Set System Closed under Symmetric Difference is Abelian Group

Theorem
Let $$\mathcal S$$ be a system of sets.

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

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

Then $$\left({\mathcal S, *}\right)$$ is an abelian group.

G0: Closure
By definition (above), $$\left({\mathcal S, *}\right)$$ is closed.

G1: Associativity

 * $$\forall A, B, C \in \mathcal S: \left({A * B}\right) * C = A * \left(\right)$$ as Symmetric Difference is Associative.

G2: Identity
From Symmetric Difference Self Null, we have that:
 * $$\forall A \in \mathcal S: A * A = \varnothing$$

So it is clear that $$\varnothing$$ is in $$\mathcal S$$, from the fact that $$\left({\mathcal S, *}\right)$$ is Closed.

Then we have:


 * $$\forall A \in \mathcal S: A * \varnothing = A = \varnothing * A$$ from Symmetric Difference with Null and Symmetric Difference is Commutative.

Thus $$\varnothing$$ acts as an identity.

G3: Inverses
From the above, we know that $$\varnothing$$ is the identity element of $$\left({\mathcal S, *}\right)$$.

We also noted that
 * $$\forall A \in \mathcal S: A * A = \varnothing$$

From Symmetric Difference Self Null.

Thus each $$A \in \mathcal S$$ is self-inverse.

Commutativity

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

We see that $$\left({\mathcal S, *}\right)$$ is closed, associative, commutative, has an identity $$\varnothing$$, and each element has an inverse (itself), so it satisfies the criteria for being an abelian group.