Commutativity of Group Direct Product

Theorem
Let $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$ be groups.

Let $\struct {G \times H, \circ}$ be the group direct product of $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$, where the operation $\circ$ is defined as:


 * $\tuple {g_1, h_1} \circ \tuple {g_2, h_2} = \tuple {g_1 \circ_g g_2, h_1 \circ_h h_2}$

Let $\struct {H \times G, \star}$ be the group direct product of $\struct {H, \circ_h}$ and $\struct {G, \circ_g}$, where the operation $\star$ is defined as:


 * $\tuple {h_1, g_1} \star \tuple {h_2, g_2} = \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}$

The group direct product $\struct {G \times H, \circ}$ is isomorphic to the $\struct {H \times G, \star}$.

Proof
The mapping $\theta: G \times H \to H \times G$ defined as:
 * $\forall g \in G, h \in H: \map \theta {g, h} = \tuple {h, g}$

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

Injective
Thus $\theta$ is injective by definition.

Surjective
Thus $\theta$ is surjective by definition.

Group Homomorphism
Let $\tuple {g_1, h_1}, \tuple {g_2, h_2} \in G \times H$.

Then:

thus proving that $\theta$ is a homomorphism.