A CENTRAL SERIES ASSOCIATED with V (G) a Dissertation Submitted

A CENTRAL SERIES ASSOCIATED WITH V (G)

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

Nabil Mlaiki

August 2011 Dissertation written by

Nabil Mlaiki

B.S., Kent State University, 2004

M.A., Kent State University, 2006

Ph.D., Kent State University, 2011

Approved by

Mark L. Lewis, Chair, Doctoral Dissertation Committee

Stephen M. Gagola, Jr., Member, Doctoral Dissertation Committee

Donald L. White, Member, Doctoral Dissertation Committee

David Allender, Member, Doctoral Dissertation Committee

Mike Mikusa, Member, Doctoral Dissertation Committee

Accepted by

Andrew Tonge, Chair, Department of Mathematical Sciences

Timothy S. Moerland, Dean, College of Arts and Sciences

ii TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... iv

INTRODUCTION ...... 1

1 BACKGROUND ABOUT FINITE GROUPS ...... 8

2 BACKGROUND ABOUT V (G) ...... 13

3 GENERAL LEMMAS FOR THE CENTRAL SERIES ASSOCIATED WITH V (G) 16

4 PROOF OF THE FIRST TWO THEOREMS AND THEOREM 4 ...... 21

5 CAMINA PAIRS AND CAMINA TRIPLES ...... 23

6 FUTURE RESEARCH QUESTIONS ...... 31

BIBLIOGRAPHY ...... 32

iii ACKNOWLEDGEMENTS

I would like to express my sincere gratude to Dr. Lewis for being a good advisor and mentor. I enjoyed our sessions and thank him for guidance, support and encouragement he has given me throughout this process. THANK YOU!

I also, like to thank Dr. White and Dr. Gagola for all their help. I thank my mother, and my brother Neji Mlaiki for all their support.

iv INTRODUCTION

Throughout this dissertation, G is a ﬁnite group. Denote by C the set of complex numbers. Deﬁne the general linear group, GL(n, C), to be all invertible n×n matrices over

C. A C-representation of G is a homomorphism < : G → GL(n, C). For many purposes, the set of representations has too much information about the group. So, it will be enough to consider the traces of the representations. Recall that the trace of a matrix is the sum of the diagonal entries of the matrix. The character χ aﬀorded by < is the function given

by χ(g) = tr<(g) for g ∈ G, where tr is the trace of <(g). Thus, χ is a function from G

to C. However, except when n = 1, this function χ is not a homomorphism. We will be

interested in the set of all characters that can occur for a group G. One can show that the

sum of characters is a character. A character is said to be irreducible if it cannot be written

as a sum of other characters of G. Every character is a sum of irreducible characters, and

if one knows the irreducible characters of G, then one can determine all characters of G.

Hence, we focus on the irreducible characters of G. We write Irr(G) for the set of irreducible

characters of G.

Let g be an element of G. The conjugacy class of g is the set denoted by cl(g) and

deﬁned by cl(g) = {xgx−1 | x ∈ G}. The conjugacy classes of G partition G. That is, if

x, y ∈ G, then either cl(x) ∩ cl(y) is empty or cl(x) = cl(y). One can show that characters

are constant on the conjugacy classes of a group. Hence, characters are what we call class

functions. A class function of G is any function from G to C that is constant on conjugacy

classes. A class function is a character if and only if it is aﬀorded by some representation.

One can deﬁne addition ϕ + γ of class functions by (ϕ + γ)(x) = ϕ(x) + γ(x), and scalar

multiplication cϕ by (cϕ)(x) = cϕ(x). Note that with this addition and scalar multiplication

the set of class functions of a group G forms a vector space, and one can show that the set

1 2

of irreducible characters forms a basis for this vector space. There is another basis for this vector space. Let C be a conjugacy class for G. Deﬁne the class function fc by fc(g) = 1 if g ∈ C and fc(g) = 0 if g 6∈ C. The set {fc | C is a conjugacy class of G} is also a basis whose size is the number of conjugacy classes of G. This implies that the number of irreducible characters equals the number of conjugacy classes. As proved in Corollary 2.7

P 2 in [4], χ∈Irr(G) χ(1) = |G|. Given a character χ of G, note that χ(1) = n, where n is the integer, so that < : G →

GL(n, C) is a representation aﬀording χ. Let χ be a character of a group G. We say that

χ(1) is the degree of χ. Characters of degree 1 are called linear characters. Linear characters are exactly the characters that are homomorphisms from G to C∗ =∼ GL(1, C). If λ is a linear character, then λ(g) is invertible for all g ∈ G. Hence, λ(g) 6= 0 for all g ∈ G. An irreducible character χ of G is nonlinear if χ(1) 6= 1. For nonlinear characters the situation is diﬀerent. The following result is Theorem 3.15 in [4], and is due to Burnside: if χ is an irreducible character of G with χ(1) > 1, then χ(g) = 0 for some g ∈ G. We write nl(G) for the set of nonlinear irreducible characters of G.

Let X be a subset of G. Define hXi to be the subgroup generated by X (i.e., the smallest subgroup of G containing X). Now, define the vanishing off subgroup V (χ) of a character χ to be the subgroup generated by all the elements g ∈ G where χ(g) is not 0. Note that the set of elements where χ(g) is not 0 tends not to be a subgroup. That is why we consider the subgroup V (χ) = hg ∈ G | χ(g) 6= 0i. If λ is a linear character, then observe that V (λ) = G.

We mention that V (χ) is the smallest subgroup of G such that χ vanishes on G \ V (χ).

Also, if χ is nonlinear, then we know χ must vanish on some elements of G, but it is possible

to have the set of elements where χ does not vanish generate all of G. Hence, we can have

V (χ) = G for χ ∈ nl(G). In the groups we are interested in, we will have V (χ) < G for χ ∈

nl(G).

In this dissertation, we are going to study a related subgroup. We deﬁne the vanishing 3

oﬀ subgroup of G, denoted by V (G), by V (G) = hg ∈ G | there exists χ ∈ nl(G) such that

χ(g) 6= 0i. It was ﬁrst introduced by Lewis in [8], which appeared in 2008. We exclude linear characters so that it is possible to have V (G) < G. Also, note that V (G) is the smallest subgroup of G such that all nonlinear irreducible characters vanish on G \ V (G). In this

dissertation we are interested in the case where V (G) is proper in G. Note that V (G) may

have some elements g such that for every χ ∈ nl(G), χ(g) = 0.

A series of normal subgroups Ni C G such that N0 ≤ N1 ≤ · · · ≤ Nn, is called a

central series in G if Ni+1/Ni ≤ Z(G/Ni) for all i = 0, ··· , n − 1. We need to consider

the commutator of two elements in a group. Let x, y ∈ G. The commutator of x and y

is denoted by [x, y] and is deﬁned by [x, y] = x−1y−1xy. Also, if H,K are two subgroups

of G, the commutator subgroup of H and K is [H,K] = hh−1k−1hk | h ∈ H, k ∈ Ki.

Now, consider the term Gi as the i-th term in the lower central series, which is deﬁned by

0 G1 = G, G2 = G = [G, G], and Gi = [Gi−1,G] for i ≥ 2. We are going to study a central

series associated with the vanishing oﬀ subgroup, deﬁned by V1 = V (G), and Vi = [Vi−1,G]

for i ≥ 2. Lewis, in [8], proved that Gi+1 ≤ Vi ≤ Gi. Our main focus is on the case when

Vi < Gi for some i > 3. In [8], Lewis showed that in the case when Vi < Gi, we have Vj < Gj

for all j such that 1 ≤ j ≤ i. An abelian group is a group with the property that for every

x ∈ G we have |cl(x)| = 1. If, in such group, every nonidentity element has the same prime

order p, then the group is called an elementary abelian p-group. In [8], Lewis proved that

if V2 < G2, then there exists a prime p such that Gi/Vi is an elementary abelian p-group

for all i ≥ 1. Also, he proved that if V3 < G3 there exists a positive integer n such that

0 2 2n |G : V1| = |G : V2| = p . The main questions that we are trying to answer in Chapters

3 and 4 are can we generalize the results in [8] to the case where Vi < Gi for i > 3? What

can we say about the index of Vi−1 in Gi−1?

