Torsion element of a non-elementary Hyperbolic group Let $\Gamma$ be a non elementary hyperbolic group acting on the Gromov boundary $\partial\Gamma$. Let $a \in \Gamma$ be a torsion element i.e $\langle a\rangle$ is finite. Does $a$ fix every element of $\partial\Gamma$?
I know that that every element $a$ is either a parabolic element in which case it has one fixed point, or a hyperbolic element in which case it has two fixed points or an elliptic element in which case it has finite order. For non-elliptic elements $a$ also there is a north pole-south pole property: for each such $a \in \Gamma$ there are two points $x_a^+$ and $x_a^{-}$ on the boundary, not necessarily distinct such that $$a^nx\xrightarrow{n \to \infty} x_a^+, \forall x \ne x_a^-$$
and $ax_a^+=x_a^+$ and $ax_a^-=x_a^-$. 
This convergence property is common to all convergence groups, where I looked in as well but found no information about the fixed points of elliptic elements. Can anything be said about the fixed points of elliptic elements? Am I missing something?
Thanks for the help!!
 A: A finite-order element of a hyperbolic group may have no fixed point at infinity. Consider for instance the free product $\mathbb{Z}_2 \ast \mathbb{Z}_2$ (or $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$ is you want a non-elementary example). You have an action on a simplicial tree and the finite-order elements inverse edges. 
In fact, to answer your first question, it is not too difficult to show that the kernel of the action $\Gamma \curvearrowright \partial \Gamma$ is the unique maximal finite normal subgroup of $\Gamma$. A possible sketch of proof goes as follows:


*

*Let $K$ denote the kernel of the action $\Gamma \curvearrowright \partial \Gamma$. Because $\Gamma$ is non-elementary, the boundary contains three pairwise distinct points $\alpha,\beta,\gamma \in \partial \Gamma$. Notice that, for every $N \geq 1$, the intersection $$[\alpha,\beta]^{+N} \cap [\beta,\gamma]^{+N} \cap [\alpha,\gamma]^{+N}$$ is bounded, where $[\cdot,\cdot]$ denotes the union of all the geodesics between two points at infinity. Taking $N$ large enough so that the intersection is non-empty, it follows that $K$ stabilises a bounded subset, proving that $K$ must be finite. Moreover, as a kernel, it is automatically normal.

*Next, let $F \lhd \Gamma$ be a finite normal subgroup of $\Gamma$. Notice that, for every $x \in \Gamma$, the orbit $F \cdot x$ has the same diameter as the the orbit $F \cdot 1$. Indeed, $$F \cdot x = xx^{-1}Fx \cdot 1= x (F \cdot 1).$$ It follows that $F$ fixes pointwise $\partial \Gamma$, hence $F \leq K$. 
