Product of Commuting Idempotent Elements is Idempotent

Theorem

Let $\struct {S, \circ}$ be a semigroup.

Let $a, b \in S$ be idempotent elements of $S$.

Let $a$ and $b$ commute:

$a \circ b = b \circ a$

Then $a \circ b$ is idempotent.

Proof

From Semigroup Axiom $\text S 0$: Closure we take it for granted that $\struct {S, \circ}$ is closed under $\circ$.

Hence:

\(\ds \paren {a \circ b} \circ \paren {a \circ b}\)

\(=\)

\(\ds \paren {a \circ \paren {b \circ a} } \circ b\)

Semigroup Axiom $\text S 1$: Associativity

\(\ds \)

\(=\)

\(\ds \paren {a \circ \paren {a \circ b} } \circ b\)

$a$ and $b$ commute

\(\ds \)

\(=\)

\(\ds \paren {a \circ a} \circ \paren {b \circ b}\)

Semigroup Axiom $\text S 1$: Associativity

\(\ds \)

\(=\)

\(\ds a \circ b\)

$a$ and $b$ are idempotent by the premise

Thus $a \circ b$ is idempotent.

$\blacksquare$