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

Transcription

1 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

2 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, Stephen M. Gagola, Donald L. White, Chair, Doctoral Dissertation Committee Members, Doctoral Dissertation Committee Joanne Caniglia, Hassan Peyravi, Accepted by Andrew Tonge, Chair, Department of Mathematical Sciences James L. Blank, Dean, College of Arts and Sciences

3 Table of Contents Dedication iv Acknowledgments v 1 Introduction 1 2 Background Group Theory Character Theory 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

4 To my mother, for her continued love and kindness, and to my father, who gave me a love of mathematics.

5 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

6 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/F 2 (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 F 2 (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

7 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 = 1 G + m 1 α 1 + m 2 α 2, where α 1, α 2 Irr(G) # are distinct characters and m 1 and m 2 are strictly positive integers. Then dl(g) 8. Throughout, G is a finite solvable group with center Z(G), and the set of irreducible 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 = 1 G + m 1 α 1 + m 2 α 2, where α 1, α 2 Irr(G) # are distinct characters and m 1, m 2 are strictly positive integers. Then both α 1 and α 2 are real-valued characters. The next natural question to consider is the situation where χχ has three nonprincipal irreducible constituents. This will be broken into two cases. The first to 2

8 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 = 1 G + m 1 α 1 + m 2 α 2 + m 2 α 2, where α 1, α 2 Irr(G) # are distinct characters and m 1 and m 2 are strictly positive integers. Then, 1. the order of G is even, 2. dl(g) 6, 3. both ker(α 1 ) and ker(α 2 ) are abelian groups with either ker(α 1 ) = Z or ker(α 2 ) = 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 = 1 G + m 1 α 1 + m 2 α 2 + m 3 α 3, where α i Irr(G) # are distinct characters and m i are strictly positive integers for all i = 1, 2, 3. Then dl(g) 16. 3

9 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 χ E Irr(E). Then χχ = 1 G + n α i, i=1 where α i Irr(G) # are distinct and ker(α i ) = Z for all i, 1 i n. 4

10 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 = {g n 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

11 discussion. The first is the normal subgroup. A subgroup N G is called normal when N g = g 1 Ng = 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 C G (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 1 y 1 xy and the commutator subgroup is the subgroup G = [G, G]. This subgroup is generated by all commutators of two elements in G, i.e., G 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 = 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 = Φ(P ). Let G be a group. Then Sylow p-subgroups are subgroups P of G with p-power 6

12 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 O p (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 G 0, G 1,...,G n such that 1 = G 0 G 1 G n = G and G i+1 /G i 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), G = G (1), G = (G ) = 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 N i such that 1 = N 0 N 1 N m = G with the property that N i /N i 1 Z(G/N i 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

13 we let F = F(G), the second Fitting subgroup, denoted by F 2 (G), is defined by F 2 (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 G is the wreath product of G and H, and is often denoted by W = H 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 S n. We say that a transitive group G S n 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

14 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 [χ, ϕ] = 1 G χ(g)ϕ(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

15 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 χ = k n i χ i, i=1 with n i 0 and χ i Irr(G). Then those χ i with positive coefficients n i 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 1 G. In our situation, 1 G is an irreducible constituent of χχ since 1 = [χ, χ] = [χχ, 1 G ], 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 χ H is a character of H. Likewise, if θ is a character of G then the induced character θ G is given by θ G (g) = 1 H θ (xgx 1 ), x G 10

16 where θ (h) = θ(h) for h H and θ (y) = 0 for y / H. Moreover, θ G (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 χ N and suppose that θ = θ 1, θ 2,..., θ t are the distinct conjugates of θ in G. Then t χ N = e θ i, i=1 where e = [χ N, θ]. Proof. See Theorem 6.2 of [9]. We say θ extends to G if there is some χ Irr(G) such that χ H = θ and θ induces irreducibly if θ G Irr(G). If χ is the unique irreducible constituent of θ G 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., θ g, where θ Irr(N). The stabilizer of θ is the subgroup G θ = {g G θ g = θ} and θ is G-invariant if G θ = 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

