External Direct Product Commutativity

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

If $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ are commutative, then $\left({S \times T, \circ}\right)$ is also commutative.

Generalized Result
Let $\displaystyle \left({S, \circ}\right) = \prod_{k=1}^n S_k$ 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 $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ be commutative.

and we see that $\left({S \times T, \circ}\right)$ is commutative.

Proof of Generalized Result
Suppose that, for all $k \in \N^*_n$, $\circ_k$ is commutative.

Let $\left({s_1, s_2, \ldots, s_n}\right)$ and $\left({t_1, t_2, \ldots, t_n}\right)$ be elements of $\left({S_1, \circ_1}\right) \times \left({S_2, \circ_2}\right) \times \cdots \times \left({S_n, \circ_n}\right)$.

... hence the result.