On O-Basis Groups and Generalizations
Except where reference is made to the work of others, the work described in this
dissertation is my own or was done in collaboration with my advisory committee. This
dissertation does not include proprietary or classifled information.
Jason Ervin
Certiflcate of Approval:
Gary Gruenhage
Professor
Mathematics and Statistics
Randall Holmes, Chair
Professor
Mathematics and Statistics
Huajun Huang
Assistant Professor
Mathematics and Statistics
Joe F. Pittman
Interim Dean
Graduate School
On O-Basis Groups and Generalizations
Jason Ervin
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
August 4, 2007
On O-Basis Groups and Generalizations
Jason Ervin
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon the request of individuals or institutions and at
their expense. The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Dissertation Abstract
On O-Basis Groups and Generalizations
Jason Ervin
Doctor of Philosophy, August 4, 2007
(M.S., Mississippi State University, 2002)
(B.S., Mississippi State University, 2000)
(B.S., Mississippi State University, 2000)
(A.A., Mississippi Delta Community College, 1998)
79 Typed Pages
Directed by Randall Holmes
A class of flnite groups which we call o-basis groups is generalized and explored. One
reason for interest in these groups lies with the concept?s origins. The notion of o-basis
group arose from the study of the existence, in the n-fold tensor product of a complex inner
product space, of an orthogonal basis consisting entirely of \standard symmetrized tensors".
We call such a basis an o-basis. The term \symmetrized" refers to the action on the tensor
product of a subgroup of the symmetric group Sn. Given a subgroup of Sn, one may ask if
the corresponding symmetrized tensor space has an o-basis. The answer will depend in part
on the structure of the given group. Since any group can be homomorphically embedded
onto a subgroup of the symmetric group, arbitrary flnite groups may be considered. It has
already been shown that if G is an o-basis group and ? : G ! Sn is a homomorphism, then
the symmetrized space corresponding to ?(G) has an o-basis. The study of these groups
therefore may well be of interest to those working with o-bases of symmetrized spaces. Our
focus, however, is on the group structure and character theory of o-basis groups themselves
with a view toward using the o-basis property as a means of distinguishing between abstract
iv
flnite groups. The tools come from flnite group theory and the character theory of flnite
groups. Field theory appears very brie y. In previous work, some interesting classes of
groups have been shown to be o-basis, and so far all groups identifled as o-basis are nilpotent.
Particularly compelling are the dihedral groups. It has been shown that the o-basis dihedral
groups are precisely those that are 2-groups. These are also precisely the nilpotent dihedrals.
With this in mind, we ask whether or not all o-basis groups are nilpotent. We consider
this question for a restricted class of groups. Conversely, there are examples of nilpotent
groups that are not o-basis leading us to explore conditions on a nilpotent group which will
guarantee that the group is o-basis. The results obtained indicate a possible connection
between the o-basis property and the nilpotence class of a group.
The second main division of the present work is an exploration of a generalization of
o-basis groups. While the following deflnitions contain technicalities, the reader should be
able, without preliminary preparation, to understand the nature of the generalization. A
group is o-basis if for each subgroup H ? G and ? 2 Irr(G) for which (?;1)H 6= 0, there are
a certain number of \orthogonal cosets" of H in G. We generalize by relaxing the subgroup
condition as follows. Let K ? G. We say G is K-o-basis if for each ? 2 Irr(G) and each
subgroup H containing K where (?;1)H 6= 0, there are the required number of \orthogonal
cosets" of H. The o-basis groups, therefore, are the hei-o-basis groups, where hei denotes
the identity subgroup. Note that to apply this notion to a given class of groups, K must
be deflned for all groups in that class. Some results are obtained for the case when K is
a member of the lower central series and when it is a member of the upper central series.
Finally, the still open question of whether a direct product of o-basis groups is o-basis is
brie y discussed.
v
Acknowledgments
For the many and patient efiorts on my behalf, I would like to express my most sincere
gratitude to Dr. Randall Holmes. To Dr. Gary Gruenhage and Dr. Huajun Huang, I would
also like to earnestly convey my thanks. To the professors of the Auburn Department of
Mathematics who have given their time and energy toward making my program successful
and to the Department itself for supplying the resources that have made it possible, I am
truly grateful.
vi
Style manual or journal used Journal of Approximation Theory (together with the style
known as \aums"). Bibliograpy follows van Leunen?s A Handbook for Scholars.
