Given a torsion-free hyperbolic group $G$, does there exist a number $n(G)$ such that for any $x,y,z\in G$, $x^n y^n z^n =1$ implies that $x$, $y$, and $z$ commute pairwise?

Some musings/questions... When $G$ is free, the result is true for $n=2$.

Clearly, this does not generalize to hyperbolic groups (e.g. non-orientable surface groups of genus 3).

Somewhat related: given any 3 non-commuting elements $x,y,z\in G$ there is a number $n(x,y,z)$ such that $\langle x^n, y^n , z^n \rangle $ is free. Is this true??..definitely in case of two elements (Gromov).

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    $\begingroup$ 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. $\endgroup$ – YCor Jun 24 '13 at 21:25
  • $\begingroup$ 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 $W(G)$". $\endgroup$ – YCor Jun 24 '13 at 21:30
  • $\begingroup$ Yves, the $G$ is assumed torsion-free.... $\endgroup$ – Alexey Kvashchuk Jun 25 '13 at 0:43
  • $\begingroup$ but thanks for confirming the answer to the second question. the argument with disjoint ends works. $\endgroup$ – Alexey Kvashchuk Jun 25 '13 at 0:44
  • $\begingroup$ 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). $\endgroup$ – YCor Jun 25 '13 at 18:24

The second statement is true if $x,y,z$ are not torsion as proved by Arzhantseva and, independently, by Kapovich and Weidmann (see Arzhantseva, Goulnara N. A dichotomy for finitely generated subgroups of word hyperbolic groups. Topological and asymptotic aspects of group theory, 1–10, Contemp. Math., 394, Amer. Math. Soc., Providence, RI, 2006 and Kapovich, Ilya; Weidmann, Richard Nielsen methods and groups acting on hyperbolic spaces. Geom. Dedicata 98 (2003), 95–121. This also implies the first statement (about the equation $x^ny^nz^n=1$) in the torsion-free case.

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  • $\begingroup$ It's not clear why the positive answer to the 2nd question implies a positive question to the first, since in the second statement $n$ depends on $(x,y,z)$ and in the first statement there is a single $n(G)$. Unless the given references give $n(x,y,z)$ independent of $(x,y,z)$. $\endgroup$ – YCor Jun 24 '13 at 21:36
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    $\begingroup$ $n(x,y,z)$ depends only on $\delta$, the hyperbolicity constant. It is something like $12\delta$. $\endgroup$ – user6976 Jun 24 '13 at 21:56
  • $\begingroup$ Mark, thanks for the answer. I'm familiar with that result (i.e. an alternative that either a given f.g. subgroup of $G$ is free or a tuple of it's generator is Nielsen equivalent to a tuple of "short" elements). However, like Yves, I fail to see how it implies the answer to the first question. $\endgroup$ – Alexey Kvashchuk Jun 25 '13 at 0:42
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    $\begingroup$ Let $G$ be torsion-free. Take any three pairwise non-commuting elements $x,y,z$. Then $x^n,y^n,z^n$ are not equivalent to short elements where $n=n(\delta)$ does not depend on $x,y,z$ (this is a standard hyperbolic argument). Then $x^n,y^n,z^n$ freely generate free subgroup. Hence if $x^{2n}y^{2n}z^{2n}=1$, then $x^{n}, y^n, z^n$ pairwise commute, a contradiction (the centralizer of an element in a torsion-free hyperbolic group is cyclic, so if $x,y$ do not commute, no powers of $x,y$ commute either). $\endgroup$ – user6976 Jun 25 '13 at 3:10
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    $\begingroup$ Yes, that is correct. It is no difficult to prove using asymptotic cones for example. $\endgroup$ – user6976 Jun 25 '13 at 18:55

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