Set System Closed under Symmetric Difference is Abelian Group

Theorem
Let $\SS$ be a system of sets.

Let $\SS$ be such that:
 * $\forall A, B \in \SS: A \symdif B \in \SS$

where $A \symdif B$ denotes the symmetric difference between $A$ and $B$.

Then $\struct {\SS, \symdif}$ is an abelian group.

By presupposition on $\SS$, $\struct {\SS, \symdif}$ is closed.


 * $\forall A, B, C \in \SS: \paren {A \symdif B} \symdif C = A \symdif \paren {B \symdif C}$ as Symmetric Difference is Associative.

So $\symdif$ is associative.

From Symmetric Difference with Self is Empty Set, we have that:
 * $\forall A \in \SS: A \symdif A = \O$

So it is clear that $\O$ is in $\SS$, from the fact that $\struct {\SS, \symdif}$ is closed.

Then we have:


 * $\forall A \in \SS: A \symdif \O = A = \O \symdif A$ from Symmetric Difference with Empty Set and Symmetric Difference is Commutative.

Thus $\O$ acts as an identity.

From the above, we know that $\O$ is the identity element of $\struct {\SS, \symdif}$.

We also noted that
 * $\forall A \in \SS: A \symdif A = \O$

From Symmetric Difference with Self is Empty Set.

Thus each $A \in \SS$ is self-inverse.

Commutativity

 * $\forall A, B \in \SS: A \symdif B = B \symdif A$ as Symmetric Difference is Commutative.

So $\symdif$ is commutative.

We see that $\struct {\SS, \symdif}$ is closed, associative, commutative, has an identity element $\O$, and each element has an inverse (itself), so it satisfies the criteria for being an abelian group.

Also see

 * Set System Closed under Union is Commutative Semigroup
 * Set System Closed under Intersection is Commutative Semigroup