Composition of Commuting Idempotent Mappings is Idempotent

Theorem

Let $S$ be a set.

Let $f, g: S \to S$ be idempotent mappings from $S$ to $S$.

Let:

$f \circ g = g \circ f$

where $\circ$ denotes composition.

Then $f \circ g$ is idempotent.

Proof 1

 $\displaystyle \left({f \circ g}\right) \circ \left({f \circ g}\right)$ $=$ $\displaystyle f \circ \left({g \circ f}\right) \circ g$ Composition of Mappings is Associative $\displaystyle$ $=$ $\displaystyle f \circ \left({f \circ g}\right) \circ g$ by hypothesis $\displaystyle$ $=$ $\displaystyle \left({f \circ f}\right) \circ \left({g \circ g}\right)$ Composition of Mappings is Associative $\displaystyle$ $=$ $\displaystyle f \circ g$ $f$ and $g$ are idempotent by hypothesis

$\blacksquare$

Proof 2

By Set of All Self-Maps is Semigroup, the set of all self-maps on $S$ forms a semigroup under composition.

The result follows from Product of Commuting Idempotent Elements is Idempotent.

$\blacksquare$