normalizer of a cyclic subgroup in a torsion-free hyperbolic group

Question

How one can show that in a torsion-free hyperbolic group if elements $x$ and $y$ (edit: $y\ne1$) satisfy: $$ xy^mx^{-1}=y^n $$ then $m=n$ and $x$ and $y$ belong to the same cyclic subgroup?

Answer 1

Here is one argument. You will need some basic facts about hyperbolic groups which you can find, for instance, in the book by Bridson and Haefliger "Metric Spaces of Non-Positive Curvature".

I will assume that $y$ is nontrivial, otherwise there is nothing to prove (except $n=m$ will fail). Consider the boundary $\partial G$ of the hyperbolic group $G$. Then, since $G$ is torsion-free, $y$ has exactly two fixed points $p, q$ in $\partial G$. The same holds for the powers $y^m, y^n$. Hence, $x$ preserves the set $S=\{p, q\}$: The points $p, q$ are either both fixed by $x$ or are swapped by it. In either case, the subgroup $H$ generated by $x$ and $y$ preserves the set $S$ and, hence, is elementary. Every elementary subgroup of a hyperbolic group is virtually cyclic, i.e., contains a cyclic subgroup of finite index. Since $G$ is torsion-free, the subgroup $H$, therefore, is infinite cyclic (see below). The fact that $n=m$ follows immediately from commutativity of the group $H$.

Lemma. Every nontrivial torsion-free virtually cyclic group $H$ is infinite cyclic.

Proof. Let $C$ be an infinite cyclic subgroup of finite index in $H$. Consider the discrete action of $C$ on the real line by translations. Using the induced representation construction, we obtain a discrete isometric action of $H$ on a Euclidean space $E$. Then, by a Bieberbach's theorem, there exists an affine subspace $A\subset E$ invariant under the action of $H$, on which $H$ acts cocompactly. Since $H$ is virtually cyclic, the subspace $A$ has to be 1-dimensional. Since the action of $H$ on $E$ is properly discontinuous, the kernel of the action of $H$ on $A$ is finite. Since $H$ is torsion-free, the kernel is trivial. The action of $H$ on the line $A$ has to be by translations (since $H$ is torsion-free). Hence, $H$ acts on $A$ as an infinite cyclic group. Since the kernel of the action is trivial, the group $H$ is infinite cyclic. QED

One can probably give a purely algebraic proof of this, avoiding affine actions, but I am not sure how to do so.