Internal Direct Product Theorem/General Result

Theorem
Let $G$ be a group whose identity is $e$.

Let $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ be a sequence of subgroups of $G$.

Then $G$ is the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ iff:


 * $(1): \quad G = H_1 H_2 \cdots H_n$
 * $(2): \quad \left \langle {H_k} \right \rangle_{1 \le k \le n}$ is a sequence of independent subgroups
 * $(3): \quad$ For each $k \in \left[{1 \,.\,.\, n}\right]$, $H_k \triangleleft G$.

Proof
By definition, $G$ is the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ iff the mapping:


 * $\displaystyle C: \prod_{k \mathop = 1}^n H_k \to G: C \left({h_1, \ldots, h_n}\right) = \prod_{k \mathop = 1}^n h_k$

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

Necessary Condition
Let $G$ be the internal group direct product of $\left \langle {H_k} \right \rangle_{1 \le k \le n}$.


 * $(1): \quad$ From Internal Group Direct Product Surjective, $G = H_1 H_2 \cdots H_n$.
 * $(2): \quad$ From Internal Group Direct Product Injective, $\left \langle {H_k} \right \rangle_{1 \le k \le n}$ is a sequence of independent subgroups.
 * $(3): \quad$ From Internal Group Direct Product Isomorphism, $\forall k \in \left[{1 \,.\,.\, n}\right]: H_k \triangleleft G$.

Sufficient Condition
Now suppose the three conditions hold.


 * $(1): \quad$ From Internal Group Direct Product Surjective, $C$ is surjective.
 * $(2): \quad$ From Internal Group Direct Product Injective, $C$ 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 $\left \langle {H_k} \right \rangle_{1 \le k \le n}$.