External Direct Product Identity

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:
 * $$e_S$$ is the identity for $$\left({S, \circ_1}\right)$$, and:
 * $$e_T$$ is the identity for $$\left({T, \circ_2}\right)$$;

then $$\left({e_S, e_T}\right)$$ is the identity for $$\left({S \times T, \circ}\right)$$.

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

If $$e_1, e_2, \ldots, e_n$$ are the identities of $$\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$$ respectively, then $$\left({e_1, e_2, \ldots, e_n}\right)$$ is the identity of $$\left({S, \circ}\right)$$.

Proof
$$ $$

So the identity is $$\left({e_S, e_T}\right)$$.

Proof of Generalized Result
This follows directly from the above.