Noether's problem for abelian extensions of cyclic $p$-groups

0.26 MB·English

by  Ivo M. Michailov

NOETHER’S PROBLEM FOR ABELIAN EXTENSIONS OF CYCLIC
p-GROUPS
IVO M. MICHAILOV
Abstract. Let K be a field and G be a finite group. Let G act on the rational
3
1 function field K(x(g) : g ∈ G) by K automorphisms defined by g·x(h) = x(gh) for
0 any g,h∈ G. Denote by K(G) the fixed field K(x(g) : g ∈ G)G. Noether’s problem
2 then asks whether K(G) is rational (i.e., purely transcendental) over K. The first
n main result of this article is that K(G) is rational over K for a certain class of p-
a groupshavinganabeliansubgoupofindexp. ThesecondmainresultisthatK(G)is
J rationaloverK for anygroupoforderp5 orp6 (p isanoddprime)havinganabelian

0 normalsubgroupsuchthatits quotientgroupiscyclic. (Inboththeoremsweassume
3 that if charK 6= p then K contains a primitive pe-th root of unity, where pe is the

exponent of G.)
]
G
A
.
h
t
a
m 1. Introduction
[

Let K be any field. A field extension L of K is called rational over K (or K-rational,
1
v
for short) if L ≃ K(x ,...,x ) over K for some integer n, with x ,...,x algebraically
4 1 n 1 n
8
independent over K. Now let G be a finite group. Let G act on the rational function
2
7
field K(x(g) : g ∈ G) by K automorphisms defined by g·x(h) = x(gh) for any g,h ∈ G.
.
1
0 Denoteby K(G) thefixed field K(x(g) : g ∈ G)G. Noether’s problemthenasks whether
3
1 K(G) is rational over K. This is related to the inverse Galois problem, to the existence
:
v
of generic G-Galois extensions over K, and to the existence of versal G-torsors over
i
X
K-rational field extensions [Sw, Sa1, GMS, 33.1, p.86]. Noether’s problem for abelian
r
a
groups was studied extensively by Swan, Voskresenskii, Endo, Miyata and Lenstra,
etc. The reader is referred to Swan’s paper for a survey of this problem [Sw]. Fischer’s
Theorem is a starting point of investigating Noether’s problem for finite abelian groups
in general.
Date: January 31, 2013.
2000 Mathematics Subject Classification. primary 14E08 14M20;secondary 13A50,12F12.
Key words and phrases. Noether’s problem, Rationality problem, Meta-abelian group actions.
This work is partially supported by a project No RD-05-275/15.01.2013of Shumen University.
1
2 IVO M. MICHAILOV
Theorem 1.1. (Fischer [Sw, Theorem 6.1]) Let G be a finite abelian group of exponent
e. Assume that (i) either char K = 0 or char K > 0 with char K ∤ e, and (ii) K
contains a primitive e-th root of unity. Then K(G) is rational over K.
On the other hand, just a handful of results about Noether’s problem are obtained
when the groups are non-abelian. This is the case even when the group G is a p-group.
The reader is referred to [CK, HuK, Ka1, Ka2, Ka3] for previous results of Noether’s
problem for p-groups. The following theorem of Kang generalizes Fischer’s theorem for
the metacyclic p-groups.
Theorem 1.2. (Kang[Ka1, Theorem 1.5]) Let G be a metacyclic p-group with exponent
pe, and let K be any field such that (i) char K = p, or (ii) char K 6= p and K contains
a primitive pe-th root of unity. Then K(G) is rational over K.
The next job is to study Noether’s problem for meta-abelian groups. Three results
due to Haeuslein, Hajja and Kang respectively are known.
Theorem 1.3. (Haeuslein [Ha]) Let K be a field and G be a finite group. Assume that
(i) G contains an abelian normal subgroup H so that G/H is cyclic of prime order p,
(ii) Z[ζ ] is a unique factorization domain, and (iii) ζ ∈ K where e is the exponent
p pe
of G. If G → GL(V) is any finite-dimensional linear representation of G over K, then
K(V)G is rational over K.
