TITLE: If $(|G|, |H|) > 1$, does it follow that $\operatorname{Aut}(G \times H) \neq \operatorname{Aut}(G) \times \operatorname{Aut}(H)$?
QUESTION [16 upvotes]: Let $G$ and $H$ be finite groups. If $|G|$ and $|H|$ are coprime, then 
$$\operatorname{Aut}(G \times H) \cong \operatorname{Aut}(G) \times \operatorname{Aut}(H)$$
holds. What about when $(|G|, |H|) > 1$? In this case we know that $\operatorname{Aut}(G) \times \operatorname{Aut}(H)$ is contained in $\operatorname{Aut}(G \times H)$, but the isomorphism above might not hold. For example $\operatorname{Aut}(C_2 \times C_2)$ has order $6$ but $\operatorname{Aut}(C_2) \times \operatorname{Aut}(C_2)$ is  trivial. 
Is the isomorphism possible at all when $(|G|, |H|) > 1$?

REPLY [3 votes]: No, it does not follow.
Let $G,H$ be two finite non-cyclic simple groups with $|G|,|H|$ both even (this in always holds) but not multiple of one another (for instance one can take $|G|=60$ and $|H|=168$). Clearly $2\mid\gcd(|G|,|H|)$, but the subgroups $G,H$ of the product group $G\times H$ are characteristic subgroups (stable under all automorphims), so that every automorphism of $G\times H$ comes from the automorphisms it induces in the factors $G,H$ separately. To see that the factor subgroups are characteristic, it suffices to consider the image of one factor, say $G$, by an automorphism, and then projected to the other factor, $H$. The result is  a subgroup of$~H$ isomorphic to a quotient of$~G$. But $H$ has no subgroup isomorphic to all of $G$ (consider the orders), and the only other quotient of $G$ is the trivial group (by simplicity of$~G$), so that is what the result is. This means that $G$ was stable under the automorphism. The same goes with $G$ and $H$ interchanged.