External Direct Product Associativity

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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 $\circ_1$ and $\circ_2$ be associative.


Then $\circ$ is also associative.


General Result

Let $\displaystyle \left({S, \circ}\right) = \prod_{k \mathop = 1}^n S_k$ be the external direct product of the algebraic structures $\left({S_1, \circ_1}\right), \left({S_2, \circ_2}\right), \ldots, \left({S_n, \circ_n}\right)$.


If $\circ_1, \ldots, \circ_n$ are all associative, then so is $\circ$.


Proof

\(\displaystyle \paren {\tuple {s_1, t_1} \circ \tuple {s_2, t_2} } \circ \tuple {s_3, t_3}\) \(=\) \(\displaystyle \tuple {\paren {s_1 \circ_1 s_2} \circ_1 s_3, \paren {t_1 \circ_2 t_2} \circ_2 t_3}\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {s_1 \circ_1 s_2 \circ_1 s_3, t_1 \circ_2 t_2 \circ_2 t_3}\) Definition of Associative Operation
\(\displaystyle \) \(=\) \(\displaystyle \tuple {s_1 \circ_1 \paren {s_2 \circ_1 s_3}, t_1 \circ_2 \paren {t_2 \circ_2 t_3} }\)
\(\displaystyle \) \(=\) \(\displaystyle \tuple {s_1, t_1} \circ \paren {\tuple {s_2, t_2} \circ \tuple {s_3, t_3} }\)

Thus $\circ$ is associative.

$\blacksquare$


Sources