0 To answer these questions, we deﬁne some subgroups. First, set D3/V3 = CG/V3 (G /V3). p Lewis, in [8], proved that if V3 < G3, then either |G : D3| = |G : V1| or D3 = V1. To 4

study the case when i > 3, we will add an additional hypothesis and a new deﬁnition. First, we need to deﬁne some more subgroups. For each integer i ≥ 3, set Yi/Vi = Z(G/Vi) and

Di/Vi = CG/Vi (Gi−1/Vi).

The new deﬁnition we need is the deﬁnition of H1. We say Gk is H1 if for every normal subgroup N of G where Vk ≤ N < Gk, we have Vk−1/N = Gk−1/N ∩ Yk(G/N). In [8], it

0 was proved that G3 is H1. Under the additional hypothesis that G /Vi is abelian we are

able to show that Gi is H1 for all i > 3. We are also interested in computing the index of

Vi in Gi. We will see that this index depends on the size of D3. In other words, it depends

0 on the size of the centralizer of G modulo Vi.

We now come to our ﬁrst theorem. When Vk < Gk, and Gi is H1 for i = 3, ··· , k, we are able to prove that Dk = D3, which is very useful to prove some of the results of this

dissertation.

0 Theorem 1. If Vk < Gk, G /Vk is abelian, and Gi is H1 for all i = 3, ··· , k, then Dk = D3.

Our second theorem should be considered to be the main result of this dissertation. We

0 are able to prove that |Gi−1 : Vi−1| = |G : D3|, for every i ≥ 4, where Vi < Gi, and G /Vi is

0 abelian. Hence, for a nilpotent group of class c, if Vc < Gc, and G /Vc is abelian, then we have |Gi−1 : Vi−1| = |G : D3| for all 4 ≤ i ≤ c, and |Gc : Vc| ≤ |G : D3|.

0 Theorem 2. If Vk < Gk, G /Vk is abelian, for some k ≥ 3, then

1. |Gk−1 : Vk−1| = |G : D3| if k ≥ 4,

2. Dk = D3,

3. Gk is H1,

4. |Gk : Vk| ≤ |G : D3|.

If 3 ≤ i ≤ k − 1, then Vi will satisfy the same hypothesis. So, Di = D3,Gi is H1 and when i ≥ 4, |Gi−1 : Vi−1| = |G : D3|. Note that the above result was motivated from 5

the bound of subgroups by MacDonald in [9], where he proved for a Camina group G that

0 |G3| ≤ |G : G |. Our motivation for adding the hypothesis G/Vk abelian is that the results in [9] were under the hypothesis that G is metabelian (i.e. G0 is abelian). Hence, proving this conclusion under a similar metabelian hypothesis seems like a reasonable ﬁrst step.

In the Camina group case, removing the metabelian hypothesis required totally diﬀerent techniques.

In the second part of this dissertation, we will study Camina triples. Camina triples are a generalization of Camina pairs. Camina pairs were ﬁrst introduced in 1978 by A.R.

Camina in [1]. Camina’s work in [1] was inspired by the study of Frobenius groups, which we will deﬁne in the ﬁrst chapter.

Throughout this dissertation, we say that (G, N) is a Camina pair when N is a normal subgroup of a group G, and for all x ∈ G \ N, x is conjugate to all of xN. Chillag and

MacDonald proved in [2] two equivalent conditions of a pair (G, N) to be a Camina pair.

They showed that (G, N) is a Camina pair if and only if for every x ∈ G \ N we have

|CG(x)| = |CG/N (xN)|. Also, they proved that (G, N) is a Camina pair if and only if for all x ∈ G \ N and z ∈ N, there exists an element y ∈ G so that [x, y] = z. In [9], MacDonald showed that if (G, N) is a Camina pair where G is a p-group, then N is a term in both the lower and the upper central series. As was proved in [2], if χ ∈ Irr(G) where N ker(χ), then χ vanishes on G \ N. Camina proved in [1] that if (G, N) is a Camina pair, then either

N is a p-group or G/N is a p-group for some prime p, or G is Frobenius group with kernel

Let 1 < M ≤ N be two nontrivial normal subgroups of a ﬁnite group G. We say that

(G, N, M) is a Camina triple (CT ) if for every g ∈ G \ N, g is conjugate to all of gM.

Notice that Camina pairs are special cases of Camina triples when M = N. Note that we

can construct an example of a Camina triple from a Camina pair as follows: let (G, M) be

any Camina pair, and N be any group. Deﬁne the group H = G × N by {(x, y) | x ∈ G 6

and y ∈ N}. Now consider the subgroups K = M × 1N and L = M × N. We claim that

(H, L, K) is a Camina triple. To see this, we only need to show that every element (x, y) in H \ L is conjugate to all the elements of K(x, y) in H. Hence it is enough to show that x is conjugate to all of Mx in G, but if (x, y) is in H \ L, then x ∈ G \ M. Since (G, M) is a Camina pair, x is conjugate to all of Mx in G as desired. It turns out that there is a connection between Camina triples and our ﬁrst topic, the vanishing oﬀ subgroup. Note that (G, V (G),G0) is a Camina triple. Also, Camina triples and V (G) are both motivated by studying Camina groups and Camina pairs.

In our third theorem, we will prove some facts about Camina triples. We were able to prove the following theorem. First, deﬁne for a subgroup M of G, Irr(G | M) = {χ ∈

Irr(G) | M ker(χ)}

Theorem 3. If (G, N, M) is a Camina triple, then the following are true:

1. M is solvable.

2. M has a normal π-complement Q with M/Q nilpotent, where π is the set of primes

that divide | G : N |.

3. If x ∈ M, then there exists χ ∈ Irr(G | M) such that χ(x) 6= 0.

4. If x ∈ G \ N, then χ(x) = 0 for all χ ∈ Irr(G | M).

The following corollary is an immediate consequence of Theorem 3.

Corollary 1. If (G, N) is a Camina pair and x ∈ N, then there exists χ ∈ Irr(G | N) such that χ(x) 6= 0.

In closing, as an application of our techniques we answer an open question about Camina groups. Let G be a ﬁnite group. We say that G is a Camina group if for every x ∈ G \ G0, then cl(x) = xG0. In [9], MacDonald conjectured that if G is a Camina group of nilpotence 7

n 2n 0 class 3, then |G3| ≤ p , where p = |G : G |. He gave a sketch of a proof. But, Dark and

Scoppola in [3] observed that MacDonald’s proof was not conclusive. So, they proved that

3n if G is a Camina group of nilpotence class 3, then |G3| ≤ p 2 . In our fourth theorem we were able to give a conclusive proof of MacDonald’s conjecture.

n Theorem 4. If G is a Camina group of nilpotence class 3, then |G3| ≤ p .

