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

Some books use this property to characterized the finite nilpotent group:

A finite group $G$ is nilpotent if and only if whenever $a,b\in G$ are elements of finite order with $\gcd(\mathrm{ord}(a),\mathrm{ord}(b))=1$ then $ab=ba$.

I wonder what happen when $G$ is infinite. Does it still hold, become just an "if" statement, just an "only if" statement, or neither?

Edit: what I meant is that the property holds whenever $a$ and $b$ are of finite order (even when $G$ is infinite)

share|cite|improve this question
How do you define the $\gcd$ if either $\text{ord}(a)$ or $\text{ord}(b)$ is infinite? – Qiaochu Yuan Apr 30 '11 at 17:59
If $a$ and $b$ are elements of finite order in any nilpotent group with $gcd(ord(a),ord(b))=1$, then they commute. – user641 Apr 30 '11 at 18:09
@Quaichu Yuan : I guess an element of finite order exists in an infinite group, for example, infinite direct product of a finite group. @Steve D: so the "only if" part also true for infinite nilpotent group? – Ajat Adriansyah Apr 30 '11 at 18:38
The upper triangular matrices in $\mathrm{GL}_2(\mathbb{R})$ with positive diagonal coefficients form a solvable, non-nilpotent group, which has no finite order elements other than the identity, so the "if" part is not true in the infinite case. – Plop Apr 30 '11 at 18:54
up vote 6 down vote accepted

If $G$ is nilpotent, and $a$ and $b$ are element of relatively prime finite order, then $a$ and $b$ commute; this holds whether $G$ is finite or not. (That is, the condition is necessary, so "only if" holds in all cases).

Note. Plop's argument below is much easier than mine here. This is probably a result of When-You-Have-a-Hammer Syndrome. I've used commutator calculus a lot in my work, so it tends to be one of the first tools I reach for when dealing with these kinds of problems.

Since an element of finite order can be written as a product of elements of prime power order, it suffices to consider the case where $a$ and $b$ are of prime power order, $\mathrm{ord}(a) = p^n$, $\mathrm{ord}(b) = q^m$, $p$ and $q$ distinct primes, $n,m$ positive integers.

The key is to use the following consequences of P. Hall's collections process (you can see the basic results in Marshall Hall's Group Theory book; I'm quoting these theorems from R.R. Struik's Nilpotent products of cyclic groups, Canad. J. Math. 12 (1960), 447-462.

Theorem. Let $x$ and $y$ be any two elements of a group. Ldt $u_1,\ldots,u_n,\ldots$ be a fixed sequence of commutators in $x$ and $y$ of non-decreasing wight; that is, $u_1 = [x,y]$, $u_2=[[x,y],x]$, $u_3 = [[x,y],y]$, etc. Then $$(xy)^n = x^ny^n u_1^{f_1(n)}u_2^{f_2(n)}\cdots u_t^{f_t(n)}\cdots$$ where $$f_i(n) = a_1\binom{n}{1}+a_2\binom{n}{2}+\cdots+a_{w_i}\binom{n}{w_i},$$ $a_i$ are rational integers, and $w_ii$ is the weight of $u_i$ as a commutator in $x$ and $y$. The first formula is an identity if the group is nilpotent; otherwise, it can be considered as giving a series of approximations to $(xy)^n$ modulo successive terms of the lower central seres.

Theorem. Let $\alpha$ be a fixed integer, and let $G$ be a group such that the $n$th term of the loweer central series of $G$ is trivial. Then if $b_j\in G$ and $r\lt n$, $$[b_1,\ldots,b_{i-1},b_i^{\alpha},b_{i+1},\ldots,b_r] = [b_1,\ldots,b_r]^{\alpha} v_1^{f_1(\alpha)}v_2^{f_2(\alpha)}\cdots$$ where the $v_k$ are commutators in $b_1,\ldots,b_r$ of weight greater than $r$, and every $b_j$, $1\leq j\leq r$ appears in each commutator $v_k$. The $f_i$ are of the form $$f_i(n) = a_1\binom{n}{1}+a_2\binom{n}{2}+\cdots+a_{w_i}\binom{n}{w_i}$$ where $w_i$ is the weight of $v_i$ minus $r-1$.

