Equality of norm groups of subextensions of S n ðnp5þ extensions of algebraic number fields

http://www.elsevier.com/locate/jnt Journal of Number Theory 102 (2003) 257–277

Equality of norm groups of subextensions of S _n ðn p5Þ extensions of algebraic number fields

Leonid Stern

Department of Mathematics, Towson University, Baltimore, MD 21252, USA Received 30 April 2002; revised 7 February 2003

Communicated by J.S. Hsia

Abstract

Let E=k be a Galois extension of algebraic number fields with the Galois group isomorphic to the symmetric group S_non np5 letters. For any field extensions kDK; LDE a necessary and a sufficient condition is given for the equality N_K=kK^¼ N_L=kL^to hold, where N_K=kK^is the group of norms from K to k of the elements of the multiplicative group K^of K:

r2003 Elsevier Inc. All rights reserved.

Let k be an algebraic number field. The main theorem in[9]states that for any finite Galois extensions K and L of k

N_K=kK^DN_L=kL^ iff LDK:

In particular, the equality of norm groups of finite Galois extensions of k is equivalent to the equality of the extensions. For finite non-Galois extensions K and L of k the equality of norm groups holds provided that K and L are conjugate over k: The converse, however, is false in general. To investigate the equality of norm groups in the case of non-Galois extensions, we defined in[10]a ‘‘weaker’’ equality of norm groups which is called almost equal norm groups. This relation is equivalent to the equality of certain subsets of a finite group that are defined as follows. Let H be a subgroup of a finite group G: We define a subset of G by

PGðHÞ ¼ fg^1hg=hAH of prime power order; gAGg:

We note that in[6,7]the relation of almost equal norm groups was characterized in terms of a decomposition type of ideals in finite extensions of algebraic number

ARTICLE IN PRESS

E-mail address:[email protected].

0022-314X/03/$ - see front matter r 2003 Elsevier Inc. All rights reserved.

doi:10.1016/S0022-314X(03)00082-9

fields. In[11, Theorem 3, p. 343]we proved that the relation of almost equal norm groups is equivalent to the equality of norm groups of idele groups. More specifically, for any algebraic number field k we denote by Jk the idele group of k:

Also, for any finite extension of algebraic number fields L=k we set NðL=kÞ ¼ k^-NL=kJL; where N_L=k is the norm operator. It follows that NðL=kÞ ¼ T

vðk^-½NL=kL^ _vÞ (see for instance[10, Proposition (1.3), p. 113]), where v ranges over all primes of k; and ½N_L=kL^ _v is the topological closure of N_L=kL^ in k^_v (the multiplicative group of the completion of k at v). So if N_K=kK^¼ NL=kL^ for two finite extensions K=k and L=k of algebraic number fields, then NðK=kÞ ¼ NðL=kÞ:

Let K=k and L=k be finite extensions of algebraic number fields. Let E be a finite Galois extension of k containing K and L: Suppose that G¼ GðE=kÞ and HK¼ GðE=KÞ; HL¼ GðE=LÞ: In [11]we proved that the inclusion PGðHKÞDPGðHLÞ is equivalent to each of the inclusions

N_K=kJ_KDN_L=kJ_L and NðK=kÞDNðL=kÞ: ð1Þ We thus obtain a necessary group theoretic condition for the equality of norm groups,

N_K=kK^¼ N_L=kL^ implies PGðHKÞ ¼ PGðHLÞ: ð2Þ We wish to determine sets of extensions with the same norm group that are not pairwise conjugate over the base field. The necessary condition (2) leads us to make the following definitions.

Definition 1. Let G be a finite group. A sequence of subgroups fHig^m_i¼1 of G is called a full system of subgroups if for any 1pi; jpm the following conditions are satisfied:

(a) Hi and Hj are not conjugate in G for iaj;

(b) PGðHiÞ ¼ PGðHjÞ;

(c) for any subgroup H of G; if PGðHÞ ¼ PGðHsÞ for some 1pspm; then H is conjugate to Ht in G for some 1ptpm:

We will write SðH1; y; H_mÞ; if H1; y; H_m is a full system of subgroups. The length of a full system of subgroups SðH1; y; H_mÞ is the number m of subgroups in the sequence.

Two full systems of subgroups SðH1; y; H_nÞ and SðN1; y; N_mÞ of a finite group G are equal, if they are of the same length, and the subgroups in the systems can be reordered, if necessary, in such a way that H_i and N_i are conjugate in G for all 1pipm:

Definition 2. Let G be a finite group. We defineSðGÞ to be the set of all distinct nontrivial full systems of subgroups of G; i.e. the set of all distinct full systems of length greater than or equal to two.

The symmetric group S2 has only two conjugacy classes of subgroups, and SðS2Þ ¼|: There are four conjugacy classes of subgroups of S3

with representatives: the two trivial subgroups, and Sylow 2 and 3-subgroups of S3: It follows that SðS3Þ ¼|: By (2) we therefore obtain the following theorem.

