# Automorphism group of finitary symmetric group equals symmetric group

## Statement

Let $S$ be an infinite set. Let $K = \operatorname{Sym}(S)$ be the Symmetric group (?) on $S$, i.e., the group of all permutations on $S$, and $G = \operatorname{FSym}(S)$ be the subgroup of $K$ that comprises the finitary permutations. In other words, $G$ is the Finitary symmetric group (?) on $S$.

Then, $G$ is normal in $K$. Further consider the map $\varphi: K \to \operatorname{Aut}(G)$ sending an element $a \in K$ to conjugation by $a$ on $G$. This map is an isomorphism.

## Proof

1. By fact (1), the map $\varphi: K \to \operatorname{Aut}(G)$ is well-defined, and by fact (2), $\varphi$ is injective.
2. By fact (3) every element of $\operatorname{Aut}(G)$ sends transpositions to transpositions. Combining this with fact (4), we obtain that every element of $\operatorname{Aut}(G)$ comes from an inner automorphism in $K$. Thus, $\varphi$ is surjective.