# 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

 $\displaystyle \map \theta {g_1, h_1}$ $=$ $\displaystyle \map \theta {g_2, h_2}$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {h_1, g_1}$ $=$ $\displaystyle \tuple {h_2, g_2}$ Definition of $\theta$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {g_1, h_1}$ $=$ $\displaystyle \tuple {g_2, h_2}$ Equality of Ordered Pairs

Thus $\theta$ is injective by definition.

$\Box$

### Surjective

 $\displaystyle \tuple {h, g}$ $\in$ $\displaystyle H \times G$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {g, h}$ $\in$ $\displaystyle G \times H$ $\displaystyle \leadsto \ \$ $\, \displaystyle \exists \tuple {g, h} \in G \times H: \,$ $\displaystyle \tuple {h, g}$ $=$ $\displaystyle \map \theta {g, h}$

Thus $\theta$ is surjective by definition.

$\Box$

### Group Homomorphism

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

Then:

 $\displaystyle \map \theta {\tuple {g_1, h_1} \circ \tuple {g_2, h_2} }$ $=$ $\displaystyle \map \theta {g_1 \circ_g g_2, h_1 \circ_h h_2}$ Definition of $\circ$ $\displaystyle$ $=$ $\displaystyle \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}$ Definition of $\theta$ $\displaystyle$ $=$ $\displaystyle \tuple {h_1, g_1} \star \tuple {h_2, g_2}$ Definition of $\star$ $\displaystyle$ $=$ $\displaystyle \map \theta {g_1, h_1} \star \map \theta {g_2, h_2}$ Definition of $\theta$

thus proving that $\theta$ is a homomorphism.

$\blacksquare$