A subset $S$ of a group $G$ is called an Engel set if, for all $x,y\in S$,
there is a non-negative integer $n=n(x,y)$ such that $[x,\,_n y]=1$. In this
paper we are interested in finding conditions for a group generated by a finite
Engel set to be nilpotent. In particular, we focus our investigation on groups
generated by an Engel set of size two.Comment: to appear in Journal of Algebr

Abdollahi, Alireza

Brandl, Rolf

Tortora, Antonio

English

arXiv

A subset S of a group G is called an Engel set if, for all x, y ∈ S, there is a non-negative integer n = n(x, y) such that [x,_n y] = 1. In this paper we are interested in finding conditions for a group generated by a finite Engel set to be nilpotent. In particular, we focus our investigation on groups generated by an Engel set of size two

Alireza Abdollahi

Rolf Brandl

TORTORA, ANTONIO

Archivio della Ricerca - Università di Salerno

Groups generated by a finite Engel setAlireza Abdollahi∗Department of Mathematics, University of Isfahan81746-73441, Isfahan, Iran;and School of Mathematics, Institute for Research in Fundamental Sciences (IPM)P.O.Box: 19395-5746, Tehran, IranEmail: a.abdollahi@math.ui.ac.irRolf BrandlMathematisches Institut, Am Hubland 1297074 Würzburg, GermanyAntonio Tortora†Dipartimento di Matematica, Università di SalernoVia Ponte don Melillo, 84084 - Fisciano (SA), ItalyE-mail: antortora@unisa.itAbstractA subset S of a group G is called an Engel set if, for all x, y ∈ S, thereis a non-negative integer n = n(x, y) such that [x, ny] = 1. In thispaper we are interested in finding conditions for a group generated bya finite Engel set to be nilpotent. In particular, we focus our investi-gation on groups generated by an Engel set of size two.2010 Mathematics Subject Classification: 20F45; 20F19Keywords: Engel set, nilpotent group1 IntroductionA subset S of a group G is called an Engel set if, for all x, y ∈ S, there is anon-negative integer n = n(x, y) such that [x, ny] = 1. It is known that, fora group G satisfying Max-ab, a normal subset S ⊆ G is an Engel set if andonly if it is contained in the Fitting subgroup of G (see [7], Theorem 7.23; see∗The first author’s research was in part supported by a grant from IPM (No. 90050219)and by the Center of Excellence for Mathematics, University of Isfahan.†The third author is grateful to Professor A. Rhemtulla for interesting discussionsand useful suggestions. He also wishes to thank the Department of Mathematical andStatistical Sciences at the University of Alberta for its fine hospitality while part of thiswork was being carried out.1also [1]) and so, in this case, 〈S〉 is nilpotent whenever S is finite. However,a group generated by a finite Engel set is not necessarily nilpotent: Golod’sexamples show that there exist infinite non-nilpotent groups generated byan Engel set with three or more elements (see [5]). Furthermore, if S is anEngel set of size three, then an easier example of a non-nilpotent groupgenerated by S is the wreath product of the alternating group of degree 5with the cyclic group of order 3: it has a presentation of type (r, s, t) (see[3]), i.e. S = {a, b, c} where 〈a, b〉 is nilpotent of class r, 〈a, c〉 is nilpotent ofclass s and 〈b, c〉 is nilpotent of class t. All these groups are not soluble, butthe nilpotency does not hold even in the soluble case. In [3] it was shownthat every group with a presentation of type (1, 2, 2) is soluble of length atmost 3 and that there are non-nilpotent groups of this type.In this paper, we first get that any nilpotent-by-abelian group generatedby a finite Engel set is nilpotent and then we focus on groups generatedby an Engel set of size two. In particular, we prove that such a group isnilpotent whenever it is abelian-by-(nilpotent of class 2). This is the bestpossible result in the soluble case. In fact, we construct by GAP (see [4]) anon-nilpotent counterexample which is abelian-by-(nilpotent of class 3). Onthe other hand, some of the counterexamples in [3], mentioned above, areabelian-by-(nilpotent of class 2) and generated by an Engel set of size three.2 Groups that are Nilpotent-by-AbelianWe start with a result that is certainly already known. It generalizes, formetabelian groups, two basic properties of commutators.Lemma 2.1. Let G be a metabelian group and x, y, z be elements of G. Forall