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Let $\Gamma$ be a nonelementary hyperbolic group. Then is it true that $\Gamma^{p}:=<\gamma^{p}| \gamma \in \Gamma>$ is a finite index subgroup of $\Gamma$? Here $p$ is a prime number. What is known about this problem? What can we say when $p$ is just an odd number(not necessarily prime) or just a postive integer?

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    $\begingroup$ Since free groups are hyperbolic, surely the index can be infinite for sufficiently large $p$? The Burnside group $B(2,n)$ is known to be infinite for $n \ge 8000$ - I don't know if that is still the best known. $\endgroup$ – Derek Holt Nov 29 '11 at 9:03
  • $\begingroup$ What if p=2 or 3 or some small numbers? is there known results? $\endgroup$ – user9552 Nov 29 '11 at 10:26
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    $\begingroup$ For $p=2,3,4$ the group $G/G^p$ is finite for any finitely generated group $G$, because it is a quotient of the free Burnside group of exponent $p$, which is known to be finite for such small values of $p$. The smallest non-trivial value for which this is not yet known is $p=5$. $\endgroup$ – Ashot Minasyan Nov 29 '11 at 10:30
  • $\begingroup$ To Ashot Minasyan:Is there a reference for that for p=2,3,4, the group $G/G^{p}$ is the quotient of the free Burnside group of exponent $p$? Thanks. $\endgroup$ – user9552 Nov 29 '11 at 12:11
  • $\begingroup$ This web-page contains an overview of the results together with the references: www-gap.dcs.st-and.ac.uk/~history/HistTopics/… $\endgroup$ – Ashot Minasyan Nov 29 '11 at 13:07
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A. Yu. Olshanskii in the paper "Periodic quotient groups of hyperbolic groups." ((Russian) Mat. Sb. 182 (1991), no. 4, 543--567; translation in Math. USSR-Sb. 72 (1992), no. 2, 519–541) proved that for every torsion-free non-elementary hyperbolic group $G$ there is a number $N \in \mathbb{N}$ such that for any odd $n \ge N$ the quotient $G/G^n$ is infinite.

In a more recent article, Ivanov and Olshanskii ("Hyperbolic groups and their quotients of bounded exponents". Trans. Amer. Math. Soc. 348 (1996), no. 6, 2091–2138) proved a similar statement for an arbitrary non-elementary hyperbolic group $G$ (torsion is allowed): there is $n=n(G) \in \mathbb{N}$ such that $G/G^n$ is infinite. In this case one cannot say that $G/G^k$ is infinite for any sufficiently large odd $k$, because if the group $G$ is generated by elements of, say, order $3$, then for any $k$ not divisible by $3$, $G^k=G$.

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  • $\begingroup$ @Ashot: there are also "even" versions of these results. $\endgroup$ – Mark Sapir Nov 29 '11 at 11:13
  • $\begingroup$ @Mark, surely the second result (of Ivanov-Olshanskii) cannot be improved? In the torsion-free case you are probably thinking about the paper by Delzant and Gromov? But Ivanov-Olshanskii already prove this for all $n$ diisible by $2^5$. $\endgroup$ – Ashot Minasyan Nov 29 '11 at 11:32
  • $\begingroup$ I meant a paper by Ivanov and Olshanskii. In your answer you always put $n$ odd. For the OP, it does not matter probably. $\endgroup$ – Mark Sapir Nov 29 '11 at 13:47
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    $\begingroup$ I would like to use this opportunity to advertise the work of a young colleague: more than infinite, these quotients $G/G^n$ ($G$ hyperbolic, $n$ odd large enough) have exponential growth, arbitrarily close to the growth of $G$. See math.vanderbilt.edu/~coulonrb/papiers/growth.pdf $\endgroup$ – Benoît Kloeckner Nov 29 '11 at 17:24
  • $\begingroup$ @Benoit: Thanks for pointing out. This is in fact the most geometrically intuitive proof. Remi is giving a series of lectures about it in Vanderbilt this semester. It is based on ideas of Gromov (and a paper by Delzant-Gromov), but it is the first place where the ideas are implemented clearly and cleanly. Also Remi's work applies to Gromov's random groups, etc. $\endgroup$ – Mark Sapir Nov 29 '11 at 20:04

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