Theorem 3. Let E=k be a finite Galois extension of algebraic number fields with the Galois group G: Let kDK; LDE be subfields of E: If G is isomorphic to S2or S3; then N_K=kK^ ¼ N_L=kL^ if and only if K and L are conjugate over k:

The symmetric group S₄ has 11 conjugacy classes of subgroups, and SðS4Þ ¼ fSðZ=2Z; V4Þg; where V4is the Klein 4-group, and Z=2Z is an arbitrary subgroup of V4of order two. If E=k is a Galois extension with the Galois group isomorphic to S4; and kCKCLCE are the fixed fields of V4*Z=2Z; then by (1) the equality NðK=kÞ ¼ NðL=kÞ holds. We will show that the Hasse Norm Principle (HNP) holds for K=k; i.e. NðK=kÞ ¼ N_K=kK^:

In general, for any finite extension X =Y of algebraic number fields, the factor group of NðX =Y Þ by N_{X =Y}X^is finite, and it is called the total obstruction to HNP for X =Y : The order of this factor group is denoted by iðX =Y Þ: If iðX =Y Þ ¼ 1; then we say that HNP holds for X =Y : It is well known that HNP holds for cyclic finite Galois extensions. Also, HNP holds for any extension of prime degree[1, Lemma 4];[8, Proposition 10.11].

To prove the above equality NðK=kÞ ¼ NK=kK^;we note that K is the compositum of a quadratic extension of k (the fixed field of A4) with a cubic extension of k (the fixed field of a Sylow 2-subgroup of S4). By Proposition 2[5, p. 315]HNP holds for K=k: So the equality NK=kK^¼ NL=kL^holds if and only if HNP holds for L=k: We will determine in this article when HNP holds for L=k:

Let L=k and T =k be finite extensions of algebraic number fields. In[12]we defined the first obstruction to HNP for L=k corresponding to T =k as the factor group of k^-NL=kJ_LN_T=kJ_T by NðT=kÞNL=kL^:We note that if T is a finite Galois extension of k containing L; then the first obstruction to HNP for L=k corresponding to T=k is equal to k^-NL=kJ_L=NðT=kÞNL=kL^:This factor group, called the first obstruction to HNP for L=k corresponding to the tower kDLDT ; was introduced and investigated in [4]. Let E=k be a finite Galois extension containing L and T ; and suppose that G¼ GðE=kÞ; H ¼ GðE=LÞ; and N ¼ GðE=TÞ: For each prime v of k we fixa k- embedding of E into the algebraic closure ˜kvof the completion kvof k at v: This will also fixa decomposition group Gv¼ res_Ev=E½GðEv=k_vÞ of v in E: There is a one-to- one correspondence between the primes o of L above v and the distinct double cosets HxoGv of H and Gv in G: Furthermore, Ho¼ H-xoGvx^1_o is a decomposition group of o in E: Similarly, there is a one-to-one correspondence between the primes n of T above v and the distinct double cosets NynGv of N and Gv in G; and Nn¼ N-ynG_vy^1_n is a decomposition group of n in E: Let v be an arbitrary prime of k:

Suppose that G¼S

ojvHx_oG_v¼S

njvNy_nG_vare decompositions of G into the unions of distinct double cosets of H; G_vand N; G_vin G; respectively. From now on in this article, unless stated otherwise, we will denote by X⁰ the commutator subgroup

½X ; X of a group X : For each prime v of k we define a group homomorphism lv:Q

ojv Ho=H_o⁰-H=H⁰ by the rule

lv½ðhoHo0Þ_ojv ¼ Y

ojv

ho

0

@

1 AH⁰;

and a group homomorphism c_v:Q

ojv Ho=H_o⁰-Gv=Q

njv ðN_n^ynGv0Þ by the rule

c_v½ðhoHo0Þ_ojv ¼ Y

ojv

x^1_o hoxo

0

@

1

AY

njv

ðN_n^ynGv0Þ;

where N_n^yn ¼ y^1_n Nnyn:It follows that the diagram Q

ojv

Ho=H_o⁰ !^lv H=H⁰

kcv kp

Gv=Q

njv

ðN_n^ynGv0Þ !^pv G=NG⁰;

ð3Þ

is commutative for each prime v of k; where p and pvare canonical homomorphisms.

This commutative diagram can be extended to a commutative diagram in a natural way

‘

v

Q

ojv

H_o=H_o⁰

! ! Qlv

H=H⁰ kQ

c_v kp

‘

v

Gv=Q

njv

ðN_n^ynGv0Þ ! Qpv

G=NG⁰:

By Theorem 1.7 of [12] the first obstruction to HNP for L=k corresponding t o the extension T =k is isomorphic to Ker p=ðQ

l_vÞ½KerQ

c_v : We note that Ker p¼ ðH-NG⁰Þ=H⁰; and ðQ

l_vÞ½Ker Q

c_v is a subgroup of this factor group generated by subgroups l_v½Ker c_v ; where v ranges over all primes of k: In order to describe the group ðQ

l_vÞ½Ker Q

c_v we will use a subgroup of G that was introduced in[4]to investigate first obstructions to HNP corresponding to towers of field extensions.

