Set of Subsemigroups of Commutative Semigroup form Subsemigroup of Power Structure

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:

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.