Theorem 1.4. (Hajja [Haj]) Let K be a field and G be a finite group. Assume that (i)
G contains an abelian normal subgroup H so that G/H is cyclic of order n, (ii) Z[ζ ]
n
is a unique factorization domain, and (iii) K is algebraically closed with charK = 0. If
G → GL(V) is any finite-dimensional linear representation of G over K, then K(V)G
is rational over K.
Theorem 1.5. ([Ka3, Theorem 1.4]) Let K be a field and G be a finite group. Assume
that (i) G contains an abelian normal subgroup H so that G/H is cyclic of order n,
(ii) Z[ζ ] is a unique factorization domain, and (iii) ζ ∈ K where e is the exponent of
n e
G. If G → GL(V) is any finite-dimensional linear representation of G over K, then
K(V)G is rational over K.
NOETHER’S PROBLEM FOR ABELIAN EXTENSIONS OF CYCLIC p-GROUPS 3
Note that those integers n for which Z[ζ ] is a unique factorization domain are
n
determined by Masley and Montgomery.
Theorem 1.6. (MasleyandMontgomery[MM]) Z[ζ ] is a unique factorization domain
n
if and only if 1 ≤ n ≤ 22, or n = 24,25,26,27,28,30,32,33,34,35,36,38,40,42,45,48,
50,54,60,66,70,84,90.
Therefore, Theorem 1.3 holds only for primes p such that 1 ≤ p ≤ 19. One of
the goals of our paper is to show that the this condition can be waived, under some
additional assumptions regarding the structure of the abelian subgroup H.
Consider the following situation. Let G be a group of order pn for n ≥ 2 with an
abelian subgroup H of order pn−1. Bender [Be] determined some interesting properties
of these groups. We study further the case when the p-th lower central subgroup G
(p)
is trivial. (Recall that G = G and G = [G,G ] for i ≥ 1 are called the lower
(0) (i) (i−1)
central series.) For our purposes we need to classify with generators and relations these
groups. We achieve this in the following lemma.
Lemma 1.7. Let G be a group of order pn for n ≥ 2 with an abelian subgroup H
of order pn−1. Choose any α ∈ G such that α generates G/H, i.e., α ∈/ H,αp ∈ H.
Denote H(p) = {h ∈ H : hp = 1,h ∈/ Hp}, and assume that [H(p),α] ⊂ H(p). Assume
also that the p-th lower central subgroup G is trivial. (Recall that G = G and
(p) (0)
G = [G,G ] for i ≥ 1 are called the lower central series.) Then H is a direct
(i) (i−1)
product of normal subgroups of G that are of the following four types:
(1) (C )s for some s ≥ 1. There exist generators α ,...,α of (C )s, such that
p 1 s p
[α ,α] = α for 1 ≤ j ≤ s−1 and α ∈ Z(G).
j j+1 s
(2) C for some a ≥ 1. There exists a generator β of C such that [β,α] = βbpa−1
pa pa
for some b : 0 ≤ b ≤ p−1.
(3) C × C × ··· × C × (C )s for some k ≥ 1,a ≥ 2,s ≥ 1. There exist
pa1 pa2 pak p i
generators α11,α21,...,αk1 of Cpa1 × Cpa2 × ··· × Cpak such that [αi,1,α] =
αpai+1−1 ∈ Z(G) for i = 1,...,k−1. There also exist generators α ,...,α
i+1,1 k,2 k,s+1
of (C )s, such that [α ,α] = α for 1 ≤ j ≤ s and α ∈ Z(G).
p k,j k,j+1 k,s+1
4 IVO M. MICHAILOV
(4) Cpa1 × Cpa2 ×···× Cpak for some k ≥ 2,ai ≥ 2. For any i : 1 ≤ i ≤ k there
exists a generator α of the factor C , such that [α ,α] = αpai−1 ∈ Z(G)
i,1 pai i,1 i+1,1
and [α ,α] ∈ hαpa1−1,...,αpak−1i.
k,1 1,1 k,1
ThefirstmainresultofthispaperisthefollowingtheoremwhichgeneralizesTheorem
1.3.
