Set System Closed under Symmetric Difference is Abelian Group

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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.


Proof

G0: Closure

By presupposition on $\mathcal S$, $\left({\mathcal S, *}\right)$ is closed.

$\Box$


G1: Associativity

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

So $*$ is associative.

$\Box$


G2: Identity

From Symmetric Difference with Self is Empty Set, we have that:

$\forall A \in \mathcal S: A * A = \O$

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


Then we have:

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

Thus $\O$ acts as an identity.

$\Box$


G3: Inverses

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

We also noted that

$\forall A \in \mathcal S: A * A = \varnothing$

From Symmetric Difference with Self is Empty Set.

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

$\Box$


Commutativity

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

So $*$ is commutative.

$\Box$


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

$\blacksquare$


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