Connections Between the Number of Constituents and the Derived Length of a Group a Dissertation Submitted to Kent State Universi

Connections Between the Number of Constituents and the Derived Length of a

Group

A dissertation submitted to Kent State University in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy

by,

Lisa Rose Hendrixson

May, 2017 Dissertation written by

Lisa Rose Hendrixson

B.S., Kent State University, 2011

M.A., Kent State University, 2013

Ph.D., Kent State University, 2017

Approved by

Mark L. Lewis , Chair, Doctoral Dissertation Committee

Stephen M. Gagola , Members, Doctoral Dissertation

Donald L. White , Committee

Joanne Caniglia ,

Hassan Peyravi ,

Accepted by

Andrew Tonge , Chair, Department of Mathematical

Sciences

James L. Blank , Dean, College of Arts and Sciences Table of Contents

Dedication iv

Acknowledgments v

1 Introduction 1

2 Background 5

2.1 Group Theory ...... 5

2.2 Character Theory ...... 9

3 Known Results 13

4 Two Nonprincipal Irreducible Constituents 22

5 Three Nonprincipal Irreducible Constituents-The Special Case 33

6 Three Nonprincipal Irreducible Constituents-The General Case 58

7 Examples 85

Bibliography 93

iii To my mother, for her continued love and kindness, and to my father, who gave me

a love of mathematics. Acknowledgments

I would like to thank my advisor, Dr. Mark Lewis, for his assistance and support throughout my education. I would like to thank Dr. White and Dr. Gagola for their many helpful suggestions that have improved this document and the works and talks that came out of it. Lastly, I would like to thank my parents for their unwavering support and understanding.

v Chapter 1

Introduction

In this dissertation, we discuss the relationship between the structure of a ﬁnite solvable group and the number of nonprincipal irreducible constituents belonging to a particular product of characters. Our main focus is to answer, at least in part, the following question: Is the derived length bounded in terms of the number of nonprincipal irreducible constituents of a particular product of characters, namely

χχ?

Both Adan-Bante and Keller have made strides in answering this question. In [1],

Adan-Bante proves that G/ ker(χ) has derived length bounded by a linear function involving the number of nonprincipal irreducible constituents of χχ. Also, Keller was able to prove in [12] that G/F2(G) is bounded in terms of the number of character degrees, although he notes that the bound proved in that paper is not the best possible and suggests what it should be. The notation F2(G) will be deﬁned later.

In [2], Adan-Bante completely classiﬁes the solvable groups which have a faithful irreducible character χ such that χχ has one nonprincipal irreducible constituent, and gives some group structure of those solvable groups having an irreducible faithful character χ such that χχ has two nonprincipal irreducible constituents. We begin

1 by proving that the best possible bound when χχ has two nonprincipal irreducible

constituents is 8. In fact, we are able to prove that the two constituents in question

must be real characters. Then we consider the situation when χχ has three such

constituents. This includes the special case where two of the nonprincipal irreducible

constituents are complex conjugates, and the general case where none of them are

related.

Theorem A. Let G be a ﬁnite solvable group and let χ ∈ Irr(G) be a faithful character. Assume that

χχ = 1G + m1α1 + m2α2,

# where α1, α2 ∈ Irr(G) are distinct characters and m1 and m2 are strictly positive

integers. Then dl(G) ≤ 8.

Throughout, G is a ﬁnite solvable group with center Z(G), and the set of irre-

ducible characters of G is denoted by Irr(G). Also, for a group V , we denote the set

of nonidentity elements of V by V #.

Theorem B. Let G be a ﬁnite solvable group with a faithful character χ ∈ Irr(G)

such that

χχ = 1G + m1α1 + m2α2,

# where α1, α2 ∈ Irr(G) are distinct characters and m1, m2 are strictly positive inte-

gers. Then both α1 and α2 are real-valued characters.

The next natural question to consider is the situation where χχ has three non-

principal irreducible constituents. This will be broken into two cases. The ﬁrst to

2 be considered will be when χχ has complex-valued irreducible constituents, which is handled by Theorem C. When we are ﬁnished with Theorem C, we will turn our attention to Theorem D, the case when χχ has only real nonprincipal irreducible constituents. We will also see that Theorem C is a consequence of Theorem B.

Theorem C. Let G be a ﬁnite solvable group with center Z = Z(G) and let χ ∈ Irr(G) be a faithful character. Assume that

χχ = 1G + m1α1 + m2α2 + m2α2,

# where α1, α2 ∈ Irr(G) are distinct characters and m1 and m2 are strictly positive integers. Then,

1. the order of G is even,

2. dl(G) ≤ 6,

3. both ker(α1) and ker(α2) are abelian groups with either ker(α1) = Z or ker(α2) =

Z, and χ(1) is a power of a prime.

Theorem D. Let G be a ﬁnite solvable group and let χ ∈ Irr(G) be a faithful character. Assume that

χχ = 1G + m1α1 + m2α2 + m3α3,

# where αi ∈ Irr(G) are distinct characters and mi are strictly positive integers for all i = 1, 2, 3. Then dl(G) ≤ 16.

3 It should be mentioned that the largest example we have has derived length 6, and

is presented at the end of this paper. This suggests that there is some improvement

that could be made to the bound in Theorem D, as well as a possibly larger example

that has yet to be found. It seems likely that both of these could be true.

The next theorem provides a starting point when looking for more examples. In

[1], Adan-Bante only proves an implicit bound for the derived length of groups with

the desired property. Therefore, it seems likely that the following theorem could also

provide some information on ﬁnding an explicit bound for the groups in question.

Theorem E. Let G be a ﬁnite group with E ≤ G a nonabelian subgroup. Assume

E/Z is an abelian chief factor of G with Z = Z(G), and that G/E acts with n ≥ 2

# orbits on (E/Z) . Let χ ∈ Irr(G) be faithful and suppose that χE ∈ Irr(E). Then

n X χχ = 1G + αi, i=1

# where αi ∈ Irr(G) are distinct and ker(αi) = Z for all i, 1 ≤ i ≤ n.

4 Chapter 2

Background

2.1 Group Theory

Since our discussion will center on groups, it is necessary to begin by introducing those

topics that will play an important role. The references that will be used are [10] and

[11]. A group is a set G together with an associative binary operation ◦ deﬁned on G such that there exists e ∈ G with the properties that for each x ∈ G, x ◦ e = e ◦ x = x and for each x ∈ G there exists some y ∈ G such that x ◦ y = y ◦ x = e. If the binary

operation is also commutative, then the group is called an abelian group. A group is said to be cyclic if there exists some g ∈ G with G = {gn|n ∈ Z}; these groups are automatically abelian. We say that H is a subgroup of G if H is a subset of G which is also a group. A proper subgroup H is called maximal if whenever H ≤ K ≤ G then either H = K or K = G. The Frattini subgroup, denoted Φ(G), is the intersection of all maximal subgroups. Also, we denoted the index of H in G by |G : H|, which can be thought of as |G|/|H| when the groups under consideration are ﬁnite. As all of the groups in this paper will be ﬁnite, this is appropriate.

There are several particular types of subgroups which will play a role in our

5 discussion. The ﬁrst is the normal subgroup. A subgroup N ≤ G is called normal

when N g = g−1Ng = N for all g ∈ G, and is denoted by N/G. The socle of G is the subgroup of G generated by all of the minimal normal subgroups of G. The centralizer of X ⊆ G is denoted CG(X) and is the set of elements in G which commute with all elements of X. Likewise, the center of G is the set of elements which commute with all elements of G, and is denoted by Z(G). Also relevant to our topics is the commutator subgroup. The commutator of two elements x, y ∈ G is denoted by

[x, y] = x−1y−1xy and the commutator subgroup is the subgroup G0 = [G, G]. This subgroup is generated by all commutators of two elements in G, i.e., G0 is the set of

all ﬁnite products of elements of the form [x, y] for x, y ∈ G. Let H be a subgroup of

G. Then G/H is the set {gH|g ∈ G}. If H is a normal subgroup of G, then G/H is

referred to as the factor group, or quotient group. In particular, a chief factor L/K

of a group G is a factor group, where K < L are normal subgroups of G and there

are no normal subgroups M such that K < M < L.

A subgroup P of G is called a p-group if the order |P | is equal to a power of some

prime p. This group is also elementary abelian if every nonidentity element of P has

the same prime order p and P is an abelian group. We also say that P is extra-special

if P 0 = Z(P ) has order p. The following lemma will be used frequently, without proof

(it appears as Exercise 4A.4 in [11]).

Lemma 2.1. If P is extra-special, then P/Z(P ) is elementary abelian. Equivalently,

Z(P ) = P 0 = Φ(P ).

Let G be a group. Then Sylow p-subgroups are subgroups P of G with p-power

6 order such that p does not divide |G : P |. Now, let H be a subgroup of G and π a set of primes. We say that H is a Hall π-subgroup of G if every prime divisor of |H| is in π and none of the primes in π divide |G : H|. The notation Op(G) will be used to denote the unique largest normal p-subgroup of G.

Other types of groups will be important throughout the discussion. For all of our results, we will assume that G is solvable, i.e., there exists a ﬁnite collection of normal subgroups G0, G1,. . . ,Gn such that

1 = G0 ≤ G1 ≤ · · · ≤ Gn = G

and Gi+1/Gi is abelian for 0 ≤ i < n. In particular, solvable groups have what is referred to as a derived series, in which the normal subgroups are the commutator subgroups. Write G = G(0), G0 = G(1), G00 = (G0)0 = G(2),...,G(n) = 1. Then we have

G = G(0) ≥ G(1) ≥ G(2) ≥ · · · ≥ G(n) = 1 and the derived series is comprised of these groups. The derived length is the integer n. Also of note is the fact that in a solvable group, the chief factors are elementary abelian p-groups. A group G is nilpotent if G has a series of normal subgroups Ni such that

1 = N0 ≤ N1 ≤ · · · ≤ Nm = G

with the property that Ni/Ni−1 ≤ Z(G/Ni−1) for 1 ≤ i ≤ m. The Fitting subgroup of G is the largest normal nilpotent subgroup of G, and is denoted by F(G). If

7 we let F = F(G), the second Fitting subgroup, denoted by F2(G), is deﬁned by

F2(G)/F = F(G/F ).

A group G is a semi-direct product if G = NH with N/G, H ≤ G, and N ∩H = 1.

If H is also normal in G, then we call this product a direct product. When discussing such products the following lemma is important and will play a role in several of the results in this paper.

Lemma 2.2 (Diamond Lemma). Let N/G and H ≤ G. Then H ∩ N/H and

H/(H ∩ N) ∼= NH/N.

Another type of product is the wreath product. Let G and H be groups and let B be the group of functions f : G → H. Then W = B o G is the wreath product of G and H, and is often denoted by W = H o G.

There are a few other types of groups that we will need in our discussion. Let

Ω = {1, 2, . . . , n} for some positive integer n. Then the set of permutations of these

“letters” is denoted by Sn. We say that a transitive group G ≤ Sn is a primitive permutation group if it is not possible to decompose Ω into proper subsets of size greater than 1 such that these sets are permuted by the action of G. A transitive action is deﬁned such that any α, β ∈ Ω, α can be mapped to β using some permutation from

G. If a group is not primitive, we call it imprimitive, and the proper subsets of Ω that are permuted by G are called the imprimitivity spaces. Another type of action is a faithful action, which means that the only element of G that ﬁxes every element in Ω is the identity element of G.

A module is an abelian additive group (meaning that the binary operation is

8 addition) that is being acted on by a group. A module is called irreducible if it has

no proper nontrivial submodules.

Lastly, if we have a map ϕ : G → H, then the map is called a homomorphism if

ϕ(xy) = ϕ(x)ϕ(y). If the map is also one-to-one and onto, i.e., a bijection, then we

call it an isomorphism. In the next section, speciﬁc homomorphisms will be described.

2.2 Character Theory

In this section, most of what will be described can be found in [9], which is a standard

reference for character theory. We begin by describing matrices and certain groups

of matrices. The set of complex numbers is denoted by C. Also, the general linear

group, denoted by GL(n, C), is the group of invertible n × n matrices with entries in

C. The trace of a matrix A ∈ GL(n, C) is the sum of all the diagonal entries of A.

We denote this by tr(A).

Now, let G be a group and C the complex numbers. A representation X of G is

a homomorphism X : G → GL(n, C) for some positive integer n.A character χ of

G aﬀorded by X is given by the formula χ(g) = tr(X(g)). Characters are constant

on conjugacy classes, which means that all characters are class functions. The inner

product of two class functions χ and ϕ is deﬁned by

1 X [χ, ϕ] = χ(g)ϕ(g), |G| g∈G

where ϕ(g) is the complex conjugate of ϕ(g). A character is irreducible precisely

when [χ, χ] = 1, and homogeneous when it is the integer multiple of some irreducible

9 character (the corresponding module will also be called homogeneous). The set of irreducible characters of G is denoted by Irr(G), and there are only ﬁnitely many of them by Corollary 2.7 of [9]. In particular, the characters in Irr(G) form a basis for the set of class functions. Suppose

k X χ = niχi, i=1

with ni ≥ 0 and χi ∈ Irr(G). Then those χi with positive coeﬃcients ni are called the irreducible constituents of χ. Also, the degree of the character χ is χ(1), which is the dimension of X and a positive integer. A character is called linear if χ(1) = 1, and a group is abelian if and only if all of its characters are linear, by Corollary 2.6 of [9].

Furthermore, linear characters are the only characters which are homomorphisms; in general, characters are not homomorphisms. Every group has at least one linear character, namely the principal character, which sends all the elements of G to 1 and is denoted by 1G. In our situation, 1G is an irreducible constituent of χχ since

1 = [χ, χ] = [χχ, 1G], where the second equality follows by comments made on page

48 of [9].

Next, let H ≤ G be a subgroup. If χ is a character of G, then its restriction to H denoted by χH is a character of H. Likewise, if θ is a character of G then the induced character θG is given by

1 X θG(g) = θ◦(xgx−1), |H| x∈G

10 where θ◦(h) = θ(h) for h ∈ H and θ◦(y) = 0 for y∈ / H. Moreover, θG(1) = |G :

H|θ(1).

For the work here, there are several results which will be important. The ﬁrst is

Cliﬀord Theory. The main result of Cliﬀord, which will be used many times, is the

following:

Lemma 2.3. Let N/G and let χ ∈ Irr(G). Let θ be an irreducible constituent of χN and suppose that θ = θ1, θ2, . . . , θt are the distinct conjugates of θ in G. Then

t X χN = e θi, i=1

where e = [χN , θ].

Proof. See Theorem 6.2 of [9].

We say θ extends to G if there is some χ ∈ Irr(G) such that χH = θ and θ induces irreducibly if θG ∈ Irr(G). If χ is the unique irreducible constituent of θG and θ is

G-invariant, then we say that χ and θ are fully ramiﬁed. Also, G acts on the set

Irr(N) by conjugation, i.e., θg, where θ ∈ Irr(N). The stabilizer of θ is the subgroup

g Gθ = {g ∈ G | θ = θ} and θ is G-invariant if Gθ = G.

There are a few other theorems which will be crucial to our discussion.

Theorem 2.4 (Ito’s Theorem). Let A/G be abelian. Then χ(1) divides |G : A| for

all χ ∈ Irr(G).

Proof. See Theorem 6.15 of [9].

11 Theorem 2.5 (Gallagher’s Theorem). Let N/G and let χ ∈ Irr(G) be such that

χN = θ ∈ Irr(N). Then the characters βχ for β ∈ Irr(G/N) are irreducible and distinct for distinct β and are all of the irreducible constituents of θG.

Proof. See Corollary 6.17 of [9].

Theorem 2.6. Let K/L be an abelian chief factor of G. Suppose θ ∈ Irr(K) is

invariant in G. Then one of the following occurs:

(a) θL ∈ Irr(L);

2 (b) θL = eϕ for some ϕ ∈ Irr(L) and e = |K : L|;

Pt (c) θL = i=1 ϕi where the ϕi ∈ Irr(L) are distinct and t = |K : L|.

Proof. See Theorem 6.18 of [9].

12 Chapter 3

Known Results

In this chapter, we present a number of results which will be frequently used, but which are more speciﬁc than those found in a standard text. Much of what appears in this chapter can be found in [13]. The ﬁrst is a well-known result stated in [13], with a proof given in the preceding paragraph:

Lemma 3.1. If an abelian group A has a faithful irreducible module W , then A is

cyclic.

The next results will be used many times when dealing with the diﬀerent derived

lengths of groups acting on modules of various sizes.

Lemma 3.2. Suppose G 6= 1 is solvable and every normal abelian subgroup of G is cyclic. Let F = F(G) and let Z be the socle of the cyclic subgroup Z(F ). Set

A = CG(Z). Then there exist E,T/G with

(i) F = ET , Z = E ∩ T , and T = CF (E).

(ii) A Sylow q-subgroup of E is cyclic of order q or extra-special of exponent q or 4.

13 (iii) E/Z = E1/Z × · · · × En/Z for chief factors Ei/Z of G with Ei ≤ CG(Ej) for

i 6= j.

2ni (iv) For each i, Z(Ei) = Z, |Ei/Z| = pi for a prime pi and an integer ni, and

2ni+1 E = O 0 (Z) · F for an extra-special group F = O (E ) /G of order p . i pi i i pi i i

(v) There exists U ≤ T of index at most 2 with U cyclic, U/G and CT (U) = U.

(vi) G is nilpotent if and only if G = T .

(vii) T = CG(E) and F = CA(E/Z).

(viii) E/Z ∼= F/T is a completely reducible G/F -module and faithful A/F module.

(ix) A/CA(Ei/Z) . Sp(2ni, pi).

(x) If every normal abelian subgroup of G is central in F , then T = Z(F ) is cyclic.

