Maximal Subgroups of Finite Groups

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MAXIMAL SUBGROUPS OF FINITE GROUPS

By LINDSEY-KAY LAUDERDALE

A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY

UNIVERSITY OF FLORIDA 2014 c 2014 Lindsey-Kay Lauderdale ⃝ To Mom, Dad and Lisa ACKNOWLEDGMENTS I would like to thank my advisor Dr. Alexandre Turull, who ﬁrst sparked my interest in group theory. His endless patience and support were invaluable to the completion of this dissertation. I am also grateful for the support, questions and suggestions of my committee members Dr. Kevin Keating, Dr. David Mazyck, Dr. Paul Robinson and Dr. Peter Sin.

4 TABLE OF CONTENTS page ACKNOWLEDGMENTS ...... 4 LIST OF TABLES ...... 6 ABSTRACT ...... 7

CHAPTER 1 DISSERTATION OUTLINE ...... 9 2 A BRIEF BACKGROUND OF FINITE GROUP THEORY ...... 11 2.1 Introduction ...... 11 2.2 Maximal Subgroups ...... 14 2.3 Notation and Some Deﬁnitions ...... 17 2.4 On the Number of Maximal Subgroups ...... 18

3 UPPER BOUNDS FOR THE NUMBER OF MAXIMAL SUBGROUPS .... 20 3.1 Introduction ...... 20 3.2 Upper Bounds Proven by Others ...... 21 3.3 Preliminary Results ...... 22 3.4 Proof of Main Theorem ...... 23 4 LOWER BOUNDS FOR THE NUMBER OF MAXIMAL SUBGROUPS .... 25 4.1 Introduction ...... 25 4.2 Preliminary Results ...... 25 4.3 Proof of Main Theorem ...... 29 4.4 A Special Case ...... 35

5 SOME FURTHER LOWER BOUNDS FOR THE NUMBER OF MAXIMAL SUBGROUPS ...... 38 5.1 Introduction ...... 38 5.2 Deﬁnitions ...... 40 5.3 Preliminary Results ...... 41 5.4 m(G) for p-Solvable Groups ...... 43 5.5 |m(G)| for Nonsolvable Groups ...... 46 5.6 Proof| | of Main Theorems ...... 51 5.7 Remarks on α(G) ...... 52 5.8 Remarks on Property B ...... 54 REFERENCES ...... 56

BIOGRAPHICAL SKETCH ...... 57

5 LIST OF TABLES Table page 5-2 Values of α(G)...... 54

6 Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulﬁllment of the Requirements for the Degree of Doctor of Philosophy MAXIMAL SUBGROUPS OF FINITE GROUPS By Lindsey-Kay Lauderdale May 2014 Chair: Alexandre Turull Major: Mathematics Finite group theory is a topic that has held the attention of mathematicians for over ahundredyears.Thisisinpartduetoitsapplicationsthroughoutmultiplebranchesof science including biology, chemistry and physics. Investigating ﬁnite groups is a classical problem, with many open questions remaining. A major component to understanding

ﬁnite groups is comprehending their structure. One way to investigate the structure of ﬁnite groups is to study their maximal subgroups.

In this dissertation, we study the number of maximal subgroups in a finite group G,denoted m(G) . When G is a cyclic group, an elementary calculation proves that | | m(G) = π(G) ,whereπ(G)denotesthesetofprimeswhichdivide G .Wefurtherprove | | | | | | that if G = P P P ,then m(G) = m(P ) + m(P ) + + m(P ) , where 1 × 2 ×···× n | | | 1 | | 2 | ··· | n | π(G)= p ,p ,...,p and P Syl (G)foreachi 1, 2,...,n . { 1 2 n} i ∈ pi ∈{ } We proceed by turning our attention to bounding the number of maximal subgroups in an arbitrary finite group. First, we consider the upper bound for the number of maximal subgroups in a finite group. Upper bounds for the number of maximal subgroups have gradually been sharpened by other mathematicians through making further and further assumptions on the group. We make more assumptions on the structure of a finite group and then improve the existing upper bound for the number of the maximal subgroups in a finite solvable group.

7 To continue to narrow the range of the possible number of maximal subgroups in a ﬁnite group, we must consider the lower bound. We prove multiple lower bounds for the number of maximal subgroups in an arbitrary noncyclic ﬁnite group. In general, for a noncyclic group G, m(G) π(G) + p where p π(G)isthesmallestprimethatdivides | |≥| | ∈ G .IfG has a noncyclic Sylow subgroup and q π(G)isthesmallestprimesuchthat | | ∈ Q Syl (G)isnoncyclic,then m(G) π(G) + q.Weconcludebyproducingtwonew ∈ q | |≥| | lower bounds for m(G) ,bothofwhichconsideralloftheprimesinπ(G). | |

8 CHAPTER 1 DISSERTATION OUTLINE This first chapter gives an outline of the remaining chapters in this dissertation. The main content begins in Chapter 2 with some basic definitions and a natural way of classifying groups by their structural properties. We proceed with some results which provide a foundation for the study of maximal subgroups of finite groups, the focus of this dissertation. After a brief introduction to maximal subgroups, we calculate the number of maximal subgroups of some basic finite groups. However, with minimal assumptions on a group we cannot always calculate the exact number of maximal subgroups. Thus we attempt to narrow the range for the possible number of maximal subgroups in a finite group by considering both upper and lower bounds.

Chapter 3 focuses on the upper bounds for the number of maximal subgroups. Here we only consider the number of maximal subgroups in ﬁnite solvable groups. Several upper bounds for the number of maximal subgroups have been calculated by other mathematicians, each one improving the upper bound that was proven before. In the last section of Chapter 3,wemakesomefurtherassumptionsonastructureofaﬁnitesolvable group to improve the upper bound for the number of the maximal subgroups.

To continue to narrow the range for the possible number of maximal subgroups of agivenfinitegroup,wenextconsiderthelowerbound.BothChapter4 and Chapter 5 consider the lower bounds for the number of maximal subgroups in a finite group. In 2013, the main content of Chapter 4 appeared in Volume 101 of Archiv der Mathematik and was titled, ‘Lower bounds on the number of maximal subgroups in a finite group’ (see [6]). This chapter focuses on producing a lower bound for the number of maximal subgroups in afinitegroupandthenproceedstoimprovetheboundbymakingminimalassumptionson the structure of the group. We conclude this chapter by providing an example of a group which achieves the stated lower bounds.

9 In Chapter 5,weimprovethepreviouslystatedlowerboundsforthenumberof maximal subgroups by making further assumptions on the structure of the group. In particular, the lower bound is improved by partitioning the set of primes which divide the order of the group into three sets. Again, we conclude the chapter by providing examples of groups which achieve the stated lower bounds. The main content of Chapter 5 is currently submitted.

10 CHAPTER 2 A BRIEF BACKGROUND OF FINITE GROUP THEORY 2.1 Introduction

Groups are mathematical objects which measure symmetry. They are studied throughout many areas of mathematics, as well as multiple branches of science. A group is deﬁned below: Deﬁnition 2.1.1. Agroupisanorderedpair(G, ), where G is a set and is a binary ∗ ∗ operation on G satisfying the following axioms:

(i) (a b) c = a (b c), for all a, b, c G ∗ ∗ ∗ ∗ ∈ (ii) there exists an element e G such that for all a G we have a e = e a = a ∈ ∈ ∗ ∗ 1 1 1 (iii) for each a G there is an element a− of G such that a a− = a− a = e. ∈ ∗ ∗ This simple deﬁnition of a group can yield extremely diﬃcult questions which arise

in group theory. In this dissertation, we study finite groups only and part of finite group theory attempts to classify finite groups. One of the simplest ways of classifying finite

groups is by their order, or the number of elements in the group. A list of small groups, classiﬁed by order, can be found in the program Groups, Algorithms, and Programming

(GAP) [4]. A more natural way to classify groups is by their structural properties. These

structural properties form a hierarchy, as seen below: Cyclic Groups Abelian Groups Nilpotent Groups Solvable Groups All Groups ⊂ ⊂ ⊂ ⊂ The simplest structural property is seen in cyclic groups. A cyclic group is a group

which is generated by a single element. Every finite cyclic group is isomorphic to Zn, where n is a positive integer. All cyclic groups are abelian groups, that is for all a, b G ∈ we have a b = b a. Abelian groups are completely classified and the classification can be ∗ ∗ seen in the following theorem: Theorem 2.1.2. Let G be a finite abelian group. Then

G = Zn1 Zn2 Znm , ∼ × ×···×

11 for some integers n1,n2,...,nm satisfying the following conditions:

(1) (i) n 2 for all i 1, 2,...,m i ≥ ∈{ } (ii) n n for j 1, 2,...,m 1 j+1| j ∈{ − } (2) The expression in (1) is unique, up to isomorphism.

Proof. See for example [3,Theorem3.22].

The next category of groups is nilpotent groups. Nilpotent groups enjoy the following equivalent properties:

Theorem 2.1.3. Let G be a ﬁnite group. Suppose p1,p2,...,pn are the distinct primes dividing the order of G and let P be a Sylow p -subgroup of G for i 1, 2,...,n . Then i i ∈{ } the following are equivalent:

(1) G is nilpotent

(2) G = P P P ∼ 1 × 2 ×···× n (3) For all i 1, 2,...,n , P ✂ G ∈{ } i (4) Every maximal subgroup is normal

(5) For every proper subgroup H of G, NG(H) >H.

Proof. See for example [5,Theorem1.26].

The penultimate category of groups is solvable groups. A solvable group is a group with a composition series, all of whose composition factors are cyclic of prime order. The Jordan-Hölder Theorem provides information about the composition factors of a nontrivial finite group. Theorem 2.1.4. (Jordan-Hölder Theorem) Let G be a nontrivial finite group. Then

(1) G has a composition series

(2) The composition factors in a composition series are unique, up to isomorphism.

Proof. See for example [3,Theorem3.22].

12 The composition series of a group assists in breaking down the group into smaller pieces. There are many results which dictate situations when a given group has composition factors which are cyclic of prime order. For example, in 1911, W. Burnside proved the following result: Theorem 2.1.5. (Burnside Theorem) For primes p and q, every group of order paqb is solvable.

