# If $H \unlhd G$ is CC-closed, then $H$ is a Hall subgroup

I'm stuck with this problem.

Let $H$ be a normal subgroup of a finite group $G$ such that $C_G(x)\subseteq H$ for every non-identity element $x\in H$ (that is, $H$ is a normal CC-subgroup of $G$). Then $\textrm{gcd}(|H|,|G/H|)=1$ (i.e. $H$ is a Hall-subgroup of $G$).

I think that Sylow's Theorems would be helpful here. I spent hours thinking and I couldn't solve it.

Thanks in advance.

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Just to help people search: "Why is the Frobenius kernel a Hall subgroup?" – Jack Schmidt Jan 31 '12 at 23:50
And here is a generalization of the statement, with more or less the same proof: math.stackexchange.com/questions/76413 – Jack Schmidt Jan 31 '12 at 23:57

## 1 Answer

Hint: Consider a p-element of H and an element of the center of a Sylow p-subgroup containing it. Centers and centralizers unite!

If p divides H, then let Q be a Sylow p-subgroup of H, and let P be a Sylow p-subgroup of G containing Q. Let z be an element of the center of P, and let x be a non-identity element of Q. Since x and z commute, $z \in C_G(x) \leq H$, and so z in H. However, z is in the center of P, and so $P \leq C_G(z) \leq H$ and P is actually a Sylow p-subgroup of H as well. Hence $[G:H]$ is not divisible by p.

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Thanks a lot Jack! – fiorerb Feb 1 '12 at 0:00