Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
You haven't said what $b$ is - presumably you intended $b=a^m$. Then you need to replace $a$ with $b$ in your statements of both Step 1 and Step 2. – Zev Chonoles Jun 1 '12 at 20:44
I have edited the question to reflect your corrections. Sorry that was quite a blunder... :P – roo Jun 1 '12 at 21:58
I'm reading Theorem 6.14 of Fraleigh's book right now, which I assume you're referencing. What part of his proof do you find confusing? Maybe someone here can clarify it. – yunone Jun 1 '12 at 22:09
I'm actually reading an earlier edition to the one you are referring to. I've actually figured it out, and I'll answer my own question in a moment. Thanks very much for checking though! – roo Jun 1 '12 at 22:26
up vote 3 down vote accepted

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?

share|cite|improve this answer
@Dylan: I sure do; thanks. – Brian M. Scott Jun 1 '12 at 22:52

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 $|a^m|$?

share|cite|improve this answer
This is what I meant to say, sorry for the error. – roo Jun 1 '12 at 21:59

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!

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.