ON FINITELY GENERATED NILPOTENT GROUPS and THEIR SUBGROUPS by BRYAN SANDOR MARTIN EVANS, COMMITTEE CHAIR MARTYN DIXON BRUCE TRAC

Home , Nilpotent group, Polycyclic group

ON FINITELY GENERATED NILPOTENT GROUPS AND THEIR SUBGROUPS

by BRYAN SANDOR MARTIN EVANS, COMMITTEE CHAIR MARTYN DIXON BRUCE TRACE ALLEN STERN JON CORSON

A DISSERTATION

Submitted in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in the Department of Mathematics in the Graduate School of The University of Alabama

TUSCALOOSA, ALABAMA

2017 Copyright Bryan Sandor 2017 ALL RIGHTS RESERVED ABSTRACT In this work we investigate nilpotent groups G in which all proper subgroups (or all subgroups of inﬁnite index) have class smaller than the class of G. Our main results are obtained by considering analagous questions for Lie algebras and using the Lazard correspondence and the Mal’cev correspondence. Among other things, for each n ≥ 3, we prove the existence of nilpotent groups of class 2n in which every proper subgroup (or subgroup of inﬁnite index) has class at most n.

ii DEDICATION I would like to dedicate this work to the memory of Professor Bettina Richmond, my thesis advisor at Western Kentucky University, who started me on my road to success.

iii ACKNOWLEDGMENTS Certainly I would never have made it this far without the patient and unconditional guidance of my advisor, Professor Martin Evans. He has the best temperament and per- sonality anyone could hope for in a mentor. I esteem our collaboration and will forever be grateful for his tutelage. My parents, Daniel and Dolly, have been my bedrock of support as I pursued this degree. I do not believe they ever doubted that I could earn this, even when I doubted myself. I lack the vocabulary and elegance of a poet to truly express my love for them. They are my heart. My brother, Adam, is the true artist between us and his passion for his craft served as an inspiration for me. My sister-in-law, Gina, worked very hard for her Doctorate in Veterinary Medicine. Her dedication and determination to settle for nothing less motivated me. My cousins Tom and Kay Bloodworth know the importance of being able to laugh, lending every measure of levity to even the gravest of situations, and there were plenty throughout school that they could always help me laugh through. Their parents, Tay “Scott” and Martha, are spiritual parents to me and were integral to my obedience to the Gospel. My grandfather, William Glenn “Poppy” Bloodworth, is nothing short of the patriarch to our family and has a heart bigger and stronger (despite medical evidence to the contrary) than even the Grinch on Christmas. Halie “Nana” was taken from him and us too soon and I wish she were here to see her eldest grandchild graduate with a doctorate as she was there to see me graduate high school. My best friends, Jen Manchenko and Rachel and Trey Ward, never let me get away with beating myself up and helped me maintain my sanity throughout my studies here.

iv My friends Anne Duffee and Erin Watley were wonderful confidantes, helping me realize that it really is normal to feel like I am out of my educational league, even though it is not true. We laughed, complained, cried, and celebrated together throughout the program. I thank the Department of Mathematics at the University of Alabama for giving me financial and professional support, fostering my teaching and researching skills. Finally, I thank G-d, who fashioned me in HIS image and gave me all the skills I possess. Psalm 148 is my favorite to read and to sing but I close with the following: “Let everything that hath breath praise the LORD.” -Psalm 150:6

v CONTENTS

ABSTRACT ...... ii

DEDICATION ...... iii

ACKNOWLEDGMENTS ...... iv

LIST OF FIGURES ...... vii

CHAPTER 1. INTRODUCTION ...... 1

CHAPTER 2. RESULTS MOTIVATED BY THE WORK OF C. K. GUPTA . . . . 11

CHAPTER 3. EXISTENCE THEOREMS ...... 19

CHAPTER 4. ASSOCIATIVE ALGEBRAS ...... 34

REFERENCES ...... 38

vi LIST OF FIGURES

3.6 Young tableau associated with anti-symmetrizer Θ ...... 24

vii CHAPTER 1

INTRODUCTION

In this chapter we collect together various deﬁnitions and basic results we will need in the sequel.

Deﬁnition 1. Let G be a group. The center of G, denoted Z(G), is the subgroup of G consisting of the elements which commute with every element of G.

Deﬁnition 2. A group G is called nilpotent if it has a central series, that is, a normal series

1 = G0 ≤ G1 ≤ ... ≤ Gn = G such that Gi+1/Gi is contained in the center of G/Gi for all i. The length of a shortest central series of G is the nilpotent class of G.

Deﬁnition 3. Let g1, g2, g3, . . . , gn,... be elements of a group G. The commutator of g1

−1 −1 and g2 is [g1, g2] = g1 g2 g1g2.A simple commutator of weight n ≥ 2 is inductively deﬁned

as [g1, g2, . . . , gn] = [[g1, g2, . . . , gn−1], gn], where [g1] = g1.

Recall that the conjugate of g by h is gh := h−1gh for all g, h ∈ G. Our ﬁrst lemma can be proved by direct calculation.

Lemma 1.1. Let G be a group and let x, y, z ∈ G. Then

1. [x, y]−1 = [y, x],

2. [x, zy] = [x, y][x, z]y and [xz, y] = [x, y]z[z, y], and

3. [x, y−1, z]y[y, z−1, x]z[z, x−1, y]x = 1 (The Hall-Witt identity).

Deﬁnition 4. Let X be a nonempty subset of a group G. Deﬁne the subgroup generated by X, denoted by hXi, to be the intersection of all subgroups of G which contain X.

1 For nonempty subsets X1 and X2 of a group G, let

[X1,X2] = h[x1, x2] | x1 ∈ X1, x2 ∈ X2i.

We similarly deﬁne [X1,X2,...,Xn] inductively by

[X1,X2,...,Xn] = [[X1,X2,...,Xn−1],Xn].

With this notation we introduce two important central series.

Deﬁnition 5. The series

G = γ1(G) ≥ γ2(G) ≥ γ3(G) ≥ ...

in which γn+1(G) = [γn(G),G] is called the lower central series of G.

Notice that every γn(G) is fully-invariant in G and that γn(G)/γn+1(G) lies in the center of G/γn+1(G).

Deﬁnition 6. The upper central series is given by

1 = ζ0(G) ≤ ζ1(G) ≤ ζ2(G) ≤ ...,

where ζn+1(G)/ζn(G) = Z(G/ζn(G)).

Each ζn(G) is characteristic but not necessarily fully-invariant in G. Notice that

ζ1(G) = Z(G).

Lemma 1.2. [16, 5.1.9] Let 1 = G0 ≤ G1 ≤ ... ≤ Gn = G be a central series in a nilpotent group G. Then

(i) γi(G) ≤ Gn−i+1, so that γn+1(G) = 1;

2 (ii) Gi ≤ ζi(G), so that ζn(G) = G;

(iii) the nilpotent class of G equals the length of the upper central series, which itself equals the length of the lower central series.

In particular, a group G is nilpotent if and only if the lower central series of G reaches the identity subgroup after ﬁnitely many steps or, equivalently, the upper central series of G reaches the group itself after ﬁnitely many steps. It is worth explicitly mentioning that if G is nilpotent of class c, then γc+1(G) = 1 (but γc(G) 6= 1) and ζc(G) = G (but ζc−1(G) 6= G).

Theorem 1.3. Let G be a nilpotent group of class c. Then all of its subgroups are nilpotent and have class at most c. Furthermore, if N ¡ G, then G/N is also nilpotent of class at most c.

We now begin a brief discussion of torsion in nilpotent groups.

Deﬁnition 7. Let π be a nonempty set of prime numbers. A π-number is a positive integer whose prime divisors belong to π. An element of a group G is called a π-element if its order is a π-number. If every element in G is a π-element, then the group is called a π-group.

In particular, when π = {p}, we speak of p-elements and p-groups. Note that if G is ﬁnite, by Lagrange’s theorem every element is a p-element if and only if the order of G is a power of p.

Deﬁnition 8. The commutator subgroup of G (or derived subgroup of G), denoted by G0, is the subgroup of G generated by all the commutators.

Note that G/G0 is the largest abelian quotient group of G. We call it the abelianization

of G and sometimes denote it by Gab.

Lemma 1.4. [16, 5.2.6] Let P be a group-theoretical property which is inherited by images of tensor products and by extensions. If G is an nilpotent group such that G/G0 has P, then G has P.

3 For example, let P be the property of being finite. Then we obtain the result: if G is a nilpotent group and G/G0 is finite, then G too is finite.

Theorem 1.5. [16, 5.2.7] Let G be a nilpotent group. Then the elements with ﬁnite order in G form a fully-invariant subgroup, T (G), called the torsion subgroup of G, such that G/T (G) Q is torsion-free and T (G) = p Tp(G) where Tp(G) is the unique maximum p-subgroup of G.

Proof. Let π be a nonempty set of prime numbers and let Tπ(G) denote the subgroup generated by all π-elements of G. Now (Tπ(G))ab is also generated by π-elements and, being abelian, is a π-group. Since G is nilpotent and (Tπ(G))ab is a π-group, the subgroup Tπ(G) is also a π-group. Let π equal the set of all primes and we conclude T (G) = Tπ(G) consists of elements of ﬁnite order, hence T (G) has torsion. Now, taking π = {p}, we have that Tp Q is a p-group. Then Tp ¡ G and so T (G) = p Tp(G).

