# 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.

$\blacksquare$