TITLE: Amenability of $S^{\infty}$
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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.