Internal Direct Product Theorem/Proof 2

Sufficient Condition
Let $\phi: H_1 \times H_2 \to G$ be the mapping defined as:
 * $\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$

Let $\phi$ be an isomorphism.


 * $(1): \quad$ 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 $\phi: H_1 \times H_2 \to G$ be the mapping defined as:
 * $\forall \tuple {h_1, h_2} \in H_1 \times H_2: \map \phi {h_1, h_2} = h_1 \circ h_2$

Suppose the three conditions hold.


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

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

So by definition, $G$ is the internal group direct product of $H_1$ and $H_2$.