17 Theorem 2.5 (Gallagher s Theorem). Let N G and let χ Irr(G) be such that χ N = 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 θ G. 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) θ L Irr(L); (b) θ L = eϕ for some ϕ Irr(L) and e 2 = K : L ; (c) θ L = t i=1 ϕ i where the ϕ i Irr(L) are distinct and t = K : L. Proof. See Theorem 6.18 of [9]. 12

18 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 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 = C G (Z). Then there exist E, T G with (i) F = ET, Z = E T, and T = C F (E). (ii) A Sylow q-subgroup of E is cyclic of order q or extra-special of exponent q or 4. 13

19 (iii) E/Z = E 1 /Z E n /Z for chief factors E i /Z of G with E i C G (E j ) for i j. (iv) For each i, Z(E i ) = Z, E i /Z = p 2n i i for a prime p i and an integer n i, and E i = O p i (Z) F i for an extra-special group F i = O pi (E i ) G of order p 2n i+1 i. (v) There exists U T of index at most 2 with U cyclic, U G and C T (U) = U. (vi) G is nilpotent if and only if G = T. (vii) T = C G (E) and F = C A (E/Z). (viii) E/Z = F/T is a completely reducible G/F -module and faithful A/F module. (ix) A/C A (E i /Z) Sp(2n i, p i ). (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 (q m )/GF (q)). Then the semi-linear group is given by Γ(V ) = { x ax σ a GF (q m ) #, σ G }, and the affine semi-linear group is given by AΓ(V ) = { x ax σ + b a GF (q m ) #, σ G, b V }. 14

20 In particular, AΓ(V ) = V Γ(V ), and Γ(V ) is a point-stabilizer for zero in AΓ(V ), which is a doubly transitive permutation group. The notation Γ(q m ) 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 V F 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 dim F (V ) = te dim F (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 V U. If char(f) = 0, assume that F is algebraically closed. Then dim F (V ) = e dim F (W ) with e 2 = H : U. 15

21 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 Lemma 3.6. Let V be a faithful irreducible F[G]-module and suppose V = V 1 V n (n > 1) is a system of imprimitivity for G. Let γ : G S n be the homomorphism induced by the permutation action of G on the V i. Set S = γ(g), which is a transitive subgroup of S n. Finally, let H = N G (V 1 )/C G (V 1 ). Then G is isomorphic to a subgroup of H 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

22 Lemma 3.8. Let G be solvable and let V 0 be a faithful and completely reducible F[G]-module over an arbitrary field F. Set n = dim F (V ). Then dl(g) log 2 3(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(q n ), or one of the following occurs: (a) q n = 3 4, F(G) is extra-special of order 2 5, F 2 (G)/F(G) = 5 and G/F 2 (G) Z 4. (b) q n = 3 2, 5 2, 7 2, 11 2, or Here F(G) = QT, where T = Z(G) Z(GL(V )) is cyclic, Q 8 = 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: In particular, dl(g) 4. q n T G/F(G) Z 3 or S or 4 Z S or 6 S Z 3 or S S 3 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

23 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 = G α. Then G has a unique minimal normal subgroup V such that G = GV, G V = 1, C G (V ) = V, and V acts regularly on Ω. This means that Ω = V = q n 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 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 = q n for a prime q and G is permutation-isomorphic to a subgroup of AΓ(q n ), the affine semi-linear group. In particular, dl(g) 2. (ii) G is an imprimitive linear group with imprimitivity spaces V 1, V 2, where V = 18

