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 $h \in H$, $k \in K$.

We have:

and:

Thus:
 * $\sqbrk {h, k} \in H \cap K$

But as $H \cap K = \set e$, it follows that:
 * $\sqbrk {h, k} = e$

It follows from Commutator is Identity iff Elements Commute that:
 * $h k = k h$

and the result follows.