Symmetric groups on infinite sets are complete

Statement

Let $S$ be an infinite set. The symmetric group on $S$, denoted $\mathrm{S}\mathrm{y}\mathrm{m}(S)$, is a complete group: it is centerless and every automorphism of it is inner.

Facts used

1. Finitary symmetric group is characteristic in symmetric group
2. Automorphism group of finitary symmetric group equals symmetric group
3. Finitary symmetric group is automorphism-faithful in symmetric group

Proof

Given: $S$ is an infinite set, $K=\mathrm{S}\mathrm{y}\mathrm{m}(S)$, $\sigma$ is an automorphism of $K$.

To prove: $\sigma$ is inner.

Proof: Let $G=\mathrm{F}\mathrm{S}\mathrm{y}\mathrm{m}(S)$ be the subgroup of $K$ comprising the finitary permutations.

1. By fact (1), $\sigma$ restricts to an automorphism, say $\tau$ of $G$.
2. By fact (2), the automorphism $\tau$ of $G$ arises from some inner automorphism, say ${\sigma }^{\prime }$, of $K$.
3. Consider the ratio ${\sigma }^{\prime }{\sigma }^{-1}$. The restriction of this automorphism to $G$ is $\tau {\tau }^{-1}$ which is the identity map. By fact (3), ${\sigma }^{\prime }{\sigma }^{-1}$ is the identity map on $K$, so $\sigma ={\sigma }^{\prime }$. Thus, $\sigma$ is inner.