P-GROUP CODEGREE SETS and NILPOTENCE CLASS A

Home , Class of groups

p-GROUP CODEGREE SETS AND NILPOTENCE CLASS

A dissertation submitted to

Kent State University in partial

fulﬁllment of the requirements for the

degree of Doctor of Philosophy

Sarah B. Croome

May, 2019 Dissertation written by

Sarah B. Croome

B.A., University of South Florida, 2013

M.A., Kent State University, 2015

Ph.D., Kent State University, 2019

Approved by

Mark L. Lewis , Chair, Doctoral Dissertation Committee

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

Donald L. White

Robert A. Walker

Joanne C. Caniglia

Accepted by Andrew M. Tonge , Chair, Department of Mathematical Sciences

James L. Blank , Dean, College of Arts and Sciences TABLE OF CONTENTS

Table of Contents ...... iii

Acknowledgments ...... iv

1 Introduction ...... 1

2 Background ...... 6

3 Codegrees of Maximal Class p-groups ...... 19

4 Inclusion of p2 as a Codegree ...... 28

5 p-groups with Exactly Four Codegrees ...... 38

Concluding Remarks ...... 55

References ...... 55

iii Acknowledgments

I would like to thank my advisor, Dr. Mark Lewis, for his assistance and guidance. I would also like to thank my parents for their support throughout the many years of my education. Thanks to all of my friends for their patience.

iv CHAPTER 1

Introduction

The degrees of the irreducible characters of a ﬁnite group G, denoted cd(G), have often

been studied for their insight into the structure of groups. All groups in this dissertation

are ﬁnite p-groups where p is a prime, and for such groups, the degrees of the irreducible

characters are always powers of p. Any collection of p-powers that includes 1 can occur as

the set of irreducible character degrees for some group [11]. In fact, the group can always be

chosen so that its nilpotence class is at most 2. The question of when a set of character degrees

bounds the nilpotence class is also of interest, and the answer is hardly straightforward. Even

for groups with just two character degrees, there are starkly contrasting possibilities. In [13],

it is shown that {1, p} can occur as the character degree set of groups with arbitrarily large

class, while Theorem 3.10 of [14] states that if cd(G) = {1, pe} for e > 1, then the class of G

is at most p.

The purpose of this dissertation is to investigate the set of codegrees of a group and its

relationship to nilpotence class. The codegree of an irreducible character χ of a ﬁnite group

G is defined by cod(χ) = |G : ker(χ)|/χ(1), the index of the kernel of χ in G, divided by the degree of χ. The set of codegrees of the irreducible characters of G is denoted cod(G). This definition for codegrees first appeared in [17] in 2007, where the authors use a graph-theoretic approach to compare the structure of a group with its set of codegrees. Earlier use of the term codegree appears in [6] in 1989, with the slightly different definition |G|/χ(1). In their

paper, Chillag and Herzog examine the structure of groups whose codegrees are divisible by

at most two primes. This deﬁnition, without the actual term codegree, appeared even earlier

1 in [9] in 1981. In his paper, Gagola showed that for a ﬁnite group G with normal subgroup

N, if χ ∈ Irr(N) is invariant in G, then χ is extendible to G when (|G : N|, |N|/χ(1)) = 1.

Returning to the modern defintion of codegrees, in 2016, Du and Lewis proved in [7] that when G is a finite p-group and |cod(G)| = 3, the nilpotence class of G, denoted c(G), is at most 2, suggesting that groups with few codegrees may have bounded nilpotence class. Our original purpose was to find a sharp bound for the nilpotence class of groups with exactly four codegrees. When a group has exactly 4 codegrees and some additional restrictions we can show that the nilpotence class of the group is at most 4.

Theorem 1. Let G be a ﬁnite p-group such that cod(G) = {1, p, pb, pa}, where 2 ≤ b < a.

If any of the following hold, then G has nilpotence class at most 4:

(i) |cd(G)| = 2,

(ii) cd(G) = {1, p, p2},

(iii) |G : G0| = p2.

The coclass of a p-group G with nilpotence class c is given by logp |G| − c. If G has four codegrees and coclass at most 3, then G has nilpotence class at most 4, bounding the order of G for a given prime p.

Theorem 2. Let G be a p-group such that cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has coclass at most 3, then G has nilpotence class at most 4, and |G| ≤ p7.

With the following additional hypothesis, we can extend the result of Theorem 2 to p-groups with coclass at most 7.

2n Hypothesis (∗). If G is a p-group with nilpotence class n such that |G| ≥ p , then |Z2(G)|= 6 p2.

2 Theorem 3. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If

G has coclass at most 7, and G and all of its quotients satisfy Hypothesis (∗), then the nilpotence class of G is at most 4 and |G| ≤ p11.

In [7], Du and Lewis were able to bound c(G) in terms of the largest member of cod(G).

They showed that if pa, (where a > 1), is the largest codegree of G, then c(G) ≤ 2a − 2, and in some speciﬁc cases, c(G) ≤ 2a − 3. When |cod(G)| = 4, we can improve this bound.

Theorem 4. If G is a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a, then

c(G) ≤ a + 1.

This bound can be improved slightly when the two largest codegrees are consecutive

powers of p and the group does not have p2 as a codegree. Notice that in Theorem 5,

p2 ∈/ cod(G). The question of when p2 must be included in the set of codegrees of a group

motivates the results in chapter 4.

Theorem 5. If G is a ﬁnite p-group such that cod(G) = {1, p, pa−1, pa} for a ≥ 4, then

c(G) ≤ a.

It is also interesting to ask about the nilpotence class of G when cod(G) contains as

many codegrees as possible for the group’s order. Surprisingly, if G has order pn and cod(G)

contains every power of p up to pn−1, it turns out that the nilpotence class of G can either

be quite small or as large as possible, but not in between. This is the result of Theorem 6.

Theorem 6. If G is a p-group such that |G| = pn ≥ p2, then cod(G) = {1, p, p2, . . . , pn−1}

if and only if one of the following occurs:

∼ (i) G = Zpn−1 × Zp,

∼ (ii) G = Zpn−1 o Zp and has nilpotence class 2,

(iii) G has maximal class and |cd(G)| = 2.

3 While examining maximal class p-groups, it became apparent that the codegrees of these

groups are often consecutive powers of p. From Theorem 6, we know that a maximal class

group with order pn+1 will only have all powers of p up to pn as codegrees if the group

has exactly two character degrees. With three character degrees, the codegrees are still

consecutive powers of p, but the largest power may vary.

Theorem 7. Let G be a maximal class p-group for some prime p such that cd(G) = {1, p, pb}.

If |G| = pn, then cod(G) = {pi | 0 ≤ i ≤ c}, for some integer n − b ≤ c ≤ n − 2.

Any metabelian maximal class group can have at most three character degrees, so we know from Theorem 7 that the codegrees of such a group will be consecutive powers of p.

In this case, however, there are only two possibilities for the largest power.

Corollary 8. Let G be a metabelian maximal class p-group for some prime p. If |G| = pn,

then cod(G) = {pi | 0 ≤ i ≤ c}, where c = n − 1 or n − 2.

The codegrees of a normally monomial maximal class p-group will also be consecutive

p-powers. This result is similar to that of Theorem 7, but does not depend on the number

of character degrees of the group.

Theorem 9. Let G be a normally monomial maximal class p-group for some prime p. Let

n i |G| = p , b(G) = max{χ(1) | χ ∈ Irr(G)}, and b = logp(b(G)). Then cod(G) = {p | 0 ≤ i ≤ c} for some integer c ≥ n − b.

It is easily shown that p2 and p3 are always included among the codegrees of maximal

class p-groups (when the order of the group is large enough for the codegree to occur, e.g.,

p3 ∈ cod(G) when |G| ≥ p4). If the order of G is at least p6, G also has p4 as a codegree.

This evidence, including Theorems 6, 7, 9, and Corollary 8, leads us to ask whether the

codegrees of maximal class p-groups are always consecutive powers of p.

4 In the case of character degrees of p-groups, there is one (non-trivial) degree that stands out as more inﬂuential than the rest. In [15], it is shown that sets of p-power character degrees that include p are not class bounding. The set of codegrees of a p-group always includes p [7, Corollary 2.3], but the next smallest power, p2, plays an important role when determining nilpotence class. When p2 is missing from the set of codegrees of a group G, the quotient G/G0 is elementary abelian [7, Corollary 2.5]. In addition, if |G| = p5 or p6, then

G has nilpotence class at most 2, and if |G| = p7, then G has nilpotence class at most 3.

As p-groups with large nilpotence class relative to their order have a more predictable structure, it is often possible to characterize groups with small coclass in ways that are impossible for groups with a ﬁxed nilpotence class but arbitrarily large order. The following theorem describes a feature shared by large enough p-groups of coclass 2 and coclass 3.

Theorem 10. Let G be a p-group.

(i) If G has coclass 2 and order at least p5, then p2 ∈ cod(G).

(ii) If G has coclass 3 and order at least p6, then p2 ∈ cod(G).

With the addition of Hypothesis (∗), we can broaden this result to p-groups of arbitrary

ﬁnite order that have large enough nilpotence class.

Theorem 11. Let a p-group G and all of its quotients satisfy Hypothesis (∗). If G has coclass n ≥ 3 and |G| ≥ p2n, then p2 ∈ cod(G).

5 CHAPTER 2

Background

A group G is a set together with a binary operation ∗ such that the following hold:

(i) G is closed under the operation ∗,

(ii) the operation ∗ is associative,

(iii) there is an element 1 in G, called the identity, such that for every g in G, g∗1 = 1∗g = g,

(iv) for every element x ∈ G, there is an element y ∈ G such that x ∗ y = y ∗ x = 1.

A homomorphism from a group (G, ∗) to a group (H, ·) is a map ϕ such that for elements x and y of G, ϕ(x ∗ y) = ϕ(x) · ϕ(y). If ϕ is a bijection, then the map is an isomorphism.

Isomorphic groups are generally regarded as identical.

A group that is also commutative is called abelian. The number of elements of a ﬁnite group G is called the order of G, and is denoted |G|. An element g ∈ G has order n if this is the smallest positive integer such that gn = 1, and this number always divides |G|. The smallest integer n such that for every g ∈ G, gn = 1 is called the exponent of G. If the order of G is a power of a prime p, then G is called a p-group. An elementary abelian p-group is an abelian p-group for which every non-identity element has order p.

A subgroup H is a subset of G that is also a group under the operation of G. This relationship is usually denoted as H ≤ G, rather than H ⊆ G, which is more commonly used to indicate a subset. If H is properly contained in G, we write H < G. The index of the subgroup H in G is given by |G : H| = |G|/|H|, and is always a positive integer.

6 The center of G, Z(G), is the subgroup consisting of all elements that commute with

every element in G. The upper central series of G is the series 1 ≤ Z = Z1 ≤ Z2 ≤ · · ·, where Z(G/Zi) = Zi+1/Zi. If this series terminates at G, then G is said to be nilpotent.

The nilpotence class of G is given by the least integer n such that Zn+1 = G, and is denoted c(G).

For an element x ∈ G, conjugation by another element y ∈ G is given by xy = y−1xy.

The subgroup H is normal in G if H is closed under conjugation by all elements in G, and we denote this relationship by H E G. For a normal subgroup H of G, the quotient G/H is also a group. The conjugacy class of an element x ∈ G is the set Cl(x) = {xg | g ∈ G}.

Conjugacy classes form a partition of G, and a central element is always the sole member of its conjugacy class. As a result, the number of conjugacy classes of an abelian group is equal to the order of the group.

The commutator subgroup, or derived subgroup, is denoted G0, and is generated by the commutators of G. For x and y in G, the commutator of x with y is given by [x, y] = x−1y−1xy. It is important to notice that not all elements in G0 are necessarily commutators.

0 The lower central series of G is G = G1 ≥ G = G2 ≥ ..., where Gi+1 = [Gi,G]. For a nilpotent group, the length of the lower central series is the same as that of the upper central

th series. As their names suggest, the i term of the lower central series, Gi, is contained in

th the (n − i + 1) term of the upper central series, Zn−i+1.

If for every i, 1 ≤ i ≤ c(G), Gi is the only normal subgroup of a p-group G of order |Gi|, then G is said to be normally constrained, a deﬁnition given in [4]. When p is odd and G is normally constrained, we have the following lemma.

Lemma 2.1. [4, Theorem 3.5] Let p be an odd prime and let G be a normally constrained p-group of nilpotence class at least 3 such that |G : G0| = p2n. Then for 2 ≤ i < c(G), we

n 2n have p ≤ |Gi : Gi+1| ≤ p .

7 The commutator subgroup is the smallest normal subgroup of G such that the quotient of G by the subgroup is abelian. For any normal subgroup N of G, if G/N is abelian, then N contains G0. A nonabelian group is called metabelian if its commutator subgroup is abelian.

When the derived subgroup of a p-group G has index p2 in G, some restrictions are placed on the structure of G, including those in the following lemma.

0 2 3 Lemma 2.2. Let G be a p-group such that |G : G | = p . Then |G : G3| = p and

5 |G : G4| ≤ p .

0 2 0 Proof. Since |G : G | = p , G is two-generated, and we can write G = hx, yi. Then G /G3 =

0 0 h[x, y]i/G3, and since G/G is elementary abelian, |G : G3| = p. For G3/G4, we have

2 G3/G4 = h[x, y, x], [x, y, y]i/G4, and hence |G3 : G4| = p or p .

For a p-group G, if h is an element of Zi, then [h, g] is an element of Zi−1 for all g ∈ G.

Consequently, we have the following lemma.

p p Lemma 2.3. Let G be a p-group and let h ∈ Z2(G). Then [h , g] = [h, g] for all g ∈ G.

As a particular example, let h ∈ Z2. Then [h, g] ∈ Z for all g ∈ G, and

[h2, g] = h−2g−1h2g

= h−1h−1g−1h(gg−1)hg

= h−1(h−1g−1hg)g−1hg

= [h, g]h−1g−1hg

= [h, g]2.

