Prove: The symmetric group $\mathfrak{S}_4$ does not have a normal subgroup of order 2.

I found the following way of showing this, which does not make me happy. My questions are: Is this proof correct? And: Is there a nicer way of showing this, by using the more abstract concepts of algebra, for example isomorphism theorems and sylow theorems? For example, how could I determine, whether there is a normal subgroup of order 4?

My attempt: Suppose there is a normal subgroup $S \unlhd \mathfrak{S}_4$. For satisfying the group axioms, $S$ must have the structure $S = \{e, g\}$, where $e \in \mathfrak{S}_4$ is the neutral permutation and $g \in \mathfrak{S}_4$ any selfinverse element, that means $gg = e$. For permutations, this means, that $g$ has cycles of length 1 and 2 only, but at least one cycle of length 2, since it is not the neutral permutation. Now I can inspect the two possible cases:

Case 1: g consist of two cycles of length 2

Case 2: g constst of one cycle of length 2 and two cycles of length 1

For both cases, it is easy to find a $h \in \mathfrak{S}_4$ so that $hgh^{-1} \neq g$ (I don't want to step into messy details here). But it is also $hgh^{-1} \neq e$, since $hgh^{-1} = e$ would lead to $g=e$. So we have $hgh^{-1} \notin S$, which shows that $S$ is not a normal subgroup.

Thank you for your help.