Composition of Commuting Idempotent Mappings is Idempotent

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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.

$\blacksquare$