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Let $G$ be a word-hyperbolic group and let $\partial G$ be its (Gromov) boundary. Do there exist criteria that imply that all non-trivial finite order elements of $G$ act fixed-point freely on $\partial G$?

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  • $\begingroup$ Surely not, because the boundary is a tree and so every finite order element fixes a point? (I believe it forms a tree as this is where JSJ-decompositions come from, if I have interpreted Bowditch's paper on cut points and canonical splitting's correctly). $\endgroup$ – user1729 Aug 9 '13 at 14:45
  • $\begingroup$ The boundary is certainly not always a tree. If the group acts properly and cocompactly on $\mathbb{H}^{n}$, the boundary is homeomorphic to $S^{n-1}$. $\endgroup$ – user68316 Aug 9 '13 at 14:48
  • $\begingroup$ Hmm. Can what you say happen in the one-ended case? $\endgroup$ – user1729 Aug 9 '13 at 14:49
  • $\begingroup$ If $\partial G$ is connected, then $G$ has one end. $\endgroup$ – user68316 Aug 9 '13 at 15:01
  • $\begingroup$ Oh yeah, Bowditch mentions that! $\endgroup$ – user1729 Aug 9 '13 at 15:37

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