Deﬁnition 9. The subgroup H of the group G is called subnormal in G if there exists a

ﬁnite chain {Hi} of subgroups Hi of G with

H = H0 ≤ H1 ≤ ... ≤ Hn = G

such that Hi is a normal subgroup of Hi+1 for every i = 0, 1, . . . , n − 1. The length of a minimal such chain is the defect of H in G.

The following well known lemma is fundamental and can be proved quite easily by induction on the class of G.

Lemma 1.6. Let H ≤ G where G is a nilpotent group of class c. Then H is subnormal in G and its defect is at most c.

Deﬁnition 10. If X is a nonempty subset of a group G, then a word on X is an element w ∈ G of the form

e1 e2 en w = x1 x2 . . . xn ,

where xi ∈ X, ei = ±1, and n ≥ 0. If n = 0, then w := 1.

4 It is easy to see that hXi is precisely the set of all words on X.

Definition 11. Let n be a positive integer. A group G is said to be an n-generator group if it can be generated by some n-subset {x1, x2, . . . , xn} ⊆ G. In this case, each such xi is called a generator of G. A group is finitely generated if it is n-generator for some n ∈ N. If G is finitely generated, we let d(G) denote the cardinality of a smallest generating set of G.

The next lemma includes routine results; their proofs may be found, for example, in [5, Lemma 3.1–3.2].

Lemma 1.7. Let G be a group.

1. Let N be a normal subgroup of G. If {g1, g2, . . . , gr} is a family of elements of G

containing m elements of N (with m ≤ r), then [g1, g2, . . . , gr] ∈ γm(N).

2. If G = hXi and the commutator [x1, x2, . . . , xn] = 1 whenever x1, x2, . . . , xn ∈ X, then G is nilpotent of class at most n − 1.

Deﬁnition 12. A proper subgroup M of a group G is a maximal subgroup of G if M ≤ H < G implies M = H.

Deﬁnition 13. Let Λ be a partially ordered set with partial order ≤. An element, λm ∈ Λ, is a maximal element if it does not precede any other element of Λ.

Lemma 1.8. Let G be a ﬁnitely generated group and let K be a proper subgroup of G. Then G contains a maximal subgroup M such that K ≤ M. In particular, G has a maximal subgroup.

Proof. Let G = hg1, g2, . . . , gdi, where d = d(G) and let S be the set of all proper subgroups

of G that contain K. Evidently S is non-empty since K ∈ S. Let S = {Si} be a chain of

subgroups in S and let H = ∪iSi. It is easy to see that H is a subgroup of G. We claim that H ∈ S. Since obviously K ≤ H this is equivalent to claiming that H is a proper subgroup

of G. Suppose that H = G. Then for each i = 1, 2, . . . , d there exists ni such that gi ∈ Sni .

5 However the ﬁnite set {n1, n2, . . . , nd} has a largest element, nˆ say, and it now follows that

Snˆ = G, a contradiction. Therefore H is a proper subgroup of G and so H ∈ S. This shows that every chain in S has an upper bound in S and Zorn’s lemma now implies that S contains a maximal element M. Clearly M has the desired properties.

Deﬁnition 14. The Frattini subgroup of an arbitrary group G, denoted Frat(G), is deﬁned to be the intersection of all the maximal subgroups of G, with the stipulation that it shall equal G if G should prove to have no maximal subgroups.

Deﬁnition 15. An element g of a group G is called a nongenerator of G if G = hg, Xi implies G = hXi when X is a subset of G.

Lemma 1.9. Let G be a ﬁnitely generated group. Then Frat(G) is precisely the set of nongenerators of G.

Proof. Suppose g is a nongenerator of G with g∈ / Frat(G). Then there exists some maximal subgroup M of G with g∈ / M. Therefore M 6= hg, Mi. However since M is maximal, hg, Mi = G. Since g is a nongenerator, hg, Mi must be M, implying M = G, a contradiction. Thus Frat(G) contains all the nongenerators of G. Now, let X be a subset of G and suppose g ∈ Frat(G) satisﬁes G = hg, Xi but G 6= hXi. Then g∈ / X and by Lemma 1.8 there is some maximal subgroup M of G such that X ≤ M. Clearly g∈ / M. However Frat(G) ≤ M and g ∈ Frat(G), a contradiction. The proof is complete.

Lemma 1.10. Let G be a ﬁnitely generated nilpotent group. Then G0 ≤ Frat(G).

Proof. We argue by induction on the class c of G. If c = 1, then G0 = 1 and certainly Frat(G) contains the trivial subgroup. Now suppose the result holds for class c ≤ k − 1 and let G have class k. Let g ∈ G0 and suppose that X is a subset of G such that G = hg, Xi. By the induction hypothesis, G/Z(G) is generated by the set X∗ = {xZ(G) | x ∈ X} so

6 that for each h ∈ G there exists zh ∈ Z(G) such that hzh ∈ hXi. However, for all h, y ∈ G we have

zy zh [hzh, yzy] = [hzh, zy][hzh, y] = [hzh, y] = [h, y] [zh, y] = [h, y].

Consequently, hXi contains all commutators of elements of G and therefore G = hXi. This means g is a nongenerator of G and from Lemma 1.9 we have g ∈ Frat(G).

The following deﬁnes a useful class of groups.

Deﬁnition 16. A group G is polycyclic if it has a ﬁnite series

1 = G0 ¢ G1 ¢ ··· ¢ Gn = G

such that the factor groups Gi+1/Gi are cyclic for all i. Furthermore, if each Gi+1/Gi is inﬁnite, then G is said to be poly-(inﬁnite cyclic).

Lemma 1.11. Let G be a ﬁnitely generated torsion-free nilpotent group. Then G is poly- (inﬁnite cyclic).

For a proof of this see [16, 5.2.20]. The following lemma shows that the number of inﬁnite factors in a cyclic series of a polycyclic group is uniquely determined.

Lemma 1.12. [16, 5.4.13] In a polycyclic group G the number of inﬁnite factors in a cyclic series (i.e. a series of the type given in Deﬁnition 16) is independent of the series and hence is an invariant of G.

Deﬁnition 17. Let G be a polycyclic group. The Hirsch length of G, denoted h(G), is the number of inﬁnite factors in a cyclic series of G.

Lemma 1.13. Let G be a poly-(inﬁnite cyclic) group of Hirsch length h. Then G can be generated by h elements.

Proof. Let 1 = X0 ¡ X1 ¡ X2 ¡ ··· ¡ Xh = G be a series with inﬁnite cyclic factors.

Choose elements xi ∈ G such that hxiXi−1i = Xi/Xi−1 for each i = 1, 2, . . . , h. Then clearly hx1, x2, . . . , xji = Xj for j = 1, 2, . . . , h, and in particular G = hx1, x2, . . . , xhi.

7 Lemma 1.14. Let G be a polycyclic group of Hirsch length h and suppose that K is a subgroup of ﬁnite index in G. Then K has Hirsch length h.

Lemma 1.15. Let H be a ﬁnitely generated subgroup of ﬁnite index in the torsion-free nilpotent group G. Then d(G) is at most the Hirsch length of H.

Proof. It is clear that G is ﬁnitely generated and so Lemma 1.11 shows that it is poly-(inﬁnite cyclic). Lemma 1.14 shows that H and G have the same Hirsch length and the result follows from Lemma 1.13.

Deﬁnition 18. Let Λ be a partially ordered set with partial order ≤. Then Λ satisﬁes the

maximal condition if each nonempty subset Λ0 contains at least one maximal element. Equivalently, we say Λ satisﬁes the ascending chain condition if there does not exist

an inﬁnite properly ascending chain λ1 < λ2 < . . . in Λ. Let G be a group. We say that G satisﬁes the maximal condition on subgroups (or G has max) if every set of subgroups of G has a maximal element.

Note that if G has max then every subgroup of G is ﬁnitely generated.

Theorem 1.16. Let G be a polycyclic group. Then G has max. In particular, every subgroup of G is ﬁnitely generated.

Deﬁnition 19. The Möbius function, µ(n): N → {−1, 0, 1}, is deﬁned as µ(1) := 1 and,

e1 e2 es given the prime factorization of n = p1 p2 . . . ps for distinct primes p1, p2, . . . , ps,

  0 if any ei > 1 µ(n) = .  s (−1) otherwise

For example, µ(6) = 1 since 6 = 2 · 3 is the product of two distinct prime numbers, µ(7) = −1, since 7 is trivially the product of one prime number, and µ(27) = 0 since 27 = 33.

Let Fn denote the free group of rank n for n ≥ 2. For each c ∈ N the group

Nn,c = Fn/γc+1(Fn) is called the free nilpotent group of class c and rank n. It is torsion-free

8 and has class (exactly) n. It is known that each term, γi(Nn,c)/γi+1(Nn,c), of its lower central series is a free abelian group of rank [6, 11.2.2]

1 X M (r) = µ(d)nr/d. n r d|r

Consequently, the Hirsch length of Nn,c is

c   X 1 X H = µ(d)nr/d . n,c r  r=1 d|r

It is also known that every n-generator group that is nilpotent of class at most c is a homomorphic image of Nn,c. It follows that the Hirsch length of any n-generator group that is nilpotent of class at most c is less than or equal to Hn,c. Clearly this implies the following, although there are easier proofs available. (For instance it can be proved by using Lemma 1.4.)

Corollary 1.17. If G is a ﬁnitely generated nilpotent group, then G is polycyclic.

Combining this result with Theorem 1.16 gives the following.