Proof. See Corollary 1.10 of [13].

For several results in this chapter, we need some deﬁnitions. A standard reference

for this is page 37 of [13]. Let q be a prime power and take V to be a vector space over GF (q) of dimension m. Fix a ∈ V # = V − {0}, b ∈ V , and σ ∈ G =

Gal(GF (qm)/GF (q)). Then the semi-linear group is given by

Γ(V ) = x 7→ axσ|a ∈ GF (qm)#, σ ∈ G ,

and the aﬃne semi-linear group is given by

AΓ(V ) = x 7→ axσ + b|a ∈ GF (qm)#, σ ∈ G, b ∈ V .

14 ∼ In particular, AΓ(V ) = V o Γ(V ), and Γ(V ) is a point-stabilizer for zero in AΓ(V ),

which is a doubly transitive permutation group. The notation Γ(qm) is often used in

place of Γ(V ) to remind the reader of the dimension and prime power involved.

Lemma 3.3. Suppose that V is a faithful irreducible GF (q)[G]-module for a solvable group G and a prime power q. Let F = F(G).

(a) If F is abelian and VF is homogeneous, then G ≤ Γ(V ).

(b) If V is quasi-primitive and F = T , then G ≤ Γ(V ).

Proof. See Corollary 2.3 of [13].

The next two lemmas will provide the dimension of a given module. In practice,

they will yield a list of dimensions that are either impossible, or will give us a bound.

Lemma 3.4. Let F be a group with center Z. Suppose that V is a faithful irreducible

F[F ]-module, where F is a ﬁeld that has positive characteristic or is algebraically closed. Assume that char(F) - |F |. Let W be an irreducible Z-submodule of V . Then

dimF (V ) = te · dimF (W ) for integers t and e, with e = χ(1) for a faithful irreducible ordinary character χ of F.

Proof. See Lemma 2.4 of [13].

Lemma 3.5. Assume that H = EU, where U = Z(H) is cyclic, U ∩ E = Z(E), E

is nilpotent and the Sylow subgroups of E are extra-special or of prime order. Let

V be a faithful irreducible F[H]-module and W an irreducible submodule of VU . If

char(F) = 0, assume that F is algebraically closed. Then dimF (V ) = e · dimF (W )

with e2 = |H : U|.

15 Proof. See Corollary 2.6 of [13].

This lemma will allow us to bound the derived length of imprimitive linear groups

G, as will be seen in Theorems 3.10 and 3.11.

Lemma 3.6. Let V be a faithful irreducible F[G]-module and suppose V = V1 ⊕ · · ·⊕

Vn (n > 1) is a system of imprimitivity for G. Let γ : G → Sn be the homomorphism induced by the permutation action of G on the Vi. Set S = γ(G), which is a transitive

subgroup of Sn. Finally, let H = NG(V1)/CG(V1). Then G is isomorphic to a subgroup

of H o S as linear groups.

Proof. See Lemma 2.8 of [13].

The next two results will allow us to handle several dimensions at the same time.

Lemma 3.7. Let G be a solvable irreducible subgroup of GL(n, q). Then

(i) if n = 2 and q is a prime power, dl(G) ≤ 4.

(ii) if q = 2 and n = pr, where p and r are primes which are not necessarily distinct,

then dl(G) ≤ 6.

Proof. For the ﬁrst part of the lemma, see Theorem 2.11 of [13]. For the second part

of the lemma, see Theorem 2.14 of [13].

The next lemma will be useful is bounding the derived length for certain prime

powers. In some cases, this result will give bounds which are higher than those that

we can ﬁnd. When that happens, we will use the previous results to improve what

this lemma provides.

16 Lemma 3.8. Let G be solvable and let V 6= 0 be a faithful and completely reducible

F[G]-module over an arbitrary ﬁeld F. Set n = dimF (V ). Then dl(G) ≤ 8 +

5 2 log3(n/8).

Proof. See Theorem 3.9 of [13].

The next several results will involve bounding the derived length of a group G depending on how many orbits G has when acting on a vector space V . The ﬁrst is

a well-known result of Huppert.

Theorem 3.9 (Huppert). Let V be a vector space of dimension n over GF (q), q

a prime power. Suppose that G is a solvable subgroup of GL(V ) that transitively

permutes the elements of V #. Then G ≤ Γ(qn), or one of the following occurs:

n 4 5 (a) q = 3 , F(G) is extra-special of order 2 , |F2(G)/F(G)| = 5 and G/F2(G) ≤ Z4.

(b) qn = 32, 52, 72, 112, or 232. Here F(G) = QT , where T = Z(G) ≤ Z(GL(V )) is ∼ ∼ cyclic, Q8 = Q, T ∩ Q = Z(Q) and Q/Z(Q) = F(G)/T is a faithful irreducible

G/F(G)-module. We also have one of the following entries:

qn |T | G/F(G) 2 3 2 Z3 or S3 2 5 2 or 4 Z3 2 5 4 S3 2 7 2 or 6 S3 2 11 10 Z3 or S3 2 23 22 S3

In particular, dl(G) ≤ 4.

Proof. By Theorem 6.8 of [13], we know that either G is a semi-linear group or

G/F(G) is a metacyclic group with F(G) extra-special. For our purposes, it is enough

17 to realize that semi-linear groups are metacyclic, so their derived lengths are at most

2. In the second case, we see that both G/F(G) and F(G) have derived length at most 2, so G has derived length at most 4.

The next two results will also be useful in determining derived lengths for various

groups. They also inform where examples might appear. Both were proved by Foulser

in [5] and simpliﬁed by Dornhoﬀ in [4]. For these theorems, we must consider a

primitive permutation group G acting on a ﬁnite set Ω. A standard reference for this

situation is page 39 of [13]. First, let G be a solvable primitive permutation group on

Ω with α ∈ Ω having as its point stabilizer G = Gα. Then G has a unique minimal normal subgroup V such that G = GV , G ∩ V = 1, CG(V ) = V , and V acts regularly

on Ω. This means that |Ω| = |V | = qn is a prime power. Also, the mapping that takes v ∈ V to vα is a G permutation isomorphism between V and Ω. In this case, G acts

on V by conjugation. Since V is the unique minimal normal subgroup of G, it follows

that V is irreducible and faithful as a module for G. Because of the permutation

isomorphism, the rank of G on Ω is the number of orbits, including the trivial orbit,

of G on V .

Theorem 3.10. Let G be a primitive solvable permutation group of rank 3; write

G = GV where V is a minimal normal subgroup of G and G is the stabilizer of a

point in Ω, the set upon which G acts. Then one of the following holds:

(i) V has order |V | = qn for a prime q and G is permutation-isomorphic to a

subgroup of AΓ(qn), the aﬃne semi-linear group. In particular, dl(G) ≤ 2.

(ii) G is an imprimitive linear group with imprimitivity spaces V1,V2, where V =

18 V1 ⊕ V2. Here H = NG(V1) has index 2 and H/CH (V1) is a linear group that

acts transitively on V1 − {0}. Thus, dl(G) ≤ 5.

(iii) G acts as a primitive linear group on V and G has one of the degrees 72, 132,

172, 192, 232, 292, 312, 472, 34, 74, 26, or 36.

Proof. See Theorem 1.1 of [5]. To see the second conclusion of (ii), we know that dl(G) ≤ 5 since

G ≤ H/CH (V1) o C2, by the argument of Lemma 1.4 of [12]. However, this wreath product has derived length bounded above by 5, since from Theorem 3.9, dl(H/CH (V1)) ≤ 4.

We will also need the following theorem, proved by Foulser in [5] and simpliﬁed by Dornhoﬀ in [4]. This will be useful when χχ has three nonprincipal irreducible constituents.

Theorem 3.11. Let G be a primitive solvable permutation group of rank 4; write

G = GV where V is a minimal normal subgroup of G and G is the stabilizer of a point in Ω, the set upon which G acts. Then one of the following holds:

(i) V has order |V | = qn for a prime q and G is permutation-isomorphic to a

subgroup AΓ(qn), the aﬃne semi-linear group. In particular, dl(G) ≤ 2.

r (ii) G is an imprimitive linear group with V = ⊕i=1Vi where the Vi are imprimitivity

spaces and r = 2 or 3. Here H = NG(V1) and H/CH (V1)is a linear group that

# acts transitively on V1 = V1 − {0}. Thus, dl(G) ≤ 5.

19 (iii) G acts as a primitive linear group on V and G has one of the degrees qn for qn =

p2, where p is a prime and p ≤ 71, or qn ∈ {24, 34, 54, 74, 26, 36, 28, 210, 310, 212}.

Proof. See Theorem 1.2 of [5]. To see the second conclusion of (ii), we know that dl(G) ≤ 5 since

G ≤ H/CH (V1) o Cr, by the argument of Lemma 3.6. However, this wreath product has derived length bounded above by 5, since from Theorem 3.9, dl(H/CH (V1)) ≤ 4.

Lastly, the following lemma will be used when χχ has three nonprincipal irreducible constituents. In particular, we will show that if χχ = 1G+m1α1+m2α2+m2α2, then both ker(α1) and ker(α2) must be abelian.

Lemma 3.12. Let G be a ﬁnite solvable group. Let V be a symplectic vector space of dimension n over GF(q), where q is a power of a prime number. Assume that V is a faithful G-module and the action of G on V preserves the symplectic form. Also assume that G acts transitively on V #. Set e2 = |V | = qn. Let F(G) be the Fitting subgroup of G and F2(G)/F(G) be the Fitting subgroup of G/F(G). Then one of the following holds:

∼ ∼ (i) e = 2, and either G = S3 or G = Z3.

∼ ∼ (ii) e = 3, and either G = Q8 or G = SL(2, 3).

(iii) e = 5, and G ∼= SL(2, 3).

(iv) e = 7, and G ∼= GL(2f , 3).

20 5 (v) e = 9, F(G) is an extra-special group of order 2 , |F2(G)/F(G)| = 5, and

G/F2(G) is a subgroup of the cyclic group of order 2.

Proof. See Theorem 2.2.1 of [2].

21 Chapter 4

Two Nonprincipal Irreducible

Constituents

The ﬁrst lemma will be useful in all of the following situations. It is based oﬀ Lemma

4.2.1 of [2].

Lemma 4.1. Let G be a ﬁnite solvable group. Let χ ∈ Irr(G) be a faithful character.

Assume n X χχ = 1G + miαi, i=1

# where the αi ∈ Irr(G) are distinct and the mi ∈ N. Let N be a normal subgroup of

G. Then χN ∈ Irr(N) if and only if N 6≤ ker(αi) for all i = 1, 2, . . . , n.

Proof. Notice that

n n X X [χN , χN ] = [(χχ)N , 1N ] = [1N + mi(αi)N , 1N ] = 1 + mi[(αi)N , 1N ]. i=1 i=1

Thus, [χN , χN ] = 1 if and only if [(αi)N , 1N ] = 0 for all i = 1, 2, . . . , n. Therefore,

[χN , χN ] = 1 if and only if N 6≤ ker(αi) for all i. Since [χN , χN ] = 1 if and only if

22 χN ∈ Irr(N), the result follows.

For Theorem A, we will need to examine all the cases from Theorem 3.10. Fortu-

nately, most of these cases have been handled elsewhere, in particular in Section 2 of

[13]. The only remaining case that needs dealing with is the situation where V is a

G-module of order 36, which was handled in [7]. It is stated here for convenience.

Lemma 4.2. If H is a solvable group that has a faithful, primitive, and irreducible

module V of order 36, then dl(H) ≤ 5.

Proof. See Lemma 5 of [7].

Based on this lemma, we are now ready to prove Theorem A.

Theorem A. Let G be a ﬁnite solvable group and let χ ∈ Irr(G) be a faithful character. Assume that

χχ = 1G + m1α1 + m2α2,

# where α1, α2 ∈ Irr(G) are distinct characters and m1 and m2 are strictly positive

integers. Then dl(G) ≤ 8.

Proof. Let Z = Z(G). From Lemma 3.1.2 and Lemma 4.2.4 of [2], we know that T Z = ker(αi) and that when there are only two nonprincipal irreducible constituents

of χχ, then Z is equal to the kernel of at least one of these constituents. For this

proof, we must consider two cases: ker(α1) is abelian and nonabelian. First, suppose

that ker(α1) is nonabelian. This implies Z = ker(α2). Then we can take E/Z to

be a chief factor of G such that E ≤ ker(α1). Notice that E is nonabelian, since

23 we can take a character θ ∈ Irr(ker(α1)), with the property that it is nonlinear and

θE ∈ Irr(E). Note that dl(E) = 2 since E/Z is a chief factor of a solvable group and is thus abelian, and Z = Z(G). By Proposition 4.2.13 of [2], we know that G/E is isomorphic to either SL(2, 3) or the group isoclinic to GL(2, 3), which we denote by GL(2f , 3). Thus, the derived length of G/E is either 3 or 4 and thus implies that dl(G) ≤ dl(G/E) + 2, which is 5 or 6.

This leaves us with the case where ker(α1) and ker(α2) are both abelian. From

Proposition 4.2.5 and Proposition 4.2.9 of [2], we know χ(1)2 = |E : Z| and χ(1) is a power of a prime. If ker(α1) 6= ker(α2), then Proposition 4.2.9 of [2] also implies that dl(G) ≤ 6. So, we assume that ker(α1) = ker(α2) = Z. Again, let E/Z be a chief factor of G. We know from Proposition 4.2.5 of [2] that E/Z is a faithful irreducible module of G/E and that G/E has exactly two orbits on Irr(E/Z)#. Thus, G/Z is a permutation group of rank 3 and by Theorem 3.10, there are three situations. In the ﬁrst case, G/E is metacyclic, and thus dl(G) ≤ dl(G/E) + 2 ≤ 4. In the second case, G/E is an imprimitive linear group, which implies that dl(G) ≤ dl(G/E) + 2 ≤

5 + 2 = 7.

We now consider the possibility that G/Z satisﬁes conclusion (iii) of Theorem 3.10.

Thus, E/Z is primitive, irreducible, and faithful as a module for G/E and the degree of

G/Z as a permutation group is one of the degrees listed in conclusion (iii) of Theorem

3.10. As mentioned before Theorem 3.10, the degree of G/Z as a permutation group is |E : Z|. Thus, |E : Z| is one of 72, 132, 172, 192, 232, 292, 312, 472, 34, 74, 26, or

36, most of which have been handled in [13]. In particular, if |E : Z| = p2 for a prime p, then G/E is isomorphic to a subgroup of GL(2, p) and so we may apply Lemma

24 3.7 to see that dl(G/E) ≤ 4, which implies that dl(G) ≤ 6. If |E : Z| = 34 or 74, then we see that E/Z has dimension 4 over the ﬁnite ﬁeld Zq where q is either 3 or 7.

5 Thus, we may apply Lemma 3.8 to see that dl(G/E) ≤ 8+ 2 log(4/8) ≈ 6.42. We can

conclude that dl(G/E) ≤ 6 and dl(G) ≤ 8. Notice that for our purposes, we could

have also appealed to Lemma 3.8 when |E : Z| = p2. If |E : Z| = 26, then we have

that G/E is isomorphic to a subgroup of GL(6, 2), and hence we may apply Lemma

3.7 to see that dl(G/E) ≤ 6 and dl(G) ≤ 8.

The remaining possibility is that |E : Z| = 36. Unfortunately, as we mentioned

above, this case is not speciﬁcally handled in [13] and the bound that Lemma 3.8

provides is larger than the bound we want to prove. However, by Lemma 4.2, we

know that dl(G/E) ≤ 5, and so dl(G) ≤ 7. This completes the proof.

The next theorem will allow us to prove the existence of some examples.

Theorem 4.3. Let H be a group acting on an extra-special group E of order p2n+1.

Suppose that this action is faithful, irreducible, and has three orbits on Irr(E/Z(E)).

Also suppose that the action centralizes Z(E). If G is the semi-direct product of H

acting on E, then there exists a faithful character χ ∈ Irr(G) of degree χ(1) = pn

such that χχ = 1G + α1 + α2, where α1, α2 ∈ Irr(G) are distinct and the degrees of

these characters are exactly the orbit sizes.

Proof. Let Z = Z(E). Since H acts irreducibly on E/Z and Z = Z(E) = Z(G), we

have that E/Z is a chief factor of G. Let ψ ∈ Irr(E) be nonlinear. Then, since E/Z

is fully ramiﬁed, ψZ has a unique irreducible constituent λ. Since H centralizes Z,

it must be that λ extends to Z × H. We will show that Theorem 6.1 of [8] can be

25 applied, and using that theorem, we have that ψ extends to a character χ ∈ Irr(G),

where G = E o H. Then by Lemma 4.3.1 of [2], we know that this χ has degree

χ(1) = pn and χχ has the desired form.

There are two facts which make the use of Theorem 6.1 of [8] possible. The ﬁrst is that noncentral chief factors are strong. That is, they satisfy Theorem 6.1 (iii) of

[8]. In [8], Isaacs proved this fact, and we reproduce it here for the convenience of the reader. We have that G is a solvable group and E/Z is a noncentral chief factor

of G. Let C = CG(E/Z) < G and choose N such that N/C is also a chief factor of

G. Then S = N/C satisﬁes the conditions of the deﬁnition. That is, CE/Z (S) = 1,

(|S|, |E/Z|) = 1, and S is solvable. In our situation, since H acts nontrivially on

E/Z, we know that E/Z is a noncentral chief factor and is thus a strong section.

