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Suppose that $H$ is a subgroup of $G$ such that $\gcd(|G|,\lvert\operatorname{Aut}(H)\rvert)=1$.

Prove that $N_G(H)=C_G(H)$.

I want to solve the above problem. I think because there is a normalizer, we should think about inner homomorphism and use the condition to make trivial homomorphism.

But in this case, there is no assumption about $H$ being normal, so I can't use the inner homomorphism.

Help me solve this problem!

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This is easy. The $N/C$ Theorem tells you that $N_G(H)/C_G(H)$ embeds isomorphically as a subgroup of $\operatorname{Aut}(H)$, thus $\left\lvert N_G(H)/C_G(H)\right\rvert$ divides $\left\lvert \operatorname{Aut}(H)\right \rvert$. But $\left\lvert N_G(H)/C_G(H)\right\rvert$ also divides $|G|$ (since it divides $|N_G(H)|$ which in turn divides $|G|$), so it divides $\gcd(|G|,\left\lvert \operatorname{Aut}(H)\right \rvert)=1$. Therefore $|N_G(H)| = |C_G(H)|$ and since $C_G(H) \leq N_G(H)$ we get $N_G(H)=C_G(H)$.

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    $\begingroup$ Wow~ thanks a lot. From you, I can learn the N/C theorem. $\endgroup$ – Pearl Nov 26 '18 at 13:07

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