Inverses in Group Direct Product/Proof 1

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: and:
 * $g^{-1}$ is an inverse of $g \in \left({G, \circ_1}\right)$
 * $h^{-1}$ is an inverse of $h \in \left({H, \circ_2}\right)$

then $\left({g^{-1}, h^{-1}}\right)$ is the inverse of $\left({g, h}\right) \in \left({G \times H, \circ}\right)$.

Proof
Let: and:
 * $e_G$ be the identity for $\left({G, \circ_1}\right)$
 * $e_H$ be the identity for $\left({H, \circ_2}\right)$.

Also let: and
 * $g^{-1}$ be the inverse of $g \in \left({G, \circ_1}\right)$
 * $h^{-1}$ be the inverse of $h \in \left({H, \circ_2}\right)$.

Then:

So the inverse of $\left({g, h}\right)$ is $\left({g^{-1}, h^{-1}}\right)$.