Proof. See for example [5,Theorem7.8],aproofusingﬁnitegrouptheory.

It was W. Burnside who conjectured that every nonabelian simple group was of even order. This fact was proven in 1963 by W. Feit and J. Thompson.

Theorem 2.1.6. (Feit-Thompson Theorem) Let G be a group. If G has odd order, then G is solvable.

Proof. See [13].

The remaining category of groups includes all groups, or groups that may or may

not have a composition series with only cyclic composition factors. The composition factors of every finite group are simple groups. Since a group can be constructed from its compositions factors, the Jordan-Hölder Theorem illustrates the importance of understanding finite simple groups. Thus the classification of finite simple groups was

amajorachievement.Theclassificationtookover50yearsandmanymathematiciansto complete. Theorem 2.1.7. (Classification of Finite Simple Groups, 2004) Let G be a finite simple group. Then G is isomorphic to one of the following groups:

(1) A cyclic group of prime order

(2) An alternating group with degree at least 5

(3) A simple group of Lie type

(4) One of the 26 sporadic simple groups.

13 Proof. See [12].

Because the ﬁnite simple groups are known, we can study arbitrary groups through their composition factors. It is natural to now focus on studying the maximal subgroups of ﬁnite groups. 2.2 Maximal Subgroups

One way to investigate the structure of a finite group is to study its maximal subgroups. Since before the classification of finite simple groups, researchers have been studying the maximal subgroups of all finite groups. The definition of maximal subgroups is seen below: Definition 2.2.1. AsubgroupM of a group G is called a maximal subgroup if M = G ̸ and the only subgroups containing M are M and G. In nontrivial finite groups, maximal subgroups will always exist because the subgroups form a partially ordered set under inclusion. Since the set of subgroups is finite, this partially ordered set will have a maximal element. However, not all groups will have maximal subgroups. For example, the rational numbers under addition have no maximal subgroups.

We will proceed by stating some well-known results about maximal subgroups of ﬁnite groups. We may use some of these results in the later chapters without explicitly stating so. Lemma 2.2.2. Let H be a proper subgroup of a ﬁnite group G. Then there is a maximal subgroup of G containing H.

Proof. See for example [3,Exercise2.4.16].

In addition to every proper subgroup being contained in a maximal subgroup, maximal subgroups of certain ﬁnite groups enjoy some very nice properties. A few of these properties can be seen in the lemmas below:

14 Lemma 2.2.3. Let G be a ﬁnite group. If M is a maximal subgroup of G, then M is normal in G or M is self-normalizing.

Proof. See for example [3,Exercise4.3.23].

Lemma 2.2.4. Every maximal subgroup of a ﬁnite nilpotent group is normal of prime index.

Proof. See for example [3,Theorem3.22]and[5,Theorem1.20].

Lemma 2.2.5. Every maximal subgroup of a ﬁnite solvable group has prime power index.

Proof. See for example [5,Exercise3B.1].

An important subgroup of a finite group involving its maximal subgroups is called the Frattini subgroup. Definition 2.2.6. Let G be a finite group. The Frattini subgroup of G,denotedΦ(G), is the intersection of the maximal subgroups of G. Next, we state some properties of Frattini subgroups of finite groups.

Lemma 2.2.7. The Frattini subgroup of a ﬁnite group is nilpotent.

Proof. See for example [3,Exercise6.1.25].

Lemma 2.2.8. Let G be a ﬁnite group. Then Φ(G) is a characteristic subgroup of G.

Proof. See for example [3,Exercise6.1.21].

Lemma 2.2.9. Let p be a prime number. If P is a p-group, then P/Φ(P ) is elementary abelian.

Proof. See for example [5,Exercise1D.8].

Understanding the maximal subgroups of a ﬁnite group can provide a wealth of information into the group itself. For example, under certain conditions, knowing only the number of maximal subgroups in a group is enough to identify the group. The maximal subgroups of ﬁnite simple groups have been researched and they are understood. The

15 maximal subgroups of small finite simple groups are listed in the Atlas of Finite Groups [16]. More general situations have also been considered. In [1], M. Aschbacher classifies the maximal subgroups of classical groups. He proves that every maximal subgroup of classical groups belongs to one of eight classes of subgroups or has a generalized Fitting subgroup which is a nonabelian simple group. In [7], P. Kleidman and M. Liebeck study the eight classes introduced by Aschbacher. We state their results below. Set G = PGL(n, q)andletV be the underlying n-dimensional vector space over GF (q), a finite field with q elements.

C1: Stabilizers of subspaces of V

C2: Stabilizers of direct sum decompositions of V into subspaces of the same dimension

C3: Stabilizers of extension ﬁelds of GF (q)ofprimedegree

C : Stabilizers of tensor product decompositions V = V V 4 1 ⊗ 2

C5: Linear groups over subﬁelds of prime index

C6: Normalizers of r-groups of symplectic type, where r is a prime diﬀerent from p

t C7: Stabilizers of tensor product decompositions V = i=1Vi,whereVi all have the same dimension ⊗

C8: Classical groups Theorem 2.2.10. (Aschbacher’s Theorem) Let H be a subgroup of PGL(n, q) not containing PSL(n, q). Then one of the following holds:

(1) H is contained in a member of one of the classes C1,C2,...,C8

(2) H is almost simple and is induced by an absolutely irreducible subgroup modulo scalars.

Proof. See [1].

It turns out that more information about H can be stated if it is not a member of one

of the classes C1,C2,...,C8.

16 Theorem 2.2.11. Let H fall under case (2) in Aschbacher’s Theorem (Theorem 2.2.10). Then either H is or , with m = n +1or m = n +2, or H has order at most q3n. Sm Am Proof. See [7].

The O’Nan-Scott Theorem gives a classiﬁcation of the maximal subgroups of the symmetric group. Theorem 2.2.12. (O’Nan-Scott Theorem) If H is any proper subgroup of (except ), Sn An then H is a subgroup of one or more of the following subgroups:

(1) An intransitive group , where n = k + m Sk ×Sm (2) An imprimitive group , where n = km Sk ≀Sm (3) A primitive wreath product, , where n = km Sk ≀Sm d d (4) An aﬃne group AGLd(p) ∼= p : GLd(p), where n = p m (5) A group of shape T .(Out(T ) m), where T is a non-abelian simple group, acting ×S m 1 on the cosets of the diagonal subgroup Aut(T ) , where n = T − ×Sm | | (6) An almost simple group acting on the cosets of a maximal subgroup.

Proof. See [11].

Because the maximal subgroups of a ﬁnite group contain such a vast amount of

information, there are currently open questions regarding the maximal subgroups of an arbitrary ﬁnite group.

2.3 Notation and Some Deﬁnitions

One open question regarding the maximal subgroups of a finite group is the following: With limited assumptions on the structure of a finite group G,whatcanwesayaboutthe number of maximal subgroups in G?Toaidinansweringthisquestion,wewillintroduce the following definitions and notation. Definition 2.3.1. Let H be a subgroup of the finite group G.AsubgroupK of G is called a complement for H in G if G = HK and H K =1. ∩

17 Definition 2.3.2. A Hall subgroup of a finite group is a subgroup whose order is coprime to its index. Definition 2.3.3. Let π be any set of prime numbers. A finite group G is said to be

π-separable, if there exists a normal series where each factor is a π-group or a π′-group. Throughout this dissertation, we use the following notation. Below, we assume that G is a ﬁnite group and p is a prime number.

✵ (G)= xp : x G • 1 ⟨ ∈ ⟩ Aut(G)denotestheautomorphismgroupofG • C (H)denotesthecentralizerofasubgroupH in G • G 1 1 [K, L]= k− l− kl : k K, l L ,whereK and L are subgroups of G • ⟨ ∈ ∈ ⟩ m(G)denotesthesetofmaximalsubgroupsofG • m (G)denotesthesetofmaximalsubgroupsofG whose index in G is a power of p • p N (H)denotesthenormalizerofasubgroupH in G • G n (G)denotesthenumberofSylowp-subgroups of G • p Φ(G)denotestheFrattinisubgroup • π(G)denotesthesetofprimeswhichdividetheorderofG • Syl (G)denotesSylowp-subgroups of G • p Ω (G)= x G : xp =1 • 1 ⟨ ∈ ⟩ Out(G)denotestheouterautomorphismgroupofG • Op(G)denotesthesmallestnormalsubgroupofG whose index is a power of p • O (G)denotesthelargestnormalp-subgroup of G • p Z(G)denotesthecenterofG • 2.4 On the Number of Maximal Subgroups

For some groups G,wecanmakeminimalassumptionsonitsstructureandstill calculate the exact number of maximal subgroups. We discuss some such situations below.

18 When G is a ﬁnite cyclic group it is easy to see that m(G) = π(G) .Ifp is a prime | | | | number and G is a noncyclic p-subgroup, then m(G) can be calculated too. For such a | | pk 1 G,ifG/Φ(G)isanelementaryabeliangroupofrankk where k 2, then m(G) = p −1 . ≥ | | − Because the number of maximal subgroups can be calculated for all p-groups, the number of maximal subgroups can also be calculated for all nilpotent groups. We shall see the number of maximal subgroups of a nilpotent group is the sum of the number of maximal subgroups of each of its Sylow subgroups. It is natural to next consider the number of maximal subgroups for both solvable and nonsolvable groups. However, with these assumptions we cannot expect to calculate the exact number of maximal subgroups for an arbitrary group. However, if given a ﬁnite group, it is possible to calculate the number of maximal subgroups. Resources like GAP

[4]andtheAtlas of Finite Groups [16]canbeusedtoaidinsuchatask.Weseekto

find the number of maximal subgroups of a general finite group, while making minimal structural assumptions. Thus we attempt to narrow the range for the possible number of maximal subgroups of an arbitrary finite group. One way to approach this is to consider both upper and lower bounds for the number of maximal subgroups in a finite group.

