> 
> Prove that if a subgroup $H$ of a finite cyclic group $G=\langle a
>  \rangle $ of order $n$ is generated by $a^m$, then $m$ is a divisor of
>  $n$.
> 
> 
> 

It can be easily shown that $m$ is the least positive integer such that $H=\langle a^m\rangle$. But I stuck to show that $m\mid n$. If $o(a^m)=r$, then $a^{mr}=e\implies n\mid mr$ (since $o(a)=n$). I think my approaches going on wrong. How can prove that $m\mid n$?