Amenability of an “almost Hamiltonian” group

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.

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