Computer software used The document preparation package TEX (speciflcally LATEX)
together with the departmental style-flle aums.sty.
vii
Table of Contents
1 Preliminaries 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Character Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3.1 Brief Introduction to Character Theory . . . . . . . . . . . . . . . . 8
1.3.2 Basic Concepts and Related Theorems . . . . . . . . . . . . . . . . . 17
1.3.3 New Characters from Old: Products, Induction, Restriction
and Conjugation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1.3.4 Semi-Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.3.5 Frobenius Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2 O-Basis Groups 26
2.1 Construction and Deflnition . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.2 Connections with Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . 30
3 New Work 32
3.1 A Generalized Deflnition and Some Preliminary Results . . . . . . . . . . . 32
3.2 O-basis Groups and Nilpotency . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.3 The Upper and Lower Central Series . . . . . . . . . . . . . . . . . . . . . . 61
4 Conclusions 64
Bibliography 68
Notation 69
viii
Chapter 1
Preliminaries
1.1 Introduction
In this dissertation we study the notion of o-basis group. We also deflne and explore
generalizations of this notion. O-basis groups were deflned by Holmes in [Hlms], but their
origins go back to an earlier paper by Holmes and Tam, [Hlms,Tam]. In that former
paper, the authors studied the problem of the existence, in the n-fold tensor product of
a complex inner product space, of a basis consisting entirely of \standard symmetrized
tensors". Holmes would later call such a basis an o-basis. The term \symmetrized" refers
totheactionofasubgroupGofthesymmetricgroupSn onthetensorproduct. Thequestion
arises as to the extent to which the existence of an o-basis depends upon the structure of
the group G. In particular, one may seek conditions on G which will guarantee that the
corresponding symmetrized space has an o-basis. Since any group can be homomorphically
embedded into Sn, abstract flnite groups can be considered. For example, the authors
proved in [Hlms,Tam] that if G is the dihedral group of order 2n (the group of symmetries
of the regular n-gon), then the corresponding symmetrized tensor space has an o-basis if and
only if n is a power of 2. Motivated by this, Holmes deflned, in [Hlms], the o-basis groups
as a class of abstract flnite groups satisfying certain, rather technical, conditions. He then
provedthat, givenano-basisgroupG, foranyquotientofG, regardlessofhowitisembedded
into Sn, the corresponding tensor space has an o-basis. Thus, o-basis groups may well be
of interest to those working with the existence of o-bases in tensor spaces. That problem
and its connection with o-basis groups are discussed in more detail in section 2.2 below.
1
The focus in the present work is on o-basis groups themselves. In deflning o-basis groups,
Holmes? aim was to use the concept as a tool for distinguishing between abstract groups.
For example, we have noted that the o-basis property chooses, from among the dihedrals,
exactly those of prime-power order. Following the example of the dihedrals, Holmes was
able to produce a list of familiar groups that are o-basis according to the deflnition he gives
in [Hlms].
Theorem 1.1.1 ([Hlms], p. 142) The following groups are o-basis groups (p, prime,
n ? 1):
(i) any flnite abelian group,
(ii) the dihedral group D2n,
(iii) the quaternion group Q2n,
(iv) the semidehedral group S2n,
(v) the group with presentation hx;ajxp = 1 = apn?1;ax = a1+pn?2i
(vi) any group of order p3,
(vii) any extra-special p-group.
Holmes also provides a list of groups that are not o-basis.
Theorem 1.1.2 ([Hlms], p. 138) The following groups are not o-basis groups:
(i) any dihedral group Dn (order 2n) with n not a power of 2.
(ii) any 2-transitive subgroup of Sn with n ? 3 (e.g., the alter-
nating group An, n ? 4 and the symmetric group Sn n ? 3),
(iii) any flnite simple group of Lie type.
Let us state, for future reference, the dihedral group result noting that if follows from
Theorems 1.1.1 and 1.1.2 taken together.
2
Theorem 1.1.3 A dihedral group of order 2n is o-basis if and only if n is a power of 2.
Noting that all of the above examples that are o-basis are also p-groups, Holmes asked
if all p-groups are o-basis. He answered this in the negative by constructing a group of order
34 that is not.
It is at this point that the present study begins. The new work presented here can
be divided into two parts. The flrst part deals with the relationship between the property
of being o-basis and that of being nilpotent. Since all prime power groups are nilpotent,
we see that, so far, every group that has been identifled as o-basis is also nilpotent. Even
more compelling is the fact, already mentioned, that the dihedral groups that are o-basis are
precisely those that are 2-groups. These also happen to be precisely the nilpotent dihedrals.
