Proof showing power set is group using isomorphism today we discussed a proof about power set is a group with the operation symmetric difference. We thought how to show symmetric difference is associative. Our professor had the idea to use an isomorphism. 
Be power set $P(X)$ from the set $X$.
We used the isomorphism $$ \psi : \; P(X) \rightarrow \text{Map}(X; \mathbb{Z} / 2 \Bbb {Z} ) \\ A \mapsto \left( x \mapsto \begin{cases} 1 & x \in A \\
0 & x \notin A \end{cases} \right)  $$
We showed $ \text{Map}(X; \mathbb{Z} / 2 \Bbb {Z} )$ is group with the operation $ (f + g)(x) = f(x) + g(x) $.
Then we showed $ \psi (A \triangle B) = \psi (A) + \psi (B) $
I'm not feeling totally comfortable how he argued. Maybe someone could explain it further.
Thanks!
 A: This is a example of what we call "transfer of structure" in algebra. You have two sets : $P(X)$ and $\mathrm{Map}(X,\mathbb{Z}/2\mathbb{Z})$, and on the second one you have group structure (addition of maps). Then you use a bijection $\psi: P(x)\to \mathrm{Map}(X,\mathbb{Z}/2\mathbb{Z})$ to identify the two sets, and you say that since you had a group operation on the second one, you can define one on the first one through this bijection.
So generally, if you have a bijection $\psi: A\to B$ between two sets, and you have a group structure on $B$ (say the operation is denoted $+$), then you cand efine a group operation on $A$ by $\psi(a+b) = \psi(a) + \psi(b)$, or if you prefer you define $a+b := \psi^{-1}(\psi(a) + \psi(b))$.
The fact that this defines a group structure on $A$ is easily checked axiom by axiom, from the fact that it is a group structure on $B$.
Since in your case the operation it defines on $P(X)$ coincides with symmetric difference, it shows that symmetric difference is indeed a group law.
