External Direct Product Associativity/Necessary Condition
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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 $\struct {T, \circ_2}$.
Let $\circ$ be associative.
Then $\circ_1$ and $\circ_2$ are both also associative.
Proof
Let $\circ$ be associative.
Aiming for a contradiction, suppose it is not the case that both $\circ_1$ and $\circ_2$ are associative.
Without loss of generality, suppose $\circ_1$ is not associative.
Hence:
\(\ds \exists s_1, s_2, s_3 \in S: \, \) | \(\ds \paren {s_1 \circ_1 s_2} \circ_1 s_3\) | \(\ne\) | \(\ds s_1 \circ_1 \paren {s_2 \circ_1 s_3}\) | as $\circ_1$ is not associative | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall t_1, t_2, t_3 \in T: \, \) | \(\ds \tuple {\paren {s_1 \circ_1 s_2}, s_3} \circ \tuple {\paren {t_1 \circ_2 t_2}, t_3}\) | \(\ne\) | \(\ds \tuple {s_1 \circ_1 \paren {s_2 \circ_1 s_3}, t_1 \circ_2 \paren {t_2 \circ_2 t_3} }\) | Equality of Ordered Pairs | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall t_1, t_2, t_3 \in T: \, \) | \(\ds \paren {\tuple {s_1, t_1} \circ \tuple {s_2, t_2} } \circ \tuple {s_3, t_3}\) | \(\ne\) | \(\ds \tuple {s_1, t_1} \circ \paren {\tuple {s_2, t_2} \circ \tuple {s_3, t_3} }\) | Definition of Operation Induced by Direct Product |
This contradicts our assumption that $\circ$ is associative.
The same argument can be applied, mutatis mutandis, to the case where $\circ_2$ is not associative.
Hence, by Proof by Contradiction, $\circ_1$ and $\circ_2$ are both associative.
$\blacksquare$
Also see
- External Direct Product Commutativity
- External Direct Product Identity
- External Direct Product Inverses
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.5$