Using these two theorems and decreasing induction, it is easy to verify that if $G$ is nilpotent, then for a sufficiently large power of $p$ we have $$[a^{p^N},b] = [a,b^{p^N}].$$ For a sufficiently large $N$, the left hand side is trivial; on the right hand side, since $b$ is of order prime to $p$, it follows that $b$ is a power of $p^N$, so the fact that $[a,b^{p^N}]$ is trivial, which is equivalent to the fact that $a$ commutes with $b^{p^N}$, implies that $a$ also commutes with any power of $b^{p^N}$, in particular that $a$ commutes with $b$.

So if $G$ is nilpotent, whether finite or infinite, and $a,b\in G$ are two elements of finite coprime order, then $a$ and $b$ commute.

The converse is true for finite groups (a finite group is nilpotent if and only if it is a direct product of $p$-groups, which implies that two elements of coprime order will commute. But the converse is false for infinite groups. In addition to Plop's example, here's a torsion example in which the condition is satisfied nonvacuously:

Fix a prime $p$. For each $n\gt 1$, let $G(p,n)$ be a nilpotent group of order $p^n$ and maximal class (that is, of class $n-1$); such groups exist. Let $$\mathfrak{G}_p = \bigoplus_{n=2}^{\infty} G(p,n).$$ Then $\mathfrak{G}_p$ is torsion, but is not nilpotent, since the $n$th term of the lower central series of $\mathfrak{G}_p$ is the direct sum of the $n$th terms of the lower central series of the $G(p,m)$ for each $m$, so it is not trivial for every $n$. Now take two distinct primes, $p$ and $q$, and let $\mathcal{G}=\mathfrak{G}(p)\oplus\mathfrak{G}(q)$. This is torsion, has nontrivial elements of coprime order, any two elements of coprime order commute, but $\mathcal{G}$ is not nilpotent (since it has non-nilpotent sugroups).

So the condition is necessary in general, and sufficient in the finite case for nilpotency.

share|cite|improve this answer
A small refinement: Must such a (torsion) group be locally nilpotent? (No. Tarski monster or direct products of McLain's groups; p-groups need not be nilpotent, but such a group must be a direct product of its Sylow p-subgroups.) – Jack Schmidt Apr 30 '11 at 21:06
Thank you very much :) – Ajat Adriansyah May 1 '11 at 2:11

There is also a simple proof of the "only if" part by induction on the smallest $n$ such that $G^{(n)}=\{e\}$ (where $G^{(0)}=G$ and $G^{(n+1)}=[G,G^{(n)}]$). If $G$ is abelian it is obviously true. If $n>1$, let $C$ be the center of $G$. Then $G/C$ has a smaller $n$, and the order of $\bar{a} = a \mod C$ divides $o(a)$ (the order of $a$), and the same for $b$, so $ab=ba \mod C$, that is, $[a,b]=aba^{-1}b^{-1} \in C$. But then $a$ and $b$ commute to $[a,b]$, so $[a,b]^{o(a)}=a^{o(a)} \left(ba^{-1}b^{-1}\right)^{o(a)}=a^{o(a)} \left(ba^{-o(a)}b^{-1}\right)=e$, and similarly $[a,b]^{o(b)}=e$, and a Bezout relation between $o(a)$ and $o(b)$ allows us to conclude that $[a,b]=e$.

share|cite|improve this answer
Heh; that's what I get for being very worried usually about just exactly how large an $N$ I need for $[a^{p^N},b]$ to equal $[a,b^{p^N}]$. This is of course much simpler. – Arturo Magidin Apr 30 '11 at 21:02
@Plop I only have one doubt in your proof, please answer, why can we assume $G/C$ has smaller $n$, |$G/C$| is smaller than |$G$| but still their derived lengths can be equal? – Bhaskar Vashishth Sep 25 '14 at 0:07

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.