SOLVABLE GROUPS WHOSE CHARACTER DEGREE GRAPHS HAVE
DIAMETER THREE
A dissertation submitted
to Kent State University in partial
fulfillment of the requirements for the
degree of Doctor of Philosophy
by
Carrie T. Dugan
August, 2007
Dissertation written by
Carrie T. Dugan
B.S., Marshall University, 1986
M.S., Kent State University, 2003
Ph.D., Kent State University, 2007
Approved by
Mark Lewis,
Chair, Doctoral Dissertation Committee
Stephen Gagola,
Members, Doctoral Dissertation Committee
Donald White
Elizabeth Mann
Austin Melton
Accepted by
Andrew Tonge,
Chair, Department of Mathematical Sciences
Jerry Feezel,
Dean, College of Arts and Sciences
ii
TABLE OF CONTENTS
Acknowledgements
iv
Introduction
1
Related Results
6
Construction
21
Bilinear Forms
28
Commutators
40
Irreducible Characters Part I
52
Irreducible Characters Part II
56
Irreducible Characters Part III
60
A Specific Example
68
Irreducible Characters Part IV
75
iii
ACKNOWLEDGEMENTS
There are many people that deserve recognition for my ability to complete my
dissertation. I would like to begin by thanking my family. My husband gave me the
strength to persevere and the support to make it all come to fruition. He kept me
grounded when times were difficult. This could not have been completed without his
love and devotion to my dream.
There are others in my family who also stood by me throughout graduate school.
My parents were two of my biggest cheerleaders for the duration. My father finished
his doctoral work when my siblings and I were young. So he and my mother knew
how hard it was for me to do graduate work with children at home. My father knew
how difficult my work would be and my mother knew how difficult my husband’s
work would be. Knowing this helped me finish what I started.
I want to thank Steve Gagola for being a mentor during my years at graduate
school. I took abstract algebra from Dr. Gagola my first year at Kent State University.
He made the course exciting and interesting and is one of the main reasons I chose
to pursue research in group theory.
Finally, there are not enough words to express my gratitude toward my advisor,
Mark Lewis. I would not have survived the intensity of the program without his
guidance. He became more of a mentor than I thought was possible. I could always
count on him to make time for me and to answer the many questions I had. He is
not only someone I admire tremendously. He is my lifelong friend.
iv
CHAPTER 1
INTRODUCTION
In a broad sense, character theory is an alternative way to study finite groups. The
characters of a finite group G are functions, χ, from G to the complex numbers. A
representation of a group lets us think of the group in terms of matrices, and characters
are defined by the trace of a representation of G. The value of a function of this type
at 1 ∈ G is known as the character degree, or degree of χ. The irreducible characters
of G, denoted Irr(G), are the characters arising from the irreducible representations
of G. Our focus is on irreducible characters and their degrees. Within a finite group,
the conjugacy classes partition the group and the number of irreducible characters of
G is equal to the number of conjugacy classes of G. More on conjugacy classes can
be found in the next chapter. The set of all irreducible character degrees is the set
cd(G) = {χ(1) | χ ∈ Irr(G)}.
We define the set of vertices, ρ(G), to be the set of all primes that divide degrees
in cd(G). The character degree graph, or degree graph, ∆(G), associated with cd(G)
obtains its vertices from the set ρ(G). Within ∆(G) we have an edge connecting
p, q ∈ ρ(G) if pq divides some degree a ∈ cd(G). The distance between two vertices in
a graph is the minimum number of edges in any path joining those two points. The
diameter of ∆(G) is defined to be the maximum distance between any two vertices. A
graph is connected if there is a path from every vertex in ρ(G) to every other vertex.
In a disconnected graph we typically characterize each connected component of the
1
2
graph. The diameter of a disconnected graph is defined to be the maximum diameter
of its connected components.
It is known that in a solvable group the number of connected components is at
most two. This was shown in [12]. Corollary 4.5 of [13] tells us that if G is a solvable
group and ∆(G) is disconnected, at least one connected component of ∆(G) is a
complete graph and the other connected component has diameter at most two. We
mention that it appears that Pálfy was the first to prove that in a solvable group G, if
∆(G) is disconnected, then each connected component is a complete graph. Thus, in
this case, the diameter is one. The graph of the group we will construct has diameter
three, and thus, is only relevant to the case when ∆(G) is connected.
A very important and very relevant characterization of a character degree graph
of a solvable group is known as Pálfy’s condition. In [15], Pálfy proved that any
solvable group has the property that any three primes in ρ(G) must have an edge in
∆(G) that is incident to two of those vertices. We say that a graph satisfies Pálfy’s
condition if it has this property. It turns out that connected graphs satisfying Pálfy’s
condition have diameter at most three. This was proven true by Manz, Willems, and
Wolf in [13]. Following this result, Huppert [3] conjectured that the diameter of ∆(G)
was at most two in the case of a solvable group G with one connected component.
At the time of his conjecture, there were no known examples where the diameter of
∆(G) was three. All that was known was that the diameter had a bound of three.
This is the origin of the work done by Lewis in [9].
The problem solved in this paper began as a conjecture by Lewis in [9], that the
type of group he constructed could be generalized to produce a degree graph with
diameter three. Thus, we saw the possibility of a family of groups whose structure was
the same as, or similar to that of the group in [9] and in particular, whose character
3
degree graph has diameter three. We began to think of the result in [9] in more
general terms with the goal in mind of constructing a similar family of groups whose
character degree graphs yielded a diameter of three.
In [9], Lewis showed that a solvable group whose character degree graph has
diameter three does exist. In the construction of the group he used specific primes,
p = 2, q = 3, and r = 5, which produced a degree graph having diameter three. Here
we generalize that idea and choose any prime p, and primes q and r, which will have
certain divisibility conditions placed upon them. These conditions will be outlined in
Chapter 3 when we detail the construction of our group.
The main algebraic structure of our group is that of the group in [9]. However, as
we began working on this result, it became very apparent that with q > 3, the group
takes on a much more complicated structure. Many things become complicated when
q > 3. If we had instead fixed q = 3 and chosen p = 2 and r with the appropriate
parameters, we could have mimicked the result in [9]. Even if we let p be any prime
and fix q = 3, the proof and techniques of proving are not that different from the case
of Lewis. Thus, the real work involved in this problem lies in the case when q > 3.
The techniques used in [9] can be used here, however, only up to those results
needed to prove the general q = 3 case. In fact, this specific case is summarized in
Chapter 9. Only a couple of minor points needed to be changed from those of Lewis
in Chapter 9. The majority of the work comes when we move beyond this case.
One purpose of the restrictions placed on p, q, and r is to ensure coprime group
actions, which brings us to the point that the conditions we have set on our primes
are essentially those used in [6]. Also, the construction of our group is the same construction Lewis used, which can be found in [6]. The main result in [6] characterizes
the derived subgroup of a group that is being acted on coprimely as nilpotent, hence
4
solvable. Our interest was not this, necessarily, but the construction of the group in
addition to the results on coprime actions.
A relevant result is Lemma 4.1 in [6]. In short, Isaacs characterizes a particular
type of commutator. We devote the entire Chapter 5 of this paper to commutators
because understanding the commutators of certain elements within our group is essential to the end result here. Much of the work involved in this paper sits inside
Chapter 5. Though Lemma 4.1 in [6] is helpful in a few instances, it is not enough
to characterize all of the commutators we are interested in. There is more on the
definition of the commutator of two elements and why they are such an integral part
of our main goal in Chapters 2 and 5.
Arguably, one of the most helpful results in attaining our goal here is Theorem
4.4 in [16]. Riedl also constructed a group using construction techniques similar to
those of Isaacs. In [16], he focused on calculating character degrees of a particular
class of p-groups. As the reader will see in Chapter 3, within our group lies this same
class of p-groups. Thus, we rely heavily on Theorem 4.4 in [16].
As we set out to prove our group had diameter three, we knew from previous
results of others that we would have a minimum of six vertices in our graph. To see
this, notice that a graph with diameter three must have |ρ(G)| ≥ 4. Zhang proved in
[17] that there does not exist a solvable group with four vertices and diameter three.
In [10], Lewis showed the case of five vertices and diameter three does not occur.
Therefore, we knew |ρ(G)| ≥ 6. In general, we expected the cardinality of ρ(G) to be
larger than six. The group constructed in [9] is somewhat rare, since
(pqr −1)(p−1)
(pq −1)(pr −1)
pq −1 pr −1
,
,
p−1 p−1
and
turned out to be 7, 31, and 151, which are all prime numbers.
As expected, it is possible to have more than six vertices with this family of groups,
depending on the the primes used. What occurs, however, is that there are certain
5
‘families’ of primes within ρ(G) that form complete graphs and for all intents and
purposes can each be thought of as one vertex, since the diameter of a complete graph
is one. Thus, in reality, a group of this type will more than likely have |ρ(G)| larger
than six, but because of the nature of this set we still have a graph with diameter
three. In particular, our graph is isomorphic to the graph in [9].
CHAPTER 2
RELATED RESULTS
This Chapter is in place to provide the general reader with enough background
to better understand the results in this paper. All of these results are well known
among group theorists, who may wish to skip to the next chapter. We provide the
information in this chapter without proof. The majority of the results provided can
be found in [7] and [5].
We begin this chapter with one of the most fundamental results in group theory,
the First Isomorphism Theorem. The main idea to extract from this theorem is that
the normal subgroups of any group can be found in the manner outlined. They are
in fact, kernels of homomorphisms between groups.
Theorem 2.1 (First Isomorphism) Let G and H be groups, let ϕ : G −→ H be a
surjective homomorphism and let N = ker(ϕ). Then H ∼
= G/N . In fact, there exists
a unique isomorphism θ : G/N −→ H such that πθ = ϕ, where π is the canonical
homomorphism from G to G/N .
The next theorem is equally as important as the First Isomorphism Theorem. When
two groups are isomorphic they are essentially ‘equal’ in the sense that they share
group theoretic properties, as the next theorem tells us.
6
7
Theorem 2.2 (Correspondence) Let ϕ : G −→ H be surjective homomorphism and
let N = ker(ϕ). Define the following set of subgroups:
S = {U | N ≤ U ≤ G} and T = {V | V ≤ H}.
Then ϕ(·) and ϕ−1 (·) are inverse bijections between S and T . Furthermore, these
maps respect containment, indices, normality and factor groups.
With the Isomorphism Theorems at hand we can mention another well-known
theorem that is used to help understand the structure of the group. Within the
structure of the group we construct here, for that matter many groups, lie what
are called diamonds. It is always extremely helpful to be able to find isomorphic
subgroups within a group to determine part of the structure of the group. In other
words, we know that isomorphic groups share group theoretic properties. Thus, if we
can find a group, or a subgroup that is isomorphic to a group in which we know the
structure, then we answer many questions.
Theorem 2.3 (Diamond) Let N C G and H ⊆ G. Then H ∩ N C H and H/(H ∩
N) ∼
= N H/N .
Group actions help us count elements in a finite group since, for one thing, the
orbits partition the group. One very important tool used relative to group actions and
counting elements is the Fundamental Counting Principle (FCP). In order to show
the FCP, we need to mention the stabilizer of an element. Let G act on the set Ω and
α ∈ Ω. We define the stabilizer of α to be Gα = {g ∈ G | α · g = α}. As an example,
a very common group action is conjugation. Conjugation is given by αg = g −1 αg. It
is clear that for g to stabilize α we must have g −1 αg = α so that the stabilizer of α
8
will be the centralizer of α in G, written CG (α). The following theorem, Theorem 4.9
in [7], leads directly into the FCP. Thus, it is worth mentioning.
Theorem 2.4 Let G act on Ω and let O be an orbit of this action. Let α ∈ O and
write H = Gα for the stabilizer. Then there exists a bijection O ↔ {Hx | x ∈ G}.
In other words, Theorem 2.4 says that there is a bijection between the elements
of an orbit of the action and the right cosets of a stabilizer. We write the FCP as a
corollary to Theorem 2.4, as was done in [7].
Corollary 2.5 Suppose G acts on Ω, and let O be an orbit. Then |O| = |G : Gα |,
where α is any element of O. If G is finite, we have |O| = |G|/|Gα |, and in particular,
|O| divides |G|.
A corollary and an application of the FCP relative to conjugation that is important
to character theory needs mentioning. Conjugation is a very common group action
that is used and happens to be related to many results in character theory. One of
the most important results related to this within character theory is that the number
of conjugacy classes of a finite group is equal to the number of irreducible characters.
Corollary 2.6 Let g ∈ G and let cl(g) denote the conjugacy class containing g. Then
|cl(g)| = |G : CG (g)|. In particular, for a finite group , all class sizes divide the order
of the group.
Suppose H, N are groups, and H acts on N via automorphisms. An action of this
type means that for all x, y ∈ N and h ∈ H, we have (xy) · h = (x · h)(y · h). A very
useful result arises when gcd(|N |, |H|) = 1 and N is abelian. This result is known as
Fitting’s Lemma. Before we state Fitting’s Lemma we mention a couple of definitions.
9
If N and H are groups, the commutator group generated by N and H, denoted [N, H],
is defined to be the group generated by the set of all [n, h] = n−1 h−1 nh, where n ∈ N
and h ∈ H. A subgroup H ⊆ G is characteristic in G if θ(H) = H for every
automorphism θ of G.
Lemma 2.7 (Fitting) Let H act on N by automorphisms. Assume gcd(|N |, |H|) = 1
and N is abelian. Then the following hold:
˙
1. N = CN (H)×[N,
H].
2. If H and N are subgroups of a group G, with N / G, G = HN , and the given
˙
action of H on N being conjugation within G, then G = CN (H)×H[N,
H].
3. If G is as in 2, then CN (H) and [N, H] are characteristic subgroups of G.
The family of groups that we construct in this paper are solvable groups. So, we
give the definition of a solvable group and a few group theoretic results related to this
type of group. A group G is solvable if there exists a finite collection of subgroups
G0 , G1 , . . . , Gn such that
1 = G0 / G1 / · · · / Gn = G
and Gi+1 /Gi is abelian for 0 ≤ i ≤ n − 1. Abelian groups are automatically solvable
since G and 1 satisfy the definition.
Two more useful results related to group actions and solvable groups are Glauberman’s Lemma and the corollary to it. These results involve transitive actions. Suppose G acts on Ω. We say that the action is transitive if for every two elements
α, β ∈ Ω, there exists an element g ∈ G such that α · g = β.
10
Lemma 2.8 (Glauberman) Let H and N be groups and suppose H and N act on Ω.
Also, assume that H acts on N via automorphisms. Suppose these actions are related
by (α · x) · h = (α · h) · xh for α ∈ Ω, h ∈ H, and x ∈ N . If N acts transitively on
Ω, gcd(|N |, |H|) = 1, and at least one of H or N is solvable, then H fixes some point
of Ω.
Corollary 2.9 Let H, N , and Ω satisfy the hypotheses of Glauberman’s Lemma.
Then the set of H-fixed points of Ω forms an orbit under the action of CN (H).
In addition to the above definition of solvability, we can also think of solvability
in terms of derived subgroups, or commutator subgroups. The derived subgroup G0
of a group G is the subgroup generated by all commutators [x, y] = x−1 y −1 xy for
x, y ∈ G. Now, we can define G(2) = G00 = (G0 )0 and in general G(n) = (G(n−1) )0 for
n > 0. The series of subgroups G = G(0) ⊇ G(1) ⊇ · · · is the derived series of G. This
leads to Theorem 8.3 in [7].
Theorem 2.10 A group G is solvable if and only if G(n) = 1 for some n. Also,
subgroups and factors groups of solvable groups are solvable.
In the context of Theorem 2.10, the smallest integer n such that G(n) = 1 is called
the derived length. We now state Corollary 8.5 of [7], as it is relevant to understanding
some of the structure of our group.
Corollary 2.11 Let 1 = N0 C N1 C · · · C Nn = G be a composition series for G
(each factor Ni+1 /Ni is a simple group for 0 ≤ i < n). Then G is solvable if and only
if the composition factors Ni+1 /Ni all have prime order.
A Sylow p-subgroup of a finite group G is defined to be a subgroup whose order
is pa , where pa is the full power of p dividing the order of G. The set of all Sylow
11
p-subgroups of G is denoted Sylp (G). There are three main theorems attributed to
Sylow, not the least of which states that these subgroups exist in every finite group.
For our purposes, however, we mention that these theorems can be found in chapter
5 of [7] and state Corollaries 5.9 and 5.10 from [7].
Corollary 2.12 Let G be finite and let P ∈ Sylp (G). Then |SylP (G)| = |G : NG (P )|,
and in particular, this divides |G : P |.
Corollary 2.13 Let S ∈ Sylp (G). The following are equivalent:
1. S C G.
2. S is the unique Sylow p-subgroup of G.
3. Every p-subgroup of G is contained in S.
4. S is characteristic in G.
It is important for the reader to have a basic background on a special type of
group, called a Frobenius Group. Here we provide some definitions, as well as results,
involving Frobenius groups and Frobenius group actions. The best resource for more
information on Frobenius groups is [2].
Given a finite group G, let H ≤ G with 1 < H < G. Assume that whenever
g ∈ G − H, we have H ∩ H g = 1. Then G is a Frobenius group with respect to H
and H is the Frobenius complement.
Theorem 2.14 (Frobenius) Let G be a Frobenius group with Frobenius complement
H. Then the set, N , of all elements of G not conjugate to an element of H − 1,
satisfies (i) N is a normal subgroup of G, (ii) G = N H, and (iii) N ∩ H = 1.
12
Theorem 2.14 guarantees the existence of the normal subgroup N . The subgroup
N in Theorem 2.14 is called the Frobenius kernel. Notice that N can be found in
the Frobenius group G by setting N = (G − ∪g∈G H g ) ∪ {1}. It is also clear that the
size of the Frobenius kernel is equal to the index of H in G. Thus, N is uniquely
determined by H. It should be clear that all normal subgroups of G that trivially
intersect H are contained in the kernel N .
An alternative definition to the one above is to say that a finite group G is a
Frobenius group with Frobenius kernel N if N is a proper, nontrivial normal subgroup
of G and CG (n) ≤ N for all elements n 6= 1 of N . This definition, plus the above
results, can be summarized in the following lemma, which is actually problem 7.1 in
[5].
Lemma 2.15 Let N C G, H ≤ G, with N H = G and N ∩H = 1. Then the following
are equivalent:
1. CG (n) ≤ N for all 1 6= n ∈ N ;
2. CH (n) = 1 for all 1 6= n ∈ N ;
3. CG (h) ≤ H for all 1 6= h ∈ H;
4. every x ∈ G − N is conjugate to an element of H;
5. if 1 6= h ∈ H, then h is conjugate to every element of N h;
6. H is a Frobenius complement in G.
Galois theory is another very important topic in group theory that deserves to
be discussed. Part of the construction of our group involves a Galois group, which
we define here. Suppose F ≤ E is a field extension with | E : F |< ∞. The set
13
of F -automorphisms of E is the set of all automorphisms of E that fix all elements
of F . This set is a subgroup of Aut(E) and is called the Galois group of E over
F . We denote this group by Gal(E/F ). An extension of fields F ≤ E is a Galois
extension if | E : F |< ∞ and F ix(Gal(E/F )) = F , where F ix is defined by,
F ix(H) = {α ∈ E | σ(α) = α for all σ ∈ H ≤ Aut(E)}. F ix(H) is a subfield of E
and is called the fixed field of H. Galois theory is used as an aid in constructing groups
from fields. The following theorem is widely known as the Fundamental Theorem of
Galois theory. It tells us that there is a correspondence between the subgroups of the
Galois group and the intermediate fields of E.
Theorem 2.16 Let F ≤ E be a Galois extension with G = Gal(E/F ). Write
F = {K | F ≤ K ≤ E, K a field} and G = {H | H ≤ G}.
1. The maps f (·) = F ix(·) and g(·) = Gal(E/·) are inclusion reversing inverse
bijections between F and G.
2. If g(K) = H, then |E : K| = |H| and |K : F | = |G : H|. In particular,
|E : F | = |G|.