The second relevant fact, which we shall prove next, is that the complements of E in G are conjugate. This implies that H will be conjugate to the complement found in Theorem 6.1 of [8]. From the last paragraph, we know that E/Z is a strong section of G. Also, for C = CG(E/Z), we know that E ≤ C. Furthermore, Dedekind’s

Lemma yields that C = E(C ∩ H), and since C ∩ H is the kernel of the action of

H on E/Z and this action is faithful, this implies that C = E. Now, let N be a

subgroup such that N/E is also a chief factor of G. Since E/Z is strong, we know

that CE/Z (N/E) = 1 and (|N/E|, |E/Z|) = 1. Thus, E is a normal Hall-subgroup of

N and the Schur-Zassenhaus Theorem states that all the complements of E in N are

in fact conjugate. Thus, their normalizers are conjugate, i.e., if H1 and H2 are both

complements of E in N, then NG(H1) is conjugate to NG(H2).

Next, we want to prove that all complements of E in G are normalizers of com-

26 plements of E in N. Recall that G = E o H and take L = HZ ∩ N = Z(H ∩ N).

Then EL = EH ∩ N = N and E ∩ L = E ∩ HZ = Z. Also, since N C G, we know that L C HZ. Thus, HZ ≤ NG(L) and [NG(L) ∩ E,L] ≤ Z. This implies that

NG(L)∩E ≤ CN/L(L/Z) = 1. Hence, NG(L) = H(NG(L)∩E) and HZ = NG(HZ∩L).

Therefore, HZ ∩ L is a complement of the Hall-subgroup E in N and its normalizer

HZ is a complement of E in G. Thus, complements of E in G are normalizers of complements of E in N. Therefore, the complements of E in G are conjugate, as they are normalizers of conjugate subgroups, which is what we wanted to prove.

For odd primes, this yields examples of our groups whenever there exists a module

with the appropriate orbit structure.

Corollary 4.4. Let p be an odd prime. If the group H acts faithfully, irreducibly, and symplectically on a vector space V of order p2n with exactly three orbits, then there exists a group G with a faithful character χ ∈ Irr(G) satisfying χ(1) = pn and

χχ = 1G + α1 + α2 for distinct α1, α2 ∈ Irr(G).

Proof. Let E be an extra-special group of order p2n+1 and exponent p. Then by

the comments made on page 404 of [6], Sp(2n, p) is isomorphic to a subgroup of

Aut(E) which centralizes Z(E). Since H acts symplectically on V , we know that H is isomorphic to a subgroup of Sp(2n, p) and hence, H acts via automorphisms on

E. Furthermore, the action of H on E/Z(E) is isomorphic to the action of H on V .

Thus, if we let G = E o H, then we have satisﬁed the hypotheses of Theorem 4.3, and that theorem gives the conclusion.

When p = 2, it is not enough for the action of H to be symplectic. To apply

27 Theorem 4.3, we need H to be a subgroup in the appropriate orthogonal group. In particular, to obtain our example of a solvable group G with dl(G) = 8, we need a

− subgroup of O6 (2). Such an example will be presented later in Chapter 7.

To prove Theorem B, we will need two new lemmas. The ﬁrst includes more information than is needed at this point, since we will be using it again for Lemma

5.4.

Lemma 4.5. Let G be a solvable group. Let V be a vector space of order qn for qn = p2, where p is a odd prime and p ≤ 71, or qn ∈ {34, 54, 74, 36, 310}. Assume that

V is a faithful, irreducible, and primitive G-module. Finally, assume that G 6≤ Γ(V ).

Then G contains the central involution of GL(n, q).

Proof. Since V is a primitive faithful G module, it restricts homogeneously to every abelian normal subgroup of G. Thus, every abelian normal subgroup of G has a faithful, irreducible module. This implies that every normal abelian subgroup of G is cyclic, and we may apply Lemma 3.2 to G. This result will be used extensively in what follows.

Set F = F(G). First, assume that dim(V ) = 2. Then by Lemma 3.7 F = QT ∼ where Q = Q8, the quaternion group of order 8, and T ≤ Z (GL(2, p)). Also, Q∩T = ∼ Z(Q) = Z2. Thus, if dim(V ) = 2, the central involution is contained in G.

Next, assume that dim(V ) = 4. Then since V is a faithful F -module, we know that q does not divide |F |, where q ∈ {3, 5, 7}. Since G is not a subgroup of Γ(V ), we may assume that F is nonabelian by Lemma 3.3. Let Z be the socle of Z(F ). Then by the ﬁrst conclusion of Lemma 3.2 there exist normal subgroups Q and T of G such

28 that F = QT and Q ∩ T = Z. Also, the Sylow r-subgroups of Q are extra-special or

cyclic of prime order by the second conclusion of Lemma 3.2. Hence, by Lemma 3.5,

we have that there exists an integer e such that e2 = |F : T | = |Q : Z|. Notice that

e divides dimZq (W ), where V = f · W for some integer f. Since q cannot divide e,

we have that e ∈ {1, 2, 4}. Furthermore, e 6= 1, since otherwise, Lemma 3.3 implies

that G is a subgroup of Γ(V ), which we have removed by assumption. So, |F | is

even. Hence, |Z(F )| is also even. Since Z(F ) is cyclic, this implies it has a unique subgroup L of order 2. Now, since V is irreducible, the involution x ∈ L can either

fix everything in V or nothing. If x fixes everything, then x is in the kernel of the action of G on V . However, this is a contradiction since we are assuming that V is a faithful module. Therefore, x fixes nothing and sends every element v ∈ V to its inverse −v. In particular, x is the central involution of GL(V ) and x ∈ G.

Next, assume that dim(V ) = 6. Then, by a similar argument, we know that e divides 6 and q = 3 cannot divide e. Thus, we see that e = 2, and |F | is again even. Then we may apply the previous argument to F to get that G must contain the central involution of GL(6, 3).

Lastly, assume that dim(V ) = 10. The previous argument works in the case that e = 2 or 10 but fails when e = 5. However, by Theorem 1.2 of [5] and comments made on page 1 of that paper, we get that G has a minimal normal nonabelian subgroup N with even order. From this, we can use a similar argument for all possible e, which completes the proof.

We are now ready to prove the following lemma, which will make Theorem B

29 possible.

Lemma 4.6. Let G be a ﬁnite solvable group. Let V be a symplectic vector space of dimension 2n over GF(q), where q is a prime power. Assume that V is a faithful irreducible G-module and the action of G on V preserves the symplectic form. Also

# assume that G acts with two orbits on V = V − {0}. Call them O1 and O2. Set

2 2n e = |V | = q . Then if v ∈ Oi, so is −v.

Proof. Notice that our hypotheses imply the hypotheses of Theorem 3.10. Therefore, we will study the cases given there. We know that v and −v are in orbits of equal size. So, if O1 and O2 have diﬀerent sizes, we obtain the conclusion. Hence, we may assume that the orbits have equal size, which implies that |G| is divisible by

1 2 1 2n 2 (e − 1) = 2 (q − 1).

Assume Theorem 3.10(i). Then G is isomorphic to a subgroup of Γ(V ), N =

G ∩ Γ0(V ) is a normal cyclic subgroup of G, and G/N acts faithfully on N. Also,

since N is cyclic, we know that |Aut(N)| = ϕ(|N|), where ϕ is the Euler ϕ-function,

and since G/N acts faithfully on N, we have that |G| ≤ |N|ϕ(|N|).

Suppose that |N| is not a prime number and is not equal to 4. Then ϕ(|N|) ≤

|N| − 3. By Lemma 2.2.3 of [2], |N| ≤ e + 1. So ϕ(|N|) ≤ e − 2, which implies that

2 1 2 # |G| < e − 1. Therefore, |G| = 2 (e − 1), G acts Frobeniusly on V , as does N.

Also, G having even order and N being cyclic imply that N has a unique subgroup of order 2. But there is only one element in Γ(V ) that acts Frobeniusly, which is the element that sends v to −v. This implies that elements are in the same orbits as their inverses.

30 Now suppose that |N| = p, a prime. Then p ≤ e + 1. If p < e + 1, then p ≤ e, and

|G| ≤ |N|ϕ(|N|) = p(p − 1) ≤ e(e − 1) < e2 − 1,

which yields the same results as the last paragraph. So assume that p = e+1. Notice

2 1 that p(p−2) = e −1 and 2 p(p−2) in an integer dividing |G|. However, e ≥ 3 implies

1 that p > 4 is an odd prime. Hence, p(p − 2) is an odd integer, and 2 p(p − 2) is not

an integer. This is a contradiction.

1 2 So assume that |N| = 4. Then |G| divides 8, and thus 2 (e − 1) divides 8. Hence,

1 2 2 2 2 (e − 1) is 1, 2, 4, or 8. Since V is a symplectic vector space, |V | = e = r for some

# ∼ prime power r. Thus, |V | = 8 and G ≤ SL(2, 3). Therefore, G = Q8, the quaternion group of order 8. However, since Q8 acts on a vector space of order 9 transitively, this case is impossible.

Next, assume Theorem 3.10(ii). Then the vector space V has two spaces of imprimitivity V1 and V2 with V = V1 ⊕ V2. If H = NG(V1), we know that H/CH (V1)

# acts transitively on V1 and v ∈ V1 is in the same orbit as its inverse.

Finally, assume Theorem 3.10(iii). Then by Lemma 4.5, we know that G contains

the central involution of GL(2n, q) and therefore v and −v are in the same orbit, as

desired.

Using this lemma, it is now possible to prove Theorem B.

Theorem B. Let G be a ﬁnite solvable group with a faithful character χ ∈ Irr(G)

31 such that

χχ = 1G + m1α1 + m2α2,

# where α1, α2 ∈ Irr(G) are distinct characters and m1, m2 are strictly positive inte-

gers. Then both α1 and α2 are real-valued characters.

Proof. First, notice that χχ is a real-valued character. So, if ker(α1) 6= ker(α2) and αi

is a constituent of χχ, then αi is also a constituent of χχ for i = 1, 2 with ker(αi) =

ker(αi). In our case, if ker(α1) 6= ker(α2), then αi = αi, and αi is a real-valued character. So, let Z = Z(G) and assume that ker(α1) = ker(α2). Then by Lemma

4.2.4 of [2], we know that Z = ker(χχ) = ker(α1) ∩ ker(α2) = ker(α1) = ker(α2).

Let E/Z be a chief factor of G. Then by Proposition 4.2.5 of [2], E/Z is a fully

ramiﬁed section of G with respect to χE and λ ∈ Irr(Z) such that [χZ , λ] 6= 0. Also,

G/E acts symplectically and faithfully on E/Z with two orbits on (E/Z)#. Suppose

the αi are complex-valued characters that are not real-valued. Then since χχ is real-

# valued, α2 = α1. Hence the orbits of G/E on Irr(E/Z) are inverses of each other, i.e., if the nontrivial orbits are O1 and O2 and θ ∈ O1, then θ ∈ O2. However, by

Lemma 4.6, we know that θ and θ must belong to the same orbit. Thus, we have a

contradiction of the assumption that αi are complex, which completes the proof.

32 Chapter 5

Three Nonprincipal Irreducible

Constituents-The Special Case

We are now ready to consider the situation when χχ has three nonprincipal irreducible constituents. This situation has two cases. The ﬁrst, which we discuss in this chapter, is when two of the constituents are complex conjugates. The second case will be discussed in the next chapter.

Hypothesis 5.1. Let G be a ﬁnite solvable group with χ ∈ Irr(G) a faithful character.

Assume

χχ = 1G + m1α1 + m2α2 + m2α2,

# where the αi ∈ Irr(G) are distinct and the mi are strictly positive integers for all i = 1, 2. Set Z = Z(G).

Lemma 5.2. Let X be a ﬁnite group with structure X = R × S and θ a character of

33 X. Assume that θ(g) = 0 if g ∈ X − R, or if g ∈ X − S. Then

  0 if g 6= 1 θ(g) =  θ(1) otherwise.

Therefore, θ is a multiple of the regular character.

Proof. See [2], Lemma 4.2.2.

Lemma 5.3. Let G be a ﬁnite solvable group with χ ∈ Irr(G) faithful such that

χχ = 1G + m1α1 + m2α2 + m3α3,

with ker(α2) = ker(α3). Let Z = Z(G). Then either ker(α1) = Z or ker(α2) = Z. In particular, if we have Hypothesis 5.1, then ker(α1) = Z or ker(α2) = Z.

Proof. Here, we follow the style of Lemma 4.2.4 in [2]. Since α2 and α3 have the same kernel, we need only consider ker(α1) and ker(α2). Assume that Z is proper in both kernels. Then let R/Z be a chief factor of G with R ≤ ker(α1). Note that

R 6≤ ker(α2). Also, let S/Z be a chief factor of G with S ≤ ker(α2). Set T = RS.

Then R ∩ S = Z since ker(α1) ∩ ker(α2) = Z. This implies that T 6≤ ker(αi) for i = 1, 2. By Lemma 4.1, χT ∈ Irr(T ).

Notice that T/R is also a chief factor of G. Let ψ ∈ Irr(R) such that [χR, ψ] 6= 0.

2 Then by Theorem 2.6, either χR = eψ for some ψ ∈ Irr(R) and e = |T : R| or

T ψ = χT . Both of these cases imply that χ(g) = 0 if g ∈ T − R. Similarly, χ(g) = 0 if g ∈ T − S. Also, since R/Z and S/Z are chief factors of G, they are elementary

34 abelian. Since R and S are normal subgroups of G with intersection Z and product

T , we have that T/Z = R/Z × S/Z. Thus, Lemma 5.2 gives us

  0 if g ∈ T − R,    χχ(g) = 0 if g ∈ T − S,     2 χ(1) if g ∈ R ∩ S = Z.

Lemma 5.2 also tells us that χχ is a multiple of the regular character of RS/Z. Thus,

2 2 χ(1) ≥ |T : Z|. However, since χT ∈ Irr(T ) it is also true that χ(1) ≤ |T : Z|.

T Hence, equality holds and (χχ)T = 1Z is the regular character of T/Z. Since all

T characters in Irr(T/Z) are linear, they appear with multiplicity 1 in 1Z = (χχ)T and

mi = 1 for i = 1, 2.

Notice that

T [(χχ)R, 1R] = [((χχ)T )R, 1R] = [(1Z )R, 1R] = |T : R|.

Since R ≤ ker(α1) and R 66= ker(α2), we have

1 + α1(1) = |T : R|.

In a similar fashion, 1 + α2(1) + α3(1) = |T : S|. This yields the equation

2 |T : Z| = χ(1) = 1 + α1(1) + α2(1) + α3(1) = |T : R| + |T : S| − 1.

35 Also since T = RS and R ∩ S = Z, we can rewrite this as

|T : R||R : Z| − |T : R| − |R : Z| + 1 = 0.

Thus,

(|T : R| − 1)(|R : Z| − 1) = 0.

But since T/R and R/Z are chief factors of G, neither of them can have size 1.

Therefore, Z = ker(α1) or Z = ker(α2) = ker(α3), as desired. In particular, if we

consider the situation outlined in Hypothesis 5.1, we see that since α2 and α2 are complex conjugates, they have the same kernel. Thus, if we assume Hypothesis 5.1, we get the conclusion.

Thus, it remains to go through all the possible equalities between ker(α1), ker(α2),

and Z = Z(G). We will begin with the case that all three subgroups are equal. In

fact, this is impossible. Then we will consider the case when one of the kernels is

an abelian group properly containing Z(G). Lastly, we will assume that one of the

kernels is a nonabelian group, properly containing Z(G), which is equal to the other

kernel. For the case when Z(G) = ker(α1) = ker(α2), we will need Theorem 3.11.

Lemma 5.4. Let G be a ﬁnite solvable group. Let V be a symplectic vector space

of dimension 2n over GF(q) for some prime power q. Assume that V is a faithful

irreducible G-module. Suppose the action of G on V preserves the symplectic form,

# 2 2n and that G acts on V with three orbits, say O1,O2, and O3. Set e = q = |V |.

Then v ∈ Oi implies that −v ∈ Oi for 1 ≤ i ≤ 3.

36 Proof. If q = 2, then our conclusion is immediate since V is an elementary abelian

2-group and v = −v in this case. Thus, we will assume that q is an odd prime.

Also, our hypotheses imply those of Theorem 3.11, so we will examine the three

situations there. Notice that v and −v must be in orbits of equal size. So, if all three

nontrivial orbits have distinct sizes, the result is trivial. Hence, we may assume that

two orbits have the same size, which implies that the orbit sizes are a = |O2| = |O3|

2 and e − 1 − 2a = |O1|, both of which must divide |G|.

Assume Theorem 3.11(i). Then G ≤ Γ(V ) and N = G ∩ Γ0(V ) is a cyclic normal

subgroup of G and G/N acts faithfully on N. This means that |Aut(N)| = ϕ(|N|),

where ϕ is the Euler ϕ-function, and |G| ≤ |N|ϕ(|N|). First assume that |N| is not

prime and is not equal to 4. Then ϕ(|N|) ≤ |N| − 3 and |N| ≤ e + 1 by Lemma 2.2.3

of [2]. Thus, |G| < e2 − 1.

1 2 Claim 1. |G| ≥ 3 (e − 1).

Proof. Since max(a, e2 − 1 − 2a) ≤ lcm(a, e2 − 1 − 2a) ≤ |G|, it suﬃces to show that

2 1 2 1 2 max(a, e − 1 − 2a) ≥ 3 (e − 1). If a ≥ 3 (e − 1), then we are ﬁnished. So assume

1 2 2 2 2 2 2 2 1 2 a < 3 (e −1). Then −2a > − 3 (e −1) and e −1−2a > e −1− 3 (e −1) = 3 (e −1).

This completes the claim.

1 2 2 By the claim, 3 (e − 1) ≤ |G| < e − 1. Then

|G| 1 (e2 − 1) e − 1 qn − 1 2n ≥ ≥ 3 = = . |N| e + 1 3 3

Thus, we need prime powers qn such that this inequality holds. This implies that

37 2n 2 2 4 2 2 2 q = e ∈ {3 , 3 , 5 , 7 }. Now, |N| divides |Γ0(V )| = e − 1 and |G : N| divides the

size of the Galois group. Here, we may assume that |N| is odd as well since otherwise,

it contains the central involution of the ﬁeld and we are ﬁnished.

