Symmetric Difference on Power Set forms Abelian Group

Theorem
Let $$S$$ be a set such that $$\varnothing \subset S$$ (i.e. $$S$$ is not empty.

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

Let $$\mathcal{P} \left({S}\right)$$ be the power set of $$S$$.

Then the algebraic structure $$\left({\mathcal{P} \left({S}\right), *}\right)$$ is an abelian group

G0: Closure
Let $$A, B \subseteq S$$, i.e. $$A, B \in \mathcal{P} \left({S}\right)$$.

Thus we see that $$\left({\mathcal{P} \left({S}\right), *}\right)$$ is closed.

G1: Associativity
$$\forall A, B, C \subseteq S: \left({A * B}\right) * C = A * \left(\right)$$ as Symmetric Difference is Associative.

G2: Identity
$$\forall A \subseteq S: A * \varnothing = A = \varnothing * A$$ from Symmetric Difference with Null and Symmetric Difference is Commutative.

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

G3: Inverses
$$\forall A \subseteq S: A * A = \varnothing$$ from Symmetric Difference Self Null

Thus each $$A \in \mathcal{P} \left({S}\right)$$ is self-inverse.

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

We see that $$\left({\mathcal{P} \left({S}\right), *}\right)$$ 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.