Classifying Finite Subgroups of Rational Division Algebras. · Classifying Finite Subgroups of Rational Division Algebras. ... that the ﬁnite subgroups of rational division algebras

Faculty of Scienceand Bio-Engineering SciencesDepartment of Mathematics

Classifying Finite Subgroups ofRational Division Algebras.

Master thesis submitted for the degree of Master in Mathematics.

Doryan Temmerman

Promotor: Prof. E. Jespers

2014-2015

Contents

1 Introduction 1

2 The goal 32.1 The theorems of Amitsur . . . . . . . . . . . . . . . . . . . . . . . 4

3 Number theory 73.1 Dedekind domains . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Number fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113.3 Valuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.4 Local fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.5 Hensel’s lemma and applications . . . . . . . . . . . . . . . . . . . 17

4 Finite group theory 214.1 Z0- and Z-groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 Quaternion groups . . . . . . . . . . . . . . . . . . . . . . . . . . 264.3 The group SL(2, 5) . . . . . . . . . . . . . . . . . . . . . . . . . . 284.4 Fitting subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.5 Solvable groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

5 Finite-dimensional algebras 415.1 Tensor product of central simple algebras . . . . . . . . . . . . . . 425.2 Schur-index and exponent . . . . . . . . . . . . . . . . . . . . . . . 445.3 Quaternion algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 455.4 Crossed products . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

6 Proving the theorems of Amitsur 516.1 Classifying Z-groups . . . . . . . . . . . . . . . . . . . . . . . . . 516.2 Finite, non-solvable subgroup(s) of division algebras . . . . . . . . 616.3 Finite, solvable subgroups with a quaternion Sylow subgroup . . . . 65

Bibliography 71

Index 73

List of Symbols 75

I

Samenvatting

In 1953 stelde Herstein in [6] de vraag hoe de eindige deelgroepen van schevelichamen er uit zien. Hij bewees dat voor positieve karakteristiek deze groepensteeds cyclisch zijn. Later, in 1955, bewees Amitsur in [2] een classificatie stellingvoor de eindige deelgroepen van scheve lichamen in het geval van karakteristiek0. Deze classificatie zal het hoofd onderwerp van deze thesis zijn. We volgenhet bewijs van Shirvani en Wehrfritz in [16]. We zullen bewijzen dat de eindigedeelgroepen van de scheve lichamen van karakteristiek 0 de Z-groepen zijn, samenmet groepen opgebouwd rond een veralgemeende quaternionen groep en SL(2, 5).Hiervoor zullen we de volgende groepen bestuderen:

• Frobenius complementen.

• Z- and Z0-groepen.

• veralgemeende quaternionen groepen.

• oplosbare groepen.

• SL(2, 5).

Uit de classificatie zal onmiddelijk volgen dat de groep SL(2, 5) de enige eindigeonoplosbare groep is van een scheef lichaam. De Z-groepen zullen ingebed wordenin een scheef lichaam door het gebruiken van kruisproducten, en de andere groepenzullen vooral gebruik maken van tensorproducten van o.a. quaternionen algebras.

Een andere stelling dat hier in dicht verband mee staat is de classificatie stellingvan de Z-groepen. Deze stelling werd ook bewezen door Amitsur in hetzelfde ar-tikel. Dit zal het tweede grote onderwerp van deze tekst zijn. De stellingen wordenbewezen, opnieuw gebruikmakend van het boek van Shirvani en Wehrfritz [16],door middel van technieken uit, o. a. de getaltheorie (getallenlichamen, p-adischegetallen, lokale lichamen,...) en door het construeren van eindigdimensionale al-gebras gebruikmakend van tensorproducten, quaternionen algebras en de kruispro-ducten.

II

Abstract

In 1953 a question was asked by Herstein in [6] what the finite subgroups ofdivision algebras looked like. He proved that in positive characteristic these areonly cyclic groups. In 1955, Amitsur [2] proved a classification theorem for thefinite subgroups of rational division algebras. It is this classification that is the mainsubject of this thesis. We will prove, using the proof given by Shirvani and Wehrfritzin [16], that the finite subgroups of rational division algebras are Z-groups, build upfrom generalized quaternion groups or the group SL(2, 5). This leads us to studyingthe following groups:• Frobenius complements.

• Z- and Z0-groups.

• generalized quaternion groups.

• solvable groups.

• SL(2, 5).

It will follow automatically from the classification that SL(2, 5) is the only non-solvable finite subgroup of a division algebra. The Z-groups are embedded in di-vision algebras using crossed products and the other groups mainly use quaternionalgebras and tensor products of both types.

An important theorem that is closely related to this classification is the classifi-cation of the Z-groups. This theorem was also proved by Amitsur in the same articleand is the second subject of this dissertation. We will again follow the proof givenby Shirvani and Wehrfritz ([16]) using techniques from, amongst others, numbertheory (number fields, p-adic numbers, local fields,...) and by constructing finite-dimensional algebras using tensor products, quaternion algebras and crossed prod-ucts.

III

Dankwoord

Dit dankwoord gaat in eerste instantie uit naar mijn promotor, professor Jespers.Hij stond me bij tijdens onze wekelijkse afspraken om te antwoorden op mijn, somszeer eenvoudige, vragen, ook al was het de laatste tijd wat chaotisch. Ik ben onderde indruk van de intensiteit waarmee hij deze tekst heeft nagelezen en hoe hij demoed gevonden heeft om mij op zo veel mogelijk punten te wijzen, opdat dezethesis toch als degelijke wiskundige tekst kan doorgaan.

Hierbij wil ik ook graag Inneke bedanken om me te wijzen op de wiskundigehiaten en stijlfouten die in vroegere versies nog talrijker aanwezig waren. Ook zouik graag Geoffrey willen vermelden in dit dankwoord; niet alleen stond hij me bijmet mijn wiskundige vragen en gaf hij goed advies over de tekst zelf, hij was erook bij voor de meer ontspannende momenten buiten de muren van de wiskunde.

Bij uitbreiding wil ik alle professoren en assistenten bedanken die me laatstejaren begeleid hebben op de VUB en de UA. Zonder hun zou deze thesis vanzelf-sprekend nooit tot stand gekomen zijn. Zij waren een geweldige gids op tochtdoorheen het bos van de wiskunde...

Professor Becher en Sten Veraa wil ik extra bedanken voor de assistentie bijhet hoofdstuk over getaltheorie. Zonder hun hulp en aangerade lectuur zou ditonderdeel nog steeds een raadsel voor me zijn.

A special thanks goes to Jesus Hernandez Gil, who, while working on the samesubject, was able to clarify a substantial amount of my questions.

Carmen, Filip, Lies en Lucas, mijn klasgenootjes deze laatste 5 jaar. Hoewel hetnormaal is dat we in dit laatste jaar niet al te veel meer les hadden samen, vond ikhet toch fijn om jullie af en toe eens tegen te komen op de campus en dan snel watpaniekerige opmerkingen over de thesissen of examens uit te wisselen. Desondankskonden we, vooral onder leiding van Filip, ons toch amuseren met onderwerpen dieminder met wiskunde te maken hadden. Ook onze kerstfeestjes, die soms tijdens deles werden voortgezet, zal ik missen.

Natuurlijk bestaat de ondersteuning van deze thesis niet slechts uit het helpenmet wiskundige problemen, maar moet er ook aan de persoon die deze schrijftgedacht worden. In dit kader wil ik graag mijn vriendin Annaline speciaal in debloemetjes zetten: enerzijds voor het ondersteunen van de innerlijke wiskundigemet het, af en toe, brengen van de nodige koffie en versnaperingen, anderzijds voorhet omgaan met mijn aanwezigheid als ik weer eens vast zat met een probleem.Ik heb ook veel steun gehad van mijn broer Yarno met z’n geweldige uitsprakenzoals “Hey, dat symbool ken ik!” en “Ma, da is toch kei makkelijk...”. Ik wil ook

IV

CONTENTS V

mijn ouders niet vergeten in dit tekstje, die reikhalzend uitkijken naar het momentwaarop ze de titel van deze thesis voldoende onder de knie hebben om ze te herhalenvoor geınteresseerde kennissen.

Daarbuiten zijn er natuurlijk ook nog de mensen van SWAMP, waarop ik konrekenen om elke maandagavond (en occasionele avonden daarbuiten) klaar te staanmet ontspannende momenten van spel en plezier. Hierbij wil ik er graag proseniorSaskia uitpikken, die m’n hoofd onnodig, maar met plezier, volstopte met filmtitels,schilderijen en muziekstukken.

Als allerlaatste wil ik graag het koffiemachine op kot bedanken. Zonder zijnhulp en ondersteuning hadden de ochtenden en namiddagen er toch anders uitgezien.

Doryan Temmerman,Brussel, mei 2015

VI CONTENTS

1

Introduction

“A mind needs books as a sword needs a whetstone, if it is to keep itsedge.”

– George Martin, A Game of Thrones

Before starting a theorem trying to classify all the finite groups that can beembedded in division rings (also called division algebras), one should look at theorigins of this question. A group is a set that allows a multiplication with divisionand a division ring is a set that allows an addition, subtraction, multiplication anddivision (or at least almost) with some compatibility constraints between them. Onecan thus find groups inside of these division rings. The question then naturallyoccurs how much these constraints influence the form of the groups that can beretrieved herein.

The first time one proposed a classification of these subgroups, at least in the fi-nite case, was in 1953 in [6] by Herstein. He proved (as will be shown in the secondchapter) that for a division ring of positive characteristic, the finite subgroups arealways cyclic. An example of a non-cyclic group in the real quaternions is easilygiven by

{±1,±i,±j,±k},

so he knew this would not be the case in characteristic 0. Herstein was able to provethat for the real quaternions, if the group G has odd order, then it has to be cyclic.In the general case, the best results he could muster were the following.

Lemma 1.1. Let D be a division ring. If a group G is of odd prime-power order,and G is embeddable in the multiplicative group of D, then G is cyclic.

Theorem 1.2. Let D be a division ring. If a group G is of odd order and G isembeddable in the multiplicative group of D, then there exist elements a and b of Gand an r ∈ N0 such that:

• an = bm = 1 for some n and m odd.

• bab−1 = ar for some r.

1

2 1. INTRODUCTION

These results led him to conjecture that all the subgroups of odd order in adivision ring are cyclic.

It was in 1955 that Amitsur, by giving a classification of all finite subgroups ofrational division rings, disproved this conjecture in [2]. It is this classification byAmitsur (see Theorem 2.8) that will be the main subject of this master thesis. Wewill follow the proof of the classification given by Shirvani and Wehrfritz in [16].As we will mainly follow this book, most of the results can be found here, unlessstated otherwise. A theorem that is closely related to the classification theorem ofthe finite subgroups of rational division algebras is that of the classification of theso called Z-groups (see Theorem 2.12). This theorem is another major subject ofthis dissertation. It will be discussed in Section 2.1 and will eventually be proven inChapter 6. Chapter 2 is dedicated to laying out some of the new terminology usedin the rest of this thesis, as well as stating the two main theorems. Also, the theoremof Herstein is proved concerning the finite subgroups of division rings of positivecharacteristic.

The proof of the theorem concerning Z-groups relies heavily on some number-theoretical results, so, in Chapter 3, we give some background that is needed onnumber theory. As these are mainly well-known results, almost no proofs will begiven and the reader who feels confident in his or her knowledge of this subjectneed not burden him- or herself with the details.

In Chapter 4 we look closer at some of the groups that appear in this thesis,such as Z-groups, quaternion groups, solvable groups, etc. It will mainly containproperties of such groups that are needed in Chapter 5 and Chapter 6.

Chapter 5 is highly important, for it contains results on central simple algebrasand crossed products. These mathematical objects play a central role in the actualproof of the classification theorems by Amitsur.

Lastly, Chapter 6 is a synthesis of the other chapters and contains the actualproofs of our two main theorems.

I will assume the reader to have an understanding of algebra up to masters level,such as basic knowledge about (finite) groups, (division) rings, modules and Galois-theory. In certain chapters we will make assumptions on the objects we work with.These assumptions will be noted at the beginning of the chapter or section. Allrings are assumed to be with unity.

It should be noted that, due to the difficult nature of the classification theoremsdiscussed in this dissertation, we will often use results that are given without proof,for they are either well-known or would lead us too far from our topic. In thesecases, a reference is given so that the reader may look up the proofs of these results.Even though we use these results, they will still leave a fair amount of theoreticalwork to be done in the rest of the proofs in this dissertation. In other words, they donot render this work trivial.

2

The goal

“I will take the ring,” he said, “though I do not know the way.”– J.R.R. Tolkien, The Lord of the Rings

In this chapter we will give a short layout about the classification of subgroupsof division rings with positive characteristic by Herstein. Then we will focus onthe theorem by Amitsur for characteristic 0 (the case of a rational division algebra),mainly giving some definitions and some basic results. We will also always assumea group G to be finite.

Our goal here is to classify all the finite subgroups of division rings. As is clearfrom the introduction, we can make a first distinction based on the characteristicof the division ring. We start with the easy case, namely if we consider divisionrings of positive characteristic. We will need the following famous theorem fromWedderburn that can be found in any algebra book dealing with finite fields (forexample, see Cohn [4, Theorem 8.6]).

Theorem 2.1 (Wedderburn). Every finite division ring is commutative.

For a ring R denote Mn(R) the ring of n-dimensional matrices over R. Theunique field with pa elements is denoted by Fpa . We introduce a special kind ofgroup, the general linear group GL(n, F ) = {A ∈ Mn(F ) | det(A) 6= 0}. Thisgroup has the special linear group SL(n, F ) = {A ∈ GL(n, F ) | det(A) = 1} asa subgroup. For GL(n,Fpa) (respectively SL(n,Fpa)) one also writes GL(n, pa)(respectively SL(n, pa)).

Recall that, for a prime p, a finite group is called a p′-group if p is not a divisor ofthe order of the group. With N we denote the positive integers, including 0. We alsouse the notation N0 = N \ {0}. One can now easily prove the following theorem byHerstein.

Theorem 2.2 (Herstein [6]). Let G be a finite multiplicative subgroup of a divisionring D of characteristic p > 0. Then G is a cyclic p′-group.

Proof. We know that the division ring D contains the prime subfield Fp, the field

3

4 2. THE GOAL

with p elements. Consider K the Fp-subalgebra of D generated by G:

K = {n∑i=0

figi | n ∈ N, fi ∈ Fp, gi ∈ G}.

This is clearly a finite set as both G and Fp are finite. So, 0 6= r ∈ K. Then thereexist m,n ∈ N with m > n such that rn = rm, showing that 1 = r.rm−n−1, so r isinvertible. Hence K is a division algebra. By the Wedderburn theorem it has to bea field. As it is well known that the finite subgroups of fields are cyclic and G is asubgroup of K, it follows that G is cyclic.

Moreover, K is a finite field of characteristic p. We know that K×, the multi-plicative group of invertible elements of K, has pk − 1 elements, for some k ∈ N0.As G ≤ K× we get |G| | pk − 1, and thus G is a p′-group.

To see that every cyclic p′-group G is embedded in a division ring D of charac-teristic p, we use the following reasoning: denote |G| = n, so (n, p) = 1. As Znis finite, there exist positive integers k and l with k > l such that pk ≡ pl mod n.But then, as (n, p) = 1, p is invertible modulo n, so

pk−l ≡ 1 mod n.

So nm = pk−l − 1 for some m ∈ N. Denote a a generator, of order pk−l − 1,of (Fpk−l)×. Now it is obvious that G ∼= 〈am〉 ≤ (Fpk−l)×. So G can indeed beembedded in a field of characteristic p. This proves the following classification.

Theorem 2.3. A finite group G is embedded in a division ring of characteristicp > 0 if and only if G is a cyclic p′-group. In this case, we can always embed it ina finite field of characteristic p > 0.

In what follows we will consider the case where the characteristic is 0. As wasclear from the introduction, this case will not only contain cyclic groups. The restof this dissertation is dedicated to the study of this theorem in the case of a rationaldivision algebra.

2.1 The theorems of AmitsurWe will denote by Cn the multiplicative cyclic group of order n. The two maintheorems require some new terminology.

Definition 2.4. One says that a finite groupG is a Z0-group if all its Sylow-subgroupsare cyclic. Within this class, we say that a group is a Z-group if it is also embed-dable in a division ring of characteristic 0.

Recall the following definitions.

2.1. THE THEOREMS OF AMITSUR 5

Definition 2.5. The commutator subgroup G′ of a group G is defined as

G′ = 〈(a, b) = aba−1b−1 | a, b ∈ G〉.

It is the smallest normal subgroup N of G such that G/N is abelian.

Definition 2.6. A group G is called solvable if there exist some normal subgroupsG1, . . . , Gn of G, such that {1} = G0 ⊆ G1 ⊆ . . . ⊆ Gn = G and Gi+1/Gi isabelian for 0 ≤ i < n.

Remark 2.7. This definition is equivalent to saying that the derived series

G ⊇ G′ = G(1) ⊇ G′′ = G(2) ⊇ . . . ⊇ G(m) = (G(m−1))′ ⊇ . . . ,

becomes trivial, that is, there exists a positive integer n such that

G(n) = {1}.

In particular, if G is solvable and non-trivial, then G 6= G′.

It is well-known that the class of solvable groups is closed under taking quotientgroups and subgroups.

We will often need the order of some integer modulo another number, so weintroduce the following notation: for n and m coprime numbers, denote by om(n)the multiplicative order of n modulo m. This is the smallest positive integer k suchthat nk ≡ 1 mod m. We give the classification theorem by Amitsur, which is themain theorem of this thesis. It will be proven at the end of this dissertation, seeSection 6.3. We state the result in the form given by Shirvani and Wehrfritz [16].

Theorem 2.8 (Amitsur [2]). A finite group G is a subgroup of a division ring ofcharacteristic 0 if and only if G is isomorphic to one of the following groups:

a) a Z-group.

b) i) the binary octahedral group O∗ of order 48:

{±1,±i,±j,±ij, ±1± i± j ± ij2

} ∪ {±a± b√2| a, b ∈ {1, i, j, ij}}.

ii) CmoQ, where m is odd, Q is quaternion of order 2t for some t ≥ 3, anelement of order 2t−1 centralizes Cm and an element of order 4 invertsCm.

iii) M ×Q8, with M a Z-group of odd order m and 2 - om(2).

iv) M × SL(2, 3), where M is a Z-group of order m coprime to 6 and2 - om(2).

c) SL(2, 5), a group of order 120.

6 2. THE GOAL

Remark 2.9. By a quaternion group, we mean a group of the following form

Q2t = 〈x, y | x2t−2

= y2, y4 = 1, xy = x−1〉,for t ≥ 3. We call Q2t the quaternion group of order 2t. One might recognise Q8 asisomorphic to {±1,±i,±j,±k}, our example from page 1, by identifying i with xand j with y. Also remark that Z(Q2t) = {1, x2t−2} ∼= C2, where Z(Q2t) denotesthe center of the group Q2t .

Remark 2.10. The subdivision in the way this theorem is stated will come up nat-urally in the proof. Note that it follows that SL(2, 5) is the only non-solvable finitesubgroup of a division ring of characteristic 0. Under b) is listed the groups thatare solvable and have a quaternion subgroup (in particular a non-cyclic Sylow sub-group) and a) lists the groups that are solvable and whose Sylow-subgroups are allcyclic.

Remark 2.11. The binary octahedral group O∗ also has the presentation

〈r, s | r2 = s3 = (rs)4, r4 = 1〉.In order to complete the classification, Amitsur also proves the following classi-

fication of Z-groups. We state the result in the form given by Shirvani and Wehrfritz[16].

Theorem 2.12 (Amitsur [2]). A group G is a Z-group if and only if it is one of thefollowing types:

a) cyclic.

b) Cm o C4, where m is odd and C4 acts by inversion.

c) G0×G1× . . .×Gs, with s ≥ 1, the orders of the groups Gi are coprime andG0 is the only cyclic subgroup amongst them. Each of the Gi, for 1 ≤ i ≤ s,is of the form

Cpa o(Cqb11× . . .× Cqbrr

),

for p, q1, . . . , qr distinct primes. Moreover, each of the groups Cpa o Cqb isnon-cyclic (i.e. if Cqα denotes the kernel of the action of Cqb on Cpa , thenα 6= b) and satisfies the following properties:

(i) qoqα(p) - o |G||Gi|

(p).

(ii) one of the following is true:

• q = 2, p ≡ −1 mod 4, and α = 1,

• q = 2, p ≡ −1 mod 4, and 2α+1 - p2 − 1,

• q = 2, p ≡ 1 mod 4, and 2α+1 - p− 1,

• q > 2, and qα+1 - p− 1.

These cases are mutually exclusive, except for the group Cpa oC4 with p ≡ −1mod 4 and C4 acting by inversion. This group is covered by both the parts b) andc). This last theorem will also be proven in this dissertation (see Section 6.1).

3

Number theory

“42, ” said Deep Thought, with infinite majesty and calm.– Douglas Adams, The Hitchhiker’s Guide to the Galaxy

In later chapters we will rely heavily on some results from number theory. Thischapter is dedicated to an introduction to this theory and will contain enough infor-mation for readers who are new to this area to understand the rest of this dissertation.Number theory will often work with so called number fields: finite algebraic fieldextensions K of Q. Due to this fact, for the rest of this chapter, we will assume allrings and division algebras to be commutative and of characteristic 0. This chapteris loosely based on Neukirch [11] and Jacobson [7] and the results can be found inthese books unless stated otherwise.

3.1 Dedekind domainsWe begin with the definition of a domain.

Definition 3.1. A commutative ring R with unit is called a domain if it has nonon-trivial zero-divisors.

The following type of domains will be important.

Definition 3.2. A domain R is called a Dedekind domain if every non-zero idealI / R is a product of maximal ideals P1, . . . ,Pn :

I = P1 · . . . ·Pn.

In a Dedekind domain R it can be easily checked that every non-zero primeideal is a maximal ideal, implying

Spec(R) = Max(R) ∪ {(0)}.

