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

Question

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 $.

anon · Accepted Answer

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 $phiin{rm Aut}(G)times{rm Aut}(H)$.

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

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