Abstract Algebra: Theory and Applications · Problem 9.4.51

The key insight is that an internal direct product is defined precisely so that its elements can be uniquely written as products of components from each subgroup—this mirrors exactly how the external direct product works. To prove the isomorphism, construct a map sending each element $g \in G$ to its unique factorization $(h_1, h_2, \ldots, h_n)$ where $h_i \in H_i$, then verify this map is a bijective homomorphism using the defining properties of internal direct products (normality, pairwise commutativity, and trivial intersections).

The key insight is that an internal direct product is defined precisely so that its elements can be uniquely written as products of components from each subgroup—this mirrors exactly how the external direct product works. To prove the isomorphism, construct a map sending each element $g \in G$ to its unique factorization $(h_1, h_2, \ldots, h_n)$ where $h_i \in H_i$, then verify this map is a bijective homomorphism using the defining properties of internal direct products (normality, pairwise commutativity, and trivial intersections).