# 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.

$\blacksquare$

## Sources

- 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $13$: Direct products: Proposition $13.1 \ (2)$