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?



Amenability of $S^{\infty}$
The answer to [this question](https://mathoverflow.net/questions/293614/is-each-locally-compact-group-topology-on-the-permutation-group-discrete/293619#293619) 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](https://mathoverflow.net/questions/293614/is-each-locally-compact-group-topology-on-the-permutation-group-discrete/293619#293619), 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.


