Hello I have a particular question, about the proof of the following theorem:
Theorem: If $r|p-1$, with $p$ an odd prime, there are $\phi(r)$ incongruent integers which have order $r$ modulo $p$.
Proof: Suppose that $g$ has order $r$ modulo p. Then, by the previous theorem, the $r$ integers $g^0 , g^1, ... , g^{r-1}$ are incongruent modulo $p$ and satisfy the equation $x^r \equiv 1 \text{ mod } p$. Thus, the sequence $g^n \text{ mod } p$ is periodic and $n$ is defined modulo $r$, with $n=0,1,...,r-1$. Then, for $gcd(b,r)=1$, $g^{bn} \text{ mod } p$ is a simple permutation of the sequence $g^n \text{ mod } p$. When $gcd(b,r) \neq 1$ the sequence $bn \text{ mod } r$ will contain repititions and therefore the corresponding integers $g^b$ will be of order less than $r$. Thus, we have either zero or $\phi(r)$ incongruent roots of order $r$. Since all integers in the set $1,...,p-1$ have some order, the total number of roots, for all divisiors of $p-1$, is equal to $p-1$. We note, with 2.35, that $\sum_{r|p-1} \phi(r) = p-1$. Thus, there are $\phi(r)$ roots of order $r$ for each divisor $r$ of $p-1$
My question now, starting from this proof, is how to come to the conclusion that ..., we have either zero or $\phi(r)$ incongruent roots of order $r$ holds.
I think that this refers to the exponents, so that one can use $\phi(r)$ to find the number of exponents of $g$, so that, for example, $g^{bn}$ is a permutation and not a repetition. I'm not entirely sure about this, so I'm happy to ask here! Thanks a lot!