Idempotent Elements form Subsemigroup of Commutative Semigroup

Theorem

Let $\struct {S, \circ}$ be a semigroup such that $\circ$ is commutative.

Let $I$ be the set of all elements of $S$ that are idempotent under $\circ$.

That is:

$I = \set {x \in S: x \circ x = x}$

Then $\struct {I, \circ}$ is a subsemigroup of $\struct {S, \circ}$.

Proof 1

By Subsemigroup Closure Test we need only show that:

For all $x, y \in I$: $x \circ y \in I$.

That is:

$\paren {x \circ y} \circ \paren {x \circ y} = x \circ y$

We reason as follows:

\(\ds \paren {x \circ y} \circ \paren {x \circ y}\)

\(=\)

\(\ds \paren {x \circ y} \circ \paren {y \circ x}\)

$\circ$ is commutative

\(\ds \)

\(=\)

\(\ds \paren {x \circ \paren {y \circ x} } \circ y\)

$\circ$ is associative

\(\ds \)

\(=\)

\(\ds \paren {x \circ \paren {x \circ y} } \circ y\)

$\circ$ is commutative

\(\ds \)

\(=\)

\(\ds \paren {x \circ x} \circ \paren {y \circ y}\)

$\circ$ is associative

\(\ds \)

\(=\)

\(\ds x \circ y\)

$x$ and $y$ are idempotent

Hence the result.

$\blacksquare$

Proof 2

By Subsemigroup Closure Test we need only show that:

For all $x, y \in I$: $x \circ y \in I$.

As $x, y \in I$, they are idempotent.

We have that $\circ$ is commutative.

Thus, by definition, $x$ and $y$ commute.

From Product of Commuting Idempotent Elements is Idempotent, $\left({x \circ y}\right)$ is idempotent.

That is:

$x \circ y \in I$

$\blacksquare$