Theorem 1.8. Let G be a group of order pn for n ≥ 2 with an abelian subgroup H
of order pn−1, and let G be of exponent pe. Choose any α ∈ G such that α generates
G/H, i.e., α ∈/ H,αp ∈ H. Denote H(p) = {h ∈ H : hp = 1,h ∈/ Hp}, and assume
that [H(p),α] ⊂ H(p). Denote by G = [G,G ] the lower central series for i ≥ 1
(i) (i−1)
and G = G. Let the p-th lower central subgroup G be trivial. Assume that (i) char
(0) (p)
K = p > 0, or (ii) char K 6= p and K contains a primitive pe-th root of unity. Then
K(G) is rational over K.
The key idea to prove Theorem 1.8 is to find a faithful G-subspace W of the regular
representation space K · x(g) and to show that WG is rational over K. The
g∈G
subspace W is obtained as an induced representation from H by applying Lemma 1.7.
L
(A particular case we proved in the preprint [Mi].)
The next goal of our article is to study Noether’s problem for some groups of orders
p5 and p6 for any odd prime p. We use the list of generators and relations for these
groups, given by James [Ja]. It is known that K(G) is always rational if G is a p-group
of order ≤ p4 and ζ ∈ K where e is the exponent of G (see [CK]). However, in [HoK]
e
is shown that there exists a group G of order p5 such that C(G) is not rational over C.
The second main result of this article is the following rationality criterion for the
groups of orders p5 and p6, having an abelian normal subgroup such that its quotient
group is cyclic.
Theorem 1.9. Let G be a group of order pn for n ≤ 6 with an abelian normal subgroup
H, such that G/H is cyclic. Let G be of exponent pe. Assume that (i) char K = p > 0,
or (ii) char K 6= p and K contains a primitive pe-th root of unity. Then K(G) is
rational over K.
NOETHER’S PROBLEM FOR ABELIAN EXTENSIONS OF CYCLIC p-GROUPS 5
We do not know whether Theorem 1.9 holds for any n ≥ 7. However, we should not
”over-generalize” Theorem 1.9 to the case of any meta-abelian group because of the
following theorem of Saltman.
Theorem 1.10. (Saltman[Sa2]) Forany prime number p and for any field K with char
K 6= p (in particular, K may be an algebraically closed field), there is a meta-abelian
p-group G of order p9 such that K(G) is not rational over K.
We organize this paper as follows. We recall some preliminaries in Section 2 that will
be used in the proofs of Theorems 1.8 and 1.9. There we also prove Lemma 2.5 which
is a generalization of Kang’s argument from [Ka2, Case 5, Step II]. In Section 3 we
prove Lemma 1.7 which is of independent interest, since it provides a list of generators
and relations for any p-group G having an abelian subgroup H of index p, provided
that [H(p),α] ⊂ H(p) and G = 1. Our main results – Theorems 1.8 and 1.9 – we
(p)
prove in Sections 4 and 5 respectively.
2. Preliminaries
We list several results which will be used in the sequel.
Theorem 2.1. ([HK, Theorem 1]) Let G be a finite group acting on L(x ,...,x ),
1 m
the rational function field of m variables over a field L such that
(i): for any σ ∈ G,σ(L) ⊂ L;
(ii): the restriction of the action of G to L is faithful;
(iii): for any σ ∈ G,
σ(x ) x
1 1
. .
.. = A(σ) .. +B(σ)
   
σ(x ) x
m m
   
where A(σ) ∈ GL (L) and B(σ) is m × 1 matrix over L. Then there exist
m
z ,...,z ∈ L(x ,...,x ) so that L(x ,...,x )G = LG(z ,...,z ) and σ(z ) =
1 m 1 m 1 m 1 m i
z for any σ ∈ G, any 1 ≤ i ≤ m.
i
Theorem 2.2. ([AHK, Theorem 3.1]) Let G be a finite group acting on L(x), the ratio-
nal function field of one variable over a field L. Assume that, for any σ ∈ G,σ(L) ⊂ L
and σ(x) = a x+ b for any a ,b ∈ L with a 6= 0. Then L(x)G = LG(z) for some
σ σ σ σ σ
z ∈ L[x].
