External Direct Product Inverses

Theorem
Let $\left({S \times T, \circ}\right)$ be the external direct product of the two algebraic structures $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$.

Let:
 * $s^{-1}$ be an inverse of $s \in \left({S, \circ_1}\right)$

and:
 * $t^{-1}$ be an inverse of $t \in \left({T, \circ_2}\right)$.

Then $\left({s^{-1}, t^{-1}}\right)$ is an inverse of $\left({s, t}\right) \in \left({S \times T, \circ}\right)$.

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

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

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

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

Then:

Thus the inverse of $\left({s, t}\right)$ is $\left({s^{-1}, t^{-1}}\right)$.