Groups with all subgroups subnormal - Dipartimento di Matematica ...

Groups with all subgroups subnormalCarlo CasoloDipartimento di Matematica “Ulisse Dini”,Università di Firenze,I-50134 Firenze Italye-mail: casolo@math.unifi.itOtranto, 4 - 8 giugno 2007

Contents1 Locally nilpotent groups 51.1 Commutators and related subgroups . . . . . . . . . . . . . . . . 51.2 Subnormal subgroups and generalizations . . . . . . . . . . . . . 111.3 Classes of groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.4 Nilpotent groups and their generalizations . . . . . . . . . . . . . 231.5 Classes of locally nilpotent groups . . . . . . . . . . . . . . . . . 291.6 Preliminaries on N 1 . . . . . . . . . . . . . . . . . . . . . . . . . 392 Torsion-free Groups 432.1 Locally nilpotent torsion–free groups . . . . . . . . . . . . . . . . 432.2 Isolators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.3 Torsion–free N 1 -groups . . . . . . . . . . . . . . . . . . . . . . . . 493 Groups of Heineken and Mohamed 553.1 General remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.2 Basic construction . . . . . . . . . . . . . . . . . . . . . . . . . . 573.3 Developements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.4 Minimal non-N groups . . . . . . . . . . . . . . . . . . . . . . . . 674 Bounded defects 714.1 n-Baer groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 714.2 Roseblade’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 744.3 First applications . . . . . . . . . . . . . . . . . . . . . . . . . . . 815 Periodic N 1 -groups 835.1 N 1 -groups of finite exponent . . . . . . . . . . . . . . . . . . . . . 835.2 Extensions by groups of finite exponent . . . . . . . . . . . . . . 975.3 Periodic hypercentral N 1 -groups . . . . . . . . . . . . . . . . . . 1016 The structure of N 1 -groups 1056.1 Solubility of N 1 -groups . . . . . . . . . . . . . . . . . . . . . . . . 1056.2 Fitting Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1076.3 Hypercentral and Smith’s groups . . . . . . . . . . . . . . . . . . 1126.4 The structure of periodic N 1 -groups . . . . . . . . . . . . . . . . 1193

4 CONTENTS7 Beyond N 1 1257.1 Generalizing subnormality . . . . . . . . . . . . . . . . . . . . . . 1257.2 Groups with many subnormal subgroups . . . . . . . . . . . . . . 1297.3 The subnormal intersection property. . . . . . . . . . . . . . . . . 1377.4 Other classes of locally nilpotent groups . . . . . . . . . . . . . . 139

Chapter 1Locally nilpotent groupsIn this chapter we review part of the basic theory of locally nilpotent groups.This will mainly serve to fix the notations and recall some definitions, togetherwith some important results whose proofs will not be included in these notes.Also, we hope to provide some motivation for the study of groups with allsubgroups subnormal (for short N 1 -groups) by setting them into a wider frame.In fact, we will perhaps include more material then what strictly needed tounderstand N 1 -groups.Thus, the first sections of this chapter may be intended both as an unfaithfullist of prerequisites and a quick reference: as such, most of the readers mightwell skip them. As said, we will not give those proofs that are too complicate or,conversely, may be found in any introductory text on groups which includes someinfinite groups (e.g. [97] or [52], for nilpotent groups we may suggest, amongmany, [56]). For the theory of generalized nilpotent groups and that of subnormalsubgroups, our standard references will be, respectively, D. Robinson’s classicalmonography [96] and the book by Lennox and Stonehewer [64].In the last section we begin the study of N 1 -groups, starting with the firstbasic facts, which are not diffucult but are fundamental to understand the restof these notes.1.1 Commutators and related subgroupsLet x, y be elements of a group G. As customary, we denote by x y = y −1 xy theconjugate of x by y. The commutator of x and y is defined in the usual way as[x, y] = x −1 y −1 xy = x −1 x y .Then, for n ∈ N, the iterated commutator [x, n y] is recursively defined as follows[x, 0 y] = x, [x, 1 y] = [x, y]and, for 1 ≤ i ∈ N,[x, i+1 y] = [[x, i y], y].5

6 CHAPTER 1. LOCALLY NILPOTENT GROUPSSimilarly, if x 1 , x 2 , . . . x n are elements of G, the simple commutator of weight nis defined recursively by[x 1 , x 2 , . . . , x n ] = [[x 1 , . . . , x n−1 ], x n ].We list some elementary but important facts of commutator manipulations.They all follow easily from the definitons, and can be found in any introductorytext in group theory.Lemma 1.1 Let G be a group, and x, y, z ∈ G. Then(1) [x, y] −1 = [y, x];(2) [xy, z] = [x, z] y [y, z] = [x, z][x, z, y][y, z];(3) [x, yz] = [x, z][x, y] z = [x, z][x, y][x, y, z];(4) (Hall-Witt identity) [x, y −1 , z] y [y, z −1 , x] z [z, x −1 , y] x = 1.Lemma 1.2 Let G be a group, x, y ∈ G, n ∈ N, and suppose that [x, y, y] = 1;then [x, y] n = [x, y n ]. If further [x, y, x] = 1, then(xy) n = x n y n [y, x] (n 2) .If X is a subset of a group G then 〈X〉 denotes the subgroup generated byX. If U and V are non-empty subsets of the group G, we set[U, V ] = 〈[x, y] | x ∈ U, y ∈ V 〉and define inductively in the obvious way [U, n V ], for n ∈ N. Finally, if A ≤ G,and x ∈ G, we let, for all n ∈ N, [A, n x] = 〈[a, n x] | a ∈ A〉.If H ≤ G, H G denotes the largest normal subgroup of G contained in H,and H G the normal closure of H in G, i.e. the smallest normal subgroup of Gcontaining H. Clearly,H G = ⋂H g and H G = 〈H g | g ∈ G〉.g∈GMore generally, if X and Y are non-empty subsets of the group G, we denoteby X Y the subgroup 〈x y | x ∈ X, y ∈ Y 〉.The following are easy consequences of the definitions.Lemma 1.3 Let H and K be subgroups of a group. Then [H, K] 〈H, K〉.Lemma 1.4 Let X, Y be subsets of the group G. ThenIf N G, then [N, 〈X〉] = [N, X].[〈X〉, 〈Y 〉] = [X, Y ] 〈X〉〈Y 〉 .The next, very useful Lemma follows from the Hall-Witt identity.

1.1. COMMUTATORS AND RELATED SUBGROUPS 7Lemma 1.5 (Three Subgroup Lemma). Let A, B, C be subgroups of the groupG, and let N be a normal subgroup such that [A, B, C] and [B, C, A] are containedin N. Then also [C, A, B] is contained in N.The rules in Lemma 1.1, as well as others derived from those, may be appliedto get sorts of handy analogues for subgroups. For instance, if A, B, Care subgroups of G and [A, C] is a normal, then [AB, C] = [A, C][B, C]. Moregenerally, we haveLemma 1.6 Let N, H 1 , . . . , H n be subgroups of the group G, with N G, andput Y = 〈H 1 , . . . , H n 〉. Then[N, Y ] = [N, H 1 ] . . . [N, H n ].The same commutator notation we adopt for groups actions: let the groupG act on the group A. For all g ∈ G and a ∈ A, we set [a, g] = a −1 a g , and[A, G] = 〈[a, g] | a ∈ A, g ∈ G〉. With the obvius interpretations, the propertieslisted above for standard group commutators continue to hold.For a group G, the subgroup G ′ = [G, G] is called the derived subgroup ofG, and is the smallest normal subgroup N of G such that the quotient G/Nis abelian. The terms G (d) (1 ≤ d ∈ N) of the derived series of G are thecharacteristic subgroups defined by G (1) = G ′ and, inductively ,G (n+1) = (G (n) ) ′ = [G (n) , G (n) ](the second derived subgroup G (2) is often denote by G ′′ ). The group G is solubleif there exists an n such that G (n) = 1; in such a case the smallest integer n forwhich this occurs is called the derived length of the soluble group G. Of course,subgroups and homomorphic images of a soluble group of derived length d aresoluble with derived length at most d.A group is said to be perfect if it has no non-trivial abelian quotiens; thus,G is perfect if and only if G = G ′ .By means of commutators are also defined the terms γ d (G) of the lowercentral series of a group G: set γ 1 (G) = G, and inductively, for d ≥ 1,γ d+1 (G) = [γ d (G), G] = [G, d G].These are also characteristic subgroups of G. A group G is nilpotent if, for somec ∈ N, γ c+1 (G) = 1. The nilpotency class (or, simply, the class) of a nilpotentgroup G is the smallest integer c such that γ c+1 (G) = 1.Lemma 1.7 Let G be a group, and m, n ∈ N \ {0}. Then(1) [γ n (G), γ m (G)] ≤ γ n+m (G);.(2) γ m (γ n (G)) ≤ γ mn (G);From (1), and induction on n, we have

8 CHAPTER 1. LOCALLY NILPOTENT GROUPSCorollary 1.8 For any group G and any 1 ≤ n ∈ N, G (n) ≤ γ 2 n(G). In particulara nilpotent group of class c has derived length at most [log 2 c] + 1.Also, by using (1) and induction, one easily proves the first point of the followingLemma, while the second one follows by induction and use of the commutatoridentities of 1.1,Lemma 1.9 Let G be a group, and 1 ≤ n ∈ N. Then(1) γ n (G) = 〈[g 1 , g 2 , . . . , g n ] | g i ∈ G, i = 1, 2, . . . , n〉.(2) If S is a generating set for G, then γ n (G) is generated by the simplecommutators of weight at least n in the elemets of S ∪ S −1 .The upper central series of a group G is the series whose terms ζ i (G) aredefined in the familiar way: ζ 1 (G) = Z(G) = {x ∈ G | xg = gx ∀g ∈ G} is thecentre of G, and for all n ≥ 2, ζ n (G) is defined byζ n (G)/ζ n−1 (G) = Z(G/ζ n−1 (G)).A basic observation is that, for n ≥ 1, ζ n (G) = G if and only if γ n+1 (G) = 1,and so G is nilpotent of class c if and only if G = ζ c (G) and c is the smallestsuch positive integer. This follows at once from the following property.Lemma 1.10 Let G be a group, and 1 ≤ n ∈ N. Then [γ n (G), ζ n (G)] = 1.The next remark is often referred to as Grün’s Lemma.Lemma 1.11 Let G be a group. If ζ 2 (G) > ζ 1 (G) then G ′ < G.Let us recall here some elementary but more technical facts, which we willfrequently use, about commutators in actions on an abelian groups.Thus, let A be a normal abelian subgroup of a group G, F ≤ A, and letx ∈ G. It is then easy to see that, for all i ∈ N,[F, i x] = { [a, i x] | a ∈ F } and F 〈x〉 = 〈 [F, i x] | i ∈ N 〉.Lemma 1.12 Let A be a normal abelian subgroup of the group G, and H ≤ G.Suppose that H/C H (A) is abelian. Then, for all a ∈ A, x, y ∈ H:[a, x, y] = [a, y, x].Proof. Since H/C H (A) is abelian, [a, xy] = [a, yx], and, by espanding the commutatorsusing Lemma 1.1, [a, y][a, x] y = [a, x][a, y] x . Since A is abelian, we getthe desired equality [a, x] −1 [a, x] y = [a, y] −1 [a, y] x .Corollary 1.13 Let A be a normal abelian subgroup of the group G, such thatG/C G (A) is abelian. Then, for all X, Y ≤ G: [A, X, Y ] = [A, Y, X].Lemma 1.14 Let A be a a normal elementary abelian p-subgroup of a groupG. Then, for all x ∈ G, [A, p m x] = [A, x pm ] for all m ∈ N.

1.1. COMMUTATORS AND RELATED SUBGROUPS 9Proof. It is convenient to look at x as to an endomorphism, via conjugation, ofthe abelian group A. Then, for all a ∈ A, [a, x] = a −1 a x = a x−1 , whence, as Ahas exponent p,[a, p x] = a (x−1)p = a xp −1 = [a, x p ],and the inductive extension to any power x pm is immediate.Corollary 1.15 Let 1 ≠ A be a normal elementary abelian p-subgroup of thegroup G. If G/C G (A) is a finite p-group, then there exists n ≥ 1 such thatA ≤ ζ n (G).Proof. Let C = C G (A). We argue by induction on m, where |G/C| = p m . Ifm = 0, A is central in G. Thus, let m ≥ 1, N/C a maximal subgroup of G/C,and x ∈ G \ N. Then, by inductive assumption, A ≤ ζ k (N), for some k ≥ 1. LetA 0 = ζ(N) ∩ A; then A 0 ≠ 1 and C G (A 0 ) ≥ N. Now, x p ∈ N, and by Lemma1.14[A 0 , p x] = [A 0 , x p ] ≤ [A 0 , N] = 1.This means that A 0 ≤ ζ p (G). Ny repeating this same argument for all thecentral N-factors contained in A, we get [A, pk G] = 1, whence A ≤ ζ pk (G).Lemma 1.16 Let A be an abelian group, and x an automorphism of A suchthat [A, n x] = 1, for n ≥ 1.(i) If x has finite order q, then [A qn−1 , x] = 1.(ii) If A has finite exponent e ≥ 2, then [A, x en−1 ] = 1.(iii) Let the group H act on A with [A, n H] = 1 (n ≥ 1); then γ n (H) ≤ C H (A).Proof. (i) By induction on n. If n = 1 we have nothing to prove. Thus, let n ≥ 2,and set B = [A, x]. Then [B, n−1 x] = 1, whence, by inductive assumption,[A qn−2 , x, x] = [[A, x] qn−2 , x] = [B qn−2 , x] = 1.Now, let b ∈ A qn−2 . Then, since [b, x, x] = 1 = [b, x, b], by Lemma 1.2 we have[b q , x] = [b, x] q = [b, x q ] = 1. Hence, [A qn−1 , x] = [(A qn−2 ) q , x] = 1, as wanted.(ii) By induction on n. If n = 1, then 1 = [A, x] = [A, x e0 ]. Let n ≥ 2, andset B = [A, x en−2 ] ≤ [A, x]. Then, by inductive hypothesis,[A, x en−2 , x en−2 ] = [B, x en−2 ] = 1 .By Lemma 1.2, we then have [A, x en−1 ] = [A, x en−2e ] = [A, x en−2 ] e = 1.(iii) By induction on n, being the case n = 1 trivial. Let n > 1. Then Hacts on [A, H] and [A, H, n−1 H] = 1, hence, by inductive assumption[A, H, γ n−1 (H)] = 1. (1.1)Let A 0 = [A, n−1 H] and A = A/A 0 . Then H acts on A and [A, n−1 H] = 1. Byinductive assumption we have [A, γ n−1 (H)] = 1, which means [γ n−1 , A] ≤ A 0 .

10 CHAPTER 1. LOCALLY NILPOTENT GROUPSSince [A 0 , H] = 1, we get [γ n−1 (H), A, H] = 1, which, together with (1.1 andthe Three Subgroup Lemma, yields γ n (H), A] = [H, γ n−1 (H), A] = 1.Point (iii) of Lemma 1.16 is a particular case of a theorem of Kalužnin, whichwe will state later, together with an important generalization due to P. Hall.It is not difficult to extend similar remarks to the case when A is nilpotent.in which case it is to be espected that the numerical values will depend also onthe nilpotency class of A. We show only one of these possible generalizations.Lemma 1.17 Let A be a nilpotent group of class c, and x an automorphism ofA such that |x| = q and [A, n x] = 1, for n ≥ 1. Then [A qcn−1 , x] = 1.Proof. We argue by induction on the class c of A. The case c = 1 is just point(i) of the previous Lemma. Thus, we assume c ≥ 2 and write B = A q(c−1)n−1 .Then, by inductive assumption,[B, x] ≤ γ c (A) ≤ Z(A).In particular, [B, x, B] = 1, and so by Lemma 1.2, [B qn−1 , x] = [B, x] qn−1 .Also, [B, x] is abelian and so [[B, x], x] = [B, x, x]. Thus, by case c = 1,[[B, x] qn−1 , x] = 1. Hence [B qn−1 , x, x] = 1. Thus[B qn , x] = [B qn−1 , x] q = [B qn−1 , x q ] = 1.Therefore, A qcn−1 = B qn ≤ C A (x), as wanted.Let us state a handy corollary, for which we need to fix the following notation.Given a group G, and an integer n ≥ 1, we denote by G n the subgroup of Ggenerated by the n-th powers of all the elements of G, and set G ω = ⋂ n∈N Gn .Corollary 1.18 Let G be a periodic nilpotent group. Then G ω ≤ Z(G).Now a technical result (Lemma 1.21) which will be very useful. For the proofwe first need the following observationLemma 1.19 Let A be a nilpotent group of class c > 0, and let x be an autoomorphismof A. Then, for every q ≥ 1,[A qc , 〈x〉] ≤ [A, 〈x〉] q .Proof. By induction on c. If c = 1 we have equality [A q , 〈x〉] = [A, 〈x〉] q . Thus,let c ≥ 2, T = γ c (A), and set D = [A, 〈x〉] q . Then, D is normal in A and 〈x〉-invariant. By inductive assumption, [A qc−1 , 〈x〉] ≤ DT ; i.e., setting A = A/D,[A qc−1 , 〈x〉] ≤ T ≤ Z(A).If a ∈ A and u = a qc−1 , we have [Du, 〈x〉] ≤ T , and so [Du q , x] = [Du, x] q = 1,which is to say thatthus completing the proof.[a qc , 〈x〉] = [u q , 〈x〉] ⊆ D = [A, 〈x〉] q ,

1.2. SUBNORMAL SUBGROUPS AND GENERALIZATIONS 11Corollary 1.20 Let A be a nilpotent group of class c > 0, and let x 1 , . . . , x d beautoomorphisms of A. Then, for every q ≥ 1,[A qcd , 〈x 1 〉, . . . 〈x d 〉] ≤ [A, 〈x 1 〉, . . . , 〈x d 〉] q .Lemma 1.21 Let A be a nilpotent group of class c, let x 1 , x 2 , . . . , x d be automorphismsof A such that [A, n 〈x i 〉] = 1 for all i = 1, . . . , d. Let q 1 , . . . , q d beintegers ≥ 1, and q = q 1 · · · q d . Then[A qncd , 〈x 1 〉, . . . , 〈x d 〉] ≤ [A, 〈x q11 〉, . . . , 〈xq dd〉].Proof. We argue by induction on d ≥ 1. If d = 1, q = q 1 , write R = [A, 〈x q 〉].Then R 〈A, x〉, and by applying Lemma 1.17 to the action of x on A/R, wehave (since x q centralizes A/R)[A qcn , 〈x〉] ≤ Rwhich is what we want.Let then d ≥ 2. Write s = q 1 . . . q d−1 and B = [A sncd−1 , 〈x 1 〉, . . . , 〈x d−1 〉].By inductive assumptionNow, q ncdB ≤ [A, 〈x q11 〉, . . . , 〈xq d−1〉]. (1.2)d−1= s ncd q ncdd; thus, using Corollary 1.20,[A qncd , 〈x 1 〉, . . . , 〈x d 〉] ≤ [[A sncd , 〈x 1 〉, . . . , 〈x d−1 〉] qnc dBy the case d = 1 we then have[A qncd , 〈x 1 〉, . . . , 〈x d 〉] ≤ [B 〈xq〉 , 〈x q dd〉] = [B, 〈x q dd〉],from which, applying (1.2), we get the desidered inclusion., 〈xd 〉] ≤ [B qnc d , 〈x〉].1.2 Subnormal subgroups and generalizationsA subgroup H of the group G is said to be subnormal (written H ⊳ ⊳ G) if H isa term of a finite series of G; i.e. if there exists d ∈ N and a series of subgroups,such thatH = H d H d−1 . . . H 0 = G.If H ⊳⊳ G, then the defect of H in G is the shortest lenght of such a series; it willbe denoted by d(H, G). We shall say that a subgroup H of G is n-subnormal ifH ⊳ ⊳ G and d(H, G) ≤ n.Clearly, subnormality is a transitive relation, in the sense that if S ⊳ ⊳ Hand H ⊳ ⊳ G, then S ⊳ ⊳ G. Moreover, if S ⊳ ⊳ G, then S ∩ H ⊳ ⊳ H for everyH ≤ G, and SN/N ⊳ ⊳ G/N for every N G. Also, the intersection of a finiteset of subnormal subgroups is subnormal; but this is not in general true for theintersection of an infinite family of subnormal subgroups. The join 〈S 1 , S 2 〉 of

12 CHAPTER 1. LOCALLY NILPOTENT GROUPStwo subnormal subgroups S 1 and S 2 is not in general a subnormal subgroup(see [64] for a full discussion of this point).The reason why groups with all subgroups subnormal became a subject ofinvestigation lies in the following elementary facts.Proposition 1.22 (1) In a nilpotent group of class c every subgroup is subnormalof defect at most c.(2) A finitely generated group in which every subgroup is subnormal is nilpotent.Let H ≤ G; the normal closure series (H G,n ) n∈N of H in G is defined recursivleybyH G,0 = G, H G,1 = H G , and H G,n+1 = H HG,n .By definiton, H G,n+1 H G,n , and it is immediate to show that if H ⊳ ⊳ Gand H = H d H d−1 . . . H 0 = G is a series from H to G, then, for all0 ≤ n ≤ d, H G,n ≤ H n . Thus, a subgroup H is subnormal in G if and only ifH G,d ≤ H for some d ≥ 0, and the small such d is the defect of H. The followingis easily proved by induction on n.Lemma 1.23 Let G be a group, and H ≤ G. Then(1) H G,n = H[G, n H] for all n ∈ N.(2) For d ≥ 1, H is d-subnormal if and only if [G, d H] ≤ H.For our pourposes it is convenient to explicitely state also the following easyobservation.Lemma 1.24 Let H be a subgroup of the group G and suppose that, for somen ≥ 1, H G,n ≠ H. Then there exist finitely generated subgroups G 0 and H 0 ofG and H, respectively, such that [G 0 , n H 0 ] ≰ H.We recall another elemenatry and useful fact (for a proof see [64]).Lemma 1.25 Let H and K be subnormal subgroups of the group G. If 〈H, K〉 =HK, then 〈H, K〉 is subnormal in G.Series. Although we will not be directly interested in generalizations of subnormality,we will sometimes refer to them, notably to ascendancy; also, whenworking with subnormal subgroups in infinite groups, in order to have a betterunderstanding of what is going on, or to think to feasible extensions of ourresults, it may be useful to be aware of them.Our definition of a (general) subgroup series in a group is the standard oneproposed by P. Hall (which in turn includes the earlier Mal’cev’s definition).We give only a brief resume of the principal features of this basic notion, byessentially reproducing part of §1.2 of [96], to which we refer for a fuller account.Let Γ be a totally ordered set; a series of type Γ of a group G is a set{(V γ , Λ γ ) | γ ∈ Γ}

1.2. SUBNORMAL SUBGROUPS AND GENERALIZATIONS 13of pair of subgroups V γ , Λ γ of G such that(i) V γ Λ γ for all γ ∈ Γ;(ii) Λ α ≤ V β for all α < β (α, β ∈ Γ);(iii) G \ {1} = ⋃ γ∈Γ (Λ γ \ V γ ).Each 1 ≠ x ∈ G lies in one and only one of the difference sets Λ γ \ V γ .Moreover, for each γ ∈ Γ,V γ = ⋃β γ; thus the second equality inidentity (1.3) imply Λ γ = V γ+1 , and so the terms Λ γ are superfluos in definingthe ascending series. For such a series it is customary to add the term V α = Gif α is a limit ordinal. Hence, given an ordinal α, an ascending series of type αof G is a set of subgroups {V γ | γ ≤ α} such that V 0 = 1, V γ V γ+1 for γ < α,β>γ

14 CHAPTER 1. LOCALLY NILPOTENT GROUPSV α = G andV γ = ⋃β

1.3. CLASSES OF GROUPS 15if α is a limit ordinal (to be strictly adherent to our conventions on series weshould add the group G as last term, but this omission will not cause anytroubles, and will keep the exposition more linear). Clearly, it is a central seriesof G. The union of the terms of this series is a fully invariant subgroup of Gcalled the hypercentre of G. Thus, the hypercentre is the term ζ α (G) of theupper central series of G correspending to the smallest ordinal α such thatζ α (G) = ζ α+1 (G). The group G is called hypercentral if G is a term of the uppercentral series of G.Similarly, we talk also of the extended lower central series of a group G. Itsterms are inductively defined for every ordinal α, in the natural way, by settingγ 0 (G) = G, γ α+1 (G) = [γ α (G), γ α (G)] for every ordinal α, andγ β (G) = ⋂γ α (G)α

16 CHAPTER 1. LOCALLY NILPOTENT GROUPSa hyper–X–group. We will often refer in particular to hyperabelian groups; thatis, groups admitting a normal ascending series with abelian factors. Similarly, agroup G is said to be a hypo-X-group if G admits a descending normal series allof whose factors are X-groups. Thus, a hypoabelian group is a group admittinga normal descending series with all factors abelian. (Of course, we may definean extended derived series of the group G, by setting: G (1) = G ′ = [G, G],G (α+1) = [G (α) , G (α) ] and G (β) = ⋂ α

1.3. CLASSES OF GROUPS 17Lemma 1.29 Let (N λ ) λ∈Λ be a family of normal subgroups of the group G,such that ⋂ λ∈Λ N λ = 1. Let H ≤ G, and Z = C G (H). Then Z = ⋂ λ∈Λ ZN λ.Proof. Let g ∈ ⋂ λ∈Λ ZN λ. Then, for all a ∈ H, and all λ ∈ Λ,Thus [a, g] = 1 and so g ∈ C G (H) = Z.[a, g] ∈ [H, ZN λ ] ≤ [H, N λ ] ≤ N λ .Varieties. Let W be a subset of the free group F on a countable set of freegenerators X. The variety V(W ) defined by W is the class of all groups G suchthat φ(w) = 1 for every homomorphism φ : F → G and every w ∈ W . Aconvenient way to look at this is to consider every element w = w(x 1 , . . . , x n )of W (with {x 1 , . . . , x n } a subset of X) as a law that has to be satisfied bythe groups in the variety V(W ); in the sense that G ∈ V(W ) if and only if, inG, w(g 1 , . . . , g n ) = 1 for every substitution x i ↔ g i by elements g i ∈ G. Forexample, the class of abelian groups is the variety defined by the single word[x 1 , x 2 ] = x −11 x−1 2 x 1x 2 .It is clear that every variety V(W ) is a group class which is closed by subgroups,quotients and cartesian products (and thus it is r-closed too). Theconverse of this fact is also true (for a proof ee e.g. [97], 2.3.5; or [52], 15.2.1).Theorem 1.30 (Birkhoff) A class of groups is a variety if and only if it isclosed by subgroups, quotients and cartesian products.In general, given a set W ⊆ F and a group G, the subgroup W (G) generatedby all possible substitutions by elements of G in the words w = w(x 1 , . . . , x n )of W , is called the verbal subgroup of G defined by W . ThusW (G) = 〈w(g 1 , . . . , g n ) | w(x 1 , . . . , x n ) ∈ W, g i ∈ G〉.Hence G ∈ V(W ) if and only if the W -verbal subgroup of G is trivial. Forinstance, if n ≥ 2, in any group G, the n-th term of the lower central seriesγ n (G) is the verbal subgroup defined by the single law [x 1 , x 2 , . . . , x n ].It is obvious that for every group homomorphism φ : G → H we haveφ(W (G)) ≤ W (H). Therefore verbal subgrous are fully characterstic, and inparticular if N G then W (G/N) = W (G)N/N.Locally graded groups. A group G is said to be locally graded if every nontrivialfinitely generated subgroup of G has a non-trivial finite homomorphicimage. This is a rather large class of groups, containing for instance all residuallyfinite groups and all locally–(soluble by finite) groups. It is often consideredin order to avoid finitely generated simple groups, and in particular the socalledTarski monsters, i. e. infinite groups in which all proper subgroups arecyclic of the same order. Tarski monsters do exist and have been constructedby Ol’shnskii (see [88] for the periodic case, and [87] for the torsion–free case)and Rips (unpublished).Groups like the Golod-Shafarevic finitely generated infinite p-groups (for asimple approach see Ol’shnskii [88]) are locally graded (in fact they are even

18 CHAPTER 1. LOCALLY NILPOTENT GROUPSresidually finite). In the theory of locally nilpotent groups, to avoid such groupstoo, it is sometimes convenient to restrict to a proper, but still large, subclassof the class of locally graded groups, which is denoted by W and was introducedby Phillips and Wilson in [92]: a group G is in W if every non-nilpotentfinitely generated subgroup of G has a non-nilpotent finite image. A theorem ofRobinson [95] ensures that W contains all locally (hyperabelian by finite) groups.Observe that the class of locally graded groups and the class W are clearlylocal and closed by subgroups, but are not closed by quotients, as considerartionof free groups of rank at least 2 shows.Countable recognition. In many situations, it is convenient to be able to dealjust with countable groups in a certain class. Thus, the following concept is ofimportance. We say that a class of groups P is countably recognizable if a groupG belongs to P provided that all countable subgroups of G belong to P. Observethat a finitely generated group is countable.Theorem 1.31 Let 1 ≤ c ∈ N. The following classes of groups are countablyrecognizable: nilpotent groups, nilpotent groups of class at most c, soluble groups,soluble groups of derived length at most c,Proof. The claim is clearly true for the classes of nilpotent groups of class atmost c, and of soluble groups of derived length at most c. In fact for these casesit is enough to make the assumption on finitely generated subgroups.Now, suppose that all countable subgroups of the group G are nilpotent. Inparticular all finitely generated subgroups of G are nilpotent. Suppose that, forall i ≥ 1 there exists a finitely generated subgroup U i of G whose nilpotencyclass is greater that i. Then the subgroup 〈U i ; i ∈ N〉 is countable and notnilpotent, which contradicts our assumption. Hence there exists a bound on thenilpotency class of finitely generated subgroups of G, and so G is nilpotent. Theproof for the class of soluble groups is similar.In fact, it is not difficult to prove that a countable union of countably recognizablegroup classes is countably recognizable. For this and more general resulton this subject see section 8.3 in [96].Radicable groups. A property which is somehow opposite from being a finitenesscondition, in the sense that the trivial group is the only finite group thatsatisfies it, is radicability. A group G is radicable if for every 1 ≠ g ∈ G andevery 0 ≠ d ∈ N, there exists in G a d-rooth of g, i.e. an element h ∈ G suchthat h d = g. The most obviuos example of a radicable group is the additivegroup Q of the rationals.Radicable abelian groups are called divisible groups. Besides the group Q,the fundamental divisible groups are the groups of Prüfer type C p ∞ (often calledquasicyclic groups); these latter ones are defined for every prime number p: C p ∞is isomorphic to the multiplicative group of all p n -th complex roots of unity forall n ∈ N. The Prüfer group C p ∞ has the following presentationC p ∞= 〈u 0 , u 1 , u 2 , . . . | u 0 = 1, u p i+1 = u i for i ∈ N〉,

1.3. CLASSES OF GROUPS 19and the property that every proper subgroup is one of the 〈u i 〉 (and thus a cyclicp-group). An abelian group is divisible if and only if it is isomorphic to a directproduct of copies of Q and groups of Prüfer type (see for instance [97] 4.1.5).A group G is semi-radicable if, with the notation introduced in 1.18, G ω = G.It is rather strightforward to see that a semi-radicable abelian group is divisible.This is true for nilpotent groups also, but observe that Lemma 1.18 implies that aperiodic semi-radicable nilpotent group is abelian; on the other hand the groupsof upper unitriangular rational matrices UT (n, Q) are examples of torsion-freeradicable nilpotent groups that are not abelian.The following Lemma is a sort of refinement of 1.16(i).Lemma 1.32 Let A be a normal abelian divisible subgroup of G, and H ≤ Gsuch that [A, n H] = 1 for some positive integer n. If H/H ′ is periodic, then[A, H] = 1.Proof. Let B = [A, H, H]. Then B is normal in 〈A, H〉 and, by the Three-Subgroup Lemma 1.5, [H ′ , A] = [H, H, A] ≤ B. Thus H ′ ≤ C H (A/B), and,since A/B is divisible, it follows from Lemma 1.16 (i) that [A/B, H] = 1 or, inother words, [A, H] = B. From this, the result follows.Next, an interesting property of subnormal (more generally, ascendant) divisiblesubgroups.Lemma 1.33 Let A be a periodic abelian divisible subgroup of the group G. IfA is ascendant, then A G is abelian and divisible.Proof. Let A be an ascendant periodic divisible abelian subgroup of G and letA = A 0 A 1 . . . A α = G be an ascending series from A to G. For everyordinal β ≤ α let U β = A A β; then U β+1 ≤ A A β+1β= A β . Hence U β+1 normalizesU β and so the U β (1 ≤ β ≤ α) are the terms of an ascending series from A toU α = A G . Suppose, by contradiction, that A G is not abelian, and let β be theleast ordinal such that U β is not abelian. Then, clearly, 1 < β cannot be a limitordinal, so U β = A A β= U A ββ−1 . Let g ∈ A β. Then the abelian subgroups U β−1and U g β−1 are both normal in U β, hence [U β−1 , U g β−1 , U g β−1] = 1, and so, byLemma 1.32, [U β−1 , U g β−1 ] = 1. This shows that AA β= U β is abelian, againstour choice. Thus A G is abelian, and it is then clear that it is divisible.With the same arguments it is easy to see that two ascendant periodic divisibleabelian subgroups of a group generate an abelian (ascendant) subgroup. Fornon-periodic groups the situation can be very different: see, for instance, [64]2.1.7 for an example of a group generated by two subnormal torsion-free abeliandivisible group which is not hypoercentral.We continue by mentioning some important classes of groups defined byfiniteness conditions. We ecall that a finiteness condition is a property whichis satisfied by all finite groups (often for trivial reasons). So, for instance, theproperties of being periodic, finitely generated, locally finite or linear (i.e. isomorphicto a subgroup of some matrix group GL(n, K) for some field K) all arefiniteness conditions

22 CHAPTER 1. LOCALLY NILPOTENT GROUPS(ii) G/G ′ is finitely generated;(iii) G is polycyclic;(iv) G satisfies Max.We also recall a couple of well known and important results. The first is due toMal’cev, and the second to Hirsch (for proofs, see [97], or Segal [99] which is thestandard reference for polycyclic groups).Theorem 1.40 Let G be a polycyclic group. Then(i) Every subgroup of G is the intersection of subgroups of finite index of G;(ii) G is nilpotent if and only if every finite quotient of G is nilpotent.A further related and useful result is the following one.Theorem 1.41 A finitely generated torsion-free nilpotent group is resiiduallya finite p-group for every prime p.An important feature of polycyclic groups is the fact that if G is polycyclic,then in any finite series with cyclic factors of G the number of infinite factorsis an invariant, called the Hirsch length of G (and denoted by h(G)).Finite rank. There are several notions of rank of a group. When not otherwisespecified, by a group of finite rank we will always mean a group of finite Prüferrank, i.e. a group G with the property that there exists a d ∈ N such thatevery finitely generated subgroup of G can be generated by d elements. If thishappens, the smallest such d is called the (Prüfer) rank of G. This is clearly afiniteness condition. For example the additive groups Z, Q and the groups C p ∞all are abelian groups of rank 1.Amomg others, a much weaker condition is that of finite abelian subgrouprank: a group G has finite abelian subgroup rank if every abelian subgroup ofG which is either free abelian or elementary abelian is finitely generated.FC-groups. An FC-group is a group in which every element has a finite numberof conjugates. Thus, G is an FC-group if and only if |G : C G (g)| is finite forevery g ∈ G.Proposition 1.42 (R. Baer, B. Neumann) Let G be an FC-group. Then(i) G/Z(G) is periodic and residually finite;(ii) if g is an element of finite order of G, then 〈g〉 G is finite;(iii) the set T (G) of all elements of finite order of G is a characteristicsubgroup of G, and G ′ ≤ T (G).Strictly related to elements with a finite number of conjugates is Dic’manLemma.Lemma 1.43 Let U be a normal subset of the group G (i.e. x g ∈ U for everyx ∈ U and g ∈ G). If U is finite and consists of elements of finite order, then〈U〉 is a finite normal subgroup of G.

1.4. NILPOTENT GROUPS AND THEIR GENERALIZATIONS 231.4 Nilpotent groups and their generalizationsIn nilpotent groups the nature of the lower central factors, and sometimes thatof the whole group, is strictly related to the properties of the first of them. Thisis elucidated by the following result.Theorem 1.44 (Robinson [94]) Let H be a group and A = H/H ′ . Then, forevery c ≥ 1, there is an epimorphism:A ⊗ A ⊗ · · · ⊗ A} {{ }c times−→ γ c (H)/γ c+1 (H)From this, a number of facts connecting the properties of a nilpotent group Hto that of its abelianization H/H ′ , follow more or less easily; for instance, thefollowing elementary but basic observation.Proposition 1.45 Let G be a nilpotent group, and π a set of primes. If Gadmits set S of generators all of whose elements have finite bounded π-order,then G is a π-group of finite exponent; if, further. S is finite, then G is finite.The following useful property may be also duduced rather easily from 1.44.Proposition 1.46 The rank of a finitely generated nilpotent group G does notexceed a value which depends on the numeber of generators and the nilpotencyclass of G.Another handy fact that can be proved using 1.44 isLemma 1.47 Let H be a nilpotent group of class c ≥ 1; then the mapis a homomorphism in every variable.Thus, we have in particular,H × . . . × H → γ c (H)(x 1 , . . . , x c ) ↦→ [x 1 , . . . , x c ]Corollary 1.48 Let S be a generating set for the group G. Then G is nilpotentof class at most c if and only if [x 1 , x 2 , . . . , x c+1 ] = 1 for any elementsx 1 , x 2 , . . . , x c+1 ∈ S.The centre of a nilpotent group (i.e. the first factor of the upper centralseries) has also a certain influence on the whole group. Here is an importantinstance of this.Proposition 1.49 Let G be a nilpotent group of class c and suppose that thecentre of G has finite exponent e. Then G has exponent dividing e c .Proof. We let Z = Z(G), and proceed by induction on the nilpotency class c ofG. If c = 1 then G = Z and there is nothing to prove. Let c ≥ 2 and let y ∈ ζ 2 (G),g ∈ G. Then [g, y] ∈ Z and so, by Lemma 1.2, 1 = [g, y] e = [g, y e ]. Thereforey e ∈ Z, showing that ζ 2 (G)/Z has exponent dividing e. By inductive assumptionG/Z has exponent dividing e c−1 and from this the conclusion follows.

24 CHAPTER 1. LOCALLY NILPOTENT GROUPSThere are two more properties relative to the mutual behaviour of the termsof the lower and upper central series of a group, which are often useful, that welike to recall. Their proofs may be found, for instance, in §14.5 of [97].Proposition 1.50 (Baer) Let G be a group such that, for some i ≥ 1, G/ζ i (G)is finite, then γ i+1 (G) is finite.Proposition 1.51 (P. Hall) Let G be a group such that, for some i ≥ 1, γ i+1 (G)is finite, then G/ζ 2i (G) is finite.The class of all nilpotent groups whose nilpotency class does not exceed acertain integer c ≥ 1 will be denoted by N c . Needless to say, for every c ≥ 1,the class N c is closed by subgroups, quotients and cartesian products (in fact,it forms a variety). The class of nilpotent groups N = ⋃ c∈N N c is closed bysubgroups and quotients, but it is not closed under direct products (indeed, thesmallest variety containg N is the class of all groups). However, a fundamentalresult (due in its generality to Fitting), ensures that the subgroup generated bytwo normal nilpotent subgroups is still nilpotent (i.e. N =n 0 N); for the proofsee for instance [97] or [52].Theorem 1.52 (Fitting’s Thorem). Let H and K be nilpotent normal subgroupsof a group G, of nilpotency class c and d, respecyively. Then their joinHK is a nilpotent normal subgroup of G of nilpotency class at most c + d.Nilpotency criteria. For finite groups there are a number of conditions eachof those is equivalent to nilpotency. The next theorem lists some of the mostrelevant and simple of them.Theorem 1.53 Let G be a finite group. Then the following conditions are equivalentto nilpotency.(i) every chief factor of G is central;(ii) every maximal subgroup of G is normal;(iii) G is the direct product of its primary (i.e. Sylow) subgroups;(iv) for every proper subgroup H of G, N G (H) > H.For infinite groups all of these conditions are weaker than nilpotence, andimposing any of them gives rise to different classes of so-called generalized nilpotentgroups. In fact there are many other ways to define classes of generalizednilpotent groups, and those obtained by imposing any of the conditions (i) - (iii)of the theorem determine classes of groups that are rather far even from beinglocally nilpotent.However, there are relevant nilpotency criteria which work for arbitrarygroups. The following is one of the most useful, specially when dealing withgroups that are known to be soluble.Theorem 1.54 (P. Hall [36]) Let N be a normal subgroup of the group G. IfN is nilpotent of class c and G/N ′ is nilpotent of class d, then G is nilpotent ofclass at most ( ) (c+12 d − c)2 .

1.4. NILPOTENT GROUPS AND THEIR GENERALIZATIONS 25Proof. See [97], 5.2.10.As a consequence, we have that if the group class X is closed by quotientsand normal subgroups, and has the property that all metabelian groups in Xare nilpotent (of class bounded by c), then all soluble groups in X are nilpotent(of class bounded by a function of c and the derived length of the group).Another nilpotency criterion (also due to P. Hall) arises from the idea of aseries stanbilizer. Let S = {(V γ , Λ γ ) | γ ∈ Γ} be a series of the group G. Thestability group of S is the set of all automorphisms φ of G that centralize everyfactor of S; that is [Λ γ , φ] ≤ V γ for all γ ∈ Γ. It is readily seen that the stabilitygroup of a series is a subgroup of Aut(G). Obviously, a group H ≤ Aut(G) issaid to stabilize the series S of G if H is contained in the stability group of S.Theorem 1.55 (P. Hall [36]) Let H ≤ Aut(G) stabilize a finite series of lengthn of the group G. Then H is nilpotent of class at most ( n2).Proof. Let G = G 0 G 1 . . . G n = 1 be a series of length n of Gstabilized by H. We argue by induction on n, being the case n = 1 trivial. Letn ≥ 2 and Y 0 = Y = C H (G 1 ). By inductive hypothesis, H/Y is nilpotent ofclass at most ( )n−12 . For i ≥ 1, write Yi = [Y, i H] = [Y i−1 , H]; we show, byinduction on i, that [G, Y i ] ≤ G i+1 . This is clear for i = 0; let i ≥ 1, then[Y i , G] = [Y i−1 , H, G]. Let h ∈ H, y ∈ Y i−1 , g ∈ G, then, taking into accountthat [H, G, Y i−1 ] ≤ [G i , Y ] = 1, by the Hall-Witt identity 1.1 we have[y, h −1 , g] h [g, y −1 , h] y = 1.So [y, h −1 , g] ∈ [G, Y i−1 , H] H and, applying the inductive assumption[Y i , G] = [H, Y i−1 , G] ≤ [G i , H] ≤ G i+1 .For i = n − 1 we get [Y n−1 , G] ≤ G n = 1, that is Y n−1 = [Y, n−1 H] = 1. Thus,Y ≤ ζ n−1 (H). Since H/Y has class at most ( )n−12 , this completes the proof.The bound ( n2)on the nilpotency class of the stability group H has beenimproved by Hurley in [49]. For stabilizers of finite normal series, it is in factmuch stricter.Theorem 1.56 (Kalužnin) The stability group of a finite normal series oflength n of a group is nilpotent of class at most n − 1.Proof. Essentially the same of that of point (iii) of Lemma 1.16.A class of groups is a class of generalized nilpotent groups if it contains Nand every finite member of it is nilpotent (see chapter 6 in [96]).Local nilpotency. That of locally nilpotent groups is perhaps the most obviousclass of generalized nilpotent groups. We remind from section 1.3 that a groupG is locally nilpotent if every finitely generated subgroup of G is nilpotent. Thelocally dihedral 2-group is among the simplest examples of non-nilpotent locallynilpotent groups.Although it will not play a great role in tthe rest of these notes, the Hirsch–Plotkin Theorem is one of the basic results in the theory of infinite groups.

