Normal factors with trivial intersection and product $G$ form an internal direct product.Suggest a slogan
Let \(G\) be a group and let \(H,K\le G\) be normal subgroups with \(H\cap K=\theset{e}\) and \(HK=G\). Then the map \[ H\times K\to G,\qquad(h,k)\mapsto hk, \] is an isomorphism of groups.
For \(h\in H\) and \(k\in K\), the commutator \(hkh^{-1}k^{-1}\) lies in \(K\) because \(K\) is normal and in \(H\) because \(H\) is normal, so it is \(e\) and \(hk=kh\). Hence \((h,k)(h',k')=(hh',kk')\mapsto hh'kk'=hkh'k'\), and the map is a homomorphism. Its kernel consists of the pairs with \(hk=e\), so \(h=k^{-1}\in H\cap K=\theset{e}\), and the map is injective. It is surjective because \(HK=G\).
What conditions on subgroups $H, K \leq G$ give $G \cong H \times K$?