In the ﬁrst chapter, we give some general background and results in group theory that we will be using in later chapters. In the second chapter, we state some facts about the vanishing oﬀ subgroup and some important results about the centralizer of G0 and its relationship with

V (G). In the third chapter we give some general lemmas about the central series associated with the vanishing oﬀ subgroup. In the fourth chapter we prove our main theorem of this dissertation. Also, we prove the fourth theorem. In the ﬁfth chapter we prove our third theorem along with some results and facts about Camina triples. In our sixth chapter we state some open questions for future research. CHAPTER 1

BACKGROUND ABOUT FINITE GROUPS

In this chapter, we give some background deﬁnitions and results about ﬁnite groups. Let

G be a ﬁnite group and N a subgroup of G. We say that N is normal in G if for every x ∈ G,

we have xNx−1 = N. If g ∈ G and H is a subgroup of G, then the set Hg = {xg | x ∈ H}

is called the right coset of H. The set G/N = {Nx | x ∈ G} is a group if and only if N is

normal in G. The index of N in G, denoted by |G : N| or |G/N|, is |{Nx | x ∈ G}|. The following result is Theorem 2.23 in [5] and it is due to Lagrange.

Theorem 1.1. If N is a subgroup of G, then |G| = |G : N||N|. In particular if G is ﬁnite, then |N| divides |G| and |G|/|N| = |G : N|.

Let g ∈ G. The centralizer of g in G is CG(g) = {x ∈ G | xg = gx}. The following

theorem is Corollary 4.11 in [5]. It is a useful result that relates the order of the conjugacy

class of an element of a group and the index of its centralizer.

Theorem 1.2. If g ∈ G, then |cl(g)| = |G : CG(g)|.

A well deﬁned map f : G → H, where G and H are groups, is called a homomorphism

if f(xy) = f(x)f(y) for all x, y ∈ G. If a homomorphism is one-to-one and onto, then it is

called an isomorphism. An isomorphism from a group to itself is called an automorphism. A

subgroup N of a group G is called a characteristic subgroup of G if for every automorphism

f of G, we have f(N) = N. If f is a homomorphism of G, then the normal subgroup

N = {g ∈ G | f(g) = 1} is called the kernel of the homomorphism and is denoted ker(f).

Also, if f is a homomorphism from a group G to a group H, then we say that f is surjective

8 9

or onto if for every y ∈ H there exists x ∈ G such that f(x) = y. The following theorem is known as the First Isomorphism theorem.

Theorem 1.3. If f is a surjective homomorphism from a group G to a group H, then

G/ker(f) is isomorphic to H.

A ﬁnite collection of normal subgroups Ni of a group G is a normal series of G provided that 1 = N0 ≤ N1 ≤ · · · ≤ Nr = G. If such a series exists, and Ni/Ni−1 is abelian for all

1 ≤ i ≤ r, then we say that G is solvable. Every maximal group of a ﬁnite solvable group must have a prime power index. Another fact about solvable groups is that given a normal subgroup M of a group G, if M and G/M are solvable, then G is solvable. Also, subgroups and quotient subgroups of a solvable group are solvable.

The center of a group is the subgroup of G deﬁned by Z(G) = {x ∈ G | gx = xg for all g ∈ G}. We now consider the terms of the upper central series of G, deﬁned by

Z1 = Z(G) and Zi/Zi−1 = Z(G/Zi−1). A normal series is a central series if in addition, we have Ni/Ni−1 ≤ Z(G/Ni−1) for 1 ≤ i ≤ r. A group G is nilpotent if it has a central series.

There are a few equivalent conditions that a group has to have to be nilpotent. A group

G is nilpotent if and only if all Sylow subgroups are normal in the group. Also, for every normal subgroup M > 1 of G we have M ∩ Z(G) > 1. A group G is nilpotent if and only if

G is the direct product of its nontrivial Sylow subgroups.

Let Gk be the k-th term in the lower central series deﬁned in the introduction. The following result is from Theorem 8.17 in [5].

Lemma 1.4. Let G be any group. Then G is nilpotent if and only if Gn = 1 for some positive integer n.

If G = 1, then G is said to have nilpotence class 0. If G > 1 and G is nilpotent, the nilpotence class of G is the positive integer n so that Gn > 1 and Gn+1 = 1. 10

We now give the deﬁnition of Frobenius groups. Let H be a proper subgroup of a group

G. For every element g ∈ G, deﬁne Hg = {g−1xg | x ∈ H}. Assume that G has a subgroup

H > 1 that satisﬁes H ∩ Hg = {1} for every g ∈ G \ H. Then H is called a Frobenius complement in G. A group which contains a Frobenius complement is called a Frobenius group. The following result is Theorem 7.2 in [4], and it is due to Frobenius.

Theorem 1.5. If G is a Frobenius group with complement H, then there exists a normal subgroup N in G with G = HN and H ∩ N = {1}.

x As stated in Lemma 7.3 in [4], one obtains N = (G \ ∪x∈GH ) ∪ {1}. The subgroup N is called the Frobenius kernel of G.

Recall, the commutator of two elements in a group. Given x, y ∈ G, the commutator of x and y is denoted by [x, y] and deﬁned by [x, y] = x−1y−1xy. Also, let H,K be two subgroups

of G. The commutator subgroup of H and K is [H,K] = hh−1k−1hk | h ∈ H, k ∈ Ki. Deﬁne

a higher commutator by [x, y, z] = [[x, y], z], and [H,K,L] = [[H,K],L]. The following result

is Lemma 8.26 in [5] and is called P. Hall’s Identity.

Lemma 1.6. Let x, y, z ∈ G. Then

[x, y−1, z]y[y, z−1, x]z[z, x−1, y]x = 1.

The next result is an application of Hall’s Identity. It is called the Three Subgroups

Lemma, it is Lemma 8.27 in [5].

Lemma 1.7. If X,Y,Z are three subgroups of G such that [X,Y,Z] = 1 and [Y,Z,X] = 1, then [Z,X,Y ] = 1.

The following lemma is Lemma X.3 in [6], and it is due to Dedekind.

Lemma 1.8. If H ≤ K ≤ G and L ≤ G, then K ∩ HL = H(K ∩ L). 11

Now, we are going to give some background deﬁnitions and results about characters.

Let χ be a character of a group G. The kernel of χ is ker(χ) = g ∈ G | χ(g) = χ(1). The

following result is Lemma 2.22 in [4].

Lemma 1.9. Let N be a normal subgroup of G.

1. If χ is a character of G and N ≤ ker(χ), then χ is constant on cosets of N in G and

the function χb on G/N deﬁned by χb(Ng) = χ(g) is a character of G/N.

2. If χb is a character of G/N, then the function χ deﬁned by χ(g) = χb(Ng) is a character of G.

3. In both (a) and (b), χ ∈ Irr(G) if and only if χb ∈ Irr(G/N).

Based on the above lemma, we can view Irr(G/N) as Irr(G/N) = {χ ∈ Irr(G) | N ≤

ker(χ)}. We introduced the notation of Irr(G | N) in the introduction. It is not diﬃcult

to see that Irr(G) is a disjoint union of Irr(G/N)) and Irr(G | N). The following lemma is

Corollary 2.23 in [4].

Lemma 1.10. If G be a group with commutator group G0, then

1. G0 = ∩{ker(λ) | λ is linear },

2. |G : G0| is equal to the number of linear characters of G.

Let χ be a character of a group G, and let g ∈ G. Then χ(g) is the complex conjugate of χ(g). One can show that χ(g−1) = χ(g). The next result is Theorem 2.18 in [4], and it is called the Second Orthogonality Relation.

Lemma 1.11. If g ∈ G, then

X χ(g)χ(g) = |CG(g)|. χ∈Irr(G) 12

Deﬁne a function ρ of a ﬁnite group G by ρ(g) = 0 if g 6= 1 and |G| if g = 1. The

function ρ is a character, and it is called the regular character of the group G. Also, the P regular character ρ of a group G has the property that ρ = χ∈Irr(G) χ(1)χ. If N is a normal subgroup of a group G, then the regular character of the quotient group G/N is P ρG/N (N) = |G/N| and ρG/N (Ng) = 0 if g ∈ G \ N. Also, ρG/N (g) = θ∈Irr(G/N) θ(1)θ. Let P σ be a character of G such that σ(g) = ρG/N (Ng). Thus, ρ − σ = θ∈Irr(G|N) θ(1)θ, and (ρ − σ)(g) = |G| − |G/N| if g = 1, (ρ − σ)(g) = −|G/N| if g ∈ N \{1}, and (ρ − σ)(g) = 0 if g ∈ G \ N.

