$S_4$ does not have a normal subgroup of order 8?

Question

Prove that $S_4$ does not have a normal subgroup of order 8.

My arrangement is .

Assume that $S_4$ does have a normal subgroup H of order 8. Since $\mid{S_4}\mid=24$ and $\mid{H}\mid=8$. Then, by Lagrange's theorem $\mid{S_4/H}\mid=24/8=3$

would someone help me out finishing this problem.

score 1 · Answer 1 · answered Mar 10 '16 at 15:55

1

Hint:

The sizes of the conjugacy classes of $S_4$ are: $1,6,8,6,3$.

A normal subgroup is a union of conjugacy classes and contains the identity.

answered Mar 10 '16 at 15:55

drhab

153,781

Thank you so much. I know that there is no way to obtain 8 as the sum of terms. But we have not gone over conjugacy classes neither Sylow Thm – ADAM Mar 10 '16 at 15:58
A look here might help. Good luck. – drhab Mar 10 '16 at 16:03

$S_4$ does not have a normal subgroup of order 8?

1 Answers1