Power Set with Intersection is Commutative Monoid
Theorem
Let $S$ be a set and let $\powerset S$ be its power set.
Then $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.
The only invertible element of this structure is $S$.
Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cap}$ cannot be a group.
Proof
From Power Set is Closed under Intersection:
- $\forall A, B \in \powerset S: A \cap B \in \powerset S$
From Set System Closed under Intersection is Commutative Semigroup, $\struct {\powerset S, \cap}$ 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 commutative monoid.
$\blacksquare$
Also see
Sources
- 1951: Nathan Jacobson: Lectures in Abstract Algebra: Volume $\text { I }$: Basic Concepts ... (previous) ... (next): Chapter $\text{I}$: Semi-Groups and Groups: $1$: Definition and examples of semigroups: Example $7$
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 4.4$. Gruppoids, semigroups and groups: Example $77$
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.4$
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{T}$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 29$. Semigroups: definition and examples: $(3)$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $5$: Semigroups: Exercise $2$