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Let $A, B$ be finite nonabelian groups such that $(|A|, |B|) = 1$. Is $\operatorname{Aut}(A\times B) = \operatorname{Aut}(A)\times\operatorname{Aut}(B)$?

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Yes.

Let $\phi:A\times B\rightarrow A\times B$ be an automorphism. It splits into surjective homomorphisms $\phi_A:A\times B\rightarrow A$ and $\phi_B:A\times B\rightarrow B$. Now $B\subseteq \ker \phi_A$ since $\frac{|A\times B|}{|\ker\phi_A|}=|A|$ (this is where we use coprimality of the orders of $A$ and $B$. For order reasons $\phi_A|_{A\times\{1\}}$ must then be an automorphism of $A$. The same argument applies to $\phi_B$, such that after forgetting the kernels we may write $\phi=\phi_A\times \phi_B$. It is now clear that composition happens componentwise, such that the inclusion $\text{Aut}(A)\times \text{Aut}(B)\rightarrow \text{Aut}(A\times B)$ is an isomorphism.

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