F^GðHÞ ¼ /f½h; g =hAH-gHg^1; gAGgS;

where /X S denotes the subgroup of G generated by a subset X of G; and½h; g ¼ h^1g^1hg: The group F^GðHÞ contains H⁰;and is a subgroup of H-G⁰:Finally, let S be the (finite) set of primes of k whose decomposition groups in E are not cyclic. We

define a subgroup H⁰DX_L=kðT; EÞDH-NG⁰ by X_L=kðT; EÞ=H⁰¼Y

vAS

l_v½Ker c_v : ð4Þ

By [12] ðQ

lvÞ½Ker Q

c_v coincides with the factor group of F^GðHÞ /H-PGðNÞSX_L=kðT; EÞ by H⁰:It follows that

k^-NL=kJLN_T=kJT

NðT=kÞN_L=kL^ D H-NG⁰

F^GðHÞ/H-PGðNÞSXL=kðT; EÞ ð5Þ [12, Theorem 1.11]. We wish now to establish a group theoretic property of XL=kðT; EÞ:

Let H and N be subgroups of a finite group G: For any subgroup X of G we define a subgroup of H-NG⁰=H⁰ as follows. Suppose that G¼Sn

i¼1Hx_iX¼Sm j¼1Ny_jX are decompositions of G into the unions of distinct double cosets of H; X and N; X ; respectively. We define H_i¼ H-xiXx^1_i for all 1pipn; and Nj¼ N-yjXy^1_j for all 1pjpm: The following mapping cX:Qn

i¼1 H_i=H_i⁰-X=Qm

j¼1ðN_j^yjX⁰Þ given by c_Xðh1H10; y; h_nHn0Þ ¼ Yⁿ

i¼1

x^1_i hixi

!Y^m

j¼1

ðN_j^yjX⁰Þ

is a group homomorphism. We have a commutative diagram of group homomorph- isms similar to diagram (3)

Qⁿ

i¼1

Hi=H_i⁰ !^lX H=H⁰

kcX kp

X =Q^m

j¼1

ðN_j^yjX⁰Þ !^pX G=NG⁰;

ð6Þ

where lX is given by lXðh1H10; y; h_nHn0Þ ¼ ðQn

i¼1 hiÞH⁰; and p; pX are canonical homomorphisms. It follows that lX½Ker c_X D Ker p ¼ H-NG⁰=H⁰:We will show that lX½Ker c_X does not depend on the choice of representatives fxig and fyjg of double cosets H; X and N; X in G; respectively.

Lemma 4. Let

A

A B

B

C

~ ~

g





~

~

f

be a commutative diagram of group homomorphisms, i.e. ga¼*af and *gf ¼ g: Then g½Ker a D*g½Ker *a :

The proof of the lemma is a straightforward verification of the inclusion.

Proposition 5. In the notation of diagram (6) the group l_X½Ker c_X is independent of the choice of representatives of double cosets H; X and N; X in G: Moreover, for any gAG the equality l_X½Ker c_X ¼ lX^g½Ker c_X^g holds.

Proof. Suppose that G¼Sn

i¼1HxiX ¼Sn

i¼1HaiX and G¼Sm

j¼1NyjX¼ Sm

j¼1Nb_jX are decompositions of G into the unions of distinct double cosets. Let ai¼ hixiaið1pipnÞ and bj¼ njyjb_jð1pjpmÞ for some hiAH; njAN; and ai;b_jAX : We define Hi; N_j as above, and ˜Hi¼ H-aiXa_i^1; ˜Nj¼ N-bjXb^1_j :Then

H˜_i¼ H-X^a1i ¼ H-X^{a1i x1i h1i} ¼ H-X^{x1i h1i} ¼ ðH-X^x1i Þ^h1i ¼ H_i^h1i : Similarly, ˜Nj¼ njNjn^1_j ð1pjpmÞ: We define

j :Yⁿ

i¼1

Hi=H_i⁰-Yⁿ

i¼1

H˜i= ˜Hi0

by the rule j½ðziH_i⁰Þ ¼ ðhiz_ih^1_i H˜_i⁰Þ: Since ˜H_i¼ hiH_ih^1_i ð1pipnÞ; it is easy to show that j is a well-defined group homomorphism. The equalities N˜_j¼ njN_jn^1_j ð1pjpmÞ imply that

N˜^b_j^jX⁰¼ Nⁿ

1 j bj

j X⁰¼ N_j^yjbjX⁰¼ N_j^yjX⁰ (since b_jAXÞ for each 1pjpm: Let Y ¼Qm

j¼1ðN_j^yjX⁰Þ ¼Qm

j¼1ð ˜N^b_j^jX⁰Þ:

Id X

X

X 

X

X

X

X / Y



~

~

H^~_i/H^~_i′ Πⁿ_i=1

H_i/H_i′

H/H′

G/NG′ X/Y

Πⁿ_i=1

,

where *l_X and *c_X are defined similarly to l_X and c_X; respectively. It is easy to see that *lXj¼ lX: To prove *cXj¼ c_X we choose an arbitrary element

ðziH_i⁰ÞAQn

i¼1 H_i=H_i⁰:Then

*c_Xj½ðziH_i⁰Þ ¼ *cX½ðz_i^hi1H˜_i⁰Þ ¼ Yⁿ

i¼1

z^h_i^{1i ai}

! Y :