24 V 1 V 2. Here H = N G (V 1 ) has index 2 and H/C H (V 1 ) is a linear group that acts transitively on V 1 {0}. Thus, dl(g) 5. (iii) G acts as a primitive linear group on V and G has one of the degrees 7 2, 13 2, 17 2, 19 2, 23 2, 29 2, 31 2, 47 2, 3 4, 7 4, 2 6, or 3 6. Proof. See Theorem 1.1 of [5]. To see the second conclusion of (ii), we know that dl(g) 5 since G H/C H (V 1 ) C 2, 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/c H (V 1 )) 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 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 = q n for a prime q and G is permutation-isomorphic to a subgroup AΓ(q n ), the affine semi-linear group. In particular, dl(g) 2. (ii) G is an imprimitive linear group with V = r i=1v i where the V i are imprimitivity spaces and r = 2 or 3. Here H = N G (V 1 ) and H/C H (V 1 )is a linear group that acts transitively on V # 1 = V 1 {0}. Thus, dl(g) 5. 19

25 (iii) G acts as a primitive linear group on V and G has one of the degrees q n for q n = p 2, where p is a prime and p 71, or q n {2 4, 3 4, 5 4, 7 4, 2 6, 3 6, 2 8, 2 10, 3 10, 2 12 }. Proof. See Theorem 1.2 of [5]. To see the second conclusion of (ii), we know that dl(g) 5 since G H/C H (V 1 ) C r, 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/c H (V 1 )) 4. Lastly, the following lemma will be used when χχ has three nonprincipal irreducible constituents. In particular, we will show that if χχ = 1 G +m 1 α 1 +m 2 α 2 +m 2 α 2, then both ker(α 1 ) and ker(α 2 ) must be abelian. Lemma 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 e 2 = V = q n. Let F(G) be the Fitting subgroup of G and F 2 (G)/F(G) be the Fitting subgroup of G/F(G). Then one of the following holds: (i) e = 2, and either G = S 3 or G = Z 3. (ii) e = 3, and either G = Q 8 or G = SL(2, 3). (iii) e = 5, and G = SL(2, 3). (iv) e = 7, and G = GL(2, 3). 20

26 (v) e = 9, F(G) is an extra-special group of order 2 5, F 2 (G)/F(G) = 5, and G/F 2 (G) is a subgroup of the cyclic group of order 2. Proof. See Theorem of [2]. 21

27 Chapter 4 Two Nonprincipal Irreducible Constituents The first lemma will be useful in all of the following situations. It is based off Lemma of [2]. Lemma 4.1. Let G be a finite solvable group. Let χ Irr(G) be a faithful character. Assume χχ = 1 G + n m i α i, i=1 where the α i Irr(G) # are distinct and the m i N. Let N be a normal subgroup of G. Then χ N Irr(N) if and only if N ker(α i ) for all i = 1, 2,..., n. Proof. Notice that [χ N, χ N ] = [(χχ) N, 1 N ] = [1 N + n m i (α i ) N, 1 N ] = 1 + i=1 n m i [(α i ) N, 1 N ]. i=1 Thus, [χ N, χ N ] = 1 if and only if [(α i ) N, 1 N ] = 0 for all i = 1, 2,..., n. Therefore, [χ N, χ N ] = 1 if and only if N ker(α i ) for all i. Since [χ N, χ N ] = 1 if and only if 22

28 χ N Irr(N), the result follows. For Theorem A, we will need to examine all the cases from Theorem Fortunately, 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 3 6, 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 3 6, 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 = 1 G + m 1 α 1 + m 2 α 2, where α 1, α 2 Irr(G) # are distinct characters and m 1 and m 2 are strictly positive integers. Then dl(g) 8. Proof. Let Z = Z(G). From Lemma and Lemma of [2], we know that Z = ker(α i ) 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(α 1 ) is abelian and nonabelian. First, suppose that ker(α 1 ) is nonabelian. This implies Z = ker(α 2 ). Then we can take E/Z to be a chief factor of G such that E ker(α 1 ). Notice that E is nonabelian, since 23

