From: Jim HowieDate: 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.