External Direct Product Commutativity

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Theorem

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

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


Then $\circ$ is also a commutative operation.


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 commutative, then so is $\circ$.


Proof

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

\(\displaystyle \left({s_1, t_1}\right) \circ \left({s_2, t_2}\right)\) \(=\) \(\displaystyle \left({s_1 \circ_1 s_2, t_1 \circ_2 t_2}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({s_2 \circ_1 s_1, t_2 \circ_2 t_1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({s_2, t_2}\right) \circ \left({s_1, t_1}\right)\)

Thus $\circ$ is commutative.

$\blacksquare$


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