Normal factors that generate $G$ and meet the product of the others trivially form an internal direct product.Suggest a slogan
Let \(G\) be a group with subgroups \(H_1,\ldots,H_n\) such that
\(H_i \normal G\) for all \(i\),
\(G = H_1 \cdots H_n\), and
for each \(k\), \(H_k \intersect \left(H_1 \cdots H_{k-1}H_{k+1} \cdots H_n\right) = \theset{e}\).
Then the map \(\prod_{i=1}^n H_i\to G\), \((h_1,\ldots,h_n)\mapsto h_1\cdots h_n\), is an isomorphism, so \(G \cong \prod_{i=1}^n H_i\).