Proof on order of element of cyclic group must divide the order of the group. I am proving the following problem. I finished proving it but do not know is it sufficient enough.

Problem: Prove that the order of an element in a cyclic group $G$ must
divide the order of the group.

My original proof:

Let the order of $G$ be $n$, and the generator of $G$ be $a$, then $a^n=e$. Then, for each element $b\in G, \exists k, m \in \mathbb{Z}$ s.t. $|b|=k$ and $b=a^m$. It must follow that:
$b^{k}=(a^{m})^{k}$ $\implies$ $e=a^{mk}$ $\implies$$a^n=a^{mk}$ $\implies$ $n=mk$.
Since $m\in \mathbb{Z}$, then $k|n$.

My edited proof:
Let the order of $G$ be $n$, and the generator of $G$ be $a$, then $a^n=e$. Then, for each element $b\in G, \exists k, m \in \mathbb{Z}$ s.t. $|b|=k$ and $b=a^m$. It must follow that:
$k=\frac{n}{gcd(n,m)} \implies n=(k)gcd(n,m)$
Since $gcd(n,m)\in \mathbb{N}$, then $k|n$.
My question is that I have seen lots of people used Division Algorithm to prove this, but it is only when the element is the generator, here it is any element. Yet, I don't know my prove is sufficient or not. Any suggestion is appreciated.

Your proof is wrong because you state that $a^n=a^{mk}\implies n=mk$. Why do you think that this is true?
However, $a^n=a^{mk}=e\implies n\mid mk$, but this is not enough to deduce that $k\mid n$, which is what you want to prove.
