In an exercise I am asked to prove the following:
Show that if $G$ is a finite cyclic group with order $n$, the equation $x^m=e$ has $m$ solutions for all $m\mid n$, where $e$ is the neutral element.
My attempt was the following (I think that it is correct): If an element $a$ of order $m\mid n$, it generates a subgroup with order $m$, in which every element satisfies that equation. Up to this point, there is a problem: is this cyclic subgroup of that order unique? If this is not true, then there would be the double minus one solutions to that equation. Therefore, the only issue that needs to be solved would be undoubtly that uniqueness.
It would be easy to think that every cyclic group is isomorphic to a group of type $\mathbb{Z}_n$ (a fact that I actually know), and in that case I would not have any doubt, but the thing is that I have not seen that in class and I would like to avoid using it if possible.
With your reasoning you have already shown that there are at least $m$ solutions
To prove the other implication, the most straightforward is I think to use the definition of the order of an element. First consider $c$ a generator of $G$. I am going to assume that you have seen that the order of $x$ is $n$.
Now consider $x$, such that $x^m=e$. Write $x=c^k$ with $k$ between $0$ and $n-1$, then $c^{km}=e$ and so $n$ has to divide $km$, but then $\frac{n}{m}$ divides $k$ and there are only $m$ multiples of $\frac{n}{m}$ that are between $0$ and $n-1$ thus there cannot be more than $m$ solutions to the equation.