Finitary symmetric group is normal in symmetric group

This article gives the statement, and possibly proof, of a particular subgroup or type of subgroup (namely, Finitary symmetric group (?)) satisfying a particular subgroup property (namely, Normal subgroup (?)) in a particular group or type of group (namely, Symmetric group (?)).

Statement

Let $S$ be a set. Let $K = \operatorname{Sym}(S)$ be the symmetric group on $S$, and $G$ be the finitary symmetric group on $S$. In other words, $G$ is the subgroup of $K$ comprising the finitary permutations, i.e., the permutations that move only finitely many elements. Then, $G$ is a normal subgroup of $K$.

Related facts

• Finitary symmetric group is characteristic in symmetric group
• Finitary symmetric group is automorphism-faithful in symmetric group

Proof

Given: A set $S$. $G = \operatorname{FSym}(S)$ is a subgroup of $K = \operatorname{Sym}(S)$. $\sigma \in K$ and $\alpha \in G$.

To prove: $\sigma\alpha\sigma^{-1} \in G$.

Proof: If $\sigma \in K$ and $\alpha \in G$, we have:

$(\sigma\alpha\sigma^{-1})(\sigma(x)) = \sigma(\alpha(x))$.

Thus, $\sigma(x)$ is moved by $\sigma \alpha \sigma^{-1}$ if and $x$ is moved by $\alpha$. Since $\sigma$ is a permutation, this shows that the number of points moved by $\alpha$ and $\sigma\alpha\sigma^{-1}$ is equal. In particular, $\sigma\alpha\sigma^{-1}$ also moves only finitely many points, and hence is in $G$.

Thus, $G$ is normal in $K$.