Let \(H,K\trianglelefteq G\) satisfy \[ H\cap K=\{e\},\qquad HK=G. \] Prove that \(G\cong H\times K\). Generalize the criterion to finitely many normal subgroups.
Solution
1. The subgroups \(H\) and \(K\) commute elementwise.
For \(h\in H\) and \(k\in K\), the commutator \[ [h,k]=hkh^{-1}k^{-1} \] lies in \(H\) because \(H\) is normal, and lies in \(K\) because \(K\) is normal. Hence \[ [h,k]\in H\cap K=\{e\}. \] Thus \(hk=kh\).
2. Multiplication gives an isomorphism.
Define \[ \mu:H\times K\to G,\qquad (h,k)\mapsto hk. \] By 1 it is a homomorphism. It is surjective because \(HK=G\). If \(hk=e\), then \(h=k^{-1}\in H\cap K\), so \(h=k=e\). Thus \(\mu\) is injective.
3. Finite-family version.
Let \(H_1,\dots,H_r\trianglelefteq G\). If \[ G=H_1\cdots H_r \] and for every \(i\), \[ H_i\cap \prod_{j\ne i}H_j=\{e\}, \] then the subgroups commute pairwise and the multiplication map \[ H_1\times\cdots\times H_r\longrightarrow G \] is an isomorphism.
Equivalently, the internal direct-product condition is precisely that the multiplication map be surjective with trivial kernel. Pairwise trivial intersections alone are not sufficient for three or more factors.