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
1. The subgroup \(G\times\{1\}\) is characteristic in \(G\times H\).
An element \((g,h)\in G\times H\) belongs to \(G\times\{1\}\) if and only if its order divides \(\abs G\).
Indeed, if \(h=1\), then \[ \operatorname{ord}(g,1)=\operatorname{ord}(g) \] divides \(\abs G\) by Lagrange’s theorem.
Conversely, suppose \[ \operatorname{ord}(g,h)\mid\abs G. \] Then \(\operatorname{ord}(h)\) divides \(\operatorname{ord}(g,h)\), so it divides \(\abs G\). It also divides \(\abs H\). Since \[ \gcd(\abs G,\abs H)=1, \] one has \(\operatorname{ord}(h)=1\), hence \(h=1\).
Thus \(G\times\{1\}\) is characterized purely by element orders. Every automorphism preserves element orders, so every automorphism of \(G\times H\) preserves \(G\times\{1\}\).
2. The subgroup \(\{1\}\times H\) is characteristic in \(G\times H\).
Interchanging the roles of \(G\) and \(H\) in step 1 shows that \[ \{1\}\times H \] is exactly the set of elements whose orders divide \(\abs H\). Hence it too is preserved by every automorphism.
3. Every automorphism \(\varphi\) of \(G\times H\) has the form\[ \varphi(g,h)=\bigl(\alpha(g),\beta(h)\bigr) \]for uniquely determined\[ \alpha\in\operatorname{Aut}(G), \qquad \beta\in\operatorname{Aut}(H). \]
By steps 1 and 2, \(\varphi\) restricts to automorphisms of the two factors. Thus there are uniquely determined automorphisms \(\alpha\) and \(\beta\) such that \[ \varphi(g,1)=(\alpha(g),1) \] and \[ \varphi(1,h)=(1,\beta(h)). \] Since \[ (g,h)=(g,1)(1,h), \] the homomorphism property gives \[ \begin{aligned} \varphi(g,h) &= \varphi(g,1)\varphi(1,h)\\ &= (\alpha(g),1)(1,\beta(h))\\ &= (\alpha(g),\beta(h)). \end{aligned} \]
4. The map\[ \Psi:\operatorname{Aut}(G\times H) \longrightarrow \operatorname{Aut}(G)\times\operatorname{Aut}(H) \]that sends \(\varphi\) to its two restrictions is a group isomorphism.
Step 3 shows that \(\Psi\) is injective.
Given \[ (\alpha,\beta)\in \operatorname{Aut}(G)\times\operatorname{Aut}(H), \] the map \[ (g,h)\longmapsto(\alpha(g),\beta(h)) \] is an automorphism of \(G\times H\), with inverse \[ (g,h)\longmapsto(\alpha^{-1}(g),\beta^{-1}(h)). \] Hence \(\Psi\) is surjective.
Finally, restriction commutes with composition, so \(\Psi\) is a homomorphism. Therefore it is an isomorphism.
5. Q.E.D.
Step 4 gives the required isomorphism.