In summary, two questions arise.
? Which nilpotent groups are o-basis?
(not all of them, as the example of order 34 shows)
? Are all o-basis groups nilpotent?
Concerning the flrst of these questions, we begin by proving that if G0 Z(G) (a
condition implying nilpotentcy), then G is o-basis. We use this fact as a tool to obtain
some further results. It also raises an interesting question for possible future study. Any
group with G0 Z(G) has nilpotence class less than or equal to 3 (see the deflnition of
n(G) in the notation section). Noting that the example of order 34 has class 4, one might
ask whether or not an o-basis group must be of class 3. This question remains open.
The reader might recall that nilpotent groups are characterized as being direct products
of their Sylow subgroups. We use this fact to show that a nilpotent group is o-basis if and
3
only if each of its Sylow subgroups is o-basis. The method of proof does not generalize
to direct products in general, and we brie y discuss the dilemma that arises after proving
the result. From this result on Sylow subgroups, we see that in some sense the question
of which nilpotent groups are o-basis would be answered if one knew which p-groups are
o-basis. In [Hlms], Holmes has given some su?cient conditions on a p-group for it to be
o-basis. In the present study, we take the approach of considering groups of increasingly
higher prime power order. As shall be seen, it follows quickly from the deflntion of o-basis
that all abelian groups, and therefore all groups of order p2, are o-basis. Holmes has also
shown that all groups of order p3 are o-basis (Theorem 1.1.1). Arriving at p4, Holmes?
group of order 34 gives the flrst example of a prime power group this is not o-basis. In
hopes of better understanding groups of order p4, we derive some necessary conditions for
such a group to fail to be o-basis. The groups that arise from this investigation begin to
look very much like Holmes? example.
After these considerations, we turn to the second of our questions. Are all o-basis
groups nilpotent? Again taking our cue from the dihedrals, we narrow the focus of the
question by considering a class of "dihedral-like" groups. More precisely, let p be prime
and let A C G be abelian with jG : Aj = pn for some positive integer n. Is it true that,
whenever G is o-basis, G is also nilpotent? We obtain some limited results for small values
of n.
The second major division of this work begins in section 3.1 where we deflne a gen-
eralized notion of o-basis group. The reader should be able to understand the nature of
the generalization without begin concerned with the technicalities of the deflnitions given
4
below. O-basis groups could be deflned as follows (although, our working deflnition will
include some additional conditions in order to eliminate trivialities).
A flnite group G is called an o-basis group if for all H ? G, and all ? 2 Irr(G),
there exist at least ?(e)(?;1)H cosets of H in G which are mutually orthogonal relative to
B?H.
This deflnition gives conditions to be satisfled by all pairs (H;?), with H ? G and ?
an irreducible character of G. To generalize this, we require these conditions to hold for
only certain subgroups. We also make an attempt at more convenient notation.
For H ? G and ? 2 Irr(G), we say G is?H;??-o-basis if there are at least ?(e)??;1?H
cosets of H in G which are mutually orthogonal relative to B?H. Let K ? G. If G is ?H;??-
o-basis for all subgroups H with K H and all ? 2 Irr(G), we say G is K-o-basis.
Let us note that the new deflnition encompasses the old since an o-basis group is one that is
hei-o-basis, where e denotes identity element. One might try to use the notion of K-o-basis
to distinguish between groups in a given class. To do this, K must be chosen so that it
is deflned for all groups in the class. For example, it makes sense to ask for any flnite
group whether or not the group is Z(G)-o-basis, where Z(G) denotes the center of G. In
section 3.3, we explore the notion of K-o-basis, where K is an element of the lower central
series and also where K is an element of the upper central series. For example, we show
that all flnite groups are 3-o-basis, where 3 is the third term of the lower central series
(see the notation section for the deflnition of these series).
Our tools come from flnite group theory and from the character theory of flnite groups.
Field theory also appears very brie y.
5
The remainder of this preliminary chapter is devoted to material that will be needed in
the main body which the reader may need to be introduced to or at least reminded of. We
discuss some general concepts and facts from Group Theory in section 1.2 and Character
theory in sections 1.3.1 - 1.3.3. Sections 1.3.4 and 1.3.5 are devoted to the specialized topics
of Semi-Direct Products and Frobenius Groups respectively.