6 IVO M. MICHAILOV
Theorem 2.3. ([CK, Theorem 1.7]) If charK = p > 0 and G is a finite p-group, then
K(G) is rational over K.
e
The following Lemma can be extracted from some proofs in [Ka2, HuK].
Lemma 2.4. Let hτi be a cyclic group of order n > 1, acting on K(v ,...,v ), the
1 n−1
rational function field of n−1 variables over a field K such that
τ : v 7→ v 7→ ··· 7→ v 7→ (v ···v )−1 7→ v .
1 2 n−1 1 n−1 1
If K contains a primitive n-th root of unity ξ, then K(v ,...,v ) = K(s ,...,s )
1 n−1 1 n−1
where τ : s 7→ ξis for 1 ≤ i ≤ n−1.
i i
Proof. Define w = 1 + v + v v + ··· + v v ···v ,w = (1/w ) − 1/n,w =
0 1 1 2 1 2 n−1 1 0 i+1
(v v ···v /w )−1/n for 1 ≤ i ≤ n−1. Thus K(v ,...,v ) = K(w ,...,w ) with
1 2 i 0 1 n−1 1 n
w +w +···+w = 0 and
1 2 n
τ : w 7→ w 7→ ··· 7→ w 7→ w 7→ w .
1 2 n−1 n 1
Define s = ξ−ijw for 1 ≤ i ≤ n−1. Then K(w ,...,w ) = K(s ,...,s )
i 1≤j≤n j 1 n 1 n−1
and τ : s 7→ ξis for 1 ≤ i ≤ n−1. (cid:3)
i P i
Moreover, we are now going to generalize Kang’s argument from [Ka2, Case 5, Step
II], obtaining the following Lemma which plays an important role in our work.
Lemma 2.5. Let k > 1, let p be any prime and let hαi be a cyclic group of order p,
acting on K(y ,y ...,y : 1 ≤ i ≤ p − 1), the rational function field of k(p − 1)
1i 2i ki
variables over a field K such that
α : y 7→ y 7→ ··· 7→ y 7→ (y y ···y )−1, for 1 ≤ j ≤ k.
j1 j2 jp−1 j1 j2 jp−1
Assume that K(v ,v ...,v : 1 ≤ i ≤ p − 1) = K(y ,y ...,y : 1 ≤ i ≤ p − 1)
1i 2i ki 1i 2i ki
where for any j : 1 ≤ j ≤ k and for any i : 1 ≤ i ≤ p−1 the variable v is a monomial
ji
in the variables y ,y ...,y . Assume also that the action of α on K(v ,v ...,v :
1i 2i ki 1i 2i ki
1 ≤ i ≤ p−1) is given by
α : v 7→ v vp , v 7→ v 7→ ··· 7→ v 7→ A ·(v vp−1vp−2···v2 )−1,
j1 j1 j2 j2 j3 jp−1 j j1 j2 j3 jp−1
for 1 ≤ j ≤ k,
NOETHER’S PROBLEM FOR ABELIAN EXTENSIONS OF CYCLIC p-GROUPS 7
where A is some monomial in v ,...,v for 2 ≤ j ≤ k and A = 1. If K
j 1i j−1i 1
contains a primitive p-th root of unity ζ, then K(v ,v ...,v : 1 ≤ i ≤ p − 1) =
1i 2i ki
K(s ,s ...,s : 1 ≤ i ≤ p−1) where α : s 7→ ζis for 1 ≤ j ≤ k,1 ≤ i ≤ p−1.
1i 2i ki ji ji
Proof. We write the additive version of the multiplication action of α, i.e., consider the
Z[π]-module M = (⊕ Z·v ), where π = hαi. Denote the submodules
1≤m≤k 1≤i≤p−1 mi
M = (⊕ Z·v ) for 1 ≤ j ≤ k. Thus α has the following additive action
j 1≤m≤j 1≤Li≤p−1 mi
L
α : v 7→ v +pv ,
j1 j1 j2
v 7→ v 7→ ··· 7→ v 7→ A −v −(p−1)v −(p−2)v −···−2v ,
j2 j3 jp−1 j j1 j2 j3 jp−1
where A ∈ M .
j j−1
By Lemma 2.4, M is isomorphic to the Z[π]-module N = ⊕ Z · u where
1 1≤i≤p−1 i
u = v ,u = αi−1 ·v for 2 ≤ i ≤ p−1, and
1 12 i 12
α : u 7→ u 7→ ··· 7→ u 7→ −u −u −···−u 7→ u .
