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

Sources