positive integers n, we have:(i) [x−1, ny] = [x, ny]−x−1;(ii) [xy, nz] = [x, nz][x, nz, y][y, nz].Proof. Since G is metabelian, every g in G induces on G′ an endomorphism−1+g that maps u to u−1ug, and any two of these endomorphisms commute.We thus have:[x−1, ny] = ([x, y]−x−1)(−1+y)n−1= [x, y]−(−1+y)n−1x−1 = [x, ny]−x−1 .The proof of (ii) is similar.As a consequence of Lemma 2.1, we get:Lemma 2.2. If G is a metabelian group generated by an Engel set S, thenany x ∈ S is a left Engel element. In particular, G is locally nilpotent.2Proof. Take a finite subset of S, say T = {x1, . . . , xr}, and suppose [xi, nxj ] =1 for all 1 ≤ i, j ≤ r. By the previous lemma, every xi is a left n-Engelelement in G. Then (−1 +xi)n = 0. It follows that any product in the endo-morphisms −1 + xi of weight (n− 1)r + 1 is trivial. Hence 〈T 〉 is nilpotentof class at most (n− 1)r + 2. This proves that G is locally nilpotent.For a finite Engel set, we then obtain the following:Theorem 2.3. Let G be a nilpotent-by-abelian group generated by a finiteEngel set. Then G is nilpotent.Proof. If N is a normal nilpotent subgroup of G such that G/N is abelian,then G/N ′ is nilpotent by Lemma 2.2 and so G is nilpotent by a well-knownresult of P. Hall.3 Engel sets of size twoLet G = 〈x, y〉 be a group and assume that {x, y} is an Engel set. Then[x, ny] = 1 and [y, mx] = 1 for some positive integers n,m. We also say thatthe elements x and y are mutually Engel and, whenever n ≥ m, that theyare mutually n-Engel. If n = m = 2, then G is obviously nilpotent of classat most 2 and the nilpotency still holds for n = 2 and m = 3.Proposition 3.1. Let G = 〈x, y〉 be an arbitrary group such that [x, y, y] = 1and [y, x, x, x] = 1. Then G is nilpotent of class at most 3.Proof. By the Hall-Witt identity we have[[y, x], x−1, y]x[x, y−1, [y, x]]y[y, [y, x]−1, x][y,x] = 1,from which it follows[y, x, x−1, y] = 1since [x, y−1] = [x, y]−1 and [y, [y, x]−1] = [x, y, y]−1 = 1. Then [y, x, x, y] = 1and hence [y, x, x] ∈ Z(G). Now [x, y, y] = [y, x, x] = 1 modulo Z(G), soG/Z(G) is nilpotent of class ≤ 2 and G is nilpotent of class ≤ 3.However, as we will see in the next section, this is not true in general,even in the soluble case. We are therefore led to consider extra conditionsfor a group generated by an Engel set of size two to be nilpotent. In thesequel, we will turn our attention to groups which are abelian-by-(nilpotentof class 2).Let G be any abelian-by-(nilpotent of class 2) group generated by twomutually Engel elements x and y. By assumption [x, ny] = 1 and [y, nx] = 1for some n. Suppose, by way of contradiction, that G is not nilpotent. Then3G has a non-nilpotent finite image by Theorem 10.51 of [7] and so we mayassume that G is finite.Using induction on the order of the group, we may assume that G isa minimal counterexample. It follows that G contains a unique minimalnormal subgroup A such that G/A is nilpotent. As G is not nilpotent thereis a maximal subgroup H that is not normal. On the other hand G/A isnilpotent, therefore A  H (otherwise H/A  G/A implies that H  G).Thus G = AH. The group A ∩ H is normal in G and A ∩ H < A. Theminimality of A then forces A ∩H = 1.Clearly, A is an elementary abelian p-group for some prime p and H isnilpotent. Let P be the Sylow p-subgroup of H. Then AP/A  G/A andso AP is the Sylow p-subgroup of G. Since AP is nilpotent, we have that[A,AP ] < A and by the minimality of A, the normal subgroup [A,AP ] mustbe trivial. Thus [A,P ] = 1 and PG = PAH = PH = P , that is P G. ButA  P , hence P = 1 and H is a Hall p′-subgroup of G.Lemma 3.2. Every nontrivial element of Z(H) acts fixed point freely on Aby conjugation.Proof. For all z ∈ Z(H) and h ∈ H, CA(z)h = CA(z) and thus CA(z) G.As 〈z〉 cannot be normal in G, we get CA(z) = 1 by minimality of A.The next lemma shows that H is nilpotent of class 2 and that we canrestrict our attention to n = 3.Lemma 3.3. Let G = AH = 〈x, y〉 be a minimal counterexample thatis abelian-by-(nilpotent of class 2). Then A = γ3(G), [x, y, y, y] = 1 and[y, x, x, x] = 1.Proof. Of course, A ⊆ γ3(G) by minimality of A. Let q 6= p be a prime.Then any q-subgroup of γ3(G) is necessarily trivial. But G/A is a p′-group,therefore A = γ3(G) and H is nilpotent of class 2.Assuming now [x, n−1y] 6= 1, we will prove that n ≤ 3. Let y = ah wherea ∈ A, h ∈ H, and