Is there a short way to prove that $(\mathcal{P}(X),\Delta)$ is a commutative group

Question

Is there a way to prove that $(\mathcal{P}(X),\Delta)$ is a commutative group, where $\Delta$ denotes the symmetric difference, without doing tedious computations?

Answer 1

Define for $A\in\mathcal P(X)$ the characteristic function $\chi_A $ on $X$ by $$\chi(x)=1\;\text{if}\; x\in A\;\text{and $0$ otherwise}$$ and prove that $$\chi_A\chi_B=\chi_{A\cap B}$$ $$\chi_A+\chi_B-\chi_{A\cap B}=\chi_{A\cup B}$$ $$\chi_{A^c}=1-\chi_A$$ and use this function to answer your question.

Answer 2

If you think of $\Delta$ as follows:

$A\Delta B$ is the set of things that are in exactly one of $A$ and $B$

things flow rather nicely. Associativity and commutativity are immediate, and once you identify the identity and complement, they are straightforward too.

Answer 3

To continue the thought I made in a comment on Sami Ben Romdhanes answer, first prove that $\mathbb Z_2$ is a additive group. Then consider the set $$ \operatorname{Maps}(X,\mathbb Z_2)=\left\{\, f\colon X\to\mathbb Z_2\, \right\} $$ of maps from $X$ to $\mathbb Z_2$. This is again a commutative group with pointwise addition. Now consider the map \begin{align} \chi : \mathcal P(X) &\longrightarrow \operatorname{Maps}(X,\mathbb Z_2) \\ A &\longmapsto \chi_A \end{align} where $\chi_A$ is the characteristic function of $A\subseteq X$. This map is a bijection and we have $$ \chi_{A\Delta B} = \chi_A + \chi_B, $$ thus $(\mathcal P(X),\Delta)$ and $(\operatorname{Maps}(X,\mathbb Z_2), +)$ are isomorphic groups.