3. If g(K) = H and σ ∈ G, then g(σ(K)) = H σ , the conjugate of H by σ in G.
Also, H C G if and only if K is Galois over F , and in this case, Gal(K/F ) ∼
=
G/H.
Another result needed for our work in this paper is Theorem 21.10 in [7]. As the
reader will see in Chapter 3, our Galois group is cyclic, as is determined by the next
theorem. To be a cyclic group means that the entire group is generated by some
element within the group.
14
Theorem 2.17 Let F ≤ E be finite fields with | F |= q, a prime power. Then E is
Galois over F and G = Gal(E/F ) is cyclic. In fact, G = hσi, where σ : E −→ E is
defined by σ(α) = αq for α ∈ E.
We mention briefly that finite fields are typically called Galois fields and the Galois
field with q elements is denoted GF (q), where q is any prime power. The next three
results about Galois fields are Theorem 21.7 and Corollaries 21.8 and 21.9 from [7].
Additional information on finite fields can be found in Chapter 21 of [7].
Theorem 2.18 Let E = GF (q n ), where q is a prime power. If m ≥ 1 is an integer,
then the following are equivalent.
1. E has a subfield of order q m .
2. m divides n.
3. q m − 1 divides q n − 1.
According to Theorem 2.18, we now have a very useful tool to find certain subfields
within a particular finite field. This fact will be used in the construction of our group,
as well as the determination of certain character degrees. The following corollary also
aids in determining subfields.
Corollary 2.19 Let E, F ≤ L be fields, where |E| = q n and |F | = q m for some prime
power q. Then |E ∩ F | = q d where d = gcd(m, n). In particular, F ≤ E if and only
if m|n.
We end this series of results on finite fields with a corollary that is used to determine all subfields of a finite field.
15
Corollary 2.20 Let F ≤ E be fields with |F | = q < ∞ and |E : F | = n < ∞. Then
for each divisor m of n, there exists a unique subfield K with F ≤ K ≤ E and such
that |K| = q m . There are no other intermediate fields between F and E.
Part of the construction of our group involves rings and ideals. So, we turn our
attention briefly to related ring theory and definitions. Since a ring forms an abelian
group with respect to addition, all results related to this type of group can be applied
to the additive group within a ring. An ideal in a ring is a special subgroup of the
additive group of a ring that is closed under multiplication by ring elements. An ideal
is also the kernel of a ring homomorphism. If R is a ring and a ∈ R, we set (a) as
the principal ideal generated by a. We call a ring where all ideals are principal a
principal ideal ring (PIR). Another ideal we are interested in is the Jacobson radical.
This ideal is the intersection of all maximal right ideals in the ring. More will be said
about the Jacobson radical in the next chapter. A recommended resource for more
on rings and ideals is Chapter 12 in [7].
We now introduce some background on results in character theory. The best
known resource for more information on character theory is [5]. Character theory is
an alternative way to study finite groups and we very briefly point out that ordinary
characters of a group are functions into the complex numbers. Throughout this discussion we will always assume the group G is finite. The set Irr(G) = {θ1 , . . . , θk } of
all irreducible characters of G is the set of characters afforded by irreducible represenP
tations. The degree of an irreducible character, θ, is denoted θ(1). If θ = ki=1 ni θi
is a character, then those θi with ni > 0 are called the irreducible constituents of θ.
We mentioned earlier that the number of irreducible characters of a group is equal to
the number of conjugacy classes of the group. If θ is a character of G, then θ denotes
the complex conjugate.
16
Let ϕ and θ be class functions (complex valued functions that remain constant
on conjugacy classes) of a group G. Then the inner product of ϕ and θ is defined
P
1
to be [ϕ, θ] = |G|
g∈G ϕ(g)θ(g). If ϕ and θ are characters, then [ϕ, θ] = [θ, ϕ] is a
nonnegative integer. Also, ϕ is irreducible if and only if [ϕ, ϕ] = 1.
We provide the definition of an induced class function. Let H ≤ G and let ϕ be
a class function of H. Then ϕG , the induced class function on G, is given by
ϕG (g) =
1 X ◦
ϕ (xgx−1 ),
|H| x∈G
where ϕ◦ is defined by ϕ◦ (h) = ϕ(h) if h ∈ H and ϕ◦ (y) = 0 if y ∈
/ H. An important
point to make here is that if ϕ ∈ Irr(H), then ϕG is a character of G, but it is not
true in general that ϕG is irreducible.
The next lemma is an important tool used quite frequently in character theory
and deserves mentioning.
Lemma 2.21 (Frobenius Reciprocity) Let H ≤ G and suppose that ϕ is a class
function on H and that θ is a class function on G. Then [ϕ, θH ] = [ϕG , θ], where θH
denotes the restriction of θ to H.
A corollary to Frobenius reciprocity states that if ϕ ∈ Irr(H), then there exists
χ ∈ Irr(G) such that ϕ is a constituent of χH .
The main character theory results that we use focus on are those related to normal
subgroups. Let H C G. If θ is a class function of H and g ∈ G, we define θg : H −→ C
by θg (h) = θ(ghg −1 ). We say that θg is conjugate to θ. The following is Lemma 6.1
in [5].
Lemma 2.22 Let H C G and let ϕ, θ be class functions of H and x, y ∈ G. Then
17
1. ϕx is a class function;
2. (ϕx )y = ϕxy ;
3. [ϕx , θx ] = [ϕ, θ];
4. [χH , ϕx ] = [χH , ϕ] for any class function χ of G;
5. ϕx is a character if ϕ is.
We are now ready to introduce what is widely called Clifford theory. The first
very important result for our work, which is Theorem 6.2 in [5], is the following.
Theorem 2.23 (Clifford) Let H C G and let χ ∈ Irr(G). Let θ be an irreducible
constituent of χH and suppose θ = θ1 , θ2 , . . . , θt are the distinct conjugates of θ in G.
P
Then χH = e ti=1 θi , where e = [χH , θ].
The next theorem is arguably the most important tool in character theory. Before
we state the theorem we need a definition. Let H C G and let θ ∈ Irr(H). Then
IG (θ) = {g ∈ G|θg = θ} is called the stabilizer, or inertia group, of θ in G. The
stabilizer is a subgroup of G that contains H. From the FCP, | G : IG (θ) | is the size
of the orbit of θ in the action of G on the irreducible characters of H. So, from the
P
formula χH = e ti=1 θi in Theorem 2.23, we have that t = |G : IG (θ)|.
Theorem 2.24 (Clifford) Let H C G, θ ∈ Irr(H), and T = IG (θ). Let
A = {ψ ∈ Irr(T )|[ψH , θ] 6= 0}andB = {χ ∈ Irr(G)|[χH , θ] 6= 0}.
Then
1. if ψ ∈ A, then ψ G is irreducible;
18
2. the map ψ 7→ ψ G is a bijection of A onto B;
3. if ψ G = χ, with ψ ∈ A, then ψ is the unique irreducible constituent of χT that
lies in A;
4. if ψ G = χ, with ψ ∈ A, then [ψH , θ] = [χH , θ].
Using the notation of Theorem 2.24, we say that θ is invariant in G if T = G. If
P
θ is invariant in G, then we have θG =
ei χi , where χi ∈ Irr(G) and (χi )H = ei θ.
We say that θ is extendible if (χj )H = θ for some j. In other words, ej = 1. If θ
is extendible, it is automatically invariant. This leads to the next theorem and a
corollary attributed to Gallagher.
Theorem 2.25 Let N C G and let ϕ, θ ∈ Irr(N ) be invariant in G. Assume ϕθ is
irreducible and that θ extends to χ ∈ Irr(G). Let S = {ν ∈ Irr(G)|[ϕG , ν] 6= 0} and
T = {ψ ∈ Irr(G)|[(ϕθ)G , ψ] 6= 0}. Then ν 7→ νχ defines a bijection of S onto T .
Next is Gallagher’s Theorem, which is Theorem 2.25 with ϕ = 1N .
Corollary 2.26 (Gallagher) Let N C G and let χ ∈ Irr(G) be such that χN = θ ∈
Irr(N ). Then the characters νχ for ν ∈ Irr(G/N ) are irreducible, distinct for distinct
ν, and are all of the irreducible constituents of θG .
Chapter 6 in [5] is by far the most used resource within this paper, not the least of
which are some of the problems at the end of the chapter. We state these as lemmas
here, again without proof. In order below are problems 6.1, 6.3, and 6.12 from [5].
Lemma 2.27 Let N C G and θ ∈ Irr(N ). Then θG ∈ Irr(G) if and only if IG (θ) =
N.
19
Problem 6.1, or Lemma 2.27 as it is labeled here, asserts that induced characters
of normal subgroups N are irreducible if and only if the stabilizer is only N .
Next, problem 6.3 introduces a new type of character known as a fully ramified
character. These types of characters are interesting to study and occur throughout
[8].
Lemma 2.28 Let N C G and let χ ∈ Irr(G) and θ ∈ Irr(N ) with [χN , θ] 6= 0. Then
the following are equivalent:
1. χN = eθ, with e2 = |G : N |;
2. χ vanishes on G − N and θ is invariant in G;
3. χ is the unique irreducible constituent of θG and θ is invariant in G.
If the hypotheses within Lemma 2.28 are met, we say that χ and θ are fully
ramified with respect to G/N .
The next lemma, problem 6.12, says that a particular character in a particular
type of group either induces irreducibly, is fully ramified, or extends. It is sometimes
referred to as the ‘going up’ theorem and is directly related to Theorem 6.18 in [5],
which is sometimes called the ‘going down’ theorem. We will see later that this result
will be relevant to our work, as it was in [9].
Lemma 2.29 Let K/L be an abelian chief factor of G (i.e., K, L C G and no M C G
exists with L < M < K). Let ϕ ∈ Irr(L) and T = IG (ϕ). Assume that KT = G.
Then one of the following occurs:
1. ϕK ∈ Irr(K).
2. ϕK = eθ for some θ ∈ Irr(K), where e2 = |K : L|.
20
3. ϕK =
Pt
i=1 θi ,
where θi ∈ Irr(K) are distinct and t = |K : L|.
We feel this chapter provides an ample background for the reader to understand
the terminology, notation, and most of the relevant theory used within this paper.
As mentioned above, other references provide proofs of the results presented, as well
as more detail on the subject matter.
CHAPTER 3
CONSTRUCTION
To begin the construction of our group, we need to place the necessary restrictions
on the primes p, q, and r. The nth cyclotomic polynomial, denoted Φn (x), is the monic
polynomial in C[x] whose roots are precisely the primitive nth roots of unity. Consider
the monic polynomial xn − 1. According to Lemma 20.4 in [7], we know that
xn − 1 =
Y
Φd (x),
d|n
where the product runs over all positive divisors d of n. Notice that Φ1 (x) = x − 1
and, since xp̂ − 1 = Φ1 (x)Φp̂ (x) for any prime p̂, we have Φp̂ (x) =
xp̂ −1
.
x−1
We now choose p to be any prime, primes q < r, and consider certain cyclotomic polynomials in p. Because q and r are primes, we know Φq (p) =
and Φr (p) =
pr −1
.
p−1
pq −1
p−1
Since the only divisors of qr are 1, q, r, and qr, we can write
pqr − 1 = Φ1 (p)Φq (p)Φr (p)Φqr (p). Thus,
pqr −1
p−1
= Φq (p)Φr (p)Φqr (p). We choose the
distinct primes q and r so that Φq (p), Φr (p), and Φqr (p) are pairwise relatively prime.
As was mentioned in Chapter 1, we use the method found in Chapter 4 of [6], as
well as in [9], to construct our group G. We begin by letting F be the field of order
pqr and take F {X} to be the skew polynomial ring. A skew polynomial ring gets its
name from the slightly unorthodox method of multiplication. In particular, within
21
22
the skew polynomial ring the definition of multiplication on the right by a scalars is
given by Xα = αp X for every element α ∈ F . A point to make here is that this type
of multiplication is one of the major contributors to complicating the commutator of
elements within our group.
Next we define R to be the ring obtained from the quotient of F {X} by the ideal
P
generated by X q+1 , written (X q+1 ). This ideal is the set of all i≥q+1 αi X i , where
αi ∈ F . Let x be the image in R of the indeterminate X. Clearly xR = Rx is a
nilpotent ideal since (xR)q+1 = xq+1 R = 0. Also R/xR ∼
= F . Hence, xR is a maximal
ideal of R. In fact, xR is the unique maximal ideal of R and therefore, is the Jacobson
radical of R. We will write J for the Jacobson radical of R. Notice that elements of
J have the form α1 x + α2 x2 + · · · + αq xq where the αi ∈ F .
The Jacobson radical of a ring has some very interesting properties. One such
property is that every element of J is quasiregular. This means that if z ∈ J, then
1 − z has a multiplicative inverse in R. This is equivalent to the assertion that
(1 − z)J = J(1 − z) = J. A direct consequence of this is that 1 + J is a subgroup of
the group of units of R. By the fact that 1 + J is in and of itself then a group, we
know that inverses exist for every element within this structure.
We now define P to be the group 1 + J. It is clear from the structure of J that
the elements of P have the form 1 + α1 x + α2 x2 + · · · + αq xq where the αi ∈ F . Due
to the size of F there are pqr choices for each of the αi . Thus, it is easy to see that
2
the order of P is pq r . Also, P is the unique Sylow p-subgroup of our group G that
we are constructing.
As was done in [9], we define P i = 1 + J i . Observe that P 1 = P and P q+1 = 1,
and P i /P i+1 is isomorphic to the additive group of F for i = 1, 2, . . . q. It is easy to
see that P i is normal in G for every positive integer i. Another important note is that
23
P
q+1
2
is abelian and P k is not abelian for k = 1, 2, . . . , q−1
. This family of p-groups
2
has within it quotient groups that behave in such a manner that we can apply results
from [16] to factor groups within this family, as we will see later.
We now introduce some group actions. Since F × is cyclic of order divisible by
p − 1, we know from Theorem 2.9 of [7] that there is a unique subgroup C of F ×
qr
−1
of order pp−1
. There is a natural action of C on R via automorphisms defined by
P
P
i
( qi=1 αi xi ) ∗ c = qi=1 αi cp xi . However, we define a different action on R.
Consider the subgroup F0× of F × of order p − 1. From Lemma 4.6 in [6], we know
˙ 0× . This fact plays a crucial role in determining the action of C on
that F × = C ×F
R. As in [9], we define the action of C on R via ring automorphisms by
pq −1
p−1
(α0 + α1 x + · · · + αq xq ) · c = α0 c0 + α1 c p−1 x + · · · + αq c p−1 xq .
To see that this is a group action, first notice that if t = α0 + α1 x + · · · + αq xq , then
t · 1 = t. Also, if c, d ∈ C, then
pq −1
p−1
(t · c) · d = (α0 + α1 c p−1 x + · · · + αq c p−1 xq ) · d
p−1
p−1
pq −1
pq −1
= α0 + α1 c p−1 d p−1 x + · · · + αq c p−1 d p−1 xq
p−1
pq −1
= α0 + α1 (cd) p−1 x + · · · + αq (cd) p−1 xq = t · (cd).
To see that C acts via ring automorphisms, let r1 = α0 + α1 x + · · · + αq xq , r2 =
β0 + β1 x + · · · + βq xq ∈ R, and c ∈ C. Then
(r1 + r2 ) · c = ((α0 + β0 ) + (α1 + β1 )x + · · · + (αq + βq )xq ) · c.
24
Using the definition of the action, we extend this to
pq −1
pq −1
pq −1
(α0 +β0 )+(α1 +β1 )cx+· · ·+(αq +βq )c p−1 xq = (α0 +β0 )+· · ·+(αq c p−1 +βq c p−1 )xq .
Distributing xi within each term and combining like terms we obtain
pq −1
pq −1
(α0 + α1 cx + · · · + αq c p−1 xq ) + (β0 + β1 cx + · · · + βq c p−1 xq ) = (r1 · c) + (r2 · c).
Now that we know the action of C on R preserves ring addition, we can verify the
preservation of products using monomials as representative elements of R. Let r1 =
i
αi xi , r2 = αj xj ∈ R, c ∈ C, and assume i + j ≤ q. Then (r1 r2 ) · c = (αi αjp xi+j ) · c.
The definition of the action gives the expression
i
αi αjp c
pi+j −1
p−1
xi+j .
(3.1)
We now consider
pj −1
pi −1
(r1 · c)(r2 · c) = αi c p−1 xi αj c p−1 xj .
Multiplication within the skew polynomial ring and combining exponents of c gives
us
i
pi −1
pj −1
i
i
pi −1
αi αjp c p−1 +( p−1 )p xi+j = αi αjp c p−1 +
pi+j −pi
p−1
xi+j .
Lastly, by simplifying the exponent of c we have the expression
i
αi αjp c
pi+j −1
p−1
xi+j ,
which equals (3.1). Hence, C acts via ring automorphisms on R.
Let G be the Galois group of F over the subfield of order p. Theorem 2.17 tells
25
us that G is cyclic of order qr. For any element σ ∈ G we define the action of G on R
to be
(α0 + α1 x + · · · + αq xq ) · σ = α0σ + α1 σ x + · · · + αq σ xq ,
where the αi ∈ F , and the action of σ on C is just the action of σ on F × . It
is not difficult to see that G acts via automorphisms on R. Let σ ∈ G and set
r1 = α0 + α1 x + · · · + αq xq and r2 = β0 + β1 x + · · · + βq xq . Then
(r1 + r2 ) · σ = ((α0 + · · · + αq xq + β0 + · · · + βq xq )) · σ = ((α0 + β0 ) + · · · + (αq + βq )xq ) · σ.
The definition of the Galois action and the fact that σ is a subgroup of Aut(F ) gives
us
(α0 + β0 )σ + · · · + (αq + βq )σ xq = (α0σ + β0σ ) + · · · + (αqσ + βqσ )xq .
By distributing each xi and combining like terms, we have
(α0σ + · · · + αqσ xq ) + (β0σ + · · · + βqσ xq ) = (r1 · σ) + (r2 · σ).
Again, to show the action of G on R preserves ring multiplication, we can use monomials as representative elements of R. Let σ ∈ G, and r1 = αi xi , r2 = βj xj ∈ R.
Then
i
(r1 r2 ) · σ = (αi xi βj xj ) · σ = (αi βjp xi+j ) · σ.
Using the definition of the action and that σ is an automorphism of F , we have
i
i
i
(αi βjp )σ xi+j = αiσ (βjp )σ xi+j = αiσ (βjσ )p xi+j .
26
Reassociating this product yields
(αiσ xi )(βjσ xj ) = (r1 · σ)(r2 · σ).
Therefore, G acts via ring automorphisms on R. Clearly, G also acts on C via automorphisms. Hence, we form the semidirect product of G with C. Suppose αi xi ∈ R, c ∈ C,
and σ ∈ G. To show that GC acts via automorphisms on R, we need to show
((αi xi ) · c) · σ = ((αi xi ) · σ) · cσ . We begin by noticing from the left side of the equation
pi −1
pi −1
((αi xi ) · c) · σ = (αi c p−1 xi ) · σ = αiσ (c p−1 )σ xi .
pi −1
Again, σ is an automorphism. So, the right hand side becomes αiσ (cσ ) p−1 xi . Separating this product gives us
(αiσ xi ) · cσ = ((αi xi ) · σ) · cσ .
Therefore, GC acts on R via automorphisms. Notice P is invariant under this action.
We now construct our group G from the semidirect product of GC with P .
For the appropriate fixed primes, the order of the group is very large. The order
2
qr
−1
of G is qrpq r ( pp−1
). Even in the case presented by Lewis, his group had order in the
neighborhood of 1019 . Whether or not a smaller example of a solvable group with a
degree graph of diameter three exists is not known, as of yet.
CHAPTER 4
BILINEAR FORMS
The results in this chapter do not involve the characters of our group directly but
will play an important role in helping us understand the kernels of our characters
better. Also, many complicated strings of sums occur later when we look at commutators of elements inside of P . The lemmas here eliminate some of the work later.
