Idempotent Elements form Submonoid of Commutative Monoid

Theorem
Let $\left({S, \circ}\right)$ be a commutative monoid.

Let $e \in S$ be the identity element of $\left({S, \circ}\right)$.

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

That is:


 * $I = \left\{{x \in S: x \circ x = x}\right\}$

Then $\left({I, \circ}\right)$ is a submonoid of $\left({S, \circ}\right)$ with identity $e$.

Proof
By Idempotent Elements form Subsemigroup of Commutative Semigroup, $\left({I, \circ}\right)$ is a subsemigroup of $\left({S, \circ}\right)$.

By Identity Element is Idempotent, $e \in I$.

By Identity of Algebraic Structure is Identity of Substructure Containing It, $e$ is an identity of $(I, \circ)$.

Since $\left({T, \circ}\right)$ is a semigroup and has an identity, $\left({T, \circ}\right)$ is a monoid.

Since $T \subseteq S$, $\left({T, \circ}\right)$ is a submonoid of $\left({S, \circ}\right)$ with identity $e$.