External Direct Product Commutativity

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Theorem

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

Then:

$\circ_1$ and $\circ_2$ are commutative

if and only if:

$\circ$ is commutative.


Proof

Sufficient Condition

Let $\circ_1$ and $\circ_2$ be commutative operations.

\(\ds \tuple {s_1, t_1} \circ \tuple {s_2, t_2}\) \(=\) \(\ds \tuple {s_1 \circ_1 s_2, t_1 \circ_2 t_2}\) Definition of Operation Induced by Direct Product
\(\ds \) \(=\) \(\ds \tuple {s_2 \circ_1 s_1, t_2 \circ_2 t_1}\) Definition of Commutative Operation
\(\ds \) \(=\) \(\ds \tuple {s_2, t_2} \circ \tuple {s_1, t_1}\) Definition of Operation Induced by Direct Product

Thus $\circ$ is commutative.

$\Box$


Necessary Condition

Let $\circ$ be commutative.

Aiming for a contradiction, suppose it is not the case that both $\circ_1$ and $\circ_2$ are commutative.

Without loss of generality, suppose $\circ_1$ is not commutative.

Hence:

\(\ds \exists s_1, s_2 \in S: \, \) \(\ds s_1 \circ_1 s_2\) \(\ne\) \(\ds s_2 \circ_1 s_1\) as $\circ_1$ is not commutative
\(\ds \leadsto \ \ \) \(\ds \forall t_1, t_2 \in T: \, \) \(\ds \tuple {s_1 \circ_1 s_2} \circ \tuple {t_1 \circ_2 t_2}\) \(\ne\) \(\ds \tuple {s_2 \circ_1 s_1} \circ \tuple {t_2 \circ_2 t_1}\) Equality of Ordered Pairs
\(\ds \leadsto \ \ \) \(\ds \forall t_1, t_2 \in T: \, \) \(\ds \tuple {s_1, t_1} \circ \tuple {s_2, t_2}\) \(\ne\) \(\ds \tuple {s_2, t_2} \circ \tuple {s_1, t_1}\) Definition of Operation Induced by Direct Product

This contradicts our assumption that $\circ$ is commutative.

The same argument can be applied, mutatis mutandis, to the case where $\circ_2$ is not commutative.


Hence, by Proof by Contradiction, $\circ_1$ and $\circ_2$ are both commutative.

$\blacksquare$


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