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

I want to show the following statement:

Let $G$ be a nilpotent group and $a,b\in G$ such that there exist $m,n\in\mathbf{N}_{>0}$ such that $\text{gcd}(m,n)=1$ and $a^m=b^n=1$. Then $ab=ba$.

If $G$ is finite, this is clear to me, since I know that then $G$ is the direct product of its Sylow subgroups.

I have found a sketch of a proof in Hall's Theory of groups: If $G=G_1,G_2,\ldots$ is the lower central series of $G$, show that $[a,b]\in G_i$ for any $i$, which then implies $[a,b]=e$. Hall even gives another hint: If $[a,b]\in G_i$, show that $[a,b]^m\in G_{i+1}$ and $[a,b]^n\in G_{i+1}$. Unfortunately I can't do that. I managed the case $m=2$: $$a^{-1}b^{-1}aba^{-1}b^{-1}ab=a^{-1}(b^{-1}a^{-1}ba)a(a^{-1}b^{-1}ab)=[a,[a,b]],$$ but I don't see how to generalize this.

Another hint would be nice.

share|cite|improve this question
Are you familiar with the commutator identities, like $[ab,c] = b^{-1}[a,c]b[b,c]$? They imply that, if $[a,b]$ is in the centre of the group, then $[a^k,b] = [a,b^k] = [a,b]^k$ for any $k$. Now you know that the image of $[a,b]$ in $G/G_{i+1}$ is in the centre of $G/G_{i+1}$ ... – Derek Holt Nov 21 '11 at 22:55
@Stefan Walter This question is very close to this one… – Plop Nov 22 '11 at 0:02
@Plop: And your answer there is the perfect answer to this question. Why I didn't realize this before, I don't know. Should this question be closed as a duplicate? Alternatively, I would also accept an answerification of Derek's comment. Thank you both! – Stefan Nov 22 '11 at 19:58
@Derek: It seems like I can't notify two people in one comment. So I added this one. – Stefan Nov 22 '11 at 20:00

Since you are okay in the finite case, can you show that the subgroup generated by $a$ and $b$ is finite?

share|cite|improve this answer
I'd be interested in how you would prove this. Do you have a concrete idea? – j.p. Nov 22 '11 at 12:54
@James, I think you misunderstood, the theorem only holds for finite groups. I thought I was clear in stating that. So the subgroup generated by $a$ and $b$ is automatically finite. – Nicky Hekster Nov 23 '11 at 22:22

You might want to note that the finite groups $G$ having the property that for all $a, b \in G$ with $gcd(order(a), order(b))=1$, one has $order(ab)=order(a).order(b)$, are the nilpotent groups.
In other words, for a finite group $G$ the order function is "multiplicative" iff $G$ is nilpotent.

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.