I am reading I.N.Herstein-Topics in Algebra
Question:
For $G$ a finite group. Let $|G|=n$, then $O(a)|n \enspace \forall a\in G$.
Note:
I have already proved that if $G$ is a finite group, then $\exists N\in \mathbb{N}$ such that $a^N=e \enspace \forall a \in G$ and $N\leq n$.
Proof:
Suppose $\exists a\in G$ and a minimal $m\in \mathbb{N}$ such that $a^m=e$ and $m\nmid n$.
Given $G$ is finite. Therefore, we have $$a^n=e\\\text{By using division algorthim,}\enspace n=km+r\enspace\text {for some k}\in \mathbb{N} \\ \implies a^{n-km}.\{a^m\}^k=e \\ \implies a^{n-km}.e=e \\ \implies a^{n-km}=e$$ This Contradicts minimality of $m$.
Therefore, $O(a)|n \enspace \forall a\in G$.
This also proves that $N|n$.
Is My proof Correct? Any Improvements and suggestions are appreciated.