1.4. NILPOTENT GROUPS AND THEIR GENERALIZATIONS 27Theorem 1.61 Let G be a locally nilpotent group. Then(a) (Baer [3]) Every maximal subgroup of G is normal.(b) (Mal’cev, McLain [71]) Every chief factor of G is central. Thus, everychief series of a locally nilpotent group is central and it is a composition series.Proof. (a) Let G be locally nilpotent and suppose by contradiction that M is amaximal subgroup of G which is not normal. Then N ≱ G ′ , and so there existsg ∈ G ′ \M. Since G = 〈M, g〉, there exists a finitely generated subgroup X of Msuch that g ∈ 〈X, g〉 ′ . Let H = 〈X, g〉; then X ≤ M ∩ H, and H = (M ∩ H)H ′ .Since H is nilpotent, this forces M ∩ H = H and the contradiction g ∈ M ∩ H.(b) It is enough to show that a minimal normal subgroup A of the locallynilpotent group G is central. If this is not the case there exist a ∈ A and g ∈ Gsuch that b = [a, g] ≠ 1. Since, by minimality of A, A = 〈b〉 G , we getthus contradicting Lemma 1.60.〈a〉 ⊆ 〈b〉 G ⊆ [〈a〉, G] G = [〈a〉, G],Note that locally nilpotent groups need not admit maximal subgroups: forexample, the wreath product C p ∞ ≀ C p ∞ is a locally finite p-group with nomaximal subgroups.Lemma 1.62 Let G be a group. Then all subgroups of G are serial if and onlyif for every H ≤ G all maximal subgroups of H are normal.Proof. Suppose that every subgroup of G is serial; let H ≤ G and M a maximalsubgroup of H. By intersecting with H every term of a series of G containingM, we get a series of H containing M. But M is maximal in H, so M H.Conversely, suppose that for every H ≤ G all maximal subgroups of H arenormal. Let L ≤ G and let C be the family of all chains of subgroups of G thatcontain L as a term and satisfiy conditions (ii) and (iii) (but not necessarily (i))of the definition of a series. By standard application of Zorn’s Lemma, C hasa maximal element, which, because of the assumption on G, must also satisfynormality condition (i), and it is therefore a series of G with L as a term.Now, by point (a) of Theorem 1.61, we have:Corollary 1.63 (Baer [3]) In a locally nilpotent group every subgroup is serial.This Corollary, as well as Theorem 1.61, follows also as an application ofMal’cev’s general (and by now classical) method for proving local theorems inalgebraic systems. For this important method we refer to the Appendix of [52]or Section 8.2 of [96].Seriality of all subgroups does not imply local nilpotence. In fact, in [121]J. Wilson constructs finitely generated infinite p-groups - hence not (locally)nilpotent - in which every subgroup is serial (and every chief factor is central).On the other hand groups in which every subgroup is ascendant (called N-groups) are locally nilpotent and, as such, will be considered more at length inthe next section.

28 CHAPTER 1. LOCALLY NILPOTENT GROUPSEngel conditions. Along with that of locally nilpotent groups, the class ofEngel groups is the class of generalized nilpotent groups that have received mostattention through the years.An element g of the group G is said to be left Engel if, for any x ∈ G, thereexists a positive integer n = n(g, x) such that [x, n g] = 1. If further such aninteger n does not depend on x, then g is called a left n-Engel element. A groupG is called an Engel group if every element of G is left Engel, and it is calledan n-Engel group if every element of G is left n-Engel, for a fixed n. A groupwhich is n-Engel for some n is called a bounded Engel group. For a given n ≥ 1,the class of all n-Engel groups is a variety.A classical result of Zorn (see [97], 12.3.4) ensures that finite Engel groups arenilpotent. Observe that every locally nilpotent group is an Engel group. In fact,if G is locally nilpotent, and x, g ∈ G, then 〈x, g〉 is nilpotent, and this impliesthat, for some n ∈ N, [x, n g] = 1. On the other hand, the celebrated examplesdue to Golod are finitely generated Engel groups that are not nilpotent (in fact,for every d ≥ 2 and any prime p, Golod constructs d-generated infinite p-groupsin which all (d−1)–generated subgroup are nilpotent - an thus finite). However,it appears to be still an open question whether bounded Engel groups are locallynilpotent. Although no counterexample is known, and there are important recentrusults that prove this in some relevant cases, it seems unlikely that to be truein general.The general theory of Engel groups is well beyond the scope of these notes.In fact, we will restrict to a few facts, that are more closely connected to oursubject. A few results on n-Engel groups with small n will be recalled in Section4.1, while, moving to general bounded Engel conditions, we like to mention herea couple of recent and deep theorems.Theorem 1.64 (J. Wilson [122]) A residually finite bounded Engel group islocally finite.Theorem 1.65 (Zel’manov [123]) A torsion–free locally nilpotent n-Engel groupis nilpotent of nilpotency class depending only on n.By using these and Zel’manov solution of the restricted Burnside problem, it ispossible to show that locally graded bounded Engel groups are locally nilpotent(see [54]), and then the following general statement (see, for instance, [12]).Theorem 1.66 For every n ≥ 1 there exist integers e(n) and c(n) such that ifG is a locally graded n-Engel group then γ c(n) (G) e(n) = 1.These represent the reaching point of the work of many authors, and the proofscannot be included here; what will be enough for most of our pourposes is amuch earlier version, first due to Gruenberg (and whose proof can be found in[96], 7.36).Proposition 1.67 For n, d ≥ 1 there exist integers e = e(n, d) and c = c(nd, )such that if G is a soluble n-Engel group of derived length d, then γ c (G) e = 1.

1.5. CLASSES OF LOCALLY NILPOTENT GROUPS 29e(n, d) and c(n, d) may be given explicit upper bounds; in particular one hasCorollary 1.68 A torsion-free soluble n-Engel group of derived length d isnilpotent of class bounded by n d−1 .1.5 Classes of locally nilpotent groupsIn this section we give a brief account of some relevant classes of locally nilpotentgroups. Our approach follows that of D. Robinson in the second volume of [96],in the sense that most of the classes that we will point out are defined in termsof embedding properties of all their (finitely generated or arbitrary) subgroups.Baer and Gruenberg groups. The following basic result is due to Baer [4]for the case of subnormal subgroups and to Gruenberg [32] for that of ascendantones.Theorem 1.69 Let H and K be finitely generated nilpotent subgroups of thegroup G. If H and K are subnormal (ascendant), then J = 〈H, K〉 is a subnormal(ascendant) nilpotent subgroup of G.For the proof we need a Lemma which will be useful on other occasions.Lemma 1.70 (Gruenberg [32]) Let G be a locally nilpotent group and X afinitely generated subgroup of G. If A is an ascendant (subnormal) subgroup ofG normalized by X then there exists an ascending (finite) series containing Aall of whose terms are normalized by X.Proof. Let A = A 0 A 1 . . . A α = G be an ascending series from A to G.For each ordinal β ≤ α putB β = ⋂A x β.x∈XClearly the B β (β ≤ α) are normalized by X and are the terms of a chain ofsubgroups of G with B 0 = A 0 = A and B α = A α = G; we show that theyform an ascending series. For every β < α it is clear that B β B β+1 , sowhat we have to prove is that for every limit ordinal β ≤ α, ⋃ λ

30 CHAPTER 1. LOCALLY NILPOTENT GROUPSand there is nothing to prove. Let d ≥ 1; then H J is finitely generated and soit is generated by a finite number of conjugates of H. Let H x be such a conjugate;then, like H, H x is ascendant and the defect of H in 〈H, H x 〉 is at mostd − 1, so that 〈H, H x 〉 is ascendant in G by inductive assumption. Repeatingthis argument a finite number of times, we conclude that H J is ascendant inG. Since K normalizes H J , by Lemma 1.70 there exists an ascending seriesH J = T 0 T 1 . . . T α = G all of whose terms are normalized by K.For each ordinal β ≤ α, let J β = T β K. As, clearly, ⋃ λ

1.5. CLASSES OF LOCALLY NILPOTENT GROUPS 31(n ≥ 1). Let g be the automorphism of A that fixes every direct summand A nand acts on it as a unitriangular matrix whose non-diagonal entries are 1 overthe main diagonal and 0 everywhere else (thus, g is the linear extension of themap e g 1,n = e 1,n and e g i,n = e i,n + e i−1,n if 0 < i ≤ n). Let G = A⋊〈g〉 bethe semidirect product defined by this action. Then G is clearly torsion-free. Toprove that G is a Gruenberg group it is enough to show (by Theorem 1.71) that〈g〉 is ascendant in G. But this is clear: for every n ≥ 1, let B n = A 1 × . . . × A nand B 0 = 1; then B n+1 〈g〉/B n ≃ A n+1 〈g〉 and so B n 〈g〉 is subnormal in B n+1 〈g〉for all n ≥ 0. By refining each these intermediate finite series we get an ascendingseries (of type ω) from 〈g〉 to G (more formally, assign the inverse lexicographicorder to the base {e i,n | 1 ≤ i ≤ n, 0 ≠ n ∈ N} of A, and for each (i, n) letB (i,n) = 〈e j,k | (j, k) ≤ (i, n)〉; then the subgroups H (0,0) = 〈g〉 and H (i,n) =B (i,n) 〈g〉, for 1 ≤ i ≤ n, are the terms of an ascending series). However, G isnot a Baer group. In fact, for each n ≥ 2, [A n , n−1 〈g〉] ≠ 1, and so 〈g〉 cannotbe subnormal in G.Not all locally nilpotent groups are Gruenberg groups (see [96] for an example).On the other hand, by observing that a countable locally nilpotent groupis the union of an ascending chain of (finitely generated) nilpotent groups, oneeasily proves that every countable locally nilpotent group is a Gruenberg group.Thus, in particular, the class of Gruenberg groups is not countably recognizable;while it easily follows from Lemma 1.24 that the class of Baer groups iscountably recognizable.We now give another characterization of Gruenberg groups inside the class oflocally nilpotent groups. Following Mal’cev we say that a group G is a SN ∗ -group if G admits an ascending series with abelian factors. Since subgroups andquotients of abelian groups are abelian, it is easy to see that every subgroupand every quotient of a SN ∗ -group is an SN ∗ -group.Lemma 1.72 A group G has a unique maximal normal SN ∗ -subgroup, whichcontains every ascendant SN ∗ -subgroup of G.Proof. Suppose that N 1 N 2 N 3 . . . is a chain of SN ∗ -subgroups of thegroup G. Then, for every n ≥ 1, N n /N n−1 is a SN ∗ -group. So, if we start fromthe terms of an abelian ascending series of N 1 and successively add the inverseimages modulo N n−1 of the terms of an abelian ascending series of N n /N n−1 ,we eventually get an abelian ascending series of N = ⋃ n∈N N n; therefore, Nis a SN ∗ -subgroup of G. If we further assume that all the subgroups N n arenormal in G, we get that ⋃ n∈N N n is a normal SN ∗ -subgroup of G. Thus, byZorn’s Lemma every group G admits maximal normal SN ∗ -subgroups. A similarargument shows that if N and K are normal SN ∗ -subgroups of G, then NK isa normal SN ∗ -subgroup of G. This proves that G has a unique maximal normalSN ∗ -subgroup, which we may call the SN ∗ -radical of G.Just for this proof, let us denote by Θ(G) the SN ∗ -radical of a group G. LetH be an ascending SN ∗ -subgroup of G, and H = H 0 ≤ H 1 ≤ . . . ≤ H α = Gan ascending series from H to G. We prove that H ≤ Θ(G) by induction onthe ordinal α. Let α = β + 1; since Θ(H β ) is characteristic in H β , Θ(H β ) is anormal SN ∗ -subgroup of G, and so it is contained in Θ(G). Now, H ≤ Θ(H β )by inductive assumption, and we are done. Thus, let α be a limit ordinal. Then

32 CHAPTER 1. LOCALLY NILPOTENT GROUPSthe inductive assumption ensures that Θ(H λ ) ≤ Θ(H µ ), for all λ ≤ µ < α.Hence S = ⋃ β

1.5. CLASSES OF LOCALLY NILPOTENT GROUPS 33abelian factors, and let d = 1+p+. . .+p n−1 . Then for every g ∈ G, [G, d 〈g〉] = 1.Thus, a soluble p-group of finite exponent is a Baer group and a bounded Engelgroup.Proof. Fixed a prime p, for n ≥ 1 we write d(n) = 1 + p + . . . + p n−1 . We thenargue by induction on n. If n = 1 our claim is trivial, so let n ≥ 2, g ∈ G and letA by the first non trivial term of a normal series of G with elementary abelianfactors. Then G/A has a series of this kind with n − 1 factors, therefore, byinductive assumption [G, d(n−1) 〈g〉] ≤ A. Now, observe that certainly g pn−1 ∈ A.Thus, since A is an elementary abelian p-group, by Lemma 1.14 we have[A, p n−1 〈g〉] = [A, p n−1 g] = [A, g pn−1 ] = 1.Therefore 1 = [G, d 〈g〉] , where d = d(n − 1) + p n−1 = d(n).Corollary 1.77 A p-group with a normal nilpotent subgroup of finite index andfinite exponent is nilpotent.Proof. A group satisfying the assumptions in the statement is certainly soluble,so the result follows immediately from 1.76 and 1.75.As we are interested in subnormal subgroups, let us also mention the following.Lemma 1.78 A perfect subnormal subgroup of a Baer group is normal.Proof. See [64], 2.5.10.Let G be a group. Then the subgroup B(G) generated by all cyclic subnormalsubgroups of G is called the Baer radical of G; the subgroup Γ(G) generated byall cyclic ascendant subgroups of G is called the Gruenberg radical of G.Clearly, Baer and Gruenberg radicals are characteristic subgroups containedin the Hirsch-Plotkin radical, and, by Theorem 1.71 they contain, respectively,every subnormal (ascendant) Baer (Gruenberg) subgroup of the group. We remarkthat, even in a locally nilpotent group, the subgroup generated by twosubnormal nilpotent subgroups need not be nilpotent (see [64] for more details).Finally, observe the following consequence of Theorem 1.73.Corollary 1.79 Let G be a locally nilpotent group and let Γ(G) be its Gruenbergradical. Then Γ(G/Γ(G)) = 1.Fitting groups. A group G is called a Fitting group if every finitely generatedsubgroup of G is contained in a normal nilpotent subgroupBy Fitting’s Theorem, a group G is a Fitting group if and only if, for everyelement x of G, the normal closure 〈x〉 G = 〈{x g |g ∈ G}〉 is nilpotent. The classof Fitting groups is contained in the class of Baer groups, and it is closed bysubgroups and homomorphic images, but not by normal products (see Theorem2.1.2 in [64]). The following remark is easy to prove.Proposition 1.80 A Fitting group is hyperabelian.

34 CHAPTER 1. LOCALLY NILPOTENT GROUPSThe Fitting radical F (G) of a group G is the subgroup generated by all x ∈ Gsuch that 〈x〉 G is nilpotent.In other terms, the Fitting radical of G is the subgroup generated by all normalnilpotent subgroups of G. Clearly, F (G) is a Fitting group and is containedin the Baer radical B(G), but in general it does not contain all normal Fittingsubgroups of G. On the other hand, examples constructed by Dark show thatthere exist Baer groups with no non-trivial normal abelian subgroup.Lemma 1.81 A nilpotent by abelian Baer group is a Fitting group.Proof. Let G be a Baer group such that G ′ is nilpotent, and let x ∈ G. Then〈x〉 G = 〈x〉[G, 〈x〉] ≤ 〈x〉G ′ , which is nilpotent by Lemma 1.75.Wreath products (and wreath powers) are a very useful tool when consructinggroups with particular features. The next example is a simple issue of that.Before, let us recall an easy property of standard restricted wreath products.Lemma 1.82 Let A, H be groups and G = A ≀ H the standard wreath product.We look at H as a subgroup of G (complementing the base group). Let K be aninfinite subggroup of H. Then N G (K) = N H (K) (in particular, if H is infinite,N G (H) = H).Proof. Let A, H, G and K be as in the statement, and let G = BH, where B isthe base group. Then, N G (K) = N B (K)N H (K). Now [N B (K), K] ≤ B ∩ K = 1and so N B (H) = C B (H). Now, an element f ∈ B centralizes K if and only iff is constant on all orbits of K. Since H is taken in its regular permutationrepresentation and K is infinite, such orbits are all infinite, and so (being ourproduct the restricted one) C B (K) = 1. Thus N G (K) = N H (K).Example. An abelian by nilpotent Baer group that is not Fitting. Let p be afixed prime and let A be a vector space over the field GF (p) with base indexedon N, {a i | i ∈ N}. Let x be the automorphism of A defined bya x i = a i + a i−1 if i ≢ 0 (mod p) and a x np = a np (∀n ∈ N).Observe that x has order p. We look at x as an automorphism of the additivegroup A (which is an elementary abelian p-group) and consider the semidirectproduct H = A⋊〈x〉. Then, being A abelian, an easy computation shows thatζ(H) = C A (x) = 〈a i | i ≡ 0 (mod p)〉, and for 1 ≤ n ≤ p − 1,ζ n (H) = 〈a i | i ≡ 0, 1, . . . , n − 1 (mod p)〉.So, H/ζ p−1 (H) is abelian, and thus H is nilpotent of class p (and exponentp 2 ). Consider now the wreath product G = C p ≀ H = BH, where C p is a cyclicgroup of order p. Then G is soluble of exponent p 3 and so, by Proposition 1.76,it is a Baer group. On the other hand, 〈x〉 H contains all elements [a i , x] andso 〈x〉 H = H ′ 〈x〉 is an infinite subgroup of H. By 1.82, N G (〈x〉 H ) = H. But〈x〉 G ∩ H = 〈x〉 H , and so 〈x〉 H is self-normalizing in 〈x〉 G which therefore is notnilpotent (for, clearly, 〈x〉 G > 〈x〉 H ). Thus G is not a Fitting group.

1.5. CLASSES OF LOCALLY NILPOTENT GROUPS 35Hypercentral groups. Hypercentral groups, often called ZA-groups, are anatural generalization of nilpotent groups. We recall their definition.A group G is hypercentral if it admits an ascending central series.Arguing as in the finite case, it is easy to show that a group G is hypercentralif and only if G coincides with its hypercentre, or, in other words, if there existsan ordinal α such that ζ α (G) = G. If G is hypercentral, then the least ordinalα such that ζ α (G) = G is called the (hypercentral) length of G. One has thefollowing easy characterization of hypercentral groups.Proposition 1.83 A group G is hypercentral if and only if every non-trivialhomomorphic image of G has non-trivial centre.Proof. Since the quotient of any group modulo its hypercentre has obviouslytrivial center, one implication is clear. Conversely, let G be hypercentral of lengthα, and let N be a proper normal subgroup of G. Then there exists a smallestordinal β < α such that N does not contain ζ β (G). Clearly β is not a limitordinal, and is not 0. Thus, ζ β−1 (G) ≤ N, and so it follows that ζ β (G)N/N isa non-trivial central subgroup of G/N.Thus, the class of hypercentral groups is closed by subgroups and quotients;we leave to the reader the exercise of proving that it is countably recognizable.A simple way to constract hypercentral groups of length ω is to take directproducts of nilpotent groups with unbounded nilpotency class. The locally dihedral2-group is hypercentral of length ω+1, and, similarly, all Černikov p-groupsare hypercentral. The group in the example above is a torsion-free hypercentralgroup of length ω + 1.In fact, for every ordinal α there exist hypercentral groups of length α.It is not difficult to show directly that every hypercentral group is locallynilpotent. However, we take a different approach.Proposition 1.84 Every subgroup of a hypercentral group is ascendant.Proof. Let H be a subgroup of the hypercentral group G, and suppose thatG has hypercentral length α. Then, by setting H λ = ζ λ (G)H, for all ordinalsλ ≤ α, one clearly obtains an ascending series from H to G.Therefore a hypercentral group is a Gruenberg group and so it is locallynilpotent. The locally dihedral 2-group is the simplest example of a hypercentralgroup which is not a Baer group (on the other hand it is clear that a hypercentralgroup of length ω is a Fitting group). Hypercentral groups share with nilpotentgroups a number of useful properties, that may be proved by adjusting in aneasy way the proof for the nilpotent case;.Lemma 1.85 Let G be a hypercentreal group. Then(i) If 1 ≠ N G then N ∩ ζ(G) ≠ 1;(ii) if A is a maximal normal abelian subgroup of G, then A = C G (A).With the aid of 1.85 it is easy to prove that the class of hypercentral groups isclosed by normal products.

36 CHAPTER 1. LOCALLY NILPOTENT GROUPSProposition 1.86 (P. Hall [37]) Let H, K be two normal hypercentral subgroupsof a group; then HK is hypercentral.Proof. If a group G is the product of two normal hypercentral subgroups, and Wis the hypercentre of G, then G/W is also a product of two normal hypercentralsubgroups and, by definition of hypercentre, it has trivial centre. Thus, in orderto prove that W = G, it will suffice to show that a product 1 ≠ G = HK oftwo normal hypercentral subgroups H and K necessarily has non-trivial centre.Thus, we may assume H ≠ 1, and let Z = ζ(H). Clearly, Z ∩ ζ(K) ≤ ζ(G),so we suppose Z ∩ ζ(K) = 1. But then, since Z ∩ K K, Lemma 1.85 forcesZ ∩ K = 1. Hence [Z, K] ≤ Z ∩ K = 1, and so Z ∈ ζ(G).For further reference, we also observe the following fact.Lemma 1.87 Let N be a normal subgroup of the locally nilpotent group G. IfN is hypercentral and G/N is finitely generated, then G is hypercentral.Proof. Let S be a finite subset of G that generates G modulo N, and let H be thehypercentre of G. Assume, by contradiction, that H ≠ G. Then, since G/N isnilpotent, H ≱ N, and so K = H ∩ N < N. Clearly K G and, by Proposition1.83, A/K = ζ(N/K) ≠ 1. Let a ∈ A \ K, and U = 〈a, S〉K. Then, beingfinitely generated, U/K is nilpotent and (A ∩ U)/K is a non-trivial subgroup ofit. Hence V/K = ζ(U/K) ∩ (A ∩ U)/K is not trivial. Now [V, N] ≤ K becauseV ≤ A, and [V, 〈S〉] ≤ K because V/K ≤ ζ(U/K). Thus, since G = N〈S〉, weget V/K ≤ ζ(G/K) and the contradiction V ≤ H ∩ N = K.We add some considerations about the dual and much more intricate caseof groups with a lower (i.e. descendant) central series. Such groups are calledhypocentral. A simple example of a hypocentral group which is not locally nilpotentis the infinite dihedral group D ∞ . If G is hypocentral and α is the smallestordinal such that γ α (G) = 1, we say that G has hypocentral type length α. Forinstance, the infinite dihedral group has hypocentral type length ω. Hypocentralgroups form a class of generalized nilpotent groups; however, this class is toolarge to be considered in general. For instance, by a famous theorem of Magnus,it includes every free group.In fact, free groups (and the infinite dihedral group as well) belong to thenarrower class of residually nilpotent groups. A group G is residually nilpotentif for each 1 ≠ x ∈ G there exists a normal subgroup N of G such that G/N isnilpotent and x ∉ N. It is immediate to prove that G is residually nilpotent ifand only if γ ω (G) = ⋂ n∈N γ n(G) = 1. Notice also that a locally nilpotent groupwhich is residually finite is residually nilpotent (but not the converse).Golod examples prove the existence of finitely generated residually finite p-groups that are not finite (observe that a residually finite p-group is residuallynilpotent). Also, we have already mentioned (Theorem 1.34) that the recent solutionby Zelmanov of the Restricted Burnside Problem implies that a residuallyfinite p-group of finite exponent is locally nilpotent.Let us give a simple example of a locally nilpotent hypocentral group thatis not residually nilpotent.

1.5. CLASSES OF LOCALLY NILPOTENT GROUPS 37Example. Let K be a field and, for every 1 ≤ n ∈ N, let T n be the unitriangularmatrix group UT (n, K). Let W = Dir n≥1 T n ; then γ k (W ) = Dir n≥1 γ k (T n ) forall k ∈ N. Thus, γ ω (W ) = 1 and W is residually nilpotent (and hypercentral oflength ω). Let Z = ζ(W ) = Dir n≥1 ζ(T n ). Now, for all n ≥ 1, ζ(T n ) is isomorphicto the additive group of K via, say, the isomorphism φ n . Let N be the kernel ofthe homomorphismZ → (K, +)(x 1 , x 2 , . . .) ↦→ ∑ n≥1 φ n(x n ).Then N W , Z/N ≃ (K, +), and Nζ(T n ) = Z for all n ≥ 1. Finally, letG = W/N. Since G/Z ≃ W/Z ≃ Dir n≥1 (T n /ζ(T n )), we clearly have that G/Zis residually nilpotent, i.e. γ ω (G) ≤ Z/N. On the other hand, for all k ∈ N,γ k (G) = γ k(W )NN≥ ζ k(T k+1 )NN= ζ(T k+1)NN= Z N .Thus γ ω (G) = Z/N and G is not residually nilpotent (but γ ω+1 (G) = 1.)Observe that this example incidentally shows that, even in the class of locallynilpotent groups, homomorphic images of residually nilpotent groups need notbe residually nilpotent. In fact, we will show in section 3.5 that every locallynilpotent group is a homomorphic image of a suitable residually finite locallynilpotent group.The normalizer condition. A group G is said to satisfy the normalizer conditionif H ≠ N G (H) for all proper subgroups H of G. Following [96], we denoteby N the class of all groups satisfying the normalizer condition.Proposition 1.88 A group G satisfies the normalizer condition if and only ifevery subgroup of G is ascendant. Thus N-groups are Gruenberg groups.Proof. Since, clearly, a proper ascendant subgroup of a group cannot be selfnormalizing,in one direction the implication is obvious. Conversely, let G be anN-group, and H a proper subgroup of it. Then one defines an ascending series ofsuccessive normalizers by setting N 0 (H) = H, N α+1 (H) = N G (N α (H)) for anyordinal α, and N β (H) = ⋃ α

38 CHAPTER 1. LOCALLY NILPOTENT GROUPSTo prove that the class of hypercentral groups is properly contained in Nis not that easy. The first examples of N-groups with trivial centre are dueto Heineken and Mohamed [47] and appeared in 1968. These groups, whoseconstruction we will report in chapter 3, are extensions of an elementary abelianp-group by a Prùfer group C p ∞ (for any fixed prime p), and have the propertythat all of their proper subgroups are nilpotent and subnormal.Example. Let G = C p wr C p ∞, and let B be the base group of G. If H is asubgroup of G such that BH ≠ G, then BH is nilpotent and normal in G (inparticular H is nilpotent and subnormal of G). On the other hand, if H = C p ∞,by Lemma 1.82 we have H = N G (H), and so ζ(G) = 1. Thus, G is a Fittinggroup but it is not hypercentral.Example. The group of the example at page 32 is a Baer group that is not Fitting,and that also does not satisfy the normalizer condition. Another example withthese properties is the group G = C p ≀(C p ≀A), where A is an infinite elemenatryabelian p-group. G is a soluble group of exponent p 3 , and so it is a Baer groupby Proposition 1.76. But G is not a Fitting group, and does not satisfy thenormalizer condition.Let us add some more comments on radicable groups. For torsion-free groupsthis property is not very decisive: a theorem of Mal’cev ensures that everynilpotent torsion-free group N may be embedded in torsion-free radicable groupwhich is still nilpotent of the same nilpotency class of N. For periodic groupsthe situation is different: a periodic hypercentral semi-radicable group is abelian(and radicable). Nevertheless, radicable locally finite p-groups may also be rathercomplicated: in fact, a consequece of a result of Baumslag [6] is that every p-group may be embedded in a radicable p-group. Here is a sketch of the argument.Let P be any group, C n a cyclic group of order n, and embed P as the diagonalsubgroup δ(P ) in the base group of the standard wreath product P 1 = P ≀ C n ;then every element of δ(P ) has a n-th root in P 1 . If we start from a p-groupP = P 0 , take n = p, and iterate the process (the embedding is in the diagonalsubgroup P i ↦→ δ(P i ) ≤ P i ≀ C p = P i+1 ), we get a direct limit group P , whichis a radicable p-group, and contains a copy of the original P as a subgroup.Observe that if P is locally finite then such is P ; moreover if P is nilpotent thenP is subnormal in P (of defect equal to its nilpotency class) and P is a Fittinggroup. On the other hand we have:Proposition 1.89 A radicable periodic group satisfying the normalizer conditionis abelian.Proof. Let G be a radicable periodic N-group. We may clearly assume that Gis a p-group for a prime p. Let x = x 0 ∈ G. Then there exists x 1 ∈ G suchthat x p 1 = x 0, and for i ≥ 2, inductively we find x i ∈ G with the property thatx p i = x i−1 . Let U = 〈x i | i ∈ N〉; then U ≃ C p ∞. Since G is a N-group, Uis ascendant in G and so, by Lemma 1.33, U G is abelian. This means that xcommutes with all of its conjugates. Hence [y, x, x] = 1 for all x, y ∈ G. Now,let x, y ∈ G with m = |x|, and let t ∈ G such that t m = y; then by Lemma 1.2we have [x, y] = [x, t m ] = [x m , t] = 1, thus proving that G is abelian.

1.6. PRELIMINARIES ON N 1 39This does not hold for semi-radicable groups: in fact, let U be one of the p-groups constructed by Heineken and Mohamed. Then U/U ′ ≃ C p ∞ (see, in fact,Chapter 3), and it is not difficult to see that U is semi-radicable not radicable.Finiteness conditions. Locally nilpotent groups satisfying various finitenessconditions have been largely studied in the past, and much is known aboutthme. While refering to the first volume of Robinson’s monograph [96] for a fullaccount, we restrict to mentioning, for further reference, just a special case of aresult of PlotkinTheorem 1.90 Let G be a locally nilpotent group. Then(1) G satisfies the maximal condition on abelian subgroups if and only if G isa finiltely generated nilpotent group;(2) G satisfies the minimal condition on abelian subgroups if and only if G isa direct product of finitely many Černikov p-groups.1.6 Preliminaries on N 1The class of groups in which every subgroup is subnormal, which we denote byN 1 , represents a case for which it is difficult to make any immediate but nottrivial observation.Among the natural classes of generalized nilpotent groups, N 1 is perhapsthe closest to nilpotency, as it will be seen in these notes. Indeed, it may beuseful to warn that, although Lemma 1.60 seems to confirm the idea that locallynilpotent groups are plenty of normal (and subnormal) subgroups, this is notquite true in general. For instance, F. Leinen (see [62]) has shown that, given aprime p, in the unique countable existentially closed locally finite p-group (whichwas discovered by P. Hall, and contains as a subgroup every finite p-group) allsubnormal subgroups are normal and form a unique chain of subgroups.Besides groups of Heineken–Mohamed type (non-nilpotent groups with allof their proper subgroups nilpotent and subnormal), another way of explicitelyconstructing non–nilpotent N 1 -groups (which we treat in Chapter 6), was discoveredby H. Smith. It produces in particular hypercentral, residually finite,N 1 -groups of finite rank. Thus, none of these properties: hypercentrality, finiterank, residual finiteness, associated to N 1 is enough to ensure nilpotency.Clearly, a N 1 -group is a Baer group satisfying the normalizer condition,but not viceversa, as the direct product of infinitely many nilpotent groupswith unbounded classes shows. It is also obvoius that the class N 1 is closed bysubgrops and quotients; but it is not closed by direct products. In fact, takingfor granted the existence of a N 1 -group H with trivial centre, then the diagonalsubgroup D = {(x, x) | x ∈ H} of the direct power H × H, is self-normalizing.Indeed, it is not difficult to prove the following fact.Lemma 1.91 Let H be a group and let D be the diagonal subgroup of H × H.Then:(a) D ≠ N H×H (D) if and only if ζ(H) ≠ 1;

40 CHAPTER 1. LOCALLY NILPOTENT GROUPS(b) D is subnormal in H × H if and only if H is nilpotent;(c) D is ascendant in H × H if and only if H is hypercentral;Observe that point (c) gives a sort of ’outer’ characterization of hypercentralgroups inside the class N: a group G is hypercentral if and only if the directproduct G × G satisfies the normalizer condition.Let us repeat another elemantary but basic fact (in fact, Lemma 1.24). LetH be a subgroup of the group G. Then H is subnormal of defect at most n if andonly if [G, n U] ≤ H for any finitely generated subgroup U of H. In particular,if all finitely generated subgroups of H are subnormal of defect at most n, thenH is subnormal of defect at most n.We now start proving something. The first result is indeed one of the mostuseful arguments in studying N 1 -groups. In essence it was firstly observed byC. Brookes in [7].Theorem 1.92 (Brookes). Let G be a group in N 1 , and let Θ be a family ofsubgroups of G such that G ∈ Θ. Then there exists a subgroup H ∈ Θ, a finitelygenerated subgroup F of H, and a positive integer d, such that every F ≤ K ≤H, with K ∈ Θ, has defect at most d in H.Proof. Let G be a counterexample. By an inductive procedure we construct twochains of subgroups{1} = F 0 ≤ F 1 ≤ . . . ≤ F i ≤ F i+1 ≤ . . .G = H 0 ≥ H 1 ≥ . . . ≥ H i ≥ H i+1 ≥ . . .such that, for each i, j ∈ N, F i is finitely generated, H i ∈ Θ, F i ≤ H j and[H i , i F i+1 ] ≰ H i+1 .Set F 0 = {1}, H 0 = G, and suppose we have already defined F 0 , . . . , F iand H 0 , . . . , H i . Since F i ≤ H i ∈ Θ, and G is a counterexample, there existsa subgroup Θ ∋ H i+1 ≤ H i with F i ≤ H i+1 , and d(H i+1 , H i ) = i + 1. Thisimplies that there exists a finitely generated subgroup K of H i+1 such that[H i , i K] ≰ H i+1 . We put F i+1 = 〈F i , K〉 . Then F i+1 is finitely generated,F i ≤ F i+1 ≤ H i+1 , and [H i , i F i+1 ] ≰ H i+1 .By induction, we thus construct the two chains {F i } i∈N , {H i } i∈N with thedesired proprties. We then putF = ⋃ i∈NF i .Then F ≤ ⋂ i∈N H i is subnormal in G. So there exists an integer k such that[G, k F ] ≤ F . In particular we havewhich contradicts the choice of F k+1 .[G, k F k+1 ] ≤ [G, k F ] ≤ F ≤ H k+1Next proposition generalizes a result appearing in [101], where its proof iscredited to D. Robinson.

1.6. PRELIMINARIES ON N 1 41Proposition 1.93 Let G ∈ N 1 , and A a normal nilpotent periodic subgroup ofG. Let A ω = ⋂ n∈N An . Then there exists d ≥ 1 such that A ω ≤ ζ d (G).Proof. Write D = A ω . We may clearly suppose that G/D is countable; thus letG/D = {a 1 D, a 2 D, a 3 D, . . .}. Assume that, for a 1 ≤ n ∈ N we have integersm 1 , m 2 , . . . , m n such that, if U n = 〈a m11 , am2 2 , . . . , amn n 〉, then A ∩ U n = 1. Now,U n is a subgroup of the finitely generated nilpotent group 〈U n , a n+1 〉. Also, sinceA is periodic, A ∩〈U n , a n+1 〉 is finite. then, by Theorem 1.40, there exists a subgroupof finite index of 〈U n , a n+1 〉 that contains U n and has trivial intersectionwith A. In particular, there exists a m n+1 ≥ 1 such that U n+1 = 〈U n , a mn+1n+1 〉has trivial intersection with A. In this way we get, by induction, a sequence(m n ) n≥1 of integers such that, for all n, A ∩ 〈a m11 , . . . , amn n 〉 = 1. We now setU = 〈a mnn | 1 ≤ n ∈ N〉. Then A ∩ U = 1, and for each x ∈ G there exists a1 ≤ k ∈ N such that x k ∈ U.Now, G ∈ N 1 , so U is a subnormal subgroup; let d be the defect of U in G.Then [A, d U] ≤ A ∩ U = 1. Let x 1 , . . . , x d ∈ G, and let m 1 , . . . , m d ∈ N withx mii∈ U. Then Lemma 1.21 yields[D, x 1 , . . . , x d ] ≤ [A, 〈x m11 〉, . . . , 〈xm dd 〉] ≤ [A, d U] = 1.This proves that D ≤ ζ d (G).It is convenient to state explicitely an immediate corollary of this.Corollary 1.94 Let G ∈ N 1 , and D be a normal abelian divisible periodic subgroupof G. If G/D is nilpotent (hypercentral), then G is nilpotent (hypercentral).

42 CHAPTER 1. LOCALLY NILPOTENT GROUPS

Chapter 2Torsion-free GroupsThe proof that torsion–free N 1 -groups are nilpotent is relatively simple and doesnot require a lot of preparation. Thus, inverting the historical development, wepresent it before anything else in this short chapter. The price will be that, inorder to be as self consistent as possible, we will state and prove for a specialcase some results that will be later (and with much more effert) shown to holdin general (notably Proposition 2.17 and Lemma 2.19); the proofs in the torsionfreecase are considerably shorter, and we hope that the repetition will not annoythe reader.2.1 Locally nilpotent torsion–free groupsLet us begin with some simple properties of the ascending central series of atorsion-free group.Lemma 2.1 Let G be an Engel group, and a, b ∈ G with 〈a〉 G torsion-free.Assume that there exists 1 ≤ n ∈ N such that [a, b n ] = 1. Then [a, b] = 1.Proof. Since G is an Engel group there exists an integer k such that [a, k b] = 1.We make induction on k (for k = 1 there is nothing to prove). Our assumptionimplies that b n is in the centre of 〈a, b〉, and so [[a, b], b n ] = 1. Now, [a, b] ∈ 〈a〉 Gand by inductive assumption we then have [a, b, b] = 1, whence, by Lemma 1.2,[a, b] n = [a, b n ] = 1. Since [a, b] ∈ 〈a〉 G , which is torsion-free, we conclude that[a, b] = 1.Corollary 2.2 Let G be a locally nilpotent group, and a, b ∈ G. Suppose that[a n , b m ] has finite order, for some n, m ≥ 1. Then [a, b] has finite order.Proof. Apply Lemma 2.1 to G/T , where T is the torsion subgroup of G.Another immediate application of this Lemma is the following useful fact.Proposition 2.3 Let G be a locally nilpotent group.(1) If N is a normal torsion-free subgroup of G then G/C G (N) is torsion-free;43

44 CHAPTER 2. TORSION-FREE GROUPS(2) if G is torsion-free then G/ζ α (G) is torsion-free for every ordinal α.Proof. (1) Let b ∈ G and 1 ≤ n ∈ N be such that b n ∈ C G (N). Then, byLemma 2.1, b ∈ C G (N). This shows that G/C G (N) is torsion-free.(2) Let G be torsion-free. Then point (1) applied to N = G yields G/ζ 1 (G)torsion-free; and the same argument, applied to any ordinal of type α + 1 showsthat G/ζ α+1 (G) is torsion-free if such is G/ζ α (G). To complete the proof byinduction on α, it remains to consider the case of a limit ordinal β. Thus, takeζ β (G) = ⋃ α

2.2. ISOLATORS 45Corollary 2.6 Let G be a locally nilpotent torsion-free group, and H a finitelygenerated normal subgroup of G. Then H ≤ ζ h (G), where h is the Hirsch lengthof H.Proof. This follows easily from Lemma 2.5 and induction on the Hirsch lengthh(H), keeping in mind that 1 ≠ Z(H) is normal in G, H/Z is torsion free, andh(H/Z(H)) + h(Z(H)) = h(H).2.2 IsolatorsThe basic aspects of root extraction in locally nilpotent groups are subsumedin the elegant P. Hall’s theory of isolatorsl [38], which is a fondamental tool inwhat follows, and that we introduce in its simpler form.Recall that if π is a set of primes, an integer n ≠ 0 is a π-number if all of itsprime divisors belong to π.Definition 2.1 Let π be a set of primes and let H be a subgroup of a group G.The π-isolator of H in G is the setI π G(H) = { x ∈ G | x n ∈ H for some π-number n ≥ 1} .If π is the set of all primes, we then omit it in the notation and speak aboutthe isolator I G (H) of H ≤ G; thusI G (H) = {x ∈ G | x n ∈ H for some 1 ≤ n ∈ N}.The results we prove thereafter are stated for the full isolator, since it is this casethat we will need, although some of them (in particular Lemmas 2.7 and 2.8)admit a ’local’ version which may be proved by specializing the same arguments.Lemma 2.7 Let G be a locally nilpotent group. Then, for all H ≤ G, I G (H) isa subgroup of G.Proof. Let x, y ∈ I G (H), where H is a subgroup of the locally nilpotent groupG. Then there exists 1 ≤ m ∈ N, such that 〈x m , y m 〉 ≤ H. Now, U = 〈x, y〉is nilpotent, of class, say, c. We prove, by induction on c, that |U : 〈x m , y m 〉|is finite, from which U ⊆ I G (H) clearly follows. If U is abelian, this fact isclear. Otherwise, by inductive assumption, we have that Y = γ c (U)〈x m , y m 〉has finite index in U. Now, γ c (U) is generated by the simple commutators oflength c whose entries are x and y. If w = [u 1 , . . . , u c ] is such a commutator, then(see Lemma 1.47) w mc = [u m 1 , . . . , u m c ] ∈ γ c (U) ∩ 〈x m , y m 〉). Thus, the abeliangroup Y/〈x m , y m 〉) ≃ γ c (U)/(γ c (U)∩〈x m , y m 〉) is finitely generated by elementsof bounded exponent, and is therefore finite. Hence, as wanted, |U : 〈x m , y m 〉|is finite.Needless to say, if H is a subgroup of the locally nilpotent group G, thenI G (I G (H)) = I G (H), and I G (H) G if H G. We say that a subgroup H ofthe group G is isolated (respectively π-isolated) if H = I G (H) (H = I π G (H)).

46 CHAPTER 2. TORSION-FREE GROUPSLemma 2.8 Let G be a locally nilpotent group, and let H, K ≤ G. Then, forevery 1 ≤ n ∈ N,(1) [G, I G (H)] ≤ I G ([G, H]), thus if U/V is a central factor of G, then alsoI G (U)/I G (V ) is a central factor;(2) γ n (I G (H)) ≤ I G (γ n (H));(3) I G (H) (n) ≤ I G (H (n) ).Proof. (1) Let M = I G ([G, H]). Then M G, since [G, H] G, and G/M istorsion–free. Let b ∈ I G (H), and n ∈ N such that b n ∈ H. Then, for any g ∈ G,[g, b n ] ∈ [G, H] ≤ M. Now, G/M is torsion-free and thus from Lemma 2.1 itfollows [g, b] ∈ M. This shows that [G, I G (H)] ≤ M.(2) We proceed by induction on n. If n = 1, then the inclusion reduces toI G (H) = I G (H). Let now n ≥ 2, and set K = γ n (H). Let x 1 , . . . , x n ∈ I G (H);then there exists 1 ≤ m ∈ N such that x m i ∈ H for all 1 ≤ i ≤ n. By inductivehypothesis, y = [x 1 , . . . , x n−1 ] ∈ I G (γ n−1 (H)), and so there exists 1 ≤ t ∈ Nsuch that y t ∈ γ n−1 (H). Hence, [y t , x m n ] ∈ K ≤ I G (K). By Lemma 2.1, thisimplies [y, x n ] ∈ I G (K), which is what we wanted.(3) For n = 1, H (1) = γ 2 (H) and we have proved the inclusion in point (1).Thue, let n ≥ 2. Applying the induction hypothesis and again point (1), we get:I G (H) (n) = γ 2 (I G (H) (n−1) ) ≤ γ 2 (I G (H (n−1) )) ≤ I G (γ 2 (H (n−1) )) = I G (H (n) ),which is our assertion.Lemma 2.8 is an instance of a more general result established by P. Hall in[38]: if H 1 , . . . , H, n are subgroups of a group G and θ is any word in n variables,we define θ(H 1 , . . . , H n ) to be the subgroup of G generated by all the elementsof the form θ(h 1 , . . . , h n ) where h i ∈ H i for all i = 1, . . . , n. If G is locallynilpotent, then θ(IG π (H 1), . . . , IG π (H n)) ≤ IG π (θ(H 1, . . . , H n )). To prove this, webegin with a lemma.Lemma 2.9 Let A 1 , . . . , A n be subgroups of the locally nilpotent group G, andfor each i = 1, . . . , n, let B i ≤ A i with |A i : B i | finite. Let π be the set of allprime divisors of the indices |A i : B i | and θ(x 1 , . . . , x n ) a word. Then the index|θ(A 1 , . . . , A n ) : θ(B 1 , . . . , B n )| is finite and a π-number.Proof. Let H = θ(A 1 , . . . , A n ), K = θ(B 1 , . . . , B n ), and suppose by contradictionthat |H : K| is not a π-number. Then, since G satisfies the maximalcondition on subgroups (Proposition 1.39), there exists N G maximal suchthat |HN : KN| is either infinite or diveded by a prime not in π. We may wellassume N = 1. Let Z be the centre of G. Then Z ∩K G and so (by our choiceof N) Z ∩ K = 1. Suppose that Z contains an infinite cyclic subgroup Y . Then|HY : KY | is a π-number, and therefore 1 ≠ Y ∩ H. Thus C = Y ∩ H is aninfinite cyclic group. Let q be a prime with q ∉ π. Then 1 ≠ C q G, whence|C q H : C q K| is a π-number. But, as K ∩ C = 1, we have the contradiction|C q H : C q K| = |H : C q K| = |H : CK||CK : C q K| = q|H : CK|.

