# Finitary symmetric group equals center of symmetric group modulo finitary alternating group

## Statement

Let $S$ be an infinite set, $G = \operatorname{Sym}(S)$ be the Symmetric group (?) on $S$, $K = \operatorname{FSym}(S)$ be the Finitary symmetric group (?), and $H$ be the Finitary alternating group (?). Then, $K/H$ is the Center (?) of $G/H$.

## Proof

We need to pove that if $\sigma \in G$ is such that $[G,\sigma] \le H$, then $\sigma \in K$. We do this by picking any $\sigma \notin K$, and show that $[G,\sigma]$ is not contained in $H$.

If $\sigma \notin K$, $\sigma$ must move infinitely many elements. There are two cases, that are collectively exhaustive:

• $\sigma$ contains infinitely many finite cycles.
• $\sigma$ contains an infinite cycle.