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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.

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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
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@Stefan Walter This question is very close to this one math.stackexchange.com/questions/36056/… –  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 Walter 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 Walter Nov 22 '11 at 20:00

2 Answers 2

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

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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.

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