Furthermore, it can be proven that the product in the definition is unique, up tothe order of the maximal ideals. Also, fields and unique factorisation domains are

7

8 3. NUMBER THEORY

examples of Dedekind domains since ideals generated by irreducible elements areprime. In this way, a Dedekind domain can be seen as a generalisation of the notionof a unique factorisation domain. An important example of a Dedekind domain thatwill be used later is Z.

Recall that one says that a domain R is integrally closed if R = {r ∈ K |r is integral over R}, where K is the field of fractions of R and an element is calledintegral over R if it is a zero of a monic polynomial over R.

Theorem 3.3. A domain R is a Dedekind domain if and only if R is noetherian,integrally closed and all non-trivial prime ideals are maximal.

Corollary 3.4. A localisation of a Dedekind domain with respect to a multiplica-tively closed set is again a Dedekind domain.

For the remainder of the section, if R is a domain then we denote by K its fieldof fractions.

Definition 3.5. Let R be a domain. An R-submodule of K of the form s−1I , withs ∈ R \ {0} and I a non-zero ideal of R, is called a fractional ideal of R. By F(R)one denotes the set of fractional ideals.

One can define a product on such fractional ideals by setting

(s−1I) · (l−1J) = (sl)−1IJ,

and F(R) becomes a monoid for this product. The neutral element is 1R.If R is a ring and I a set, then one denotes the direct product of R, indexed

by I as RI = {(ai)i∈I | ai ∈ R ∀i ∈ I}. This is again a ring with point-wise multiplication and addition. One also writes R(I) = {(ai)i∈I ∈ RI | ai 6=0 only finitely many times}. This is a subring of RI .

Theorem 3.6. If R is a Dedekind domain, then F(R) is a group. Moreover, everyfractional ideal has a unique representation as a product∏

P∈Max(R)

PvP ,

with (vP)P∈Max(R) ∈ Z(Max(R)).

If we think about Dedekind domains as a generalisation of unique factorisationdomains, then this theorem is the generalisation of extending the unique factori-sation of elements in a unique factorisation domain to the elements in its field offractions by allowing inverses of irreducible elements.

Corollary 3.7. For a Dedekind domain R, the map

ϕ : Z(Max(R)) → F(R) : (vP)P∈Max(R) 7→∏

P∈Max(R)

PvP ,

3.1. DEDEKIND DOMAINS 9

is an isomorphism of groups. It is used to define a map v : K× → Z(Max(R)) wherev is obtained by the following diagram:

K× −→ F(R) −→ Z(Max(R))

a 7−→ aR 7−→ ϕ−1(aR).

Thus a ∈ K× is mapped to (vP)P∈Max(R) ∈ Z(Max(R)) such that aR =∏

P∈Max(R)

PvP .

This map is called the total valuation map of K.

The following theorem is an important one in this section as it will give rise tothe notions of ramification index and residue degree.

Theorem 3.8. Let R be a Dedekind domain with field of fractions K. Let L/K bea finite separable extension, S the integral closure of R in L and p a prime ideal ofR. Then S is a Dedekind domain and if we write

pS = Pe11 · . . . ·Per

r ,

for the decomposition in different primes P1, . . . ,Pr of S, then S/Pi is a finitedimensional R/p-vector space for each 1 ≤ i ≤ r and we denote its dimensionwith fi. Furthermore

[L : K] =r∑i=1

eifi.

If L/K is a Galois extension, then e1 = . . . = er, and this number is denotede(p, L/K), and f1 = . . . = fr, and this number is denoted f(p, L/K). In particular

[L : K] = e(p, L/K)f(p, L/K)r.

In the above theorem, we say that the primes P1, . . . ,Pr lie over p and thisproperty is denoted by Pi|p. The number e(p, L/K) is called the ramification indexof the prime p in the extension L/K and f(p, L/K) is called the residue degree.The number r is often denoted as r(p, L/K). Intuitively, the numbers r(p, L/K)and e(p, L/K) give an idea how far pS is from being a prime or prime-power. Ife(p, L/K) = 1, we call the prime p unramified.

Corollary 3.9 (Multiplicativity of the ramification index and the residue degree).LetR be a Dedekind domain with field of fractionsK and F/L and L/K are (finite)Galois extensions. Let moreover P be a prime ideal of S, the integral closure of Rin L, lying over p, a prime ideal of R. Then

δ(p, F/K) = δ(P, F/L)δ(p, L/K),

where δ is a symbol in {e, f, r}.

10 3. NUMBER THEORY

We introduce the notion of a P-adic valuation. Because of Corollary 3.7, con-sider the following homomorphism for P ∈ Max(R):

vP : K× −→ Z : a 7→ vP(a),

where vP(a) is the exponent of P in the decomposition of aR. This map is thecomposition of v and the natural projection map onto the P-component ofZ(Max(R)).We extend this map to vP : K −→ Z ∪ {∞} by setting vP(0) =∞. This is the socalled P-adic valuation on K.

A local Dedekind domain (i.e. a Dedekind domain with a unique non-zeromaximal ideal) that is not a field is called a discrete valuation ring.

Proposition 3.10. Let R be a Dedekind domain and let P ∈ Max(R). The locali-sation RP of R at P is a discrete valuation ring and considering RP as a subringof K we have

RP = {x ∈ K | vP(x) ≥ 0} and PRP = {x ∈ K | vP(x) > 0}.The valuation map vP also has the following nice property that will allow us to

associate a norm to such a valuation (see also Section 3.3).

Proposition 3.11. Let R be a Dedekind domain, P a maximal ideal and K the fieldof fractions of R. The P-adic valuation map satisfies the following properties forelements a, b ∈ K:

• vP(a) =∞⇔ a = 0.

• vP(ab) = vP(a) + vP(b).

• vP(a+ b) ≥ min{vP(a), vP(b)}.For any 1 < r ∈ R, define the P-adic norm as follows:

| . |P: K → R : a 7→ r−vP(a).

Note that this norm satisfies a stronger property than the triangular inequality:

|a+ b|P ≤ max{|a|P, |b|P}.These properties allow to define a metric on K. The completion of K with respectto this metric is denoted KP and is again a field. Also, in this case the integralclosure S of R in KP is a local ring with unique prime PS. With this notation, onehas for a Galois extension K/F and P | p:

[KP : Fp] = e(p, K/F )f(p, K/F ),

and[K : F ] =

∑P|p

[KP : Fp].

The equalities are intuitively easy to see: if we remember that KP has a uniqueprime P that lies over p, then obviously r(p, KP/Fp) = 1 and the ramificationindex and residue degree do not change from e(p, K/F ) and f(p, K/F ) becausewe do not add any other prime ideals. These formulas are thus an easy consequenceof Theorem 3.8.

3.2. NUMBER FIELDS 11

3.2 Number fieldsA very important example of a Dedekind domain is Z with field of fractions Q.Here, the prime ideals correspond to the prime numbers, so the Dedekind property(every ideal is the product of some prime ideals) corresponds to the prime decom-position of the integers: if a =

∏paii is the prime decomposition of a, then

(a) =∏

(pi)ai .

Also keep in mind that Z is a principal ideal domain, so every ideal is of this form.For notation’s sake we will often write vp when we mean v(p) and similar adapta-tions will be used throughout. For a fraction a

b∈ Q× with a and b coprime integers,

the p-adic valuation becomes vp(ab

)= vp(a) if p is a prime divisor of a and else

vp(ab

)= −vp(b). Also, in the definition of a p-adic norm | . |p, we will take the

ground element to be p:|a|p = p−vp(a),

for every a ∈ Q×.Important fields are the so called number fields; these are finite algebraic exten-

sions K of the field Q. For these fields, we write OK for the algebraic closure of Zin K and call this the ring of integers of K.

We recall the definition of Euler’s totient function.

Definition 3.12. The function

ϕ : N0 → N0 : n 7→ |{m ∈ N0 | 0 < m ≤ n and m is coprime to n}|,

is called Euler’s totient function or Euler’s phi function.

From basic commutative group theory we can deduce that ϕ(n) = |{x ∈Z/nZ | x is a generator of Z/nZ}| = |{x ∈ Z/nZ | o(x) = n}|. It is well knownthat for an n ∈ N0 we get the following formula

ϕ(n) = n∏p|n

(1− 1

p

),

where p is a prime.Denote ζk = e

2πki ∈ C, a primitive k-th root of unity. A number field is called

cyclotomic if it is isomorphic to Q(ζk) for some positive integer k. The follow-ing proposition gives us insight in the ramification constants of cyclotomic numberfields.

Proposition 3.13. Let p be a prime, m a positive integer and consider the cyclo-tomic extension Q(ζm)/Q. Write m = pkm′ with (p,m′) = 1 and k ∈ N, thenp factors in OQ(ζm) into r different primes (ideals) of residue degree f and withramification index e such that

rf = ϕ(m′), f = om′(p) and e = ϕ(pk).

In particular, if p - m, then p is unramified in the extension.

12 3. NUMBER THEORY

Proof. See Weiss [17, Theorem 7.2.4 and Theorem 7.4.3].

Corollary 3.14. Let p be a prime number and n,m arbitrary positive integers withm and p coprime. If p is a prime of OQ(ζm) lying over p then

r(p,Q(ζmpn)/Q(ζm)) = 1

Proof. By Proposition 3.13, r(p,Q(ζm)/Q) = ϕ(m)om(p)

and r(p,Q(ζmpn)/Q) = ϕ(m)om(p)

.

Because of Corollary 3.9, r(p,Q(ζmpn)/Q(ζm)) =r(p,Q(ζmpn )/Q)r(p,Q(ζm)/Q) = 1.

Another field that is often studied in number theory is the so called field of p-adic numbersQp; it is the completion ofQwith respect to the p-adic norm. As suchwe have constructed infinitely many new fields.

Proposition 3.15. If p and q are primes in Z, then

Qp∼= Qq ⇔ p = q.

The closure Zp of Z in Qp with respect to this metric is called the ring of p-adic integers. The ring Zp also has the following other descriptions which could beuseful for intuition:

• {x ∈ Qp | |x|p ≤ 1} (compare to Proposition 3.19),

• The ring consisting of the formal series

∞∑n=0

aipi,

where ai ∈ {0, . . . , p− 1},

• Z[[X]]/(X−p), where Z[[X]] denotes the ring consisting of the formal powerseries over Z,

• lim←−n

Z/pnZ, the inverse limit of the rings Z/pnZ.

For the second type described above, we can consider Qp to be the formal sums

∞∑n=−∞

aipi,

where again ai ∈ {0, . . . , p− 1}. In any case, Qp is the field of fractions associatedto Zp. The units of Zp are exactly those that have p-adic norm 1.

3.3. VALUATIONS 13

3.3 Valuations

We will now generalise the concept of a valuation so that we do not always need toconstruct a field starting from a Dedekind domain. There are two types of valua-tions, namely a multiplicative and an exponential valuation, defined in the followingdefinitions.

Definition 3.16. Let K be a field. A map

| . |: K −→ R+,

is called a (multiplicative) valuation if it satisfies the following properties for alla, b ∈ K:

• |a| = 0⇔ a = 0,

• |ab| = |a||b|,

• |a+ b| ≤ |a|+ |b|.

To avoid trivial cases we will exclude the valuation |a| = 1 if a 6= 0 and |0| = 0.One calls two multiplicative valuations| . |1 and | . |2 equivalent if they generate thesame topology, or equivalently if | . |1=| . |s2 for some 0 < s ∈ R.

The P-adic norm is an example of a multiplicative valuation.

Definition 3.17. Let K be a field. A map

v : K −→ R ∪ {∞},

is called an (exponential) valuation if it satisfies the following properties for alla, b ∈ K:

• v(a) =∞⇔ a = 0,

• v(ab) = v(a) + v(b),

• v(a+ b) ≥ min{v(a), v(b)}.

For an exponential valuation we will avoid the case where v(a) = 0 if a 6= 0 andv(0) = ∞. One calls two exponential valuations v1 and v2 equivalent if v1 = sv2for some 0 < s ∈ R.

The P-adic valuation is an example of an exponential valuation.

Definition 3.18. The valuation | . | is called non-archimedean if the set {|n| | n ∈N} is bounded. If not, then we call the valuation archimedean.

14 3. NUMBER THEORY

One can prove that

| . | is non-archimedean ⇔ |x+ y| ≤ max{|x|, |y|} for all x, y in K.

The P-adic norm is an example of a non-archimedean valuation. An example of anarchimedean valuation is the absolute value on Q.

Given a multiplicative non-archimedean valuation | . |, for each 1 < q one candefine v(x) := − logq |x| for x ∈ K× and v(0) := ∞ the associated exponentialvaluation for a non-archimedean multiplicative valuation.

On the other hand, given an exponential valuation v, one can define a non-archimedean multiplicative valuation for each 1 < q by setting |x| := q−v(x) forx ∈ K× and |0| := 1. These constructions are each others inverse up to equivalenceof valuations. In the case of a non-archimedean valuations one often uses the wordvaluation to mean either the exponential or multiplicative variant.

Within the field K one can find a ring that will act as a local Dedekind domain,as described in Proposition 3.10.

Proposition 3.19. Let K be a field with a valuation v. The set

O = {x ∈ K | v(x) ≥ 0} = {x ∈ K | |x| ≤ 1},

is a subring of K with units

O× = {x ∈ K | v(x) = 0} = {x ∈ K | |x| = 1},

and unique maximal ideal

P = {x ∈ K | v(x) > 0} = {x ∈ K | |x| < 1}.

The ring O is called a valuation ring of K and κ = O/P a residue field of K.Remark that these structures are invariant under equivalence of valuations.

The P-adic valuations are examples of so called discrete normalised valuations.

Definition 3.20. A valuation v is called discrete if it admits a smallest positive values. One can see that in this case v(K) = sZ ∪ {∞}.

It is called a normalised discrete valuation if s = 1. Clearly every discretevaluation is equivalent to a normalised valuation.

This definition is compatible with the notion of discrete valuation ring definedearlier: K is a field with a discrete valuation v if and only if the associated valuationringO is a discrete valuation ring. The discrete normalised valuations have a usefulproperty.

Theorem 3.21. Let v be a discrete normalised valuation. An element π ∈ O withv(π) = 1 is called a uniformising parameter. For every uniformising parameter πand every a ∈ K× there exists a unique decomposition

a = uπm,

with u a unit of O and m = v(a).In particular, O is a local principal ideal domain and the ideals are of the form

Pn = πnO = {x ∈ K | v(x) ≥ n}.

3.4. LOCAL FIELDS 15

3.4 Local fieldsAs we have done in Section 3.1, we can take the completion of a field K in ac-cordance with some arbitrary valuation | . |. It is easy to see we can extend thisvaluation to the completion. We have seen completions of Q with respect to thep-adic valuation in Section 3.2 and called them the p-adic numbers Qp. It appearswe now know all the completions of Q with respect to valuations.

Theorem 3.22 (Ostrowski). Any valuation on Q is equivalent to | . |p for someprime number p or | . |, the absolute value. In particular, the absolute value is theonly archimedean valuation on Q and all the completions are given by

R,Q2,Q3,Q5, . . . ,Qp, . . .

where no two in this collection are isomorphic.

We introduce local fields.

Definition 3.23. A discrete valued field K is called a local field if it is completewith respect to its valuation and κ, its residue field, is finite.

There is a more general notion of local field, but as we will only need this fordiscrete valued fields of characteristic 0, this definition suffices. The more generaldefinition would allow for local fields on a non-discrete valued field and in the casesof positive characteristic. Equivalently we can say that the topology defined by thevaluation is locally compact, hence the name of a local field.

Theorem 3.24. The local fields of characteristic 0 are exactly the finite extensionsof the fields Qp. Moreover, their valuation rings are local.

If we allow for non-discrete valuations in our definition of local fields, then RandC are added to this list. If we also allow for fields of positive characteristic, thenthe finite extensions of Fp((x)), the field of fractions of Fp[[X]], are also added.

Because of Theorem 3.24 one also calls the local fields of characteristic 0 p-adicnumber fields. If L/K are both local fields, with unique prime ideals P | p of theirvaluation rings, then we can unambiguously define the ramification index and theresidue degree of the extensions:

e(L/K) := e(p, L/K) and f(L/K) := f(p, L/K).

Remark 3.25. Every finite extension of Qp has a residue field isomorphic to Fprfor some r ∈ N. Indeed, one can easily prove that the residue field of Qp is iso-morphic to Fp. Clearly now the residue field κ of a finite extension K/Qp is finitedimensional over Fp, proving the claim.

Definition 3.26. A Galois extension L/K is called abelian (cyclic) if Gal(L/K) isan abelian (cyclic) group.

16 3. NUMBER THEORY

Recall that for a finite Galois extension L/K we can define a norm map from Lto K:

NL/K : L→ K : x 7→∏

σ∈Gal(L/K)

σ(x).

This map is multiplicative and defines a group homomorphism between L× andK×. For example in the case of C/R one has NC/R(z) = zz = |z|2, motivatingthe name ”norm map”. For a tower of finite, abelian local field extensions we havefollowing useful proposition that allows us to pull back information of the norms.

Proposition 3.27. Let F/L/K be a tower of local field extensions such that F/Kis a finite abelian extension. Then we have the following commuting diagram:

Gal(F/L) L×/NF/L(F×)

Gal(F/K) K×/NF/K(F×)

∼=

NL/K

∼=

In particular, if x ∈ L×, then

x ∈ NF/L(F×)⇔ NL/K(x) ∈ NF/K(F×).

Proof. For the first statement, see Serre [15, XIII, §4, Corollary and XIII, §4, Propo-sition 10]. The second statement is equivalent to saying that NL/K is injective onL×/NF/L(F×), which follows from the commutativity of the diagram.

One last definition in this chapter defines unramified and totally ramified exten-sions. We need this for a theorem concerning the norms of such extensions, seeTheorem 3.33.

Definition 3.28. A finite field extension L/K of local fields is said to be unramifiedif the extension of residue fields λ/κ is seperable and

[L : K] = [λ : κ] = f(L/K).

Similarly one defines a totally ramified extension if

[L : K] = e(L/K).

Proposition 3.29. A subextension of an unramified extension is again unramifiedand the composite of two unramified extensions is again unramified.

Example 3.30. It follows from Proposition 7.12 in Neukirch [11] that for a p-adicnumber field K with residue field κ ∼= Fpa , the extension L := K(ζn) with n andp coprime is unramified of degree on(pa). The group Gal(L/K) is generated byσ : ζ 7→ ζp

a.

3.5. HENSEL’S LEMMA AND APPLICATIONS 17

Example 3.31. It follows from Proposition 7.13 in Neukirch [11] that the extensionQp(ζpn)/Qp is totally ramified of degree ϕ(pn) = (p− 1)pn−1.

Corollary 3.32. The residue field κ of K = Qp(ζn) with n = pam and (p,m) = 1is isomorphic to Fpo with o = om(p) and [K : Qp] = om(p)ϕ(pa).

Proof. Write K = Qp(ζpa)(ζm). Then the extension Qp(ζpa)/Qp is totally ramifiedby Example 3.31 and Qp(ζpa) has residue field Fp. The extension K/Qp(ζpa) isnow unramified by Example 3.30 and om(p) = [K : Qp(ζpa)] = [κ : Fp].

The last statement follows from

[K : Qp] = [K : Qp(ζpa)][Qp(ζpa) : Qp] = om(p)ϕ(pa).

Theorem 3.33. Suppose that L/K is an unramified extension. Then a unit inU(OK) is a norm of a unit in U(OL) if and only if its residue class modulo P,a prime of K, is a norm of the residue field of L. In particular, if the residue field ofK is finite, then

NL/K(U(OL)) = U(OK).

Proof. See Cassels and Frolich [3, Corollary pg. 29].

3.5 Hensel’s lemma and applicationsIn this section we work towards some results that are specifically used in the clas-sification of Z-groups in Section 6.1.

Lemma 3.34. In Qp with p 6= 2, all the roots of unity are (p − 1)-th roots of unityin Zp and they are all different in Zp/pZp. In Q2, the roots of unity are exactly ±1.

To prove this lemma, we state Hensel’s Lemma, a proof of which can be foundin Neukirch [11, pg. 129].

Lemma 3.35 (Hensel’s Lemma). If f(X) ∈ Zp[X] and an a ∈ Zp exists such that

f(a) ≡ 0 mod p, f ′(a) 6≡ 0 mod p,

then there exists a unique α ∈ Zp such that f(α) = 0 and a ≡ α mod p.

Example 3.36. Let p be a prime and k an integer such that 0 ≤ k ≤ p−1. Considerthe polynomial f(X) = Xp − X . As kp ≡ k mod p by Fermat’s little theorem,we have

f(k) ≡ 0 mod p and f ′(k) = pkp−1 − 1 ≡ −1 6≡ 0 mod p.

So, by Hensel’s Lemma 3.35, we obtain elements αk ∈ Zp satisfying αpk = αkand αk ≡ k mod p. Note that these p elements α0, . . . , αp−1 are all differentbecause they are different modulo p. This shows that f(X) = X(Xp−1 − 1) splitscompletely over Zp and we have exactly p− 1 p-adic (p− 1)-th roots of unity.

18 3. NUMBER THEORY

Proof of Lemma 3.34. Suppose x ∈ Qp is a root of unity: xn = 1, for some positiveinteger n. Then |x|np = 1 and |x|p = 1, showing that x ∈ Z×p (see Section 3.2). Wemay thus assume in the sequel that all roots of unity are in Z×p .

Let us first consider roots of unity of order relatively prime to p and let ζ1 andζ2 be such roots. The integer m, being the product of their orders, is still relativelyprime to p. The elements ζ1 and ζ2 are roots of the polynomial f(X) = Xm − 1.Considering the fact that |f ′(ζi)|p = |mζm−1i |p = |ζi|m−1p = 1 (which moreovershows that f ′(ζi) 6= 0) , the uniqueness part of Hensel’s Lemma 3.35 implies that

ζ1 ≡ ζ2 mod p⇒ ζ1 = ζ2.

