Let G be a word-hyperbolic group with torsion and let ∂G be its boundary. Do there exist criteria that imply that all non-trivial finite order elements of G act fixed-point freely on ∂G?



Fixed points on boundary of hyperbolic group
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 fixed-point set of $F$. 
I will assume that $F$ is the maximal finite subgroup with the fixed-point set $L$. 


Lemma. The normalizer $H$ of $F$ in $G$ has the property that:


1. $H$ is quasiconvex in $G$.
2. 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 1-point 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 
quasi-preserves 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 fixed-point sets in $\partial G$, it suffices to assume that every infinite quasiconvex subgroup in $G$ has trivial centralizer. 


