Set System Closed under Symmetric Difference is Abelian Group

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

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

where $A * B$ denotes the symmetric difference between $A$ and $B$.

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

$\text G 0$: Closure
By presupposition on $\SS$, $\struct {\SS, *}$ is closed.

$\text G 1$: Associativity

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

So $*$ is associative.

$\text G 2$: Identity
From Symmetric Difference with Self is Empty Set, we have that:
 * $\forall A \in \SS: A * A = \O$

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

Then we have:


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

Thus $\O$ acts as an identity.

$\text G 3$: Inverses
From the above, we know that $\O$ is the identity element of $\struct {\SS, *}$.

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

From Symmetric Difference with Self is Empty Set.

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

Commutativity

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

So $*$ is commutative.

We see that $\struct {\SS, *}$ 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