Prove or disprove:


Let $G$ be a countable group. Let $H < G$ be an amenable subgroup with a finite conjugacy class. Then the normal closure of $H$ is also amenable. 


Thanks!



Amenable normal closure
It's true. Indeed, by your assumption, the normalizer of $H$ has finite index and hence contains a finite index normal subgroup $N$. For every $g\in G$, the intersection $H\_g=N\cap gHg^{-1}$ is normal in $N$, and only depends on $g\in G/N$. Since a subgroup generated by amenable normal subgroups is amenable, we deduce that the subgroup $M$ generated by the $H\_g$ is amenable. $M$ is normal in $G$ and contains a finite index subgroup of $H$, and is contained in the normal closure of $H$.


So, working in $G/M$, we can reduce to the case when $H$ is finite. In this case, $H$ is contained in the FC-center of $G$ (the union of all finite conjugacy classes), which is well-known (easy exercise) to be a locally virtually abelian normal subgroup, and hence amenable.