29 we can take a character θ Irr(ker(α 1 )), with the property that it is nonlinear and θ E 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 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(2, 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(α 1 ) and ker(α 2 ) are both abelian. From Proposition and Proposition of [2], we know χ(1) 2 = E : Z and χ(1) is a power of a prime. If ker(α 1 ) ker(α 2 ), then Proposition of [2] also implies that dl(g) 6. So, we assume that ker(α 1 ) = ker(α 2 ) = Z. Again, let E/Z be a chief factor of G. We know from Proposition 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) In the second case, G/E is an imprimitive linear group, which implies that dl(g) dl(g/e) = 7. We now consider the possibility that G/Z satisfies conclusion (iii) of Theorem 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 As mentioned before Theorem 3.10, the degree of G/Z as a permutation group is E : Z. Thus, E : Z is one of 7 2, 13 2, 17 2, 19 2, 23 2, 29 2, 31 2, 47 2, 3 4, 7 4, 2 6, or 3 6, most of which have been handled in [13]. In particular, if E : Z = p 2 for a prime p, then G/E is isomorphic to a subgroup of GL(2, p) and so we may apply Lemma 24

30 3.7 to see that dl(g/e) 4, which implies that dl(g) 6. If E : Z = 3 4 or 7 4, then we see that E/Z has dimension 4 over the finite field Z q where q is either 3 or 7. Thus, we may apply Lemma 3.8 to see that dl(g/e) log(4/8) We can 2 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 = p 2. If E : Z = 2 6, 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 = 3 6. 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 p 2n+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) = p n such that χχ = 1 G + α 1 + α 2, where α 1, α 2 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, ψ Z 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

31 applied, and using that theorem, we have that ψ extends to a character χ Irr(G), where G = E H. Then by Lemma of [2], we know that this χ has degree χ(1) = p n 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 = C G (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, C E/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 = C G (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 C E/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 H 1 and H 2 are both complements of E in N, then N G (H 1 ) is conjugate to N G (H 2 ). Next, we want to prove that all complements of E in G are normalizers of com- 26

32 plements of E in N. Recall that G = E H and take L = HZ N = Z(H N). Then EL = EH N = N and E L = E HZ = Z. Also, since N G, we know that L HZ. Thus, HZ N G (L) and [N G (L) E, L] Z. This implies that N G (L) E C N/L (L/Z) = 1. Hence, N G (L) = H(N G (L) E) and HZ = N G (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 p 2n with exactly three orbits, then there exists a group G with a faithful character χ Irr(G) satisfying χ(1) = p n and χχ = 1 G + α 1 + α 2 for distinct α 1, α 2 Irr(G). Proof. Let E be an extra-special group of order p 2n+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 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

33 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 O 6 (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 q n for q n = p 2, where p is a odd prime and p 71, or q n {3 4, 5 4, 7 4, 3 6, 3 10 }. Assume that V is a faithful, irreducible, and primitive G-module. Finally, assume that G 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 = Q 8, the quaternion group of order 8, and T Z (GL(2, p)). Also, Q T = Z(Q) = Z 2. 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

34 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 e 2 = F : T = Q : Z. Notice that e divides dim Zq (W ), where V = f W for some integer f. Since q cannot divide e, we have that e {1, 2, 4}. Furthermore, e 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

35 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 O 1 and O 2. Set e 2 = V = q 2n. Then if v O i, so is v. Proof. Notice that our hypotheses imply the hypotheses of Theorem Therefore, we will study the cases given there. We know that v and v are in orbits of equal size. So, if O 1 and O 2 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 (e2 1) = 1 2 (q2n 1). Assume Theorem 3.10(i). Then G is isomorphic to a subgroup of Γ(V ), N = G Γ 0 (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 of [2], N e + 1. So ϕ( N ) e 2, which implies that G < e 2 1. Therefore, G = 1 2 (e2 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

36 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) < e 2 1, which yields the same results as the last paragraph. So assume that p = e+1. Notice that p(p 2) = e 2 1 and 1 p(p 2) in an integer dividing G. However, e 3 implies 2 that p > 4 is an odd prime. Hence, p(p 2) is an odd integer, and 1 p(p 2) is not 2 an integer. This is a contradiction. So assume that N = 4. Then G divides 8, and thus 1 2 (e2 1) divides 8. Hence, 1 2 (e2 1) is 1, 2, 4, or 8. Since V is a symplectic vector space, V = e 2 = r 2 for some prime power r. Thus, V # = 8 and G SL(2, 3). Therefore, G = Q 8, the quaternion group of order 8. However, since Q 8 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 V 1 and V 2 with V = V 1 V 2. If H = N G (V 1 ), we know that H/C H (V 1 ) acts transitively on V # 1 and v V 1 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

