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here is another interesting question that I can't answer on my own.

Let G be a countable, discrete group such that for any subgroup H of G and any element s of G we have: [H : sHt] is finite and [H : tHs] is finite where st=ts=1 i.e. t is the inverse of s. Then G is amenable.

Note that in the case when the indexes above are exactly equal 1 then G is Hamiltonian which is solvable and therefore amenable.

Thanks in advance.

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Another way to state this condition is that every subgroup $H<G$ has commensurator $\text{Comm}_G(H)$ equal to all of $G$. The commensurator in $G$ of a subgroup $H$ consists of all those $g$ for which $H\cap gHg^{-1}$ has finite index in $H$ and in $gHg^{-1}$.) (Note that in your question you need to intersect $H$ with $gHg^{-1}$ to ensure you get a subgroup of $H$.) Do you have examples of not-virtually-solvable groups for which this condition holds? – Tom Church May 31 '10 at 18:57
A classic is Pedersen's book, unfortunately not readable here: – Steve Huntsman May 31 '10 at 19:18
Pedersen's book is available here in electronic form: – Dmitri Pavlov May 31 '10 at 19:25
I think it is a bad idea to completely replace a question like this, in particular because Tom Church spent time writing a comment for the original question. If you figured out the first question as it says in the edit history, you could post the answer. – Jonas Meyer May 31 '10 at 19:36
Quasi-finite groups (groups in which every proper subgroup is finite) fulfill the requirements (every subgroup is commensurated). Some known examples of such groups are infinite with Kazhdan's Property T and hence are non-amenable. Possibly the question for torsion-free groups is more subtle. – YCor Oct 6 '15 at 17:23

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