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 $$\left({S, \circ}\right) = \prod_{k=1}^n S_k$$ 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 $$k \in \mathbb{N}^*_n$$, 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.