Define the binary operation $\bigtriangleup$ on $P(A)$ by $X \triangle Y = (X-Y) \cup (Y-X)$.Prove $\langle P(A), \triangle\rangle$ is group.

Question

Let A be a set.Define the binary operation $\bigtriangleup$ on $P(A)$ by $X \bigtriangleup Y = (X-Y) \cup (Y-X)$ for all $X,Y \in P(A)$.Prove that $\langle P(A), \bigtriangleup\rangle$ is an abelian group.

In this question I stuck in showing that the binary operator $\bigtriangleup$ is associative.

I wrote what are $X \bigtriangleup (Y \bigtriangleup Z)$ and $(X \bigtriangleup Y) \bigtriangleup Z$, and then by drawing Venn diagrams, I concluded that these two operations gives different results.I checked my calculations, but couldn't find anything.

I can write here my calculations, but the way that I have tried to solve is pretty tedious way to prove such kind of thing, so I am asking that how can I prove that this binary operator is, in fact, associative ?

Answer 1

$1_{A \Delta B} = (1_A + 1_B) \mod 2$. Now, $$1_{(A \Delta B) \Delta C} = (1_{A \Delta B} + 1_C) \mod 2 = (1_A + 1_B + 1_C) \mod 2$$ From the symmetry, it follows that $(A \Delta B) \Delta C = A \Delta (B \Delta C)$

Answer 2

The easiest way to prove that this operator is associative is probably to note that $\triangle$ is commutative, and to prove that $x \in X \triangle (Y \triangle Z)$ if and only if it lies in all three of $X,Y,Z$ or it lies in precisely one of $X,Y,Z$. It follows that the same is true for the statement $x \in Z \triangle (X \triangle Y) = (X \triangle Y) \triangle Z$.

Answer 3

Note that: $$1_{A\triangle B}=1_A+1_B-2.1_{A}1_{B}$$

Now work out $1_{X\triangle(Y\triangle Z)}$ and $1_{(X\triangle Y)\triangle Z}$ and observe that the outcomes are the same.