$S_4$ does not have a normal subgroup of order $8$ 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 :)
 A: If you have the Sylow theorems available, then you can argue as follows. The dihedral group $D_4$ (symmetries of a square) has order $8$ and can be viewed as a subgroup of $S_4$ by regarding a symmetry of the square as a permutation of the four vertices. The permutation $p$ that interchanges two adjacent vertices of the square while fixing the other two vertices is in $S_4$ but not in $D_4$. Since it has order $2$, this $p$ must be in a $2$-Sylow subgroup.  So you have at least two distinct $2$-Sylow subgroups of $S_4$, namely $D_4$ and one containing $p$.  But all the $2$-Sylow subgroups are conjugate to each other, so, as soon as there's more than one of them, none of them can be normal.
A: Another approach : Two elements in $S_n$ are conjugate to each other iff they have the same cycle decomposition. Hence, all the transpositions must be in $H$ if one of them is (because $H$ is normal). 
This leaves one non-identity element, which must have order 2 (since all the transpositions are their own inverses). That element must be a product of transpositions. Such a thing would be conjugate to something else, which is not in $H$.
If you assume that $H$ does not contain any transpositions, it still contains an element of order 2 ...
A: 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.
A: You approach is OK, but here is a simpler one. Since $S_4/H \cong C_3$ is abelian, it follows that $[S_4,S_4]=A_4 \subseteq H$. Can you see this leads to a contradiction?
A: Yet another approach: Normal subgroups are comprised of entire conjugacy classes, one of which must be the identity's conjugacy class, but the class equation for $S_4$ is $$24=1+3+6+6+8,$$ and there is no way to obtain $8$ as a sum of terms from the right hand side including $1$.
A: Suppose $N$ is a normal subgroup of $S_4$ of order $8$. The $2$-cycles are: $(12), (13), (14), (23), (24)$ and $(34)$. So we have $6$ elements of order $2$.
Claim: There exists an element of order $2$ that is not in $N$.
Proof: If $N$ contains all the $2$-cycles of $S$, then $1, (12), (13), (14), (23), (24)$ and $(34)$ are all in $N$. Since $|N|=8$, then $N$ contains an element of order $3$ or $4$. So $N$ must also contain the inverse of that element (and the elements of order $3$ or $4$ are not self-inverse) so that $N$ will contain at least $9$ elements. This contradicts the order of $N$. Therefore there exists a $2$-cycle $(ab)\notin N$. 
Now take $H=\{1,(ab)\}$. $H$ is a subgroup of $G$. Since $N$ is normal in $G$, $HN$ is a subgroup of $G$. Clearly, $H\cap N=1$ since  $(ab)\notin N$. It follows that
$|HN|=\frac{|H||N|}{|H\cap N|}=|H||N|=16$. But $16$ is not a divisor of $24$, contradiction Lagrange's Theorem. Therefore $S_4$ does not have a normal subgroup of order $8$.
