how do I prove that $S_4$ has no normal subgroup of order 6 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?
 A: A normal subgroup is a union of conjugacy classes, one of which must be the identity element.
The non-identity conjugacy classes in $S_4$ have sizes $3$, $6$, $6$, and $8$, and no subset of these numbers sums to $5$.
A: 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.
A: 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. 
A: There are only two groups of order $6$: $\mathbb{Z}/6\mathbb{Z}$ and $S_3$.  Since $S_4$ doesn't have any elements of order $6$, it cannot contain a copy of $\mathbb{Z}/6\mathbb{Z}$.  Thus if $|N| = 6$, it must be isomorphic to a copy of $S_3$; relabel the elements in $S_4$ s.t. $N = \{e,(12),(13),(23),(123),(132)\}$.   However, this isn't normal as $(14)(13)(14) = (34) \notin N$.
