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
\(\ds \map \theta {g_1, h_1}\) | \(=\) | \(\ds \map \theta {g_2, h_2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {h_1, g_1}\) | \(=\) | \(\ds \tuple {h_2, g_2}\) | Definition of $\theta$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {g_1, h_1}\) | \(=\) | \(\ds \tuple {g_2, h_2}\) | Equality of Ordered Pairs |
Thus $\theta$ is injective by definition.
$\Box$
Surjective
\(\ds \tuple {h, g}\) | \(\in\) | \(\ds H \times G\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \tuple {g, h}\) | \(\in\) | \(\ds G \times H\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists \tuple {g, h} \in G \times H: \, \) | \(\ds \tuple {h, g}\) | \(=\) | \(\ds \map \theta {g, h}\) |
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:
\(\ds \map \theta {\tuple {g_1, h_1} \circ \tuple {g_2, h_2} }\) | \(=\) | \(\ds \map \theta {g_1 \circ_g g_2, h_1 \circ_h h_2}\) | Definition of $\circ$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}\) | Definition of $\theta$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \tuple {h_1, g_1} \star \tuple {h_2, g_2}\) | Definition of $\star$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \theta {g_1, h_1} \star \map \theta {g_2, h_2}\) | 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 $13.1 \ (2)$