19 CHAPTER 3 UPPER BOUNDS FOR THE NUMBER OF MAXIMAL SUBGROUPS 3.1 Introduction

There have been several mathematicians who estimated the number of maximal subgroups in finite groups. In particular, their work focuses on the upper bound for the number of maximal subgroups in finite solvable groups. The upper bound for the number of maximal subgroups has gradually been sharpened by making further and further assumptions on the group. The first upper bound, proven by G. E. Wall, only considered the order of the group. Together R. J. Cook, J. Wiegold and A. G. Williamson also considered the smallest prime that divides the order of the group. Later M. Herzog and O. Manz studied the smallest prime and largest prime which divide the order of the group as well as the order of the group. Finally, B. Newton used all of the primes that divide the order of the group, along with their multiplicities, to state his upper bound for the number of maximal subgroups of finite group. All of the aforementioned results are explicitly stated in Section 3.2 below. In this chapter, we make more assumptions on the structure of a finite group to further improve the upper bound for the number of maximal subgroups in a finite solvable group. For a finite group G,recallwedenoteOp(G)asthesmallestnormalsubgroupofG whose index in G is a power of p and denote mp(G)asthesetofmaximalsubgroupsofG whose index in G is a power of p.Ourmainresultofthischapteristhefollowing: Theorem 3.1.1. Let G be a finite solvable group and suppose p π(G).IfP Syl (G) is ∈ ∈ p cyclic, then m (G) 1,p . More precisely, | p |∈{ }

p, if Op(G)=G mp(G) = ⎧ . | | p ⎨⎪ 1, if O (G)

⎩⎪

20 3.2 Upper Bounds Proven by Others

For solvable groups, during the last fifty years estimations on the upper bound for the number of maximal subgroups have been gradually sharpened. The earliest such estimate was proven by G. E. Wall. He computed an upper bound for the number of maximal subgroups in a finite solvable group in [14]. Theorem 3.2.1. (Wall) Let G be a finite solvable group. Then m(G) < G . | | | | Proof. See [14].

Almost thirty years later, R. J. Cook, J. Wiegold and A. G. Williamson improved Wall’s result in [15].

Theorem 3.2.2. (Cook, Wiegold, Williamson) Let G be a ﬁnite solvable group and suppose p the smallest prime in π(G). Then

G 1 m(G) | |− . | |≤ p 1 − This bound is achieved if and only if G is elementary abelian.

Proof. See [15].

The largest possible number of maximal subgroups in a solvable group was narrowed further in [8] by M. Herzog and O. Manz. Theorem 3.2.3. (Herzog-Manz) Let G be a ﬁnite solvable group with Frattini subgroup Φ(G). We denote by q the largest prime in π(G) and by p the smallest prime in π(G). Then q G/Φ(G) p m(G) | |− . | |≤ p(q 1) − Proof. See [8].

The most accurate result to date was proven by B. Newton in [10], assuming that the

finite group is solvable. For a prime p,hefirstprovesaboundforthenumberofmaximal subgroups in a finite solvable group with p-power index.

21 Theorem 3.2.4. (Newton) Let p be a prime number and let G be a ﬁnite solvable group of order pkm, where p ! m. Suppose that [G : Op(G)] = pr. Then

r k r+1 p 1 p − p m (G) − + − . | p |≤ p 1 p 1 − − pk+1 p Moreover, m (G) is never greater than − , and if Op(G)

B. Newton uses his theorem above to bound the number of maximal subgroups for a ﬁnite solvable group.

Corollary 3.2.5. (Newton) Let G be a ﬁnite solvable group with G = pr1 pr2 prm for | | 1 2 ··· m distinct primes p ,p ,...,p , where pr1 =min pri :1 i m . Then 1 2 m 1 { i ≤ ≤ } m pr1 1 pri+1 p m(G) 1 − + i − i . | |≤ p 1 p 1 1 i=2 i − % − Proof. See [10].

3.3 Preliminary Results

In this section, we collect some results to aid in the proof of Theorem 3.1.1.

Lemma 3.3.1. (Newton) Let G be a ﬁnite solvable group and let p be a prime number dividing G . Suppose that M and M are inconjugate maximal subgroups of G, both | | 1 2 of which have p-power index in G, and neither of which is normal in G. Then (M 1 ∩ M ) Op(G)=G. Furthermore, if P Syl (Op(G)), then (M P )(M P )=P . 2 0 ∈ p 1 ∩ 0 2 ∩ 0 0 Proof. See [10].

Lemma 3.3.2. Let G be a ﬁnite solvable group and suppose that p π(G). Let M be a ∈ maximal subgroup of G such that [G : M]=pn.IfP Syl (G) is cyclic, then n =1. ∈ p

g Proof. Let U = g G M .ThenU is the largest normal subgroup of G contained in M. ∈ Let V/U be a chief& factor of G.ThenMV G because V is normal in G.SinceM is ≤

22 maximal and M

3.4 Proof of Main Theorem

We now prove the main result. Proof of Theorem 3.1.1. Let be the set of maximal subgroups of G with p-power index. M Partition into the following two sets: = M : M ✂ G and = M : M N { ∈M } L { ∈M M G .Tobound m (G) ,wewillbound and separately. ̸! } | p | |N | |L| Let M contain Op(G). Denote M/Op(G)byM¯ and G/ Op(G)byG¯.ThenM¯ ∈M is a maximal subgroup of G¯.SinceG¯ is a p-group, [G¯ : M¯ ]=p and M¯ ✂ G¯.Therefore M ✂ G and M .SothesetofmaximalsubgroupsofG which each contain Op(G)is ∈N .SinceG/ Op(G)isacyclicp-group, N

0, if Op(G)=G = ⎧ . |N | p ⎨⎪ 1, if O (G)

Now suppose that M ,M are⎩⎪ not conjugate. If P Syl (Op(G)) then 1 2 ∈L 0 ∈ p G = M P .Therefore[G : M ]=[P : M P ]. Since P P , P is cyclic. So at least one i 0 i 0 i ∩ 0 0 ⊆ 0 of M P or M P must be P ,asP =(M P )(M P )byTheorem3.3.1. Without 1 ∩ 0 2 ∩ 0 0 0 1 ∩ 0 2 ∩ 0 loss of generality, suppose M P = P .ByLemma3.3.2,wehave[G : M ]=p and so 1 ∩ 0 0 1

p =[G : M ]=[P : M P ]=1, 1 0 1 ∩ 0 acontradiction.Thereforeallsubgroupsin are conjugate in G. L If Op(G)=G,then = and so = as G is nontrivial. Therefore m (G) = N ∅ L̸ ∅ | p | =[G : M ]=p. Finally, suppose that Op(G)isapropersubgroupofG.Fora |L| 1 p contradiction, assume =0andletM .ThenO (G) M = G.IfQ Sylp(M ), |L| ̸ L ∈L L ∈ L then p p O (G) p M p O (G) p Q p L G p = | p | | | = | p | | | = O (G) Q p. | | O (G) M p O (G) Q | | | ∩ L| | ∩ |

23 Therefore G = Op(G) Q.LetR Syl (G)withR Q and assume R = Q.ThenQ Rp, ∈ p ⊇ ̸ ⊆ so that (Q Op(G))/ Op(G) (G/ Op(G))p,acontradiction.ThusR = Q,contradicting ⊆ our choice of M .Therefore = and mp(G) = =1,asdesired. L L ∅ | | |N |

24 CHAPTER 4 LOWER BOUNDS FOR THE NUMBER OF MAXIMAL SUBGROUPS 4.1 Introduction

The aforementioned results regarding upper bounds for the number of maximal subgroups of finite solvable groups are valuable in bounding the number of maximal subgroups in a finite group. To continue to narrow the range for the possible number of maximal subgroups of a given finite group, we now consider the smallest possible number of maximal subgroups. In this chapter, we produce three lower bounds for the number of maximal subgroups in any finite group. Fix a positive integer m.Itisnaturaltosuspectthatamongallgroupsoforderm, m(G) is smallest when G is a cyclic group. If G is the cyclic group, then the number of | | maximal subgroups is equal to the number of distinct primes which divide the order of the group. We confirm both facts below (Corollary 4.1.2).

Given a set of prime numbers τ with τ = n,weinvestigatelowerboundsfor m(G) | | | | when G is a finite noncyclic group with π(G)=τ.Weobtainthefollowingmaintheorem: Theorem 4.1.1. Let n be a positive integer and suppose that G is a finite noncyclic group such that π(G) = n. Assume that p π(G) is the smallest prime which divides G . Then | | ∈ | | m(G) n + p. Furthermore, if G has a noncyclic Sylow subgroup and q π(G) is the | |≥ ∈ smallest prime such that Q Syl (G) is noncyclic, then m(G) n + q. ∈ q | |≥ The lower bounds in Theorem 4.1.1 are best possible and this fact is confirmed in Proposition 4.3.5 below. An immediate consequence of this theorem is the following lower bound on the number of maximal subgroups in an arbitrary finite group. Note that it also yields a new characterization of cyclic groups. Corollary 4.1.2. If G is a finite group G, then m(G) π(G) . Furthermore, equality | |≥| | holds if and only if G is cyclic. 4.2 Preliminary Results

Here, we collect some results needed to prove Theorem 4.1.1.

25 Deﬁnition 4.2.1. AgroupG is supersolvable if there exists normal subgroups Ni with

1=N N N = G, 0 ⊆ 1 ⊆···⊆ r where each factor Ni/Ni 1 is cyclic for 1 i r. − ≤ ≤ Theorem 4.2.2. If all the Sylow subgroups of a ﬁnite group G are cyclic, then G is supersolvable.

Proof. By [5,Theorem5.16],bothG′ and G/G′ are cyclic. Thus G is supersolvable.

Theorem 4.2.3. Every minimal normal subgroup M of a ﬁnite group is a direct product of isomorphic simple groups. Furthermore, if M is solvable then it is an elementary abelian p-group for some prime p.

Proof. See for example [5, Exercise 2A.5 and Lemma 3.11].

Theorem 4.2.4. Let G be a ﬁnite group. If G is not solvable and P Syl (G), then P is ∈ 2 noncyclic.

Proof. Suppose P Syl (G)iscyclic.ThenG has a normal 2-complement N (see for ∈ 2 example [5,Corollary5.14]).SinceN is of odd order, it is solvable by the Feit-Thompson Theorem. And because G/N is cyclic, G is also solvable.

Theorem 4.2.5. (Schur-Zassenhaus) Every normal Hall subgroup of a ﬁnite group has a complement.

Proof. See for example [5,Theorem3.8].

