TITLE: $S_4$ does not have a normal subgroup of order $8$
QUESTION [16 upvotes]: Question is to prove that :
$S_4$ does not have a normal subgroup of order $8$ 
I do not have any specific idea how to proceed but: 
Assuming there exists a normal subgroup $H$ of order $8$ in $S_4$, 
As $H\unlhd G$, $HK\leq G$ for any subgroup $K\leq G$ and $|HK|=\frac{|H||K|}{|H\cap K|}$
what i am trying to do is try getting an element $ x $ of order $2$ which is not in $H$ and set $K=\{1,x\}$ then $HK$ would be a group of order $16$ which is a contradiction as $S_4$
 can not have a subgroup of order $16$.
as there are six $2-cycles$ and three  products of disjoint $2$ cycles but $|H|=8$, there does exists an element of order $2$ which is not in $H$ and thus we are done.
I am sure this would be the nicest way or the stupidest way one can ever do :P
I would be thankful if someone can help me to see if anything is wrong in my approach.
I would be thankful if someone can give me a hint for an alternate approach.
Thank You :)

REPLY [2 votes]: I know this question is old, but there is an approach that uses one of the other problems in the same section as this one in "Abstract Algebra" by Dummit and Foote. However, it appears a little later . I thought I would just add it.

If $N$ is a normal subgroup of a finite group $G$ and $GCD(|N|, |G : N|) = 1$, then $N$ is the unique subgroup of $G$ of order $|N|$.

Using this result, assume $S_4$ has a normal subgroup, $N$, of order $8$. $GCD(8, 24/8) = 1$, thus it must be the only subgroup of order $8$. $S_4$ also has a subgroup, $K$,  isomorphic to $D_8$ (dihedral group of order 8/symmetries of a square), by associating $1,2,3,4$ with vertices of square. Thus $K=N$ by the result above.
However we can show $K$ is in fact not normal. Let's say our labeling of the vertices is such that $(1,2,3,4) \in K, (1,2, 3) \notin K$ , then $(2,1,3)(1,2,3,4)(1,2,3) = (1,2,4,3) \notin K$. Thus a contradiction.
Or simply note that $S_4$ has more than one subgroup of order $8$ leading to a contradiction as well.