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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.

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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.

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  • $\begingroup$ 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 $\endgroup$ – ADAM Mar 10 '16 at 15:58
  • $\begingroup$ A look here might help. Good luck. $\endgroup$ – drhab Mar 10 '16 at 16:03

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