Theorem 4.2.6. (Maschke) Let G be a ﬁnite group and let F be a ﬁeld whose characteristic does not divide G .IfV is any FG-module and U is any submodule of V , then V has | | a submodule W such that V = U W . ⊕ Proof. See for example [3,Sec.18.1Theorem1].

The following lemmas in this section are useful in the proof of Theorem 4.1.1.

26 Lemma 4.2.7. Suppose that G is a ﬁnite group, and let π(G)= p ,p ,...,p . For each { 1 2 n} i 1, 2,...,n , let P Syl (G).IfG = P P P , then ∈{ } i ∈ pi 1 × 2 ×···× n

m(G) = m(P ) + m(P ) + + m(P ) . | | | 1 | | 2 | ··· | n |

Proof. Note that because G is nilpotent, all maximal subgroups of G will have prime index in G.Foreachi 1, 2,...,n ,letm (G)denotethesetofmaximalsubgroupsofG with ∈{ } pi index p .Deﬁneϕ : m (G) m(P )byϕ (M)=P M,forallM m (G). Since i i pi → i i i ∩ ∈ pi

G = PiM, [P : P M]=[G : M ]=p . i i ∩ i i Then P M ✂ P and so P /(P M)isofprimeorder.ThusP M m(P ). i ∩ i i i ∩ i ∩ ∈ i Now deﬁne ψ : m(P ) m (G)by i i → pi

ψ (N )=P N P , i i 1 ×···× i ×···× n for all N m(P ). Since N is a maximal subgroup of a p -group, N ✂ P and [P : N ]= i ∈ i i i i i i i p .Thusψ (N ) ✂ G and [G : ψ (N )] = p .Soψ (N ) m (G). i i i i i i i i ∈ pi Since ϕ and ψ are inverses of each other, ϕ is a bijection and m (G) = m(P ) . i i i | pi | | i | Observe that

m(G) = m (G) + m (G) + + m (G) , | | | p1 | | p2 | ··· | pn | because every maximal subgroup of G has prime index. Thus

m(G) = m(P ) + m(P ) + + m(P ) , | | | 1 | | 2 | ··· | n | as desired.

Lemma 4.2.8. Let N be a nonabelian minimal normal subgroup of a ﬁnite group G. Let p be the largest prime in π(N). Then there are at least p maximal subgroups of G which do not contain N.

27 Proof. Let P be a Sylow p-subgroup of N.SetN0 = NG(P ). Then G = NN0 by the Frattini Argument. Because P N and N is a minimal normal subgroup of G, N = G ⊂ 0 ̸ and there exists a maximal subgroup M of G containing N .ObservethatN M 0 ̸⊆ because otherwise G = NM M.ThereforeM is a maximal subgroup of G which ⊆ does not contain N.Infact,sinceN is the direct product of isomorphic simple groups by Theorem 4.2.3,thereexistsasimplenormalsubgroupS of N which is not contained in M. Set R = S M and n =[S : R]. Since S acts by left multiplication on the set of ∩ left coset of R in S,thereexistsanontrivialhomomorphismfromS to the symmetric group on n elements. Then, because S is a simple group, S must be isomorphic to a subgroup of the alternating group on n elements. Thus p n.SinceS = R =1,R ≤ ̸ ̸ is not a normal subgroup of S.ItfollowsthatM is not a normal subgroup of G.Then

M is self-normalizing because it is a maximal subgroup of G.Thusthereareatleastp

S-conjugates of M,yieldingatleastp conjugates of M in G.ThereforeG has at least p maximal subgroups which do not contain N,asdesired.

Lemma 4.2.9. Let n 3 be an integer. Suppose S is a set of n distinct primes. Assume ≥ that p and q are the largest and smallest elements of S, respectively. Then p n + q. ≥ Proof. Observe that all elements of S are at least 2 units apart from each other, except for possibly the two smallest elements of S which are 1 unit apart. Therefore

p q 2(n 2) + 1, − ≥ − which is at least n as n 3. So p q n and p n + q,asdesired. ≥ − ≥ ≥ Lemma 4.2.10. Let G be a ﬁnite group with a noncyclic Sylow subgroup. Suppose that p π(G) is the smallest prime such that P Syl (G) is noncyclic. If N is a solvable ∈ ∈ p minimal normal subgroup of G and all Sylow subgroups of G/N are cyclic, then N is a p-group.

28 Proof. Suppose that N is a q-group, where q π(G)andp = q.ThenP N =1and ∈ ̸ ∩ P = PN/N G/N,bytheSecondIsomorphismTheorem.SincePN/N Syl (G/N), ∼ ≤ ∈ p PN/N is cyclic and cannot be isomorphic to the noncyclic group P ,acontradiction. Therefore, N is a p-group.

Lemma 4.2.11. Let N be a solvable normal subgroup of a ﬁnite group G. Suppose that H is a Hall subgroup of N. Then G = NG(H)N.

Proof. Let g G.SinceHg N g = N and Hg has the same order of a Hall subgroup of ∈ ⊆ N, Hg is a Hall subgroup of N.ThereforeH and Hg are N-conjugate, say Hgn = H for some n N.Thengn N (H)and ∈ ∈ G

1 g N (H)n− N (H)N. ∈ G ⊆ G

Therefore G = NG(H)N,asdesired.

Lemma 4.2.12. Let G be a ﬁnite solvable group. If H is a Hall subgroup of G and K is a subgroup which contains NG(H), then K is self-normalizing.

Proof. Let g G satisfy Kg = K.ThenHg is a Hall subgroup of K. Since all Hall ∈ subgroups of K are conjugate, there exists n K such that Hgn = H.Sogn N (H) ∈ ∈ G and

1 g N (H)n− K. ∈ G ⊆ Therefore K is self-normalizing.

4.3 Proof of Main Theorem

With the goal of proving Theorem 4.1.1,weﬁrstinvestigate m(G) in a cyclic group | | G. Theorem 4.3.1. For a ﬁnite cyclic group G, m(G) = π(G) . | | | |

29 α1 α2 αn Proof. Let G be a ﬁnite cyclic group. Then G is isomorphic to Zp Zp Zpn , 1 × 2 ×···× where G = pα1 pα2 pαn .ByLemma4.2.7, | | 1 2 ··· n

α1 α2 αn m(G) = m(Zp ) + m(Zp ) + + m(Zpn ) . | | | 1 | | 2 | ··· | |

αi α 1 For each pi π(G), Zp has a unique maximal subgroup isomorphic to Zp i− and so ∈ i i

Next, we prove Theorem 4.1.1 when G is a p-group and then in the case where all

Sylow subgroups of G are cyclic. Lemma 4.3.2. Let p be a prime number and suppose that P is a noncyclic p-subgroup. Let P/Φ(P ) be the elementary abelian group of rank k, where k 2 and Φ(P ) denotes the ≥ Frattini subgroup of P . Then pk 1 m(P ) = − . | | p 1 − In particular, m(P ) p +1. | |≥ Proof. Since P/Φ(P )isanelementaryabeliangroupofrankk,wehave

pk 1 m P/Φ(P ) = − . p 1 − ' ( )' ' ' Because Φ(P )iscontainedineverymaximalsubgroupofP ,

pk 1 m(P ) = m P/Φ(P ) = − . | | p 1 − ' ( )' pk 1 ' ' pk 1 Observe that p −1 is smallest when k =2.Inthiscase, p −1 = p +1,sothat m(P ) − − | |≥ p +1.

Theorem 4.3.3. Suppose G is a ﬁnite noncyclic group such that π(G) = n and let | | p π(G) be the smallest prime dividing G . If all Sylow subgroups of G are cyclic, then ∈ | | m(G) n + p. | |≥

30 Proof. Suppose this is false. Let G be a counterexample of minimum order. Then m(G)

Therefore π(G/N)=π(G) q and N is a cyclic Sylow q-subgroup. Let H be −{ } acomplementofN in G,whichexistsbytheSchur-ZassenhausTheorem.ThenH is a maximal subgroup of G.Lett π(G/N)bethesmallestprimedividing G/N ,andnotice ∈ | | that t p.IfG/N is noncyclic, then ≥

m(G/N) n 1+t n 1+p | |≥ − ≥ − by induction. Therefore G has at least n 1+p maximal subgroups that contain N. − Thus m(G) n + p because H is maximal, a contradiction. Hence G/N is cyclic. | |≥ Observe that H is not normal in G as G is noncyclic. So H is self-normalizing and has [G : H]conjugatesinG,allofwhicharemaximalanddonotcontainN.SinceH is not normal in G and p is the smallest prime that divides G ,[G : H] >p.ByTheorem4.3.1, | | m(G/N) = n 1. These n 1maximalsubgroupsofG/N will correspond to at least | | − − n 1maximalsubgroupsofG which contain N.Thus m(G) >n 1+p,aﬁnal − | | − contradiction.

Finally, we prove the remaining case of Theorem 4.1.1.

31 Theorem 4.3.4. Let G be a ﬁnite noncyclic group such that π(G) = n, where n 2. | | ≥ Assume that G has a noncyclic Sylow subgroup and p π(G) is the smallest prime such ∈ that P Syl (G) is noncyclic. Then m(G) n + p. ∈ p | |≥ Proof. Suppose this is false. Let G be a counterexample of minimum order. Then m(G)

m(G) n π(N) + s n π(N) + π(N) +2=n +2, | |≥ −| | ≥ −| | | | acontradiction.

Therefore, N must be solvable. Since m(G) pand

m(G/N) n 1+q n + p. | |≥ − ≥ In either case, we obtain n + p maximal subgroups of G/N,whichwillcorrespondtoat least n + p maximal subgroups of G which contain N,acontradiction.ThereforeallSylow

32 subgroups of G/N are cyclic. So G/N is supersolvable and thus G is solvable. Then G has a Hall p′-subgroup, say H.

For each prime q π(G)withq = p,letH be a Hall q′-subgroup containing P . ∈ ̸ q Let M be a maximal subgroup which contains H ,yieldingn 1distinctmaximal q q − subgroups of G.IfH ✂ G,thenG/H is a noncyclic p-group and Lemma 4.3.2 implies m(G/H) p +1.Thesep +1maximalsubgroupsofG/H will correspond to at least | |≥ p +1maximalsubgroupsofG that contain H,andarethereforedistinctfromthen 1 −

Mq’s. Thus m(G) (n 1) + (p +1)=n + p, | |≥ − acontradiction.

