# If $\gcd(|G|,\lvert\operatorname{Aut}(H)\rvert) =1$ then $N_G(H)=C_G(H)$

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!

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