Since ai¼ hixiaið1pipnÞ and X⁰DY ; it follows that z^h_i^{1i ai}Y ¼ z^x_i^iaiY¼ ðz^x_iⁱÞ^aiY ¼ z^x_iⁱY : So *c_Xj½ðziH_i⁰Þ ¼ ðQn

i¼1 z^x_iⁱÞY ¼ c_X½ðziH_i⁰Þ : By Lemma 4 l_X½Ker c_X D

*l_X½Ker *cX : By symmetry the opposite inclusion holds. So lX½Ker c_X ¼ *lX½Ker *cX : We will now prove that the equality lX½Ker c_X ¼ lX^g½Ker c_X^g holds for any gAG: Since G¼Sn

i¼1Hx_iX is the decomposition of G into the union of distinct double cosets of H; X in G; it follows that G¼Sn

i¼1Hx_igX^gis the decomposition of G into the union of distinct double cosets of H; X^g in G: For each 1pipn

H_i¼ H-xiXx^1_i ¼ H-xigX^gðxigÞ^1: Similarly, G¼Sm

j¼1Ny_jX¼Sm

j¼1Ny_jgX^g:For each 1pjpm N-yjgX^gðyjgÞ^1¼ N-yjXy^1_j ¼ Nj: We define w : X =Qm

j¼1ðN_j^yjX⁰Þ-X^g=Qm

j¼1ðN_j^yjgðX^gÞ⁰Þ in a natural way by conjuga- tion by g:

Xg

Xg

X

X





Xg

X

H_i/H_i′ Πⁿ_i=1

(N_j (X^g)′) X^g/Π^m_j=1

(N_j X′) X/Π^m_j=1

y_jg

y_j H_i/H_i′

H/H′

G/NG′ Id

Πⁿ_i=1

.

It is easy to verify that lX ¼ lX^g and wc_X¼ c_Xg: So by Lemma 4 lX½Ker c_X DlX^g½Ker c_Xg : By symmetry the opposite inclusion holds. So lX½Ker c_X ¼ lX^g½Ker c_Xg : &

By Proposition 5 in the notation of diagram (3) lv½Ker c_v is determined by a decomposition group Gvof a prime v of k in E; and it is independent of the choice of a decomposition group of v in E: The following proposition shows that XL=kðT; EÞ can be determined in certain cases by fewer decomposition groups Gv than it was stated in definition (4).

Proposition 6. Let H and N be subgroups of a finite group G: Then for any subgroups X DY of G in the notation of diagram (6) the inclusion lX½Ker c_X DlY½Ker c_Y holds.

Proof. Let G¼Sn

i¼1Ha_iY ¼Sm

j¼1Na_jY be decompositions of G; respectively, into the unions of distinct double cosets of H; Y and N; Y in G: Since X is a subgroup of Y ; it follows that HaiY ¼Ssi

r¼1HbirX ð1pipnÞ and NajY ¼Stj

r¼1Nb_jrXð1pjpmÞ are the unions of distinct double cosets of H; X and N; X ; respectively. It follows that Sn

i¼1ðSsi

r¼1Hb_irXÞ ¼Sm j¼1ðStj

r¼1Nb_jrXÞ are decompositions of G; respectively, into the unions of distinct double cosets of H; X and N; X in G: Let

H_i¼ H-aiYa^1_i and H_ir¼ H-birXb^1_ir ð1pipn; 1prpsiÞ;

and

Nj¼ N-ajY a^1_j and Njr¼ N-bjrX b^1_jr ð1pjpm; 1prptjÞ:

Since birAHaiY ; it follows that there are hirAH and yirAY such that

b_ir¼ hira_iy_ir ð7Þ

for all 1pipn; 1prpsi:Then for any 1pipn; 1prpsi

H_ir^hir¼ ðH-X^b1ir Þ^hir ¼ H-X^b1irhir¼ H-X^y1ira1i

DH-Y^{y1ir a1i} ¼ H-Y^a1i ¼ Hi: ð8Þ Furthermore, for any 1pjpm; 1prptj; b_jrANajY : Suppose that b_jr¼ ujrajvjr for some u_jrAN and v_jrAY ð1pjpm; 1prptjÞ: Then

N_jr^bjrX⁰¼ ðN-X^bjr1Þ^bjrX⁰¼ ðN^bjr-XÞX⁰¼ ðN^ajvjr-XÞX⁰DðN^ajvjr-Y^vjrÞY⁰

¼ ðN^aj-YÞ^vjrY⁰¼ ðN^aj-YÞY⁰¼ ðN-Y^a1j Þ^ajY⁰¼ N_j^ajY⁰: ð9Þ By (8) we can define a homomorphism

j_i:Y^si

r¼1

H_ir=H_ir⁰-Hi=H_i⁰

by the rule j_i½ðxirHir0Þ_1prpsi ¼ ðQsi

r¼1 h^1_ir xirhirÞHi0for each 1pipn: By (9) we have a canonical homomorphism

w : X =Y^m

j¼1

Y^tj

r¼1

N_jr^bjrX⁰

!

-Y=Y^m

j¼1

ðN_j^ajY⁰Þ:

The above-defined homomorphisms yield a commutative diagram

X

Y

Y

X



H_i/H_i′ Πⁿ

Πi

i=1

H_ir/H_ir)

H/H′, Πⁿ _i=1 (Π^s_{r =1}ⁱ

N_jr X′) X/Π^m _j=1 (Π_{r =1}^{tj jr}

Y′) Y/Π^m _j=1 (N_j
^j

′

ð10Þ

where c_X; c_Y and lX; lY are defined similarly to c_X and lX; respectively, in diagram (6).

