Identity of Group Direct Product

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Theorem

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

If:

$e_G$ is the identity for $\left({G, \circ_1}\right)$

and:

$e_H$ is the identity for $\left({H, \circ_2}\right)$

then $\left({e_G, e_H}\right)$ is the identity for $\left({G \times H, \circ}\right)$.


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