The title says it all. I am trying to solve a homework problem, the ultimate goal of which is to prove that $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic for prime $p$. (This is from Herstein's book, if you are interested). I already solved the previous parts, one of which is if $G$ is a finite abelian group such that $x^m=e$ has at most $m$ solutions for any $m$ dividing $n$, then $G$ is cyclic. I am trying to apply this to $(\mathbb{Z}/p\mathbb{Z})^{\times}$ with order $p-1$. This would require showing that $x^m=1$ has at most $m$ solutions, which would prove that $(\mathbb{Z}/p\mathbb{Z})^{\times}$. One thing I noticed is the set $\{x\in (\mathbb{Z}/p\mathbb{Z})^{\times}: x^m=e\}$ is a subgroup of $(\mathbb{Z}/p\mathbb{Z})^{\times}$. I was thinking that I can show that this subgroup cannot have more than $m$ elements using Lagrange's Theorem. The problem is, I do not know anything about the divisors of $p-1$, nor do I ever use the fact that $p$ is prime. I cannot use anything other than Lagrange's Theorem and basic facts about groups and subgroups, as well as Euler's Theorem and Fermat's Little Theorem.
Any suggestions? Please do not give full solutions, but hints are welcome.
Edit: I do not have much experience with abstract algebra so if I used something that you had mentioned not to please comment.
Hint-1
Think about the number of $x \pmod{p}$ having some order $d$ where $d$ divides $p-1$.This is well known.
Hint-2
If there exists an $ord_p(x) = z$ where $z|m$, then $x^m \equiv 1 \pmod{p}$
Proceed iff hints fail
Solution :
Answer to hint one is $\phi(d)$, the proof can be found here -
Prove that exactly $\phi(d)$ elements have order $d$
When $d \mid n$, the number of elements of order $d$ in a cyclic subgroup of order $n$ is $\phi (d)$
The total solutions to $x^m \equiv 1 \pmod{p}$ will be the sum of all $\phi(d)$ where $d|m$ and $d|(p-1)$ (why?,comment if you have a problem with this statement).
This means that $\sum_{d|m}\phi(d)= m$(This is also well known and is $m$ itself) is an upper bound, as it may or may not satisfy $d|p-1$.
It will satisfy when $m|p-1$.
Therefore, Number of solutions $\leq m$