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.

Facts used

 * 1) uses::Finitary symmetric group is characteristic in symmetric group
 * 2) uses::Automorphism group of finitary symmetric group equals symmetric group
 * 3) uses::Finitary symmetric group is automorphism-faithful in symmetric group

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.