Let $G$ be a cyclic subgroup of order $n$, generated by say $a\in G$ where the identity of $G$ is labelled $e$. Let $H$ be the cyclic subgroup of $G$ generated by some $a^{m}\in G$. Then I want to show that the order of $H$ is equal to $n/d$ where $d$ is the greatest common divisor of $n$ and $m$.
So far I've got:
This can be reduced to the statement: If $b = a^{m}$, the smallest positive integer $k$ such that $b^{k} = e$ is $n/d$, where $d$ is the greatest common divisor of $n$ and $m$.
Step 1: I've shown that $b^{n/d} = e$ (the easy part).
But this step is giving me problems.
Step 2: Show that there is no $k < n/d$ such that $b^{k} = e$.
I'm reading this from the book "A First Course in Abstract Algebra" by John B. Fraleigh, but I cannot follow his proof. I can supply his argument if anyone requests it.