To prove that diagram (10) is commutative we will first prove the equality wc_X ¼ c_YðQ

j_iÞ: Let xirAHir ð1pipn; 1prpsiÞ be arbitrary elements.

By (8) h^1_ir x_irh_irAH_i; and therefore a^1_i h^1_ir x_irh_ira_iAY : In (7) we defined yirAY ð1pipn; 1prpsiÞ: Now using the fact that the commutator

½a^1_i h^1_ir xirhirai; y_ir is an element of Y⁰ we obtain

b^1_ir x_irb_irY⁰¼ a^1_i h^1_ir x_irh_ira_iY⁰ ð11Þ for any 1pipn; 1prpsi:It follows now by (11) that

wc_X½ððxi1Hi10; y; x_isiHisi

0ÞÞ_1pipn

¼ w Y

1pipn 1prpsi

b^1_ir xirbir

0 B@

1 CA Y

1pjpm 1prptj

ðN_jr^bjrX⁰Þ 2

66 4

3 77 5

¼ Y

1pipn 1prpsi

b^1_ir x_irb_ir 0

B@

1 CAY^m

j¼1

ðN_j^ajY⁰Þ

¼ Y

1pipn 1prpsi

a^1_i h^1_ir x_irh_ira_i 0

B@

1 CAY^m

j¼1

ðN_j^ajY⁰Þ: ð12Þ

On the other hand c_YðY

j_iÞ½ððxi1H_i1⁰; y; x_isiH_isi⁰ÞÞ_1pipn

¼ c_Y Y^s1

r¼1

h^1_1rx_1rh_1rH₁⁰; y;Y^sn

r¼1

h^1_nrx_nrh_nrH_n⁰

!

" #

¼ Y

1pipn 1prpsi

a^1_i h^1_ir x_irh_ira_i 0

B@

1 CAY^m

j¼1

ðN_j^ajY⁰Þ: ð13Þ

By (12) and (13) we obtain the desired equality wc_X¼ c_YðQ

j_iÞ: To complete the proof that diagram (10) is commutative it remains to show that the equality lYðQ

j_iÞ ¼ lX holds. Let xirAHir ð1pipn;

1prpsiÞ be arbitrary elements. Since xirAH_irDH; it follows that h^1_ir x_irh_irH⁰¼ x_irH⁰:Then

lYðY

j_iÞ½ððxi1Hi10; y; x_isiHisi

0ÞÞ_1pipn

¼ lY

Y^si

r¼1

h^1_ir xirhirHi0

!

1pipn

" #

¼ Y

1pipn 1prpsi

h^1_ir x_irh_ir 0

B@

1 CAH⁰

¼ Y

1pipn 1prpsi

x_ir 0

B@

1

CAH⁰¼ lX½ððxi1H_i1⁰; y; x_isiH_isi⁰ÞÞ_1pipn :

Since diagram (10) is commutative, it follows by Lemma 4 that l_X½Ker c_X DlY½Ker c_Y : &

We wish now to generalize Proposition 6.

Proposition 7. Let G be a subgroup of a finite group G; and let H; N be a˜ subgroups of G: In the notation of diagram (6) for any subgroup Y of ˜G we obtain two subgroups of H=H⁰:lY½Ker c_Y is defined by G; H; N; and Y ;˜ lX½Ker c_X is defined by G; H; N; and X¼ G-Y: Then lX½Ker c_X D lY½Ker c_Y :

Proof. Suppose that G˜ ¼Sn

i¼1DiðHÞ and G˜¼Sr

j¼1DjðNÞ be decompositions of ˜G into the unions of distinct double cosets of H; Y and N; Y ; respectively.

Let fDiðHÞg_1pipm and fDjðNÞg_1pjps be all double cosets for which D_iðHÞ-Ga| and DjðNÞ-Ga|: Suppose that DiðHÞ ¼ HxiY for 1pipn and D_jðNÞ ¼ NyjY for 1pjpr; and x_i; y_iAG for 1pipm and 1pjps: It follows that G ¼Sm

i¼1Hx_iX¼Ss

j¼1Ny_jX are decompositions of G into the unions of distinct double cosets of H; X and N; X in G; respectively.

We define

H˜_i¼ H-xiYx_i^1 ð1pipnÞ and Hi¼ H-xiXx^1_i ð1pipmÞ;

N˜j¼ N-yjYy^1_j ð1pjprÞ and Nj¼ N-yjXy^1_j ð1pjpsÞ

Diagram (6) yields the following diagram of group homomorphisms.

(N_j X′) X /Π^s_j=1 H_i/H_i′

H/H′

G/NG′,

G/NG′, H/H′

Πⁿ_i=1

H_i/H_i′ Π^m_i=1

~ ~

~ ~ ~

y_j

(N_j Y′) Y /Π^r_j=1 ^yj

X

Y





 

X

Y

X

Y

Id

~

where j; Z; w are the canonical homomorphisms. The remaining homomorphisms in the above diagram are defined similarly to the corresponding homomorphisms in diagram (6). It is easy to show that the equalities l_X ¼ lYj and c_Yj¼ Zc_Xhold. It follows by Lemma 4 that l_X½Ker c_X DlY½Ker c_Y : &

By Propositions 6 and 7 we immediately obtain the following theorem.

