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 $$\left({S, \circ}\right) = \prod_{k=1}^n S_k$$ 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 $$k \in \mathbb{N}^*_n$$, $$\circ_k$$ 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.