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 $\left({T, \circ_2}\right)$.

If: and:
 * $s^{-1}$ is an inverse of $s \in \left({S, \circ_1}\right)$
 * $t^{-1}$ is 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: and:
 * $e_S$ be the identity for $\left({S, \circ_1}\right)$
 * $e_T$ be the identity for $\left({T, \circ_2}\right)$.

Also let: and
 * $s^{-1}$ be the inverse of $s \in \left({S, \circ_1}\right)$
 * $t^{-1}$ be the inverse of $t \in \left({T, \circ_2}\right)$.

Then:

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