2
$\begingroup$

Let $G$ be a hyperbolic group acting faithfully on $\mathbb{Z}$ such that:

  1. The action is highly transitive - it is $k$-transitive for each $k \in \mathbb{N}$.

  2. For every quasiconvex subgroup $H \leq G$ of infinite index, the orbits of the action of $H$ on $\mathbb{Z}$ are finite.

I want to know of any special properties $G$ must have.

Must $G$ have a finite index subgroup?

An example of a special property: $G$ does not have a finite normal subgroup.

$\endgroup$
  • $\begingroup$ Algebraically, what does the highly transitive hypothesis mean about a point stabilizer? $\endgroup$ – HJRW Jan 7 '15 at 13:38
  • 1
    $\begingroup$ @HJRW Let $L$ be the point stabilizer. It means that for every two $k$-tuples of distinct cosets $(Lx_1, \dots, Lx_k)$ and $(Ly_1, \dots, Ly_k)$ there is some $a \in G$ such that $Lx_ia = Ly_i$ for each $1 \leq i \leq k$. That is, $$\bigcap_{i=1}^k x_i^{-1}Ly_i \neq \emptyset.$$ $\endgroup$ – Pablo Jan 7 '15 at 13:45
  • $\begingroup$ Could you give an example of such an action when $G$ is free? $\endgroup$ – HJRW Jan 7 '15 at 14:25
  • $\begingroup$ @HJRW Yes. See arxiv.org/pdf/1308.3192v1.pdf and references therein. In fact, I suspect that I can construct such an action for EVERY hyperbolic group without finite normal subgroups (this is clearly a necessary condition). It seems that it is even possible to add the condition that every element has a fixed point. $\endgroup$ – Pablo Jan 7 '15 at 14:34
  • 1
    $\begingroup$ that seems plausible. Of course, your question would then be equivalent to a famous open problem. $\endgroup$ – HJRW Jan 7 '15 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.