Therefore H is not normal in G.LetMp be a maximal subgroup of G which contains

NG(H). Then Mp is self-normalizing by Lemma 4.2.12.ThereforeMp has at least p conjugates all of which are maximal. Since m(G)

We claim that P ✂ G and thus Op(G)=P .SinceMp is self-normalizing and Mq ✂ G for all q π(G)withq = p we have m(G) = n 1+p,namelyM for q = p and the ∈ ̸ | | − q ̸ conjugates of M .SupposeN (P ) = G.LetM be a maximal subgroup of G.Observe p G ̸ that P is not contained in the p conjugates of Mp,eachhasindexp in G.SoM = Mq for some q π(G)withq = p.ButifM N (P )thenitisself-normalizing,acontradiction. ∈ ̸ q ⊇ G

Therefore P ✂ G and Op(G)=P . We also claim that P must be elementary abelian. First, suppose that P is not abelian. Then P/P′ is not cyclic and P ′ is trivial by minimality. So P is abelian. Next, assume that P is not elementary abelian. Observe that P/Ω1(P )mustbecyclicby minimality. Hence ✵ (P )iscyclicandP/ Ω (P ) ✵ (P ) is not cyclic. By minimality, 1 1 ∩ 1 this forces Ω (P ) ✵ (P )tobetrivial.So( P is elementary) abelian. 1 ∩ 1

33 Suppose that H acts irreducibly on P .ThenN (H) 1 ,P .SinceH is not P ∈ { } normal in G, N (H) = P .IfN (H)istrivial,thenH is maximal* and+ P ̸ P

[G : H]= P p2. | |≥

So H has at least p2 conjugates, all of which are maximal and do not contain P .Therefore

m(G) (n 1) + p2 >n+ p, | |≥ − acontradiction.ThusH does not act irreducibly on P .ThenP has a nontrivial proper

H-invariant subgroup P1.ByMaschke’sTheorem,thereexistsanontrivialproper H-invariant subgroup P of P such that P = P P . Now H must act non-trivially 2 1 × 2 on at least one of P1 or P2. Without loss of generality, assume H acts non-trivially on P1.

Let M be a maximal subgroup which contains NG(H). Then the p conjugates of M will not contain P1.ThereforeifMˆ is a maximal subgroup which contains P1H,itisdistinct from the aforementioned n 1+p maximal subgroups of G. Hence −

m(G) (n 1+p)+1=n + p, | |≥ − aﬁnalcontradiction.

Theorem 4.1.1 follows immediately from Lemma 4.3.2,Theorem4.3.3 and Theorem 4.3.4. We conclude this section by proving the lower bound in Theorem 4.1.1 is best possible.

Proposition 4.3.5. Let n be a positive integer. Given any set of distinct primes p ,p ,...,p where p is the smallest, there exists a ﬁnite group G with π(G)= { 1 2 n} 1 p ,p ,...,p and m(G) = n + p . { 1 2 n} | | 1 Proof. Consider the noncyclic group

G = Zp1 Zp1 Zp2 Zpn . × × ×···×

34 By Lemma 4.3.2, m(Zp1 Zp1 ) = p1 + 1. Also, m(Zpi ) =1fori 2,...,n by | × | | | ∈{ } Lemma 4.3.1.Thus m(G) =(n 1) + (p +1)=n + p | | − 1 1 by Lemma 4.2.7,andthelowerboundinTheorem4.1.1 is best possible.

4.4 A Special Case

We seek to improve the lower bound in Theorem 4.1.1.Thusweconsideraﬁnite group G such that π(G) =2. | | Theorem 4.4.1. Let G be a ﬁnite noncyclic group such that π(G)= p, q , where p

Proof. See for example [3,Corollary4.2.5].

Lemma 4.4.3. (Dedekind’s Lemma) Let H and K be subgroups of a group G, and let H U G, where U is also a subgroup. Then ⊆ ⊆

HK U = H(K U). ∩ ∩

Proof. See for example [5, Lemma X.3].

Now, we can prove Theorem 4.4.1.

Proof. Suppose this is false and let G be a counterexample of minimum order, say G = paqb. Assume P is a Sylow p-subgroup of G and Q is a Sylow q-subgroup of | | G. First, suppose G is nilpotent. Then G is the direct product of its nontrivial Sylow subgroups, G = P Q,byTheorem2.1.3.BecauseG is noncyclic at least one of P or Q is ×

35 also noncyclic. If P is noncyclic, then p2 divides the order of G.Then, m(P ) p +1and | |≥ m(Q) 1sothat m(G) p +2,acontradiction.ThereforeQ must be noncyclic and | |≥ | |≥ possibly p2 ! G .Eitherway, m(Q) q +1and m(P ) 1, forcing m(G) q +2,a | | | |≥ | |≥ | |≥ contradiction. Therefore G is not nilpotent. Then G has a maximal subgroup M which is not normal by Theorem 2.1.3.ThusM has [G : NG(M)] = [G : M]conjugatesinG.Byatheorem of Burnside (Theorem 2.1.5), G is solvable and so [G : M]isaprimepower.Ifp [G : M], | then by Lemma 4.4.2 p2 [G : M]becauseM is not a normal subgroup of G.Thusp2 | divides the order of G and m(G) p2 p +2,acontradiction.Thereforep2 does not | |≥ ≥ divide the order of G and p ! [G : M]. But [G : M] =1,sothatq [G : M]. Hence ̸ | m(G) q. Notice that Q is not contained in any conjugate of M.ThereforeifK is any | |≥ maximal subgroup of G containing Q,thenK is diﬀerent from all the conjugates of M.

Thus m(G) q +1,forcingp ! (q 1). | |≥ −

Let np(G)denotethenumberofSylowp-subgroups of G and let nq(G)denotethe number of Sylow q-subgroups of G.BySylow’sthirdtheorem,n (G) p and n (G) qb. q | p | Then Q ✂ G and n (G) qb.SinceG is not nilpotent n (G) = 1. Also n (G) = q because p | p ̸ p ̸ 2 p ! (q 1). Therefore np(G) q .SinceΦ(Q)isacharacteristicsubgroupofQ and every − ≥ characteristic subgroup of a normal subgroup is normal Φ(Q) ✂ G.SupposeΦ(Q) =1, ̸ Then G/Φ(Q) < G and | | | |

q +2 m(G/Φ(Q)) m(G) , ≤| |≤| | where the ﬁrst inequality holds by minimality, a contradiction. Therefore, Φ(Q)=1and Q is elementary abelian because Q is nilpotent. Let L be a maximal P -invariant subgroup of Q.ThenL ✂ G and by minimality L is trivial. Therefore P acts irreducible on Q and

CQ(P )istrivial.Thus N (P )=P C (P )=P. G × Q

36 Suppose that NG(P )isnotamaximalsubgroupofG.LetM0 be a maximal subgroup of

G containing NG(P ). By Dedekind’s Lemma (Lemma 4.4.3),

M = M N (P )Q = N (P )(M Q), 0 0 ∩ G G 0 ∩ acontradiction.ThereforeNG(P )=P is a maximal subgroups of G,whichhas

[G : N (P )] = [G : P ]=qn q +2, G ≥ aﬁnalcontradiction.

Proposition 4.4.4. Given any two primes p and q such that p

Proof. Observe that m(Zpq Zp) = p +2 and m(Fpq) = q +1 and m(Zpq Zq) = q +2. | × | | | | × | Therefore the bound in Theorem 4.4.1 is best possible.

Naturally, the next situation to consider is π(G) =3,whereG is a ﬁnite group. | | However, this case is much more complicated than the previously discussed situation where π(G) =2.ThusweproceedinChapter5 by not only considering the cardinality of | | π(G), but also partitioning the primes in π(G)intothreesets.

37 CHAPTER 5 SOME FURTHER LOWER BOUNDS FOR THE NUMBER OF MAXIMAL SUBGROUPS 5.1 Introduction

Some lower bounds for the number of maximal subgroups were investigated in the previous chapter. In general, m(G) π(G) and equality holds if and only if G is cyclic. | |≥| | Under further assumptions on the structure of G,thislowerboundwasimproved.IfG is noncyclic then m(G) π(G) + p,wherep π(G)isthesmallestprimethatdivides G . | |≥| | ∈ | | Additionally, if G has a noncyclic Sylow subgroup and q π(G)isthesmallestprimesuch ∈ that Q Syl (G)isnoncyclic,thenitisproven ∈ q

m(G) π(G) + q. (5-1) | |≥| |

Observe that given any set of distinct primes q ,q ,...,q where q is the smallest, there { 1 2 n} 1 exists a ﬁnite group G with π(G)= q ,q ,...,q and { 1 2 n}

m(G) = π(G) + q . | | | | 1

For example, G = Zq1 Zq1q2 qn achieves the lower bound in Inequality 5-1. However, the × ··· previously stated lower bounds can be restrictive because they consider at most one prime in π(G). For instance, let G = 10 Z11 Z143,where 10 is the symmetric group on 10 S × × S letters. Then π(G) =6andG has a noncyclic Sylow 2-subgroup. Therefore q =2and | | Inequality 5-1 yields m(G) 8, however m(G) =4002.ThelowerboundinInequality | |≥ | | 5-1 can be drastically improved with minimal assumptions on the structure of the group. In the present chapter, we improve the existing lower bound for m(G) by partitioning | | π(G)intothreesets.Weconsiderthesetofprimesp for which G is not p-solvable, denoted τ(G), and set of primes q for which G is q-solvable. We further subdivide the latter set of primes into the set of primes which correspond to cyclic Sylow subgroups of G,denotedλ(G), and the set of primes which correspond to noncyclic Sylow subgroups of G,denotedκ(G). Under these considerations we obtain two new lower bounds for

38 m(G) .Inadditiontothesetsλ(G), κ(G)andτ(G), the first lower bound includes a | | further assumption on G.WeassumethatthenonabeliancompositionfactorsofG satisfy atechnicalproperty,namelyPropertyB,whichisdefinedinDefinition5.2.6. Theorem 5.1.1. Let G be a finite group. If the nonabelian composition factors of G satisfy Property B, then

m(G) π(G) + p + p. (5-2) | |≥| | p κ(G) p τ(G) ∈% ∈% Property B is not a very restrictive property and we conclude this chapter by providing some examples of simple groups that satisfy Property B (see Section 5.8). However, not all finite simple groups satisfy Property B and knowing the composition factors of a group may increase the possible lower bound. For a group G,weintroduce below the invariant α(G)whichtakesintoaccountmorefinelytheinfluenceofthe composition factors into our problem. The following theorem does not require any special condition on the composition factors.

