Given a group $G$ acting on a $CAT(0)$ complex $X$ by isometries can $G$ contain a divisible element, i.e. an element $g\in G$ such that $\forall n\in\mathbb N$ there is $h\in G$ such that $g=h^n$.
Let $G$ be a group acting geometrically on a CAT(0) space $X$.
- Show that there exists some $N \geq 0$ such that any torsion element of $G$ has order $\leq N$.
- Deduce that $G$ cannot contain a divisible element of finite order.
- Let $g \in G$ be an infinite order element. In particular, it must be loxodromic, ie., there exists a bi-infinite geodesic $\gamma$ on which it acts by translation. If $g=h^n$, then $h$ also acts on $\gamma$ by translation. Moreover, $\ell(g)=\ell(h)^n$, where $\ell(\cdot)$ denotes the translation length.
- Show that, if $g$ is a divisible element and if we fix a point $x \in \gamma$, then there exist infinitely many elements $h \in G$ satifying $d(x,hx) \leq 1$. Hence a contradiction with the fact that $G$ acts geometrically on $X$.
In fact, the argument essentially holds in a more general setting, for instance for semihyperbolic groups.