My question is related to this question I asked last year. I am just getting around to picking up the problem again, and normally, I would just ask a follow-up question to the answer that was given, but the person who wrote it is away from MSE due to medical reasons, and I am in urgent need of an answer.
My original question was:
Let $p$ be a prime number and $\gcd(p,n)=1$. Define an equivalence relation on $\mathbb{Z}_{p}$ as follows: $x \sim y$ iff $n^{r}x = n^{t}y$ for some $r,t \geq 0$. Let $m$ be the number of equivalence classes of this equivalence relation. Prove that $m-1$ is a divisor of $p-1$.
I followed the answerer's advice and showed that $x \sim 0$ iff $x = 0$, and now am working on the next step,which is to show that any two equivalence classes of non-zero elements of $\mathbb{Z}_{p}$ have the same number of elements, call it $k$.
However, I am not sure how to do this part. Could someone please let me know how I am supposed to do this?
Thank you.
Consider the multiplicative group $\mathbb{Z}_p^\times$ (discard $0$). Then the equivalence relation can be written, for nonzero elements, $$ xy^{-1}\in\langle n\rangle $$ so the equivalence relation is just the one defining the quotient group $\mathbb{Z}_p^\times\big/\langle n\rangle$. In particular the equivalence classes have the same number of elements. In the whole set $\mathbb{Z}_p$ there's also the class of $0$, with just one element.