Order of a subgroup of a finite cyclic group

Question

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.

Answer 1

Let $b=a^m$, and let $d=\gcd(m,n)$, $m'=\frac{m}d$, and $n'=\frac{n}d$. Suppose that $b^k=e$, where $k>0$. Then $a^{km}=e$, so $km$ is a multiple of $n$; say $km=rn$ for some $r>0$. Divide through by $d$: $km'=rn'$, so $km'$ is a multiple of $n'$, where $m'$ and $n'$ are relatively prime. Can you finish it from here?

Answer 2

If $a$ is the generator of $G$ and $|G|=n$, then $a^{n/d}\neq e$. You want to find the smallest $k$ such that $(a^m)^k=e$, that is, what is $|a^m|$?

Answer 3

I ended up relying on a lemma from number theory: If $a$ divides $bc$, then $\frac{a}{d}$ divides $c$, where $d = gcd(a,b)$ (a tedious but simple verification using Fundamental Theorem of Arithmetic).

Then if $(a^{s})^{k} = e$, then $n|sk$. Therefore by the lemma, $\frac{n}{d} | k$, where $d = gcd(n,s)$. This is the remaining step to prove in my OP.

Thanks for your attention all. Sorry if I wasted anyone's time!