Internal Direct Product Theorem/Proof 2

Sufficient Condition
Let $\phi: H \times K \to G$ be the mapping defined as:
 * $\forall \tuple {h, k} \in H \times K: \map \phi {h, k} = h \circ k$

Let $\phi$ be an isomorphism.


 * $(1): \quad$ From Codomain of Internal Direct Isomorphism is Subset Product of Factors, $G = H \circ K$.
 * $(2): \quad$ From Internal Group Direct Product is Injective, $H$ and $K$ are independent subgroups of $G$.
 * $(3): \quad$ From Internal Group Direct Product Isomorphism, $H$ and $K$ are normal subgroups of $G$.

Necessary Condition
Let $\phi: H \times K \to G$ be the mapping defined as:
 * $\forall \tuple {h, k} \in H \times K: \map \phi {h, k} = h \circ k$

Suppose the three conditions hold.


 * $(1): \quad$ From $G = H \circ K$, $\phi$ 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 $H$ and $K$.