If there exists a group H such that G is isomorphic to H/Z(G), then G is called capable.

For p-groups, we have the following lemma.

Lemma 2.4. [8, Lemma 2.5] If G is a p-group such that |G0| = p and |G : Z(G)| > p2, then

G is not capable.

8 A character χ of the group G is a map from G to C which, among other properties, is constant on the conjugacy classes of G. Sums of characters are also characters, but those characters that cannot be written as a sum are called irreducible. The set of irreducible characters of G is written Irr(G). The degree of a character χ is given by χ(1) (a positive

integer), where 1 is the identity element of G, and the degrees of the irreducible characters

of G form the set cd(G) = {χ(1) | χ ∈ Irr(G)}. A fundamental equation that is satisﬁed by

the irreducible characters of a group G is the following.

X |G| = χ(1)2 (2.1) χ∈Irr(G)

Characters of degree 1 are called linear, and these are always irreducible. A linear

character is a homomorphism from the group into the multiplicative group of C, the ﬁeld of complex numbers, that is, linear characters take values that are roots of unity. A special

linear character is the principal character, which takes the value 1 on all elements of the

group. The set of linear characters of G is denoted Lin(G). Lemma 2.5 is a consequence of

Equation 2.1.

Lemma 2.5. A group G is abelian if and only if Irr(G) = Lin(G).

For χ ∈ Irr(G), the kernel of χ is given by ker(χ) = {g ∈ G | χ(g) = χ(1)}. This is a

normal subgroup of G, and in fact, every normal subgroup of G is either the kernel of some

irreducible character of G, or the intersection of kernels. The intersection of the kernels

of all linear characters of G is exactly G0, and the intersection of all kernels of irreducible

characters of G is the trivial group, h1i. An irreducible character with trivial kernel is said

to be faithful. The center of an irreducible character χ is Z(χ) = {g ∈ G | |χ(g)| = χ(1)}.

This is also a normal subgroup of G, and the intersection of the centers of the irreducible

characters of G is exactly Z(G). Lemma 2.27 of [12] contains two remarkable facts, stated

below as Lemma 2.6, which will be crucial to most of the proofs in this dissertation.

9 Lemma 2.6. For χ ∈ Irr(G), let K = ker(χ) and let Y = Z(χ). Then Y/K is cyclic, and

Z(G/K) = Y/K.

Since linear characters take values that are roots of unity, the center of a linear character contains the whole group. Consequently, the quotient of a group by the kernel of a linear character is always cyclic, and a linear character is faithful if and only if the group is cyclic.

For an abelian group G, the set of linear characters also forms a group, called the dual group, which is denoted by Gˆ as in exercise 2.7 of [12]. The dual group Gˆ is actually isomorphic to

G. If λ is a linear character of G, then the order of λ in Gˆ is given by |G/ker(λ)|. This is because λ is a faithful character of the cyclic group G/ker(λ).

The index of the center of χ in G is always at least χ(1)2, a consequence of Corollary

2.30 of [12].

Lemma 2.7. [12, Corollary 2.30] Let χ ∈ Irr(G). Then χ(1)2 ≤ |G : Z(χ)|.

When G/Z(χ) is abelian, we can say more.

Lemma 2.8. [12, Theorem 2.31] Suppose that χ ∈ Irr(G) and that G/Z(χ) is abelian. Then

|G : Z(χ)| = χ(1)2.

When G is a p-group with |G0| = p, the center of any irreducible character of G will contain G0. This yields the following lemma.

Lemma 2.9. Let G be a p-group such that |G0| = p. Then |G : Z(G)| is a square.

Proof. Let χ ∈ Irr(G) be nonlinear. As Z(χ) ≥ Z(G) ≥ G0, we have G/Z(χ) is abelian, and by Lemma 2.8, |G : Z(χ)| = χ(1)2. Let h ∈ Z(χ) and g ∈ G. Then [g, h] ∈ ker(χ). Notice that G0 is not contained in ker(χ), as this would imply G/ker(χ) is abelian, contradicting that χ is nonlinear. Hence G0 ∩ ker(χ) = {1}, which shows that [g, h] = 1. Since g was arbitrary, h ∈ Z(G). Now Z(χ) is a subgroup of Z(G) but also contains Z(G), so they are equal. Hence |G : Z(G)| = χ(1)2.

10 For a normal subgroup N of G, there is a natural bijection between irreducible characters of G/N and the irreducible characters of G whose kernels contain N. This is stated formally as Lemma 2.10.

Lemma 2.10. [12, Lemma 2.22] Let N C G.

(i) If χ is a character of G and N ⊆ ker(χ), then χ is constant on cosets of N in G and

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

(ii) If χˆ is a character of G/N, then the function χ deﬁned by χ(g) =χ ˆ(Ng) is a character

of G.

(iii) In both (i) and (ii), χ ∈ Irr(G) if and only if χˆ ∈ Irr(G/N).

We can therefore freely consider χ ∈ Irr(G/N) as a character of G whose kernel contains

N, and ker(χ) will usually refer to the kernel of χ as a character of G. Members of the set Irr(G|N) are those irreducible characters of G whose kernels do not contain the normal subgroup N. An important consequence of Lemma 2.10 is that the codegree of an irreducible character of G whose kernel contains the normal subgroup N will be the same as the codegree of χ regarded as an irreducible character of G/N. As a result, we have Lemma 2.11.

Lemma 2.11. Let G be a group and N E G. Then cod(G/N) ⊆ cod(G).

If H is a subgroup of a group G, and χ is a character of H, then χG, the character χ induced to G, is deﬁned in Deﬁnition 5.1 of [12] as

1 X χG(g) = χ◦(xgx−1), (2.2) |H| x∈G

where χ◦(y) = χ(y) if y ∈ H and 0 otherwise. For the kernel of χG, we have the following

lemma.

11 Lemma 2.12. [12, Lemma 5.11] Let χ be a character of H ≤ G. Then

\ ker(χG) = (ker(χ))x. (2.3) x∈G The degree of an irreducible character divides the order of the group, and consequently the character degrees of p-groups are p-powers. From Theorem 2.32 of [12], we have the

following lemma.

Lemma 2.13. For a p-group G, the center of G is cyclic if and only if G has a faithful irreducible character.

All p-groups have a non-trivial center, and as a result are nilpotent. Every normal subgroup of a p-group intersects the center nontrivially. Of particular interest is the following lemma.

Lemma 2.14. [2, Lemma A.6.2] Let G be a p-group. Then Zi(G) contains all normal subgroups of G of order at most pi. In particular, if N is a normal subgroup of G of order

i i greater than p , then |N ∩ Zi(G)| ≥ p .

Itˆo’sTheorem [12, Theorem 6.15] is extremely useful in the study of group characters.

Here, it will be used to prove Lemma 2.16, which in turn will be used to prove many of the results in this dissertation.

Theorem 2.15. (Itˆo’sTheorem) Let A C G be abelian. Then χ(1) divides |G : A| for all χ ∈ Irr(G).

An irreducible character χ of a group G that is induced from a linear character of a

subgroup of G is called monomial. If this subgroup is normal in G then χ is called normally monomial. If every irreducible character of G is monomial, we say G is an M-group, and if

every χ ∈ Irr(G) is normally monomial, then G is also said to be normally monomial. All

p-groups are nilpotent, and by Corollary 6.14 of [12], they are therefore M-groups.

12 Lemma 2.16. If a ﬁnite p-group G has a faithful irreducible character of degree p, then G

has a normal abelian subgroup of index p and cd(G) = {1, p}.

Proof. Let χ ∈ Irr(G) be a faithful character of degree p. As G is an M-group, there exists a subgroup A ≤ G and λ ∈ Lin(A) such that λG = χ. From [12], Deﬁnition 5.1, we have

1 X p = χ(1) = λG(1) = λ◦(1) = |G : A|, |A| g∈G

so A is a maximal subgroup and hence is normal in G. Since A0 is characteristic in A, which

0 is normal in G, we have A E G. The core in G of a subgroup H ≤ G is the smallest normal

0 T x subgroup of G containing H, thus A ≤ coreG(ker λ) = x∈G(ker λ) . By Lemma 2.12, this is equal to ker(λG) = ker(χ) = 1, which shows that A0 is trivial and hence A is abelian. By

Theorem 2.15, the degree of every irreducible character of G divides the index of any normal abelian proper subgroup of G, and hence cd(G) = {1, p}.

A group with cd(G) = {1, p} must satisfy one of two conditions ﬁrst given by Isaacs and

Passman in [13] as Theorem C 4.8. We state this theorem below as it appears in [2].

Lemma 2.17. [2, Theorem 22.5] If G is a nonabelian p-group with cd(G) = {1, p}, then one

and only one of the following holds:

(i) G has an abelian subgroup of index p,

(ii) G/Z(G) is of order p3 and exponent p.

Frequently, we will show through Lemma 2.16 or Lemma 2.17 that a group has an abelian

subgroup of index p, and then use the following lemma to obtain a contradiction about the

possible orders of the group, the derived subgroup, and the center.

Lemma 2.18. [2, Lemma 1.1] Let A be an abelian subgroup of index p of a nonabelian p-group G. Then |G| = p · |G0| · |Z(G)|.

13 A result similar to Lemma 2.18 holds for an arbitrary group G with abelian subgroup A

when G/A is cyclic.

Lemma 2.19. [12, Lemma 12.12] Let A C G with A abelian and G/A cyclic. Then |A| = |G0||A ∩ Z(G)|.

A p-group is called special if G0 = Z(G). A special group is extraspecial if the center has

order p. If an extraspecial group has order pm, then m must be odd, and for each m, there are two extraspecial p-groups. These groups have exactly two irreducible character degrees, as described in [10, Example 7 (b)].

Lemma 2.20. Let G be extraspecial with |G| = p2n+1. Then cd(G) = {1, pn}.

In [3], the authors give the following characterization of a p-group that is both capable and extraspecial.

Lemma 2.21. [3, Corollary 8.2] An extraspecial p-group is capable if and only if it has order p3, and is either dihedral of order 8 when p = 2 or has exponent p when p is odd.

If G is a p-group such that for every maximal subgroup N of Z(G) the quotient G/N is

extraspecial, then G is called semi-extraspecial. In [18], it is noted that G/G0 is elementary

abelian. In Lemma 1 and Satz 1 of [1], Beisiegel proves the following fact for semi-extraspecial

groups.

Lemma 2.22. If G is semi-extraspecial, then G is special and |G0|2 ≤ |G : G0|.

Semi-extraspecial groups have exactly two irreducible character degrees, as described in

Theorem 5.5 of [16].

Lemma 2.23. If G is semi-extraspecial, then cd(G) = {1, p|G : G0|}.

We will also make use of the following lemma.

14 Lemma 2.24. (M. Lewis) Let G be a p-group with nilpotence class n ≥ 2. Then one of the

following occurs:

(i) G has maximal class,

(ii) G is extraspecial,

n (iii) |Zn−1| ≥ p .

In [8], the authors investigate two conditions on p-groups that relate to a particular

restriction of the group’s irreducible character degrees. Two of their results are given in the

following theorems.

Theorem 2.25. [8, Theorem B] Let G be a p-group such that |G : Z(G)| = p2n is a square.

The following statements are equivalent:

(i) cd(G) = {1, pn},

(ii) the normal subgroups of G either contain G0 or are contained in Z(G).

Theorem 2.26. [8, Theorem C] Let G be a p-group such that |G : Z(G)| = p2n+1 is a not square. The following statements are equivalent:

(i) cd(G) = {1, pn},

0 (ii) for any N E G, either G ≤ N or |NZ(G): Z(G)| ≤ p.

The condition given in part (ii) of Theorem 2.25 is referred to in [8] as the strong condition, and the condition in part (ii) of Theorem 2.26 is referred to as the weak condition. The next lemma, part of Theorem F of [8], draws a conclusion about p-groups that satisfy the weak condition.

Lemma 2.27. If G is a 2-group of nilpotence class 3 which satisﬁes the weak condition on normal subgroups, then |G : Z(G)| = 23 or 24.

15 Further characterizations of p-groups satisfying the weak condition are given in the fol-

lowing lemma, a combination of Theorem G and Theorem 5.2 of [8].

Lemma 2.28. Let G be a p-group that satisﬁes the weak condition on normal subgroups.

∼ (i) If G has nilpotence class 3, then Z2(G)/Z(G) = Zp or Zp × Zp.

(ii) If G has nilpotence class 4, then |G| ≤ p6. Furthermore, if p = 2, then |G| = 25.

The paper [7] by Du and Lewis provides much of the necessary background of character

codegrees. A primary fact that is used often in this dissertation is that the degree of a

nonprincipal irreducible character is always strictly less than its codegree. (The principal

character of a group G, denoted 1G, is the one that takes value 1 on all elements of G.)

Lemma 2.29. [7, Lemma 2.1] Let G be a p-group and let χ ∈ Irr(G). If χ 6= 1G, then

χ(1) < cod(χ).

A direct consequence of this fact is the following lemma.

Lemma 2.30. [7, Lemma 3.1] Let G be a group and suppose that χ ∈ Irr(G) is faithful.

Then |G| = χ(1)cod(χ) < χ(1)2.

The derived subgroup of an abelian group is always trivial. Combined with the following lemma, this fact implies that the codegrees of abelian p-groups are consecutive p-powers.

Lemma 2.31. [7, Lemma 2.2] Let G be a group, let e be the exponent of G/G0, and let d be a divisor of e. Then d ∈ cod(G/G0).

Before proving their main results, Du and Lewis make three key observations about the possible codegrees of p-groups. First, 1 and p are always included as codegrees of nontrivial p-groups, shown in Corollary 2.3 of [7]. Second, cod(G) = {1, p} if and only if G is a nontrivial elementary abelian p-group. Third, if p2 is not a codegree of G, then G/G0 is elementary abelian.

