Associative Idempotent Anticommutative

Theorem
Let $$\circ$$ be a binary operation on a set $$S$$.

Let $$\circ$$ be associative.

Then $$\circ$$ is anticommutative iff:


 * $$\circ$$ is idempotent, and
 * $$\forall a, b \in S: a \circ b \circ a = a$$.

Proof
Let $$\circ$$ be an associative operation on $$S$$.


 * Suppose $$\circ$$ is anticommutative.

Let $$a \circ a = x$$ for some $$a \in S$$. Then:

$$ $$ $$ $$

So $$\circ$$ being associative and anticommutative implies that $$\circ$$ is idempotent.

Now, let $$a \circ b \circ a = x$$ for some $$a, b$$.

$$ $$ $$ $$ $$ $$

So $$\circ$$ being associative and anticommutative implies, via the fact (also proved) that $$\circ$$ is idempotent, that $$a \circ b \circ a = a$$.


 * Now, suppose $$\circ$$ (which we take to be associative) is idempotent and $$a \circ b \circ a = a$$. We now need to show that $$\circ$$ is anticommutative.

Suppose $$a \circ b = b \circ a$$. Then:

So, $$\circ$$ being idempotent and $$a \circ b \circ a = a$$ implies that $$\circ$$ is anticommutative.

And we're done.

Reference
Seth Warner: Modern Algebra, 1965, Exercise 2:17 (a).