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I have made several attempts at various times to understand the many equivalent definitions of an amenable group. Is the following statement correct?

A group $G$ is amenable if and only if, for any finite subset $X$ of $G$ and any $\epsilon > 0$, there is a finite subset $A$ of the subgroup $\langle X \rangle$ of $G$ generated by $X$, such that $|xA \, \Delta\, A|/|A| < \epsilon$ for all $x \in X$.

Thanks!

Derek Holt
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  • $(\forall \epsilon >0)(\exists X\subseteq ^{ f }G)(\exists A\subseteq ^{ f }\left< X \right> )(\forall x\in X)\left(\frac { |Ax\setminus A| }{ |A| } +\frac { |A\setminus Ax| }{ |A| } <\epsilon\right)$!? I think there is a relation between this and big/small/skinny sets in a group. –  Feb 24 '13 at 17:44
  • The standard Følner definition of amenable exactly what you write with "of the subgroup $\langle X\rangle$" erased. So what you write is a priori slightly stronger. But if $G$ is amenable, given finite $X$, given that $\langle X\rangle$ is amenable, you get the property as you write. – YCor Feb 25 '13 at 04:28

1 Answers1

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Yes, this is correct.

As Yves Cornulier notes, your condition implies the Følner condition which is the same as yours, but with the condition $A \subseteq \langle X \rangle$ removed.

Conversely, suppose $G$ is amenable. Then the subgroup $H = \langle X \rangle$ is amenable and if $(F_n)_{n \in \mathbb{N}}$ is a Følner sequence for $H$ then choosing $n$ large enough allows us to take $A = F_n$.

An advantage of your condition is that it is immediately clear that a subgroup of a group satisfying your condition also satisfies your condition. To show that the usual Følner condition passes to subgroups requires an additional argument.

Martin
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    Thanks to you and Yves for your help! – Derek Holt Feb 25 '13 at 08:06
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    By the way, that the Følner condition passes to subgroups can be seen directly. Its negation is the isoperimetric inequality: $\exists S$ finite and $e>0$ such that for every finite $F$ there exists $s\in S$ such that $#(sF\Delta F)\ge e#(F)$. Then it's easy that this negative condition passes to overgroups, just partitioning $F$ according to the left cosets of the smallest group. – YCor Mar 06 '13 at 00:36