Throughout this chapter, unless otherwise noted, we will let F = GF (pf ), where
p is any prime and let m, n ∈ Z+ . In keeping with methods used in [9], we generalize
the definition of maps used in that paper to characterize the coefficients in P that
occur.
We begin by defining the map h·, ·im,n from F × F −→ F by
m
n
ha, bim,n = abp − ap b.
We define ha, F im,n = {ha, bim,n | b ∈ F }. Note that if a = 0, then ha, F im,n = {0}.
This map has several useful and interesting properties. Those properties that are
needed for the purposes of this paper are outlined in this chapter. In general, as we
will see below, the nature of these forms will depend on gcd(m, n), gcd(m, f ), and
gcd(m + n, f ).
27
28
Lemma 4.1 Let d = gcd(m, n) and suppose d | f . Let Fd be the subfield of F of
m
n
order pd . Then the map h·, ·im,n from F × F −→ F defined by ha, bim,n = abp − ap b
is an Fd -bilinear map. In particular, if we fix a ∈ F , the map ha, ·im,n : F −→ F is
an additive homomorphism.
Proof. Let α, β ∈ Fd and a1 , a2 , b ∈ F . To show linearity in the first coordinate of
h·, ·im,n we will need to show that
hαa1 + βa2 , bim,n = αha1 , bim,n + βha2 , bim,n .
By definition of h·, ·im,n , we see that
m
n
hαa1 + βa2 , bim,n = (αa1 + βa2 )bp − (αa1 + βa2 )p b.
k
k
k
Since the characteristic of F is p, we have that for all γ and δ in F , (γ ±δ)p = γ p ±δ p
for any integer k. We use this fact to expand the last expression and get
m
n
m
m
n
n
(αa1 + βa2 )bp − (αa1 + βa2 )p b = αa1 bp + βa2 bp − (αa1 )p b − (βa2 )p b
m
n
n
m
n
n
= αa1 bp − αp ap1 b + βa2 bp − β p ap2 b.
Now, o(α) and o(β) both divide pd − 1 since α and β are elements of Fd . As a result
n
n
αp = α and β p = β. Hence,
m
n
m
n
m
n
m
n
αa1 bp − (αa1 )p b + βa2 bp − (βa2 )p b = αa1 bp − αap1 b + βa2 bp − βap2 b.
29
And again by definition of h·, ·im,n , we have
m
n
n
m
n
m
m
n
αa1 bp − αap1 b + βa2 bp − βap2 b = α(a1 bp − ap1 b) + β(a2 bp − ap2 b)
= αha1 , bim,n + βha2 , bim,n .
Linearity in the second coordinate can be shown using a similar argument. The last
statement of the lemma follows immediately from h·, ·im,n being Fd -bilinear.
q.e.d.
We should mention some easy, yet relevant, facts regarding ha, bim,n . First, notice
that
m
n
n
m
ha, bim,n = abp − ap b = −(ap b − abp ) = −hb, ain,m .
This result will be used frequently without mention. Secondly, this clearly implies
the fact that ha, F im,n = hF, ain,m . Formal proofs of these results are omitted.
The importance of calculating stabilizers of the various characters is to help compute character degrees by applying well-known Clifford theory. The first few results
in this chapter help us get a handle on the behavior that the commutators exhibit.
Of equal importance are the kernels of our characters. Later results in this chapter
will be used to help us describe the kernels of certain characters. The next two lemmas
will be used directly and quite frequently when computing coefficients of the various
powers of x within the commutators.
m
m
Lemma 4.2 Let a, b ∈ F . Then a(ha, bim,n )p = ha, abp im,m+n .
Proof. By definition of h·, ·im,n we have that
m
m
n
m
a(ha, bim,n )p = a(abp − ap b)p .
30
Recall from the proof of Lemma 4.1 that since the characteristic of F is p, we know
m
m
m
that for any α and β in F , (α − β)p = αp − β p . Hence,
m
n
m
m
m
m+n
m+n
bp ] = a(abp )p − ap
a(abp − ap b)p = a[(abp )p − ap
m
bp ].
By distributing a, we have
m
m
a[(abp )p − ap
m
m
m
m+n
m
(abp ).
And finally, by definition of h·, ·im,n we have
m
m
m+n
a(abp )p − ap
m
m
(abp ) = ha, abp im,m+n .
q.e.d.
Continuing on the same topic, this next lemma will handle a slightly more complicated calculation that appears later. This lemma helps us pair two separate forms
together that have the appropriate parameters. Notice the similarities and differences
between Lemma 4.2 and this next lemma.
Lemma 4.3 Let a, b, c ∈ F . Then
k
m
m
k
a(hb, cim,n )p + b(ha, cik,n )p = ha, bcp ik,m+n + hb, acp im,k+n .
Proof. We begin as usual and notice that
k
m
m
n
k
k
n
m
a(hb, cim,n )p + b(ha, cik,n )p = a(bcp − bp c)p + b(acp − ap c)p .
31
Recall that F has characteristic p, so
m
n
k
k
n
m
a[(bcp − bp c)p ] + b[(acp − ap c)p ]
m
k
n
k
k
m
n
m
= a[(bcp )p − (bp c)p ] + b[(acp )p − (ap c)p ].
Expanding this further we have
m
k
n
k
k
m
n
m
a[(bcp )p − (bp c)p ] + b[(acp )p − (ap c)p ]
k
k+m
= abp cp
k+n
− abp
k
m
k+m
cp + bap cp
m+n
− bap
m
cp .
If we group the first term from this with the fourth term and the second term with
the third, we obtain
k
k+m
abp cp
k
k+n
− abp
k+m
= abp cp
k
m
k+m
cp + bap cp
m+n
− bap
m
m
m+n
− bap
k+m
cp + bap cp
cp
k+n
− abp
m
k
cp .
The result now follows since, by definition,
k
k+m
abp cp
m+n
− bap
m
m
m
k+m
cp + bap cp
k+n
− abp
k
cp
k
= ha, bcp ik,m+n + hb, acp im,k+n .
q.e.d.
We now move on to information regarding the general maps. In the next few
lemmas we determine much needed information regarding the kernels of these maps.
We also compare the kernels of these maps to the kernel of the trace map. As we will
32
see, there is a direct relationship between the kernels of these maps and the kernel of
the trace map when gcd(m, f ) = 1.
f
−1
Lemma 4.4 Suppose d = gcd(m, f ) and let 0 6= a ∈ F . Assume gcd(pm − 1, ppd −1
)=
n −1
1. Define the map am,n : F −→ F by am,n (b) = ha, bim,n for all b ∈ F . If o(ap
pf −1
,
pd −1
not divide
) does
then ker(am,n ) = 0 and ha, F im,n = F . Otherwise, | ker(am,n ) |= pd
and | ha, F im,n |= pf −d .
Proof. Note that am,n is an additive homomorphism and that
Im(am,n ) = ha, F im,n ⊆ F.
By the First Isomorphism Theorem, we know that
| ha, F im,n |=| Im(am,n ) |=
|F |
.
| ker(am,n ) |
So we work to compute | ker(am,n ) |.
Suppose b ∈ ker(am,n ). Now, 0 ∈ ker(am,n ), so without loss we can assume b 6= 0.
m
n
Thus, 0 = ha, bim,n = abp − bap . Since a and b are not zero, this implies that
m −1
bp
n −1
= ap
.
Since d | m, we can write pm −1 = (pd −1)h(p) for some polynomial h. Notice that
m −1
(bp
n −1
o(ap
pf −1
f −1
) pd −1 = (bh(p) )p
n −1
), if o(ap
m −1
= 1. So, o(bp
) does not divide
pf −1
,
pd −1
) |
pf −1
.
pd −1
m −1
Since we know that o(bp
)=
this is a contradiction, and we conclude
that ker(am,n ) ={0}. Hence, | ha, F im,n |=| F | and so F = ha, F im,n .
n −1
Now assume o(ap
)|
pf −1
.
pd −1
Our field F is finite and therefore F × is cyclic by
Lemma 17.12 in [5]. According to Theorem 2.9 in [5], for each divisor of pf − 1,
there is exactly one subgroup of F × . In particular, F × has a unique subgroup, B × ,
33
of order
pf −1
.
pd −1
f
−1
Since gcd(pm − 1, ppd −1
) = 1, raising the elements of B × to the power
pm − 1 induces a bijection on B × , so there exists a unique element b0 ∈ B × such that
m −1
bp0
n −1
= ap
pf −1
pd −1
. Since
m −1
c ∈ GF (pd ) where b0 p
and pd − 1 are relatively prime, we can write b = b0 c for
n −1
= ap
m −1
. Hence, bp
m −1
= b0 p
m −1
cp
.
Recall that we can write pm − 1 = (pd − 1)h(p) for some polynomial h. So,
m −1
b0 p
n −1
ap
cp
m −1
m −1
= b0 p
d −1)(h(p))
c(p
m −1
. But this forces bp
= ap
m −1
. Our choices of b0 and c yield b0 p
n −1
d −1)(h(p))
c(p
=
, which is true if b ∈ ker(am,n ). Thus, the elements
of ker(am,n ) are in one to one correspondence with the elements of GF (pd ). Therefore,
| ker(am,n ) |= pd and via the First Isomorphism Theorem, | ha, F im,n |= pf −d . q.e.d.
Before we begin the next few lemmas, we take a moment to mention our definition
and a few of the properties of trace maps. Our definition is actually taken from
Theorem 23.5 in [5].
Suppose E is a finite field extension of K and suppose α is an element of E. Let
m =| E : K[α] |. Finally, let s denote the sum of the roots of the minimal polynomial
of α over K, counting multiplicities. We define T rE/K (α) = ms.
We also mention two corollaries from [5], 23.10 and 23.11, that have relevance to
our purposes here. Let L ≤ K ≤ E, where E is Galois over L, and let U be a set
of representatives for the right cosets of Gal(E/K) in Gal(E/L). Then T rK/L (α) =
P
σ
σ∈U α for all α ∈ K. The second corollary is a special case of Corollary 23.10
where K is Galois over L. Here we can take E = K. The corollary states that if we
P
let L ≤ K be Galois with G = Gal(K/L), then T rK/L (α) = σ∈G ασ for all α ∈ K.
One important property of the trace map is the L-linearity of the map. In other
words, for elements α, β ∈ K and scalars a, b ∈ L, we have T rK/L (aα + bβ) =
aT rK/L (α) + bT rK/L (β).
If the characteristic of L is p and L = GF (p), then
34
T rK/L (αp ) = T rK/L (α) for all α ∈ K. Unless otherwise stated, we will use the
notation Tr to denote the trace T rF/GF (p) : F −→ GF (p).
A subspace K, of a vector space E over GF (p), is called a hyperplane for E if it
has dimension one less than E. From Lemma 4.4, we saw that | ha, F im,n |= pf −d .
Thus, the subspace ha, F im,n is a hyperplane in F if and only if d = 1. Notice that
the First Isomorphism Theorem tells us that | ker(T r) |= pf −1 , hence ker(T r) will
also be a hyperplane. The hyperplanes within our group correspond to the kernel of
the trace map. This correspondence may, or may not be one-to-one, as we will see
below.
Lemma 4.5 Let d = gcd(m + n, f ). If a ∈ GF (pd ), then ha, F im,n ⊆ ker(T r). In
particular, ha, F im,n = ker(T r) if and only if gcd(m, f ) = 1.
Proof. Let a ∈ GF (pd ) and b ∈ F. Using the definition of h·, ·im,n and the linearity
of the trace map, we have
m
n
m
n
T r(ha, bim,n ) = T r(abp − ap b) = T r(abp ) − T r(ap b).
The property that T r(αp ) = T r(α) for all α ∈ F gives us
m
n
m
n
m
m
m+n
T r(abp ) − T r(ap b) = T r(abp ) − T r((ap b)p ) = T r(abp ) − T r(ap
m
bp ).
By linearity we obtain
m
m+n
T r(abp ) − T r(ap
m
m
m+n
bp ) = T r(abp − ap
m
m+n
bp ) = T r((a − ap
m
)bp ).
35
n+m
Since by assumption d | m + n and a ∈ GF (pd ), we have that ap
T r((a − ap
n+m
m
= a, and so,
m
)bp ) = T r((a − a)bp ) = 0.
Hence, ha, F im,n ⊆ ker(T r).
As discussed above, ha, F im,n = ker(T r) if and only if ha, F im,n is a hyperplane in
F . Thus, ha, F im,n = ker(T r) if and only if gcd(m, f ) = 1.
q.e.d.
Corollary 4.6 In general we have h1, F im,n ⊆ ker(T r). In particular, h1, F im,n =
ker(T r) if and only if gcd(m, f ) = 1.
This is an immediate consequence of Lemma 4.5 with a = 1.
Lemma 4.7 Let d = gcd(m, f ). Assume gcd(pd − 1, fd ) = 1. Write a = a0 c where
f
m+n −1)k
−1
o(a0 ) | (pd − 1) and o(c) | ( ppd −1
). Then ha, F im,n ⊆ c(p
f
m+n −1)k
−1
). In particular, ha, F im,n = c(p
k ≡ (pm − 1)−1 mod( ppd −1
n −1
gcd(m, f ) = 1 and o(ap
)|
ha0 , F im,n where
ker(T r) if and only if
pf −1
.
pd −1
n −1)k
Proof. Let b ∈ F. Since F is a field, we can write b = b0 c(p
for some b0 ∈ F .
Then substituting appropriately for a and b and by definition of h·, ·im,n , we have
n −1)k
ha, bim,n = ha0 c, b0 c(p
n −1)k
im,n = (a0 c)(b0 c(p
m
n
n −1)k
)p − (a0 c)p (b0 c(p
).
From here we see that
n −1)k
(a0 c)(b0 c(p
m
n
)p − (a0 c)p (b0 c(p
n −1)k
m
m (pn −1)k
) = a0 cbp0 cp
n
n
− ap0 cp b0 c(p
n −1)k
.
36
By combining the exponents of c in each term, we are left with
m
m (pn −1)k
a0 cbp0 cp
n
n
n −1)k
− ap0 cp b0 c(p
m (pn −1)k
= c1+p
m
n +(pn −1)k
a0 bp0 − cp
n
a0 p b0 .
We now compare the exponents of c in each term in the last expression. We begin
with the exponent of c in the the first term. Recalling that the choice of k gives
1 ≡ k(pm −1), we have 1+pm (pn −1)k = 1+kpm+n −kpm ≡ k(pm −1)+k(pm+n −pm ) =
k(pm −1+pm+n −pm ) = k(pm+n −1). Similarly, working with the exponent of c in the
second term, we have pn + kpn − k ≡ k(pm − 1)pn + kpn − k = k(pm+n − pn + pn − 1) =
k(pm+n − 1). Now, we have
m (pn −1)k
c1+p
m
n +(pn −1)k
a0 bp − cp
n
m+n −1)
ap0 b0 = ck(p
m
n
(a0 bp0 − ap0 b0 ).
Again, by definition of h·, ·im,n we have
ck(p
m+n −1)
m
m+n −1)k
Therefore, ha, F im,n ⊆ c(p
n −1
divides
pf −1
pd −1
m+n −1)
)|
ha0 , b0 im,n .
ha0 , F im,n and the first part of the lemma is proven.
m+n −1)k
Suppose ha, F im,n = c(p
Lemma 4.4, we have o(ap
n
(a0 bp0 − ap0 b0 ) = ck(p
ker(T r). Thus, | ha, F im,n |=| ker(T r) |= pf −1 . By
pf −1
pd −1
and gcd(m, f ) = 1. Conversely, suppose o(ap
n −1
)
and gcd(m, f ) = 1. Then by Lemma 4.4, | ha, F im,n |= pf −1 . By Lemma
4.5, ha0 , F im,n = ker(T r), so ha, F im,n ⊆ c(p
m+n −1)k
ker(T r). Since | ha, F im,n |=
pf −1 =| ker(T r) |, we see that equality follows.
q.e.d.
Corollary 4.8 Let f = qr where q < r are primes. If 0 < i < q and α ∈ F , then
hα, F ii,q−i = c(p
q −1)k
ker(T r) for some element c ∈ F so that o(c) divides
pf −1
.
pq −1
37
q −1)k
Furthermore, for each such c, there is an α so that hα, F ii,q−i = c(p
ker(T r).
Therefore, the set of hyperplanes of the form hα, F ii,q−i is the set of hyperplanes
q −1)k
c(p
ker(T r) as c runs through the elements of F of order
pf −1
.
pq −1
In particular, this
second set of hyperplanes is independent of i.
Proof. In the situation of Lemma 4.7, we can let d = q, m = i, and n = q−i. We then
q −1)k
have hα, F ii,q−i = c(p
ker(T r), where o(c) |
pf −1
pd −1
f
−1
and k ≡ (pi − 1)−1 mod( ppq −1
). As
in Lemma 4.7, for each such c we can write α = α0 c where o(α0 ) | pq − 1. As a result,
q −1)k
the set of hyperplanes of the form hα, F ii,q−i is the set of hyperplanes c(p
as c runs through the elements of F of order
pf −1
,
pq −1
ker(T r)
and this set is independent of i.
q.e.d.
The following two lemmas are special cases of Lemma 4.7 where in each case m
and f are relatively prime. The purpose of these lemmas will become clear in the last
chapter of this paper.
Lemma 4.9 Let a ∈ F such that o(a) |
pf −1
.
p−1
Choose c ∈ Z+ so that
c ≡ (pm − 1)−1 mod(o(a)).
Set b ≡ c(pn − 1)mod(o(a)). Then ha, F im,n = hF, ab im,n .
Proof. By Lemma 4.7 we know that ha, F im,n = a
pm+n −1
pm −1
ker(T r). On the other hand,
we also know from Lemma 4.7 that hF, ab im,n = hab , F in,m = ab(
pm+n −1
)
pn −1
ker(T r). Our
choice of b gives the desired result since
µ
b
pm+n − 1
pn − 1
¶
µ
≡
pn − 1
pm − 1
¶µ
pm+n − 1
pn − 1
¶
=
pm+n − 1
.
pm − 1
q.e.d.
38
Lemma 4.10 Let a ∈ F such that o(a) |
pf −1
.
p−1
Choose b ∈ Z+ such that b ≡
n
−1
( pp−1
)mod(o(a)). Then ha, F i1,m−1 = hab , F in,m−n .
Proof. As in the proof of Lemma 4.9, we know from Lemma 4.7 that ha, F i1,m−1 =
a
pm −1
p−1
pm −1
ker(T r). Similarly, we have that hab , F in,m−n = ab( pn −1 ) ker(T r). The given
choice of b yields
µ
b
pm − 1
pn − 1
¶
µ
≡
pn − 1
p−1
¶µ
pm − 1
pn − 1
¶
=
pm − 1
.
p−1
q.e.d.
The last two lemmas of this chapter characterize all of the hyperplanes in our field
F . For these two lemmas we assume we are within our group G.
Lemma 4.11 Assume a ∈ F and a 6= 0. Then aker(T r) = ker(T r) if and only if
a ∈ Zp .
Proof. If a ∈ Zp , then aker(T r) = ker(T r) since multiplication by elements in Zp
does not affect whether or not the trace is zero.
Conversely, suppose aker(T r) = ker(T r) and assume a ∈
/ Zp . Thus, a − ap 6= 0.
Let d ∈ F with T r(d) 6= 0. Let c ∈ F so that (a − ap )c = d. Since multiplication by
elements in Zp does affect whether or not the trace is zero, we may assume o(c) divides
pqr −1
.
p−1
Because aker(T r) = ker(T r), it follows that a(γ − γ p ) ∈ ker(T r) for all γ ∈ F .
By the definition of the kernel of the trace map, we know that aγ − (aγ)p ∈ ker(T r).
Subtracting these terms, we have aγ p − ap γ p − (a − ap )γ p ∈ kerT r for all γ ∈ F . Now,
raising to the p-power is a bijection for the cyclic group of order
pqr −1
,
p−1
so there is a
γ ∈ F so that γ p = c. This yields (a − ap )γ p = (a − ap )c = d ∈
/ ker(T r). This is a
contradiction. Therefore, a ∈ Zp .
q.e.d.
