MSC 20F45 Recently Havas and Vaughan-Lee proved that 4-Engel groups are locally nilpotent. Their proof relies on the fact that a certain 4-Engel group T is nilpotent and this they prove using a computer and the Knuth-Bendix algorithm. In this paper we give a short hand-proof of the nilpotency of T 

Gunnar Traustason

CiteSeerX

A note on the local nilpotence of 4-Engel
groups
Gunnar Traustason
Centre for Mathematical Sciences, Lund University,
Box 118, SE-221 00 Lund, Sweden
MSC 20F45
Recently Havas and Vaughan-Lee proved that 4-Engel groups are lo-
cally nilpotent. Their proof relies on the fact that a certain 4-Engel
group T is nilpotent and this they prove using a computer and the
Knuth-Bendix algorithm. In this paper we give a short hand-proof of
the nilpotency of T .
1 Introduction
Recently, Havas and Vaughan-Lee have proved that 4-Engel groups are locally
nilpotent [1]. Their proof uses repeatedly the fact that a certain 3-generator
4-Engel group T is nilpotent. This group T = 〈a, b, c〉 is the largest 4-Engel
group satisfying the extra properties that
〈a, b〉 is abelian,
〈a, c〉 is nilpotent of class at most two,
〈b, c〉 is nilpotent of class at most three.
The largest quotient of exponent 5 of this group also played a role in the
proof of Vaughan-Lee that 4-Engel groups of exponent 5 are locally nilpo-
tent [5]. The proof of Havas and Vaughan-Lee is a computer proof using the
1
2
Knuth-Bendix algorithm. In Section 2 we will give a short hand proof of the
nilpotence of T . This then implies that we have essentially a computer-free
proof of the local nilpotence of 4-Engel groups. (Havas and Vaughan-Lee
also use the nilpotent quotient algorithm to calculate the nilpotency class of
a number of 4-Engel groups, but the class is never very large and so these
calculations could be done by hand using commutator calculus or Lie ring
methods. Even so these calculations are without doubt better done by com-
puter as this is both quicker and more reliable).
As in the proof of Havas and Vaughan-Lee, we will use the fact that any
2-generator 4-Engel group is nilpotent [4]. In this paper we will denote by
R(G), the Hirsch-Plotkin radical of G. It is well known that for any 4-Engel
group G, R(G/R(G)) = {1}. To prove that T is nilpotent we can thus with-
out loss of generality assume that R(T ) = {1}. As any 2-generator 4-Engel
group is nilpotent it follows that the torsion elements of T form a normal
subgroup that is a direct product of p-groups. By [5], every 4-Engel 2-group
and every 4-Engel 3-group is locally nilpotent. Since we are assuming that
R(T ) = {1} we thus have that T has no elements of order 2 and 3. In fact
T has no elements of order 5 by Vaughan-Lee’s result but we don’t want to
assume this as we also want to give a computer-free proof of the finiteness of
the largest exponent 5 quotient of T .
In turns out to be the case that if one adds the extra condition that [b, c, c] = 1
then one doesn’t need the 4-Engel identity to prove the nilpotence of T . One
only needs the radical properties above. This we will prove in Section 3. We
have not been able to do this without this extra condition and it may well
be the case that the 4-Engel property is needed for the stronger version.
2 The nilpotence of T
As in the proof of Havas and Vaughan-Lee we use the fact [4] that any 2-
generator 4-Engel group is nilpotent. We will also need the following lemmas.
Lemma 2.1 Let G be any 4-Engel group without 2-elements and let x, y ∈ G.
(1) If [y, x, x] = 1 then 〈x〉〈y〉 is abelian.
(2) If [y, x, x, x] = 1 then 〈x〉〈y〉 is nilpotent of class at most 2.
3
Proof. By [4] the group 〈x, y〉 is nilpotent. The two statements can now be
read from a polycyclic presentation of the free 2-generator 4-Engel group. See
for example [2]. In fact part (1) follows also directly from the next lemma.
2
Lemma 2.2 Let G be any 4-Engel group and x, y, z ∈ G. If z commutes
with x, xy and xy
−1
then z commutes with all elements in 〈x〉〈y〉.