suppose n > 3. We have [x, y, y] ∈ A and n − 2 ≥ 2, sothat [x, n−2y] and [x, n−2y, y] lie in A. It follows that[x, n−2y, yp] = [x, n−2y, y]p = 1.Notice that yp = a1hp with a1 ∈ A and h = hαp for some integer α. Thus1 = [x, n−2y, yp] = [x, n−2y, a1hp] = [x, n−2y, hp]and1 = [x, n−2y, hαp] = [x, n−2y, h].But then1 = [x, n−2y, ah] = [x, n−2y, y],that is a contradiction.4We need one more preliminary lemma before proving our main result.Lemma 3.4. Let x = ah, y = bk where a, b ∈ A and h, k ∈ H. If [x, y] =[h, k], then[a, k−1] = [b, h−1], [a, h] = 1 and [b, k] = 1,with a 6= 1 and b 6= 1.Proof. We have[h, k] = [x, y] = [ah, bk] = [a, k]h[h, k][h, b]k.This implies [a, k]h[h, b]h−1kh = 1 and then [a, k]k−1= [b, h]h−1, or equiva-lently [a, k−1] = [b, h−1].As G 6= H we must have that one of a, b is nontrivial. Without loss ofgenerality, we may assume a 6= 1. Clearly, [y, x, x] ∈ A and 1 6= [y, x] ∈Z(H). Then 1 = [y, x, x, x] = [y, x, x, h] and[x, h][y,x] = [x[y,x], h] = [[y, x, x]−1x, h] = [x, h].Thus 1 = [x, h, [y, x]] = [[a, h]h, [y, x]] = [a, h, [y, x]]h, so [a, h] is fixed by[y, x]. By Lemma 3.2 it follows that [a, h] = 1. As a consequence b 6= 1,otherwise [a, k] = 1 and [a, [h, k]] = 1. Arguing as for a, we then concludethat [b, k] = 1.Theorem 3.5. Let G be any abelian-by-(nilpotent of class 2) group gener-ated by two mutually Engel elements x and y. Then G is nilpotent.Proof. Put x = ah, y = bk where a, b ∈ A and h, k ∈ H. Then [x, y] = [h, k]cwith [h, k] ∈ Z(H) and for some c ∈ A. By Lemma 3.3 we know that[x, y, y], [y, x, x] ∈ A and [x, y, y, y] = [y, x, x, x] = 1.This gives[x, y, yp] = 1 and [x, y, xp] = 1.If 〈xp, yp〉∩A 6= 1, the commutator [x, y] commutes with a nontrivial elementof A. Thus [h, k] = 1 by Lemma 3.2, and [x, y] ∈ A. Indeed G′ ≤ A and Gis nilpotent by Lemma 2.2. Therefore A ∩ 〈xp, yp〉 = 1 and we may assumeH = 〈xp, yp〉, since 〈h, k〉 ' 〈h, k〉A/A = 〈xp, yp〉A/A ' 〈xp, yp〉. It followsthat c must be trivial. Then 1 6= [x, y] = [h, k] and, by Lemma 3.4, we have[a, k−1] = [b, h−1] and [a, h] = 1,with a 6= 1.Now, the Hall-Witt identity[a, k−1, h]k[k, h−1, a]h[h, a−1, k]a = 15implies[a, k−1, h]k = [k, h−1, a]−h.But [k, h−1, a] commutes with h, so [[a, k−1], h] = [[b, h−1], h] commutes withhk−1. Then [b, h, h]h−1= [b, h−1, h]−1 commutes with hk−1, in particular[b, h, h] commutes with hk−1h = hk−1. Hence [b, h, h] ∈ CA(hk−1).Let B = CA(hk−1) and K = 〈h, hk−1〉A. Then B K because [h−1, k] ∈Z(H). If q is the order of h, we also have B = [b, hq]B = [b, h]qB. However,the order of [b, h] is coprime with q, thus [b, h] ∈ B and [a, k−1] = [b, h−1] ∈B. So [a, k−1, hk−1] = 1 and [k, a, h] = 1. Finally, from[a, k, h]k−1[k−1, h−1, a]h[h, a−1, k−1]a = 1,it follows [k, h, a] = 1 which contradicts Lemma 3.2.When x and y are mutually 3-Engel elements, we get thanks to GAPthat the group G in Theorem 3.5 is nilpotent of class at most 8. In fact,using the ANU Nilpotent Quotient package of W. Nickel (see [6]), wecan construct the largest nilpotent quotient of G which is isomorphic to G.Also notice that the theorem above can be extended to a group gener-ated by more than two mutually Engel elements, provided that none of thegenerators has order divisible by 2 or 3.Corollary 3.6. Let S be a finite Engel set and assume that G = 〈S〉 isabelian-by-(nilpotent of class 2). If every element in S has order that is notdivisible by 2 or 3, then G is nilpotent.Proof. For all x, y ∈ S, the subgroup 〈x, y〉 is nilpotent by Theorem 3.5.Thus the claim follows by Proposition 1 of [3].Using Theorem 3.5, we now present a criterion for nilpotency of a finitesoluble group depending on information on its Sylow subgroups.Corollary 3.7. Let G = 〈x, y〉 be a finite soluble group with x and y mutu-ally Engel elements. If all Sylow subgroups of G are nilpotent of class ≤ 2,then G is nilpotent.Proof. Let G be a counterexample of least possible order and let N bea minimal normal subgroup of G. Then G/N is nilpotent by minimality.Moreover, all Sylow subgroups of G/N are nilpotent of class ≤ 2, so thatG/N is nilpotent of class ≤ 2. On the other hand N is abelian, because G issoluble. Hence G is abelian-by-(nilpotent of class 2) and thus nilpotent byTheorem 3.5: a contradiction.64 ExamplesOur first example shows that, for any positive integer n, there exists a groupgenerated by two mutually n-Engel elements