Associative Idempotent Anticommutative

== Theorem ==
Let $\circ$ be a binary operation on a non-empty set $S$.
Let $\circ$ be associative.
Then $\circ$ is anticommutative :
:$(1): \quad \circ$ is idempotent
and:
:$(2): \quad \forall a, b \in S: a \circ b \circ a = a$
== Proof ==
Let $\circ$ be an associative operation on $S$.
=== Necessary Condition ===
Suppose $\circ$ is anticommutative.
Since $S$ is non-empty, let $a \in S$ be arbitrary.
By definition, $\circ$ is anticommutative implies that:
:$\forall a \in S: a \circ a = a \circ a \iff a = a$
By $\circ$ is a well-defined operation, $a \circ a$ is well-defined.
It remains to be shown that $a \circ a \in S$.
Let $a, w \in S$ be arbitrary.
Since $\circ$ is anticommutative, then:
:$\paren {a \circ a} \circ w = w \circ \paren {a \circ a} \iff a \circ a = w$
Thus $a \circ a \in S$.
That is:
:$\forall \paren {a \circ a}, w \in S: \paren {a \circ a} \circ w = w \circ \paren {a \circ a} \iff a \circ a = w$
Let $w = a \in S$. 
Then:
:$\forall \paren {a \circ a}, a \in S: \paren {a \circ a} \circ a = a \circ \paren {a \circ a} \iff a \circ a = a$
Hence:
:$\forall a \in S: \paren {a \circ a} \circ a = a \circ \paren {a \circ a} \iff a \circ a = a$
So $\circ$ being associative and anticommutative implies that $\circ$ is idempotent.
Now, it remains to be shown that for any $a, b \in S$, if $\circ$ is associative and anticommutative, then:
:$ a \circ b \circ a = a$.
In particular, it remains to be shown that:
:$a \circ b \circ a \in S$.
By assumption:
:$\forall a, b \in S: a \circ b = b \circ a \iff a = b$
By $\circ$ is a well-defined operation, $a \circ b$ and $b \circ a$ are well-defined.
Let $a, b, x \in S$, then:
:$\paren {a \circ b} \circ x = x \circ \paren {a \circ b} \iff a \circ b = x$
This implies that:
:$\forall \paren {a \circ b}, x \in S: \paren {a \circ b} \circ x = x \circ \paren {a \circ b} \iff a \circ b = x$
Hence there exists $\paren {a \circ b} \in S$ for some $a, b \in S$. 
Similarly, let $b, a, y \in S$, then:
:$\paren {b \circ a} \circ y = y \circ \paren {b \circ a} \iff b \circ a = y$
This implies that:
:$\forall \paren {b \circ a}, y \in S: \paren {b \circ a} \circ y = y \circ \paren {b \circ a} \iff b \circ a = y$
Hence there exists $\paren {b \circ a} \in S$ for some $a, b \in S$. 
Let $a, b, z \in S$, then:
:$\paren {a \circ b \circ a} \circ z = z \circ \paren {a \circ b \circ a} \iff a \circ b \circ a = z$
Hence there exists $\paren {a \circ b \circ a} \in S$ for some $a, b \in S$.
Hence:
So $\circ$ being associative and anticommutative implies, via the fact (also proved) that $\circ$ is idempotent, that $a \circ b \circ a = a$.
=== Sufficient Condition ===
Now, suppose $\circ$ (which we take to be associative) is idempotent and:
:$\forall a, b \in S: a \circ b \circ a = a$
It remains to be shown that $\circ$ is anticommutative.
Suppose:
:$a \circ b = b \circ a$.
Then:
Similarly:
Also:
Then:
Hence:
So:
:$(1): \quad \circ$ is idempotent
and:
:$(2): \quad \forall a, b \in S: a \circ b \circ a = a$
together imply that $\circ$ is anticommutative.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.17 \ \text{(b)}$
Category:Abstract Algebra
Category:Anticommutativity
Category:Associativity
Category:Idempotence