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 $g=h^n$.

  • $\begingroup$ You need to add the condition that the action is properly discontinuous and cocompact, as required by the definition of a CAT(0) group, otherwise there are easy counter-examples. $\endgroup$ – Moishe Kohan Sep 19 '16 at 19:00

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.

  • $\begingroup$ why if $g$ has to be an infinite order element it has to be loxodromic? could it not be parabolic? $\endgroup$ – Spotty Oct 16 '16 at 19:14
  • $\begingroup$ If a group acts cocompactly on a CAT(0) space, then it contains no parabolic. This statement should be available in Bridson and Haefliger's book; I should be able to find a precise reference if needed. $\endgroup$ – Seirios Oct 17 '16 at 7:59

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