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 😛
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 🙂
Best Answer
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?