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Publicly Available Published by De Gruyter December 21, 2019

On (𝑛 + ½)-Engel groups

  • Enrico Jabara EMAIL logo and Gunnar Traustason
From the journal Journal of Group Theory

Abstract

Let n be a positive integer. We say that a group G is an (n+12)-Engel group if it satisfies the law [x,yn,x]=1. The variety of (n+12)-Engel groups lies between the varieties of n-Engel groups and (n+1)-Engel groups. In this paper, we study these groups, and in particular, we prove that all (4+12)-Engel {2,3}-groups are locally nilpotent. We also show that if G is a (4+12)-Engel p-group, where p5 is a prime, then Gp is locally nilpotent.

1 Introduction

Let G be a group and g,hG. The commutator of g and h is the element

[g,h]=g-1h-1ghG.

We recursively define [g,hn], where n is a positive integer, as [g,h1]=[g,h] and [g,hn+1]=[[g,hn],h] for n1. A subset SG is an Engel set of G if, for every g,hS, there is a positive integer k=k(g,h) such that [g,hk]=1. If k is bounded above by some positive integer n, we say that S is an n-Engel subset, and if furthermore G=S, then G is an n-Engel group. Recall that every 2-Engel group is nilpotent of class at most 3. By a classic result of Heineken [4], every 3-Engel group is locally nilpotent, and this result was later generalized to include all 4-Engel groups [3] (see also [9]).

Recall that an element aG is said to be left n-Engel if [x,an]=1 for all xG and right n-Engel if [a,xn]=1 for all xG. We denote the subset of left n-Engel elements by Ln(G) and that of right n-Engel elements by Rn(G).

Definition.

Let G be a group and n a positive integer.

  1. We say that aG is a left (n+12)-Engel element if [x,an,x]=1 for all xG.

  2. We say that aG is a right (n+12)-Engel element if [a,xn,a]=1 for all xG.

  3. We say that G is an (n+12)-Engel group if it satisfies the law [x,yn,x]=1.

We denote the subset of left (n+12)-Engel elements by Ln+12(G) and the right (n+12)-Engel elements by Rn+12(G). Thus G is an (n+12)-Engel group if and only if Ln+12(G)=G or equivalently Rn+12(G)=G. We denote the variety of m-Engel groups by m.

Remark.

It is not difficult to prove that

L1+12(G)=R2(G)andR1+12(G)=L2(G).

Thus, in particular, 1+12=2.

Lemma 1.1.

Let G be a group and n a positive integer. We have

Ln(G)Ln+12(G)Ln+1(G).

In particular, EnEn+12En+1.

Proof.

That Ln(G)Ln+12(G) is obvious. To see that Ln+12(G)Ln+1(G), let aLn+12(G). Then, for any xG, we have

1=[ax,an,ax]=[x,an,x][x,an+1]x=[x,an+1]x.

Thus [x,an+1]=1 and aLn+1(G). ∎

Our main results on (n+12)-Engel groups are the following.

Theorem B.

Let G be a (4+12)-Engel {2,3}-group. Then G is locally nilpotent.

Theorem C.

Let G be a (4+12)-Engel p-group, where p is a prime and p5. Then Gp is locally nilpotent.

A major ingredient to the proofs is a result on Engel sets that is also of independent interest. Let en(x,y) be the n-Engel word [x,yn].

Theorem A.

Let R=a,b be the largest 2-generator group satisfying the relations e3(a,b)=e3(b,a)=e3(a-1,b-1)=e3(b-1,a-1)=1. Then R is nilpotent of class 4.

We will see later that these relations imply that S={a,b,a-1,b-1} is a 3-Engel subset of R.

2 Proof of Theorem A

Consider the n-Engel word en(x,y)=[x,yn]=1. As we will focus in particular on the 3-Engel word, we will often use e(x,y) instead of e3(x,y).

Lemma 2.1.

Suppose G is a group with elements a,b, where

e(a,b)=e(a-1,b-1)=1.

Then b,ba=b,[a,b] is nilpotent of class at most 2.

Proof.

From the equations

1=[a,b,b,b]=[b-ab,b,b]=[b-a,b,b]b,
1=[a-1,b-1,b-1,b-1]=[ba-1,b-1,b-1]b-1,

we see that 1=[b-a,b,b] and 1=[ba-1,b-1,b-1]a=[b,b-a,b-a]. Thus we have that b,ba is nilpotent of class at most 2. ∎

Lemma 2.2.

