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?

3$\begingroup$ 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

2$\begingroup$ 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 nontrivial 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 webpage contains an overview of the results together with the references: wwwgap.dcs.stand.ac.uk/~history/HistTopics/… $\endgroup$ – Ashot Minasyan Nov 29 '11 at 13:07
A. Yu. Olshanskii in the paper "Periodic quotient groups of hyperbolic groups." ((Russian) Mat. Sb. 182 (1991), no. 4, 543567; translation in Math. USSRSb. 72 (1992), no. 2, 519–541) proved that for every torsionfree nonelementary 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 nonelementary 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 $3$, $G^k=G$.

$\begingroup$ @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 IvanovOlshanskii) cannot be improved? In the torsionfree case you are probably thinking about the paper by Delzant and Gromov? But IvanovOlshanskii 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

1$\begingroup$ 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 DelzantGromov), 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