Inverses in Group Direct Product/Proof 1

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

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

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

and
 * $h^{-1}$ be the inverse of $h \in \left({H, \circ_2}\right)$.

Then:

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