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S Apr 19, 2015 at 6:56 history suggested Seirios CC BY-SA 3.0
Improved formating and edited tags
Apr 19, 2015 at 6:28 review Suggested edits
S Apr 19, 2015 at 6:56
S Apr 19, 2015 at 4:06 history suggested Duchamp Gérard H. E. CC BY-SA 3.0
I corrected theee title (I am a one of Schützenberger's student)
Apr 19, 2015 at 3:49 review Suggested edits
S Apr 19, 2015 at 4:06
Jun 25, 2013 at 20:56 vote accept Alexey Kvashchuk
Jun 25, 2013 at 18:24 comment added YCor ah ok you're right. But anyway for a discrete hyperbolic group, "torsion-free" is a very strong hypothesis while "trivial finite radical" is a very weak hypothesis (as we can always boil down to it by modding out).
Jun 25, 2013 at 0:44 comment added Alexey Kvashchuk but thanks for confirming the answer to the second question. the argument with disjoint ends works.
Jun 25, 2013 at 0:43 comment added Alexey Kvashchuk Yves, the $G$ is assumed torsion-free....
Jun 24, 2013 at 21:30 comment added YCor PS: for these question we need the hyperbolic group to have a trivial finite radical (the finite radical $W(G)$ is the largest finite normal subgroup, which in a discrete hyperbolic group $G$ does exist). Otherwise the last question has stupid counterexamples (take semidirect products $M\rtimes F$ with $M$ finite and $F$ free). Or alternatively replace everywhere "commute" by "commute modulo $W(G)$".
Jun 24, 2013 at 21:25 comment added YCor for the last question, two non-commuting elements of infinite order in a discrete hyperbolic group have disjoint ends (the ends of a hyperbolic isometry are the two fixed ends). So Gromov's argument of freeness (for some suitable powers) works with no change.
Jun 24, 2013 at 20:32 answer added user6976 timeline score: 8
Jun 24, 2013 at 19:46 history asked Alexey Kvashchuk CC BY-SA 3.0