37 such that χχ = 1 G + m 1 α 1 + m 2 α 2, where α 1, α 2 Irr(G) # are distinct characters and m 1, m 2 are strictly positive integers. Then both α 1 and α 2 are real-valued characters. Proof. First, notice that χχ is a real-valued character. So, if ker(α 1 ) ker(α 2 ) and α i is a constituent of χχ, then α i is also a constituent of χχ for i = 1, 2 with ker(α i ) = ker(α i ). In our case, if ker(α 1 ) ker(α 2 ), then α i = α i, and α i is a real-valued character. So, let Z = Z(G) and assume that ker(α 1 ) = ker(α 2 ). Then by Lemma of [2], we know that Z = ker(χχ) = ker(α 1 ) ker(α 2 ) = ker(α 1 ) = ker(α 2 ). Let E/Z be a chief factor of G. Then by Proposition of [2], E/Z is a fully ramified section of G with respect to χ E and λ Irr(Z) such that [χ Z, λ] 0. Also, G/E acts symplectically and faithfully on E/Z with two orbits on (E/Z) #. Suppose the α i are complex-valued characters that are not real-valued. Then since χχ is realvalued, α 2 = α 1. Hence the orbits of G/E on Irr(E/Z) # are inverses of each other, i.e., if the nontrivial orbits are O 1 and O 2 and θ O 1, then θ O 2. 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 α i are complex, which completes the proof. 32

38 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 = 1 G + m 1 α 1 + m 2 α 2 + m 2 α 2, where the α i Irr(G) # are distinct and the m i 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

39 X. Assume that θ(g) = 0 if g X R, or if g X S. Then 0 if g 1 θ(g) = θ(1) otherwise. Therefore, θ is a multiple of the regular character. Proof. See [2], Lemma Lemma 5.3. Let G be a finite solvable group with χ Irr(G) faithful such that χχ = 1 G + m 1 α 1 + m 2 α 2 + m 3 α 3, with ker(α 2 ) = ker(α 3 ). Let Z = Z(G). Then either ker(α 1 ) = Z or ker(α 2 ) = Z. In particular, if we have Hypothesis 5.1, then ker(α 1 ) = Z or ker(α 2 ) = Z. Proof. Here, we follow the style of Lemma in [2]. Since α 2 and α 3 have the same kernel, we need only consider ker(α 1 ) and ker(α 2 ). Assume that Z is proper in both kernels. Then let R/Z be a chief factor of G with R ker(α 1 ). Note that R ker(α 2 ). Also, let S/Z be a chief factor of G with S ker(α 2 ). Set T = RS. Then R S = Z since ker(α 1 ) ker(α 2 ) = Z. This implies that T ker(α i ) for i = 1, 2. By Lemma 4.1, χ T Irr(T ). Notice that T/R is also a chief factor of G. Let ψ Irr(R) such that [χ R, ψ] 0. Then by Theorem 2.6, either χ R = eψ for some ψ Irr(R) and e 2 = T : R or ψ T = χ T. 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

40 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, χ(1) 2 if g R S = Z. Lemma 5.2 also tells us that χχ is a multiple of the regular character of RS/Z. Thus, χ(1) 2 T : Z. However, since χ T Irr(T ) it is also true that χ(1) 2 T : Z. Hence, equality holds and (χχ) T = 1 T Z is the regular character of T/Z. Since all characters in Irr(T/Z) are linear, they appear with multiplicity 1 in 1 T Z = (χχ) T and m i = 1 for i = 1, 2. Notice that [(χχ) R, 1 R ] = [((χχ) T ) R, 1 R ] = [(1 T Z) R, 1 R ] = T : R. Since R ker(α 1 ) and R ker(α 2 ), we have 1 + α 1 (1) = T : R. In a similar fashion, 1 + α 2 (1) + α 3 (1) = T : S. This yields the equation T : Z = χ(1) 2 = 1 + α 1 (1) + α 2 (1) + α 3 (1) = T : R + T : S 1. 35

