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 $\left({s, t}\right) \in \left({S \times T, \circ}\right)$.

Generalized Result
Let $\displaystyle \left({S, \circ}\right) = \prod_{k=1}^n S_k$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$.

Let $\left({x_1, x_2, \ldots, x_n}\right) \in S$.

If $y_k$ is an inverse of $x_k$ in $\left({S_k, \circ_k}\right)$ for each of $k \in \N^*_n$, then $\left({y_1, y_2, \ldots, y_n}\right)$ is the inverse of $\left({x_1, x_2, \ldots, x_n}\right) \in S$ in $\left({S, \circ}\right)$.

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 $\left({s^{-1}, t^{-1}}\right)$.

Proof of Generalized Result
This follows directly from above and can be proved explicitly using the same technique.