2.2. ISOLATORS 47Thus Z does not have any elements of infinite order. Let R by a cycli subgroupof prime order q of Z. As before we have R ≤ H, R∩K = 1, and |H : RK|a π-number. Hence|H : K| = |H : RK||RK : K| = |H : RK||R : R ∩ K| = |H : RK|q.Since we are assuming that |H : K| is not a π-number, this forces q ∉ π.Therefore Z is a finite π ′ -group. But then, by Proposition 1.49, G is a π ′ -group,which is clearly a contradiction.We may now prove Hall’s result.Theorem 2.10 (P. Hall) Let θ(x 1 , . . . , x n ) be a word, π a set of primes, andH 1 , . . . , H n subgroups of a locally nilpotent group G, thenθ(I π G(H 1 ), . . . , I π G(H n )) ≤ I π G(θ(H 1 , . . . , H n )).Proof. Let U = θ(IG π (H 1), . . . , IG π (H n)), V = θ(H 1 , . . . , H n ), and take an elementg = θ(g 1 , . . . , g n ) with g i ∈ IG π (H 1). For any i = 1, . . . , n we thenhave g mii ∈ H i for some π-number m i ; we write A i = 〈g i 〉 and B i = 〈g mii 〉.Since 〈A 1 , . . . , A n 〉 is nilpotent, we can apply Lemma 2.9 and deduce that|θ(A 1 , . . . , A n ) : θ(B 1 , . . . , B n )| is a π-number. As θ(B 1 , . . . , B n ) is subnormalin θ(A 1 , . . . , A n ) and g ∈ θ(A 1 , . . . , A n ), it follows that g m ∈ θ(B 1 , . . . , B n ) ≤ Vfor some π-number m. Thus g ∈ IG π (V ). Since the elements like g generate U,we have U ≤ IG π (V ), as wanted.Corollary 2.11 Let H, K be subgroups of a locally nilpotent group G, then[I G (H), I G (K)] ≤ I G ([H, K]).Remarks. Let H be a subgroup of a locally nilpotent group G. Observe that theCorollary implies that I G (N G (H)) ≤ N G (I G (H)); in particular, the normalizerof an isolated subgroup is also isolated. Another immediate consequence is thatif H is subnormal of defect d, then I G (H) is subnormal of defect at most d.We now move to torsion-free groups, for which the results are stronger.Lemma 2.12 Let G be a locally nilpotent, torsion-free group, and let H ≤ G.Then, for every ordinal α,ζ α (I G (H)) = I G (ζ α (H))Proof. We make induction on α. If α = 0, then the equality reduces to 1 = I G (1)which is satisfied since G is torsion-free. Assume now that α = β + 1 for someordinal β, and let K = ζ β (H). Let x ∈ I G (ζ α (H)), and let g ∈ I G (H). Thenthere exists 1 ≤ m ∈ N such that [g m , x m ] ∈ K. By Lemma 2.1, it follows that[g, x] ∈ I G (K), and this holds for all g ∈ I G (H). Now, by inductive assumption,I G (K) = ζ β (I G (H)), and so x ∈ ζ α (I G (H)). Conversely, let y ∈ ζ α (I G (H)).Then y n ∈ H for some 1 ≤ n ∈ N. Hence,[H, y n ] ≤ [I G (H), y n ] ∩ H ≤ ζ β (I G (H)) ∩ H = I G (K) ∩ H = I H (K).Now, by Lemma 2.3, I H (K) = K. Thus, y n ∈ ζ α (H) and so y ∈ I G (ζ α (H)).

48 CHAPTER 2. TORSION-FREE GROUPSSyuppose now that α is a limit ordinal, i.e. α = ⋃ β

2.3. TORSION–FREE N 1 -GROUPS 492.3 Torsion–free N 1 -groupsIn this section we show that a torsion-free group with all subgroups subnormalis nilpotent. In [78], W. Möhres proved that such a group is soluble and hypercentral,and later H. Smith [108] was able to establish nilpotency, Here, wewill follow the proof given in [15], which in turn makes a heavy use of Möhres’sideas. Let us begin with a general observation.Lemma 2.16 Let H be a torsion-free nilpotent group of class c, and assumethat H/H ′ can be generated by r elements. Then the Hirsch length of H isbounded by r + r 2 + . . . + r c .Proof. Let A = H/H ′ . Then, for every 1 ≤ i ≤ c, there is an epimorphism:A ⊗ A ⊗ · · · ⊗ A} {{ }i times−→ γ i (H)/γ i+1 (H)(Theorem 1.44). Now, A is a r-generated abelian group, and so it has Hirschlength at most r. Similarly, for each i ≥ 1, the i-th tensor power A ⊗ · · · ⊗ A hasHirsch length at most r i . Hence, for each 1 ≤ i ≤ c, γ i (H)/γ i+1 (H) has Hirschlength at most r i . Since γ c+1 (H) = 1, it plainly follows that H has Hirsch lengthat most r + r 2 + . . . + r c .We first deal with groups with all subgroups subnormal of bounded defect.Thus, for each 1 ≤ n ∈ N, let us denote by U n the class of groups in whichevery subgroup is subnormal of defect at most n. It is clear that every U n groupis locally nilpotent and (n + 1)-Engel. We observe that a torsion-free group Gin U n is in fact a n-Engel group. Let x ∈ G and Y = 〈x〉 G,n−1 ; then 〈x〉 Y .Since Y is torsion-free and locally nilpotent, it follows from Lemma 2.5 thatx ∈ Z(Y ); in particular [g, n x] = [g, n−1 x, x] ∈ [Y, x] = 1 for all g ∈ G.The next Proposition is a special case of Roseblade’s Theorem (see section4.2), and, of course of Zel’manov theorem 1.65.Proposition 2.17 There exists a function ρ 0 : N → N, such that a torsion-freegroup in which every subgroup is subnormal of defect at most n, is nilpotent ofnilpotency class not exceeding ρ 0 (n).Proof. We will define by recursion on n a value ρ 0 (n), such that, if G is atorsion-free U n -group, then γ ρ0(n)+1(G) = 1.A U 1 -group is a group in which every subgroup is normal, and it is wellknown since Dedekind that a torsion-free such group is abelian. Thus ρ 0 (1) = 1.Let n ≥ 1, and assume we have defined ρ 0 (i) for 1 ≤ i ≤ n − 1. Let G be atorsion-free U n -group. Then, for each H ≤ G, we have a seriesH = H G,n H G,n−1 . . . H G,1 = H G G.Now, if H ≤ K ≤ H G , then clearly K G = H G . It follows that H G,1 /H G,2belongs to U n−1 . Similarly, we have, for all i = 1, . . . , n − 1,H G,iH G,i+1 ∈ U n−i.

50 CHAPTER 2. TORSION-FREE GROUPSFor 1 ≤ i ≤ n − 1, we put H i+1 = I H G,i(H G,i+1 ). Then H i+1 H G,i , andH G,i /H i+1 is a torsion-free U n−i -group. By inductive assumption, H G,i /H i+1is nilpotent of class at most ρ 0 (n − i) and so it is solvable of derived length atmost [log 2 (ρ 0 (n − i))] + 1. Let c(n) = ∑ n−1i=1 ([log 2(ρ 0 (i))] + 1). then(H G ) (c(n)) ≤ I G (H)and this holds for every H ≤ G. Write M = (H G ) (c(n)) . Then from M ≤ I G (H),we cleary get I H G(M) ≤ I G (H). Now, H G /I H G(M) is a soluble torsion-free n-Engel group, hence by Corollary 1.68, it is nilpotent of class at mostα(n) = n c(n) .Thus γ α(n)+1 (H G ) ≤ I H G(M) ≤ I G (H), and this holds for every H ≤ G. Inparticular, for all x ∈ G, 〈x〉 G is nilpotent of class at most α(n).Now, let x 1 , x 2 , . . . , x α(n) be elements of G, and let H = 〈x 1 , x 2 , . . . , x α(n) 〉.Then, by Fitting Theorem, H G is nilpotent of class at most α(n) 2 . In particular,H has nilpotency class at most α(n) 2 . Since H is generated by α(n) elements,it follows from Lemma 2.16 that its Hirsch length is bounded byg(n) = α(n) + α(n) 2 + . . . + α(n) α(n)2 ≤ α(n) α(n)2 +1 .Hence, γ α(n)+1 (H G ) has Hirsch length at most α(n) α(n)2 +1 , and so, by Corollary2.6,γ α(n)+1 (H G ) ≤ ζ α(n) α(n) 2 +1(G).This yields that G is nilpotent of class at most α(n) + α(n) α(n)2 +1 .The exact values of ρ 0 (n) (in the torsion–free case) are known only for n ≤ 4,and in these cases we have ρ 0 (n) = n. For n = 2 this follows from Levi’s resultson 2-Engel groups, while for n = 3, 4 it has been established, respectively, byTraustason [118] and Smith and Traustason [114] (see also Section 4.2).Question 1 (see [114]) Is the nilpotency class of every torsion–free group withall subgroups n-subnormal bounded by n?We now drop the assumption of bounded defects.Proposition 2.18 (Möhres [78]) Let G be a non-nilpotent torsion-free N 1 -group. Then there exist a n ∈ N, a non-nilpotent subgroup H of G and a finitelygenerated subgroup F of H, such that all subgroups U with F ≤ U ≤ H have defectat most n in H. If G is countable, then H can be taken such that I G (H) = G.Proof. Without loss of generality, we may assume that G is countable counterexample.Set H 0 = 1, and suppose that, for a 1 ≤ n ∈ N, we have found afinitely generated subgroup H n−1 of G, and elements x 1 , . . . , x n−1 , such that{x 1 , . . . , x n−1 } ∩ H n−1 = ∅. Then, by Lemma 2.15, there exists a K n ≤ G,with I G (K n ) = G and such that {x 1 , . . . , x n−1 } ∩ K n = ∅¿ Since G is a counterexampleto the proposition, there exists a finitely generated subgroup H n ofK n , containing H n−1 , that has defect at least n + 1 in G. Hence, there exists a

2.3. TORSION–FREE N 1 -GROUPS 51x n ∈ [G, n H n ] \ H n . Then {x 1 , . . . , x n−1 , x n } ∩ H n = ∅. Let now H = ⋃ n∈N H n.H is subnormal in G of defect, say, d. Thus,x d ∈ [G, d H d ] ≤ [G, d H] ≤ H = ⋃ n∈NH n ,whence x d ∈ H j for some j > d, which contradicts the choice of H j .Lemma 2.19 (Möhres). A torsion-free group in which all subgroups are subnormalis soluble.Proof. Let G be a torsion-free group N 1 -group. Since solubility is a countablyrecognizable property (see 1.31), we may assume that G is countable and notnilpotent. Then by Proposition 2.18 there exist a n ∈ N, a non-nilpotent subgroupH of G and a finitely generated subgroup F of H, such that I G (H) = G,and all subgroups U with F ≤ U ≤ H have defect at most n in H. We nowproceed by induction on n to prove that H is soluble. If n = 1, then F G andH/F is Hamiltonian. Hence |(G/F ) ′ | ≤ 2, and, as F is nilpotent, we have inparticular that H is soluble. Let now n ≥ 1, and observe that if F ≤ U ≤ F H ,then U H = F H . Hence, all subgroups of F H containing F have defect at mostn − 1 in F H . By inductive assumption, F H is solvable, and by Lemma 2.8,N = I H (F H ) is a normal soluble subgroup of H. Finally, H/N is solvable byProposition 2.17, and so H is soluble. As G = I G (H), by Lemma 2.8 we concludethat G is soluble.The next Lemma is indeed a key argument. Given a prime p, and positiveintegers k, n, we definef p (k, n) = (n + 2)p [logpk(n+2)]+1 .Lemma 2.20 Let G = A〈x〉 be a nilpotent group, where A G is an elementaryabelian p-group. Assume also that there exists a subgroup F of A, and an ∈ N, such that |F | = p k , and every subgroup H of G with F ≤ H is subnormalof defect at most n in G. Then [A, fp(k,n)−1 x] = 1.Proof. Set s = f p (k, n), and m = [log p k(n + 2)] + 1. Then p m > k(n + 2).Assume, by contradiction, that [A, s−1 x] ≠ 1. By obvious induction we maythen assume [A, s x] = 1. Also, the subgroupsA, [A, x], [A, 2 x], [A, 3 x], . . . , [A, s−1 x], [A, s x] = 1are all distinct. In particular, we have|[A, (n+1)p m x]| ≥ p s−(n+1)pm = p (n+2)pm −(n+1)p m = p pm > p k(n+2) = |F | n+2 .Now, by Lemma 1.14,[A, n+2 x pm ] = [A, (n+2)p m x] = [A, s x] = 1whence [F, n+2 x pm ] = 1. As F 〈xpm 〉 is generated by the subgroups [F, i x pm ], itthere follows that|F 〈xpm 〉 | ≤ |F | n+2 .

52 CHAPTER 2. TORSION-FREE GROUPSLet now H = 〈A, x pm 〉. Since A is normal abelian and F 〈xpm 〉 ≤ A, we haveF H = F 〈xpm 〉 . Now, H/F H = (A/F H )(〈x pm 〉F H /F H ), where A/F H is normalabelian, and 〈x pm 〉F H /F H is a cyclic subgroup of defect at most n. By Lemma1.59, A/F H is nilpotent of class at most n + 1. In particular we have[A, n+1 x pm ] ≤ F H = F 〈xpm 〉 .Since, by Lemma 1.14, [A, (n+1)p m x] = [A, n+1 x pm ], we finally have|[A, (n+1)p m x]| ≤ |F 〈xpm 〉 | ≤ |F | n+2contradicting what we had obtained above.Lemma 2.21 Let G be a torsion free locally nilpotent group. Let A be an abeliannormal subgroup of G such that G/A is abelian. Suppose that there exist a finitelygenerated subgroup F of A and n ∈ N such that all subgroups of G containingF are subnormal of defect at most n. Then G is nilpotent (and its nilpotencyclass is bounded by a function of (n, rk(F ))).Proof. Assume the hypothesis of the Lemma, and let k be the rank of F .Let x ∈ G and X = F 〈x〉 . Then X ≤ A because F ≤ A G. Since G islocally nilpotent, 〈F, x〉 is nilpotent and X is a finitely generated torsion freeabelian group. Let r be the rank of X. Now set Y = X 2 . Then Y 〈F, x〉and X/Y is an elementary abelian group of order 2 r . Let F = F Y/Y . Then|F | = 2 k and all subgroups of 〈F, x〉/Y that contain F have defect at most n.Also, X = X/Y = F 〈x〉 . By Lemma 2.20, [X, s x] = 1, where s = f 2 (k, n) − 1.Let 2 h the smallest power of 2 larger than s. Then [X, x 2h ] = [X, 2 h x] = 1, so Fhas at most 2 h conjugates in 〈F, x〉/Y . Since X is an abelian group generatedby the conjugates of F , we get 2 r = |X/Y | ≤ |F | 2h = 2 k2h and thus r ≤ k2 h(observe that h does not depend on x, but only on k and n).We have then obtained that, for all x ∈ G, F 〈x〉 is a torsion free abeliangroup of rank at most u = k2 h . Since 〈F, x〉 is torsion free and nilpotent, itfollows that, for all x ∈ G,[〈F, x〉, u x] = 1 .Now, F ≤ 〈F, x〉, so 〈F, x〉 is subnormal of defect at most n in G. Thus we have,for all g, x ∈ G[g, n+u x] = [[g, n x], u x] ∈ [〈F, x〉, u x] = 1.Then G is a metabelian torsion free (n + u)-Engel group and so by Corollary1.68, G is nilpotent of class at most n + u.The following variant of Theorem 1.54 appears in W. Möhres doctoral dissertation.Lemma 2.22 Let N be a nilpotent normal subgroup of the locally nilpotenttorsion-free group G. If G/I G (N ′ ) is nilpotent, then G is nilpotent.

2.3. TORSION–FREE N 1 -GROUPS 53Proof. Observe that, by Lemma 2.8, I G (I G (N) ′ ) ≥ I G (N ′ ) = I G (I G (N ′ )) ≥I G (I G (N) ′ ). Thus, I G (I G (N) ′ ) = I G (N ′ ). Since I G (N) is nilpotent by Lemma2.12, we may assume that N = I G (N).We now proceed by induction on the nilpotency class c of N; the case c = 1being just our assumption. Let c ≥ 2, and let K = I G (γ c (N)). Then K G, andG/K is torsion-free. Moreover, K ≤ Z(N), and N/K has class at most c − 1.Thus, by inductive hypothesis, G/K is nilpotent. Let K/K = K 0 /K ≤ K 1 /K ≤. . . ≤ K d /K = N/K be the intersection of the upper central series of G/K withN/K; and for s = 0, 1, . . . , 2d let T s = 〈 [K i , K j ] | 0 ≤ i, j ≤ d, i + j = s〉. Now,if 1 ≤ i, j ≤ d, by the three subgroup Lemma 1.5, we have[K i , K j , G] ≤ [K j , G, K i ][G, K i , K j ] ≤ [K j−1 , K i ][K i−1 , K j ] ≤ T i+j−1 ,showing that [T s , G] ≤ T s−1 for all s ≥ 1. In other words, G centralizes theseries 1 = [K, K] = T 0 ≤ T 1 ≤ . . . ≤ T 2d = N ′ . By Lemma 2.8, G centralizesthe series of the isolators 1 = I G ({1}) ≤ I G (T 1 ) ≤ . . . ≤ I G (N ′ ). As G/I G (N ′ )is nilpotent by assumption, it follows that G is nilpotent.We are finally in a position to prove the main result.Theorem 2.23 (H. Smith [108]). A torsion-free group in which all subgroupsare subnormal is nilpotent.Proof. Let G be a torsion free group with all subgroups subnormal. By a Lemma2.19, G is soluble. We argue by induction on the derived length d of G.Suppose first that G is metabelian and, by contradiction, that G is notnilpotent. Then, by Proposition 2.18 we may assume that there exists a finitelygenerated subgroup F of G and a n ∈ N such that all subgroups of G containingF are subnormal of defect at most n. Let H = F G ′ and L = I G (H ′ ). H isa normal subgroup of G, so L is normal, and G/L is torsion-free. Since G ′ isabelian and F is finitely generated and subnormal, H is nilpotent. Since G isnot nilpotent, it follows from 2.22 that G/L is not nilpotent. So we may assumethat H is abelian. By Lemma 2.21, G is nilpotent.The general case is now an immediate application of Lemma 2.22. Let d bethe derived length of G and let N = G ′ . By inductive hypothesis, N is nilpotent.By the metabelian case G/I G (N ′ ) is nilpotent, and so G is nilpotent by 2.22.

54 CHAPTER 2. TORSION-FREE GROUPS

Chapter 3Groups of Heineken andMohamedIn the literature two quite different methods for constructing non-nilpotent N 1 -groups are known. The first goes back to a celebrated 1968 paper by H. Heinekenand I. J. Mohamed, and produces p-groups with trivial centre and no propersubgroup of finte index, while the second was discovered by H. Smith in 1982,and gives rise to mixed groups that are hypercentral and residually finite. Wedescribe Smith’s contructions later in Chapter 6, when we will specifically dealwith hypercentral N 1 -groups, while to the Haineken-Mohamed groups, whichhave been much more investigated, we devote the present Chapter.3.1 General remarksIn their mentioned paper [47], H. Heineken and I. J. Mohamed provided thefirst examples of N 1 -groups with trivial centre. The groups they constructed are(locally finite) p-groups for a prime p, and the extension of an infinite elementaryabelian group by a Prüfer group; furthermore, all their proper subgroups aresubnormal and nilpotent.Heineken and Mohamed construction was studied and extended by manyauthors (see e.g. [9], [40], [41], [73], [75]) and it became customary to call agroup G of Heineken-Mohamed type if G is not nilpotent and all of its propersubgroups are nilpotent and subnormal. In particular, in [48] the same authorsshow that there exist 2 ℵ0 non-isomorphic groups sharing these properties, Brunoand Phillips [9] and Möhres [75] studied, respectively, the Schur multiplier andthe automorphisms group of certain groups of Heineken-Mohamed type, andHartley [41] showed that, for every n ≥ 1, there exist p-groups of Heineken-Mohamed type G such that G ′ is an abelian group of exponent p n . For sometime all groups thus constructed were metabelian, and the question as to whethera soluble group G of Heineken-Mohamed type may have arbitrary derived lengthwas eventually solved in the affermative by Menegazzo in [72]. In the same paper,Menegazzo gave a very general method for constructing groups of Heineken-55

56 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMEDMohamed type, which was in turn inspired by Hartley approach ([40]), andwhich is the one that we will present here.Before the actual construction, let us prove the following fact.Proposition 3.1 (Heineken and Mohamed [47]) Let p be a prime and let G bea p-group of Heineken-Mohamed type such that G ≠ G ′ . Then(i) G is countable;(ii) G/G ′ ≃ C p ∞and (G ′ ) p ≠ G ′ = γ 3 (G);(iii) for every H ≤ G, G ′ H = G implies H = G.Conversely, if G is a non-nilpotent p-group with a normal nilpotent subgroup Nof finite exponent such that G/N ≃ C p ∞ and NH ≠ G for every proper subgroupH of G, then G is a group of Heineken-Mohamed type.Later we shall prove Möhres Theorem that every N 1 -group is soluble; thusthe extra condition G ≠ G ′ in the statement of Proposition 3.1 is redundant,and all groups of Heineken-Mohamed type have the properties listed.For further reference, we isolate part of the proof of 3.1 in a separate andelementary Lemma.Lemma 3.2 (Newman and Wiegold [86]) Let G be a non-trivial group such thatUV ≠ G for all pairs of proper normal subgroups U and V . Then there existsa prime number p such that G/G ′ is either a cyclic p-group (possibly trivial) orG/G ′ ≃ C p ∞ and G ′ = γ 3 (G).Proof. Let first assume that G is abelian. Let 1 ≠ x ∈ G; then there exists aprime p such that 〈x p 〉 ≠ 〈x〉. Let U be a subgroup of G maximal such thatx p ∈ U but x ∉ U (it exists by Zorn’s Lemma). Then all subgroups of G/Ucontain xU. Since G/U is abelian, we have that G/U is either a non-trivialcyclic p-group or isomorphic to C p ∞. If G is not a p-group there exists a y ∈ Gand a prime q ≠ p such that 〈y q 〉 ≠ 〈y〉. Arguing as before, we then get a propersubgroup V of G such that G/V is a q-group. But then, clearly, G = UV ,contradicting the assumptions on G. Thus, G is a p-group, and from this iteasily follows that G is either cyclic or of type C p ∞. Now for the general casewe are left to show that G ′ = γ 3 (G). But this is immediate, since H = G/γ 3 (G)is a nilpotent group and H/H ′ ≃ G/G ′ is cyclic or a Prüfer group, and so H isabelian.Proof of Proposition 3.1. Since, by definition, G is not nilpotent but all ofits proper subgroups are nilpotent, G must be countable by Theorem 1.31. Byassumption, G ′ ≠ G and so G ′ is nilpotent. It thus follows from Lemma 1.75that G/G ′ is not finitely generated. Also, by Fitting’s Theorem, G cannot be theproduct of two proper normal subgroups and therefore G/G ′ ≃ C p ∞ by Lemma3.2. Finally, suppose that (G ′ ) p = G ′ . Then G ′ is an abelian divisible groupby Lemma 1.18. Now, every cyclic subgroup X of G is subnormal and so G ′ iscentralized by X by Lemma 1.32. It follows that G ′ ≤ ζ(G), which contradictsthe non-nilpotence of G. Hence (G ′ ) p ≠ G ′ .For the converse, suppose that the non-nilpotent p-group G satisfies theconditions of the second part of the statement, and let H be a proper subgroup

3.2. BASIC CONSTRUCTION 57of G. Then NH ≠ G and so, since G/N ≃ C p ∞, NH/N is finite. Thus, NHis nilpotent by Corollary 1.77. Therefore H is nilpotent and subnormal in NH.Since NH is normal in G, it follows that H is subnormal in G. Hence G is agroup of Heineken-Mohamed type.3.2 Basic constructionAs mentioned before, our approach follows closely Menegazzo [72].For the rest of this section, we fix a prime p and denote by U the Prüfergroup C p ∞, which we take with a fixed set of standard generators u 1 , u 2 , u 3 , . . .:U = 〈u 1 , u 2 , . . . | u p 1 = 1, up i+1 = u i (i ≥ 1)〉.For each i ≥ 1, we write U i = 〈u i 〉. Also, we denote by R = F p [U] the groupalgebra of U over the field F p = Z/pZ, and by U its augmentation ideal. Thismeans that U is the kernel of the (ring) epimorphism ɛ : R → F p defined by( ∑ɛu∈U)a u u = ∑ a u .u∈UThen, U is the ideal of R generated by all the elements of type u − 1 for u ∈ U.Similarly, for each i ≥ 1, we put R i = F p [U i ] and let U i denote the augmentationideal of R i . Then, clearly, R = ⋃ i≥1 R i, U = ⋃ i≥1 U i, andU i = (u i − 1)R ifor every i ≥ 1. Moreover (u i − 1) pi = u pii − 1 = 0; hence all elements of Uare nilpotent and therefore, by elementary ring theory, all elements of R \ U areinvertible. Our first Lemma is standard and not difficult to prove.Lemma 3.3 The ideals of R i are exactly the principal ideals(u i − 1) k R i for 0 ≤ k ≤ p i .These are all distinc and form a totally ordered set with respect to inclusion.An immediate consequence isLemma 3.4 The set of ideals of R is totally ordered.Proof. It is enough to show that, for all u, v ∈ R, if u does not belong to vRthen v belongs to uR. Now, given u, v ∈ R, there clearly exists i ≥ 1 such thatu, v ∈ R i . But then, by Lemma 3.3, either uR i ≤ vR i or vR i ≤ uR i . Thus, theLemma is proved.We observe a consequence of this, which will be used in the next section.Corollary 3.5 Let M be a (right) R-module and y ∈ M, r ∈ R with 0 ≠ x =yr. Then Ann R (y) = rAnn R (x).

58 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMEDProof. Clearly, rAnn R (x) ⊆ Ann R (y). Since 0 ≠ x = y(r1), Ann R (y) ⊈ rR,and so (by Lemma 3.4) Ann R (y) ⊆ rR. From this the claim easily follows.Lemma 3.3 suggests also a convenient way to parametrize the set of allideals of R. In fact, let I be an ideal of R; then, for each i ≥ 1, there is a unique0 ≤ k i ≤ p i such thatI ∩ R i = (u i − 1) ki R i .We thus associate to I the sequence (k 1 , k 2 , . . .). Since R = ⋃ i≥1 R i, this sequenceuniquely determines I. Observe also that, since (I ∩ R i+1 ) ∩ R i = I ∩ R i ,the sequence is such that, for every i ≥ 1,p(k i − 1) < k i+1 ≤ pk i . (3.1)Conversely, a sequence (k 1 , k 2 , . . .) of integers 0 ≤ k i ≤ p i satisfying (3.1) is thesequence associated to the ideal ∑ i≥1 (u i − 1) ki R of R.Lemma 3.6 Let I be a non-zero ideal of R, and let (k 1 , k 2 , . . .) be the sequenceassociated to I. Then IU = I if and only if for every i ≥ 1 there exists j ≥ isuch that pk j > k j+1 .Proof. Suppose that the sequence for I satisfies the condition in the statementand let i ≥ 1. Choose j ≥ i such that pk j > k j+1 . Then, for some t > 0,(u j − 1) kj = (u j+1 − 1) pkj = (u j+1 − 1) kj+1 (u j+1 − 1) t .This implies (u j − 1) kj ∈ IU and, consequently, (u i − 1) ki ∈ IU. Therefore IUhas the same sequence as I and so IU = I.Conversely, assume IU = I, and let i ≥ 1 with (u i − 1) ki ≠ 0. Then itis easy to see that there exists t ≥ i such that (u i − 1) ki ∈ (I ∩ R t )U t . SinceI∩R t = (u t −1) kt R t and U t = (u t −1)R t , we have that (u i −1) ki = (u t −1) kt+1 vfor some v ∈ R t . Hence (u i − 1) ki R < (u t − 1) kt R, and so in the chain(u i − 1) ki R ≤ (u i+1 − 1) ki+1 R ≤ . . . ≤ (u t − 1) kt Rat least one of the inclusions is proper, say (u j −1) kj R < (u j+1 −1) kj+1 R, whichmeans j j+1 < pk j .We now come to the key definition of an HM-system. Let V be a right R-module which is generated by a sequence of elements a = (a i ) i≥l (for somepositive integer l). For any sequence v = (v i ) i≥l of elements of V , we setτ i,k (v) = −v i +k∑a i+s (u i+s − 1) ps−1 + v i+k+1 (u i+k+1 − 1) pk+1 −1s=0for all i ≥ l, k ≥ 0. We then say that a is a HM-system in V iffor every sequence v = (v i ) i≥l .V = 〈τ i,k (v) | i ≥ l, k ≥ 0〉

3.2. BASIC CONSTRUCTION 59Proposition 3.7 (Menegazzo [72]) Let G be a p-group with a normal elementaryabelian subgroup N ≠ 1 such that [G, N] = N and G/N ≃ C p ∞ = U. Letη : U → G/N be an isomorphism, and make N into a R-module in the obviousway. For each i ≥ 1, let g i ∈ R i such that g i N = u η i , and let a i = g −1i g p i+1 (thusa i ∈ N). Suppose further that G = 〈g i | i ≥ l〉 for some l ≥ 1. If the sequencea = (a i ) i≥l is a HM-sequence for N then G is a group of Heineken-Mohamedtype.Proof. Since [G, N] = N ≠ 1, G is not nilpotent. Hence, by proposition 3.1 itsufficies to show that HN = G forces H = G for every H ≤ G. For n ∈ N andu ∈ U we write n u = n (uη) , and for all i ≥ l, k ≥ 0, we setσ i,k =k∏s=0a (ui+s−1)ps −1i+s .We show, by induction on k ≥ 0, that g pk+1i+k+1 = g iσ i,k for all i ≥ l. For k = 0this is trivial since σ i,0 = a i . Thus, let k ≥ 1 and assume g pki+k = g iσ i,k−1 . Theng pk+1i+k+1 = (gp i+k+1 )pk = (g i+k a i+k ) pk = g pk+...+u i+k+1i+k=−1i+k aupk i+k= g i σ i,k−1 a (u i+k−1) pk −1i+k= g i σ i,k .Now, let H ≤ G with NH = G. Then, for every i ≥ l, H contains an elementof the form g i v i with v i ∈ N. Let v be the sequence (v i ) i≥l . For every i ≥ l,k ≥ 0, writing τ i,k = τ i,k (v), and using the identities established above, we have(g i+k+1 v i+k+1 ) pk+1 = g pk+1i+k+1 v(u i+k+1−1) pk+1 −1i+k+1= g i σ i,k v (u i+k+1−1) pk+1 −1()= g i v i v −1i σ i,k v (u i+k+1−1) pk+1 −1i+k+1= g i v i τ i,k .i+k+1=Hence, τ i,k ∈ H for every i ≥ l and k ≥ 0, and thus H contains the subgroupgenerated by the elements τ i,k , which is N, since a is a HM-system. ThereforeH ≥ NH = G, and so H = G as wanted.Our next task is then to find R-modules admitting HN-systems. We de thatwith the aid of Lemmas 3.4 and 3.6.Proposition 3.8 Let I be a non-zero ideal of R such that I = IU < U, and let(k 1 , k 2 , . . .) be the sequence associated to I. Fix l ≥ 1 with 0 < k l < p l , and foreach i ≥ l set { (ui − 1)c i =ki if k i+1 = pk i(u i+1 − 1) pki−1 if k i+1 < pk i .Then c = (c i ) i≥l is a HM-system for I as a R-module.Proof. We first make sure that c is a generating set for I. Thus, let J be theideal (i.e. R-submodule) generated by c. Then J ≤ I: in fact c i ∈ I by definitionif k i+1 = pk i , and, if k i+1 < pk i , c i = (u i+1 − 1) pki−1 ∈ (u i+1 − 1) ki+1 R ≤ I. For

60 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMEDthe reverse inclusion, consider first i ≥ l. If k i+1 = pk i then R i ∩ I = c i R i ≤ J;if k i+1 < pk i ,R i ∩ I = (u i − 1) ki R i = (u i+1 − 1) pki R i = c i (u i+1 − 1)R i ≤ J.If 1 ≤ i < l, then (u i − 1) ki ∈ (u l − 1) k lR ≤ J. Hence J = I.We now prove that c satisfies the requirements of a HM-system for I as aR-module. Let v = (v i ) i≥l be a sequence of elements of I, and for every i ≥ l,k ≥ 0, write τ i,k = τ i,k (v). We prove that for every i > l there exists k ≥ 0 suchthat(u i+1 − 1) ki−1 ∈ τ i,k R. (3.2)This of course will imply that I is generated by the set {τ i,k | i ≥ l, k ≥ 0}.therefore assuring that c is a HM-system for I.Thus, let i ≥ l. If k i = pk i−1 then, by Lemma 3.6, there is a j ≥ i suchthat(u i−1 − 1) ki−1 = (u j−1 − 1) kj−1 and k j < pk j−1 . Hence we may assumek i < pk i−1 . Now, there exists h > 0 such that v i ∈ I ∩ R i+h , and there existsk ≥ h such that k i+k+1 < pk i+k . Then c i+k = (u i+k+1 − 1) pk i+k−1 , andτ i,k = −v i + c i + . . . + c i+k+1 (u i+k+1 − 1) pk−1 −1 + w (3.3)where w = c i+k (u i+k − 1) pk −1 + v i+k+1 (u i+k+1 − 1) pk−1 −1 . We then havew = (u i+k+1 − 1) pk i+k−1 (u i+k − 1) pk −1 + v i+k+1 (u i+k+1 − 1) pk+1 −1 == (u i+k+1 − 1) pk i+k−1+p k+1 −p + v i+k+1 (u i+k+1 − 1) pk+1 −1 == (u i+k+1 − 1) pk+1 −1 ( (u i+k+1 − 1) p(k i+k−1) + v i+k+1)== (u i+k+1 − 1) pk+1 −1 ( (u i+k − 1) k i+k−1 + v i+k+1).Now, v i+k+1 ∈ I and (u i+k − 1) k i+k−1 ∉ I, and so it follows from Lemma 3.4that (u i+k − 1) k i+k−1 and (u i+k − 1) k i+k−1 + v i+k+1 generate the same ideal ofR. Therefore, there exists an invertible element ɛ ∈ R such that(u i+k − 1) k i+k−1 + v i+k+1 = (u i+k − 1) k i+k−1 ɛ.Thus, w = (u i+k+1 − 1) pk+1 −1+p(k i+k −1) ɛ. All other summands in the rightterm of (3.3) belong to I ∩ R i+k ; hence, denoting by w ′ their sum, we havew ′ = (u i+k − 1) m η = (U i+k+1 − 1) pm η for some m ≥ n i+k and some invertibleelement η of R i+k . By observing that the exponents of u i+k+1 − 1 in w andin w ′ are not congruent modulo p, we deduce that the ideals w ′ R and wR aredistinct. Therefore, τ i,k = w ′ + w generates the largest of the two ideals w ′ Rand wR. In particular,(u i+k+1 − 1) pk+1 −1+p(k i+k −1) = wɛ −1 ∈ τ i,k R. (3.4)Now, taking into account that pk i−1 ≥ k i + 1, we havep k+2 k i−1 ≥ p k+1 (k i + 1) ≥ pk i+k + p k+1 > p k+1 − 1 + p(k i+k − 1),and therefore, by (3.4), (u i−1 − 1) ki−1 = (u i+k+1 − 1) pk+2 k i−1belongs to τ i,k R.This proves (3.2) and the Proposition.We can now proceed to the construction of Heineken-Mohamed groups.

3.2. BASIC CONSTRUCTION 61Theorem 3.9 (Menegazzo [72]) To every non-zero ideal I of R such that I =IU < U there corresponds a group of Heineken-Mohamed type G = G(I) suchthat G/G ′ ≃ U and G ′ ≃ I (as R-modules). Moreover, if J is another ideal ofR with J = JU < U and I ≠ J, then G(I) and G(J) are not isomorphic.Proof. Let I be as in the statement and let (k 1 , k 2 , . . .) be the associated sequence.Choose l ≥ 1 such that 1 < k l < p l and for every i ≥ l define theelement c i as in Proposition 3.8. We will inductively define a sequence (a i ) i≥lof elements of R satisfying the following conditions:a i ∈ (u i − 1)R and a i+1 (u i+1 − 1) p−1 = a i + c i (3.5)for every i ≥ l. Set a l = 0, and assume that, for i ≥ l, we have found a l , . . . , a iwith the desired properties. Now, k i ≥ k l > 1 and c i is either (u i − 1) ki or(u i+1 − 1) pk I −1 ; in any case c i ∈ (u i − 1)R and so there exists b ∈ R such thatc i + a i = (u i − 1)b = (u i+1 − 1) p b. By setting a i+1 = (u i+1 − 1)b we get a newelement in the sequence that satisfies (3.5).Consider now the semidirect product W = R⋊U, where R is meant tobe the additive group of the ring (thus the multiplication in W is given by(r, u)(r ′ , u ′ ) = (ru ′ +r ′ , uu ′ )), and for every i ≥ l, let g i = (a i , u i ). Let G = G(I)be the subgroup of W generated by all the g i ’s:Then, for every i ≥ l,G = 〈 (a i , u i ) ∈ W | i ≥ l 〉.g p i+1 = (a i+1(u i+1 − 1) p−1 , u p i+1 ) = (a i + c i , u i ) = g i (c i , 1),and therefore G∩(R×1) contains the U-invariant subgroup N generated by theset {(c i , 1) | i ≥ l}, which, as a U-module, is isomorphic to I. Clearly G/N =〈g i N | i ≥ l〉 ≃ U; moreover, since IU = I, we have N = [N, U] = [N, G].Finally, the sequence (g −1i g p i+1 ) i≥l = ((c i , 1)) i≥l is a HM-system for N ≃ U I,and so we may apply Proposition 3.7 to conclude that G is a group of Heineken-Mohamed type.Now, for the second part of the statement, let J be another ideal of Rwith J = JU < U, write G 1 = G(I), G 2 = G(J), and assume that there is agroup isomorphism α : G 1 → G 2 . By construction, there are canonical isomorphismG ′ 1 ≃ R I and G ′ 2 ≃ R J (as R-modules). Now, α induces an isomorphismG 1 /G ′ 1 → G 2 /G ′ 2, which, combined with the natural isomorphisms with U, givesan isomorphism of U, which we extend by linearity to an isomorphism θ of R.Then, for every x ∈ I = G ′ 1 and u ∈ R:(xu) α = x α u θ .It follows that Ann R (x α ) = Ann R (x), for every x ∈ I. Now, if x = (u i −1) mpi−k ,with 1 ≤ k ≤ i and (m, p) = 1, it is easy to see thatAnn R (x) = (u k − 1) pk −m R.Therefore, for all i ≥ l, Ann R (c α i ) = Ann R(c i ) implies c α i R = c iR. Thus weconclude that I = I α = J.

62 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMEDComments. (1) The groups G constructed in Theorem 3.9 are certainly notnilpotent as G ′ = [G, G ′ ]. A similar behaviour has the upper central series ofany G = G(I). In fact, if 0 ≠ r ∈ R, there exists u ∈ U such that ru ≠ r.This implies (with the notetion used in the proof of 3.9) that ζ(G) ∩ N = 1,and therefore [ζ 2 (G), G] ≤ ζ(G) ∩ G ′ ≤ ζ(G) ∩ N = 1, forcing ζ 2 (G) = ζ(G).Factoring G by ζ(G) we thus obtain groups of Heineken-Mohamed type withtrivial centre. Observe also that ζ(G(I)) is contained in U; hence ζ(G(I)) is nottrivial if and only if I(u 1 − 1) = 0.(2) There are 2 ℵ0 distinct ideal-sequences (k 1 , k 2 , . . .) that satisfy the conditionsof Lemma 3.6, each of those is associated to a different ideal of R,. Therefore,by the second part of Theorem 3.9, we haveCorollary 3.10 For every prime p there are 2 ℵ0 non-isomorphic groups G ofHeineken-Mohamed type such that G/G ′ ≃ C p ∞ and G ′ elementary abelian.A result which was also proved by Heineken and Mohamed [48], Hartley [40]and Meldrum [73].3.3 DevelopementsIn [72] Menegazzo is able to exploit the tecniques reported above to establish theexistence, for every prime p, of a p-group of Heineken-Mohamed type G whosederived subgroup is abelian of infinite exponent (as we are dealing with p-groups,this means that G ′ contains elements of order p n for every n ≥ 0). Since Hartleyhad previously proved in [41] that there exist Heineken-Mohamed groups withderived subgroup of arbitrary finite exponent p n , we have the following result,whose proof we do not include here.Theorem 3.11 For every prime p and any e ∈ {p n | n ∈ N}∪{∞} there existsa p-group G of Heineken-Mohamed type such that G ′ is abelian of exponent e.Another important result from [72] is the following one.Theorem 3.12 (Menegazzo) For every prime p and every n ≥ 1 there existp-groups of Heineken-Mohamed type whose derived length is exactly n.We try at least to indicate the ideas used in the proof of this. We start bydescribing a method of lifting an action on an abelian gruop to an action on anilpotent one, which we will soon specialize to the extend the action of U onR = F p [U].Let A be a commutative ring (with identity) of prime characteristc p, andlet 1 ≤ n ∈ N. To each odered n-tuple (a 1 , . . . , a n ) ∈ A n we associate a unitriangular(n + 1) × (n + 1)-matrix⎛1 a 1 a 2 . . . a n⎜ 0 1 a p 1 . . . a p n−1⎟Σ(a 1 , . . . , a n ) =⎜⎝⎞⎟. . . 1 a pn−1 ⎠110 0 1 . . . a p2n−2

3.3. DEVELOPEMENTS 63We then setΣ n (A) = {Σ(a 1 , . . . , a n ) | a 1 , . . . , a n ∈ R}.It is easily checked that Σ(A) is a subgroup of the group of all upper unitriangularA-matrices of order n + 1. In particular, Σ n (A) is a nilpotent p-groupof finite exponent. Also, Q = Q n (A) = {Σ(0, a 2 , . . . , a n ) | a 2 , . . . , a n ∈ A} is anormal subgroup of Σ = Σ n (A), Σ/Q is isomorphic to the additive group of A,and the set of matrices {Σ(a 1 , a 2 , . . . , a n ) | a 2 = · · · = a n = 0} is a set of cosetrepresentatives of Σ modulo M (all these facts are not hard to chek by directcomputations). For every 1 ≤ i ≤ n, we define π i : Σ n (A) → A as the naturalprojection Σ(a 1 , . . . , a n ) ↦→ a i .The following observatrion may be easily proved by matrix computations,and we omit the details.Lemma 3.13 Let α = Σ(a 1 , . . . , a n ) and β = Σ(b 1 , . . . , b n ) be elements ofΣ n (A), and suppose that 1 ≤ t, s ≤ n are such that a i = 0 for all i < t andb i = 0 for all i < s. Let [α, β] = Σ(q 1 , . . . , q n ). Then q i = 0 for all i < t + s, andq t+s = a t b pts− b s a pst .Let now X be a group of multiplications of A. Then X acts on Σ n (A) in thefollowing wayΣ(a 1 , a 2 , . . . , a n ) x = Σ(a 1 x, a 2 x p+1 , . . . , a n x pn−1 +...+p+1 ). (3.6)That this defines a group action may be seen immedaitely by observing that(3.6) coincides with conjugating Σ(a 1 , . . . , a n ) (in the group of all invertibleA-matrices of order n + 1) by the diagonal matrix⎛⎞1xx p+1D(x) =⎜⎟⎝⎠x pn−1 +...+p+1and that x ↦→ D(x) clearly defines a group isomorphism. Under this action thenormal subgroup Q = Q n (A) defined before is X-invariant, and the action of Xon the factor group Σ n (A)/Q is equivalent to the natural action by multiplicationof X on A.Now, using the notations of section 3.2, we specialize to the case X = U =〈u 1 , u 2 , . . .〉 ≃ C p ∞ and A = F p [U] = R. Recall, in particular, that U denotesthe augmentation ideal of R.Lemma 3.14 Let I be an ideal of R with I = IU < U.(i) Let H be a subgroup of Σ n (R) such that for 1 ≤ s ≤ n, Hπ i = 0 for alli < s, and I ps−1 +...+1 ⊆ Hπ s ; then [H, H]π j = 0 for j < 2s, andI p2s−1 +...+1 ⊆ [H, H]π 2s .