Theorem 8. Let G be a subgroup of a finite group ˜G; and let H; N be subgroups of G: In the notation of diagram (6) for any subgroup Y of ˜G and for any subgroup X of G that is contained in Y we obtain two subgroups of H=H⁰: lY½Ker c_Y is defined by G; H; N; and Y ; l˜ _X½Ker c_X is defined by G; H; N; and X : Then lX½Ker c_X DlY½Ker c_Y :

Definition 9. LetA be a set of subgroups of a finite group G: We define MðAÞ to be a subset of A such that

(a) for any X AA there are Y AMðAÞ and gAG such that X^gDY ; (b) for any X ; Y AMðAÞ and gAG; X^gaY:

By Theorem 8 we obtain the following corollary.

Corollary 10. Let H and N be subgroups of a finite group G: Let A be a set of subgroups of G: Then in the notation of diagram (6) we have equality of subgroups of H-NG⁰=H⁰

Y

X AA

lX½Ker c_X ¼ Y

X AMðAÞ

lX½Ker c_X :

The set SðS4Þ of distinct nontrivial full systems of subgroups of S4 is fSðZ=2Z; V4Þg; where Z=2Z ¼ /ð1; 2Þð3; 4ÞS and V4¼ /ð1; 2Þð3; 4Þ; ð1; 3Þð4; 2ÞS is the Klein 4-group.

Lemma 11. Let E=k be a Galois extension with the Galois group G¼ GðE=kÞ isomorphic to S4: Let kCKCLCE be the fixed fields of V₄*Z=2Z: If there is a prime v of k whose decomposition group in E contains the Sylow 2-subgroup of A4; then N_K=kK^ ¼ N_L=kL^:

Proof. Let v be a prime of k whose decomposition group G_vin E contains the Sylow 2-subgroup V4 of A4: Let F be the fixed field of A4:Then there is a prime o of F above v whose decomposition group A4-Gv in E contains V4:By Proposition 1.1 [10, p. 111]N_K=FK^¼ N_L=FL^:It follows that

N_K=kK^¼ N_{F =k}ðN_K=FK^Þ ¼ N_{F =k}ðN_L=FL^Þ ¼ N_L=kL^: &

The set of distinct nontrivial full systems of subgroups of S5 is SðS5Þ ¼ fSðH1; H₂Þ; SðN1; N₂Þ; SðR1; R₂Þg; where

H₁ ¼ /ð1; 2; 3Þð4; 5ÞSD Z=6Z and H2¼ /ð3; 4; 5Þ; ð4; 5ÞSD S3

N1 ¼ /ð2; 3Þð4; 5ÞSD Z=2Z and N2¼ /ð2; 3Þð4; 5Þ; ð2; 4Þð3; 5ÞSD V4

R1 ¼ /ð1; 2Þð4; 5Þ; ð3; 4; 5ÞSD S3 and R2¼ /ð2; 3Þð4; 5Þ; ð2; 4Þð3; 5Þ; ð3; 4; 5ÞSD A4:

2 2

2

3

3 3

6

10

4 2

5

2 5

20

k T

F E

L₁

L₂

K₂ K₁K₂

K₁ M₁

M₂

Fig. 1.

Let E=k be a Galois extension with the Galois group isomorphic to S5: Suppose that Ki; L_i and Mi ði ¼ 1; 2Þ are the fixed fields of Hi; N_i and Ri; respectively. The compositum K1K2 is fixed by H1-H2¼ /ð4; 5ÞS: We identify S4

with the symmetric group on the setf2; 3; 4; 5g; and A4with R2:Let F ; T be the fixed fields of A5 and S4; respectively. Field extensions of k contained in E are shown above inFig. 1.

The proof of the following lemma is similar to the proof of Lemma 11.

Lemma 12. In the notation ofFig. 1 if there is a prime v of k whose decomposition group in E contains a Sylow 2-subgroup of A₅; then N_L1=kL₁^¼ NL2=kL^₂ and NM1=kM₁^¼ NM2=kM₂^:

Proof. We first note that N2 is a Sylow 2-subgroup of A5: Let v be a prime of k whose decomposition group Gvin E contains a Sylow 2-subgroup of A5;say N₂^g for some gAA₅: There is a prime of F and there is a prime of T above v whose decomposition groups in E are A5-Gv and S4-gGvg^1; respectively. The first decomposition group contains the Sylow 2-subgroup N₂^gof A₅:By Proposition 2.4 of [12, p. 258] N_M1=FM₁^¼ N_M2=FM₂^: It follows as in the proof of Lemma 11 that N_M1=kM₁^¼ NM2=kM₂^: The second decomposition group contains the Sylow 2- subgroup N2 of A4:By Lemma 11 NL1=TL^₁¼ NL2=TL₂^:So NL1=kL^₁¼ NL2=kL^₂: &

To prove the next proposition we will use a theorem that was proved in[4]. We will include the statement of the theorem here for the convenience of the reader.

