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Abstract
We give an infinite family of examples that generalize the construction given by Noce, Tracey and Traustason of a locally finite 2-group G containing a left 3-Engel element x where the normal closure of x in G, is not nilpotent. The construction is based on a family of Lie algebras that are of interest in their own right and make use of a classical theorem of Lucas, regarding when
is even.
1. Introduction
Let G be a group. An element is a left Engel element in G, if for each
there exists a non-negative integer n(x) such that
If n(x) is bounded above by n then we say that a is a left n-Engel element in G. Throughout this paper we will assume that, when dealing with commutators or Lie products, these are left normed. Recall that the Hirsch-Plotkin radical of a group G is the subgroup generated by all the normal locally nilpotent subgroups of G and that this is also locally nilpotent. It is straightforward to see that any element of the Hirsch-Plotkin radical HP(G) of G is a left Engel element and the converse is known to be true for some classes of groups, including solvable groups and finite groups (more generally groups satisfying the maximal condition on subgroups) [Citation1, Citation3]. The converse is however not true in general and this is the case even for bounded left Engel elements. In fact whereas one sees readily that a left 2-Engel element is always in the Hirsch-Plotkin radical this is still an open question for left 3-Engel elements. Recently there has been a breakthrough and in [Citation7] it is shown that any left 3-Engel element of odd order is contained in HP(G). From [Citation12] one also knows that in order to generalize this to left 3-Engel elements of any finite order it suffices to deal with elements of order 2.
It was observed by William Burnside [Citation2] that every element in a group of exponent 3 is a left 2-Engel element and so the fact that every left 2-Engel element lies in the Hirsch-Plotkin radical can be seen as the underlying reason why groups of exponent 3 are locally finite. For groups of 2-power exponent there is a close link with left Engel elements. If G is a group of exponent then it is not difficult to see that any element a in G of order 2 is a left
-Engel element of G (see the introduction of [Citation14] for details). For sufficiently large n we know that the variety of groups of exponent
is not locally finite [Citation6, Citation8]. As a result one can see (for example in [Citation14]) that it follows that for sufficiently large n we do not have in general that a left n-Engel element is contained in the Hirsch-Plotkin radical. Using the fact that groups of exponent 4 are locally finite [Citation11], one can also see that if all left 4-Engel elements of a group G of exponent 8 are in HP(G) then G is locally finite.
Swapping the role of a and x in the definition of a left Engel element we get the notion of a right Engel element. Thus an element is a right Engel element, if for each
there exists a non-negative integer n(x) such that
If n(x) is bounded above by n, we say that a is a right n-Engel element. By a classical result of Heineken [Citation5] one knows that if a is a right n-Engel element in G then is a left
-Engel element.
In [Citation9] M. Newell proved that if a is a right 3-Engel element in G then and in fact he proved the stronger result that
is nilpotent of class at most 3. The natural question arises whether the analogous result holds for left 3-Engel elements. In [Citation10] it is shown that this is not the case by giving an example of a locally finite 2-group with a left 3-Engel element a such that
is not nilpotent. Moreover in [Citation4] an example is given, for each odd prime p, of a locally finite p-group containing a left 3-Engel element x where
is not nilpotent.
In this paper we extend the example above, of a 2-group, to an infinite family of examples. The construction will be based on a family of Lie algebras that generalize the Lie algebra given in [Citation13]. These algebras are of interest in their own right and will make use of a classical theorem of Lucas. Before stating Lucas’s Theorem we need some notation.
Let p be a prime and consider non-negative integers m and n written in base p
where
We introduce a partial order
where
if and only if
for
Theorem 1.1
(Lucas’ Theorem). The binomial coefficient is divisible by p if and only if
Remark.
Notice that when p = 2 we get that the binomial coefficient is odd if and only if
2. The Lie algebra L
In this section we construct a family of Lie algebras that extend the example given in [Citation13]. The construction makes an interesting use of Lucas’ Theorem.
Let be the field of order 2 and let
for
Consider the
-dimensional vector space
We equip L with a binary product where
such that
and
and where
We then extend the product linearly on L. The next theorem is our first main result.
Theorem 2.1.
L is a Lie algebra over
Proof.
Let and suppose that
where
for
In order to show that the product is alternating, it only remains to see that
and
(recall that the characteristic is 2). Firstly
In order for the product to be zero we know by Lucas’ Theorem that we need Assume for contradiction that
Then
for all
which implies that
This gives
that contradicts
Now, for we need
But,
Having established that the product is alternating we turn to the Jacobi identity . If we have basis elements e, f then, as and
we get
Therefore, we only need to deal with the cases when the three basis elements are different. We will divide our analysis into few cases. Notice first, that any Jacobi relation involving one occurrence of x and two v(i)’s, for
is clearly 0. There is one remaining type of a Jacobi relation that involves x. This is the one for a triple
where
First, let We have,
since
and thus
If
then
Let us next consider triples of the type where
Case 1:
Let and
Then,
Case 2:
Let Then,
where the last equality follows from Pascal’s Rule.
