Power Set with Intersection is Commutative Monoid

Theorem
Let $$S$$ be a set and let $$\mathcal P \left({S}\right)$$ be its power set.

Then $$\left({\mathcal P \left({S}\right), \cap}\right)$$ is a commutative monoid whose identity is $$S$$.

The only invertible element of this structure is $$S$$.

Thus (except in the degenerate case $$S = \varnothing$$) $$\mathcal P \left({S}\right)$$ cannot be a group.

Proof

 * First, from Power Set Closed under Intersection, we have that $$\forall A, B \in \mathcal P \left({S}\right): A \cap B \in \mathcal P \left({S}\right)$$.


 * Next, we have that, from Set System Closed with Intersection is Semigroup, $$\left({\mathcal P \left({S}\right), \cap}\right)$$ is a commutative semigroup.


 * From Identity of Power Set with Intersection, we have that $$S$$ acts as the identity.


 * Finally, we show that only $$S$$ has an Inverse:

For $$T \subseteq S$$ to have an inverse under $$\cap$$, we require $$T^{-1} \cap T = S$$. From this it follows that $$T = S = T^{-1}$$.

The result follows by definition of monoid.

Also see

 * Power Set with Union is a Monoid