Theorem 13. Let kDT DF be a tower of finite extensions of an algebraic number field k with F Galois over k: Let G¼ GðF =kÞ and H ¼ GðF =TÞ: If G contains subgroups G1; y; G_n; and subgroups H_sDGs-Hðs ¼ 1; y; nÞ such that

Yⁿ

s¼1

Cor^G_G^s: Yⁿ

s¼1

Hˆ^3ðGs;ZÞ- ˆH^3ðG; ZÞ

is a surjective homomorphism, and

iðTs=k_sÞ ¼ 1 for each s¼ 1; y; n;

where GðF =ksÞ ¼ Gs and GðF =TsÞ ¼ Hs ðs ¼ 1; y; nÞ; then NðF =KÞDNT=kT^: In the above theorem ˆH^3ðGs;ZÞ is a Tate cohomology group, and ˆH^3ðGs;ZÞ ¼ H2ðGs;ZÞ: Similarly, Hˆ^3ðS5;ZÞ ¼ H2ðS5;ZÞ: We will show in the following proposition that the equality N_K1=kK₁^¼ N_K2=kK₂^ (see Fig. 1) holds for any Galois extension E=k with the Galois group isomorphic to S₅:

Proposition 14. In the notation ofFig. 1, HNP holds for both extensions K1=k; K₂=k;

and N_K1=kK₁^ ¼ N_K2=kK₂^:

Proof. To prove the equality of norm groups we will use Theorem 13 to show that NðE=kÞDN_Ki=kK_i^ for each i¼ 1; 2: Let G1 be a Sylow 2-subgroup of S₅ containing H₁-H2¼ /ð4; 5ÞS: Let G2 and G₃ be a Sylow 3-subgroup and a Sylow 5-subgroup of S₅; respectively. By Proposition 3.1.15 of [13, p. 92]

Y³

i¼1

Cor^G_S5ⁱ: Y³

i¼1

Hˆ^3ðGi;ZÞ- ˆH^3ðS5;ZÞ

is a surjective homomorphism. Let Hij¼ Gj-Hi ð1pip2; 1pjp3Þ; and let kj and Tij be the fixed fields of Gj and Hij; respectively. For ja1;

T_ij=k_j is a cyclic extension of degree 1, 3, or 5 for any 1pip2: So HNP holds for all extensions Tij=k_j for which i¼ 1; 2 and ja1: The group G1 is isomorphic to the dihedral group D₈ of order 8. Since D₈ contains only one nontrivial central element, which is the square of an element of order four, it follows that the nontrivial central element of G₁ is an even permutation in S₅: So the cyclic group /ð4; 5ÞS is not normal in G1: Since H_i1¼ /ð4; 5ÞS for i ¼ 1; 2; it follows by Satz 1[2]that HNP holds for Ti1=k₁for i¼ 1; 2:

So by Theorem 13, NðE=kÞDNKi=kK_i^ for each i¼ 1; 2: By (1) the equality PS5ðH1Þ ¼ PS5ðH2Þ implies NðK1=kÞ ¼ NðK2=kÞ: By (5) the first obstruction NðKi=kÞ=NðE=kÞNKi=kK_i^ ði ¼ 1; 2Þ to HNP for Ki=k corresponding to E=k is a homomorphic image of Hi-S50=F^S5ðHiÞ; where S50¼ A5 is the commutator subgroup of S5:Since Hi-A5¼ F^S5ðHiÞ is the cyclic subgroup of order 3 of Hi; it follows that the obstruction NðKi=kÞ=NðE=kÞNKi=kK_i^is trivial for each i¼ 1; 2: So for each i¼ 1; 2

NðKi=kÞ ¼ NðE=kÞN_Ki=kK_i^¼ N_Ki=kK_i^:

It follows that HNP holds for both extensions K1=k and K2=k; and therefore the equality N_K1=kK₁^¼ N_K2=kK₂^ holds. &

In Lemmas 11 and 12 we discussed the equality of norm groups of certain extensions in the case when there is a prime of the base field whose decomposition group in an extension contains a Sylow 2-subgroup of A4 or A5: We wish now to consider the case when a decomposition group of an arbitrary prime does not contain a Sylow 2-subgroup of A4 or A5: Since A4¼ R2; the Sylow 2-subgroup V4 of A4 is also a Sylow 2-subgroup of A5:

Lemma 15. Let E=k be a finite Galois extension of an algebraic number field k with the Galois group GDS5: In the notation ofFig. 1, if for any prime of k its decomposition group in E does not contain Sylow 2-subgroups of A₅; then N_L1=kL₁^aNL2=kL^₂ and N_M1=kM₁^aNM2=kM₂^:

Proof. The commutator subgroup of S₅ is A5: So for each i¼ 1; 2 the first obstructions to HNP for Li=kand Mi=k corresponding to E=k are isomorphic to

Ni-A5=F^GðNiÞX_Li=kðE; EÞ and Ri-A5=F^GðRiÞX_Mi=kðE; EÞ;

respectively. It follows that F^GðN1Þ ¼ 1; F^GðN2Þ ¼ N2; and F^GðR1Þ ¼ R10¼ /ð3; 4; 5ÞS; F^GðR2Þ ¼ R2: Since all four groups Ni; R_i ði ¼ 1; 2Þ are subgroups of A5;it follows that