Theorem 1.18. Let G be a ﬁnitely generated nilpotent group. Then G has max. In particular, every subgroup of G is ﬁnitely generated.

Theorem 1.19. Let G be a finitely generated nilpotent group. Then every subgroup H of infinite index in G lies in a subgroup M that is maximal with respect to having infinite index in G.

Proof. Let ΩH be the set of all subgroups of G which contain H and have inﬁnite index in

G. Since G has max, there exists a maximal element M of ΩH . Now H ≤ M, by deﬁnition, and if K ≤ G is such that M K we have [G : K] < ∞. The result follows.

Theorem 1.20. Let H be a maximal subgroup of a ﬁnitely generated nilpotent group G. Then H ¡ G.

9 Proof. Since H is a maximal subgroup of G, we have Frat(G) ≤ H and since G is a ﬁnitely generated nilpotent group, it follows that G0 ≤ Frat(G), by Lemma 1.10. Therefore G0 ≤ H. Thus H/G0 ≤ G/G0 and since G/G0 is abelian, H/G0 is normal in G/G0, hence H is normal in G.

Deﬁnition 20. Let G be a group. The higher commutator subgroups of G are deﬁned inductively: G(0) = G; G(i+1) = (G(i))0;

that is, G(i+1) is the commutator subgroup of G(i). The series

G = G(0) ≥ G(1) ≥ G(2) ≥ ...

is called the derived series of G.

Note that the factors G(n)/G(n+1) are abelian groups. Note if G00 = 1, then G is said to be metabelian.

10 CHAPTER 2

RESULTS MOTIVATED BY THE WORK OF C. K. GUPTA

One of the fundamental results in the theory of nilpotent groups was proved by Fitting.

Theorem 2.1. (Fitting’s Theorem.) Let M and N be normal nilpotent subgroups of a group G. If c and d are the nilpotent classes of M and N, then L = MN is a nilpotent subgroup of G with class at most c + d.

For our purposes the following consequence of Fitting’s theorem is particularly important.

Corollary 2.2. Let G be a nilpotent group and suppose that every proper subgroup of G has class at most n. Then G has class at most 2n.

Proof. Suppose ﬁrst that G is not ﬁnitely generated. Then for every g1, g2, . . . , gn+1 ∈ G the group hg1, g2, . . . , gn+1i is a proper subgroup of G and so it has class at most n. Therefore

[g1, g2, . . . , gn+1] = 1. Lemma 1.7 now shows that G has class at most n and the result holds. It follows that we may now assume that G is ﬁnitely generated. If G/G0 is cyclic then G is cyclic by Lemma 1.4 and again the result holds. So suppose that d(G/G0) = d ≥ 2 and let

0 0 0 0 x1, x2, . . . , xd ∈ G be such that hx1G , x2G , . . . , xdG i = G/G .

0 0 Let M = hx2, x3, . . . , xd,G i and N = hx1, x2, . . . , xd−1,G i. These are both normal proper subgroups of G and so each has class at most n. Moreover we see that MN =

0 hx1, x2, . . . , xd,G i = G so Fitting’s theorem shows that G has class at most 2n. The proof is complete.

11 In her paper [5], C.K. Gupta studied nilpotent groups G of class c in which every proper subgroup has class less than c. Notice that the first part of the preceding proof shows that such a group must be finitely generated. In fact considerably more is true: such groups must be finite, a result that first appeared in [17]. Thus Gupta’s results are about finite groups. Among other things, she proved the following interesting theorem.

Theorem 2.3. [5, Theorem 1.1] Let n be a positive integer greater than 1. If G is a nilpotent group whose proper subgroups are all nilpotent of class at most n, then the class of G is at

nd most m, where m ≤ d−1 < m + 1 and d is the minimal number of generators of G.

Retaining the notation in this theorem, we note that Corollary 2.2 shows that m ≤ 2n but it now follows that if G does have class 2n then d(G) = 2. We record this surprising fact as

Corollary 2.4. Let G be a nilpotent group of class 2n in which every proper subgroup has class at most n. Then G is a 2-generator ﬁnite group.

There are examples of such groups in the literature [11]. We will discuss them in the next chapter. Our goal in this chapter is to prove results like Gupta’s but for torsion-free nilpotent groups. Some modifications are surely necessary since if H is a subgroup of finite index in a torsion-free nilpotent group G then the class of H is equal to the class of G, (see Lemma 2.12 below). The appropriate modification is to restrict attention to subgroups of infinite index. Thus we investigate properties of nilpotent groups of class m in which all subgroups of infinite index have class at most n. We begin with a result about abelianizations.

Theorem 2.5. Let G be an inﬁnite nilpotent group such that every subgroup of inﬁnite index in G has class at most n. Suppose that the class of G is greater than 2n. Then G/G0 has torsion-free rank 1.

12 Proof. By Lemma 1.7 there exist x1, x2, . . . , x2n+2 ∈ G such that [x1, x2, . . . , x2n+2] 6= 1. Let

H = hx1, x2, . . . , x2n+2i. Since H has class greater than n it follows that H has ﬁnite index in G, [G : H] = k, say. Consequently G is the union of its cosets H, y2H, . . . , ykH for some

y2, y3, . . . , yk ∈ G and it follows that G = hH, y2, . . . , yki. Hence G is finitely generated. If G/G0 is finite, then the remark following Lemma 1.4 shows that G is finite, a contradiction. If G/G0 has torsion-free rank greater than 1 then there exists K ¡ G such that ∼ G/K = C∞ ×C∞. Let aK and bK together generate G/K. Set M = hK, ai and N = hK, bi. Thus M and N are normal subgroups of G, each has infinite index in G (and so has class at most n), and G = MN. Fitting’s theorem applies and shows that G has class at most 2n, a contradiction. The proof is complete.

0 ∼ The following examples show that in the exceptional case G/G = C∞ that occurs in the previous theorem, there is no bound on the class of G. We introduce a deﬁnition ﬁrst.

Deﬁnition 21. If W is a set of words in x1, x2,... , the class of all groups G such that W (G) = 1 is called the variety Var(W ) determined by W .

As an example, if W = {[x1, x2]}, then Var(W ) is the class of all abelian groups and similarly if W = {[x1, x2, . . . , xc+1]}, then Var(W ) is the class of all nilpotent groups of class at most c.

Theorem 2.6. (compare with [2, Proposition 3]) Let n and k be positive integers with n < k and let p be a prime. There exists a 3-generator nilpotent group Gn,k of class k that is an extension of a finite p-group Q of class n by an infinite cyclic group. Consequently, every subgroup of infinite index Gn,k has class at most n.

Proof. Let P be a 2-generator ﬁnite p-group with class n and let pl denote its exponent. (For instance we may take P = ha, b | apn = bpn−1 = 1, ab = a1+pi, a ﬁnite metabelian p-group of

2n−1 n class n, order p , and exponent p .) Now let V2k represent the free group of rank 2k in

Var(P ), the variety of groups generated by P , freely generated by a1, a2, . . . , ak, b1, b2, . . . , bk.

13 Note that each non-cyclic ﬁnitely generated free group in Var(P ) is nilpotent of class n, has exponent equal to pl, and is ﬁnite [13, Theorem 15.71].

The function θ : {a1, a2, . . . , ak, b1, b2, . . . , bk} → V2k, given by θ(ai) = aiai−1, θ(bi) =

∗ bibi−1, for i = 2, 3, . . . , k, and θ(a1) = a1, θ(b1) = b1, extends to an automorphism θ of V2k.

Let G = V2k o hyi, the semi-direct product of V2k with the inﬁnite cyclic group hyi, in which

y ∗ the action of y on V2k is given by q = θ (q) for all q ∈ V2k.

Since [ai, y] = ai−1 and [bi, y] = bi−1 for i = 2, 3, . . . , k, it follows that G = hak, bk, yi.

0 ∼ pl pl Moreover, G/V2k = hy, a1, a2, . . . , ak, b1, b2, . . . , bk | [ai, aj] = [bi, bj] = [ai, bj] = ai = bi = 1, for i, j = 1, 2, . . . , k, [ai, y] = ai−1, [bi, y] = bi−1 for i, j = 2, 3, . . . , k, [a1, y] = [b1, y] = 1i and

0 we see that G/V2k is nilpotent of class k. Since V2k is also nilpotent, a well-known result of P. Hall (see, for instance, [16, 5.2.10, page 134]) now implies G is nilpotent. Evidently the class of G is at least k.

Let N be the normal closure of {a1, a2, . . . , ak−1, b1, b2, . . . , bk−1} in V2k so that we have

∼ x V2k/N = V2 where, of course, V2 is the free group of rank 2 in Var(P ). Note that N = N ∼ so that N ¡ G. It is easy to see that G/N = V2 × C∞ and it follows that γn+1(G) ≤ N. Now set Gn,k = G/γk+1(G) and note that Gn,k has class k. Finally, let Q = V2k/γk+1(G) and note ¯ ∼ ¯ that Q has class n since Q/N = V2 where N = N/γk+1(G).

Let us now rephrase Theorem 2.5 in the following way.

Theorem 2.7. Let G be an inﬁnite nilpotent group such that G/G0 has torsion-free rank at least 2 and suppose that every subgroup of inﬁnite index in G has class at most n. Then G has class at most 2n.

Now to make further progress we will need a number of lemmas. Some of these are “folklore.”

Lemma 2.8. Let G be a nilpotent group, let H ≤ G have index n in G and suppose that g ∈ G. Then gn ∈ H.

