Aut(GxH) isomorphic to Aut(G)xAut(H)

Can anyone help with the following proof?

Let H,G be final groups.

$(|G|,|H|) = 1$, prove that $Aut(G×H) ≅ Aut(G) ×Aut(H)$

I found this question Show that if $ \gcd(|G|,|H|) = 1 $, then $ \text{Aut}(G \times H) \cong \text{Aut}(G) \times \text{Aut}(H) $. but didnt really understand the solution, would appreciate if anyone could help with that.

thank you

Solution 1:

HINT: There's two things you need to show.

• $(i)$ $\vert Aut(G)\times Aut(H)\vert\le \vert Aut(G\times H)\vert$, and

• $(ii)$ $\vert Aut(G)\times Aut(H)\vert\ge \vert Aut(G\times H)\vert$.

Let's start with $(i)$. Suppose I have an automorphism $\alpha$ of $G$, and an automorphism $\beta$ of $H$. Do you see how to build a "product automorphism" $\alpha\times \beta$ of $G\times H$? Do you see how to show that this is injective - that is, $\alpha_0\times\beta_0=\alpha_1\times\beta_1\implies \alpha_0=\alpha_1, \beta_0=\beta_1$?

As for $(ii)$, this is harder. We basically want to show that every automorphism of $G\times H$ is of the form $\alpha\times \beta$ for some $\alpha\in Aut(G)$ and $\beta\in Aut(H)$. This is where the condition on the gcd of the orders of $G$ and $H$ comes in. HINT: suppose the orders of $G$ and $H$ are relatively prime. Can $G$ and $H$ have nonidentity elements $a$ and $b$ with the same order?