Symmetric Difference on Power Set forms Abelian Group

Theorem

Let $S$ be a set such that $\O \subset S$ (that is, $S$ is non-empty).

Let $A \symdif B$ be defined as the symmetric difference between $A$ and $B$.

Let $\powerset S$ denote the power set of $S$.

Then the algebraic structure $\struct {\powerset S, \symdif}$ is an abelian group.

Proof

From Power Set is Closed under Symmetric Difference, we have that $\struct {\powerset S, \symdif}$ is closed.

The result follows directly from Set System Closed under Symmetric Difference is Abelian Group.

$\blacksquare$

Sources

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): Chapter $4$. Groups: Exercise $12$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.4$
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.1 \ \text{(b)}$
• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Exercise $\text{T}$
• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Definition of Group Structure: $\S 26 \kappa$