Theorem 5.1.2. If G is a ﬁnite group, then

m(G) λ(G) + (p +1)+α(G) (p +1). (5-3) | |≥| | p κ(G) p τ(G) ∈% ∈% Notice that Theorem 5.1.2 implies Theorem 5.1.1 when α(G) 1. The invariant α(G) ≥ is closely connected to the values α( )as runs over the nonabelian composition factors C C of G.Inthepenultimatesection,(seeSection5.7), we analyze α(G)andseethatα( ) C tends to infinity for many families of finite simple groups . Although, we provide many C examples of finite simple groups which satisfy both α( ) > 1andPropertyB,ourresults C do not use the classification of finite simple groups. Both Theorem 5.1.1 and Theorem 5.1.2 improve the lower bound in Inequality

5-1.ReconsideringthegroupG = 10 Z11 Z143,thelowerboundinInequality S × × 5-3 yields m(G) 4001, a vast improvement over the lower bound in Inequality 5-1. | |≥ In most cases Theorem 5.1.2 gives a stronger result than Theorem 5.1.1. However, it

39 is possible for the lower bound in Inequality 5-2 to be sharper than the lower bound in Inequality 5-3 because not all ﬁnite groups G satisfy α(G) 1. For instance, α( ) < 1 ≥ A6 where is the alternating group on 6 letters, and α PSL(2, 7) < 1. Both and A6 A6 PSL(2, 7) satisfy Property B, thus Theorem 5.1.1 is applicable.( ) In such a case, say

G = PSL(2, 7) Z11 Z143,thelowerboundinInequality5-3 yields m(G) 21, but the × × | |≥ lower bound in Inequality 5-2 yields m(G) 28. | |≥ 5.2 Deﬁnitions

We begin with a basic definition. Definition 5.2.1. Let G be a finite group and let p be a prime number. Then G is p-solvable if all of the nonabelian composition factors of G do not have p dividing their order. To improve the existing lower bounds for m(G) ,firstpartitionπ(G)intothreesets: | | λ(G), κ(G)andτ(G). Definition 5.2.2. Let G be a finite group.

Deﬁne λ(G)asthesetofp π(G)suchthatG is p-solvable and P Sylp(G)is • cyclic. ∈ ∈

Deﬁne κ(G)asthesetofp π(G)suchthatG is p-solvable and P Sylp(G)is • noncyclic. ∈ ∈

Define τ(G)asthesetofp π(G)suchthatp divides the order of a nonabelian chief • factor of G. ∈ To aid in counting some of the maximal subgroups of G with index a τ(G)-number, we make the following definitions: Definition 5.2.3. Let S be a nonabelian simple group and assume M is a maximal subgroup of S.WesaythatM satisfies Property A if NAut(S)(M)S = Aut(S). Let mA(S) denote the set of maximal subgroups of S that satisfy Property A. Definition 5.2.4. Let G be a finite group and suppose is a nonabelian composition C factor of G.Defineγ( )= p π( )(p +1)andγ(G)= p τ(G)(p +1). C ∈ C ∈ , ,

40 Deﬁnition 5.2.5. Let G be a ﬁnite group and suppose is a nonabelian composition C m ( ) | A C | factor of G.Setα( )= γ( ) .IfG is not solvable, set C C

α(G)=min α( ): is a nonabelian composition factor of G { C C } and if G is solvable, then set α(G)=0. Definition 5.2.6. Let S be a nonabelian finite simple group. We say that S satisfies Property B if m(G) γ(S), for all subgroups G such that S G Aut(S). | |≥ ≤ ≤ 5.3 Preliminary Results

In this section, we collect some number theoretic results which will aid us in bounding both the number of maximal subgroups of a group that satisfy Property A as well as groups which satisfy Property B. Lemma 5.3.1. Let n 3 be an integer. Suppose S is a set of n distinct primes. Assume ≥ that p and q are the largest and smallest elements of S, respectively. Then p n + q. ≥ Proof. Observe that all elements of S are at least 2 units apart from each other, except for possibly the two smallest elements of S which are 1 unit apart. Therefore p q − ≥ 2(n 2) + 1, which is at least n as n 3. So p q n and p n + q,asdesired. − ≥ − ≥ ≥ Lemma 5.3.2. Let l 3 be an integer. If p ,p ,...,p are prime numbers such that ≥ 1 2 l p1 p +1foralli 1, 2,...,l 1 . Hence ≥ l i ∈{ − } l 1 l 1 − − p l>p (p +1) p + (p +1) l l i ≥ l i i=1 i=1 - % and thus l p l (p +1), l ≥ i i=1 % as desired.

41 Lemma 5.3.3. Let q =2f 8 and suppose n q 1,q+1 .Ifn is not a prime number, ≥ ∈{ − } then n (p +1)+2. ≥ p n %| Proof. Suppose n = p p p ,wherep p for all i 2, 3,...,l .Ifn =9,theresult 1 2 ··· l 1 ≥ i ∈{ } is clear. Thus we assume n>9andproceedbyinductiononl.Sincen is odd and q is a power of two, p 5. Then (p 1)(p 1) 5andsothat 1 ≥ 1 − 2 − ≥ 2 p p (p +1)+2. 1 2 ≥ i i=1 % Thus we suppose l 1 − p1p2 pl 1 (pi +1)+2. ··· − ≥ i=1 % Because (p1p2 pl 1 1)(pl 1) 2, ··· − − − ≥ n = p p p 1 2 ··· l

p1p2 pl 1 + pl +1 ≥ ··· − l 1 − (p +1)+2+p +1 ≥ i 1 i=1 %l

= (pi +1)+2 i=1 % (p +1)+2, ≥ i p n %| as desired.

Lemma 5.3.4. If q =2f 8, then ≥

2(q +1) (p +1). ≥ p (q3 q) |%−

42 Proof. To prove this result, we divide the proof into three cases. First suppose q +1is prime. Then q 1isnotaprimenumberand −

q 1 (p +1)+2 − ≥ p (q 1) |%− by Lemma 5.3.3. Adding (q +1)+2tobothsidesoftheinequalityaboveyields

2(q +1) (p +1). ≥ p (q3 q) |%− Thus assume that q 1isprime.Thenq +1isnotprimeand −

q +1 (p +1)+2 ≥ p (q+1) |% by Lemma 5.3.3. Adding (q 1) + 2 to both sides of the inequality above yields −

2(q +1) (p +1). ≥ p (q3 q) |%− Finally, suppose that q +1norq 1isprime.ThenLemma5.3.3 implies −

q +1+q 1 (p +1)+4 − ≥ p (q2 1) |%− and 2(q +1) (p +1), ≥ p (q3 q) |%− as desired.

5.4 m(G) for p-Solvable Groups | | For a ﬁnite solvable group, we will make use of the well known fact that every maximal subgroup has prime powered index. For p λ(G), we count the number of ∈ maximal subgroups of G by counting the number of maximal subgroups whose index in G is a power of p.Recalldenotesmp(G)thesetofmaximalsubgroupsofG whose index in G is a power of p.Theﬁrsttwolemmasofthissectionarestandardresultswhichwillaid us in bounding m (G) from below. | p |

43 Lemma 5.4.1. Let G be a ﬁnite p-solvable group, where p π(G).IfH is a Hall ∈ p′-subgroup of G and K is a subgroup which contains NG(H), then K is self-normalizing.

g g Proof. Let g G satisfy K = K.ThenH is a Hall p′-subgroup of K. Since all Hall ∈ gn p′-subgroups of K are conjugate, there exists n K such that H = H.Sogn N (H) ∈ ∈ G and

1 g N (H)n− K. ∈ G ⊆ Therefore K is self-normalizing.

Lemma 5.4.2. Let N be a solvable normal subgroup of a ﬁnite group G. Suppose that H is a Hall subgroup of N. Then G = NG(H)N.

1 g N (H)n− N (H)N. ∈ G ⊆ G

Therefore G = NG(H)N,asdesired.

Let p π(G)foraﬁnitegroupG.RecallthatO (G)isthelargestnormalp-subgroup ∈ p of G.Thefollowingtheoremgivesalowerboundon m (G) . | p | Theorem 5.4.3. Suppose G is a ﬁnite group and assume p λ(G) κ(G). Then ∈ ∪ m (G) 1 and m (G) p +1when p κ(G). | p |≥ | p |≥ ∈ Proof. Suppose this is false. Let G be a counterexample of minimum order and assume

P Syl (G). Since G has a maximal subgroup which contains a Hall p′-subgroup, ∈ p m (G) 1. Therefore m (G)

First, suppose that P ✂ G.ThenΦ(P ) ✂ G and P/Φ(P )isnoncyclic.Byminimality,

Φ(P )=1andsoP is elementary abelian. Let H be a Hall p′-subgroup of G.Supposethat

44 H acts irreducibly on P .ThenN (H) 1 ,P .Byminimality,O (G)=1andso P ∈ { } p′ H G.ThenN (H)= 1 and H is maximal.* Therefore+ ̸! P { }

[G : H]= P p2 | |≥

2 and H has at least p conjugates, all of which are maximal and lie in mp(G), a contradiction. Thus H does not act irreducibly on P .ThenP has a nontrivial proper H-invariant subgroup P1.ByMaschke’sTheorem(Theorem4.2.6), there exists a nontrivial proper H-invariant subgroup P of P such that P = P P . Now H must act non-trivially on 2 1 × 2 at least one of P1 or P2. Without loss of generality, assume H acts non-trivially on P1.

