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It is a consequence of the theorem of M. Gromov on groups with polynomial growth and a result of P. Pansu(*) that if a finitely generated group $\Gamma = \langle S\rangle,\, S=S^{-1}$ satisfies $(\dagger)\quad |S^n| \le C\cdot n^c$ for all $n\in\mathbb{Z}_{>0}$ and some $C,c\in[0,+\infty[$ then the group actually satisfies that $|S^n| \sim C_S\cdot n^{d_\Gamma}$ for some constants $C_S\in]0,+\infty[$ and $d_\Gamma\in\mathbb{Z}_{\ge 0}$. In particular for any group $\Gamma$ the limit in $[0,+\infty]$ of the sequence $\lim \log|S^n|/\log n$ exists (**), and if finite it is an integer. Is there a known proof of the latter result which does not use Gromov's theorem?

(*) To be self contained: the former states that a group $\Gamma$ satisfying $(\dagger)$ must contain with finite index a nilpotent group, and the latter studies the growth of nilpotent groups and establishes the stated conclusion for such.

(**) To deal with all groups maybe you need an extension of Gromov's result that covers the case where you have the polynomial bound$(\dagger)$ only for an infinite sequence of integers $n$.

The question might appear a bit gratuitous, so I'll say that the motivation for asking this comes from the hope to extend the result about the integral exponent to the generic volume growth of so-called "unimodular random hyperbolic surfaces" (a generalization of infinite covers of compact hyperbolic surfaces).

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    $\begingroup$ My feeling is that any structural analysis of groups (or other group-like objects) of polynomial growth that is refined enough to give this claim would also be powerful enough to prove Gromov's theorem. For your application, there may be some recent work of Hrushovski on the structural theory of approximate equivalence relations that may be relevant (see ma.huji.ac.il/~ehud/approx-eq.pdf ), though it may not be so easy to translate his general formalism to a specific concrete setting such as yours. $\endgroup$
    – Terry Tao
    Feb 26, 2015 at 19:48
  • $\begingroup$ That's probably right. It may have been better to phrase the question with the motivation more in the spotlight, in any case I am grateful for the reference. $\endgroup$ Feb 27, 2015 at 12:49

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