Proof Let u = xy
−1
. Then z commutes with u, uy and uy
2
and thus with
u, [u, y], [u, y, y]. As G is 4-Engel, we have
1 = [u, zy, zy, zy, zy]
= [u, y, zy, zy, zy]
= [u, y, y, zy, zy]
= [u, y, y, y, zy]
= [u, y, y, y, z]y.
Thus z commutes also with uy
3
and as 〈x〉〈y〉 = 〈u〉〈y〉 is generated by u, uy, uy
2
and uy
3
, the result follows. 2
We now prove that the group T = 〈a, b, c〉 is nilpotent. Without loss of
generality we can assume that T has trivial Hirsch-Plotkin radical. Thus in
particular T has a trivial centre and no elements of order 2 or 3. As well
as the lemmas above the following lemma is going to play a key role in the
proof.
Lemma 2.3 Let u be an element in T . If
H = 〈u〉〈[c,a]〉 ≤ CT (〈b, c〉)
then u = 1.
Proof First consider any h ∈ H that commutes with [c, a]. Then h commutes
with c, [c, a] and b. We next show that the same holds for the elements
[h, a], [h, a, a] and [h, a, a, a]. Firstly, as a commutes with b and [c, a] it is
clear that these elements commute with b and [c, a]. We show by an inductive
argument that they also commute with c. But
[h, a]c = [h, a[a, c]]
= [h, [a, c]][h, a][a,c]
= [h, a],
4
and thus [h, a] commutes with c. In fact this argument shows that if v in T
commutes with c and [c, a] then the same is true for [v, a]. Thus the elements
h, [h, a], [h, a, a] and [h, a, a, a] all commute with c and b. As T is 4-Engel it
follows that [h, a, a, a] ∈ Z(T ) and thus trivial. But then [h, a, a] is in the
centre and also trivial. Continuing like this, we see that h = 1.
We use this now to deduce that u = 1. As [u, [c, a], [c, a], [c, a]] commutes
with [c, a] it is trivial by the previous paragraph. But then [u, [c, a], [c, a]]
commutes with [c, a] and is also trivial. Continuing like this we see that
u = 1. 2
We divide the rest of the proof into few steps.
Step 1. We show that [c, b, b] = 1.
The 2-generator subgroup 〈ac, b〉 is nilpotent and
[ac, b, b, b] = [c, b, b, b] = 1.
It follows from Lemma 2.1 that 〈b〉〈ac〉 is nilpotent of class at most two. In
particular [ac, b, b, ac] commutes with b. Then
[ac, b, b, ac] = [c, b, b, ac] = [c, b, b, a][c, b, b, a, c],
and as [c, b, b, a] commutes with b (as [c, b, b] and a do) we conclude that
[c, b, b, a, c, b] = 1. (1)
The Hall-Witt identity gives us
1 = [[c, b, b], c−1, a]c[c, a−1, [c, b, b]]a[a, [c, b, b]−1, c][c,b,b]
= [[c, a]−1, [c, b, b]]a[c, b, b, a, c].
By (1) the latter commutator commutes with b. Thus [[c, b, b], [c, a]−1] com-
mutes with b and replacing a by a−1 we see that [[c, b, b], [c, a]] also commutes
with b. We have now seen that
[c, b, b], [c, b, b][c,a], [c, b, b][c,a]
−1
5
all commute with b and by Lemma 2.2 we can then deduce that all the el-
ements in H = 〈[c, b, b]〉〈[c,a]〉 commute with b. These elements also clearly
commute with c. Now Lemma 2.3 gives that [c, b, b] = 1.
Step 2. We show that [b, c, c] = 1.
By Step 1 we have that
[ac, b, b] = [c, b, b] = 1.
It follows from Lemma 2.1 that 〈b〉〈ac〉 is abelian. In particular it follows that
the element
[ac, b, ac] = [c, b, ac] = [c, b, c][c, b, a][c, b, a, c]
commutes with b and as [c, b, a], [c, b, c] commute now with b, we conclude
that
[c, b, a, c, b] = 1. (2)
We again use the Hall-Witt identity. This time we have
1 = [[c, b], a, c]a
−1
[a−1, c−1, [c, b]]c[c, [c, b]−1, a−1][c,b].
By (2) the first commutator commutes with b and it is clear that the last one
also commutes with b as [b, c, c], a and [c, b] all commute with b. It follows
that the second commutator also commutes with b. That is
[[c, b]c, [c, a]−1] commutes with b. (3)
And replacing a with a−1 we see that [[c, b]c, [c, a]] also commutes with b.
This means that for u = [c, b][c, b, c] we have that [u, [c, a]] commutes with
b. Replacing c, a by c−1, a−1 we see that for v = [c, b]−1[c, b, c]2, [v, [c, a]]
commutes with b. Then
[uv, [c, a]] = [u, [c, a]]v[v, [c, a]]
commutes with b as v also commutes with b by Step 1. But uv = [b, c, c]3.
So [b, c, c]3, [[b, c, c]3, [c, a]] and [[b, c, c]3, [c, a]−1] all commute with b. Lemma
2.2 and Lemma 2.3 give just as in the proof of Step 1 that [b, c, c]3 = 1. As
T has no elements of order 3 it follows that [b, c, c] = 1.
6
Step 3. We show that T = {1}.
Now by Step 1 and Step 2, [b, c] commutes with b and c. By (3) we have as
before that all elements in
〈[b, c]〉〈[c,a]〉
commute with b and c and we conclude as before, using Lemmas 2.2 and 2.3,
that [c, b] = 1.
Now b is in the centre of T and thus trivial. Then T = 〈a, c〉 which was
nilpotent by our assumption. Hence T is trivial. 2
3 A general criterion for the nilpotence of a
3-generator group
Let S = 〈a, b, c〉 be any group where
〈a, b〉 is abelian (4)
〈a, c〉 is nilpotent of class at most 2 (5)
〈b, c〉 is nilpotent of class at most 3. (6)
In the last section we proved that S is nilpotent if S also satisfies the 4-
Engel identity. We now drop this last condition. We still want S to keep the
property that
R(S/R(S)) = {1} and S/R(S) has no elements of order 2, 3. (7)
Notice that this property (7) is satisfied by a large class of groups. This holds
for example in any group of exponent p or any n-Engel p-group, p 6= 2, 3. To
this we add the following extra condition
[b, c, c] = 1. (8)
We have not been able to prove the nilpotence of S without the extra property
(8) and maybe the 4-Engel property is needed to prove the stronger version.
Proposition 3.1 S is nilpotent.
7
Proof By property (7) we can assume without loss of generality that S has
trivial Hirsch-Plotkin radical. From the property that 〈a, c〉 is nilpotent of
class at most 2 it follows that 〈c〉〈a〉 is abelian and 1 = [c, a, a] gives
ca
2
= c2ac−1.
We now use this together with the property that 〈c〉〈b〉 is abelian and [a, b] =
1. We have for any r ∈ Z
1 = [ca
2br , ca
2
]
= [c2ab
r
c−b
r
, c2ac−1]
= [c2ab
r
, c2ac−1]c
−b
r
[c−b
r
, c2ac−1]
= [c2ab
r
, c−1]c
−b
r
[c−b
r
, c2a]c
−1
= [c2ab
r
, c−1][c−b
r
, c2a].
It follows that
[c−1, c2ab
r
] = [c−b
r
, c2a].
By our conditions, it is clear that the elements c, cab
r
commute with the
elements ca, cb
r
. Therefore the LHS commutes with cb
r
, ca and the RHS
commutes with c, cab
r
. Thus both sides (being equal) commute with all these
four conjugates. In particular we get
[c, cab
r
]−2 = [cb
r
, ca]−2
and as [c, cab
r
] commutes with [cb
r
, ca] and S has no elements of order 2, it
follows that [c, cab
r
] = [cb
r
, ca]. In particular we have
[c, cab] = [cb, ca] commutes with c, cb, ca, cab (9)
[c, cab
2
] = [cb
2
, ca] commutes with c, cb
2
, ca, cab
2
(10)
[cb, cab
2
] = [cb
2
, cab] commutes with cb, cb
2
, cab, cab
2
(11)
where (11) follows from (9) taking a conjugate with b. The next aim is to
show that H = 〈c〉S is abelian. Clearly
〈c〉S = 〈c, cb, cb
2
, ca, cab, cab
2
〉.