If q2n = 34, then |N| divides 80, and in particular, |N| = 5. This implies that

80 1 2 |G : N| divides 4 and |G| ≤ 20. But we know that 20 < 3 = 3 (e − 1) ≤ |G|. This

is a contradiction. If q2n = 32, then |N| divides 8. Since N cannot be the trivial

group, it has even order in this case and we are ﬁnished. If q2n = 52, then |N| = 3

and |G| ≤ 6. However, our claim implies that |G| ≥ 8, another contradiction. Lastly,

if q2n = 72, then |N| = 3 and |G| ≤ 6. But since our claim implies that |G| ≥ 16, we

have a contradiction here as well.

Next assume that |N| = p for a prime p. Then |N| = p ≤ e+1 and ϕ(|N|) = p−1.

Assume ﬁrst that p < e + 1, i.e., p ≤ e. Then we get that |G| < e2 − 1 and the claim

1 2 2 yields that 3 (e −1) ≤ |G| < e −1. This yields the same results as the last paragraph.

So, assume that p = e + 1. Then p is a Fermat prime and e = qn is a power of 2, which is ﬁnished by our earlier remark.

Now assume that |N| = 4. Then since |N| is even, it contains the central involution of Γ0(V ) and this element sends elements to their inverses. This completes this case.

r Assume Theorem 3.11(ii). Then G is an imprimitive linear group and V = ⊕j=1Vj where the Vj are imprimitivity spaces and r = 2 or 3. We know that H = NG(V1)

# has index r in G, and that H = H/CH (V1) acts transitively on V1 . Thus, Lemma

3.6 implies that G ≤ H o Zr, and we may apply Theorem 3.9 to H.

First, suppose that r = 2 and that Theorem 3.9(i) holds, i.e., H ≤ Γ(V1). Since q

# is an odd prime, we know that |V1| is odd, and |V1 | is even. Hence, H is also even and

38 contains some involution which inverts some element of V1. This handles the orbit

# # which has form V1 ∪ V2 . Now, let v = (a, 0) ∈ V − (V1 ∪ V2) with a 6= 0. Then, for an involution g ∈ G, we know that v · g = (0, b) = w. Now, consider v − w = (a, −b).

Then (v − w) · g = v · g − w · g = w − v = − (v − w). Thus, the orbit containing v − w ∈ V − (V1 ∪ V2) contains inverses. Since v − w is in one of the orbits that make up V − (V1 ∪ V2), it follows that the second orbit contains inverses as well.

2 2 2 2 2 4 Next, assume Theorem 3.9(b),(c). Then, H ≤ GL(V1) where |V1| = 3 , 5 , 7 , 11 , 23 , 3 , and Lemma 4.5 yields that G contains the central involution. Therefore, the orbits contain inverses, as desired.

So assume that there are three imprimitivity spaces, i.e., V = V1 ⊕ V2 ⊕ V3.

Hence, elements of V are 3-tuples, meaning they have form (v1, v2, v3), where vi ∈ Vi for 1 ≤ i ≤ 3. Since 0 ∈ Vi is centralized by all g ∈ G, if v · g = w, it is impossible to v to have a diﬀerent number of nonzero coordinates than w. Hence, we obtain the following orbits: {(0, 0, 0)}, {(a, b, c)|exactly one of a, b, c is nonzero},

{(a, b, c)|exactly two of a, b, c are nonzero}, and {(a, b, c)|a, b, c 6= 0}. By their deﬁni- tion, v and −v must be in the same orbit, since they will have the same number of nonzero coordinates. Thus, we are ﬁnished in this case.

Finally, assume Theorem 3.11(iii). Then by Lemma 4.5, we have that G contains the central involution, and we are ﬁnished.

We will use the last lemma to prove the following proposition.

Proposition 5.5. Assume Hypothesis 5.1. Then ker(α1) and ker(α2) are distinct.

Proof. Suppose that Z = ker(α1) = ker(α2) and let E/Z be a chief factor of G. Let

39 λ ∈ Irr(Z) such that [χZ , λ] 6= 0. By Lemma 4.1, χE ∈ Irr(E) since E 6≤ ker(αi) for

i = 1, 2. Then Theorem 2.6 implies that E/Z is a fully ramiﬁed section with respect to

χE and λ. By appealing to Lemma 3.2.1 of [2], we see that CG(E/Z) = ECG(E) and

E/Z is a symplectic vector space. Notice that since χ ∈ Irr(G) is faithful, Z(χ) = Z.

Also, χE ∈ Irr(E) implies that CG(E) ≤ Z(χ) = Z by Lemma 3.2.2 of [2]. Hence,

CG(E/Z) = E and G/E acts faithfully on E/Z. Furthermore,

E (χχ)E = 1Z ,

by Lemma 3.1.1 of [2].

Now, we have that

E 1Z = (χχ)E = 1E + m1 (α1)E + m2 (α2)E + m2 (α2)E .

By Cliﬀord Theory, the irreducible constituents of (α1)E form an orbit. If the irre-

ducible constituents of (α1)E,(α2)E, and (α2)E are all the same, then there is a single

orbit under the action of G/E on Irr(E/Z)# and on the action of G/E on (E/Z)# by

Brauer’s Theorem (Theorem 6.32 of [9]). However, by Theorem A of [2], this means

# that χχ = 1G + α for some α ∈ Irr(G) , which is a contradiction. So, assume that two of the orbits given by (α1)E,(α2)E, and (α2)E are the same. Then by Proposition

# 4.3.3 of [2], this implies that χχ = 1G + α + β for some α, β ∈ Irr(G) , which is also

a contradiction. So, we suppose that the orbits given by (α1)E,(α2)E, and (α2)E are

all distinct, namely the orbits of α2 and α2 are not the same. Then, their orbits must

40 be inverses of each other. However, by Lemma 5.4, we know that since we have three

orbits, if x ∈ O for some orbit, then x−1 ∈ O also. This is a contradiction. Thus, ker(α1) 6= ker(α2), and the proof is complete.

Next, we will discuss the option that one of the kernels is an abelian group strictly

containing the center of G.

Lemma 5.6. Let G be a ﬁnite solvable group with χ ∈ Irr(G) faithful and

n X χχ = 1G + miαi, i=1

# where the αi ∈ Irr(G) are distinct, mi ∈ N, and n ≥ 2. Fix j such that K = ker(αj) is maximal among the ker(αi). Suppose that K is abelian, with K > Z(G) = Z. Let

θ ∈ Irr(K) such that [χK , θ] 6= 0. Denote the inertia group of θ by Gθ. Let L/K be a

chief factor of G. Then

(i) G = GθL and K = Gθ ∩ L.

(ii) G/L acts faithfully and irreducibly on L/K.

(iii) χ(1) = |L : K|, and thus is a prime power.

Proof. Since L > ker(αi) for all i, χL ∈ Irr(L) by Lemma 4.1. Since χK is reducible,

L K is abelian, and Z(χ) = Z 6= ker(α1), Theorem 2.6 implies that χL = θ and

Gθ is proper in G. By Exercise 5.7 of [9], we have that GθL = G. Also, we know

L that θ = χL ∈ Irr(L) and so by Exercise 6.1 of [9], we know that Lθ = K. But

Lθ = Gθ ∩ L, and thus K = Gθ ∩ L.

41 Now, observe that L ≤ CG(L/K) /G. If we let C = CG(L/K), then G = GθL

implies that C = (C ∩ Gθ)L. Also notice that C ∩ Gθ is normal in Gθ and that

[C ∩ Gθ,L] ≤ [C,L] ≤ K. Thus, C ∩ Gθ is normal in G. If L < C then K is proper

in the normal subgroup C ∩ Gθ and χC∩Gθ is an irreducible character. However, since

χ is induced from some character of Gθ and C ∩ Gθ ≤ Gθ < G, it is impossible for

χC∩Gθ to be irreducible. So it must be the case that C = L. This proves that the action is faithful.

L Since χL = θ and θ(1) = 1,

χ(1) = θL(1) = |L : K|θ(1) = |L : K|,

which is a prime power since L/K is a chief factor of a solvable group. Thus, χ(1) is

also a prime power. Lastly, since K is maximal among the kernels, Lemma 3.1.3 of

[2] yields that mj = 1.

Proposition 5.7. Assume Hypothesis 5.1. Assume that K = ker(α1) is an abelian

group properly containing Z(G) = ker(α2). Then dl(G) ≤ 6.

Proof. Let θ ∈ Irr(K) such that [χK , θ] 6= 0 and let L/K be a chief factor of G. Then

by Lemma 5.6, we know that G = GθL, where Gθ is the inertia group of θ in G, and

G/L acts faithfully and irreducibly on L/K. Also by that lemma, m1 = 1.

L By Lemma 3.1.1 of [2], (χχ)L = 1K +Φ where Φ is a character of L and [ΦK , 1K ] =

L 0. Also, since (χχ)L = 1L +(α1)L +m2(α2)L +m2(α2)L, we have that 1K = 1L +(α1)L

# and Φ = m2(α2)L + m2(α2)L. This implies that G/L acts transitively on Irr(L/K)

and hence on (L/K)# by Brauer’s Theorem (Theorem 6.32 of [9]). Therefore, G/L is

42 one of the groups in Theorem 3.9. Hence, dl(G/L) ≤ 4. Since L/K is a chief factor of

G and K is abelian, we have that dl(L) = 2. Therefore, dl(G) ≤ dl(G/L) + dl(L) ≤

4 + 2 = 6.

Before we begin the next proposition, we need the following lemma.

Lemma 5.8. Let G be a ﬁnite solvable group. Let V be a vector space of dimension

n over GF(q), where q is a prime power. Assume that V is a faithful G-module and that G acts on V # = V − {0} with two orbits of equal size. Also assume one of the following situations:

(i) G is an imprimitive linear group.

(ii) G acts as a primitive linear group on V and G has one of the degrees 72, 132,

172, 192, 232, 292, 312, 472, 34, 74, 26, or 36.

Then v and −v belong to the same orbit.

Proof. Suppose case (i). Then V = V1 ⊕ V2 ⊕ · · · ⊕ Vm. Now, if m ≥ 3, we have orbits

of the following forms: an orbit containing n-tuples with only one coordinate equal

to zero, an orbit of n-tuples with two coordinates equal to zero, an orbit of n-tuples

with three coordinates equal to zero, etc. However, we are assuming that there are

# # # # only two orbits. Thus, m = 2 and the orbits are {0}, V1 ∪ V2 , and V − V1 ∪ V2 .

In all these cases, if v belongs to one of them, so does its inverse −v. So, suppose case (ii). Then |V | = qn is one of the prime powers listed in that case, and by Lemma

4.5, G has the central involution of GL(V ). This completes the proof.

43 Proposition 5.9. Assume Hypothesis 5.1. Assume also that K = ker(α2) is an

abelian group properly containing Z. Then dl(G) ≤ 6.

Proof. Let L/K be a chief factor of G. By Lemma 5.6, we know that if θ ∈ Irr(K)

such that [χK , θ] 6= 0, and L/K is a chief factor of G then G = GθL, K = Gθ ∩L, and

G/L acts faithfully and irreducibly on L/K. It remains to show how many orbits are in the action of G/L on L/K. By Lemma 3.1.1 of [2], we know that

L (χχ)L = 1K + Φ,

where Φ is either the zero function or a character of L with [ΦK , 1K ] = 0. Also,

(χχ)L = 1L + m1(α1)L + m2(α2)L + m2(α2)L.

Since ker(α1) = Z < ker(α2), we know that [(α1)K , 1K ] = 0. This means that

L Φ = m1(α1)L and 1K = 1L + m2(α2)L + m2(α2)L. So the irreducible constituents of

# (α2)L + (α2)L form Irr(L/K) . If the orbits of (α2)L and (α2)L are the same, then

G/L acts transitively on Irr(L/K)#, and hence on (L/K)# by Brauer’s Theorem,

implying that dl(G/L) ≤ 4 by Theorem 3.9. If they are diﬀerent then G/L acts on

(L/K)# with two orbits of equal size that are inverses of each other. By Lemma

5.8 and Theorem 3.10, this means that G/L is isomorphic to a subgroup of Γ(L/K).

However, these subgroups are metacyclic, which implies that dl(G/L) ≤ 2. Finally,

since L/K is a chief factor of G and K is abelian by assumption, we know that dl(L) = 2 and thus dl(G) ≤ 6.

44 We are now ready to consider the case when one of the kernels is a nonabelian

group which properly contains the other. First, we will need several lemmas which

extend directly from claims made in [2].

Lemma 5.10. Let G be a ﬁnite solvable group with χ ∈ Irr(G) a faithful character.

Suppose also that n X χχ = 1G + miαi, i=1

# where the αi ∈ Irr(G) are distinct, the mi ∈ N, and n ≥ 2. Also, for all i, deﬁne

Ki = ker(αi). Let I be a set of indices such that I < {1, 2, . . . , n} and such that T Z(G) < L, where L = i∈I Ki. Let θ ∈ Irr(L). Then for E ≤ L such that E 6≤ L∩Kj for j∈ / I , θE ∈ Irr(E).

Proof. Suppose that θE is reducible. Then [θE, θE] > 1. Let χL = fθ + ∆ for some character ∆ of L. Then since f = [χL, θ] 6= 0 we have

[χL, χL] = [fθ + ∆, fθ + ∆]

= f 2[θ, θ] + [∆, ∆]

= f 2 + [∆, ∆]

2 < f [θE, θE] + 2f[θE, ∆E] + [∆E, ∆E]

= [χE, χE].

Thus, [(χχ)E, 1E] = [χE, χE] > [χL, χL] = [(χχ)L, 1L]. But since E and L are contained in precisely the same Ki, it must be the case that [(χχ)E, 1E] = [(χχ)L, 1L].

Therefore, θE ∈ Irr(E), as desired.

45 Lemma 5.11. Let G be a ﬁnite solvable group with χ ∈ Irr(G) a faithful character.

Suppose also that n X χχ = 1G + miαi, i=1

# where the αi ∈ Irr(G) are distinct, the mi ∈ N, and n ≥ 2. Also, for all i, deﬁne

Ki = ker(αi). Let K ≤ G be a nonabelian subgroup of G. Suppose that for all Ki, either Ki = K or Ki = Z, where Z = Z(G). Let E/Z be a chief factor of G such that E ≤ K. Then CG(E/Z) = E and G/E acts faithfully on the symplectic vector space E/Z.

Proof. See Claim 4.2.17 of [2].

The next lemma is a generalization of two claims from [2]. In this situation, we

can use Lemmas 5.10 and 5.11 to get that G/E acts transtively on (E/Z)#, as Adan-

Bante does in [2]. Then, it is possible to use Lemma 3.12 to restrict the possible

values of e = |E : Z|1/2.

Lemma 5.12. Let G be a ﬁnite solvable group with χ ∈ Irr(G) a faithful character.

Suppose also that n X χχ = 1G + miαi, i=1

# where the αi ∈ Irr(G) are distinct, the mi ∈ N, and n ≥ 2. Also, for all i, deﬁne

Ki = ker(αi). Let K ≤ G be a nonabelian subgroup of G. Suppose that for all Ki, either Ki = K or Ki = Z, where Z = Z(G). Suppose also that E/Z is a chief factor of G such that E ≤ ker(α1) = K and G/E acts transitively on E/Z. Let

2 θ ∈ Irr(K) be such that [χK , θ] 6= 0 and ϕ = θE ∈ Irr(E). Also let e = |E : Z|. Then

46 e ∈ {2, 3, 5, 7, 9} and the following occur:

(a) If e ∈ {2, 3, 5, 7}, then ϕ extends to G, i.e., there exists some δ ∈ Irr(G) such

that δE = ϕ.

(b) If e = 9, then E < K and α1 is not a faithful character.

Proof. For part (a), see Claim 4.2.20 of [2]. For part (b), see Claim 4.2.39 of [2].

It is now possible to prove that both ker(α1) and ker(α2) must be abelian sub-

groups of G.

Proposition 5.13. Assume Hypothesis 5.1. Then ker(α1) is abelian.

Proof. Assume that ker(α1) is not abelian and let E/Z be a chief factor of G such that

E ≤ K = ker(α1). Let θ ∈ Irr(K) be such that [χK , θ] 6= 0. Since K is nonabelian

and χ is faithful, we know that χK is a sum of G-conjugate nonlinear characters, one of which is θ. Hence, θ(1) > 1. By Proposition 5.3, Z = ker(α2) and by Lemma 5.10,

θE = ϕ ∈ Irr(E). Since ϕ(1) > 1, E/Z is a chief factor of G, and Z = Z(G), we

have that E/Z is a fully ramiﬁed section with respect to ϕ and λ ∈ Irr(Z) such that

1/2 [θZ , λ] 6= 0. Thus, ϕ is G-invariant and ϕ(1) = |E : Z| . Also by Lemma 5.11, G/E

acts faithfully and symplectically on E/Z.

Claim 1. G/E acts transitively on (E/Z)#.

Proof of Claim 1. Since E/Z is a fully ramiﬁed section of G and Z = Z(G),

E (θθ)E = 1Z ,

47 E by Lemma 3.1.1 of [2]. Moreover, since [θE, χE] 6= 0, we know that (χχ)E = 1Z + Φ,

where Φ is a character of E. Since K = ker(α1) ≥ E, our assumption implies that

(χχ)E = (1G + m1α1 + m2α2 + m2α2)E

= (1 + m1α1(1))1E + m2(α2)E + m2(α2)E

E = 1Z + Φ.

