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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.


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I think it is unclear (to me at least) what you mean. Do you mean that there are only finitely many $G$-conjugates of $H$, or do you mean that there is an element of $H$ which only has finitely many conjugates ( and in that case, I assume you mean to exclude the identity)? –  Geoff Robinson Jun 9 '13 at 19:29
I mean that there are only finitely many groups of the form $gHg^{-1}$. In any case, the identity is the only element in its conjugacy class. –  Vladimir Jun 9 '13 at 20:00
@Vladimir: it may not be the case, but your phrasing of the question makes it look like homework. Can you motivate the question a bit? –  Alain Valette Jun 10 '13 at 4:33

1 Answer 1

up vote 5 down vote accepted

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.

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Thanks! But I'm not sure I understand. The normalizer of $H$ has finite index in what? And why? –  Vladimir Jun 9 '13 at 15:36
Please note that $H$ does not necessarily have finite index in $G$, but only a finite congujacy class. –  Vladimir Jun 9 '13 at 15:56
The set of subgroups conjugate to $H$ can be identified with $G/A$ where $A$ is the normalizer of $H$. So your assumption on the number of conjugates is precisely that $A$ has finite index in $G$. –  YCor Jun 9 '13 at 16:39

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