Let M be a maximal subgroup which contains NG(H)P2.ThenM is self-normalizing by

Lemma 5.4.1,andsothep conjugates of M will not contain P1 and are elements of mp(G). If L is a maximal subgroup which contains P H,thenL m (G)andisdistinctfromthe 1 ∈ p aforementioned p maximal subgroups of m (G). Hence m (G) p +1,acontradiction. p | p |≥ Therefore, P G and O (G)isapropersubgroupofP .LetJ be a Hall p′-subgroup ̸! p of O (G). Then J =1and p,p′ ̸

G = N (J) O (G) N (J) O (G), G p,p′ ⊆ G p where the equality holds by Lemma 5.4.2. Notice that NG(J)isapropersubgroupofG

because Op′ (G)=1byminimality.IfM is a maximal subgroup of G which contains

NG(J), it is self-normalizing by Lemma 5.4.1.SoM has at least p conjugates in mp(G), all of which do not contain O (G). By minimality, m (G/ O (G)) 1andsothereisat p | p p |≥ least one maximal subgroup in m (G)whichcontainsO (G). Therefore m (G) p +1, p p | p |≥ aﬁnalcontradiction.

Observe that the theorem above proves both Theorem 5.1.1 and Theorem 5.1.2 for all ﬁnite solvable groups. Suppose G is a ﬁnite solvable group. Then α(G)=0andthe

45 nonabelian composition factors G satisfy Property B. Thus, theorem 5.4.3 yields at least

λ(G) + (p +1)= π(G) + p | | | | p κ(G) p κ(G) ∈% ∈% distinct maximal subgroups of G. 5.5 m(G) for Nonsolvable Groups | | Throughout this section we will make use of the fact that every minimal normal subgroup of a finite group is the direct product of isomorphic simple groups. For a nonsolvable group, we will count the maximal subgroups by separating them into those that contain a minimal normal subgroup N and those that do not contain N. Lemma 5.5.1. Let N be a nonabelian minimal normal subgroup of a finite group G and suppose that N = S S S , where S is a nonabelian simple group and S = S for 1 × 2 ×···× l i i ∼ j all i, j 1, 2,...,l . Let S = S and fix some ϕ : S S to be an isomorphism. Let M ∈{ } ∼ i i → i be a maximal subgroup of S and define Q = ϕ (M) ϕ (M) ... ϕ (M).IfM satisfies M 1 × 2 × × l

Property A, then G = N NG(QM ).

1 Proof. Since N ✂ G, N N (Q )isasubgroupofG.Letg G and observe that gQ g− G M ∈ M 1 is a proper subgroup of N.Deﬁnethemapσ : N N by σ (n)=gng− for all n N. g → g ∈ Then σ Aut(N). So σ will send minimal normal subgroups of N to minimal normal g ∈ g subgroups of N,thusσ permutes the minimal normal subgroups of N.Let be the g Sl 1 symmetric group on l objects. Then there exists ψ such that gS 1 g− = S ,forall ∈Sl ψ− (i) i i 1, 2,...,l . ∈{ } 1 1 Deﬁne the map τ : S S by τ (s)=ϕ− gϕ 1 (s)g− for all s S.Then g,i → g,i i ψ− (i) ∈ τ Aut(S)foralli 1, 2,...,l . Since Aut(S()=N (M))S,thereexistssome g,i ∈ ∈{ } Aut(S) s 1 S such that g,ψ− (i) ∈

1 1 1 sg,ψ 1(i)Ms− 1 = ϕ− gϕψ 1(i)(M)g− . − g,ψ− (i) i − ( )

46 Applying ϕi,weobtaintheequality

1 1 ϕi(sg,ψ 1(i)Ms− 1 )=gϕψ 1(i)(M)g− . − g,ψ− (i) −

Set n = ϕ (s 1 ),ϕ (s 1 ),...,ϕ(s 1 ) N.Then g 1 g,ψ− (1) 2 g,ψ− (2) l g,ψ− (l) ∈ ( ) 1 1 1 gQ g− = gϕ 1 (M)g− gϕ 1 (M)g− M ψ− (1) ×···× ψ− (l) 1 1 = ϕ1(sg,ψ 1(1)Ms− 1 ) ϕl(sg,ψ 1(l)Ms− 1 ) − g,ψ− (1) ×···× − g,ψ− (l) 1 = n ϕ (M) ϕ (M) n− g 1 ×···× l g ( 1 ) = ngQM ng− .

1 1 1 So n− gQ g− n = Q and therefore n− g N (Q )andg n N (Q ). Since g M g M g ∈ G M ∈ g G M n N, g N N (Q )andG = N N (Q ). g ∈ ∈ G M G M Lemma 5.5.2. Let N be a nonabelian minimal normal subgroup of a finite group G and suppose that N = S S S , where S is a nonabelian simple group and 1 × 2 ×···× l i S = S for all i, j 1, 2,...,l . Let S = S and assume M is a maximal subgroup of i ∼ j ∈{ } ∼ i S which satisfies Property A. Fix some ϕ : S S to be an isomorphism and define i → i Q = ϕ (M) ϕ (M) ... ϕ (M). Then N (Q ) is a maximal subgroup of G. M 1 × 2 × × m G M Proof. Since N is a minimal normal subgroup of G and Q N, N (Q )isaproper M ⊂ G M subgroup of G.SupposeK is a proper subgroup of G which contains NG(QM ). Then N is not contained in K by Lemma 5.5.1.Sothereexistssomei 1, 2,...,l such ∈{ } that K does not contain S ,butK S = ϕ (M). Observe that G acts transitively i ∩ i i on S ,S ,...,S and N acts trivially on S ,S ,...,S ,soK acts transitively on { 1 2 l} { 1 2 l} 1 S ,S ,...,S .Foreachj 1, 2,...,l ,thereissomek K such that k S k− = S { 1 2 l} ∈{ } j ∈ j i j j 1 and K S = k (K S )k− .ThereforeK S = S and so K S = ϕ (M)as ∩ j j ∩ i j ∩ j ̸ j ∩ j j ϕ (M) K S .SoK N = Q and N (Q ) N = Q .So j ⊆ ∩ j ∩ M G M ∩ M

G = KN = NG(QM )N

47 and K = N (Q )(K N)=N (Q ). G M ∩ G M

Therefore NG(QM )isamaximalsubgroupofG.

The following two lemmas count the number of maximal subgroups which do not contain N,theﬁrstlemmausingPropertyAandthesecondlemmausingPropertyB. Lemma 5.5.3. Suppose N is a nonabelian minimal normal subgroup of a ﬁnite group

G and assume N = S S S , where S is a simple group and S = S for all 1 × 2 ×···× l i i ∼ j i, j 1, 2,...,l . Then there are at least m (S ) maximal subgroups of G which do not ∈{ } | A 1 | contain N.

Proof. Let S = S for i 1, 2,...,l . Fix some ϕ : S S to be an isomorphism. ∼ i ∈{ } i → i Set n = m (S) ,sothatS has n maximal subgroups, say M ,M ,...,M ,suchthat | A | 1 2 n N (M )S = Aut(S)forallj 1, 2,...,n .DeﬁneQ = ϕ (M ) ϕ (M ) Aut(S) j ∈{ } Mj 1 j × 2 j ×···×

ϕl(Mj). Then G = N NG(QMj )byLemma5.5.1 and NG(QMj )isamaximalsubgroupof G by Lemma 5.5.2.ThereforeG has at least n maximal subgroups which do not contain

N,asdesired.

Lemma 5.5.4. Suppose N is a nonabelian minimal normal subgroup of a ﬁnite group G and assume N = S S S , where S = S for all i, j 1, 2,...,l .IfS satisﬁes 1 × 2 ×···× l i ∼ j ∈{ } 1

Property B, then there are at least γ(S1) maximal subgroups of G which do not contain N.

Proof. Let p be the largest prime in π(N)andsupposeP is a Sylow p-subgroup of N.ThenG = N N (P ) by the Frattini Argument. Because P N and N is a G ⊂ minimal normal subgroup of G, N (P ) = G and there exists a maximal subgroup L G ̸ of G containing N (P ). Observe that N L because otherwise G ̸⊆

G = NL L. ⊆

Therefore L is maximal subgroup of G which does not contain N.

48 For i 1, 2,...,l ,deﬁneR = S L.ObservethatL acts transitively on ∈{ } i i ∩ S ,S ,...,S .ThusR and R are L-conjugate, for i, j 1, 2,...,l and have the same { 1 2 l} i j ∈{ } order. Since N is the direct product of isomorphic simple groups, Si is not contained in L for all i 1, 2,...,l . Also S = R =1andR is not a normal subgroup of S .Itfollows ∈{ } i ̸ i ̸ i i that L is not a normal subgroup of G,andthusL is a self-normalizing maximal subgroup of G.

Because Si acts by left multiplication on the set of left coset of Ri in Si,thereexists anontrivialhomomorphismfromSi to the symmetric group on [Si : Ri]elementsandSi must be isomorphic to a subgroup of the alternating group on [Si : Ri]elements.Thus p [S : R ]. Let s S and deﬁne g =(s ,s ,s ,...,s) N.ConjugatingL by g,we ≤ i i i ∈ i 1 2 3 l ∈ obtain at least pl distinct maximal subgroups of G which do not contain N.

If l 2, then G has at least p π(N) maximal subgroups which do not contain N as ≥ ·| | p> π(N) by Lemma 5.3.1.Since | |

p π(N) = p π(S ) γ(S ) ·| | ·| 1 |≥ 1 by Lemma 5.3.2,thereareatleastγ(S1)maximalsubgroupswhichdonotcontainN.So assume that l =1.ThenN = S and N C (N) ✂ G.ThenG/C (N)isisomorphicto 1 × G G H,where

N H Aut(N). ≤ ≤ Since N satisﬁes Property B, there are at least γ(H) γ(S )maximalsubgroupsofG ≥ 1 which do not contain N,asdesired.

We conclude this section, with two theorems that allow us to count some of the maximal subgroups of a nonsolvable group by their index. Theorem 5.5.5. If G be a ﬁnite group, then it has at least γ(G)α(G) maximal subgroups of G with index a τ(G)-number.