By Cliﬀord theory, it must be that the irreducible constituents of (α2)E form a G-

orbit, as do those of (α2)E. If they are distinct orbits, then they must be inverses of

each other. However, this contradicts Lemma 4.6. Therefore, they are the same orbit

and G/E acts transtively on Irr(E/Z)# and hence on (E/Z)# by Brauer’s Theorem

(Theorem 6.32 of [9]).

Thus we can apply Lemma 3.12, which means that e = |E : Z|1/2 ∈ {2, 3, 5, 7, 9}.

Claim 2. e∈ / {2, 3, 5, 7}.

Proof of Claim 2. Since [ϕ, χE] 6= 0 and ϕ extends to δ ∈ Irr(G) by Lemma 5.12(a),

Theorem 2.5 states that there exists some ψ ∈ Irr(G/E) such that χ = δψ.

Since δ extends ϕ and G/E acts faithfully on E/Z, we have that ker(δ) ≤ E. Also since χ is faithful, ker(δ) ≤ E, and χ = δψ, we have that ker(δ) = ker(χ) ∩ E = 1, which implies that δ is a faithful character of G, E/Z is a chief factor of G where

Z = Z(G), and G/E acts transitively on (E/Z)# by assumption. So Lemma 3.3.2 of

[2] gives that δδ = 1G + c1γ for some γ ∈ Irr(G) and an integer c1. By Theorem A of

[2], c1 = 1 and by Lemma 3.1.2 of [2], Z = ker(γ).

48 Pn Now let ψψ = 1G + i=1 aiβi for some positive integers ai and distinct characters

βi ∈ Irr(G) for i = 1, 2, . . . , n. Since ψ ∈ Irr(G/E), we have that βi ∈ Irr(G/E) also

for all i. Then,

1G + m1α1 + m2α2 + m2α2 =χχ

=(δψ)(δψ)

=(δδ)(ψψ)

n X =(1G + γ)(1G + aiβi) i=1 n n X X =1G + γ + aiβi + aiβiγ. i=1 i=1

Now ker(β1) ≥ E and ker(γ) = Z < E. Also, notice that Z = ker(α2). So, γ and α2 have the same kernel. But since γ is real and α2 is complex, this is a contradiction.

Thus, e∈ / {2, 3, 5, 7}.

Claim 3. e 6= 9.

Proof of Claim 3. Recall that E ≤ K = ker(α1). Thus, χE is reducible. Also, E/Z is

a fully ramiﬁed section with respect to θE, which implies that χ(1) = em for m ≥ 2.

Also, since χ(g) = 0 for all g ∈ E − Z, we have that

0 = χχ(g) = 1 + α1(1) + m2α2(g) + m2α2(g).

This means that

1 + α1(g) α2(g) + α2(g) = − m2

49 is both rational and an algebraic integer. Thus, α2(g) + α2(g) ∈ Z. So m2 divides

1 + α1(1) and in particular 2m2 divides 1 + α1(1).

Now, our hypothesis implies that

2 χ(1) 1 + α1(1) α2(1) = − , 2m2 2m2

which yields that

E 1 + α1(1) (α2)E = s1Z − 1E 2m2

for some integer s ≥ 1+α1(1) . Thus, 2m2

2 2 2 1 + α1(1) e m 1 + α1(1) se − = α2(1) = − , 2m2 2m2 2m2

m2 2 and s = . Therefore, 2m2 divides m . 2m2

2 Also, since α2 ∈ Irr(G) and ker(α2) = Z, we have α2(1) < |G : Z| and α2(1)

divides |G : Z|. So assume that e = 9 and recall that χ(1) = em. We will obtain a

contradiction. By Lemma 3.12(v), we have that |G : Z| divides 25920 = 10 × 32 × 81.

By Claim 1 and Lemma 3.12(v) above and Lemma 4.2.12 of [2], we know that α1(1) ∈

{1, 2, 4, 5, 8, 10}. By Lemma 5.12(b) and Lemma 4.2.12 of [2], we can shorten this

1+α1(1) 11 list to α1(1) ∈ {1, 2, 4, 5, 10}, which puts m2 ∈ {1, 3} and gives that < . 2m2 2

1+α1(1) Suppose that m2 = 1. Then since 2 is an integer, we have that α1(1) ∈ {1, 5}.

2 Also, 2m2 divides m , which implies that 2 divides m. If m ≥ 4, then

χ(1)2 1 + α (1) α (1) = − 1 2 2 2

50 4292 1 + 5 ≥ − 2 2 =81 × 8 − 3 = 645.

2 2 Thus, α2(1) ≥ 645 > 25920 ≥ |G : Z|, which is a contradiction since we know

2 that α2(1) must be smaller than |G : Z|. So, assume that m = 2. Then either

α2(1) = 2 × 81 − 1 = 161 or α2(1) = 2 × 81 − 3 = 159. However, neither of these divides |G : Z|. Since α2(1) must divide |G : Z|, this is a contradiction. Therefore, m2 6= 1.

2 2 Now, assume m2 = 3. We know that 2m2 divides m , so 6 divides m and m ≥ 36.

This implies that χ(1)2 e2m2 9236 = ≥ = 486, 2m2 6 6

1+5 2 2 and thus α2(1) ≥ 486 − 6 = 485. So, α2(1) ≥ 485 > 25920 ≥ |G : Z|, which also

2 contradicts the fact that α2(1) < |G : Z|. So m2 6= 3 and e 6= 9.

Notice that we have considered all the cases given in Lemma 3.12, which completes

the proof of the proposition.

It remains to show that ker(α2) must also be abelian.

Proposition 5.14. Assume Hypothesis 5.1. Then ker(α2) is abelian.

Proof. Suppose that ker(α2) is not abelian and let E/Z be a chief factor of G such that

E ≤ K = ker(α2). Let θ ∈ Irr(K) be such that [χK , θ] 6= 0. Since K is nonabelian

and χ is faithful, we have that χK is a sum of G-conjugate nonlinear characters, one

of which is θ. Hence θ is nonlinear. By Lemma 5.3, Z = ker(α1). Also by Lemma

51 5.10, θE = ϕ ∈ Irr(E). Since ϕ(1) > 1, E/Z is a chief factor of G, and Z = Z(G), we have that E/Z is a fully ramiﬁed section of G with respect to ϕ and λ ∈ Irr(Z) such

1/2 that [θZ , λ] 6= 0. Thus, ϕ is G-invariant and ϕ(1) = θE(1) = |E : Z| . Finally, by

Lemma 5.11, CG(E/Z) = E and G/E acts faithfully on the symplectic vector space

E/Z.

Claim 1. G/E acts transitively on (E/Z)#.

Proof of Claim 1. We know that E and Z are fully ramiﬁed with respect to ϕ and λ.

E Since Z = Z(G), we have that (θθ)E = 1Z . Also, since [χE, ϕ] 6= 0,

E (χχ)E = 1Z + Φ,

where Φ is a character of E. Since K = ker(α2) ≥ E, our hypothesis implies that

(χχ)E =(1G + m1 + m2α2 + m2α2)E

=1E + m1(α1)E + 2m2α2(1)1E

=(1 + 2m2α2(1))1E + m1(α1)E

E =1Z + Φ.

By Cliﬀord Theory, the irreducible constituents of (α1)E are G-conjugate. Thus,

Irr(E/Z)# is a G-orbit and by Brauer’s Theorem (Theorem 6.32 of [9]), we have that

G/E acts transitively on (E/Z)#.

Now by Lemma 3.12, we have that e ∈ {2, 3, 5, 7, 9} and G/E is isomorphic to

52 diﬀerent groups for each value of e.

Claim 2. e∈ / {2, 3, 5, 7}

Proof of Claim 2. Since [χE, ϕ] 6= 0 and ϕ extends to δ ∈ Irr(G) by Lemma 5.12(a),

there exists some ψ ∈ Irr(G/E) such that χ = δψ by Gallagher’s Theorem. Since δ extends ϕ and G/E acts faithfully on (E/Z)#, we know that ker(δ) ≤ E. Also, since

χ = δψ is faithful and ker(δ) ≤ E, it must also be the case that ker(δ) = ker(χ)∩E = 1 and δ is a faithful character. Thus, by Lemma 3.3.2 and Theorem A of [2],

δδ = 1G + γ,

for some γ ∈ Irr(G). By Lemma 3.1.2 of [2], Z = Z(G) = ker(γ).

Pn Let ψψ = 1G + i=1 ciβi for some positive integers ci and βi ∈ Irr(G). Since

ψ ∈ Irr(G/E), we know that βi ∈ Irr(G/E) for all i. Also,

1G + m1α1 + m2α2 + m2α2 =χχ n n X X =1G + γ + ciβi + ciγβi. i=1 i=1

Since ker(βi) ≥ E and ker(γ) = Z < E, we know that γ is distinct from the βi. Also, n ≤ 2 since we know that we have exactly three nonprincipal irreducible constituents of χχ. If n = 1, then we get that ψψ = 1G + α2. But since ψψ is a real character and

α2 is not, this is impossible. Thus, n = 2.

Since m2 = 1, ker(α1) = Z, and ker(α2) ≥ E, we get that α1 = γ, α2 = β1,

α2 = β2, and γβ1 + γβ2 = 2α2(1)α1. This implies that ψψ = 1G + α2 + α2.

53 2 Let H = ker(ψ) and ZH /H = Z(G/H). We know that ψ(1) = 1 + 2α2(1) > 1.

Hence ψ(1) > 1 and since ψ is a faithful irreducible character of G/H, it follows that

G/H is nonabelian. Thus, there exists some L/G such that L/ZH is a chief factor of

G. Notice that ψψ has two nonprincipal irreducible constituents, which are complex conjugates. Thus, Theorem B ﬁnishes the claim.

Claim 3. e 6= 9.

Proof of Claim 3. Recall that E ≤ K = ker(α2). Thus, χE is reducible. Also, since

E/Z is fully ramiﬁed with respect to ϕ, we have that χ(1) = em for m ≥ 2. Now,

χ(g) = 0 for all g ∈ E − Z. Thus, χχ(g) = 0 = 1G + m1α1(g) + α2(1) + α2(1), and

1 + 2α2(1) α1(g) = − . m1

1+2α2(1) Since α1 ∈ Irr(G), α1(g) is an algebraic integer for all g ∈ G. Since is m1

a rational algebraic integer, it must be in Z. Thus, m1 divides 1 + 2α2(1). By

hypothesis, 2 χ(1) 1 + 2α2(1) α1(1) = − , m1 m1

which implies that

E 1 + 2α2(1) (α1)E = s1Z − m1 for some integer s ≥ 1+2α2(1) . Therefore, m1

2 2 2 1 + 2α2(1) m e 1 + 2α2(1) se − = α1(1) = − m1 m1 m1

54 m2 2 and s = . In particular, m1 divides m . m1

2 Also, since α1 ∈ Irr(G) and Z = ker(α1), we have that α1(1) < |G : Z| and

α1(1) divides |G : Z|. So assume that e = 9 and recall that χ(1) = em. Then by

Lemma 5.12(b), we know that E < K and α2 is not a faithful character of G/E.

Also, we have that |G : Z| divides 10 × 32 × 81 = 25920 by Lemma 3.12(v). By the above statements and Lemma 4.2.12 of [2], we know that α1 ∈ {1, 2, 4, 5, 10} and m1 ∈ {1, 3, 5, 7, 9, 11, 21}. Thus,

1 + 2α (1) 2 ≤ 21. m1

Now, assume m1 = 1. Then

2 χ(1) 1 + 2α2(1) α1(1) = − m1 m1

2 =χ(1) − (1 + 2α2(1))

≥4 × 81 − 21 = 303.

2 2 But then α1(1) ≥ 303 > 25920 ≥ |G : Z|, and this is a contradiction. So m1 6= 1.

2 So assume that m1 ∈ {3, 5, 7, 9, 11, 21}. Then m ≥ 9 and

2 χ(1) 1 + 2α2(1) α1(1) = − m1 m1 9 × 92 ≥ − 21 3 =3 × 81 − 21 = 222.

55 2 2 However, α1(1) ≥ 222 > 25920 ≥ |G : Z|, and we again have a contradiction. So, e 6= 9.

As we have examined all possible values of e, the proof is complete.

Lastly, it remains to show that |G| is even.

Proposition 5.15. Assume Hypothesis 5.1. Then |G| is even.

Proof. If χ(1) is even, then, we are ﬁnished since χ(1) divides |G| by Theorem 3.11 of [9]. So assume χ(1) is odd. If the αi are real, then G has conjugacy classes with elements of order two, which implies that |G| is even. So, assume that the αi are complex. Since χχ is a real character, we have that 1G + m1α1 + m2α2 + m2α2 is also real. Notice that both 1G and m2α2 + m2α2 are real-valued. Thus, m1α1 = χχ −

1G − m2α2 − m2α2 is real-valued, a contradiction. This completes the argument.

We are now ready to prove Theorem C.

Theorem C. Let G be a ﬁnite solvable group with center Z = Z(G) and let χ ∈ Irr(G) be a faithful character. Assume that

χχ = 1G + m1α1 + m2α2 + m2α2,

# where α1, α2 ∈ Irr(G) are distinct characters and m1 and m2 are strictly positive integers. Then,

1. the order of G is even,

2. dl(G) ≤ 6,

56 3. both ker(α1) and ker(α2) are abelian groups with either ker(α1) = Z or ker(α2) =

Z, and χ(1) is a power of a prime.

Proof. Suppose our hypotheses. Then, the ﬁrst part of the theorem follows by Propo- sition 5.15. Also, by Propositions 5.5, 5.7, 5.9, 5.13, and 5.14, we know that both ker(α1) and ker(α2) must be abelian, with one of them properly containing Z = Z(G).

Therefore, by Propositions 5.7 and 5.9, we know that dl(G) ≤ 6.

This completes all possible situations in the special case and provides a bound of dl(G) ≤ 6. In the last chapter, we will provide an example of the special case.

At this point, it is unclear whether there is an example satisfying the hypotheses of

Proposition 5.7, and it would be interesting to learn if this proposition is impossible. Attempting to ﬁnd such a group has only led to cases were χχ has three real nonprincipal irreducible constituents and dl(G) = 6. While this is not helpful in our special situation, it does show that if we generalize to the situation with χχ having three nonprincipal irreducible constituents, then dl(G) ≥ 6.

57 Chapter 6

Three Nonprincipal Irreducible

Constituents-The General Case

We are now ready to discuss the situation where χχ has precisely three nonprincipal

irreducible constituents, all of which are unique and real. Notice that we are allowed

to assume that they are real characters, since χχ is a real character. Otherwise, one of the nonprincipal irreducible constituents is complex, and one of the other two must be its complex conjugate to ensure that χχ is real. This puts us back into the special

case of Theorem C, with a possible renumbering of the αi. For simplicity, we make

the following hypothesis.

Hypothesis 6.1. Let G be a ﬁnite solvable group with χ ∈ Irr(G) a faithful character.

Assume

χχ = 1G + m1α1 + m2α2 + m3α3,

# where the αi ∈ Irr(G) are distinct, real-valued characters and the mi are strictly

positive integers for all i = 1, 2, 3. Set Z = Z(G) and deﬁne Ki = ker(αi) for

i = 1, 2, 3.

58 In order to prove Theorem D, we will need several propositions of the many

diﬀerent cases that can occur. We will apply Theorem 3.11 repeatedly in those

propositions. Many of the examples listed in Theorem 3.11 (iii) have been handled

in Sections 2 and 3 of [13]. The exceptions to this are when |V | ∈ {36, 28, 310, 212}.

When V is a faithful, primitive, irreducible G-module and |V | = 36, then dl(G) ≤ 5.

This is handled by Lemma 4.2, and uses the same arguments found in Sections 2 and

3 of [13]. We will use those same techniques again to handle the remaining cases.

Lemma 6.2. Let G be a solvable group and V a primitive, irreducible, and faithful

G-module. Also suppose that |V | ∈ {28, 310, 212}. Then dl(G) ≤ 7.

Proof. Since V is primitive and faithful in all cases, V restricts homogeneously to

every normal subgroup of G, and so every abelian normal subgroup of G has a faithful

irreducible module. Thus, by Lemma 3.1, every normal abelian subgroup of G is

cyclic. Hence, the conditions of Lemma 3.2 are satisﬁed, and we will use it repeatedly

in the rest of this argument. Also, set F = F(G), the Fitting subgroup of G. Notice

that if F is abelian, then by Lemma 3.3 G is isomorphic to a subgroup of Γ(V )

and dl(G) ≤ 2. Thus, we will assume throughout the following claims that F is

nonabelian.

Claim 1. If |V | = 28, then dl(G) ≤ 5.

Proof. Note that V is a faithful F -module and as such, 2 does not divide |F |. Let

Z be the socle of Z(F ). By the ﬁrst conclusion of Lemma 3.2, there exist normal

subgroups Q and T of G such that F = QT , Q ∩ T = Z, and T = CF (Q). By the

59 second conclusion of Lemma 3.2, the Sylow q-subgroups of Q are either extra-special or cyclic of prime order.

Next, by the ﬁfth part of Lemma 3.2, there exists a subgroup U of T such that

|T : U| ≤ 2. Now, by Lemma 3.2(iii), we have that Q/Z is the direct product of chief factors of G. Hence, Q/Z is abelian, which implies that F 0 ≤ TZ = T . Since

|F : QU| ≤ 2 as well, we know that F 0 ≤ QU. Thus, F 0 ≤ T ∩QU = U (T ∩ Q) = UZ.

Also, since Z is central in F and U is cyclic, this yields that UZ is abelian. Thus, dl(F ) = 2.

