TITLE: Discrete uniform probability on a sample space of prime cardinality
QUESTION [7 upvotes]: My question is to show that if I have a fair die with $p$ faces, where $p$ is prime, and the experiment consists of rolling it once, no two proper events can be independent.
Here is my approach:
Suppose $A$ and $B$ are independent events. Then $$P(A \cap B) = P(A) * P(B) = \frac{|A|}{p} * \frac{|B|}{p} = \frac{|A||B|}{p^{2}} < 1$$
Not sure where a contradiction would happen to know how to proceed.
Any help?

REPLY [4 votes]: Neat question!
The outcomes of the die are $\{1,2,\cdots,p\}$. Suppose we have a pair of proper events $A,B$ and we assume they are independent. I'm assuming here that "proper" means neither event is empty and neither event is equal to the whole space. If $A\cap B=\{\emptyset\}$ we immediately get a contradiction, so suppose that $A\cap B=C$, for some proper event $C$. We have:
$\frac{|C|}{p}=P(A\cap B)=P(A)P(B)=\frac{|A||B|}{p^2}$. 
Rearranging this gives:
$$p|C|=|A||B|.$$
Since neither $A$ nor $B$ is the full space, $0<|A|<p$ and $0<|B|<p$. Since $|C|>0$ and $p$ is prime, $p$ must divide either $|A|$ or $|B|$. This is clearly a contradiction. Note that if $p$ was not prime (say, $p=4$), then you would only need the factors of $p$ to divide either $|A|,|B|$, so $p$ being prime is necessary.