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
 A: 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?
A: Consider the map $\psi\colon \operatorname{Aut}(G) \times \operatorname{Aut}(H) \to \operatorname{Aut}(G\times H)$ given by $(\alpha,\beta)\mapsto \alpha\times\beta$ where $\alpha\times\beta\colon G\times H \to G\times H$ is given by $(g,h)\mapsto(\alpha(g),\beta(h))$.
I skip the step of showing that $\psi$ is well-defined, which means that $\alpha\times\beta$ is indeed an automorphism of $G\times H$. This can be easily checked.

*

*We show that $\psi$ is a group homomorphism. First we have
$$
((\alpha_1\circ\alpha_2)\times(\beta_1\circ\beta_2))(g,h)=(\alpha_1(\alpha_2(g)),\beta_1(\beta_2(h))) = ((\alpha_1\times\beta_1)\circ(\alpha_2\times\beta_2))(g,h),
$$
which means $(\alpha_1\circ\alpha_2)\times(\beta_1\circ\beta_2) = (\alpha_1\times\beta_1)\circ(\alpha_2\times\beta_2)$. Thus we get
\begin{align*}
\psi((\alpha_1,\beta_1)(\alpha_2,\beta_2)) &= \psi(\alpha_1\circ\alpha_2,\beta_1\circ\beta_2)\\&=(\alpha_1\circ\alpha_2)\times(\beta_1\circ\beta_2) \\&= (\alpha_1\times\beta_1)\circ(\alpha_2\times\beta_2) \\&= \psi(\alpha_1,\beta_1)\circ\psi(\alpha_2,\beta_2).
\end{align*}


*Too see that $\psi$ is injective, we determine $\ker(\psi)$:
\begin{align*}
(\alpha,\beta)\in\ker(\psi) &\iff \alpha\times\beta=\operatorname{id}_{G\times H} \\&\iff \forall g\in G, h\in H\colon (\alpha(g),\beta(h))=(g,h) \\&\iff \forall g\in G, h\in H\colon \alpha(g)=g \ \land \ \beta(h)=h \\&\iff \alpha=\operatorname{id}_G \ \land \ \beta=\operatorname{id}_H.
\end{align*}
Thus the kernel is trivial, which means that $\psi$ is injective.


*Too see that $\psi$ is surjective, consider $\gamma\in\operatorname{Aut}(G\times H)$ and $g\in G$. Assume $\gamma(g,e_H)=(x,y)$ with $y\ne e_H$. Thus we have $n:=\operatorname{ord}(y)>1$, where $n\mid|H|$. If $g=e_G$, then $y(g,e_H)=(e_G,e_H)$, so $g\ne e_G$. Thus we have $m:=\operatorname{ord}(g)>1$ as well, where $m\mid|G|$. From
$$
(e_G,e_H)=\gamma(e_G,e_H)=\gamma(g^m,e_H)=\gamma(g,e_H)^m = (x,y)^m = (x^m,y^m)
$$
it follows $n\mid m$ and hence $n\mid|G|$. So $n>1$ is common divisor of $|G|$ and $|H|$, which is a contradiction to our condition $\gcd(|G|,|H|)=1$. This means $\gamma(g,e_H)$ has the form $(x,e_H)$, and analogously, $\gamma(e_G,h)$ has the form $(e_G,y)$. Let $\pi_G$ denote the projection homomorphism $G\times H \to G$ given by $(g,h)\mapsto g$. Thus the map $\alpha\colon G\to G$ given by $g\mapsto\pi_G(\gamma(g,e_H))$ is an automorphism (with inverse $g\mapsto\pi_G(\gamma^{-1}(g,e_H))$). Analogously, the map $\beta\colon H\to H$ given by $h\mapsto\pi_H(\gamma(e_G,h))$ is an automorphism as well. In other words: $\gamma(g,e_H)=(\alpha(g),e_H)$ and $\gamma(e_G,h)=(e_G,\beta(h))$. It remains to check $\gamma = \alpha\times\beta$, which can be easily done. Thus we have $\gamma=\alpha\times\beta=\psi(\alpha,\beta)$, which means that $\psi$ is surjective.
Summary: $\psi$ is a group isomorphism.