16 Lemma 2.32. Let G be a ﬁnite p-group. Then {1, p} ⊆ cod(G).

Proof. The principal character 1G takes the value 1 on all elements of G, and hence its kernel is all of G. This shows that cod(1G) = 1. The existence of p in cod(G) follows from Lemma

2.31, as p divides the exponent of G/G0.

Lemma 2.33. [7, Lemma 2.4] Let G be a p-group. Then cod(G) = {1, p} if and only if G

is elementary abelian.

Lemma 2.34. [7, Corollary 2.5] Let G be a nontrivial p-group. If p2 ∈/ cod(G), then G/G0

is elementary abelian.

The following lemma can be inferred from [7].

Lemma 2.35. Let G be a p-group with p2 ∈/ cod(G). If χ is an irreducible character of G

such that cod(χ) > p, then χ is non-linear.

Proof. Let χ be an irreducible character of G such that cod(χ) = pa for some a > 2,

and suppose that χ is linear. Then ker(χ) ≥ G0, so G/ker(χ) is abelian. Since χ is a

faithful irreducible character of G/ker(χ), this quotient must be cyclic, and |G/ker(χ)| =

χ(1)cod(χ) = pa > p2. By Lemma 2.34, G/G0 is elementary abelian, which is impossible

since ker(χ) ≥ G0 and G/ker(χ) is cyclic with order greater than p2.

The ﬁrst main result of Du and Lewis in [7] is the following theorem.

Theorem 2.36. [7, Theorem 1.1] If G is a p-group and p < pa = max(cod(G)), then the

following are true.

(i) The nilpotence class of G is at most 2a − 2.

(ii) If either p = 2 and a > 2 or p = 3 and a > 3, then G has nilpotence class at most

2a − 3.

17 The second main result of Du and Lewis provided the original motivation for this dis-

sertation. They give a constant bound for the nilpotence class of p-groups with exactly 3

codegrees, leading to the question whether such a bound also exists for p-groups with exactly

4 codegrees.

Theorem 2.37. [7, Theorem 1.2] If G is a p-group and |cod(G)| = 3, then G has nilpotence class at most 2.

18 CHAPTER 3

Codegrees of Maximal Class p-groups

The structure of a maximal class p-group is restricted by the central series. In these

groups, the upper and lower central series coincide, and the order of each successive term

increases or decreases respectively by a factor of p. As G/Zi has a cyclic center of order

p for each Zi in the upper central series of G, each nonlinear irreducible character of G can be viewed as a faithful character of G/Zi for some i. Since cod(G/Zi) ⊆ cod(G/Zi−1), many facts that are easily proven for small maximal class p-groups can be extended to those with arbitrarily large order. In this section, we ﬁnd families of maximal class p-groups with consecutive p-power codegrees, and we show that if the group is large enough, all powers of p up to p4 will be included as codegrees.

Notice the restriction n ≥ 4 in Theorem 6 (ii). If G has class 2 and |G| = p3, then G is not isomorphic to Zp2 × Zp, instead, G will be as in (iii). In this case, note that G is extraspecial.

Theorem 6. If G is a p-group such that |G| = pn ≥ p2, then cod(G) = {1, p, p2, . . . , pn−1} if and only if one of the following occurs:

∼ (i) G = Zpn−1 × Zp,

∼ (ii) G = Zpn−1 o Zp, G has nilpotence class 2, and n ≥ 4,

(iii) G has maximal class and |cd(G)| = 2.

∼ Proof. Assume ﬁrst that G = Zpn−1 × Zp. As G is abelian, it is isomorphic to Gb, the group

of irreducible characters of G. If λ ∈ Irr(G), the order of λ in Gb is given by |G : ker(λ)| =

19 cod(λ). Hence the orders of characters in Gb correspond to the codegrees of G, and we have cod(G) = {pi | 0 ≤ i ≤ n − 1}. ∼ ∼ Next, assume that G = Zpn−1 o Zp and G has nilpotence class 2. Let hxi = Zpn−1

1+pn−2 and choose ϕ ∈ Aut(Zpn−1 ) as an automorphism of order p such that ϕ(x) = x . Taking the semidirect product as {(a, b) | a ∈ hxi, b ∈ hϕi}, with multiplication given by

b−1 1 ∼ (a1, b1)(a2, b2) = (a1(a2 ), b1b2), we have G = h(x, 1), (1, ϕ)i. Notice that [(x, 1), (1, ϕ)] = (xpn−2 , 1), which has order p, and hence G0 = h(xpn−2 , 1)i has order p. As (xp, 1) ∈ Z, and

|G : h(xp, 1)i| = p2, we have Z = h(xp, 1)i. Now G/G0 ∼= h(x, 1)i × h(1, ϕ)i/h(xpn−2 , 1)i ∼=

0 i 0 Zpn−2 × Zp, which implies cod(G/G ) = {p | 0 ≤ i ≤ n − 2}. Since Z is cyclic, G is the unique subgroup of G of order p, and hence any nonlinear character of G must be faithful. Let

χ ∈ Irr(G) be nonlinear. Since |G : Z| = p2, by Lemma 2.7, χ(1) = p. Hence cod(χ) = pn−1, and we have cod(G) = {pi | 0 ≤ i ≤ n − 1}.

Finally, assume that G has maximal class and |cd(G)| = 2. The quotient G/Zn−2 has

3 order p and class 2. A nonlinear irreducible character of G/Zn−2 must have degree p, as

2 the center of G/Zn−2 has index p . Hence cd(G) = {1, p}. For i ≤ n − 2, the quotient

G/Zi is non-abelian with a cyclic center, and hence has a faithful nonlinear irreducible character with codegree pn−i−1. From Lemma 2.32, we always have 1 and p in cod(G). Thus cod(G) = {pi | 0 ≤ i ≤ n − 1}.

For the forward direction, assume ﬁrst that G is abelian with cod(G) = {pi | 0 ≤ i ≤ n − 1}. Since pn ∈/ cod(G), G cannot be cyclic. As G is isomorphic to Gb, there is an element

n−1 ∼ of order p in G. Since G is not cyclic, we must have G = Zpn−1 × Zp. Now suppose G has class 2, n ≥ 4, and cod(G) = {pi | 0 ≤ i ≤ n − 1}. Let χ ∈ Irr(G)

have codegree pn−1. If χ is linear, then the kernel of χ equals G0 and has order p. By Lemma

2.6, G/G0 is cyclic, which implies G/Z is cyclic, contradicting that G has class 2. Hence

χ(1) = p and χ is faithful. By Lemmas 2.6 and 2.8, Z is cyclic and has index p2 in G. Hence

20 G has two noncentral generators, which implies |G0| = p. Since Z is cyclic, G0 is the unique normal subgroup of G of order p, so an irreducible character of G is nonlinear if and only if it is faithful. Thus cod(G/G0) = {pi | 0 ≤ i ≤ n − 2}. As an abelian group, G/G0 is isomorphic to G/G[0. Hence G/G[0 (and therefore G/G0), has an element of order pn−2. Since

0 0 ∼ G/G is not cyclic, we must have G/G = Zpn−2 × Zp. Write G = ha, bi, where apn−2 and bp are elements of G0, but no smaller power of either element is in G0. Suppose apn−2 = 1. Since |G : Z| = p2 and G/Z is not cyclic, it must be elementary abelian. Thus ap ∈ Z, and since Z has a unique subgroup of order p, G0 is contained in hapi. As ahapi and bhapi have order p, and hahapi, bhapii = G/hapi, we have

|G/hapi| = p2. On the other hand, since pn−2 is the smallest power of a that equals 1, the order of hapi is pn−3. Hence |G : hapi| = p3, a contradiction. Thus apn−2 6= 1, and a has order pn−1. Now |G : hai| = p. Suppose there is no element g ∈ G − hai that has order p. Then G has a unique subgroup of order p, implying G is either cyclic or a generalized quaternion 2-group. Since generalized quaternion 2-groups have maximal class and |G| ≥ p4, either case contradicts that G has class 2. Hence such an element g exists, which shows that ∼ ∼ G = hai o hgi = Zpn−1 o Zp. Claim. If |G| = pn and cod(G) = {pi | 0 ≤ i ≤ n − 1}, then G has maximal class or c(G) ≤ 2.

Proof of claim. Induct on |G|. Clearly the result holds when |G| ≤ p4, so let |G| = pn and cod(G) = {pi | 0 ≤ i ≤ n − 1}. Assume c(G) > 2. If χ ∈ Irr(G) such that cod(χ) = pn−1, then χ is faithful and has degree p. Hence Z is cyclic, and G has a unique normal subgroup of order p, say N. The kernel of any non-faithful irreducible character must contain N, and hence such a character will also be a character of G/N. By Lemma 2.16, cd(G) = {1, p}. If µ is a faithful irreducible character of G then µ is non-linear and has degree p, so cod(µ) = pn−1.

This shows that an irreducible character of G has codegree pn−1 if and only if it is faithful.

21 Thus cod(G/N) = {pi | 0 ≤ i ≤ n − 2}, and by the inductive hypothesis, c(G/N) ≤ 2 or c(G/N) = n − 2.

Suppose G/N has class n − 2. Then G has either class n − 2 or n − 1. If c(G) = n − 1,

2 0 n−2 we are done, so assume c(G) = n − 2. Then |Gn−2| = p , Z = Gn−2, and |G | = p . By

Lemma 2.18, pn = |G| = p|Z||G0| = pn+1, a contradiction.

Now suppose c(G/N) ≤ 2. If G/N is abelian, then c(G) ≤ 2 and we are done, so let c(G/N) = 2, and assume c(G) = 3. This implies N = G3. Let γ ∈ Irr(G/G3) such that

n−2 0 cod(γ) = p . If γ is linear then |ker(γ)/G3| = p, so ker(γ) = G /G3, and by Lemma 2.6,

G/G0 is cyclic, which is impossible. Hence γ(1) = p and γ must be a faithful character of

0 G/G3. Put Z(G/G3) = X/G3 and notice that X/G3 is cyclic. Since Z/G3 and G /G3 are both contained in X/G3, and X/G3 has a unique subgroup of each possible index, we must

0 n 0 n−1 0 have Z < G . Recalling Lemma 2.18, we have p = p|Z||G | and p = p|X/G3||G /G3|,

0 p hence |X : Z| = p and G = X. Put X = ha, G3i, so Z = ha ,G3i. Since Z is cyclic, we

p p p either have a ∈ G3 or G3 ≤ ha i. If a ∈ G3, then Z = G3. Again applying Lemma 2.18,

4 p this shows that |G| = p , so G has maximal class. Suppose instead G3 ≤ ha i. The faithful

0 character γ ∈ Irr(G/G3) has center Z(γ) = X = G . Since G/Z(γ) is abelian, we have

2 2 0 n 0 |G : X| = γ(1) = p . This shows that G = Z2. By Lemma 2.18, p = |G| = p|Z||G | = p · pn−3 · pn−2 = p2(n−2), which shows that in this case as well we have |G| = p4. This proves the claim.

We may now assume G has maximal class and cod(G) = {pi | 0 ≤ i ≤ n − 1}. Let

χ ∈ Irr(G) have codegree pn−1. If χ is linear, then |ker(χ)| = p, so ker(χ) = Z. Since

Z(χ) = G and Z(χ)/ker(χ) is cyclic, this implies G/Z is cyclic, which is impossible. Hence

χ is nonlinear, and must be faithful with degree p. By Lemma 2.16, cd(G) = {1, p}.

An abelian group of order pn can have any number of codegrees, from 2, in the case of an elementary abelian group, up to n + 1 if the group is cyclic. Notice that the codegrees of an

22 abelian group are always consectutive powers of p, a consequence of Lemma 2.31. The set of codegrees of a group of class 2, on the other hand, may not always contain consecutive powers of p. The comments after Lemma 4.2 mention some examples of this fact for groups of order p5. The groups listed in the the small groups library of Magma [5] as SmallGroup(35, i) for

i = 28, 29, and 30 are examples of maximal class p-groups where a codegree with power one

less than the power of the order of the group (in this case p4) is missing from the set of

codegrees.

The next lemma considers maximal class groups with p2 as a character degree. The result

is similar to that of Theorem 6 (iii), with the largest codegree now reduced to pn−2.

Theorem 3.1. Let G be a maximal class p-group with |G| = pn and cd(G) = {1, p, p2}.

Then cod(G) = {pi | 0 ≤ i ≤ n − 2}.

Proof. A maximal class p-group with an irreducible character of degree p2 must have order

at least p5, since |G : Z| ≥ p4 by Lemma 2.7. Let G be a maximal class p-group such that

|G| = p5 and cd(G) = {1, p, p2}. Suppose χ ∈ Irr(G) has codegree p4, and notice that χ must

be a faithful character with degree p. Lemma 2.16 implies cd(G) = {1, p}, a contradiction,

so cod(χ) ≤ p3 for all χ ∈ Irr(G). Since c(G) = 4, Theorem2.37 and Lemma 2.33 imply that

|cod(G)| ≥ 4, and hence cod(G) = {1, p, p2, p3}.

Now let |G| = pn, where n ≥ 6. Suppose cod(µ) = pn−1 for some µ ∈ Irr(G). If

µ is linear, then |ker(µ)| = p, which is impossible since the kernel of a linear character

contains G0, and G is maximal class. Hence µ(1) = p and µ is faithful, a contradiction by

Lemma 2.16. Let χ ∈ Irr(G) be faithful. Then χ(1) = p2 and cod(χ) = pn−2 is the largest

codegree of G. If cd(G/Z) = {1, p}, then by Theorem 6, cod(G/Z) = {1, p, p2, . . . , pn−2}. If

cd(G/Z) = {1, p, p2}, then by the inductive assumption cod(G/Z) = {pi | 0 ≤ i ≤ n − 3}.

In either case, cod(G) = {pi | 0 ≤ i ≤ n − 2}.

