From: Jim Howie 
Date: Mon, 21 Oct 1996 11:07:11 +0100
 
Let $G$ be the free nilpotent group of class two on 4 generators a,b,c,d.
Then [a,b][c,d] is not a commutator.
 
Jim Howie.

------------------------------------

From: Derek Holt 
Date: Mon, 21 Oct 1996 12:13:44 +0100

Jim Howie gave an simple counterexample to this.
One can also build finite counterexamples as follows.
 
Let  Q  be an elementary abelian p-group of order p^n with generators
q_1,q_2,...,q_n.
 
Then one can form a class two nilpotent group G, with generators
g_1, ..., g_n (corresponding to the q_i), with G/G' ~= Q, where G' has
order p^x, for x=n(n-1)/2, and G' is generated by the commutators
[g_i,g_j] (i 5.
 
Derek Holt.

--------------------------------------

From: R M Bryant 
Date: Mon, 21 Oct 96 13:00:58 BST

Dear Juergen,
One can show that for every natural number n there is a group G such
that G is nilpotent of class 2 and there are elements of G' which
cannot be written as the product of n commutators. One example of
this (which is quite widely known) is to take G to be the largest nilpotent
group on m generators (m sufficiently large) in which every element has
order dividing 4.  From the known structure of this group, G/Z(G)
has order 2^m, but G' has order 2^{m(m-1)/2}.  It follows that the
number of elements of G' which can be written as a product of n or
fewer commutators is at most 2^{2mn}  -- I hope I have got all these
numbers right.  The result then follows.
Roger Bryant

---------------------------

Date: Thu, 24 Oct 1996 09:50:21 +0100
Andreas Caranti 

Just a few comments (written in Plain \TeX\ notation) to the messages of Jim
Howie, Derek Holt and Roger Bryant.
 
There is a four-generator example in Huppert's Endliche Gruppen I, III.1.12.
 
Let $Q$ be an elementary abelian $p$-group of order $p^{n}$, with $n > 1$, and
take $p > 2$, say. Then the Schur multiplier $M$ of $Q$ has order $p^{n \choose
2}$ and is naturally isomorphic to the exterior square $Q \wedge Q$ (see
Robinson's ``A Course in the Theory of Groups'', 11.4.16). Among all the
possible covering groups of $Q$, one may choose one of exponent $p$, as $p > 2$.
 
Commutators in the covering group correspond to decomposable tensors in $M$,
that is, to elements of the form $u \wedge w$, for $u, w \in Q$. Such tensors
are well-known to form an algebraic variety, which in this case is the
Grassmann manifold of lines in the
$n-1$-dimensional projective space on $Q$. Equations for this variety are
well-known, just ask your local geometer. There are no equations for $n = 2$,
and for
$n = 3$ (by the point-line duality in projective planes), so all elements are
commutators here. For
$n = 4$, there is the single equation
$$
 x_{1,2} x_{3,4} - x_{1,3} x_{2,4} + x_{1,4} x_{2,3}  = 0,
$$                             
for $x = \sum x_{i, j} v_{i} \wedge v_{j} \in M$, with the $v_{i}$ a base for
$Q$. This is mentioned in Huppert's example refered to above.
 
For every element $x \in M$ there exists a unique natural number $m$, and $2m$
linearly independent elements of $Q$, $u_{i}, w_{i}$, for $1 \le i \le m$, such
that
$x = u_{1} \wedge w_{1} + \dots + u_{m} \wedge w_{m}$. In particular, there is
an element of $M$ that cannot be written as the sum of less than $\left\lfloor
{n
\over 2} \right\rfloor$ decomposable tensors.
 
All this may be actually useful in some group-theoretic contexts. You may want
to see the following two papers, in which I've been involved:
 
\item{$\bullet$} Ann.\ Mat.\ Pura Appl.\ 134 (1983), 93--146
 
\item{$\bullet$} Bull.\ Austral.\ Math.\ Soc.\ 30 (1984), 67--71
 
Best,
 
Andreas Caranti

---------------

Subject: class 2 commutators
Date: Sat, 26 Oct 1996 21:11:44 PDT
From: Bill Bogley 

Based on investigation of some examples, Juergen Ecker began a
discussion of commutators in nilpotent groups of class 2, and asked whether a
product of pure commutators in such a group is always a pure commutator.
Examples by Jim Howie, Derek Holt, Roger Bryant, and Andreas Caranti have shown
that this is not always the case. More than one four-generator example was
given.   
 
It happens that a group supporting such an example cannot be generated by
fewer than four elements:  In a three-generator nilpotent group G of class
two, every element of the commutator subgroup G' is a pure commutator.
 
For suppose that G is generated by x,y, and z. Then G' is abelian, generated
by [x,y], [x,z], and [y,z], and the commutator product [-.-]:GxG->G' is
bilinear. 
 
Given u in G', there are integers a,b,c such that
 
                       u = [x,y]^a * [x,z]^b * [y,z]^c
 
If a=0, then u = [x^b,z][y^c,z] = [x^b*y^c,z] is a pure commutator, and
similarly if b or c is 0.
 
Suppose abc is nonzero and let g=gcd(a,c). Write 
 
                               a=a'g and c=c'g;
 
then gcd(a',c')=1 and so there are integers k,m such that
                               
                                a'k + c'm = b.
 
Then the pure commutator
 
   [x^a'*z^(-c'),x^m*y^g*z^k] = [x,y]^(a'g) * [x,z]^(a'k+c'm) * [y,z]^(c'g)
 
is equal to u.