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Let $H$ be a subgroup of a finite group $G$. Is it true that $\lvert \operatorname{Aut}(H)\rvert$ divides $\lvert \operatorname{Aut}(G)\rvert$? What if we also assume $G$ is abelian? (I know that $\lvert \operatorname{Aut}(H)\rvert \space \big| \space \lvert \operatorname{Aut}(G)\rvert$ if $G$ is cyclic).

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It's not even true for abelian groups in general. Take $H=C_2\times C_2$ as a subgroup of $G=C_4\times C_2$. Then $\lvert \operatorname{Aut}(G)\rvert=8$, while $\lvert \operatorname{Aut}(H)\rvert=6$.

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