This theorem can be extended to the case when the character degree p2 is replaced by

23 an arbitrarily large power of p. In this case, however, the largest codegree may vary.

Theorem 7. Let G be a maximal class p-group such that such that cd(G) = {1, p, pb}. If

|G| = pn, then cod(G) = {pi | 0 ≤ i ≤ c}, for some integer n − b ≤ c ≤ n − 2.

Proof. First, notice that n ≥ 2b+1, as the square of the degree of every irreducible character

of G must divide |G : Z|. If cod(χ) = pn−1 for χ ∈ Irr(G), then χ is a faithful character of

degree p. By Lemma 2.16, this contradicts |cd(G)| = 3. Hence pn−2 is the largest possible codegree of G. We proceed by induction on |G|, with the base case |G| = p5 (and hence

b = 2) established by Theorem 3.1. Now let |G| = pn, and let χ ∈ Irr(G) be faithful. If

χ(1) = p, then by Lemma 2.16, cd(G) = {1, p}, a contradiction. Hence χ(1) = pb and

cod(χ) = pn−b.

There are two possibilities for cd(G/Z), either |cd(G/Z)| = 2, or G/Z and G have the

same three character degrees. In the ﬁrst case, we can apply Theorem 6 to get cod(G/Z) =

{pi | 0 ≤ i ≤ n−2}. Since b ≥ 2, we have cod(χ) = pn−b ≤ pn−2, so cod(G) = cod(G/Z) and

we are done. In the second case, we have cod(G/Z) = {pi | 0 ≤ i ≤ c0}, where c0 ≥ n − 1 − b.

If c0 = n − 1 − b, then cod(G) = {pi | 0 ≤ i ≤ n − b}. If c0 > n − 1 − b, then c0 ≥ n − b, and

cod(G/Z) = cod(G).

Corollary 8 is an easy corollary of Theorem 6 and Theorem 3.1.

Corollary 8. Let G be a metabelian maximal class p-group. If |G| = pn, then cod(G) =

{pi | 0 ≤ i ≤ c}, where c = n − 1 or n − 2.

Proof. If G is a metabelian maximal class p-group, then G0 is abelian and has index p2 in

G. By Lemma 2.15, if χ is an irreducible character of G, then χ(1) | p2. Hence |cd(G)| = 2

or cd(G) = {1, p, p2}, and we can apply Theorem 6 or Theorem 3.1, respectively.

From Lemma 2.32, we know that 1 and p are always present in the set of codegrees of a

p-group G. When G has maximal class, we also have p2 ∈ cod(G).

24 Lemma 3.2. If G is a p-group that has maximal class, then p2 ∈ cod(G).

n+1 Proof. Let G have order p and class n. The quotient G/Zn−2 is an extraspecial group of

3 order p . If χ ∈ Irr(G/Zn−2) is nonlinear, then χ must be faithful. Also notice that χ(1) = p,

2 2 since |G/Zn−2 : Z(G/Zn−2)| = p . Hence cod(χ) = p , and since cod(G/Zn−2) ⊆ cod(G), we

have p2 ∈ cod(G).

Following this trend, we ﬁnd that whenever G has maximal class and |G| ≥ p4, we have

p3 ∈ cod(G).

Lemma 3.3. If G is a maximal class p-group such that |G| ≥ p4, then p3 ∈ cod(G).

n Proof. Let |G| = p , where n ≥ 4. Since G has maximal class, Z(G/Zn−4) is cyclic and

hence there exists a faithful character χ ∈ Irr(G/Zn−4). As G/Zn−4 is not abelian, χ(1) > 1,

2 3 3 3 but also χ(1) ≤ |G : Zn−3| = p , so χ(1) = p. Thus cod(χ) = p , so p ∈ cod(G/Zn−4) ⊆

cod(G).

We would like to be able to continue increasing the order of the group to obtain each

next largest codegree, however our results so far are limited to codegrees up to size p4, as in

the following lemma.

Lemma 3.4. Let G be a maximal class p-group such that |G| ≥ p6. Then p4 ∈ cod(G).

n 5 Proof. Let |G| = p , where n ≥ 6. Since |G : Zn−5| = p , the largest possible degree of

2 an irreducible character of G/Zn−6 is p . By Theorem 6 and Theorem 3.1, we must have

4 p ∈ cod(G/Zn−6) ⊆ cod(G).

This bound is sharp, and the groups listed in the the small group library of Magma [5]

as SmallGroup(35, i) for i = 28, 29, and 30 are examples of maximal class groups of order p5 with p4 not included among their codegrees. When |G| = pn and c(G) = n − 1, it is not surprising that pn−1 is not always included in the set of codegrees, as this would imply

25 G has a faithful character of degree p, which in turn implies that cd(G) = {1, p}, and the existence of groups of maximal class with other sets of character degrees is well known.

Whether this pattern continues is an interesting question. We have not been able to prove that p5 must be a codegree for maximal class groups of order at least p8, nor have we found a counterexample. If a counterexample G of order p8 exists, then G must have p3 ∈ cd(G), but no faithful irreducible character of G can have degree p3.

Our ﬁnal theorem in this chapter is restricted to normally monomial maximal class p- groups. In a group of this type, a faithful irreducible character will always have the largest degree in cd(G), allowing us to apply induction to G/Z.

Theorem 9. Let G be a normally monomial maximal class p-group. Let |G| = pn, b(G) =

i max{χ(1) | χ ∈ Irr(G)}, and b = logp(b(G)). Then cod(G) = {p | 0 ≤ i ≤ c} for some integer c ≥ n − b.

Proof. Let G be nonabelian with order p3 and let χ ∈ Irr(G). Since χ(1)2 ≤ |G : Z| = p2, we have cd(G) = {1, p}. If χ(1) = p, then by Lemma 2.29, cod(χ) ≥ p2. An irreducible character of G with codegree p3 must be faithful and linear, which is impossible since G is nonabelian. Hence cod(χ) = p2, and cod(G) = {1, p, p2}.

Now let G be normally monomial with |G| = pn and c(G) = n − 1. Since quotients of normally monomial groups are also normally monomial, we can apply induction and assume

i 0 0 0 0 cod(G/Z) = {p | 0 ≤ i ≤ c }, where c ≥ n − 1 − b and b = logp(b(G/Z)). Notice that b0 ≤ b. If b0 < b, then b0 ≤ b − 1, so c0 ≥ n − 1 − b0 ≥ n − 1 − (b − 1) = n − b. Hence cod(G/Z) = {pi | 0 ≤ i ≤ c0} ⊆ cod(G).

Let χ ∈ Irr(G). If χ is not faithful, then χ can be equated with an irreducible character of G/Z and hence cod(χ) ∈ cod(G/Z). Thus assume χ is faithful. As G is normally

G monomial, there exists a subgroup H E G and λ ∈ Lin(H) such that χ = λ and χ(1) =

G 0 0 λ (1) = |G : H|. Since H is characteristic in H, which is normal in G, we have H EG. Thus

26 0 x G H ≤ coreG(ker(λ)) = ∩x∈G(ker(λ)) . By Lemma 2.12, this is equal to ker(λ ) = ker(χ) = 1,

which shows that H0 is trivial and hence H is abelian. If θ ∈ Irr(G), then by Lemma 2.15,

θ divides |G : H| = χ(1). Thus χ(1) = b(G), and cod(χ) = pn−b ∈ cod(G/Z).

Assume now that b0 = b, so cod(G/Z) = {pi | 0 ≤ i ≤ c0}, where c0 ≥ n − 1 − b.

Let χ ∈ Irr(G) and note again that if χ is not faithful then cod(χ) ∈ cod(G/Z) so we may assume χ is faithful. Since G is normally monomial, we have (as in the previous paragraph) that b(G) = χ(1), and therefore cod(χ) = pn−b. Thus either cod(χ) ∈ cod(G/Z), in which case cod(G) = cod(G/Z) = {pi | 0 ≤ i ≤ c}, where c ≥ n − b, or cod(G) =

{1, p, . . . , pn−1−b, pn−b}.

27 CHAPTER 4

Inclusion of p2 as a Codegree

This chapter contains the ﬁrst steps toward a more general question regarding the inclusion of p2 in a group’s codegrees. While the evidence is limited, it seems to suggest that p2

might be among the codegrees of G whenever |G| = p2n or p2n−1 and c(G) ≥ n (assuming n ≥ 3).

The next lemma gives our ﬁrst indication of when p2 will, or will not, be included among

a group’s codegrees.

Lemma 4.1. Let G be a p-group. Then p2 ∈ cod(G) if and only if either the exponent of

0 2 3 G/G is at least p or there exists N C G such that G/N is extraspecial of order p .

Proof. If G/G0 has exponent at least p2, then p2 ∈ cod(G), since otherwise G/G0 is ele-

3 mentary abelian by Lemma 2.34. If N C G such that G/N is extraspecial of order p , then c(G/N) = 2 implies there exists χ ∈ Irr(G/N) such that χ(1) = p. Since (G/N)0 = Z(G/N)

and |Z(G/N)| = p, χ must be faithful and hence cod(χ) = p2.

Now assume p2 ∈ cod(G). Let χ ∈ Irr(G) have codegree p2. If χ is linear, then p2 =

|G : ker(χ)| and by Lemma 2.6, G/ker(χ) is cyclic. Since the kernel of any linear character

contains G0, we see that the exponent of G/G0 is at least p2. Now assume χ is not linear. By

Lemma 2.29, χ(1) = p. Hence, p3 = cod(χ)χ(1) = |G : ker(χ)|, which shows that G/ker(χ)

is an extraspecial group of order p3.

We are aware of the existence of groups of order p4 with class 2 that do not have p2 as a codegree. For a particular example, we have the group listed in the Small Groups database

28 of Magma [5] as SmallGroup(34, 14). For an arbitrary prime p, let G ∼= hx, y, z | ap = bp = p2 p 3 c , [a, b] = c i. Here G is the central product of an extraspecial group of order p and Zp2 . This group has order p4, and G0 is the unique normal subgroup of order p. Any non-linear irreducible character must be faithful of degree p, and hence has codegree p3. As G/G0 is elementary abelian, any linear character must have kernel of order at least p3, and hence has codegree at most p.

As p2 is not always a codegree of G when |G| = p5 and G has class 2, it will be useful to know something about the structure of G in that case.

Lemma 4.2. Let G be a p-group with order p5 and nilpotence class 2. If p2 ∈/ cod(G), then cod(G) = {1, p, p3}, and G is either extraspecial or has no faithful irreducible characters.

Proof. Let χ ∈ Irr(G) with cod(χ) > p, and notice that by Lemma 2.35 χ is non-linear.

Then χ(1)cod(χ) ≤ |G| = p5 implies p3 ≤ cod(χ) ≤ p4. If cod(χ) = p4, then χ is faithful and Z is cyclic. Since G has class 2, by Lemma 2.8 we have |G : Z| = χ(1)2 = p2, and hence

|Z| = p3. Also notice that G0 is contained in Z, and p2 ∈/ cod(G) implies G0 is elementary abelian. Since Z is cyclic and contains the elementary abelian subgroup G0, we have |G0| = p.

Then G/G0, which is also elementary abelian, contains the cyclic subgroup Z/G0 with order p2, which is impossible, so p4 ∈/ cod(G). Since G is not elementary abelian, cod(G) 6= {1, p} by Lemma 2.33. Hence we must have p3 ∈ cod(G), so cod(G) = {1, p, p3}.

The linear characters of G are not faithful, as G has nilpotence class 2 and hence |G0| > 1.

Suppose χ ∈ Irr(G) is faithful and note that cod(χ) = p3. Then χ(1)cod(χ) = |G| implies

χ(1) = p2 and |G : Z| = p4. Since h1i < G0 ≤ Z, we have G0 = Z and G is extraspecial.

Both cases of Lemma 4.2 can occur: the extraspecial groups are well known, and the groups identiﬁed by Magma [5] as SmallGroup(35, i) for i = 44, 45, 64, 65, and 66 have no ∼ faithful irreducible characters. In general, consider H = G × Zp, where G is the central product described in the discussion preceding Lemma 4.2. This group has order p5, and no

29 faithful irreducible characters. Nonlinear irreducible characters will have degree p and kernel

of size p, giving p3 as a codegree. Linear characters will have kernel of size at least p4, as

H/H0 is elementary abelian.

Lemma 4.3 is the next step toward investigating the connection between the order of a p-group, its nilpotence class, and the presence of p2 as a codegree.

Lemma 4.3. If G is a group with order p5 and nilpotence class at least 3, then p2 ∈ cod(G).

Proof. If c(G) = 4, then by Lemma 3.2, p2 ∈ cod(G). Thus we may assume c(G) = 3.

Suppose p2 ∈/ cod(G). Since G does not have maximal class and is not extraspecial, we

3 2 know by Lemma 2.24 that |Z2| = p . Suppose |Z| = p . Let χ ∈ Irr(G/Z) be non-linear.

By Lemma 2.29, χ(1) < cod(χ), which implies cod(χ) ≥ p3. Thus p4 ≤ χ(1)cod(χ) = |G :

ker(χ)| ≤ |G : Z| = p3, which is impossible. Therefore |Z| = p, and p4 ≤ χ(1)cod(χ) = |G :

4 ker(χ)| ≤ |G : Z| = p shows that χ is faithful and hence Z2/Z is cyclic. As |Z| = p, we

0 0 0 0 have Z = [G ,G] < G ≤ Z2. Since Z2/Z is cyclic, while G /[G ,G] is elementary abelian, we

have |G0| = p2.

p p p Let Z2 = ha, Zi. For any g ∈ G,[a, g] ∈ Z, so 1 = [a, g] = [a , g], which implies a ∈ Z.

As G0 = hap,Zi, this implies G0 = Z, a contradiction. Hence p2 ∈ cod(G).

We can now prove the ﬁrst half of Theorem 10 using induction, with Lemma 4.3 as the

base case.

