TITLE: how do I prove that $S_4$ has no normal subgroup of order 6
QUESTION [6 upvotes]: Let $N$ be a normal subgroup of $S_4$.
I have proven that $|N|\ne 2,3,8$.
Yet, I don't know how to prove that $|N|\neq 6$.
Should I compute all subgroups and check this?

REPLY [3 votes]: Suppose $N$ is normal subgroup of order $6$, then $S_4/N \cong C_4$ or $S_4/N \cong C_2\times C_2$ both are abelian. It fallows that $[S_4,S_4]=A_4 \subseteq N$
which Leads to contradiction.

REPLY [3 votes]: Suppose $S_4$ has a normal subgroup of order $6$ and call it $N$. Then since $|N|=6~(even)$, Hence it contains an element of order $2$ and elements of order $2$ in $S_4$ are either a cycle of length $2$ or product of two cycles each of length $2$.
Case-1 If it contains element of cycle $2$ then it will contain $\{(1~2),(1~3),(1~4),(2~3),(2~4),(3~4)\}$ because all are conjugate and $N$ is normal. So order of $N$ must be greater than $7$. which  is a contradiction.
Case-2 If it contains an element which is product of two cycles each of length $2$
then it will contain $\{(1~2)(3~4),(1~3)(2~4),(1~4)(2~3)\}$ because all are conjugate and $N$ is normal. So $V_4$ is normal subgroup of $N$. So from $Lagrange ~theorem$ $4\mid6$ which is a contradiction.