Symmetric group $\mathfrak{S}_4$ does not have a normal subgroup of order 2. 
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.

Every subgroup of order two in $S_4$ is generated by either a transposition $\sigma = (ij)$ or a double transposition $\rho = (ij)(kl)$. If $\tau = (il)$ then 
$$\tau\sigma\tau^{-1} = (jl)$$
$$\tau\rho\tau^{-1} = (lj)(ki)$$
are not in this subgroup of order two. It seems a bit unsatisfactory to claim that "there is $h$ such that $hgh^{-1}\neq g$" without proof, because that's the crux of the argument. You must exhibit such $h$. 