1 2 p−1 1 2 p−1 1
Let Φ (T) ∈ Z[T] be the p-th cyclotomic polynomial. Since Z[π] ≃ Z[T]/(Tp −1),
p
we find that Z[π]/Φ (α) ≃ Z[T]/Φ (T) ≃ Z[ω], the ring of p-th cyclotomic integer. As
p p
Φ (α) · x = 0 for any x ∈ N, the Z[π]-module N can be regarded as a Z[ω]-module
p
through the morphism Z[π] → Z[π]/Φ (α). When N is regarded as a Z[ω]-module,
p
N ≃ Z[ω] the rank-one free Z[ω]-module.
We claim that M itself can be regarded as a Z[ω]-module, i.e., Φ (α)·M = 0.
p
Returntothemultiplicative notations. Notethatallv ’saremonomialsiny ’s. The
ji ji
action of α on y given in the statement satisfies the relation αm(y ) = 1
ji 0≤m≤p−1 ji
for any 1 ≤ j ≤ k,1 ≤ i ≤ p−1. Using the additive notations, we get Φ (α)·y = 0.
Q p ji
Hence Φ (α)·M = 0.
p
DefineM′ = M/M . ItfollowsthatwehaveashortexactsequenceofZ[π]-modules
k−1
(2.1) 0 → M → M → M′ → 0.
k−1
Since M is a Z[ω]-module, (2.1) is a short exact sequence of Z[ω]-modules. Proceeding
by induction, we obtain that M is a direct sum of free Z[ω]-modules isomorphic to N.
Therefore, M ≃ ⊕ N , where N ≃ N is a free Z[ω]-module, and so a Z[π]-module
1≤j≤k j j
also (for 1 ≤ j ≤ k).
8 IVO M. MICHAILOV
Finally, we interpret the additive version of M ≃ ⊕ N ≃ Nk it terms of the
1≤j≤k j
multiplicative version as follows: There exist w that are monomials in v for 1 ≤ j ≤
ji ji
k,1 ≤ i ≤ p−1 such that K(w ) = K(v ) and α acts as
ji ji
α : w 7→ w 7→ ··· 7→ w 7→ (w w ...w )−1 for 1 ≤ j ≤ k.
j1 j2 jp−1 j1 j2 jp−1
AccordingtoLemma2.4,theaboveactioncanbelinearizedaspointedinthestatement.
(cid:3)
Now, let G be any metacyclic p-group generated by two elements σ and τ with
relations σpa = 1,τpb = σpc and τ−1στ = σε+δpr where ε = 1 if p is odd, ε = ±1 if
p = 2, δ = 0,1 anda,b,c,r ≥ 0 are subject to some restrictions. For thethe description
of these restrictions see e.g. [Ka1, p. 564].
Theorem 2.6. (Kang [Ka1, Theorem 4.1]) Let p be a prime number, m,n and r are
positive integers, k = 1 + pr if (p,r) 6= (2,1) (resp. k = −1 + 2r with r ≥ 2). Let
G be a split metacyclic p-group of order pm+n and exponent pe defined by G = hσ,τ :
σpm = τpn = 1,τ−1στ = σki. Let K be any field such that charK 6= p and K contains
a primitive pe-th root of unity, and let ζ be a primitive pm-th root of unity. Then
K(x ,x ,...,x )G is rational over K, where G acts on x ,...,x by
0 1 pn−1 0 pn−1
σ : x 7→ ζkix ,
i i
τ : x 7→ x 7→ ··· 7→ x 7→ x .
0 1 pn−1 0
3. Proof of Lemma 1.7
It is well known that H is a normal subgroup of G. We divide the proof into several
steps.
