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

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