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Let $G$ be the group of all permutations of $\mathbb{N}$. If I am not mistaken, this group is denoted by $S^{\infty}$.

Is there a precise locally compact topology on $G$ such that $G$ would be an amenable group? Or is $G$ isomorphic to a dense subgroup of an amenable group?

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    $\begingroup$ I am afraid that the unique locally compact group topology on $S^\infty$ is the discrete topology. $\endgroup$
    – Taras Banakh
    Commented Feb 22, 2018 at 14:03
  • $\begingroup$ @TarasBanakh And the discrete one is not amenable. But is it isomorphic to a dense subgroup of an amenable group? $\endgroup$
    – Ali Taghavi
    Commented Feb 22, 2018 at 14:46
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    $\begingroup$ I think that your two questions are equivalent: $S_\omega$ admits a non-discrete locally compact topology if and only if $S_\omega$ is isomorphic to a dense subgroup of non-discrete locally compact topological group. This can be shown using the fact that the topology of point-wise convergence is the smallest group topology on $S_\omega$. So, now I am thinking on the existence of a non-discrete locally compact group topology on $S_\omega$ and cannot find a quick answer (neither my collegues - Ravsky, Gutik -- that work in this field). This is an interestng question. $\endgroup$
    – Taras Banakh
    Commented Feb 22, 2018 at 14:57
  • $\begingroup$ @AliTaghavi In the second question, you probably mean a subgroup of a _locally _compact amenable group. With its usual Polish topology, $G$ is amenable. $\endgroup$
    – user95282
    Commented Feb 22, 2018 at 18:59

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The answer to this question implies that each locally compact group topology $\tau$ on the permutation group $S^\infty$ is discrete. Since the discrete group $S^\infty$ is known to be non-amenable (it contains a free group with two generators), the locally compact topological group $(S^\infty,\tau)$ is not amenable.

By Corollary in the answer to this question, the permutation group $S^\infty$ is not isomorphic to a dense subgroup of a non-discrete locally compact group. This implies that $S^\infty$ is not isomorphic to a dense subgroup of an amenable locally compact group.

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  • $\begingroup$ Thank you very much for your attention to my question and your great answer. $\endgroup$
    – Ali Taghavi
    Commented Feb 23, 2018 at 7:34

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