Let G be a ﬁnite group and p a prime. A Sylow p-subgroup of G is a subgroup P ≤ G, such that |P | = pa is the full power of p that divides |G|. The set of all Sylow p-subgroups of G is denoted by Sylp(G). For any ﬁnite group, Sylow p- subgroups exist for every prime.

Also, all the subgroups in Sylp(G) are conjugate. Hence, if for a prime p, there exists a normal Sylow p-subgroup, then that Sylow p-subgroup is unique. We say that a subgroup

N is a Hall p-complement of G for some prime p if p does not divide |N| and |G : N| is a power of p. Similarly, if we let π be a ﬁnite set of prime numbers, a subgroup L is said to be a Hall π-complement of G if |L| is not divisible by any prime in π and |G : L| is only divisible by primes in π. In general, groups are not guaranteed to have a Hall p-complement, but for solvable groups they are.

Deﬁne Op(G) to be the unique minimal normal subgroup of p-power index, and Ap(G) to be the unique minimal normal subgroup of G such that G/Ap(G) is an abelian p-group.

The groups Op(G) and Ap(G) are guaranteed to exist for any group. Now, we are going to state Tate’s Theorem, and it is Theorem 6.31 in [4].

p p p Theorem 1.12. Let P ∈ Sylp(G) and P ≤ N. If N ∩ A (G) = A (N), then N ∩ O (G) =

Op(N). CHAPTER 2

BACKGROUND ABOUT V (G)

In this chapter we state some definitions and results about the subgroup V (G). Define the vanishing off subgroup V (χ) of a character χ to be the subgroup generated by all the elements g ∈ G. Thus, V (χ) = hg ∈ G | χ(g) 6= 0i. Note that V (χ) is a normal subgroup of

G. Also, |G : V (χ)| divides χ(1)2, which was proved on page 200 of [4]. As mentioned by

Lewis in [8], observe that if H is any subgroup so that χ vanishes on G\H, then V (χ) ≤ H.

In other words, V (χ) is the smallest subgroup K of G such that χ vanishes on G \ K. Note

that if λ is a linear character of G, then V (λ) = G. That is why in this dissertation, we

are only interested in studying V (χ) where χ is not linear. In other words, it is possible to

have V (χ) proper in G.

Deﬁne the vanishing oﬀ subgroup V (G) of G to be the subgroup generated by all the

elements g ∈ G where there is some nonlinear character χ ∈ Irr(G) so that χ(g) 6= 0,

that is, V (G) = hg ∈ G | there exists χ ∈ nl(G) such that χ(g) 6= 0i. Also, note that

V (G) is the smallest subgroup of G such that all nonlinear irreducible characters vanish on

G \ V (G). In particular, we have V (χ) ≤ V (G) for all χ ∈ nl(G). In fact, one can show that Q V (G) = χ∈nl(G) V (χ). Notice that V (G) is a characteristic subgroup of G. Lewis showed in [8], if G is nonsolvable, then V (G) = G. Thus, we only consider the case where G is

solvable. The following result is Theorem 1 in [8].

Theorem 2.1. If G is a nonabelian solvable group, then G/V (G) is either cyclic or an

elementary abelian p-group for some prime p.

Our goal in this chapter is to state some facts about the central series associated with

V (G), which was deﬁned in the introduction of this dissertation. The following lemma is a

13 14

nice result which was proved in Lemma 4.1 of [8]. Note that conclusion (1) was mentioned in the introduction.

Lemma 2.2. If G is a group, then the following are true:

1. If Vn < Gn for some n, then Vi < Gi for all i with 1 ≤ i ≤ n.

2. If Vn < Gn, then G/Vn is nilpotent of nilpotence class n and Vi(G/Vn) = Vi/Vn for

all 1 ≤ i ≤ n.

As mentioned in the introduction, the idea of trying to ﬁnd the index of Vi−1 in Gi−1 when Vi < Gi was inspired by Lewis’s work and techniques in [8]. So, in the next few lemmas we state some results from [8]. The next lemma, Lemma 4.6 from [8], is a result

0 that deals with the case V3 < G3. First, recall that D3/V3 = CG/V3 (G /V3).

Lemma 2.3. Assume that V3 < G3. Suppose that b ∈ D3 \ V1, and let B/V2 = CG/V2 (bV2).

Then B ≤ D3.

The next result is Lemma 4.7 in [8], which gives us an interesting relation between V1 and two special centralizers.

Lemma 2.4. Assume that V3 < G3. Suppose that a ∈ G \ D3, and let A/V2 = CG/V2 (aV2).

Then A ∩ D3 = V1.

Now, we are ready to state a deeper result about the index of V1 in G and its relationship

0 with the index of V2 in G , which was proved by Lewis as Lemma 4.10 in [8].

Lemma 2.5. If V3 < G3, then there exist a positive integer n and a prime p such that

0 2 2n |G : V1| = |G : V2| = p .

In Lemma 4.11 in [8], Lewis was able to determine the structure of D3.

0 p Lemma 2.6. If V3 < G3, then either |G : D3| = |G : V2| = |G : V1| or D3 = V1. 15

Lewis presented examples in [8] that show both possibilities occur in the above Lemma.

We now show that if Gk is H1, then Vk−1 = Gk−1 ∩ Yk.

Lemma 2.7. Assume that Vk < Gk for k ≥ 3. If there exists N such that Vk ≤ N < Gk with Vk−1/N = Gk−1/N ∩ Z(G/N), then Vk−1 = Gk−1 ∩ Yk.

Proof. Observe that Yk/N ≤ Z(G/N). We have

Vk−1/N ≤ (Yk ∩ Gk−1)/N = Yk/N ∩ Gk−1/N ≤ Z(G/N) ∩ Gk−1/N = Vk−1/N.

Thus, we have equality throughout. So, Vk−1 = Gk−1 ∩ Yk as desired.

As an immediate consequence, note that if Gk is H1, then Vk−1 = Gk−1 ∩ Yk. We now

give the lemma that was proved as Lemma 4.12 in [8], which shows that G3 is H1.

0 Lemma 2.8. Assume that V3 < G3. If Y/V3 = Z(G/V3), then V2 = G ∩ Y.

The next theorem gives us information regarding the terms of the central series associated with V (G). In particular, all the quotients Gi/Vi are elementary abelian p-groups for some prime p, which was proved by Lewis as Lemma 4.4 in [8].

0 Theorem 2.9. Let G be a group. If V2 < G , then there is a prime p so that Gi/Vi is an elementary abelian p-group for all i ≥ 1.

This next result describes the relationship between V (G) and nonabelian quotients of

G, which was proved by Lewis as Lemma 3.3 in [8].

Lemma 2.10. Let N be a normal subgroup of G so that G/N is nonabelian. Then N ≤

V (G) and V (G/N) ≤ V (G)/N. CHAPTER 3

GENERAL LEMMAS FOR THE CENTRAL SERIES ASSOCIATED WITH V (G)

In this chapter, we prove some lemmas that are useful for the proofs of our theorems.

Also, some of these facts give us a good idea about the relation between the lower central series and the central series associated with the vanishing oﬀ subgroup that we deﬁned in the introduction. Lewis showed in [8] that both series are related by proving that Vi ≤ Gi ≤

Vi−1.