Finally, consider for
Clearly, if all coefficients αi are even then the Jacobi identity holds. So, assume without loss of generality that α1 is odd. Then, as
we get
for
Then,
so
Case 1:
Consider the case where Then,
Notice that
is odd if and only if
Hence
and
Thus,
as required.
Case 2:
Consider the case where We want to show that if α1 is odd then exactly one of
is odd. We shall consider each term separately. We have
We assumed that α1 must be odd, hence we need both binomial coefficients to be odd. We have is odd if and only if
for all i. Let t be the smallest index such that
Therefore,
In order for to be odd we must have that
So,
hence,
and
for all
Notice that it follows in particular that
for all i except possibly i = t.
Notice that the assumption implies that
That is
Then we must have that
and
If there is an equality, for example
then, by symmetry, we have already shown in Case 1 that the Jacobi identity holds. So, we may assume that
Hence, similarly as above,
and
We have two cases:
ct = 0: Assume without loss of generality that at = 0 and bt = 1. Then,
and
hence
is odd and
is even. Notice that this implies that the smallest index s such that
is s = t. Hence,
is odd, since
for all
and
This shows that the Jacobi identity holds.
ct = 1: Assume without loss of generality that at = 0 and bt = 1. Then, both
hence both binomials coefficients
and
are odd. Now, similarly as in case (a), if s is the smallest index such that
then s = t and so
is odd. It only remains to show that
is even. But,
so the smallest index l such that
is
Then, in order for
to be odd we require
which contradicts our assumption, hence
must be even and the Jacobi identity holds.□
Lemma 2.2.
The Lie algebra L has trivial center.
Proof.
Take an element of L say where
that lies in the center of L. Multiplying by x gives
therefore μ = 0. Then, multiplying by w gives
and therefore
Lemma 2.3.
is a simple ideal of L.
Proof.
Consider the ideal I generated by y, where and
are not all zero. We first show that
If
this is clear. If not, take the smallest i such that
where
Taking y and mulitiplying n – i times by w gives us
that implies that
Having established that
we can multiply it by
to see that
Hence I = W.□
Let As Z(L) is trivial, E is the associative enveloping algebra of L.
Lemma 2.4.
The associative enveloping algebra E is finite-dimensional.
Proof.
This follows from the fact that hence dim (End (L)) = (n + 2)2, thus E must be of finite dimension.□
We will use L to construct a locally nilpotent Lie algebra over of countably infinite dimension. This will then help us to construct a locally finite group G with a left 3-Engel element y where
is not nilpotent. We now introduce a notation that was used in [Citation13] of modified unions of subsets of
We let
For each non-empty subset A of we let WA be a copy of the vector space
that is
with addition
We then take the direct sum of these
that we turn into a Lie algebra with multiplication
when
and
and extend linearly on
The interpretation here is that
Finally, we extend this to the semidirect product with
induced from the action
Notice that
has basis
and that
for all
and
Lemma 2.5.
is locally nilpotent.
Proof.
Notice that any finitely generated subalgebra of is contained in some
Thus it suffices to show that S is nilpotent. Observe first that any Lie product with a repeated entry of or wB is trivial and thus a non-trivial Lie product of the generators of S can include in total at most
such elements. As
we have
for all
Thus we see that S is nilpotent of class at most
Therefore,
is locally nilpotent.
3. The group G
For an element we denote by
the linear operator on
induced by multiplication by y on the right. In this section we find a group G inside
containing
where
is a left 3-Engel element in G, but where
is not nilpotent. The next Lemma is a preparation for this.
Lemma 3.1.
The adjoint linear operator on
satisfies:
, for all
Proof.
This follows from our earlier observation that
for all
Follows from
□
Now let y be any of the generators for
Since,
for all y it follows that
Thus, is an involution in
Notice also that the following are elementary abelian 2-groups of countably infinite rank.
We will be working with the group
Lemma 3.2.
The following commutator relations hold in G:
, if
Proof.
We have
where we have used Lemma 3.1. Part (b) is proven similarly. For (c) we have
Parts (d) and (e) are proved similarly.
Remark.
Notice that as is locally nilpotent it follows from Lemma 3.2 that G is locally nilpotent. Next proposition clarifies the structure of G.
Proposition 3.3.
We have In particular, for every element
there exists an expression
with
and
Proof.
Suppose that
where
are products of elements of the form
From Lemma 3.2 we know that
and
commutes with all products of the form
for
We can thus collect the
’s to the left starting with the leftmost occurrence. This may introduce more elements of the form
but no new elements
We thus have that
where gi is of the form
for
or of the form
Notice also that we can assume that
where
This reduces our problem to the case when
Suppose
where the terms gi are
in some order. By Lemma 3.2 we have that
and
if
or
if
We can thus collect the terms so that
where hi are of the form
with
or of the form
where D is a modified union of at least two sets from
Thus,
where
for
and h is a product of elements of the form
with
or of the form
where D is a modified union of at least two sets from
Repeating this collection process we get
where
for
and k is a product of elements of the form
where
or of the form
where E is a modified union of at least three sets from
Continuing in this manner we conclude that after k steps
where
for
and f is a product of elements of the form
where
or of the form
where H is a modified union of at least k + 1 sets from
However, any modified union of at least
sets from
is trivial and thus f = 1 when
This completes the proof.