14 Proof. Let m be the defect of H in G. We argue by induction on m. If m = 1 then H ¡ G and we may form the factor group G/H. Since [G : H] = n it follows from Lagrange’s theorem that (gH)n = gnH = H and so gn ∈ H as required. Suppose that m ≥ 2 and that the result holds whenever the defect is less than m.

Let H = Hm ¡ Hm−1 ¡ ··· ¡ H0 = G be a subnormal series and set q = [G : Hn−1] and

r = [Hm−1 : Hm]. Now n = [G : Hm] = [G : Hm−1][Hm−1 : Hm] = qr and our inductive

q q r n q r hypothesis shows ﬁrst that g ∈ Hm−1 and then that (g ) ∈ Hm. Thus g = (g ) ∈ Hm = H and the result follows.

Lemma 2.9. Let G be a group, let a ∈ γk(G) for some k ≥ 1, and let c ∈ G. Then for each integer n ≥ 1 we have that

n n (i) [a , c] ≡ [a, c] mod γk+2(G), and

n n (ii) [a, c ] ≡ [a, c] mod γk+2(G).

Proof. (Throughout, all congruences are to be read modulo γk+2(G).)

n−1 n−1 Since [a , c] and [a, c ] are elements of γk+1(G), they are central modulo γk+2(G). Therefore [an, c] = [an−1a, c] = [an−1, c]a[a, c] ≡ [an−1, c][a, c] and [a, cn] = [a, cn−1c] = [a, c][a, cn−1]c ≡ [a, c][a, cn−1]. Both parts of the lemma now follow easily by induction on n.

Lemma 2.10. Let G be a group, let r be a positive integer and let g1, g2, . . . , gn ∈ G for some n ≥ 2. Then

r r r rn [g1, g2, . . . , gn] ≡ [g1, g2, . . . , gn] mod γn+1(G).

Proof. The proof is by induction on n, beginning with n = 2.

r r r r Since g1 ∈ γ1(G) = G we have that [g1, g2] ≡ [g1, g2] mod γ3(G) by Lemma 2.9(i).

r r Moreover, it follows from Lemma 2.9(ii) that [g1, g2] ≡ [g1, g2] mod γ3(G) and conse-

15 r r r r r r r2 quently, working modulo γ3(G), we have that [g1, g2] ≡ [g1, g2] ≡ ([g1, g2] ) ≡ [g1, g2] . Therefore induction starts. Suppose now that the result is true for some positive integer

n ≥ 2 and let g1, g2, . . . , gn, gn+1 ∈ G. Note that our inductive hypothesis yields that

r r r rn r r r rn [g1, g2, . . . , gn] ≡ [g1, g2, . . . , gn] mod γn+1(G) and so [g1, g2, . . . , gn] = [g1, g2, . . . , gn] h

r r r r rn r for some h ∈ γn+1(G). Thus we have [g1, g2, . . . , gn, gn+1] = [[g1, g2, . . . , gn] h, gn+1] =

rn r h rn r [g1, g2, . . . , gn] , gn+1] [h, gn+1] ≡ [[g1, g2, . . . , gn] , gn+1] mod γn+2(G) since h is central

modulo γn+2(G) and the result now follows easily from Lemma 2.9 since [g1, g2, . . . , gn] ∈

γn(G). (This last part is similar to the argument at the beginning of this proof of the case n = 2.)

Deﬁnition 22. Let βn(G) denote the subgroup generated by nth powers of elements of the group G, i.e.

n βn(G) = hg | g ∈ Gi.

Lemma 2.11. Let G be nilpotent of class n. Then γn(βr(G)) = βrn (γn(G)) for all integers r ≥ 1.

Proof. It follows from Lemma 1.7(2) that γn(G) is generated by simple commutators of the

form [g1, g2, . . . , gn] for if all of these are trivial G has class at most n − 1. Consequently,

rn since βrn (γn(G)) is abelian, we have that βrn (γn(G)) = h[g1, g2, . . . , gn] | g1, g2, . . . , gn ∈

r r r Gi = h[g1, g2, . . . , gn] | g1, g2, . . . , gn ∈ Gi = γn(βr(G)) as required.

Lemma 2.12. Let G be a ﬁnitely generated torsion-free nilpotent group of class m and let H be a subgroup of ﬁnite index in G. Then H has class m.

Proof. Let [G : H] = r and note that Lemma 2.8 implies that βr(G) ≤ H. It now follows from Lemma 2.11 that γm(H) ≥ γm(βr(G)) = βrm (γm(G)) 6= 1, where the last inequality holds since G is torsion-free. The result follows.

Lemma 2.13. Let G be a ﬁnitely generated nilpotent group and suppose that H ≤ G is such that [G : HG0] < ∞. Then [G : H] < ∞.

16 Proof. We argue by induction on c, the class of G, the case c = 1 being clear. By induction we

may assume that [G/γc(G): Hγc(G)/γc(G)] < ∞ and so [G : Hγc(G)] = n, say. Clearly the

result will follow once we show that [Hγc(G): H] is finite. To this end note that H ¡Hγc(G) ∼ since γc(G) ≤ Z(G) and so it suffices to show that γc(G)/(H ∩ γc(G)) = Hγc(G)/H is finite.

Appealing to Lemma 2.11 and Lemma 2.8, we obtain that βnc (γc(G)) = γc(βn(G)) ≤

γc(Hγc(G)) = γc(H), where the last equality follows since γc(G) is central in G. It follows

that (the finitely generated nilpotent group) γc(G)/(H ∩ γc(G)) has finite exponent and is therefore finite. The result is proved.

Lemma 2.14. Let G be a ﬁnitely generated nilpotent group such that G/G0 has torsion-free rank r. Then there exists a subgroup H of G such that [G : H] < ∞ and H/H0 is free abelian of rank r.

0 0 0 Proof. Let hh1G , h2G , . . . , hrGi be a free abelian subgroup of G/G that has rank r and let

0 H = hh1, h2, . . . , hri. Now [G : HG ] < ∞ so Lemma 2.13 shows that H is of ﬁnite index in G. Moreover, HG0/G0 ∼= H/(G0 ∩ H) is free abelian of rank r and, since H is r-generator, we deduce that H/H0 is free abelian of rank r. The proof is complete.

Theorem 2.15. Let G be a ﬁnitely generated nilpotent group of class c such that G/G0 has torsion-free rank r ≥ 2 and suppose that every subgroup of inﬁnite index has class at most

n ∈ N. Suppose further that either

(i) G/G0 is torsion-free, or

(ii) G is torsion-free.

Then c ≤ bnr/(r − 1)c. Consequently, if r ≥ n2 + 1 then G has class at most n, and if G has class 2n then r = 2.

Proof. Let G be as stated. Suppose ﬁrst that G/G0 is torsion-free so that G/G0 is free

0 abelian of rank r. Let y1, y2, . . . , yr ∈ G be such that hy1, y2, . . . , yr,G i = G. Then, since

0 G ≤ Frat(G), it follows that Y = {y1, y2, . . . , yr} is a set of generators of G. Observe that,

17 G for i = 1, 2, . . . , r, the subgroup Ni = (Y \{yi}) , (the normal closure of Y \{yi} in G), has inﬁnite index in G and therefore has class at most n.

By Lemma 1.7(ii), there exists a commutator C = [x1, x2, . . . , xc] 6= 1 where xi ∈ Y for i = 1, 2, . . . , c. Let ti denote the (exact) number of occurences of yi in C. Then, by Lemma

1.7(i), we have that γc−ti (Ni) 6= 1 for i = 1, 2, . . . , r, and so c − ti ≤ n. Thus, for each i = 1, 2, . . . , r we have c ≤ n + ti, and on summing these inequalities over i = 1, 2, . . . , r, we P obtain rc ≤ rn + ti = rn + c. After rearranging this inequality we see that c ≤ nr/(r − 1). The proof of (i) is complete. Suppose now that G is torsion-free and let H be a subgroup of finite index in G such that H/H0 is free abelian of rank r; such a subgroup exists by Lemma 2.14. Each subgroup of infinite index in H also has infinite index in G and therefore has class at most n. Part (i) now implies that the class of H is at most bnr/(r − 1)c and part (ii) follows since the class of G is equal to the class of H by Lemma 2.12.

18 CHAPTER 3

EXISTENCE THEOREMS

In this chapter we answer two questions. The ﬁrst is related to the work of C.K. Gupta we discussed in Chapter 2.

Question 3.1. Let n be a positive integer. Does there exist a nilpotent group Hn of class 2n in which every proper subgroup has class at most n?

As we have seen, if such a group Hn exists it must be ﬁnite and 2-generator.

Clearly the answer to Question 3.1 is ‘yes’ if n = 1: for each prime p we can let H1 be a non-abelian p-group of order p3, although there are many other possibilities. Redei [14] has classified all finite non-abelian groups in which all proper subgroups are abelian; for each prime p his classification contains infinitely many p-groups. In contrast the answer is ‘no’ for n = 2 [11, Corollary 1 to Theorem 1]. As far as we are aware the only other value of n for which the answer is known is n = 3: I.D. Macdonald has constructed, for each prime p ≥ 5, a finite p-group of class 6 in which every proper subgroup has class at most 3 [11, Theorem 4]. For infinite groups we ask the following question which is related to our work in Chapter 2.

Question 3.2. Let n be a positive integer. Does there exist a torsion-free nilpotent group

Kn of class 2n in which every subgroup of inﬁnite index has class at most n?