Theorem 10 (i). If G is a p-group with coclass 2 and order at least p5, then p2 ∈ cod(G).

Proof. Induct on |G|. Lemma 4.3 establishes the base case where |G| = p5, so assume

|G| = pn+2. Since G has coclass 2, the nilpotence class of G is n, the class of G/Z is n − 1, and Z has order at most p2. If |Z| = p2, then |G/Z| = pn and by Lemma 3.2, p2 ∈ cod(G/Z). If |Z| = p, then |G/Z| = pn+1, so G/Z has coclass 2 and by the inductive hypothesis, p2 ∈ cod(G/Z).

30 If we increase |G| in Lemma 4.3 to p6, the result p2 ∈ cod(G) will still hold. There are examples of groups of order p6 with class 2 where p2 ∈/ cod(G), e.g., semi-extraspecial groups. In these groups, G0 = Z and by Lemma 2.22 we know |G0|2 ≤ |G : G0|. Thus if

|G| = p6 and c(G) = 2, we have |Z| = |G0| = p2. Recall that G/G0 is elementary abelian for

semi-extraspecial groups. If a linear character λ ∈ Irr(G) has codegree 2, then G/ker(λ) is a cyclic quotient of order p2, which is impossible since the kernel of a linear character contains

G0. If χ ∈ Irr(G) is nonlinear, then by Lemma 2.23, χ(1) = p2, so cod(χ) > p2, and we have p2 ∈/ cod(G).

The proof of Lemma 4.4 makes use of the strong and weak conditions described in the second part of Theorems 2.25 and 2.26.

Lemma 4.4. If G is a p-group with |G| = p6 and c(G) ≥ 3, then p2 ∈ cod(G).

Proof. If G has nilpotence class 4 or 5, we have p2 ∈ cod(G) by Theorem 10 (i) or Lemma

class 3. If G/ker(µ) has class 3, then Lemma 4.3 implies p2 ∈ cod(G/ker(µ)), contradicting

p2 ∈/ cod(G). On the other hand, if G/ker(µ) has class 2, then p4 ∈ cod(G/ker(µ)) is

impossible by Lemma 4.2. Thus µ must be faithful. At least one of p4 or p5 must be in

cod(G), as |cod(G)| ≥ 4 by Theorem 2.37. Thus, G has a faithful irreducible character, and

hence Z is cyclic.

Since G has class 3, we know that G0 6≤ Z, so Z cannot be realized as the intersection of

kernels of only linear characters of G. Therefore Z must be contained in the kernel of one

or more non-linear irreducible characters of G. Since such a character is clearly not faithful,

31 it must have codegree p3. Let χ ∈ Irr(G) be one such character. Then p4 ≤ χ(1)cod(χ) =

|G : ker(χ)| ≤ |G : Z| ≤ p5, which shows that |Z| = p or p2.

Case 1. Assume |Z| = p2. Put K = ker(χ) and notice that by the above inequality we now have K = Z. Since G3 is elementary abelian and contained in Z, which is cyclic, we

4 have |G3| = p. By Lemmas 2.6 and 2.8, Z(χ) = Z2, and |Z2| = p .

Let Z(G/G3) = X/G3 and notice that Z ≤ X ≤ Z2. If G/G3 is extraspecial, then

0 0 0 2 X/G3 = (G/G3) = G /G3 implies that X = G has order p , and thus X = Z. This is

0 impossible as G has class 3 and hence G 6= Z. By Lemma 4.2, we now have that G/G3 has

2 3 no faithful irreducible characters, thus X/G3 cannot be cyclic, so |X/G3| = p or p .

3 2 If |X/G3| = p , then X = Z2, and |G : X| = p shows that G/G3 has exactly two

0 2 noncentral generators. Thus |G | = p . Let Z2 = ha, zi, where Z = hzi, and notice that a and z both have order p2. Since G/G0 is elementary abelian, gp ∈ G0 for every g ∈ G. Hence

0 p p p p G = ha , z i. As a ∈/ Z, there exists some g ∈ G such that 1 6= [a , g], and since a ∈ Z2, we have [a, g] ∈ Z which implies [ap, g] = [a, g]p. Thus [a, g]p 6= 1, so [a, g] is an element of Z with order greater than p and therefore generates Z. Since [a, g] is also an element of

G0, this shows that Z ≤ G0, which is impossible as they have the same order but cannot be

2 equal. Thus we may assume that |X/G3| = p .

0 0 By Lemma 2.9, we know that |G /G3| 6= p, so G = X. Recall that since Z is cyclic, G3 is the unique normal subgroup of G with order p, and hence the kernel of any nonfaithful irreducible character of G must contain G3. Any such character that is also nonlinear must

3 have codegree p , and since G/G3 has no faithful irreducible characters, its kernel must have order p2. Thus any nontrivial normal subgroup of G that does not contain G0 is either the

3 2 kernel of an irreducible character of G with codegree p , having order p and containing G3, or an intersection of such kernels and therefore equaling G3. Hence G satisﬁes the weak

2 condition, and by Lemma 2.28 (i), Z2/Z cannot be cyclic of order p . This is a contradiction

32 since Z(χ)/K = Z2/Z is cyclic by Lemma 2.6.

Case 2. Assume |Z| = p. Let χ ∈ Irr(G) have codegree p3 and put K = ker(χ). If

|K| = p then by Lemma 4.2, G/Z is extraspecial. The only capable extraspecial group has order p3 (Lemma 2.21), so this is impossible, and we may assume that |K| = p2. By Lemma

3 4 2.24, we have |Z2| ≥ p . Suppose |Z2| = p . Since Z(χ) ≥ Z2, Lemma 2.7 implies Z(χ) = Z2.

Since |K : Z| = p, and K/Z must intersect nontrivially with Z2/Z, we have that K ≤ Z2.

2 By Lemma 2.6, Z(χ)/ker(χ) = Z2/K is cyclic. Let Z2 = ha, Ki. Since |Z2 : K| = p , we

p p have a ∈/ K > Z, and hence there exists some g ∈ G such that [a , g] 6= 1. As a ∈ Z2, we have [a, g] ∈ Z, so 1 6= [ap, g] = [a, g]p ∈ Z. This is impossible since [a, g] ∈ Z and |Z| = p,

3 thus we may assume |Z2| = p .

The order of G0 is now either p2 or p3. Let Z(χ) = ha, Ki. Observe that [Z(χ),G] is contained in both K and G0. If |G0| = p2, then K ∩ G0 = Z, giving [Z(χ),G] ≤ Z, and hence

[a, g] ∈ Z for all g ∈ G. Since ap ∈/ Z, we can ﬁnd some g ∈ G such that [ap, g] 6= 1. As before, we have 1 6= [ap, g] = [a, g]p, which is impossible since |Z| = p. Thus |G0| = p3 and

0 we have G = Z2.

Recall that the only non-faithful irreducible characters of G are either linear, or have codegree p3 and kernel of order p2. Again, G satisﬁes the weak condition and Lemma 2.26 implies cd(G) = {1, p2}. Since G has at least one irreducible character χ with codegree p3 and |ker(χ)| = p2, p ∈ cd(G), which is a contradiction.

Lemma 4.4 provides the base case for the induction used to prove the second half of

Theorem 10.

Theorem 10 (ii). If G is a p-group with coclass 3 and order at least p6, then p2 ∈ cod(G).

Proof. Induct on |G|. The base case |G| = p6 is established by Lemma 4.4, so we may

assume |G| = pn+3. As G has coclass 3, the order of Z is at most p3. When |Z| = p3, G/Z

33 has maximal class and p2 ∈ cod(G/Z) by Lemma 3.2. When |Z| = p2, G/Z has coclass 2 and p2 ∈ cod(G/Z) by Theorem 10 (i). The ﬁnal possibility is |Z| = p, in which case G/Z has coclass 3, and by the inductive hypothesis, p2 ∈ cod(G/Z).

The following is an easy corollary of Lemma 3.2 and Theorem 10.

Corollary 4.5. If G is a group with order p7 and nilpotence class at least 4, then p2 ∈ cod(G).

The question of whether p2 is in the set of codegrees for groups of order p8 with class 4 remains unsettled. If there exists a group G with p2 ∈/ cod(G), we can make certain claims about the group’s structure. These claims are detailed in Lemma 4.6.

Lemma 4.6. Let G be a p-group with |G| = p8, c(G) = 4, and p2 ∈/ cod(G). Then the following hold:

(i) Z = G4 is the unique normal subgroup of G of order p,

2 (ii) Z2 = G3 is the unique normal subgroup of G of order p ,

3 (iii) cod(G/Z2) = {1, p, p },

4 0 2 5 4 0 5 (iv) either |Z3| = p , Z3 = G , and cd(G/Z2) = {1, p, p }, or |Z3| = p , p ≤ |G | ≤ p , and

cd(G/Z2) = {1, p}.

Proof. Let G be as stated. If |Z| ≥ p2, then c(G/Z) = 3 and p4 ≤ |G/Z| ≤ p6. Suppose

|G/Z| = p4. Notice that G/Z has maximal class, and by Theorem 3.2, p2 ∈ cod(G/Z), which is a contradiction. If p5 ≤ |G/Z| ≤ p6, then p2 ∈ cod(G/Z) by Lemmas 4.3 and 4.4. Hence

|Z| = |G4| = p, which proves (i).

3 2 Suppose |G3| ≥ p , and let N be a normal subgroup of G of order p such that N

6 2 G3. Then G/N has order p and class 3, and by Lemma 4.4, p ∈ cod(G/N), which is a

2 2 contradiction. Hence |G3| = p . Let N C G with |N| ≥ p , N G3. Then c(G/N) = 3,

34 p4 ≤ |G/N| ≤ p6, and as before, we have p2 ∈ cod(G/N), a contradiction. Hence we may

2 assume that all normal subgroups of G with order at least p contain G3.

Suppose G/G3 has a faithful character. Then Z(G/G3) = X/G3 is cyclic. Since G/G3

0 0 2 0 has class 2, we have X/G3 ≥ (G/G3) = G /G3. As p ∈/ cod(G), G /G3 is elementary

0 0 abelian, and hence |G /G3| = p. Since G/G is also elementary abelian while X/G3 is cyclic,

2 0 we have |X/G3| = p , and G /G3 is the unique normal subgroup of G/G3 of order p. Every nontrivial normal subgroup of a p-group intersects the center nontrivially, and hence contains

0 2 G /G3, so G/G3 satisﬁes the strong condition. Thus cd(G/G3) = {1, p } by Lemma 2.25.

Also notice that G/G3 has no nonfaithful nonlinear characters, as each nontrivial kernel

0 K/G3 contains G /G3, which implies G/K is abelian and hence has no nonlinear irreducible characters. Thus, no kernel of a nonlinear character of G can properly contain G3 and the kernel of any nonfaithful nonlinear irreducible character of G is either G3 or G4. Any normal subgroup of G with order at least p3 must contain G0, which shows that G satisﬁes the weak

6 condition, and by Lemma 2.28 (ii), |G| ≤ p . This is a contradiction, and therefore G/G3 cannot have a faithful character.

Now, the kernel of a nonlinear character of G cannot have order p2, and |G0| ≥ p3. Since every normal subgroup is the intersection of one or more kernels of irreducible characters, in order to realize G3 as such a kernel or intersection of kernels, there must be some nonlinear

If G/Z has no faithful characters, then cod(G/Z) = {1, p, p3}, implying c(G/Z) ≤ 2, a contradiction. Hence G/Z has a faithful character, and Z2/Z is cyclic. Let Z2 = ha, Zi and

35 3 p p suppose |Z2| ≥ p . Then a ∈/ Z, and there exists some g ∈ G such that [a , g] 6= 1. For all

x ∈ G,[a, x] ∈ Z, so [ap, x] = [a, x]p = 1 (since |Z| = p). Hence 1 6= [ap, g] = [a, g]p = 1, a

2 contradiction. Thus |Z2| = p and hence Z2 = G3, proving (ii).

To see (iv), consider |Z3|. As G/Z2 has no faithful irreducible characters, Z3/Z2 is not

4 6 4 cyclic, so |Z3| ≥ p . Suppose |Z3| = p . Then cd(G/Z2) = {1, p}, which implies |ker(χ)| = p

2 for all nonlinear χ ∈ Irr(G) such that |ker(χ)| ≥ p . By Theorem 2.25, G/Z2 satisﬁes the

strong condition. Hence ker(χ) ≤ Z3, and since Z(χ) ≥ Z3, we have that Z3/ker(χ) is cyclic.

p Put ker(χ) = K, G = G/Z, and Z3 = ha, Ki. For all g ∈ G,[a, g] ∈ Z2. Since a ∈/ Z2, there is some x ∈ G such that [ap, x] 6= 1. Then [a, x]p = [ap, x] 6= 1, but this is a contradiction

p 6 since |Z2| = p, and hence [a, x] = 1. Therefore |Z3|= 6 p .

4 0 3 Suppose |Z3| = p . If |G | = p , then by Lemma 2.4, G/Z2 is not capable, a contradiction.

0 4 0 Hence |G | = p , that is, G = Z3. By Lemma 2.18, none of G, G/Z, or G/Z2 has an

abelian subgroup of index p. By Lemma 2.17, neither cd(G) nor cd(G/Z2) is {1, p}. Hence

2 3 p ∈ cd(G/Z2), and we have some χ ∈ Irr(G) with |ker(χ)| = p .

3 Suppose p∈ / cd(G/Z2). Then |ker(χ)| = p for all nonlinear χ ∈ Irr(G/Z2). By Lemma

2.14, Z3 ≥ ker(χ) for all such χ, and hence G is normally constrained. If p is odd, then by

0 4 2 Lemma 2.1, |G : G | = p implies p ≤ |G3 : G4|, a contradiction since |G3 : G4| = p. If

3 p = 2, then since G/Z satisﬁes the weak condition, Lemma 2.27 implies that |G : Z2| = p

4 2 or p , a contradiction. Hence cd(G/Z2) = {1, p, p }.

