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.