64 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMED(ii) Let d = [log 2 (n)], m = p 2d−1 + . . . + p + 1, and suppose further thatI m ≠ 0. Let H ≤ Σ n (R) with I ⊆ Hπ 1 ; then H has derived length d.Proof. Let (k 1 , k 2 , . . .) be the sequence associated to I. Then k i+1 ≤ pk i for alli ≥ 0, and, by Lemma 3.6, I is generated by the set {(u i − 1) ki | k i+1 < pk i }.For j ≥ 1, we write I j = I pj−1 +...+1 .(i) By Lemma 3.13, [H, H]π i = 0 for every j < 2s, and [H, K]π s+t containsall elements of the formx(a, b) = ab pt − ba pswith a, b ∈ I s . Observe that x(a, b) ∈ I 2s . For i ≥ 1 such that k i+1 < pk i , wetakea = (u i − 1) ki(ps−1 +...+1) = (u i+1 − 1) ki(ps +...+p)b = (u i+1 − 1) ki+1(ps−1 +...+1).Then, we havek i+1 (p 2s−1 +. . .+p s )−k i+1 (p s−1 +. . .+1) < k i (p 2s +. . .+p s+1 )−k i (p s +. . .+p).which implies that ba ps ∈ ab ps R, and ba ps R < ab ps R. By the total ordering ofR–ideals, it follows that x(a, b)R = ab ps R containsba ps (u i+1 − 1) (pki−ki+1)(ps−1 +...+1) = (u i − 1) ki(ps+t−1 +...+1) .Since the set of all elements (u i − 1) ki(ps+t−1 +...+1) , generates I 2s as an ideal,the proof of point (i) is completed by observing the [H.H]π 2s contains the idealgenerated by all the elements x(a, b) for a, b ∈ I s . Now, under our assumptionson H, (xy)π 2s = xπ 2s + yπ 2s for all x, y ∈ [H, H]; moreover, if a, b ∈ I s , andu ∈ U, thenx(a, b)u ps +1 = ab s u 1+ps − a s bu 1+ps = x(au, bu) ∈ [H, H]π 2s ;since the power p s + 1 is an automorphism of the group U, we conclude that[H, H]π 2s contains the ideal I 2s .(ii) Let H ≤ Σ n (R) be such that Hπ 1 ⊇ I. Then, point (i) and an obviousinduction shows that H (r) π 2 r ⊇ I 2 r, for every 0 ≤ r ≤ d. ThereforeH (d) π 2 d ⊇ I 2 d = I m ≠ 0,and thus H has derived length d (it cannot be more).Observe that if the sequence (k 1 , k 2 , . . .) assoociated the ideal I = IU < U,satisfies k j = 1 for j ≤ m, then I m ≠ 0; thus, there exist 2 ℵ0 ideals I thatsatisfy the condition in point (ii) of the Lemma.From now on, we suppose n ≥ 2 to be fixed, and simply write Σ = Σ n (R).Let W be the semidirect product W = Σ⋊U, and let I be a fixed ideal of Rsuch that I = IU < U. We then refer to the notations used in section 3.2;in particular l is an integer choosen as in Proposition 3.8, and (c i ) i≥l is theHM-system for the R-module I defined in the same Proposition.

3.3. DEVELOPEMENTS 65Let (k i ) i≥1 be the sequence associate to I, we define integers r i (for i ≥ l),as follows:{ pr l =l+1 − pk l if k l+1 = pk lp l+1 (3.7)− pk l + 1 if k l+1 < pk land, for i > l,⎧0 if k i+1 = pk i and k i = pk i−1⎪⎨p(pkr i =i−1 − k i − 1) if k i+1 = pk i and k i < pk i−1(3.8)1 if k i+1 < pk i and k i = pk i−1⎪⎩p(pk i−1 − k i − 1) + 1 if k i+1 < pk i and k i < pk i−1These numbers are singled out because of the following fact.Lemma 3.15 With the notations of Proposition 3.8, and definitions (3.7) and(3.8), set, for every i ≥ l, w i = (u i+1 − 1) ri . Then the following hold.(i) c l w l = 0; and c i w i = c i−1 for all i > l.( ∏i)(ii) Fore every i ≥ l, Ann R (c i ) =s=l w s R.Proof. Point (i) follows easily from the definitions of the elements c i and of thenumbers r i . Now, using point (i), Corollary 3.5 and an obvious induction, wesee that, in order to prove (ii), it is enough to observe that Ann R (c l ) = w l R,which is again clear by the definition.The relevance of this is in turn motivated by the following Lemma.Lemma 3.16 Let M be a R-module, which is generated by the sequence (d i ) i≥l ,such that d l w l = 0 a, and d i w i = d i−1 for all i > l. Then there exists a R-homomorphism σ : I −→ M, with σ(c i ) = d i , for every i ≥ l. In particular,(d i ) i≥l is a HM-system for M.Proof. Since w l ∈ Ann R (d l ), by Lemma 3.5, Lemma 3.15 and an obviousinductive argument, we have Ann R (c i ) ⊆ Ann R (d i ), for every i ≥ l. Thus,for every i ≥ l, there is the natural projection R/Ann R (c i ) → R/Ann R (d i ),which in turn yields a homomorphism of R-modules σ i : c i R → d d R (sincec i R ≃ R R/Ann R (c i ) and d i R ≃ R R/Ann R (d i )). Now, by Lemma 3.15σ j (c i ) = σ j((j∏s=i+1w s)cj)=(j∏s=i+1w s)dj = d i = σ i (c i )for every l ≤ i < j. Hence the maps σ i are compatible, and so the positionc i ↦→ d i (for i ≥ l), may be extended to a R-homomorphism σ : I −→ M. Thelast assertion follows easily from the definition of HM-system.Next step is to prove the existence of elements of Σ that will allow to applyLemma 3.16 (in suitable abelian factors). Thus, Menegazzo extablishes the followingcrucial fact, whose proof (by induction on n, being the case n = 1 partof the proof of Theorem 3.9) is rather long; and we refer to the original paper[72] for it.

66 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMEDLemma 3.17 There exist elements x i , y i in Σ, for all i ≥ l, such that:(i) x i π j , y i π j ∈ (u i − 1)R for every i ≥ l and every j = 1, . . . , n;(ii) y i π 1 = c i for every i ≥ l; and x l = 1;(iii) [y l , rl u l+1 x l+1 ] = 1, and [y i , ri u i+1 x i+1 ] = y i−1 for every i > l;(iv) x up−1 i+1i+1 . . . xui+1i+1 x i+1 = x i y i for every i ≥ l.Now, for the proof of Theorem 3.12, we set g iconsider the subgroup G of W given by= u i x i for every i ≥ l, andG = 〈g i | i ≥ l〉.By property (iv) in Lemma 3.17, we have, for every i ≥ l.g p i+1 = (u i+1x i+1 ) p = u p i+1 xup−1i+1i+1 . . . xui+1x i+1 = u i x i y i = g i y i . (3.9)Write N = 〈y i | i ≥ l〉 G . Then (3.9) shows that G ′ ≤ N ≤ Σ∩G, and G/N ≃ U.In fact, as G ∋ g l = u l , and G/N ≃ U, we have Σ ∩ G = Σ ∩ N〈u l 〉 = N. Also,for i ≥ l, let j > i minimal such that r j > 0; then, by point (iii) of 3.17,y i = y j−1 = [y i , ri g i+1 ] ∈ [N, G]. Hence N = [N, G] = G ′ .Now, let D = N ′ N p and write N = N/D; let also η denote the iomorphismU → G/N which maps u i ↦→ g i N for all i ≥ l. Then, N becomes a R-moduleby letting, for all u ∈ U and yD ∈ N,(yD) u = y uη D.As an R-module, N is generated by the sequence (y i D) i≥l . Now, point (iii) inLemma 3.17, yields(y l D) w l= [y l , rl g l+1 ]D = 1and, for every i > l,(y i D) wi = [y i , ri g i+1 ]D = y i−1 D.Thus, by Lemma 3.16, (y i D) i≥l is a HM-system for the R-module N. It thenfollows from Proposition 3.7 that G/D is a group of Heineken–Mohamed type.To deduce that G is also a group of Heineken–Mohamed type it is now easy,and requires only the following observation.Lemma 3.18 Let G be a p-group of finite exponent, and let N be a normalnilpotent subgroup G. If G/N p N ′ is a Heineken–Mohamed group, then G is aHeineken–Mohamed group.Proof. Let G and N be as in the assumptions, write K = N p N ′ , and let S bea proper subgroup of G. If SK < G, then SK/K is nilpotent and subnormal,whence in particular NS/K is also nilpotent by Lemma 1.59. Since N/N ′ hasfinite exponent, it is easy to deduce that SN/N ′ is nilpotent. Thus, by P. Hall’snilpotency criterion 1.54, NS is nilpotent. In particular, S is nilpotent, andS ⊳ ⊳ NS ⊳ ⊳ G.

3.4. MINIMAL NON-N GROUPS 67Thus, let KS = G. In such a case, KS = G by Proposition 3.1, and soK(N ∩ S) = N ∩ KS = N. Since N is nilpotent, it follows N ∩ S = N. ThusN ≤ S, and consequently S = G.The proof of Theorem 3.12 will be completed once we prove that the groupG constructed above may have arbitrary derived length. As G ′ = [G, N] = N,we have to show that n ≥ 2 and ideal I may be chosen such that N has arbitraryderived length. This is easily achieved by first observing that, by point (ii) ofLemma 3.17, I = Nπ 1 : in fact (ab)π 1 = aπ 1 + bπ 1 for every a, b ∈ N, andif u i ∈ U (i ≥ l), Nπ 1 ∋ (a gi )π 1 = (aπ 1 )u i , for every a ∈ N. Now, givend ≥ 1, we take n ≥ 2 d , and I an ideal with I = IU < U and I m ≠ 0, wherem = p 2d−1 +. . .+p+1. Then Lemma 3.14 yields the desired conclusion. Observealso that, using the remark following the proof of Lemma 3.14, it is not difficultto show that there exists 2 ℵ0 pairwise non-isomorphic Heineken–Mohamed p-groups of a given derived length (we recall also that we will see in Chapter 6that every Heineken-mohamed group is in fact soluble).Another construction which somehow extends that of Heineken and Mohamed,and we like to mention, appears in W. Möhres doctoral thesis [75].Proposition 3.19 For every prime number p and every integer n ≥ 1, thereexists a group G ∈ N 1 such that(1) Z(G) = 1;(2) G ′ is an elementary abelian p-group;(3) G/G ′ is isomorphic to the direct product of n copies of the C p ∞.Clearly (see Proposition 3.1), if n ≥ 2, the groups obtained by this Propositionare not of Heineken-Mohamed type. Nevertheless the existence of N 1 -groupswith the properties described in 3.19 becomes relevant in view of the content ofour final result on periodic N 1 -groups (Theorem 6.23).3.4 Minimal non-N groupsDespite of its simplicity, Möhres’ Lemma 2.15 (and its variations, see e.g. [82]) isoften useful in reducing certain problems to the periodic (or to the finitely generated)case. We now leave for a while our main theme to treat just a particularcase, somehow related to HM-groups, in which this occurs.Let P be a class of groups. A group G is called minimal non-P if G doesnot beleng to P, but all its proper subgroups are P-groups. We are interested inminimal non-nilpotent groups (minimal non-N). Finite minimal non-N groupsare very well understood by a result of O. J. Schmidt (see [97] 9.19). Infiniteexamples are the Heineken-Mohamed groups and the infinite dihedral 2-group.We showProposition 3.20 Let G be a minimal non-nilpotent group. Then, either G isfinitely generated or it is a countable locally finite p-group (for some prime p)of one of the following types:

68 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMED(i) a perfect group;(ii) a Černikov p-group;(iii) a (soluble) group of Heineken-Mohamed type.Proof. Let G be a minimal non-nilpotent group, and assume that G is notfinitely generated. Then G is locally nilpotent and it is countable by Theorem1.31. Let T be the torsion subgroup of G.Suppose T ≠ G. Then G/T is a countable locally nilpotent torsion-freegroup, so, by Lemma 2.15, it admits a proper subgroup H/T with I G/T (H/T ) =G/T . Now H (and H/T ) is nilpotent by minimality of G, whence G/T is nilpotentby Corollary 2.13. Let N/T be the derived subgroup of G/T . Since G/T isnot trivial, G/N cannot be a p-group (for any prime p), so by Lemma 3.2 thereexist two proper subgroups U/N and V/N of it such that UV = G. Now, Uand V are then normal nilpotent subgroups of G, and it follows from Fitting’sTheorem that G is nilpotent, contradicting our assumption.Thus T = G, and so, being locally nilpotent, G is the direct product of itsprimary components. If there are two of more such components, then G is thedirect product of two proper subgroups and so it is nilpotent. Therefore onlyone primary component may exist, and so G is a locally finite p-group for someprime p.Suppose that G is not perfect (which is case i)), and let N = G ′ . Then byLemma 3.2 G/N is either cyclic or C p ∞.Assume firts that G/N is cyclic, and let x ∈ G such that G = N〈x〉. Observethat G/N p is nilpotent by Corollary 1.77; in particular X = 〈x〉N p is subnormalin G. Now, if N p ≠ N = G ′ , X is a proper subgroup, and so X G is also a propersubgroup of G. But then G = NX G is nilpotent by Fitting’s Theorem. Thus,N p = N or, in other words, N is semi-radical, and it follows from Lemma 1.18that N is an abelian divisible p-group, a direct prodoct of groups of type C p ∞.Let A ≤ N be such a subgroup; then A has a finite number of conjugates inG, so A G is the product of finitely many copies of A. If A G ≠ N then A G 〈x〉is nilpotent, forcing [A G , x] = 1, which is a contradiction. Thus, A G = N hasfinite rank, and G is a Černikov p-group.Assume finally that G/N ≃ C p ∞. Let H be a proper subgroup of G, IfNH ≠ G then NH is nilpotent and normal in G and so H is subnormal inG. Thus G is a group of Heineken-Mohamed type if we show that no propersubgroup H of G exists such that NH = G. Suppose, by contradiction, that His such a subgroup. Then H ∩ N is a proper subnormal subgroup of N, and it isnormal in H; hence, being N nilpotent, M = (H ∩ N) N N ′ is a proper subgroupof N which is normalized by NH = G. It follows that MH is a proper subgroupof G. We may than assume M = 1. Hence N is abelian, and N ∩H = 1 (this lastcondition imply H ≃ C p ∞). Since N is not centralized by H (otherwise H Gand G is nilpotent), there exists an element x ∈ H such that C N (x) ≠ N.Now, as H is abelian, C N (x) is normalized by H, so C N (x)H is a proper, andhence nilpotent, subgroup of G. But also [N, x] ≠ N, as N〈x〉 is nilpotent andN ∩ 〈x〉 = 1; whence [N, x]H is nilpotent. It follows that there exists n ∈ N suchthat [[N, X], n H] = 1, where X = 〈x〉. Then, by 1.13,1 = [N, X, H, . . . , H] = [N, H, . . . , H, X]

3.4. MINIMAL NON-N GROUPS 69which means that [N, n H] ≤ C N (X). Since we observed above that C N (X)H isnilpotent, we conclude that H is subnormal, and this implies that G is nilpotent,a contradiction.Clearly, not every Černikov p-group is minimal non-N, and we leave to thereader to work out a more precise description for this case. More relevant is toreport that Asar [1] has proved that case (i) cannot occur. It is also important tonote that finitely generated minimal non-N groups appear to be very difficultto understand: the finitely generated groups with all proper subgroups cyclic(the so-called Tarski monsters), constructed by Ol’shanskii [87] and Rips, are,obviously, of this kind (and they can be torsion-free). In view of these examples,it is common in the literature on the argument to restrict investigations to classesof groups that are large enough to comprise important cases but exclude Tarskimonsters and objects alike. The usual restriction is to locally graded groups.Now, let G be a locally graded finitely generated group with all propersubgroups nilpotent, and assume that G is not finite. Then G is a finite extensionof a nilpotent group N; since finite minimal non-N groups are soluble, we maytake N ≥ G ′ . We know that (being finitely generated) G/N is a cyclic p-group forsome prime p. Also, as a subgroup of finite index of a finitely generated group, Nis finitely generated nilpotent infinite group; hence the torsion subgroup T (N) isfinite and, by 1.41, N/T (N) admits a characteristic subgroup X/N with N/Xa finite non-trivial p-group. But then X G and G/X is a finite p-group,contradicting N = γ 3 (G) (which in turn follows from Lemma 3.2).Thus, together with the aforementioned result of Asar, we have:Theorem 3.21 Let G be a locally graded minimal non-nilpotent group. Then,G is either finite, or a Černikov p-group, or a p-group of Heineken-Mohamedtype (in particular - as we will see later - G is soluble).In fact, Heineken-Mohamed groups are nilpotent-by-Černikov, and with similar(but more elaborated) methods it is possible to prove the following Theorem.Theorem 3.22 Let G be a locally graded group in which every proper subgroupis nilpotent-by-Černikov. Then G is nilpotent-by-Černikov.A result that, as well as Theorem 3.21, is due to the combined efforts of a numberof people; see Newman and Wiegold [86], Bruno [8], Otal and Peña [90], Brunoand Phillips [10], H. Smith [105], Napolitani and Pegoraro [82] and Asar [1].

70 CHAPTER 3. GROUPS OF HEINEKEN AND MOHAMED

Chapter 4Bounded defectsThe main result to be proved in this chapter (at least in view of its subsequentapplications in these notes) is a fundamental theorem of Roseblade, stating thata group in which every subgroup is subnormal of defect at most n (for n ≥ 1)is nilpotent of nilpotency class not exceeding a value depending only on n. Wealso include some related material (mostly without proofs).4.1 n-Baer groupsFor every n, r ≥ 1, we denote by U n,r the class of all groups in which everysubgroup that can be generated by r elements is n-subnormal (i.e. subnormalof defect at most n). By definition, U n,r+1 ⊆ U n,r for every n, r ≥ 1; we setU n = ⋂ r≥1U n,r .Then, from Lemma 1.24 it immediately follows,Proposition 4.1 for every n ≥ 1, U n is the class of groups in which everysubgroup is n-subnormal.Given n ≥ 1, U n,1 is the class of groups in which every cyclic subgroup is n-subnormal; such groups are usually called n–Baer groups. Occurencies of groupsof this kind we have already encountered. For instance, Proposition 1.76 statesthat a soluble p-group of finite exponent is an n-Baer group, where n depends onthe exponent and on the derived length of the group. However, not many generalresults are known about n-Baer groups, and we have precise informations onlyfor small values of n, which we will briefly report.Before, let us notice the obvious fact that every n-Baer group G is (n + 1)-Engel, that is it satisfies the identity x n+1 y] = 1. Thus, as a first step in treatingn-Baer groups we recall some known facts about n-Engel groups (for n small).Clearly, 1-Engel groups are just the abelian groups. 2-Engel groups are also wellunderstood; their description is essentially due to Levi [65], who also provedthat every group of exponent 3 is 2-Engel.71

72 CHAPTER 4. BOUNDED DEFECTSTheorem 4.2 (Levi [65]). Let G be a 2-Engel group. Then γ 4 (G) = 1, andγ 3 (G) has exponent dividing 3. Thus, a torsion–free 2-Engel group is nilpotentof class at nost 2.3-Engel groups are much more complicated. They need not be nilpotent: thestandard wreath product G = C ≀ A of a cyclic group C of order 2 by an infiniteelementary abelian 2-group A is not nilpotent (for example Z(G) = 1) but it is3-Engel, as it is easily cheched. The fact that 3-Engel groups are locally nilpotentis not at all immediate and was established in [43] by Heineken, who also provedthat if G is a 3-Engel group with no elements of order 2 or 5, then γ 5 (G) = 1.On the other hand, Bachmuth and Mochizuki showed in [2] that there exists a3-Engel group of exponent 5 that is not even soluble (while 3-Engel 2-groupsare soluble, see [33]). The following stetements collect the most relevant knownfacts about 3-Engel groups.Theorem 4.3 (N. Gupta, M. Newman [35]) Let G be a 3-Engel group. Then1. if G is n-generated, with n > 2, then it is nilpotent of class at most 2n − 1,if, further, G does not have elements of order 5, then G has class at mostn + 2;2. γ 5 (G) has exponent dividing 20, and this is best possible;3. the subgroup G 5 generated by the fifth powers of elements of G satisfies thelaw [[a, b, c], [d, e]] = 1To complete the statement of point 1. we mention that if G is a 2-generated3-Engel group, then |γ 4 (G)| ≤ 2 (Heineken [43]), and this is best possible (C.K. Gupta, see [34]).Proposition 4.4 (L. C. Kappe and W. P. Kappe [50]). Let G be a group. Thefollowing are equivalent:1. G is a 3-Engel group;2. 〈x〉 G is a 2-Engel group for every x ∈ G;3. γ 3 (〈x〉 G ) = 1 for every x ∈ G.Recently Havas and Vaughan-Lee [42] succeeded in proving that 4-Engel groupsare locally nilpotent (see also Traustason [117] for a mostly computer–free approach).Of course, for n-Baer groups local nilpotency is not in question. We alreadyobserved that a n-Baer group is (n+1)-Engel group (but not necessarily n-Engel,se e.g. [67]). However, if G is not periodic then G is in fact n-Engel.Lemma 4.5 Let G be a non-periodic group in which the set T or(G) of torsionelements is a finite subgroup. Suppose that, for some n ≥ 1, all elements ofinfinite order in G are left n-Engel. Then G is n-Engel.

4.1. N-BAER GROUPS 73Proof. Write T = T or(G). Since T G is finite and G is not periodic, thecentralizer C G (T ) contains an element x of infinite order.Let g ∈ G. If |g| = ∞ then g is left n-Engel by assumption. Thus, let g ∈ T .Then, for any y ∈ G, [y, xg] = [y, g][y, x] g = [y, g][y, x], as 〈x〉 G ≤ C G (T ).Continuing by induction on i ≥ 1, we haveand, since [y, i g, xg] ∈ T ,[y, i+1 xg] = [[y, i g][y, i x], xg] = [y, i g, xg] [y,i x] [y, i x, xg][y, i+1 xg] = [y, i g, xg][y, i x, xg] = [y, i+1 g][y, i+1 x].Thus, for every k ≥ 1, [y, k xg] = [y, k g][y, k x]. Now, both x and xg have infiniteorder and so are left n-Engel. Hence, in particular,1 = [y, n xg] = [y, n g][y, n x] = [y, n g].This proves that g is left n-Engel. Therefore, G is n-Engel.Proposition 4.6 (see [51]) Let n ≥ 1. Every non-periodic n-Baer group is n-Engel.Proof. Let G be a non-periodic n–Baer group. To show that G is n-Engel, wemay clearly assume that G is finitely generated. Then G is nilpotent, and so, inparticular, T = T or(G) is a finite normal subgroup of G. By Lemma 4.5, it isthen sufficient to show that all elements of infinite order of G are left n-Engel.Thus, let x ∈ G have infinite order, and let g ∈ G. Then [g, n x] ∈ 〈x〉, and sothere exists m = m(g) ≥ 0 such that [g, n x] = x m . Hence x m ∈ γ n+1 (G), andx m2= [g, n−1 x, x] m = [g, n−1 x, x m ] ∈ γ 2n+1 (G).Proceeding in this way, we have, for any r ≥ 1, x mr ∈ γ rn+1 (G). But G isnilpotent, so there exists r ≥ 1 such that x mr = 1. Since |x| = ∞, the onlypossibility is then m = 0. Thus, [g, n x] = 1, and we are done.Clearly, 1-Baer groups are just those groups in which every subgroup isnormal. These are the well-known Dedekind groups (see [97] 3.5.7)Proposition 4.7 (Dedekind). U 1 = U 1,1 , and G ∈ U 1 if and only if G is eitherabelian or the direct product G = Q × D of a quaternion group of order 8 and aperiodic abelian group D that does not have elements of order 4. In particular,if G ∈ U 1 , then |γ 2 (G)| ≤ 2. Torsion–free U 1 are abelian.The class of 2-Baer groups was first studied by Heineken, who proved thatif G is a 2-Baer group then G/ζ(G) is 2-Engel;. from Theorem 4.2 nilpotency ofG follows, togheter with informations on the lower central factors. These werelater completed by Mahdavianary. The combined result isTheorem 4.8 (Heineken [44], Mahdavianary [66]) Let G be a 2-Baer group,then γ 4 (G) = 1.

74 CHAPTER 4. BOUNDED DEFECTSSpecial classes of 2-Baer p-groups (p = 2, 3) are classified in further papersof Mahdavianary ([67], [68]), while in [89], E. Ormerod describes all 2-Baerp-groups for p ≥ 5.It follows immediately from Proposition 4.4 that every 3-Engel is a 3-Baergroup; thus, 3-Baer groups need not be even be soluble (in fact the Bachmuth-Mochizuki group shows that 3-Baer groups of finite exponent need not be soluble(cfr. Proposition 1.76)). Also, by 4.4 and 4.6, a non-periodic group is 3-Baer ifand only if it is 3-Engel. Some positive results on arbitrary 3-Baer groups are tobe found in Traustason [118]; in particular, Traustason proves that every 3-Baergroup G admits a normal subgroup N, which is nilptent of class at most 3 (andin fact abelian if G does not have 2-elements) such that G/N is a 3-Engel group.Finally, metabelian n-Baer groups are the subject of a paper by L. C. Kappeand Garrison [29].4.2 Roseblade’s TheoremAs mentioned before, Roseblade’s Theorem says (in particular) that, for everyn ≥ 1, there exists a positive integer ρ(n) such that a group in which everysubgroup is n-subnormal (thus, a U n -group) is nilpotent of nilpotency classbounded by ρ(n). Thus (recalling that N denotes the class of all nilpotent groups,N = ⋃ n∈NU n .Theorem 4.9 (Roseblade [98]) There exist functions f, ρ : N → N such that,for every n ≥ 1, a group in which every f(n)-generated subgroup is subnormalof defect at most n is nilpotent of nilpotency class not exceeding ρ(n). ThusU n,f(n) ⊆ N ρ(n) .Before coming to it, let we mention that the value of ρ(n) that one obtains fromthe proof of Roseblade’s Theorem is quite likely far larger than the real bound.The actual bound has been determined only for n = 1 and n = 2. Clearlyρ(1) = 1, while ρ(2) = 3 follows from Mahdavianary Theorem 4.8 (althoughit is not hard to see that the class U 2 is strictly smaller than the class of 2-Baer groups). For n = 3, we have the following proposition (to be considered inconnection to the metioned results on 3-Baer groups in [118])Proposition 4.10 (Traustason [117]) Let G be a 3-Engel U 3 -group with noelements of order 2. Then γ 5 (G) = 1.(Thus, a U 3 -group with no elements of order 2 has derived length at most 4.)We already mentioned in Chapter 2, that if n ≤ 4, and the group G ∈ U n istrosion-free, then γ n+1 (G) = 1; this is due Stadelmann [115] for n = 2 (but thisfollows at once from 4.2 and 4.6), Traustason [117] (the previous Propositionplus 4.6) for n = 3, and Smith and Traustason [114] for n = 4.For the proof of 4.9 we follow [64], and start with a preliminary result dealingwith Engel groups.

4.2. ROSEBLADE’S THEOREM 75Lemma 4.11 Let A be a normal abelian subgroup of the group G. Suppose thatG/C G (A) is abelian and there exists n ≥ 1 with [a, n x] = 1 for all a ∈ A. Thenthere exists 0 < β(n) ∈ N such thator, eqiuivalently, A β(n) ≤ ζ 2 n−1(G).Proof. See e.g. [64], Lemma 6.1.6.[A, 2 n−1 G] β(n) = 1This Lemma is in fact the key ingredient (together with an inductive argumentusing Hall”s nilpotency criterion) in the proof of Proposition 1.67. As wementioned in Chapter 1, nowadays (thanks to Zelmanov’s solution of the restrictedBurnside Problem plus some tools from the theory of profinite groups)it is possible to say much more (at least for locally graded groups) as seen inTheorem 1.66. However, for the proof of Roseblade Theorem, we need only thosefacts (like. 4.11) that can be proved without invoking such deep results, and sowe proceed along this line.Lemma 4.12 There exists a function c : N × N → N such that if G ∈ U n,n issoluble of derived length d, then G ∈ N c(n,d) .Proof. Clearly c(n, 1) = 1 for every n ≥ 1 and c(1, d) = 12. Now, assume d = 2,n ≥ 2, set t = 2 n−1 . Let A = G ′ ; then A is a normal abelian subgroup of Gand G/C G (A) is abeliian; since G is (n + 1)-Engel it follows from Lemma 4.11that A β(n+1) ≤ ζ 2 n(G). Thus, we may assume that A has exponent dividingb = β(n). By Lemma 1.16 we then have that G/C G (A) is abelian of exponentdividing b n . Let x 0 , x 1 , . . . , x n ∈ G and set H = 〈x 0 , x 1 , . . . , x n 〉. Then H ′ isgenerated by the H-conjugates of the commutators [x i , x j ], 0 ≤ i < j ≤ n.Since H ′ ≤ A and H/C H (A) is a (n + 1)-generated abelian group of exponentdividing b n , it follows that the number of generators of H ′ does not exceedc = c(n) = ( )n+12 b n(n+1) , whence |H ′ | ≤ b c since H ′ has exponent dividing b.Now, by Lemma 1.12,[A, x 0 , x 1 , . . . , x n , g] = [a, g, x 0 , x 1 , . . . x n ]for every g ∈ G, and a ∈ A, showing that K = [A, x 0 , x 1 , . . . , x n ] is a normalsubgroup of G. Since G ∈ U n,n , [A, x 0 , x 1 , . . . , x n−1 ] ≤ H and so K ≤ H ′ hasorder bounded by b c . As G is locally nilpotent, this implies that K is containedin the b c -th term of the upper central series G. Thusγ n+3 (G) = [A, n+1 G] ≤ ζ b c(G).Recalling that we worked modulo A b , we conclude that G is nilpotent of classat mostc(n, 2) = 2 n + b c + n + 2.We now fix n ≥ 1 and proceed by induction on the derived length d ofG ∈ U n,n . Then, by inductive assumption, N = G ′ ∈ N c(n,d−1) , while G/N ′is metabelian and so, by what proved above, it is nilpotent of class at most

76 CHAPTER 4. BOUNDED DEFECTSc(n, 2). By Theorem 1.54, we conclude that G is nilpotent of class at mostc(n, d) = ( ) (c(n,d−1)+12 c(n, 2) − c(n,d−1))2 .The following property of the automorphims group of an abelian p-groupwas first proved by P. Hall for the finite case, and later extended by Baer andHeineken [5].Lemma 4.13 Let A be a abelian p-group of rank r. Then any p-subgroup ofAut(A) can be generated by at most r(5r − 1)/2 elements.Proof. See [5], or [64] page 178.We are now ready to prove Roseblade’s Theorem.Proof of Theorem 4.9 By Dedekind’s Theorem 4.7, f(1) = 1 and ρ(1) = 2. Wethen let n ≥ 2 and proceed by inductione on n.We set d = (n − 1)([log 2 (ρ(n − 1))] + 1), and definef(n) = c(n, d) + f(n − 1) + 1where c(n, d) is the value obtained in Lemma 4.12 (observe that f(n) ≥ n). LetG ∈ U n,f(n) ; we have to show that G is nilpotent of bounded class.Let X be a s-generated subgroup of G, with s ≤ c(n, d) + 1, and denoteas usual by X G,i the i-th term of the normal closure series of X in G. Sinces ≤ f(n), X G,n ≤ X. For i = 1, . . . , n − 1, and let Y be a f(n − 1)-generatedsubgroup of H G,i , then V = 〈X, Y 〉 is generated by s+f(n−1) ≤ f(n) elementsand so it is subnormal of defect at most n in G, whence it is subnormal of defectat most n − i in V G,i = X G,i . Thus (since f(n − 1) ≥ f(n − i)), we have thefollowing:X G,iX G,i+1 ∈ U n−i,f(n−i). (4.1)for all 1 ≤ i ≤ n − 1. Now, by inductive assumption, X G,i /X G,i+1 is nilpotentof class at most ρ(n − i) ≤ ρ(n − 1), and so its derived length is at most[log 2 (ρ(n − 1))] + 1 by 1.8. Thus, by the definition of d given above,(X G ) (d) ≤ X G,n ≤ X. (4.2)Let c = c(n, d) + 1 (as we will write from now on). By applying Lemma 4.12 to(4.2), we have that for any c-generated subgroup X of Gγ c (X G ) ≤ X (4.3)In particular, it follows from 4.2 that for every x ∈ G, 〈x〉 G is soluble ofderived length at most d + 1, and therefore it is nilpotent of class not exceedingl = c(n, d + 1). By Fitting’s Theorem it follows that, for every s ≥ 1, and anyx 1 , . . . , x s ∈ Gγ sl+1 (〈x 1 , . . . , x s 〉 G ) = 1. (4.4)By Proposition 1.46, we conclude that there exists r = r(n) such thatevery c−generated subgroup of G has rank at most r. (4.5)

4.2. ROSEBLADE’S THEOREM 77We now observe that we may assume that G is a p-group for some primep. In fact, in order to prove that G has bounded nilpotency class it is enoughto prove this for a finitely generated (and thus nilpotent) G; but then G isresidually finite; thus we may suppose that G is finite and, consequently, a p-group for some prime p.LetA be a normal abelian subgroup of G. Let x 1 , . . . , x c ∈ G and writeH = 〈x 1 , . . . , x c 〉. Then, by (4.3),[A, c H G ] ≤ γ c (H G ) ≤ A ∩ H. (4.6)Thus, by (4.5) we have that B = [A, c H G ] is a normal abelian subgroup of Gof rank at most r. We may then apply Lemma 4.13 to conclude that G/C G (B)is generated by at most r(5r − 1)/2 elements. Then (4.4) tels us that G/C G (B)has nilpotency class at most r 1 = (r(5r − 1)/2)l, and so [B, γ r1+1(G)] = 1. Inparticular, setting C = γ r1+1(G),[A, x 1 , x 2 , . . . , x c ] ≤ C G (C). (4.7)Since A is normal and abelian, this yields [A, c G, C] = 1. Since c ≤ r 1 + 1 and[A, γ c (G)] ≤ [A, c G], we obtain[A, C ′ ] ≤ [A, C, C] = 1 (4.8)for any abelian normal subgroup A of G. Now, let x ∈ G. If K = 〈x〉 G , thenK/K ′ is an abelian normal subgroup of G/K ′ , and so [K, C ′ ] ≤ K ′ , by (4.8).Since, by the remark following (4.3), K has class at most l, a simple inductiveargument using the Three Subgroups Lemma, shows that[K, l C ′ ] = 1.This holds for every x ∈ G; in particular we have γ l+1 (C ′ ) = 1. Therefore, G issoluble of derived length bounded by [log 2 r 1 ] + [log 2 l] + 3. By Lemma 4.12, weconclude that G is nilpotent of class at mostρ(n) = c(n, [log 2 r 1 ] + [log 2 l] + 3).This completes the proof of the Theorem.For torsion-free groups, it follows form Zel’manov deep result on boundedEngel groups (Theorem 1.65) that groups in U n,1 are nilpotent of bounded class.Obviously, this is not in general the case: for instance, let G = C≀A be the wreathproduct of a group of order 2 by an infinite elementary abelian 2-group, andlet B denote its base group; then for every x ∈ G, 〈x〉 G ≤ B〈x〉 is nilpotentof class at most 2 and from Fitting theorem it follows that, for every n ≥ 1and x 1 , . . . x n ∈ G ,〈x 1 , . . . x n 〉 G has class at most 2n; hence, every n-generatedsubgroup of G has defect at most 2n in its normal closure, and so G ∈ U 2n+1,n(for every n ≥ 1), but G is not nilpotent . Indeed groups in U 2n+1,n need noteven be soluble: using the same argument (via Proposition 4.4) one shows thatthe mentioned Bachmuth–Mochizuki group ([2]), which is not soluble, belongsto U 2n+1,n for every n ≥ 1. In his original paper, Roseblade asks the following:

78 CHAPTER 4. BOUNDED DEFECTSQuestion 2 Is U n,n ⊆ N ρ1(n), for some positove integer ρ 1 (n)?(in view of the examples given above, the feeling is that U n,r ⊆ N for somen/2 ≤ r ≤ n). Also, it follows from Roseblade’s Theorem that, for every n ≥ 1,there exists r(n) ≥ 1, such that U n,r(n) = U n (it is plain that r(1) = 1 and notdifficult to see that r(2) = 2); thus, a related question isQuestion 3 Find a reasonable bound for r(n).An Engel-type version of these questions could be the following.Question 4 Let G be a group, n ≥ 1, and suppose that[g, x 1 , . . . , x n ] ∈ 〈x 1 , . . . , x n 〉for every g, x 1 , . . . , x n ∈ G. Is it true that G is nilpotent of class bounded by afunction of n?For locally nilpotent torsion groups, Roseblade’s Theorem has been generalizedby E. Detomi ([24]) along a direction which is clearly suggested by Brookes’trick (Theorem 1.92).Theorem 4.14 Let G be a periodic locally nilpotent group. Assume that thereexist a finite subgroup F of G, and n ∈ N, such that every subgroup of Gcontaining F is subnormal of defect at most n in G. Then γ β(n)+1 (G) is finitefor a positive integer β(n) depending only on n. In particular, G is nilpotent.We begin the proof with a rather simple observation.Lemma 4.15 Let A be a normal subgroup of the group G, such that G/C G (A)is abelian. Suppose that there exists 1 ≤ n, m ∈ N such that |[A, x 1 , . . . , x n ]| ≤ mfor all x 1 , . . . , x n ∈ G. Then |[A, 2n G]| ≤ g(n, m), where g(1, m) = (m!) 2 , andg(n + 1, m) = (g(n, m)!) 2 .Proof. Observe first that, since G/C G (A) is abelian, [A, xy] = [A, yx] for everyx, y ∈ G. Hence, for all x, y ∈ G [A, x] is normal in G, and [A, x, y] = [A, y, x].Assume first n = 1 and proceed by induction on m. If m = 1, we have nothingto prove. Thus, let m ≥ 2. If [A, x, y] = 1 for all x, y ∈ G then we are done.Otherwise, there exist x, y ∈ G such that [A, x, y] ≠ 1. Let N = [A, x][A, y], andG = G/N.Now, if z = zN ∈ G, then [A, z] = [A, z]N/N ∼ = [A, z]/([A, z] ∩ N). Supposethat there exists an element z ∈ G such that |[A, z]| ≥ m. Then [A, z] ∩ N = 1.In particular, [N, z] ≤ N ∩ [A, z] = 1, which in turn implies[A, x, z] = [A, y, z] = 1.Also, as [A, x] = 1, we get [A, xz] = [A, x][A, z][A, x, z] = [A, z], and so, by thesame argument used above, N ∩ [A, xz] = 1 and [a, y, xz] = 1. Hence, for alla ∈ A, we have[a, y, xz] = [a, xz, y] = [[a, x][a, x, z][a, z], y] = [a, x, y][a, z, y] = [a, x, y].

4.2. ROSEBLADE’S THEOREM 79Thus, we reach the contradiction 1 = [A, y, xz] = [A, x, y] ≠ 1. Therefore,|[A, z]| ≤ m − 1 for all z ∈ G. By inductive hypothesis, we then haveand consequently,|[A, 2 G]| ≤ ((m − 1)!) 2 ,|[A, 2 G]| ≤ |[A, 2 G]||N| ≤ ((m − 1)!) 2 m 2 = (m!) 2 .Thus, the Lemma is proved for n = 1, and we now continue the proof byinduction on n. If we fix x ∈ G, then [A, x] G and G/C G ([A, x]) is abelian, asC G ([A, x]) ≥ C G (A). Moreover, for all x 2 , . . . , x n ∈ G, |[[A, x], x 2 , . . . , x n ]| ≤ m.Hence, by inductive assumption,|[[A, x], 2(n−1) G]| ≤ g(n − 1, m).This holds for every x ∈ G. But [[A, x], 2(n−1) G] = [[A, 2(n−1) G], x]. It thenfollows by the case n = 1 that|[A, 2n G]| = |[[A, 2(n−1) G], 2 G]| ≤ g(1, g(n − 1, m)) = (g(n − 1, m)!) 2 = g(n, m),thus completing the proof.To shorten the notation, let us denote by U + n the class of all locally nilpotentgroups which admit a finite subgroup F such that all subgroups F ≤ H ≤ Gare subnormal of defect at most n in G.Lemma 4.16 Let G ∈ U + n , and suppose that G has a nilpotent subgroup N,with finite index in G and nilpotency class c. Then γ cn+1 (G) is finite.Proof. Since N has finite index, we may possibly replace it by its normal coreN G . Thus, we assume N G. Let T be the torsion subgroup of G. Then G/Tis a locally nilpotent torsion-free group with a subgroup of finite index NT/T ,which is nilpotent of class at most c; by Corollary 2.13, G/T is nilpotent of classat most c; thus γ c+1 (G) ≤ T .Let F be a finite subgroup of G such that all subgroups of G containing Fare subnormal of defect at most n. Let T be a transversal of G modulo N, andset H = 〈F, T 〉. Then H is finitely generated (hence nilpotent) and subnormalof defect d ≤ n in G. If d = 1, then G/H = NH/H ≃ N/N ∩ H is nilpotent ofclass at most c; hence γ c+1 (G) ≤ H ∩ T is finite (because H ∩ T = T or(H)),and we are done. Continuing by induction on d, let d ≥ 2. Now, H G ∈ U + nand N ∩ H G is a finite-index subgroup of H G ; so γ c(d−1)+1 (H G ) is finite byinductive assumption. Therefore, by Fitting’s Theorem applied to G = NH G ,we conclude|γ cd+1 (G)| ≤ |γ (d−1)c+1 (H G )| · |γ c+1 (N)| < ∞which is what we wanted.This allows to prove the specific Hall–type reduction needed.Lemma 4.17 There exists a function f(d, c; n) with the following property. LetG ∈ U + n and N a normal subgroup of G; if γ c+1 (N) and γ d+1 (G/N ′ ) are finite,then γ f(d,c;n) (G) is finite.

