Internal Group Direct Product Commutativity/Proof 2

Proof
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
 * $\sqbrk {x, y} := x^{-1} y^{-1} x y$

We have that:

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

We have:

and:

Thus:
 * $\sqbrk {h_1, h_2} \in H_1 \cap H_2$

But as $H_1 \cap H_2 = \set e$, it follows that:
 * $\sqbrk {h_1, h_2} = e$

It follows from Commutator is Identity iff Elements Commute that:
 * $h_1 h_2 = h_2 h_1$

and the result follows.