# Commutator Length of Finitely Generated Linear Groups

Authors: Mahboubeh Alizadeh Sanati
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

---

## Abstract

The commutator length “cl(G)” of a group G is the least natural number c such that every element of the derived subgroup of G is a product of c commutators. We give an upper bound for cl(G) when G is a d-generator nilpotent-by-abelian-by-finite group. Then, we give an upper bound for the commutator length of a soluble-by-finite linear group over C that depends only on d and the degree of linearity. For such a group G, we prove that cl(G) is less than
k(k+1)/2+12d3+o(d2), where k is the minimum number of generators of (upper) triangular subgroup of G and o(d2) is a quadratic polynomial in d. Finally we show that if G is a
soluble-by-finite group of Prüffer rank r then cl(G)≤r(r+1)/2+12r3+o(r2), where o(r2) is a quadratic polynomial in r.


---

## Body

## 1. Introduction and Motivation

We recall that ifg is a nonidentity element  of the  commutator
subgroup of an arbitrary nonabelian group G,
denote by cl(g) that the least
integer such that g can be written as a product of cl(g) commutators. The commutator length of G is defined as(1.1)cl(G)=sup{cl(g)∣g∈G}.It is assumed that the
commutator length of the identity element is zero.In the present
paper, we studycl(G) when G is a soluble-by-finite linear group. In
general, the commutator length of a linear group
need not  be finite.
For example, the group G,
as a subgroup GL(2,Q),
generated by(1.2)(1021),(1201)is free and cl(G) is infinite [1]. Also, the rank of a group is d
if the minimum number of its generators is d.
In this case, G is called d-generator, too.In 1966, Stroud [2] proved that the commutator
length of a free nilpotent group of rank n is at most n.
After about twenty years, analogous result was obtained for free metabelian
nilpotent groups of rank n by Allambergenov and Roman'kov [3].
Using geometric methods, Goldstein and Turner  [4] and Culler  [5],
extended by Bardakov [1] using permutational method, obtained an algorithm
for computing the commutator length of elements of free groups. Many other
people have worked on the following: “when is cl(G) finite, if G is free soluble group of rank k and solvability length d?”
In some special cases the answer is positive;  one can see [6–8].In this paper, we have restricted our attention to
groups which are nilpotent by abelian by finite. In the case whereG is a linear group over a field of
characteristic zero, the upper bound depends only on the number of generators of
the group and the degree of linearity. Recall that the Prüffer rank of a group G is the least cardinal r such that every finitely generated subgroup of G can be generated by r elements. Our main results are as follows.Theorem A.
LetG=〈x1,…,xd〉 be a group with a nilpotent-by-abelian  normal k-generator subgroup H of finite index. Then,(1.3)cl(G)≤k(k+1)2+12d3+o(d2),where o(d2) is a quadratic polynomial in d.Theorem B.
LetG=GL(n,F) be a d-generator linear group, where charF=0 and G does not contain a nonabelian free group.
Then,(1.4)cl(G)≤m(m+1)2+12d3+o(d2),where m is the number of generators of a
triangularizable normal subgroup T of G, and o(d2) is a quadratic polynomial in d.
Moreover, the number m is bounded by a function of n and d.Corollary 1.1.
LetG be a soluble-by-finite group of finite Prüffer
rank r.
Then,(1.5)cl(G)≤r(r+1)2+12r3+o(r2),where o(r2) is a quadratic polynomial in r.Corollary 1.2.
LetG be a linear group of finite Prüffer rank r.
Then,(1.6)cl(G)≤r(r+1)2+12r3+o(r2),where o(r2) is a quadratic polynomial in r.


