Looking for a proof that if a normal subgroup of $Sym(\mathbb N)$ has an element of infinite order then it is the whole group

Question

I know that $Sym (\mathbb N)$ , the group of all permutations on $\mathbb N$ , has exactly two non-trivial normal subgroups and both of these subgroups are torsion , so it follows that if a normal subgroup of $Sym(\mathbb N)$ has an element of infinite order then it is the whole group . But I am looking for a simpler way to prove that "if a normal subgroup of $Sym(\mathbb N)$ has an element of infinite order then it is the whole group " without having to go through determining all the normal subgroups of $Sym(\mathbb N)$ . Is it possible ? I can get that if $N$ is a normal subgroup of $Sym(\mathbb N)$ containing an element of infinite order then $N$ contains an element of order $2$ with exactly countably infinitely many two-cycles and countably infinitely many fixed points . Is this fact any important in proving a simple way my required statement ? Please help . Thanks in advance

Answer 1

My comment wasn't quite right, so I will try again.

By considering the standard permutation representation of a (finite or infinite) dihedral group in which a point stabilizer is a reflection, we see that any finite or infinite cycle can be written as the product of two involutions (i.e. elements of order $2$). So, by decomposing an arbitrary permutation $h$ of ${\mathbb N}$ as a product of cycles, it follows that $h$ is the product of two involutions.

Now, if $g$ is an involution in ${\rm Sym}({\mathbb N})$ with infinitely many transpositions and fixed points, then any involution in $G$ is either conjugate to $g$ or, if it has only finitely many transpostions or fixed points, a product of two conjugates of $g$. It follows than any $h \in {\mathbb N}$ is the product of at most four conjugates of $g$.