# Show that if $gcd(|G|,|H|) = 1$, then $text{Aut}(G times H) cong text{Aut}(G) times text{Aut}(H)$.

Mathematics Asked by La Rias on December 24, 2020

Let $G$ and $H$ be finite groups. If $gcd(|G|,|H|) = 1$, then I want to show that
$$text{Aut}(G times H) cong text{Aut}(G) times text{Aut}(H).$$
In my attempt, I first defined the map $f: text{Aut}(G) times text{Aut}(H) to text{Aut}(G times H)$ by
$$forall (phi_{G},phi_{H}) in text{Aut}(G) times text{Aut}(H): quad f(phi_{G},phi_{H}) stackrel{text{def}}{=} phi_{G} times phi_{H},$$
but I don’t know how to use the fact that $gcd(|G|,|H|) = 1$.

Automatically ${rm Aut}(G)times{rm Aut}(H)subseteq{rm Aut}(Gtimes H)$. The gcd condition forces this to be an equality.

Let $phiin{rm Aut}(Gtimes H)$. Show $phi(Gtimes1)= Gtimes1$; argue $subseteq$ by contradiction (show the order of an element outside of $Gtimes1$ is divisible by a factor of $|H|$...). Symmetrically, $phi(1times H)=1times H$.

After that it is straightforward to check $phi=(phi|_G,phi|_H)$, hence $phiin{rm Aut}(G)times{rm Aut}(H)$.

Correct answer by anon on December 24, 2020

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