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Update: I've reversed the inequality.. sorry for the mistake!

Let $G$ be a finitely generated group with a fixed symmetric generating set $S$. Let $d_S$ be the word distance with respect to $S$ and let $B(e,R)$ and $S(e,R)$ denote the open ball and the sphere around the unit element $e$ with radius $n\in\mathbb N$.

Question: Does there exist a non-amenable group such that $|B(e,n)|>|S(e,n)|$ for all but finitely many $n$'s?

I hope it is not too easy... I'm trying to use $\mathbb F_2\times(\mathbb Z_2)^s$, with $s$ big enough, but I'm quite confused at the moment and I'm not convinced one can do the job with a group containing $\mathbb F_2$. The balls increase too quickly and even if you try to make their growth slower using a hugely generated amenable group, it seems to me that at a certain point $\mathbb F_2$ wins.. mmm sorry for the informal language of this sentence. I hope you have understood what I meant.

Thanks in advance for any comment/suggestion

Valerio

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    $\begingroup$ Yes. The group $(\mathbb Z/2\mathbb Z)\ast (\mathbb Z/3 \mathbb Z)$ should give an example. $\endgroup$
    – Andreas Thom
    Commented May 30, 2011 at 19:25
  • $\begingroup$ yes, it works! you double just a part of the previous sphere.. :) I guess one can modify the example in such a way that the double condition is verified with arbitrarily big constant. I mean, it is know that non-amenability is equivalent to the existence of a costant $k$ such that $|N_k(F)|\geq2|F|$, for any finite subset $F\subseteq G$, where $N_k(F)$ is the set of elements whose distance from $F$ is less then $k$. The question is: fixed $k_0\in\mathbb N$, is there a non-amenable semigroup such that the best choice of $k$ is exactly $k_0$? $\endgroup$
    – Valerio Capraro
    Commented May 30, 2011 at 21:18
  • $\begingroup$ (mmm.. I'm thinking that maybe it is not considered polite to ask another question in a comment, because it seems that one demands an answer from that particular person.. so I don't cancel because you might have already read, but no worry.. I'm not demanding anything..) $\endgroup$
    – Valerio Capraro
    Commented May 30, 2011 at 21:29
  • $\begingroup$ @Valerio: in your question you mean: "does there exist a non-amenable group and a finite generating subset such that..." $\endgroup$
    – YCor
    Commented Jul 1, 2013 at 19:07

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unless I am missing something... The free group on 2 generators already has this p-ty. Since $|B(e,n)|=2 \cdot 3^n-1$ and $S(e,n)=4\cdot 3^{n}$.

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  • $\begingroup$ You hit first, Kate (:-) $\endgroup$
    – Alain Valette
    Commented May 30, 2011 at 16:41
  • $\begingroup$ ahahah I think I wrote the inequality in the wrong way!! Sorry :(( (I was sad that you were not at my talk today.. I also quoted you a couple of times! :( ) $\endgroup$
    – Valerio Capraro
    Commented May 30, 2011 at 18:33

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