which are not (n − 1)-Engel.This is the dihedral group of order 2n+1.Example 4.1. Let consider G = 〈x, y |x2 = y2 = 1, (xy)2n = 1〉. If z = xy,then [x, y] = z2 and zx = zy = z−1. For any k ≥ 1, we get by induction[x, ky] = z−(−2)k and [y, kx] = z(−2)k . Therefore [x, n−1y], [y, n−1x] 6= 1whereas [x, ny] = [y, nx] = 1. Thus x and y are mutually n-Engel elements.Furthermore, we have G = 〈y, z〉 and [y, 2z] = [z, ny] = 1, so even y and zare mutually n-Engel elements. The following is an example obtained by GAP of a non-nilpotent groupG generated by two mutually 3-Engel elements, for which γ4(G) is abelian.Example 4.2. Let W = S3wrZ4 be the wreath product of the symmetricgroup of degree 3 with the cyclic group of order 4. Thus, |W | = 2634. Wehave W = QnN , where N is an elementary abelian group of order 34 andQ ' Z2wrZ4. Moreover, Q is nilpotent of class 4. With the notation of GAP,let ele:=Elements(W ), x := ele[4] and y := ele[228]. Then o(x) = o(y) = 4and [x, 3y] = [y, 3x] = 1. As o(xy−1) = 6, the subgroup G = 〈x, y〉 of W isnot nilpotent. Finally, one can check that G = S nN where S is a group oforder 25 which is nilpotent of class 3.For completeness reasons, we point out that W = 〈x, y′〉 with y′ :=ele[509] of order 6 and [x, 3y′] = [y′, 4x] = 1. Hence, W is a generated bytwo mutually 4-Engel elements and is not nilpotent. Notice that some more non-nilpotent groups generated by two mutuallyn-Engel elements can be found in the literature. For instance, Corollary 0.2of [2] says that, for n ≥ 26, the group G(n) = 〈x, y | [x, ny] = [y, nx] = 1〉is not nilpotent. We can improve upon this. In fact, we show below thatG(4) is not soluble, because it has a quotient isomorphic to the symmetricgroup S8.Example 4.3. Let S8 be the symmetric group of degree 8, and let x =(1, 2, 3, 4)(5, 6)(7, 8) and y = (1, 3)(2, 5)(4, 7, 6, 8). Put xn = [x, ny] andyn = [y, nx], for any n ≥ 0 (so x0 = x, y0 = y). We then have:x1 = (1, 6)(2, 7)(3, 8)(4, 5) y1 = (1, 6)(2, 7)(3, 8)(4, 5)x2 = (1, 5)(4, 6) y2 = (2, 4)(5, 7)x3 = (1, 5)(2, 3)(4, 6)(7, 8) y3 = (1, 3)(2, 4)(5, 7)(6, 8)x4 = (1) y4 = (1) .In particular, [x, 4y] = [y, 4x] = 1. However x and y are of order 4, butxy = (1, 5, 8, 6, 2)(3, 7, 4) is of order 15. The subgroup G = 〈x, y〉 is thusnon-nilpotent. Using GAP, it is easy to see that |G| = 8!, so G = S8. 7We now discuss the situation of Example 4.3. Clearly, if the pair (x, y) ∈G×G satisfies the condition[x, 4y] = [y, 4x] = 1, (∗)then all conjugates (xg, yg), for all g ∈ G, satisfy the analogous property.Therefore it is sensible to consider classes under conjugation.It turns out by GAP that the only pairs (x, y) ∈ G×G satisfying (∗), thatgenerate a non-nilpotent subgroup of G, have both x and y with cycle struc-ture of type (4)(2)(2) and, in addition, x, y necessarily generate the wholegroup G. Without loss of generality, we may assume x = (1, 2, 3, 4)(5, 6)(7, 8). For this x, we calculated all solutions y ∈ G of (∗). We ended up withprecisely 64 solutions. Of course, the group CG(x) of order 32 acts on thepairs of solutions. The stabilizer of this action is CG(x) ∩CG(y) = Z(G) = 1,so that we obtain two essentially distinct solutions.Other examples? Suppose that in some finite group we can find Sylowp-subgroups P,Q and elements x ∈ P, y ∈ Q such that [x, y] ∈ P ∩ Q. Letc be the nilpotency class of P . Thus, [x, c+1y] = [y, c+1x] = 1. If xy is nota p-element, then 〈x, y〉 is non-nilpotent. The groups in Examples 4.2 and4.3 are of this form for p = 2. It would be very interesting to find analogousexamples for all odd primes p.References[1] A. Abdollahi, Engel graph associated with a group, J. Algebra 318(2007), 680–691.[2] D. J. Collins and A. Juhász, Engel groups III, Israel J. Math. 174(2009), 73-91.[3] G. Endimioni and G. Traustason, Groups that are pairwise nilpotent,Comm. Algebra 36 (2008), 4413–4435.[4] The GAP Group, GAP – Groups, Algorithms, and Programming, Version4.4.12, 2008; http://www.gap-system.org.[5] E. S. Golod, Some problems of Burnside type, 1968 Proc. Int. Congr.Math. (Moscow 1966), 284–289; Amer. Math. Soc. Transl. (2) 84(1969), 83-88.[6] W. Nickel, Computation of nilpotent Engel groups, J. Austral. Math.Soc. Ser. A 67 (1999), 214–222.[7] D. J. S. Robinson, Finiteness conditions and generalized soluble groups,Part 1 and 2, Springer-Verlag, Berlin, 1972.8

