# Identity of Group Direct Product

## 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.

$\blacksquare$