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Let $G$ be a totally disconnected Polish topological group (e.g., a closed subgroup of the homeomorphism group of the Cantor set). If $G$ is amenable, is every closed subgroup of $G$ also amenable?

Notes: This is apparently not true if we remove the "totally disconnected" requirement. It is true if we require $G$ to be locally compact.

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    $\begingroup$ The extremely amenable totally disconnected Polish group $\mathrm{Aut}(\mathbf{Q},\le)$ contains a free subgroup of rank 2 as (closed) discrete subgroup. (Fix a left-invariant total ordering on $F_2$, and let $F_2$ act on itself $X=F_2\times\mathbf{Q}$ with lexicographic order. This action is free and $X$ is isomorphic to $\mathbf{Q}$ as chain. So this realizes $F_2$ as discrete subgroup of $\mathrm{Aut}(\mathbf{Q},\le)$. $\endgroup$
    – YCor
    Commented Jan 31, 2022 at 17:46
  • $\begingroup$ Thanks a lot! I'm not sure I understand what you mean by "$F_2$ act on itself $X=F_2\times Q$" $\endgroup$
    – Vladimir
    Commented Jan 31, 2022 at 18:04
  • $\begingroup$ oops: $F_2$ acts on itself, and trivially on $\mathbf{Q}$, and consider the product action on $X=F_2\times\mathbf{Q}$. $\endgroup$
    – YCor
    Commented Jan 31, 2022 at 18:05
  • $\begingroup$ Thanks again! Is it easy to see that this subgroup is discrete and closed? $\endgroup$
    – Vladimir
    Commented Jan 31, 2022 at 18:14
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    $\begingroup$ Discrete implies closed. Yes any subgroup of Sym(N) acting freely is clearly discrete. $\endgroup$
    – YCor
    Commented Jan 31, 2022 at 19:35

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