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Let $G$ and $H$ be two finite groups. We know if $\alpha\in \mathrm{Aut}(G)$ and $\beta\in \mathrm{Aut}(H)$, then $\alpha\times\beta\in \mathrm{Aut}(G\times H)$. Hence, $| \mathrm{Aut}(G\times H)|\geq| \mathrm{Aut}(G)\times \mathrm{Aut}(H)|$. Also we know if $(|G|,| H|)=1$, then $| \mathrm{Aut}(G\times H)|=| \mathrm{Aut}(G)\times \mathrm{Aut}(H)|$.

Now, can we say that if $(| G|,| H|)\neq1$, then $| \mathrm{Aut}(G\times H)|>| \mathrm{Aut}(G)\times \mathrm{Aut}(H)|$?


merged by Alexander Gruber Jun 15 '13 at 20:02

This question was merged with If $(|G|, |H|) > 1$, does it follow that $\operatorname{Aut}(G \times H) \neq \operatorname{Aut}(G) \times \operatorname{Aut}(H)$? because it is an exact duplicate of that question.