39
Lemma 4.12 Every hyperplane in F has the form aker(T r) for some 0 6= a ∈ F .
Proof. Lemma 4.11 implies that aker(T r) = bker(T r) if and only if b = az for some
z ∈ Zp . This implies that there are
pqr −1
p−1
total number of hyperplanes in F is
hyperplanes of the form aker(T r). But, the
pqr −1
,
p−1
so every hyperplane has this form. q.e.d.
CHAPTER 5
COMMUTATORS
As was mentioned earlier, commutators will play a large role in the understanding
of stabilizers of the characters of our group G. To see if an element is in the stabilizer
of a character, we need to conjugate that element by another element. The reason
that commutators arise is the fact that if we have elements s and t, for example,
then ts = s−1 ts = t(t−1 s−1 ts) = t[t, s]. Thus, conjugation can be thought of as
multiplication on the right by the commutator [t, s].
Rather than attempting to calculate t−1 and s−1 for use within [t, s], we use the
identity st[t, s] = ts. With this we will calculate the products st, ts, and st[t, s]. At
that point we solve for coefficients of each xi in [t, s].
Recall that R is the skew polynomial ring with Jacobson radical J and that
P = 1 + J and P i = 1 + J i . Because J is the Jacobson radical of R, which is the skew
polynomial ring, right multiplication within P is determined by the multiplication
within R. This is one of the complicating factors of computing [t, s]. The other, more
complicating factors, are the forms h·, ·ii,j that appear within the coefficients and how
they combine or do not combine with each other.
With all of this in mind we begin this chapter with a result that helps us begin
to understand the commutator [t, s] where s ∈ P and t ∈ P
40
q+1
2
.
41
Lemma 5.1 Let s = 1 + α1 x + α2 x2 + · · · + αq xq ∈ P and t = 1 + β q+1 x
q+1
2
2
β q+3 x
q+3
2
2
+ · · · + βq xq ∈ P
q+1
2
. Then [t, s] is of the form
1 + δ1 x
where
δj =
j−1
X
q+3
2
+ δ2 x
pi
−αi δj−i
q+5
2
+ · · · + δ q−1 xq ,
2
−
j
X
i=1
and j ∗ = j +
+
hαi , βj ∗ −i ii,j ∗ −i
i=1
q+1
.
2
Proof. We begin by considering the product st. Write t = 1 + y, where y =
β q+1 x
2
q+1
2
+ β q+3 x
q+3
2
2
+ · · · + βq xq . We can now think of the product st as
st = s(1 + y) = s + sy.
Thus, we will consider the equation
st = s + y + (α1 x)y + (α2 x2 )y + · · · + (αq−1 xq−1 )y + (αq xq )y.
Define j ∗ = j +
q+1
2
as in the statement of the lemma, where 1 ≤ j ≤
∗
analyze the term from st that will involve xj for
q+3
2
q−1
.
2
We now
≤ j ∗ ≤ q. Notice that this term
∗
will result from terms βj ∗ xj added to sums of products of the form αi xi βj ∗ −i xj
In other words, all of y except for the first term β q+1 x
q+1
2
2
∗
∗
∗ −1
+ α2 x2 βj ∗ −2 xj
∗
2
∗ −2
+ · · · + α q−1 x
2
q−1
2
βj ∗ − q−1 xj
2
∗
∗
= βj ∗ xj + α1 βjp∗ −1 xj + α2 βjp∗ −2 xj + · · · + α q−1 βj ∗ − q−1 xj .
2
.
will combine with the
remaining terms of sy to form the term we are interested in. Thus, we consider
βj ∗ xj + α1 xβj ∗ −1 xj
∗ −i
2
∗ − q−1
2
42
∗
Hence, for the term involving xj we can write
j∗
β x +
j∗
∗
α1 βjp∗ −1 xj
+
2
∗
α2 βjp∗ −2 xj
q−3
+ ··· + α
j
X
∗
= β j ∗ xj +
q−3
2
∗
2
βjp∗ − q−3 xj
2
i
q−1
+α
q−1
2
∗
2
βjp∗ − q−1 xj
2
∗
αi βjp∗ −i xj .
i=1
Summing over all values of j we have that
q−1
q−1
st = s + β q+1 x
q+1
2
2
+
2
X
j∗
βj ∗ x +
j=1
j
2 X
X
i
∗
αi βjp∗ −i xj .
j=1 i=1
Now consider the product ts. The details provided in the calculation of st should
help the reader understand that we can begin by writing
ts = s + y + y(α1 x) + y(α2 x2 ) + · · · + y(αq−1 xq−1 ) + y(αq xq ).
∗
Again, we consider the term involving xj for
q+3
2
≤ j ∗ ≤ q. This term will come from
∗
βj ∗ xj in y added to sums of products of the form βj ∗ −i xj
∗
βj ∗ xj + βj ∗ −1 xj
∗
∗ −1
α1 x + βj ∗ −2 xj
j ∗ −1
= βj ∗ xj + βj ∗ −1 α1p
∗ −2
∗
∗ −i
α2 x2 + · · · + βj ∗ − q−1 xj
j ∗ −2
q−1
j∗ − 2
∗
2
∗
βj ∗ x +
i=1
j ∗ −i
αi p
α q−1 x
∗
βj ∗ −i xj .
q−1
2
2
xj + · · · + βj ∗ − q−1 αpq−1
So for the term involving xj in ts we have the summation
j
X
∗ − q−1
2
2
xj + βj ∗ −2 α2p
j∗
αi xi . We see that
2
∗
xj .
43
If we sum over all possible values for j we obtain
q−1
ts = s + β q+1 x
q+1
2
2
+
2
X
q−1
j∗
βj ∗ x +
j=1
j
2 X
X
j ∗ −i
αi p
∗
βj ∗ −i xj .
j=1 i=1
We now move on to calculate st[t, s]. First, it is straightforward to see that [t, s]
can be written as 1 + δ1 x
q+3
2
[t, s] = 1 + z where z = δ1 x
+ δ2 x
q+3
2
q+5
2
+ · · · + δ q−1 xq . As above, we begin by writing
2
+ δ2 x
q+5
2
+ · · · + δ q−1 xq . Thus, from this we see that
2
st[t, s] can be written
q−1
st[t, s] = st + stz = st + (s + β q+1 x
2
q+1
2
+
2
X
q−1
j∗
βj ∗ x +
j=1
j
2 X
X
i
∗
αi βjp∗ −i xj )z.
j=1 i=1
Much of the expression on the right side will collapse to zero because the exponent
of x in many terms will exceed q. Consider the expression (β q+1 x
q+1
2
2
)z. The smallest
power of x here is q + 2. Thus, the products in that term disappear. Similarly,
Pj
P q−1
P q−1
∗
pi
j∗
2
2
βj ∗ xj )z = 0 and ( j=1
( j=1
i=1 αi βj ∗ −i x )z = 0.
At this point, we now have a much shorter equation to analyze. In other words,
we now have
st[t, s] = st + sz.
We now simplify sz in our equation.
sz = z + α1 xz + α2 x2 z + · · · + αq−1 xq−1 z + αq xq z.
∗
Consider the term involving xj for
q+3
2
∗
≤ j ∗ ≤ q. This term will come from δj xj in
z and sums of products of the form αi xi δj−i xj
∗ −i
in the pieces αk xk z for appropriate
44
values of k. As a result, we see that the term we are interested in is the sum
∗
δj xj + α1 xδj−1 xj
j∗
= δj x +
∗ −1
+ α2 x2 δj−2 xj
∗
p
α1 δj−1
xj
+
∗ −2
+ · · · + α q−3 x
q−3
2
2
δj− q−3 xj
∗ + q+3
2
2
q−3
∗
p2
α2 δj−2
xj
+ ··· +
p 2
j∗
α q−3 δj−
q−3 x .
2
2
Thus, we can summarize our term with the following:
j∗
δj x +
j−1
X
i
∗
p
αi δj−i
xj .
i=1
Summing over all values of j we obtain
q−1
2
X
q−1
j∗
δj x +
j−1
2 X
X
j=1
i
∗
p
αi δj−i
xj .
j=1 i=1
Combining all of these results for st[t, s], we see that
q−1
st[t, s] = s + β q+1 x
q+1
2
2
+
2
X
q−1
q−1
j∗
βj ∗ x +
j=1
2
X
j∗
δj x +
j
2 X
X
j=1
q−1
pi
j∗
αi βj ∗ −i x +
j=1 i=1
j−1
2 X
X
i
∗
p
αi δj−i
xj .
j=1 i=1
∗
We now want to compare coefficients of xj in both st[t, s] and ts. Recall that st[t, s] =
ts. First notice that since both the left and right sides of this equation have the
P q−1
q+1
∗
2
identical terms s + β q+1 x 2 + j=1
βj ∗ xj , these terms cancel. Thus, fixing j and
2
equating the x
j∗
terms gives us
q−1
j∗
δj x +
j
2 X
X
j=1 i=1
q−1
pi
j∗
αi βj ∗ −i x +
j−1
2 X
X
j=1 i=1
q−1
pi
αi δj−i x
j∗
=
j
2 X
X
j=1 i=1
j ∗ −i
αi p
∗
βj ∗ −i xj .
45
If we only compare coefficients from this equation, we see that
δj +
j
X
i
αi βjp∗ −i
+
i=1
j−1
X
pi
αi δj−i
=
j
X
j ∗ −i
αi p
βj ∗ −i .
i=1
i=1
Solving for δj we obtain
δj = −
j−1
X
i
p
αi δj−i
−
i=1
j
X
i
αi βjp∗ −i +
i=1
j
X
j ∗ −i
αi p
βj ∗ −i .
i=1
For our result, we combine the last two sums and simplify this to
δj = −
j−1
X
pi
αi δj−i
−
i=1
=−
j−1
X
j
X
i
αi βjp∗ −i − αi p
j ∗ −i
βj ∗ −i
i=1
pi
αi δj−i
i=1
−
j
X
hαi , βj ∗ −i ii,j ∗ −i .
i=1
q.e.d.
Though Lemma 5.1 is helpful to the reader to begin to understand the coefficients
that arise in the commutator [t, s], we would like to simplify them even further. The
main purpose of Lemma 5.1 is to aid in the very technical proof of the next lemma.
The next lemma does, in fact, condense the general formula for the coefficients. In
other words, by dissecting Lemma 5.1 we are able to recombine the coefficients into
a more concise formula. To do this, we certainly incorporate the result from Lemma
5.1. However, we rely very heavily on the results of Lemmas 4.2 and 4.3 for our proof.
Lemma 5.2 Let s ∈ P and t ∈ P
q+1
2
be defined as in Lemma 5.1. Inductively, we
46
define d1 = β q+1 . For k > 1 define
2
dk = −βk∗ −1 +
k−1
X
l
αl (dk−l )p ,
l=1
where k ∗ = k +
q+1
.
2
If we write
[t, s] = 1 + δ1 x
then δk =
q+3
2
+ δ2 x
q+5
2
+ · · · + δ q−1 xq ,
2
Pk
i=1 hαi , dk−i+1 ii,k∗ −i .
Proof. We proceed by induction on k. In the initial case when k = 1, we know from
Lemma 5.1 that δ1 = −hα1 , β q+1 i1, q+1 . Hence, the result is true in this case. Suppose
2
2
that the result holds for all k = 1, . . . , m − 1. By Lemma 5.1 and the induction
hypothesis we have the following:
δm = −
m−1
X
i=1
=−
m−1
X
pi
αi δm−i
m
X
−
hαi , βm∗ −i ii,m∗ −i
i=1
m−i
m
X
X
i
−αi ( hαj , dm−i−j+1 ij,m∗ −i−j )p −
hαi , βm∗ −i ii,m∗ −i .
i=1
j=1
i=1
We can move αi inside the inner sum to get
m−1
m−i
XX
i=1 j=1
i
αi (hαj , dm−i−j+1 ij,m∗ −i−j )p −
m
X
hαi , βm∗ −i ii,m∗ −i .
(5.1)
i=1
For the moment we will ignore the single summation portion of (5.1) and only work
with the double summation. At this point there are two main cases to consider. The
two cases are when i = j and when i 6= j. We will first consider the easier case when
i
i = j. Here we have a sum of products of the form αi (hαi , dm−2i+1 ii,m∗ −2i )p . We can
47
apply Lemma 4.2 to write
i
i
αi (hαi , dm−2i+1 ii,m∗ −2i )p = hαi , αi dpm−2i+1 ii,m∗ −i .
Notice that for i = j, we must have i ≤ m − i or i ≤ b m2 c. Set v = b m2 c. So, if we
sum over all possible values of i, for each i we can apply this result to exactly one j.
Thus, for our general case we obtain
v
α1 (hα1 , dm−1 i1,m∗ −2 )p + · · · + αv (hαv , d1 iv,m∗ −2v )p
v
= hα1 , α1 dpm−1 i1,m∗ −1 + · · · + hαv , αv dp1 iv,m∗ −v .
(5.2)
We will come back to (5.2) later in the proof when we combine all of our results. For
now we move to the case when i 6= j. In order to begin analyzing this more difficult
case we will consider when i = 1 and in essence we argue inductively on i. The reason
we choose i = 1 is not because this is a formal inductive argument, but because this
value of i gives us the maximum number of components for j. The reader will then
be able to recognize the pattern more quickly when we sum over all values of i. When
i = 1, the double summation of (5.1) becomes
m−1
X
α1 (hαj , dm−j ij,m∗ −j−1 )p .
j=1
From this, when j = 1 we pick up the first term in (5.2). Thus, we will do an
analysis summing over j > 1. If we isolate one term and fix j = j0 , then within this
sum we have α1 (hαj0 , dm−j0 ij0 ,m∗ −j0 −1 )p . We would like to apply Lemma 4.3 here.
j
To do this we need to combine this term with the term αj0 (hα1 , dm−j0 i1,m∗ −j0 −1 )p 0 .
Notice that this term comes from the outer summation in (5.1) when i = j0 . Hence,
48
in general we will obtain m − 2 terms of this form that can be combined with the
term α1 (hαj0 , dm−j0 ij0 ,m∗ −j0 −1 )p using Lemma 4.3. In the specific case that we are
considering, if we apply Lemma 4.3, we have
j
α1 (hαj0 , dm−j0 ij0 ,m∗ −j0 −1 )p + αj0 (hα1 , dm−j0 i1,m∗ −j0 −1 )p 0
j0
= hα1 , αj0 dpm−j0 i1,m∗ −1 + hαj0 , α1 dpm−j0 ij0 ,m∗ −j0 .
(5.3)
We can sum over all possible values for j0 > 1 and we get
m−1
X
j0
(hα1 , αj0 dpm−j0 i1,m∗ −1 + hαj0 , α1 dpm−j0 ij0 ,m∗ −j0 ).
j0 =2
Notice that had we fixed i = j0 and j = 1, we would have considered the equation
j
αj0 (hα1 , dm−j0 i1,m∗ −j0 −1 )p 0 + α1 (hαj0 , dm−j0 ij0 ,m∗ −j0 −1 )p
j0
= hαj0 , α1 dpm−j0 ij0 ,m∗ −j0 + hα1 , αj0 dpm−j0 i1,m∗ −1 .
(5.4)
Of course, this term was accounted for already in (5.3). In other words, in general
there is some repetition occurring. For example, if we give the label (a, b) to the term
b
a
hαa , αb dpm−b ia,m∗ −a + hαb , αa dpm−b ib,m∗ −b ,
then we will always have (a, b) = (b, a), and hence, a repeated term. Now consider
the double summation in (5.1) with i = 2. We see that summing over all j we will
repeat (1, 2) that was accounted for in (5.3). Hence, fixing j = j0 and combining the
49
appropriate terms we obtain
m−2
X
j0
2
(hα2 , αj0 dpm−j0 i2,m∗ −2 + hαj0 , α2 dpm−j0 ij0 ,m∗ −j0 ).
j0 =3
In general for i = i0 , to avoid repeating prior terms we would sum from j0 = i0 + 1
to m − i0 . Now, as i0 increases, so does the starting value for j0 in this sum. Also,
as i0 increases, the maximum value for j0 decreases. Thus, the number of terms is
decreasing as i0 increases. We must have i0 < j0 ≤ m − i0 . So, i0 ≤ m − i − 1 and
2i0 ≤ m − 1. Thus, we need i0 ≤ b m−1
c for these sums to make sense in general. Set
2
u = b m−1
c and notice that in general we have sums of pairs
2
j0
(hα1 , αj0 dpm−j0 i1,m∗ −1 + hαj0 , α1 dpm−j0 ij0 ,m∗ −j0 )+
j0
u
· · · + (hαu , αj0 dpm−j0 iu,m∗ −u + hαj0 , αu dpm−j0 ij0 ,m∗ −j0 ).
So, for each of these pairs if we sum over all possible values of j0 , we get
m−1
X
j0
(hα1 , αj0 dpm−j0 i1,m∗ −1 + hαj0 , α1 dpm−j0 ij0 ,m∗ −j0 )+
j0 =2
··· +
m−u
X
j0
u
(hαu , αj0 dpm−j0 iu,m∗ −u + hαj0 , αu dpm−j0 ij0 ,m∗ −j0 ).
(5.5)
j0 =u+1
Notice that in the last summation m − u = u + 1, so there is only one pair to consider
there. Hence, adding (5.2) and (5.5) yields
v
hα1 , α1 dpm−1 i1,m∗ −1 + · · · + hαv , αv dp1 iv,m∗ −v
+
m−1
X
j0
(hα1 , αj0 dpm−j0 i1,m∗ −1 + hαj0 , α1 dpm−j0 ij0 ,m∗ −j0 )+
j0 =2
50
u
u+1
· · · + hαu , αu+1 dpm−2u iu,m∗ −u + hαu+1 , αu dpm−2u iu+1,m∗ −u−1 .
(5.6)
Again, we fix i = i0 and combine the appropriate terms to form hαi0 , ·ii0 ,m∗ −i0 . From
(5.6) we see that
i0
i0 +1
m−i0
hαi0 , ·ii0 ,m∗ −i0 = hαi0 , αi0 dpm−i0 + αi0 +1 dpm−i0 + · · · + αm−i0 dip0
ii0 ,m∗ −i0 .
We can now generalize this and sum over all values of i to get
m
X
i
i+1
m−i
hαi , αi dpm−i + αi+1 dpm−i + · · · + αm−i dpi
ii,m∗ −i =
m
X
i=1
i=1
hαi ,
m−i
X
j
αj dpm−j ii,m∗ −i .
j=1
At this point, if we combine the single summation from (5.1) and invoke the induction
hypothesis again, we have our desired result
δm =
m
X
hαi , −βm∗ −i +
i=1
m−i
X
j
αj dpm−j ii,m∗ −i
j=1
m
X
=
hαi , dm−i+1 ii,m∗ −i .
i=1
q.e.d.
The purpose of the next lemma is to consider a specific case that appears later.
The point of the lemma is to show the reader which coefficients disappear within the
commutator in this case.
Lemma 5.3 Assume the notation of Lemma 5.1. Suppose that s = 1 + αxm . Then
1. δj = 0 for 1 ≤ j ≤ m − 1.
2. δj = −hα, βj ∗ −m im,j ∗ −m + hα, eim,j ∗ −m where e is a polynomial in α and
β q+1 , . . . , βj ∗ −m−1 when m ≤ j ≤
2
q−1
.
2
51
Proof. Since the coefficients of x, x2 , . . . , xj in s are 0 when 1 ≤ j ≤ m − 1, we see
from Lemma 5.1 that δj = 0 when 1 ≤ j ≤ m − 1. Suppose that m ≤ j ≤
q−1
.
2
Then
m
by Lemma 5.2, we have δj = hα, dj−m+1 im,j ∗ −m where dj−m+1 = −βj ∗ −m +α(dj−2m )p .
(We mention here that we write dj−2m = 0 if j − 2m ≤ 0.) It is easy to see that dj−2m
is a polynomial in α and β q+1 , . . . , βj ∗ −m−1 , and so, the result follows.
2
q.e.d.
