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15.4 (2002)

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Is it true that large growth implies non-amenability? More precisely, consider a number $\epsilon > 0$, an integer $k \geqslant 2$, a group $\Gamma$ generated by a set $S$ of $k$ elements, and the corresponding exponential growth rate $\omega(\Gamma, S)$ defined as in 14.7. For $\epsilon$ small enough, does the inequality $\omega(\Gamma, S) \geqslant 2k - 1 - \epsilon$ imply that $\Gamma$ is non-amenable?

Contributor: P. de la Harpe

Progress

If $\omega(\Gamma, S) = 2k - 1$, it is easy to show that $\Gamma$ is free on $S$, and in particular non-amenable; see Section 2 in (R. I. Grigorchuk, P. de la Harpe, J. Dynam. Control Syst., 3, no. 1 (1997), 51–89).

In general, no, it does not. Counterexamples can be found even in the classes of abelian-by-nilpotent and metabelian-by-finite groups. (G. N. Artzhantseva, V. S. Guba, L. Guyot, J. Group Theory, 8 (2005), 389–394).

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