49 Proof. Suppose this is false. Let G be a counterexample of minimum order. Assume N is a minimal normal subgroup of G such that N = S S ... S ,whereS is a ∼ 1 × 2 × × l i simple group and S = S for all i, j 1, 2,...,l . First, suppose that N is solvable. i ∼ j ∈{ } Then τ(G)=τ(G/N)andα(G)=α(G/N). By minimality, G/N has at least γ(G)α(G) maximal subgroups with index a τ(G)-number. Thus G has at least γ(G)α(G)maximal subgroups which contain N and have index a τ(G)-number, a contradiction. Therefore N is not solvable. By minimality, G/N has at least γ(G/N) α(G/N) maximal subgroups with index a τ(G/N)-number. So G has at least γ(G/N)α(G/N) maximal subgroups which contain N and have index a τ(G)-number. In addition, there are at least m (S ) maximal subgroups which do not contain N by Lemma 5.5.3. | A 1 | Observe that

mA(S1) mA(S1) α(N)=α(S1)=| | = | |, γ(S1) γ(N) and thus m (S ) = γ(N)α(N). In total, G has at least | A 1 |

γ(G/N)α(G/N)+γ(N)α(N) γ(G)α(G) ≥ maximal subgroups with index a τ(G)-number, a ﬁnal contradiction.

Theorem 5.5.6. Let G be a ﬁnite group. If the nonabelian composition factors of G satisfy Property B, then there exists at least γ(G) maximal subgroups of G with index a

τ(G)-number.

Proof. Suppose this is false. Let G be a counterexample of minimum order and assume N is a minimal normal subgroup of G. First, suppose that N is solvable. Then τ(G)= τ(G/N)andG/N has at least γ(G)maximalsubgroupswithindexaτ(G)-number, by minimality. Thus G has at least γ(G)maximalsubgroupswhichcontainN and have index a τ(G)-number, a contradiction. Therefore N is not solvable. Suppose N = S S S ,whereS is a nonabelian 1 × 2 ×···× l i simple group and S = S for all i, j 1, 2,...,l .Byminimality,G/N has at least i ∼ j ∈{ }

50 γ(G/N)maximalsubgroupswithindexaτ(G/N)-number. So G has at least γ(G/N) maximal subgroups which contain N and have index a τ(G)-number. By Lemma 5.5.4, G has at least γ(S1)maximalsubgroupswhichdonotcontainN.Intotal,G has at least

γ(G/N)+γ(S )=γ(G/N)+γ(N) γ(G) 1 ≥ maximal subgroups with index a τ(G)-number, a ﬁnal contradiction.

5.6 Proof of Main Theorems

Proof of Theorem 5.1.1. We obtain at least

λ(G) + (p +1) | | p κ(G) ∈% distinct maximal subgroups of G with index a λ(G) κ(G)-number by Theorem 5.4.3.By ∪ Theorem 5.5.6, G has at least γ(G)maximalsubgroupsofG with index a τ(G)-number. Therefore,

Proof of Theorem 5.1.2. We obtain at least

λ(G) + (p +1) | | p κ(G) ∈% distinct maximal subgroups of G with index a λ(G) κ(G)-number by Theorem 5.4.3. ∪ By Theorem 5.5.5, G has at least γ(G)α(G)maximalsubgroupsofG with index a τ(G)-number. Therefore,

m(G) λ(G) + (p +1)+γ(G)α(G), | |≥| | p κ(G) ∈% as desired.

51 We conclude this section by proving the bound in Theorem 5.1.2 is best possible. The same example proves that the lower bound in Theorem 5.1.1 is best possible. Proposition 5.6.1. Given any ﬁnite set of primes Ω and any subset Ω¯ of Ω, there exists a ﬁnite solvable group G with π(G)=Ω, κ(G)=Ω¯ and

m(G) = λ(G) + (p +1)+γ(G)α(G). | | | | p κ(G) ∈% Proof. Suppose that Ω= p ,p ,...,p and Ω¯ = p ,p ,...,p where 0 l n. { 1 2 n} { 1 2 l} ≤ ≤ Consider the noncyclic group

G = Zp1 Zp1 Zp2 Zp2 Zpl Zpl Zpl+1 Zpl+2 Zpn , × × × ×···× × × × ×···× which is of order p2p2 p2p p p .ThenΩ=π(G), Ω¯ =κ(G)andα(G)=0.By 1 2 ··· l l+1 l+2 ··· n

Lemma 4.3.2, m(Zpi Zpi ) = pi +1fori 1, 2,...,l and observe that m(Zpi ) =1for | × | ∈{ } | | i l +1,...,n .Thus ∈{ } l m(G) = λ(G) + (p +1), | | | | i i=1 % by Lemma 4.2.7,andthelowerboundinTheorem5.1.2 is best possible.

There exists nonsolvable groups which demonstrate the lower bound in Theorem 5.1.2 is best possible. For example, suppose G is a ﬁnite simple group such that m(G)= mA(G). Then λ(G)= = κ(G) ∅ and m(G) = m (G) = γ(G)α(G) | | | A | and thus G achieves the lower bound in Theorem 5.1.2.

5.7 Remarks on α(G)

In this section, we give some groups G which satisfy α(G) > 1. Lemma 5.7.1. Let denote the simple alternating group on n letters. If n 9, then An ≥ α( ) >n. Otherwise α( )= 21 , α( )= 10 , α( )=3and α( )= 127 . An A5 13 A6 13 A7 A8 21

52 Proof. If 5 n 8, the result holds by the Atlas of Finite Groups [16]. So we assume ≤ ≤ that n 9. Let denote the symmetric group on n letters. For each integer k with ≥ Sn n 1 n,wehaveatleastn( 2) = − distinct A S ×S − ∩A 2 − 2 maximal subgroups of .Observethattheaforementionedmaximalsubgroupsare An distinct from the n conjugates of the maximal subgroup n 1 in n.LetM be one of A − A n(n 4) the aforementioned − + n maximal subgroups of . Since Aut( )= and M is 2 An An Sn normalized by an element of n n,itfollowsthatNAut( n)(M) n = Aut( n). Therefore S \A A A A n(n 4) n2 2n m ( ) − + n and −2 /γ( n) >n,asdesired. | A An |≥ 2 A Lemma 5.7.2. Let q =2f .Ifq 8, then ≥ q2 + q +1 α(PSL(2,q)) . ≥ 2(q +1)

21 If q =4, then α(PSL(2,q)) = 13 .

Proof. Let S = PSL(2,q). If q =4,theresultholdsbytheresultholdsbytheAtlas

f of Finite Groups [16]. So assume that q 8. Then D2(q 1), D2(q+1) and Z2 "Zq 1 are ≥ − − maximal subgroups of S by Dickson’s Theorem (see [2]). Since S is simple, D2(q 1) has − 3 f 1 3 f 1 [q q :2(q 1)] = 2 − (q +1)conjugates,D has [q q :2(q +1)]=2 − (q 1) − − 2(q+1) − − 3 f 2 conjugates and [q q : Z2 "Zq 1]=q +1. Thus S has at least q + q +1 distinctmaximal − − subgroups. The outer automorphism group of S is generated by a ﬁeld automorphism ϕ of order f,whichﬁxesallthreeoftheaforementionedconjugacyclassesofmaximal subgroups. Therefore m (S) q2 + q +1.Since S = q3 q,Lemma5.3.4 implies | A |≥ | | − q2 + q +1 q2 + q +1 α(S) , ≥ p τ(S)(p +1) ≥ 2(q +1) ∈ as desired. ,

53 Table 5-2. Values of α(G). Gα(G) α(G) Gα(G) α(G) ⌊ ⌋ ⌊ ⌋ 309 703,131,313 M11 25 12 Fi22 47 14, 960, 240 2,850 10 10 M12 25 114 Fi23 > 10 > 10 1,948 10 10 M22 33 59 Fi24′ > 10 > 10 44,414 c 495,529 M23 57 779 M L 33 15, 016 1,507,537 6,862,629 M24 57 26, 448 He 39 175, 964 13,016 367,882,130 J1 53 245 Ru 65 5, 659, 725 16,964 291,965,788 J2 21 807 Suz 47 6, 212, 038 68,410 8,935,530,772 J3 33 2, 073 HN 53 168, 594, 920 10 10 83,897,408 J4 > 10 > 10 O′N 85 987, 028 424,818,005 10 10 Co3 57 7, 452, 947 Th > 10 > 10 3,581,796,533 10 10 Co2 57 62, 838, 535 Ly > 10 > 10 10 10 10 10 Co1 > 10 > 10 B>10 > 10 68,410 10 10 HS 33 2, 073 M>10 > 10

Lemma 5.7.3. Let q = pf , where p is an odd prime number. If q 13, then ≥ q2 α(PSL(2,q)) . ≥ 2(q +1)

8 10 67 Otherwise α(PSL(2, 7)) = 15 , α(PSL(2, 9)) = 13 and α(PSL(2, 11)) = 25 .

Proof. This proof is omitted due to its similarities to the proof of Lemma 5.7.2.

We conclude this section by giving values of α(G), where G is a sporadic simple group. The following values given in the table below were calculated from the Atlas of Finite Groups [16].

5.8 Remarks on Property B

In this section, we given some examples of ﬁnite simple groups which satisfy Property B.

Lemma 5.8.1. If n 5, then satisﬁes Property B. ≥ An

54 Proof. If n =6,theresultholdsbytheresultholdsbytheAtlas of Finite Groups [16]. So assume that n = 6. Since Aut( )= ,theresultfollowsfromtheproofof ̸ An Sn Lemma 5.7.1.

Lemma 5.8.2. Suppose q = pf , where p is a prime number. If q 4, then PSL(2,q) ≥ satisﬁes Property B.

Proof. If n 7, 9 ,thentheresultholdsbytheAtlas of Finite Groups [16]. So assume ∈{ } that n 7, 9 .SinceOut(PSL(2,q)) = Z(2,q 1) Zd,theresultfollowsfromtheproofsof ̸∈ { } ∼ − × Lemma 5.7.2 and Lemma 5.7.3.

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56 BIOGRAPHICAL SKETCH Lindsey-Kay Lauderdale was born in Champaign, Illinois. She graduated from Centennial High School and completed a Bachelor of Science in mathematics from the University of Illinois at Urbana-Champaign in 2009. In 2011, Lindsey-Kay earned a Master of Science in mathematics from the University of Florida. In her free time she enjoys shopping, watching television and nail art.