In other words, distinct roots of unity of power relatively prime to p must be incon-gruent modulo p. But in Example 3.36 we already found p− 1 different (p− 1)-throots of unity, all belonging to a non-zero class in Zp/pZp. It is clear that there cannot be a root of unity congruent to 0 modulo p, so indeed all the roots of unity oforder relatively prime to p in Zp (and thus Qp) are (p − 1)-th roots of unity. In thecase where p = 2, this means that only the p-adic number 1 is such a root of unity.

We will now consider p-power roots of unity. For p 6= 2, we will prove that theonly p-th root of unity is 1 in Zp, showing that all p-power roots of unity are trivial:if ζpk = 1, then (ζp

k−1)p = 1 and thus by assumption ζpk−1

= 1. By induction thiswould show that all p-power roots of unity are trivial. For p = 2, we will show thatthe only 4th roots of unity are±1, implying similarly that all 2-power roots of unityare ±1.

Assume p 6= 2 and suppose ζp = 1 for some ζ ∈ Zp. This implies that ζ ≡ζp = 1 mod p using Fermat’s Little Theorem. By the uniqueness part in Hensel’sLemma 3.35 on the polynomial f(X) = Xp − 1 with |f ′(ζ)|p = |pζp−1|p = 1

pwe

have that the set

{x ∈ Zp | |x− ζ|p < |f ′(ζ)|p} = {x ∈ Zp | |x− ζ|p ≤1

p2} = ζ + p2Zp,

contains a unique p-th root of unity, namely ζ . We will show that ζ ≡ 1 mod p2,which proves that 1 ∈ ζ + p2Zp. This implies that 1 = ζ .

As ζ ≡ 1 mod p, write ζ = 1 + py for a y ∈ Zp. Then

1 = ζp = (1 + py)p = 1 + p(py) +

p−1∑i=2

(p

i

)(py)i + (py)p.

The binomial coefficients in the sum over i are all at least divisible by p and becausei ≥ 2, the whole sum over i is divisible by p3. Also (py)p is divisible by p3 sincep ≥ 3. This yields

1 ≡ 1 + p2y mod p3,

so 0 ≡ p2y mod p3. Therefore p divides y and ζ ≡ 1 mod p2.Now let p = 2. We prove that if ζ is a 4th root of unity, then ζ = ±1. Suppose

for the sake of contradiction that ζ 6= ±1, then ζ2 = −1. This implies that ζ2 ≡ −1mod 4, but we also have

ζ ∈ Z×2 ⇒ ζ ≡ 1 or 3 mod 4⇒ ζ2 ≡ 1 mod 4,

3.5. HENSEL’S LEMMA AND APPLICATIONS 19

a contradiction.In general, every root of unity in Zp is a product of a p-power root and a root of

order relatively prime to p. For p = 2 this is±1. For a different p, this is just ζ withζp−1 = 1.

Now that we know what the roots of unity look like in the p-adic numbers, weuse this to prove the following proposition.

Proposition 3.37. Let k = Qp and K = k(ζmp), with p an odd prime and m apositive integer. Suppose α is a root of unity in k such that α = NK/k(β) for someβ ∈ K. Then α = 1.

Proof. As the norm satisfies the following rule:

NK/k = NQp(ζp)/k ◦NK/Qp(ζp),

we may assume m = 1. As K/k is an extension of local fields, there is a uniqueprime ideal P lying over p. Now denote v the p-adic valuation on k and also itsextension to K. Then for every s ∈ Aut(K/k), we have that v(s(β)) = v(β). AsP = {x ∈ K | v(x) ≥ 1}, we see that s(β) ≡ β mod P. If we write out thedefinition of the norm then

α = NK/k(β) =∏

s∈Aut(K/k)

s(β) ≡P βp−1 ≡P 1,

where in the last equivalence we used that the residue field of K is isomorphic toFp (see Corollary 3.32). We have also used that [K : k] = p − 1, which is clearfrom Example 3.31. But then α ≡ 1 mod (k ∩P) = (p). By Lemma 3.34, as 1 isthe only root of unity with residue 1 modulo p, α = 1.

Theorem 3.38 (The Hasse Norm Theorem). Let L/K be a cyclic extension of anumber field K. Let P be a prime of K and P a prime of L lying over P , then

a ∈ NL/K(L)⇔ a ∈ NLP/KP (LP) for every pair of primes P | P.

Proof. See Cassels & Frohlich [3, pg. 185].

20 3. NUMBER THEORY

4

Finite group theory

“It’s an impossible puzzle, and I love puzzles!”– Ron Swanson, Parks and Recreation

Obviously, this dissertation requires a chapter on finite group theory. In thischapter we will further discuss some types of groups that appear throughout thisthesis, such as Z0-groups, Z-groups, quaternion groups, solvable groups, etc. Theproperties listed here are to be seen as an extension of our group theoretical knowl-edge.

For a group G with subgroup H the subgroup NG(H) = {g ∈ G | g−1Hg ⊆H} is called the normalizer of H in G. It is the biggest subgroup of G of which His a normal subgroup. The subgroup of all elements of G that commute with H isdenoted by CG(H) = {g ∈ G | gh = hg ∀h ∈ H} and is called the centralizer ofH in G. In this notation, CG(G) = Z(G).

We will sometimes use the following argument, called the Frattini argument,after Giovanni Frattini, an Italian mathematician who first used it in a paper in 1885while defining the Frattini subgroup. It correlates a normal subgroup H of G and aSylow subgroup of H to the whole group G.

Lemma 4.1 (Frattini’s Argument). Let G be a finite group, H a normal subgroupof G and P a Sylow p-subgroup of H , then

G = NG(P )H.

Proof. We take g ∈ G arbitrary. As P ≤ H and H is normal in G, g−1Pg ≤ H .But g−1Pg is also a Sylow p-subgroup of H . By the Sylow theorems, it has to beH-conjugate to P :

g−1Pg = h−1Ph,

for some h in H . Clearlyhg−1P (hg−1)−1 = P,

and thus gh−1 ∈ NG(P ). The lemma follows.

21

22 4. FINITE GROUP THEORY

There is one occasion, namely in the proof of Theorem 6.8, where we will needan advanced version of this argument for solvable groups. This will be Lemma 4.4.Beforehand, we introduce the notion of a Hall subgroup.

Definition 4.2. Let G be a finite group. We call a subgroup H of order n a Halln-subgroup of G if |H| and [G : H] are coprime.

Theorem 4.3 (P. Hall). LetG be a finite solvable group of order∏i∈I

paii for different

primes pi. If we take J ⊆ I arbitrary, then Hall nJ -subgroups exists for nJ =∏i∈J⊆I

paii and all the Hall nJ -subgroups are conjugate.

Proof. See Robinson [14, Theorem 9.1.7].

In the proof of Lemma 4.1 we only used that Sylow p-subgroups are conjugate,so using Theorem 4.3, a proof of the next lemma is obvious.

Lemma 4.4 (Frattini’s argument for solvable groups). Let G be a finite solvablegroup, H a normal subgroup of G and P a Hall subgroup of H , then

G = NG(P )H.

The following lemmas are used in Shirvani and Wehrfritz [16] without proof butfor simplicity’s sake they are proved here. The homomorphism IdG is the identityhomomorphism on the group G.

Lemma 4.5. Let N /G. If Aut(N) is a 2-group and CG(N) is a 2-group, then G isa 2-group.

Proof. Let g ∈ G arbitrary. Denote the action of g on N by αg ∈ Aut(N). SinceAut(N) is a 2-group, write 2n = o(αg) for some n. Then αg2n = α2n

g = IdN , sog2

n ∈ CG(N), showing that the order of g2n is also a 2-power since CG(N) is a2-group. This proves that g itself is of order a 2-power.

Lemma 4.6. Aut(Cn) ∼= (Z/nZ)×.

Proof. In a cyclic group Cn = 〈x〉, every automorphism is given by mapping thegenerator x to another generator. As an element xk is a generator of Cn if and onlyif (n, k) = 1, we obtain ϕ(n) choices for k, proving that |Aut(Cn)| = ϕ(n). As aninjection in : (Z/nZ)× → Aut(Cn) is now easily given by i(k + nZ)(x) = xk, thelemma is proven.

Corollary 4.7. Let pa be an odd prime power. The only automorphism of Cpa oforder 2 is the inversion.

Proof. By Lemma 4.6, Aut(Cpa) ∼= (Z/paZ)× ∼= (Fpa)×. We know this is a cyclicgroup of order pa − 1. Such a cyclic group clearly has only one element of order 2.It is obvious this has to be the inversion.

4.1. Z0- AND Z-GROUPS 23

Lemma 4.8. Let f ∈ Aut(Cpk). If (o(f), p) = 1 and f |Cp = idCp , then f = IdCpk

.

Proof. Consider the following commutative diagram for any m and d positive inte-gers such that d divides m:

(Z/mZ)× Aut(Cm)

(Z/dZ)× Aut(Cd)

im

π res

id

Here, im and id are the isomorphisms constructed in the proof of Lemma 4.6, π isthe canonical surjection and res is the restriction map.

For the proof, let m = pk and d = p and let α be an automorphism of Cpk oforder coprime to p such that res(α) = IdCp . It follows that s ∈ (Z/pkZ)×, thepre-image of α under im, is in the kernel of π. This kernel is of order

ker(π) =|(Z/pkZ)×||(Z/pZ)×|

=ϕ(pk)

ϕ(p)= pk−1,

so s has order a divisor of pk−1. Due to the isomorphism im, s has the same orderas α, which is coprime to p. It follows that s ≡ 1 mod pk, so IdC

pk= im(s) =

α.

4.1 Z0- and Z-groupsFrom now on, let D denote a division ring of characteristic 0. If R is a subring of Dand G ⊆ D× is a group, we will denote R[G] for the subring of D generated by Rand G. This is the smallest subring of D that contains both R and G. We will oftenuse this in the case R = Q, the prime subfield of D.

If a group G acts on a set X , we will often denote the action of a g ∈ G on ax ∈ X by xg. This notation is also used for the conjugation action of G on itself.

For the next propositions, we need the notion of a Frobenius complement.

Definition 4.9. A groupG is called a Frobenius complement if it is a group of fixed-point-free automorphisms of a finite group K, so if kg = k for g ∈ G, k ∈ K, theng = eG or k = eK .

It appears that all the groups G we deal with are of this form.

Proposition 4.10. IfG is a finite subgroup ofD, thenG is a Frobenius complement.

Proof. The subring R := Z[G] of D can be seen, as an additive group, as the directsum of n copies of Z, with n = |G|. The action of G on R by right multiplicationis clearly faithful, and also fixed-point-free. Indeed, if rg = r, then, in D, wewould have r(g − 1) = 0, so r = 0 or g = 1. We can then see G as a subgroup of


GL(n,Z) via this action. The fixed-point-free property translates to: if g ∈ G is notthe trivial element, then 1 is not an eigenvalue of g. In other words det(g − 1) 6= 0for g 6= 1. If we take a prime p ∈ N and using reduction modulo p, we obtain arepresentation of G in GL(n, p) . The canonical map associated to this reductionis denoted by g 7→ g. Since G is finite, it is clear we can find a prime p such thatthis representation is faithful (by taking a large prime for example) and such thatp - det(g − 1) for all 1 6= g ∈ G. Clearly then det(g − 1) 6= 0 for all non-trivialg ∈ G, meaning that G acts fixed-point-freely on the finite group Fnp . This showsthat G is a Frobenius complement.

We recall a famous result for Frobenius complements (see Passman [12, Theo-rem 18.1]).

Theorem 4.11. The Sylow subgroups of a group G that is a Frobenius complementare either cyclic or quaternion groups. In particular, the subgroups of order pq, forp and q primes, are cyclic and abelian subgroups of Frobenius complements arecyclic.

We give some examples. Clearly the group C7oC2, where C2 acts by inversion,is a Z0-group, but not a Frobenius complement (so not a Z-group) by Theorem 4.11.Indeed, it is itself a subgroup of order pq for p and q primes, but not cyclic. C7oC4,where C4 acts by inversion, is a Z-group because it is isomorphic to the groupgenerated by ζ7, a primitive 7-th root of unity inside of C ⊆ R+Ri+Rj +Rij =H(R), the real quaternions and j ∈ H(R).

Taking into account Proposition 4.10 and Theorem 4.11, we see that the struc-ture of a finite subgroup G of D is determined by its Sylow 2-subgroups. Thesecan either be cyclic or quaternion. In the former case, all Sylow subgroups of Gare cyclic, and so G would be a Z0-group and even a Z-group. This gives us thedifference between a group of the form a) and b) in Theorem 2.8 as described inthe remark afterwards. One has a decent structure theorem for Z0-groups, a prooffor which comes from Hall [5, Theorem 9.4.3].

Theorem 4.12. If G is a Z0-group, then G is of the form Cm o Cn for coprimeintegers m and n. If moreover G is a Z-group, then for all primes p | n, thesubgroup Cp of Cn centralizes Cm.

We will prove this theorem, but first some auxiliary results.

Theorem 4.13 (Frobenius). If d divides the order of a finite group G, then thenumber of solutions for the equation xd = 1 in G is a multiple of d.

Proof. An elementary proof can be found in Khurana and Khurana [9, page 222].

Proposition 4.14. Subgroups and epimorphic images of Z0-groups are again Z0-groups.

4.1. Z0- AND Z-GROUPS 25

Proof. Suppose G is a Z0-group and H ≤ G. A Sylow p-subgroup of H is a sub-group of a Sylow p-subgroup of G. As subgroups of cyclic groups are themselvescyclic, H is again a Z0-group.

Now assume H is normal in G. For a prime p dividing the order of G, let ggenerate a Sylow p-subgroup of order pa and write |H| = pbm with (p,m) = 1.Denote by g the natural image of g in G/H . The order of this element is stilla p-power and divides pa−b (this is the highest power of p that divides |G/H|).Suppose o(g) = pc for c < a − b. Then gpc = 1 in G/H , so gpc ∈ H . Since gpc

is an element of order pa−c > pb, this yields a contradiction for pa−c - |H|. Weconclude that g generates a group of order pa−b in G/H . This subgroup is a cyclicSylow p-subgroup of G/H . All Sylow p-subgroups are then cyclic since they areconjugate. Due to the fact that pwas chosen arbitrary, this proves thatG/H is againa Z0-group.

Proposition 4.15. If G is a Z0-group, then G is solvable.

Proof. Write |G| =s∏i=1

peii its prime decomposition, with p1 < . . . < ps. We will

first show that, for an integer of the form m = pfii pei+1

i+1 . . . pess , where fi ≤ ei, the

equation xm = 1 has exactly m solutions in G. This is obvious for m = |G|. Via aninduction argument, it will then suffice to prove the following: if p is a prime factorof |G|, such that it is the smallest prime dividing mp and xmp = 1 has exactly mpsolutions, then xm = 1 has exactly m solutions.

Suppose pf+1 is the highest power of p dividing mp. Since the Sylow p-subgroup of G is cyclic, there exists an element in G of order exactly pf+1. So,not all solutions of xpm = 1 are solutions of xm = 1. The number of solutionsfor xm = 1 is a multiple of m by Theorem 4.13, say km, and the number of solu-tions for xpm = 1 is pm by hypothesis. We obtain that 1 ≤ k < p and there arepm− km = (p− k)m elements satisfying xmp = 1 but not xm = 1.

Now, an element satisfying xmp = 1 but not xm = 1 has order exactly mp.Thus, there are ϕ(mp) elements of this form (see the remark after Definition 3.12).

The integer ϕ(mp) = mp∏q|mp

= m(p−1)∏p 6=q|m

(1− 1

q) is divisible by p−1. Hence,

because of the reasoning above, (p − k)m is divisible by p − 1. Since p was thesmallest prime dividing mp, p − 1 will not divide m, so p − 1 has to divide p − k.As 1 ≤ k < p, this is only possible if k = 1, showing that xm = 1 has exactly msolutions and proving the claim.

This is particularly so for m = pess . A Sylow ps-subgroup of G has order m.Since conjugation keeps the order of an element intact, this shows that this Sylowps-subgroup is a normal subgroup (xp

ess − 1 has exactly pess solutions). Since it is

cyclic (remember, G was a Z0-group), it is of course solvable.We have shown that a Z0-group G has a solvable normal subgroup H . But by

Lemma 4.14, H and G/H are also Z0-groups, so by induction we may assume Hand G/H to be solvable. Then G is also solvable.


Theorem 4.16. If two consecutive factor groups G(n−1)/G(n) and G(n)/G(n+1) ofthe derived series G ≥ G(1) ≥ G(2) ≥ G(3) ≥ . . . of a group G are cyclic, thenG(n+1) = {1}.

Proof. See Hall [5, Theorem 9.4.2].

Proof of Theorem 4.12. Using the classification of finite abelian groups, it is easyto see that an abelian Z0-group is in fact cyclic. Hence, as G is already solvable byProposition 4.15, the factor groups of G ⊃ G′ ⊃ G′′ ⊃ . . . are cyclic. But then, byTheorem 4.16, G′′ = 1. Suppose first that G′ = 1. Then G is abelian, so cyclic.Hence, it is of the desired form with n = 1.

Now suppose G′ 6= 1. Since G′′ = 1, it is abelian. According to Proposi-tion 4.14, it is also a Z0-group, so it is cyclic. Let a be a generator of G′ of orderm = |G′|. Let b an element that yields a coset bC ′ that is a generator for G/G′ (thisis possible because G/G′ is cyclic). Hence, a and b generate G and b−1ab = ar forsome r 6= 1 (otherwise G would be abelian, a contradiction with G′ 6= 1). Let n bethe order of G/G′, so bn ∈ G′ which implies that bn commutes with a. The actionof bn on a is then given by

a = b−nabn = arn

,

showing that rn ≡ 1 mod m. Every element of G is of the form biaj . The generalcommutators are then of the form (biaj, buaw), but these can be rewritten in termsof (ak, bl). By calculating

(ak, bl) = a−kb−lakbl = a−kakrl

= ak(rl−1) = (a(r−1))k(1+...+r

l−1),

we see that they themselves can be written in terms of ar−1. Thus G′ is also gen-erated by ar−1, showing that (r − 1,m) = 1. Write bn = aj , for some integer j.Thus, arj = (ar)j = (b−1ab)j = b−jajbj = b−jbnbj = bn = aj , or equivalently(ar−1)j = 1. Since (r − 1,m) = 1, this implies j ≡ 0 mod m. This shows thatbn = 1 and hence G ∼= Cm o Cn. Moreover, if m and n would have a commonprime factor p, then a

mp and b

np would generate a non-cyclic group of order p2, a

contradiction with the fact that G is a Z0-group and Theorem 4.11.If G was already a Z-group, then by Theorem 4.11 the subgroups of order pq

are cyclic, and the second part follows.

4.2 Quaternion groupsIt is already clear that the quaternion groups play an important role in this thesis.This section is dedicated to the study of some properties of these quaternion groups.These properties are used without proof in Shirvani and Wehrfritz [16], but areproven here.

Recall that a quaternion group is of the following form

Q2t = 〈x, y | x2t−2

= y2, y4 = 1, xy = x−1〉,

for t ≥ 3.

4.2. QUATERNION GROUPS 27

Lemma 4.17. Aut(Q8) ∼= S4.

Proof. Indeed, Q8 has exactly 3 cyclic subgroups of order 4: 〈x〉, 〈y〉, 〈xy〉. SinceAut(Q8) acts on the set of these three subgroups, it induces a homomorphism Φ :Aut(Q8)→ S3. This map is surjective because the automorphisms f : x 7→ y, y 7→x and g : y 7→ xy, xy 7→ y give us two transpositions in S3, and we know S3 isgenerated by these transpositions.

Furthermore, the kernel of Φ contains the maps k such that k(〈x〉) = 〈x〉 andk(〈y〉) = 〈y〉 (automatically we will have k(〈xy〉) = 〈xy〉). If k(〈x〉) = 〈x〉then k(x) = x or x3 and similarly k(y) = y or y3. It is clear these choices areindependent of each other and that they are all of order 2 or 1. We have shown thatker Φ ' K4, Klein’s 4-group.

Clearly the two automorphisms f and g from earlier are not in the kernel of Φand Aut(Q8) ≥ K :=< f, g >∼= S3 with Φ(K) = S3. As ker Φ∩K = 1 we get thatAut(Q8) = ker ΦoK ∼= K4oS3 where the first equality follows from an argumentbased on the order of the groups. One can check that non-trivial elements of ker Φdo not commute with K. The latter thus acts faithfully on K4 by conjugation, andup to isomorphism there is only one such action. Thus Aut(Q8) = K4 o S3

∼= S4.This last isomorphism is obtained by mapping K4

∼= Z2 × Z2 to the subgroup〈(1, 2)(3, 4), (1, 3)(2, 4)〉 and by embedding S3 in S4 in the canonical way.

Lemma 4.18. Q8 o C3∼= SL(2, 3)

Proof. In SL(2, 3) we find Q8 as a normal subgroup

Q8∼= 〈(

0 12 0

),

(2 22 1

)〉.

The group C3 can be found in SL(2, 3) as the subgroup

〈(

1 10 1

)〉.

In SL(2, 3) these groups have a trivial intersection (Q8 has no element of order 3).

Furthermore |〈(

0 12 0

),

(2 22 1

)〉o 〈

(1 10 1

)〉| = 8 ·3 = 24 = |SL(2, 3)|, showing

that indeed Q8 o C3∼= SL(2, 3).

Remark 4.19. If we denote

c =

(1 10 1

), y =

(0 12 0

)and x =

(2 22 1

),

then we can see that in SL(2, 3):

xc = y, yc =

(1 22 2

)= xy and (xy)c = x.

This means that the action of C3 on Q8 is given by cyclically permuting x, y andxy.


Lemma 4.20. Aut(Q2t) is a 2-group for t > 3.

