Let \(G,H\) be finite groups of relatively prime orders. Prove that \[ \operatorname{Aut}(G\times H) \cong \operatorname{Aut}(G)\times\operatorname{Aut}(H). \]
Solution
Write \[ G_0\coloneqq G\times\{1\}, \qquad H_0\coloneqq\{1\}\times H. \]
1. Every homomorphism \(G\to H\) and every homomorphism \(H\to G\) istrivial.
Let \[ \psi:G\longrightarrow H \] be a homomorphism. The order of \(\im\psi\) divides \(\abs{G}\) because \(\im\psi\) is a quotient of \(G\), and it divides \(\abs{H}\) because it is a subgroup of \(H\). Since \[ \gcd(\abs{G},\abs{H})=1, \] one has \(\abs{\im\psi}=1\). Thus \(\psi\) is trivial. The same argument with \(G\) and \(H\) interchanged proves the other assertion.
2. Every automorphism\[ \varphi\in\Aut(G\times H) \]preserves both \(G_0\) and \(H_0\).
For \(g\in G\), write \[ \varphi(g,1)=(\alpha(g),\beta(g)). \] The maps \[ \alpha:G\to G, \qquad \beta:G\to H \] obtained by composing \(\varphi|_{G_0}\) with the two projections are homomorphisms. By step 1, \(\beta\) is trivial, so \[ \varphi(g,1)=(\alpha(g),1)\in G_0. \] Thus \(\varphi(G_0)\subseteq G_0\). The same argument gives \[ \varphi(H_0)\subseteq H_0. \]
Apply the same reasoning to \(\varphi^{-1}\). It gives \[ \varphi^{-1}(G_0)\subseteq G_0, \qquad \varphi^{-1}(H_0)\subseteq H_0. \] Applying \(\varphi\) to these inclusions yields the reverse inclusions. Hence \[ \varphi(G_0)=G_0, \qquad \varphi(H_0)=H_0. \]
3. Every \(\varphi\in\Aut(G\times H)\) has a unique expression\[ \varphi(g,h)=(\alpha(g),\delta(h)) \]with\[ \alpha\in\Aut(G), \qquad \delta\in\Aut(H). \]
By step 2, the restrictions of \(\varphi\) to \(G_0\) and \(H_0\) are automorphisms of those factors. Thus there are unique \[ \alpha\in\Aut(G), \qquad \delta\in\Aut(H) \] such that \[ \varphi(g,1)=(\alpha(g),1), \qquad \varphi(1,h)=(1,\delta(h)). \] Since \[ (g,h)=(g,1)(1,h), \] the homomorphism property gives \[ \begin{aligned} \varphi(g,h) &= \varphi(g,1)\varphi(1,h)\\ &= (\alpha(g),\delta(h)). \end{aligned} \] Uniqueness follows by evaluating at \((g,1)\) and \((1,h)\).
4. The map\[ \Phi:\Aut(G\times H)\longrightarrow\Aut(G)\times\Aut(H), \qquad \varphi\longmapsto(\alpha,\delta), \]where \((\alpha,\delta)\) is supplied by step 3, is a group isomorphism.
Composition is componentwise in the formula from step 3, so \(\Phi\) is a homomorphism. It is injective because that formula determines \(\varphi\) uniquely. It is surjective because, for every \[ (\alpha,\delta)\in\Aut(G)\times\Aut(H), \] the map \[ (g,h)\longmapsto(\alpha(g),\delta(h)) \] is an automorphism of \(G\times H\) whose image under \(\Phi\) is \((\alpha,\delta)\).
5. Therefore\[ \boxed{ \Aut(G\times H) \cong \Aut(G)\times\Aut(H) }. \]
This is the isomorphism constructed in step 4.
6. Q.E.D.
Step 5 is the required conclusion.