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
- $\circ$ 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.
$\blacksquare$