Lemma 3.1. If G is nilpotent and |Gi| = p, then for every x ∈ Gi−1 \ (Gi−1 ∩ Yi), we have cl(x) = xGi.

Proof. Because G is nilpotent, we can write G = P × Q where P is a p-group and Q is

0 a p -group. Hence, Gi−1 = Pi−1 × Qi−1. As |Gi| = p, we have Gi = Pi. In particular,

Qi−1 ≤ Z(G). Observe that Gi−1/Gi is central in G/Gi. Thus, it follows that cl(x) ⊆ xGi.

We deduce that |cl(x)| ≤ p. Recall that x ∈ Gi−1 \ Yi, which implies Q ≤ CG(x). Now,

|cl(x)| = |G : CG(x)| divides |G : Q| = |P |. Therefore, |cl(x)| is either 1 or p. Since x is not central, we must have |cl(x)| = p = |xGi|. We conclude that cl(x) = xGi.

Now, we get a relationship between the central series associated with the vanishing oﬀ subgroup of the whole group and a quotient group of that group.

Lemma 3.2. If Vk < Gk for some k ≥ 3, then for every N < Gk, we have Vi(G/N) = Vi/N for every 2 ≤ i ≤ k.

Proof. We are going to prove this by induction. In Lemma 2.2 in [8], we have V1(G/V2) =

V (G)/V2. Let X/N = V (G/N). By Lemma 2.10, X ≤ V (G). On the other hand, V2/N

16 17

is normal in G/N. By Lemma 2.10 applied to G/N, we have V (G)/V2 = V1(G/V2) =

V1((G/N)/(V2/N)) ≤ V (G/N)/(V2/N) = (X/N)/(V2/N) = X/N. So, V (G) ≤ X ··· X =

V (G). Hence, V2(G/N) = V2/N. That will be the initial case of the induction. Now, suppose

that i > 2 and assume that Vi−1(G/N) = Vi−1/N. Hence, Vi(G/N) = [Vi−1(G/N), G/N] =

[Vi−1/N, G/N] = [Vi−1,G]N/N = Vi/N as desired.

Now, we see the importance of the H1 hypothesis.

Lemma 3.3. If Vi = 1 and Gi is H1, then for every x ∈ Gi−1 \ Vi−1, we have cl(x) = xGi.

Proof. Since Vi = 1, we have Gi is central in G. Thus, [x, G] is central. This implies that g → [x, g] is a homomorphism so that [x, G] = {x−1xg | g ∈ G}. It follows that the map

−1 a 7→ x a is a bijection from cl(x) to [x, G]. Hence, cl(x) = xGi if and only if [x, G] = Gi.

Since x ∈ Gi−1, it follows that [x, G] ≤ Gi. Suppose that [x, G] < Gi, and we want to

ﬁnd a contradiction. We can ﬁnd N such that [x, G] ≤ N < Gi, where |Gi : N| = p.

Since x 6∈ Yi, [x, G] 6= 1. Thus, N > 1. Applying Lemma 2.10, it is not diﬃcult to see that Vi−1(G/N) ≤ Vi−1/N. Notice that xN ∈ Yi(G/N). On the other hand, we have xN ∈ Gi−1/N = (G/N)i−1. Thus, xN ∈ Yi(G/N) ∩ Gi−1/N = Vi−1(G/N) ≤ Vi−1/N.

Therefore, x ∈ Vi−1, which contradicts the choice of x.

The following lemma is a nice result about the irreducible characters in Irr(G|Gk).

Lemma 3.4. If Vk = 1 and Gk is H1 where k ≥ 3, then all the characters in Irr(G|Gk) vanish on Gk−1 \ Vk−1.

Proof. Consider x ∈ Gk−1 \Vk−1. By Lemma 3.3 we have cl(x) = xGk. Applying the second orthogonality relation we get,

X 2 X 2 X 2 |G|/|Gk| = |G|/|cl(x)| = |CG(x)| = |χ(x)| = |χ(x)| + |χ(x)| .

χ∈Irr(G) χ∈Irr(G/Gk) χ∈Irr(G|Gk) 18

Since Gk−1/Gk is central in G/Gk we can use the second orthogonality relation in G/N to see that,

X 2 X 2 |G : Gk| = |χ(xGk)| = |χ(x)| .

χ∈Irr(G/Gk) χ∈Irr(G/Gk) Hence, X |χ(x)|2 = 0.

Deﬁne Ei/(Gi−1 ∩ Yi) = CG/Gi−1∩Yi (Gi−2/(Gi−1 ∩ Yi)). We know that Vi−1 ≤ Gi−1, and since Vi = [Vi−1,G],Vi−1 ≤ Yi, and hence Vi−1 ≤ Gi−1 ∩ Yi. Since [Gi−1,Di−1] ≤ Vi−1 ≤

Gi−1 ∩ Yi, it follows that Di−1 ≤ Ei.

Recall that as a consequence of Lemma 2.7, if Gi is H1, then Vi−1 = Gi−1 ∩ Yi. Hence,

Di−1/Vi−1 = CG/Vi−1 (Gi−2/Vi−1) = CG/Gi−1∩Yi (Gi−2/Gi−1 ∩ Yi) = Ei/(Gi−1 ∩ Yi). In particular, Di−1 = Ei.

0 Notice that only in our next lemma we do use the hypothesis that G /Vi is abelian.

0 Lemma 3.5. Let Vi < Gi, suppose that i ≥ 4, and assume that G /Vi is abelian. Then

Di ≤ Ei.

0 Proof. We may assume that Vi = 1. Hence, Di = CG(Gi−1) and G is abelian. Since

0 0 0 G is abelian, we obtain [G, Di,Gi−2] ≤ [G ,G ] = 1. On the other hand, we obtain

[Gi−2, G, Di] = [Gi−1,Di] = 1. Hence, by the Three Subgroups lemma we get [Di,Gi−2,G] =

1. Therefore, [Di,Gi−2] ≤ Yi. Now, we know that [Di,Gi−2] = [Gi−2,Di] ≤ Gi−1. Hence,

[Di,Gi−2] ≤ Gi−1 ∩ Yi. We deduce that Di ≤ Ei, as desired.

In the next lemma we obtain an upper bound for the index of Di in G.

Lemma 3.6. Assume that Vi = 1. If |Gi| = p, then |G : Di| ≤ |Gi−1 : Gi−1 ∩ Yi|. 19

Proof. By Theorem 2.1, we know that Gi−1/Vi−1 is an elementary abelian p-group. Hence, we can ﬁnd x1, ··· , xt ∈ Gi−1 \ Yi, such that Gi−1 = hx1, ··· , xt,Gi−1 ∩ Yii, where |Gi−1 :

j = 1, ··· , t. Thus,

t t \ Y t |G : Di| = |G : CG(xj)| ≤ |G : CG(xj)| = p = |Gi−1 : Gi−1 ∩ Yi|. j=1 j=1

In our next lemma, we prove a very interesting isomorphism that will be a key to get the index of Vi in Gi.

Lemma 3.7. Assume that Gi is H1. Let a ∈ Gi−1 \ Vi−1 and set K/Vi = CG/Vi (aVi). Then ∼ G/K = Gi/Vi.

Proof. Without loss of generality, we may assume that Vi = 1. Consider the map from

G to Gi deﬁned by g → [g, a]. Since a ∈ Gi−1, we have [g, a] ∈ Gi for every g ∈ G.

Hence, this map is well deﬁned. Also, we know that Gi is central in G. Thus, this map is a homomorphism with kernel K. By Lemma 3.3 this map is onto. Therefore, by the First ∼ Isomorphism theorem, we conclude that G/K = Gi.

Now, we prove the following result.