Now, by Lemma 3.5, there exists an integer e such that e2 = |F : T | = |Q : Z|. So, all faithful characters of Q must have degree e. Suppose that χ ∈ Irr(F ) is faithful with θ ∈ Irr(Q) such that [χQ, θ] 6= 0. Notice that since Z is central in F , λ will be the unique irreducible constituent of χZ . Thus, λ is faithful, as is θ. Since θ(1) = e, we know that e divides χ(1), and so e also divides the degree of every faithful irreducible character of F .

Next, by Lemma 3.4 we know that e divides dimZ2 (W ), where W is the irreducible

F -submodule of V such that V = f · W for some integer f. Now, dim(W ) divides dim(V ) = 8. Since 2 cannot divide |F |, we have that 2 does not divide e either. This implies that e = 1 and F = T . Hence, by Lemma 3.3, G must be isomorphically contained in Γ(V ) and have derived length dl(G) ≤ 2.

Claim 2. If |V | = 310, then dl(G) ≤ 7.

Proof. In this case, V being a faithful F -module implies that 3 does not divide |F |.

Again, we let Z be the socle of Z(F ), and by Lemma 3.2, there exist normal subgroups

60 Q and T of G such that F = QT with T ∩ Q = Z. Also by Lemma 3.2, the Sylow

q-subgroups of Q are either extra-special or cyclic of prime order. As before, we

can show that dl(F ) = 2. Also, by Lemma 3.5, there exists an integer e so that

e2 = |F : T | = |Q : Z| and with e dividing the degree of every faithful irreducible

constituent of χZ , where χ ∈ Irr(F ). By Lemma 3.4, e divides dimZ3 (W ) where W is

once again the irreducible F -submodule of V with V = f · W and f ∈ Z. Also, the

dimension of W divides the dimension of V . Thus, e ∈ {1, 2, 5, 10}. If e = 1, then as

before, G is isomorphically contained in Γ(V ) and dl(G) ≤ 2.

Now, let A = CG(Z). Since G/A is isomorphic to a subgroup of the automorphism

group of Z denoted by Aut(Z), and Z is cyclic, we have that G/A is abelian with

derived length 1. If e = 2, then we have that A/F is isomorphic to a subgroup of

GL(2, 2) and dl(A/F ) ≤ 2. Thus, dl(G) ≤ dl(G/A)+dl(A/F )+dl(F ) ≤ 1+2+2 = 5.

If e = 5, then A/F is isomorphically contained in GL(2, 5) and dl(A/F ) ≤ 4. Thus,

dl(G) ≤ dl(G/A) + dl(A/F ) + dl(F ) ≤ 1 + 4 + 2 = 7. Lastly, if e = 10, then by

Conclusion (iii) of Lemma 3.2, we know that we can write Q/Z = (Q1/Z) × (Q2/Z),

where the Qi/Z are chief factors of G. By the fourth conclusion of Lemma 3.2, and since |Q : Z| = 100, we also know that one of them has order 22 and the other has

2 ∼ order 5 . This yields that A/F is contained in Aut(Q/Z) = S3 × GL(2, 5), which

implies that dl(A/F ) ≤ 4 and dl(G) ≤ 7 as before. This completes the claim.

Claim 3. If |V | = 212, then dl(G) ≤ 7.

Proof. Since V is a faithful F -module, 2 cannot divide |F |. Using the ﬁrst conclusion

of Lemma 3.2, we can ﬁnd normal subgroups Q and T of G such that F = QT and

61 Q ∩ T = Z, the socle of Z(F ). As in the last claim, we can show that dl(F ) = 2.

By the second conclusion of Lemma 3.2, we know that the Sylow q-subgroups of Q are either extra-special or cyclic of prime order. Thus, we may apply Lemma 3.5 to obtain an integer e such that e2 = |F : T | = |Q : Z|. As before, e divides the degree

of every faithful irreducible character of F . Also, by Lemma 3.4, e divides dimZ3 (W ), where W is an irreducible F -submodule of V such that V = f ·W and f is an integer.

Thus, since dim(W ) divides dim(V ), we have that e divides 12. Since 2 cannot divide

|F |, we have that e must be an odd integer. Thus, e = 1 or 3.

If e = 1, then as before, F = T and G ≤ Γ(V ) by Lemma 3.3. This means that dl(G) ≤ 2. So, assume that e = 3. Then for A = CG(Z), we obtain that A/F is isomorphically embedded in GL(2, 3), meaning that dl(A/F ) ≤ 4. Since dl(G/A) = 1 as in the last claim, we get that dl(G) ≤ dl(G/A) + dl(A/F ) + dl(F ) ≤ 1 + 4 + 2 = 7.

This completes the proof.

Our ﬁrst consideration will be the possibility that all the kernels equal Z(G).

Proposition 6.3. Assume Hypothesis 6.1. Suppose that ker(α1) = ker(α2) = ker(α3) =

Z. Let E/Z be a chief factor of G. Then E/Z is a fully ramiﬁed section of G. Also,

G/E acts faithfully and symplectically on E/Z with three orbits on (E/Z)#. Thus,

2 dl(G) ≤ 9. Also, m1 = m2 = m3 = 1 and χ(1) = |E : Z| is a prime power.

Proof. By Lemma 4.1, χE ∈ Irr(E) since E 6≤ ker(αi) for all i = 1, 2, 3. Since χE

is not a linear character, we have that E/Z is a fully ramiﬁed section with respect

to χE and λ ∈ Irr(Z) such that [χZ , λ] 6= 0. By Lemma 3.2.1 of [2], we know that

62 CG(E/Z) = E · CG(E) and that E/Z is a symplectic vector space with the conjugate

action of G on E/Z preserving the symplectic form. By Lemma 3.2.2 of [2], we also

have that CG(E) ≤ Z(χ) = Z, which implies that CG(E/Z) = E. Thus, G/E acts

faithfully and symplectically on E/Z.

E 2 Since E/Z is fully ramiﬁed, we know that (χχ)E = 1Z . Hence, χ(1) = |E : Z|,

which is a prime power since G is a solvable group. Therefore, χ(1) is also a prime

power. Now,

E 1Z = (χχ)E = 1E + m1(α1)E + m2(α2)E + m3(α3)E.

By Cliﬀord theory, the irreducible constituents of (α1)E form a G-orbit, as do those

of (α2)E and (α3)E. If they all have the same irreducible constituents, then G/E

acts transitively on Irr(E/Z)#, and hence Lemma 3.3.2 and Theorem A of [2] imply

that χχ = 1G + α, where α ∈ Irr(G), contradicting our assumption that χχ has three nonprincipal irreducible constituents. If the irreducible constituents of (α2)E

# and (α3)E are the same, then G/E acts on Irr(E/Z) with two orbits and by Lemma

4.3.1 of [2], χχ = 1G + α + β, for α, β ∈ Irr(G), again contradicting our assumption that χχ has three distinct nonprincipal irreducible constituents. Thus, the number

of orbits of G/E on Irr(E/Z)# is three. By Brauer’s Theorem (Theorem 6.32 of [9]),

there are also three orbits under the action of G/E on (E/Z)#.

Now, by Theorem 3.11, we have three cases. If G/E ≤ Γ(E/Z), then dl(G/E) ≤ 2,

since subgroups of the semi-linear group are metacyclic. If G/E is an imprimitive

linear group acting on E/Z, then dl(G/E) ≤ 5. Our ﬁnal possibility is that G/E

63 is one of the exceptional cases listed in Theorem 3.11. If |E : Z| = p2 for a prime p ≤ 71, Lemma 3.7 yields that dl(G/E) ≤ 4. If |E : Z| ∈ {24, 34, 54, 74, 26, 36}, then

Lemma 3.8 yields that dl(G/E) ≤ 7. If |E : Z| = 210, then Lemma 3.7 implies that dl(G/E) ≤ 6. Lastly, if |E : Z| ∈ {28, 310, 212}, Lemma 6.2 implies that dl(G/E) ≤ 7.

Now, since E/Z is a chief factor of a solvable group, it is abelian, which means that dl(E) = 2 and dl(G) ≤ dl(G/E) + 2 ≤ 9. Lastly, observe that m1 = m2 = m3 = 1 by

Lemma 3.1.3 of [2].

Our next goal will be the situation where we have two distinct kernels. Recall that Lemma 5.3 implies that one of the kernels in question must be Z = Z(G). Thus, we will consider four diﬀerent cases depending on whether one or two characters have kernels equal to the center, and whether or not the third kernel is abelian or not. The

ﬁrst case we consider is when two of the kernels equal the center and the third kernel is abelian.

Proposition 6.4. Assume Hypothesis 6.1. Assume also that K = ker(α1) is abelian and properly containing Z and ker(α2) = ker(α3). Then, Z = ker(α2) = ker(α3) and dl(G) ≤ 6.

Proof. Since K > Z and Z = ker(α1) ∩ ker(α2) ∩ ker(α3) = ker(α1) ∩ ker(α2), we know that Z = ker(α2) = ker(α3) by Lemma 5.3.

Let θ ∈ Irr(K) such that [χK , θ] 6= 0. Let L/K be a chief factor of G. By Lemma

5.6, we know that G = GθL, K = Gθ ∩ L, and G/L acts faithfully and irreducibly on

64 L/K. Now, Lemma 3.1.1 of [2] implies that

L (χχ)L = 1K + Φ,

where Φ is a character of L such that [ΦK , 1K ] = 0. Also,

(χχ)L = 1L + m1(α1)L + m2(α2)L + m3(α3)L.

Since ker(α2) = ker(α3) = Z < ker(α1), we know that [(α2)K , 1K ] = 0 = [(α3)K , 1K ].

L Thus, Φ = m2(α2)L+m3(α3)L and 1K = 1L+m1(α1)L. So, the irreducible constituents

# # of (α1)L form the set Irr(L/K) and G/L acts transitively on Irr(L/K) , and on

(L/K)#, by Theorem 6.32 of [9]. Finally, Theorem 3.9 implies that dl(G/L) ≤ 4 and since L/K is a chief factor of a solvable group and K is abelian, dl(L) = 2. Therefore, dl(G) ≤ dl(G/L) + 2 ≤ 6.

The next case we consider is when one of the characters has Z(G) as its kernel and the other two have the same abelian subgroup as their kernels.

Proposition 6.5. Assume Hypothesis 6.1. Also assume that K = ker(α1) = ker(α2) >

Z and that K is abelian. Then Z = ker(α3) and dl(G) ≤ 8.

Proof. Since ker(α1) = ker(α2) > Z and Z = ker(α1) ∩ ker(α2) ∩ ker(α3) = ker(α1) ∩ ker(α3), it must be the case that Z = ker(α3) by Lemma 5.3. This proves the ﬁrst statement.

Let θ ∈ Irr(K) such that [χK , θ] 6= 0 and let L/K be a chief factor of G. From

Lemma 5.6, we know that G/L acts faithfully and irreducibly on L/K. Now, by

65 Lemma 3.1.1 of [2],

L (χχ)L = 1K + Φ,

where Φ is a character of L such that [ΦK , 1K ] = 0. Also,

(χχ)L = 1L + m1(α1)L + m2(α2)L + m3(α3)L.

L Since ker(α3) = Z < K, [(α3)K , 1K ] = 0. Thus, Φ = m3(α3)L and 1K = 1L +

m1(α1)L + m2(α2)L. So, the irreducible constituents of (α1)L and (α2)L form the

set Irr(L/K)#. In particular, G/L acts with two orbits on Irr(L/K)#, and hence it

also acts with two orbits on (L/K)#, by Theorem 6.32 of [9]. By Theorem 3.10, we

have three cases. If G/L ≤ Γ(L/K), then dl(G/L) ≤ 2. If G/L is imprimitive, then

dl(G/L) ≤ 5. Lastly, if we have one of the exceptional cases, Lemma 3.8 and Lemma

4.2 yield that dl(G/L) ≤ 6. Also, dl(L) = 2, and so dl(G) ≤ 8.

Next, we examine the nonabelian case, where one of the characters has Z(G) as

its kernel and the other two have a nonabelian subgroup K as their kernels. In fact, we will show that this case is impossible.

Proposition 6.6. Assume Hypothesis 6.1. Assume also that K = ker(α1) = ker(α2)

properly contains Z. Then K is abelian.

Proof. Assume not, that is, assume that K is a nonabelian subgroup of G. Choose

E ≤ K such that E/Z is a chief factor of G, and let θ ∈ Irr(K) that [χK , θ] 6= 0.

Since K is nonabelian and χ is a faithful character, χK is the sum of G-conjugate nonlinear characters, which means that θ(1) > 1. Also, since the intersection of the

66 kernels of the αi is Z, we have that Z = ker(α3) by Lemma 5.3. Furthermore, Lemma

5.10 yields that θE ∈ Irr(E). Notice also that E/Z is a fully ramiﬁed section with respect to θE and λ ∈ Irr(Z) with [θZ , λ] 6= 0, since θE is a nonlinear character, E/Z is a chief factor of G, and Z = Z(G). Thus, θE is also a G-invariant character and

2 θE(1) = |E : Z|. Hence, Lemma 5.11 shows that G/E acts faithfully on E/Z. Also,

Lemma 3.1.3 of [2] implies that m1 = m2 = 1.

Claim 1. G/E acts transitively on (E/Z)#.

E E Proof. Now, (θθ)E = 1Z . Since [χE, θE] 6= 0, we have that (χχ)E = 1Z + Φ, where Φ is a character of E. Since K ≥ E,

E 1Z + Φ = (χχ)E = (1 + α1(1) + α2(1))1E + m3(α3)E.

Hence, Cliﬀord Theory implies that the irreducible constituents of (α3)E are G- conjugates. Thus, Irr(E/Z)# is a G-orbit, and by Brauer’s Theorem (Theorem 6.32 of [9]), so is (E/Z)#.

Thus, if e2 = |E : Z|, then e ∈ {2, 3, 5, 7, 9} by Lemma 3.12. Also, if e ∈

{2, 3, 5, 7}, then θE is extendable to G by Lemma 5.12.

Claim 2. Let e ∈ {2, 3, 5, 7}. Let ϕ ∈ Irr(G) extend θE ∈ Irr(E). Then χ = ϕψ for some ψ ∈ Irr(G/E) and

1. ϕϕ = 1G + α3

2. ψψ = 1G + α1 + α2

67 3. m3 = 1 + α1(1) + α2(1)

4. αiα3 = αi(1)α3 for i = 1, 2.

Proof. Since [χE, δ] 6= 0 and θE extends to ϕ ∈ Irr(G), by Theorem 2.5, there exists

ψ ∈ Irr(G/E) such that χ = ϕψ. Observe now that ker(ϕ) ≤ E since ϕ is an extension of θE and G/E acts faithfully on E/Z. Since χ is faithful and ker(ϕ) ≤ E, we have that ker(ϕ) = ker(χ) ∩ E = 1. Thus, ϕ is a faithful character of G, E/Z is a chief factor of G, and G/E acts transitively on the nonidentity elements of E/Z. Thus, we conclude that ϕϕ = 1G + γ for some γ ∈ Irr(G) by Lemma 3.1.2 and Theorem A of

[2]. Notice also that Z = ker(γ).

Pn Next, let ψψ = 1G + i=1 aiβi, for integers ai > 0 and βi ∈ Irr(G/E). Observe that

1G + α1 + α2 + m3α3 = χχ

= (ϕϕ)(ψψ)

n X = (1G + γ)(1G + aiβi) i=1 n n X X = 1G + γ + aiβi + aiβiγ. i=1 i=1

Since ker(βi) ≥ E and ker(γ) = Z does not contain E, we have that γ 6= βi. Since the left hand side has three irreducible constituents, we know that n = 2. Also, since ker(βi) ≥ E and ker(γ) = Z, we have that ker(βiγ) does not contain E. Since m1 = m2 = 1, ker(α1) = ker(α2) ≥ E, and ker(α3) = Z, we have that β1 = α1,

β2 = α2, and γ = α3, which gives us the ﬁrst two parts of the claim. Also, both

68 α1α3 and α2α3 are multiples of α3. Thus, α1α3 = α1(1)α3, α2α3 = α2(1)α3, and

m3 = 1 + α1(1) + α2(1). This completes the claim.

Now, let H = ker(ψ), where ψ ∈ Irr(G/E) is as in Claim 2, and let ZH /H =

Z(G/H). Then, Theorem 3.1.2 of [2] implies that

Z ker(α ) ker(α ) K H = 1 ∩ 2 = , H H H H

2 where the last equality follows from our assumption. Notice that ψ(1) = 1 + α1(1) +

α2(1) > 1. Thus, G/H is not abelian and there exists a normal subgroup L of G such

that L/ZH is a chief factor of G. Also, since

ker(α ) ker(α ) Z 1 = 2 = H , H H H

and L/ZH = L/K is a chief factor of G, Proposition 4.2.5 of [2] yields that G/L acts

faithfully and symplectically on L/K with two orbits on the nonidentity elements of

L/K. Further, ψ(1)2 = |L : K|.

Our next goal is to examine all the possible groups that could appear as G/E.

First, we state some facts that will help narrow our focus. Observe that E < L. Thus,

when we take a speciﬁc G/E, we will consider what possible L/E could appear. In

particular, since L/K is a symplectic vector space, p2a must divide L/E for some

prime p and positive integer a. These prime powers p2a will yield the vector spaces

L/K since p2a = |L/K|. Once we have the order of L/E, we can ﬁnd the speciﬁc

69 G/L for this case, since G G/E ∼= . L L/E

Then, we check to see if the G/L acts with the appropriate number of orbits on L/K.

Claim 3. e 6= 2.

∼ Proof. Suppose that e = 2. Then by Lemma 3.12, we know that either G/E = S3, the

∼ ∼ symmetric group on three letters, or G/E = Z3. Suppose ﬁrst that G/E = Z3. Then since L/E is a normal subgroup of G/E, it must be the case that L/E is either the trivial subgroup or the whole group. However, neither of these has order divisible by

2a ∼ a prime power of the form p . So, G/E = S3. Thus, L/E, being a normal subgroup of G/E, must be the trivial subgroup, Z3, or the whole group. Again, none of these has order divisible by a prime to an even power. Thus, e 6= 2.