Proof. Take t > 3 arbitrary and fixed, write Q2t = Q and let ϕ ∈ Aut(Q).Now ϕ(x) = xry or ϕ(x) = xr for some r. Suppose for the sake of contradic-tion that ϕ(x) = xry for some r. As we have xyx = y, we get that ϕ(x4) =(xryxr)y(xryxr)y = y4 = 1. This yields a contradiction due to the fact thato(x) = 2t−1 6= 4. We proved that that ϕ(x) is of the form xr and since the or-der needs to be preserved, (r, 2t−1) = 1. Furthermore, ϕ(y) = xiy for somei arbitrary because we need surjectivity. So we can choose r in φ(2t−1) differ-ent ways, and the image of y in 2t−1 different ways. It is also clear that ev-ery choice of such r and i yields an automorphism by extending the images ofx and y multiplicatively to the whole group. This shows that |Aut(Q)| dividesφ(2t−1)2t−1 = 2t−2φ(2)2t−1 = 22t−2, so Aut(Q) is a 2-group.

4.3 The group SL(2, 5)It is known that SL(2, 5) is generated by

x =

(−1 1−1 0

)and y =

(1 01 1

).

They also give rise to the following presentation for SL(2, 5):

〈x, y, z | x3 = y5 = z2 = 1, z = (xy)2, [x, z] = [y, z] = 1〉.

Lemma 4.21. SL(2, 5)/〈z〉 ∼= A5, the alternating group on 5 elements.

Proof. The group A5 has a presentation 〈a, b | a2 = b3 = (ab)5 = 1〉. Let a = xyand b = x in SL(2, 5)/〈z〉. Then o(a) = 2, o(b) = 3 and ab = xyx = (xy)2y−1 =y−1 is of order 5. As the order of the two groups are the same, we have thatSL(2, 5)/〈z〉 ∼= A5.

Corollary 4.22. SL(2, 5) is non-solvable.

Proof. We know thatA5 is a simple group. As it is non-abelian, A5 is non-solvable.Moreover, solvability is a characteristic that is inherited on quotients, so SL(2, 5)can not be solvable.

We introduce a new group: the orthogonal group O(n, F ) = {A ∈ GL(n, F ) |AAt = AtA = In} and its associated special orthogonal group SO(n, F ) =SL(n, F ) ∩ O(n, F ). A matrix A of O(n, F ) is also called an orthogonal matrix.The group SO(3,R) can be seen as the group of rotations of the 3-dimensional realspace around a line.

We need a new kind of algebra, called a real quaternion algebra

H(R) := R⊕ Ri⊕ Rj ⊕ Rij,

4.4. FITTING SUBGROUP 29

with multiplication rules i2 = −1, j2 = −1 and ij = −ji. This algebra is a divisionring (see also Theorem 5.12). Compare this to the definition given in Section 5.3.

In the following proposition we use the following norm on H(R):

|.| : H(R)→ R : a+ bi+ cj + dij 7→ a2 + b2 + c2 + d2.

Proposition 4.23. The subgroup of H(R)× consisting of the quaternions of norm1 maps homomorphically onto SO(3,R) with kernel {±1}.

Proof. See Lam [10, Theorem 3.1].

Proposition 4.24. SL(2, 5) is embedded in H(R).

Proof. The group A5 can be seen as the orientation preserving symmetries of aregular dodecahedron (or by duality the regular icosahedron), soA5 is a subgroup ofSO(3,R). As SL(2, 5)/〈z〉 ∼= A5, and 〈z〉 ∼= C2, it follows from Proposition 4.23that SL(2, 5) is a subgroup of H(R).

An explicit embedding in H(R) is given by

y = ζ5 ∈ C ⊆ H(R) and x =(y2 − y) + (y2 − 1)j√

5.

These x and y can then be interpreted as the generators for the presentation ofSL(2, 5) given earlier.

It appears SL(2, 5) is the only non-solvable group that is embedded in a divisionalgebra (as will be shown in Theorem 6.5), and so it occupies a special place in thisthesis.

4.4 Fitting subgroupIn this section we define the Fitting subgroup of a group, but we will first recallsome results concerning nilpotent groups.

Definition 4.25. A group G is said to be nilpotent if there exist normal subgroupsG1, . . . , Gn of G, such that

{1} = G0 ⊆ G1 ⊆ . . . ⊆ Gn = G,

and Gi+1/Gi ⊆ Z(G/Gi) for each 0 ≤ i < n.

The following theorem is well known (see for example Robinson [14, Theorem5.2.4]).

Theorem 4.26. If G is a finite group, then the following properties are equivalent:

1. G is nilpotent.

2. For every proper subgroup H < G, H 6= NG(H).


3. Every maximal subgroup of G is normal.

4. Every Sylow subgroup of G is normal.

5. G is the direct product of its Sylow subgroups.

Definition 4.27. The Fitting subgroup of a finite group G is the maximal normalnilpotent subgroup of G and is denoted by Fit(G).

One can prove that there is indeed a unique maximal normal nilpotent subgroupfor a finite group G, using the fact that if M and N are normal and nilpotent sub-groups of G, then MN is again normal and nilpotent. An elementary p-group is agroup in which every non-trivial element as order p.

Now we prove that for solvable groups, the Fitting subgroup somehow lies rigidin the whole group. More precisely, CG(Fit(G)) ≤ Fit(G). First we need thefollowing lemma.

Lemma 4.28. If G is a solvable group and {1} 6= N is a minimal normal subgroupof G, then N is an elementary p-group.

Proof. N ′ is a characteristic subgroup of N , so N ′ is normal in G. Moreover, asN is solvable, N ′ 6= N (see Remark 2.7) and by the minimality of N this impliesN ′ = {1}. So N is abelian.

Consider the subgroup P = {a ∈ N | ap = 1} of N , for some p | |N |. ThenP 6= {1} and it is a characteristic subgroup of N . Therefore, P is normal in G, soby the minimality of N we obtain P = N .

Proposition 4.29. If G is a solvable group, then

CG(Fit(G)) ≤ Fit(G).

Proof. Suppose that CG(Fit(G)) � Fit(G) and write A = Z(Fit(G)). Then A �CG(Fit(G)). Let H be a minimal normal subgroup of G strictly containing A andsuch that H ≤ CG(Fit(G)). Such a group exists, for CG(Fit(G)) itself satisfies theconditions except maybe the minimality. Then H/A is a minimal normal subgroupof G/A, a solvable group. By Lemma 4.28, H/A is abelian, and so H ′ ≤ A.

As A is abelian, H(2) = {1} which implies that H is nilpotent. This means thatH ≤ Fit(G), but also H ≤ CG(Fit(G)). We conclude that H ≤ A, a contradictionwith the fact that H strictly contains A.

4.5 Solvable groupsIn this section, if we talk about a group G, we will always assume it is embeddedin a division ring D of characteristic 0. We will use some results that are obtainedin Section 5.1 and Section 5.2, so it should be noted that both those sections do notrely on results obtained in this section. As the path of non-solvable subgroups ofa division ring will come to a dead end with SL(2, 5) (see Section 6.2), we will

4.5. SOLVABLE GROUPS 31

focus on studying the solvable case. A first small result deals with cyclic groups.To make notations easier, we say that a prime power pα divides an integer n exactlyif pα divides n and pα+1 does not divide n.

Lemma 4.30. Let n, r and s be positive integers.

1. If n, r and s are mutually coprime, then

ors(n) = lcm{or(n), os(n)}.

2. Let q be a prime dividing n− 1 and let i be a positive integer.

(a) If q = 2 and n ≡ −1 mod 4, write n2 − 1 = 2dt with (2, t) = 1, theno = o2(n) = 1 and no − 1 is exactly divisible by 2,o = o22(n) = . . . = o2d(n) = 2 and no − 1 is exactly divisible by 2d,o = o2i(n) = 2i−d+1 if i > d, and no − 1 is exactly divisible by 2i.

(b) In all cases not covered by (a), write n − 1 = qdt with (q, t) = 1, theno = oq(n) = . . . = oqd(n) = 1 and no − 1 is exactly divisible by qd,o = oqi(n) = qi−d if i > d, and no − 1 is exactly divisible by qi.

3. If q is a prime and m, i are positive integers with (q,m) = 1, then

oqi(m) = oq(m) · qa,

where a ∈ N and depends on i.

Proof. 1. This can be easily seen by considering at the multiplicative order of nin Z/rZ× Z/sZ.

2. First of all, in (a), it is clear that o2(n) = 1 if n ≡ −1 mod 4 and that fori ≥ 2 we have o2i(n) 6= 1. But as n2 − 1 is exactly divisible by 2d, we geto2i(n) = 2 for i = 2, . . . , d. In the case of (b), it is also clear that oqi(n) = 1for i = 1, . . . d if n−1 is exactly divisible by qd. In all of these cases, the factthat no − 1 is divisible by the mentioned prime power is also trivial to check.It remains to prove that no higher power can divide it.

Remark the following: suppose we have for some k and q that nk − 1 isexactly divisible by qs, where either q > 2 or s ≥ 2. So nk = 1 + qsm, withq - m. Then

nkq = (1 + qsm)q ≡ 1 + qs+1m 6≡ 1 mod qs+2.

This shows that nqk − 1 is exactly divisible by qs+1.

Consider the case where k = oqs(n). Then k divides oqs+1(n), which in itselfdivides kq. We get that either oqs+1(n) = k or oqs+1(n) = kq. We assumedthat qs+1 - nk − 1 whence oqs+1(n) 6= k, implying that oqs+1(n) = kq. Thisproves part (b) and partially (a). The only thing that remains to check is thecase where n − 1 is exactly divisible by 2, say n = 4m − 1. But then theargument above can be applied on n2 − 1 = 8(2m2 −m), proving the result.


3. If q | (m − 1), then this is just part 2. Suppose q - (m − 1), in particularq 6= 2. Write k = oq(m). If qd is the highest power of q dividing mk − 1,then by the previous parts we have k = oq(m) = oq2(m) = . . . = oqd(m) andoqi(m) = qi−dk if i > d. In both cases, the power of q that is multiplied withoq(m) = k is indeed only dependent of i.

As said in the remark just after our main theorem (Theorem 2.8), the differencebetween the Z-groups and the groups listed in b) lies in the existence of a quaternionsubgroup. An important result in relation to this remark is Theorem 4.32. To provethis theorem, we first prove following proposition concercing the centralizer of aquaternion subgroup in a group G.

Proposition 4.31. Suppose G is a solvable group that is embeddable in a rationaldivision ring D with a quaternion subgroup Q of order 2t.

(i) If t = 3, thenCG(Q) ∼= C2 ×M,

where M is a Z-group and both the order m of M and om(2) are odd.

(ii) If t > 3, thenCG(Q) ∼= C2.

Proof. First, let t = 3. It is easy to see that, ifQ is a subgroup of a quaternion groupQ1 ≤ G, then CQ1(Q) = {±1} = Z(Q1). As the Sylow 2-subgroups of G are allquaternion (G is a Frobenius complement and contains a quaternion subgroup, seeTheorem 4.11), we get that 〈−1〉 is the Sylow 2-subgroup of CG(Q). Indeed, LetP be a Sylow 2-subgroup of CG(Q). Then the group generated by P and Q is a2-group, so it is contained in a quaternion group Q1. Then

P ≤ CQ1(Q) = {±1},

proving that P = {±1}. CG(Q) is of the form

CG(Q) ∼= 〈−1〉 ×M,

for M a subgroup of odd order.By Theorem 5.13, Q[Q] ∼= H. Let g ∈ M be arbitrary of odd prime order p.

Using Theorem 5.3, we get that

Q[Q, g] ∼= H⊗Q CQ[Q,g](Q[Q]) = H⊗Q Q[g].

This is a quaternion sub-algebra of D, so it has to be a division algebra. Moreover,since g is a p-th root of unity,Q[g] has no real embedding. In view of Theorem 5.12,we need the residue degree of 2 in Q[g]/Q ∼= Q(ζp)/Q to be odd. By Proposition3.13, this residue degree is exactly op(2). Doing this for every p | m = |M |, andusing the first part of Lemma 4.30, we obtain that om(2) is odd.


Now, we assume t > 3. It is enough to prove this for t = 4, since CG(Q2t) ⊆CG(Q2t−1) and clearly 〈−1〉 ⊆ CG(Q2t) for every t. Also, by the first part and thisreasoning, it suffices to prove that CG(Q16) is a 2-group.

Write Q = Q16. Similarly to the above, take g ∈ CG(Q) of odd prime order p.In this case, Q[Q] ∼= H(Q(

√2)) (see Theorem 5.13), so we obtain again a division

ringQ[Q, g] ∼= Q[Q]⊗Q(√2) Q(

√2)[g] ∼= H⊗Q Q(

√2)[g]

in D. Remark that Q[g] ⊆ Q(√

2)[g] has no real embedding. Moreover, 2 is asquare in Q(

√2)[g] so the ramification index of 2 in Q(

√2)[g]/Q is even. These

facts yield a contradiction with Theorem 5.12. We conclude that CG(Q) is a 2-group.

By O2(G) we denote the maximal normal 2-subgroup of G. We can now provethe following theorem.

Theorem 4.32. Let G be solvable group that is embedded in a rational divisionring D and suppose Q = O2(G) is quaternion of order 8. Then G is isomorphic toone of the following groups:

1. Q×M , with M a Z-group of order m and om(2) is odd.

2. SL(2, 3)×M , with M a Z-group of order m, (m, 6) = 1 and om(2) is odd.

3. The binary octahedral group O∗ of order 48, 〈r, s | r2 = s3 = (rs)4, r4 = 1〉.

Proof. Assume the finite group G is solvable and Q = O2(G) ∼= Q8. Consider thenormal subgroup K := QCG(Q) of this group G. As Q is normal in G, we have agroup homomorphism

G→ Aut(Q) : g 7→ (·g : G→ G : x 7→ xg).

Clearly, the kernel of this morphism isCG(Q), soG/CG(Q) embeds into Aut(Q) ∼=S4 (see Lemma 4.17). Also |Q ∩ CG(Q)| = |Z(Q)| = |〈−1〉| = 2, so |K| =|Q||CG(Q)||Q∩CG(Q)| = |Q||CG(Q)|

2.

Now, |G/K| = 2|G||Q||CG(Q)| = |G/CG(Q)|

4, so G/K can only be of order 1, 2, 3 or

6 since G/CG(Q) embeds into a group of order 4! = 23 · 3. We will look at thesepossibilities and deduce that they correspond to the cases given in the theorem. Firstremark that by Proposition 4.31

CG(Q) ∼= 〈−1〉 ×M, (4.1)

with M a Z-group of order m such that both m and om(2) are odd. This impliesthatK ∼= Q×M because the 〈−1〉 subgroup of CG(Q) comes from the intersectionbetween Q and CG(Q).

If G/K is trivial (so of order 1), G ∼= K ∼= Q ×M , and we are in case 1 ofthe theorem. If |G/K| = 3, then since o3(2) = 2 (indeed, 22 = 3 + 1), we have


that 3 - m. Indeed, assume m = 3k. Using Lemma 4.30 parts 1 and 3, we obtainthat o3(2) | om(2), but this is not the case since om(2) is odd. Hence, the Sylow 3-subgroup ofG has order 3 and does not centralizeQ (by the construction ofK). LetC ∼= C3 be such a Sylow subgroup and denote L = 〈Q,C〉 = Q o C ∼= SL(2, 3)(see Lemma 4.18). Clearly G = LM . If we can now prove that Q[Q] = Q[L],then M centralizes Q[L], in particular M would centralize L. We would then haveG ∼= SL(2, 3)×M , and G is of the second type.

We now go on to prove Q[Q] = Q[L]. If we look at Remark 4.19, we can finda generator c of C and generators x and y of Q such that the action of c cyclicallypermutes x, y and xy. On the other hand, if we define

t = −1 + x+ y + xy

2∈ Q[Q],

then conjugation by t has the same effect on x, y and xy. The element ct−1 thencentralizes Q[Q]. Using this, we calculate

tct−1 = ct−1t = c,

showing that c and t commute. In this case Q[c, t] is a field. Moreover, in thisfield, c and t are both cube roots of unity. There are only at most 3 cube roots ofunity in any field and 〈t〉 consists of 3 cube roots. We conclude that c ∈ 〈t〉 andQ[L] = Q[Q].

We consider the last 2 cases (|G/K| being 2 and 6) together. In both these cases,a Sylow 2-subgroup of G has to be quaternion (as we already have that O2(G) isquaternion, a Sylow 2-subgroup can not be cyclic so it has to be quaternion byTheorem 4.11). The order of G is exactly divisible by 24 since |K| is exactly di-visible by 23 and the index of K in G is 2 or 6. This implies further that a Sylow2-subgroup of G is quaternion of order 16. Let Q1 be such a group. If we writeΓ = CQ[G](Q) = CQ[G](Q[Q]) and identify H and Q[Q] (see Theorem 5.13), thenby Theorem 5.3

Q[G] = H⊗Q Γ.

Remark that also M ⊆ Γ by equation 4.1. Denote Z = Z(Γ), then

Q[G] ∼= H⊗Q Γ ∼= H⊗Q Z ⊗Z Γ ∼= H(Z)⊗Z Γ.

This is still a division ring, so using Proposition 5.8, m(H(Z)) and m(Γ) have tobe coprime. Since m(H(Z)) is even, m(Γ) should be odd.

Using again Theorem 5.13, there exists a central element g that is a square rootof 2 in Q[Q1] and such that Q[Q1] = H⊗QQ[g]. As Q ⊂ Q1, g has to centralize Qtoo, so g ∈ Γ. Now look at the inclusions

Z ⊆ Z[g] ⊆ E ⊆ Γ,

where E is the maximal subfield of Γ containing the element g. On one hand,as g2 = 2, Z[g]/Z is either a quadratic or trivial extension. On the other hand,


[Z[g] : Z] divides [E : Z] = m(Γ) (see Proposition 5.9), which is odd. This showsthat [Z[g] : Z] = 1, and g ∈ Z. As M ⊆ Γ, it now commutes with both g and Q.This shows that it also centralizes Q[Q1], implying that M is the trivial group (seeProposition 4.31).

If we look back at what this means, thenK = Q andG/K can not have order 2.Indeed, if it would, then G is a 2-group of order 16 and O2(G) would be the wholegroup, a contradiction since O2(G) ∼= Q8. Thus G/K has order 6. This is onlypossible if G/CG(Q) ∼= S4. We have an extension

1→ C2 → G→ S4 → 1,

such that the Sylow 2-subgroup of G is quaternion. G is a group of order 48, andwe will prove it is isomorphic to the binary octahedral group.

For this argument, take C = 〈c〉 a Sylow 3-subgroup of G and P a Sylow2-subgroup of NG(C). We may now take Q1 lying over P . Moreover, CQ is anormal subgroup because it is of index 2 in our groupG. Using Frattini’s Argument(Lemma 4.1) on this group and taking C to be the Sylow 3-subgroup, we obtain

G = NG(C)CQ = PCCQ = PCQ = CPQ.

We have used the fact that NG(C) = PC. This is clear as P is the Sylow 2-subgroup of NG(C) and |NG(C)| divides |G| = 24 · 3. As obviously C ≤ NG(C),we get that PC = NG(C). Since Q is a 2-subgroup of G, it lies in a Sylow 2-subgroup of G. Moreover, all the Sylow 2-subgroups of G are conjugate and Qis normal, so we may assume Q being a subgroup of Q1. This further impliesPQ ≤ Q1. Considering the order of Q1 and the fact that G = CPQ, we find that

PQ = Q1.

This clearly implies that |P ∩ Q| = 2 and that P is of order 4. There are only 2groups of order 4: the Klein group or C4. As Q1

∼= Q16 does not contain a groupisomorphic to the Klein group and P ⊂ Q1, we obtain that P is cyclic of order 4.

Now let x ∈ Q1 be of order 8 and y a generator for P . Since the index of Q inQ1 is 2, Q1/Q ∼= C2. This means that x2 ∈ Q. Look at the inclusions

〈x〉 ≤ 〈x, y〉 ≤ Q1.

The group 〈x〉 has order 8 and Q1 has order 16, so 〈x, y〉 has to be equal to one ofboth. For contradiction’s sake, assume that 〈x, y〉 = 〈x〉. Then y ∈ 〈x〉 and sincethe order of y is 4

y = x2 or y = x6.

In the latter case, y3 = x18 = x2 and thus since P = 〈y〉 = 〈y3〉 we obtain

P = 〈x2〉,

in both cases. This is a contradiction because |P ∩ Q| = 2, but a generator of P ,namely x2, lies in both groups. We find that Q1 = 〈x, y〉. A similar reasoning can


be applied to see that 〈x2, x−1y〉 = Q. Indeed, as y ∈ Q1 \ Q = xQ, the elementx−1y is an element of Q. We have an inclusion

〈x2〉 ≤ 〈x2, x−1y〉 ≤ Q.

In this case x−1y /∈ 〈x2〉, because otherwise y ∈ 〈x〉 and we get the same contra-diction as before.

By Remark 4.19, either c or c−1 conjugates x2 to x−1y so

(x2)c = x−1y or y−1x.

Then again, y−1x = x−1y−1, so replacing y by y−1 (remark that this changes noth-ing in the above) we may assume that

(x2)c = x−1y.

One can calculate that the only elements of order 4 are x2, x−1y, xy and xy−1. Asconjugation by c does not change the order of an element, (x−1y)c has to be amongthis list. Because the order of c is 3, we can not have that (x−1y)c = x2 or x−1ysince we already have (x2)c = x−1y. Suppose that (x−1y)c = xy. Then

cx = x−1ycy−1 = x−1c2 = (cx)−1,

and so cx is of order 2, implying cx ∈ 〈x4〉. This would mean that c ∈ 〈x3〉 = 〈x〉,clearly a contradiction. We deduct that

(x−1y)c = xy−1.

As a conclusion, G is given by the presentation

〈x, y, c | x4 = y2, y4 = 1, xy = x−1, (x2)c = x−1y, (x−1y)c = xy−1, cy = c2, c3 = 1〉,

which can be seen as the following presentation

〈r, s | r2 = s3 = (rs)4, r4 = 1〉,

by setting r = x2y and s = −c. The group G is indeed the binary octahedralgroup.