The answer is ‘yes’ if n = 1 for we can set K1 = hx, y | [x, y, x] = 1 = [x, y, y]i, the free nilpotent group of rank 2. Moreover, using the ﬁnite p-groups of class 6 referred to above, it is possible to construct a torsion-free nilpotent group of class 6 in which every subgroup of

19 inﬁnite index has class at most 3 [2, Proposition 4]. Therefore the answer to Question 3.2 is ‘yes’ if n = 3. In this work we will answer Questions 3.1 and 3.2 by ﬁrst establishing the following theorem which deals with the analogous situation for Lie algebras.

Theorem 3.3. Let n ≥ 3 be a positive integer and let K be a ﬁeld of characteristic 0 or a (n2−1)n ﬁeld of characteristic p > 0 where p does not divide 2 . Then there exists a nilpotent Lie K-algebra L of class 2n in which every proper subalgebra has class at most n.

Deﬁnition 23. Let K be a ﬁeld. A Lie algebra over K (or a Lie K-algebra) is a vector space L over K together with a bilinear operation

[·, ·]: L × L → L

called the Lie bracket that satisﬁes

(i) [x, x] = 0 for all x ∈ L and

(ii) [[x, y], z] + [[y, z], x] + [[z, x], y] = 0 for all x, y, z ∈ L (The Jacobi Identity).

As a consequence of property (i), we have that the Lie bracket is anticommutative, i.e. [y, x] = −[x, y] for all x, y ∈ L.

Deﬁnition 24. A subalgebra M of the Lie K-algebra L, denoted M ≤ L is a subspace with the properties that [x, y] ∈ M for all x, y ∈ M. A subalgebra M is an ideal of L, written M ¡L, if M satisﬁes the additional property that [x, z] ∈ M whenever x ∈ M and z ∈ L.

Note that, by the anticommutativity of the Lie bracket, there is an obvious sense in which we do not need to distinguish “right ideals” and “left ideals” in Lie algebras. Much of our terminology from group theory carries over to the theory of Lie algebras in a straightforward manner. We may talk of maximal subalgebras and maximal ideals of

20 a Lie K-algebra L. These are respectively the natural analogs of maximal subgroups and maximal normal subgroups of a group G.

Similarly, if M is an ideal of a Lie K-algebra L we may form the quotient Lie algebra L/M: L/M is already a vector space over K and we deﬁned the Lie bracket here by [x + M, y + M] := [x, y] + M for all x + M, y + M ∈ L/M. The center of the Lie K-algebra L is Z(L) = {x ∈ L | [x, y] = 0 for all y ∈ L} and a central series in L is a chain of ideals

0 ¡ L1 ¡ L2 ¡ ··· ¡ Ln = L

such that Li+1/Li ≤ Z(L/Li) for i = 2, 3, . . . , n. If L has a central series and the shortest one has length c we say that L is nilpotent of class c. The reader will notice that all of this is strictly analogous to the group theoretic situation considered in Chapter 1. Continuing in this vein, if L is a Lie K-algebra we deﬁne

γ1(L) = L and γi+1(L) = [γi(L),L], the subalgebra generated by all products [x, y] where x ∈ γi(L) and y ∈ L. It is not diﬃcult to see that γi(L) ¡ L for all i and that

... ≤ γn(L) ≤ γn−1(L) ≤ ... ≤ γ1(L) = L is a central series in L. Naturally this is called the lower central series of L. Just as in the group case, L is nilpotent of class c if and only if γc+1(L) = 0 and γc(L) 6= 0.

Our proof of Theorem 3.3 provides an explicit non-zero element of γ2n(L), the (2n)th term of the lower central series of L. The existence of such an element can be deduced from ‘Schur-Weyl duality’, particularly [4, Section 9.3.4], and a celebrated theorem of Klyachko [9], at least in case K = C. In this work we do not pursue this approach although we point out that the operator Θ introduced below is the anti-symmetrizer of a particular Young tableau. Note that the obvious analogue of Theorem 3.3 for n = 1 is true without any restric- tion on K since one can take L to be the Lie K-algebra of smallest dimension that has class 2; evidently this dimension is 3. On the other hand, Theorem 3.15 below shows that Theorem

21 3.3 does not extend to the case n = 2. An interesting consequence of Theorem 3.15 is that the answer to Question 3.2 is ‘no’ for n = 2. With Theorem 3.3 in hand we can prove our main group-theoretic results.

Theorem 3.4. Let n ≥ 3 be a positive integer and let p be a prime number such that

p > n + 1. Then there exists a ﬁnite p-group Hn of class 2n in which every proper subgroup has class at most n.

Theorem 3.5. Let n ≥ 3 be a positive integer. Then there exists a torsion-free nilpotent group Kn of class 2n in which every subgroup of inﬁnite index has class at most n.

Theorem 3.4 follows very quickly from Theorem 3.3 and some well-known properties of the Lazard correspondence. Similarly, after discussing torsion-free radicable nilpotent groups, we will use Theorem 3.3 and the Mal’cev correspondence to deduce Theorem 3.5. A brief discussion of these correspondences is given below. If x, y ∈ L for some Lie algebra L so that [x, y] denotes the Lie product of x and y,

[x, y, z] means [[x, y], z], and we write [x, ry] for [x, y, y, . . . , y], where there are r occurrences of y. Whether [x, y] denotes a group commutator or a Lie product will always be clear from the context.

Throughout the sequel K denotes a ﬁeld. Moreover, for each prime p, we write Fp for the ﬁeld that has exactly p elements. Before we prove Theorem 3.3 we recall some facts from multilinear algebra and the representation theory of general linear groups. Two good general references for this material are [3] and [4].

Let V denote a vector space over K and let T = T (V ) denote the tensor algebra of L∞ r 0 r V . We view T (V ) as a graded K-algebra: T (V ) = r=0 T (V ) where T (V ) = K and T =

r T (V ) is the subspace of T (V ) spanned by {v1 ⊗ v2 ⊗ ... ⊗ vr | vi ∈ V for i = 1, 2, . . . , r }, the set of simple tensors of degree r. Let B be a basis of V and note that T r is spanned by the set of simple tensors of the form v1 ⊗ v2 ⊗ ... ⊗ vr where vi ∈ B for i = 1, 2, . . . , r.

22 r There is an action of GL(V ) on each T given by α(v1 ⊗ v2 ⊗ ... ⊗ vr) = α(v1) ⊗

α(v2) ⊗ ... ⊗ α(vr) for all α ∈ GL(V ) and all vi ∈ V and so we have an action of GL(V ) on the graded algebra T (V ) that preserves degree. Note that the eﬀect of α on each T r is completely determined by its eﬀect on the simple tensors v1 ⊗ v2 ⊗ ... ⊗ vr with vi ∈ B for i = 1, 2, . . . , r.

For each positive integer r, let Sr denote the symmetric group of degree r. Then,

r for each r ≥ 1, there is a right action of Sr on T given by (w1 ⊗ w2 ⊗ ... ⊗ wr)ρ =

r wρ(1) ⊗ wρ(2) ⊗ ... ⊗ wρ(r) for all ρ ∈ Sr and all wi ∈ V . Therefore we may view T as a right

r KSr-module. The (left) action of GL(V ) on T commutes with this right action of KSr

α(w)Φ = α(wΦ)

r for all α ∈ GL(V ), w ∈ T , and Φ ∈ KSr, and so each element of KSr can be viewed as (inducing) a GL(V )-module homomorphism of T r.

Let n ≥ 3 be an integer and let E denote the subgroup of S2n generated by the transpositions

(1 2), (3 4), (5 n + 3),..., (i n − 2 + i),..., (n + 2 2n), where i = 5, 6, . . . , n + 2, an elementary abelian 2-group of order 2n, and let

X Θ = sgn(σ)σ, (∗) σ∈E an element of KS2n. It is not diﬃcult to see that

n+2 Y Θ = (1 − (1 2))(1 − (3 4)) (1 − (i n − 2 + i)). (∗∗) i=5 We remark in passing that Θ is the anti-symmetrizer associated with the Young tableau

23 1 3 5 . . . i . . . n + 2

2 4 n + 3 . . . n − 2 + i ... 2n

Figure 3.6: Young tableau associated with anti-symmetrizer Θ

in which i = 5, 6, . . . , n + 2.

2n Lemma 3.7. Let n ≥ 3 be a positive integer and let v = v1 ⊗ v2 ⊗ ... ⊗ v2n ∈ T be such

that at least n + 1 of the elements v1, v2, . . . , v2n are equal. Then vΘ = 0.

Proof. At least one of the ordered pairs

{v1, v2}, {v3, v4}, {v5, vn+3},..., {vi, vn−2+i},..., {vn+2, v2n},

for i = 5, 6, . . . , n + 2, contains two equal entries. In each case v is annihilated by one of the factors in the description of Θ given at (∗∗). Since these factors commute with each other, the result follows.

We next recall some well-known connections between tensor algebras, polynomial rings and Lie algebras. (A good general reference for this material is [7, Chapter V.4].) Let

A be an associative K-algebra and recall that A becomes a Lie K-algebra on deﬁning Lie

− multiplication by [a1, a2] = a1a2 − a2a1 for all a1, a2 in A. We write A for A viewed as a Lie algebra in this way.