Corollary 3.8. If Gi is H1, then |Gi : Vi| ≤ |G : Di|.

Proof. Let K be as in Lemma 3.7. We know since a ∈ Gi−1 and Di/Vi = CG/Vi (Gi−1/Vi) that Di ≤ K. Hence, |Gi : Vi| = |G : K| ≤ |G : Di|.

The following result is very useful to prove our main theorem.

0 Lemma 3.9. Assume that Vi < Gi, G /Vi is abelian, and Gi−1 is H1, for i ≥ 4. Let

a ∈ Gi−2 \ Vi−2 and set K/Vi−1 = CG/Vi−1 (aVi−1). Then K ≤ Di. 20

0 Proof. We may assume that Vi = 1. Hence, Vi−1 is central in G, G is abelian, Yi = Z(G), and Di = CG(Gi−1). Fix x ∈ K, and let w ∈ G be arbitrary. Notice that [a, x] ∈ Vi−1 ≤ Yi.

0 0 0 0 Thus, [a, x, w] = 1. Also, [x, w] ∈ G . Because i ≥ 4, Gi−2 ≤ G , a ∈ G . Since G is abelian,

[x, w, a] ≤ [G0,G0] = 1. Therefore, by Lemma 1.6, we obtain [w, a, x] = 1. This implies that x centralizes [w, a]. Since a 6∈ Vi−2 and Gi−1 is H1, we deduce by Lemma 3.3 that as w runs through all of G,[w, a] runs through all of Gi−1. Hence, x centralizes Gi−1. Thus, x ≤ Di.

Therefore, K ≤ Di.

As a consequence of the previous lemma, we get the following corollary.

0 Corollary 3.10. If Vi < Gi, G /Vi is abelian, and Gi−1 is H1 for i ≥ 4, then Di−1 ≤ Di.

Proof. Let a ∈ Gi−2 \Vi−2 and set K/Vi−1 = CG/Vi−1 (aVi−1). Then, by Lemma 3.9, we have

K ≤ Di. Also, we know that Di−1 ≤ K. Thus, Di−1 ≤ Di.

We now get an upper bound for |Gi−1 : Gi−1 ∩ Yi|.

Lemma 3.11. If Vi < Gi and Gi−1 is H1, then |G : Ei| ≥ |Gi−1 : Gi−1 ∩ Yi|.

Proof. Fix a ∈ Gi−2 \ Vi−2, and consider the map f from G to Gi−1/Vi−1 deﬁned by f(g) =

[a, g]Vi−1. As in the proof of Lemma 3.7, we know that f is an onto homomorphism. It follows that f maps G/Ei onto Gi−1/f(Ei). Thus, |Gi−1 : f(Ei)| ≤ |G : Ei|. Since a ∈ Gi−2,

[Ei, a] ≤ Gi−1 ∩ Yi, and thus f(Ei) ≤ (Gi−1 ∩ Yi). Then |Gi−1 : Gi−1 ∩ Yi| ≤ |Gi−1 : f(Ei)|.

Hence, |G : Ei| ≥ |Gi−1 : Gi−1 ∩ Yi| as required. CHAPTER 4

PROOF OF THE FIRST TWO THEOREMS AND THEOREM 4

In this chapter, we prove the ﬁrst two Theorems along with Theorem 4 using the general lemmas that we proved in the previous chapter.

Now, we prove Theorem 1.

Proof of Theorem 1. We have Di = D3 for i = 3. That will be our initial case of induction.

Assume that the theorem is true for i − 1. We are going to prove it for i. By hypothesis, we have Gi is H1. Hence, by Lemma 2.7, we have Vi−1 = Gi−1 ∩ Yi. Thus, Ei = Di−1. We know by the inductive hypothesis that Di−1 = D3. Thus, Ei = D3. By Lemma 3.5, we know that

Di ≤ Ei. By Corollary 3.10, we have Di−1 ≤ Di. Thus, Di−1 ≤ Di ≤ Ei = Di−1. Therefore, we deduce that Di = Ei = Di−1 = D3.

Now, we are ready to prove our second theorem.

Proof of Theorem 2. We are going to prove this theorem by induction. Notice that the initial case of induction (i = 3) is done by Lewis in [8]. Now, assume that the theorem is true for k = i − 1. We are going to prove it for k = i. Also, in this proof, without loss

of generality, we may assume that Vi = 1. We also know by the inductive hypothesis that

Di−1 = D3 and Gi−1 is H1. Now, by Lemma 3.5 we have that Di ≤ Ei. By Corollary 3.10, we have Di ≤ Di−1. First we assume that |Gi| = p. Thus, we obtain

|G : Di| ≥ |G : Di−1| ≥ |Gi−1 : Vi−1| ≥ |Gi−1 : Gi−1 ∩ Yi|.

But, by Lemma 3.6, we have |G : Di| ≤ |Gi−1 : Gi−1 ∩ Yi|. Hence, we have equality throughout the above inequality. Hence, Vi−1 = Gi−1 ∩ Yi, and |Gi−1 : Vi−1| = |G : Di−1|.

21 22

Now, assume that |Gi| > p. Consider a normal subgroup N, such that Vi ≤ N < Gi and |Gi : N| = p. The above argument shows that Vi−1(G/N) = Yi(G/N) ∩ Gi−1/N. Thus,

Gi satisﬁes H1. By strong induction we have G4, ··· ,Gi−1 satisfy H1. Thus, we may apply

Theorem 1 to see that Di = D3. First deﬁne DiN /N = CG/N (Gi−1/N). Note that Di ≤ DiN ,

and so DiN = D3. The above argument yields |G : D3| = |G : Di−1| = |Gi−1 : Vi−1|. To

prove part (d), since Gi is H1, by Corollary 3.8 we obtain |Gi : Vi| ≤ |G : Di| as desired.

Now, we prove Theorem 4, which is a conclusive proof of MacDonald’s conjecture in [9] about the order of G3, in the case when G is a Camina group of nilpotence class 3.

Proof of Theorem 4. Note that, V1 = G2. Hence, V2 = G3. We deduce that V3 = G4 = 1. ∼ Let a ∈ G2 \V2 and set K = CG(a). Thus, by Lemma 3.7, we have G3 = G/K. But, D3 ≤ K.

n By MacDonald in [9], we know that |G : D3| = p . Thus,

n |G3| = |G : K| ≤ |G : D3| = p . CHAPTER 5

CAMINA PAIRS AND CAMINA TRIPLES

In this chapter, we prove Theorem 3 along with some facts about Camina triples and

Camina pairs. Recall, if N is a normal subgroup of a group G, then we say that (G, N) is

a Camina pair if and only if for every x ∈ G \ N, x is conjugate to all of xN. Notice that

means the conjugacy class of x is a union of cosets of N in G. In [9], MacDonald proved that

if (G, N) is a Camina pair where G is a p-group, then N is a term in both the upper and

the lower central series of G. In the next lemma, we state some known equivalent conditions

for a pair (G, N) to be a Camina pair.

Lemma 5.1. Let 1 < N be a proper normal subgroup of a group G. Then the following are

equivalent:

1. (G, N) is a Camina pair.

2. If x ∈ G \ N, then |CG(x)| = |CG/N (xN)|.

3. If xN and yN are conjugate in G/N and nontrivial, then x is conjugate to y in G.

4. For x ∈ G \ N and z ∈ N, there exists an element y ∈ G so that [x, y] = z.

The second part of Lemma 5.1, was proved in [1] and [2]. The third and fourth parts of

Lemma 5.1 were only proved in [2].