Theorem 4.32 gives us a way of identifying groups that are of type b)i), iii) andiv) in the classification Theorem 2.8. We are still missing a means of identifyingtype b)ii). The rest of this chapter will be dedicated to this problem.

Lemma 4.33. Suppose that O is the ring of integers of a cyclotomic extensionof Q and that P is the only prime ideal of O lying over the prime 2. If G is amultiplicative cyclic subgroup of O of order coprime to 2, then

G ∩ (1 + P) = {1}.


Proof. Suppose we have 1 6= x ∈ G ∩ 1 + P. We may assume without loss ofgenerality that x has order p, for p a prime dividing |G|. Write x = 1 + α forα ∈ P.

As Z ⊆ O and P is the only prime lying over 2, we get

Z(2) ⊆ OP,

the localisation of Z (respectivelyO) to the ideal (2) (respectively P). The prime pis thus invertible in OP.

We calculate:

1 = xp = (1 + α)p = 1 + pα +

(p

2

)α2 + . . .+ pαp−1 + αp,

and this is true if and only if

p

(1 + p−1

(p

2

)α + . . .+ αp−2

)α + αp = 0,

or equivalently

p

(1 + p−1

(p

2

)α + . . .+ αp−2

)= −αp−1.

This forms a contradiction: the right hand side of this equation is not invertible inOP, but the left hand side is a product of 2 units. To see that y = 1 + p−1

(p2

)α +

. . . + αp−2 is a unit, remark that y − 1 ∈ OPP, the only ideal of OP. Now y cannot be in OPP as well since this would imply 1 ∈ OPP. As every element outsideof OPP is invertible, y is a unit.

Theorem 4.34. If G = Cm o Q is the split extension of a cyclic group Cm of oddorder by a quaternion group of order 2t and it is embeddable in a rational divisionalgebra D, then one of the following properties holds:

(i) G = Cm ×Q8.

(ii) an element of order 2t−1 of Q centralizes Cm and an element of order 4 of Qinverts Cm.

Proof. Let P be a quaternion subgroup of order 8 in D×. Then, as we have donebefore, using Theorem 5.3 and Theorem 5.13 we find

D = H⊗Q CD(Q[P ]) = H⊗Q F ⊗F Γ = H(F )⊗F Γ,

where Γ = CD(P ) and F is the center of D. Because of Proposition 5.8 and thefact that m(H(F )) = 2, we obtain that m(Γ) is odd. As a consequence, m(D) =m(H(F ))m(Γ) is not divisible by 4. By Proposition 5.9, neither is [E : F ∩ E] =[EF : F ], for E a subfield of D such that E/(E ∩ F ) is a Galois extension. To


see this, extend E to a maximal subfield K. Then m(A) = [K : F ] according toProposition 5.9 and m(A) = [K : E][E : F ] is not divisible by 4. Thus

4 - |Gal(E/(E ∩ F ))|. (4.2)

The actual proof will be split into two cases, namely the cases t > 3 and t = 3.The easy part is when t > 3. Remember that in this caseQ = Q2t is of the followingform

Q2t = 〈x, y | x2t−2

= y2, y4 = 1, xy = x−1〉.

Define E := Q[Cm] ⊆ D. Clearly E is a field. Moreover, x acts on E by conjuga-tion and it induces an automorphism of order a 2-power. Because this action is triv-ial on E ∩F (F is the center of D), we find this automorphism in Gal(E/(E ∩F )).Because 4 is not a divisor of the order of this group (see (4.2)), the automorphismis of order 1 or 2. Suppose for contradiction’s sake that x does not centralize Cm.Then the order of the action is 2 and x2 centralizes E. Now x and y induce 2 dif-ferent automorphisms on E(x2) (independent of how y acts on E and due to thefact that t > 3). These automorphisms are trivial on E(x2) ∩ F . We have found 2different elements of order 2 in Gal(E(x2)/(E(x2) ∩ F )), a contradiction with thefact that 4 does not divide the order of this group. We conclude that x centralizesCm.

Now we need to prove that y inverts Cm. Suppose there is a Sylow p-subgroupof Cm that is centralized by y. Then this Sylow subgroup, which is moreover cyclic,would be a subgroup of CG(Q). According to Proposition 4.31, this is a contra-diction since this Sylow subgroup is of odd order. The action of y on the Sylowsubgroup can not be faithful (of order 4) because of Theorem 4.12, so it is of order2. Using Corollary 4.7, the action of y is by inversion on the Sylow subgroup. Inconclusion, y acts by inversion on all of Cm since p was chosen arbitrary.

Now consider the case t = 3. If neither x nor y centralizes Cm, by the samereasoning as above, x and y induce the same automorphism of order 2 on Q[Cm].Then xy centralizes Q[Cm]. We may assume by a change in notation that x central-izes Q[Cm] (since mapping x to xy and y to y is an automorphism of Q8, see theproof of Lemma 4.17). If y also centralizes Cm, then obviously

G ∼= Cm ×Q8.

If y does not centralize Cm, then it has to invert at least one Sylow subgroup ofCm. This is due to a similar reasoning as we did at the end of the case t > 3. Wewill call this Sylow subgroup P = 〈p〉 and we have to prove that y inverts everyother Sylow subgroup too. Write

pa = |P |, D1 = Q[PQ] and D1 ⊇ E1 = Q[p, x] ∼= Q[ζpa , ζ4] = Q[ζ4pa ].

Clearly E1 is a maximal subfield of D1. Also write

Z = Z(D) (with D = Q[G]), Z1 = Z(D1) and E = Q[Cm],


as before. As E1 is a maximal subfield, Z1 is going to be a subfield of E1. It is clearthat Z1 6= E1 (since x does not commute with y for example) and one can easilycheck that

Z ′1 := Q[px+ (px)−1] ⊆ Z1 ⊆ E1.

As [E1 : Z ′1] = 2, we find that Z1 = Z ′1. We will prove that Z is a real field usingour knowledge of Z1. This will indeed suffice for the proof: if there is a Sylowsubgroup of Cm that y centralizes, then the generator of this Sylow subgroup wouldcorrespond to a primitive odd order root of unity in Z. This yields a contradictionwith the fact that Z is a real field.

We claim that the prime 2 is unramified in the extension E1/Z1. Indeed, if 2would have a ramification index different from 1, then e(2, E1/Z1) = 2 becausee(2, E1/Z1) | [E1 : Z1] = 2. This means that for O = OE1

∼= Z[ζ4pa ] we have

2O = P2,

for some prime P. The element y acts on O since it fixes Z and inverts ζ4pa . Thenthe ideal 2O is fixed under the action of y since it fixes a generator. This yields thatPy is again a prime ideal in the decomposition of 2O, showing that y normalizes P.Then we can let y act onO/P. Using Lemma 4.33, P ∩ (1 +P) = {1}. But, y actsnon-trivially on P with at least 3 elements, so it acts non-trivially on O/P. This isof course only possible if f(2, E1/Z1) > 1. A contradiction is found within the factsthat 2 < f(2, E1/Z1)e(2, E1/Z1) and f(2, E1/Z1)e(2, E1/Z1) | [E1 : Z1] = 2,eventually showing that e(2, E1/Z1) = 1.

By Proposition 3.13, e(2, E1/Q) = ϕ(4) = 2 (remember that E1∼= Q[ζ4pa ]), so

using the multiplicativity of the ramification index

2 = e(2, E1/Q) = e(2, E1/Z1)e(2, Z1/Q) = e(2, Z1/Q).

Since y /∈ Z1, one can see that Z1 ⊆ E(x). We have the following inclusions

Z1 ⊆ CE(x)(y) ⊆ Z.

The last inclusion follows from the fact that D = E(x)(y) and clearly E(x) iscommutative. Now the number

e(2, Z/Q) = e(2, Z/Z1)e(2, Z1/Q) = 2e(2, Z/Z1),

is even. Also, D contains the subalgebra Q[G] ⊗ Z ∼= H(Z), which thereforebecomes a division algebra (quaternion algebras are division or split, and it can notbe split because it is a subring of a division algebra). Using Theorem 5.12 and theabove, we conclude that Z has to be a real field, proving the theorem.

We will use the Schur-Zassenhaus Theorem, a proof of which can be found inJacobson [7, Theorem 6.16].

Theorem 4.35 (Schur-Zassenhaus Theorem). IfG is a finite group withH a normalHall subgroup, then G is a semi-direct product of H and another subgroup K.


To prove Corollary 4.39 elegantly, we introduce now the notion of supersolv-ability.

Definition 4.36. A group G is called supersolvable if there exist normal subgroupsG1, . . . , Gn ofG such that {1} = G0 ⊆ G1 ⊆ . . . ⊆ Gn = G andGi+1/Gi is cyclicfor 0 ≤ i < n.

A result on supersolvable groups we will use is the following theorem.

Theorem 4.37. The elements of odd order in a supersolvable group form a charac-teristic subgroup.

Proof. See Robinson [14, Theorem 5.4.9].

Proposition 4.38. If G is a finite solvable group, embeddable in a division ring Dof characteristic 0, with O2(G) cyclic, then G is supersolvable.

Proof. We will prove that in this case Fit(G) is cyclic andG/Fit(G) is abelian. Thiswould suffice because we can expand the series 1 E Fit(G) E G to a series

1 E Fit(G) E H1 E . . . E Hn E G,

for H1, . . . , Hn normal subgroups of G and each factor cyclic using the Classifica-tion of Finite Abelian Groups. Indeed, as G/Fit(G) is abelian, let K1 be a cyclicsubgroup of G/Fit(G). This K1 is automatically normal and thus it is of the formH1/Fit(G) for a normal subgroup H1 of G containing Fit(G) (follows from theisomorphism theorems for groups). The other groups Hi can now be found byinduction as G/H1 is also abelian.

The group F := Fit(G), by its definition is finite nilpotent. Using Theorem4.26, F is the direct product of its Sylow subgroups:

F =∏p||F |

Sp.

These Sylow subgroups Sp are subgroups of the Sylow subgroups of G. For oddprimes, these are cyclic (see Theorem 4.11). The group S2 on the other hand isnormal in G since for every g ∈ G:

Sg2 ⊂ F g ⊆ F and Sg2 is a 2-group,

proving that Sg2 ⊆ S2 by the factorisation of F . Then S2 ⊆ O2(G), so S2 is cyclic.The group F is thus a direct product of cyclic groups of relative prime orders,proving that F is itself cyclic.

By Proposition 4.29, CG(F ) ⊆ F , implying that G/F is embedded in theabelian group Aut(F ). We conclude that G/F is also abelian, proving the proposi-tion.

Corollary 4.39. If G is a finite solvable group, embeddable in a division ring D ofcharacteristic 0, with O2(G) cyclic, then G is a split extension of a 2′-group by a2-group.

Proof. This is direct consequence of Proposition 4.38, Theorem 4.37 and the Schur-Zassenhaus Theorem 4.35.

5

Finite-dimensional algebras

“Don’t they teach recreational math anymore?’– The Doctor, Doctor Who

This chapter is dominated by a certain type of algebra, namely the central simplealgebras.

Definition 5.1. Let F be a field. An F -algebra A is called central if it is a finite-dimensional F -algebra with center F . A central F -algebra A is called simple if ithas no non-trivial two-sided ideals (A is simple as a ring). The algebra A is calleda central division algebra if it is central simple algebra and a division algebra.

Using the well-known Wedderburn Theorem, we know that such a central sim-ple F -algebra A is always either a central division algebra or an n-dimensionalmatrix ring A ∼= Mn(D) over D, a central division algebra. One denotes the di-mension of A (as a vector space) over F by dimF (A) or [A : F ]. We also remindthe reader of the following lemma.

Lemma 5.2. If A is a finite dimensional algebra over a field F , then A is a divisionring if and only if A has no zero-divisors.

If we have a central simple F -algebra A, then we can always create a newcentral simple F -algebra Aop with it. As a vector space, Aop is the same as A butthe multiplicative operation is defined as

a · b = ba.

It is clear this algebra is again a central simple F -algebra. We define the envelopingalgebra of a central simple F -algebra A to be

Ae := A⊗F Aop.

Obviously [Ae : F ] = [A : F ]2. Consider the morphism

ψ : Ae → EndF (A) :∑

ai ⊗ a′i 7→ (a 7→∑

aiaa′i),

41

42 5. FINITE-DIMENSIONAL ALGEBRAS

where ai, a′i and a are elements of the vector space A. If we let {e1, . . . , en} bean F -basis of A, then every element of Ae can be uniquely written as

∑ei ⊗ ai

for some ai ∈ Aop. We will prove that ψ is an injection. For this, take elementsα =

∑ei ⊗ ai and β =

∑ei ⊗ bi and suppose that ψ(α) = ψ(β). For every

1 ≤ j ≤ n we obtain∑

i eiejai =∑

i eiejbi. This implies that for every i and jwe have an equality ejai = ejbi. Since the ej’s form a basis, we may conclude thatai = bi for every i, proving injectivity. Moreover, since the dimension of EndF (A)and Ae over F are the same, this map is an isomorphism. This shows that Ae is acentral simple F -algebra since it is well known that EndF (A) ∼= Mn(A) is a centralsimple F -algebra.

5.1 Tensor product of central simple algebrasIn our proofs we will often use the tensor product of F -algebras, so in this sectionwe will give some results considering such constructions. A first one deals withfactorising an algebra as a tensor product of some subalgebra.

Theorem 5.3. Let B be a finite-dimensional central simple F -algebra contained inan F -algebra A. Then A = B ⊗F CA(B), where CA(B) is the centralizer of Bin A. The map I 7→ B ⊗ I is a bijection between the set of ideals of CA(B) andA. Moreover the center of A coincides with the center of CA(B), considered as asubalgebra of A by 1⊗ CA(B).

Proof. The proof given below is based on Jacobson [7, Theorem 4.7].Regard A as a Be-module by considering the following action:

(∑

bi ⊗ βi)a =∑

biaβi,

with bi, βi ∈ B for all i’s and a ∈ A. Remark that such an action is present on anyalgebra containing B, in particular on B itself. Moreover, for this action, B is anirreducible Be-module.

All irreducible Be-modules are isomorphic. In particular, they are isomorphicto B. In any irreducible Be algebra we can find an element c such that

(b⊗ 1)c = (1⊗ b)c

and(b⊗ 1)c = 0⇒ b = 0,

since we can do this in B (for example, take c = 1). As Be is simple, A is adirect sum of irreducible Be-modules. Take such a ci, as described above, in everyirreducible component of A, then A =

⊕Bci and ci ∈ CA(B). Moreover, every

c ∈ CA(B) can be written in a unique way as c =∑bici and the fact that cb = bc

for every b ∈ B implies that bi ∈ Z(B) = F . This shows that CA(B) =∑Fci

and it is now clear that B ⊗F CA(B) ∼= A.

5.1. TENSOR PRODUCT OF CENTRAL SIMPLE ALGEBRAS 43

To see that the map for the ideals given in the theorem is surjective, take J anideal of A. Then it is a Be-submodule of A and so J =

∑Bdi for some di similar

to what we did before. Clearly di ∈ J ∩ CA(B) =: I . This I is an ideal of CA(B)and J = B ⊗ I . For the injectivity, we will show that CA(B) ∩ (B ⊗ I) = I . Takean F -basis {1 = b1, b2, . . . , bn} of B. An element of B ⊗ I can then be written ina unique way as

∑biki with ki ∈ I . An element of CA(B) = 1⊗ CA(B) is of the

form c1b1. The intersection then contains only elements of the form k1b1 = k1 ∈ I ,proving that CA(B) ∩ (B ⊗ I) = I .

The center of A is contained in CA(B) and so it needs to be contained in thecenter of CA(B). On the other hand, if an element is contained in the center ofCA(B) then it commutes with both CA(B) (it is an element of its center) and B (itis contained in CA(B)). We conclude that the element is contained in the center ofA.

Corollary 5.4. Let B be a finite-dimensional central simple F -algebra, and C anarbitrary F -algebra. Then the map I 7→ B ⊗F I is a bijection between the set ofideals of C and A = B ⊗F C. Moreover, the center of A coincides with the centerof C, considered as the subalgebra 1⊗F C.

Proof. We consider B and C as subalgebras of A by identifying them with B ⊗ 1and 1⊗ C respectively. We now claim that C is the centralizer of B in A, for thenwe can use the previous theorem. Denote by {cγ} an F -basis of C. Then everyelement a ∈ A can be written in a unique way as a =

∑γ

bγ ⊗ cγ for some bγ ∈ B.

This implies that, for an arbitrary b ∈ B, the equation ab = ba is equivalent to theequation bγb = bbγ for every bγ . So, claiming that an a ∈ A commutes with everyb ∈ B is equivalent to claiming that a =

∑γ

βγ ⊗ cγ with βγ ∈ Z(B) = F . Since

the tensor product is over F , this is moreover equivalent to claiming that a is in C.This proves C = CA(B).

Corollary 5.5. Let B be a finite-dimensional central simple F -algebra, and C anarbitrary F -algebra. Put A = B ⊗F C. Then

• A is simple if and only if C is simple,

• A is central if and only if C is central.

Proof. Immediate from the previous corollary.

In particular, the tensor product of two central simple F -algebras is again acentral simple F -algebra. On the set of central simple F -algebras we can alsodefine an equivalence relation: we call A and B, both central simple F -algebras,equivalent (denoted A ∼ B) if there exist strictly positive integers m and n suchthat

Mn(A) ∼= Mm(B).


On the set

Br(F ) = {A | A is a central simple F -algebra}/ ∼,

we define a multiplication as follows

[A][B] = [A⊗B].

This set becomes a group for the multiplication, the so called Brauer group of F .The unit in this group is [F ] and [A]−1 = [Aop].

5.2 Schur-index and exponentThe fact that a central simple F -algebraA is always isomorphic toMn(D) for somepositive integer n 6= 0 and a central division F -algebraD gives rise to the followingdefinition.

Definition 5.6. With notation as above, the Schur-index m(A) of A is defined to bethe square root of [D : F ]. The exponent e(A) of A is defined as the least integer ksuch that A⊗k = A⊗F A⊗F . . .⊗F A ∼= Mr(F ) for some r. In other words, e(A)is the order of A in Br(F ).

Remark that the Schur-index is always an integer. Indeed, if we denote Fby the algebraic closure of F , then D ⊗F F is a central simple F -algebra usingCorollary 5.5. Moreover, [D : F ] = [D ⊗F F : F ] and as F is algebraicallyclosed, D ⊗F F is isomorphic to Ml(F ) for some positive integer l. But then[D : F ] = [D ⊗F F : F ] = l2. From the definition it follows that the Schur-indexof an algebra A can be interpreted as a measure of how far the representations of Alie from being defined over F .

It can be proved that e(A) | m(A) where we have equality if F is an algebraicnumber field (see Theorem 5.7).

Let F be an algebraic number field. By a finite prime P of F we mean anon-zero prime ideal of the ring of integers OF of F , and an infinite prime P isan equivalence class of embeddings of F in C. In this context, two embeddingsof F in C are called equivalent if they are equal up to complex conjugation. If anembedding of F inC in such a prime maps F inR, then we call the prime an infinitereal prime, otherwise an infinite complex prime. For infinite primes, the valuationdetermined by the prime is the modulus of the image under an embedding in theprime. In all cases, we denote the completion of F with respect to (the valuationdetermined by) P by FP.

Theorem 5.7. Let F be an algebraic number field and A a finite-dimensional cen-tral simple F -algebra. Let P be a (finite or infinite) prime of F . Then

AP = A⊗F FP,

5.3. QUATERNION ALGEBRAS 45

is a central simple FP-algebra, m(AP) = e(AP) and

m(A) = lcm{m(AP) | P a prime of F},e(A) = lcm{e(AP) | P a prime of F}.

In particular, e(A) = m(A) and if m(A) is a prime power, then there exists aprime P such that m(A) = m(AP).

Proof. See Reiner [13, Theorem 32.17 and Theorem 32.19].

The Schur-index appears when we tensor 2 finite-dimensional central divisionalgebras. It determines when such a tensor product is again a division algebra.

Proposition 5.8. Let D1 and D2 be central division algebras over F , where F isan algebraic number field. Then D1 ⊗F D2 is a division algebra if and only if theintegers m(D1) and m(D2) are coprime.

Proof. First of all, by Corollary 5.5, we know that A := D1 ⊗F D2 is a centralsimple algebra over F . By the definition of the Schur-index, it is clear that A isa division algebra if and only if m(A) =

√[A : F ] =

√[D1 : F ] · [D2 : F ] =

m(D1)m(D2). But, as the Schur-index and the exponent coincide in these threecases (F is an algebraic number field), this is also equivalent to e(A) = e(D1)e(D2).This, being the orders in the commutative Brauer group of F , andA being the prod-uct of D1 and D2 in this group, happens only if and when e(D1)(= m(D1)) ande(D2)(= m(D2)) are coprime.

We will also use the following proposition.

Proposition 5.9. Let A be a finite dimensional central simple F -algebra and La subfield of A containing F . Then L is a maximal subfield in A if and only if[L : F ] = m(A).

Proof. See Reiner [13, Theorem 7.15].

5.3 Quaternion algebrasLater, when considering quaternion groups Q in D, we will often work with thealgebra generated by Q and Q. It will come as no surprise that this algebra will bea quaternion algebra over a field (see Theorem 5.13). We say that

H = Q⊕Qi⊕Qj ⊕Qij,

is the rational quaternion algebra with multiplication rules i2 = −1, j2 = −1 andij = −ji. It is easy to see that H is a division algebra; clearly Q is in the center ofH and [H : Q] = 4. As the dimension of H over its center is a square, this meansthat either H is commutative or Q is the center. Since H is not commutative, H has


to be central and has Schur-index 2. A quaternion algebra over a field F is definedas

H(F ) = H⊗Q F.Using Corollary 5.5, we see that H(F ) is a central simple F -algebra. Taking

into account its dimension over F , then it is either a division algebra or isomorphicto M2(F ). In this last case, we call the quaternion algebra split.

The question when this algebra is a division algebra for some finite Galois ex-tension F of Q is answered in Theorem 5.12.