Claim 4. e 6= 3, 5.

∼ Proof. Suppose that e = 3. Then by Lemma 3.12, G/E = Q8, the quaternion group,

∼ ∼ or G/E = SL(2, 3). If G/E = Q8, then L/E must be isomorphic to the trivial group,

Z2, Z4, or the whole group. Since the order of L/E must be divisible by a prime to ∼ an even power, we can rule out the ﬁrst two options. If L/E = Z4, then L/K is a

∼ # vector space of four elements and G/L = Z2 acts with two orbits on (L/K) of sizes

1 and 2. These sizes are the degrees of α1 and α2, which implies that ψ(1) = 2. Also,

since e = 3, we have that ϕ(1) = 3, which implies that χ(1) = 6 and α3(1) = 2 by

70 Claim 2. Thus,

2 36 = χ(1) = 1 + α1(1) + α2(1) + m3α3(1) = 4 + 2m3.

∼ Thus, m3 = 16. However, by Claim 2, m3 = 4. So, L/E = Q8, L/K is a vector space of size 4, and G/L is the trivial group. However, since the trivial group ﬁxes every element of L/K, we have four orbits of size 1, contradicting the requirement that we have three orbits.

So, G/E ∼= SL(2, 3). Since the order of L/E must also be divisible by a prime to an even power, L/E is isomorphic to either Q8 or the whole group, with |L/K| = 4 in

∼ ∼ # both cases. If L/E = Q8, then G/L = Z3, which acts transitively on (L/K) . Since this is impossible, it must be the case that L/E ∼= SL(2, 3). Then G/L is the trivial group and acts on L/K with four orbits of size 1, another contradiction. So e 6= 3.

Lastly, if e = 5, then G/E ∼= SL(2, 3). By the previous paragraph, this is impossible, and so e 6= 5.

Claim 5. e 6= 7.

Proof. Suppose e = 7. By Lemma 3.12, G/E ∼= GL(2f , 3), the isoclinic group to

GL(2, 3). Then L/E must be Q8, SL(2, 3), or the whole group. If L/E is one of the

ﬁrst two groups, then |L/K| = 4. If L/E is the whole group, then L/K has order 4 or 16, and G/L ∼= {1} acts with either four orbits of size 1 or 16 orbits of size 1, both a contradiction of our having three orbits.

∼ ∼ # If L/E = Q8, then G/L = S3, which acts transitively on (L/K) . So, we must

71 ∼ ∼ have L/E = SL(2, 3). Then G/L = Z2, which acts with two orbits of sizes 1 and

2. Again, these are the degrees of α1 and α2 respectively, and Claim 2 implies that

ψ(1) = 2. Also, since e = 7, ϕ(1) = 7 and Claim 2 implies that α3(1) = 6 and

χ(1) = 14. Thus,

2 196 = χ(1) = 4 + 6m3.

This implies that m3 = 32. However, by Claim 2, m3 = 4. Thus, e 6= 7.

The last situation that we need to deal with is the case when e = 9. Before we begin, there are some facts that need to be addressed. Since E ≤ K = ker(α1) = ker(α2), we have that χE is a reducible character. Since E/Z is a fully ramiﬁed section of G with respect to θE, we have that χ(1) = ef, where f ≥ 2 is an integer. Also,

χ(g) = 0 for all g ∈ E − Z. Therefore,

0 = χχ(g) = 1 + α1(1) + α2(1) + m3α3(g)

for all g ∈ E − Z. Thus,

1 + α1(1) + α2(1) α3(g) = − . m3

Since α3 ∈ Irr(G), we know that α3(g) is an algebraic integer. Since it is also a rational number, α3(g) ∈ Z and m3 divides 1 + α1(1) + α2(1). Also, by our hypothesis,

2 χ(1) 1 + α1(1) + α2(1) α3(1) = − . m3 m3

72 Since

E 1 + α1(1) + α2(1) (α3)E = s · 1Z − 1E m3 for some integer s ≥ 1+α1(1)+α2(1) , we get that m3

1 + α (1) + α (1) e2f 2 1 + α (1) + α (1) se2 − 1 2 = − 1 2 m3 m3 m3

2 2 and s = f /m3. In particular, m3 divides f . Also, since α3 ∈ Irr(G) with Z =

2 ker(α3), we have that α3(1) < |G : Z| and α3(1) divides |G : Z|.

Claim 6. e 6= 9.

Proof. Assume that e = 9 and recall that χ(1) = ef. By part (b) of Lemma 5.12,

E < K and α1 and α2 are not faithful characters of G/E. Now, by Lemma 3.12, |G :

Z| divides 10 × 32 × 81 = 25290. By Lemma 4.2.12 of [2], we have that α1(1), α2(1) ∈

{1, 2, 4, 5, 10} , which implies that

m3 ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 21} .

Thus, 1 + α (1) + α (1) 1 2 ≤ 21. m3

If m3 = 1, then

2 χ(1) 1 + α1(1) + α2(1) α3(1) = − m3 m3

> 4 × 81 − 21

73 = 303.

2 2 However, this implies that α3(1) ≥ 303 = 91809 > 25920 ≥ |G : Z|. So, m3 6= 1.

2 If m3 ≥ 3, then f ≥ 9 and

93 α (1) ≥ − 21 = 222, 3 3

2 2 which means that α3(1) ≥ 222 = 49284 > 25920 ≥ |G : Z|, another contradiction.

2 2 So, m3 = 2. This implies that 2 divides f. If f > 2, then f > 4 and χ(1) /m3 ≥

2 8 × 81. Thus, α3(1) ≥ 8 × 81 − 21 = 627, again implying that α3(1) > |G : Z|. So,

2 f = 2. Then χ(1) /m3 = 162. Also, since 2 divides 1 + α1(1) + α2(1), exactly one of

α1(1) and α2(1) is odd. Recall that

2 χ(1) − (1 + α1(1) + α2(1)) 1 α3(1) = = 162 − (1 + α1(1) + α2(1)). m3 2

Therefore,

α3(1) ∈ {154, 156, 157, 158, 159, 160} ,

none of which divides 25290. Hence, α3(1) cannot divide |G : Z|, a contradiction.

Therefore, e 6= 9.

Since we have examined all possible values of e, we must conclude that K is

abelian.

The last of these cases to consider is when two of the characters have Z(G) as

74 their kernels and the third character has a nonabelian subgroup of G as its kernel.

Proposition 6.7. Assume Hypothesis 6.1. Assume also that K = ker(α1) is a nonabelian subgroup properly containing Z = ker(α2) = ker(α3). Then dl(G) ≤ 7.

Proof. First, we choose E ≤ K such that E/Z is a chief factor of G. Then, let

θ ∈ Irr(K) with [χK , θ] 6= 0. As before, this implies that θ(1) > 1. By Lemma 5.10, we know that ϕ = θE ∈ Irr(E). Also, E/Z is a fully ramiﬁed section of G with respect to ϕ and λ ∈ Irr(Z) such that [ϕZ , λ] 6= 0, since E/Z is a chief factor of G, ϕ is a nonlinear character, and Z = Z(G). Hence, ϕ is G-invariant and ϕ(1)2 = |E : Z|.

Also, by Lemma 5.11, G/E acts faithfully on E/Z.

Now, G/E acts with at most two orbits on (E/Z)#. To see this, notice that

E E (θθ)E = ϕϕ = 1Z . Since [χE, ϕ] 6= 0, we have that (χχ)E = 1Z + Φ, where Φ is a character of E. Now, since K ≥ E, we know that

E 1Z + Φ = (χχ)E = (1 + m1α1(1))1E + m2(α2)E + m3(α3)E.

By Cliﬀord Theory, the constituents of (α2)E form an orbit, as do those of (α3)E.

Thus, G/E acts on (E/Z)# with at most two orbits. If G/E acts transitively on

(E/Z)#, then we may apply Theorem 3.9 to obtain that dl(G/E) ≤ 4. Since E/Z is a chief factor of G and Z = Z(G), dl(E) = 2. Therefore, dl(G) ≤ 6 in this case.

We may now assume that G/E acts with two orbits on (E/Z)#. Here, we may apply Theorem 3.10 to get three subcases. In the ﬁrst, G/E is metacyclic and as such has dl(G/E) ≤ 2. In the second, we have shown that dl(G/E) ≤ 5. In the last, we must consider several special cases. However, by Lemma 3.7, Lemma 3.8, Lemma

75 3.7, and Lemma 4.2, we know that dl(G/E) ≤ 5. Thus, dl(G) ≤ 7, as desired.

We are now ready to examine the cases with three distinct kernels. The ﬁrst

possibility that we consider is the stacked case, where K1 > K2 > K3 = Z.

Proposition 6.8. Assume Hypothesis 6.1. Let K1 > K2 > K3 = Z with K1 abelian.

Then dl(G) ≤ 6.

Proof. We know by Lemma 5.6 that if L > K1 such that L/K1 is a chief factor of G,

and θ ∈ Irr(K1) with the property that [χK1 , θ] 6= 0, then G = LGθ, K1 = L∩Gθ, and

G/L acts faithfully and irreducibly on L/K1. Recall that χL ∈ Irr(L) since L 6≤ Ki for all i. Thus, by Lemma 3.1.1 of [2], we have that

L (χχ)L = 1K1 + Φ,

where Φ is either the zero function or a character of L such that [ΦK1 , 1K1 ] = 0. Also,

(χχ)L = 1L + m1(α1)L + m2(α2)L + m3(α3)L.

Since K1 > K2 > K3, we know that [(α2)K1 , 1K1 ] = 0 = [(α3)K1 , 1K1 ]. Thus,

L # 1K1 = 1L + m1(α1)L. Hence, G/L acts transitively on Irr(L/K1) , and by Brauer’s

# Theorem (Theorem 6.32 of [9]), transitively on (L/K1) . Therefore, by Theorem 3.9,

dl(G/L) ≤ 4. Lastly, since L/K1 is a chief factor of G and K1 is abelian, dl(L) = 2

and dl(G) ≤ 6, as desired.

Proposition 6.9. Assume Hypothesis 6.1 and assume K1 > K2 > K3 = Z(G) = Z with K2 nonabelian. Then dl(G) ≤ 6.

76 Proof. Let E ≤ K2 such that E/Z is a chief factor of G. Let θ ∈ Irr(K2) such that

[χK2 , θ] 6= 0. Since K2 is nonabelian and χ is a faithful character, we know that

θ(1) 6= 1. Now, by Lemma 5.10, θE ∈ Irr(E). Thus, E/Z is a fully ramiﬁed section

2 of G with respect to θE and λ. Hence, θE is G-invariant and θE(1) = |E : Z|. Now,

by Lemma 3.2.1 of [2], we know that CG(E/Z) = E · CG(E). Let C = CG(E/Z) and

assume that CG(E) 6= Z. Then we have three cases.

In the ﬁrst case, we assume that CG(E) ≤ K2. Then θCG(E) ∈ Irr(CG(E)) and we

have that

E ≤ CK2 (CG(E)) ≤ Z(θ),

where the second containment follows from Lemma 3.2.2 of [2]. However, since θE ∈

Irr(E), this implies that θ(1) = 1, a contradiction. Thus, CG(E) 6≤ K2.

Next, we assume that CG(E) 6≤ K1. Then CG(E) 6≤ Ki for all i. Then by Lemma

4.1, this means that χCG(E) ∈ Irr(CG(E)) and by Lemma 3.2.2 of [2], we have

E ≤ CG(CG(E)) ≤ Z(χ) < E.

But this is another contradiction. So CG(E) ≤ K1 but CG(E) 6≤ K2.

Notice now that E ∩CG(E) is a normal subgroup of G and that Z ≤ E ∩CG(E) ≤

E. Since E/Z is a chief factor of G, this means that either E ∩ CG(E) = Z or

E ∩ CG(E) = E. If the latter occurs, this means that E ≤ CG(E), and E is abelian.

Since this is impossible, we have that Z = E ∩ CG(E) and so K2 ∩ C = E.

Now, let γ ∈ Irr(K1) such that [χK1 , γ] 6= 0. Also, let H ≤ C such that H/E is a

77 chief factor of G.

K2C

C K2 H

CG(E) E CH (E) Z

Claim 1. Let N C G such that N ≤ K1 but N 6≤ Ki for i = 2, 3. Then γN ∈ Irr(G).

Proof. This follows from a proof similar to that of Lemma 5.10.

In particular, this implies that γH ∈ Irr(H). Next, we want to determine if

γE ∈ Irr(E) or if γH is fully ramiﬁed.

2 Claim 2. γE = fθE, where f = |H : E|.

Proof. Suppose that γE ∈ Irr(E). Then θ = γK2 ∈ Irr(K2) and γE = θE is G-

Pn invariant. Let χK1 = g i=1 γi, where the γi are the conjugates of γ and g = [χK1 , γ].

Pn Then since γK2 ∈ Irr(K2), we know that χK2 = g i=1(γi)K2 as well. Then since γE

is irreducible and G-invariant, we have that χE = gnγE. Now,

2 2 2 g n = [(χχ)E, 1E] = 1 + m1α1(1) + m2α2(1) = [(χχ)K2 , 1K2 ] = g n.

78 Hence, n2 = n and n = 1. Thus,

2 1 + m1α1(1) = [(χχ)K1 , 1K1 ] = g = [(χχ)K2 , 1K2 ] = 1 + m1α1(1) + m2α2(1).

But this implies that m2α2(1) = 0, which is a contradiction. Thus, γE = fθE as desired.

Now, by the ﬁrst claim above, we know that γCH (E) ∈ Irr(CH (E)). We want to know how this character behaves with respect to the chief factor CH (E)/Z. Since

it is impossible to induce up from Z, it must be that either γZ ∈ Irr(Z) or γCH (E) is fully ramiﬁed. Notice that γ(1)2/θ(1)2 = |H : E| and θ(1)2 = |E : Z|. Thus,

γ(1)2 = |H : Z|. Since all irreducible characters of Z are linear, and |H : Z| > 1, it is

impossible for γZ to be an irreducible character. Therefore, γCH (E) is fully ramiﬁed, making CH (E)/Z a fully ramiﬁed section. Also, we have that

2 |H : Z| = γ(1) = |CH (E): Z| = |H : E|.

But this implies that |E : Z| = 1, which is impossible since E/Z is a chief factor of G.

Since we have exhausted our cases, it must be that CG(E) = Z and C = E · CG(E) =

Now, since E/Z is a fully ramiﬁed section of G with respect to θE and λ, and

E Z = Z(G), we have that (θθ)E = 1Z . Since [χE, θE] 6= 0, we also have that (χχ)E =

79 E 1Z + Φ, for some character Φ of E. Hence, E ≤ K2 < K1, yields that

E 1Z + Φ = (χχ)E = (1 + m1α1(1) + m2α2(1))1E + m3(α3)E.

Thus, by Cliﬀord Theory, the irreducible constituents of (α3)E form an orbit. Also,

Theorem 6.32 of [9] yields that G/E acts transitively on (E/Z)#. Hence, by Theorem

3.9, dl(G/E) ≤ 4. Since dl(E) = 2, we get our desired conclusion of dl(G) ≤ dl(G/E) + dl(E) ≤ 4 + 2 = 6.

We are ﬁnally ready to consider the case when all three characters have distinct kernels that are not stacked. We allow for the possibility that one of the kernels is

Z(G), but it is not required for the proof. For the last few results, we must borrow some notation from [1]. Let L ≤ G. Then (N, θ) ≤ (L, ϕ) is a chain if N ≤ L,

θ ∈ Irr(N), ϕ ∈ Irr(L), and [ϕN , θ] 6= 0. Now, let X be a family of normal subgroups of G with G ∈ X. The chain

(N0, θ0) > (N1, θ1) > ··· > (Nk, θk),

with N0 = G and θ0 = χ, is a (X, χ)-reducing chain if Ni ∈ X and (θi)Ni+1 is reducible for all i = 0, 1, . . . , k. Furthermore, this chain is maximal if it also has the following properties:

(i) For any i with 0 < i ≤ k, the group Ni is maximal in the set

{M ∈ X|M ≤ Ni−1 and (θi−1)M is reducible} .

80 (ii) For any M ∈ X such that M < Nk, the restriction (θk)M is irreducible.

The following lemma is motivated by results in [1]. We also mention that we use the standard notation to denote the vanishing-oﬀ subgroup of a character. In particular, if ϕ ∈ Irr(G), then V (ϕ) = hg ∈ G|ϕ(g) 6= 0i is the smallest subgroup of

G such that ϕ vanishes on G − V (ϕ). Also, we deﬁne

( ) \ Ω = ker(αi)|S ⊆ {1, 2, 3} , i∈S

T where i∈S ker(αi) = G when S is empty.

Lemma 6.10. Let (G, χ) = (N0, θ0) > (N1, θ1) > ··· > (Nk, θk) be a maximal (Ω, χ)- reducing chain. For each i, let Li/Ni be a chief factor of G such that Li ≤ Ni−1.

(a) If Ci = CNi−1 (Li/Ni), then Ci/Ni is abelian.

(b) Deﬁne ψi = (θi−1)Li ∈ Irr(Li). Then for Mi ≤ G such that NiV (ψi) ≤ Mi < Li

and Li/Mi is a chief factor of Ni−1, the subgroup Hi = NG(Mi) acts faithfully

# on Irr(Li/Mi) with at most ri orbits, where

ri = | {αj|Ni ≤ ker(αj) and Ni−1 6≤ ker(αj)} |.

well.

Proof. The ﬁrst conclusion of the lemma is Lemma 6.10 of [1]. The other conclusions follow from the proof of Lemma 6.4 of [1].