80 CHAPTER 4. BOUNDED DEFECTSProof. Since γ c+1 (N) is a finite normal subgroup of G, we may well assumethat γ c+1 (N) = 1. Now, by a result of P. Hall (Proposition 1.51), we have thatA/N ′ = ζ 2c (G/N ′ ) has finite index in G/N ′ . Since A/N ′ has nilpotency classat most 2d, Fitting Theorem yields that AN/N ′ has nilpotency class at most2d + 1. Then, by Hall criterion (Theorem 1.54), AN is nilpotent of class at mostm = ( ) (c+12 (2d + 1) − c)2 . As AN has finite index in G, we finally apply Lemma4.16 to get the desired conclusion.Proof of Theorem 4.14. We proceed by induction on n ≥ 1. If n = 1, F G andG/F is a Dedekind group; thus β(1) = 1.Assume n ≥ 2, and let N = F G . Then N ∈ U + n−1 , and so, by inductiveassumption, γ β(n−1)+1 (N) is finite. By Lemma 4.17 we are done if we show thatγ k (G/N ′ ) is finite for some k depending only on n. Thus, we may assume thatN = F G is abelian.By Roseblade Theorem, G/N is nilpotent of class bounded by ρ(n); in particularthe derived length l of G/C G (N)) is bounded by log 2 (ρ(n)). Fixed n, weargue by induction on l.Thus, assume first that l = 1, i.e. G ′ cetralizes N. Let π = π(F ) be the setof all prime divisors of |F |. Then N is an abelian π-group. Given p ∈ π, let X pbe the product of all q-components of N with q ≠ p; then X p G and N/X p isa p-group. If we prove that γ µ (G/X p ) is finite for a uniform µ = µ(n), then weare done because π is a finite set of primes. Thus, we may suppose that N = F Gis an abelian p-group for some prime p. Let |F | = p k , let p r be the exponent ofF , and observe that p r is also the exponent of N. Write G = G/N p , N = N/N p ,and so on. Let x ∈ G, and x = xN p . By assumption, 〈F , x〉 is nilpotent andsubnormal; hence, by 1.59, 〈N, x〉 is nilpotent. Also, every subgroup of 〈N, x〉containing F has defect at most n, so by Lemma 2.20 we have[N, fp(k,n)−1 x] = 1.Let s be the smallest power of p grater than f p (k, n) − 1. As N is an elementaryabelian p-group, it follows from 1.14 that [N, x s ] = 1; i. e.[N, x s ] ≤ N p for all x ∈ G. (4.9)Write now t = sp r(log p r+1) . Since N has exponent p r , (4.9) yields[N, x t ] = 1 for all x ∈ G. (4.10)Thus, the exponent of G/C G (N) is at most t. Now, take x 1 , x 2 , . . . , x ρ(n) ∈ G,and let H = 〈F, x 1 , . . . , x ρ(n) 〉 = F H 〈x 1 , . . . , x ρ(n) 〉. Since H/CH(N) is abelian(as such, by assumption, is G/C G (N)), its order is at most t ρ(n) , and consequently|F H | ≤ |F | tρ(n) .Now, F ≤ F H NH, and all subgroups of NH/F H are subnormal of defect atmost n. By Roseblade’s Theorem, NH/F H is nilpotent of class at most ρ(n);hence [N, ρ(n) H] ≤ F H , and, in particular,∣∣[N, x 1 , . . . , x ρ(n) ] ∣ ∣ ≤ ∣ ∣F H∣ ∣ ≤ |F | tρ(n) . (4.11)

4.3. FIRST APPLICATIONS 81We may then apply Lemma 4.15 and have that [N, 2ρ(n) G] is finite. Now, G/Nis nilpotent of class at most ρ(n), and so we conclude thatγ 3ρ(n)+1 (G) = [γ ρ(n)+1 (G), 2ρ(n) G] ≤ [N, 2ρ(n) G]is finite. Thus, the case in which G/C G (N) is abelian is done.Suppose now l ≥ 2. Let K = C G (N)G ′ . Then, by inductive assumnption,γ ν (K) is finite for some ν which depends only on n and l − 1. Also, N ≤ K, andC G (NK ′ /K ′ ) ≥ K ≥ G ′ . It is now easy to see that we may apply the previouscase to the group G/K ′ , concluding that γ 3ρ(n)+1 (G/K ′ ) is finite. ApplyingLemma 4.17 we thus conclude that γ k (G) is finite, for some k that ultimatelydepends oinly on n. By the remarks made at the beginning, this completes theproof.In her paper, Detomi also shows that Theorem 4.14 does not hold whendropping the assumption of G being locally nilpotent (an example in which|F | = 2 and γ 2 (G) = γ 3 (G) is infinite is given), nor it is true for locally nilpotentnon-periodic groups; although she proves that, in this case, G is hypercentral.Proposition 4.18 Let F be a finitely generated subgroup of the locally nilpotentgroup G, and suppose that there exists n ≥ 1 such that every subgroup of Gcontaining F has defect at most n. Then G is hypercentral (and soluble).Proof. By assumption F is nilpotent and subnormal in G, and, by RosebladesTheorem, each section of the normal closure series of F in G is nilpotent; thus,G is soluble.Now, it is clearly enough to prove that G has non-trivial centre. If n = 1,then F G, G/F is nilpotent of class at most two, and F , as a finitely generatednormal subgroup of a locally nilpotent group, is contained in some term of theupper central series of G, which is then nilpotent. Thus, letting n ≥ 2, andassuming that the claim is true for smaller values, we may suppose Z(F G ) ≠ 1;in particular, F is contained in a normal subgroup N of G which has non-trivialcentre. We now proceed by induction on the derived length d of G/N. If d = 0,then G = N has non-trivial centre. Let d ≥ 1, set G 1 = G ′ N and A = Z(G 1 ).Then A ≠ 1, by inductive hypothesis. Also, A is a normal subgroup of G. LetU be a finitely generated subgroup of G containing F . By assumption, U issubnormal of defect at most n in G, hence [A, n U] ≤ U and [A, n U] is finitelygenerated,. As G ′ ≤ C G (A), for every g ∈ G we get[A, n U] g = [A, n U g ] ≤ [A, n U[U, g]]] = [A, n U],and so [A, n U] is normal in G. As [A, n U] is finitely generated, [A, n U] ≤ ζ k (G)for some k ≥ 1. Thus, if [A, n U] ≠ 1, then Z(G) ≠ 1. Otherwise, if [A, n U] = 1for any finitely generated subgroup U of G (containing F ), then [A, n G] = 1,i.e. A ≤ ζ n (G), and again Z(G) ≠ 1.4.3 First applicationsA first immediate application of Roseblade’s Theorem allows to reduce the studyof periodic N 1 -groups to the case of p-groups.

82 CHAPTER 4. BOUNDED DEFECTSLemma 4.19 Let G be periodic N 1 -group. Then there exists 1 ≤ m ∈ N, suchthat all but finitely many primary components of G are nilpotent of nilpotencyclass at most m. in particular, G is nilpotent if and only if all of its primarycomponents are nilpotent.Proof. Since G is locally nilpotent, it is isomorphic to the direct product of itsprimary components. In one sense, the implication is trivial. Conversely, supposethat all primary components of G are nilpotent. If the nilpotency class of thecomponents is not bounded, then by Roseblade’s Theorem, for each positiveinteger n there is a primary component P n of G and a subgroup H n of P n ofdefect n (and P n ≠ P k if n ≠ k). But then, the subgroup H = 〈H n |n ∈ N〉 cannotbe subnormal in G. Thus, the nilpotency class of the primary components of Gis bounded, and therefore G is nilpotent.Then, in conjunction with Brookes’ trick, a first step towards the proof ofsolubility of N 1 -groups.Lemma 4.20 Let G be a N 1 -group. Then there exists a 1 ≤ n ∈ N such thatG (n) = G (n+1) .Proof. Let G be a N 1 -group which we may clearly assume not to be soluble.Then, by Theorem 1.92 applied to the family on non-soluble subgroups of G,there exists a non-soluble subgroup H of G, a finitely generated subgroup F ofH, and a positive integer d, such that for every F ≤ K ≤ H, if K is not soluble,then K has defect at most d in H. Let ρ = ρ(d) be the bound in Roseblade’sTheorem 4.9, and let d = [log 2 (ρ)] + 1. Now, if K is a non-soluble subgroup ofH containing F , then K H is not soluble, and so all subgroups of H/K H havedefect at most n. By Roseblade’s Theorem, H/K H is nilpotent of class at mostρ, and so it is soluble of derived length at most d.Corollary 4.21 ([13]). A residually soluble N 1 -group is soluble.Now, an application of Detomi’s Theorem.Proposition 4.22 (H.Smith [107]). A periodic residually finite N 1 -group isnilpotent.Proof. Let G be a periodic residually finite N 1 -group. Since every subgroup ofG is residually finite, we may assume that G is countable. By Theorem 1.92there exists a subgroup H of finite index, a finitely generated subgroup F ofH, and a positive integer d, such that every F ≤ K ≤ H, such that |H : K| isfinite, has defect at most d in H. Now, let K be a finitely generated subgroupof H containing F . Since G is periodic, K is finite, whence, by Lemma 1.28, Kis a intersection of subgroups of finite index of H. It follows that K has defectat most d in H. This implies that every subgroup of H containing F has defectat most d in H. By Theorem 4.14, H is nilpotent. Since |G : H| is finite, weconclude that G is nilpotent.Both Corollary 4.21 and Proposition 4.22 will be later superseded (respectively,by Theorem 6.4 and Theorem 5.29).

Chapter 5Periodic N 1 -groups5.1 N 1 -groups of finite exponentIn this section we prove that a soluble N 1 -group of finite exponent is nilpotent;a most important result, due to W. Möhres, which lies at the core of the wholetheory of N 1 -groups. Möhres proof is based on a delicate analysis ([76] and [77])of p-groups which are the extension of two (infinite) elementary abelian groups,and we rather closely follow his approach.For the next results, up to Proposition 5.9, we fix the following notation: p isa given prime, A an elementary abelian p-group, and B an elementary abelianp-group acting on A.We recall a couple of elementary facts. From Lemma 1.12, we have that ifn ≥ 1, x 1 , . . . , x n ∈ B and σ is a permutation of {1, . . . , n} then[a, x 1 , . . . , x n ] = [a, x σ(1) , . . . , x σ(n) ]for any a ∈ A, while from Lemma 1.14 it follows that [A, p x] = 1 for all x ∈ B.We set Z 0 = {1} and, for every n ∈ N, Z n+1 /Z n = C A/Zn (B). Then, forevery a ∈ A and n ≥ 1, a ∈ Z n if and only if [a, x 1 , . . . , x n ] = 1 for everyx 1 , . . . , x n ∈ B. Observe also that if U is a finite B-invariant subgroup of A,then U ≤ Z logp |U|. Finally, if B is finite then, by Corollary 1.77, the naturalsemidirect product AB is nilpotent, so there exists n ∈ N such that [A, n B] = 1.The first Lemma we prove is a standard tool in the theory of (soluble) p-groups of finite exponent and Lie algebras in characteristic p.Lemma 5.1 Let 0 ≤ n ≤ p − 1, a ∈ A and x 1 , . . . , x n ∈ B. Suppose that[a, x 1 , . . . , x n ] ≠ 1. Then there exists x ∈ 〈x 1 , . . . , x n 〉, such that [a, n x] ≠ 1.Proof. We argue by induction on n. If n = 1 the claim is trivial. Thus, letn ≥ 2, and let X = 〈x 1 , . . . , x n 〉. Since X is finite, there exists k ∈ N such that[A, k X] = 1. So, in order to prove the Lemma, we may well assume [a, n+1 X] = 1.By inductive assumtion there exists y ∈ 〈x 2 , . . . , x n 〉 such that [[a, x 1 ], n−1 y] ≠ 1.For every i ∈ {0, 1, . . . , n} let b i = [a, n−i x 1 , i y] (n i) . Now, since [A,n+1 X] = 1 thesubstitution of elements from X in commutators of type [a, t 1 , . . . , t n ] is linear83

84 CHAPTER 5. PERIODIC N 1 -GROUPSin every component (see Lemma 1.47). From this it easily follows that, for every0 ≤ k ≤ n,n∏∏[a, n x 1 y k ] = [a, n−i x 1 , i y k ] i) n (n = b kii .i=oNow, the reduction modulo p of the (n + 1) × (n + 1) matrix (k i ) k,i=0,...,nis a Vandermonde matrix on Z/pZ, and so its determinant is not zero. Sinceb n−1 = [a, x 1 , n−1 y] ≠ 1, it thus follows that there exists 0 ≤ k ≤ n such that[a, n x 1 y k ] ≠ 1. As x 1 y k ∈ X, this is what we wanted to show.The case of this Lemma that we will use frequently is when n = p−1. Observethat if a ∈ A and x ∈ B are such that [a, p−1 x] ≠ 1 then M = 〈a〉 〈x〉 has order p p(in fact, if |M| ≤ p p−1 then, as M is 〈x〉-invariant, [M, p−1 x] = 1). Thus, both{a, a x , . . . , a xp−1 } and {a, [a, x], . . . , [a, p−1 x]} are independent generating setsfor M (in general, if, for some 1 ≤ n ≤ p − 1, [a, n x] ≠ 1, then a, a x , . . . , a xn areindependent). In other words, M is the regular F p [〈x〉]–module. Observe alsothat, for every a ∈ A and x ∈ B, [a, p−1 x] = aa x · · · a xp−1 .These remarks are further extended in the next Lemma.Lemma 5.2 Let n ≥ 1, and a ∈ A.(i) If a ∉ Z n(p−1) , there exist x 1 , . . . , x n ∈ B with [a, p−1 x 1 , . . . , p−1 x n ] ≠ 1.(ii) If x 1 , . . . , x n ∈ B are such that [a, p−1 x 1 , . . . , p−1 x n ] ≠ 1, then x 1 , . . . , x nare independent in B (whence 〈x 1 , . . . , x n 〉 = p n ).(iii) If x 1 , . . . , x n ∈ B are such that [a, p−1 x 1 , . . . , p−1 x n ] ≠ 1, then the setof all elements [a, t1 x 1 , . . . , tn x n ], for every (t 1 , . . . , t n ) ∈ {0, 1, . . . , p − 1} n islinearly independent.Proof. (i) For n = 1 the claim follows from Lemma 5.1. Let n ≥ 2 and assumethe property holds for n − 1. If a ∈ A \ Z n(p−1) then, by inductive assumption,there exists x 1 , . . . , x n−1 such that [a, p−1 x 1 , . . . , p−1 x n−1 ] ∉ Z p−1 , whence bycase n = 1, we find x n ∈ G, with [a, p−1 x 1 , . . . , p−1 x n−1 , p−1 x n ] ≠ 1.(ii) The fact is trivial for n = 1. Thus, arguing by induction on n, we supposethat x 1 , . . . , x n−1 are linearly independent. Now, [a, p−1 x 1 , . . . , p−1 x n−1 , x i ] = 1for every i = 1, . . . , n − 1. Hence b = [a, p−1 x 1 , . . . , p−1 x n−1 ] is centralized byY = 〈x 1 , . . . , x n−1 〉. If x n ∈ Y we have a contradiction. Therefore x n ∉ Y andx 1 , . . . , x n−1 , x n are linearly independent.(iii) By induction on n. For n = 1 this fact has already been observed.Thus, let n ≥ 2, ∆ a non-empty subset of {0, . . . , p − 1} n , and for each letbe given an integer k t with t ∈ ∆ let 1 ≤ k t ≤ p − 1. We have to show thatb = ∏ t∈∆ [a, t 1x 1 , . . . , tn x n ] kt ≠ 1. Let m = min{t n | t ∈ ∆}, s = p − 1 − m, and∆ 0 = {t ∈ ∆ | t n = m}. If c = [a, p−1 x n ], then[b, s x n ] = ∏[a, t1 x 1 , . . . , tn−1 x n−1 , p−1 x n ] kt= ∏[c, t1 x 1 , . . . , tn−1 x n−1 ] kt .t∈∆ 0 t∈∆ 0By inductive assumption [b, s x n ] ≠ 1, whence b ≠ 1.In the hypothesis of point (iii) of the previous Lemma, let X = 〈x 1 , . . . , x n 〉.It then follows from (ii) and (iii) that |X| = p n and |〈a〉 X | = p pn . Hence C X (a) =i=o

5.1. N 1 -GROUPS OF FINITE EXPONENT 851 and {a x | x ∈ X} is a set of independent generators of 〈a〉 X . After theseremarks one easily deduce the following Lemma.Lemma 5.3 Let n ∈ N, a ∈ A \ Z n(p−1) , and let X = 〈x 1 , . . . , x n 〉 ≤ B, with[a, p−1 x 1 , . . . , p−1 x n ] ≠ 1; then(i) if y 1 , . . . , y m ∈ X are independent, then [a, p−1 y 1 , . . . , p−1 y m ] ≠ 1;(ii) X ∩ C B (a) = 1, and so C B (a) has index at least p n in B.We now move to some more specific facts.Lemma 5.4 Let n, s ∈ N with n ≥ 1 and p s > n. If a 1 , . . . , a n ∈ A \ Z s(p−1)then there exists x ∈ B such that [a a , x] ≠ 1 for every i = 1, . . . , n.Proof. By point (i) of Lemma 5.2 and point (ii) of Lemma 5.3, we have that|B : C B (a i )| ≥ p s for every i = 1, . . . , n. Since p s > n, a result of B. H. Neumann[84] implies B ≠ ⋃ ni=1 C B(a i ), and the claim follows.Lemma 5.5 Let n ≥ 1, t = t n = (p − 1) 2n−1 , a 1 , . . . , a n ∈ A, x 1 , . . . x t ∈ B,and suppose that [a i ,x 1 , . . . ,x t ] ≠ 1, for every i = 1, . . . n. Then there existsy ∈ 〈x 1 , . . . , x t 〉 such that [a i , p−1 y] ≠ 1 for every i = 1, . . . , n.Proof. By induction on n. Case n = 1 follows from Lemma 5.1. Thus, let n ≥ 2,t = (p − 1) 2n−1 , and asssume the claim true for n − 1. Let s = (p − 1) 2n−2 ; thent = (p − 1)s = (p − 1) 2 t n−1 . We show that for each j = 1, . . . , p − 1, there existsy j ∈ X j = 〈x (j−1)s+1 , . . . , x js 〉, such that{[an , y 1 , . . . , y j , x js+1 , . . . , x t ] ≠ 1(5.1)[a i , p−1 y 1 , . . . , p−1 y j x js+1 , . . . , x t ] ≠ 1 for i = 1, . . . , n − 1.We start by finding y 1 . For each i = 1, . . . , n, we set b i = [a i , x s+1 , . . . , x t ].Then, by assumption, [b i , x 1 , . . . , x s ] ≠ 1 for all i = 1, . . . , n. Now, by Lemma5.3, C X1 (b n ) has index at least p s/(p−1) = p tn−1 in X 1 . Thus, there is a lineralyindependent subset {z 1 , . . . , z tn−1 } of {x 1 , . . . , x s }, such that Y = 〈z 1 , . . . , z tn−1 〉intersects trivially C X1 (b n ). By the inductive assumption on n, we then findy 1 ∈ Y such that [b i , p−1 y 1 ] ≠ 1 for for all j = 1, . . . , n − 1. Then[a i , p−1 y 1 , x s+1 , . . . , x t ] = [b i , p−1 y 1 ] ≠ 1for i = 1, . . . , n − 1; and [a n , y 1 , x s+1 , . . . , x t ] = [b n , y 1 ] ≠ 1. So conditions 5.1are satisfied for j = 1.Suppose that, for 1 ≤ k < p − 1, we are given y 1 , . . . , y k with the requiredproperties. Then, by setting b i = [a i , p−1 y 1 , . . . , p−1 y k , x (k+1)s+1 , . . . , x t ]for i = 1, . . . , n − 1, b n = [a n , y 1 , . . . , y k , x (k+1)s+1 , . . . , x t ], and repeating thesame argument used for j = 1 we find y k+1 ∈ 〈x ks+1 , . . . , x (k+1)s 〉 that togetherwith y 1 , . . . , y k satisfies 5.1.Thus, we eventually get elements y 1 , . . . , y p−1 ∈ 〈x 1 , . . . , x t 〉 such that{[an , y 1 , . . . , y p−1 ] ≠ 1[a i , p−1 y 1 , . . . , p−1 y p−1 ] ≠ 1 for i = 1, . . . , n − 1.

86 CHAPTER 5. PERIODIC N 1 -GROUPSBy the first of these inequalities and Lemma 5.1 it follows that there existsy ∈ 〈y 1 , . . . , y p−1 〉 such that [a n , p−1 y] ≠ 1. But from the remaining inequalitiesand Lemma 5.3 we also have [a i , p−1 y] ≠ 1 for all i = 1, . . . , n − 1, thus finishingthe proof.Proposition 5.6 There exists a function α : N \ {0} → N, with the propertythat if U is a subgroup of A of order at most p n and U ∩ Z α(n) = 1, then thereexists y ∈ B such that [a, p−1 y] ≠ 1 for all 1 ≠ a ∈ U.Proof. For 1 ≤ n ∈ N, we set α(n) = n(p − 1) 2pn−2 . Let U be a subgroupof A with |U| ≤ p n and U ∩ Z α(n) = 1. Let s = (p − 1) 2pn−3 ; thus α(n) =n · s · (p − 1). Since p n > |U| \ {1}, it follows from Lemma 5.4 that there existelements x 1 , . . . , x s in B such that [a, x 1 , . . . , x s ] ≠ 1 for all a ∈ U \ {1}. Buts ≥ (p − 1) 2|U\{1}|−1 , and so, by Lemma 5.5, there exists y ∈ B such that[a, p−1 y] ≠ 1 for all 1 ≠ a ∈ U.Observe that, if U and y are as in the statement of 5.6, then |U 〈y〉 | = |U| p .In fact, by the Jordan canonical form, U 〈y〉 = 〈u 1 〉 〈y〉 × . . . × 〈u s 〉 〈y〉 for suitableu 1 , . . . , u s ∈ U with U = 〈u 1 , . . . , u s 〉. By the remark following Lemma 5.1,|〈u i 〉 〈y〉 | = p p for every i = 1, . . . , s. Hence |U 〈y〉 | = p ps = |U| p .Lemma 5.7 For every n ≥ 1 there exists a function f n : N × N → N, such thatthe following holds:if |B| ≥ p fn(r,s) , U ≤ Z n , z ∈ Z 1 \U, and |U| ≤ p r , then there exists H ≤ B,with |H| = p s and z ∉ U H .Proof. We argue by induction on n. Clearly, f 1 is given by f 1 (r, s) = s, for all(r, s) ∈ N × N. We then assume that, for n ≥ 2, functions f i have been foundfor 1 ≤ i ≤ n − 1, and procced to define the values f n (r, s).Trivially, for any r, s ∈ N, f n (0, s) = s and f n (r, 0) = 0.To provide f n (1, 1) let us introduce an auxiliary function h : N \ {0} → N, bysetting h(1) = f n−1 (1, 1) and, for t ≥ 2, h(t) = f n−1 (n − 2, h(t − 1)) + 1. Thenlet f n (1, 1) = h((p − 1) 2 + 1).Suppose that, for the given A and B, the conclusion of the statementt failsfor r = s = 1 (and holds for n − 1). Then, there exist1 ≠ a ∈ Z n \ Z n−1 , z ∈ Z 1 \ 〈a〉 such that z ∈ 〈a〉 〈y〉 for all 1 ≠ y ∈ B. (5.2)If n ≥ p + 1 then a ∉ Z p and so, by 5.1 there exists y ∈ B with [a, p−1 y] ∉ Z 1 ,whence Z 1 ∩ 〈a〉 〈y〉 = Z 1 ∩ 〈a, [a, x], . . . [a, p−1 x]〉 = 1. Thus, n ≤ p.For every 1 ≠ y ∈ B we denote by d(y) the smallest positive integer such that[a, d(y) y] ≠ 1. Thus 〈[a, d(y) y]〉 = C A (y) ∩ 〈a〉 〈y〉 . By our assumptions, for every1 ≠ y ∈ B, we have 1 ≤ d(y) ≤ n−1 and 〈z〉 = 〈[a, d(y) y]〉. So there is a uniquelydetermined 1 ≤ m(y) ≤ p − 1 with [a, d(y) y] = z m(y) .We say that a subset {y 1 , . . . , y t } of B is stable (with respect to a and z) if– y 1 , . . . , y t are independent;– for every ∅ ≠ {i 1 , . . . , i s } ⊆ {1, . . . , t}, d(x i1 · · · x is ) = n − 1 andm(x i1 · · · x is ) ≡s∑m(x ij ) (mod p).j=1

5.1. N 1 -GROUPS OF FINITE EXPONENT 87Let K ≤ B; we show, by induction on t ≥ 1, thatif |K| ≥ h(t) then K admits a stable subset of cardinality t. (5.3)For t = 1, |K| ≥ p fn−1(1,1) . Then, by the inductive assumption on n and theassumption (5.2), a ∉ Z n−1 (K), and so, by Lemma 5.1, there exists x ∈ K suchthat d(x) = n − 1.Thus, let t ≥ 2, and suppose |K| ≥ p h(t) . As above, there exists x 1 ∈ K suchthat d(x 1 ) = n − 1. Let V = 〈[a, x 1 ], . . . , [a, n−2 x 1 ]〉. Since [a, x 1 ], . . . , [a, n−1 x 1 ]are linearly independent, we have |V | = p n−2 and z = [a, n−1 x] −m(x1) ∉ V . LetK = K 1 × 〈x 1 〉; then |K 1 | ≥ p fn−1(n−2,h(t−1)) . Since V ≤ Z n−1 , the inductiveassumption on n implies that there exists H ≤ K 1 with |H| = p h(t−1) andz ∉ V H . Then z ∉ V H (〈a〉 ∩ Z n−1 ) = V H 〈a〉 ∩ Z n−1 , whence z ∉ V H 〈a〉. Wework with A/V H acted on by H. If there exists y ∈ H such that z ∉ V H 〈a〉 〈y〉 ,then obviously z ∉ 〈a〉 〈y〉 , which is in contrast with (5.2). Thus, H satisfies(5.2) on A/V H with respect to aV H and zV H . By induction on t it follows thatH ≤ K 1 admits a subset {x 2 , . . . , x t } of cardinality t − 1, which is stable withrespect to aV H and zV H (observe that, since z ∈ Z 1 , this means, in particular,that {x 2 , . . . , x t } is stable with respect to a and z).Now, {x 1 , x 2 , . . . , x t } is an independent subset of K, and d(x i ) = n−1 for everyi = 1, . . . , t. Let 1 ≠ y ∈ 〈x 2 , . . . , x t 〉. Then, as a ∈ Z n , for 1 ≤ k ≤ n − 1,k−1∏[a, k x 1 y] = [a, k x 1 ][a, k y] [a, i x 1 , k−i y] (n−1 i ) ∈ [a,k x 1 ][a, k y]V H . (5.4)i=1Let 1 ≤ s ≤ t, {i 1 , . . . , i s } a subset of {2. . . . , t}, y = x i1 · · · x is , and d = d(x 1 y).By (5.4),1 ≠ z m(x1y) = [a, d x 1 ][a, d y]v,with v ∈ V H , and so [a, d+1 y] ∈ V H . Since the set {x 2 , . . . , x t } is stable withrespect to aV H and zV H , necessarily we have d = n − 1. Moreover, by applyingagain (5.4) with k − n − 1, we have [a, n−1 x 1 y] = [a, n−1 x 1 ][a, n−1 y]w withw ∈ V H ; but then w ∈ V H ∩ 〈z〉, i.e. w = 1. Hencez m(x1y) = [a, n−1 x 1 y] = [a, n−1 x 1 ][a, n−1 y] = z m(x1) z m(y) ,and, since {x 2 , . . . , x t } is stable with respect to a and z,m(x 1 x i1 · · · x is ) = m(x 1 y) ≡ m(x 1 ) + m(y) ≡ m(x 1 ) +s∑m(x ij ) (mod p).This completes the proof of claim (5.3).Now, letting t = (p−1) 2 +1, if we suppose (by contradiction) that |B| ≥ p h(t) ,then by (5.3), B admits a stable subset {x 1 , . . . , x t } with respect to a and z,of cardinality t. Since, for each 1 ≤ i ≤ t, m(x i ) ∈ {1, 2, . . . , p − 1}, thereexists a subset {i 1 , . . . , i p } of {1, . . . , t} such that m = m(x i1 ) = m(x ij ) for allj = 1, . . . , p. But then stability of {x 1 , . . . , x t } implies the contradiction.0 ≠ m(x i1 · · · x is ) ≡j=1p∑m(x ij ) = pm ≡ 0 (mod p).j=1

88 CHAPTER 5. PERIODIC N 1 -GROUPSTherefore, if |B| ≥ p h((p−1)2 +1) , then B, in its action on A, cannot verify (5.2).Thus we may define f n (1, 1) = h((p − 1) 2 + 1).Now, for s ≥ 1, let f n (1, s) = max{f n−1 (n − 1, f n (1, s − 1)) + 1, f n (1, 1)}; weprove by induction on s that this setting satisfies the desired property.For s = 1 this has already been established. Thus, let s ≥ 2, |B| ≥ p fn(1,s) ,1 ≠ a ∈ Z n and z ∈ Z 1 \ 〈a〉. By the inductive assumption on n we may wellsuppose a ∈ Z n \ Z n−1 . Let x ∈ B such that z ∉ 〈a〉 〈x〉 (it exists by cases = 1) and write D = [〈a〉, 〈x〉] = 〈[a, x], . . . , [a, n−1 x]〉. Then, D ≤ Z n−1 and|D| ≤ p n−1 . Let B = B 1 × 〈x〉; then |B 1 | ≥ p fn−1(n−1,fn(1,s−1)) , and so, bythe inductive assumption on n, there exists V ≤ B 1 , with |V | = p fn(1,s−1) andz ∉ D V . Thus z ∉ D V (〈a〉 ∩ Z n−1 ) = D V 〈a〉 ∩ Z n−1 , and, in particular, z ∉D V 〈a〉. Therefore, by the inductive assumption on s, there exists W ≤ V with|W | = p s−1 and sD V ∉ 〈a〉 W D V /D V ; thus z ∉ 〈a〉 W D V . Let then H = 〈W, x〉.Since W ≤ V ≤ B 1 , H = W × 〈x〉. Thus |H| = p s , and〈a〉 H = (〈a〉 〈x〉 ) W = (D〈a〉) W = D W 〈a〉 W ∌ z.This completes the discussion of the case r = 1.To conclude the proof we put, for every r, s ≥ 1,f n (r, s) = max{f n−1 (r, s), f n (r − 1, f n (1, s))},and show by induction on r that this satisfies the property in the statement.For r = 1 this has been proved above. Thus, let r ≥ 2. |B| ≥ p fn(r,s) , U ≤ Z nwith |U| ≤ p r , and let z ∈ Z 1 \ U. By induction on n we may also assumeU ∉ Z n−1 . Then, let a ∈ U \ Z n−1 , and let U = 〈a〉 × U 1 , with U ∩ Z n−1 ≤ U 1 .Now, |U 1 | ≤ p r−1 and so, by the inductive assumption on r and the definitionof f n (r, s), there exists V ≤ B, with |V | = p fn(1,s) and z ∉ U V 1 . Thenz ∉ U 1 [U 1 , V ] ≥ [U 1 , V ](U ∩ Z n−1 ) = [U 1 , V ]U ∩ Z n−1and so z ∉ [U 1 , V ]U = U V 1 〈a〉. Considering the action of V on A/U 1 V , we have,by case r = 1, that there exists H ≤ V , with |H| = p s and zU V 1 ∉ 〈a〉 H U V 1 /U V 1 .Then z ∉ 〈a〉 H U V 1 ≥ 〈a〉 H U H 1 = U H . This completes the proof of the inductivestep on r, and thus the proof of the Lemma.We go on by eliminating the role of the parameter n in Lemma 5.8.Lemma 5.8 There exists a function α 1 : N → N, such that, for every r ∈ N,the following holds:if |B| ≥ p α1(r) , U ≤ A with |U| ≤ p r , and z ∈ Z 1 \ U, then there exists1 ≠ x ∈ B, with z ∉ U 〈x〉 .Proof. We set α 1 (0) = 1 and, inductively, α 1 (n) = f α(n)+1 (n, α 1 (n − 1)), whereα and f k are the functions of Proposition 5.6 and Lemma 5.7. We prove byinduction on n that α 1 has the desired properties.Thus, let n ≥ 1, and |B| ≥ α 1 (n). Let U ≤ A with |U| ≤ p n , and z ∈ Z 1 \ U.Write U = U 1 × U 2 , where U 1 = U ∩ Z α(n)+1 .

5.1. N 1 -GROUPS OF FINITE EXPONENT 89If U 1 = 1, then UZ 1 /Z 1 ∩ Z α(n) (A/Z 1 ) = 1 and so, by Proposition 5.6(since α 1 (n) ≥ α(n)), there exists x ∈ B with |U 〈x〉 Z 1 /Z 1 | = |U| p . But thenU 〈x〉 ∩ Z 1 = 1, and we are done.Assume now U 1 ≠ 1; then |U 2 | ≤ p n−1 . By the definition of α 1 (n) andLemma 5.7, there exists H ≤ B with |H| = p α1(n−1) and z ∉ K = U1 H . Now,K ≤ Z α(n)+1 , and so KU 2 ∩Z α(n)+1 = K(U 2 ∩Z α(n)+1 ) = K; hence z ∉ KU 2 . Byconsidering the action of H on A/K, we know, by the inductive assumption, thatthere exists 1 ≠ x ∈ H such that zK ∉ (KU 2 /K) 〈x〉 = KU 〈x〉2 /K.. Thereforez ∉ KU 〈x〉2 ≥ U 〈x〉 , and we are done.We are ready to prove the main result of this part.Proposition 5.9 (Möhres [76], Satz 3.5) There is a function β : N × N → Nsuch that for every r, s ∈ N the following holds:if |B| ≥ p β(r,s) , U ≤ A with |U| ≤ p r and a ∈ A \ U, then there exists H ≤ Bwith |H| = p s and a ∉ U H .Proof. Trivially, β(r, 0) = 0 for every r ≥ 0. For s ≥ 1 and all r ≥ 0, we setβ(r, s) = α 1 (rp s−1 ) + s − 1,where α 1 is the function of Lemma 5.8, and proceed by induction on s to provethat such function β satisfies the desired property.Thus, let s ≥ 1 and |B| ≥ p β(r,s) . Let U ≤ A with |U| ≤ p r , and a ∈ A \ U.We may clearly assume that B is finite. Hence A = Z m for some m ≥ 1. Letd ≥ 0 be minimal such that a ∈ Z d+1 U. Then a = zu for some u ∈ U andz ∈ Z d+1 \ U. Since a ∉ Z d U, also z ∉ Z d U. Now zZ d /Z d ∈ Z 1 (A/Z d ) and so,by Lemma 5.8, since (as s ≥ 1) β(r, s) ≥ a 1 (n), there exists x ∈ B such thatzZ d ∉ (UZ d /Z d ) 〈x〉 . Thus z ∉ U 〈x〉 , and consequently a ∉ U 〈x〉 .If s = 1 we are done. Otherwise, let Y be a complement of 〈x〉 in B. Then|Y | ≥ p β(r,s)−1 . Now, β(r, s) − 1 = α 1 (rpp s−2 ) + (s − 1) − 1 = β(rp, s − 1). Since|U 〈x〉 | ≤ |U| p ≤ p rp , there exists, by the inductive assumption, W ≤ Y with|W | = p s−1 and a ∉ (U 〈x〉 ) W . Then H = W 〈x〉 = W × 〈x〉 has order p s anda ∉ (U 〈x〉 ) W = U H . This completes the proof.We now move to actual group extensions. Given a prime number p, we denoteby Φ the set of all pairs (G, A) where G is a p-group, A a normal elementaryabelian subgroup of G, and G/A is elementary abelian. In this case, by lettingB = G/A we may apply the results proved so far.We begin with a couple of elementary observations.Lemma 5.10 Let (G, A) ∈ Φ, n ≥ 1, x 1 , . . . , x n ∈ G, and X = 〈x 1 , . . . , x n 〉.Then(i) AX is nilpotent of class at most n(p − 1) + 2;(ii) |X| ≤ p γ(n) , where γ(1) = 2, and γ(n) = 2n + p n( n2)for n ≥ 2.Proof. (i) This follows easily from the fact that, for every x ∈ G, [A, p x] = 1,and elementary commutator calculus.

90 CHAPTER 5. PERIODIC N 1 -GROUPS(ii) Let S = 〈[x i , x j ] | i, j = 1, . . . , n〉. Then, by Lemma 1.4, X ′ = S X ≤ A.Now, X has exponent dividing p 2 , and so |X/X ′ | ≤ p 2n . Also, |S| ≤ p (n 2) and[X : N X (S)] ≤ [X : X ∩ A] ≤ p n . Thuswhich is what we wanted.|X| ≤ p 2n |X ′ | ≤ p 2n · |S| pn ≤ p 2n+pn ( n 2) ,The next fundamental result (Proposition 5.12) is somehow more generalthan we actually need in the present contest, but in this form it will be usefulin later applications. For its proof, we need a special variation of the Chevalley–Warning Theorem (see e.g. [100]).Lemma 5.11 ([77]) For every m, d, n ∈ N\{0}, there exists a value α(m, d, n),such that if s ≥ α(m, d, n), and f 1 , . . . , f m ∈ Z(x 1 , . . . x s ) are homogeneouspolynomials of degree at least 1, with ∑ mi=1 deg f i ≤ d, and p is a prime, thenthere exists (a 1 , . . . a s ) ∈ Z s with at least one entry a j not a multiple of p, andfor all i = 1, . . . , m.Proof. See Möhres [77], Lemma 1.5f i (a 1 , . . . , a s ) ≡ 0 (mod p n )Proposition 5.12 Let G be a nilpotent p-group of class c ≥ 2, and suppose thatγ c (G) has rank 1. Let F be subgroup of G with |F | ≤ p n and γ c (F ) = 1. Let Hbe a normal subgroup of G such that G/H is elementary abelian of order at leastα((n + 1) c , (n + 1) c c, n). Then there exists y ∈ G \ H such that γ c (〈F, y〉) = 1.Proof. Let s = α((n + 1) c , (n + 1) c c, n), and let {Hy 1 . . . , Hy s } be a set of sindependent elements of G/H. Let also {x 1 , . . . , x n } be a set of generators of F(which certainly exists since |F | ≤ p n ).Denote by S be the set of all functions σ : {1, . . . , c} → {0, 1, . . . , n}, suchthat 0 ∈ Im(σ) ≠ {0}. Observe that |S| < (n + 1) c .For σ ∈ S, let q = q σ = |σ −1 (0)| (then 1 ≤ q ≤ c − 1), and write σ −1 (0) ={σ(1), . . . σ(q)} where σ(1) < . . . < σ(q). We define a map φ σ : G q → γ c (G) bysetting, for all g 1 , . . . , g q ∈ G, φ σ (g 1 , . . . , g q ) = [z 1 , . . . , z c ], where z i = x σ(i) ifσ(i) ≠ 0, and z i = g l if i ∈ σ −1 (0) and i = σ(l). Finally, for all g ∈ G, we setω σ (g) = φ σ (g, . . . , g).Since G has class c ≥ 2, γ c (G) is locally cyclic and γ c (F ) = 1, it follows fromCorollary 1.48 that, for every g ∈ G,γ c (〈F, g〉) = {ω σ (g) | σ ∈ S}. (5.5)Let z ∈ γ c (G) be a generator of the unique subgroup of order p n of γ c (G). Now,in any of the commutators φ σ (g 1 , . . . , g q ) (with σ ∈ S and g 1 , . . . , g q ∈ G) thereappears at least one element of F , and therefore (Lemma 1.44)φ σ (g 1 , . . . , g q ) ∈ 〈z〉. (5.6)

5.1. N 1 -GROUPS OF FINITE EXPONENT 91Given σ ∈ S, let us write q = q σ , and denote by J σ the set of all q-tuplesj = (j(1), . . . j(q)) of elements in {1, . . . , s}. By (5.6), for every σ ∈ S and everyj ∈ J σ , there exists a unique element a σ,j ∈ I = {0, 1, . . . , p n − 1} such thatφ σ (y j(1) , . . . , y j(q) ) = z aσ,j . (5.7)Now, let t 1 , . . . t s be independent indeterminates over Z, and for every σ ∈ S letf σ = ∑j∈J σa σ,j t j(1) · · · t j(q) ∈ Z[t 1 , . . . , t s ]. (5.8)Then each such f σ is homogeneous of degree q = q σ ≤ c − 1.By Lemma 1.47, the commutators of weight c in G are homomorphisms ineach component; moreover, since |〈z〉| = p n , commutators that involve elementsfrom F (like the φ σ ), behave linearly modulo p n Z in each component. Thus, if(m 1 , . . . , m s ) ∈ Z/p n Z, we have, for every σ ∈ S,ω σ (y m11 · · · ys ms) = φ σ (y m1= ∏1 · · · ysmsφ σ (y m j(1)j(1)j∈J σ= ∏, . . . , y m11 · · · y mss ) =, . . . , y m j(q)j(q)) =j∈J σφ σ (y j(1) , . . . , y j(q) ) m j(1)···m j(q)== ∏j∈J σz aσ,jm j(1)···m j(q)= z fσ(m1,...ms) . (5.9)Now, since ∑ σ∈S deg f σ ≤ |S|(c − 1) < (n + 1) c c, by Lemma 5.11, there existsa s-tuple (k 1 , . . . , k s ) ∈ Z such that not all the entries k i are multiples of p,and f σ (k 1 , . . . , k s ) ≡ 0 (mod p n ) for every σ ∈ S. Thus, if y = y k11 · · · yks s , theny ∉ H, as at least one of the k i ’s is not zero (mod p), and γ c (〈F, y〉) = 1 by(5.5) and (5.9).Remark. We will use Proposition 5.12 in its full force in the next section. Atthe moment, for groups in the class Φ, one may well suppose |γ c (G)| = p. Inthis case, the polynomials in (5.8) induce F p -multilinear maps, and the standardChevalley–Warning Theorem (see e.g. [100] p. 5, or [83] p. 50) may be appliedinstead of Proposition 5.12, with the smaller bound s = (n + 1) c c to get thedesired conclusion (we leave the details to the reader). ThusLemma 5.13 Let G be a nilpotent p-group of class c ≥ 2, with |γ c (G)| = p,and let F be subgroup of G with |F | ≤ p n and γ c (F ) = 1. Let H be a normalsubgroup of G such that G/H is elementary abelian and |G/H ≥ (n+1) c c. Thenthere exists y ∈ G \ H such that γ c (〈F, y〉) = 1.Repeated applications of this Lemma easily yield the following.Corollary 5.14 Let (G, A) ∈ Φ, with G nilpotent of class c ≥ 2, let n ≥ 0 andsuppose that |G/A| ≥ p nc c+n−1 . Then there exists Y ≤ G such that γ c (Y ) = 1and |AY : A| = p n .We immediately apply this.

92 CHAPTER 5. PERIODIC N 1 -GROUPSLemma 5.15 Let n ≥ 1. There exists a function g n : N → N, such that if(G, A) ∈ Φ and |G/A| ≥ p gn(c) , then⋂{X ≤ G | |AX/A| = p n } ≤ γ c+1 (G).Proof. We may clearly put g n (0) = n, and g n (1) = n + 1. Let c ≥ 2 andsuppose we have already found g n (c − 1) with the desired property. We setg n (c) = g n (c − 1) c c + n, and show that it sartisfies our requirement.Let (G, A) ∈ Φ, with |G/A| ≥ p gn(c) ; let W = ⋂ {X ≤ G | |AX : A| = p n },and K = γ c (G). Since g n (c) > g n (c − 1), we have, by inductive assumption,W ≤ K. If γ c+1 (G) = [K, G] = K, there is nothing more to prove. Thus,let γ c+1 (G) < K. Take γ c+1 (G) ≤ T < K with |K : T | = p. Then T G andγ c (G/T ) = K/T is cyclic of order p. By Corollary 5.14 and the choice of g n (c+1),there exists a subgroup H/T of G/T with γ c (H) ≤ T and |AT/A| = p gn(c−1) .By inductive assumption we have⋂{X ≤ H | |(A ∩ H)X/(A ∩ H)| = p n } ≤ γ c (H) ≤ T.But, for X ≤ H, |(A ∩ H)X/(A ∩ H)| = |X/A ∩ X| = |AX/A|, and so W ≤ T .Now, this holds for every maximal subgroup T/γ c+1 (G) of K/γ c+1 (G). SinceK/γ c+1 (G) is elementary abelian, we conclude that W ≤ γ c+1 (G).We now look to a kind of opposite situation, that is when G admits ’long’non–trivial commutators.Lemma 5.16 Let (G, A) ∈ Φ, n ≥ 1. Let x 1 , . . . , x n , y ∈ G and X = 〈x 1 , .., x n 〉.Suppose that A∩X = 1 and |[A, p−1 x 1 , . . . , p−1 x n , p−1 y]| ≥ p pn+1 . Then for every1 ≠ a ∈ A there exists c ∈ A such that a ∉ 〈X, yc〉.Proof. Let ∆ be the set of all n-tuples t = (t 1 , . . . , t n ) of integers 0 ≤ t i ≤ p − 1,with t i ≠ 0 for at least one i ∈ {1, . . . , n}.For every t = (t 1 , . . . , t n ) ∈ ∆, 0 ≤ j ≤ p − 2, and g ∈ X, we defineω t,j (g) = [g, t1 x 1 , . . . , tn x n , j g] and τ t (g) = [g p , t1 x 1 , . . . , tn x n ]. We show that,for every g ∈ X,A ∩ 〈X, g〉 = 〈g p , τ t (g), ω t,j (g) | t ∈ ∆, 0 ≤ j ≤ p − 2〉 . (5.10)Observe that A ∩ X = 1 implies X elementary abelian. It is thus clear thatA ∩ 〈X, g〉 = 〈g p 〉〈X, g〉 ′ . So it will suffice to show that〈X, g〉 ′ = 〈τ t (g), ω t,j (g) | t ∈ ∆, 0 ≤ j ≤ p − 2〉 . (5.11)By Lemma 5.1, 〈X, g〉 ′ is generated by all the conjugates of the elements [g, x i ](i = 1, . . . , n). Since 〈X, g〉 ′ ≤ A is abelian, we deduce that 〈X, g〉 ′ is generatedby the set of all the elements [g, x s ] xk 11 ···xkn n gj , and so it is generated by the setof all commutators[g, x s , k1 x 1 , . . . , kn x n , j g] (5.12)with s ∈ {1, . . . n}, 0 ≤ k i ≤ p − 1 for i = 1, . . . , n, and 0 ≤ j ≤ p − 1. Now,since x i x j = x j x i and the commutators [g, x i ], [g, x j ] also commute, we see

5.1. N 1 -GROUPS OF FINITE EXPONENT 93that [g, x i , x j ] = [g, x j , x i ] for sll i, j = 1, . . . , n,; moreover, as x p i = 1, we have[g, p x i ] = [g, x i ][g, x i ] xi · · · [g, x i ] xp−1 i = 1, and similirly [g, x i , p−1 g] = [g p , x i ],for all i = 1, . . . , n. These observations allow to freely rearrange the elementsx 1 , . . . , x n in (5.12), to deduce that k s ≤ p − 2, and eventually to rewrite (5.12)as a commutator of type ω t,j (g) (if j ≠ p − 1), or of type τ t (g) (if j = p − 1);thus proving identity (5.11), and consequently establishing (5.10).Let x 1 , . . . , x n , y be as in the statement of the Lemma, and I = {0, . . . , p−2}.Let S be the set of all functions from ∆ × I ∪ ∆ ∪ {0} in {0.1, . . . , p − 1}. Thenlog p |S| = |∆ × I ∪ ∆ ∪ {0}| = p n+1 − p + 1.For every σ ∈ S and every c ∈ A, letb σ (c) = (yc) pσ(0) · ∏∏τ t (yc) σ(t) · ω t,j (yc) σ(t,j) . (5.13)t∈∆Then, by (5.12), for every c ∈ A, we haveLet K be the kernel of the linear map on A,(t,j)∈∆×IA ∩ 〈X, yc〉 = {b σ (c) | σ ∈ S}. (5.14)a ↦→ [a n−1 x 1 , . . . , n−1 x n , n−1 y].Then, by hypothesis, |A/K| ≥ p pn+1 > |S|.Let now a ∈ A with a ∈ 〈X, yc〉, for every c ∈ A. Then, by (5.14) there existc, c ′ ∈ A with Kc ≠ Kc ′ , and σ ∈ S, such thatNow, for every t ∈ ∆, j ∈ I, and every c ∈ A, we haveb σ (c) = a = b σ (c ′ ). (5.15)ω t,j (yc) = ω t,j (y)[c, t1 x 1 , . . . , tn x n , j y]τ t (yc) = τ t (y)[c, t1 x 1 , . . . , tn x n , p−1 y](yc) p = y p [c, p−1 y].Thus, setting b = c −1 c ′ , (5.15) and (5.13) entail1 = [b, p−1 y] ∏ t∈∆[b, t1 x 1 , . . . , tn x n , p−1 y] σ(t) ∏ (t,j)[b, t1 x 1 , . . . , tn x n , j y] σ(t,j)But, as Kc ≠ Kc ′ , b = c −1 c ′ ∉ K, whence, by Lemma 5.2, all the commutatorsthat appear in the above product are linearly independent. It then follows thatσ is the zero–constant, and so a = 1. This proves the Lemma.We may now complete Lemma 5.15.Lemma 5.17 There exists a function α 3 : N \ {0} → N such that, for everyn ≥ 1, if (G, A) ∈ Φ and |G/A| ≥ p α3(n) , then⋂{X ≤ G | |AX/A| = p n } = 1.

