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

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.