Commutativity of Group Direct Product

Theorem
The group direct product $G \times H$ is isomorphic to $H \times G$.

Proof
The mapping $\theta: G \times H \to H \times G$ defined as:
 * $\forall g \in G, h \in H: \theta \left({\left({g, h}\right)}\right) = \left({h, g}\right)$

is to be shown to be a group homomorphism, and that $\theta$ is bijective, as follows:

Injective
If $\theta \left({\left({g_1, h_1}\right)}\right) = \theta \left({\left({g_2, h_2}\right)}\right)$, then $\left({h_1, g_1}\right) = \left({h_2, g_2}\right)$.

Thus by Equality of Ordered Pairs, $h_1 = h_2$ and $g_1 = g_2$.

Thus $\left({g_1, h_1}\right) = \left({g_2, h_2}\right)$ and $\theta$ is injective.

Surjective
Let $\left({h, g}\right) \in H \times G$.

Then $h \in H$ and $g \in G$, so $\left({g, h}\right) \in G \times H$ and thus $\theta$ is surjective.

Group Homomorphism
Let $\left({g_1, h_1}\right), \left({g_2, h_2}\right) \in G \times H$. Then:

thus proving that $\theta$ is a homomorphism.