# nilpotent group is almost abelian - counterexample for infinite order case

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)

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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

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.

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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 $[a,b]=e$.

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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 at 0:07