Let G be a group with elements a,b, where

e(a,b)=e(a-1,b-1)=1.

Then [aϵ,bϵ1,bϵ2,bϵ3]=1 for all ϵ,ϵ1,ϵ2,ϵ3{1,-1}.

Proof.

By symmetry, it suffices to deal with the case when ϵ=1. As

[a,bϵ1,bϵ2,b-1]=[a,bϵ1,bϵ2,b]-b-1,

we can also assume that ϵ3=1. Then, from

[a,bϵ1,b-1,b]=[[a,bϵ1,b]-1,b]b-1=[a,bϵ1,b,b]-[a,bϵ1,b]-1b-1,

we can also assume, without loss of generality, that ϵ2=1. We are thus only left with showing that [a,b-1,b,b]=1, but this follows from Lemma 2.1 and the fact that [a,b-1,b,b]=[bab-1,b,b]γ3(b,ba). ∎

Remark.

It follows from Lemma 2.2 that if

e(a,b)=e(a-1,b-1)=e(b,a)=e(b-1,a-1)=1,

then S={a,b,a-1,b-1} is a 3-Engel subset.

Lemma 2.3.

Let G be a group with elements a,b satisfying

e(a,b)=e(a-1,b-1)=1.

Then [[a,b,b]-1,a] commutes with ba.

Proof.

From the Hall–Witt identity and Lemma 2.1, we have

1=[[a,b,b],a-1,b]a[a,b-1,[a,b,b]]b[b,[a,b,b]-1,a][a,b,b]
=[a,b,b,a-1,b]a=[[a,b,b,a]-1,ba].

It follows that [a,b,b,a] commutes with ba, and thus, using Lemma 2.1 again, [[a,b,b]-1,a]=[a,b,b,a]-[a,b,b]-1 commutes with ba as well. ∎

Proof of Theorem A.

Let R=a,b be the largest group satisfying the relations

e(a,b)=e(b,a)=e(a-1,b-1)=e(b-1,a-1)=1.

By the remark after Lemma 2.2, we know that S={a,b,a-1,b-1} is a 3-Engel subset of R.

In order to show that R is nilpotent of class at most 4, we need to show that a,bZ4(R). This is equivalent to showing that [b,a]Z3(R). As a,[a,b] and b,[a,b] are nilpotent of class at most 2, we see that [a,b,a]=[b,a,a]-1 and [b,a,b]=[a,b,b]-1. In order to show that [b,a]Z3(R), we need to show that [b,a,a] and [b,a,b]=[a,b,b]-1 are in Z2(R). As [b,a,a,a]=[a,b,b,b]=1, it suffices to show that [[b,a,a]-1,b],[[a,b,b]-1,a]Z(R). In the following calculations, we again use the fact that b,[a,b] and a,[a,b] are nilpotent of class at most 2. We have

[b,a,a][b,a,a,b]=[b,a,a]b=[[b,a]b,ab]=[[b,a,b][b,a],a[a,b]]=[[b,a,b][b,a],a][a,b]=[b,a,b,a][b,a][a,b][b,a,a][a,b]=[[a,b,b]-1,a][b,a,a].

Thus

(2.1)[[a,b,b]-1,a]=[b,a,a,b][b,a,a]-1=[[b,a,a]-1,b]-1.

From (2.1), we thus see that it suffices to show that [[a,b,b]-1,a]Z(R), and in fact, it suffices to show that [[a,b,b]-1,a] commutes with b as then, by symmetry, the RHS of (2.1) commutes with a, and thus [[a,b,b]-1,a] commutes then with a as well.

From Lemma 2.2, we know that

e(aα,bβ)=e(bβ,aα)=1for allα,β{1,-1}.

In particular, equation (2.1) holds if we replace a by a-1 or b by b-1. Calculating in the group a,ab that is nilpotent of class at most 2, we see that [b,a-1,a-1]=[ab,a-1]=[a-b,a]=[b,a,a]. From this and (2.1), it follows that [[a,b,b]-1,a] is invariant under replacing a by a-1. Notice also that

[a-1,b,b]=[b-a-1,b]=[b-1,ba]a-1=[b-a,b]-a-1=[a,b,b]-a-1.

Thus

[[a,b,b]-1,a]=[[a-1,b,b]-1,a-1]=[[a,b,b],a-1]a-1=[a,b,b,a]-a-2=[[a,b,b]-1,a][a,b,b]a-2.