Theorem 3.4.
The element is a left 3-Engel element in G such that
is not nilpotent.
Proof.
Showing that is a left 3-Engel element in G is equivalent to showing that
commutes with
for all
Let
be an arbitrary element in G, where
We want to show that
Let Then
Notice that since by Lemma 3.2 we know that
commutes with all elements of the form
for
Then, by induction we obtain that
where
Since the commutator of
with
is
Then,
using the fact that
However, the normal closure of in G is not nilpotent, as for
we have
where
One might wonder if the nilpotency class of the subgroup generated by r conjugates is unbounded for the family we have constructed. That turns out not to be the case. Our next aim is to show that the subgroup generated by any r conjugates of
in G will be nilpotent of r-bounded class.
We first work in a more general setting. For each and
let
where
for
Then let E* = ⟨ad(x),e(A): e ∈ E and ∅≠A ⊆ ℕ⟩ . As is locally nilpotent, one sees readily that the elements of
are nilpotent and thus
is a subgroup of
Despite the fact that the normal closure of in G is not nilpotent, it turns out that the nilpotency class of the subgroup generated by any r conjugates grows linearly with respect to r. In order to see this we must first introduce some more notation. In relation with the r conjugates we let
be any r subsets of
For each r-tuple
of non-negative integers and each
we let
Notice that
Consider the r conjugates of in G. Recall that each conjugate is of the from
Without loss of generality one can assume that each Ck is a singleton set. The following argument also works for the more general case. Let
and
Then we have seen (see the proof of Theorem 3.4) that
Let
where i is the k-th coordinate and Let
Our aim is to find an upper bound for the nilpotence class of F. For this we need to understand better the two aspects of multplying and
These are
Under which conditions the binomial coefficients are non-trivial;
Under which conditions is the Lie product
non trivial.
The next two lemmas will help clarify these questions.
Lemma 3.5.
If , then
, where
Proof.
Suppose that
We have that is odd if and only if
The latter happens if and only if there exists no l such that
where
In particular, for the binomial coefficient to be non-zero we need
□
Lemma 3.6.
If and
, then
Proof.
As a preparation we first show that if
To see this let
where
for
Then
In particular for to be odd we need
That is we need
Thus if
we have
Having established this preliminary result we turn to
where
Firstly,
and so it immediately follows from the result above that
if
that is if
Then, for
we have
From our preliminary result above we know that if
We can therefore assume that
and hence
Therefore,
which, by our preliminary result again, is trivial when
that is if
Given that
we have in particular that the product is trivial when
Hence,
if
□
From these two lemmas we get the following result.
Lemma 3.7.
Let . If
is nonzero, then
We are now ready for establishing the linear upper bound for the nilpotence class of F.
Lemma 3.8.
F is nilpotent of a class at most
Proof.
Notice that by Lemma 3.5 we have that for any
We thus only need to consider products of elements
where
Take any such product of even length 2u where u is going to be determined later. Suppose the product is
(1)
(1)
where
and
We want to determine u so that this product becomes 0. From Lemma 3.7 we know that for this product to be non-trivial we need
for all
For
let tl be the sum of all the lth coordinates of the superfixes of the 2 u elements in the product. For this to be non-zero we need
If one of is greater than n + 1 then Lemma 3.5 implies that the product is zero. We need to find how big u has to be so that this happens. Notice that the largest value of
is greater or equal to the mean value and this is at least
Therefore, it suffices that
This holds when Hence,
□
From this it is not difficult to obtain a linear upper bound for nilpotence class of the group generated by r conjugates of First we extend the analysis of F to the subalgebra
of
If no element
occurs then any product of elements of Q of length
is trivial. Here we are using the fact that
Lemma 3.8 and the fact that
commutes with elements of the form
when
Suppose therefore that at least one element
is involved in a product of elements of Q. If an element of the form
included, where
we can pick
and
that are of closest distance within the product. This reduces things to the following situations:
These products are trivial, as commutes with
and the products of
and
are trivial by Lemma 3.5. We can therefore assume that no element
is involved, where
Then we only have a product in
’s and elements of the form
By Lemma 3.5 and the fact that
the product needs to alternate between
and elements of the form
We will make use of the fact that
We have
where the last equality follows from Lemma 3.5. Similarly any product of the form
is 0. To conclude we have seen that
Now let H be a subgroup of G generated by any r conjugates
of
in G. Then
Thus we have the following result.
Proposition 3.9.
Let be any r conjugates of
in G. Then
is nilpotent of class at most
Acknowledgments
The first author is partially supported by ‘The Norton Scholarship’. We acknowledge the EPSRC (grant number 16523160) for support. Moreover, we would like to thank Marialaura Noce for drawing our attention to Lucas’ Theorem.
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