Anticommutativity of External Direct Product

Theorem
Let $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ be algebraic structures where $S$ and $T$ both have at least two distinct elements.

Let $\left({S \times T, \circ}\right)$ be their external direct product.

Then $\left({S \times T, \circ}\right)$ is anticommutative iff at least one of $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ is.

Sufficient Condition
Suppose $\left({S \times T, \circ}\right)$ is anticommutative and neither $\left({S, \circ_1}\right)$ nor $\left({T, \circ_2}\right)$ are.

Then for some distinct $a, b \in S$:


 * $a \circ_1 b = b \circ_1 a$

Similarly, for some distinct $c, d \in T$:


 * $c \circ_2 d = d \circ_2 c$

Then we compute by definition of $\circ$ in the external direct product:


 * $\left({a,c}\right) \circ \left({b,d}\right) = \left({a\circ_1 b, c \circ_2 d}\right)$
 * $\left({b,d}\right) \circ \left({a,c}\right) = \left({b\circ_1 a, d \circ_2 c}\right)$

So:


 * $\left({a,c}\right) \circ \left({b,d}\right) = \left({b,d}\right) \circ \left({a,c}\right)$

which contradicts our assumption that $\left({S \times T, \circ}\right)$ was anticommutative.

Hence at least one of $\left({S, \circ_1}\right)$ and $\left({T, \circ_2}\right)$ are anticommutative.

Necessary Condition
Suppose $\left({S \times T, \circ}\right)$ is not anticommutative.

Then for some $\left({a, b}\right), \left({c, d}\right) \in S \times T$:


 * $\left({a, b}\right) \circ \left({c, d}\right) = \left({c, d}\right) \circ \left({a, b}\right)$

which by the definition of $\circ$ implies:


 * $\left({a \circ_1 c, b \circ_2 d} \right) = \left({c \circ_1 a, d \circ_2 b}\right)$

Therefore we conclude:


 * $a \circ_1 c = c \circ_1 a$
 * $b \circ_2 d = d \circ_2 b$

Hence, neither $\left({S, \circ_1}\right)$ nor $\left({T, \circ_2}\right)$ can be anticommutative.

Hence if one of $\left({S, \circ_1}\right)$ or $\left({T, \circ_2}\right)$ is anticommutative, then $\left({S \times T, \circ}\right)$ must be as well.