I am looking for the proof in the title because I am trying to prove that if $G$ is a cyclic group of order n with generator $a$, then the subgroup generated by $a^m$ has order $n/(m,n)$. I already understand that the order of the group is the same as the order of the generator, but I fail to grasp how the order can be related to gcd. 

I know there is a related question in : [Let $G$ be a cyclic group of order $n$. Prove that every subgroup $H$ of $G$ is of the form $<a^m>$ where $m$ is a divisor of $n$.](https://math.stackexchange.com/questions/1451997/let-g-be-a-cyclic-group-of-order-n-prove-that-every-subgroup-h-of-g-is). But I am interested in the evaluation of the order of the subgroup.