NðL1=kÞ=NðE=kÞNL1=kL^₁DN1=X_L1=kðE; EÞ and

NðL2=kÞ ¼ NðE=kÞN_L2=kL^₂; ð14Þ

and

NðM1=kÞ=NðE=kÞNM1=kM₁^DR1=X_M1=kðE; EÞ and

NðM2=kÞ ¼ NðE=kÞN_M2=kM₂^: ð15Þ We will prove that

X_L1=kðE; EÞ ¼ N10 ð¼ 1Þ ð16Þ and

XM1=kðE; EÞ ¼ R10 ð¼ /ð3; 4; 5ÞSÞ: ð17Þ If equalities (16) and (17) hold, then by (14) and (15)

N_L1=kL^₁aNL2=kL^₂ and N_M1=kM₁^aNM2=kM₂^:

If the decompositions groups Gv of all primes v of k in E are cyclic, then by the definition of the group X_L1=kðE; EÞ equality (16) holds. Similarly, equality (17) holds if G_vis cyclic for all primes v of k: We assume, therefore, that there is a prime v of k for which G_vis not cyclic. Let S be the (finite) set of all primes v of k for which G_vis not cyclic. We will show that l_v½Ker c_v is the trivial subgroup of order one of N₁=N₁⁰ ðR1=R₁⁰Þ for each prime vAS: We note that for each prime vAS; Gv is a noncyclic solvable subgroup of G¼ S5;and by the assumption Gvdoes not contain a Sylow 2-subgroup of A5: Let A be the set of all decomposition groups of primes from S in E; and let ADB be the set of all solvable noncyclic subgroups of G that do not contain a Sylow 2-subgroup of A5:We will show that lX½Ker c_X is trivial for all X AMðBÞ: By Theorem 8 this will imply that lv½Ker c_v is the trivial subgroup of order one of N1=N₁⁰ ðR1=R₁⁰Þ for each prime vAS: The set MðBÞ contains two groups: a semidirect product of two cyclic groups C6 C2¼ /ð1; 2; 3Þð4; 5Þ; ð2; 3ÞS ¼ /a; bS with a⁶¼ b²¼ 1; b^1ab¼ a^1; and a semidirect product of cyclic groups C₅ C4 ¼ /ð1; 2; 3; 4; 5Þ; ð2; 4; 5; 3ÞS ¼ /a; bS with a⁵¼ b⁴¼ 1; b^1ab¼ a³:

We first consider the case when X ¼ C6 C2: The group G is the union of sixdistinct double cosets of N1 and X in G with repre- sentatives f1; ð3; 4Þ; ð3; 4; 5Þ; ð2; 4Þð3; 5Þ; ð1; 4; 3; 2Þ; ð1; 4; 2Þð3; 5Þg: In the notation of diagram (6) the corresponding subgroups of N1 are ðN1Þ₁¼ ðN1Þ₄¼ N₁; and the remaining four subgroups are trivial. It follows that the kernel of

c_X:Y⁶

i¼1

ðN1Þ_i=ðN1Þ_i⁰-X=X⁰

is the cyclic group of order 2 generated by the 6-tuple in which the first and the fourth components are (2,3)(4,5), and the remaining components are trivial. So lX½Ker c_X is the trivial subgroup of order one of N1=N₁⁰:

Suppose now that X is equal to the second group C5 C4 in MðBÞ: Then G is the union of four distinct double cosets of N₁ and X in G with repre- sentatives f1; ð4; 5Þ; ð3; 5; 4Þ; ð3; 5Þg: In the notation of diagram (6) the corresponding subgroups of N₁ are ðN1Þ₁¼ ðN1Þ₂¼ 1; and ðN1Þ₃¼ ðN1Þ₄¼ N1: The kernel of

c_X:Y⁴

i¼1

ðN1Þ_i=ðN1Þ_i⁰-X=X⁰

is the cyclic group of order 2 generated by ð1; 1; ð2; 3Þð4; 5Þ; ð2; 3Þð4; 5ÞÞ:

It follows that lX½Ker c_X is the trivial subgroup of order one of N1=N₁⁰: This completes the proof of the equality XL1=kðE; EÞ ¼ N10: So N_L1=kL^₁aNL2=kL^₂:

We wish now to prove that N_M1=kM₁^aNM2=kM₂^: The group G is the union of three distinct double cosets of R1 and X ¼ C6 C2 in G with representativesf1; ð2; 4; 3Þ; ð1; 4; 2; 5; 3Þg: In the notation of diagram (6) correspond- ing subgroups of R1 are ðR1Þ₁¼ /ð1; 2Þð4; 5ÞS; ðR1Þ₂¼ 1; and ðR1Þ₃ ¼ R1: The kernel of

c_X :Y³

i¼1

ðR1Þ_i=ðR1Þ_i⁰-X=X⁰

is the cyclic group of order 2 generated by the ordered triple in which the second component is trivial, and the remaining two components contain ð1; 2Þð4; 5Þ: It follows that lX½Ker c_X is the trivial subgroup of order one of R1=R₁⁰:Finally, we assume that X¼ C5 C4:Then G is the union of two distinct double cosets of R1and X in G with representatives f1; ð4; 5Þg: In the notation of diagram (6) the corresponding subgroups of R1 are ðR1Þ₁ ¼ /ð1; 2Þð3; 5ÞS; and ðR1Þ₂¼ /ð1; 2Þð3; 4ÞS: The kernel of

c_X:ðR1Þ₁ ðR1Þ₂-X=X⁰