5 2 Finally, suppose |Z3| = p . Recall that for χ ∈ Irr(G), χ(1) ≤ |G : Z|. Since |G : Z3| =

3 0 3 p , we have cd(G/Z2) = {1, p}. By Lemma 2.9, |G |= 6 p , since |G : Z3| is not a square.

If we consider only p-groups satisfying Hypothesis (∗), these results can be extended to

groups that are arbitrarily large. Theorem 4.7 restates Theorem 11 in terms of nilpotence

class.

Theorem 4.7. Let a group G and all of its quotients satisfy Hypothesis (∗). If |G| = p2n or

36 p2n−1, where n ≥ 3, and the nilpotence class of G is at least n, then p2 ∈ cod(G).

Proof. Induct on n. The base case when n = 3 is established by Lemmas 4.3 and 4.4. Now

assume |G| = p2n or p2n−1 and c(G) = c ≥ n, where n ≥ 4. Suppose |Z| ≥ p2. Then

c(G/Z) = c−1 and |G/Z| ≤ p2n−2 or p2n−3. If |G/Z| ≥ p5 then we are done by the inductive

assumption. Since c − 1 ≥ 3, we know |G/Z| ≥ p4, and if |G/Z| = p4 then p2 ∈ cod(G/Z)

by Lemma 3.2.

2 We may now assume |Z| = p. By Hypothesis (∗), |Z2 : Z| ≥ p . Suppose Z2/Z has

p exponent greater than p, and let a ∈ Z2 such that a ∈/ Z. There exists g ∈ G such that

[ap, g] 6= 1. Since [a, g] ∈ Z, we have 1 6= [ap, g] = [a, g]p, which is trivial, as |Z| = p.

This contradiction shows that Z2/Z is elementary abelian, and we can ﬁnd N C G such that Z < N < G, |N : Z| = p, and c(G/N) = c − 1. Now |G/N| = p2n−2 or p2n−3, and we are

done by the inductive assumption.

Theorem 11 now follows as a corollary of Theorem 4.7.

Theorem 11. Let a p-group G and all of its quotients satisfy Hypothesis (∗). If G has coclass n ≥ 3 and |G| ≥ p2n, then p2 ∈ cod(G).

Proof. Let G have coclass n, and |G| = p2m or p2m+1, where m ≥ n. If |G| = p2m, then c(G) = 2m − n ≥ m. If |G| = p2m+1, then c(G) = 2m + 1 − n ≥ m. In either case, we are done by Theorem 4.7.

37 CHAPTER 5

p-groups with Exactly Four Codegrees

Our original purpose was to ﬁnd a sharp bound for the nilpotence class of groups with

exactly four codegrees. Ideally, this bound would be a simple constant. At this time, our

best result for arbitrary p-groups depends on the largest codegree, but in the cases when

|cd(G)| = 2 or |G : G0| = p2, we have attained the predicted bound of nilpotence class at

most 4.

Lemma 5.1. Let G be a ﬁnite p-group such that cod(G) = {1, p, pb, pa}, where 2 ≤ b < a.

If |G| is minimal such that G has nilpotence class n ≥ 3, then G has a faithful irreducible

character.

Proof. Suppose G has no faithful irreducible characters. Let cod(G) = {1, p, pb, pa} and let

|G| be minimal such that G has class n. Let K be the kernel of an irreducible character of

G. Either cod(G/K) ( cod(G), in which case c(G/K) ≤ 2, or cod(G/K) = cod(G), and since |G| is minimal, c(G/K) ≤ n − 1. Hence, there is a central series of G with the n − 1

term contained in K for all kernels of irreducible characters of G, and since the intersection

of these kernels is trivial, G has class at most n − 1, a contradiction. Thus G must have a

faithful irreducible character.

The upper bound on the nilpotence class of G when pa = max(cod(G)) is given as 2a − 2 or 2a − 3 in Theorem 2.36. Restricting |cod(G)| to 4 yields the stronger bound of Theorem

38 Theorem 4. If G is a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a, then

c(G) ≤ a + 1.

Proof. Let G be a minimal counterexample and notice that c(G) = a + 2, for otherwise

c(G) ≥ a + 3 implies c(G/Z) ≥ a + 2 ≥ 5, and by Theorem 2.37, cod(G/Z) = {1, p, pb, pa},

contradicting that |G| is minimal. By Lemma 5.1, G has a faithful character of codegree at

a 2a−1 most p and by Lemma 2.30, |G| ≤ p . Using Theorem 2.37 and that c(G/Za−1) = 3,

we have cod(G/Za−1) = cod(G). Let χ be an irreducible character of G/Za−1 such that

a a a a cod(χ) = p . Then p ≤ χ(1)p = |G : ker(χ)| ≤ |G : Za−1| ≤ p so χ is a faithful linear

character of G/Za−1, which is impossible since G/Za−1 has class 3. Hence c(G) ≤ a + 1.

Theorem 5 further improves the bound in Theorem 4 when the two largest codegrees are

consecutive powers of p and p2 ∈/ cod(G).

Theorem 5. If G is a ﬁnite p-group such that cod(G) = {1, p, pa−1, pa} for a ≥ 4, then c(G) ≤ a.

Proof. Let G be a minimal counterexample. Then c(G) = a + 1 and by Lemma 5.1, G has

a 2a−1 a faithful character with codegree at most p , giving |G| ≤ p . The quotient G/Za−2

has class 3 and therefore has the same set of codegrees as G. Let χ be an irreducible

a character of G/Za−2 such that cod(χ) = p . By Lemma 2.35, χ is nonlinear, so we have

a+1 a a+1 p ≤ χ(1)p = |G : ker(χ)| ≤ |G : Za−2| ≤ p , which shows that χ is faithful, χ(1) = p

and by Lemma 2.16, cd(G/Za−2) = {1, p}. Let µ be an irreducible character of G/Za−2 with

a−1 a a+1 codegree p . Then p = µ(1)cod(µ) = |G : ker(µ)| ≤ |G : Za−2| = p , which shows that

|ker(µ): Za−2| = p and hence ker(µ) ≤ Za−1.

Suppose |Za−1 : Za−2| = p. Then µ is a faithful character of G/Za−1, and Za/Za−1 is

0 2 0 cyclic, so Za−1 ≤ G ≤ Za. Since p ∈/ cod(G), we have that G/G is elementary abelian and

0 2 hence |Za : G | ≤ p. Notice that Z(µ) = Za, and by Lemma 2.8, |G : Z(µ)| = p . We also

39 0 0 0 0 have |G : Za−1| = p, as G ≥ Za−1 ≥ G3 and G /G3 is elementary abelian while G /Za−1

0 4 is cyclic. If G = Za, then |G/Za−2| = p , which contradicts that χ is a faithful nonlinear

a 4 0 5 character with codegree p ≥ p . Thus |Za : G | = p and we have |G/Za−2| = p . By Lemma

4.3, this contradicts p2 ∈/ cod(G).

We may now assume that |Za−1 : Za−2| > p. Let G/Za−2 = H, so H has class 3,

|Z(H)| > p, and µ is an irreducible character of H with |ker(µ)| = p. Since χ is a faithful

irreducible character of H, Z(H) is cyclic and hence ker(µ) is the unique subgroup of H of

order p. Notice that H3 is elementary abelian, and since it is contained in the cyclic subgroup

0 Z(H), we have H3 = ker(µ). As H/H3 has class 2, H /H3 ≤ Z(H/H3) = Z(µ)/H3, which

2 shows that H/Z(µ) is abelian and hence |H : Z(µ)| = p . We then have that Z(µ) = Z2(H),

as Z(µ) ≤ Z2(H). By Lemma 2.6, Z(µ)/ker(µ) = Z2(H)/H3 is cyclic. We also have

0 0 0 H3 ≤ H ≤ Z2(H), so H /H3 is cyclic as well as elementary abelian, and hence |H : H3| = p.

0 This is impossible since |Z(H): H3| = p and |H : Z(H)| ≥ p. Thus c(G) ≤ a.

The next lemma can be used to shorten proofs bounding the nilpotence class when a

group has exactly four codegrees. With this lemma, we can assume that such a group

with nilpotence class greater than 4 has at least one faithful irreducible character, and the

codegree of any such character must be as large as possible.

Lemma 5.2. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has nilpotence class 5, then faithful irreducible characters of G have codegree pa.

Proof. Let χ ∈ Irr(G) be faithful and suppose cod(χ) = pb. If b = 2, then |G| ≤ p5 by Lemma 2.30, which is impossible since G has nilpotence class 5. Hence b ≥ 3 and

2 r p ∈/ cod(G). Let ϕ ∈ Irr(G/Z3) be nonlinear, and let cod(ϕ) = p . Note that since ϕ

is nonlinear, r is equal to either b or a. Put ker(ϕ) = K and Z(ϕ) = Y . By deﬁnition,

r 2 ϕ(1)p = |G : K|, and since G/Z3 has nilpotence class 2, |G : Y | = ϕ(1) by Lemma 2.8.

Combining these equations yields prϕ(1) = ϕ(1)2|Y : K|, and hence pr = ϕ(1)|Y : K|. Now

40 |G| = χ(1)pb ≤ χ(1)pr = χ(1)ϕ(1)|Y : K|, so χ(1)ϕ(1) ≥ |G : Y ||K| = ϕ(1)2|K|. This

shows that pr > χ(1) ≥ ϕ(1)|K|, so ϕ(1)|K| < pr = ϕ(1)|Y : K|. Finally, this implies

|K| < |Y : K|.

Since G/K has class 2, G/Y is abelian, and hence G0 ≤ Y . Thus Y ≥ G0K ≥ K, and

since Y/K is cyclic, the quotient Y/G0K is also cyclic. Recalling that G/G0 is elementary

abelian since p2 ∈/ cod(G), this also shows that Y/G0K is elementary abelian. Being both

cyclic and elementary abelian, the order of Y/G0K is at most p. Using the fact that Y

0 0 0 contains G , we have G3 = [G ,G] ≤ [Y,G] ≤ K. This shows that G ∩ K contains G3.

0 0 0 Since G /G3 is elementary abelian, G /G ∩ K is also elementary abelian. By the Diamond

Isomorphism Theorem, G0/G0 ∩K is isomorphic to G0K/K, which is then elementary abelian,

but also cyclic, since Y/K is cyclic. This shows that the order of G0K/K is at most p. We

0 0 2 now have |Y : K| = |Y : G K||G K : K| ≤ p . Since ϕ is an irreducible character of G/Z3,

3 we know that p ≤ |Z3| ≤ |K|, but the previous paragraph showed that |K| < |Y : K|,

which is at most p2, a contradiction. Hence, any faithful characters of G must have codegree

pa.

Recall that p-groups with exactly two character degrees may have unbounded nilpotence class if cd(G) = {1, p}, and otherwise have nilpotence class at most p. If the group also has

exactly four codegrees, then it has nilpotence class at most 4.

Theorem 1 (i). If G is a ﬁnite p-group such that |cd(G)| = 2 and cod(G) = {1, p, pb, pa},

where 2 ≤ b < a, then G has nilpotence class at most 4.

Proof. Let G be a minimal counterexample. By Lemmas 5.1 and 5.2, G has at least one

faithful irreducible character χ, and such a character must have codegree pa. Let |G| = pn,

and notice that χ(1) = pn−a. Since c(G/Z) = 4, Theorem 2.37 implies that cod(G/Z) =

cod(G). Let ϕ ∈ Irr(G/Z) have codegree pa. If ϕ is nonlinear, then ϕ(1) = pn−a, and hence

|G : ker(ϕ)| = cod(ϕ)ϕ(1) = papn−a = pn = |G|. This shows that ϕ is a faithful irreducible

41 character of G, contradicting that the kernel of ϕ contains Z. Hence ϕ must be linear. By

Lemma 2.35, p2 ∈ cod(G), and by a similar argument, pa = p3. By Lemma 2.30, we have

|G| < cod(χ)2 = p6, implying that |G| ≤ p5, and hence G has nilpotence class at most 4.

Theorem 1 (ii). If G is a ﬁnite p-group such that cd(G) = {1, p, p2} and cod(G) =

{1, p, pb, pa}, where 2 ≤ b < a, then G has nilpotence class at most 4.

Proof. Let G be a minimal counterexample. Notice that for the kernel K of an irreducible character of G, if cod(G/K) = cod(G) then cd(G/K) = cd(G), for otherwise we have a contradiction with Theorem 1 (i). Since |G| is minimal, we must have c(G/K) < c(G), and we can apply Lemmas 5.1 and 5.2. Now c(G) = 5 and G has a faithful irreducible character

χ with codegree pa. By Lemma 2.16, χ must have degree p2, for otherwise |cd(G)| = {1, p}, a contradiction. This implies that pa+2 = χ(1)cod(χ) = |G|, and since G has class 5, we see that a is at least 4.

The quotient G/Z2 has nilpotence class 3, hence cod(G/Z2) = cod(G) and we can ﬁnd

a 2 3 γ ∈ Irr(G/Z2) with codegree p . Since a ≥ 4 and |cod(G)| = 4, either p or p is not in

a+1 a cod(G). Hence γ is nonlinear, and we have p ≤ γ(1)cod(γ) ≤ |G : Z2| ≤ p , which is impossible. Thus G can have nilpotence class at most 4.

Groups with derived subgroup of index p2 have many special properties, including re-

strictions on the index of G3 and G4 in G. In Theorem 1 (iii), we use these facts to show

that if such a group has exactly four codegrees, then it must have nilpotence class at most

Theorem 1 (iii). Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a.

If |G : G0| = p2, then G has nilpotence class at most 4.

Proof. Let G be a minimal counterexample. Then G has nilpotence class 5, and by Lemmas

a 5.1 and 5.2, G has a faithful character χ with codegree p . Let ϕ ∈ Irr(G/G3) be nonlinear.