## 2. Theorem A

One begins with known
results on commutator length that one needs in the proof of  Theorem A.Lemma 2.1 (Hartley [6]).
LetG=〈x1,…,xd〉 and let H be a nilpotent normal subgroup of G. Suppose that H=〈y1G,…,ykG〉 is generated by the conjugacy classes in G of elements y1,…,yk. Then, every element of [H,G] can be expressed in the form(2.1)∏i=1d[ai,xi]∏j=1k[bj,yj],where ai's and bj's are in H and the product is taken for
definiteness in order of increasing suffices.Lemma 2.2 (Stroud [2]).
LetG=〈x1,…,xd〉 be a nilpotent group. Then, every element of G′ can be expressed as the product of d commutators [x1,g1],…,[xd,gd] for suitable gi∈G(1≤i≤d).Now, we can prove the following lemma that is based on
an argument of Rhemtulla [8]. Note that a weaker bound can be obtained from the
theorem of Akhavan-Malayeri [9].Lemma 2.3.
LetG=〈x1,…,xd〉 be a nilpotent-by-abelian group, then(2.2)cl(G)≤d(d+1)2.Proof.
We know thatG′=〈[xr,xs]G∣1≤r,⁢s≤d〉 is nilpotent, since G is nilpotent by abelian. So, by Hartley's
theorem, every element of [G′,G] can be expressed in the form ∏i=1d[ai,xi]∏j=1k[bj,yj],
where ai's and bj's are in G′ and yj∈{[xr,xs]:1≤r,⁢s≤d} and k=d(d−1)/2.
But G/[G′,G] is always nilpotent and by Stroud's theorem
every element of G′ has the form ∏i=1d[xi,gi] modulo [G′,G] for some gi∈G.
Hence, every element of g∈G′ can be expressed in the form(2.3)g=∏i=1d[xi,gi](∏i=1d[ai,xi]∏j=1k[bj,yj]).We can write ∏i=1d[xi,gi]∏i=1d[ai,xi]=[xd,gd]⋯[x1,g1][a1,x1]⋯[ad,xd].
By severally useing the equality [a,b][c,a]=[cb−1,a]ba,
one sees that every element of G′ has the following form:(2.4)g=∏i=1d[aigi−1,xi]aixici∏i=1k[bj,yj],for suitable elements ci∈G′.
Hence, the assertion holds.The proof of Theorem A requires the following lemma
that has been recently proved by Nikolov and Segal [10].Lemma 2.4 (Nikolov and Segal [10]).
LetG be any d-generator finite group and let H be a normal subgroup of G.
There exists a function α=α(d) such that every element of [H,G] is equal to a product of α=α(d) commutators [u,v] with u∈H and v∈G.
In particular cl(G)≤12d3+o(d2), where o(d2) is a quadratic polynomial in d.Proof  of Theorem A.
We knowcl(G)≤cl(G/H′)+cl(H).
By Lemma 2.3, we conclude cl(G)≤cl(G/H′)+k(k+1)/2.
We now aim to find an upper bound for cl(G/H′).
Without loss of generality, we can assume that H′=1.
So H is abelian and by Lemma 3 from [7], every
element of [H,G] can be expressed as a product of at most d commutators, Πi=1d[xi,hi],
where hi∈H.
Hence, we can assume that [H,G]=1.
Then, H is central, and G is a central-by-finite group. Therefore, G′ is finite. Since H is finitely generated abelian group, Hn is torsion free for some positive integer n.
So G′∩Hn=1 and G′≃(G/Hn)′.
But G/Hn is a finite group and by Segal's theorem,
every element of its derived subgroup (and so G′) is a product of at most 12d3+o(d2) commutators, where o(d2) is a quadratic polynomial in d.
That is, every element of G′ modulo H′ is a product of 12d3+o(d2) commutators. Now, the proof is completed.Proof.
We can assume thatG is generated by r elements. According to Robinson's theorem (see [11, Section 10.3]), G is nilpotent-by-abelian-by-finite group. Now,
the result holds from Theorem A.Lemma 2.5 (Platonov [12]).
LetG be a subgroup of GL(n,F) of finite Prüffer rank r.
Then, G is soluble by finite, and if charF=p>0 then G contains an abelian normal subgroup of finite
index bounded in terms of r,⁢n, and p.Proof.
Without loss of
generality, we can assume thatG is r-finitely generated. Then, according to
Platonov's theorem G is soluble by finite. Now, the result follows
from Corollary 1.1.In the caseG is d-generator soluble group, we have the
following result which was obtained by Akhavan-Malayeri [9, Theorem 1].Theorem 2.6 (Akhavan-Malayeri [9]).
LetH be a nilpotent-by-abelian normal subgroup of
finite index of a d-generator soluble group G.
Then,(2.5)cl(G)≤k(k+1)2+72d2+47d,where k is the number of generators of H.


## 3. Theorem B

The  next set of results deals with soluble-by-finite
linear groups over a field of characteristic zero and one tries to find the
number of generators of the upper triangular subgroups of these groups.Lemma 3.1 (Wehrfritz [13]).
LetG be a soluble subgroup of GL(n,F) and let N be a unipotent (nilpotent) normal subgroup of G.
Then, G contains a triangularizable normal subgroup T containing N such that the index of T in G is finite and divides(3.1)f(n)=(max⁡Σi=1tni=n{∏i=1t(ni2(ni2!))nini!})!.Lemma 3.2 (Tits [14]).
IfG is a linear group over a field of
characteristic p≥0 that does not contain nonabelian free
subgroups, then if p=0, G contains a soluble normal subgroup of finite
index and if p>0, G contains a soluble normal locally finite
subgroup.In Tits's article, it was proved that every finitely
generated linear group which does not contain a nonabelian free group has a
soluble subgroup of finite index. If one can calculate an upper bound for this
index, then by using the method of the next lemma one can obtain an upper bound
for the commutator length of a finitely generated linear group which does not
contain a nonabelian free group, in general case. In the casecharF=0,
one can use the following theorem.Lemma 3.3 (Jordan [13]).
IfcharF=0, and G is a soluble-by-finite subgroup of GL(n,F), then G contains a soluble normal subgroup of finite
index at most β(n)=(8n+1)2n2−(8n−1)2n2.Proof  of Theorem B.
By Tits's theorem,G is soluble by finite and so by Jordan's
theorem, G has a soluble normal subgroup H of finite index at most β(n).
Wehrfritz's theorem says that H has a nilpotent-by-abelian subgroup T such that the index of T in H divides f(n).
So we have(3.2)[G:T]=[G:H][H:T]≤β(n)f(n).We consider g(n)=β(n)⋅f(n).
We know by Nielsen-Schrier's theorem that the rank of T is at most m=dg(n)+1−g(n).
Now, the result holds by Theorem A.



---

*Source: 281734-2008-08-18.xml*