41 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(α 1 ) or Z = ker(α 2 ) = ker(α 3 ), as desired. In particular, if we consider the situation outlined in Hypothesis 5.1, we see that since α 2 and α 2 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(α 1 ), ker(α 2 ), 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(α 1 ) = ker(α 2 ), we will need Theorem 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, and that G acts on V # with three orbits, say O 1, O 2, and O 3. Set e 2 = q 2n = V. Then v O i implies that v O i for 1 i 3. 36

42 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 = O 2 = O 3 and e 2 1 2a = O 1, both of which must divide G. Assume Theorem 3.11(i). Then G Γ(V ) and N = G Γ 0 (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 of [2]. Thus, G < e 2 1. Claim 1. G 1 3 (e2 1). Proof. Since max(a, e 2 1 2a) lcm(a, e 2 1 2a) G, it suffices to show that max(a, e 2 1 2a) 1 3 (e2 1). If a 1 3 (e2 1), then we are finished. So assume a < 1 3 (e2 1). Then 2a > 2 3 (e2 1) and e 2 1 2a > e (e2 1) = 1 3 (e2 1). This completes the claim. By the claim, 1 3 (e2 1) G < e 2 1. Then 2n G 1 N 3 (e2 1) e + 1 = e 1 3 = qn 1. 3 Thus, we need prime powers q n such that this inequality holds. This implies that 37

43 q 2n = e 2 {3 2, 3 4, 5 2, 7 2 }. Now, N divides Γ 0 (V ) = e 2 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 q 2n = 3 4, then N divides 80, and in particular, N = 5. This implies that G : N divides 4 and G 20. But we know that 20 < 80 = (e2 1) G. This is a contradiction. If q 2n = 3 2, then N divides 8. Since N cannot be the trivial group, it has even order in this case and we are finished. If q 2n = 5 2, then N = 3 and G 6. However, our claim implies that G 8, another contradiction. Lastly, if q 2n = 7 2, 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 < e 2 1 and the claim yields that 1 3 (e2 1) G < e 2 1. This yields the same results as the last paragraph. So, assume that p = e + 1. Then p is a Fermat prime and e = q n 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 Γ 0 (V ) and this element sends elements to their inverses. This completes this case. Assume Theorem 3.11(ii). Then G is an imprimitive linear group and V = r j=1v j where the V j are imprimitivity spaces and r = 2 or 3. We know that H = N G (V 1 ) has index r in G, and that H = H/C H (V 1 ) acts transitively on V # 1. Thus, Lemma 3.6 implies that G H Z r, 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(V 1 ). Since q is an odd prime, we know that V 1 is odd, and V # 1 is even. Hence, H is also even and 38

44 contains some involution which inverts some element of V 1. This handles the orbit which has form V # 1 V # 2. Now, let v = (a, 0) V (V 1 V 2 ) with a 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 (V 1 V 2 ) contains inverses. Since v w is in one of the orbits that make up V (V 1 V 2 ), it follows that the second orbit contains inverses as well. Next, assume Theorem 3.9(b),(c). Then, H GL(V 1 ) where V 1 = 3 2, 5 2, 7 2, 11 2, 23 2, 3 4, 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 = V 1 V 2 V 3. Hence, elements of V are 3-tuples, meaning they have form (v 1, v 2, v 3 ), where v i V i for 1 i 3. Since 0 V i 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 0}. By their definition, 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(α 1 ) and ker(α 2 ) are distinct. Proof. Suppose that Z = ker(α 1 ) = ker(α 2 ) and let E/Z be a chief factor of G. Let 39