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

Generalized Result
Let $$G$$ be a group whose identity is $$e$$.

Let $$\left \langle {H_n} \right \rangle$$ be a sequence of subgroups of $$G$$.

Let $$\phi: \prod_{k=1}^n H_k \to G$$ be a mapping defined by $$\phi \left({\left({h_1, h_2, \ldots, h_n}\right)}\right) = \prod_{k=1}^n h_k$$.

Then $$\phi$$ is injective iff $$\bigcap_{k=1}^n H_k = \left\{{e}\right\}$$.

Proof

 * First we show that if $$\phi$$ is injective, then $$H_1 \cap H_2 = \left\{{e}\right\}$$.

So, suppose $$\phi$$ is 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 $$H_1 \cap H_2 = \left\{{e}\right\}$$.


 * Now, let $$H_1 \cap H_2 = \left\{{e}\right\}$$.

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

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.