Accessible Published by De Gruyter December 21, 2019

On (𝑛 + ½)-Engel groups

Enrico Jabara 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 + 1 2 ) -Engel group if it satisfies the law [ x , y n , x ] = 1 . The variety of ( n + 1 2 ) -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 + 1 2 ) -Engel { 2 , 3 } -groups are locally nilpotent. We also show that if G is a ( 4 + 1 2 ) -Engel p-group, where p 5 is a prime, then G p is locally nilpotent.

1 Introduction

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

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

We recursively define [ g , h n ] , where n is a positive integer, as [ g , h 1 ] = [ g , h ] and [ g , h n + 1 ] = [ [ g , h n ] , h ] for n 1 . A subset S G is an Engel set of G if, for every g , h S , there is a positive integer k = k ( g , h ) such that [ g , h k ] = 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 a G is said to be left n-Engel if [ x , a n ] = 1 for all x G and right n-Engel if [ a , x n ] = 1 for all x G . We denote the subset of left n-Engel elements by L n ( G ) and that of right n-Engel elements by R n ( G ) .

Definition.

Let G be a group and n a positive integer.

  1. (1)

    We say that a G is a left ( n + 1 2 ) -Engel element if [ x , a n , x ] = 1 for all x G .

  2. (2)

    We say that a G is a right ( n + 1 2 ) -Engel element if [ a , x n , a ] = 1 for all x G .

  3. (3)

    We say that G is an ( n + 1 2 ) -Engel group if it satisfies the law [ x , y n , x ] = 1 .

We denote the subset of left ( n + 1 2 ) -Engel elements by L n + 1 2 ( G ) and the right ( n + 1 2 ) -Engel elements by R n + 1 2 ( G ) . Thus G is an ( n + 1 2 ) -Engel group if and only if L n + 1 2 ( G ) = G or equivalently R n + 1 2 ( G ) = G . We denote the variety of m-Engel groups by m .

Remark.

It is not difficult to prove that

L 1 + 1 2 ( G ) = R 2 ( G ) and R 1 + 1 2 ( G ) = L 2 ( G ) .

Thus, in particular, 1 + 1 2 = 2 .

Lemma 1.1.

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

L n ( G ) L n + 1 2 ( G ) L n + 1 ( G ) .

In particular, E n E n + 1 2 E n + 1 .

Proof.

That L n ( G ) L n + 1 2 ( G ) is obvious. To see that L n + 1 2 ( G ) L n + 1 ( G ) , let a L n + 1 2 ( G ) . Then, for any x G , we have

1 = [ a x , a n , a x ] = [ x , a n , x ] [ x , a n + 1 ] x = [ x , a n + 1 ] x .

Thus [ x , a n + 1 ] = 1 and a L n + 1 ( G ) . ∎

Our main results on ( n + 1 2 ) -Engel groups are the following.

Theorem B.

Let G be a ( 4 + 1 2 ) -Engel { 2 , 3 } -group. Then G is locally nilpotent.

Theorem C.

Let G be a ( 4 + 1 2 ) -Engel p-group, where p is a prime and p 5 . Then G p is locally nilpotent.

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

Theorem A.

Let R = a , b be the largest 2-generator group satisfying the relations e 3 ( a , b ) = e 3 ( b , a ) = e 3 ( a - 1 , b - 1 ) = e 3 ( 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 e n ( x , y ) = [ x , y n ] = 1 . As we will focus in particular on the 3-Engel word, we will often use e ( x , y ) instead of e 3 ( 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 , b a = b , [ a , b ] is nilpotent of class at most 2.

Proof.

From the equations

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

we see that 1 = [ b - a , b , b ] and 1 = [ b a - 1 , b - 1 , b - 1 ] a = [ b , b - a , b - a ] . Thus we have that b , b a 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 [ a ϵ , b ϵ 1 , b ϵ 2 , b ϵ 3 ] = 1 for all ϵ , ϵ 1 , ϵ 2 , ϵ 3 { 1 , - 1 } .

Proof.

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

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

we can also assume that ϵ 3 = 1 . Then, from

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

we can also assume, without loss of generality, that ϵ 2 = 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 ] = [ b a b - 1 , b , b ] γ 3 ( b , b a ) . ∎

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 b a .

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 , b a ] .

It follows that [ a , b , b , a ] commutes with b a , and thus, using Lemma 2.1 again, [ [ a , b , b ] - 1 , a ] = [ a , b , b , a ] - [ a , b , b ] - 1 commutes with b a 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 , b Z 4 ( R ) . This is equivalent to showing that [ b , a ] Z 3 ( 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 ] Z 3 ( R ) , we need to show that [ b , a , a ] and [ b , a , b ] = [ a , b , b ] - 1 are in Z 2 ( 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 , a b ] = [ [ 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 ( a α , b β ) = e ( b β , a α ) = 1 for all α , 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 , a b that is nilpotent of class at most 2, we see that [ b , a - 1 , a - 1 ] = [ a b , 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 , b a ] 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 b a and a b . Replacing a by a - 1 for the RHS, we see that the common element also commutes with b a - 1 . Likewise, it commutes with a b - 1 . As

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

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 R 4 . ∎

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 S 12 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 2 5 3 4 , so, in particular, G is not nilpotent.

