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
From Power Set is Closed under Union:
 * $\forall A, B \in \mathcal P \left({S}\right): A \cup B \in \mathcal P \left({S}\right)$

From Set System Closed under Union is Commutative Semigroup, $\left({\mathcal P \left({S}\right), \cup}\right)$ is a commutative semigroup.

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

It remains to be shown 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 commutative monoid.

Also see

 * Power Set with Intersection is Commutative Monoid