Find all finite sets $A$ so that $A\times\mathcal P(A) =\mathcal P(A)\times A$, where $\mathcal P(A)$ is the power set of $A$. 
Find all finite sets $A$ so that $A\times\mathcal P(A) =\mathcal P(A)\times A$, where $\mathcal P(A)$ is the power set of $A$.

Now I'm a beginner at Discrete Math so I'm not sure how to tackle this problem. I was expecting that a set with elements would provided to prove this. My best hunch is that all sets should work since its the same on both sides of the equal sign. If you could help me find out how work through the question, instead of just an answer that would be great!
 A: Hint 1: $\mathcal P (A)$ always has more elements than $A$. In particular $A \neq \mathcal P(A)$.
Hint 2: If $A \times B =B \times A$ then you can prove that either $A=B$ or one of $A,B$ is empty. 
Indeed, if you assume by contradiction that you have $A \neq B$ and that neither $A$ nor $B$ is empty, then you can find an element $x$ which is in exactly one of the sets $A$ or $B$. Now, by using that neither $A$ nor $B$ is empty, you can pick some $y$ in the other set. Look at $(x,y)$.
Hint 3 Conclude that $A=\emptyset$ is the only solution.
A: Assuming that "x" is the Cartesian product of two sets, we have the definition 

For two sets $A$ and $B$, the cartesian product is given by 
  $$A\times B=\{ (a,b):a\in A, b\in B\}$$

Suppose that $A\times P(A)$ for some set $A$. 
Then, each element of $P(A)$ must be in $A$. 
However, by Cantor's theorem, we have that $|A|<|P(A)|$ for any $A$, thus there is at least one element of $P(A)$ that is not in $A$.
Thus it cannot be true that $A\times P(A)=P(A)\times A$ for any set $A$. 