But, from Lemma 2.3 and (2.1), we know that [[a,b,b]-1,a]=[[b,a,a]-1,b]-1 commutes with ba and ab. Replacing a by a-1 for the RHS, we see that the common element also commutes with ba-1. Likewise, it commutes with ab-1. As

[a,b,b]a-2=[b,a]b-1[a,b]ba-2=a-bab-1a-1aba-2b-1b=a-bb-a-1aba-2b-1b,

it follows that [[a,b,b]-1,a] commutes with b.

As R is nilpotent of class at most 4, it follows that R is metabelian, and using the nilpotent quotient algorithm nq of Nickel [8] (which is implemented in GAP [10]), one can see that the class is exactly 4. It turns out that R is torsion-free with R4. ∎

Remark.

An interesting related result [1, Proposition 3.1] states that if S={a,b} with [a,b,b]=[b,a,a,a]=1, then S is nilpotent of class at most 3.

The following examples show that the hypotheses of Theorem A cannot be weakened.

Example 1 ([1, Example 4.2]).

Let x and y be elements of S12 defined by

x=(1,2)(3,4)(5,6)(7,9,10,8)(11,12),
y=(1,3)(2,4,5,7)(6,8)(9,11)(10,12).

Then o(x)=4=o(y), [x,y,y,y]=1=[y,x,x,x] and G=x,y has order 2534, so, in particular, G is not nilpotent.

Example 2.

Let x and y be elements of S12 defined by

x=(1,2,3,4)(5,6,8,10)(7,9,11,12),
y=(1,3)(2,4,5,7)(6,9)(8,11)(10,12).

Then o(x)=4=o(y),

[x,y,y,y]=1=[x-1,y,y,y],
[y,x,x,x,x]=1=[y-1,x,x,x,x]

and G=x,y has order 2634, so, in particular, G is not nilpotent.

3 Proofs of Theorem B and Theorem C

Lemma 3.1.

Let G=x,y be a group, yLn+12(G). Then [x,yn]Z(G).

Proof.

That [x,yn] commutes with x is a direct consequence of yLn+12(G). Then [yx,yn,yx]=[x,yn,yx]=[x,yn,x][xn+1y]x=[x,yn+1]x shows that [x,yn] commutes also with y. ∎

Lemma 3.2.

Let G be a group, and let a,bG. Suppose that, for some n2, we have that {a,b,a-1,b-1} is an n-Engel subset of G. Then

en-1(b-a,b)=en-1(b,b-a)=en-1(ba,b-1)=en-1(b-1,ba)=1.

Proof.

We have

1=[a,bn]=[b-ab,bn-1]=[b-a,bn-1]b

and therefore [b-a,bn-1]=1. Replacing b by b-1, we see that [a,b-1n]=1 implies [ba,b-1n-1]=1. Next use [a-1,bn]=1, which implies [b-a-1,bn-1]=1 and thus, after conjugation by a, that [b-1,ban-1]=1. Replacing b by b-1, we see that [a-1,b-1n]=1 implies [b,b-an-1]=1. ∎

Proposition 3.3.

Let G be a (4+12)-Engel 2-group. Then G is locally nilpotent.

Proof.

Taking the quotient of G by the Hirsch–Plotkin radical, we can assume that HP(G)=1, and we want to show that G=1. We argue by contradiction and suppose G1. As groups of exponent 4 are locally finite, there must be an element gG of order 8. We get a contradiction by showing that g4G is abelian and thus g4HP(G)=1.

Let hG, and consider the subgroup H=g,g1, where g1=g-h. Let

H¯=H/Z(G)=g¯,g¯1,

where g¯=gZ(H) and g¯1=g1Z(H). By Lemma 3.2, we know that

e(g,g1)=e(g1,g)=e(g-1,g1-1)=e(g1-1,g-1)=1.

By Theorem A, we then know that H¯ and therefore H is finite. Using GAP or MAGMA, one can then check that [g-4h,g4]=[g14,g4]=1, and thus we have shown that g4G is abelian. ∎

Proposition 3.4.

Let G be a (4+12)-Engel 3-group. Then G is locally nilpotent.

Proof.

As before, we can assume that HP(G)=1, and the aim is then to show that G=1. We argue by contradiction and suppose that G1. As groups of exponent 3 are locally finite, there must be an element gG of order 9. Let hG and g1=g-h. As in the proof of Proposition 3.3, one sees that H is finite and then, with the help of GAP or MAGMA, that [g13,g3,g3]=1. Thus [h,g3,g3,g3]=1 for all hG, and thus g3 is a left 3-Engel element of G. By the main result of [5], we then know that g3HP(G)=1, which contradicts the fact that o(g)=9. ∎

Lemma 3.5.

