Internal Direct Product Theorem

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $H_1, H_2 \le G$.

Then $G$ is the internal group direct product of $H_1$ and $H_2$ :


 * $(1): \quad G = H_1 \circ H_2$
 * $(2): \quad H_1 \cap H_2 = \left\{{e}\right\}$
 * $(3): \quad H_1, H_2 \lhd G$

where $H_1 \lhd G$ denotes that $H_1$ is a normal subgroup of $G$.

Sufficient Condition
Let $G$ be the internal group direct product of $H_1$ and $H_2$.


 * $(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $G = H_1 \circ H_2$.
 * $(2): \quad$ From Internal Group Direct Product is Injective, $H_1$ and $H_2$ are independent subgroups of $G$.
 * $(3): \quad$ From Internal Group Direct Product Isomorphism, $H_1 \lhd G$ and $H_2 \lhd G$.

Necessary Condition
Let $C: H_1 \times H_2 \to G$ be the mapping defined as:
 * $\forall \left({h_1, h_2}\right) \in H_1 \times H_2: C \left({\left({h_1, h_2}\right)}\right) = h_1 \circ h_2$

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, $C$ 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 $H_1$ and $H_2$.