Internal Group Direct Product Commutativity

Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.

Let $H_1, H_2 \le G$.

Let $\left({G, \circ}\right)$ be the internal group direct product of $H_1$ and $H_2$.

Then:
 * $\forall h_1 \in H_1, h_2 \in H_2: h_1 \circ h_2 = h_2 \circ h_1$

Proof
Let $G$ be the internal group direct product of $H_1$ and $H_2$.

Then by definition the mapping:
 * $C: H_1 \times H_2 \to G: C \left({\left({h_1, h_2}\right)}\right) = h_1 \circ h_2$

is a (group) isomorphism from the cartesian product $\left({H_1, \circ \restriction_{H_1}}\right) \times \left({H_2, \circ \restriction_{H_2}}\right)$ onto $\left({G, \circ}\right)$.

Let the symbol $\circ$ also be used for the operation induced on $H_1 \times H_2$ by $\circ \restriction_{H_1}$ and $\circ \restriction_{H_2}$.

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

Then: