Groups of prime-power order with a small second derived quotient

16 Pages·0.17 MB·English

by  Csaba Schneider

Groups of prime-power order with a small second
derived quotient
3
0
0
Csaba Schneider
2

n School of Mathematics and Statistics
a
J
The University of Western Australia
9

35 Stirling Highway 6009 Crawley
]
R
Western Australia
G
h. www.maths.uwa.edu.au/∼csaba
t
a [email protected]
m
[
9 January 2003

1
v
8
7
0 Abstract
1
0 Foroddprimesweprovesomestructuretheoremsforfinitep-groupsG,suchthat
3 ′′ ′ ′′ 3
0 G 6= 1 and |G/G | = p . Building on results of Blackburn and Hall, it is shown
/ ′
h that γ3(G) is a maximal subgroup of G, the group G has a central decomposition
t ′
a into two simpler subgroups, and, moreover, G has one of two isomorphism types.
m
: Keywordsandphrases: finitep-groups,derivedsubgroup,secondderivedsubgroup.
v
Xi 2000 Mathematics Subject Classification: 20D15, 20-04.
r
a
1 Introduction
′′
It is well-known that in a finite p-group G the condition G 6= 1 implies that
|G′/G′′| > p3; see for example Huppert [10] III.7.10. In this article we prove a num-
′′
ber of results about groups in which equality holds; that is, we assume that G 6= 1
′ ′′ 3
and |G/G | = p . Such groups have already been investigated by, among others,
′ ′′ 3
N. Blackburn and P. Hall. Blackburn [3] proved that the condition |G/G | = p
1
′′
implies that G is abelian generated by two elements and itis nearly homocyclic. In
the same article healso publisheda result, which heattributed to Hall, that for odd
′′
primes the same condition implies that |G |6 p. Here we mostly consider p-groups
for odd p, and our main results are concerned with such groups.
Let G be a finite p-group and γ (G) the i-th term of the lower central series, so
i
′ ′′
that γ (G) = G, γ (G) = G, etc. If G 6= 1 then we have the following chain of
1 2
normal subgroups:
′ ′′
G > G = γ (G) >γ (G) > γ (G) > G > 1. (1)
2 3 4
′ ′′ 3
If, in addition, we assume that |G/G | = p , then it easily follows that the order
′ 2
of G/γ (G) is at most p . The result of this simple argument is improved by the
3
following theorem.
Theorem 1.1 Let p > 3 and G be a finite p-group, such that |G′/G′′| = p3 and
′′ ′ ′′
G 6= 1. Then |G/γ (G)| = p and G = γ (G).
3 5
TheproofofthisresultisgiveninSection3. Oursecondtheorem,whoseproofis
in Section 4, is that G can bewritten as acentral productof two simpler subgroups.
Theorem 1.2 Let p > 3 and G be a finite p-group, such that |G′/G′′| = p3 and
′′
G 6= 1. Then G can be factorised as G= HU, where
(i) H is a normal subgroup of G generated by at most 5 generators;
(ii) γ (H) = γ (G) for all i > 2;
i i
′
(iii) U is a normal subgroup of G, such that U 6 γ (G);
5
(iv) H and U centralise each other.
An example is given after the proof of this theorem to show that the number
“5” is, in general, best possible, and that there are, in some cases, other central
decompositions of G in which the subgroups can have different isomorphism types.
Our proofs are based on commutator calculus. To simplify notation, we
write long commutators according to the left-normed convention; for example
[a,b,c] = [[a,b],c]. We use the well-known commutator identities that can be found
in most group theory textbooks (see for instance Huppert [10] III.1.2-III.1.3). In
additiontothese,weneedthecollection formula,whichisprovedasLemmaVIII.1.1
2
byHuppertandBlackburn [11]. Wemainly usethisresultinthesimplestcasewhen
it can be stated as
p p ′ p
[x ,y]≡ [x,y] mod(N ) γ (N) where N = (cid:10)x,[x,y](cid:11).
p
The Hall-Witt identity will occur in a lesser known form which can be found in
Magnus, Karrass & Solitar [13] on page 290:
x z y
[x,y,z ][z,x,y ][y,z,x ] = [x,y,z[z,x]][z,x,y[y,z]][y,z,x[x,y]] = 1.
We often manipulate generating sets of groups. In order to avoid cumbersome
repetitions, we introduceapiece of notation. LetGbea group,g asymbolreferring
to a group element, and x an element in G. After the occurrence of the expression
x ; g, the name g will refer to the element x. For example, let G be the cyclic
group of order two and let g denote its non-identity element. If we perform the
replacement g2 ; g, then the symbol g will refer to the identity element of G.
One can naturally ask whether it is possible for a fixed prime to give a classi-
fication of groups which satisfy the conditions of the previous two theorems. It is
6
conceivablethatBlackburn’s[2]descriptionofgroupsofmaximalclasswithorderp
and degree of commutativity 0 is a good starting point. However, increasing the
number of generators and allowing the abelian factor to have exponent higher than
p led to complications which could not be resolved within the research presented
here.
Our results can also be viewed in a wider context. It was first shown by
Hall [8] (Theorem 2.57) that the conditions i > 1 and G(i+1) 6= 1 imply that
i i+1
|G(i)/G(i+1)| > p2 +1, and |G| > p2 +i+1 (see also Huppert [10] III.7.10 and
III.7.11). The lower bound for the order of G has recently been improved by
Mann [12] and the author [15]. Both of these improvements are, however, mi-
(i+1)
nor, and the order of the smallest p-group G such that G 6=1 is still unknown;
the smallest known examples were constructed by Evans-Riley, Newman, and the
author [5]. If p > 3 then we also do not know how sharp Hall’s lower bound is for
(i) (i+1)
|G /G |. As the example of the Sylow 2-subgroup of the symmetric group with
i+2
degree 2 shows, this result is best possible for p = 2; it is not known otherwise.
Our research was originally motivated by these questions, and it is hoped that a
more detailed understanding of groups with a small second derived quotient will
give us a hint of the solution to some of the above problems. Some partial results
can be found in the author’s PhD thesis [15].
3
Our results are inspired by Lie algebra calculations, and it is possible to prove
some of them using the Lie ring method. In fact, Theorem 1.1 can be proved
by first verifying the corresponding result for Lie algebras and then using the Lie
ring associated with the lower central series. This approach would lead to some
interesting new results for Lie algebras, which are beyond the scope of the present
article.
The paper is structured as follows. In Section 2 we prove a lemma which is a
generalisation of Blackburn’s Theorem 1.3 [2]. A consequence of this result is that
we can often restrict our interest to groups which are generated by two or three
elements. In Sections 3 and 4 we prove Theorems 1.1 and 1.2, respectively. In
Section 5 we characterise the commutator subgroup of G, and show that it has one
of two isomorphism types.
2 A general lemma and some consequences
We have seen in the introduction that in a finite p-group G, the conditions
′ ′′ 3 ′′ ′ 2
|G/G | = p and G 6= 1 imply that G/γ (G) has order at most p . The aim
3
of this section is to show that G has a subgroup H with a small generating set,
such that, apart from the first term, the lower central series of H coincides with the
lower central series of G. This result generalises Blackburn’s Theorem 1.3 [2] and
Slattery’s Lemma 2.1 [16].
Lemma 2.1 Let G be a nilpotent group and H a subgroup of G, such that
′ ′
G = H γ (G). Then γ (G) = γ (H) for all i > 2. Moreover, H is a normal
3 i i
subgroup of G.
Proof. First we prove by induction on i that γ (G) = γ (H)γ (G) for all i > 2.
i i i+1
By the conditions of the lemma, this is true for i = 2. Supposethat our claim holds
for some i−1 > 2, and let us show that it holds for i as well. As it is obvious that
γ (H)γ (G) 6 γ (G), we only have to prove that γ (G) 6 γ (H)γ (G). Using
i i+1 i i i i+1
the induction hypothesis and III.1.10(a) of Huppert [10], we compute
γi(G) = [γi−1(G),G] = [γi−1(H)γi(G),G]
= [γi−1(H),G][γi(G),G] = [γi−1(H),G]γi+1(G).
Thereforeitisenoughtoprovethat[γi−1(H),G] 6 γi(H)γi+1(G). Firstwenotethat
γ (G) > γ (H)γ (G) > γ (G), and hence γ (H)γ (G) is a normal subgroup of
i i i+1 i+1 i i+1
4
G. Using the induction hypothesis we obtain
[G,γi−2(H),H] 6 [γi−1(G),H] = [γi−1(H)γi(G),H]
= [γi−1(H),H][γi(G),H] 6 γi(H)γi+1(G)
and
′ ′
[H,G,γi−2(H)] 6 [G,γi−2(H)] = [H γ3(G),γi−2(H)]
′
= [H ,γi−2(H)][γ3(G),γi−2(H)] 6 γi(H)γi+1(G).
Using the Three Subgroups Lemma (see [10] III.1.10(b)), we obtain
[γi−1(H),G] = [γi−2(H),H,G] 6 γi(H)γi+1(G),
and hence our statement is correct.
Let us prove that γ (G) = γ (H) for all i > 2. If the nilpotency class of G is c,
i i
that is γ (G) = 1, then γ (G) = γ (H) = 1. If γ (G) = γ (H) for some
c+1 c+1 c+1 i+1 i+1
i, such that 36 i+1 6 c+1, then, by the result of the previous paragraph,
γ (G) = γ (H)γ (G) = γ (H)γ (H) = γ (H).
i i i+1 i i+1 i
Using induction, we obtain γ (G) = γ (H) for all i > 2. The normality of H is an
i i
easy consequence of the fact that G′ = H′ 6 H. 2
Corollary 2.2 Let G be a finite p-group.
′
(i) If G/γ (G) is cyclic of order p, then G has a 2-generator normal subgroup
3
H, such that γ (G) = γ (H) for all i> 2.
i i
′ 2
(ii) IfG/γ (G)iselementaryabelianoforderp ,thenGhasa3-generatornormal
3
subgroup H, such that γ (G) = γ (H) for all i > 2.
i i
Proof. (i) Suppose that G′/γ (G) = (cid:10)[a,b]γ (G)(cid:11) for some a, b ∈ G, and set
3 3
′ ′
H = (cid:10)a, b(cid:11). As we have H γ (G) = G, Lemma 2.1 implies that H is a normal
3
subgroup and γ (G) = γ (H) for all i> 2.
i i
′ 2
(ii) Suppose that G/γ (G) is elementary abelian of order p , and suppose that
3
′
G/γ (G) = (cid:10)[a,b]γ (G), [c,d]γ (G)(cid:11) for some a, b, c, d ∈ G. Select a subgroup H
3 3 3
in G as follows. If [a,c], [a,d], [b,c], [b,d] are all in γ (G) then let H = (cid:10)a, bc, d(cid:11).
3
α β
Otherwise suppose without loss of generality that [a,c] ≡ [a,b] [c,d] modγ (G)
3
5
for some α and β, such that 0 6 α, β 6 p − 1, and at least one of α and β is
non-zero. If α 6= 0, then set H = (cid:10)a, c, d(cid:11), otherwise set H = (cid:10)a, b, c(cid:11). It is easy
′ ′
to see that H γ (G) = G, and so, using Lemma 2.1, we obtain that H is a normal
3
subgroup and γ (G) = γ (H) for all i> 2. 2
i i
3 Proof of Theorem 1.1
In this section we prove Theorem 1.1.
′ ′′ 3 ′′
Suppose first that G is a finite p-group, such that |G/G | = p and G 6= 1. If
′
the quotient G/γ (G) is cyclic, then Lemma 2.1 of Blackburn [2] implies that
3
′′ ′ ′ ′
G = [G,G] = [G,γ (G)] 6 γ (G), (2)
3 5
and there is a chain
′ ′′
G> G = γ (G) > γ (G) > γ (G) > γ (G) > G > 1 (3)
2 3 4 5
′
of normal subgroups. In particular, if |G/γ (G)| = p, then (2) and (3) imply that
3
′′ ′ 2 ′
G = γ (G); similarly if |G/γ (G)| = p , then (2) implies that G/γ (G) must be
5 3 3
elementary abelian.
′ 2
Now assume that |G/γ (G)| = p ; we show that this can only happen when
3
p = 2. By Corollary 2.2, there is a 3-generator subgroup H of G, such that
γ (G) = γ (H) for all i > 2. After replacing G by H, we may assume without
i i
loss of generality that G = (cid:10)a, b, c(cid:11) for some a, b, c ∈ G. Moreover, from (1) it
′′
follows that G = γ (G), and hence we may suppose that G has nilpotency class 4.
4
′ 2
As G/γ (G) is elementary abelian of order p , we have that there are some α, β,
3
and γ not all zero, such that 0 6 α, β, γ 6 p−1 and
α β γ
[a,b] [a,c] [b,c] ≡ 1 modγ (G).
3
γ
If α = β = 0, then [b,c] ≡ 1 modγ (G), that is [b,c] ∈ γ (G). If α = 0 and β 6= 0,
3 3
then we obtain [aβbγ,c] ≡ 1 modγ (G). If we replace aβbγ ; a, then we obtain
3
that in the new generating set [a,c] ∈ γ (G). Similarly, if α 6= 0 and β = 0, then
3
we replace a−αcγ ; a, and obtain that after the substitution [a,b] ∈ γ (G). If
3
α 6= 0, and β 6= 0, then we replace aβ/αbγ/α ; a and bcβ/α ; b. Then it is easy
to see that in the new generating set [a,b] ∈ γ (G). After possibly reordering the
3
generators, we may suppose without loss of generality that G is generated by three
6
′
elements a, b, and c, such that G/γ (G) = (cid:10)[a,b]γ (G), [a,c]γ (G)(cid:11), and moreover
3 3 3
[b,c] ∈ γ (G). Note that in this case [a,b,c] ≡ [a,c,b] mod γ (G) also holds. Then
3 4
′′
G 6= 1 and γ (G) = 1 imply that
5
−1
[[a,b],[a,c]] = [a,b,a,c][a,b,c,a] 6= 1
and
−1
[[a,c],[a,b]] = [a,c,a,b][a,c,b,a] 6= 1.
If [a,b,a] ∈ γ (G), then [a,b,a,c] ∈ γ (G), and hence [a,b,a,c] = 1. Similarly
4 5
[a,b,c] ∈ γ (G), implies that [a,b,c,a] = 1; therefore at least one of the elements
4
[a,b,a]and[a,b,c] doesnotlieinγ (G). Similarly, atleastoneof[a,c,a] and[a,b,c]
4
must also lie outside γ (G).
4
First we assume that [a,b,c] ∈ γ (G). In this case we must have [a,c,a] 6∈γ (G)
4 4
and [a,b,a] 6∈ γ (G). As γ (G)/γ (G) is cyclic of order p, there is some α, such
4 3 4
that 0 6 α 6 p−1 and [a,bcα,a] ≡ 1 modγ (G), and we carry out the replacement
4
bcα ; b. In the new generating set [b,c] ∈ γ (G) still holds, and, in addition, we
3
obtain [a,b,a] ∈ γ (G).
4
Sowithoutlossofgeneralityweassumethat[a,b,a] ∈ γ (G)and[a,b,c] 6∈ γ (G).
4 4
In this case [a,b,c,a] = [a,c,b,a] 6= 1, in other words a 6∈ C (γ (G)). On the
G 3
other hand, [a,b,b,a] = [a,b,a,b], and hence [a,b,b,a] = 1. If [a,b,b] 6∈ γ (G),
4
then γ (G) = (cid:10)[a,b,b],γ (G)(cid:11), and so a ∈ C (γ (G)), which is impossible; there-
3 4 G 3
fore [a,b,b] ∈ γ (G). If [a,c,a] 6∈ γ (G) then there is some α 6= 0, such that
4 4
[a,c,abα] ∈ γ (G); in this case we let abα ; a and obtain [a,c,a] ∈ γ (G). In the
4 4
new generating set [b,c] ∈ γ (G) and [a,b,a], [a,b,b] ∈ γ (G) still hold. Then
3 4
1 = [[a,b],a,c][[a,c],[a,b]][c,[a,b],a]
−1 −1 −2
= [a,b,a,c][a,c,a,b][a,b,c,a] [a,b,c,a] = [a,b,c,a] ,
2
that is, [a,b,c,a] = 1, and hence p = 2. This completes the proof of Theorem 1.1.
4 The proof of Theorem 1.2
In the previous section we proved Theorem 1.1, and hence we know that in a group
′
G the conditions of Theorem 1.2 imply that |G/γ (G)| = p. Thus, according to
3
Corollary 2.2, G has a 2-generator subgroup H, such that for all i > 2 we have
γ (G) = γ (H). We use this subgroup to obtain the desired factorisation. First we
i i
show that we can choose a generating set which satisfies some extra conditions.
7
Lemma 4.1 Let G be a 2-generator, finite p-group, such that |G′/G′′| = p3,
′ ′′
|G/γ (G)| = p, and G 6= 1. Then generators a and b of G can be chosen, such
3
that the following hold:
(i) γ (G)/γ (G) = (cid:10)[b,a]γ (G)(cid:11);
2 3 3
(ii) γ (G)/γ (G) = (cid:10)[b,a,a]γ (G)(cid:11) and [b,a,b] ∈ γ (G);
3 4 4 4
(iii) γ (G)/γ (G) = (cid:10)[b,a,a,a]γ (G)(cid:11) and [b,a,a,b] ∈ γ (G);
4 5 5 5
(iv) γ (G)/γ (G) = (cid:10)[b,a,a,a,b]γ (G)(cid:11) and [b,a,a,a,a] ∈ γ (G).
5 6 6 6
Proof. Wemaysupposewithoutlossofgenerality thatGhasclass5. Asnoticedin
′
the introduction, our conditions imply that the factors G/γ (G), γ (G)/γ (G), and
3 3 4
γ (G)/γ (G)arecyclicwithorderp. UsingtheargumentpresentedbyBlackburn[2]
4 5
in Lemma 2.9, we can choose the generating set {a, b}, so that properties (i)-(iii)
′′ ′′
hold. It follows from (2) and (3) that G = γ (G), and G = (cid:10)[[b,a,a],[b,a]](cid:11).
5
As the element [[b,a,a],[b,a]] is central and has order p, we have |γ (G)| = p, and
5
using Blackburn’s argument on page 89, the set {a, b} can be chosen so that the
additional property (iv) also holds. 2
Lemma 4.2 Let p > 3 and G be a finite p-group, such that |G′/G′′| = p3 and
′′
G 6= 1. Then G has a minimal generating set {a,b,u ,u ,...,u }, such that
1 2 r
(i) H = (cid:10)a,b(cid:11) is a normal subgroup of G, such that γ (H) = γ (G) for all i > 2;
i i
further, a and b are as in Lemma 4.1;
(ii) [a,u ]∈ γ (G) for all u ;
i 5 i
(iii) [b,u ]∈ γ (G) for all u ;
i 4 i
(iv) [u ,u ] ∈ γ (G) for all u and u .
i j 5 i j
In particular, u ,...,u ∈ C (cid:0)G′(cid:1).
1 r G
Proof. First recall Hall’s theorem that |G′′| = p, and so (3) implies that G has
class 5. Select a, b ∈ G, such that the subgroup H = (cid:10)a,b(cid:11) and its generators
are as in Lemma 4.1. It is easy to see that a, b are linearly independent modulo
the Frattini subgroup of G. Therefore they can be viewed as elements of a min-
imal generating set {a,b,u ,...,u }. Now suppose that for each i ∈ {1,...,r},
1 r
[u ,a] ≡ [b,a]αi and [u ,b] ≡ [b,a]βi moduloγ (G) with some α , β ∈ {0,...,p−1}.
i i 3 i i
8
Then [u b−αiaβi,b] ∈ γ (G) and also [u b−αiaβi,a] ∈ γ (G). If we perform the re-
i 3 i 3
placement u b−αiaβi ; u , then it is easy to see that {a,b,u ,...,u } is also a
i i 1 r
minimal generating set for G and
[a,u ], [b,u ]∈ γ (G) for all i∈ {1,...,r}.
i i 3
Now suppose that for all i∈ {1,...,r} we have
[u ,a] ≡ [b,a,a]αi[b,a,a,a]βi mod γ (G)
i 5
for some α , β ∈ {0,...,p−1}. Then computing modulo γ (G) we obtain
i i 5
[u [b,a]−αi[b,a,a]−βi,a]
i
= [u [b,a]−αi,a][u [b,a]−αi,a,[b,a,a]−βi][[b,a,a]−βi,a]]
i i
≡ [u ,a][u ,a,[b,a]−αi][[b,a]−αi,a][[b,a,a]−βi,a]
i i
≡ [u ,a][b,a,a]−αi[b,a,a,a]−βi ≡ 1.
i
If we replace u [b,a]−αi[b,a,a]−βi ; u , then [u ,a] ∈ γ (G). Since the images of
i i i 5
the u over the Frattini subgroup did not change, the set {a,b,u ,...,u } is still
i 1 r
a minimal generating system for G. We show that this generating set satisfies the
properties required by the lemma.
We claim that [u ,b] ∈γ (G) for all i∈ {1,...,r}. To prove this weobserve that
i 4
1 = [b,a,u [u ,b]][u ,b,a[a,u ]][a,u ,b[b,a]] = [b,a,[u ,b]][b,a,u ][u ,b,a],
i i i i i i i i
and thus
−1 −1 −1 −1
[b,a,u ] = [u ,b,a] [b,a,[u ,b]] = [[b,u ] ,a] [[u ,b],[b,a]]
i i i i i
= [b,u ,a][[u ,b],[b,a]]. (4)
i i
In particular, [b,a,u ]∈ γ (G). Now consider
i 4
1 = [[b,a],a,u [u ,[b,a]]][u ,[b,a],a[a,u ]][a,u ,[b,a][b,a,a]] =[b,a,a,u ].
i i i i i i
We can obtain similarly [b,a,a,a,u ] = 1. As [u ,b] ∈ γ (G), [u ,b] ≡ [b,a,a]εi
i i 3 i
modulo γ (G) for some ε ∈ {0,...,p−1}. The Hall-Witt identity implies that
4 i
1 = [[b,a],u ,b[b,[b,a]]][b,[b,a],u [u ,b]][u ,b,[b,a][b,a,u ]]
i i i i i
= [[b,a],u ,b][[u ,b],[b,a]].
i i
9
Using (4) we get [b,a,u ,b] =[b,a,a,a,b]−εi. Moreover,
i
[[u ,b],[b,a]] = [[b,a,a]εi,[b,a]] = [b,a,a,a,b]−εi,
i
and thus [b,a,a,a,b]−2εi = 1, from which it follows that ε = 0, in other words
i
[u ,b] ∈ γ (G).
i 4
We now prove that u ,...,u ∈ C (cid:0)G′(cid:1). We have already seen that [b,a,a],
1 r G
[b,a,a,a] are centralised by the u , so it suffices to prove that [b,a,u ] = 1 for all
i i
i∈ {1,...,r}. This is clear because
1= [b,a,u [u ,b]][u ,b,a[a,u ]][a,u ,b[b,a]] = [b,a,u ].
i i i i i i
It remains to show that [u ,u ] lies in γ (G) for all i, j ∈ {1,...,r}. It easily
i j 5
follows using the Hall-Witt identity that [u ,u ,a] = 1 and [u ,u ,b] = 1, therefore
i j i j
[u ,u ] ∈ Z(H)∩H′ = γ (H) = γ (G). The proof is complete. 2
i j 5 5
Proof of Theorem 1.2. Choose a generating set {a,b,u ,u ...,u } for G as in
1 2 r
the previous lemma. In the first stage of the proof we show that this generating set
can be modified so that, in addition to the properties required by Lemma 4.2, one
of the following holds:
(a) u ,...,u ∈ C (a); or
1 r G
(b) u ,...,u ∈ C (cid:0)(cid:10)a,u (cid:11)(cid:1).
2 r G 1
If u ,...,u ∈ C (a) then (a) holds and we are done. Suppose that there is at
1 r G
least one u which does not centralise a. Without loss of generality we may assume
i
that [u ,a] = [b,a,a,a,b]. If [u ,a] = [b,a,a,a,b]αi for some i ∈ {2,...,r}, then let
1 i
u u−αi ; u . In this way we obtain a generating set {a,b,u ,...,u }, such that
i 1 i 1 r
[u ,a] = [b,a,a,a,b] and (cid:10)u ,...,u (cid:11) 6C (a).
1 2 r G
If u ,...,u centralise u , then (b) holds and we are done. We assume without
2 r 1
loss of generality that [u ,u ] = [b,a,a,a,b]. If [u ,u ] = [b,a,a,a,b]βi for some
2 1 i 1
i ∈ {3,...,r}, then let u u−βi ; u . In this way we obtain a generating set, such
i 2 i
that u ,...,u centralise a, and u ,...,u centralise u . Repeating this process, we
2 r 3 r 1
construct a generating set {a,b,u ,...,u ,...,u }, such that
1 k r
1. [u ,a] = [b,a,a,a,b];
1
2. u ,...,u centralise a;
2 r
3. [u ,u ] =[b,a,a,a,b] for all i ∈ {1,...,k−1};
i+1 i
10