Step I. Let β be any element of H that is not central. Since G = {1}, there exist
1 (p)
β ,...,β ∈ H for some k : 2 ≤ k ≤ p such that [β ,α] = β , where 1 ≤ j ≤ k−1 and
2 k j j+1
β 6= 1 is central. We are going to show now that the order of β is not greater than
k 2
p. In particular, from the multiplication rule [a,α][b,α] = [ab,α] (for any a,b ∈ H) it
follows that all p-th powers are contained in the center of G.
From [β ,α] = β it follows the well known formula
j j+1
(p) (p) ( p )
(3.1) α−pβ αp = β β 1 β 2 ···β p−1 β ,
1 1 2 3 p p+1
NOETHER’S PROBLEM FOR ABELIAN EXTENSIONS OF CYCLIC p-GROUPS 9
where we put β = ··· = β = 1. Since αp is in H, we obtain the formula
k+1 p+1
(p) (p) ( p )
β 1 β 2 ···β k−1 = 1.
2 3 k
Hence (β · βaj)p = 1 for some integers a . It is not hard to see that this identity is
2 j6=2 j j
impossible if the order of β is greater than p. Indeed, if ℓ = max{j : βp 6= 1}, then βp
Q 2 j ℓ
is in the subgroup generated by βp,...,βp . Therefore [βp,α] = [βb2p···βbℓ−1p,α] =
2 ℓ−1 ℓ 2 ℓ−1
βb2p···βbℓ−1p 6= 1 for some b ,...,b ∈ Z . On the other hand, [βp,α] = βp = 1,
3 ℓ 2 ℓ−1 p ℓ ℓ+1
which is a contradiction.
Step II. Let us write the decomposition of H as a direct product of cyclic subgroups
(not necessarily normal in G): H ≃ (C )t × C × C × ···C for 0 ≤ t,2 ≤
p pa1 pa2 pas
a ≤ a ≤ ··· ≤ a . Choose a generator α ∈ C . Since G = {1}, there exist
1 2 s 11 pa1 (p)
α ,...,α ∈ H for some k : 2 ≤ k ≤ p such that [α ,α] = α , where 1 ≤ j ≤ k−1
12 1k 1j 1j+1
and α 6= 1 is central. From Step I it follows that the order of α is not greater than
1k 12
p. We are going to define a normal subgroup of G which depends on the nature of the
element α . We will denote it by hhα ii, and call it the commutator chain of α .
12 11 11
Simultaneously, we will define a complement in H denoted by hhα ii.
11
Case II.1. Let α = αpa1−1c1 for some c : 0 ≤ c ≤ p −1. Define hhα ii = hα i.
12 11 1 1 11 11
Clearly, hhα ii is a normal subgroup of type (2). Define hhα ii = (C )t·hα ,...,α i.
11 11 p 21 s1
Case II.2. Let α ∈/ Hp. According to the assumption in the statement of our
12
Lemma, [H(p),α]∩Hp = {1},wehaveα ∈/ Hpforallj. Definehhα ii = hα ,...,α i.
1j 11 11 1k
Therefore hhα ii ≃ C ×(C )k−1 is a normal subgroup of type (3). Define hhα ii =
11 pa1 p 11
(C )t−k+1 ·hα ,...,α i, where (C )t−k+1 is the complement of (C )k−1 in (C )t.
p 21 s1 p p p
Case II.3. Let α ∈ Hp. Then α = αpai−1di, where A ⊂ {1,2,...,s},1 ≤
12 12 i∈A i1
d ≤ p−1. Put i = min{i ∈ A}.
i 0 Q
If i = 1, then α = (αd1 αpai−a1di)pa1−1. Now, we can replace the generator
0 12 11 i∈A,i6=1 i1
α with α′ = αd1 αpai−a1di. Clearly, ord(α′ ) = ord(α ) and [α′ ,α] ∈ hα′ i,
11 11 11 i∈A,i6=1Qi1 11 11 11 11
so this case is reduced to Case I.
