# Symmetric groups on infinite sets are complete

## Statement

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

## Proof

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

To prove: $\sigma$ is inner.

Proof: Let $G = \operatorname{FSym}(S)$ 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 $\sigma'$, of $K$.
3. Consider the ratio $\sigma'\sigma^{-1}$. The restriction of this automorphism to $G$ is $\tau\tau^{-1}$ which is the identity map. By fact (3), $\sigma'\sigma^{-1}$ is the identity map on $K$, so $\sigma = \sigma'$. Thus, $\sigma$ is inner.