Identity of Group Direct Product

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Theorem

Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.

Let $e_G$ be the identity for $\struct {G, \circ_1}$.

Let $e_H$ be the identity for $\struct {H, \circ_2}$.


Then $\tuple {e_G, e_H}$ is the identity for $\struct {G \times H, \circ}$.


Proof 1

\(\displaystyle \left({g, h}\right) \circ \left({e_G, e_H}\right)\) \(=\) \(\displaystyle \left({g \circ_1 e_G, h \circ_2 e_H}\right) = \left({g, h}\right)\)
\(\displaystyle \left({e_G, e_H}\right) \circ \left({g, h}\right)\) \(=\) \(\displaystyle \left({e_G \circ_1 g, e_H \circ_2 h}\right) = \left({g, h}\right)\)


So the identity is $\left({e_G, e_H}\right)$.

$\blacksquare$


Proof 2

A specific instance of External Direct Product Identity, where the algebraic structures in question are groups.

$\blacksquare$


Sources