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

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.

It remains to be shown 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 Monoid