Idempotent Elements form Subsemigroup of Commutative Semigroup/Proof 1
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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 $x, y \in I$: $x \circ y \in I$.
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.
$\blacksquare$