42 2 As G/G3 has class 2, G/Z(ϕ) is abelian and |G : Z(ϕ)| = ϕ(1) . Since Z(ϕ) contains

G0 which has index p2 in G, we see that ϕ(1) = p. Put cod(ϕ) = pr. Then pr+1 =

3 cod(ϕ)ϕ(1) = |G : ker(ϕ)| ≤ |G : G3|. By Lemma 2.2, |G : G3| = p , which implies r = 2

2 and cod(G/G3) = {1, p, p }.

2 a The nilpotence class of G/G4 is 3, which implies that cod(G/G4) = {1, p, p , p }. Let

a a+1 θ ∈ Irr(G/G4) be nonlinear with codegree p . Then p ≤ cod(ϕ)ϕ(1) = |G : ker(ϕ)| ≤

5 |G : G4|. By Lemma 2.2, |G : G4| ≤ p , which implies a ≤ 4. Recalling that G has a faithful

character with codegree pa, we have that |G| ≤ p7. If |G| ≤ p6 , then by Lemma 3.3, a = 3,

implying that |G| is actually at most p5, which contradicts that G has nilpotence class 5.

Hence |G| = p7 and a = 4.

0 2 3 4 5 2 We now have |G : G | = p , |G : G3| = p , and |G : G4| = p or p . If |G5| = p , then G/G5

3 has maximal class, and by Lemma 3.3, p ∈ cod(G/G5), a contradiction. Hence |G5| = p.

4 Suppose |G : G4| = p and let N be a normal subgroup of G such that G5 < N < G4. Then

G/N has order p5 and class 4, and we can again obtain a contradiction from Lemma 3.3.

5 Now let |G : G4| = p and let N be a normal subgroup of G such that G4 < N < G3. Then

G/N has order p4 and class 3, and Lemma 3.3 yields the desired contradiction. Thus, no

minimal counterexample exists.

This proves Theorem 2 in the case when G has coclass 1. Notice that coclass 1 is equivalent to having maximal class, and a p-group with nilpotence class 4 that is also maximal class has order p5.

Lemma 5.3. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has coclass 2, then the nilpotence class of G is at most 4 and |G| ≤ p6.

Proof. Let G be a minimal counterexample and suppose c(G) = c > 5. The nilpotence class

of G/Z is c − 1 > 4, so G/Z must be maximal class since |G| was minimal with coclass 2,

but this contradicts Theorem 1 (iii). Hence we may assume G has nilpotence class 5.

43 Suppose G has no faithful irreducible character. If K is the kernel of an irreducible

character of G, then either c(G/K) ≤ 4 or c(G/K) = 5 and G/K has maximal class. Since

the latter contradicts Theorem 1 (iii), we must have c(G/K) ≤ 4 for all irreducible characters

of G, which contradicts that G has class 5. Hence G has a faithful irreducible character χ, and by Lemma 5.2, χ has codegree pa.

7 3 As G has class 5 and coclass 2, we see that |G| = p . If |Z2| = p , then G/Z2 has class 3

and order p4, so it has maximal class, and by Lemmas 3.2 and 3.3, cod(G) = {1, p, p2, p3}.

Since G has a faithful character χ with codegree pa = p3, Lemma 2.30 implies |G| ≤ p5, a

2 5 contradiction. Hence |Z2| = p and |Z| = p. By Lemma 2.24, we have |Z4| = p .

a a+1 Let θ ∈ Irr(G/Z2) have codegree p and notice that θ is nonlinear. Thus p ≤

5 θ(1)cod(θ) ≤ |G : Z2| = p . Since a ≥ 4, this shows that θ is a faithful character of

G/Z2 and has degree p. By Lemma 2.16, cd(G/Z2) = {1, p}. Now Lemmas 2.17 and 2.18

3 5 0 3 imply that either |G : Z3| = p , or p = |G : Z2| = p|Z3 : Z2||G : Z2|. Suppose |G : Z3|= 6 p .

3 0 3 3 Then |Z3| = p and |G : Z2| = p , contradicting Theorem 1 (iii). Hence |G : Z3| = p . As

p G/Z2 has a faithful character, Z3/Z2 is cyclic, and we can write Z3 = hx, Z2i, where x ∈/ Z2.

Let g ∈ G such that [xp, g] ∈/ Z. Putting G for G/Z, we have 1 6= [xp, g] = [x, g]p, and since

p [x, g] ∈ Z2, we have [x, g] = 1, a contradiction, and hence G must have nilpotence class at

most 4. Let G have order pn and nilpotence class c ≤ 4. Since G has coclass 2, we have

n − 2 = c ≤ 4, and hence |G| ≤ p6.

This proves Theorem 2 in the case when G has coclass 2. The next lemma completes the proof of Theorem 2.

Lemma 5.4. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has coclass 3, then the nilpotence class of G is at most 4 and |G| ≤ p7.

Proof. Let G be a minimal counterexample and suppose c(G) = c > 5. The nilpotence class

of G/Z is c − 1 > 4, so G/Z must be maximal class or have coclass 2 since |G| was minimal

44 with coclass 3, but this contradicts Theorem 1 (iii) and Lemma 5.3. Hence we may assume

G has nilpotence class 5.

Suppose G has no faithful irreducible character. If K is the kernel of an irreducible

character of G, then either c(G/K) ≤ 4, or c(G/K) = 5 and G/K has maximal class or

coclass 2, contradicting Theorem 1 (iii) and Lemma 5.3, respectively. Hence we must have

c(G/K) ≤ 4 for all irreducible characters of G, which contradicts that G has class 5. Thus

G has a faithful irreducible character χ, and by Lemma 5.2, χ has codegree pa.

As G has coclass 3 and nilpotence class 5, we have |G| = p8, and by Lemma 2.30,

5 ≤ a ≤ 7. If a = 7, then χ(1) = p. By Lemma 2.16 and Theorem 1 (i), this is impossible, so a is at most 6. Theorem 10 (ii) implies that cod(G) = {1, p, p2, pa}.

Suppose p2 ≤ |Z| ≤ p3. Let θ ∈ Irr(G/Z) have codegree pa. Since a is at least 5, θ is nonlinear, and pa+1 ≤ θ(1)cod(θ) ≤ |G : Z| ≤ p6. This shows that a = 5 and |Z| = p2. Since

5 6 G/Z2 has class 3, p ∈ cod(G/Z2). This implies |G : Z2| ≥ p , which is impossible since

6 |G : Z| = p . Hence Z = G5 has order p.

a 6 a 5 The quotient G/Z2 has class 3 and hence p ∈ cod(G/Z2). As |G : Z2| ≤ p , and p = p

6 6 or p is the codegree of a nonlinear character of G/Z2, we must have |G : Z2| = p and a = 5.

5 6 6 Let γ ∈ Irr(G/Z2) have codegree p . Then γ is nonlinear, so p ≤ γ(1)cod(γ) ≤ |G : Z2| = p ,

which shows that γ is a faithful character of G/Z2 with degree p. By Lemma 2.16, this implies

that cd(G/Z2) = {1, p}.

4 Suppose |Z3| ≥ p . As G/Z2 has a faithful character, Z3/Z2 is cyclic, and we can write

p p Z3 = hx, Z2i, where x ∈/ Z2. Let g ∈ G such that [x , g] ∈/ Z. Putting G for G/Z, we

p p p have 1 6= [x , g] = [x, g] , and since [x, g] ∈ Z2, we have [x, g] = 1, a contradiction. Hence

3 |Z3| = p .

5 Since cd(G/Z2) = {1, p} and |G : Z3| = p , Lemma 2.17 implies that G/Z2 has an abelian

6 0 subgroup of index p. By Lemma 2.18, we have p = |G/Z2| = p|Z3/Z2||G /Z2|, which shows

45 0 6 6 5 3 3 that |G | = p , and hence |Z4| = p . Applying Lemma 2.18 to G/Z3 gives p = p · p · p , which is impossible. Therefore G must have nilpotence class at most 4. Since G has coclass

3, we have n − 3 = c ≤ 4, and hence |G| ≤ p7.

For the remainder of this chapter we will consider only those groups that satisfy Hypoth-

esis (∗). Together, Lemmas 5.5–5.8 prove Theorem 3.

Lemma 5.5. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has coclass 4, and G and all of its quotients satisfy Hypothesis (∗), then the nilpotence class of G is at most 4 and |G| ≤ p8.

Proof. Let G satisfy Hypothesis (∗) and let |G| be minimal such that G has coclass 4 and

|cod(G)| = 4. Suppose c(G) = c > 5. The nilpotence class of G/Z is c − 1 > 4, so G/Z has coclass at most 3 since |G| was minimal with coclass 4, but this contradicts Theorem 1 (iii) and Lemmas 5.3 and 5.4. Hence we may assume G has nilpotence class 5.

Suppose G has no faithful irreducible character. If K is the kernel of an irreducible

character of G, then either c(G/K) ≤ 4, or c(G/K) = 5 and G/K has coclass at most 3,

contradicting Theorem 1 (iii) if G has maximal class, Lemma 5.3 if G has coclass 2, and

Lemma 5.4 if G has coclass 3. Hence we must have c(G/K) ≤ 4 for all irreducible characters

of G, which contradicts that G has class 5. Thus G has a faithful irreducible character χ,

a and by Lemma 5.2, χ has codegree p . By Lemma 2.30, a ≥ 5. As c(G/Z2) = 3, we have

a 6 p ∈ cod(G/Z2), and since G satisﬁes Hypothesis (∗), the order of G/Z2 is at most p , which

3 shows that a ≤ 5. Hence a = 5, which forces Z2 to have order p .

p Suppose Z2/Z has exponent greater than p. Then there exists x ∈ Z2 such that x ∈/ Z,

and g ∈ G such that [xp, g] 6= 1. Note that this also implies |Z| = p. Since [x, g] ∈ Z, we

p p have [x , g] = [x, g] = 1, as |Z| = p, a contradiction. Hence Z2/Z is elementary abelian.

4 Suppose |Z3| > p . As G/Z2 has a faithful character, Z3/Z2 is cyclic, and we can ﬁnd

p p x ∈ Z3 such that x ∈/ Z2. Also, there is some g ∈ G such that [x , g] ∈/ Z. Letting G = G/Z,

46 p p 4 we have 1 = [x, g] = [x , g] 6= 1, a contradiction. Hence |Z3| = p . Applying Lemma 2.18

0 6 0 4 0 6 to G/Z2, we have |G/Z2| = p|Z3/Z2||(G Z2)/Z2|. This yields p ≤ |G ∩ Z2|p = |G | ≤ p ,

since |G0| ≤ p6 by Theorem 1 (iii). Hence |G0| = p6.

7 0 If |Z4| = p , we may apply Lemma 2.18 to G/Z3, which yields |G/Z3| = p|Z4/Z3||(G Z3)/Z3|.

0 5 4 6 This forces |G ∩ Z3| to have order p , which is impossible since |Z3| = p . If |Z4| = p ,

0 0 then Z4 = G , and hence G ≥ Z2. Applying Lemma 2.18 again to G/Z2, we now have

0 6 5 |G/Z2| = p|Z3/Z2||G /Z2|, which implies p = p , a contradiction. Hence G must have

nilpotence class at most 4. Since G has coclass 4, we have n − 4 = c ≤ 4, and hence

|G| ≤ p8.

Lemma 5.6. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has coclass 5, and G and all of its quotients satisfy Hypothesis (∗), then the nilpotence class of G is at most 4 and |G| ≤ p9.

Proof. Let G satisfy Hypothesis (∗) and let |G| be minimal such that G has coclass 5 and

|cod(G)| = 4. Suppose c(G) = c > 5. The nilpotence class of G/Z is c − 1 > 4, so G/Z has coclass at most 4 since |G| was minimal with coclass 5, but this contradicts Theorem 1 (iii) and Lemmas 5.3, 5.4, and 5.5. Hence we may assume G has nilpotence class 5.

Suppose G has no faithful irreducible character. If K is the kernel of an irreducible character of G, then either c(G/K) ≤ 4, or c(G/K) = 5 and G/K has coclass at most 4, contradicting Theorem 1 (iii) and Lemmas 5.3, 5.4, and 5.5. Hence we must have c(G/K) ≤ 4

for all irreducible characters of G, which contradicts that G has class 5. Thus G has a faithful

irreducible character χ, and by Lemma 5.2, χ has codegree pa. Since G has class 5 and coclass

10 a 5, we have |G| = p , and by Lemma 2.30, a ≥ 6. As c(G/Z2) = 3, we have p ∈ cod(G/Z2),

7 and since G satisﬁes Hypothesis (∗), the order of G/Z2 is at most p , which shows that a ≤ 6.

3 Hence a = 6, which forces Z2 to have order p .

p Suppose Z2/Z has exponent greater than p. Then there exists x ∈ Z2 such that x ∈/ Z,

47 and g ∈ G such that [xp, g] 6= 1. Note that this also implies |Z| = p. Since [x, g] ∈ Z,

p p we have [x , g] = [x, g] = 1, as |Z| = p, a contradiction. Hence Z2/Z is elementary abelian. Repeating this argument in the quotient group G/Z, we conclude that Z3/Z2

7 6 is also elementary abelian. As |G/Z2| = p and p ∈ cod(G/Z2), G/Z2 has a faithful

character and hence Z3/Z2 is cyclic. This shows that |Z3/Z2| = p. By Lemma 2.18, |G/Z2| =

0 7 0 5 0 7 0 7 p|Z3/Z2||(G Z2)/Z2|. This yields p ≤ |G ∩ Z2|p = |G | ≤ p , since |G | ≤ p by Theorem

0 7 7 8 8 1 (iii). Hence |G | = p and |Z4| = p or p . If |Z4| = p , then Lemma 2.18 implies that

0 6 7 0 0 |G ∩ Z3| = p , which is impossible. If |Z4| = p , then G = Z4, so G ≥ Z2, and applying

7 6 Lemma 2.18 to G/Z2 now yields p = p , a contradiction. Hence G can have nilpotence class

at most 4. Since G has coclass 5, we have n − 5 = c ≤ 4, and hence |G| ≤ p9.