CHAPTER 6
IRREDUCIBLE CHARACTERS PART I
We begin this chapter by introducing some notation. Let N be a normal subgroup
of G, and X be a subset of Irr(N ). We define Irr(G|X ) to be the set of characters
χ ∈ Irr(G) such that some irreducible constituent of χN lies in X . In this spirit,
we define cd(G|X ) = {χ(1)|χ ∈ Irr(G|X )}. We set Irr(G|N ) = Irr(G | X ) for
X = Irr(N ) \ {1N } and cd(G|N ) = {χ(1)|χ ∈ Irr(G|N )}.
Recall from Chapter 3 that we defined P i to be 1 + J i . Also, recall that P i is
normal in G for any positive integer i. Another important fact that comes from
Theorem 2.5(i) in [16] is that the group CP/P q is a Frobenius group with respect
to C with Frobenius kernel P/P q . The next lemma provides an alternative proof, to
that of Riedl, of this relevant fact.
Lemma 6.1 The group CP/P q is a Frobenius group with complement C and Frobenius kernel P/P q .
Proof. Let 1 6= s ∈ P/P q and 1 6= c ∈ C. Notice that the order of c divides pq − 1.
pi −1
Hence, c p−1 6= 1 for i = 1, 2, . . . , q −1. If we write s = 1+α1 x+α2 x2 +· · ·+αq−1 xq−1 ,
then the previous statement tells us that there exists an i such that αi 6= 0. Therefore,
s · c 6= s. Thus, CCP/P q (c) ≤ C for all c ∈ C.
q.e.d.
There are a few observations that should be mentioned with regard to CP/P q
relative to the fact that it is a Frobenius group. One characteristic is that all proper
52
53
nontrivial subgroups of C will complement P/P q . In other words, all subgroups of
C are Frobenius complements to P/P q . We also have that for 1 < P i /P q < P/P q
and P i /P q normal in CP/P q , (CP/P q )/(P i /P q ) is a Frobenius group with Frobenius
kernel (P/P q )/(P i /P q ). Also, all normal subgroups of CP/P q are either contained
in P/P q or contain P/P q as a subgroup.
Now, the fact that P/P q is a nontrivial finite p-group tells us that P/P q is nilpotent. In general, though, in any Frobenius group the Frobenius kernel is nilpotent.
The character degrees of G are all contained in the union of the sets cd(G/P q ) and
cd(G|P q ). This follows from the fact that every irreducible character of G either has
P q in its kernel or does not have P q in its kernel. We begin by calculating irreducible
character degrees of G/P q and move on to cd(G|P q ) in the chapters following this
one. Most of the work for the proof of the following Lemma can be found in Theorem
4.4 of [16].
Lemma 6.2 For P q defined as above, we have
½
µ qr
¶
¾
qr−1
p −1
i
cd(G/P ) = 1, q, r, qr, (p 2 )
| i = 0, 1, . . . , q − 2 .
p−1
q
Proof. To begin computing irreducible character degrees of G/P q , we compute the
irreducible character degrees of G/P ∼
= CG. The Fundamental Theorem of Galois
Theory tells us that the intermediate fixed fields correspond to the subgroups of G
via inclusion reversing bijections. So the subgroup of G of order q corresponds to the
fixed field of order pr and the subgroup of order r in G corresponds to the fixed field
of order pq . In addition, G corresponds to the fixed field of order p. This leads to a
correspondence between the subgroups of F × and the subgroups of C. Viewed as a
54
multiplicative group, C is complemented by the subgroup of F × of order p − 1. Now,
F × is mapped to C via the map that sends an element α ∈ F × to αp−1 in C. Hence,
F × = C × CF × (G). As a result we see that the subgroup of order q in G centralizes
pr −1
p−1
the subgroup of order
in C and thus nonidentity elements of this subgroup lie
in orbits of size r under the action of G. The subgroup of order r in G centralizes the
subgroup of order
pq −1
p−1
in C, and the elements in this subgroup lie in orbits of size q.
The remaining elements lie in orbits of size qr.
Now the character degrees of G/P correspond to the sizes of the orbits of G acting
on C. From this, we see that cd(G/P ) = {1, q, r, qr}.
Using Theorem 4.4 of [16] we see that Irr(P/P q ) contains pi (pqr − 1) nonprincipal
characters of degree (p
qr−1
2
)i for i = 0, 1, . . . , q − 2. Under the action of C these
characters lie in orbits of size
pqr −1
.
p−1
It follows that
½³
¾
´ µ pqr − 1 ¶
qr−1 i
2
p
cd(CP/P |P/P ) =
| i = 0, 1, . . . , q − 2 .
p−1
q
q
Lemma 5.3 in [16], along with Theorem 6.34 in [5] tell us that the conjugacy classes of
P/P q are fixed by G. Thus, each C-orbit contains a G-invariant character. So, each Corbit is stabilized by G. Since G is cyclic, we have cd(G/P q |P/P q ) = cd(CP/P q |P/P q ).
Therefore, at this stage we have shown that
½
q
cd(G/P ) =
³
1, q, r, qr, p
qr−1
2
´i µ pqr − 1 ¶
p−1
¾
| i = 0, 1, . . . , q − 2 .
q.e.d.
We mentioned above that cd(G) = cd(G/P q ) ∪ cd(G|P q ). Since we have computed cd(G/P q ), to determine the structure of ∆(G) we must find cd(G|P q ). The
55
character degrees in cd(G|P q ) have not been previously computed. As a result, it
will require more work and insight than it did to determine the degrees in cd(G/P q ).
The remaining chapters are comprised of the results needed to calculate the character
degrees in cd(G|P q ).
CHAPTER 7
IRREDUCIBLE CHARACTERS PART II
Before we start the main focus of this chapter, we need a small lemma that will
be used repeatedly throughout the remainder of this paper.
Lemma 7.1 Let N be a normal subgroup of a finite group G. Let λ be a linear
character of N . Fix an element s ∈ G. Then s stabilizes λ if and only if λ([t, s]) = 1
for all t ∈ N .
Proof. Clearly, s stabilizes λ if and only if s−1 stabilizes λ. Also, s stabilizes λ if and
−1
−1
only if (λ)s (t) = λ(t) for all t ∈ N . We know (λ)s (t) = λ(ts ). Thus, s stabilizes λ if
and only if λ(ts ) = λ(t) for all t ∈ N . We observe that λ(ts ) = λ(t[t, s]) = λ(t)λ([t, s]),
as λ is linear. Therefore, s stabilizes λ if and only if λ([t, s]) = 1 for all t ∈ N . q.e.d.
Moving on to looking at stabilizers of certain characters of P q , for i ≥
q+1
2
we
define the subgroup Bi of P i by Bi = {1 + αxi |α ∈ F }. Note that
P
q+1
2
= B q+1 × B q+3 × · · · × Bq−1 × Pq .
2
2
Also, B q+1 ×B q+3 ×· · ·×Bq−1 is a CG - invariant subgroup. Characters in Irr(P
2
2
q+1
2
|P q )
have the form µ q+1 × µ q+3 × · · · × µq−1 × ϕ, where each µi ∈ Irr(Bi ) and 1P q 6= ϕ ∈
2
2
q
Irr(P ).
56
57
Let B be the set of characters ϕ ∈ Irr(P q ) whose kernels do not correspond to
hα, F ii,q−i for all α ∈ F and for i = 1, 2, . . . , q−1
. Recall that in Corollary 4.8, we saw
2
that the subspaces generated by hα, F ii,q−i are independent of i. In other words, the
sets hα, F i1,q−1 as α runs through F are the same as hβ, F ii,q−i as β runs through F .
The next lemma calculates the stabilizer of µ q+1 × µ q+3 × · · · × µq−1 × ϕ in P when
2
2
ϕ ∈ B.
Lemma 7.2 For ϕ ∈ B, the stabilizer of γ = µ q+1 × µ q+3 × · · · × µq−1 × ϕ in P is
2
P
q+1
2
2
. As a result, ϕ is fully ramified with respect to P/P q and γ P is irreducible.
Proof. Fix an element s ∈ P − P
q+1
2
. We need to show that s does not stabilize γ.
From Lemma 7.1, we see that it suffices to show that there exists an element t ∈ P
so that γ([t, s]) 6= 1. We can write s = 1 + αxi + y, where 0 6= α ∈ F , 1 ≤ i ≤
q+1
2
q−1
,
2
and y ∈ J i+1 . Since the kernel of ϕ does not correspond to hα, F ii,q−i , we can find
β ∈ F , so that 1 + hα, βii,q−i xq ∈
/ ker(ϕ). Let t = 1 + βxq−i + z, where z ∈ J q−i+1 .
From Lemma 4.1 in [6], we know that [t, s] = 1 + hα, βii,q−i xq . Hence,
γ(1 + hα, βii,q−i xq ) = ϕ(1 + hα, βii,q−i xq ) 6= 1.
Thus, when ϕ ∈ B we have that P
q+1
2
is the stabilizer of γ in P. In particular, by
Theorem 6.11 of [5], γ P is irreducible. As a result, ϕ is fully ramified with respect to
P/P q .
Now, define Cq to be the subgroup of C of order
q.e.d.
pq −1
.
p−1
This subgroup will be used
throughout the remainder of the paper.
Lemma 7.3 For B defined as above, we have cd(G|B) = { qp
q−1
qr
2
qr
−1
( ppq −1
) }.
58
Proof. From Lemma 7.2, we conclude that the characters in Irr(P |B) all have degree
p
q−1
qr
2
. Let Cq be the subgroup defined above and let Dq be the complement to Cq in
C. Now, Cq centralizes P q , and since the characters in Irr(P |P q ) are fully ramified
with respect to P/P q , it follows that Cq centralizes these characters. On the other
hand, Dq acts Frobeniously on P q , so nonprincipal characters in Irr(P q ) all lie in
orbits of Dq . Hence, characters in Irr(P |P q ) all lie in regular orbits of size
pqr −1
pq −1
under the action of C. We deduce that Irr(CP |B) consists of characters of degree
p
q−1
qr
2
qr
−1
( ppq −1
).
We know from Lemma 4.12 every hyperplane in F has the form aker(T r) for some
0 6= a ∈ F . For c ∈ C, the action of C on the hyperplanes of P q is
pq −1
{1 + βxq | β ∈ aker(T r)} · c = {1 + βc p−1 xq | β ∈ aker(T r)}.
Similarly, for σ ∈ G, the action of G on the hyperplanes in P q is
{1 + βxq | β ∈ aker(T r)} · σ = {1 + β σ xq | β ∈ aker(T r)}.
Thus, the orbits of hyperplanes under the action of C correspond to cosets of Dq in
C. Since the elements of order r in G centralize Cq , this implies the elements of order
r in G will stabilize each orbit of hyperplanes. The only hyperplanes fixed by the
elements of order q in G will be those of the form {1 + βxq | β ∈ aker(T r)}, where
o(a) divides
pr −1
.
p−1
It is not difficult to see that none of these correspond to characters
in B. Thus, G will permute the hyperplanes corresponding to characters in B in orbits
of size q. Different characters in the same CG-orbit must have different kernels, so
the action on the hyperplanes will translate to the action on the characters. Thus, q
divides the size of every orbit under the action of G on Irr(CP |B). Since the subgroup
59
of order r in G centralizes Cq , we conclude that cd(G|B) = {qp
q−1
qr
2
qr
−1
( ppq −1
)}.
q.e.d.
CHAPTER 8
IRREDUCIBLE CHARACTERS PART III
We now focus our attention on those characters ϕ ∈ Irr(P q ) whose kernels correspond to hα, F ii,q−i for some α ∈ F and some 1 ≤ i ≤
q−1
.
2
characters. Notice that A is a union of p − 1 orbits of size
Define A to be this set of
pqr −1
pq −1
under the action of
C. Each C-orbit contains a character whose kernel corresponds to ker(T r).
Fix ϕ ∈ Irr(P q ) so that ker(ϕ) = {1 + αxq | α ∈ ker(T r)}. Thus, there is a oneto-one correspondence between ker(ϕ) and ker(T r). The reader at times may find it
to be more convenient to think of ker(ϕ) as the ker(T r), or in other words, to view ϕ
as a character of F .
Now set A q−1 to be the P -orbit in Irr(P
2
q+1
2
) that contains 1B q+1 × 1B q+3 × · · · ×
2
2
1Bq−1 × ϕ.
As we will see later there are other characters of this form to consider. Specifically,
we need to analyze characters in other P -orbits where certain µi 6= 1Bi . Many of the
calculations that emerge begin to get complicated and require various approaches to
determine stabilizers. The characters in this chapter, however, are relatively speaking
much easier to work with.
To begin our lemmas we need a couple of definitions. First, we define Fq be the
field of order pq . Next, define Q to be the set of all elements 1+α1 x+α2 x2 +· · ·+αq xq
where each αi lies in Fq . We will show in the following lemma that this subgroup plays
60
61
a role in the stabilizers of certain representative characters of A q−1 .
2
Lemma 8.1 Set γ = µ q+1 × 1B q+3 × · · · × 1Bq−1 × ϕ, where µ q+1 ∈ Irr(B q+1 ). Then
2
2
2
q+1
2
the stabilizer in P of γ is the subgroup QP
.
Proof. First, we will show that elements of QP
s ∈ QP
P
q+1
2
q+1
2
2
q+1
2
stabilize γ. Choose an element
. We saw in Lemma 7.1 that it suffices to show that γ([t, s]) = 1 for all t ∈
. Write s = 1+α1 x+α2 x2 +· · ·+αq xq and t = 1+β q+1 x
q+1
2
2
+β q+3 x
q+3
2
2
+· · ·+βq xq ,
where α1 , . . . , α q−1 ∈ Fq and α q+1 , . . . , αq , β q+1 , . . . , βq ∈ F . We know from Lemma
2
2
2
j− q+1
2
5.2 that the coefficients of xj in [t, s] can be written
dk = −βk∗ +
Pk−1
l=1
l
αl (dk−l )p , and k ∗ = k +
q+1
.
2
X
hαj− q+1 , dj− q+1 −i ii,j ∗ −i where
2
i=1
2
Thus, we can write
q+1
[t, s] = 1 +
j− 2
q
X
X
j= q+3
2
hαj− q+1 , dj− q+1 −i ii,j ∗ −i xj .
2
i=1
2
q−1
2
X
Now, γ([t, s]) = 1 if and only if ϕ(1 +
hα q−1 , d q−1 −i ii,q−i xq ) = 1. By definition of
2
i=1
h·, ·ii,q−i , we have
q−1
2
q−1
2
2
X
X
i
q−i
T r( hα q−1 , d q−1 −i ii,q−i ) = T r( (α q−1 dpq−1 −i − αpq−1 d q−1 −i )).
2
i=1
2
i=1
2
2
2
2
Using the linearity of the trace map, we obtain
q−1
q−1
2
2
X
X
q−i
i
pq−i
pi
(T r(α q−1 dpq−1 −i ) − T r(αpq−1 d q−1 −i )).
T r( (α q−1 d q−1 −i − α q−1 d q−1 −i )) =
i=1
2
2
2
2
2
i=1
2
2
Since T r(ap ) = T r(a) for all a ∈ F , we have
q−i
q−i
i
q
i
2
2
T r(αpq−1 d q−1 −i ) = T r((αpq−1 d q−1 −i )p ) = T r(αpq−1 dpq−1 −i ).
2
2
2
2
2
62
q
q−i
Because α q−1 ∈ Fq , we know that αpq−1 = α q−1 , so we have T r(αpq−1 d q−1 −i ) =
2
2
2
2
2
pi
T r(α q−1 d q−1 −i ). This yields
2
q−1
2
X
2
q−1
2
X
i
i
T r(α q−1 d q−1 −i − α q−1 d q−1 −i ) =
(T r(α q−1 dpq−1 −i ) − T r(α q−1 dpq−1 −i )) = 0.
pi
2
i=1
pq−i
2
2
2
We conclude that
i=1
Thus, QP
q+1
2
find t ∈ P
2
2
2
2
P q−1
2
, d q−1 −i ii,q−i x
i=1 hα q−1
2
2
q
∈ ker(ϕ). Therefore, γ([s, t]) = 1 as desired.
is contained in the stabilizer of γ.
q+1
2
q+1
2
2
hα q−1 , d q−1 −i ii,q−i ∈ ker(T r)
We now work to show that QP
s ∈ P − QP
2
q−1
2
X
and hence, 1 +
i=1
q+1
2
is the stabilizer of γ. Consider an element
. We will show that s does not stabilize γ. By Lemma 7.1, it suffices to
so that γ([t, s]) 6= 1. Now ϕ was chosen so that ker(ϕ) = {1 + αxq | α ∈
ker(T r)}. We know from Lemma 4.5 that ker(T r) = hα, F ii,q−i if and only if α ∈ Fq .
There are two cases to consider. First consider the case when s = 1 + αxi + y, where
α∈
/ Fq and y ∈ J i+1 . Then we can choose β ∈
/ Fq so that hα, βii,q−i ∈
/ ker(T r). Let
t = 1+βxq−i , so that [t, s] = 1+hα, βii,q−i xq . Since hα, βii,q−i ∈
/ ker(T r), [t, s] ∈
/ ker(ϕ)
as desired.
Secondly, consider the case when s = 1 + α1 x + α2 x2 + · · · + αi xi + y where
α1 , . . . αi−1 ∈ Fq , αi ∈
/ Fq , and y ∈ J i+1 . Write r = 1 + α1 x + α2 x2 + · · · + αi−1 xi−1 for
r ∈ QP
q+1
2
and z = r−1 (y − (r − 1)αi xi ) ∈ J i+1 . Then
r(1 + αi xi + z) = r + rαi xi + (y − (r − 1)αi xi )
= r + αi xi + y = s.
63
Also, notice that z = r−1 (y + αi xi ) − αi xi ∈ J i+1 . Thus, we can write s = rw where
r ∈ QP
q+1
2
and w = 1 + αi xi + z. Since αi ∈
/ Fq , the previous paragraph shows w
cannot be in the stabilizer of γ. Hence, s is not in the stabilizer of γ. Therefore, the
only elements in P that stabilize γ are the elements in QP
q+1
2
.
q.e.d.
Using a technique similar to that found in [9], the next lemma helps us understand
the characters in A q−1 . The main difference between the argument below and the one
2
used by Lewis is that we now use induction since we have more than one factor group
to consider.
Lemma 8.2 The character 1B q+1 × 1B q+3 × · · · × 1Bq−1 × ϕ has a unique Cq -invariant
2
extension to QP
q+1
2
2
.
Proof. Set Qi = P i ∩ QP
q+1
2
. Observe that Q1 = QP
q+1
2
and Q
q+1
2
= P
q+1
2
. We
will use reverse induction to prove that γ q+1 = 1B q+1 × · · · × 1Bq−1 × ϕ has a unique
2
2
1
Cq -invariant extension to Q . Since γ q+1 is the only extension of γ q+1 to Q
2
2
is Cq -invariant, the base case holds. Now suppose that 1 ≤ i ≤
a unique Cq -invariant extension γi+1 to Q
q+1
2
i+1
i
and
q−1
2
and γ q+1 has
i+1
is irreducible
. We know that Q /Q
2
under the action of Cq , so by Problem 6.12 of [5], we know that either γi+1 extends
to Qi or is fully ramified with respect to Qi /Qi+1 . Since |Qi : Qi+1 | = pq is not a
square, γi+1 is not fully ramified with respect to Qi /Qi+1 . Thus, γi+1 extends to Qi .
By Glauberman’s Lemma, Lemma 13.8 in [5], γi+1 has a Cq -invariant extension to
Qi . On the other hand, by the corollary to Glauberman’s Lemma, Corollary 13.9
in [5], we know that CQi /Qi+1 (Cq ) acts transitively on the Cq -invariant extensions of
γi+1 to Qi . Since CQi /Qi+1 (Cq ) = 1, it follows that γi+1 has a unique Cq -invariant
extension γi to Qi . It follows that γ q+1 will extend to Qi , and since any Cq -invariant
2
64
extension to Qi will lie over a Cq -invariant extension to Qi+1 , we see that γi is the
unique Cq -invariant extension of γ q+1 to Qi . Taking i = 1, we have the desired result.
