Internal Group Direct Product of Normal Subgroups

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

Let $$H_1, H_2$$ be subgroups of $$G$$.

Let $$\phi: H_1 \times H_2 \to G$$ be a mapping defined by $$\phi \left({\left({h_1, h_2}\right)}\right) = h_1 h_2$$.

Let $$H_1$$ and $$H_2$$ be normal subgroups of $$G$$, and let $$H_1 \cap H_2 = \left\{{e}\right\}$$.

Then $$\phi$$ is a (group) homomorphism.

Proof

 * Let $$H_1$$ and $$H_2$$ be normal subgroups of $$G$$.

Let $$h_1 \in H_1, h_2 \in H_2$$.

Consider $$x \in G: x = h_1 h_2 h_1^{-1} h_2^{-1}$$.

$$ $$

As $$H_2$$ is normal, we have $$h_1 h_2 h_1^{-1} \in H_2$$ and thus $$x \in H_2$$.

Similarly, we can show that $$x \in H_1$$ and so $$x \in H_1 \cap H_2$$ and thus $$x = e$$.

From Commutation with Inverses: Theorem 5, $$h_1 h_2 h_1^{-1} h_2^{-1} = e$$ iff $$h_1$$ and $$h_2$$ commute.

Thus $$h_1 h_2 = h_2 h_1$$.

As $$h_1$$ and $$h_2$$ are arbitrary elements of $$H_1$$ and $$H_2$$, it follows that every element of $$H_1$$ commutes with every element of $$H_2$$.

Thus from Mapping from Cartesian Product Homomorphism iff Abelian, $$\phi$$ is a homomorphism.