Lemma 5.7. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has coclass 6, and G and all of its quotients satisfy Hypothesis (∗), then the nilpotence class of G is at most 4 and |G| ≤ p10.

Proof. Let G satisfy Hypothesis (∗) and let |G| be minimal such that G has coclass 6 and

|cod(G)| = 4. Suppose c(G) = c > 5. The nilpotence class of G/Z is c − 1 > 4, so G/Z

has coclass at most 5 since |G| was minimal with coclass 6, but for each coclass we have a

contradiction with Theorem 1 (iii) or one of Lemmas 5.3–5.6. Hence we may assume G has

nilpotence class 5.

Suppose G has no faithful irreducible character. If K is the kernel of an irreducible

character of G, then either c(G/K) ≤ 4, or c(G/K) = 5 and G/K has coclass at most

5, contradicting either Theorem 1 (iii) or one of Lemmas 5.3–5.6. Hence we must have

c(G/K) ≤ 4 for all irreducible characters of G, which contradicts that G has class 5. Thus

G has a faithful irreducible character χ, and by Lemma 5.2, χ has codegree pa. Since G has

11 class 5 and coclass 6, we have |G| = p , and by Lemma 2.30, a ≥ 6. As c(G/Z2) = 3, we

a 8 have p ∈ cod(G/Z2), and since G satisﬁes Hypothesis (∗), the order of G/Z2 is at most p ,

48 which shows that a ≤ 7.

3 7 Suppose a = 7. Then |Z2| = p . Let γ ∈ Irr(G/Z2) have codegree p and notice that γ is a faithful character of G/Z2 with degree p. Suppose Z2/Z has exponent greater than p. Then

p p there exists x ∈ Z2 such that x ∈/ Z, and g ∈ G such that [x , g] 6= 1. Note that this also implies |Z| = p. Since [x, g] ∈ Z, we have [xp, g] = [x, g]p = 1, as |Z| = p, a contradiction.

Hence Z2/Z is elementary abelian. Repeating this argument in the quotient group G/Z, we conclude that Z3/Z2 is also elementary abelian, and since Z3/Z2 is also cyclic, we have

0 0 8 |Z3/Z2| = p. By Lemma 2.18, |G/Z2| = p|Z3/Z2||(G Z2)/Z2|, which implies that |G | = p ,

8 9 9 and hence Z4 has order p or p . If |Z4| = p , then applying Lemma 2.18 to G/Z3 implies

0 7 8 0 0 that |G ∩ Z3| = p , which is impossible. If |Z4| = p , then Z4 = G , and hence G ≥ Z2.

8 7 Applying Lemma 2.18 to G/Z2 now yields p = p , a contradiction. Hence a = 6.

Let χ ∈ Irr(G) be faithful with codegree p6. Then χ(1) = p5, so |G : Z| ≥ χ(1)2 implies

|Z| = p. As before, this shows that Z2/Z is elementary abelian, and similarly, Zi/Zi−1 is

6 7 3 elementary abelian for 1 ≤ i ≤ 5. Since p ∈ cod(G/Z2), we have |G : Z2| ≥ p , so |Z2| = p

4 4 or p . Suppose |Z2| = p . Then G/Z2 has a faithful character of degree p and Z3/Z2 is cyclic, showing that |Z3/Z2| = p. Let N C G such that |Z2/N| = p and put X/N = Z(G/N). If

X = Z2, then Z2(G/N) = Z3/N, contradicting Hypothesis (∗), so we must have X = Z3.

This shows that [Z3,G] ≤ N for all such N, and hence [Z3,G] ≤ Z, which is impossible.

3 Hence |Z2| = p .

4 Suppose |Z3| = p . Let N C G such that Z < N < Z2. Put Z(G/N) = X/N and notice that Z2 ≤ X ≤ Z3. If X = Z2, then Z(G/N) = Z2/N and Z2(G/N) = Z3/N. Since G/N has class 3 or 4, this contradicts Hypothesis (∗). Hence X = Z3. Now Z2(G/N) = Z4/N,

2 2 so G/N has class 3, which implies N ≥ G4. Since G4 > Z, we have p ≤ |G4| ≤ |N| = p , which implies N = G4 for all normal subgroups of G lying between Z and Z2. This shows

2 5 that Z2/Z is cyclic, contradicting that it is elementary abelian of order p . Hence |Z3| ≥ p .

49 4 Now let N C G such that Z2 < N < Z3, |N| = p , and assume N 6≥ G3. Then G/N has class 3, and p6 ∈ cod(G/N). Since |G : N| = p7, there exists a faithful character of

G/N with degree p. Put X/N = Z(G/N) and notice that X/N is cyclic. The center of

G/N lies between Z3 and Z4, and since Z4/Z3 is elementary abelian, this shows that |Y : N|

2 5 6 is at most p and |Z3| = p . If |Z4| = p , then X = Z3, and G/N violates Hypothesis

(∗). By Lemma 2.17 G/N has an abelian subgroup of index p, therefore G/Z3 does as well,

8 9 and the same lemma implies that |Z4|= 6 p . If |Z4| = p , then by Lemma 2.18, we have

0 0 0 0 5 |G/Z3| = p|Z4/Z3||(G Z3)/Z3|, which shows that p|G ∩ Z3| = |G |. Since |G ∩ Z3| ≤ p and

0 6 0 6 0 |G | ≥ p , we have |G | = p . Now G > Z3 > N, and applying Lemma 2.18 to G/N implies

4 7 |X : N| = p , which is impossible. Hence |Z4| = p .

7 0 6 7 0 7 0 With |Z4| = p , we have |G | = p or p . If |G | = p , then G = Z4, and applying Lemma

0 6 0 2.18 to G/Z3 yields a contradiction. Now |G | = p . If G ≥ Z3, then Lemma 2.18 applied to

0 0 G/Z3 again provides a contradiction. If G 6≥ Z3, then G Z3 = Z4, and Lemma 2.18 yields

the necessary contradiction. Thus G must have class at most 4. Since G has coclass 6, we

have n − 6 = c ≤ 4, and hence |G| ≤ p10.

Lemma 5.8. Let G be a ﬁnite p-group with cod(G) = {1, p, pb, pa}, where 2 ≤ b < a. If G has coclass 7, and G and all of its quotients satisfy Hypothesis (∗), then the nilpotence class of G is at most 4 and |G| ≤ p11.

Proof. Let G satisfy Hypothesis (∗) and let |G| be minimal such that G has coclass 7 and

|cod(G)| = 4. Similarly to previous lemmas, we may assume G has class 5 and G has a

faithful irreducible character χ with codegree pa. The order of G is p12, so a ≥ 7 by Lemma

9 2.30. As G/Z2 has order at most p , we also have a ≤ 8.

9 Suppose ﬁrst that a = 8, and hence |G : Z2| = p . Then G/Z2 has a faithful irreducible

character of degree p, and by Lemma 2.16, cd(G/Z2) = {1, p}. Suppose Z2/Z has exponent

p p greater than p. Then there exists x ∈ Z2 such that x ∈/ Z, and g ∈ G such that [x , g] 6= 1.

50 Note that this also implies |Z| = p. Since [x, g] ∈ Z, we have [xp, g] = [x, g]p = 1, as |Z| = p, a contradiction. Hence Z2/Z is elementary abelian. Repeating this argument in the quotient group G/Z, we conclude that Z3/Z2 is also elementary abelian, and since it is also cyclic,

4 we have |Z3| = p . Since G/Z must satisfy Hypothesis (∗), we have |Z| = p. Let N C G such that Z < N < Z2, and put X/N = Z(G/N). Then Z2 ≤ X ≤ Z3, and since G/N must

satisfy Hypothesis (∗), we have X = Z3. This shows that [Z3,G] ≤ N, and since this holds

for all such N, we have [Z3,G] ≤ Z, which is impossible.

8 9 8 Now a = 7 and |G : Z2| = p or p . Suppose |G : Z2| = p . Then G/Z2 has a faithful

irreducible character of degree p, so cd(G/Z2) = {1, p} and Z3/Z2 is cyclic. Notice that G

has a faithful irreducible character χ with codegree p7, so χ(1) = p5, and hence |Z| ≤ p2.

2 p2 Suppose |Z| = p . Put Z3 = hx, Z2i, and let G = G/Z. Since x ∈/ Z2, there exists some

p2 p2 2 g ∈ G such that 1 6= [x , g] = [x, g] = 1, since [x, g] ∈ Z2/Z, which has order p . Hence

|Z| = p. Suppose Z2/Z has exponent greater than p. Then there exists y ∈ Z2 such that

p p p y ∈/ Z, and g ∈ G such that 1 6= [y , g] = [y, g] = 1, a contradiction. Hence Z2/Z is

elementary abelian. Repeating this argument in G/Z, we see that Z3/Z2 is also elementary

5 abelian and hence |Z3| = p . Let N C G such that |Z2/N| = p and put X/N = Z(G/N). If

X = Z2, then Z2(G/N) = Z3/N, contradicting Hypothesis (∗), so we must have X = Z3.

This shows that [Z3,G] ≤ N for all such N, and hence [Z3,G] ≤ Z, which is impossible.

3 Thus Z2 must have order p .

p Suppose Z2/Z has exponent greater than p. Then there exists x ∈ Z2 such that x ∈/ Z,

and g ∈ G such that [xp, g] 6= 1. Note that this also implies |Z| = p. Since [x, g] ∈ Z, we

p p have [x , g] = [x, g] = 1, as |Z| = p, a contradiction. Hence Z2/Z is elementary abelian,

and similarly, Zi/Zi−1 is elementary abelian for 1 ≤ i ≤ 5.

4 2 Suppose |Z3| = p . If |Z| = p , then G/Z would violate Hypothesis (∗), so we have

|Z| = p. Let H C G such that Z < H < Z2 and c(G/H) = 4. Put Z(G/H) = Y/H and

51 notice that Z2 ≤ Y ≤ Z3. If Y = Z3, then G/H would have class 3, a contradiction, so

Y = Z2. Now Z2(G/H) = Z3/H, which contradicts Hypothesis (∗).

5 4 Now |Z3| ≥ p . Let N C G such that |N| = p , Z2 < N < Z3, and c(G/N) = 3. Since G/N has class 3, p7 ∈ cod(G/N), and hence G/N has a faithful irreducible character of

degree p. Put X/N = Z(G/N) and notice that X/N is cyclic. The center of G/N lies

between Z3 and Z4, and since Z4/Z3 is elementary abelian, this shows that |Y : N| is at

2 5 6 most p and |Z3| = p . If |Z4| = p , then X = Z4, which contradicts c(G/N) = 3, or X = Z3,

7 and G/N violates Hypothesis (∗). Hence |Z4| ≥ p . By Lemma 2.17, G/N has an abelian

subgroup of index p. Thus G/Z3 also has such a subgroup, and the same lemma implies

9 |Z4|= 6 p .

Applying Lemma 2.18 to G/N shows that p7 ≤ |G0|, and by Theorem 1 (iii), we have

0 9 0 9 10 0 7 |G | ≤ p . If |G | = p , then |Z4| = p , and Lemma 2.18 implies |G ∩ Z3| = p , which is

5 0 8 8 10 10 impossible as |Z3| = p . If |G | = p , then |Z4| = p or p . When |Z4| = p , Lemma 2.18

0 7 8 0 0 implies |G ∩ Z3| = p , which is impossible. When |Z4| = p , then Z4 = G , and G ≥ N.

Lemma 2.18 applied to G/N implies that |X/N| = p3, which is impossible.

0 7 7 0 Now |G | = p . If |Z4| = p , then G = Z4, and applying Lemma 2.18 to G/Z3 yields

10 0 6 a contradiction. If |Z4| = p , then applying Lemma 2.18 to G/Z3 implies |G ∩ Z3| = p ,

8 2 0 2 which is impossible. Now assume |Z4| = p . By Lemma 2.18, p ≤ |G ∩ N| = |X/N| ≤ p ,

which shows that |G0 ∩ N| = p2. Then |G0N| = (|G0||N|)/|G0 ∩ N| = p9. Since G0 and N

8 are both contained in Z4 which has order p , this is impossible. Hence G must have class at

most 4. Since G has coclass 7, we have n − 7 = c ≤ 4, and hence |G| ≤ p11.

52 Concluding Remarks

Many of the theorems in this dissertation suggest the possibility of more general results.

The questions that remain open include the following.

• Does every p-group with exactly four codegrees have nilpotence class at most 4?

• If G is a p-group with coclass n ≥ 3 and |G| ≥ p2n, is p2 a codegree of G?

• Are the codegrees of maximal class p-groups always consecutive p-powers?

The conditions of the ﬁrst two questions are met by all groups in the Small Groups database of Magma, although Hypothesis (∗) is not true for all of these groups. This suggests that

Hypothesis (∗) could be replaced by some other condition that may help generalize our results even further.

In Chapter 3, we identiﬁed several conditions on maximal class groups that guarantee consecutive p-power codegrees. In the Small Groups database, we were unable to identify any maximal class group without consecutive p-power codegrees. In addition to searching the Small Groups database, we also examined the maximal class groups of orders 59, 511, 513, and 515. The list describing these groups was provided by Eamonn O’Brien (2013, personal communication to Mark Lewis). As part of our investigation we recorded the set of faithful character degrees of G/Zi for each Zi in the upper central series of G. It was noticed that the largest degree of a faithful character of G/Zi was never less than the largest degree of a faithful character of G/Zi+1, a property seen in normally monomial groups. Since not all p- groups are normally monomially, if this is the reason the observed codegrees are consecutive p-powers then it seems unlikely that all maximal class p-groups will share this property.

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