External Direct Product Identity

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {S \times T, \circ}$ be the external direct product of the two monoids $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.

Let:

$e_S$ be the identity for $\struct {S, \circ_1}$

and:

$e_T$ be the identity for $\struct {T, \circ_2}$.


Then $\tuple {e_S, e_T}$ is the identity for $\struct {S \times T, \circ}$.


General Result

Let $\displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k$ 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 $e_1, e_2, \ldots, e_n$ be the identity elements 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 element of $\left({S, \circ}\right)$.


Proof

\(\displaystyle \tuple {s, t} \circ \tuple {e_S, e_T}\) \(=\) \(\displaystyle \tuple {s \circ_1 e_S, t \circ_2 e_T}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {s, t}\)
\(\displaystyle \tuple {e_S, e_T} \circ \tuple {s, t}\) \(=\) \(\displaystyle \tuple {e_S \circ_1 s, e_T \circ_2 t} = \left({s, t}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {s, t}\)


Thus $\tuple {e_S, e_T}$ is the identity of $\struct {S \times T, \circ}$.

$\blacksquare$


Sources