2
q.e.d.
We now have enough information to produce the character degrees of cd(G | A q−1 ).
2
Lemma 8.3 For A q−1 defined as above, we have cd(G | A q−1 ) =
2
³
³ qr ´ q−1
³2 qr ´
n q−1
´
o
p
−1
p −1
q−3
)(qr−q)
)(qr−q+i)
( 2 )(qr−q) pqr −1
( q−1
(
2
2
p
,
qp
,
p
|
i
=
0,
1,
.
.
.
.
q
q
p −1
p −1
p−1
2
Proof. Recall, Cq is the subgroup of C of order
pq −1
.
p−1
From Lemma 8.2 we know there
is a unique Cq -invariant extension of γ = 1B q+1 × 1B q+3 × . . . × 1Bq−1 × ϕ to QP
2
Call this character ϕ̂ ∈ Irr(QP
q+1
2
q+1
2
.
2
). By Theorem 6.11 in [5], ϕ̂ induces irreducibly to
P . Since ϕ is linear, it has degree 1. Now, ϕ̂P is a Cq -invariant irreducible character
of P whose degree is equal to | P : QP
q+1
2
|, which is p(
q−1
)(qr−q)
2
ϕ̂P is stabilized by P Cq , and using Corollary 6.28 from [5], it has
to P Cq . These characters also have degree p(
induce irreducibly to CP with degree p(
q−1
)(qr−q)
2
q−1
)(qr−q)
2
. In particular,
pq −1
p−1
extensions
. From here, our characters
qr
−1
( ppq −1
). Now, from earlier comments
on the action of G, we know that G stabilizes one of these characters and permutes
the rest in orbits of size q. As a result, we see that this yields character degrees in
cd(G | A q−1 ) of p(
2
q−1
)(qr−q)
2
qr
−1
( ppq −1
) and qp(
q−1
)(qr−q)
2
For the remaining characters in Irr(QP
invariant extension ϕ̂ in QP
q+1
2
q+1
2
, notice that QP
qr
−1
( ppq −1
).
| γ) that are not the unique Cq -
q+1
2
/P
q+1
2
is one of the groups studied
in [16]. Gallagher’s Theorem tells us that there is a bijection from Irr(QP
Irr(QP
q+1
2
| γ). The degrees of the characters in Irr(QP
q+1
2
/P
q+1
2
q+1
2
/P
q+1
2
) to
) were computed in
Theorem 4.4 of [16]. From there, we see that, excluding the unique Cq -invariant extension, in Irr(QP
q+1
2
) we have pi (pq − 1) characters of degree pi(
q−1
)
2
for i = 0, 1, . . . , q−3
.
2
Again, via Theorem 6.11 in [5], each of these characters induces irreducibly to P
65
with degree pi(
q−1
)
2
p(
q−1
)(qr−q)
2
= p(
these characters in orbits of size
q−1
)(qr−q+i)
2
pqr −1
,
p−1
for i = 1, 2 . . . , q−3
. Since C permutes
2
these characters induce irreducibly to CP .
Thus, we see in Irr(CP | A q−1 ), these characters have degrees pi(
q−1
)(qr−q+i)
2
2
qr
−1
( pp−1
)
for i = 1, 2, . . . , q−3
. Notice that G stabilizes these characters, so each extends to G.
2
Therefore, we have cd(G | A q−1 ) =
2
³ qr ´
³
´ q−1
³
´
q−1
q−1
p
−1
( 2 )(qr−q)
( 2 )(qr−q) pqr −1
( 2 )(qr−q+i) pqr −1
{p
, qp
,p
| i = 0, 1, . . . q−3
}.
pq −1
pq −1
p−1
2
q.e.d.
Now, A q−1 was defined to be the P -orbit in Irr(P
2
q+1
2
) that contains 1B q+1 ×1B q+3 ×
2
2
q
· · ·×1Bq−1 ×ϕ, where ϕ is a fixed irreducible character of P whose kernel corresponds
to the kernel of the trace map. We continue with ϕ fixed and we now define A q+1 to
2
be the P -orbits in Irr(P
q+1
2
) that contain characters of the form µ q+1 × 1B q+3 × . . . ×
2
2
1Bq−1 × ϕ but nothing in A q−1 .
2
Before we move on to the next lemma, we first provide a definition of certain
characters in Bi . Recall that P q ∼
= Bi ∼
= F . Thus, characters of Bi can be thought
of as characters of F or characters of P q . With this in mind, given α ∈ F and an
integer i ≥ 1, we define ϕi,q−i
∈ Irr(Bq−i ) by
α
ϕi,q−i
(1 + βxq−i ) = ϕ(1 + hα, βii,q−i xq ).
α
We will show in Lemma 8.5 that this map is, in fact a homomorphism. We need an
additional definition for the next lemma. We define Bq−i,q = {1 + βxq−i | β ∈ Fq }.
The importance of this set is its relationship to the kernel of the map above.
Lemma 8.4 Fix α ∈ F and i ≥
α ∈ F.
q+1
.
2
) for all i and for all
Then Bq−i,q ⊆ ker(ϕi,q−i
α
66
Proof. Let β ∈ Fq . Using the linearity of the trace map,
i
q−i
T r(hα, βii,q−i ) = T r(αβ p − αp
i
q−i
β) = T r(αβ p ) − T r(αp
β).
Since T r(ap ) = T r(a) for all a ∈ F , we have
i
i
q−i
q−i
T r(αβ p ) = T r((αβ p )p
) = T r(αp
q−i
q−i
q
β p ).
Hence,
i
T r(αβ p − αp
q−i
β) = T r(αp
q
β p − αp
q−i
β) = T r(αp
q
q−i
β p − αp
β).
q
Because β ∈ Fq , we obtain β p = β and thus,
q−i
T r(αp
q
q−i
β p − αp
q−i
β) = T r(αp
q−i
β − αp
β) = T r(0) = 0.
We combine these to obtain T r(hα, βii,q−i ) = 0. We now see that ϕi,q−i
(1 + βxq−i ) =
α
ϕ(1 + hα, βii,q−i xq ) = 1 since ker(ϕ) = {1 + αxq | α ∈ ker(T r)} and hα, βii,q−i ∈
ker(T r). We conclude that Bq−i,q ⊆ ker(ϕi,q−i
).
α
q.e.d.
In order to determine any character degrees in cd(G | A q+1 ), it will be useful to
2
have the following lemma. In this lemma, we will show that the map α 7→ ϕi,q−i
α
is a homomorphism, and we will determine its image and kernel. This will help us
identify the degrees in cd(G | A q+1 ).
2
. Then ∗ is a homoLemma 8.5 Define the map ∗ : F −→ Irr(Bq−i ) by α 7→ ϕi,q−i
α
morphism with kernel Fq and image Irr(Bq−i /Bq−i,q ).
67
Proof. For α, β ∈ F we have
q−i
ϕi,q−i
) = ϕ(1 + hα + β, γii,q−i xq )
α+β (1 + γx
= ϕ(1 + hα, γii,q−i xq )ϕ(1 + hβ, γii,q−i xq )
since h·, ·ii,q−i is bilinear and ϕ is a homomorphism, and
(1 + γxq−i )ϕi,q−i
(1 + γxq−i )
ϕ(1 + hα, γii,q−i xq )ϕ(1 + hβ, γii,q−i xq ) = ϕi,q−i
α
β
by definition of ϕi,q−i
. Thus ∗ is a homomorphism.
α
By Lemma 8.4, Im(∗) ⊆
Irr(Bq−i /Bq−i,q ). If α ∈ Fq , then
i
q−i
T r(hα, βii,q−i ) = T r(αβ p − αp
We know that T r(αp
T r(αp
q−i
q−i
q−i
β) = T r((αp
i
q−i
β) = T r(αβ p ) − T r(αp
i
q
β).
i
q
β)p ) = T r(αp β p ). Since α ∈ Fq , αp = α so
i
β) − T r(αβ p ) = 0. Thus, α ∈ ker(∗), and Fq ⊆ ker(∗).
If α ∈ ker(∗), then ϕi,q−i
= 1. This implies ϕi,q−i
(1 + βxq−i ) = 1 for all β ∈ F .
α
α
We see that ϕ(1 + hα, βii,q−i xq ) = 1 and hence hα, βii,q−i ∈ ker(T r) for all β ∈ F ,
which implies α ∈ Fq by Lemma 4.5. Therefore, Fq = ker(∗), so | ker(∗) |= pq , and
| Im(∗) |=
pqr
|F |
= q = pqr−q =| Irr(Bq−i /Bq−i,q ) | .
| ker(∗) |
p
We conclude that ∗ maps onto Irr(Bq−i /Bq−i,q ).
q.e.d.
CHAPTER 9
A SPECIFIC EXAMPLE
In this chapter we will demonstrate some of the structure of the group we will call
G3 that will be constructed based on fixing the prime q = 3. The primes p and r will
be general, but will satisfy the necessary parameters outlined in the beginning of this
paper. In the case when q = 3, the character degrees arise relatively easily. When
we move up from this case to q > 3 the character degrees become more difficult to
obtain, as we will see in the next chapter. In this chapter, we show that one character
degree in addition to previous ones appears for irreducible characters in A q+1 , where
2
A q+1 is defined as in the previous chapter. Specifically, with q = 3 this will be A2 .
2
The reader will see that this additional degree will not change the diameter of the
character degree graph of our group G.
In order to obtain the additional character degree, we need one more lemma before
we can demonstrate what occurs when q = 3. This next lemma will be extended later
to general Ai , but is needed here before we go further. Recall B q+1 ,q = {1 + βx
2
q+1
2
|
β ∈ Fq }. This is a special subgroup that is needed again in the next chapter. Notice
in the case when q = 3, we have B2,3 = {1 + βx2 | β ∈ F3 }.
Lemma 9.1 For A q+1 defined as above,
2
µ qr
¶¾
½
p −1
)(qr−q+ q−1
)
( q−1
2
2
.
cd(G | A q+1 ) = p
2
p−1
68
69
Proof. Fix 0 6= α ∈ F . Suppose s = 1 + αx
P
q+1
2
q−1
2
and t = 1 + β q+1 x
q+1
2
2
+ · · · + β q xq ∈
−1
. Let γ = µ q+1 × 1B q+3 × · · · × 1Bq−1 × ϕ. We next consider γ s (t) = γ(ts ) =
2
2
γ(t)γ([t, s]). Thus, it suffices to determine γ([t, s]). Also, from Lemma 5.3 we know
that [t, s] = 1 + hα, β q+1 i q−1 , q+1 xq . Thus, γ([t, s]) = ϕ(1 + hα, β q+1 i q−1 , q+1 xq ). Now,
2
by definition of ϕα
2
q−1 q+1
, 2
2
2
2
2
2
, we have that
γ(1 + hα, β q+1 i q−1 , q+1 xq ) = ϕα
2
2
q−1 q+1
, 2
2
2
(1 + β q+1 x
q+1
2
2
).
Then
(γ)s
−1
= (µ q+1 × 1B q+3 × · · · × 1Bq−1 × ϕ) · (ϕα
2
q−1 q+1
, 2
2
2
× 1B q+3 × · · · × 1P q ).
2
Thus, we have
−1
(γ)s
Recall B q+1 = {1 + βx
= ϕα
q+1
2
2
q−1 q+1
, 2
2
µ q+1 × 1B q+3 × · · · × 1Bq−1 × ϕ.
2
2
| β ∈ F }. By Lemma 8.5, we know that going through all
possible choices of α ∈ F , then ϕα
q−1 q+1
, 2
2
runs through all characters in Irr(B q+1 /B q+1 ,q )
2
in Irr(B q+1 ). Define D q+1 ,q = {1 + βx
2
2
q+1
2
2
| β ∈ [F, σ]}. Since F = Fq u [F, σ],
B q+1 ∼
= F , and B q+1 ,q ∼
= Fq , we have that B q+1 = B q+1 ,q u D q+1 ,q . Hence Irr(B q+1 ) =
2
2
2
2
2
2
˙
q+1 /D q+1 ), and the coset µ q+1 Irr(B q+1 /B q+1 ) contains a
Irr(B q+1 /B q+1 ,q )×Irr(B
,q
,q
2
2
2
2
2
2
2
unique character µ̂ q+1 ∈ Irr(B q+1 /D q+1 ,q ), where D q+1 ,q ⊆ ker(µ̂ q+1 ). Observe that
2
2
2
2
2
A q+1 consists of the orbits identified with the cosets µ q+1 Irr(B q+1 /B q+1 ,q ) where
2
2
2
2
B q+1 ,q * ker(µ q+1 ). Hence, γ is conjugate to γ̂ = µ̂ q+1 × 1B q+3 × · · · × 1Bq−1 × ϕ
2
2
2
2
in A q+1 .
2
We define Qi = P i ∩ Q, which is different from the definition of Qi in Lemma
70
8.2. From Lemma 8.1, we know that the stabilizer in P of γ̂ is QP
1B q+3 × · · · × 1Bq ∈ Irr(P
q+1
2
q+3
2
/(D q+1 ,q u P
2
2
1B q+3 × · · · × 1Bq−1 × ϕ extends to QP
q+1
2
2
P
q+1
2
P
q+3
2
=
B q+1 ,q u P
2
P
q+3
2
Q/Q
q+3
2
. Now, µ̂ q+1 ×
2
)) and by Lemma 8.2, γ̂
Q
q+1
2
= 1B q+1 ×
2
. Since B q+1 = B q+1 ,q u D q+1 ,q , we have
2
q+3
2
u
2
D q+1 ,q u P
2
q+3
2
.
2
P
q+3
2
We also know that Q ∩ B q+1 = B q+1 ,q . Thus, we have QP
2
q+1
2
q+1
2
2
/(D q+1 ,q u P
. With this in mind, we apply Gallagher’s Theorem to see Irr(Q/Q
bijection with Irr(QP
q+1
2
as characters in Irr(Q/Q
| γ̂
q+3
2
q+1
Q 2
q+3
2
). Hence, we can view characters in Irr(QP
| µ̂ q+1 ). Notice that Q/Q
2
q+3
2
q+3
2
2
) ∼
=
) is in
q+1
2
| γ̂)
is one of the groups studied
by Riedl in [16].
Using Riedl’s results, we have that the characters in Irr(Q/Q
all have degree p(
degree p(
p(
q−1
)( q−1
)
2
2
q−1
)(qr−q+ q−1
)
2
2
q−1
)( q−1
)
2
2
. Thus, the characters in Irr(QP
q+1
2
q+1
2
| Q
q+1
2
/Q
q+3
2
)
| γ̂) will all have
. Using Clifford theory, these induce irreducibly to P of degree
. As we saw in the proof of Lemma 8.3, under the action of C, these
characters are permuted in orbits of size
pqr −1
p−1
and hence, induce irreducibly to CP .
So, for characters in Irr(CP | A q+1 ), the degree of each is p(
q−1
)(qr−q+ q−1
)
2
2
2
qr
−1
( pp−1
).
Now, G stabilizes each of these characters. Thus, we have
½
µ qr
¶¾
q−1
p −1
( q−1
)(qr−q+
)
2
cd(G | A q+1 ) = p 2
.
2
p−1
q.e.d.
At this point we have enough results to calculate the character degrees for the
case when q = 3. We choose p to be any prime and choose r > 3 so that 3 and r do
not divide
p3r −1
.
p−1
71
We begin by recalling the construction of these groups. To construct G3 we see
that our field F now has order p3r and the ring R is obtained by factoring out the
ideal generated by X 4 . The subgroup P is the set of all elements 1+α1 x+α2 x2 +α3 x3 ,
where αi ∈ F for i = 1, 2, 3. There are p3r choices for each αi so that | P |= p9r . We
take C to be the subgroup of F × of order
p3r −1
p−1
and T to be the semidirect product
of C acting on P , by the somewhat nonstandard action by automorphisms defined in
Chapter 3. Lastly, our group G3 is then completed by taking the semidirect product
of G, the Galois group of F over the prime subfield, acting on T . In the end, G3 has
3r
−1
) which, as was mentioned earlier, is very large even if p = 2.
order 3rp9r ( pp−1
A point to make here is that when we consider the groups P i = 1+J i for purposes
of calculating character degrees, we have the same number of subgroups P i as in [9].
In particular, the stabilizers of the characters in [9] were found to be the groups P 2
and QP 2 . The key is the group P
q+1
2
and where it lies relative to P , which in turn
depends on the value of q. In other words, the number of groups P i that contain
P
q+1
2
and are contained in P directly affects the difficulty of calculating the character
degrees. The reader will be able to see this more clearly in the next chapter.
We proceed in the order of previous chapters and we now look to the character
degrees of G3 /P 3 . Using the result from Lemma 6.2, we have that
½
3
cd(G3 /P ) =
p3r − 1 3r−1
1, p, 3, r, 3r,
,p 2
p−1
µ
p3r − 1
p−1
¶¾
.
We now partition the nonprincipal characters of Irr(P 3 ) into two sets. We will let
A be the set of ϕ ∈ Irr(P 3 ) whose kernels correspond to hα, F i1,2 for some 0 6= α ∈ F ,
and we will let B be the remaining nonprincipal characters in Irr(P 3 ).
72
Define B2 = {1 + βx2 | β ∈ F }, which is a subgroup of P 2 . Also notice that,
as in [9], P 2 = B2 × P 3 . Characters in Irr(P 2 | P 3 ) have the form µ2 × ϕ where
µ2 ∈ Irr(B2 ) and 1P 3 6= ϕ ∈ Irr(P 3 ).
From Lemma 7.2, with ϕ ∈ B, we know that the stabilizer in P of µ2 × ϕ is P 2 .
In particular, ϕ is fully ramified with respect to P/P 3 and (µ2 × ϕ)P is irreducible.
3r
−1
Applying Lemma 7.3, we have that cd(G3 |B) = {3p3r ( pp3 −1
)}.
Recalling the definition of A, we set A1 to be the P -orbit in Irr(P 2 ) that contains
1B × ϕ, where ϕ ∈
/ B. It is worthwhile to remind the reader that by Lemma 8.1, we
know the stabilizer of 1B2 × ϕ is the subgroup QP 2 , as in [9], and as we mentioned
above. The character degrees in cd(G3 |A1 ), using Lemma 8.3, are known to be
½
cd(G3 | A1 ) =
µ
3r−3
p
p3r − 1
p3 − 1
¶
µ
3r−3
, 3p
p3r − 1
p3 − 1
¶
µ
,p
3r−3
p3r − 1
p−1
¶¾
.
Lastly, we define A2 to be the P -orbits in Irr(P 2 ) that contain characters of the
form µ2 × ϕ but contain nothing from A1 . Again, we mention by Lemma 8.1, we
know that the stabilizer in P of µ2 × ϕ is QP 2 . By Lemma 9.1, we have
½
µ 3r
¶¾
p −1
3r−2
cd(G3 | A2 ) = p
.
p−1
We need to confirm one more piece of necessary information before we combine
our character degrees. We claim that cd(G | A) = cd(G | A1 ) ∪ cd(G | A2 ). Given
−1
the character µ2 × ϕ, let s = 1 + αx and t = 1 + β2 x2 + β3 x3 . Consider (µ2 × ϕ)s (t).
−1
We know (µ2 × ϕ)s (t) = (µ2 × ϕ)(ts ) = (µ2 × ϕ)(t[t, s]) = (µ2 × ϕ)(t)(µ2 × ϕ)([t, s]).
73
Using an argument similar to that in [9], we have
(µ2 × ϕ)(t)(µ2 × ϕ)([t, s]) = (µ2 × ϕ)(1 + β2 x2 + (β3 + hα, β2 i1,2 )x3 )
= µ2 (1 + β2 x2 )ϕ(1 + (β3 + hα, β2 i1,2 )x3 ).
On the other hand
2
3
2,1
(µ2 ϕ2,1
α × ϕ)(t) = µ2 ϕα (1 + β2 x )ϕ(1 + β3 x )
= µ2 (1 + β2 x2 )ϕ(1 + (β3 + hα, β2 i1,2 )x3 ).
