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Let $G$ be an infinite hyperbolic group, and let us further assume that it is residually finite (or even LERF/GFERF) so that we have plenty of subgroups of finite index.

I would like to know if one can impose some reasonable conditions on $G$ under which the number of generators of finite index subgroups will have to grow (asymptotically) with the index. This happens, for example, if $G$ is free nonabelian (but this is too restrictive).

Conditions under which there is a bound from below on the subgroup growth are of course also of interest

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    $\begingroup$ The existence of hyperbolic HNN extensions (eg some free-by-cyclic groups) makes this tricky. Does it help to assume that the hyperbolic group has trivial abelianization? $\endgroup$
    – Sam Nead
    Commented Dec 23, 2014 at 21:16
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    $\begingroup$ The obvious comment is that the free group argument generalises to any 2-dimensional group with negative Euler characteristic. On the other hand, as Sam points out, with zero Euler characteristic this is usually false. $\endgroup$
    – HJRW
    Commented Dec 23, 2014 at 22:45
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    $\begingroup$ On the other hand, some of the arguments of this paper may be applicable: arxiv.org/abs/1212.4192. $\endgroup$
    – HJRW
    Commented Dec 23, 2014 at 22:51
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    $\begingroup$ More generally, if the 1st $l_2$-betti number is non-zero. A related concept is cost/rank gradient. front.math.ucdavis.edu/0701.5361 $\endgroup$
    – Ian Agol
    Commented Dec 24, 2014 at 4:29
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    $\begingroup$ I suspect that locally quasi-convex residually finite hyperbolic groups have the property that the rank grows with the index. However, I also think that the only known examples of this type might have non-zero 1st $l_2$-betti number. $\endgroup$
    – Ian Agol
    Commented Dec 24, 2014 at 4:42

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