Finitary symmetric group is normal in 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 $$K$$.