From this we see that s stabilizes µ2 × ϕ if and only if ϕ2,1
α = 1B2 . This occurs only
if ker(ϕ) = {1 + βx3 |β ∈ hα, F i1,2 }. If µ2 = 1B2 then µ2 × ϕ ∈ A1 . Otherwise
µ2 × ϕ ∈ A 2 .
Combining all of these results, we have
½
cd(G3 ) =
p3r − 1 3r−1
1, 3, r, 3r,
,p 2
p−1
µ
3p
3r−3
p3r − 1
p3 − 1
µ
¶
p3r − 1
p−1
µ
,p
3r−3
¶
p3r − 1
p−1
µ
, 3p
3r
p3r − 1
p3 − 1
¶
µ
,p
3r−2
¶
µ
,p
p3r − 1
p−1
3r−3
p3r − 1
p3 − 1
¶
,
¶¾
.
Though we do not show the character degree graph here, there are some points
worth mentioning. Notice that p, q, and r are prime divisors of degrees in cd(G3 ). In
order to determine the remaining primes in ρ(G3 ), we need to determine the primes
that divide
p3r −1
p−1
and
p3r −1
.
p3 −1
In Chapter 10 we will discuss these sets of primes in the
general case, as well as which vertices they represent in the graph ∆(G). The graph
74
we show there can be easily applied to the case when q = 3.
Many of the techniques used to conquer the case when q = 3 can be attributed
to Lewis in [9], who partly employed results of Riedl’s in [16]. Thus, this specific
example discussed here is relatively straightforward, as was expected. Moving beyond
this example to q > 3, however, takes much more effort to prove. Patterns do emerge
that make it possible to prove the general case. However, alternative techniques are
required to identify these patterns and get the desired result.
CHAPTER 10
IRREDUCIBLE CHARACTERS PART IV
For this chapter we continue to focus on characters in A. However, we will study
characters not in A q−1 or A q+1 . The characters we will look at will be characters of
2
2
the form
θ = µ q+1 × µ q+3 × · · · × µq−1 × ϕ,
2
2
where µi ∈ Irr(Bi ) and 1 6= ϕ ∈ Irr(P q ) is such that ker(ϕ) = {1+αxq | α ∈ ker(T r)}.
Recall we chose σ ∈ G such that the order of σ is r. We should also remind the reader
that the majority of the results of this chapter only apply to cases when the prime q
is larger than 3.
Lemma 10.1 Fix i, j ∈ Z+ so that i + j < q. If α ∈ Fq , then hα, [F, σ]ii,j = [F, σ]
and hα, F ii,j = hα, Fq ii,j u [F, σ].
Proof. Note that by Fitting’s lemma, F = CF (σ) u [F, σ] and CF (σ) = Fq , so
F = Fq u [F, σ]. Consider
hα, F ii,j = hα, Fq u [F, σ]ii,j = hα, Fq ii,j + hα, [F, σ]ii,j .
75
76
q
Now, hα, Fq ii,j ⊆ Fq since α ∈ Fq . Also, [F, σ] = {γ − γ p | γ ∈ F }. We now show
that hα, [F, σ]ii,j ⊆ [F, σ]. Using the bilinearity of h·, ·ii,j we have
q
q
hα, γ − γ p ii,j = hα, γii,j − hα, γ p ii,j .
Since α ∈ Fq , we can write
q
q
q
hα, γii,j − hα, γ p ii,j = hα, γii,j − hαp , γ p ii,j .
And finally, since F has characteristic p, we have
q
q
q
hα, γii,j − hαp , γ p ii,j = hα, γii,j − (hα, γii,j )p ∈ [F, σ].
Since hα, Fq ii,j ⊆ Fq , hα, [F, σ]ii,j ⊆ [F, σ], and Fq ∩ [F, σ] = 1, we know that
hα, F ii,j = hα, Fq ii,j u hα, [F, σ]ii,j .
Next, notice that
hα, F ii,j = hα, Fq ii,j u hα, [F, σ]ii,j ⊆ hα, Fq ii,j u [F, σ].
Since hα, Fq ii,j is a hyperplane in Fq , by Lemma 4.5, it follows that hα, Fq ii,j u [F, σ]
is a hyperplane in Fq u [F, σ] = F . We also know by Lemma 4.5 that hα, F ii,j is a
hyperplane in F . Hence, we must have equality. Therefore, hα, F ii,j = hα, Fq ii,j u
[F, σ] and hα, [F, σ]ii,j = [F, σ].
q.e.d.
77
Given α ∈ F , positive integers i and j, and µi+j ∈ Irr(Bi+j ) we define (µi+j )i,j
α ∈
Irr(Bj ) by
j
i+j
(µi+j )i,j
).
α (1 + βx ) = µi+j (1 + hα, βii,j x
Notice the differences between this definition and ϕi,q−i
defined in Chapter 8.
α
Lemma 10.2 Fix i, j ∈ Z+ so that
q+1
2
≤ i+j ≤ q −1 and fix α ∈ Fq . Choose c ∈ Z+
so that c ≡ (pi − 1)−1 mod(o(α)). Set b ≡ c(pj − 1)mod(o(α)). Let µi+j ∈ Irr(Bi+j )
such that ker(µi+j ) = {1 + βxi+j | β ∈ hα, F ii,j }. Define ∗ : Fq −→ Irr(Bj ) by
∗(γ) = (µi+j )γ i,j . Then ∗ is a homomorphism with kernel αZp and image Irr(Bj /E),
where E = {1 + βxj | β ∈ αb Zp u [F, σ]}.
Proof. Let γ1 , γ2 ∈ Fq . By definition,
∗(γ1 + γ2 )(·) = (µi+j )γ1 +γ2 i,j (·) = µi+j (1 + hγ1 + γ2 , ·ii,j xi+j ).
Since µi+j is itself a homomorphism, we have
µi+j (1 + hγ1 + γ2 , ·ii,j xi+j ) = µi+j (1 + hγ1 , ·ii,j xi+j )µi+j (1 + hγ2 , ·ii,j xi+j ).
Again, by definition of ∗, we have
µi+j (1 + hγ1 , ·ii,j xi+j )µi+j (1 + hγ2 , ·ii,j xi+j ) = ∗(γ1 ) ∗ (γ2 ).
Thus, ∗ is a homomorphism.
We now determine the kernel of ∗. Let β ∈ αZp . Then
∗(β)(·) = (µi+j )β i,j (·) = µi+j (1 + hβ, ·ii,j xi+j ).
78
Since β ∈ αZp , we have that 1 + hβ, ·ii,j xi+j ∈ ker(µi+j ). Hence αZp ⊆ ker(∗).
On the other hand, let 0 6= β ∈ ker(∗). Then
∗(β)(·) = (µi+j )β i,j (·) = µi+j (1 + hβ, ·ii,j xi+j ) = 1.
This implies that hβ, F ii,j ⊆ hα, F ii,j . Since these are both hyperplanes in F , we
must have equality. Thus, β ∈ αZp . With containment both directions we have
ker(∗) = αZp .
We know by Lemma 4.6 that hα, F ii,j = α
know hF, αb ii,j = hα, F ij,i = α
i+j
b( p j −1 )
p −1
pi+j −1
pi −1
ker(T r). By the same lemma we
ker(T r). Thus, there exists γ ∈ F such that
hγ, αb ii,j = hα, γii,j . As a result,
b j
b
i+j
(µi )i,j
) = µi (1 + hα, γii,j xi+j ) = 1.
γ (1 + α x ) = µi (1 + hγ, α ii,j x
Hence, {1 + βxi+j | β ∈ αb Zp } ⊆ ker((µi+j )i,j
γ ). Also, we have chosen µi+j so that
ker(µi+j ) = {1+βxi+j | β ∈ hα, F ii+j }. Notice that by Lemma 10.2, this is equivalent
to saying that {1 + βxi+j | β ∈ [F, σ]} ⊆ ker(µi+j ). By Lemma 10.1, this implies that
{1 + βxi+j | β ∈ [F, σ]} ⊆ ker((µi+j )α i,j ).
Comparing the sizes of Fq /αZp and Irr(Bj /E), we see that | Fq : αZp |= pq−1 . On
the other hand
¯
¯
¯
¯
F
¯.
| Irr(Bj /E) |=| Bj : E |= ¯¯
αZp u [F, σ] ¯
Recall that Fq ∼
= F/[F, σ] which, of course, has size pq . This combined with the
fact that | αZp |= p, we have | F : αb Zp u [F, σ] |= pq−1 . Therefore, by the First
Isomorphism Theorem, our map ∗ is an isomorphism with image Irr(Bj /E).
q.e.d.
79
The importance of this next lemma is to be able to characterize Fq in terms of
hyperplanes. This will become a useful tool when it comes to finding the character
degrees.
Lemma 10.3 Fix i ∈ Z+ so that
pq −1
.
p−1
Let b =
pi −1
.
p−1
q+1
2
≤ i ≤ q − 1 and fix α ∈ Fq so that o(α) divides
Then Fq = αb Zp u hα, Fq i1,i−1 .
Proof. Within this proof we now denote T r for the trace map from Fq to Zp . Since
hα, Fq i1,i−1 is a hyperplane in Fq and αb Zp has order p, it suffices to prove that αb does
pi −1
not lie in hα, Fq i1,i−1 . By Lemma 4.7, hα, Fq i1,i−1 = α p−1 ker(T r) = αb ker(T r), where
we consider T r to be the trace map on Fq . We need to show that αb is not in αb ker(T r),
which is equivalent to showing that 1 is not in ker(T r). We see that T r(1) = q and
since p does not divide q, we have 1 6∈ ker(T r). Therefore, αb 6∈ αb ker(T r).
q.e.d.
Alternatively, in the proof of Lemma 10.3 we could have compared the size of Fq
to that of αb Zp + hα, Fq ii,j−i to show that this is a direct sum.
In order to determine stabilizers of the characters we are studying in this chapter,
we need to calculate conjugates of those characters. Thus, we need a way to characterize these conjugates. The next two lemmas rely on results in Chapter 5 to make
this calculation much simpler.
Lemma 10.4 Let γ = µ q+1 × µ q+3 × · · · × µq−1 × ϕ, where µi ∈ Irr(Bi ) for i =
2
q+1
,...,q
2
− 1, and let s = 1 + αx
−1
γs
2
m
with α ∈ F and 1 ≤ m ≤
q−1
.
2
Then
= γ · (ν q+1 × ν q+3 × · · · × νq−m−1 × ϕm,q−m
× 1Bq−m+1 × · · · × 1P q ),
α
where νi ∈ Irr(Bi ).
2
2
80
Proof. Consider t ∈ P
q+1
2
−1
. We know that γ s (t) = γ(ts ) = γ(t[t, s]) = γ(t)γ([t, s]),
since γ is linear. Thus, it suffices to compute γ([t, s]). We write t as in Lemma 5.1.
Using notation from Lemma 5.1 for [t, s], we have
γ([t, s]) = µ q+1 (1 + δ1 x
q+1
2
2
)µ q+3 (1 + δ2 x
q+3
2
2
) · · · µq−1 (1 + δ q−3 xq−1 )ϕ(1 + δ q−1 xq ).
2
2
By Lemma 5.3, we see for i = 1, 2, . . . , q − 1 that δi has the form hα, dq−i+1 ii,q−i
where dq−i+1 is a polynomial in α and β q+1 , . . . , βq−m−1 . Also by Lemma 5.3, δ q−1 =
2
2
hα, βq−m im,q−m + e where e is also a polynomial in α and β q+1 , . . . , βq−m−1 . Thus,
2
γ([t, s]) will be the product of various characters defined over β q+1 , . . . , βq−m−1 times
2
ϕ(1 + hα, βq−m im,q−m xq ). Gathering together the like terms, we obtain
γ([t, s]) = ν q+1 (1 + β q+1 x
2
q+1
2
2
) · · · νq−m−1 (1 + βq−m−1 xq−m−1 )ϕαm,q−m (1 + βq−m xq−m ),
−1
for some characters νi ∈ Irr(Bi ). Since γ s (t) = γ(t)γ([t, s]), we have
−1
γs
= γ · (ν q+1 × ν q+3 × · · · × νq−m−1 × ϕm,q−m
× 1Bq−m+1 × · · · × 1P q ),
α
2
2
where νi ∈ Irr(Bi ).
q.e.d.
Lemma 10.5 Let γ = µ q+1 × · · · × µi × 1Bi+1 × · · · × 1Bq−1 × ϕ, where µj ∈ Irr(Bj )
2
for j =
q+1
, . . . , i,
2
γs
−1
and let s = 1 + αxm with α ∈ Fq and 1 ≤ m ≤
q−1
.
2
Then
= γ · (ν q+1 × ν q+3 × · · · × νi−m−1 × (µi )αm,i−m × 1Bi−m+1 · · · × 1P q ),
2
2
for some characters νj ∈ Irr(Bj ).
81
Proof. Consider t ∈ P
q+1
2
. We saw in the proof of Lemma 10.4 that it suffices to
−1
compute γ([t, s]) since γ s (t) = γ(t)γ([t, s]). Write t as in Lemma 5.1. Using the
notation from Lemma 5.1 for [t, s], we have
γ([t, s]) = µ q+1 (1 + δ1 x
q+1
2
2
)µ q+3 (1 + δ2 x
2
q+3
2
) · · · µi (1 + δi− q−1 xi )ϕ(1 + δ q−1 xq ).
2
2
By Lemma 5.3,we have that δ q−1 = hα, dq−m+1 im,q−m . Since α ∈ Fq , we know that
2
1 + hα, dq−m+1 im,q−m xq ∈ ker(ϕ). We have γ([t, s]) = µ q+1 (1 + δ1 x
2
q+1
2
) · · · µi (1 +
δi− q−1 xi ). Now, δ1 , . . . , δi− q−3 are polynomials in α and β1 , . . . , βi−m−1 and δi− q−1
2
2
2
is hα, βi−m im,i−m + e, where e is a polynomial in the same variables. Thus, γ([t, s])
will be the product of various characters defined over β q+1 , . . . , βi−m−1 times µi (1 +
2
hα, βi−m im,i−m xi ). Gathering together the like terms we obtain
γ([t, s]) = ν q+1 (1 + β q+1 x
2
q+1
2
2
) · · · νi−m−1 (1 + βi−m−1 xq−m−1 )µαm,i−m (1 + βi−m xi−m ),
−1
for some characters νi ∈ Irr(Bi ). Again, since γ s (t) = γ(t)γ([t, s]), we have
γs
−1
= γ · (ν q+1 × ν q+3 × · · · × νi−m−1 × (µi )αm,i−m × 1Bi−m+1 · · · × 1P q ),
2
2
for some characters νj ∈ Irr(Bj ).
q.e.d.
Recall by the definitions of Bq−i and Bq−i,q , we can say that Bq−i ∼
= F and
Bq−i,q ∼
= Fq . We also know that F = Fq u [F, σ]. Thus, Bq−i,q is complemented in
Bq−i . This leads us to define Dq−i,q = {1 + βxq−i | β ∈ [F, σ]}. Using this we can
write Bq−i = Bq−i,q u Dq−i,q . We use this fact in proving the next lemma.
Lemma 10.6 Fix i ∈ Z+ with
q+1
2
≤ i ≤ q − 1 and α ∈ Fq so that o(α) |
pq −1
.
p−1
Suppose µi ∈ Irr(Bi ) such that ker(µi ) = {1+βxi | β ∈ hα, F i1,i−1 }. Let θ ∈ Irr(P
q+1
2
)
82
with
θ = µ q+1 × · · · × µi × 1Bi+1 × · · · × 1Bq−1 × ϕ.
2
Then θ is conjugate to
µ̂ q+1 × · · · × µ̂i × 1Bi+1 × · · · × 1Bq−1 × ϕ,
2
where {1 + βxl | β ∈ hα, F i1,l−1 } ⊆ ker(µ̂l ) for l =
q+1 q+3
, 2 ,...,i
2
− 1, and µ̂i = µi .
Proof. We find a sequence of characters θ1 = θ, θ2 , . . . , θi so that
θj = µjq+1 × · · · × µji−j × µ̂i−j+1 × · · · × µ̂i × 1Bi+1 · · · × 1Bq−1 × ϕ,
2
where µjl ∈ Irr(Bl ) for l =
q+1
,...,i
2
− j. For j = 1, . . . , i − 1, we show that θj
is P -conjugate to θj+1 . By the transitivity of conjugacy, this will show that θ is
P -conjugate to θi , which has the desired form.
We consider θj , and we look at s1 = 1 + γxq−i+j for any element γ ∈ F . We will
fix γ later. By Lemma 10.4, we have that
(θj )s1
−1
= θj · (ν q+1 × · · · × νi−j−1 × ϕγq−i+j,i−j × 1Bi−j+1 × · · · × 1P q ).
2
Writing out the components of θj , we see that
(θj )s1
−1
= µj q+1 ν q+1 × · · · × µji−j−1 νi−j−1 × µji−j ϕγq−i+j,i−j × µ̂i−j+1 ×
2
2
· · · × µ̂i × 1Bi+1 × · · · × 1Bq−1 × ϕ.
83
In the i − j coordinate, we have the character µji−j ϕγq−i+j,i−j . By Lemma 8.5, we know
that going through all possible choices of γ, ϕγq−i+j,i−j runs through all characters in
Irr(Bi−j /Bi−j,q ). Recall Di−j,q = {1 + βxi−j | β ∈ [F, σ]}. We know that Bi−j =
˙
Bi−j,q u Di−j,q . Thus, Irr(Bi−j ) = Irr(Bi−j /Bi−j,q )×Irr(B
i−j /Di−j,q ), and hence, the
coset µji−j Irr(Bi−j /Bi−j,q ) contains a unique character µ∗i−j ∈ Irr(Bi−j /Di−j,q ).
Fix γ ∈ F so that µji−j ϕq−i+j,i−j
= µ∗i−j . Notice that this fixes s1 . For l =
γ
q+1
,...,i
2
(θj )s1
−1
− j − 1, set µ0l = µjl νl so that
= µ0q+1 × · · · × µ0i−j−1 × µ∗i−j × µ̂i−j+1 × · · · × µ̂i × 1Bi+1 × · · · × 1Bq−1 × ϕ.
2
Let s2 = 1 + δxj where δ is any element in Fq . As before, we mention that we will fix
δ later. By Lemma 10.5, we have
−1
(θjs1 )s2
−1
= θjs1
−1
· (κ q+1 × · · · × κi−j−1 × (µ̂i )δj,i−j × 1Bi−j+1 × · · · × 1P q ).
2
s−1
−1
We see that the i − j coordinate of (θj 1 )s2 will contain the character µ∗i−j (µ̂i )δj,i−j .
We know ker(µ̂i ) = {1 + βxi | β ∈ hα, F i1,i−1 }, and by Lemma 4.10, we see
j
−1
)(mod o(α)). Let c ≡
that ker(µ̂i ) = {1 + βxi | β ∈ hαb , F i1,i−1 } where b ≡ ( pp−1
i−j
( p p−1−1 )(mod o(α)) and E = {1+βxi−j | β ∈ αc Zp +[F, σ]}. By Lemma 10.2, we know
that as δ runs over Fq , (µ̂i )j,i−j
runs through the set Irr(Bi−j /E). By Lemma 10.3, we
δ
know that Fq = αc Zp + hα, Fq i1,i−j−1 . Let E 0 = {1 + βxi−j | β ∈ hα, Fq i1,i−j−1 }. We
see that Bi−j = E + E 0 and E ∩ E 0 = Di−j,q , so Bi−j /Di−j,q = E/Di−j,q u E 0 /Di−j,q .
0
˙
Hence, we have Irr(Bi−j /Di−j,q ) = Irr(Bi−j /E)×Irr(B
i−j /E ), and thus, the coset
µ∗i−j Irr(Bi−j /E) has a unique character µ̂i−j ∈ Irr(Bi−j /E 0 ). Notice that µ̂i−j has the
property needed in the conclusion.
84
Fix δ ∈ Fq so that µ∗i−j (µ̂i )j,i−j
= µ̂i−j . It now follows that
δ
(s2 s1 )−1
= µj+1
q+1 × · · · × µ̂i × 1Bi+1 × · · · × 1Bq−1 × ϕ,
θj
2
as desired.
q.e.d.
