How to show $F(S)$ is normal in Sym$(S)$

Question

As a follow-up of this question On non-trivial normal subgroup(s) of $A(S)$ , where $S$ is infinite , how do we show that the finitary symmetric group $F(S)$ is a subgroup and normal in Sym $(S)$ , where $S$ is an infinite set ?

Answer 1

Let me attempt the hardest part first, which is to show that $F(S)$ is a subgroup. Let $f, g: S\to S$ be permutations with finite support then observe that

$$ \mathrm{Fix}(g^{-1}\circ f) = \{ x \mid g(x) = f(x)\} \supseteq \{ x \mid g(x)= f(x) = x \} = \mathrm {Fix}(g) \cap \mathrm{Fix}(f), $$ taking complements: $$ \mathrm{supp}(g^{-1}\circ f) \subseteq \mathrm{supp}(f) \cup \mathrm{supp}(g),$$

the right hand side is a finite set, and so is the left hand side.

Normality is actually easier: assume $f\in F(S)$ and $g\in \mathrm{Sym}(S)$ then we must show that $g\circ f \circ g^{-1}\in F(S)$. We have $$ \mathrm{Fix}(g\circ f \circ g^{-1}) = \{ x \mid f(g^{-1}(x)) = g^{-1}(x) \} = \{ g(x) \mid f(x) = x \} = g(\mathrm{Fix}(f)), $$ and taking complements $$ \mathrm{supp}(g\circ f \circ g^{-1}) = g(\mathrm{supp}(f)),$$ so since $g$ is a bijection and $\mathrm{supp}(f)$ is finite, $ \mathrm{supp}(g\circ f \circ g^{-1})$ is finite, i.e. $g\circ f \circ g^{-1}\in F(S)$.