Product of Commuting Idempotent Elements is Idempotent

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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

\(\displaystyle \paren {a \circ b} \circ \paren {a \circ b}\) \(=\) \(\displaystyle \paren {a \circ \paren {b \circ a} } \circ b\) $\circ$ is associative by definition of semigroup
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \circ \paren {a \circ b} } \circ b\) $a$ and $b$ commute
\(\displaystyle \) \(=\) \(\displaystyle \paren {a \circ a} \circ \paren {b \circ b}\) $\circ$ is associative
\(\displaystyle \) \(=\) \(\displaystyle a \circ b\) $a$ and $b$ are idempotent by the premise

Thus $a \circ b$ is idempotent.

$\blacksquare$