External Direct Product Commutativity/Necessary Condition

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 commutative.

Then $\circ_1$ and $\circ_2$ are both also commutative.

Proof
Let $\circ$ be commutative.

it is not the case that both $\circ_1$ and $\circ_2$ are commutative.

, suppose $\circ_1$ is not commutative.

Hence:

This contradicts our assumption that $\circ$ is commutative.

Mutatis mutandis, the same argument can be applied to the case where $\circ_2$ is not commutative.

Hence, by Proof by Contradiction, $\circ_1$ and $\circ_2$ are both commutative.

Also see

 * External Direct Product Associativity
 * External Direct Product Identity
 * External Direct Product Inverses