Idempotent Elements form Subsemigroup of Commutative Semigroup/Proof 2
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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
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:
- $x \circ y \in I$
$\blacksquare$