# Composite State Space For Probability

I want to go about the proof that, given state space $\Omega$ and events $A$ and $B$, if $A$ and $B$ are nontrivial events ($|A| \neq 0 \wedge |A| \neq |\Omega| \wedge |B| \neq 0 \wedge |B| \neq |\Omega|$) and are independent, than we know that $|\Omega|$ must be a composite number. I considered building this proof by starting off with the assumption of independence, i.e. $$P(A \cap B) = P(A)P(B).$$ I can easily expand this out to be the case that $$P(A \cap B) = \frac{|A \cap B|}{|\Omega|} = P(A)P(B) = \frac{|A|}{|\Omega|} \frac{|B|}{|\Omega|}$$ $$\implies \frac{|A \cap B|}{|\Omega|} = \frac{|A||B|}{|\Omega|^2}.$$ I have a feeling that I can apply some sort of result about the properties of $\Omega,$ but I am not sure where to move from here. Any assistance would be greatly appreciated.

I assume we are under the assumption that $|\Omega|$ is finite.
Additionally, your computations ($P(A)=|A|/|\Omega|$) assume that every state has equal probability, which you did not state explicitly.
Your computations show $|\Omega|=|A||B|/|A \cap B|$. Note that $|A \cap B| \le \min(|A|,|B|)$. So, if $|\Omega|$ is prime, then either $|A \cap B|=|A|$ and $|\Omega|=|B|$; or $|A \cap B| = |B|$ and $|\Omega|=|A|$, contradicting your assumption that $A$ and $B$ are nontrivial.