Q
Ifi > 1,thenα = (αdi0 αpai−ai0di)pai0−1. Wecanreplacethegeneratorα
0 12 i01 i∈A,i6=i0 i1 i01
with α′ = αdi0 αpai−ai0di. Clearly, ord(α′ ) = ord(α ) and α′pai0−1 = α .
i01 i01 i∈A,i6=i0 i1Q i01 i01 i01 12
For abuse of notation we will assume henceforth that i = 2 and αpa2−1 = α .
Q 0 21 12
Consider α = [α ,α]. We have three possibilities now.
22 21
10 IVO M. MICHAILOV
Subcase II.3.1. α ∈ hαpa1−1,αpa1−1i. Define hhα ii = hα ,α i. Therefore
22 11 21 11 11 21
hhα ii ≃ C ×C is a normal subgroup of type (4).
11 pa1 pa2
Subcase II.3.2. α ∈/ Hp. Then there exist α ,...,α ∈ H for some ℓ : 2 ≤ ℓ ≤ p
22 22 2ℓ
such that [α ,α] = α , where 1 ≤ j ≤ ℓ − 1 and α 6= 1 is central. Define
2j 2j+1 2ℓ
hhα ii = hα ,α ,α ,...,α i. Therefore hhα ii ≃ C ×C ×(C )ℓ−1 is a normal
11 11 21 22 2ℓ 11 pa1 pa2 p
subgroup of type (3).
Subcase II.3.3. α ∈ Hp. According to the observations we have just made, this
22
subcase leads to the following two final possibilities.
Sub-subcase II.3.3.1. α = αpa3−1,...,α = αpar−1,α ∈ hαpa1−1,...,αpar−1i.
22 31 r−12 r1 r2 11 r1
Define hhα ii = hα ,α ,...,α i. Therefore hhα ii ≃ C ×C ×···×C is a
11 11 21 r1 11 pa1 pa2 par
normal subgroup of type (4). Define hhα ii = (C )t ·hα ,...,α i.
11 p r+11 s1
Sub-subcase II.3.3.2. α = αpa3−1,...,α = αpar−1,α ∈/ Hp. Then there exist
22 31 r−12 r1 r2
α ,...,α ∈ H for some ℓ : 2 ≤ ℓ ≤ p such that [α ,α] = α , where 1 ≤ j ≤ ℓ−1
r2 rℓ rj rj+1
and α 6= 1 is central. Define hhα ii = hα ,α ,...,α ,α ,...,α i. Therefore
rℓ 11 11 21 r1 r2 rℓ
hhα ii ≃ C × C × ···C × (C )ℓ−1 is a normal subgroup of type (3). Define
11 pa1 pa1 par p
hhα ii = (C )t−ℓ+1 ·hα ,...,α i, where (C )t−ℓ+1 is the complement of (C )ℓ−1 in
11 p r+11 s1 p p
(C )t.
p
Step III. Put H = hhα ii and H = hhα ii. Note that H ∩H = {1}. However,
1 11 2 11 1 2
H may not be a normal subgroup of G. That is why we need to show that there exist a
2
commutator chain H and a normal subgroup H of G such that H = H ×H . In this
1 2 1 2
Step, we will describe a somewhat algorithmic approach which replaces the generators
of H until the desired result is obtained.
Assume henceforth that H is not normal in G. Then there exists a generator β ∈ H
2 2
such that α−1βα = hh for some h ∈ H ,h ∈ H ,h ∈/ H . Since h = βh for some
1 2 1 1 1 2 2
h ∈ H , we get [β,α] = h h .
2 2 1 2
Let us assume first that ord(β) = p. If h ∈ Hp, then h ∈/ [H(p),α], otherwise
1 2
[H(p),α] ∩ Hp 6= {1}. In other words, h does not appear in similar chains, so we
2
can simply put h h , instead of h , as a generator of H . In this way we obtain a
1 2 2 2
group that is G-isomorphic to H . Thus we get that [β,α] is in this new copy of H .
2 2
Similarly, if h ∈ H(p) and h ∈/ [H(p),α], we can obtain a new copy of H such that
1 2 2
[β,α] is in H . If h ∈ [H(p),α], we may assume that [β,α] ∈ H . In this case hhα ii
2 2 1 11