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

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.

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