Recall, given two normal subgroups 1 < M ≤ N of a group G, we say that (G, N, M) is a

Camina triple if for every x ∈ G\N, x is conjugate to all of xM. Also, recall that Irr(G | M) =

{χ ∈ Irr(G) | M * ker(χ)}. Motivated by V (G), we deﬁne V (G | M) = hg ∈ G | there exists χ ∈ Irr(G | M) such that χ(g) 6= 0i. Observe that V (G | G0) = V (G), and V (χ) ≤ V (G | M)

23 24

Q for all χ ∈ Irr(G | M). It is not diﬃcult to see that V (G | M) = χ∈Irr(G|M) V (χ). Now, we prove a couple of results about V (G | M).

Lemma 5.2. Let N be a normal subgroup of G. If x ∈ N, then there is a χ ∈ Irr(G | N) so

that χ(x) 6= 0. In particular, N ≤ V (G | N).

P Proof. Let τ = χ∈Irr(G|N) χ(1)χ. Now, τ is a character of G. Observe that τ = ρ − σ where ρ is the regular character of G and σ is deﬁned by σ(g) = ρG/N (gN) where ρG/N is the regular character of G/N. It follows that τ(g) = |G| − |G : N| if g = 1, −|G : N| if g ∈ N \ 1, and equals 0 if g ∈ G \ N. Since |G : N| 6= 0, it follows that there is some

χ ∈ Irr(G | N) so that χ(g) 6= 0.

In our next lemma, we show that if (G, N, M) is a Camina triple and M is not inside

G0, then V (G | M) is the whole group.

0 Lemma 5.3. Let (G, N, M) be a Camina Triple. If M * G , then V (G | M) = G.

0 Proof. If M * G , then there exists a linear character λ of G such that M * kerλ. Hence

V (λ) 6 V (G | M), and since V (λ) = G, V (G | M) = G.

Inspired by the results in Lemma 5.1, we prove in the next theorem some equivalent conditions for a triple (G, N, M) to be a Camina triple, this also includes the proof of part

(4) of Theorem 3.

Theorem 5.4. If 1 6= M < N are two normal subgroups of a ﬁnite group G, then the following are equivalent:

1. (G, N, M) is a Camina triple.

2. |CG(g)| = |CG/M (Mg)| for every g ∈ G \ N.

3. For every g ∈ G \ N, we have χ(g) = 0 for all χ ∈ Irr(G | M). 25

4. V (G|M) ≤ N.

5. For all g ∈ G \ N and z ∈ M, there exists y ∈ G such that [g, y] = z.

Proof. First, we show that (1) implies (2). Assume that (G, N, M) is a Camina triple and let

x g ∈ G \ N. Notice that cl(g) = ∪x∈G(Mg) . Hence, |G : CG(g)| = |G/M : CG/M (Mg)||M|, and so |CG(g)| = |CG/M (Mg)| as desired. We now show that (2) implies (3). Assume (2) and let g ∈ G \ N. By the Second Orthogonality Relation, we have

X 2 X 2 X 2 |CG(g)| = |χ(g)| = |χ(g)| + |χ(g)| . χ∈Irr(G) χ∈Irr(G|M) χ∈Irr(G/M)

P 2 But we know by (2) that |CG(g)| = |CG/M (Mg)| = χ∈Irr(G/M) |χ(g)| . Hence, we obtain

χ(g) = 0 for all χ ∈ Irr(G | M). Hence, all the generators of V (G | M) are contained in

N. Thus, V (G | M) ≤ N as desired. Now, we show that (4) implies (1). Assume that

V (G|M) ≤ N and let x ∈ G \ N and y ∈ M. Hence, yx 6∈ N. Thus yx 6∈ V (G|M). So,

for any χi ∈ Irr(G|M), χi(x) = χi(yx) = 0. Recall that Irr(G) \ Irr(G|M) = Irr(G/M).

Write Irr(G/M) = {ϕ1, ··· , ϕr}. For each ϕi there exists ϕi ∈ Irr(G) \ Irr(G | M) such

that ϕi(x) = ϕi(Mx) = ϕi(yx). Hence, x and yx have the same character values for

all irreducible characters of G. Since the irreducible characters form a basis for the class

functions, all class functions have the same value on x and xy. This implies that x and xy

are in the same class. Hence, x is conjugate to all of xM. We conclude that (G, N, M) is a

Camina triple. Thus, (4) implies (1).

To ﬁnish the proof of the theorem, it is enough to show that (1) is equivalent to (5).

First assume that (G, N, M) is a Camina triple; that is, if g ∈ G \ N, then g is conjugate to

all of gM. Hence, if z ∈ M, then there exists y ∈ G such that y−1gy = gz. It follows that

g−1y−1gy = z. Conversely, suppose that for all g ∈ G \ N and z ∈ M there exists y ∈ G 26

such that [g, y] = z. Fix g ∈ G \ N and z ∈ M. We need to show that g is conjugate to gz. we know there exists y such that g−1y−1gy = z. This implies that y−1gy = gz. Hence, g is

conjugate to every element in gM, and (G, N, M) is a Camina triple as required.

The following lemma describes the relationship between two Camina triples in the same group.

Lemma 5.5. If (G, N1,M) and (G, N2,M) are Camina triples, then (G, N1 ∩ N2,M) is a

Camina triple.

Proof. Notice that 1 < M ≤ N1 ∩ N2. If g ∈ G \ N1 ∩ N2, then either g ∈ G \ N1 or

g ∈ G \ N2. In either case, g is conjugate to all of gM. Hence, (G, N1 ∩ N2,M) is a Camina

triple as desired.

We now show that Camina pairs are special cases of Camina triples.

Lemma 5.6. The triple (G, N, N) is a Camina triple if and only if (G, N) is a Camina

pair.

Proof. Observe that (G, N) is a Camina pair if and only if for every g ∈ G \ N, we have cl(g) = gN. This occurs if and only if (G, N, N) is a Camina triple.

We now prove a fact about the center of a group G in the case when (G, N, M) is a

Camina triple. Note that it is not diﬃcult to see that the intersection of Z(G) and the set of elements in G \ N has to be the empty set.

Lemma 5.7. If (G, N, M) is a Camina triple, then the following are true:

1. Z(G) ≤ N and

2. if K C G and K < M, then (G/K, N/K, M/K) is a Camina triple. 27

Proof. If g ∈ Z(G), then g is only conjugate to itself. Hence g is not conjugate to all of gM, and so g ∈ N. Therefore Z(G) ≤ N. Now, let K C G, with K < M. Hence, 1 < M/K ≤ N/K < G/K. Since every χ ∈ Irr(G|M) vanishes on G \ N, every χ ∈

Irr(G/K|M/K) vanishes on G/K \ N/K. It follows that (G/K, N/K, M/K) is a Camina triple as desired.

Now consider the terms of the upper central series of G when (G, N, M) is a Camina

triple. Let Z1 = Z(G) and Zi/Zi−1 = Z(G/Zi−1) for i > 1.

Lemma 5.8. If (G, N, M) is a Camina triple, and Zm < M, then Zm+1 ≤ N.

Proof. By Lemma 5.7 part 2, (G/Zm, N/Zm, M/Zm) is a Camina triple. So applying Lemma

5.7 part 1 to G/Zm, we get Z(G/Zm) ≤ N/Zm. Hence Zm+1 ≤ N as desired.

Now, we need to state this very useful theorem, which is Theorem D in [7], due to

Berkovich.

Theorem 5.9. Let N be a normal subgroup of G and suppose that every member of cd(G |

N 0) is divisible by some ﬁxed prime p. Then N is solvable and has a normal p-complement.

We need the next lemma to prove the remaining parts of our third theorem.

Lemma 5.10. If (G, N, M) is a Camina triple, then M is solvable and has a normal p-

complement for every prime p that divides | G : N |.

