Proof that the $(\mathcal P (\mathbb N),\triangle)$ is an abelian group? I'm stuck trying to figure out how to prove that $(\mathcal P (\mathbb N) \space , \space \triangle)$ is an abelian group? I know the definition, but I'm confused how to incorporate the fact that I'm dealing with a powerset instead of just the integers or natural numbers, so I'm seeking help to understand how proofs on sets works.
Also, I'm using $\triangle$ to mean symmetric difference. 
 A: Possibly the best approach is to note that if $A$ is an arbitrary set, and you associate to each element of $B \in \mathcal P ( A)$ its characteristic function
$$
\textbf{1}_{B} : {A} \to \mathbf{Z}_{2},
f(a) = \begin{cases} 1 & \text{if $a \in B$},\\ 0 & \text{if $a \notin B$,}\end{cases}
$$
then symmetric difference becomes (pointwise) sum of characteristic functions,
$$
\textbf{1}_{B} + \textbf{1}_{C} = \textbf{1}_{B \Delta C}
$$
A: Just check the axioms. Prove that symmetric difference is commutative (which is clear from its definition) and associative (which takes a little work). Prove that there’s an identity, i.e., a set $Z$ such that $Z\triangle A=A$ for all $A\in\wp(\Bbb N)$; there’s exactly one set that has that property, and it’s not hard to find. Finally, check that every $A\in\wp(\Bbb N)$ has an inverse. A hint for both of those last parts: what’s $A\triangle A$?
A: In fact $(\mathcal P (X) \space , \triangle)$ is a group for any $X$:


*

*closure is immediate since $\triangle : \mathcal P (X) \times \mathcal P (X) \to \mathcal P (X)$

*identity: $x \triangle \{\} = x$

*inverses: $x \triangle x = \{\}$ so everything is its own inverse.

*associativity: $(x \triangle y) \triangle z = x \triangle (y \triangle z)$ is obvious from a venn diagram.


and it's abelian because $\triangle$ is symmetric.