Example 2.

Let x and y be elements of S 12 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 2 6 3 4 , 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, y L n + 1 2 ( G ) . Then [ x , y n ] Z ( G ) .

Proof.

That [ x , y n ] commutes with x is a direct consequence of y L n + 1 2 ( G ) . Then [ y x , y n , y x ] = [ x , y n , y x ] = [ x , y n , x ] [ x n + 1 y ] x = [ x , y n + 1 ] x shows that [ x , y n ] commutes also with y. ∎

Lemma 3.2.

Let G be a group, and let a , b G . Suppose that, for some n 2 , we have that { a , b , a - 1 , b - 1 } is an n-Engel subset of G. Then

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

Proof.

We have

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

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

Proposition 3.3.

Let G be a ( 4 + 1 2 ) -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 G 1 . As groups of exponent 4 are locally finite, there must be an element g G of order 8. We get a contradiction by showing that g 4 G is abelian and thus g 4 HP ( G ) = 1 .

Let h G , and consider the subgroup H = g , g 1 , where g 1 = g - h . Let

H ¯ = H / Z ( G ) = g ¯ , g ¯ 1 ,

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

e ( g , g 1 ) = e ( g 1 , g ) = e ( g - 1 , g 1 - 1 ) = e ( g 1 - 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 - 4 h , g 4 ] = [ g 1 4 , g 4 ] = 1 , and thus we have shown that g 4 G is abelian. ∎

Proposition 3.4.

Let G be a ( 4 + 1 2 ) -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 G 1 . As groups of exponent 3 are locally finite, there must be an element g G of order 9. Let h G and g 1 = 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 [ g 1 3 , g 3 , g 3 ] = 1 . Thus [ h , g 3 , g 3 , g 3 ] = 1 for all h G , and thus g 3 is a left 3-Engel element of G. By the main result of [5], we then know that g 3 HP ( G ) = 1 , which contradicts the fact that o ( g ) = 9 . ∎

Lemma 3.5.

Let G be a group, and let a , b G 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

S 1 = { b a , b , b - a , b - 1 } and S 2 = { a b , a , a - 1 , a - b }

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

H 1 = a , a b and H 2 = b , b a

are nilpotent. As these groups are nilpotent, we know that every prime divisor of | H 1 | divides o ( a ) and every prime divisor of | H 2 | divides o ( b ) . Now we have [ a , b ] H 1 H 2 , and thus o ( [ a , b ] ) divides | H 1 | and | H 2 | 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 + 1 2 ) -Engel. Let H 2 be the set consisting of all elements in G whose order is a power of 2 and H 3 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 H 2 and H 3 are subgroups and that G is a direct product of H 2 and H 3 . Now take any two elements a , b G of coprime orders, and let T = a , b . By Lemma 3.1, we know that e ( a α , b β ) = e ( b β , a α ) 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 a H 2 , and let b H 2 be an element that has odd order. By the argument above, we know that [ a , b ] = 1 , and as a H 2 was arbitrary, we see that b Z ( H 2 ) . Thus H 2 / Z ( H 2 ) is a 2-group, and by Proposition 3.3, it is locally nilpotent and thus so is H 2 . As H 2 is generated by 2-elements, it is then a 2-group, and thus H 2 = H 2 , and thus H 2 is a subgroup. The proof that H 3 is a subgroup is similar, using Proposition 3.4. Now let a H 2 and b H 3 . Then [ a , b ] H 2 H 3 and is thus trivial. Hence G is a direct product of H 2 and H 3 and thus locally nilpotent. ∎

Lemma 3.6.

Let p 5 be a prime, and consider the group G = x , y , where x p 2 = y p 2 = 1 and { x , y , x - 1 , y - 1 } is a 3-Engel set. Then G has exponent p 2 , and G p 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 G p 2 = 1 and then that [ G p , G p ] = 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 p 5 be a prime, and let G be a ( 4 + 1 2 ) -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 p 2 . Let h G , and consider the subgroup H = g , g 1 , where g 1 = g - h . Let H ¯ = H / Z ( H ) = g ¯ , g ¯ 1 , where g ¯ = g Z ( H ) and g ¯ 1 = g 1 Z ( H ) . By Lemma 3.2, we know that

e ( g , g 1 ) = e ( g 1 , g ) = e ( g - 1 , g 1 - 1 ) = e ( g 1 - 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 ¯ 1 p , g ¯ p ] = 1 , that is, [ g 1 p , g p ] Z ( H ) , and thus, in particular, [ h , g p , g p , g p ] = [ g 1 p , g p , g p ] = 1 . Thus g p is a left 3-Engel element of odd order in G. By the main result of [5], it follows that g p HP ( G ) = 1 , which contradicts our assumption that o ( g ) = p 2 . ∎

Remark.

The variety n + 1 2 seems to be the “engelization” of the variety of groups satisfying the law [ y , x 1 , x 2 , , x n , y ] = 1 studied by Macdonald in [6, 7].

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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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