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:
 * $\phi \left({h_1, h_2}\right) = h_1 h_2$

Then $\phi$ is injective iff:
 * $H_1 \cap H_2 = \left\{{e}\right\}$

Necessary Condition
Let $\phi$ be an injection.

Let $\phi \left({\left({h_1, h_2}\right)}\right) = \phi \left({\left({k_1, k_2}\right)}\right)$.

As $\phi$ is injective, this means that:
 * $\left({h_1, h_2}\right) = \left({k_1, k_2}\right)$

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 $H_1 \cap H_2 = \left\{{e}\right\}$.

Sufficient Condition
Let $H_1 \cap H_2 = \left\{{e}\right\}$.

Let:
 * $\phi \left({\left({h_1, h_2}\right)}\right) = \phi \left({\left({k_1, k_2}\right)}\right)$

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:
 * $k_1^{-1} h_1 = k_2 h_2^{-1} \in H_1 \cap H_2 = \left\{{e}\right\}$

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

and thus:
 * $\left({h_1, h_2}\right) = \left({k_1, k_2}\right)$

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