Commutativity of Group Direct Product

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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