94 CHAPTER 5. PERIODIC N 1 -GROUPSProof. For a given n ≥ 1, let s = max{2γ(n − 1), β(γ(n − 1), 1)} + n, whereγ and β are, respectively, the functions defined in 5.10 and 5.9; then take c =s(p − 1) + p n + 1, and finally define α 3 (n) = g n (c), where g n is the functiondetermined in Lemma 5.15.Then, let (G, A) ∈ Φ, with |G/A| ≥ p α3(n) , and 1 ≠ a ∈ A. We prove thatthaere exists X ≤ G, with |AX/A| = p n and a ∉ X.If γ c+1 (G) = 1, the claim follows at once by Lemma 5.15. Thus, assumeγ c+1 (G) ≠ 1. Then A ≰ ζ c−1 (G) = ζ s(p−1)+p n(G), hence, if W = ζ p n(G) ∩A, A/W ≰ ζ s(p−1) (G/W ). Then, by Lemma 5.2, there exist y 1 , . . . , y s ∈ Gsuch that [A, p−1 y 1 , . . . , p−1 y s ] ≰ W . By 5.2, Ay 1 , . . . , Ay s are independentin G/A, and we may well suppose G = A〈y 1 , . . . y s 〉. Now, for l ≤ s, let{Az 1 , . . . , Az l } be a set of independent elements in G/A; we may completeit to a base Az 1 , . . . , Az l , Az l+1 , . . . , Az s of G/A. Then, by Lemma 5.3, we have[A, p−1 z 1 , . . . , p−1 z s ] ≰ W , and so [A, p−1 z 1 , . . . , p−1 z l ] ≰ ζ p n +(s−l)(p−1)(G) ∩ A,which in turn yields (as [A, p−1 z 1 , . . . , p−1 z l ] is nornal in G),|[A, p−1 z 1 , . . . , p−1 z l ]| ≥ p pn +(s−l)(p−1) . (5.16)Let x 0 = 1; we show that, for every 0 ≤ i ≤ n, there exist x 0 , x 1 , . . . , x i ∈ Gsuch that Ax 1 , . . . , Ax i are independent in G/A and a ∉ 〈x 0 , . . . , x i 〉. Supposethat, for some 0 ≤ i ≤ n − 1, we have already found x 0 , . . . , x i with theseproperties, and let U = 〈x 0 , . . . , x i 〉. Then, by 5.10, |A ∩ U| ≤ p γ(i) ≤ p γ(n−1) .Let A ≤ H ≤ G such that G/A = H/A × AU/A. Then H/A has rank s − i and,by choice of s, s − i ≥ s − (n − 1) ≥ β(γ(n − 1), 1). By Theorem 5.9 there existsy ∈ H \ A such that a ∉ (A ∩ U) 〈y〉 = D. Now, DU ∩ A = D(U ∩ A) = D, andso A/D ∩ DU/D = 1, and aD ≠ 1. Also, Ax 1 , . . . , Ax i , Ay are independent inG/A. Let K = [A, p−1 x 1 , . . . , p−1 x i , p−1 y]; then, by (5.16) and the choice of s,|K| ≥ p pn +(s−(i+1))(p−1) ≥ p pn +(s−n)(p−1) ≥ p pn +γ(n−1)p ≥ p pn |D|. (5.17)Now, D is 〈U, y〉-invariant; passing modulo D, (5.17) yields[A/D, p−1 Dx 1 , . . . , p−1 Dx i , p−1 Dy] ≥ p pn ≥ p pi+1 .Then, by Lemma 5.16, there exists Db ∈ A/D such that Da ∉ 〈XD/D, byD〉.By letting x i+1 = by, we have that Ax 1 , . . . , Ax i , Ax i+1 are independent, anda ∉ 〈x 1 , . . . , x i , x i+1 . The inductive proof is now complete, hence we eventuallyfind x 1 , . . . , x n ∈ G independent modulo A, such that a ∉ X = 〈x 1 , . . . , x n 〉.We are now in a position to deduce a major step in the proof.Theorem 5.18 (Möhres [77], Satz 2.2) There exists a function µ : N → N suchthat, for all n ≥ 1, the following holds:if (G, A) ∈ Φ is such that |G/A| ≥ p µ(n) , and U ≤ G has order at most p n , then⋂〈U, x〉 = U.x∈G\AUProof. Let n ≥ 1 be fixed, and let β and α 3 the functions defined, respectively,in 5.9 and 5.15. We put τ n (0) = β(n, α 3 (1)), and inductively, for d ≥ 1, τ n (d) =β(n, α 3 (τ n (d − 1))).

5.1. N 1 -GROUPS OF FINITE EXPONENT 951) We first consider the case in which U ∩ A = 1.Then U is elementary abelian. Let |U| = p n , and let (x 1 , . . . , x n ) be anordered set of independent generators of U. Let m 1 be the smallest positiveinteger such that [A, m1 x 1 ] ≠ 1, and for every 1 < i ≤ n, let m i ; be the smallestpositive integer such that [A, m1 x 1 , . . . , mi−1 x i−1 , mi x i ] ≠ 1. Finally, let d =d U (A) = ∑ ni=1 m i (observe that 0 ≤ d ≤ n(p − 1)).Arguing by induction on d, we show that if |G/A| ≥ p τn(d)+n , thenW =⋂〈U, x〉 = U. (5.18)x∈G\AULet first d = 0. Then U centralizes A and (G, AU) ∈ Φ. Let 1 ≠ a ∈ A. Bydefinition of τ n (0), we have |G/AU| ≥ p β(n,α3(1)) , so, by Proposition 5.9, thereexists a subgroup AU ≤ H ≤ G, |H/AU| = p α3(1) and a ∉ U H , Thus, byLemnma 5.17, the intersection of all subgroups U H 〈x〉 with x ∈ H \ AU is U H .In particular, there exists x ∈ H \ AU such that a ∉ U H 〈x〉, and, a fortiori,a ∉ 〈U, x〉. This shows that A ∩ W = 1. But thenW = W ∩ U H ≤ W ∩ AU = (W ∩ A)U = U,Assume now d ≥ 1. Then there exists a largest index 1 ≤ t ≤ n such thatm t ≠ 0. LetN = [A, m1 x 1 , . . . , mt x t ].Then N G, A/N ∩ NU/N = 1 and d U (A/N) ≤ d − 1. Since τ n (d) ≥ τ n (d − 1),by inductive assumption we have W ≤ NU.Let K be the kernel of the surjective homomorphism φ : A → N given byφ(v) = [v, m1 x 1 , . . . , mt x t ]. Let 1 ≠ a ∈ N and take b ∈ A such that φ(b) = a.Then, K G, and for every 1 ≤ i ≤ n and v ∈ A, φ([v, x i ]) = 1. Hence[A, U] ≤ K. Also A∩KU = K(A∩U) = K, so AU/K is elementary abelian and(G/K, AU/K) ∈ Φ. By definition of τ n (d), |G/AU| ≥ p β(n,α3(τn(d−1))) . Thus,by Propositon 5.9, there exists a subgroup AU ≤ H ≤ G, with |H/AU| =p α3(τn(d−1)) and bK ∉ KU H /K. In turn, by Lemma 5.17 (working in H/KU H ),there exists a subgroup KU H ≤ Y ≤ H, with |AY : AU| = p τn(d−1) , and b ∉ Y .Now, |Y : A ∩ Y | = |AY : A| = |AY : AU||AU : A| = p τn(d−1)+n , whence, byinductive assumption, W ≤ [Y ∩ A, m1 x 1 , . . . , mt x t ]U = φ(Y ∩ A)U. If a = φ(c)for some c ∈ A ∩ Y , then bc −1 ∈ K, and so b ∈ K〈c〉 ≤ Y , a contradiction.Thus, a ∉ φ(Y ∩ A) and, consequently, a ∉ W . This holds for every 1 ≠ a ∈ N.Hence, W ∩ A = W ∩ NU ∩ A = W ∩ N(U ∩ A) = W ∩ N = 1. This, as above,yields the desired conclusion (5.18).Now, as observed before, d U (A) ≤ n(p − 1); so, by letting, for every n ≥ 1,µ(n) = τ n (n(p − 1)) + n, we have the following :if |G/A| ≥ p µ(n) and U ≤ G, with |U| = p n and A ∩ U = 1, then (5.18)holds.2) Now, for the general case, let, for each n ≥ 1,µ(n) = β(n, µ(n)).Let |G/A| ≥ p µ(n) , and U ≤ G with |U| ≤ p n . Let also a ∈ A \ U.

96 CHAPTER 5. PERIODIC N 1 -GROUPSSince |A ∩ U| ≤ p n , Proposition 5.9 guarantees the existence of a subgroupA ≤ H ≤ G, such that |H/A| = p µ(n) and a ∉ (A ∩ U) H = D. Clearly, wemay let H ≥ AU. Then (U/D, A/D) ∈ Φ, aD ≠ 1, and A/D ∩ UD/D = 1 (asA ∩ DU = D(A ∩ U) = D). Therefore, by the case discussed in point 1), thereexists x ∈ H \ AU such that aD ∉ 〈UD/D, xD〉, and so, a fortiori, a ∉ 〈U, x〉.This proves that 1 = A ∩ W , where W = ⋂ x∈G\AU〈U, x〉. But then, as usualW = W ∩ AU = U(W ∩ A) = U. The proof is thus complete.Let us extend this theorem in a rather obvious way, and in a form that wewiil be able to apply more directly.Proposition 5.19 For every n, m, k ≥ 1, there exists ψ(n, m, k) ∈ N, such thatthe following holds:let (G, A) ∈ Φ with |G/A| ≥ ψ(n, m, k); if U is a n-generated subgroup of G andX a subset of A of order k, with X ∩ U = ∅, then there exist y 1 , . . . , y m ∈ G,such that ∅ = A ∩ V = 〈U, y 1 , . . . , y m 〉 and |AV : AU| = p m .Proof. We begin with defining ψ for k = 1 and all n, m. Thus, for n, m ≥ 1, weset ψ(n, m, 1) = µ(γ(n + m − 1)) (where γ(i) is as in Lemma 5.10, and µ is thefunction determined in Theorem 5.18), and show that it satisfies the requiredproperty, arguing by induction on m.Let x 1 , . . . , x n ∈ G, U = 〈x 1 , . . . , x n 〉, and let a ∈ A \ U. For m = 1, wehave ψ(n, 1, 1) = µ(γ(n)) and the claim follows from 5.18. Let m ≥ 2. Then, asψ(n, m, 1) ≥ ψ(n, m − 1, 1), by inductive assumption there exist y 1 , . . . , y m−1such that a ∉ T = 〈U, y 1 , . . . , y m−1 〉 and |AT : AU| = p m−1 . By Lemma5.10, |T | ≤ p γ(n+m−1) , and so Theorem 5.18 again implies the existence ofy m ∈ G with a ∉ V = 〈T, y m 〉 = 〈U, y 1 , . . . , y m−1 , y m 〉 and |AV : AT | = p. Thus|AV : AU| = |AV : AT ||AT : AU| = p m , and we are done.Thus, we have ψ for all cases in which k = 1. Its extension to all k ≥ 1 is byinduction: for n, m, ≥ 1, k ≥ 2, we set ψ(n, m, k) = ψ(n, ψ(n, m, 1), k − 1). Toshow that this satisfies the desired property is now an easy induction.The analisys of the case Φ now comes to an end.Proposition 5.20 Let (G, A) ∈ Φ. If G ∈ N 1 then G is nilpotent.Proof. Let (G, A) in Φ. We prove that if G is not nilpotent then it has a subgroupwhich is not subnormal.Thus, let G be not nilpotent. For every 1 ≤ n ∈ N write σ(n) = n(n + 1)/n.We prove, iductively on n ≥ 1, the existence of sequence of subgroups U n of Gand of elements a n of A, such that, for every n ≥ 1, U n is σ(n)–generated, andfor every 1 ≤ i ≤ j, U i ≤ U j and a i ∈ [A, i(p−1) U i ] \ U j .Since G is not nilpotente, there exists, by 5.1, an element y ∈ G such that[A, p−1 y] ≠ 1. Let 1 ≠ a 1 ∈ [b, p−1 y] (for some b ∈ A), by possibly replacing ywith b k y (for a suitable 0 ≤ k ≤ p − 1) we have a 1 ∉ 〈y〉; so let U 1 = 〈y〉.Now, for n ≥ 2, suppose we have already established the existence of achain of subgroups U 1 ≤ . . . ≤ U n−1 of G, and of elements a 1 , . . . , a n−1 of Awith the prescribed properties. Write U = U n−1 and X = {a 1 , . . . , a n }. LetG/A = AU/A × K/A. Since U is finite and G is not nilpotent, K/A is notnilpotent and, in particular, it isinfinite.. Let s = ψ(γ(σ(n −1)), n, n −1), where

5.2. EXTENSIONS BY GROUPS OF FINITE EXPONENT 97ψ is the function defined in 5.19 and γ that defined in 5.10. Since K is notnilpotent, by Lemma 5.2 there exist elements y 1 , . . . , y s ∈ K \ A such that[A, p−1 y 1 , . . . , p−1 y s ] ≥ p γ(σ(n))+1 (5.19)(see also the proof of 5.16). Let H = 〈A, y 1 , . . . , y s 〉; then |H/A| = p s , asAy 1 , . . . , Ay s are independent by 5.2. In fact, |HU/AU| = p s , and by inductiveassumpiton, |U| ≤ p γ(σ(n−1)) . Then, bt Lemma 5.19, there exists a subgroupV ≤ HU such that U ≤ V , X ∩ V = ∅, and |AV : AU| = p n . Now, asV = V ∩ HU = (V ∩ H)U, we may take elements x 1 , . . . , x n ∈ H such that,setting U n = 〈x 1 , . . . , x n 〉, we have U n ≤ V (hence X ∩U n = ∅) and AU n = AV .This defines U n ; observe, in fact, that, as U = U n−1 is σ(n − 1)–generated, U nis generated by σ(n − 1) + n = σ(n) elements.As Ax 1 , . . . , Ax n are independent in H/A, by Lemma 5.3 and condition 5.19 wehave [A, p−1 x 1 , . . . , p−1 x n ] > p γ(σ(n)) ≥ p |Un| . Therefore, there exists b ∈ A suchthat a n = [b, p−1 x 1 , . . . , p−1 x n ] ∉ U n . This completes the inductive step.We then find in this way the desired infinite sequence U 1 ≤ U 2 . . . of finitelygenerated subgroups of G and elements a n ∈ A such that a i ∈ [A, i(p−1) U i ] \ U j ,for all 1 ≤ i ≤ j. Now, let S = ⋃ n≥1 U n. Then S ≤ G and S ∩ {a 1 , a 2 , . . .} = ∅.It follows that S is not subnoprmal in G; for, if it were, there existed a positiveinteger d such that [A, d S] ≤ S, whence, as a d ∈ [A, d(p−1) U d ] ≤ [A, d S], thecontradiction a d ∈ S. This completes the proof.The main result of this section follows by standard arguments. We need thefollowing variation on P. Hall nilpotency criterion (1.54); the easy proof (using1.54 and the elementary observations at the end of section 1.1) we leave to thereader.Lemma 5.21 Let G be group and N a normal p-subgroup of finite exponent. IfN and G/N ′ N p are nilpotent then G is nilpotentTheorem 5.22 (Möhres [77]) A soluble N 1 -group of finite exponent is nilpotent.Proof. Let G be a soluble N 1 -group of finite exponent. Then G is the directproduct of p-groups for a finite set of primes p. Thus, we may well suppose thatG is a p-group for some prime p. Since G is soluble and has finite exponent, itadmits a finite normal series with p-elementary abelian factors. We let d be theshortest length of such a series, and argue by induction on d.If d = 1, G is abelian. Thus, let d ≥ 2 and write N = G ′ G p . Then G/N is thelargest elementary abelian quotient of G, whence by inductive assumption N isnilpotent. Let K = N ′ N p ; then G/K is an extension of the elementary abelianp-group N/K by the elementary abelian p-group G/N; hence, by Proposition5.20, G/K is nilpotent. By Lemma 5.21, G is nilpotent.5.2 Extensions by groups of finite exponentIn this section we prove another important Theorem of Möhres, saying that aperiodic N 1 -group which is the extension of a nilpotent group by a (soluble)group of finite exponent, is nilpotent.

98 CHAPTER 5. PERIODIC N 1 -GROUPSWe start with a fundamental result, which finds applications also in othercontexts.Theorem 5.23 (Möhres [79]) Let G be a nilpotent p-group, and N a normalsubgroup such that G/N is an infinite elementary abelian group. Then, for everyfinite subgroup U of G and any a ∈ G \ U, there exists a subgroup V of G withU ≤ V , a ∉ V and NV/N infinite.Proof. We procced by induction on the class c of G. If c = 1 the claim followseasily from the basic thory of abelian groups. Thus, suppose c ≥ 2, and assumethe statement true for all p-groups of class less or equal to c − 1.Let U be a finite subgroup of G, and a ∈ G \ U. Let K = γ c (G). If a ∉ KU,then we are done by inductive assumption (observe tha, since c ≥ 2, N ≥ G ′ ≥K). Hence, we may assume a ∈ KU, that is a = bu for some b ∈ K and u ∈ U:clearly, we may now replace a by b if necessary, and so suppose a ∈ K. Let Mbe a subgroup of K maximal subject to K ∩ U ≤ M and a ∉ M. Then, sinceK is central in G (in particular, it is abelian). M G and K/M has rank 1.Then a ̸ inMU (for, otherwise, a ∈ MU ∩ K = M(U ∩ K) = M), and so wemay assume M = 1, i.e. K = γ c (G) is abelian of rank 1.In this setting, we have U ∩ K = 1, and so γ c (〈U, a〉) = 1. Then, by repeatedapplications of Proposition 5.12, we conclude that there exists a subgroup Hof G, containing 〈U, a〉, with HN/N infinite and γ c (H) = 1. Now, H/(H ∩N) = NH/N is an infinite elementary abelian group, and so we may apply theinductive assumption and conclude that there exists V ≤ H such that U ≤ V ,a ∉ V and V (H ∩ N)/(H ∩ N) infinite. Clearly then V N/N is infinite and weare done.Let us state an immediate consequence, specialized to our pourposes.Lemma 5.24 Let G be a nilpotent p-group, N G with G/N is an infiniteelementary abelian group. Let F be a finitely generated subgroup and c ≥ 1 aninteger such that every H ≤ G with F ≤ H and NH/N infinite is subnormalof defect at most c in G. Then, every subgroup of G containing F has defect atmost c (whence γ β(c)+1 (G) is finite by 4.14)..Proof. Let F ≤ U ≤ G. In order to show that it has defect at most c, we mayassume that U is finitely generated. Suppose that there exists a ∈ [G, c U] \ U.Then. by Theorem 5.23, here exists V ≤ G with NV/N infinite, U ≤ V , anda ∉ V . By hypothesis, [G c U] ≤ [G, c V ] ≤ V ;hence the contradiction a ∈ V .Now we consider nilpotent–by–(finite exponent) N 1 -groups. As in the previoussection, the basic case is that of a metabelian p-group. We need a fewpreparatory lemmas (see [79]).Lemma 5.25 Let G ∈ N 1 be the extension of an abelian group A by a solublegroup of finite exponent. If G satisfies an Engel condition, then G is nilpotent.Proof. Let G and A be as in the statement, and suppose that there exists n ≥ 1such that [x, n y] = 1 for every y, x ∈ G. Hence [A, n x] = 1 for every x ∈ G. Lete be the exponent of G/A. Then, since A ≤ C G (A), by applying point (i) of 1.16we get [ A en−1 , x ] for every x ∈ G, that is B = A en−1 ≤ Z(G). Now, G/B is a

5.2. EXTENSIONS BY GROUPS OF FINITE EXPONENT 99soluble N 1 -group of finite exponent, so it is nilpotent by Theorem 5.22. Thus,G is nilpotent.Lemma 5.26 Let G ∈ N 1 be the extension of an abelian group A by an elemataryabelian p-group (p a prime). If G is not nilpotent then there is a nonnilpotentsubgroup K of G, with A ≤ K and such that a subgroup H of K isnilpotent if and only if HA/A is finite.Proof. Since G is a Baer group, every element of G is a bounded left Engelelement. In particular, for each x ∈ G, there is a largest positive integer n(x)such that [A, n x] ≠ 1. SInce G is not nilpotent, and [G, x] ≤ A for every x ∈ G,by Lemma 5.25 we have sup{n(x) | x ∈ G} = ∞. Thus, there is an infinitesequence (x i ) i≥1 of elements of G, such that n(x 1 ) ≥ 1 andn−1∑n(x n ) ≥ n + n(x i ) (5.20)i=1for all n ≥ 2. Let K = A〈x i | i ≥ 1〉.Let x, y ∈ G and m = n(x) + n(y) + 1. Then,∏[A, m xy] ≤ [A, i x, j y] = 1.m≤i+j≤2mThus, n(xy) ≤ m−1 = n(x)+n(y). From this it follows that, for every x, y ∈ G,n(xy) ≥ n(x) − n(y −1 ) = n(x) − n(y).Now, for some n ≥ 1, take x ∈ K \ A〈x 1 , . . . , x n 〉. Then, there exist t > n, and0 ≤ m j ≤ p − 1 (j = 1, 2, . . . , t), with m t ≠ 0, such that xA = x m11 . . . x mtt A.Hence, recalling (5.20),∑t−1∑t−1n(x) ≥ n(x t ) − n(x mjj ) ≥ n(x t ) − n(x j ) ≥ t > n. (5.21)j=1Let H ≤ K and suppose that H is nilpotent. Then, since H is subnormal, AHis nilpotent, say of class c. But then n(x) ≤ c for every x ∈ AH, and thus itfollows from (5.21) that AH ≤ A〈x 1 , . . . , x c 〉. In particular, |AH/A| ≤ p c .Conversely, if H ≤ K is such that AH/A is finite, then AH is nilpotent byLemma 1.75.Lemma 5.27 Let G be a p-group in N 1 , and let A be a normal abelian subgroupof G, such that G/A is elementary abelian. Then G is nilpotent.Proof. By Proposition 1.93, we may suppose thatA ω = ⋂ m≥1A pm = 1.j=1For every m ≥ 1, write K m = A pm . By Theorem 6.5, G/K m is nilpotent forevery m ≥ 1.

100 CHAPTER 5. PERIODIC N 1 -GROUPSSuppose, by contradiction, that G is not nilpotent. Then, by 1.75, G/A isinfinite. By Theorem 1.92 and by Lemma 5.26, we may also assume that thereis a finite subgroup F of G and a n ≥ 1 such that all subgroups H of G withF ≤ H and AH/A infinite, have defect at most n in G.Now, let m ≥ 1; then, G = G/K m is nilpotent, and (G/K m )/(A/K m ) isan infinite elementary abelian p-group. Also, every subgroup U/K m = U of Gcontaining F = K m F/K m and such that UK/K is infinite has defect at most nin G. Thus, by Lemma 5.24, every subgroup of G containing F has defct at mostn in G. This holds for every m ≥ 1. Now, let H be a finitely generated subgroupof G with F ≤ H. Then, by what we have just observed, G, c H] ≤ HK m forevery m ≥ 1. But H is finite, hence, by Lemma 1.27,H = ⋂ m≥1K m H.This shows that H has defect at most c in G. Then, every subgroup of G containgF has bounded defect, and so G is nilpotent by 4.14.Theorem 5.28 (Möhres [79]) A N 1 -group which is the extension of a periodicnilpotent group by a soluble group of finite exponent is nilpotent.Proof. Let G be a periodic N 1 group, with a nilpotent normal subgroup N suchthat G/N is soluble of finite exponent. By Lemma 4.19 we may assume that Gis a p-group for some prime p. As G/N is soluble of finite exponent, it admitsa finite normal series all of whose factors are elementary abelian. Proceedingby induction on the shortest length d of such a series, we reduce to the case inwhich G/N is elementary abelian. Now, by P. Hall criterion 1.54, we may alsoassume that N is abelian. So we are in a position to apply Lemma 5.26, andconclude that G is nilpotent.As a first application of Theorem 5.28, we prove a result of H. Smith [111](see also [17]).Let H be a subgroup of the group G. We write H ≤ b G if there exists aninteger m ≥ 1 such that g m ∈ H for all g ∈ G. This is equivalent to say thatG/H G is a group of finite exponent. Observe that if K ≤ b H ≤ b G then K ≤ b G.Theorem 5.29 A residually nilpotent periodic group in N 1 is nilpotent.Proof. Let G ∈ N 1 be a periodic residually nilpotent group. By Lemma 4.19,we may assume that G is a p-group for some prime p. LetG ω = ⋂ n∈NG pn .By Lemma 1.92, there exist a subgroup H ≤ b G, a finitely generated subgroupF of H, and a positive integer d, such that every F ≤ K ≤ b H has defect atmost d in H. If H is nilpotent, then G ∈ N 1 is the extension of the normalnilpotent subgroup H G by a group of finite exponent. By Theorem 5.28, G isnilpotent. Thus, we may assume that H = G.

5.3. PERIODIC HYPERCENTRAL N 1 -GROUPS 101For n ≥ 1, let G n = G pn . Then K ≤ b G, for all subgroups G n ≤ K ≤ G. Itfollows that all subgroups of G/G n that contain the finite subgroup F G n /G nhave defect at most d in G/G n . By Theorem 4.14,is finite. By Proposition 1.51,γ β(d)+1 (G/G n ) = γ β(d)+1(G)G nG nZ n /G n = ζ 2β(d) (G/G n )has finite index in G/G n . Let Y = ⋂ n∈N Z n; then G/Y is a periodic residuallyfinite N 1 -group. By Proposition 4.22, G/Y is nilpotent, of nilpotency class c, say.It follows that, for all n ≥ 1, G/G n is nilpotent of class at most m ≤ 2β(d) + c.Hence, G/G ω is nilpotent of class m. Now,γ m+1 (G)γ m+3 (G) ≤ G ω ( ) ωγ m+3 (G) = Gγ m+3 (G)is contained in the centre of G/γ m+3 (G) by Lemma 1.18. Then γ m+3 (G) =γ m+2 (G), whence, since G is residually nilpotent, γ m+2 (G) = 1, thus provingthat G is nilpotent.This Theorem does not hold in the non-periodic case, as the groups of H.Smith (section 6.3) show (which indeed are non-nilpotent residually finite N 1 -groups). However we shall later prove (Theorem 6.19) that a residually nilpotentN 1 -group is hypercentral.5.3 Periodic hypercentral N 1 -groupsHeineken–Mohamed groups have trivial centre. We show in this section thatthis is not an accident; in fact (Theorem 5.33) every non-nilpotent periodic N 1 -group must have a centreless non-nilpotent quotient. Needless to say, this alsois due to W. Möhres.Lemma 5.30 Let G ∈ N 1 be p-group, and G ′ be elementary abelian. Then(1) C G (G ′ )/Z(G) is an elementary abelian p-group;(2) a subgroup H of G is nilpotent if and only if HZ(G)/Z(G) has finiteexponent.Proof. (1) Let a ∈ C = C G (G ′ ), and x ∈ G; then [a, x, a] = 1, whence[a p , x] = [a, x] p , Thus C p ≤ Z(G). Let now a, b ∈ C and x ∈ G; then [a, b, x] =[b, xa] −1 [x, a, b] −1 = 1, showing that C ′ ≤ Z(G).(2) Let Z = Z(G) and let H ≤ G. If HZ/Z has finite exponent, then it isnilpotent by Theorem 5.22. Thus H is nilpotent. Conversely, let H be nilpotent.Then G ′ H is nilpotent, whence, by Lemma 1.14, there exists n ≥ 1 such that[G ′ , H pn ] = 1. Thus, for x ∈ G and y ∈ H pn , [x, y p ] = [x, y] p = 1, showing thatH pn+1 ≤ Z.

102 CHAPTER 5. PERIODIC N 1 -GROUPSLemma 5.31 Let G be a hypercentral p-group in N 1 , such that G ′ is elementaryabelian. Let C = C G (G ′ ) and, for every i ≥ 1, let K i = 〈x ∈ G | x pi ∈ C〉.Suppose that G is not nilpotent; then, for every i ≥ 1, K i+1 /K i is an infiniteelementary abelian p-group.Proof. Observe that K 1 ≥ G ′ , hence all factors K i+1 /K i are elementary abelianp-groups. Observe also that, by Lemma 5.30 (1) and Theorem 5.22, K i is nilpotentfor every i ≥ 1. Assume that, for some i ≥ 1, K i+1 /K i is finite. Then theabelian group G/K i has finite rank, and so it is the direct product of a finitegroup T/K i by a divisible group (of finite rank) R/K i . Now, T is a finite extensionof the nilpotent group K i , and so T is nilpotent. Since R/C is abelian andK i /C has finite exponent, a standard fact of abelian groups implies that thereexists a divisible subgroup D/C of R/C such that R = DK i . Write Z = Z(G).Now, D is hypercentral; let W/Z(G) = ζ 2 (D/Z) ∩ C/Z. Then [W, D, D] ≤ z.Since, by 5.30, C/Z is elementary abelian, we haveZ ≥ [W, D] p = [W, D p ] = [W, D].This shows that C/Z ≤ ζ(D/Z). Hence, D is a normal nilpotent subgroup ofG. Therefore, G = T D is nilpotent. This contradiction shows that K i+1 /K i isinfinite.Lemma 5.32 Let G be a hypercentral p-group in N 1 , such that G ′ is elementaryabelian. Then G is nilpotent.Proof. Suppose that G is not nilpotent. Then by 1.92 we may assume thatthere is a finite subgroup F of G, and a n ≥ 1, such that all non-nilpotentsubgroups of G containing F have defect at most n in G. We show that everysubgroup V with F ≤ V has defect at most n. As in Lemma 5.31, let Z = Z(G),C = C G (G ′ ), and, for every i ≥ 1, K i = 〈x ∈ G | x pi ∈ C〉.Let V be a finitely generated subgroup of G containing F , and suppose bycontradiction that V has defect larger than n. Then there exists a ∈ [G, n V ]\V .Clearly a ∈ G ′ and F ≤ V ≤ K m for some m ≥ 1. We construct a series ofsubgroups V ≤ V m ≤ V m+1 ≤ V m+2 ≤ . . . such that, for every j ≥ m, V j ≤ K j ,V j ≰ K j−1 , and a ∉ V j . Now, for every j ≥ m, as observed in the proof of5.31, K j is nilpotent, and K j /K j−1 is an infinite elementary abelian p-group byLemma 5.31. Thus, the existence of the subgroups V j with the desired propertiesis guaranteed by repeated applications of Theorem 5.23. LetH = ⋃V j ;j≥mthen a ∉ H ≥ V . On the other hand, H is not contained in any of the K j ’s,thus the exponent of HC/C is infinite, and so, by Lemma 5.30 (2), H is notnilpotent. Since F ≤ H it follows that H has defect at most n in G, and thisyields the contradiction a ∈ [G, n V ] ≤ [G, n H] ≤ H.This shows that all subgroups V of G, with F ≤ V B are subnormal of defectat most n in G; since G is locally nilpotent and F finite, Theorem 4.14 impliesthat G is nilpotent.

5.3. PERIODIC HYPERCENTRAL N 1 -GROUPS 103Theorem 5.33 (Möhres [81]) A periodic hypercentral N 1 -group is nilpotent.Proof. Let G be a periodic hypercentral N 1 -group. By 4.19, we may assumethat G is a p-group for some prime p.By Lemma 1.11, non-trivial hypercntral groups cannot be perfect, and so,by Lemma 4.20, G is soluble. By Theorem 1.54 and the remark which follows,we may then assume that G is metabelian. Let N = G ′ , andK = N ω = ⋂ n≥1N pn .Now, G/N p is nilpotent by Lemma 5.32. It then follows from Lemma 5.21 thatG/N pn is nilpotent for every n ≥ 1. Thus, G/K is residually nilpotent andtherefore it is nilpotent by Theorem 5.29. Since K ≤ Z(G) by Lemma 1.17, weconclude that G is nilpotent.In the next chapter we will describe examples of H. Smith which show thatthis result too does not extend to arbitrary N 1 -groups.

104 CHAPTER 5. PERIODIC N 1 -GROUPS

Chapter 6The structure of N 1 -groups6.1 Solubility of N 1 -groupsIn this section we prove what is perhaps the most relevat result on N 1 -groups;i.e. that they are soluble; a fact that has been established by W. Möhres andappears in print in [80]. We follow his approach, that my well have applicationsto other problems.Let x 1 , x 2 , . . . be an alphabet. We define the set of all outer commutatorwords inductively as follows:(i)every x i is an outer commutator word;(ii) let m, n ∈ N; if φ(x 1 , . . . , x n ), ψ(x 1 , . . . , x m ) are outer commutatorwords, then [φ(x 1 , . . . , x n ), ψ(x 1 , . . . , x m )] is an outer commutator word.Lemma 6.1 ([80]) Let G be a perfect locally finite p-group, such that for everyproper subgroup T of G, T is soluble and T G ≠ G. Then there exist a finitesubgroup U and a proper normal subgroup N of G, such that Z(G/N) = 1 and⋂〈U, x〉 ≠ U.x∈G\NProof. We assume the Lemma to be false. Then, let T be a proper subgroup ofG, and let Z/T G = Z(G/T G ). Since T G < G, and G is perfect, we have G ≠ Z,and Z(G/Z) = 1 (by Grün’s Lemma 1.11). If U be a finite subgroup of G, anda ∈ G \ U, then by our assumption there exists y ∈ G \ Z with a ∉ 〈U, y〉.Arguing by induction on n ≥ 1, we show that given any finite subgroup Uof G, any a ∈ G \ U, any proper subgroup T of G, and any outer commutatorword φ(x 1 , . . . , x n ), there exist elements y 1 , . . . , y n ∈ G, such thatφ(y 1 , . . . , y n ) ∉ T and a ∉ 〈U, y 1 , . . . , y n 〉. (6.1)For n = 1, (6.1) means that there is an elemnt y ∈ G \ T , such that a ∉ 〈U, y〉,and this is what we had above.Thus, let n ≥ 2, and assume that the claim holds for smaller integers. Let U,T , and a as above, and φ(x 1 , . . . , x n ) an outer commutator word. Since n ≥ 2,105

106 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSwe may suppose that there is a 1 ≤ k ≤ n − 1, and there are outer commutatorwords φ 1 (x 1 , . . . , x k ) and φ 2 (x k+1 , . . . , x n ) such thatφ(x 1 , . . . , x n ) = [φ 1 (x 1 , . . . , x k ), φ 2 (x k+1 , . . . , x n )].Let Z/T G = Z(G/T G ); then, as before, Z ≠ G and Z(G/Z) = 1. By inductiveassumption, there exist elements y 1 , . . . , y k ∈ G withφ 1 (y 1 , . . . , y k ) ∉ Z and a ∉ 〈U, y 1 , . . . , y k 〉,and there exist elements y k+1 , . . . , y n ∈ G such thatφ 2 (y k+1 , . . . , y n ) ∉ C G (φ 1 (y 1 , . . . , y k )Z)andThereforea ∉ 〈U, y 1 , . . . , y k , y k+1 . . . , y n 〉, (6.2)φ(y 1 , . . . , y n ) = [φ 1 (y 1 , . . . , y k ), φ 2 (y k+1 , . . . , y n )] ∉ Z,which, since Z ≥ T , together with (6.2) is what we wanted. Thus, the claimleading to (6.1) is proved.Now, write φ 1 (x 1 ) = x 1 , and, for each j ≥ 1φ j+1 (x 1 , . . . , x 2 j ) = [φ j (x 1 , . . . , x 2 j−1), φ j (x 2 j−1 +1, . . . , x 2 j )]. (6.3)Take 1 ≠ a ∈ G, and set U 0 = 1. Suppose that, for i ≥ 0, we have found finitesubgroups U 0 ≤ U 1 ≤ . . . ≤ U i , with a ∉ U i . Then, by what we had before,there exist elements y i,1 . . . , y i,2 i ∈ G such that a ∉ 〈U i−1 , y 1,i . . . , y i,2 i〉 = U iand φ i+1 (y i,1 , . . . , y i,2 i) ≠ 1.Let U = ⋃ i∈N U i. Then a ∉ U, and so U is a proper subgroup of G. Hence,by hypothesis, U is soluble, of derived length, say, d ≥ 1. But this contradicts1 ≠ φ d+1 (y d,1 , . . . , y d,2 d) ∈ U (d) .Lemma 6.2 Let G be a locally finite p-group, such that for every proper subgroupT of G, T is soluble and T G ≠ G. If G is a Fitting group, then G issoluble.Proof. Let G be as in the assumptions, and suppose by contradiction that G isnot soluble. Then G is perfect and, by Lemma 6.1, there exist a finite subgroupU and a proper normal subgroup N of G, such that Z(G/N) = 1 and thereis an element a ∈ ⋂ x∈G\N〈U, x〉 \ U. Now, G is a Fitting group, and N is aproper normal subgroup; thus there exists an element g ∈ G, with 〈g〉 G N/N anon-trivial elementary abelian p-group. Since, moreover, Z(G/N) = 1, we havethat 〈g〉 G N/N ≃ 〈g〉 G /〈g〉 G ∩ N is infinite. Now, 〈g〉 G is nilpotent, and so wemay apply Theorem 5.23 to conclude that there exists z ∈ 〈g〉 G \ N, such thata ∉ 〈U, z〉. As z ∉ N, this is a contradiction.Lemma 6.3 Let G be p-group in N 1 , and assume that all proper subgroups ofG are soluble. Then G is soluble.

6.2. FITTING GROUPS 107Proof. By Lemma 6.2 it is enough to show that G is a Fitting group. Thus,let x ∈ G; then K = 〈x〉 G is soluble because it is a proper subgroup of G.We prove that K is nilpotent arguing by induction on the derived length d ofK. If d = 1, K is abelian. Thus, let d ≥ 2, and A = K (d−1) . Then A G,and K/A = 〈xA〉 G/A ; so, by inductive assumption, K/A is nilpotent. Since it isgenerated by conjugates of x (hence by elements of bounded order), K/A hasfinite. Thus, K is a periodic N 1 -group which is an extension of an abelian groupby a soluble group of finite exponent, and so, by Theorem 5.28, K = 〈x〉 G isnilpotent. Therefore, G is is a Fitting group, and we are done.We are now in a position to prove the main Theorem.Theorem 6.4 (Möhres [80]) Every N 1 -group is soluble.Proof. Let G be N 1 -group. By 2.19 and 4.19, we may assume that G is ap-group, for some prime p. Suppose that G is not soluble; then, by 1.92, thereexists a non-soluble subgroup H of G, a finitely generated subgroup F of H, anda positive integer d, such that every non-soluble sugroup K of H with F ≤ Khas defect at most d in H. Let H = H 0 ≥ H 1 ≥ . . . ≥ H d = F be the normalclosur series of F in H, and let B = H i be the smallest non -solble term of it.Then F B is soluble, and so K = B/F B is not soluble. Furthermore, all nonsolublesubgroups of K have defect at most d. It then follows from Roseblade’sTheorem that all non-soluble subgroups of K contain the limit D of the derivedseries of K. Then, all proper subgroups of D are soluble and so, by Lemma 6.3,D is soluble. But then, Lemma 4.20, D is soluble, which is a contradiction.Having proved that every N 1 -group is soluble makes of course redundantthis assumption in Theorems like 5.22, 5.28 or in proposition 3.1. Specifically,for further reference, we restate as a Proposition, an arguemnt used in the proofof Lemma 6.3.Proposition 6.5 Let G ∈ N 1 , and suppose that G is generated by elements offinite bounded order. Then G is nilpotent of finite exponent.6.2 Fitting GroupsProposition 6.5 implies that in a N 1 group every element of finite order belongsto the Fitting radical. In this section, we generalize this by showing that everyN 1 -group is a Fitting group. This answers a question of D. Robinson, and completesthe information about the inclusion relations among some relevant classesof locally nilpotent groups, as mentioned in the second volume of [96].In fact, we shall prove something more, i.e. that in a N 1 -group every nilpotentsubgroup is contained in a normal nilpotent subgroup.We start with an observation that is certainly well known.Lemma 6.6 Let G be a nilpotent group such that its torsion subgroup T hasfinite exponent. Then there exists a 1 ≤ k ∈ N such that G k ∩ T = 1.

108 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSProof. Let q be the exponent of T .We first assume that G/T is abelian, and proceed by induction on the minimalinteger m such that T ≤ ζ m (G). If m = 1, then G ′ ≤ T ≤ ζ(G). Now, forall x, y ∈ G, Lemma 1.2 yields(xy) 2q = x 2q y 2q [y, x] q(2q−1) = x 2q y 2q .Thus G 2q = {x 2q | x ∈ G}. Also, if a = x 2q ∈ G 2q ∩ T , then 1 = a q = x 2q2 . Sox ∈ T and, consequently, a = x 2q = 1. Hence G 2q ∩ T = 1.Let now m ≥ 2, and set X = ζ(G) ∩ T . Then T/X is the torsion subgroupof G/X, and is contained in ζ m−1 (G/X). By inductive hypothesis, there is as ≥ 1 such that G s ∩ T ≤ X. Now, G s ∩ T is the torsion subgroup of G s and iscontained in its centre. By the case m = 1, we have that G 2sq ∩ X = 1 and soG 2sq ∩ T = G 2sq ∩ T ∩ G s = G 2sq ∩ X = 1.We now prove the general case by proceeding by induction on the nilpotencyclass c of G/T . Let Z/T be the centre of G/T . As Z/T is abelian, there exists, bythe case c = 1 discussed above, an s ≥ 1 such that Y = Z s has trivial intersectionwith T . Now, G/Z is torsion-free (see Proposition 2.3). Thus, Z/Y is the torsionsubgroup of G/Y , and has finite exponent. Since the nilpotency class of G/Z isc − 1, by inductive assumption there exists k ≥ 1 such that (G/Y ) k = G k Y/Yhas trivial intersection with Z/Y . In other words, G k ∩ Z ≤ G k Y ∩ Z = Y ,which in turn gives G k ∩ T ≤ Y ∩ T = 1.Now, we prove a technical but useful Lemma. Recall (see 5.29) that, forH ≤ G, H ≤ b G means that there exists an integer m ≥ 1 such that g m ∈ Hfor all g ∈ G.Lemma 6.7 Let G ∈ N 1 be such that the torsion subgroup A of G is nilpotent,and G/A n is nilpotent for every n ≥ 1. Assume that there exists a finitelygenerated subgroup F of G, and an integer d ≥ 1, such that every subgroup H,with F ≤ H ≤ b G, has defect at most d in G. Then there exists c ≥ 1, whichdepends only on d and the nilpotency class of G/A, such that every subgroup ofG containing F has defect at most c.Proof. By Proposition 1.93 we may assume A ω = ∩ n≥1 A n = 1. As A is thetorsion subgroup of the locally nilpotent group G, G/A is torsion–free and so itis nilpotent by Theorem 2.23. Let r be the nilpotency class of G/A, let β(d) asdefined by Theorem 4.14, and set m = max{r, β(d)} + 1.Let n ≥ 1. Then G/A n is nilpotent by assumption, whence, by Lemma 6.6,there exists a normal subgroup M n of G such that M n ∩ A = A n , and G/M nhas finite exponent; in particular, H ≤ b G for any H/M n ≤ G/M n . Thus, byassumption, all subgroups of G/M n containing the finite subgroup F M n /M nhave defect at most d. By Theorem 4.14, γ m (G/M n ) = γ m (G)M n /M n is finite.Also, by choice of m, γ m (G) ≤ A, so that γ m (G) ∩ M n = γ m (G) ∩ A ∩ M n =γ m (G) ∩ A n . It follows thatγ m (G)A n∼γ m (G)=A n A n ∩ γ m (G) = γ m (G)M n ∩ γ m (G) ∼ = γ m(G)M nM nis finite. By Proposition 1.51, ζ 2m (G/A n ) has finite index in G/A n .