Groups generated by a finite Engel set

AbstractA subset S of a group G is called an Engel set if, for all x,y∈S, there is a non-negative integer n=n(x,y) such that [x,yn]=1. In this paper we are interested in finding conditions for a group generated by a finite Engel set to be nilpotent. In particular, we focus our investigation on groups generated by an Engel set of size two

Elsevier - Publisher Connector 

Journal of Algebra 347 (2011) 53–59Contents lists available at SciVerse ScienceDirect
Journal of Algebra
www.elsevier.com/locate/jalgebra
Groups generated by a ﬁnite Engel set
Alireza Abdollahi a,b,1, Rolf Brandl c, Antonio Tortora d,∗,2
a Department of Mathematics, University of Isfahan, 81746-73441, Isfahan, Iran
b School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran
c Mathematisches Institut, Am Hubland 12, 97074 Würzburg, Germany
d Dipartimento di Matematica, Università di Salerno, Via Ponte don Melillo, 84084, Fisciano (SA), Italy
a r t i c l e i n f o a b s t r a c t
Article history:
Received 4 March 2011
Available online 5 October 2011
Communicated by E.I. Khukhro
MSC:
20F45
20F19
Keywords:
Engel set
Nilpotent group
A subset S of a group G is called an Engel set if, for all x, y ∈ S ,
there is a non-negative integer n = n(x, y) such that [x,n y] = 1.
In this paper we are interested in ﬁnding conditions for a group
generated by a ﬁnite Engel set to be nilpotent. In particular, we
focus our investigation on groups generated by an Engel set of size
two.
© 2011 Elsevier Inc. All rights reserved.
1. Introduction
A subset S of a group G is called an Engel set if, for all x, y ∈ S , there is a non-negative integer
n = n(x, y) such that [x, n y] = 1. It is known that, for a group G satisfying Max-ab, a normal subset
S ⊆ G is an Engel set if and only if it is contained in the Fitting subgroup of G (see [7, Theorem 7.23];
see also [1]) and so, in this case, 〈S〉 is nilpotent whenever S is ﬁnite. However, a group generated
by a ﬁnite Engel set is not necessarily nilpotent: Golod’s examples show that there exist inﬁnite
non-nilpotent groups generated by an Engel set with three or more elements (see [5]). Furthermore,
if S is an Engel set of size three, then an easier example of a non-nilpotent group generated by
* Corresponding author.
E-mail addresses: a.abdollahi@math.ui.ac.ir (A. Abdollahi), antortora@unisa.it (A. Tortora).
1 The author’s research was in part supported by a grant from IPM (No. 90050219) and by the Center of Excellence for
Mathematics, University of Isfahan.
2 The author is grateful to Professor A. Rhemtulla for interesting discussions and useful suggestions. He also wishes to thank
the Department of Mathematical and Statistical Sciences at the University of Alberta for its ﬁne hospitality while part of this
work was being carried out.0021-8693/$ – see front matter © 2011 Elsevier Inc. All rights reserved.
doi:10.1016/j.jalgebra.2011.09.013
54 A. Abdollahi et al. / Journal of Algebra 347 (2011) 53–59S is the wreath product of the alternating group of degree 5 with the cyclic group of order 3: it
has a presentation of type (r, s, t) (see [3]), i.e. S = {a,b, c} where 〈a,b〉 is nilpotent of class r, 〈a, c〉 is
nilpotent of class s and 〈b, c〉 is nilpotent of class t . All these groups are not soluble, but the nilpotency
does not hold even in the soluble case. In [3] it was shown that every group with a presentation of
type (1,2,2) is soluble of length at most 3 and that there are non-nilpotent groups of this type.
In this paper, we ﬁrst get that any nilpotent-by-abelian group generated by a ﬁnite Engel set is
nilpotent and then we focus on groups generated by an Engel set of size two. In particular, we prove
that such a group is nilpotent whenever it is abelian-by-(nilpotent of class 2). This is the best possible
result in the soluble case. In fact, we construct by GAP (see [4]) a non-nilpotent counterexample
which is abelian-by-(nilpotent of class 3). On the other hand, some of the counterexamples in [3],
mentioned above, are abelian-by-(nilpotent of class 2) and generated by an Engel set of size three.
2. Groups that are nilpotent-by-abelian
We start with a result that is certainly already known. It generalizes, for metabelian groups, two
basic properties of commutators.
Lemma 2.1. Let G be a metabelian group and x, y, z be elements of G. For all positive integers n, we have
(i) [x−1,n y] = [x, n y]−x−1 ;
(ii) [xy, nz] = [x, nz][x, nz, y][y, nz].
