Set System Closed under Symmetric Difference is Abelian Group

From ProofWiki
Jump to navigation Jump to search

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.


Proof

Group Axiom $\text G 0$: Closure

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

$\Box$


Group Axiom $\text G 1$: Associativity

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

So $\symdif$ is associative.

$\Box$


Group Axiom $\text G 2$: Existence of Identity Element

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.

$\Box$


Group Axiom $\text G 3$: Existence of Inverse Element

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.

$\Box$


Commutativity

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

So $\symdif$ is commutative.

$\Box$


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.

$\blacksquare$


Also see