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 $n$.

Question

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 $m\mid n$?

Answer 1

It's not true, even for proper subgroups. $\langle 8 \rangle \subsetneq \Bbb Z / 12 \Bbb Z$ but $8 \not | 12$. The best you can hope for is $(m, n) \gt 1.$

Answer 2

That statement is wrong, as Robert Shore's answer shows. What is true is that if $a^m$ generates a subgroup $H$ of $G=\{1,a,a^2,\dots,a^{n-1}\}$, then the order of this subgroup must divide the order of the group $G$ (i.e. $|H|$ divides $n$). But in this case we do not even need that $G$ is cyclic--the order of every subgroup of any group always has order dividing the order of the group. This fact is known as Lagrange's Theorem.