Proof. Since G is metabelian, every g in G induces on G ′ an endomorphism −1 + g that maps u to
u−1ug , and any two of these endomorphisms commute. We thus have
[
x−1,n y
] = ([x, y]−x−1)(−1+y)n−1 = [x, y]−(−1+y)n−1x−1 = [x,n y]−x−1 .
The proof of (ii) is similar. 
As a consequence of Lemma 2.1, we get:
Lemma 2.2. If G is a metabelian group generated by an Engel set S, then any x ∈ S is a left Engel element. In
particular, G is locally nilpotent.
Proof. Take a ﬁnite subset of S , say T = {x1, . . . , xr}, and suppose [xi,nx j] = 1 for all 1 i, j  r. By
the previous lemma, every xi is a left n-Engel element in G . Then (−1 + xi)n = 0. It follows that any
product in the endomorphisms −1+ xi of weight (n−1)r+1 is trivial. Hence 〈T 〉 is nilpotent of class
at most (n − 1)r + 2. This proves that G is locally nilpotent. 
For a ﬁnite Engel set, we then obtain the following:
Theorem 2.3. Let G be a nilpotent-by-abelian group generated by a ﬁnite Engel set. Then G is nilpotent.
Proof. If N is a normal nilpotent subgroup of G such that G/N is abelian, then G/N ′ is nilpotent by
Lemma 2.2 and so G is nilpotent by a well-known result of P. Hall. 
3. Engel sets of size two
Let G = 〈x, y〉 be a group and assume that {x, y} is an Engel set. Then [x, n y] = 1 and [y,mx] = 1
for some positive integers n, m. We also say that the elements x and y are mutually Engel and, when-
ever n  m, that they are mutually n-Engel. If n = m = 2, then G is obviously nilpotent of class at
most 2 and the nilpotency still holds for n = 2 and m = 3.
A. Abdollahi et al. / Journal of Algebra 347 (2011) 53–59 55Proposition 3.1. Let G = 〈x, y〉 be an arbitrary group such that [x, y, y] = 1 and [y, x, x, x] = 1. Then G is
nilpotent of class at most 3.
Proof. By the Hall–Witt identity we have
[[y, x], x−1, y]x[x, y−1, [y, x]]y[y, [y, x]−1, x][y,x] = 1,
from which it follows
[
y, x, x−1, y
] = 1
since [x, y−1] = [x, y]−1 and [y, [y, x]−1] = [x, y, y]−1 = 1. Then [y, x, x, y] = 1 and hence [y, x, x] ∈
Z(G). Now [x, y, y] = [y, x, x] = 1 modulo Z(G), so G/Z(G) is nilpotent of class  2 and G is nilpotent
of class  3. 
However, as we will see in the next section, this is not true in general, even in the soluble case.
We are therefore led to consider extra conditions for a group generated by an Engel set of size two
to be nilpotent. In the sequel, we will turn our attention to groups which are abelian-by-(nilpotent of
class 2).
Let G be any abelian-by-(nilpotent of class 2) group generated by two mutually Engel elements x
and y. By assumption [x, n y] = 1 and [y, nx] = 1 for some n. Suppose, by way of contradiction, that
G is not nilpotent. Then G has a non-nilpotent ﬁnite image by Theorem 10.51 of [7] and so we may
assume that G is ﬁnite.
Using induction on the order of the group, we may assume that G is a minimal counterexample.
It follows that G contains a unique minimal normal subgroup A such that G/A is nilpotent. As G is
not nilpotent there is a maximal subgroup H that is not normal. On the other hand G/A is nilpotent,
therefore A  H (otherwise H/AG/A implies that HG). Thus G = AH . The group A∩H is normal
in G and A ∩ H < A. The minimality of A then forces A ∩ H = 1.
Clearly, A is an elementary abelian p-group for some prime p and H is nilpotent. Let P be the
Sylow p-subgroup of H . Then AP/A  G/A and so AP is the Sylow p-subgroup of G . Since AP is
nilpotent, we have that [A, AP ] < A and by the minimality of A, the normal subgroup [A, AP ] must
be trivial. Thus [A, P ] = 1 and PG = P AH = P H = P , that is P  G . But A  P , hence P = 1 and H is
a Hall p′-subgroup of G .
Lemma 3.2. Every nontrivial element of Z(H) acts ﬁxed point freely on A by conjugation.
Proof. For all z ∈ Z(H) and h ∈ H , CA(z)h = CA(z) and thus CA(z) G . As 〈z〉 cannot be normal in G ,
we get CA(z) = 1 by minimality of A. 
The next lemma shows that H is nilpotent of class 2 and that we can restrict our attention to
n = 3.
Lemma 3.3. Let G = AH = 〈x, y〉 be a minimal counterexample that is abelian-by-(nilpotent of class 2). Then
A = γ3(G), [x, y, y, y] = 1 and [y, x, x, x] = 1.
Proof. Of course, A ⊆ γ3(G) by minimality of A. Let q 	= p be a prime. Then any q-subgroup of γ3(G)
is necessarily trivial. But G/A is a p′-group, therefore A = γ3(G) and H is nilpotent of class 2.
Assuming now [x, n−1 y] 	= 1, we will prove that n 3. Let y = ah where a ∈ A, h ∈ H , and suppose
n > 3. We have [x, y, y] ∈ A and n − 2 2, so that [x, n−2 y] and [x, n−2 y, y] lie in A. It follows that
