External Direct Product Identity/Sufficient Condition

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Theorem

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

Let:

$\struct {S, \circ_1}$ have identity element $e_S$

and:

$\struct {T, \circ_2}$ have identity element $e_T$.


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


General Result

Let $\ds \struct {\SS, \circ} = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$.


Let $e_1, e_2, \ldots, e_n$ be the identity elements of $\struct {S_1, \circ_1}, \struct {S_2, \circ_2}, \ldots, \struct {S_n, \circ_n}$ respectively.

Then $\tuple {e_1, e_2, \ldots, e_n}$ is the identity element of $\struct {\SS, \circ}$.


Proof

Let $e_S$ and $e_T$ be the identity elements of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ respectively.

\(\ds \tuple {s, t} \circ \tuple {e_S, e_T}\) \(=\) \(\ds \tuple {s \circ_1 e_S, t \circ_2 e_T}\)
\(\ds \) \(=\) \(\ds \tuple {s, t}\)
\(\ds \tuple {e_S, e_T} \circ \tuple {s, t}\) \(=\) \(\ds \tuple {e_S \circ_1 s, e_T \circ_2 t}\)
\(\ds \) \(=\) \(\ds \tuple {s, t}\)


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

$\blacksquare$


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