8
As H is normal in S and R(S) = 1, it follows that R(H) = 1. The only
commutators of weight 2 in these generators that are not obviously trivial
are those in (9)-(11). Now 〈c〉〈b〉 is abelian and 1 = [c, b, b, b] implies that
cb
3
= c3b
2
c−3bc.
Using this together with the property that 〈c〉〈a〉 is abelian and [a, b] = 1, we
see that
1 = [cb
3
, cab
3
]
= [c3b
2
c−3bc, c3ab
2
c−3abca]
= [c3b
2
c−3b, c3ab
2
c−3abca]c[c, c3ab
2
c−3abca]
= [c3b
2
c−3b, ca]c[c3b
2
c−3b, c3ab
2
c−3ab]c
ac[c, cac3ab
2
c−3ab]
= [c3b
2
c−3b, ca][c3b
2
c−3b, c3ab
2
c−3ab]c
ac[c, c3ab
2
c−3ab].
This implies that
[c3b
2
c−3b, c3ab
2
c−3ab]c
ac = [ca, c3b
2
c−3b][c3ab
2
c−3ab, c]. (12)
We work now for a moment with the two commutators on the RHS of (12).
We have
[ca, c3b
2
c−3b] = [ca, cc3b
2
c−3b] = [ca, cb
3
]
[c3ab
2
c−3ab, c] = [cac3ab
2
c−3ab, c] = [cab
3
, c].
By the result obtained above we thus have that the two commutators are
equal and that they commute with c and ca. From this and (12) we get
[ca, c3b
2
c−3b]2 = [c3b
2
c−3b, c3ab
2
c−3ab] (13)
and that both sides of (13) commute with c and ca. Next we consider the
RHS of (13). We have
[c3b
2
c−3b, c3ab
2
c−3ab] = [c3b
2
, c3ab
2
c−3ab]c
−3b
[c−3b, c−3abc3ab
2
]
= [c3b
2
, c−3ab]c
−3b
[c−3b, c3ab
2
]
= [c3b
2
, c−3ab][c−3b, c3ab
2
]
(11)
= [cb
2
, cab]−9[cb, cab
2
]−9
(11)
= [cb, cab
2
]−18.
9
By (11) this last element commutes with cb, cab
2
, cb
2
, cab and as we had already
seen that the RHS of (13) commutes with c and ca, it follows that this element
is in the centre of H and thus trivial. As S has no elements of order 2 and
3 it follows that [cb, cab
2
] = 1. As this was the RHS of (13) we also get that
the LHS is trivial. It remains now to see that [ca, cb
2
] = [ca, cb] = 1. We use
the fact that the LHS of (13) is trivial to obtain this. We have
1 = [ca, c3b
2
c−3b]
= [ca, c−3b][ca, c3b
2
]c
−3b
and thus
[ca, c3b
2
] = [ca, c−3b]−c
−3b
(9)
= [ca, cb]3
which gives (using (10)) that
[ca, cb
2
]3 = [ca, cb]3. (14)
From (10) we have that the LHS commutes with ca, cb
2
, c, cab
2
and from (9)
we have that the RHS commutes with cb and cab. Thus both sides of (14)
are in the centre of H and thus trivial. As S has no elements of order 3 it
follows that [ca, cb
2
] = [ca, cb] = 1. Thus H is abelian and therefore trivial.
This implies that S = 〈a, b〉 is abelian and therefore trivial. 2
References
[1] G. Havas and M. R. Vaughan-Lee, 4-Engel groups are locally nilpotent,
Int. J. Alg. and Comp., to appear.
[2] W. Nickel, Computation of nilpotent Engel groups, J. Austral. Math.
Soc. Ser. A 67 (1999), 214-222.
[3] G. Traustason, On 4-Engel groups, J. Algebra 178 (1995), 414-429.
[4] G. Traustason, Two generator 4-Engel groups, Int. J. Alg. and Comp.,
to appear.
[5] M. R. Vaughan-Lee, Engel-4 groups of exponent 5, Proc. London Math.
Soc. 74 (1997), 306-334.


A note on the local nilpotence of 4-Engel groups

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A note on the local nilpotence of 4-Engel groups

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