6.2. FITTING GROUPS 109Let H be a subgroup of G be such that A n F ≤ H for some n ≥ 1. Then,setting Z/A n = ζ 2m (G/A n ), we have, by what just proved, that HZ has finiteindex in G, and so, by assumption, that its defect is at most d in G. Now, clearly,H/A n has defect at most 2m in ZH/A n . Hence H has defect at most 2m inZH, and so H has defect at most c = d + 2m in G. (this holds for all n ≥ 1).Now, to show that every subgroup H ≥ F has defect at most c in G, we maywell assume that H is finitely generated.By what proved before, for every n ≥ 1, A n H has defect at most c = 2m + din G. Also, by the definition of c, we have that [G, c H] ≤ A. Thus,[G, c H] ≤ ⋂ n≥1(A n H ∩ A) = ⋂ n≥1A n (H ∩ A).But H is finitely generated nilpotent group,, and so A ∩ H ≤ tor(H) is finite.Since we are assuming ∩ n≥1 A n = 1, we conclude by Lemma 1.27 that[G c H] ≤ H ∩ A ≤ H.This proves that H has defect at most c in G.We now generalize Theorem 5.28.Theorem 6.8 (H. Smith [107]) Let G be a N 1 -group. If G is the extension ofa nilpotent group by a group of finite exponent, then G is nilpotent.For the proof, we need the following observation.Lemma 6.9 Let A be an abelian p-group, and X an elementary abelian p-groupof automorphisms of A. Then, for every n ≥ 1,[A, n X] pn≤ [A, 2n X].Proof. By induction on n. When n = 1, set A = A/[A, X, X]. Then, for everya ∈ A and every x ∈ X, [a, x, x] = 1, whence, by 1.2, [a, x] p = [a, x p ] = 1,showing that [A, X] p ≤ [A, X, X]. Let now n ≥ 2, then[A, n X] pn= [[A, n−1 X] pn−1 , X] pand so, by the inductive assumpiton and case n = 1,[A, n X] pn≤ [[A, 2(n−1) X], X] p ≤ [A, 2n−2 X, 2 X] = [A, 2n X],thus proving the Lemma.Proof of Theorem 6.8. Suppose that G is a counterexample to the theorem.Then, by an obvious inductive argument (using the fact that a N 1 -group offinite exponent is nilpotent) we may assume that G admits a normal nilpotentsubgroup N such that G/N is an elementary abelian p-group for some primep. Also, by P. Hall’s nilpotency criterion, we may reduce to the case in whichN is abelian. Let A be the torsion subgroup of N. Since G is locally nilpotentand G/C G (A) is a p-group, it follows that the p ′ -component of A is central in

110 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSG; thus we may assume that A is a p-group. If T is the torsion subgroup of G,then T ∩ N = A and T is nilpotent by Theorem 5.28; if R is a subgroup of Gsuch that G/N = T N/N × R/N, then R is not nilpotent and T ∩ R = A. Wemay therefore replace G by R, and assume that A is the torsion subgroup of G;in particular C G (A) ≥ N.By Lemma 5.26 we may furthermore suppose that a subgroup H of G isnilpotent if and only if HN/N is finite. Thus, by Brookes’ trick 1.92, we mayfinally assume that there are a finitely generated subgroup F of G and a positiveinteger d such that every subgroup H of G which contains F and such thatHN/N is infinite has defect at most d in G. Since F N/N ≃ F/(F ∩ N) isfinite, F N is nilpotent and normal in G; by invoking again P. Hall’s nilpotencycriterion, we may reduce to the case (F N) ′ = 1; in particular, F G ≤ F N isabelian (and it is easy to see that all other assumptions on A may be mantained).For n ≥ 1, let A n = A pn . by Proposition 1.93, we may suppose⋂A n = 1. (6.4)n≥1Now, since A p [A, p x] = A p [A, x p ] = A p , for every x ∈ G (Lemma 1.14), wededuce that, for every n ≥ 1, G/A n is a bounded Engel group, and so it isnilpotent by Lemma 5.25. By Lemma 6.6 there exists a normal subgroup M nof G, with A ∩ M n = A n , M n ≤ N, and G/M n a p-group of finite exponent.By Lemma 5.24, applyed to the group G/M n , its normal subgroup N/M n andF M n /M n , we deduce that every subgroup of G containing F M n has defect atmost d. This holds for any n ≥ 1, and so we may apply Lemma 6.7 and concludethat there exists c ≥ 1 such that every subgroup of G containing F has defectat most c.Let x 1 , . . . , x c ∈ G, and X = 〈x 1 , . . . , x c 〉. Then[A, x 1 , . . . , x c ] ≤ [A, c X] ≤ 〈F, X〉. (6.5)Also, C X (F ) ≥ X ∩ N X, whence |X/C X (F )| ≤ p c . Since F G is abelian,the rank of F X is bounded by rk(F )p c . Moreover, all subgroups of 〈F, X〉/F Xhave defect at most c, and so 〈F, X〉/F X is nilpotent of class at most ρ(c) byRoseblade’s Theorem. Since 〈F, X〉/F X is generated by c elements, it followsfrom Proposition 1.46 that its rank is bounded by a function of c. Therefore,for all choices of x 1 , . . . , x c ∈ G, the rank of 〈F, X〉 is bounded uniformely by avalue l (depending on c and rk(F )). In particular, from (6.5) we getrk[A, x 1 , . . . , x c ] ≤ l. (6.6)Let now D = [A, 2c G] p = Φ([A, 2c G]), and write A = A/D. By Lemma 6.9, forany x 1 , . . . , x c ∈ G we have[A, x 1 , . . . , x c ] pc≤ [A, 2c G] = [A, 2c G]/D;hence [A, x 1 , . . . , x c ] has exponent dividing p c+1 . From (6.6), we therefore deducethat|[A, x 1 , . . . , x c ]| ≤ p (c+1)l

6.2. FITTING GROUPS 111for every x 1 , . . . x c ∈ G. Since G/C G (A) is (elemetary) abelian, we may applyLemma 4.15 obtaining that [A, 2c G] = [A, 2c G]/D is finite. Since D = [A, 2c G] pand [A, 2c G] is reduced (by (6.4)), we conclude that [A, 2c G] is a normal finitesubgroup of G. Thus, [A, 2c G] ≤ ζ t (G) for some t ≥ 1. As G/A is nilpotent, wefinally obtain that G is nilpotent.Corollary 6.10 Let G ∈ N 1 . If G admits a nilpotent subgroup H with H ≤ b G,then G is nilpotent.Proof. Let G, H be as in the hypotheses. Then G/H G has bounded exponent(argue by induction on the defect of H in G). Thus, by Theorem 6.8, G isnilpotent.We are now ready to prove the main result of this section.Theorem 6.11 Let G be a group with all subgroups subnormal, and let S be anilpotent subgroup of G. Then S G is nilpotent.Proof. Let G ∈ N 1 , S a nilpotent subgroup of G, and W = S G . We fix thenotation A for the torsion subgroup of W . Then W/A is a torsion-free N 1 -group,and so it is nilpotent by Theorem 2.23. We want to prove that W is nilpotent.Arguing by induction on the defect of S in G, we may assume that S is normalin W . We begin by proving(1) The torsion subgroup A of W is nilpotent.By Lemma 4.19, it is enough to show that every primary component of A isnilpotent. By factoring modulo the product of all p ′ -components, we may assumethat A is a p-group for some prime p. By induction on the derived lenght of A,we may also assume that A has a characteristic abelian subgroup X such thatA/X is nilpotent. Now, let c be the nilpotency class of S, and let x ∈ S. Then[X, c+1 x] = [[X, x], c x] ≤ [S, c x] = 1 .Let q be a power of p greater than c+1. Then, by Lemma 1.14, every G-invariantelementary abelian section of X is centralized by x q . This holds for every x ∈S. It follows that K = 〈(x g ) q | x ∈ S, g ∈ G〉 centralizes every G-invariantelementary abelian section of X. Now, X is normal in G and has an ascendingcharacteristic series with elementary abelian factor groups, and so it follows thatX ∩ K is hypercentral in K. Since A ∩ K/X ∩ K ∼ = X(A ∩ K)/X ≤ A/X isnilpotent, A∩K is a hypercentral periodic N 1 -group. By Theorem 5.33, A∩K isnilpotent. Finally, as W is generated by the conjugates of S, W/K is nilpotentof finite exponent by Proposition 6.5. In particular, A/A ∩ K is nilpotent offinite exponent. Hence, by Theorem 5.28, A is nilpotent.(2) A is abelian.If W/A ′ is nilpotent, then, since A is nilpotent, W is nilpotent by a Theorem1.54. Hence we may assume A to be abelian.Let N = [G, S]. Then W = NS and, by Fitting’s Theorem, W is nilpotentif and only if N is such. We then prove that N is nilpotent. Suppose that Nis not nilpotent. By Corollary 6.10 and Theorem 1.92, there exists a subgroupH ≤ b N, a finitely generated subgroup F of H, and a positive integer d suchthat all subgroups L of H with F ≤ L ≤ b H have defect at most d in H.

112 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSNow, [S, G] is generated by all the commutators [x, g], with x ∈ S, g ∈ G, andso there exist a finitely generated subgroup S 1 of S, and a finitely generatedsubgroup G 1 of G, such that, writing V = 〈S 1 , G 1 〉F ≤ [S 1 , G 1 ] ≤ V ′ .(3) H satisfies the hypotheses of Lemma 6.7.Let n ≥ 1. Let B = A ∩ N be the torsion subgroup of N, and let n ≥ 1.Then B/B n has finite exponent and is invariant for W . Let x ∈ S; then, as inpoint (1), [B, c+1 x] = 1 so, by Lemma 1.16, there exists a q ≥ 1 such that x qcentralizes B/B n . Arguing as in point (1), we have that, if K = C W (B/B n ),then W/K has finite exponent, and so N/N ∩K has finite exponent. Since N/Bis nilpotent (because it is torsion-free), we have that (K ∩ N)/B n is nilpotent.Thus, by Theorem 6.8, N/B n is nilpotent. This holds for every n ≥ 1, and soN satisfies the hypotheses of Lemma 6.7. Observe now that, since H ≤ b N,H satisfies these same hypotheses and so, by Lemma 6.7, every subgroup of Hcontaining F has defect at most c in H, for some c ≥ 1.Let U = BV , Y = NU = NV , and U = U 0 U 1 . . . U n = Y be thenormal closure series of U in Y . For each j, let R j = U j ∩ N. Notice that R jis normal in U j and contains BF . Given a j, let Q j = F Rj∩H . By Roseblade’sTheorem, (R j ∩H)/Q j is nilpotent. Since F Rj ∩H ≥ Q j , (R j ∩H)/(F Rj ∩H) isnilpotent. As R j ∩ H has finite index in R j , it follows that R j /F Rj is nilpotent.Now, by induction on i, we prove that U i is nilpotent. This is trivial for i = 0,as U = BV is a Baer group, and an extension of an abelian group by a finitelygenerated group. Thus, assume that U i is nilpotent. Then R i is nilpotent andis normalized by U i+1 . Now, F ≤ V ′ ≤ U i ′ U i+1, whence F Ui+1 ≤ U i ′ ∩ R i+1.Thus, a fortiori, F Ri+1 ≤ U i ′ ∩ R i+1, and, by what we have proved above, wehave thatR i+1U i ′ ∩ R ∼= R i+1U i′i+1 U i′is nilpotent. Since U i+1 /R i+1 = U i+1 /(U i+1 ∩ N) is isomorphic to a subgroupof the finitely generated group Y/N, U i+1 /R i+1 is finitely generated, and soU i+1 /(R i+1 ∩ U i ′) is nilpotent. In particular, U i+1/U i ′ is nilpotent, and so, byP. Hall’s criterion, U i+1 is nilpotent. This completes the induction. Thus weconclude that Y = U n is nilpotent, which forces N to be nilpotent.As a particular case of the previous Theorem, we have the followingCorollary 6.12 A N 1 -group is a Fitting group.Another immediate corollary of 6.11 answers a question of H. Smith [102].Corollary 6.13 Let G be group with all subgroups subnormal. If G = 〈H, K〉where H, K are nilpotent subgroups, then G is nilpotent.6.3 Hypercentral and Smith’s groupsWe have already seen that periodic hypercentral N 1 -groups are nilpotent (Theorem5.33); thus, this section on hypercentral groups will focus on non-periodic

6.3. HYPERCENTRAL AND SMITH’S GROUPS 113(indeed, mixed) groups, beginning with H. Smith’s construction of non-nilpotenthyperabelian N 1 -groups, which we have already mentioned on several occasions.The relevance of the hypercentral case in the study of N 1 -groups (in particular,for non–periodic groups) may be for instance gathered from Theorem 6.26.Smith’s method constructs mixed N 1 -groups which have some common features,but may be adapted to produce hypercentral N 1 -groups with additionalproperties (see [101] and [112]). I will restrict to a full presentation of one singlecase (the first produced by Smith).Theorem 6.14 (H. Smith [101]) There exists a non-nilpotent group G with thefollowing properties:(1) all subgroups of G are subnormal;(2) G is hypercentral of length ω + 1;(3) if G is locally metacyclic and residually finite;(4) every subgroup H of G has finite index in the second term H G,2 of itsnormal closure series.Proof. Let p 1 , p 2 , p 3 , . . . be an infinite sequence of distict prime numbers. Forevery n ≥ 1, letH n = 〈x n , y n | x pn n n= 1 = y pn−1 nn , x ynn = x pn+1n 〉.Thus, each H n is the semidirect product of normal a cyclic group X n = 〈x n 〉 bya cyclic group 〈y n 〉, where y n acts by conjugation on X n as an automorphismof order p n−1n . Let F be the cartesian product of the groups H n :F = Car n≥1 H n .Then F is metabelian and residually finite. Also, clearly, X n ≤ ζ n (F ) for eachn ≥ 1, and so F is hypercentral of length ω + 1.For every pair n, m ≥ 1 with n ≠ m let u n,m ∈ N be such thatu n,m p m−1m ≡ 1 (mod p n n). (6.7)Let ¯z be the element of F defined by ¯z(i) = x −1in ≥ 1, let ¯x n , ȳ n ∈ F such thatfor every i ≥ 1; and, for each¯x n (i) ={xn if i = n1 if i ≠ n{yn if i = nȳ n (i) =x −ui,ni if i ≠ nNotice the following commutator relations; for every n, m ≥ 1:[¯x n , ȳ m ] = ¯x pnn if m = n[¯x n , ȳ m ] = 1 if m ≠ n[ȳ n , ȳ m ] = ¯x npnun,mxm −pmum,nif m ≠ n[ȳ n , z] = ¯x pnn(6.8)(6.9)

114 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSAlso, from (6.7), for every n ≥ 1 we haveWe then consider the subgroup G of F :ȳ pn−1 nn = ¯x n z. (6.10)G = 〈¯x n , ȳ n , z | n ≥ 1〉.Let X = Dir n≥1 X n = 〈¯x n | n ≥ 1〉. Then X is normal in G, it is periodic, locallycyclic, and contained in ζ ω (G). By the relations (6.9) we also have that X ≥ G ′ ,and that G/X is an abelian group of rank 1 (a subgroup of the additive group ofthe rationals). Thus, X is the torsion subgroup of G, and G is locally metacyclic.Furthermore G is residually finite because such is F . Hence G satisfies property(3) in the statement.Then observe that the fourth relation in (6.9) implies that, for every n ≥ 1,[z, n−1 ȳ n ] ≠ 1. Hence z ∉ ζ ω (G), and so G is hypercentral of length ω + 1, i.e.property (2) in the statement is satisfied by G.We now prove that every subgroup of G is subnormal and satisfies (4). LetA = X〈x〉; then A is an abelian normal subgroup of G, and G/A is a directproduct of cyclic p n -groups.Let S ≤ G. If S ∩ A = S ∩ X, thenSS ∩ X = SS ∩ A = SA Ais periodic, hence S is periodic and so S ≤ X, which implies that S is subnormalof defect at most 2 in GSuppose S ∩A > S ∩X. Then there exist x ∈ X and r > 0 such that xz r ∈ S.Since x has finite order, we get that there exists s > 0 such that z s ∈ S. LetX ∗ = 〈¯x n | (p n , s) = 1〉.We prove that X ∗ normalizes S. Let g ∈ S; then there exist an element a ∈ Aand integers t ∈ N, β 1 , . . . β t ≥ 1, such thatg = aȳ β1i 1· · · ȳ βti t.Let n ≥ 1 with (p n , s) = 1. Then [¯x n , g] = [¯x n , ȳ β1i 1 · · · ȳ βti t.]. Hence [¯x n , g] = 1 ∈ Sif n ∉ {i 1 , . . . , i t }; otherwise, n = j j for some j ∈ {1, . . . , t} and, letting β = β j ,[¯x n , g] = [¯x n , ȳ β n]. (6.11)Now, since G/A is abelian, by (6.10) there is a p ′ n-number k such that g k = a ′ ȳn.βThen S ∋ [z s , g k ] = [z, g k ] s = [z, ȳn] β s . Now, by (6.9), [z, ȳn] β = [¯x −1n , ȳn]βbelongs to 〈¯x n 〉, and so has order coprime to s. It then follows that[¯x n , g] = [¯x n , ȳ β n] = [z, ȳ β n] −1 ∈ 〈[z s , g k ]〉 ≤ S.Thus, we have proved that X ∗ normalizes S. Let the X ∗ = 〈¯x n | p n s〉. ThenX ∗ is a finite normal subgroup of G, and X = X ∗ X ∗ . HenceS G,2 = S[G, 2 S] ≤ S[X, S] = S[X ∗ X ∗ , S] = S[X ∗ , S] ≤ SX ∗ ,

6.3. HYPERCENTRAL AND SMITH’S GROUPS 115so |S G,2 : S| is finite, and property (4) is satisfied. Finally, X ∗ is contained insome term ζ m (G) of the upper central series of G; therefore[G, m+1 S] ≤ [X, m S] ≤ S[X ∗ , m S] ≤ S.This shows that S is subnormal and completes the proof.H. Smith’s method, in all of its occurencies in papers, gives groups of hypercentrallength ω + 1, and it is not immediate how it could be implemented inorder to obtain hypercentral N 1 -groups of different type. In particular, we askQuestion 5 For every integer n ≥ 1 construct a hypercentral N 1 -group oflength ω + n (or prove that there are not any).The above question is also motivated by the fact that there do not existhypercentral N 1 -groups of length exactly ω; this was proved by H. Smith [113](see also [16]).Theorem 6.15 Let G be a hypercentral group of hypercentral length at most ω.If all subgroups of G are subnormal, then G is nilpotent.Proof. Let G ∈ N 1 be hypercentral of hypercentral length at most ω. Thismeans that G = ⋃ n∈N ζ n(G).By Theorem 1.92 and Corollary 6.10, there exist a subgroup H ≤ b G, afinitely generated subgroup F of H, and a positive integer d, such that allsubgroups K of H, with F ≤ K ≤ b H have defect at most d in H. SinceH ≤ b G and F ≤ ζ n (G) for some n ∈ N, we may assume F = 1 and H = G.Let A be the torsion subgroup of G. By Theorem 5.33, A is nilpotent, hence,by 1.54, we may also assume that A is abelian. If G/A n is nilpotent for all n ≥ 1,then G is nilpotent by Lemma 6.7 and Roseblade’s Theorem 4.9. So, we are leftwith the case in which A is an abelian group of finite exponent. Let C = C G (A).Then, by Lemma 1.16, G/C is periodic. Now A ≤ C and C/A is torsion-freeand thus nilpotent. Hence, C is nilpotent and, by Lemma 6.6, there exists aninteger k ≥ 1 such that C k ∩ A = 1. Now, C k G, and G/C k is periodic andhypercentral. Thus G/C k is nilpotent. Since G/A is also nilpotent, we concludethat G = G/(A ∩ C k ) is nilpotent.Recently, Martinelli ([70]) gave a more complete statement, which furthermotivates Question 5.Theorem 6.16 Let G be a hypercentral non–nilpotent group in N 1 . Then Ghas hypercentral length ω + n for some 1 ≤ n ∈ N.In the same work, Martinelli provides an extension of Theorem 5.29, byshowing that a residuially nilpotent N 1 -group is hypercentral. To approach theproof of this, let us first introduce the class X 0 of all locally nilpotent groups Gsuch that A = tor(G) is nilpotent and G/A n is also nilpotent for every n ≥ 1.Lemma 6.17 Let G ∈ N 1 and H ≤ b G. If H ∈ X 0 then G ∈ X 0 .

116 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSProof. Let G ∈ N 1 , H ≤ b G with H ∈ X 0 , and write N = H G . Then G/Nhas finite exponent, and in particular, if B = tor(N), B ≥ (tor(H)) m for somem ≥ 1. From this it easily follows that N ∈ X 0 .Let A = tor(G); then B = A ∩ N and A/B ≃ AN/N has finite exponent.Since B is nilpotent, we have that A is nilpotent by Theorem 5.28.Now, let n ≥ 1. Then A n ≥ B n , and so, by assumption, A n N/A n is nilpotent.Thus, G/A n is the extension of the nilpotent normal subgroup of A n N/A n by agroup of finite exponent. From Theorem 6.8 we conclude that G/A n is nilpotent,thus proving that G belongs to X 0 .We also need to strenghten Proposition 1.93.Lemma 6.18 Let G ∈ N 1 , and let A be a normal nilpotent and periodic subgroupof G. Then there exists an integer m ≥ 0 such that(A/N) ω ≤ ζ m (G/N)for any normal subgroup N of G contained in A.Proof. We may assume that G/A is countable (this is slightly less immediatethan in the proof of 1.93: if the property fails for some G, then for every positiveinteger n there exist a normal subgroup N n ≤ A and a finitely generatedsubgroup X n of G such that [(A/N n ) ω , n X n ] ≠ 1; then consider the subgroupof G generated by A and X n for every n ≥ 0). Thus, let {Ax 1 , Ax 2 , Ax 3 , . . .} bean enumeration of the elements of G/A.Now, as in the proof of 1.93, using the chain of finitely generated nilpotentgroups 〈x 1 〉 ≤ 〈x 1 , x 2 〉 ≤ . . ., one shows that there exists a subgroup U of G,with A ∩ U = 1, and the property that for each x ∈ G there exists 1 ≤ k ∈ Nsuch that x k ∈ U (this last property follows from the fact it holds modulo A byconstruction of U, A is normal and periodic, and U is subnormal in AU). Letm be the defect of U in G; and let N be a normal subgroup of G contained inA. Then A ∩ NU = N(A ∩ U) = N, and so, writing A = A/N, U = UN/N,[A, d U] ≤ A ∩ U = 1.Let x 1 , . . . , x m ∈ G, and let k 1 , . . . , k m ∈ N with x kii ∈ U. By Lemma 1.21[A ω , Nx 1 , . . . , Nx m ] ≤ [A, 〈Nx k11 〉, . . . , 〈Nxkm m 〉] ≤ [A, d U] = 1,and this proves the Lemma.Recall from Chapter 1 (section 1.2), that a group G is hypocentral if {1} isa term of the extended lower central series of G. For a group G we also writeγ ω (G) =⋃γ n (G);1≤n∈Nthus G is residually nilpotent if and only if γ ω (G) = 1.Theorem 6.19 Let G ∈ N 1 . Then the following conditions are equivalent.(1) G ∈ X 0 ;

6.3. HYPERCENTRAL AND SMITH’S GROUPS 117(2) G is hypercentral;(3) γ ω (G) ≤ ζ m (G) for some m ∈ N;(4) G is hypocentral.Proof. Let G be a N 1 -group with G ∈ X 0 , and let A = tor(A). We first provethe following claim:there exists m ≥ 1 such thatγ m (G)A nA n is finite for all n ≥ 1. (6.12)Clearly, we may assume that G is not nilpotent. However, G/A is nilpotentby Theorem 2.23; let c be the nilpotency class of G/A. We first assume thatthere is a finitely generated subgroup F of G, and a positive integer d, suchthat every H ≤ b G containing F has defect at most d. Let β(d) as defined inTheorem 4.14, and let m = max{c, β(d)}+1. Fix n ≥ 1. Then G/A n is nilpotentby hypothesis, and so, by Lemma 6.6, there exists a normal subgroup M n ofG such that M n ∩ A = A n , and G/M n has finite exponent. Now, F M n /M nis finite, and for each H/M n ≤ G/M n , H ≤ b G. Thus, by our assumption,all subgroups of G/M n containing the finite subgroup F M n /M n have defect atmost d. By Theorem 4.14, γ m (G/M n ) = γ m (G)M n /M n is finite. Also, by choiceof m, γ m (G) ≤ A, so that γ m (G) ∩ M n = γ m (G) ∩ A ∩ M n = γ m (G) ∩ A n . Itthere follows thatγ m (G)A n∼γ m (G)=A n A n ∩ γ m (G) = γ m (G)M n ∩ γ m (G)is finite. For the general case, by Theorem 1.92 we know that there exists H ≤ b Gwhich satisfies the condition we have assumed above. Since G/H G has finiteexponent, B = A ∩ H ≥ A k for some k ≥ 1. This implies that for each n ≥ 1,B n ≥ A kn and H/B n , being a section of G/A kn , is nilpotent by hypothesis.Thus, there is a m ≥ 1 such that γ m (H)B n /B n is finite for all n ≥ 1. So,γ m (H)A nA n ∼ =γ m (H)A n ∩ γ m (H)being a factor of γ m (H)/(γ m (H) ∩ B n ) it is finite. Let H H 1 . Then γ m (H)is normal in H 1 . Since H 1 /H has finite exponent, Theorem 6.8 yields thatH 1 /γ m (H) is nilpotent, that is γ s (H 1 ) ≤ γ m (H) for some s ∈ N. Then wehave that γ s (H 1 )A n /A n is a subgroup of γ m (H)A n /A n and so it is finite. Byrepeating this argument along the normal closure series of H in G, we finallyget claim (6.12).Now we prove implication (1) ⇒ (2). We suppose that G ∈ X 0 is a counterexample,and thus that it is not hypercentral. By Lemma 1.87 and Brookes’trick 1.92, we then have that there exist a (non-hypercentral) subgroup H offinite index in G, a finitely generated subgroup F of H, and a positive integerd, such that every finite index subgroup of H containing F has defect at mostd in H. By Lemma 1.87, we may assume H = G. As above, let A be the torsionsubgroup of G. By 1.93, we may assume A ω = 1. Let m ≥ 1 as definied

118 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSin claim (6.12); then, arguing as in the second half of the proof of Lemma 6.7(using claim (6.12) in place of the first half), one shows that every subgroup ofG containing F has defect at most c = 2m + 1. But then, by Theorem 4.18 wehave that G is hypercentral.(2) ⇒ (1). Let G be a hypercentral N 1 -group. Then its torsion subgroup Ais nilpotent by Theorem 5.33. Let n ≥ 1 and C n = C G (A/A n ); G/C n is periodicby Corollary 1.21, and since AC n /A n is nilpotent, it follows from Lemma 6.6that there exists a normal subgroup M n of AC n such that M n ∩ A = A n andAC n /M n has finite exponent. Hence, G/M n is periodic and therefore nilpotentby Theorem 5.33. Since G/A is nilpotent (being torsion–free) we get that G/A nis nilpotent for every n ≥ 1, and so G is a X o -group.(1) ⇒ (3). Let G ∈ N 1 be a X o -group, and let A = tor(G). Then, by thedefinition of X 0 , and the fact that, as a torsion–free N 1 -group, G/A is nilpotent,it follows thatγ ω (G) = ⋂ n≥0γ n (G) ≤ A ω .Since A is nilpotent by assumption, 1.93 implies that A ω ≤ ζ m (G) for somem ∈ N.(3) ⇒ (4). This is clear by the definition of extended lower central series.(4) ⇒ (1). Let G be a hypocentral N 1 -group. Then its torsion subgroup Ais nilpotent by Theorem 5.29, and so there exists a positive integer m whichsatisfies the conclusion of Lemma 6.18.Suppose that G does not belong to X 0 ; then, by Lemma 6.17 and Theorem1.92, we may assume that there exists a finitely generated subgroup F of G anda positive integer d such that all subgroups H ≤ b G that contain F have defectat most d in G. G/A is nilpotent of class, say, r.Let t ≥ r (so that γ t (G) ≤ A), and write D/γ t (G) = (A/γ t G)) ω . Now,G/γ t (G) is trivially a X 0 -group, and application of Lemma 6.7 to it yields thatall subgroups of G/D that contain F D have defect at most c, where c dependsonly on r and d. As, by Lemma 6.18, [D, m G] ≤ γ t (G), we conclude that everysubgroup of G that contains F γ t (G) has defect at most c + m in G. Now, thisholds (with the same c and m) for every t ≥ r; then, if U is a finitely generatedsubgroup of G containing F , we have[G, c+m U] ≤ ⋂ t≥r(Uγ t (G) ∩ A) = ⋂ t≥rγ t (G)(U ∩ A) = γ ω (G)(U ∩ A)where the last equality holds because U ∩ A is finite. This implies that everysubgroup of G that contains F γ ω (G) has defect at most c + m in G. ThusG/γ ω (G) is hypercentral by Theorem 4.18.Let K = γ ω+m+1 (G); then G/K is also hypercentral and so it is a X 0 -groupby the already proved implication (2) ⇒ (1). But then, if Y/K = (A/K) ω ,γ ω (G) ≤ Y (by definition of X 0 ) and Y/K ≤ ζ m (G/K) (by Lemma 6.18). HenceK ≥ γ ω+m (G). Since G is hypocentral, it follows thatK = γ ω+m+1 (G) = γ ω+m (G) = 1.Hence G is hypercentral and a X 0 -group.

6.4. THE STRUCTURE OF PERIODIC N 1 -GROUPS 1196.4 The structure of periodic N 1 -groupsIn this final section, we prove that a N 1 -group is metanilpotent, and, in particular,that a periodic N 1 -group is the extension of a nilpotent group by an abeliandivisible group of finite rank.In a Heineken–Mohamed group G, G ′ is nilpotent and the factor group G/G ′is a Prüfer group C p ∞. As we have seen in Chapter 3, this has to be the caseif all proper subgroups of G are nilpotent and subnormal. Here, we prove thata similar conditon is satisfied in general by periodic groups with all subgroupssubnormal.Let A be an abelian p-group. For i ∈ N we setΩ i (A) = {a ∈ A | a pi = 1}.Then Ω i (A) ≤ A and, for all i ∈ N, Ω i+1 (A)/Ω i (A) is an elementary abelianp-group. We say that an abelian p-group A is large if Ω i+1 (A)/Ω i (A) is infinitefor all i ∈ N; otherwise we say that A is small. It is easy to see that an abelianp-group is small if and only if it is the direct product of a divisible group offinite rank by a group of finite exponent.Lemma 6.20 Let G ∈ N 1 be a p-group, and A a normal elementary abeliansubgroup of G, such that G ′ ≤ A. Then G/C G (A) is small.Proof. Let G be a counterexample, and let C = C G (A). Observe that G/Cis abelian. Let Θ be the family of all subgroups X of G such that XC/C islarge. By Lemma 1.92, there exists a Θ-subgroup H of G, a finitely generatedsubgroup F of H, and a positive integer d, such that every Θ-subgroup of Hcontaining F has defect at most d in H. For each i ≥ 0 we setH i /(H ∩ C) = Ω i (H/(H ∩ C)) = 〈g ∈ H | g pi∈ C〉.Then, as H/H ∩ C ∼ = HC/C is large, H i+1 /H i is an infinite elementary abelianp-group for all i ≥ 0. Also H i is nilpotent, by Theorem 5.28, since H i ∈ N 1is the extension of the normal nilpotent subgroup H ∩ C by a group of finiteexponent.Now, as G is a locally nilpotent p-group, F is finite. If all subgroups of Hcontaining F have defect at most d in H, then H is nilpotent by Theorem 4.14.But in that case, by Lemma 5.30, HZ/Z has finite exponent. Since Z ≤ C, itfollows that HC/C has finite exponent, contradicting the choice of H ∈ Θ.Thus, there exists a subgroup K ≥ F of H, such that d(K, H) ≥ d + 1.Then [H, d K] ≰ K. It follows that there exists a finitely generated subgroupV = V 0 of K, such that [H, d V ] ≰ K. Clearly, we may assume that F ≤ V .Let a ∈ [H, d V ] \ V , and let m be the smallest integer such that V ≤ H m . Byinduction on i we construct a seriesV = V 0 ≤ V 1 ≤ . . . ≤ V i ≤ . . .of finite subgroups of H such that, for all i ∈ N, a ∉ V i , V i ≤ H m+i , and∣ V i+1 ∣∣∣∣= p i+1 .V i+1 ∩ H m+i

120 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSSuppose we have already found V 0 , . . . , V i . Then, by Theorem 5.23, applied tothe nilpotent group H m+i+1 modulo H m+i , there exists a subgroup X of H m+i+1such that V i ≤ X, a ∉ X, and X/(X∩H m+i ) ∼ = XH m+i /H m+i is infinite. Hence,we may choose elements x 0 , x 1 , . . . , x i in X such that〈x 0 , . . . , x i 〉H m+iH m+ihas order p i+1 . We put V i+1 = 〈V i , x 0 , . . . x i 〉 ≤ X ≤ H m+i+1 . Then a ∉ V i+1 ,and V i+1 /(V i+1 ∩ H m+i ) ∼ = V i+1 H m+i /H m+i has order p i+1 .We now consider the subgroupY = ⋃ i∈NV i .Then, by construction, F ≤ Y ≤ H, and a ∉ Y . We show that Y ∈ Θ. Suppose,by contradiction, that Y = Y C/C is small. Then there exist positive integersn, k such that |Ω n+1 (Y )/Ω n (Y )| ≤ p k . By elementary facts on abelian p-groups,it follows that |Ω j+1 (Y )/Ω j (Y )| ≤ p k for all j ≥ n. For all i ∈ N, let Y i /(Y ∩C) =Ω i (Y/(Y ∩ C)). Then Y i /(Y ∩ C) ∼ = Ω i (Y ), and Y i = H i ∩ Y . Let t ≥ max{n, k}.Then, we have∣ ∣ p k ≥Ω t+m+1 (Y )∣∣∣ ∣Ω t+m (Y )∣ = Y t+m+1 ∣∣∣.Y t+mBut, by construction of Y ,Y t+m+1Y t+m= H t+m+1 ∩ YH t+m ∩ Y∼ = (H t+m+1 ∩ Y )H t+mH t+m≥ V t+1H t+mH t+mhas order at least p k+1 , and this gives a contradiction.Hence Y ∈ Θ, and so, by the choice of H, Y has defect at most d in H. Butthen,a ∈ [H, d V ] ≤ [H, d Y ] ≤ Ywhich is the final contradiction.Lemma 6.21 Let G be a p-group, and D a divisible subgroup of G of finite ranksuch that G ′ ≤ D ≤ Z(G).(1) If G/D is small, then G is the extension of a group of finite exponent byan abelian divisible group of finite rank.(2) If G/D is large, then there exists a large abelian subgroup X of G such thatD ∩ X = 1.Proof. (1) Suppose that G/D is small. Then G/D is the direct product of adivisible group D 1 /D of finite rank by a group of finite exponent. Since D 1 isthen a divisible subgroup of the nilpotent p-group G, D 1 ≤ Z(G) by Lemma1.18. Thus, we may assume that G/D has finite exponent p n . Let g, x ∈ G.Then g pn and [x, g] belong to D ≤ Z(G), whence[x, g] pn = [x, g pn ] = 1.

6.4. THE STRUCTURE OF PERIODIC N 1 -GROUPS 121Hence G ′ is a subgroup of finite exponent of D. Now, D/G ′ is a divisible subgroupof the abelian group G/G ′ , so there exists a direct summand H/G ′ ofD/G ′ in G/G ′ . Then, H G has finite exponent, and G/H ∼ = D/G ′ is divisibleof finite rank.(2) Suppose that G/D is large. Since D has finite rank, A = Ω 1 (D) isfinite. Now, the same argument used to construct Y in the proof of the previousLemma, can be employed to find a subgroup H of G such that H/H ∩D is largeand H ∩ A = 1. But, trivially, this forces H ∩ D = 1, whence H is abelian, large,and has trivial intersection with D.Lemma 6.22 Let G ∈ N 1 be a p-group, such that G ′ is nilpotent. Then thereexists a normal nilpotent subgroup N of G such that G/N is an abelian divisiblep-group of finite rank.Proof. Let H = G ′ , and C the centralizer in G of H/H ′ H p . Then H ≤ Cand C/H ′ H p is nilpotent. By Lemma 1.1, it follows that C/H ′ H pn is nilpotent,for all i ∈ N. Thus, if K/H ′ = (H/H ′ ) ω , C/K is residually nilpotent. ByTheorem 5.29, C/K is nilpotent. We then have, by Lemma 1.93, that C/H ′ isnilpotent. Since H is nilpotent by assumption, Theorem 1.54 allows to concludethat C is nilpotent. Now, by Lemma 6.20 applied to G/H ′ H p , G/C is a smallabelian p-group. By what we have observed earlier, G/C is the direct product(D/C) × (N/C) where D/C is a divisible p-group of finite rank, and N/C isa group of finite exponent. By Theorem 5.28, N is nilpotent. Since G/N isisomorphic to D/C, the proof is done.Theorem 6.23 Let G be a periodic group with all subgroups subnormal. ThenG has a normal nilpotent subgroup N such that G/N is an abelian divisiblegroup of finite rank.Proof. Let G be a periodic N 1 -group. By Lemma 4.19, we may assume that G isa p-group, for some prime p. G is soluble by Möhres Theorem 6.4. We proceed byinduction on the derived length of G. Then, by inductive assumption, H = G ′ isthe extension of a normal nilpotent subgroup by a divisible abelian subgroup offinite rank. Among such normal nilpotent subgroups of H, choose K such thatthe rank r of the divisible group H/K is as small as possible (possibly r = 0).By Theorem 6.11, K G is nilpotent. Also, H/K G is divisible of rank at most r,so we may take K to be normal in G.Now, G/K is nilpotent by Lemma 1.93, and H/K is central in G/K byLemma 1.18. Thus, we are in a position to apply Lemma 6.21 to the groupG/K. Assume first that G/H is small. Then G/K is the extension of a normalsubgroup N/K of finite exponent, by an abelian divisible group of finite rank.By Theorem 5.28, N is nilpotent, and we are done.Thus, assume that G/H is large. Let W/K be a normal subgroup of G/Kmaximal such that W ∩ H = K. We claim that G/HW is small. Suppose not,then by Lemma 6.21 there exists a large abelian subgroup X/W of G/W suchthat X ∩ HW = W . Then X ∩ H = K and X/K ∼ = XH/H is abelian. ByLemma 6.22, X admits a normal nilpotent subgroup U ≥ K such that X/U isdivisible of finite rank. ByTheorem 6.11, U G is nilpotent, U G ≤ HU and, by

122 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPSthe choice of K, as H/(H ∩ U G ) is divisible, the rank of H/(H ∩ U G ) is r. Itfollows that (H ∩ U G )/K ∼ = U G /U is finite. Now, XU G /U G ∼ = X/(U G ∩ X) isa divisible subgroup of the nilpotent p-group G/U G . By Lemma 1.18, XU G isnormal in G, i.e. XU G = X G . Moreover,X G /X = XU G /X ∼ = U G /(U G ∩ X) = U G /U ∼ = (H ∩ U G )/Kis finite. Then, there exists an integer n ≥ 0 such that, if M = (X G ) pn , thenM ≤ X. Now, X ≥ W M G, and W M ∩ H = K. By the choice of W , weget M ≤ W , which implies in particular X pn ≤ W , contradicting the fact thatX/W is large.Thus G/HW is small. Again by Lemma 6.22, W has a normal nilpotentsubgroup U ≥ K (which we may assume to be normal in G by [13]) such thatW/U is divisible of finite rank. Since HW/W ∼ = H/K is divisible of finite rankand G/HW is small, we have that G/U is the extension of a divisible abeliansubgroup of finite rank by an abelian group of finite exponent. By applying thesame argument used in the case G/H small, we get the desired conclusion.Remark. It follows from examples constructed by W. Möhres (Proposition3.19), that the rank of G/N in the above statement cannot be bounded further.In fact, let n ≥ 1, llet G be a p-group as in the statement of 3.19, and supposethat N is a nilpotent subgroup containing G ′ . It is then a standard argument,since Z(G) = 1 and G ′ is elementary abelian, to show that N/G ′ does notcontain any copy of C p ∞, and so that the rank of G/N cannot be less that n.Let us also mention a curious corollary of 6.23, that maybe confirms theimpression that periodic N 1 -groups do not differ much from Heineken-Mohamedgroups. These latter have no proper non-nilpotent subgroups; for the generalcase we have:Corollary 6.24 Let G be a periodic group in N 1 . Then there exists d ≥ 1 suchthat every non-nilpotent subgroup of G has defect at most d.(Smith’s residually finite N 1 -groups show that this is not the case for nonperiodcgroups).Theorem 6.23 comprises all other results on periodic N 1 -groups that we haveincluded in these notes; as such, together with the nilpotency of the torsion-freecase (Theorem 2.23), it represents a reaching point in the effort of describingN 1 -groups. What is not yet very well understood is the mixed case; by applyingtogether Theorems 6.23 and 2.23, we have the following fact.Theorem 6.25 Let G be a group with all subgroups subnormal. Then thereexists a normal nilpotent periodic subgroup N of G such that G/N is nilpotent.Proof. Let G ∈ N 1 , and let T be the torsion subgroup of G. Then, by Theorem2.23, G/T is nilpotent. Also, by Theorem 6.23, there exists a normal nilpotentsubgroup K of T such that T/K is a periodic divisible abelian group. ByTheorem 6.11, N = K G is nilpotent. Now, T/N is a normal periodic divisibleabelian subgroup of G/N. Since G/T is nilpotent, by Lemma 1.93 we concludethat G/N is nilpotent, thus proving the Theorem.

6.4. THE STRUCTURE OF PERIODIC N 1 -GROUPS 123Question 6 Is it true that a N 1 -group is the extension of a nilpotent group bya periodic (abelian) group of finite rank?In this direction, using 6.23 and some of the techniques developed in sections6.2, 6.3, the following result can be proved.Theorem 6.26 Every N 1 -group is the extension of a hypercentral group by anabelian periodic divisible group of finite rank.I will not include here a proof of this: it will (possibly) appear elsewhere.