Let Λ = Khx, yi, the free associative K-algebra freely generated by x and y. Thus Λ is the non-commutative polynomial ring on indeterminates x and y with coeﬃcients in K, L∞ and consequently Λ is a graded algebra: Λ = r=0 Λr where Λ0 = K and Λr is the subspace consisting of all homogeneous polynomials of (total) degree r. It is well known that the Lie subalgebra L of Λ− generated by x and y is a free Lie K-algebra of rank 2 and that L is a

24 L L graded Lie algebra, L = L1 L2 ... where Lr = L ∩ Λr. Note that the graded nature of

L implies that [x,n y,n−1 x] ∈ L2n. Our next lemma shows how to write [x,n y,n−1 x] explicitly as a polynomial in x and y.

Lemma 3.8. For each integer n ≥ 1 we have

n−1 n X X n − 1n [x, y, x] = (−1)l+k xlykxyn−kxn−1−l. n n−1 l k l=0 k=0

Proof. We use induction to prove that

m X m [w, u] = (−1)j ujwum−j, m j j=0

n−1 n−1 n for all u, w ∈ Λ and all j ∈ N. (Here the familiar identity r−1 + r = r is helpful.) The result follows easily on applying this formula twice: ﬁrst with w = x, m = n and u = y, then with w = [x,n y], m = n − 1, and u = x.

Throughout the rest of this section let V be the vector space over K that has B = {x, y} as a basis. It is easy to see that T (V ) is isomorphic as a graded algebra to Λ via the

r r algebra homomorphism ψ : T (V ) → Λ that restricts on each T to the map ψr : T → Λr

that ‘forgets the ⊗ symbols’, so that ψr(v1 ⊗ v2 ⊗ ... ⊗ vr) = v1v2 . . . vr for all vi ∈ V . (Indeed, our principal reason for introducing Λ is that ‘forgetting the ⊗ symbols’ greatly simpliﬁes the notation in our proof of the following lemma.) We shall identify T (V ) and Λ via ψ whenever it is convenient. Whether the symbols x, y denote elements of Λ or elements

of T (V ) will be clear from the context. This identiﬁcation allows us to view each Λr as a

left GL(V )-module and a right KSr-module.

Lemma 3.9. Let n ≥ 3 be an integer and let Θ ∈ KS2n be the element deﬁned at (∗) above. Then (n2 − 1)n [x, y, x]Θ = xyxyn−1xn−2Θ. n n−1 2 25 Therefore, if K has characteristic 0 or K has characteristic p > 0 where p does not divide (n2−1)n 2 then [x,n y,n−1 x]Θ 6= 0.

Proof. We begin by examining the effect of (right) multiplication by Θ on the expressions u = xlykxyn−kxn−1−l that occur as summands in Lemma 3.8. We are particularly interested in finding conditions on l and k that guarantee that uΘ 6= 0. Keeping in mind that n ≥ 3, it is not difficult to see that u(1 − (1 2)) = 0 or u(1 − (3 4)) = 0 unless l = 1 and k = 1 or k = 2. Thus uΘ = 0 unless l = 1 and k = 1 or

k = 2 and so most of the summands in the expression for [x,n y,n−1 x] given in Lemma 3.8 are annihilated by Θ. More precisely it now follows that

n [x, y, x]Θ = (n − 1)nxyxyn−1xn−2 − (n − 1) xy2xyn−2xn−2 Θ. n n−1 2

Furthermore, since (3 4)Θ = −Θ, we deduce that

n [x, y, x]Θ = (n − 1) nxyxyn−1xn−2 − xyxyn−1xn−2(3 4) Θ n n−1 2 n = (n − 1) n + xyxyn−1xn−2Θ 2 (n2 − 1)n = xyxyn−1xn−2Θ 2 as claimed. To complete the proof, note that if ρ ∈ E is such that xyxyn−1xn−2ρ = xyxyn−1xn−2 then ρ = 1. It now follows easily from the deﬁnition of Θ given at (∗) that the coeﬃcient of the monomial xyxyn−1xn−2 in the polynomial xyxyn−1xn−2Θ is 1, and we are done.

The following lemma records Lie-theoretic versions of two familiar results from group theory, one of which appears in Lemma 1.7.

Lemma 3.10. Let L be a ﬁnitely generated Lie algebra over a ﬁeld K and let X be a generating set for L. Then:

26 (i) L is nilpotent of class c if and only if [y1, y2, . . . , yc+1] = 0 for all yi ∈ X. (ii) if L is nilpotent and M is a maximal subalgebra of L, it follows that M is an ideal of L and L/M has dimension 1.

Proof. Part (i) follows easily from [1, Chapter 1, Lemma 1.1]. We refer the reader to [1, Chapter 12, Lemma 1.3] for a proof of part (ii).

Proof of Theorem 3.3. Let n and K be as stated and let V denote the vector space over K with {x, y} as a basis.

∞ r We set T+ = T+(V ) = ⊕r=1T (V ) and note that T+ is an associative algebra without 1. Recall that an algebra A without 1 is said to be nilpotent of class c ≥ 1 if the product of any c elements of A is 0 but there exist c − 1 elements of A whose product is non-zero. Let I denote the ideal of T (V ) that is generated by the set S that consists of all simple tensors v1 ⊗ v2 ⊗ ... ⊗ vk where k ≥ n + 1 and at least n + 1 of the vi are equal. Note that S is closed under multiplication by simple tensors w1 ⊗ w2 ⊗ ... ⊗ wr ∈ T of arbitrary degree r, on the right and on the left. Consequently I is spanned as a vector space over K by S and it follows that I ∩ T 2n is spanned as a vector space over K by the simple tensors v1 ⊗ v2 ⊗ ... ⊗ v2n in which at least n + 1 of the vi are equal. Observe that Lemma 3.7

2n now implies that wΘ = 0 for all w ∈ I ∩ T . In other words, if we view Θ ∈ KS2n as a GL(V )-homomorphism of T (V ), then I ∩ T 2n ≤ ker Θ. Let F be the (free) Lie subalgebra of T − that is generated by x and y and let L denote the Lie subalgebra of (T/I)− generated by a = x + I and b = y + I. Evidently the natural algebra epimorphism T → T/I induces the Lie algebra epimorphism F → L such ∼ that x 7→ a and y 7→ b and so L = F/(F ∩ I). We will show that the Lie K-algebra L has the desired properties.

Note that if k ≥ 2n + 1 then each simple tensor of the form t(x, y) = v1 ⊗ v2 ⊗ ... ⊗ vk

where vi ∈ {x, y} for i = 1, 2, . . . , k has at least n + 1 components that are equal. It follows that each such t(x, y) lies in S and hence in I. Moreover, it is easy to see that T k is spanned

k by elements of the form t(x, y) and so T ≤ I for all k ≥ 2n + 1. Therefore A = T+/I

27 is nilpotent of class at most 2n as an associative algebra without 1 and, since L ≤ A−, it follows easily that L is nilpotent of class at most 2n. On the other hand, Lemma 3.9 shows

2n that [x,n y,n−1 x] ∈/ ker Θ, and so the argument above shows that [x,n y,n−1 x] ∈/ I ∩ T .

Thus [a,n b,n−1 a] 6= 0 and we deduce that L has class 2n. It remains to show that every proper subalgebra of L has class at most n and to accomplish this it suﬃces to show that every maximal subalgebra of L has class at most n. Since F and I are GL(V )-invariant, and elements of GL(V ) act on them as Lie algebra automorphism, there is an induced action of GL(V ) on L in which elements of GL(V ) act as Lie algebra automorphisms. Now it follows easily from Lemma 3.10 (ii) that the maximal subalgebras of F are precisely the ideals γ2(F )⊕K(k1x+k2y) where k1, k2 ∈ K are such that k1x + k2y 6= 0. Since γ2(F ) is GL(V )-invariant and GL(V ) acts transitively on the set of non-zero elements of V , GL(V ) acts transitively on the set of maximal subalgebras of F and therefore on the set of maximal subalgebras of L. Consequently all maximal subalgebras of L have the same class, which we proceed to show is at most n.

Let M denote the maximal subalgebra of F generated by γ2(F ) and x. Then M is generated by X = {x, [x, y], [x, y, y],... } = {x, x ⊗ y − y ⊗ x, x ⊗ y ⊗ y − y ⊗ x ⊗ y − y ⊗ x ⊗ y − y ⊗ y ⊗ x, . . . } as a Lie algebra (see, for instance, [15, Theorem 0.6]). Note that in this last description of the generating set X, each simple tensor that occurs (as a summand of a more complicated tensor), has exactly one component that is equal to x. Consequently, if w1, w2, . . . , wn+1 ∈ X then f = [w1, w2, . . . , wn+1] ∈ I since f is a linear combination of simple tensors each of which has (exactly) n + 1 components that are equal to x. It follows from Lemma 3.10 (i) that M has class at most n and we deduce that each maximal subalgebra of L has class at most n. The proof is complete.

For the proofs of Theorems 3.4 and 3.5, we will need to make use of Mal’cev com- pletions and the correspondences of Mal’cev and Lazard. We begin with a discussion of

radicable torsion-free nilpotent groups and their relationship to nilpotent Lie Q-algebras.

