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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. Thanks!

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? No …

Let $G$ be a locally compact, second countable, non-amenable group, let $X$ be a Haudorff space that is not necessarily compact, and let $G \curvearrowright X$ be a topological action that is free (i. …

Let $G$ be a topological group, and let $H < G$ be a countable subgroup that is amenable as a discrete group. Is the closure of $H$ an amenable topological group?

Let $G$ be an infinite, discrete, countable group. Can $G$ have a translation-invariant ultrafilter? An ultrafilter $\mathcal{F} \subset 2^G$ is translation-invariant if $A \in \mathcal{F}$ implies $g …

Does there exist a finitely generated discrete amenable group $G$ that acts on a separable Hilbert space $\mathcal{H}$ by unitary transformations, and where (1) $\mathcal{H}$ has no finite dimensional …