Set of Subsemigroups of Commutative Semigroup form Subsemigroup of Power Structure

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Theorem

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

Let $\struct {\powerset S, \circ_\PP}$ denote the power structure of $\struct {S, \circ}$.

Let $\TT$ be the set of all subsemigroups of $S$.


Then $\struct {\TT, \circ_\PP}$ is a subsemigroup of $\struct {\powerset S, \circ_\PP}$.


Proof

First we establish that from Power Structure of Semigroup is Semigroup:

$\struct {\powerset S, \circ_\PP}$ is a semigroup.

From Subset Product within Commutative Structure is Commutative:

$\struct {\powerset S, \circ_\PP}$ is a commutative semigroup.


Let $A$ and $B$ be arbitrary subsemigroups of $S$.

As $A$ and $B$ are subsemigroups of $S$, they themselves are closed for $\circ$.

That is:

$\forall x, y \in A: x \circ y \in A$

and:

$\forall x, y \in B: x \circ y \in B$


By definition of operation induced on $\powerset S$:

$A \circ_\PP B = \set {a \circ b: a \in A, b \in B}$

We are to show that:

$\forall x, y \in A \circ_\PP B: x \circ y \in A \circ_\PP B$

Let $x, y \in A \circ_\PP B$ such that $x = a_x \circ b_x$, $y = a_y \circ b_y$.

We have:

\(\ds x \circ y\) \(=\) \(\ds \paren {a_x \circ b_x} \circ \paren {a_y \circ b_y}\)
\(\ds \) \(=\) \(\ds a_x \circ \paren {b_x \circ a_y} \circ b_y\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds a_x \circ \paren {a_y \circ b_x} \circ b_y\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds \paren {a_x \circ a_y} \circ \paren {b_x \circ b_y}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(\in\) \(\ds A \circ_\PP B\) Semigroup Axiom $\text S 0$: Closure: both $A$ and $B$ are closed for $\circ$

That is:

$x, y \in A \circ_\PP B \implies x \circ y \in A \circ_\PP B$

and $A \circ_\PP B$ is seen to be $\struct {\TT, \circ_\PP}$.

Hence the result.

$\blacksquare$


Sources

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