Example 5.10. H(Qp) is a division ring if and only if p = 2.

Proof. See Lam [10, Theorem 6.2.5 and Theorem 6.2.24].

Theorem 5.11. Let K be a local field, D a central division K-algebra of Schur-index m and let E be a finite extension of K. Then D⊗K E is a division algebra ifand only m does not divide [E : K].

Proof. See Reiner [13, Corollary 31.10].

Theorem 5.12. Let F be a finite Galois extension of Q. Then H(F ) is a divisionalgebra if and only if one of the following properties holds

(i) F is a real field, that is, it is embedded in R,

(ii) the ramification index and the residue degree of 2 in F/Q are odd.

Proof. We will use Example 5.10. In general, if we would have a finite prime P ofF that does not lie over 2, then it has to lie over some other prime number p. Butthen H(Qp) ⊆ H(FP) and H(FP) will not be a division algebra.

Suppose that F is real. Then H(F ) ⊆ H(R) and so it has to be a divisionalgebra itself because it can not be split as a subring of a division algebra H(R).Now assume F is not real. It follows from the definition of H(F ) that [H(F ) :F ] = 4. We see that m(H(F )) ∈ {1, 2} and H(F ) is a division algebra if andonly if m(H(F )) = 2. Taking into account Theorem 5.7, this is again equivalentwith saying that m(H(FP)) = 2 for some prime P of F , or equivalent to sayingthat H(FP) is a division algebra. Clearly, this P can not be complex as we knowthat H(C) is not a division algebra, and by assumption it is not real. Looking atthe remark made above, this P has to lie over 2. Further, using Theorem 5.11, weget that H(FP) is a division algebra if and only if 2 = m(H(Q2)) does not divide[FP : Q2] = e(2, F/Q)f(2, F/Q), proving the theorem.

The following theorem is probably the most used result for the rest of the thesis.It gives structure to the subalgebra in D generated by a quaternion subgroup.

Theorem 5.13. Let Q2t be a quaternion subgroup of some division ring D. Thenthe subalgebraQ[Q2t ] ofD is isomorphic toH(F ), where F = Q(ζ2t−1 +ζ−12t−1), thefixed field of the cyclotomic field Q(ζ2t−1) under the automorphism ζ2t−1 7→ ζ−12t−1 .In particular, Q[Q8] ∼= H, Q[Q16] ∼= H(Q(

√2)). Moreover, for any subgroup Q

isomorphic Q8 of Q2t we have Q[Q2t ] = Q[Q]⊗Q F .

5.4. CROSSED PRODUCTS 47

Proof. Recall that a quaternion algebra of order 2t is of the form

Q2t = 〈x, y | x2t−2

= y2, y4 = 1, xy = x−1〉.

Set E := Q[x] a subfield of D1 := Q[Q2t ], we see that E is isomorphic to Q(ζ2t−1)and y ∈ D1 acts by conjugation on E. The relations in our quaternion group give usthat (D1 : E) ≤ 2, but as E is commutative and D1 is not, [D1 : E] = 2. Moreover,the center of D1, from now on called F , is the fixed field of E under conjugationby y, which is the automorphism described in the theorem.

One can easily verify that the elements i = x2t−3 and j = y in D1 satisfy

i2 = j2 = −1 and ij = −ji. So, i /∈ F , which implies D1 = F [i, j]. But then D1

is an epimorphic image of H(F ) with [D1 : F ] = 4, showing that H(F ) ∼= D1.When we do this for Q8, we get that E = Q(

√−1), with fixed field Q. Thus

Q[Q8] ∼= H(Q) = H. Looking at Q16, we let ζ be a primitive 8-th root of unity.So E ∼= Q(ζ), and F ′ := Q(ζ + ζ−1) ⊆ E is fixed under the automorphism with[E : F ′] = 2, showing that F ′ = F , the whole fixed field. Now

(ζ + ζ−1)2 = ζ2 + ζ−2 + 2 = ζ−2(ζ4 + 1) + 2 = ζ−2 · 0 + 2 = 2,

which shows that F = Q(ζ + ζ−1) ∼= Q(√

2).Considering Q ∼= Q8, a subgroup of Q2t , then Q[Q8] ∼= H, so we can write

(using Theorem 5.3) Q[Q2t ] ∼= H ⊗Q C, with C the centralizer of H in Q[Q2t ].Taking the multiplicity of the Schur-index into account, the fact that m(Q[Q2t ]) =2 = m(H) shows that C also has to be a field, so C = F .

Remark that the real quaternion algebra contains copies of all the quaterniongroups. Indeed, if we take ζ2t−1 ∈ C ⊆ H(R) then the subgroup 〈ζ2t−1 , j〉 ≤ H(R)×

is isomorphic to Q2t for t ≥ 3.

5.4 Crossed productsThis section is based on Shirvani and Wehrfritz [16, Section 1.4] and Jespers anddel Rıo [8].

For the proof of Theorem 2.12 we will need to construct new rings starting froma field and a group. We start with the more general notion of a crossed product andwill afterwards show how to use this for the above mentioned construction.

Let R be a ring with subring S and a group G ⊆ U(R) that normalizes S suchthat S[G] equals R. Consider N = S ∩G, a normal subgroup in G and write

R =⊕t∈T

tS,

with T some transversal of N in G (that is a minimal set T such that G =⋃t∈T

tN ,

where the union is disjoint). If we define H := G/N , then we say that (R, S,G,H)is a crossed product. We may also call R a crossed product of S by H . For an


element a =∑t∈T

tat, put Supp(a) = {t ∈ T | at 6= 0}. This set is called the support

of a.We can construct a crossed product if we have a ring S and a group G with a

normal subgroup N that is isomorphic to a subgroup of U(S). In this case, considerN itself as a subgroup of U(S). As an extra condition, we need an action α : G→Aut(S) : g 7→ αg such that, for all g ∈ G, n ∈ N and s ∈ S we have

αg(n) = g−1ng ∈ N and αn(s) = n−1sn.

Since the action of G on S extends the conjugation of G on N and the conjugationof N on S, we will often write αg(s) = sg for g ∈ G and s ∈ S. There exists up toisomorphism a unique crossed product (R, S,G,G/N) based on this information:let T be a transversal of N to G. We define a map

f : T × T → N such that t1t2 = t3f(t1, t2),

for some unique t3 ∈ T . This is well-defined because t1t2 ∈ G =⋃t∈T

tN where this

union is disjoint. We now say that the crossed product R is the free S-module withsymbols over T such that the multiplication on R is defined by linearly extendingthe following rule:

(t1a)(t2b) = t3f(t1, t2)at2b.

One can now easily check that R is a ring with subring S and G ⊆ R such thatS[G] = R. If we would take a different transversal, we get a ring that is isomorphicto the one just constructed, showing that the construction is unique up to isomor-phism.

The following lemma allows us to conclude that the crossed product is simpleif we know S is simple in some cases. It also gives a clear description of thecenter. If a group G acts on a set X , then the notation XG is used to denote the setXG = {x ∈ X | xg = x ∀g ∈ G}.

Lemma 5.14. Let (R, S,G,G/N) be a crossed product constructed as above, suchthat the action of N on S is trivial and the action αt of T on S is not inner for everynon-trivial t. If S is a simple ring, then R is simple too. Moreover, Z(R) = Z(S)T .

Proof. This proof is inspired by the proof of Lemma 2.6.1 in Jespers and del Rıo[8].

Suppose S is simple and let I be a non-zero ideal of R. Let a ∈ I be a non-zero element such that Supp(a) is of minimal cardinality. Write a =

∑t∈T

tat its

decomposition in R. By multiplying a with t−1 for some t ∈ Supp(a) if necessary,we may assume a1 6= 0. Because of the simplicity of S, 1 ∈ Sa1S, and thus for asimilar reason, we may assume that a1 = 1.

For any s ∈ S we get that |Supp(as− sa)| < |Supp(a)|, implying that as = sadue to the minimality condition on a. Following the rules for multiplication, thisalso means that for every t ∈ Supp(a) holds αt(s)at = ats. Due to fact that s

5.4. CROSSED PRODUCTS 49

was chosen arbitrary, we obtain Sat = atS. As S is simple and Sat = atS is anon-zero two-sided ideal, it follows that 1 ∈ Sat = atS and so at ∈ U(S). Fromthe equation αt(s)at = ats it follows that αt(s) = atsa

−1t , showing that αt is an

inner automorphism. By assumption, t = 1 and thus 1 = a ∈ I , showing that R issimple.

For the second statement, note that clearly Z(S)T ⊆ Z(R). To prove the otherinclusion, take a =

∑t∈T

tat ∈ Z(R) and t ∈ Supp(a). We immediately get from the

fact that a is in the center that αt(s)at = ats for each s in S, proving again that αtis inner. By assumption, t = 1 and thus a ∈ S. Because a commutes with all t ∈ T(considered as elements of R), it follows clearly that a ∈ Z(S)T .

If we are in the situation of Lemma 5.14, then the action of G/N on S is onlydependent on the action of T on S and they will often be used interchangeably.

Later we will often consider crossed products over fields, i.e. when S is a field.The following lemma will be essential.

Lemma 5.15. Let (R, S,G,H) be a crossed product with S a division ring and Hfinite. Then the following properties are equivalent:

1. R is a division ring,

2. For every Sylow subgroup P/(S ∩G) of H , S[P ] is a division ring.

Proof. Suppose that for every Sylow subgroup P/(S ∩ G) of H the ring S[P ] isdivision. Take a non-zero right ideal M of R. Note that M can be seen as a right S-module as well as a right S[P ]-module. Thus dimS(M) = dimS[P ](M)[S[P ] : S]divides dimS(R) = |H|. Furthermore, it is easy to see that [S[P ] : S] = |P/(S ∩G)|. As P/(S ∩ G) ranges over all the Sylow subgroups of H and dimS(M) isdivisible by all the |P/(S ∩ G)|, we obtain that dimS(M) = |H| = dimS(R),effectively showing thatM = R. A similar argument can be done for the left ideals,showing that R has no non-trivial left or right ideals. The ring R is a division ring.The converse is obvious (see Lemma 5.2) as S[P ] is a subring of R.


6

Proving the theorems of Amitsur

“Look at me: still talking when there’s science to do!”– GLaDOS, Still Alive

In this final chapter we will prove the two main theorems of this dissertation, theclassification of the Z-groups and the classification of the finite subgroups of ratio-nal division algebras. Let us give a small overview of what we did so far: supposeGis a finite subgroup of a rational division algebraD. By Proposition 4.10, this grouphas to be a Frobenius complement and Theorem 4.11 tells us that these groups havecyclic or quaternion Sylow-subgroups. From this result we distinguished the Z-groups (the case where all Sylow subgroups are cyclic) and the groups that have aquaternion Sylow 2-subgroup.

In the case of Z-groups, Theorem 4.12 yields that these are groups of the form

Cm o Cn,

with n and m coprime. From their form it follows that those groups are automati-cally solvable. For this situation, it remains to check what the extra conditions areon n and m to make sure that CmoCn is embeddable in a rational division algebra.This is the classification for Z-groups and will be discussed in Section 6.1.

In the case where a quaternion subgroup is present for G, we make a distinctionbetween the non-solvable and the solvable case. Section 6.2 deals with the non-solvable case; in Section 4.3 we have seen that SL(2, 5) is a non-solvable subgroupof a rational division algebra and it will be proved that, surprisingly, it is the onlysuch group. The form of a solvable group with a quaternion subgroup has alreadybeen studied in Section 4.5 and this study is finalised in Section 6.3. Eventually, theresults from Section 6.2 and Section 6.3 are combined to prove the classificationtheorem for finite subgroups of rational division algebras.

6.1 Classifying Z-groupsIn this section, we will discuss Theorem 2.12. This theorem classifies the Z-groupsand we will start off with some preliminary work. We study the Z-groups a bit more

51

52 6. PROVING THE THEOREMS OF AMITSUR

in detail. The cyclic groups form a special case and will thus be handled separately.According to Theorem 4.12, we should consider non-cyclic groups that are of theform

G = Cm o Cn,

withm and n coprime. Write n =∏qbii the prime decomposition of n. If the Sylow

qi-subgroup Cqbii

of Cn acts trivially on Cm, then it is clear that G is isomorphic toCmq

biio C∏

j 6=iqbjj

. We may thus assume that all the Sylow subgroups of Cn act non-

trivially on Cm.From now on, we decompose n = ab for Ca the kernel of the action of Cn on

Cm. If we define the number field E := Q(ζm, ζa), then E has an automorphism oforder b that fixes ζa and that simulates the action of Cb ≤ Cn on Cm. We can makethe crossed product (E[G], E,G,Cb) as discussed in Section 5.4, where G acts onE as if ζm is the generator of Cm ≤ G and ζa is the generator of Ca ≤ G. Theaction of Cb on E is non-trivial and the only automorphism that is inner on E is thetrivial automorphism because E is commutative. So according to Lemma 5.14, thiscrossed product is simple.

For the classification of Z-groups, we want to know when the group G = CmoCn is a subgroup of a division ring D. If this is the case thenQ[G] ⊆ D is a subringof E[G], hence E[G] ∼= Q[G] because E[G] is simple. Furthermore, as E is a field,G is finite andE[G] ⊆ D has no zero-divisors,E[G] is a division algebra by Lemma5.2. Clearly, ifE[G] is a division algebra, thenG is a subgroup of a division algebra.This way, we have reduced the classification problem to answering the equivalentquestion for which m,n the group G = Cm o Cn yields a division algebra E[G].

As before, let n =∏qbii be the prime decomposition of n. Write bi = αi + βi,

whereCqαii is the kernel of the action ofCqbii

onCm. It follows thatE∩G = CmoCawith a =

∏qαii and that the groups Pi ≤ G such that Pi/(E ∩ G) is a Sylow

subgroup of G/(E ∩G) ∼= Cb are

Pi = Cm o

(Cqbii×∏j 6=i

Cqαjj

).

Following Lemma 5.15, E[G] is a division algebra if and only if E[Pi] ⊆ E[G] is adivision algebra for each i. It is clear that this E[Pi] is the crossed product, in ournotation (E[Pi], E, Pi, Cqbii

), associated with the group Pi. We have further reducedthe problem to

Cm o Cn is a Z-group if and only if Cm o

(Cqbii×∏j 6=i

Cqαjj

)is a Z-group,

and this for all i’s.We will now especially consider a group of the form

Cpa o Cqb ,

6.1. CLASSIFYING Z-GROUPS 53

for p and q different prime numbers, since they will play an important role as build-ing blocks for the bigger groups. As before, write b = α + β, where Cqα is thekernel of the action of Cqb on Cpa . We have an automorphism of order qβ on Cpa ,meaning that qβ | |Aut(Cpa)| = ϕ(pa) by Lemma 4.6. As ϕ(pa) = pa−1(p− 1) and(q, p) = 1, we get that qβ | p− 1. To avoid trivial cases (the case where this groupbecomes cyclic), we will also assume 0 6= β. In particular, as qβ | p− 1, p is odd.

In the eventual proof of the classification of Z-groups, we will reduce the caseof our large group CmoCn to smaller groups of the form

(Cpa o Cqb

)×Cr, so for

an integer r coprime to the prime numbers p and q, consider henceforth the group

G =(Cpa o Cqb

)× Cr ∼= Cpa o

(Cqb × Cr

).

In this last group, the action of Cr on Cpa is trivial. We form the crossed product(E[G], E,G,Cqβ) as discussed before, where now E = Q(ζpa , ζrqα). In this case,we denote ζ = ζrqα and

δ =op(rq

α)

op(qα).

Lemma 6.1. With the notation used before, if we write Z := Z(E[G]), then Z ⊆ E(in particular Z is a number field) and

[E : Z] = qβ.

Proof. By Lemma 5.14, Z = Z(E)Cqβ = ECqβ ⊆ E. The dimension is clear as the

kernel of the action of Cqβ on E is trivial (the action is faithful).

Lemma 6.2. With the notation used before, the exponent of the crossed productE[G] is the least positive integer e such that

ζe ∈ NE/Z .

Proof. See Albert [1, Theorem 7.17].

We will explicitly distinguish the cases q = 2 and p ≡ −1 mod 4 and the rest.In the former case, we define the integer d to be p2− 1 = 2dt with (2, t) = 1. In theother case we write p − 1 = qdt with (q, t) = 1. This is similar to what we did inLemma 4.30.

Lemma 6.3. With the notation used before, the crossed product R = E[G] is adivision ring (or equivalently G is embeddable in a division ring) if and only if oneof the following cases holds:

(i) q = 2, α = β = 1 and r = 1.

(ii) q = 2, p ≡ −1 mod 4, α = 1 and 2 - δ.

(iii) α ≥ d and q - δ.


Proof. Write Z := Z(R), then by Lemma 6.1 Z is a number field. Using Theorem5.7 we see that e(R) = m(R). Due to the definition of the Schur index, R is adivision ring if and only if m(R) =

√[R : Z] =

√[R : E][E : Z]. It also follows

from Lemma 6.1 that [E : Z] = qβ . Moreover dimE R = qβ , so R is a division ringif and only if e(R) = qβ . According to Theorem 4.12, we may assume α 6= 0.

From Lemma 6.2 it follows that e(R) is the least positive integer e such thatζe is a norm of the extension E/Z. According to the Hasse Norm Theorem 3.38,ζe ∈ NE/Z if and only if ζe ∈ NEP/ZP for each prime P of Z and primes P of Elying over P . This P can be an infinite complex, infinite real or a finite prime. If itis finite, P could lie over the prime p or over some prime k 6= p. We consider allthese cases.

a) If P is an infinite complex prime, then P is too. In this case EP = ZP = Cand ζ is trivially a norm. This would imply β = 0, a contradiction.

b) If P is a real prime, then ZP = R and P is also an infinite prime. However itis not real: q | p−1 implies that pa > 2. So ζpa ∈ E, which shows that alreadyE is not real. Thus EP = C. Moreover, 2 = [EP : ZP ] | [E : Z] = qβ , soq = 2. The root of unity ζ = ζrqα is not equal to 1 since q = 2 and α 6= 0.On the other hand ζ ∈ Z ⊆ R. As the only two real roots of unity are ±1 weobtain ζ = −1. This further implies that α = 1 = r. Since NC/R = | · |2 isthe square of the modulus, e has to be even for ζe = (−1)e to be in the imageof the norm. The smallest such positive integer is of course e = 2, and this isthe case when β = 1.

Conversely, if q = 2, α = β = 1 and r = 1, then G ∼= Cpa o C4 whereC4 acts by inversion on Cpa . Such a group G is embeddable in H(R): take aprimitive pa-th root of unity in the complex part of H(R) and j ∈ H(R), thenthese elements form a group isomorphic to G, so this case does occur.

c) Let P (and thus also P) lie over an integer prime k 6= p. Consider the fol-lowing tower of extensions:

Qk(ζ)

ZP

EP

As E = Q(ζ)(ζpa), by Example 3.30, the extension EP/Qk(ζ) is unramified.In particular EP/ZP is also unramified by Proposition 3.29. We concludethat ζ , as it is a unit of OZP , is a norm of EP by Theorem 3.33. This impliesagain β = 0, a contradiction.


d) The last and most difficult case we come across is when P (and thus also P)lies over p. We now have the following diagram of extensions:

Qp

Qp(ζ) Qp(ζpa)

ZP

EP

The extension Qp(ζpa)/Qp is totally ramified of degree ϕ(pa) = (p− 1)pa−1

by Example 3.31 and EP/Qp(ζpa) is unramified by Example 3.30. The latteris of degree o := orqα(p). Moreover, as EP/ZP is a Galois extension, wehave that

[EP : ZP ] = e(P,E/Z)f(P,E/Z) = [E : Z] = qβ.

The middle equation is a consequence of r(P,E/Z) | r(p, E/Q(ζ)) = 1 (seeCorollary 3.9 and Corollary 3.14), implying that r(P,E/Z) = 1. Considering

[EP : ZP ][ZP : Qp(ζ)][Qp(ζ) : Qp] = [EP : Qp] = orqα(p)ϕ(pa),

where the last equality follows from Corollary 3.32, yields [ZP : Qp(ζ)] =(p− 1)pa−1q−β . This number will be denoted by h. Then

ζe ∈ NEP/ZP ⇔ NZP /Qp(ζ)(ζe) ∈ NEP/Qp(ζ), (See Proposition 3.27)

⇔ ζeh ∈ NEP/Qp(ζ),

⇔ NQp(ζ)/Qp(ζeh) ∈ NEP/Qp . (See Proposition 3.27)

The extension Qp(ζ)/Qp is unramified of order o and a generator σ ofGal(Qp(ζ)/Qp) maps ζ to ζp (see again Example 3.30). Thus, if we write

s =o−1∑i=0

pi then NQp(ζ)/Qp(ζ) = ζs. Together with the above, this yields

ζe ∈ NEP/ZP if and only if ζehs ∈ NEP/Qp .

But due to Proposition 3.37, this is again equivalent to ζehs = 1, or rqα | ehs.We calculate

ehs = eq−βpa−1(p− 1)(1 + p+ . . .+ po−1) = eq−βpa−1(po − 1).


By definition r divides por(p)−1, so by Lemma 4.30 it already divides po−1.Since we are looking for the smallest e such that rqα divides ehs, and (r, q) =1, we can reduce to finding the smallest e such that qα divides ehs. If we writem = oqα(p), then by the definition of δ we can write o = mδ and

po−1 = pmδ−1 = (pm−1)δ−1∑i=0

(pm)i ≡ δ(pm−1) mod q(pm−1). (6.1)

The last equation follows from the observation that for positive integers a, b, cand d we have the rule

a ≡ b mod c⇒ ad ≡ bd mod cd,

and the fact that pm ≡ 1 mod q.

