Internal Direct Product Theorem/General Result/Proof 1

Proof
By definition, $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ the mapping:


 * $\ds C: \prod_{k \mathop = 1}^n H_k \to G: \map C {h_1, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$

is a group isomorphism from the cartesian product $\struct {H_1, \circ {\restriction_{H_1} } } \times \cdots \times \struct {H_n, \circ {\restriction_{H_n} } }$ onto $\struct {G, \circ}$.

Necessary Condition
Let $G$ be the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$.


 * $(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $G = H_1 H_2 \cdots H_n$.
 * $(2): \quad$ From Internal Group Direct Product is Injective: General Result, $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups.
 * $(3): \quad$ From Internal Group Direct Product Isomorphism:
 * $\forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$

Sufficient Condition
Now suppose the three conditions hold.


 * $(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $C$ is surjective.
 * $(2): \quad$ From Internal Group Direct Product is Injective: General Result, $C$ is injective.
 * $(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $C$ is a group homomorphism.

Putting these together, we see that $C$ is a bijective homomorphism, and therefore an isomorphism.

So by definition, $G$ is the internal group direct product of $\sequence {H_k} _{1 \mathop \le k \mathop \le n}$.