28 Our results are stronger than needed to prove Theorems 3.4 and 3.5 but are of independent interest.

m Let K be a nilpotent group and let H ≤ K. Then IK (H) = {x ∈ K | x ∈ H for some positive integer m}, the isolator of H in K, is a subgroup of K, see [10, Sec-

tion 2.3] for example. We remark that if K is torsion-free then IK (H) has the same class as √ H, (see for instance, [10, 2.3.9 (iii)]. We sometimes write H for IK (H) if K is clear from the context. A group G is said to be radicable if for every g ∈ G and every positive integer m, the equation xm = g has a solution x ∈ G. In torsion-free nilpotent groups, extraction of roots, if possible, is always unique [10, 2.1.2]. Consequently, radicable torsion-free nilpotent

groups are precisely those groups referred to as Q-powered nilpotent groups in [8], which is our most important reference in the sequel. Let G be a torsion-free nilpotent group. We say that a torsion-free nilpotent group G∗ is a Mal’cev completion of G if (i) G is a subgroup of G∗, (ii) G∗ is radicable, and (iii)

∗ IG∗ (G) = G . Mal’cev [12] has shown that such a group exists and that it is unique up to

isomorphism. We write it as GQ and call it the Mal’cev completion of G. We refer the reader to [10, Section 2.1] or [8, Chapter 9] for a modern approach to Mal’cev’s theorem. Note that √ if H ≤ GQ then H, the isolator of H in GQ, is (isomorphic to) HQ. It follows from the √ remark above that H has the same class as H and so, in particular, GQ has the same class as G.

Theorem 3.11. Let G be a radicable torsion-free nilpotent group and let c and l be positive integers with l < c. Then G has class c and all proper radicable subgroups of G have class at most l if and only if G is the Mal’cev completion of a (necessarily ﬁnitely generated) torsion-free nilpotent group H of class c in which all subgroups of inﬁnite index have class at most l.

Proof. Suppose first that G = HQ where H is a finitely generated torsion-free nilpotent group of class c in which all subgroups of infinite index have class at most l and note that

29 G has class c. Let K be a proper radicable subgroup of G and let X = K ∩ H. Clearly √ √ K ≤ X and it follows easily that K = X. Thus K and X have the same class. If [H : X] √ √ is finite then K = X = H = G, a contradiction. Thus X has infinite index in H and so the class of X, and therefore the class of K, is at most l. Now suppose that G has class c and all proper radicable subgroups of G have class at most l. Let H be a finitely generated subgroup of G that has class c; clearly such a subgroup √ √ exists. Now H is a radicable subgroup of G and, since H has class c, we deduce that √ Q G = H = H . Let K be a subgroup of infinite index in H. If IG(K) = G then IH (K) = H

and so for every h ∈ H there exists m ∈ N such that hm ∈ K. Arguing by induction on the defect of K in H it is easy to show that this implies that [H : K] < ∞, a contradiction.

Therefore IG(K) is a proper radicable subgroup of G and so has class at most l. Thus K has class at most l and the proof is complete.

The Mal’cev correspondence is a one-to-one correspondence between the class of rad-

icable torsion-free nilpotent groups and the class of nilpotent Lie Q-algebras. We will not go into great detail about this correspondence since there are several excellent treatments available, [1, Chapter 6] and [8, Chapters 9 and 10] for example. Here we adopt the approach of [8] in which corresponding groups and Lie algebras have the same underlying set. We refer the reader to [8, Theorem 10.11] and the discussion that precedes it for justification for the following summary: given a nilpotent Lie Q-algebra L one uses the Baker-Campbell-Hausdorff formula to define a multiplication on L in such a way as to obtain a radicable torsion-free nilpotent group GL. Conversely, if G is a radicable torsion-free nilpotent group one uses the inverse Baker-Campbell-Hausdorff formulae to define Lie operations + and [ , ] on G which,

together with a rule for deﬁning multiplication by elements of Q, deﬁne the structure of

a nilpotent Lie Q-algebra LG on the same set G. The crucial fact is that GLG = G and

LGL = L for all radicable torsion-free nilpotent groups G and all nilpotent Lie Q-algebras L. The approach taken in [8] makes it clear that a subset H of a radicable torsion-free nilpotent

30 group G is a radicable subgroup of G if and only if H is a Lie subalgebra of LG and then H,

viewed as a subalgebra of LG, is LH [8, Theorem 10.13 (a)]. (In the terminology of [8], ‘···

then H as a Q-powered group is in the Mal’cev correspondence with H as a Lie Q-algebra’.)

Together with the fact that G and LG have the same class [8, Theorem 10.13 (d)], this allows us to deduce the following result.

Theorem 3.12. Let c and l be positive integers with l < c. (i) Let G be a radicable torsion-free nilpotent group of class c in which every proper radicable

subgroup has class at most l. Then LG has class c and every proper subalgebra of LG has class at most l.

(ii) Let L be a nilpotent Lie Q-algebra of class c in which every proper subalgebra has class

at most l. Then GL has class c and every proper radicable subgroup of GL has class at most l.

Combining Theorems 3.11 and 3.12 in an obvious way yields the following.

Corollary 3.13. Let c and l be a positive integers with l < c. There exists a torsion-free nilpotent group of class c in which every subgroup of inﬁnite index has class at most l if and

only if there exists a nilpotent Lie Q-algebra of class c in which every proper subalgebra has class at most l.

One can also use the Baker-Campbell-Hausdorff formula to define the so-called Lazard correspondence which requires rather more restrictive conditions than the Mal’cev correspondence but allows for applications to finite groups. Again we refer the reader to [8] for details. Note that the discussion on page 124 of that work shows that, for each prime p, there exists a correspondence between nilpotent p-groups of class less than or equal to p − 1 and nilpotent Lie rings of class less than or equal to p − 1 whose additive group is a p-group. In particular, the analogue of the [8, Theorem 10.3] for the Lazard correspondence allows us deduce the following result in the same way that we deduced Theorem 3.12(ii) from [8, Theorem 10.3] .

31 Theorem 3.14. Let p be a prime, let c and l be positive integers and supose that l < c < p.

If there exists a ﬁnite nilpotent Lie Fp-algebra of class c in which every proper subalgebra has class at most l, then there exists a ﬁnite p-group of class c in which every proper subgroup has class at most l.

To prove Theorems 3.4 and 3.5 it only remains to assemble some of our previous results.

Proofs of Theorems 3.4 and 3.5. Let n ≥ 3 be a positive integer. We ﬁrst prove Theorem

3.5. Now Theorem 3.3 shows that there exists a nilpotent Lie Q-algebra L of class 2n in which every proper subalgebra has class at most n. Corollary 3.13 now applies with c = 2n and l = n to provide the desired group Kn.

(n2−1)n To prove Theorem 3.4 let p be as stated and observe that p does not divide 2 = (n+1)n(n−1) 2 . Consequently Theorem 3.3 shows that there a nilpotent Lie Fp-algebra L of class 2n in which every proper subalgebra has class at most n. Clearly L is ﬁnite dimensional and therefore ﬁnite. Theorem 3.14 now applies with c = 2n and l = n to provide the desired

group Hn.

We will now settle the remaining case, n = 2, of Question 3.2.

Theorem 3.15. Let K be a ﬁeld of characteristic diﬀerent from 2. Then there does not exist a nilpotent Lie K-algebra of class 4 in which every proper subalgebra has class at most 2.

Proof. Suppose, for a contradiction, that such a Lie algebra L exist and that X is a smallest

generating set for L (as a Lie K-algebra). Now suppose further that L is not 2-generator, i.e. that |X|> 2. Then L can- not even be generated by two elements of X together with L0 (see [1, page 242]). Let

y1, y2, y3, y4 ∈ X and note that [y1, y2], y3, y4 generate a proper subalgebra M of L. It follows

that [y1, y2, y3, y4] = [[y1, y2], y3, y4] = 0 and Lemma 3.10 (i) now shows that L is of class

32 at most 3, a contradiction. We deduce that L is 2-generator and so X = {a, b} for some a, b ∈ L. Note that [a, b] and y together generate a proper subalgebra of L, for each y ∈ L, and so 0 = [a, b, a + b, a + b] = [a, b, a, a] + [a, b, a, b] + [a, b, b, a] + [a, b, b, b] = [a, b, a, b] + [a, b, b, a] and [a, b, a, b] = −[a, b, b, a]. Moreover, the Jacobi identity yields 0 = [[a, b], a, b]+[a, b, [a, b]]+ [b, [a, b], a] = [a, b, a, b] + [b, [a, b], a] = [a, b, a, b] − [[a, b], b, a] and it follows that [a, b, a, b] = [a, b, b, a]. Consequently 2[a, b, a, b] = 0 and, since L is an algebra over a ﬁeld of characteristic other than 2, we see that all weight 4 commutators involving only the symbols a and b are zero. Lemma 3.10 (i) now shows that L has class at most 3. This contradiction completes the proof.

Theorem 3.16. There does not exist a torsion-free nilpotent group of class 4 in which every subgroup of inﬁnite index has class at most 2.

Proof. Suppose that such a group exists. Then Corollary 3.13 implies that there exists a

nilpotent Lie Q-algebra of class 4 in which every proper subalgebra of has class at most 2. This contradicts Theorem 3.15 and completes the proof.

33 CHAPTER 4

ASSOCIATIVE ALGEBRAS

In this short chapter we will show that Theorems 3.3 and 3.4 are not simply special cases of some result in Universal Algebra about general nilpotent algebraic structures. We accomplish this by proving Theorem 4.4 below. Throughout this chapter our associative algebras will all be over some fixed but arbitrary field K. The exact nature of K will not play any role in our discussion. We will require the following definitions.

Deﬁnition 25. Let Λ be an associative algebra (without 1) and let n be a positive integer.

n We write Λ for the ideal of Λ generated by all elements of the form w1w2 . . . wn where

n w1, w2, . . . , wn ∈ Λ. We say that Λ is nilpotent if Λ = 0 for some positive integer n and that Λ is nilpotent of class c if Λc+1 = 0 6= Λc.