Proof. Let χ ∈ Irr(G | M). We know by Lemma 5.4 that χ(g) = 0 for all g ∈ G \ N. By the

discussion in [4] page 200, we deduce that for every prime p divisor of |G : N|, p divides

χ(1) for all χ ∈ Irr(G | M0). So by Berkovich’s theorem, M is solvable and M has a normal

p-complement.

Now, we show that if (G, N, M) is a Camina triple, then M has a normal π-complement,

where π is the set of primes that divide |G : N|. This proves the remaining parts of Theorem

3. 28

Lemma 5.11. If (G, N, M) is a Camina triple, and π = {p prime | p divides |G : N|}, then M has a normal π-complement Q such that M/Q is nilpotent.

Proof. Since (G, N, M) is a Camina triple, by Lemma 5.10, we know that M has a normal p-

complement for every p ∈ π. Now, let Q be the intersection of these normal p-complements.

Hence, Q is a normal π-complement of M. Now, to prove that M/Q is nilpotent, it will be

enough to show that any ﬁnite group having a normal p-complement for every prime p is

nilpotent. Let G be a ﬁnite group that has a normal p-complement for every prime p. We

work by induction on |G|. If |G| = 1, then the result is trivial, so we may assume G > 1.

Let p be a prime, and we show that G has a normal Sylow p-subgroup. If G is a p-group,

then this is trivial. Thus, we may assume that |G| is divisible by some prime q which is

not p. By hypothesis, G has a normal q-complement N. Observe that N < G, and so the

induction hypothesis implies that N is nilpotent. Hence, N has a normal Sylow p-subgroup

P and thus, P is characteristic in N. But G/N is a q-group, so P is a Sylow p-subgroup of

G. It follows that G has a normal Sylow p-subgroup as required.

Let (G, N, M) be a Camina triple and π the set of primes that divide | G : N |. We know that M has a normal π-complement, say Q, where M/Q is nilpotent. If M is a π-group,

then Q = 1 and hence, M is nilpotent.

We are now going to prove some results about Camina pairs using Camina triple results,

and the fact that they are special cases of Camina triples. In [1], Camina deﬁned a diﬀerent

hypothesis that is equivalent to Camina pairs. Let G be a ﬁnite group with a proper normal

subgroup N 6= 1 and a set of irreducible non-trivial characters of G, A = {χ1 ··· χn}, where

n is a natural number, such that

1. χi vanishes on G \ N and

Pn 2. there exist natural numbers α1, ··· , αn > 0 such that i=1 αiχi is constant on N \{1}. 29

We are able to identify the characters in the Camina hypothesis in [1]. First, let N be a normal subgroup of G and θ ∈ Irr(N). The inertia group of θ in G is denoted by T and deﬁned by T = {g ∈ G | θg = θ}.

Theorem 5.12. Let If (G, N) is a Camina pair, then A = Irr(G|N).

Proof. First, we show that A ⊆ Irr(G|N). To see this, suppose χj ∈ A \ Irr(G|N). This implies that χj ∈ Irr(G/N). On the other hand, since χj ∈ A, we have that χj(x) = 0 for all x ∈ G\N. This implies χj(xN) = 0 for all xN ∈ G/N \{N}, and hence, χj is a multiple of the regular character of G/N. Since N < G, we know that the regular character of G/N is not irreducible, and so we have a contradiction since it is not possible for an irreducible character to be a multiple of a reducible character. Thus no such χj exists in A. Therefore

A ⊆ Irr(G|N).

On the other hand, for every 1N 6= θ ∈ Irr(N) and by Theorem 6.11 in [4], there exists

G G χi ∈ A such that χi ∈ Irr(G|θ). Notice that θ (g) = 0 if g 6∈ N, and if g ∈ N, then θ (g) =

1 P x G 1 |N| x∈G θ (g), hence θ (g) = |N| |T |(θ1(g) + ··· + θn(g)) = |T : N|(θ1(g) + ··· + θn(g)),

where θi, i = 1 . . . n, are the distinct conjugates of θ in G. Note that χi(g) = 0 if g 6∈ N, and

G if g ∈ N, χi(g) = a(θ1(g) + ··· + θn(g)), where a is a nonnegative integer. Hence χi = cθ .

Our last result in this chapter states some new conditions for a pair (G, N) to be a

Camina pair.

Theorem 5.13. If G is a ﬁnite group and N C G, then the following are equivalent:

1. (G, N) is a Camina pair.

2. V (G|N) = N. 30

3. There is no x in N such that χ(x) = 0 for all χ ∈ Irr(G|N), and if x ∈ G \ N, then

χ(x) = 0 for all χ ∈ Irr(G|N).

Proof. Notice that 3) implies 2) is trivial. To prove 2) implies 1), assume that V (G|N) = N.

By Theorem 5.4, (G, N, N) is a Camina triple. Thus, by Lemma 5.6, (G, N) is a Camina pair. To prove 1) is equivalent to 3), by Lemma 5.6 and Theorem 5.4, (G, N) is a Camina pair if and only if V (G | N) ≤ N. CHAPTER 6

FUTURE RESEARCH QUESTIONS

In this chapter we state some future research questions, concerning both the vanishing oﬀ series and Camina triples. The study of the vanishing oﬀ series was inspired by the results

0 of Lewis in [8]. Lewis obtained the index of V2 in G without using the hypothesis that

0 0 G is abelian. To generalize Lewis’s results, we added the hypothesis that G /Vi is abelian.

However, we only used this in Lemma 3.5, which basically was needed to show that Di ≤ Ei.

So, the question here is can we prove Theorem 1 and 2 of the dissertation without assuming

0 G /Vi is abelian? If there is a counterexample, then what can we say about the index of

0 Vi in Gi? Also, what can we say about the order of the centralizer of G modulo Vi? Can we have examples with Vi < Gi for an arbitrary large value of i? Is there a bound on the values of i that can occur?

Our second question is about Camina triples. Camina in [1], showed that if (G, N) is a Camina pair, then G is a Frobenius group with kernel N, or G/N is p-group, or N is a p-group for some prime p. We would be interested in seeing if some similar conditions can be proved for Camina triples, i.e., is it the case that G/N is a p-group, M is a p-group for some prime p, or G is “related” to a Frobenius group? We think that the case where G/N and M are p-groups for some prime p occurs when G is nilpotent.

31 BIBLIOGRAPHY

[1] A. R. Camina, “Some Conditions Which Almost Characterize Frobenius Groups,” Israel Journal of Mathematics 31 (1978), 153-160.

[2] D. Chillag and I. D. MacDonald, “Generalized Frobenius Groups,” Israel Journal of Mathematics 47 (1984), 111-122.

[3] R. Dark and C. M. Scoppola, “On Camina Groups of Prime Power Order,” Journal of Algebra 181 (1996), 787-802.

[4] I. M. Isaacs, Character Theory of Finite Groups, Academic Press, San Diego, Cali- fornia, 1976.

[5] I. M. Isaacs, Algebra A Graduate Course, Academic Press, Paciﬁc Grove, California, 1993.

[6] I. M. Isaacs, Finite Group Theory, American Mathematical Society, Providence, Rhode Island, 2008.

[7] I. M. Isaacs and Greg Knutson, “Irreducible Character Degrees and Normal subgroups,” Journal of Algebra 199 (1998), 302-326.

[8] M. L. Lewis, “The vanishing-oﬀ subgroup,” Journal of Algebra 321 (2009), 1313-1325.

[9] I. D. MacDonald, “Some p-Groups of Frobenius and Extra-Special Type,” Israel Jour- nal of Mathematics 40 (1981), 350-364.

[10] I. D. MacDonald, “More on p-Groups of Frobenius Type,” Israel Journal of Mathe- matics 56 (1986), 335-344.