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 fact about::symmetric group on $$S$$, $$K = \operatorname{FSym}(S)$$ be the fact about::finitary symmetric group, and $$H$$ be the fact about::finitary alternating group. Then, $$K/H$$ is the fact about::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.