It is a standard result that closed subgroups of locally compact amenable groups are themselves amenable, so for example $F_2$, the free group on two generators, cannot be embedded as a closed subgroup of a locally compact amenable group. However, by a result of Pestov, $F_2$ embeds as a closed subgroup of $\mathrm{Aut}(\Bbb Q,\leq)$ and the latter group is (extremely) amenable.

Are there topological groups that cannot be embedded as a closed subgroup of any amenable group?

  • 3
    $\begingroup$ It would already be interesting for Polish groups. $\endgroup$
    – YCor
    Oct 20, 2020 at 13:57
  • 2
    $\begingroup$ @YCor indeed Polish groups are the case I'm most interested in, but maybe something can be said in general $\endgroup$ Oct 20, 2020 at 14:08
  • $\begingroup$ Note that in the larger generality, an embedding with closed image need not be a homeomorphism onto its image (or it's implicit in "embedding"). $\endgroup$
    – YCor
    Oct 20, 2020 at 14:28

1 Answer 1


At least for Polish groups, which was the case I was most interested in, the answer is positive.

I mentioned this question to Ola Kwiatkowska yesterday and she immediately pointed out that one of the standard universal Polish groups, the group $\mathrm{Iso}(\Bbb U)$ of isometries of the Urysohn space, is in fact not only amenable, but even extremely amenable (and Polish subgroups of Polish groups are closed).

For arbitrary groups I still expect the answer to be positive, but I have no meaningful comments to make about that case, apart from the fact that if a group embeds as a closed subgroup into an amenable group $G$, then it also embeds as a closed subgroup into an extremely amenable group, namely $L^0(G,X,\mu)$.


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