# Does every topological group embed as a closed subgroup in an amenable group?

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?

• It would already be interesting for Polish groups.
– YCor
Oct 20 '20 at 13:57
• @YCor indeed Polish groups are the case I'm most interested in, but maybe something can be said in general Oct 20 '20 at 14:08
• 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").
– YCor
Oct 20 '20 at 14:28

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 $$L^0(G,X,\mu)$$.