Clearly, e, the smallest integer such that qα divides ehs = eq−β(po − 1)pa−1,equals qβ if and only if qα exactly divides po − 1. This happens if and only ifq does not divide δ and qα exactly divides pm − 1. To see this, write

po − 1 = δ(pm − 1) + q(pm − 1)k = (pm − 1)(δ + qk),

for some k a positive integer using equation 6.1. Clearly, if the latter conditionholds, then qα exactly divides po − 1. On the other hand, by definition qα

divides pm − 1. So, if qα exactly divides po − 1, no higher power than α of qcan divide pm − 1 and q can not divide δ.

The latter condition happens, by Lemma 4.30, exactly in the cases

q = 2, p ≡ −1 mod 4 and α = 1,

andα ≥ d.

We summarize these cases by saying that the least e for which ζe ∈ NEP/ZP isqβ if and only if

• q = 2, α = β = 1 and r = 1 or,

• q = 2, p ≡ −1 mod 4, α = 1 and 2 - δ or,

• α ≥ d and q - δ,

proving the lemma.

The following theorem will help us to reduce some cases in the ultimate classi-fication proof.

Theorem 6.4. Let p, q, r and s be distinct primes. The following groups G are notembeddable into division rings of characteristic 0:


• G = (Cp ×Cr)oCqb where qb 6= 4 and Cqb acts non-trivially on Cp and Cr,

• G = (Cp × Cr × Cs) o C4 where p, r and s are odd, C4 acts on Cp and Crby inversion and acts trivially on Cs.

Proof. First, suppose ad absurdum that G = (Cp × Cr) o Cqb is embeddable in adivision ring D = Q[G] and denote its center by F . The only primes dividing theQ-dimension of D are p, r and q and the only proper prime power is of the form qt.So of course also [D : F ] can only have these primes (or prime powers) as divisors.Since this dimension needs to be a square (see the remark after Definition 5.6),we may assume m(D) = qd for some d. By Theorem 5.7, there exists a primeP of F such that for the central FP-algebra DP = D ⊗F FP we have m(DP) =m(D) = qd. In particular, qd divides [DP : FP]. On the other hand, it is clear that[DP : FP] ≤ [D : F ] = qd. Combining these facts, m(DP) = [DP : FP] and soDP is a division algebra.

Consider the subgroup H = Cp o Cqb of G and its associated subalgebra A =Q[H] of D. Denote the center of A by Z. This group H is of the form as describedin the beginning of this section and A is isomorphic to the crossed product E[H] asexplained there. In the proof of Lemma 6.3, we have shown that m(A) = qβ , wherewe use here that the action of Cqb is not trivial.

In the division algebra DP consider the subalgebra K generated by Z and thecompletion of Q with respect to the prime k such that P | k. As DP is finitedimensional overQk, K will also be finite dimensional overQk. By Theorem 3.24,this field is a local field. Denote p′ its unique ideal (which automatically lies over kbecause Qk ⊆ K). Then p := OZ ∩ p′ is a prime ideal of Z lying over k. ClearlyZp is a subfield of K that contains Qk and Z, showing that Zp = K. In DP we canthus find the subalgebra A.Zp, which is a homomorphic image of Ap = A ⊗Z Zp.Since, by Corollary 5.5, Ap is a central simple Zp-algebra, Ap

∼= A.Zp ⊆ DP whichimplies that Ap is division by Lemma 5.2. In particular, [Ap : Zp] = m(Ap) | qβ 6=1.

From the proof of Lemma 6.3 it follows that this is only possible if p is real andqb = 4 or p lies over the prime p. By the assumption that qb 6= 4 we deduce that p(and P) lies over p. In the same way, by setting H = Cr o Cqb , we obtain that Plies over r. As a prime P can only lie over one prime in the rationals, we obtain acontradiction.

For the second case where G = (Cp×Cr×Cs)oC4∼= ((Cp × Cr)o C4)×Cs

we can repeat the above proof with H = (Cp o C4) × Cs and (Cr o C4) × Csrespectively. The fact that p lies over p now follows from the observation that p isnot a real prime. Indeed, if p would be real, this would imply situation b) in theproof of Lemma 6.3, where we deduced that s = 1, a contradiction.

We can now proceed with the proof of the Classification Theorem 2.12 for Z-groups. The theorem is repeated here.

Theorem 2.12. A groupG is a Z-group if and only if it is one of the following types:

a) cyclic,


b) Cm o C4, where m is odd and C4 acts by inversion,

c) G0×G1× . . .×Gs, with s ≥ 1, the orders of the groups Gi are coprime andG0 is the only cyclic subgroup amongst them. Each of the Gi, for 1 ≤ i ≤ s,is of the form

Cpa o(Cqb11× . . .× Cqbrr

),

for p, q1, . . . , qr distinct primes. Moreover, each of the groups Cpa o Cqb isnon-cyclic (i.e. if Cqα denotes the kernel of the action of Cqb on Cpa , thenα 6= b) and satisfies the following properties:

(i) qoqα(p) - o |G||Gi|

(p).

(ii) one of the following is true:

• q = 2, p ≡ −1 mod 4, and α = 1,

• q = 2, p ≡ −1 mod 4, and 2α+1 - p2 − 1,

• q = 2, p ≡ 1 mod 4, and 2α+1 - p− 1,

• q > 2, and qα+1 - p− 1.

Proof. Let G be a Z-group. By Theorem 4.12 it is of the form

Cm o Cn,

for some coprime integers m and n. Moreover, Cn acts non-faithfully on Cm (inother words, α 6= 0). Let us assume G is not cyclic (that is, β 6= 0) and that ifn = 4, then it does not act by inversion on Cm. We will prove that then G is of typec) in the theorem. We will do this by first splitting the Gi off from the rest of thegroup and then fixing their forms.

Write G0 = {x ∈ Cm | xy = x,∀y ∈ Cn} the centralizer of Cn in Cm. Assumewe have a prime p such that pk divides m exactly and such that G0 contains anelement of order p. Let this element be x. Take a y in Cn arbitrary and denote byαy the conjugation automorphism of y on Cpk ≤ Cm. It is trivial on Cp because xis in G0. By Lemma 4.8, αy is also trivial on Cpk , thus Cpk ≤ G0. This proves thatG0 is a Hall subgroup of G. Moreover, since it is the centralizer of Cn in Cm, it is adirect factor and we can split if off from G as a direct product of order coprime tothe rest. The group G0 is also cyclic as a subgroup of a cyclic group.

Denote the following prime decompositions

n =e∏i=1

qbii andm

|G0|=

s∏i=1

paii ,

where of course qi, pi are distinct primes and ai, bi are greater or equal to 1. EverySylow subgroup of Cn now only acts non-trivially on a unique Sylow subgroup ofCm. Indeed, suppose that C

qbii

acts non-trivially on both Cpakk and Cpajj

. By Lemma

4.8 it acts non-trivially on Cpk and Cpj . The subgroup (Cpk × Cpj) o Cqbii

of G


is embeddable in a division ring. By Theorem 6.4 this would imply qbii = 4 andthe action is by inversion (indeed, for the action is never faithful by Theorem 4.12).Now either

(Cpakk

× Cpajj

)o C4 = G, which would be a contradiction with our

assumptions, or there is another prime t dividing |G|. There are several cases toconsider: in splitting up the cases we will use that C4 can only act by inversion ortrivially (again, the action is not faithful due to Theorem 4.12) on a Sylow subgroupof Cm and that a Sylow 2′-subgroup of Cn acts on exactly one Sylow subgroup ofCm (follows from the reasoning until now)

• t = pl for some l and C4 acts by inversion on Cpall . It could be that(Cpakk

× Cpajj× Cpall

)o C4 = G.

This would yield a contradiction with our assumptions that G is not the semi-direct product of a cyclic group and C4, where the action is by inversion. If(Cpakk

× Cpajj× Cpall

)o C4 is a proper subgroup of G, we can find a new

prime t′ dividing the order of G. Repeat this argument for t′. By inductionwe will find a prime t that is one of the other cases. This argument has to stopat some point because G is a finite group.

• t = pl for some l and C4 acts trivially on Cpall . It follows that C4 acts triviallyonCpl and the groupH = (Cpk×Cpj×Cpl)oC4 ≤ G is of the second type inTheorem 6.4. This yields a contradiction with the fact that G is embeddablein a division ring.

• t = ql for some l and Cqbll

acts on one of Cpakk or Cpajj

. We find the group(Cpakk

× Cpajj

)o(C4 × Cqbll

)as a subgroup of G. So G also contains the

group(Cpk × Cpj

)o (C4 × Cql). Since this group is embeddable in a divi-

sion ring, by Theorem 4.10 and Theorem 4.11, the subgroups generated byrespectively Cpk , Cql and Cpj , Cql are cyclic. Thus(

Cpk × Cpj)o (C4 × Cql) ∼=

(Cpk × Cpj × Cql

)o C4,

is embeddable in a division ring. By Theorem 6.4, this is a contradiction,since C4 acts trivially on Cql .

• t = ql for some l and Cqbll

acts trivially on both Cpakk and Cpajj

. Then we knowthere is another Sylow subgroup of Cm that is not Cpakk or C

pajj

, associatedto a different prime pe. If C4 acts trivially on Cpaee , we find a contradictionsimilar to the second case above. If C4 acts by inversion on Cpaee , then there

is a subgroup(Cpaee × Cpajj

)o(C4 × Cqbll

)of G, with C4 acting on both

groups by inversion and Cqbll

acts non-trivially on exactly one of them. Thisforms a contradiction similar to the one given in the third case.


This proves that every Sylow subgroup of Cn acts non-trivially on a unique Sylow-subgroup of Cm.

We use this fact to factor our group into a direct product of subgroups Gi gen-erated by the Sylow pi-subgroup Cpaii of Cm and the Sylow subgroups of Cn actingnon-trivially on it. Thus we get the factorisation

G = G0 ×G1 × . . .×Gs,

where al the Gi 6= G0 are of the form

Gi = Cpaii o(Cqbi(1)i(1)

× . . .× Cqbi(t)i(t)

),

for some t depending on i. Fixing such an i and 1 ≤ j ≤ t, we write paii = pa,qbi(j)i(j) = qb and consider the subgroup Cpa o Cqb . Due to Lemma 6.3, and using the

same notation, either q = 2, α = 1 and p ≡ −1 mod 4 or α ≥ d. The first caseof this lemma does not apply here since r = 1 forces Gj = {1} for j 6= i and thusG = Cpa o C4 with C4 acting by inversion. This was a case we excluded at thebeginning of the proof.

By the definition of d, the fact that α ≥ d is equivalent to saying that 2α+1 -p2 − 1 if q = 2 and p ≡ −1 mod 4 or qα+1 - p − 1 in the other cases. It onlyremains to prove that qoqα(p) - o |G|

|Gi|(p). Clearly, if we take r a prime power dividing

ki = |G||Gi| , then

(Cpa o Cqb

)× Cr is a subgroup of G as Cr appears in one of the

other Gj . This group is thus also embeddable in a division ring. Using Lemma 6.3it follows that for all such r the prime q does not divide δ =

orqα (p)

oqα (p). By Lemma

4.30 we may equivalently say that q does not divide okiqα (p)

oqα (p), so qoqα(p) does not

divide okiqα(p). Again by Lemma 4.30, okiqα(p) = lcm{oki(p), oqα(p)}, implyingthat qoqα(p) does not divide either of them. We conclude that qoqα(p) - oki(p),proving that G is indeed a group satisfying the properties of type c).

For the other implication, suppose that G is a group satisfying one of the con-ditions listed in the theorem. Clearly the cyclic groups G are embeddable in C andeven in Q(ζm) where m = |G|. If G is of type b), namely

G = Cm o C4,

with m odd and C4 acting by inversion, then G is isomorphic to the group 〈ζm, j〉in H(R), where ζm is taken in R+ iR ∼= C.

Now assume G is of type c). Write n =∏

qbii and m = |G|n

. Then G is of theform Cm o Cn. Write qαii for the order of the kernel of the action of C

qbii

on Cm.Setting

Pi = Cm o

(Cqbii×∏i 6=j

Cqαjj

),

the Sylow subgroups of G (see page 52), we can write Pi as(Cpa o C

qbii

)× Cr

for some prime power pa and r coprime to p and qi. The extra conditions listed in

6.2. FINITE, NON-SOLVABLE SUBGROUP(S) OF DIVISION ALGEBRAS 61

the description of type c), together with Lemma 4.30, yield that Pi is a Z-group byTheorem 6.3. By Lemma 5.15, it follows that the whole group G is a Z-group.

6.2 Finite, non-solvable subgroup(s) of division alge-bras

We will start with proving the theorem of Amitsur concerning the finite subgroupsof rational division algebras. First of all, we will show that the non-solvable case isnot very interesting.

Theorem 6.5. The only finite non-solvable subgroup of a division ring of charac-teristic 0 is SL(2, 5).

We will make use of Lemma 6.6 in the proof of Theorem 6.5, but in order tofacilitate the proof, we introduce a new symbol called the Legendre symbol. For arational number a relatively prime to a prime number p, one denotes(

a

p

)=

{1 if a is a square modulo p−1 if a is not a square modulo p

.

This symbol is multiplicative (ab

p

)=

(a

p

)(b

p

).

Indeed, the fact that F×p /F×2p ∼= Z/2Z (F×2p is of index 2 in F×p ) for odd primes pimplies that a a product of two (non-)squares is a square and the product of a squarewith a non-square is a non-square. In the case where p = 2, every odd number isa square modulo 2, so the Legendre symbol is trivially multiplicative. This furtherimplies that (

a

p

)=

(a−1

p

).

Lemma 6.6. The only quadratic subfield of Q(ζp) for p an odd prime is the field

Q(

√(−1p

)p).

Proof. If√(

−1p

)p is an element of Q(ζp), then the fact that Q(

√(−1p

)p) is the

unique quadratic subfield follows from the fundamental theorem of Galois Theory,since Gal(Q(ζp)/Q) ∼= (Z/pZ)× contains a unique subgroup of index 2.

Let t =∑a∈F×p

(a

p

)ζap in Q(ζp). We claim that

(−1

p

)t2 = p,


which would prove the theorem. We calculate using the rules for the Legendresymbol seen above(

−1

p

)t2 =

∑a,b∈F×p

(−abp

)ζa+bp =

∑a,b∈F×p

(ab

p

)ζa−bp ,

=∑a,b∈F×p

(ab−1

p

)ζa−bp =

∑b,c∈F×p

(c

p

)ζbc−bp ,

=∑c 6=1

(c

p

)∑b∈F×p

ζb(c−1)p +∑b∈F×p

(1

p

),

and where we used the substitution ab−1 = c. Also, let x ∈ Fp be arbitrary but fixed

with(xp

)= −1, then

(−1)∑c∈F×p

(c

p

)=

(x

p

)∑c∈F×p

(c

p

)=∑c∈F×p

(xc

p

)=∑d∈F×p

(d

p

),

showing that∑c∈F×p

(c

p

)= 0 and

∑c 6=1

(c

p

)= −

(1

p

)= −1. If we write for short

ζ = ζ(c−1)p , then ∑

b∈F×p

ζb(c−1)p = ζ + ζ2 + . . .+ ζ l−1 = −1.

If we fill this in in the calculations above we obtain(−1

p

)t2 = (−1)(−1) + p− 1 = p,

which is what we claimed.

Proof of Theorem 6.5. We will use a theorem of Zassenhaus (see Passman [12, The-orem 18.6]) which states that, if a group G is non-solvable and a Frobenius com-plement, then it has a normal subgroup H of index at most 2 with

H ∼= SL(2, 5)×M,

where M is a Frobenius complement and a Z0-group of order coprime to 30. Alsoremark that SL(2, 5) is non-solvable (see Corollary 4.22). The rest of the proofwill be in 3 steps. First, we will show that a division ring containing SL(2, 5)always has a sub-division algebra of the form H(Q(

√5)). Then we will show that

SL(2, 5) × Cp for p a prime never embeds into a division ring. In view of theZassenhaus theorem described above, it will then be sufficient to prove that a groupwith normal subgroup SL(2, 5) of index 2 never embeds into a division ring.

6.2. FINITE, NON-SOLVABLE SUBGROUP(S) OF DIVISION ALGEBRAS 63

Step 1: Assume that G ∼= SL(2, 5) is a subgroup of D. Consider Q[G] thesubalgebra of D and write D′ = Q[G]. We prove that D′ ∼= H(Q(

√5)). This

would also further imply that we may assume D = H(Q(√

5)). In Section 4.3 weintroduced the matrices

x =

(−1 1−1 0

)and y =

(1 01 1

),

as generators for SL(2, 5).Let S := {1, y, y2, y3, x, xy, xy2, xy3} and T = Q[S] ⊆ D′. Using the well-

known equation

X t − 1 = (X − 1)(1 +X + . . .+X t−1),

we see that x2 = −x − 1 and y4 = −y3 − y2 − y − 1 because x3 = 1 = y5 andneither x nor y equals 1. This implies that x2 and y4 are in T , showing that xT ⊆ T

and Ty ⊆ T . A simple calculation gives z =

[(−1 1−1 0

)(1 01 1

)]2= −1, so

yx = x−1(xy)2y−1 = x−1zy−1 = −x2y4 = (x+ 1)(−y3 − y2 − y − 1) ∈ T.

This proves that yT ⊆∑Tyi ⊆ T , so S2 is contained in T . We proved that

D′ = T .This shows that [D′ : Q] ≤ 8. Consider D′ ⊇ E = Q[y] ∼= Q(ζ5). Since

[E : Q] = φ(5) = 4 and D′ 6= E (D′ is for example not commutative) we concludethat [D′ : Q] = 8. If we denote F the center of D′, then clearly Q ⊆ F , so [D′ :F ] ∈ {8, 4, 2, 1}. Moreover, it needs to be a square of an integer (as proven whenwe introduced the Schur index, Definition 5.6). This implies that m(D′) = 2, since[D′ : F ] 6= 1. Also, F is of degree 2 over Q, and by the Galois correspondence,E has a unique quadratic subfield (Gal(E/Q) = C4 and contains one subgroup ofindex 2). This unique quadratic subfield is Q(y1) with y1 = 2(y + y4) + 1. Werecognise this field as Q(

√5), for indeed

y21 = 4(y + y4)2 + 1 + 4(y + y4) = 4(y2 + y3 + 2) + 1 + 4(y + y4)

= 4(y + y2 + y3 + y4) + 9 = −4 + 9 = 5.

Let Q be a Sylow 2-subgroup of G. Using Proposition 4.10 and Theorem 4.11,we see that Q has to be cyclic or quaternion. It can not be cyclic, for otherwise G isa Z0-group, and by Proposition 4.15 it would be solvable. So Q is quaternion, andlooking at |G| = 23 · 3 · 5, it is of order 8. By Theorem 5.13, Q[Q] ∼= H, so wecan write D′ = H ⊗ CD′(H) by Theorem 5.3. But now CD′(H) has dimension 2over Q, so it is a field, and therefore it is the center of D′ by Corollary 5.4. In otherwords, D′ = H⊗ F ∼= H(Q(

√5)).

Step 2: We now show that SL(2, 5) is indeed the only non-solvable subgroupof a division ring. First, note that G = SL(2, 5) × Cp is never a subgroup of adivision ring for a prime p. Indeed, assume that G ⊆ D× for some D. Since


SL(2, 5) contains elements of order 2, 3 and 5, it is impossible that p ≤ 5. Indeed,since Cp × Cp ≤ G and Cp × Cp is not cyclic we would get a contradiction withTheorem 4.11. For ζ = ζp, let us consider the algebra

B = H(Q(√

5))⊗Q Q(ζ) = H⊗Q Q(√

5)⊗Q Q(ζ).

Now, we claim that C := Q(√

5) ⊗Q Q(ζ) is a field. Indeed, remark that we canwrite every element of C as 1 ⊗ u +

√5 ⊗ v for u and v in Q(ζ). If u = 0 or

v = 0, then an inverse is easily found. Suppose u 6= 0 6= v. Then 1⊗ u +√

5⊗ vhas 1 ⊗ (u − 5v2u−1)−1 +

√5 ⊗ −v(u2 − 5v2)−1 as an inverse. This is under the

assumption that u − 5v2u−1 6= 0. If this element would be zero, we can deducethat (uv−1)

2= 5 in Q(ζ). Then

√5 ∈ Q(ζ), meaning that Q(

√5) is a quadratic

subfield of Q(ζ), a contradiction with Lemma 6.6 as p > 5.Using Theorem 5.13, we can see that there exists a surjective homomorphism

from B to Q[G] = Q[SL(2, 5) × Cp] ⊆ D. As B is simple (using Corollary 5.5),B ∼= Q[G]. This implies that B can be embedded in D as a division algebra. Wenow find a contradiction with Theorem 5.12 (note that B = H(C)), as Q(ζ) has noreal embedding and the prime 2 has residue degree 2 in the extension Q(

√5)/Q.

This last fact can be seen using the following reasoning. Consider the tower ofextensions

Q

Q(√

5)

Q(ζ5)

Using Proposition 3.13, we immediatly obtain e(2,Q(ζ5)/Q) = 1 = r(2,Q(ζ5)/Q).Furthermore, due to the multiplicativity for δ a symbol from {e, f, r} (see Corol-lary 3.9) we obtain δ(2,Q(

√5)/Q) | δ(2,Q(ζ5)/Q), showing that

e(2,Q(√

5)/Q) = 1 = r(2,Q(√

5)/Q).

As

2 = [Q(√

5) : Q] = e(2,Q(√

5)/Q)f(2,Q(√

5)/Q)r(2,Q(√

5)/Q),

follows from Theorem 3.8 we obtain f(2,Q(√

5)/Q) = 2.Step 3: Let G be a group, embeddable in a division ring D, with normal sub-

group SL(2, 5) of index 2 (which we will denote by S). Let Q be a fixed Sylow2-subgroup of G. Using the same reasoning as at the end of part 1 of this proof, butnow with |G| = 24 · 3 · 5, Q has to be quaternion of order 16.

6.3. FINITE, SOLVABLE SUBGROUPS WITH A QUATERNION SYLOW SUBGROUP65

By the first part and by Theorem 5.13, we have that the subalgebras Q[S] ∼=H(Q(

√5)) and Q[Q] ∼= H(Q(

√2)) of D = Q[G] are not isomorphic and of di-

mension 8 over Q. As S has index 2 in G, we have that [D : Q] is at most2 · [Q[S] : Q] = 16 and at least [Q[S] : Q] = 8. Thus [D : Q] = 16, elseD = Q[S] = Q[Q].