Let G be a group, and let a,bG be two elements of finite order such that S={a,b,a-1,b-1} is a 4-Engel set. Then every prime divisor of o([a,b]) is a divisor of o(a) and o(b). In particular, if a and b are of coprime order, then [a,b]=1.

Proof.

By Lemma 3.2 together with Lemma 2.2, we know that

S1={ba,b,b-a,b-1}andS2={ab,a,a-1,a-b}

are 3-Engel subsets of G. By Theorem A, we know that

H1=a,abandH2=b,ba

are nilpotent. As these groups are nilpotent, we know that every prime divisor of |H1| divides o(a) and every prime divisor of |H2| divides o(b). Now we have [a,b]H1H2, and thus o([a,b]) divides |H1| and |H2| and thus also o(a) and o(b) from the discussion above. ∎

Proof of Theorem B.

Let G be a {2,3}-group that is (4+12)-Engel. Let H2 be the set consisting of all elements in G whose order is a power of 2 and H3 that of those elements whose order is a power of 3. In view of Propositions 3.3 and 3.4, it suffices to show that H2 and H3 are subgroups and that G is a direct product of H2 and H3. Now take any two elements a,bG of coprime orders, and let T=a,b. By Lemma 3.1, we know that e(aα,bβ)=e(bβ,aα)Z(T) for all α,β{1,-1}. By Lemma 3.5, it follows that [a,b]Z(T). Thus T is nilpotent, and as a and b are of coprime order, it follows that [a,b]=1. Now let aH2, and let bH2 be an element that has odd order. By the argument above, we know that [a,b]=1, and as aH2 was arbitrary, we see that bZ(H2). Thus H2/Z(H2) is a 2-group, and by Proposition 3.3, it is locally nilpotent and thus so is H2. As H2 is generated by 2-elements, it is then a 2-group, and thus H2=H2, and thus H2 is a subgroup. The proof that H3 is a subgroup is similar, using Proposition 3.4. Now let aH2 and bH3. Then [a,b]H2H3 and is thus trivial. Hence G is a direct product of H2 and H3 and thus locally nilpotent. ∎

Lemma 3.6.

Let p5 be a prime, and consider the group G=x,y, where xp2=yp2=1 and {x,y,x-1,y-1} is a 3-Engel set. Then G has exponent p2, and Gp is abelian.

Proof.

By Theorem A, we know that G is nilpotent of class at most 4. Then G is regular, and it follows easily that Gp2=1 and then that [Gp,Gp]=1 (for definition and properties of regular p-groups, see [2, Section 12.4], in particular Theorem 12.4.3). ∎

Proof of Theorem C.

Let p5 be a prime, and let G be a (4+12)-Engel p-group. Consider H=G/HP(G), where HP(G) is the Hirsch–Plotkin radical of G. The aim is to show that H is of exponent p. Passing from G to H, we can thus without loss of generality assume that the Hirsch–Plotkin radical of G is trivial, and the aim is to show that G is then of exponent p. We argue by contradiction and suppose that G has an element g of order p2. Let hG, and consider the subgroup H=g,g1, where g1=g-h. Let H¯=H/Z(H)=g¯,g¯1, where g¯=gZ(H) and g¯1=g1Z(H). By Lemma 3.2, we know that

e(g,g1)=e(g1,g)=e(g-1,g1-1)=e(g1-1,g-1)=1.

By Theorem A, we then know that H¯ is finite, and thus also H is finite. By Lemma 3.6, we know that [g¯1p,g¯p]=1, that is, [g1p,gp]Z(H), and thus, in particular, [h,gp,gp,gp]=[g1p,gp,gp]=1. Thus gp is a left 3-Engel element of odd order in G. By the main result of [5], it follows that gpHP(G)=1, which contradicts our assumption that o(g)=p2. ∎

Remark.

The variety n+12 seems to be the “engelization” of the variety of groups satisfying the law [y,x1,x2,,xn,y]=1 studied by Macdonald in [6, 7].


Communicated by Evgenii I. Khukhro


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Received: 2019-07-17
Revised: 2019-11-29
Published Online: 2019-12-21
Published in Print: 2020-05-01

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