Power Set with Union is Commutative Monoid

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

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

The only invertible element of this structure is $$\varnothing$$.

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

Proof

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


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


 * From Identity of Power Set with Union, we have that $$\varnothing$$ acts as the identity of $$\left({\mathcal P \left({S}\right), \cup}\right)$$.


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

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

The result follows by definition of monoid.

Also see

 * Power Set with Intersection is a Monoid