Also, SQ = G (this is because S is of index 2 inG andQ is not contained in S).This shows that |S ∩Q| = |S||Q|

|G| = 8, so S ∩Q is quaternion of order 8. Using this,we can write D ∼= H⊗Q Γ, where H ∼= Q[S ∩Q] (from Theorem 5.13) and Γ is thecentralizer of H in D (see Theorem 5.3). Considering the dimensions, we obtain[Γ : Q] = 4. If Γ is not a field, it is central simple over Q with m(Γ) = 2 = m(H),a contradiction with Proposition 5.8. Thus, Γ is a field. Moreover, since it is thecentralizer of Q[S ∩Q] in D, it contains the centers of Q[S] and Q[Q]. This showsthat Γ contains subfields isomorphic to Q(

√5) and Q(

√2). By the dimension, we

obtain that Γ = Q(√

5,√

2).We find a contradiction in the following reasoning. Set X := 〈x〉 ∼= C3 (see

the presentation given for S in step 1). It is a Sylow 3-subgroup of S. By Frattini’sArgument (Lemma 4.1), G = SNG(X) ≥ SP , where P is a Sylow 2-subgroup ofNG(X). Now |S||P |

|S∩P | = |SP | | |G|, so S ∩ P is a 2-subgroup of P . If it is equalto P then P ≤ S. Remark that NG(X) can not contain an element of order 5 (elseit is conjugate to y, and y does not normalise X) and contains the element x. SoNG(X) ≤ S, a contradiction with |G| = |S||NG(X)|

|S∩NG(X)| = |S| as [G : S] = 2. Since

now S ∩ P < P , we have |S||P ||S∩P | ≥|S|2n2n−1 = 2|S| = |G| (for some n ∈ N), showing

that also G = SP .In A5

∼= S/〈z〉 (see Lemma 4.21), the normalizer of a 3-cycle, say (1, 2, 3) isgiven by 〈(1, 2, 3), (1, 2)(4, 5)〉, so it is of order 6. As z itself centralizes x and isof order 2, we get that S ∩ P is of order 4. Using the formula |G| = |S||P |

|S∩P | , we findthat |P | = 8. Moreover, we can say P has to be quaternion because it is not cyclic(X does not have an automorphism of order 4). As above, we obtain

D = Q[P ]⊗Q CD(P ) ⊇ Q[P ]⊗Q Q(√

5,√

2) ∼= D,

so D = Q[P ] ⊗Q Q(√

5,√

2). Now write P = 〈i, j〉 where i centralizes x andj inverts it. Then x is an element of CD(i) = Q(

√5,√

2, i) (follows from whatwe wrote). Consider this field as a subfield of the complex numbers (we may dothis: the i from P acts as

√−1). Since x3 = 1, we obtain that 3

√1 = −1±

√3i

2is

an element of Q(√

5,√

2, i). This implies that√

3 can be found in Q(√

2,√

5), acontradiction.

6.3 Finite, solvable subgroups with a quaternion Sy-low subgroup

This section is dedicated to proving that the finite solvable subgroups G, embeddedin a rational division algebra, with a quaternion Sylow subgroup are indeed the ones


listed under type b) in Theorem 2.8. As a consequence we will also give a proof ofTheorem 2.8.

Lemma 6.7. Let n and m be positive, coprime integers. In (Cm o Cn) o G, thegroup G normalizes Cm.

Proof. Write Cm = 〈x〉 and take 1 6= c ∈ Cm and g ∈ G arbitrary. As Cm o Cn isnormalized by G, the element cg lies in Cm o Cn. As o(cg) = o(c) | m, cg can notbe an element of Cn for its order does not divide n. It remains to prove that cg cannot be written as a product of an element from Cm and a non-trivial element fromCn.

Suppose for contradiction’s sake that cg = ab for an a in Cm and a b 6= b in Cn.Write a = xl for l < m and xb = xh. Then we have

ab = xlh ⇔ ab = bxlh.

This yields (ab)t = btxk for every t ∈ N and some positive integer k depending ont. Let 1 6= t = o(ab) = o(c) | m:

1 = (ab)t = btxk ⇔ (bt)−1 = xk.

As b ∈ Cn and x ∈ Cm, this is only possible if bt = 1. Of course 1 6= t | o(b),which itself is a divisor of n and t | m. This is a contradiction with the fact that mand n are coprime.

Theorem 6.8. Assume that G is a finite solvable subgroup of a division ring D ofcharacteristic 0 such that its Sylow 2-subgroups are quaternion. Then G is isomor-phic to one of the groups:

i) the binary octahedral group O∗:

{±1,±i,±j,±ij, ±1± i± j ± ij2

} ∪ {±a± b√2| a, b ∈ {1, i, j, ij}}.

ii) Cm o Q where m is odd, Q is quaternion of order 2t for some t ≥ 3, anelement of order 2t−1 centralizes Cm and an element of order 4 inverts Cm.

iii) M ×Q8, with M is a Z-group of odd order m and 2 - om(2).

iv) M ×SL(2, 3), where M is a Z-group of order m coprime to 6 and 2 - om(2).

Proof. Although G has quaternion Sylow subgroups, it is possible that these Sylowsubgroups are not normal inG. We will prove thatG belongs to a certain type in theabove list by looking at O2(G), the maximal normal 2-subgroup of G. Since O2(G)is a subgroup of a Sylow 2-subgroup, O2(G) can also only be cyclic or quaternionby Theorem 4.11.

SupposeO2(G) is quaternion. IfO2(G) ∼= Q8, then the result follows from The-orem 4.32. If |O2(G)| ≥ 16, then its automorphismgroup is a 2-group by Lemma


4.20 and CG(O2(G)) is also a 2-group by Proposition 4.31. Using Lemma 4.5, G isitself a 2-group so G = O2(G). The group G is in this case of type ii) with m = 1.

Finally, consider the case where O2(G) is cyclic. Then by Corollary 4.39

G = M oQ,

with M a 2′-group and Q a quaternion Sylow 2-subgroup of G. It suffices to showthat M is cyclic, for then we can use Theorem 4.34. Remark that M is a Z-group:M ≤ G ⊆ D×, so M is a Frobenius complement and it has no quaternion Sylow 2-subgroups, for its order is odd. By Theorem 4.12 we find coprime positive integersm and n such that

M ∼= Cm o Cn.

The subgroup Q of G normalizes Cm by Lemma 6.7. Using Frattini’s argument forsolvable groups, Lemma 4.4, on G, M and Cn as Hall n-subgroup of M yields

G = MNG(Cn).

Considering the orders |G| = |M ||NG(Cn)||M∩NG(Cn)|

we see that |NG(Cn)| contains a Sylow2-subgroup. This subgroup is also Sylow 2-subgroup of G because |M | is odd.Without loss of generality, we may assume this is the groupQ, soQ also normalizesCn.

Write ci for the generator of Ci. Suppose we can find an element z ∈ Q thatinverts both ci. Then consider the commutator

[cm, cn]−1 = [cm, cn]z = [czm, czn] = [c−1m , c−1n ] = [cm, cn]c

−1n c−1

m .

On the other hand, both [cm, cn] and c−1n c−1m have odd order and we have followingequality:

[cm, cn](c−1n c−1

m )2 =(

[cm, cn]c−1n c−1

m

)c−1n c−1

m

= ([cm, cn]−1)c−1n c−1

m = [cm, cn]. (6.2)

If we write o(c−1n c−1m ) = 2k + 1 for some k ∈ N, then we can calculate

[cm, cn] = [cm, cn](c−1n c−1

m )2k+1

,

=(

[cm, cn](c−1n c−1

m )2k)c−1

n c−1m

,

6.2= [cm, cn]c

−1n c−1

m = [cm, cn]−1.

As the order of [cm, cn] is odd this is only possible if [cm, cn] = 1. This shows thatM is abelian and M ∼= Cm × Cn is cyclic. It only remains to find such an elementz ∈ Q.

Suppose that |Q| ≥ 16. Keeping in mind that Q normalises Cm and Cn, Gcontains subgroups isomorphic to Cm o Q and Cn o Q. Following Theorem 4.34we have an element y1 of order 4 inQ that inverts cm and an element y2 of order 4 inQ that inverts cn. If y1 also inverts cn, then we have found a suitable element. If not,


then it also follows from Theorem 4.34 that y1 centralizes cn. We can say the samefor y2; either it inverts cm and then we have a suitable element, or it centralizes cm.If the latter case holds, then y1y2 is an element of Q that inverts both cm and cn.

If on the other hand |Q| = 8, then the action of Q on M (and thus on Cm andCn) is trivial or by inversion. Write K for the kernel of the action of Cn on Cm. ByTheorem 4.12, if p | n for p a prime, then Cp ∈ K. So either K 6= {1} or K = {1},but in the latter case also Cn = {1}. The group M is then cyclic and the resultfollows from Theorem 4.34.

From here on, assume K to be non-trivial. We claim that K is normalised byQ. Indeed, for c ∈ K, z ∈ Q we have

cz−1czm =

((cz−1

m )c)z

= (cz−1

m )z = cm.

We can thus consider the subgroup LoQ of G, where L = KCm. This group L isa cyclic group because K ≤ Cn and the orders of K and Cm are coprime. We maythus apply Theorem 4.34 to this group to obtain that Q centralizes L or an elementof order 4 (say y) inverts L.

In the first case, Q centralizes K ≤ L ∩ Cn. It can not be that Q acts on Cnby inversion, because if it would, then for a c 6= 1 ∈ K ≤ Cn we would obtainc−1 = cz = c for some z ∈ Q. This is not possible in a group of odd order since cwould be of order 2. The groups Cn and Cm ≤ L are both centralized by Q, furtherimplying that also M is centralized by Q. This is a contradiction with the fact thatO2(G) is cyclic.

In the second case y inverts Cm ≤ L and at least an element of Cn ≥ K. Theaction of Q on Cn is not trivial, so y also inverts Cn. The element y is the elementwe needed above, proving the theorem.

We now have everything we need to prove our main theorem, Theorem 2.8. Forsimplicity’s sake it is repeated below.

Theorem 2.8. A finite group G is a subgroup of a division ring of characteristic 0if and only if G is isomorphic to one of the following groups:

a) a Z-group.

b) i) the binary octahedral group O∗:

{±1,±i,±j,±ij, ±1± i± j ± ij2

} ∪ {±a± b√2| a, b ∈ {1, i, j, ij}}.

ii) CmoQ where m is odd, Q is quaternion of order 2t for some t ≥ 3, anelement of order 2t−1 centralizes Cm and an element of order 4 invertsCm.

iii) M ×Q8, with M is a Z-group of odd order m and 2 - om(2).

iv) M × SL(2, 3), where M is a Z-group of order m coprime to 6 and2 - om(2).


c) SL(2, 5).

Proof. Suppose that G is a finite subgroup of a division ring of characteristic 0. IfG is non-solvable, G is isomorphic to SL(2, 5) because of Theorem 6.5, resultingin a group of type c). Suppose that G is a solvable group. By Proposition 4.10 andTheorem 4.11, G has either cyclic or quaternion Sylow subgroups. If all its Sylowsubgroups are cyclic, G is by definition a Z-group, so of type a). If G is solvableand has a quaternion Sylow subgroup, then it is of type b) because of Theorem 6.8.This exhausts all the cases, so we have shown the necessary condition.

To prove that it is sufficient for G to be one of the groups given in the typesa), b) and c), we will find embeddings for these groups into a division ring of char-acteristic 0. First of all, Z-groups are by definition embeddable in such a divisionring and SL(2, 5) is embeddable into H(R) by Proposition 4.24.

First the types b)i) and b)ii). The binary octahedral group is clearly embeddedin H(R). Consider a group Cm o Q2t , with an element of order 2t−1 centralizingCm and an element of order 4 inverting Cm. If we write

H(R) ∼= R⊕ Ri⊕ Rj ⊕ Rij,

then the part R ⊕ Ri is isomorphic to C. In here, take ζm and ζ2t−1 . The groupgenerated by ζm, ζ2t−1 and j in H(R) will be isomorphic to CmoQ2t . This followsfrom the fact that j and ζ2t−1 themselves generate a group isomorphic to Q2t , whichcan be checked by calculating the relations where ζ2t−1 plays the role of x and jplays the role of y in the notation introduced on page 6. Adding ζm to the generatorsyields the full group because j acts on ζm by inversion, ζm and ζ2t−1 commute andhave no other relations between them due to the fact that m is odd.

The groups Q8 and SL(2, 3) are already embeddable in H. For Q8 this is obvi-ous (it is generated by i and j). For SL(2, 3) we will use Remark 4.19, which saysthat SL(2, 3) ∼= Q8 o C3 and the action of C3 on Q is by cyclically permuting x, yand xy. As Q8 is generated by i and j in H, it will suffice to find an element oforder 3 in H that cyclically permutes i, j and ij. Such an element is given by

−1 + i+ j + ij

2.

It indeed permutes our elements the way we want and(−1 + i+ j + ij

2

)3

= −(1 + i+ j + ij)(−2 + 2i+ 2j + 2ij)

23,

=(1 + i+ j + ij)(1 + i+ j + ij)

22

=|1 + i+ j + ij|

4= 1,

so it is of order 3.Lastly, consider a group of the form M × Q8 or M × SL(2, 3) with M a Z-

group of odd order m and 2 - om(2). In the latter group, we also assume 3 - m.


It follows from our proof of Theorem 2.12 (see Section 6.1) that such a Z-group isembeddable in a division ringD of odd Schur index dividingm and with center F ⊆Q(ζm). Following Proposition 3.13, e(2,Q(ζm)/Q) = 1 and f(2,Q(ζm)/Q) =om(2). Since these are both odd, by Theorem 5.12, H(Q(ζm)) is a division algebra.But then of course H(F ) ⊆ H(Q(ζm)) is too, because it can not be split as asubalgebra of a division algebra. Moreover m(H(F )) = 2 and m(D) is odd, so byProposition 5.8 the algebra H(F ) ⊗F D is division. This division algebra containsa subgroup isomorphic to M ×Q8 and M × SL(2, 3).

Bibliography

[1] A. A. Albert. Structure of algebras. Revised printing. American MathematicalSociety Colloquium Publications, Vol. XXIV. American Mathematical Soci-ety, Providence, R.I., 1961.

[2] S. A. Amitsur. Finite subgroups of division rings. Trans. Amer. Math. Soc.,80:361–386, 1955.

[3] J. W. S. Cassels and A. Frohlich, editors. Algebraic Number Theory: Proceed-ings of an Instructional Conference Organized by the London MathematicalSociety (A Nato Advanced Study Institute W). Academic Press, 10, 1986.

[4] P. M. Cohn. Algebra. Vol. 2. John Wiley & Sons, Ltd., Chichester, secondedition, 1989.

[5] M. Hall, Jr. The theory of groups. The Macmillan Co., New York, N.Y., 1959.

[6] I. N. Herstein. Finite multiplicative subgroups in division rings. Pacific J.Math., 3:121–126, 1953.

[7] N. Jacobson. Basic algebra. II. W. H. Freeman and Co., San Francisco, Calif.,1980.

[8] E. Jespers and A. del Rıo. Group ring groups, orders and units. De Gruyter,In Preparation.

[9] D. Khurana and A. Khurana. A theorem of Frobenius and its applications.Math. Mag., 78(3):220–225, 2005.

[10] T. Y. Lam. The algebraic theory of quadratic forms. W. A. Benjamin, Inc.,Reading, Mass., 1973. Mathematics Lecture Note Series.

[11] J. Neukirch. Algebraic number theory, volume 322 of Grundlehren der Math-ematischen Wissenschaften [Fundamental Principles of Mathematical Sci-ences]. Springer-Verlag, Berlin, 1999. Translated from the 1992 German orig-inal and with a note by Norbert Schappacher, With a foreword by G. Harder.

[12] D. Passman. Permutation groups. W. A. Benjamin, Inc., New York-Amsterdam, 1968.

71

72 BIBLIOGRAPHY

[13] I. Reiner. Maximal orders. Academic Press [A subsidiary of Harcourt BraceJovanovich, Publishers], London-New York, 1975. London Mathematical So-ciety Monographs, No. 5.

[14] D. J. S. Robinson. A course in the theory of groups, volume 80 of GraduateTexts in Mathematics. Springer-Verlag, New York, second edition, 1996.

[15] J.P. Serre. Local fields, volume 67 of Graduate Texts in Mathematics.Springer-Verlag, New York-Berlin, 1979. Translated from the French by Mar-vin Jay Greenberg.

[16] M. Shirvani and B. A. F. Wehrfritz. Skew linear groups, volume 118 of Lon-don Mathematical Society Lecture Note Series. Cambridge University Press,Cambridge, 1986.

[17] E. Weiss. Algebraic number theory. Dover Publications, Inc., Mineola, NY,1998. Reprint of the 1963 original.

Index

Amitsur, 2, 4–6

Central algebra, 41Central simple(division) algebra, 41Classification

finite subgroups of division algebrascharacteristic 0, 5positive characteristic, 4

Z-groups, 6Commutator subgroup, 5Crossed product, 47

Derived series, 5Domain, 7

Dedekind, 7

Euler’s totient (phi) function, 11Exponent, 44Extension

abelian (cyclic), 15

Fieldlocal, 15number, 7, 11p-adic, 15

residue, 14Fractional ideal, 8Frattini’s Argument, 21

for solvable groups, 22Frobenius, 24Frobenius complement, 23

Groupbinary octahedral, 5, 6Brauer, 44cyclic, 4elementary p-, 30Fitting, 30general linear, 3

nilpotent, 29orthogonal, 28p′-, 3quaternion, 5, 6, 26

order 8, 1, 5solvable, 5special linear, 3SL(2, 3), 5SL(2, 5), 5, 28

special orthogonal, 28supersolvable, 40Z-, 4, 23Z0-, 4, 23

Hall, 22Hall subgroup, 22Hasse Norm Theorem, 19Herstein, 1, 3

Legendre symbol, 61

Norm map, 16

Ostrowski, 15

p-adic integers, 12P-adic norm, 10p-adic numbers, 12P-adic valuation, 10

Quaternion algebra, 45real, 28split, 45

Ramification index, 9, 15Residue degree, 9, 15Ring of integers, 11

Schur-index, 44Schur-Zassenhaus, 39

73

74 INDEX

Shirvani, 2Support, 47

Total valuation map, 9Totally ramified extension, 16

Unramifiedextension, 16prime, 9

Valuation(non-)archimedean, 13discrete(normalised), 14exponential, 13multiplicative, 13ring, 14

discrete, 10

Wedderburn, 3Wehrfritz, 2

INDEX 75

List of Symbols

Symbol Page DescriptionZ(A) The center of the group (or algebra) A.R×,U(R) The group of invertible elements of the

ring R.EndF (V ) The ring of F -linear endomorphisms of

the F -vector space V .Fpr 3 The unique finite field of pr elements,

with p a prime and r a positive integer.Mn(R) 3 The n-dimensional matrix ring over the

ring R.GL(n, F ) 3 The invertible n×nmatrices over the field

F .GL(n, pr) 3 The invertible n × n matrices of integers

modulo pr, with p a prime and r a positiveinteger.

SL(n, F ) 3 The n × n matrices over the field F withdeterminant 1.

SL(n, pr) 3 The n× n matrices of integers modulo pr

of determinant 1, with p a prime and r apositive integer.

Cn 4 The cyclic, multiplicative group of ordern.

G′ 5 The commutator subgroup of G.om(n) 5 The multiplicative order of nmodulom if

n and m are coprime.O∗ 5 The binary octahedral group.Q2t 6 The quaternion group of order 2t for t ≥

3.Spec(R) 7 The set of prime ideals of the ring R.Max(R) 7 The set of maximal ideals of the ring R.F(R) 8 The set of fractional ideals of a domainR.P | p 9 The prime ideal P lies over the prime

ideal p.e(P, L/K) 9 The ramification index of the prime P in

the field extension L of K.f(P, L/K) 9 The residue degree of the prime P in the

field extension L of K.vP 10 The P-adic valuation map.| . |P 10 The P-adic norm.KP 10 The completion of the field K with re-

spect to the P-adic norm.

76 INDEX

OK 11 The ring of integers of the number fieldK.

ζk 11 A primitive k-th root of unity in C.ϕ(n) 11 The Euler totient function of n.Qp 12 The p-adic numbers.Zp 12 The p-adic integers.e(L/K) 15 The ramification index of the extension of

local fields L/K.f(L/K) 15 The residue degree of the extension of lo-

cal fields L/K.NL/K 16 The norm map associated to the Galois

extension L/K.NG(H) 21 The normalizer of H in G.CG(H) 21 The centralizer of H in G.IdA 22 The identity morphism on the structureA.R[G] 23 The subring generated by the ring R and

the group G in a (division) ring D ⊇ Rand G ⊆ D×.

xg 23 The action of the element g of the groupG on the element x of a set X .

O(n, F ) 28 The orthogonal n × n matrices over thefield F .

O(n, pr) 28 The orthogonal n×n matrices of integersmodulo pr, with p a prime and r a positiveinteger.

SO(n, F ) 28 The orthogonal n × n matrices over thefield F with determinant 1.

SO(n, pr) 28 The orthogonal n × n matrices of inte-gers modulo pr of determinant 1, with pa prime and r a positive integer.

A5 28 The alternating group on 5 elements.Fit(G) 30 The Fitting subgroup of the group G.O2(G) 33 The maximal normal 2-subgroup of the

group G.[A : F ] 41 The dimension of the algebra A over the

field F .m(A) 44 The Schur-index of the algebra A.e(A) 44 The exponent of the algebra A.H(F ) 45 The quaternionalgebra over the field F .Supp(a) 48 The support of the element a in the rele-

vant crossed product.XG 48 The set of fixed points of the set X by the

action of the group G.

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