124 CHAPTER 6. THE STRUCTURE OF N 1 -GROUPS

Chapter 7Beyond N 17.1 Generalizing subnormalityHaving reached a reasonably good knowledge of the class N 1 , what is perhaps themost immediate question is to ask for groups in which every subgroup satisfiesone of the natural generalizations of subnormality; like seriality, ascendancy ordescendacy.Serial subgroups. Imposing seriality to all subgroups is not a very restrictiveconditions. By Corollary 1.63, all locally nilpotent groups satisfy it, and wementioned J. Wilson’s construction in [121] of infinite finitely generated p-groupsin which every subgroup is subnormal (we notice that, following Wilson’s line,one may also construct finitely generated non-nilpotent torsion-free groups inwhich every subgroup is serial). The groups constructed by Wilson, being ofGolod-type, are also residually finite, and therefore belong to the class of locallygraded groups. On the other hand it is clear that groups in the class W and allsubgroups serial are locally nilpotent 1 .Descendant subgroups. A subgroup H of the group G is descendant if it is aterm of a descending series of G. Like seriality, for finite groups descendancy isequivalent to subnormality. Thus, the class D of groups all of whose subgroupsare descendant is a class of generalized nilpotent groups. The following is aneasy observation.Lemma 7.1 A group G belongs to the class D if and only if H K < K for allH < K ≤ G.However, it is not even clear if groups in D are locally nilpotent. Consideration ofthe infinite dihedral group D ∞ shows that (contrary to ascendancy) to assumethat all cyclic subgroups of a group G are descendant is not enough to ensurelocal nilpotency of G. More generally, we make the following remark.Proposition 7.2 Let G be a countable residually nilpotent group. Then everyfinite and every nilpotent subgroup of G is descendant.1 Recall from Cahpter 1 that a group G belongs to the class W if every finitely generatedsubgroup of G either is nilpotent or has a non-nilpotent finite image.125

126 CHAPTER 7. BEYOND N 1Proof. Let F be a finite subgroup of the residually nilpotent group G. Then allsubgroups γ n (G)F (n ∈ N) are subnormal in G, so their chain can be refinedto get a descending series of G. Now, ⋂ n∈N γ n(G) = 1, so by Lemma 1.27,F = ⋂ n∈N γ n(G)F , showing that F is descendant.Suppose now that the subgroup H of G is nilpotent; we show by inductionon its nilpotency class c that H is descendant. If c = 1 H is abelian;by Lemma 1.29 C G (H) = ⋂ n∈N γ n(G)C G (H), so C G (H) is descendant andthus H is descendant. Let c > 1 and let Y = C G (ζ(H)). By the same arguemntused before Y = ⋂ n∈N γ n(G)Y is descendant. Now, Y is residuallynilpotent and Z = ζ(Y ) = ⋂ n∈N γ n(Y )Z, so Y/Z is residually nilpotent. NowHZ/Z ≃ H/(H ∩ Z) = H/ζ(H) is a nilpotent subgroup of Y/Z of class c − 1,and by inductive assumption HZ is descendant in Y , but H HZ so H isdescendant in Y . Since Y is descendant in G, we conclude that H is descendantin G.Remembering that a free group is residually nilpotent, we have,Corollary 7.3 In a countable free group every cyclic subgroup is descendant.Apparently, it is not known whether there exits a finitely generated infinite p-group which is residually finite and such that every subgroup of it is either finiteor has finite index. If such a group exists, then, by what observed above, it willhave all subgroups descendant.Question 7 Does there exists a non-trivial perfect (locally nilpotent) group inwhich all subgroups are descendant?Ascendant subgroups. The class of groups in which every subgroup is ascendantis of course the class N of all groups satisfying the normalizer condition.Apart from the basic facts that we recalled in Chapter 1 (it is a class of Gruenberggroups that contain every hypercentral group), little more I know ingeneral about this class. The following old question is still open.Question 8 Is every group N-group hyperabelian?Now, this seems very difficult, but nevertheless I think that some of the techniquesdeveloped for studiyng N 1 -groups, in addition to other conditions (likesolubility) may prove fruitful also for the broader class N. For insatnce, Möhres,using the methods we reported in chapter 5, has proved the following.Proposition 7.4 Let G be an N-group which is the extension of a nilpotentp-group of finite exponent by an elementary abelian p-group. Then G is hypercentral.The following question is now natural.Question 9 Is a soluble N-group of finite exponent hypercentral?and its corrispective in the torsion-free case.Question 10 Does there exist a (soluble) torsion-free N-group with trivial centre?

7.1. GENERALIZING SUBNORMALITY 127Local subnormality. A class which is intermediate between N 1 and N is theclass (which we denote by N 2 ) of groups in which every subgroup is locallysubnormal; where a subgroup H of a group G is called locally subnormal ifH ⊳ ⊳ 〈H, X〉 for all finite X ⊆ G.Trivially, in a locally nilpotent group every finitely generated subgroup islocally subnormal. Thus, the existence of locally nilpotent groups with trivialGruenberg radical shows that a locally subnormal subgroup need not be ascendant.On the other hand, it is clear that a group in which every subgroup islocally subnormal satisfies the normalizer condition, and so it is locally nilpotent.Example 3.9 Let G = C p ∞ ≀ X, where X = 〈x〉 is cyclic of order p 2 . G ishypercentral by Lemma 6.5. Let C ≃ C p ∞ be one of the coordinate subgroupsin the base group of G, and H = 〈C, x p 〉. Then H ≃ C p ∞ ≀ C p and, clearly,〈H, x〉 = 〈C, x〉 = G. On the other hand, H is ascendant but not subnormal inG, so H is not locally subnormal.This example shows that the class N 2 does not contain all hypercentral groups,and so it is a proper subclass of N (and clearly contains N 1 , in particular theHeineken-Mohamed groups which are not hypercentral). Every direct productof nilpotent groups and, more generally, every hypercentral group of length ωis a N 2 -group, while the infinite dihedral 2-group is a N 2 -group which is not aFitting group. I do not know much more about this class of locally nilpotentgroups.Question 11 Is every group in N 2 hyperabelian?Of course, this will follow from a positive answer to question 8; in general, thequestions we suggested for the class N make sense for the smaller class N 2 too.Other generalizations of subnormality. A subgroup H of a group G isalmost subnormal if H has finite index in a subnormal subgroup of G, andvirtually subnormal if H is subnormal in a subgroup that has finite index inG. Both these definitions are included in that of f-subnormality, introduced byPhillips [91]: a subgroup H of G is f-subnormal if there exists a finite seriesH 0 = H ≤ H 1 ≤ . . . ≤ H n = G such that |H i : H i−1 | < ∞ or H i−1 H i forevery i ∈ {1, . . . , n}.When applied to a single subgroup, these conditions are all different, butthings change if we consider all subgroups.Proposition 7.5 (see [19]) For any group G the following are equivalent:(1) every subgroup of G is almost subnormal;(2) every subgroup of G is virtually subnormal;(3) every subgroup of G is f-subnormal.We denote by SF the class of groups in which every subgroup is f-subnormal.For finitely generated groups there is a neat characterization of such groups.

128 CHAPTER 7. BEYOND N 1Theorem 7.6 ([64], Theorem 6.3.3) A finitely generated group is finite by nilpotentif and only if every subgroup is f-subnormal.For the general case, we have the followingTheorem 7.7 (Casolo, Mainardis [19], [20]) Let G be an SF -group, and letD(G) be the subgroup generated by the nilpotent residuals of the finitely generatedsubgroups of G. Then(1) D(G) is finite by nilpotent and contained in the torsion part of the F C-centre of G;(2) G/D(G) ∈ N 1 ;(3) G is finite by solvable;(4) if G is torsion-free then G is nilpotent;(5) if G is periodic then G is finite-by-N 1 .Stronger conditions than those assumed in Theorem 7.7 have been considered. Inthese cases, the results should be viewed as generalizations both of Roseblade’sTheorem and of a Theorem of B. Neumann saying that: The derived subgroup ofa group in which every subgroup has finite index in its normal closure is finite.We mention only a couple of these results.Theorem 7.8 (Lennox [63]) Let G be a group and suppose that there existspositive integers m, n such that |H G,n : H| ≤ m, for all H ≤ G. Thenfor some integer µ(n + m).|γ µ(m+n) (G)| ≤ m!In the same paper, Lennox obtains similar results for those groups G in whichevery subgroup is subnormal of bounded defect in a subgroup of finite boundedindex in G, and for SF-groups with suitable bounds imposed on the finte–by–subnormal series (see also [64] for a fuller account of this particular topic).More recently, Detomi [25] was able to partly extend Lennox’ result.Theorem 7.9 Let G be a group, and suppose that there exists n ≥ 1 such that|H G,n : H| < ∞ for all H ≤ G. If G is either periodic or torsion–free, thenγ δ(n) (G) is finite for some δ(n) ∈ N.It shold be noted that this last Theorem does not carry over to arbitrary groups:H. Smith’s hypercentral N 1 -groups that we will describe in Section 6.3, satisfy|H G,2 : H| < ∞ for every H ≤ G but they are not finite by nilpotent; similarexamples, in which γ 3 (G) = G ′ is infinite, are constructed in [19].Groups in which every subgroup is approached from below by a subnormalsubgroup are much less tractable, even in the special case in which H/H G isfinite for every subgroup of G (Ol’shanski infinite groups in which every propersubgroup has order p are examples of groups of this kind). The many problems

7.2. GROUPS WITH MANY SUBNORMAL SUBGROUPS 129connected with this class of groups (even when suitably restricted) have stimulatedseveral people, and a number of articles have appeared on this topic,starting perhaps with a paper by Buckley, Lennox, B. H. Neumann, H. Smithand J. Wiegold [11] (this subject involves also some non-trivial questions aboutfinite p-groups, and we mention paper [22], where more complete references maybe found). Regarding the class of groups in which every subgroup contains asubgroup of finite index which is subnormal in the whole group, I only am awareof a paper by H. Heineken [45], from which I quote the following Proposition:In a locally finite group G in which every subgroup H contains a subgroup Swith |H : S| < ∞ and S ⊳ ⊳ G, the Hirsch–Plotkin radical has finite index. Theremight well be some room left for more research on this subject: for instanceQuestion 12 Is a locally nilpotent (or soluble by finite) group with all subgroupssubnormal by finite, a finite extension of a N 1 -group ?(A nilpotent torsion-free group with all subgroups subnormal by finite is certainlynilpotent, while any non-nilpotent Černikov p-group is an example of alocally nilpoytent group with this property which is not in N 1 .)7.2 Groups with many subnormal subgroupsUnder this label are denoted in the literature groups in which the set of nonsubnormalsubgroups satisfies cenrtain (usually of finitary type) restrictions;given a specific restriction to the set of non-subnormal subgroups, the usualtarget is to describe (if any) those groups that satisiy such a restriction anddo not belong to N 1 or to the class of groups in which the set of all subgroupssatisfies that restriction.This kind of investigations goes back to Černikov, who studied groups inwhich many subgroups have a prescribed property P (structural or of embedding);in particular, close to what we are going to consder here, the case whenP is the property of being ascendant (see [21] for a survey on Černikov’s work).Perhaps even closer in methods is a 1978 paper [92] by Phillips and Wilson(in which the class W was introduced), where W-groups with ”many” serialor locally nilpotent subgroups are studied; although not explicitely referring tosubnormality, we report part of the main result of [92].Theorem 7.10 Let G be a W-group. The following are equivalent:(1) the set of all non-serial non-locally nilpotent subgroups of G satisfies theminimal condition;(2) either G is anilpotent;Černikov group, or every subgroup of G is serial or locallyand in this case, if G is not afinite cyclic.Černikov group, then G is locally nilpotent byThis is a topic that has recently seen a lot of activity, its only bound being theimagination of the scholars. Therefore, I am probably not completely aware of

130 CHAPTER 7. BEYOND N 1all the developments, and in my report I will describe only a few cases, andprovide a couple of proofs. just in order to try giving a flavour of this line ofinvestigation and an idea of the arguments involved.As in Phillips–Wilson, we begin with the minimal condition.Theorem 7.11 (Franciosi, de Giovanni [27]) Let the group G satisfy the minimalcondition on non–subnormal subgroups.(1) If G is a Baer group, then G ∈ N 1 .(2) If G is not periodic, then G ∈ N 1 .(3) If G ∈ W, then G is either a Černikov group or G ∈ N 1.Proof. (1) Let G be a Baer group satisfynig the minimal condition on non–subnormal subgroup, and suppose by contradiction that G ∉ N 1 . Thus, let Hbe a minimal non-subnormal subgroup of G. Then all proper subgroups of H aresubnormal; in particular, by Möhres Theorem, K = H ′ < H. Since H cannotbe the product of two proper subgroups, H/K is either cyclic or isomorphicto C p ∞ for some prime p. Now, G is a Baer group, so if H/K were cyclic,then H = K〈x〉 would be the product of two subnormal subgroups. HenceH/K ≃ C p ∞. Let G = K 0 > K 1 > . . . > k d = K be the normal closure series ofK in G (since H normalizes K, all K j are normalized by H), and let i ≥ 1 beminimal such that HK i is not subnormal in HK i−1 . Then K i HK i−1 andHK iK i≃ HH ∩ K iis a proper quotient of H/K ≃ C p ∞. Hence HK i /K i ≃ C p ∞, and we may replaceG by HK i−1 /K i , and H by HK i /K i , and thus assume that H ≃ C p ∞ for someprime number p. Clearly we may then also suppose that G is a p-group.Let X = N G (H). Then N G (X) = X (by 1.32 and 1.33), and H ≤ Z(X).Also, X/H satisfies Min and so X is a Černikov p-group. Now, G is a Baergroup, hence all proper subgroups of H are subnormal in G; clearly, there existsa proper (cyclic) subgroup Y of H such that Y G ≰ X. Let M be the smallestterm of the normal closure series of Y in G such that M ≰ X. Since Y ≤ Z(X),M is normalized by X. Also, Y M ≤ X and so, since Y has finite exponent, Y Mis a finite p-group. Since M is generated by normal conjugates of Y M , it followsthat M is nilpotent of finite exponent. Let N = N M (M ∩ X); then N > M ∩ Xand N is normalized by X . Let A/M ∩ X be the subgroup of all elements oforder p in Z(N/M ∩ X). Then A/M ∩ X ≠ 1, because M is nilpotent, andA is normalized by X. If A/M ∩ X is finite, then 1 < |AX : X| is finite andtherefore X ⊳⊳ AX, which contradicts X = N G (X). Thus, A/M ∩X is an infiniteelementary abelian p-group normalized by X (and by H). Let B = H(X ∩ M);then B/X ∩ M ≃ C p ∞ and N G (B) = X, whence N A (B) = A ∩ B. This, inparticular, says that B is not maximal in any subgroup S with B < S ≤ AB.So there exists an infinite chain of subgroups AB > S 1 > S 2 > . . ., withB[A, B] > S i > B for all i ≥ 1. By our assumpion on G there exists t > 1 suchthat S t is subnormal in G. But S t = B(S t ∩ A) and so S t /S t ∩ A ≃ C p ∞. It

7.2. GROUPS WITH MANY SUBNORMAL SUBGROUPS 131then follows from 1.32 and 1.33 that S t /S t ∩ A is normal in BA/S t ∩ A and soS t AB. Therefore [A, B] ≤ [A, S j ] ≤ S j , which is a contradiction.(2) Suppose that G is not periodic, and let g ∈ G be an element of infiniteorder. Then there exists integers m, n ≥ 1 such that U = 〈g 2n 〉 and V = 〈g 3n 〉are subnormal in G, whence 〈g〉 = UV is subnormal in G. Thus, the Baerradical B of G contains all elements of infinite order. Our claim will be provedif we show that G is generated by elements of infinite order. This is equivalentto prove that for every pair a, b of elements of finite order of G, the producty = ab has finite order. Suppose, to the contrary that |y| = ∞. Then y belongsto the Baer radical of 〈a, b〉, and so H = 〈a, b〉 = 〈a, y〉 is the extension of thefinitely generated nilpotent group Y = 〈y〉 〈a〉 by the finite group 〈a〉. Thus His policyclic and nilpotent by cyclic. As the torsion subgroup of Y is finite, wemay well assume that Y is torsion free. Then, if p is a prime which does notdividse the order of a, by Theorem 1.41 there exists an infinite descending chainY > N 1 > N 2 > of notmal subgroups N i with Y/N i a finite p-group. As ahas finite order, we may find a chain of this kind with all N i are normal in H.Thus, by our assumption on G, there exists t ≥ 1 such that 〈N j , a〉 is subnormalin H for all j ≥ t. Then, for all i ≥ 1, H/N i /N i is a nilpotent group, and thedirect product of its p-component Y/N i and the cyclic p ′ -group 〈N j , a〉/N i . Thus[Y, a] ≤ N i for all i ≥ 1. This yields 〈a〉 H, and so H = 〈a, b〉 is finite. Thisproves that B = G and so, by point (1), that G ∈ N 1 .(3) Let G ∈ W be a group in which the set of non-subnormal subgroupssatisfies the minimal condition. By (2) we may assume that G is periodic. Now,it is easy to see that a finitely generated periodic group in W with the minimalcondition on non-subnormal subgroups is finite; therefore G is locally finite.Suppose that G is not Černikov; then, by the Šunkov, Kegel–Wehrfritz Theorem1.37, G admits non-Černikov abelian subgroups, and by our assumptionon G there exist subnormal such subgroups. Hence, the Baer radical B of Gdoes not satisfy the minimal condition on subgroups. Bu point (1) we are doneif we prove that B = G. Clearly, it is enough to prove that any element of primepower order of G belongs to B.Thus, let g ∈ G be an element of order a power of a prime p., and let Abe the p-component of B. Suppose first that A is not Černikov. By MöhresTheorem 6.4, A is soluble. Let M = A (m) , be the smallest term of the derivedseries of A which is not a Černikov group (it exists because the class of groupswith Min is closed by extensions), and let K = A (m=1) = M ′ . Observe that Kis a Černikov Baer group and so it is contained in some finite term of the uppercentral series of M; therefore M is nilpotent. If we prove that K〈g〉 is subnormalin G, then in particular M〈g〉/K is nilpotent by 1.59 and so M〈g〉 is nilpotentbt Hall’s criterion 1.54; consequntly 〈g〉 ⊳ ⊳ K〈g〉 ⊳ ⊳ G. Thus, we assume K = 1and G = M〈g〉. Since M is a non-Černikov abelian group it has an infinitecharacteristic elementary abelian subgroup X. Since g has fnite order, there isan infinite descending chain of g-invariant subgrops X i of X, with X i ≥ X ∩〈g〉,and then X m , 〈g〉 ⊳⊳ G for some m ≥ 1. But, X m 〈g〉 is a soluble p-group of finiteexponent ans so, by Proposition 1.76, 〈g〉 ⊳ ⊳ X m 〈g〉 ⊳ ⊳ G and we are done.Suppose then that the p-component A of B is Černikov. Then, since B doesnot satisfy Min, it follows that the p ′ -component U of B is not Černikov. Again,

132 CHAPTER 7. BEYOND N 1U is soluble. Arguing exactly as in the previous case, we find a characteristicsection M/K of U such that it is enough to show that K〈g〉 is contained in somesubnormal subgroup of G contained in M〈g〉. As before, we may assume K = 1.Let X be the subgroup generated by all elements of prime order of M. Since Mis not Černikov, X is infinite. Let D = [X, 〈g〉]. By a standard fact for coprimeactions on abelian groups, [D, g] = D. Now, if D is infinite, as before we find aproper 〈g〉-invariant subgroup D 0 of D such that D 0 〈g〉 ⊳ ⊳ G, which yields thecontradictionn [D, g] ≤ D 0 < D. Thus, D is finite. This means that C X (g) isinfinite. But then we find a subgroup R of C X (g) such that R〈g〉 ⊳ ⊳ G. Since〈g〉 R〈g〉 we again conclude that 〈g〉 ⊳ ⊳ G. This completes the proof that G isa Baer subgroup and therfore (3) is established.In the same paper, Franciosi and de Giovanni consider groups with only afinite number of conjugacy classes of non-subnormal subgroups, proving thatlocally graded such groups are either finite or N 1 .Moving to the maximal condition, the following has been proved.Theorem 7.12 (Kurdachenko, Smith [57]) Let the group G satisfy the maximalcondition on non–subnormal subgroups.(1) If G is locally nilpotent, or infinite locally finite, then G ∈ N 1 .(2) G is locally (soluble–by–finite) if and only if G satisfies one of the followingconditions:(i) G is polycyclic by finite;(ii) G ∈ N 1 ;(iii) G ≠ B(G), B(G) is nilpotent, G/B(G) is polycyclic–by–finite torsion–free, and for every g ∈ G \ B(G), and every N G, with N ≤ B(G), thegroup 〈N, g〉 is finitely generated.We isolate in a Lemma one of the technical arguments involved in the proof.Lemma 7.13 Let A be a normal abelian subgroup of the soluble group G, andlet g ∈ G \ A, with gA ∈ Z(G/A). Suppose that G/A is not finitely generatedwhile A is finitely generated as Z〈g〉-module. Then the centralizer of g in Gcontains a subgroup that is not finitely generatedProof. Since A is abelian, [A, 〈g〉] = [A, g] = {[a, g] | a ∈ A}. Also, [A, g] G becausegA is central in G/A. Now, by assumption, B = A〈g〉 is finitely generated,and so B/[A, g] is a finitely generated abelian group. Let C = C G (B/[A, g]); thenC ≥ B and G/C is finitely generated (indeed, it is polycyclic, see for instance[96], 3.2.7). As G/A is not finitely generated, we get that C/A is not finitelygenerated. Now, let x ∈ C; then x ∈ C, [x, g] ∈ [B, C] ≤ [A, g], and so thereexists a ∈ A such that [x, g] = [a, g]. We have[xa −1 , g] = [x, g] a−1 [a −1 , g] = [x, g][a, g] −1 = 1which means that xa −1 ∈ C G (g). This shows that C ≤ AC G (g). Since C/Ais not finitely generated, we conclude that C C (g) = C ∩ C G (g) is not finitelygenerated.

7.2. GROUPS WITH MANY SUBNORMAL SUBGROUPS 133Proof of Theorem 7.12. Let us denote by S the class of groups satisfying the maximalcondition on non–subnormal subgroups. We begin with a rather immediateobservation.(A) Let G belong to S, and let F < H ≤ G with F finitely generated and Hnot finitely generated; then there exists a finitely generated T with F ≤ T < Hand T ⊳ ⊳ G.From this, one immediately deduces,(B) A locally nilpotent group in S is a Baer group.Now, for the proof of point (1) of the statement, we may just deal with Baergroups.(C) Let G be a Baer group in S, and 1 ≠ H ≤ G; then H ′ ≠ H and H ′ ⊳⊳ G.Proof. Let U ≤ H be a maximal non-subnormal subgroup of H, or U = 1 if thereare not any. If U = 1 let N = 1; otherwise, there exists a proper and subnormalsubgroup V of H containing U, then set N = V H . In any case N is a propernormal subgroup of H, and H/N belongs to N 1 . It then it follows H ′ < H byTheorem 6.4. Now, if H ⊳ ⊳ G, then H ′ is also subnormal. Thus, assume H isnot subnormal in G. If H/H ′ is not finitely generated, then it does not satisfiesMax, and so there exists H ′ ≤ L ≤ H with L⊳⊳ G; as H ′ L, H ′ ⊳⊳ G. If H/H ′is finitely generated, then H = H ′ X for some finitely generated subgroup X ofH. Then, H ′ X H = H, and since H is a Baer group, X = H. Thus, H is finitelygenerated and so subnormal in G.(D) Let G = AH be a Baer group in S, with A, H abelian and A G. ThenH cannot be a maximal non-subnormal subgroup of G.Proof. Observe that A∩H G, whence we may suppose A∩H = 1. Assume thatH is a maximal non-subnormal subgroup of G, and let X be a cyclic subgroupof H such that [A, X] ≠ 1. Now, X ⊳ ⊳ G, and so C A (X) ≠ 1. Since H is abelianC A (X) is normalized by H, and H < C A (X)H. Thus, C A (X)H is subnormalin G, and therefore [A, m H] ≤ C A (X)H ∩ A = C A (X) for some m ∈ N, whichwe take the smallest such. Since [A, X] ≠ 1, we have m ≥ 1. But then, since Aand H are abelian[A, m−1 H, X, H] = [A, m−1 H, H, X] ≤ [C A (X), X] = 1.Thus, [A, m−1 H, X] ≤ C A (H) = 1, which means [A, m−1 H] ≤ C A (X), againstthe choice of m.(E) Let G be a Baer group in S. Then 〈x〉 G is soluble for every x ∈ G.Proof. Let x ∈ G, and K = 〈x〉 G . Arguing by induction on the defect of 〈x〉 inG, we may assume that 〈x〉 K is soluble, and so that K is generated by normalsoluble subgroups. Another obvious inductive argument reduces us to prove thata S-group K which is generated by normal abelian subgroups is soluble. Supposethat K is not in N 1 , let H be a maximal non-subnormal subgroup of K, and letN be a normal abelian subgroup of K such that N ≰ H. Then H < NH ⊳ ⊳ G.Now, H ∩ N NH; let D = H ′ (N ∩ H). Then D < H and D ⊳ ⊳ G by point(C). In particular, D ⊳ ⊳ ND (and D < ND). Let A be the last but one termof the normal closure series of D in AD; then A is normalized by H, and thegroup AH/D violates point (D). Thus, K ∈ N 1 , and so K is soluble.

134 CHAPTER 7. BEYOND N 1Proof of point (1). We first suppose that G ∈ S is locally nilpotent, and so,by point (B), a Baer group. Assume that G is not in N 1 ; then there exists amaximal non-subnormal subgroup H of G. Now, H ′ ⊳ ⊳ G by point (C); let Kbe the samllest term of the normal closure series of H ′ in G such that K ≰ H(possibly, K = G). Then K is normalized by H and H < KH, whence KH ⊳⊳ G;so, we may replace G by KH if necessary. Then H ′ ≤ H K < H, H K HK = G,and we may also assume H K = 1, in particular, that H is abelian. Let X be acyclic subgroup of H such that K = X G ≰ H. Then K is soluble by point (E);since H ∩ K KH. there is a subgroup A of K such that H ∩ K < A HK,and A/(H ∩ K) is abelian. But this again contradicts point (D). thus, G ∈ N 1 ,and we are done.Now, assume that G is an infinite locally finite group in S. Let x ∈ G, with|x| = p n for some prime p. If there is an infinite p-subgroup containing x, then〈x〉 is subnormal in G by point (A). Thus, let P be a maximal p-subgroup ofG containing 〈x〉 and assume that P is finite. Then P is a Sylow p-subgroupof every finitely generated subgroup that contains it. By point (A) there is afinite subnormal subgroup T of G with P ≤ T ; let N = P T . Then N ⊳ ⊳ G andtherefore N = P S for every finitely generated subgroup S of G, with T ≤ S(remember that in a finite group the smallest subnormal subgroup containing aSylow subgroup is its normal closure). Thus, N G. Hence G/C G (N) is finite.In particular C G (x) has finite index in G, and so it is not finitely generated.Point (A) then ensures that there is a subnormal subgroup U of G with x ∈U ≤ C G (x), and so 〈x〉 ⊳ ⊳ G. Thus, we have proved that every element of G ofprime–power order is contained in the Baer radical of G. It clearly follows thatG is a Baer group, and we are done.Proof of point (2). Let G be a locally (soluble by finite) group in S, and letB = B(G) be the Baer radical of G. By point (1), B ∈ N 1 , and in particular Bis soluble.If B is finitely generated, then it is polyciclic and so G/C G (B) is polycyclic–by–finite (because it is a locally (soluble by finite) subgroup of Aut(B); see, forinstance, [99], Ch. 8). If C G (B)B/B is not finite, it contains (by point (A)) asubnormal finitely generated subgroup, hence a non-trivial subnormal abeliansubgroup A/B; and this implies that A is contained in the Baer radical of G, acontradiction. Thus, C G (B)B/B is finite, and consequently, G/B is polycyclicby finite. We conclude that G itself is polycyclic by finite. Conversely, a polycyclicby finite group certainly belongs to S as it satisfies Max.We are left with the case in which the Baer radical B = B(G) is not finitelygenerated, and B ≠ G.Let g ∈ G \ B, and let N be a normal subgroup of G contained in B.Suppose, by contradiction, that 〈N, g〉 = N〈g〉 is not finitely generated. Then,by (A), there exists a finitely genarated X with 〈g〉 ≤ X ≤ N〈g〉 and X ⊳ ⊳ G;in particular, N〈g〉 is subnormal in G, and there exists a smallest n ∈ N, suchthat [N, n 〈g〉]〈g〉 (the n-th term of the normal closure series of 〈g〉 in N〈g〉) isfinitely generated. Since we are assuming that N〈g〉 is not finitely generated,we must have n ≥ 1. We write U = [N, n 〈g〉], and consider, D = [N, n−1 〈g〉]〈g〉.Then, D/U is soluble and, by choice of n, it is not finitely generated; also, gU ∈

7.2. GROUPS WITH MANY SUBNORMAL SUBGROUPS 135Z(D/U). Now, U/U ′ is a normal abelian subgroup of the soluble group D/U ′ ,and is finitely generated as a Z〈g〉-module. We may then apply Lemma 7.13:since D/U is not finitely generated, we obtain that the centralizer of gU ′ in D/U ′contains a non-finitely generated subgroup. By observation (A), this impliesthat 〈gU ′ 〉 ⊳ ⊳ D/U ′ . In particular, U〈g〉/U ′ is nilpotent; since it is also finitelygenerated, it follows that U/U ′ is finitely generated. But U is a Baer group,and so U is a finitely generated nilpotent group. As U〈g〉/U ′ is also nilpotent,P. Hall’s nilpotency criterion (Theorem 1.54) yield that U〈g〉 is nilpotent. Thismeans that 〈g〉⊳⊳ U〈g〉 = [N, n 〈g〉]〈g〉, and so in 〈g〉 is subnormal in N〈g〉, whichin turn is subnormal in G. Therefore 〈g〉⊳⊳ G, and the contradiction g ∈ B. Thelast assertion in the statement of the Theorem is thus extablished.Now, let g ∈ G and suppose that g n ∈ B for some n ≥ 1. If g ∉ B, then, bywhat we have just proved B〈g〉 is finitely generated, hence B, which has finiteindex in it, is finitely generated, which is against our assumptions. This provesthat G/B is torsion free.We now prove that B is nilpotent. Fix an element g ∈ G\B, and let T denotethe torsion subgroup of B. Then T 〈g〉 is finitely generated by what we proved;and since T is soluble, it easily follows that T has finite exponent. Therefore,T is nilpotent by Theorem 5.22, and B/C B (T ) is periodic by Lemma 1.16.Moreover W = T C B (T ) is nilpotent and so, by Lemma 6.6, there is a k ≥ 1such that, writing N = W k , N ∩ T = 1. Since N is a characteristic subgroupof B, N G. Now, B/N is peridodic and B〈g〉/N is finitely generated (beinga quotient of B〈g〉), and so the same argument used for T shows that B/N isnilpotent. Since B/T is nilpotent by Theorem 2.23 and T ∩ N = 1, we concludethat B is nilpotent.Let now C/B be the Baer radical of G/B. If C/B is finitely generated,then by what we observed at the beginning of the proof of point (2), G/Bis polycyclic by finite, and we are done. Thus suppose, by contradiction, thatC/B is not finitely genertaed. Then, by Theorem 1.90, C/B admits an abeliansubgroup A/B which is not finitely generated. Let g ∈ A \ B; then applicationof Lemma 7.13 to the group A/B ′ implies that the centralizer of gB ′ in A/B ′contains a non-finitely generated subgroup. It turns out that 〈gB ′ 〉 is subnormalin A/B ′ , and so that 〈B, g〉/B ′ is nilpotent. Since B is nilpotent, it follows fromP. Hall’s criterion that 〈B, g〉 is nilpotent, and in particular that 〈g〉 is subnormalin 〈B, g〉. Since 〈B, g〉 ⊳ ⊳ G, we end up with the contradiction g ∈ B.It remains to show that groups satisfying the conditions in point (2) of thestatement do belong to S. This is trivial for polycyclic by finite groups (whichsatisfy Max) and N 1 -groups. We then suppose that the group G satisfies theconditions in (iii). Then, if B is the Baer radical of G, G/B satisfies Max.Suppose, by contradictionn, that G does not belong to S, and let Z = γ c (B)be the smallest term of the lower central series of the nilpotent group B suchthat G/Z ∈ S; then, we may clearly assumeγ c+1 (B) = 1 (i.e. Z central inB). Let H = H 1 ≤ H 2 ≤ H 3 ≤ . . . be an ascending chain of non-subnormalsubgroups of G; then H ≰ B, and since G/Z belongs to S, we may supposethat ZH i ⊳ ⊳ G for every i ≥ 1. Let x ∈ H \ B; then, since 〈B, x〉 is finitelygenerated, and x ∈ ZH ⊳ ⊳ BH, we have that B/Z is finitely generated, andtherefore G/Z is polycyclic by finite and satisfies Max. Thus there exists k ≥ 1

136 CHAPTER 7. BEYOND N 1such that ZH i = ZH k for every i ≥ k. Now, 〈Z, x〉 is finitely generated, whichmeans that Z is finitely generated as a Z〈x〉-module. Since Z〈x〉 is noetherian,Z is also noetherian, i.e. it satisfies the maximal condition on Z〈x〉–submodules.This implies that there exists an index l such that Z ∩ H i = Z ∩ H l for all i ≥ l.Now let t = max{k, l}; then for every i ≥ t,H i = H i ∩ ZH t = H t (H i ∩ Z) = H t (H t ∩ Z) = H t ,and the proof is now complete. (It should be noted that, in this situation,Kurdachenko and Smith actually show that G/B must be abelian by finite;but the proof of this requires one more page, and we thus omit it).Of course, consideration of Tarski monsters shows that the conclusions ofTheorems 7.11 and 7.12 (as well as that of most of the results we will mentionin this section) do not hold without some restrictions on the class of groupsconsidered; on the other hand, the questions as to whether 7.11 and 7.12 maybe extended to larger classes (locally graded and W groups, respectively) remainopen, and seem very difficult.Weak forms of maximal and minimal conditions on non-subnormal subgroupsare considered in [58], [59]. In [30], de Giovanni and Russo show that infinitegroups with dense subnormal subgroups are N 1 (a family S of subgroups of thegroup G is dense if for every H < K ≤ G, and H not maximal in K, thereexists a S ∈ S such that H < S < K; see also Mann [69]).Groups in which non-subnormal subgroups satisfy certain embedding restrictionshave also been considered. For instance, combined results of Franciosi, deGiovanni [26], and Kurdachenko, Smith [60], yield the following.Theorem 7.14 Let G be a group in which every non-subnormal subgroup isself-normalizing.(1) If G is not periodic, then G ∈ N 1 ;(2) if G is locally nilpotent, then G ∈ N 1 ;(3) if G is locally graded and is not locally nilpotent, then G = 〈g〉⋊Q, whereg is an element of order a power of a prime p and Q a nilpotent periodicp ′ -group.The subclass of groups with all subgroups either subnormal or abnormal isdescribed by De Falco, Kurdachenko and Subbotin [23], while in [28], Franciosi,de Giovanni and Kurdachenko characterize those groups in which every (infinite)non-subnormal subgroup has a finite number of conjugates.Along another line of research (but strictly related to the previous one, asit is already evident in [92]), one imposes inner properties to non-subnormalsubgroups. We mention only a couple of relevant results. The proofs are inthese cases too long to be included.Theorem 7.15 (Smith [109] [110]) Let G be a W–group in which every subgroupis either subnormal or nilpotent. Then

7.3. THE SUBNORMAL INTERSECTION PROPERTY. 137(1) G is soluble;(2) if G is torsion-free then G is nilpotent;(3) if G is locally finite, then G admits a normal subgroup of finite index whichbelongs to N 1 .Together with Theorem 6.23 a corollary of this is an extension of 3.20;Corollary 7.16 A locally finite group in which all non-nilpotent subgroups aresubnormal is nilpotent by Černikov.We observe that locally nilpotent groups with all subgroups subnormal or nilpotentneed not belong to N 1 ; for instance, let p be a prime, and let G = A⋊〈α〉,where A ≃ C p ∞ and α the automorphism a ↦→ a p+1 (for all a ∈ A); then Gis locally nilpotent and all non-nilpotent subgroups of it contain A (and so arenormal); however, G is not even a Baer group (indeed, Smith proves that Baergroups with all subgroups nilpotent or subnormal are N 1 -groups).We mention one more result, dealing with a class of groups which may beseen as sort of opposite to that of Baer groups.Theorem 7.17 (Heineken, Kurdachenko [46]) Let G be group in which everysubgroup is either subnormal or finitely generated.(1) If G is locally finite, then either G is Černikov or G ∈ N 1;(2) if G is locally nilpotent, then either G has finite rank or G ∈ N 1 ;(3) if G is generalized radical not nilpotent, and B(G) is its Baer radical, thenG/B(G) is finitely generated and abelian–by–finite.We recall that a group is ”generalized radical” if it admits a normal ascendingseries whose factors are either locally nilpotent or locally finite, and that locallynilpotent groups with finite rank have been fully described by Mal’cev. Groupswith all subgroups either subnormal or of finite rank are studied in [61].Needless to say, many of these and similar questions may be varied by imposingconditions on the family of all subgroups that are not subnormal withdefect not exceeding a prescribed bound d ≥ 1; aiming in this case at obtainingresults that resemble Roseblade’s Theorem. This aspect is often considered inthe same articles that treat the unbounded case, and we will not say more aboutit, leaving the interested reader to check the original papers.7.3 The subnormal intersection property.A group G is said to satisfy the subnormal intersection property (abbreviateds.i.p.) if the intersection of any family os subnormal subgroups of G is subnormal.The class of all groups satisfying s.i.p. is usually denote by S ∞ .Since the s.i.p. condition does not necessarily mean the occurrence of manysubnormal subgroups (for instance, every simple group has the s.i.p.), but rather

138 CHAPTER 7. BEYOND N 1it becomes effective when there are already many subnormal subgroups, presenceof the class S ∞ in this chapter may be not fully justified; however, I decidedto include a few comments on it, in view of the fact that, at least in certainspecific cases, some of the methods developed to study N 1 -groups apply withsome success to S ∞ . Before coming to this, let me remind one of the few generalresults on S ∞ available, namely a rather old theorem of D. Robinson [93] whichstates that a finitely generated soluble group G belongs to S ∞ if and only if Gis finite-by-nilpotent.Here, we are mainly interested in S ∞ -groups that are also Baer groups(clearly, every N 1 -group is of this kind). First, one proves a version of Brookestrick 1.92 for S ∞ -groups. The not difficult adaptation is left to the reader.Lemma 7.18 Let G be a group in S ∞ , and let Θ be a family of subnormalsubgroups of G such that G ∈ Θ. Then there exist a H ∈ Θ, a finitely generatedsubgroup F of H, and a positive integer d, such that every F ≤ K ≤ H, withK ∈ Θ, has defect at most d in H.With this and Roseblade’s Theorem, we may prove the following extension of4.21.Theorem 7.19 A residually soluble Baer group with the subnormal intersectionproperty is soluble.Proof. Let G be a residually soluble Baer group in S ∞ , and suppose by contradictionthat G is not soluble. By Lemma 7.18 applied to the family Θ of allsubnormal non-soluble subgroups of G, there exist H ∈ Θ, a finitely generatedsubgroup F of H, and a positive integer d, such that all non-soluble subnormalsubgroups of H containing F have defect at most d in H. Clearly, we mayraplace G by H, and assume that d is minimal for a counterexample.Since H is not soluble, H (m) is not soluble for every n ≥ 1; so, if V be afinitely generated subgroup containing F , V H (m) is not soluble. On the otherhand, V H (m) is subnormal in H, as H is a Baer group and V is finitely generated.Therefore the defect of V H (m) in H is at most d.We have d ≠ 1. In fact, if d = 1, then, by what we have just observed, for allm ≥ 1, all subgroups of H/H (m) containing F H (m) /H (m) are normal. ThereforeH (2) ≤ ⋂ m∈NF H (m) ,Hence, if F has derived length t,H (2+t) ≤ ⋂ m∈NH (m) = 1 ,thus contradicting the choice of H.Let now d ≥ 1. Then, by minimality of d, the normal closure F H of Fis soluble, and, for any m ≥ 1, all subgroups of H/F H H (m) are subnormalof defect at most d. By Roseblade’s Theorem, there is an integer k such thatH (k) ≤ F G H (m) , for all m ≥ 1. But then, if t is the derived length of F G ,H (k+t) ≤ ⋂ m∈NH (m) = 1 ,

7.4. OTHER CLASSES OF LOCALLY NILPOTENT GROUPS 139a contradiction that concludes the proof.One cannot remove from this theorem the hypothesis that G is a Baer group.In fact (see [14]) for every prime p, there exist residually soluble, non-soluble,locally finite p-groups in which every subnormal subgroup has defect at most four(whence they belong to S ∞ ). The main result that may then be proved, usingmethods directly derived from Möhres’ arguments, is an extension of Theorem5.29 (for a proof, we refer to [18]).Theorem 7.20 A periodic residually nilpotent group with the subnormal intersectionproperty is nilpotent.(It is an easy exercise to show that residually nilpotent groups with the s.i.p. arein fact Baer groups). Indeed, as for the N 1 case, the crucial step is to prove thestatement for groups of finite exponent. However, even in this case one cannotremove the assumption of residual nilpotence: in fact, contrary to the case of N 1 ,in [18] examples are given of metabelian p-groups of exponent p 2 that belong toS ∞ but are not nilpotent. To get one more example showing that the class ofBaer S ∞ -groups is much larger that N 1 , one may consider P. Hall generalizedwreath power Wr C N p (where C p is a cyclic group of order p) which is not difficultto check being a non-soluble Baer p-group satifying s.i.p. (for the details, see[18] or Volume II of [96]).However, I believe that there is still some room left for research on Baergroups in S ∞ . For instance, the following question should not be terribly difficultto answer.Question 13 Is every residually nilpotent group in S ∞ a N 1 -group?Some more questions (which I have not really meditated on, and thus mightwell be either trivial or very diffcult).Question 14 Do there exist non-soluble torsion–free Bear groups in S ∞ ?Question 15 Do there exist non-soluble Baer p-groups of finite exponent inS ∞ ?Perhaps, more could be proved for the class S ∞ of groups in which every subgroupsatisfies s.i.p. (this class still contains N 1 ). Of course, Tarski monstersbelong to S ∞ , thus some extra conditions are required also in this case.Question 16 Are locally graded p-groups in S ∞ locally finite? (the same questionis also open for S ∞ ).7.4 Other classes of locally nilpotent groupsStrongly Baer groups. We say that G is a strongly Baer group if every nilpotentsubgroup of G is subnormal. Clearly, strongly Baer groups are Baer groups.For every n ≥ 1, let D n be the dihedral group of order 2 n ; then the direct productDir n≥1 D n is a hypercentral Fitting group with all subgroups locally subnormal,but it is not a strongly Baer group.

140 CHAPTER 7. BEYOND N 1One of the difficulties in studying strongly Baer groups might well be thefact that this class, which is obviously closed by subgroups, it is not closed byquotients, as the following example shows. It also proves that strongly Baergroups need not satisfy the normalizer condition (nor in fact belong to N 2 ).Example 3.10 Let H be one of the p-groups constructed by Heineken andMohamed. Then A = H ′ is an infinite elementary abelian p-group, H/A ≃ C p ∞and no proper subgroup of H supplements A. Let K be the wreath productC p ≀ C p ∞, and write K = BC p ∞, where B is the base group. In the directproduct H × K, let W = A × B. Then (H × K)/W = HW/W × KW/W is thedirect product of two copies of C p ∞; we take G ≤ H ×K to be such that G/W isa diagonal subgroup of (H × K)/W . Let S be a nilpotent subgroup of G. Then,since G/B ≃ H, SB/B is a proper subgroup of G/B and so SW < G. HenceSW/W is finite because G/W ≃ C p ∞. Since W is elementary abelian it followsfrom Lemma 1.14 that W S is nilpotent. Also, SW G, so S G is nilpotent andthus certainly S is subnormal in G. Therefore, G is a strongly Baer group. ButG/A ≃ K = C p ≀ C p ∞ is not a strongly Baer group, and does not satisfy N 2 .Question 17 Does there exist a strongly Baer group which is not hyperabelian?Does there exists a (soluble) strongly Baer group that is not a Fitting group?It may be worth mentioning that many of the classical non-elementary constructionsof Baer groups (like McLain groups, P. Hall’s generalized wreath powersor Dark’s examples of Bear groups with trivial Fitting radical) do not provide,except in trivial cases, any strongly Baer group.Strong normalizer condition. Let us conclude with mentioning a class ofgroups which lies strictly between N 1 and the class N of all nilpotent groups.Given a subgroup H of a group G we define the series of the metanormalizersof H by setting NG 1 (H) = N G(H) and, for n ≥ 1, N n+1G(H) = N G(NG n (H)). Wesay that H is metanormal in G if NG n (H) = G for some 1 ≤ n ∈ N. It is thenclear that every metanormal subgroup is subnormal, and that in a nilpotentgroup every subgroup is metanormal. On the other hand, it is easy to see thatsubnormality does not in general imply metanormality: in the symmetric groupS 4 the subgroup H = 〈(12)(34)〉 is subnormal but not metanormal (in factN G (H) is a Sylow 2-subgroup of S 4 and is selfnormalizing). A group satisfiesthe Strong Normalizer Condition (SNC) if all of its subgroups are metanormal.A group satisfying SNC is clearly a N 1 -group, but need not be nilpotent, asgroups constructed by H. Smith in [101] show. On the other hand the groupsconstructed by Heineken and Mohamed, as observed by J. Lennox, do not satisfySNC; so SNC is a proper subclass of N 1 . Since Smith’s group are not periodicit seems reasonable to ask the following:Question 18 Is every periodic group satisfying SNC nilpotent?Question 19 Is every SNC-group hypercentral?Question 20 Is it true that a group G is a SNC-group if and only if for eachH ≤ G there exists a positive integer n such that γ n (G) ≤ N G (H)?An affermative answer to any of these three questions will imply affermativeanswers of the previous ones.

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