[
x,n−2 y, yp
] = [x,n−2 y, y]p = 1.
56 A. Abdollahi et al. / Journal of Algebra 347 (2011) 53–59Notice that yp = a1hp with a1 ∈ A and h = hαp for some integer α. Thus
1 = [x,n−2 y, yp
] = [x,n−2 y,a1hp
] = [x,n−2 y,hp
]
and
1 = [x,n−2 y,hαp
] = [x,n−2 y,h].
But then
1 = [x,n−2 y,ah] = [x,n−2 y, y],
that is a contradiction. 
We need one more preliminary lemma before proving our main result.
Lemma 3.4. Let x = ah, y = bk where a,b ∈ A and h,k ∈ H. If [x, y] = [h,k], then
[
a,k−1
] = [b,h−1], [a,h] = 1 and [b,k] = 1,
with a 	= 1 and b 	= 1.
Proof. We have
[h,k] = [x, y] = [ah,bk] = [a,k]h[h,k][h,b]k.
This implies [a,k]h[h,b]h−1kh = 1 and then [a,k]k−1 = [b,h]h−1 , or equivalently [a,k−1] = [b,h−1].
As G 	= H we must have that one of a, b is nontrivial. Without loss of generality, we may assume
a 	= 1. Clearly, [y, x, x] ∈ A and 1 	= [y, x] ∈ Z(H). Then 1 = [y, x, x, x] = [y, x, x,h] and
[x,h][y,x] = [x[y,x],h] = [[y, x, x]−1x,h] = [x,h].
Thus 1 = [x,h, [y, x]] = [[a,h]h, [y, x]] = [a,h, [y, x]]h , so [a,h] is ﬁxed by [y, x]. By Lemma 3.2 it
follows that [a,h] = 1. As a consequence b 	= 1, otherwise [a,k] = 1 and [a, [h,k]] = 1. Arguing as
for a, we then conclude that [b,k] = 1. 
Theorem 3.5. Let G be any abelian-by-(nilpotent of class 2) group generated by two mutually Engel elements
x and y. Then G is nilpotent.
Proof. Put x = ah, y = bk where a,b ∈ A and h,k ∈ H . Then [x, y] = [h,k]c with [h,k] ∈ Z(H) and for
some c ∈ A. By Lemma 3.3 we know that
[x, y, y], [y, x, x] ∈ A and [x, y, y, y] = [y, x, x, x] = 1.
This gives
[
x, y, yp
] = 1 and [x, y, xp] = 1.
If 〈xp, yp〉∩ A 	= 1, the commutator [x, y] commutes with a nontrivial element of A. Thus [h,k] = 1 by
Lemma 3.2, and [x, y] ∈ A. Indeed G ′  A and G is nilpotent by Lemma 2.2. Therefore A∩〈xp, yp〉 = 1
A. Abdollahi et al. / Journal of Algebra 347 (2011) 53–59 57and we may assume H = 〈xp, yp〉, since 〈h,k〉 
 〈h,k〉A/A = 〈xp, yp〉A/A 
 〈xp, yp〉. It follows that c
must be trivial. Then 1 	= [x, y] = [h,k] and, by Lemma 3.4, we have
[
a,k−1
] = [b,h−1] and [a,h] = 1,
with a 	= 1.
Now, the Hall–Witt identity
[
a,k−1,h
]k[
k,h−1,a
]h[
h,a−1,k
]a = 1
implies
[
a,k−1,h
]k = [k,h−1,a]−h.
But [k,h−1,a] commutes with h, so [[a,k−1],h] = [[b,h−1],h] commutes with hk−1 . Then [b,h,h]h−1 =
[b,h−1,h]−1 commutes with hk−1 , in particular [b,h,h] commutes with hk−1h = hk−1 . Hence [b,h,h] ∈
CA(hk
−1
).
Let B = CA(hk−1 ) and K = 〈h,hk−1 〉A. Then B  K because [h−1,k] ∈ Z(H). If q is the order of h,
we also have B = [b,hq]B = [b,h]qB . However, the order of [b,h] is coprime with q, thus [b,h] ∈ B
and [a,k−1] = [b,h−1] ∈ B . So [a,k−1,hk−1 ] = 1 and [k,a,h] = 1. Finally, from
[a,k,h]k−1[k−1,h−1,a]h[h,a−1,k−1]a = 1,
it follows [k,h,a] = 1 which contradicts Lemma 3.2. 
When x and y are mutually 3-Engel elements, we get thanks to GAP that the group G in The-
orem 3.5 is nilpotent of class at most 8. In fact, using the ANU Nilpotent Quotient package of
W. Nickel (see [6]), we can construct the largest nilpotent quotient of G which is isomorphic to G .
Also notice that the theorem above can be extended to a group generated by more than two
mutually Engel elements, provided that none of the generators has order divisible by 2 or 3.
Corollary 3.6. Let S be a ﬁnite Engel set and assume that G = 〈S〉 is abelian-by-(nilpotent of class 2). If every
element in S has order that is not divisible by 2 or 3, then G is nilpotent.
Proof. For all x, y ∈ S , the subgroup 〈x, y〉 is nilpotent by Theorem 3.5. Thus the claim follows by
Proposition 1 of [3]. 
Using Theorem 3.5, we now present a criterion for nilpotency of a ﬁnite soluble group depending
on information on its Sylow subgroups.
Corollary 3.7. Let G = 〈x, y〉 be a ﬁnite soluble group with x and y mutually Engel elements. If all Sylow
subgroups of G are nilpotent of class  2, then G is nilpotent.
Proof. Let G be a counterexample of least possible order and let N be a minimal normal subgroup
of G . Then G/N is nilpotent by minimality. Moreover, all Sylow subgroups of G/N are nilpotent of
class  2, so that G/N is nilpotent of class  2. On the other hand N is abelian, because G is soluble.
Hence G is abelian-by-(nilpotent of class 2) and thus nilpotent by Theorem 3.5: a contradiction. 
58 A. Abdollahi et al. / Journal of Algebra 347 (2011) 53–594. Examples
Our ﬁrst example shows that, for any positive integer n, there exists a group generated by two
mutually n-Engel elements which are not (n − 1)-Engel. This is the dihedral group of order 2n+1.
