Internal Group Direct Product is Injective

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:
 * $\map \phi {h_1, h_2} = h_1 h_2$

Then $\phi$ is injective :
 * $H_1 \cap H_2 = \set e$

Necessary Condition
Let $\phi$ be an injection.

Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$.

As $\phi$ is injective, this means that:
 * $\tuple {h_1, h_2} = \tuple {k_1, k_2}$

and thus:
 * $h_1 = k_1, h_2 = k_2$

From the definition of $\phi$, this means:
 * $h_1 h_2 = k_1 k_2$

Thus, each element of $G$ that can be expressed as a product of the form $h_1 h_2$ can be thus expressed uniquely.

Now, suppose $h \in H_1 \cap H_2$.

We have:

Thus we see that:

Thus:
 * $H_1 \cap H_2 = \set e$

Sufficient Condition
Let $H_1 \cap H_2 = \set e$.

Let:
 * $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$

Then:
 * $h_1 h_2 = k_1 k_2: h_1, k_1 \in H_1, h_2, k_2 \in H_2$

Thus:
 * $k_1^{-1} h_1 = k_2 h_2^{-1}$

But:
 * $k_1^{-1} h_1 \in H_1$ and $k_2 h_2^{-1} \in H_2$

As they are equal, we have:
 * $k_1^{-1} h_1 = k_2 h_2^{-1} \in H_1 \cap H_2 = \set e$

It follows that:
 * $h_1 = k_1, h_2 = k_2$

and thus:
 * $\tuple {h_1, h_2} = \tuple {k_1, k_2}$

Thus $\phi$ is injective and the result follows.