81 It is now possible to ﬁnish the ﬁnal cases of this problem using the previous lemma.

Proposition 6.11. Assume Hypothesis 6.1. Suppose also that Ki 6= Kj and that

Ki ≥ Z with at most one Ki = Z. Then dl(G) ≤ 16.

Remark. Notice that this situation also handles the case where K1 > K2 > K3 with

K1 nonabelian and K2 abelian.

Proof. First we can let (G, χ) > (K1, θ1) > (K1∩K2, θ2) > (Z, θ3) be a chain. We need

to show that this is a maximal (Ω, χ)-reducing chain. First, note that in our previous

notation, N0 = G, N1 = K1, N2 = K1 ∩ K2, and N3 = Z = K1 ∩ K2 ∩ K3. By our

deﬁnition of Ω, Ni ∈ Ω for all i = 0, 1, 2, 3. By Lemma 5.10, (θi)Ni+1 is reducible for

i = 0, 1, 2, and Ni+1 is maximal in the set {M ∈ X|M ≤ Ni−1 and (θi−1)M ∈/ Irr(M)}.

Lastly, since N3 = Z is the smallest subgroup in Ω, the last condition for our chain to

be maximal is satisﬁed. Hence, we have a maximal (Ω, χ)-reducing chain, as desired.

Let L1/K1 be a chief factor of G and let M1 ≤ G such that L1/M1 is a chief factor

# of G and K1V (χL1 ) ≤ M1 < L1. Then H1 = NG(M1) acts faithfully on Irr(L1/M1) with

r1 = | {αj|K1 ≤ ker(αj) and G 6≤ ker(αj)} | = | {α1} | = 1

orbit by the second part of Lemma 6.10. Thus, Theorem 3.9 yields that dl(H/CH (L1/M1)) ≤

4. Then the third and ﬁrst parts of Lemma 6.10 imply that dl(G/K1) ≤ 5. If K1 is

abelian, then dl(K1) = 1, and dl(G) ≤ 6. So, assume that K1 is not abelian.

We can then bound dl(K1/K2) in a similar way. Deﬁne L2, H2, and M2 similarly

82 # to the previous paragraph. Then H2 acts faithfully on Irr(L2/M2) with

r2 = | {αj|K1 ∩ K2 ≤ ker(αj) and K1 6≤ ker(αj)} | = | {α2} | = 1

orbit. Again, Theorem 3.9 and Lemma 6.10 imply that dl(K1/K1∩K2) ≤ 5. If K1∩K2 is abelian, then dl(G) ≤ dl(G/K1) + dl(K1/K1 ∩ K2) + dl(K1 ∩ K2) ≤ 5 + 5 + 1 = 11.

Lastly, we assume that K1 ∩ K2 is nonabelian. Then we can once again deﬁne L3,

# M3, and H3 similarly. Here, H3 acts faithfully on Irr(L3/M3) with

r3 = | {αj|Z ≤ ker(αj) and K1 ∩ K2 6≤ ker(αj)} | = | {α3} | = 1

orbit. Again, Theorem 3.9 and Lemma 6.10 imply that dl((K1 ∩ K2)/Z) ≤ 5. Thus, dl(G) ≤ dl(G/K1) + dl(K1/K1 ∩ K2) + dl((K1 ∩ K2)/Z) + dl(Z) ≤ 5 + 5 + 5 + 1 = 16, which completes the proof.

We are now ready to prove Theorem D.

Theorem D. Let G be a ﬁnite solvable group and let χ ∈ Irr(G) be a faithful character. Assume that

χχ = 1G + m1α1 + m2α2 + m3α3,

# where αi ∈ Irr(G) are distinct characters and mi are strictly positive integers for all i = 1, 2, 3. Then dl(G) ≤ 16.

Proof. Assume our hypotheses, i.e., that G is a ﬁnite solvable group with χ ∈ Irr(G) faithful and with the property that χχ = 1G + m1α1 + m2α2 + m3α3, for distinct

83 # αi ∈ Irr(G) and mi ∈ N. If α2 is complex-valued and α3 = α2, then dl(G) ≤ 6

by Theorem C. So, suppose that αi is real-valued for all i. Then, dl(G) ≤ 16 by

Propositions 6.3, 6.4, 6.5, 6.6, 6.7, 6.8, 6.9, and 6.11.

If we consider the proof of Theorem D, we see that for G to be a group satisfying

the hypotheses of the theorem with dl(G) = 16, Theorem 3.9 implies all Bi = Ni/Ci have the property that F(Bi) is extra-special and Bi/F(Bi) is metacyclic, using the notation from Proposition 6.11. It would be interesting to see if such a group really has a character with the desired property. Furthermore, in the next chapter, we will provide Example 7.3, which has the highest derived length we have discovered. It may be possible to ﬁnd an example of a group satisfying the hypotheses of Theorem

D with a higher derived length.

84 Chapter 7

Examples

In this section, we present a number of examples of the abstract situations in order

to show that such scenarios actually exist. The ﬁrst result is an example of Theorem

A, and veriﬁes that our bound in that case is the best possible.

Example 7.1. There exists a group G of order 165, 888 and dl(G) = 8 that has a character χ ∈ Irr(G) such that χ(1) = 8 and χχ = 1G + α1 + α2 for characters

α1, α2 ∈ Irr(G) of degrees α1(1) = 27 and α2(1) = 36.

Proof. Assume p = 2 and let E be the central product of two copies of the dihedral

group of order 8 and one copy of the quaternion group of order 8. In fact, E is

an extra-special 2-group of order 27 whose outer automorphism group contains a

− subgroup isomorphic to O6 (2) that centralizes Z(E) by comments on page 406 of [6].

6 − Let V be a vector space of order 2 . Then [3] yields that O6 (2) has a subgroup H of

4 order 6 with F = F(H) = O3(H), which is an extra-special group of order 27 and

− exponent 3. The action of H on V is faithful since H is a subgroup of O6 (2), which

acts faithfully on V .

The action of H on V is irreducible as well. To see this, let v ∈ V be a non-zero

85 vector. Then its centralizer in F is either trivial or a non-central subgroup with order

3. If CF (v) = 1, then the Fundamental Counting Principle gives an orbit of size 27.

Otherwise, the centralizer has order 3 and the orbit containing v has size 9. Hence,

the sizes of the F -orbits on V are 1, 27, and 9. Now, there are either four orbits of

size 9 or a single orbit of size 9. To see that it is actually the former, recall that F is

extra-special of order 27 and exponent 3 and thus, |F : Z(F )| = 9. Let a ∈ F −Z(F ).

Then ha, Z(F )i cannot act Frobeniusly on F , and in particular, every subgroup of

F having order 9 has some noncentral element h of order 3 with the property that

CV (h) 6= 0. In particular, every v ∈ V that belongs to an orbit of size 9 must also

0 0 belong to some CV (h ), where h is some noncentral element of F .

Next, since F/Z(F ) is elementary abelian, Fitting’s Theorem implies that there are at least four such subgroups CV (h), each having size 3. Indeed, let A, B ≤ F be distinct subgroups of order 9 and let h ∈ A and h0 ∈ B be noncentral elements. Then

0 0 F = hh, h i and CV (h) ∩ CV (h ) = 1. Therefore, we know that there are four orbits

of size 9, as desired. It remains to show that these four orbits are conjugate. To that

∼ end, we consider the action of H/F = GL(2, 3) on F/Z(F ). Notice that GL(2, 3)

transitively permutes the subgroups of order 3 in F/Z(F ) and thus, the subgroups of

order 9 in F are also transitively permuted. Inside each subgroup of order 9, we have

three subgroups of order 3 that form an F -orbit. Hence, the twelve such subgroups

of order 3 form a H-orbit. But these subgroups are of the form CF (v), where v ∈ V

belongs to an orbit of size 9. Thus, 12 divides the size of the H-orbit consisting of

these elements, implying that 36 does also. Therefore, the orbit has size at least 36.

However, since the orbit size is also required to be at most 36, it must be the case

86 that the orbit size is precisely 36. Therefore, the only orbit sizes of H on a vector space V with size 26 are 1, 27, and 36. Notice that 63 is the smallest number of the form 2n − 1 which can be written as a linear combination of 27 and 36. Notice also that every F -orbit must have size divisible by 9. Thus, since the smallest integer n such that 9 divides 2n − 1 is in fact 6, it must be the case that the action of H on V is irreducible.

Now, recall that F = F(H) has exponent 3 and H/F is isomorphic to GL(2, 3) with dl(H/F ) = 4. There is an element x ∈ Z(GL(2, 3)) with order 2 that acts nontrivially and Frobeniusly on F and hence, [F, x] = F . Thus, F must be contained in the fourth commutator subgroup of H, denoted by H(4), while hF, xi = H(3), the third commutator subgroup of H. This implies that dl(H) = 6.

Our ﬁnal goal is to show that we may take G = E o H. To this end, we will prove that H is isomorphic to a subgroup of Aut(E), the automorphism group of E. Let

A = Aut(E) and I = Inn(E), the group of inner automorphisms of E. By comments

∼ − on page 403 of [6], we know that A/I = O6 (2). Notice that by Theorem 1 of [6], A does not split over I and thus, H is not automatically isomorphic to a subgroup of A.

∼ x Now, we know that I = E/Z(E). So, if we take x ∈ E and deﬁne ϕx by ϕx(g) = g , then the map x 7→ ϕx is a homomorphism from E onto I with kernel Z(E). Notice

τ that if τ ∈ A and x ∈ E, then (ϕx) = ϕτ −1(x). To see this, let g ∈ E also. Then

τ −1 (ϕx) (g) = (τ ϕxτ)(g)

−1 = τ (ϕx(τ(g)))

87 = τ −1(τ(g)x)

= τ −1(x−1τ(g)x)

= τ −1(x−1)gτ −1(x)

= (τ −1(x))−1gτ −1(x)

= gτ −1(x)

= ϕτ −1(x)(g).

− ∗ ∗ ∼ Now, since H ≤ O6 (2), let H be the preimage of H in A, i.e., H /I = H. Then since

∗ ∗ 6 3 ∗ F C H, we have that F C H . Notice that |I| = 2 and that 3 = |F | = |F : I|.

So, I is a normal Hall subgroup of F ∗. Thus, by the Schur-Zassenhaus Theorem,

I has a complement E in F ∗. The action of E on I is isomorphic to the action of

∼ F on E/Z(E) = V . In particular, |CI (E)| = |CV (F )| = 1. Next, by the Frattini

∗ ∗ argument, H = F NH∗ (E) = IENH∗ (E) = INH∗ (E) and I ∩ NH∗ (E) = NI (E).

Also, [NI (E),E] ≤ [I,E] ≤ I since I is normal in A and [NI (E),E] ≤ E since NI (E) normalizes E. Hence, [NI (E),E] ≤ I ∩E = 1. This implies that NI (E) ≤ CI (E) = 1,

∼ ∗ ∼ and so I ∩ NH∗ (E) = 1. Now, by construction, we have that NH∗ (E) = H /I = H, and since NH∗ (E) is isomorphic to a subgroup of A, so is H, which is what we wanted to prove.

Therefore, we may take G = E o H. By Theorem 4.3, we have the desired character and using an argument similar to that involving x of order 2 and F , but with y ∈ Z(F ) of order 3 and [E, y] = E, we obtain that dl(G) = 8.

The next example shows that Proposition 5.9 actually occurs, although the bound

88 given above may not be the best possible.

∼ Example 7.2. The alternating group G = A4 has a faithful irreducible character χ with the property χχ = 1G + 2α1 + α2 + α2. In this case, Z(G) = ker(α1) and ker(α2)

is an abelian group properly containing Z(G).

∼ 0 Proof. Let G = A4, the alternating group on 4 letters. Then, |G | = 4, which implies

that G has three linear characters by Corollary 2.23(b) of [9]. These are the principal

character of G and two linear characters that are complex conjugates of each other.

By comments made on page 15 of [9], we have that

k X 2 2 12 = |G| = χi(1) = 1 + 1 + 1 + χ4(1) . i=1

Thus, χ4(1) = 3. Take χ4 = χ. We will show that χχ has the desired form.

Recall that the principal character appears in the decomposition of χχ with mul-

tiplicity 1 since 1 = [χ, χ] = [χχ, 1G]. Take α2 = χ2 and α2 = χ3 to be the two

nonprincipal linear characters of G, and let α1 = χ4. Then, we can compute their

multiplicities in χχ as well:

[χχ, α1] = 2

[χχ, α2] = 1

[χχ, α2] = 1.

Thus, χχ = 1G + 2α1 + α2 + α2. In particular, this is an example of the situation

described in Proposition 5.9, as ker(α1) < ker(α2). Finally, notice that dl(G) = 2.

89 Theorem E. Let G be a ﬁnite group with E ≤ G a nonabelian subgroup. Assume

E/Z is an abelian chief factor of G with Z = Z(G), and that G/E acts with n ≥ 2

# orbits on (E/Z) . Let χ ∈ Irr(G) be faithful and suppose that χE ∈ Irr(E). Then

n X χχ = 1G + αi, i=1

# where αi ∈ Irr(G) are distinct and ker(αi) = Z for all i, 1 ≤ i ≤ n.

Proof. Let λ ∈ Irr(Z) such that [χZ , λ] 6= 0. Since χE ∈ Irr(E) is faithful, and

E is nonabelian, χ(1) > 1. Thus, χZ is reducible. Hence E/Z is a fully ramiﬁed

2 E section, χ(1) = |E : Z|, and by Lemma 3.1.1 of [2], (χχ)E = 1Z . Since G acts on

(E/Z)# with n orbits, Brauer’s Theorem (Theorem 6.32 of [9]) implies that G acts

on Irr(E/Z)# with n orbits.

By Cliﬀord Theory, we have that for all α ∈ Irr(G), the irreducible constituents

# of αE form a G-orbit. Let α1 ∈ Irr(G) be a character such that m1 = [χχ, α1] 6= 0.

By Lemma 3.1.2 of [2], we know that Z ≤ ker(α1). Since we have n orbits, there

# # exists δ ∈ Irr(E/Z) such that [(α1)E, δ] = 0. Thus, there exists α2 ∈ Irr(G) such

that [(α2)E, δ] 6= 0 and m2 = [χχ, α2] 6= 0. Thus, the irreducible constituents of (α1)E

and (α2)E form two distinct orbits. However, since we have n orbits, there exists

# # γ ∈ Irr(E/Z) such that [(α1)E, γ] = 0 = [(α2)E, γ]. Thus, there exists α3 ∈ Irr(G)

such that [(α3)E, γ] 6= 0 and m3 = [χχ, α3] 6= 0. We can repeat this until we have

# αi for 1 ≤ i ≤ n. Since G acts on Irr(E/Z) with n orbits, these must be the only

distinct G orbits of Irr(E/Z)#.

90 Pn Now, i=1 αi(1) ≥ |E : Z| − 1, which implies that

n 2 X |E : Z| = χ(1) ≥ 1 + αi(1) ≥ |E : Z|. i=1

Therefore, mi = 1 for all i. Lastly, it remains to show that ker(αi) = Z for all i. To that end, if we deﬁne Ki = ker(αi), then we know that Ki ∩ E is a normal subgroup of G contained in E. Since χE ∈ Irr(E), it must be the case that Ki ∩ E = Z for all

i. We also need Ki = Ki ∩ E to obtain the desired conclusion. So assume not. Then

Kj 6≤ E for some ﬁxed j, 1 ≤ j ≤ n. Now, for C = EKj, Lemma 4.1 implies that

χC ∈ Irr(C) and is G-invariant. We may now apply Theorem 2.6, which says that

2 C either χKi = eϕ, where e = |C : Ki| = |E : Z|, or χC = ϕ , where ϕ ∈ Irr(Ki) such

that [χKi , ϕ] 6= 0.

First, suppose χC is fully ramiﬁed. Then

n n n X X 2 X 1 + αi(1) ≥ 1 + αi(1) = [(χχ)Ki , 1Ki ] = e = [(χχ)Z , 1Z ] = 1 + αi(1), i=1 i=1 i=1 Kj ≤ker(αi)

and we have equality throughout. So, Kj ≤ ker(αi) for all i, which implies that

Tn Kj ≤ i=1 ker(αi) = Z, yielding Kj = Z.

C Pt Now we suppose that χC = ϕ . Then χKi = k=1 ϕk, with t = |E : Z|. Here,

n n n X X X 1 + αi(1) ≥ 1 + αi(1) = [(χχ)Ki , 1Ki ] = t = |E : Z| = 1 + αi(1), i=1 i=1 i=1 Kj ≤ker(αi)

and we have equality throughout. This gives us Kj ≤ ker(αi) for all i, and Kj ≤

91 Tn i=1 ker(αi) = Z. Hence, Kj = Z. Since j was arbitrary, this holds for all j.

Therefore, ker(αi) = Z for all i = 1, 2, . . . , n, and the theorem is proved.

Example 7.3. There exists a solvable group G with a faithful character χ ∈ Irr(G)

such that χχ = 1G + α1 + α2 + α3, where the αi are all real, nonprincipal, irreducible

characters of G and ker(α2) = ker(α3) = Z(G).

∼ Proof. Let V be the vector space of order 9. Take H = V o GL(2, 3). Then H

is isomorphic to a subgroup of the general linear group GL(8, 2). Thus, we can let

∼ 8 G = W oH, where W is the natural module of GL(8, 2) with order 2 . Let χ ∈ Irr(G)

have degree χ(1) = 9. Then χ is a faithful character of G and has the property

χχ = 1G + α1 + α2 + α3.

Also, α2(1) = α3(1) = 36, α1(1) = 8, ker(α2) = ker(α3) = Z(G) = 1, and ker(α1) =

W . In particular, dl(G) = 6, meaning that the bound given in the proposition is the

best possible bound.

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