Example 4.1. Let us consider G = 〈x, y | x2 = y2 = 1, (xy)2n = 1〉. If z = xy, then [x, y] = z2 and
zx = zy = z−1. For any k  1, we get by induction [x, k y] = z−(−2)k and [y, kx] = z(−2)k . Therefore
[x,n−1 y], [y, n−1x] 	= 1 whereas [x, n y] = [y, nx] = 1. Thus x and y are mutually n-Engel elements.
Furthermore, we have G = 〈y, z〉 and [y, 2z] = [z, n y] = 1, so even y and z are mutually n-Engel
elements.
The following is an example obtained by GAP of a non-nilpotent group G generated by two mu-
tually 3-Engel elements, for which γ4(G) is abelian.
Example 4.2. Let W = S3 wrZ4 be the wreath product of the symmetric group of degree 3 with the
cyclic group of order 4. Thus, |W | = 2634. We have W = Q  N , where N is an elementary abelian
group of order 34 and Q 
 Z2 wrZ4. Moreover, Q is nilpotent of class 4. With the notation of GAP,
let ele := Elements(W ), x := ele[4] and y := ele[228]. Then o(x) = o(y) = 4 and [x, 3 y] = [y, 3x] = 1.
As o(xy−1) = 6, the subgroup G = 〈x, y〉 of W is not nilpotent. Finally, one can check that G = S  N
where S is a group of order 25 which is nilpotent of class 3.
For completeness reasons, we point out that W = 〈x, y′〉 with y′ := ele[509] of order 6 and
[x, 3 y′] = [y′, 4x] = 1. Hence, W is a generated by two mutually 4-Engel elements and is not nilpotent.
Notice that some more non-nilpotent groups generated by two mutually n-Engel elements can be
found in the literature. For instance, Corollary 0.2 of [2] says that, for n 26, the group G(n) = 〈x, y |
[x,n y] = [y, nx] = 1〉 is not nilpotent. We can improve upon this. In fact, we show below that G(4) is
not soluble, because it has a quotient isomorphic to the symmetric group S8.
Example 4.3. Let S8 be the symmetric group of degree 8, and let x = (1,2,3,4)(5,6)(7,8) and y =
(1,3)(2,5)(4,7,6,8). Put xn = [x,n y] and yn = [y,nx], for any n  0 (so x0 = x, y0 = y). We then
have
x1 = (1,6)(2,7)(3,8)(4,5), y1 = (1,6)(2,7)(3,8)(4,5),
x2 = (1,5)(4,6), y2 = (2,4)(5,7),
x3 = (1,5)(2,3)(4,6)(7,8), y3 = (1,3)(2,4)(5,7)(6,8),
x4 = (1), y4 = (1).
In particular, [x, 4 y] = [y, 4x] = 1. However x and y are of order 4, but xy = (1,5,8,6,2)(3,7,4) is of
order 15. The subgroup G = 〈x, y〉 is thus non-nilpotent. Using GAP, it is easy to see that |G| = 8!, so
G = S8.
We now discuss the situation of Example 4.3. Clearly, if the pair (x, y) ∈ G × G satisﬁes the condi-
tion
[x, 4 y] = [y, 4x] = 1, (∗)
then all conjugates (xg, yg), for all g ∈ G , satisfy the analogous property. Therefore it is sensible to
consider classes under conjugation.
It turns out by GAP that the only pairs (x, y) ∈ G × G satisfying (∗), that generate a non-nilpotent
subgroup of G , have both x and y with cycle structure of type (4)(2)(2) and, in addition, x, y necessar-
ily generate the whole group G . Without loss of generality, we may assume x = (1,2,3,4)(5,6)(7,8).
A. Abdollahi et al. / Journal of Algebra 347 (2011) 53–59 59For this x, we calculated all solutions y ∈ G of (∗). We ended up with precisely 64 solutions. Of
course, the group CG (x) of order 32 acts on the pairs of solutions. The stabilizer of this action is
CG(x) ∩ CG (y) = Z(G) = 1, so that we obtain two essentially distinct solutions.
Other examples? Suppose that in some ﬁnite group we can ﬁnd Sylow p-subgroups P , Q and el-
ements x ∈ P , y ∈ Q such that [x, y] ∈ P ∩ Q . Let c be the nilpotency class of P . Thus, [x, c+1 y] =
[y, c+1x] = 1. If xy is not a p-element, then 〈x, y〉 is non-nilpotent. The groups in Examples 4.2 and
4.3 are of this form for p = 2. It would be very interesting to ﬁnd analogous examples for all odd
primes p.
References
[1] A. Abdollahi, Engel graph associated with a group, J. Algebra 318 (2007) 680–691.
[2] D.J. Collins, A. Juhász, Engel groups III, Israel J. Math. 174 (2009) 73–91.
[3] G. Endimioni, G. Traustason, Groups that are pairwise nilpotent, Comm. Algebra 36 (2008) 4413–4435.
[4] The GAP Group, GAP – Groups, Algorithms, and Programming, Version 4.4.12, 2008, http://www.gap-system.org.
[5] E.S. Golod, Some problems of Burnside type, in: Proc. Int. Congr. Math., Moscow, 1966, in: Amer. Math. Soc. Transl. Ser. 2,
1968, pp. 284–289.
[6] W. Nickel, Computation of nilpotent Engel groups, J. Aust. Math. Soc. Ser. A 67 (1999) 214–222.
[7] D.J.S. Robinson, Finiteness Conditions and Generalized Soluble Groups, Parts 1 and 2, Springer-Verlag, Berlin, 1972.


Groups generated by a finite Engel set

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