We now have enough results to determine the location of the stabilizer of these
types of characters within our group.
q+3
2
Lemma 10.7 Fix j ∈ Z+ so that
≤ j ≤ q − 1 and fix α ∈ Fq . Let
θ = µ q+1 × · · · × µj × 1Bj+1 × · · · × 1Bq−1 × ϕ
2
such that {1 + βxi | β ∈ hα, F i1,i−1 } ⊆ ker(µi ) for i = 1, . . . , j − 1 and ker(µj ) =
{1 + βxj | β ∈ hα, F i1,j−1 }. Then the stabilizer of θ in P is contained in QP
q+1
2
.
Proof. Label S as the stabilizer of θ in P . Suppose s = 1 + α1 x + · · · + αq xq ∈ S.
Then we know that for all t ∈ P
θ([t, s]) = 1 for all t ∈ P
q+1
2
q+1
2
. Applying Lemma 5.1 to [t, s] we see that
θ([t, s]) = µ q+3 (1 + δ1 x
q+3
2
2
By Lemma 5.2, δk =
−1
, θ(t) = θs (t) = θ(t)θ([t, s]). Thus, we have
) · · · µj (1 + δj− q+1 xj )ϕ(1 + δ q−1 xq ).
2
Pk
i=1 hαi , dk−i+1 ii,k∗−i .
2
Using this result, we can replace each δi
and obtain
θ([t, s]) = µ q+3 (1 + hα1 , d1 i1, q+1 x
2
q+3
2
2
j− q+1
2
) · · · µj (1 + (
X
i=1
q−1
2
X
·ϕ(1 + ( hαi , d q+1 −i ii,q−i )xq ).
i=1
2
hαi , dj− q−1 −i ii,j−i )xj )·
2
85
Notice that since β q+1 , . . . , βq−1 are independent of each other, d1 , . . . , d q−1 are also
2
2
independent. Because of this, we can choose l so that dl+1 6= 0 and dk = 0 for k 6= l+1.
This implies δj = 0 if j < l + 1 and δj = hαj−l , dl+1 ij−l,l∗ otherwise. Similarly, for
j−
q−1
2
≤l≤
q−3
,
2
we have
∗
∗
ϕ(1 + hα q−1 −l , dl+1 iq−l,l∗ xq ) = (ϕ)αq−l,l
(1 + dl+1 xl ) = 1,
q−1
2
2 −l
q−1
−(j− q−1
)
2
2
for all dl+1 ∈ F . By Lemma 8.5, this implies α q−1 −l ∈ Fq . Notice that
2
=
q − j − 1. Thus, we have α1 , α2 , . . . , αq−j−1 ∈ Fq .
, we actually show a stronger result. For each l, in addition to
For 1 ≤ l ≤ j − q−3
2
showing that α q−1 −l ∈ Fq , we show that αl0 ∈ α
0
pl −1
p−1
2
Zp , where l0 = j −
q−1
2
− l. We
will do this by induction on l0 .
When l0 = 1, we have l = j − q−1
− 1 = j − q−3
and l∗ = j − q−1
− 1 + q−1
= j − 1.
2
2
2
2
The equation to consider here is
µj (1 + hα1 , dl+1 i1,l∗ xj )ϕ(1 + hα q−1 −l , dl+1 i q−1 −l,l∗ xq )
2
∗
q−1
2
−l,l∗
∗
= (µj )1,j−1
(1 + dl+1 xl ) · (ϕ)α 2q−1 −l (1 + dl+1 xl ) = 1,
α1
2
q−1
−l,l∗
for all dl+1 ∈ F . By Lemma 8.5, we know that Bj,q ⊆ ker((ϕ)α 2q−1 −l ). From our
2
equation above, this implies Bj,q ⊆
ker((µj )1,j−1
).
α1
On the other hand, we have seen
that α1 ∈ Fq , so we may use Lemma 10.2 to see that Dj,q ⊆ ker((µj )1,j−1
). Since
α1
∗
Bj = Bj,q u Dj,q , we deduce that (µj )1,j−1
(1 + dl+1 xl ) = 1. By Lemma 10.2, this
α1
implies that α1 ∈ αZp .
q−1
−l,l∗
∗
This gives us (ϕ)α 2q−1 −1 (1 + dl+1 xl ) = 1. By Lemma 8.5, we obtain α q−1 −l ∈ Fq .
2
2
Note that
q−1
2
−l =
q−1
2
− (j −
q−1
2
− 1) = q − j, so we have αq−j ∈ Fq .
86
We now come to the inductive step and assume that l0 > 1, and that the result
pi −1
holds for 1, . . . , l0 −1. In other words, we have α1 , . . . , αq−j+l0 −1 ∈ Fq and αi ∈ α p−1 Zp
for i = 1, . . . , l0 − 1. We obtain the equation
0
l
Y
µl∗ +i (1 + hαi , dl+1 ii,l∗ xl
∗ +i
)ϕ(1 + hα q−1 −l , dl+1 i q−1 −l,l∗ xq )
2
i=1
2
0
=
l
Y
∗
q−1
∗
−l,l∗
∗
(µl∗ +i )αi,li (1 + dl+1 xl ) · (ϕ)α 2q−1 −l (1 + dl+1 xl ) = 1,
2
i=1
for all dl+1 ∈ F .
For i = 1, . . . , l0 − 1, we know that ker(µl∗ +i ) contains
{1 + βxl
∗ +i
| β ∈ hα, F i1,l∗ +i−1 } = {1 + βxl
∗ +i
pi −1
| β ∈ hα p−1 , F ii,l∗ +i−1 },
pi −1
where equality comes from applying Lemma 4.10. Since αi ∈ α p−1 Zp , this implies
∗
∗
l
0
(µl∗ −i )i,l
αi (1 + dl+1 x ) = 1 for i = 1, . . . , l − 1. Thus, our equation becomes
0 ∗
∗
q−1
−l,l∗
∗
(µj )lα,ll0 (1 + dl+1 xl ) · (ϕ)α 2q−1 −l (1 + dl+1 xl ) = 1.
2
q−1
−l,l∗
0 ∗
By Lemma 8.5, we know that Bl∗ +1,q ⊆ ker((ϕ)α 2q−1 −l ), and so, Bl∗ +1,q ⊆ ker((µj )lα,ll0 ).
2
0 ∗
On the other hand, from Lemma 10.2, we have Dl∗ +1,q ⊆ ker((µj )lα,ll0 ). This implies
0 ∗
(µj )lα,ll0 (1
l∗
+ dl+1 x ) = 1, and hence, αl0 ∈ α
0
pl −1
p−1
j
Zp by Lemma 10.2 since
j
ker(µj ) = {1 + βx | β ∈ hα, F i1,j−1 } = {1 + βx | β ∈ hα
q−1
,l∗
0
pl −1
p−1
Zp , F il0 ,l∗ }
∗
by Lemma 4.10. Finally, we obtain (ϕ)α 2q−1 −l (1 + dl+1 xl ) = 1, and by Lemma 8.5,
2
this implies α q−1 −l ∈ Fq . This proves the inductive step and that for s to stabilize θ,
2
87
s must be in QP
q+1
2
. Therefore, S ⊆ QP
In general, we define Ai for i =
in Irr(P
q+1
2
q+1
2
.
q.e.d.
q+1 q+3
, 2 ,...,q
2
− 1 to be the set of all characters
) that are P -conjugate to some character of the form θ = µ q+1 × · · · × µi ×
2
1Bi × · · · × 1Bq−1 × ϕ, where µj ∈ Irr(Bj ) for j =
q+1
,...,i
2
and ker(µi ) = {1 + βxi |
β ∈ hα, F i1,i−1 }, for some α ∈ Fq . Recall Lemma 4.7 tells us that hα, F i1,i−1 is a
hyperplane in F . Since
pi −1
p−1
is relatively prime to
bijection for the cyclic group of order
β
pqr −1
p−1
pqr −1
.
p−1
pqr −1
,
p−1
raising to the
pi −1
p−1
power is a
Hence, there is an element β ∈ F so that
= α Therefore, hβ, F i1,i−1 = αker(T r). As a result, the particular elements
α ∈ F for which ker(µi ) = {1 + βxi | β ∈ hα, F i1,i−1 } are independent of i. Notice
that the definition of A q+1 here is consistent with the previous definition of A q+1 .
2
2
Lemma 10.8 Every character in Irr(P
Proof. Suppose θ ∈ Irr(P
q+1
2
q+1
2
| ϕ) lies in some Aj for j =
q−1
, . . . , q −1.
2
| ϕ), then
θ = µ q+1 × · · · × µi × 1Bi+1 × · · · × 1Bq−1 × ϕ,
2
where µi 6= 1. We say that i is the length of θ. We prove the lemma by induction
on i. First suppose all coordinates of θ other than ϕ are 1. In this case, we have
θ ∈ A q−1 by definition of A q−1 , and we say that θ has length
2
2
q−1
.
2
We now fix i ≥
q+1
,
2
and we assume the result holds for all characters θ with length at most i − 1. Let
s = 1 + γxq−i , and by Lemma 10.4, we have
θs
−1
= θ · (ν q+1 × · · · × νi−1 × ϕq−i,i
× 1Bi+1 × · · · × 1P q ).
γ
2
. We know from Lemma 8.5 that as γ runs
In the ith coordinate, we have µi ϕq−i,i
γ
will run over all characters in Irr(Bi /Bi,q ). We have seen
through F , then ϕq−i,i
γ
88
˙
that Irr(Bi ) = Irr(Bi /Bi,q )×Irr(B
i /Di,q ). Thus the coset µi Irr(Bi /Bi,q ) has a unique
character µ∗i ∈ Irr(Bi /Di,q ). We choose γ ∈ F so that µi ϕq−i,i
= µ∗i . Observe that
γ
−1
θs
= µ0q+1 × · · · × µ0i−1 × µ∗i × 1Bi+1 × · · · × 1Bq−1 × ϕ.
2
If µ∗i = 1, then θs
hypothesis, θs
−1
−1
has length at most (i − 1) −
q+1
,
2
and so, by the inductive
lies in one of A q−1 , . . . , Ai−1 , and we are done in this case. Thus,
2
µ∗i 6= 1. We know that Di,q ⊆ ker(µ∗i ). Now, ker(µ∗i ) ∩ Bi,q is a hyperplane in Bi,q . It
follows that ker(µ∗i ) ∩ Bi,q = {1 + βxi | β ∈ hα, Fq i1,i−1 }, and by Lemma 10.1, we have
ker(µ∗i ) = {1 + βxi | β ∈ hα, F i1,i−1 }. This implies µ∗i ∈ Ai , and proves the result.
q.e.d.
Lemma 10.9 For Ai defined as above,
½
cd(G | Ai ) =
µ
( q−1
)(qr−q+i−1)
2
p
pqr − 1
p−1
¶¾
.
Proof. Let θ ∈ Ai for some i. By definition of Ai , we can assume that θ is conjugate
to a character of the form
γ = µ q+1 × · · · × µi × 1Bi+1 × · · · × 1Bq−1 × ϕ,
2
where ker(µi ) = {1 + βxi | β ∈ hα, F i1,i−1 }. By Lemma 10.6, we may assume that
{1 + βxl | β ∈ hα, F i1,l−1 } ⊆ ker(µl ) for l =
that the stabilizer of γ is contained in QP
Ql ∩ Bl = Bl,q ∼
= Bl /Dl,q ∼
= Fq for l =
hence, we have
q+1
,...,i
2
q+1
2
− 1. By Lemma 10.7, we know
. Again, we define Qj = P j ∩ Q. Now,
q+1
, . . . , i.
2
We also know Q ∩ Bl = Bl,q , and
q+1
B q+1 ,q u P i+1 D q+1 ,q u P i+1
P 2
2
2
=
u
.
P i+1
P i+1
P i+1
89
Using a similar argument to that of Lemma 9.1 we obtain QP
Q/Q
i+1
q+1
2
/(D q+1 ,q u P i+1 ) ∼
=
2
. With this in mind, we can view γ as a non-principal character of Irr(Q
by restricting γ to Q
q+1
2
q+1
2
/Qi+1 )
. Notice that γQi+1 = 1Bi+1 × · · · × 1Bq−1 × ϕ and µ q+1 × · · · ×
2
µi ×1Bi+1 ×· · ·×1P q . By Lemma 8.2, γQi+1 extends to QP
the characters in Irr(QP
q+1
2
q+1
2
. By Gallagher’s theorem,
| γQi+1 ) are in bijection with characters in Irr(Q/Qi+1 ).
Since µi 6= 1, the characters in Irr(QP
q+1
2
| γ) are in bijection with characters in
Irr(Q/Qi+1 | Qi /Qi+1 ). Using Theorem 4.4 from [16], we have that characters in
Irr(Q/Qi+1 | Qi /Qi+1 ) all have degree p(
will all have degree p(
of degree p(
q−1
)(i−1)
2
q−1
)(qr−q+i−1)
2
orbits of size
pqr −1
p−1
q−1
)(i−1)
2
. Thus, characters in Irr(QP
q+1
2
| γ)
. Using Clifford theory, these induce irreducibly to P
. Under the action of C, these characters are permuted in
and hence, induce irreducibly to CP . Thus, for Irr(CP | Ai ), the
character degree is p(
q−1
)(qr−q+i−1)
2
qr
−1
( pp−1
). Since G stabilizes these characters, we have
½
cd(G | Ai ) =
µ
( q−1
)(qr−q+i−1)
2
p
pqr − 1
p−1
¶¾
.
q.e.d.
Theorem 10.10 Set
½
∗
A =
µ
( q−1
)(qr−q+i−1)
2
p
pqr − 1
p−1
¶
¾
| i = 0, 1, . . . , q − 1 and
µ qr
½
¶
¾
qr−1
p −1
i
B = (p 2 )
| i = 0, 1, . . . , q − 2 .
p−1
∗
Then the set of all character degrees of the group G is given by
½
µ qr
¶
µ qr
¶
q−1
q−1
p −1
p −1
qr
)(qr−q)
(
cd(G) = 1, q, r, qr, qp 2
,p 2
,
pq − 1
pq − 1
90
µ
qp
)(qr−q)
( q−1
2
pqr − 1
pq − 1
¶¾
∪ A∗ ∪ B ∗ .
Proof. As was discussed in Chapter 6, we know that cd(G) = cd(G/P q )∪cd(G | P q ).
From Lemma 6.2, we have that
½
µ qr
¶
¾
qr−1
p −1
i
cd(G/P ) = 1, q, r, qr, (p 2 )
|i = 0, 1, . . . , q − 2 = {1, q, r, qr} ∪ B ∗ .
p−1
q
Recall A was defined to be the characters ϕ ∈ Irr(P q ) whose kernels correspond to
hα, F ii,q−i for some α ∈ F and 1 ≤ i ≤
q−1
2
and B was defined to be the characters ϕ ∈
Irr(P q ) whose kernels do not correspond to hα, F ii,q−i for all α ∈ F and 1 ≤ i ≤
q−1
.
2
Now, the remaining characters of G are in Irr(G|P q ), which are in Irr(G|B)∪Irr(G|A).
To see this, consider χ ∈ Irr(G|P q ), and let ϕ be an irreducible constituent of χP q .
Since χ ∈ Irr(G|P q ), we know that ϕ 6= 1. Since P q is an elementary abelian p-group,
the kernel of ϕ will be a hyperplane in P q . By Corollary 4.8, we know that some of
the hyperplanes in P q have the form ha, F ii,q−i for some a ∈ F , and the rest do not.
If ker(ϕ) does not have the form ha, F ii,q−i for any a ∈ F , then ϕ ∈ B. If ker(ϕ) =
ha, F ii,q−i for some a ∈ F , then ϕ ∈ A. This implies Irr(G|P q ) ⊆ Irr(G|B) ∪ Irr(G|A).
The other containment is trivial, so we have Irr(G|P q ) = Irr(G|B) ∪ Irr(G|A).
Lemma 7.3 tells us that the set of character degrees whose kernels do not correspond to hyperplanes is given by
½
µ qr
¶¾
q−1
p −1
qr
cd(G|B) = qp 2
.
pq − 1
91
The characters of G whose kernels correspond to hyperplanes are found in Lemmas
8.3 and 10.9. From this we have cd(G | Ai ) is the set
½
µ qr
¶
µ qr
¶
µ qr
¶
q−1
q−1
p −1
p −1
p −1
( q−1
)(qr−q)
(
)(qr−q)
(
)(qr−q+i)
p 2
, qp 2
,p 2
pq − 1
pq − 1
p−1
¾ ½
µ qr
¶
¾
q−3
p −1
q−1
( q−1
)(qr−q+i−1)
| i = 0, 1, . . . ,
∪ p 2
|i=
,...,q − 1
2
p−1
2
µ qr
¶
µ qr
¶¾
½
q−1
p −1
p −1
( q−1
)(qr−q)
(
)(qr−q)
, qp 2
∪ A∗ .
= p 2
pq − 1
pq − 1
Combining all of these degrees we have
cd(G) = cd(G/P q ) ∪ cd(G|B) ∪ cd(G | Ai ),
where i =
q+1
,...,q
2
− 1, as desired.
q.e.d.
Clearly, p, q, and r are all vertices in ρ(G). To determine the remaining vertices
in ρ(G), we need to find the sets of prime divisors of
pqr −1
p−1
and
pqr −1
.
pq −1
The easiest
way of doing this is in terms of cyclotomic polynomials. Recall from Chapter 3, we
chose our primes p, q, and r so that Φq (p), Φr (p), and Φqr (p) are pairwise relatively
prime. We also remind the reader that because q and r are primes, Φq (p) =
Φr (p) =
pq −1
p−1
and
pr −1
.
p−1
First, consider
pqr −1
.
p−1
Recall from the discussion in Chapter 3, we can write pqr −
1 = Φ1 (p)Φq (p)Φr (p)Φqr (p). Hence,
pqr −1
p−1
= Φq (p)Φr (p)Φqr (p). Similarly,
pqr − 1
Φ1 (p)Φq (p)Φr (p)Φqr (p)
=
= Φr (p)Φqr (p).
q
p −1
Φ1 (p)Φq (p)
92
Thus, we need to consider these two products and the primes that divide them. By
the choices of p, q, and r for our group, we know Φq (p), Φr (p), and Φqr (p) will be
distinct sets of primes in ρ(G). For our purposes we will label Φn to be the set of all
primes that divide Φn (p). Using this notation, we see that we have the sets Φq , Φr ,
and Φqr that will be part of the graph.
Referring to the degrees in Theorem 10.10, notice that primes in Φq , Φr , and Φqr
are all incident to each other and to p. Also notice that r is only adjacent to q.
It should also be mentioned that relative to the character degree graph, each set,
Φn , will form a complete graph ∆(G) and will not affect the diameter. Thus, we treat
these sets of primes on the graph as single vertices.
Theorem 10.11 The set of vertices of ∆(G) is
ρ(G) = {p, q, r} ∪ Φq ∪ Φr ∪ Φqr .
The character degree graph is given by the graph in Figure 10.1.
p
PP
•H
H PP
@
@ HH
PP
HH PPP
@
HH PPP
@
PP
H
@
PP
HH
@
PP q
H
Φ
PP
H•r
Φq@•
©
³³•
³
©
¡
³³
©©
¡
³³
©
³
¡
³³
©©
¡ ©© ³³³
¡©© ³³³
³
¡©
³³
¡
©
³
•
•r
Φqr
Figure 10.1: The character degree graph of G
As is clearly seen in Figure 10.1, ∆(G) is isomorphic to the graph given by Lewis
and thus, has diameter three. We see from the graph that the diameter is being
93
determined in this case by the number of edges in a path between r and Φq . We
mentioned in the beginning that fixing p = 2, q = 3, and choosing a general r appropriately would not change the argument given by Lewis. We gain insight from the
graph as to why this is so.
Are there other groups whose degree graphs are the graph in Figure 10.1? If so,
do those groups have the same structure as the family of groups constructed here?
The answers to these questions are not known, as of yet.
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