Let X be a subset of an associative algebra Λ. We write hhXii for the subalgebra of Λ generated by X. If X = {x1, x2, . . . , xn} we write hhx1, x2, . . . , xnii for hhXii (rather than hh{x1, x2, . . . , xn}ii). An associative algebra Λ is said to be cyclic if Λ = hhxii for some x ∈ Λ. Clearly cyclic associative algebras are commutative.

It is easy to see that if hhXii = Λ then each element of Λ is a K-linear combination of monomials in the symbols in X. Consequently Λ is nilpotent of class at most c if and only if w1w2 . . . wc+1 = 0 for all w1, w2, . . . , wc+1 ∈ X. The following lemma is basically Nakayama’s lemma in a non-commutative setting.

Lemma 4.1. Let Λ be an associative algebra that is nilpotent of class c for some positive integer c. Suppose that X is a subset of Λ such that hhX ∪ Λ2ii = Λ. Then hhXii = Λ.

34 Proof. We argue by induction on c. If c = 1 there is nothing to prove so suppose that c > 1 and the result holds whenever the class of Λ is smaller than c. Let Y be a generating set for Λ so that hhY ii = Λ. Our inductive hypothesis implies that X generates Λ modulo Λc and so

c for each w ∈ Λ there exist xw ∈ hhXii and zw ∈ Λ such that w = xw+zw. Now, retaining this

notation, for all u, v ∈ Λ we have uv = (xu + zu)(xv + zv) = xuxv + zuxv + xuzv + zuzv = xuxv

c+1 2 since zuxv, xuzv, zuzv ∈ Λ = 0. Thus Λ ⊆ hhXii and the result follows.

Lemma 4.2. Let Λ be a nilpotent associative algebra, let Σ be a proper subalgebra of Λ and let ∆ be the ideal of Λ generated by Σ. Then ∆ 6= Λ.

Proof. Since Λ2 ⊆ hhΣ ∪ Λ2ii it follows that hhΣ ∪ Λ2ii is an ideal of Λ. Clearly this implies that ∆ ⊆ hhΣ ∪ Λ2ii. Now Lemma 4.1 shows that hhΣ ∪ Λ2ii= 6 Λ and the result follows.

The natural analog of Fitting’s theorem for non-cyclic associative algebras is very easy to prove.

Theorem 4.3. Let Λ be a non-cyclic nilpotent associative algebra in which every proper subalgebra has class at most n. Then Λ has class at most 2n.

Proof. Let w1, w2, . . . , w2n, w2n+1 ∈ Λ. Clearly it is suﬃcient to show that

w1w2w3w4 . . . w2n−1w2nw2n+1 = 0.

2 To this end let Σ = hh{w2n+1} ∪ Λ ii and note that Lemma 4.1 shows that Σ is a proper subalgebra of Λ since Λ is non-cyclic. Thus Σ has class at most n. Now, after inserting some

2 brackets, our product is (w1w2)(w3w4) ... (w2n−1w2n)w2n+1, a product of n terms of Λ and

n+1 w2n+1, and is therefore an element of Σ = 0. The result follows.

Theorem 4.4. Let n ≥ 3 be a positive integer and let Λ be a non-cyclic nilpotent associative algebra with the property that every proper subalgebra of Λ has class at most n. Then the class of Λ is less than 2n.

35 Proof. Suppose, for a contradiction, that the class of Λ is at least 2n. If Λ is not ﬁnitely generated then each product w1w2 . . . wn+1 is zero since hhw1, w2, . . . , wn+1ii is a proper subalgebra of Λ. In this case Λ has class at most n, a contradiction. Therefore we may assume that Λ is ﬁnitely generated (as an algebra).

Let X = {x1, x2, . . . , xr} be a minimal generating set of Λ and note that r ≥ 2 since Λ is non-cyclic. For each i ∈ {1, 2, . . . , r} and each product π = y1y2 . . . y2n with y1, y2, . . . , y2n ∈ X let di(π) denote the number of occurrences of xi in π. Thus d1(π) + d2(π) + ··· + dr(π) = 2n. Let π = y1y2 . . . y2n be a ﬁxed product with y1, y2, . . . , y2n ∈ X and let k ∈ {1, 2, . . . , r} be such that dk(π) is minimal. Let Ik denote the ideal of Λ that is generated by X \{xk} and note that 2n − dk(π) of the factors y1, y2, . . . , y2n that appear in the product π belong to Ik. Now X is a minimal generating set for Λ and so Lemma 4.2 shows that Ik is a proper ideal of Λ. Therefore Ik has class at most n. It now follows that

π = 0 unless 2n − dk(π) < n + 1, i.e. unless n ≤ dk(π). We deduce that r = 2.

To ease the notation we set x1 = a and x2 = b so that {a, b} is a minimal generating set for Λ. Note that Λ/Λ2 has dimension 2 as a vector space over K. Consider a non-zero product π = (y1y2)(y3y4) ... (y2n−1y2n) in which each yi is an element of {a, b}. Obviously each of the bracketed expressions belongs to Λ2 but notice that if a bracketed expression

2 y2l−1y2l contains the same symbol, a or b, twice, then y2l−1y2l ∈ Σ where Σ is the subal-

2 n+1 gebra of Λ generated by y2l and Λ and so (y1y2)(y3y4) ... (y2n−1y2n) ∈ Σ . Since Σ is a proper subalgebra of Λ it has class at most n and so y1y2 . . . y2l = 0, a contradiction.

Therefore each of the bracketed terms (y2l−1y2l) involves one a and one b. Furthermore, if there are two consecutive brackets (y2l−1y2l)(y2l+1y2l+2) in which y2l−1 = y2l+2 we have that

3 2 (y2l−1y2l)(y2l+1y2l+2) ∈ Σ where Σ is the subalgebra of Λ generated by y2l−1 and Λ and

n+1 we again see that (y1y2)(y3y4) ... (y2n−1y2n) ∈ Σ = 0. Consequently πa = abab . . . ab or

πb = baba . . . ba where there are exactly 2n factors, a or b, in both cases. Now {a, a+b} is also a minimal generating set for Λ and so the above argument, with a + b replacing b throughout shows that a(a + b)a(a + b) . . . a(a + b)a(a + b)(a + b)(a + b) = 0.

36 Here the factors alternate a, a + b, a, a + b . . . with the exception of the penultimate term which is a + b rather that a. Now recall that the only non-zero monomial in a and b that has exactly 2n factors and begins with a is the alternating product πa = abab . . . ab. Expanding the right side of 0 = a(a + b)a(a + b) . . . a(a + b)a(a + b)(a + b)(a + b) using the distributive

2n property show that πa = 0. Similarly πb = 0 and so Λ = 0. The proof is complete.

37 REFERENCES

[1] R. K. Amayo and I. Stewart. Inﬁnite-dimensional Lie-algebras. Noordhoﬀ International Publishing, Leyden, 1974.

[2] M. Dixon, M. Evans, and H. Smith. Minimal conditions, subgroups of inﬁnite index and Frattini subgroups. In F. de Giovanni and M. L. Newell, editors, Advances in Group Theory 2002, pages 129–143, Rome, 2002. Intensive Bimester on the Theory of Groups, Aracne Editrice.

[3] W. Fulton and J. Harris. Representation Theory: A First Course. Number 129 in Graduate Texts in Mathematics. Springer-Verlag, New York, ﬁrst edition, 2004.

[4] R. Goodman and N. Wallach. Representations and Invariants of the Classical Groups. Number 68 in Encyclopedia of Mathematics and its Applications. Cambridge University Press, Cambridge, 1998.

[5] C. K. Gupta. A bound for the class of certain nilpotent groups. Journal of the Australian Mathematical Society, 5:506–511, 1965.

[6] M. Hall, Jr. The Theory of Groups. American Mathematical Society Chelsea. American Mathematical Society, Providence, second edition, 1959.

[7] N. Jacobson. Lie Algebras. Interscience, New York, 1962.

[8] E. I. Khukhro. p-Automorphisms of Finite p-Groups. Number 246 in London Mathemat- ical Society Lecture Note Series. Cambridge University Press, Cambridge, ﬁrst edition, 1998.

[9] A. Klyachko. Lie elements in the tensor algebra. Sibirsk. Mat. Zh., 15:1296–1304, 1974.

[10] J. C. Lennox and D. J. S. Robinson. The Theory of Inﬁnite Soluble Groups. Oxford Mathematical Monographs. Clarendon Press, Oxford, 2004.

[11] I. D. Macdonald. Generalisations of a classical theorem about nilpotent groups. Illinois Journal of Mathematics, 8:556–570, 1964.

[12] A. Mal’cev. Nilpotent torsion-free groups. Izv. Akad. Nauk SSSR Ser. Mat., 13:201–212, 1949.

[13] H. Neumann. Varieties of Groups. Springer, New York, 1967.

[14] L. Redei. Das schiefe Produkt in der Gruppentheorie. Comment. Math. Helv., 20:225– 264, 1947.

38 [15] C. Reutenauer. Free Lie Algebras. Number 7 in London Mathematical Society Mono- graphs, New Series. Oxford University Press, Oxford, 1993.

[16] D. J. S. Robinson. A Course in the Theory of Groups. Number 80 in Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1996.

[17] D. I. Zaicev. Stably nilpotent groups. Matematicheskie Zametki, 2(4):337–346, October 1967.