I've fixed some typos. Also, shouldn't the right hand side of the inequality $\max_{g\in F_i} g\le r$ just be some multiple of $\delta$, the hyperbolicity constant? $\endgroup$ – Lee Mosher Aug 13 '13 at 13:45

$\begingroup$ @LeeMosher: Lee: I am not sure how to get $r$ to be $\delta$. If $F$ has a fixed point in $C$ then one can take $r=2\delta$. In general, we only can say that $F$ has an orbit of diameter $\le R$ in $C$ and one can take $r=R+2\delta$. One can then estimate $R$ using $\delta$, but I do not feel like doing this. I corrected another typo in the definition of $r$ though. As for Lemma, it feels like the result should be in the literature, but I could not find it in the "standard" sources (I checked 7). $\endgroup$ – Misha Aug 13 '13 at 16:41

$\begingroup$ Actually, Noel Brady proved that every finite subgroup of $G$ has an orbit of radius $\le 2\delta+1$ ("A note on finite subgroups of hyperbolic groups"). From this, one gets an estimate on $R$. $\endgroup$ – Misha Aug 13 '13 at 16:58