In section 2.1, we state our working deflnition of o-basis groups and look in some detail
at Holmes? original development of the concept. We also state some results obtained by
Holmes in [Hlms] that we use directly in the main body of this work. In section 2.2, we take
a closer look at the tensor space problem that gave rise to the notion of o-basis groups.
1.2 Group Theory
This seems a convenient place to collect several facts and deflnitions from group theory
that we will need and that the reader may not immediately recall.
Theorem 1.2.1 ([Hun] p. 93) Let p be a prime and G be a group of order pn for some
integer n ? 0. Suppose G acts on a flnite set S and let SG denote the set of flxed points
under the action. Then jSj?jSGj (mod p).
Deflnition 1.2.2 ([Suz], p. 50) Let K be a subgroup of a group G. We say K charac-
teristic is in G, written K char G, if every automorphism of G maps K into itself. That
is, K K for all 2 Aut(G).
Theorem 1.2.3 ([Suz], p. 51) Suppose K N are subgroups of a group G and
that K char N. If N C G, then K C G.
6
Theorem 1.2.4 ([Hun] p. 94) Let p be prime. The center of a non-trivial flnite p-group
contains more than one element.
Theorem 1.2.5 ([Hun], p. 96) Let p be a prime. If G is a flnite p-group, N C G and
N 6= hei, then N \Z(G) 6= hei.
Theorem 1.2.6 ([Suz], p. 88) Let p be prime and suppose G is a p-group. Let M be a
maximal subgroup of G. Then M C G and G=M has order p.
Deflnition 1.2.7 ([Karp] p. 811) Let p be prime. A p-group is called an extra-special
p-group if G0 = Z(G), jG0j = p and G=G0 is elementary abelian.
All groups of order p3 are extra-special. To prove this, we can use the lemma below.
Lemma 1.2.8 Suppose that G=Z is cyclic. Then G is abelian.
Proof: Suppose for the sake of contradiction that G is non-abelian. Using the assumption
that G=Z is cyclic, let g 2 G?Z such that G=Z = hgZi. Observe that CG(g) contains g and
Z(G). Thus Cg(G)=Z = G=Z so that Cg(G) = G. It follows that g 2 Z(G), a contradiction.
Therefore, G is abelian as desired.
?
Proposition 1.2.9 Let G be a non-abelian group of order p3. Then G is extra-special.
Proof: We verify the conditions of Deflnition 1.2.7. Since G is non-abelian, we have
jG0j > 1 and jZj ? p2, where Z = Z(G). If jZj = p2, then G=Z is cyclic so that, contrary
to our assumption, G is abelian by Lemma 1.2.8. Thus jZj = p. By Theorem 1.2.5, Z is
contained in every normal subgroup of G. In particular, Z G0. Also, G=Z is abelian,
7
being a p-group with order p2. If follows that G0 Z so that G0 = Z. Finally, note that
if G=G0 = G=Z is cyclic then Lemma 1.2.8 again gives that G is abelian. It follows that
G=G0 ?= Zp ?Zp. That is, G=G0 is elementary abelian and the proof is complete.
?
Deflnition 1.2.10 ([Suz], p.159) Let p be prime. An abelian group A is said to be
elementary abelian if ap = 1 for all a 2 A.
If A is an elementary abelian p-group, then A is isomorphic to a direct sum of cyclic groups
of order p.
1.3 Character Theory
In this section, we introduce the basics of the character theory of flnite groups and
present a number of deflnitions and results which we will call upon throughout this work.
It is hoped that this will be an informative and enjoyable introduction to character theory
for those readers not familiar with it. Secondly, we aim to make the subsequent discussion
of o-basis groups more accessible and meaningful.
1.3.1 Brief Introduction to Character Theory
Character Theory can be developed in two alternate contexts, that of linear represen-
tations and that of modules over the group algebra. The resulting theories are essentially
equivalent. We begin with representations. Let G be a flnite group, let K be a fleld, and let
V be a flnite-dimensional vector space over K. We denote by GL(V) the group of invertible
linear transformations of V onto itself. A group homomorphism ? : G ! GL(V) is called
8
a linear K-representation (or simply a representation) of G in V. In the main body of
this work, we will always take K to be the fleld of complex numbers. This assumption, as
will be brie y explained below, simplifles the theory. We state what we can in the more
general context to make the reader aware of that theory.
We move from representations to characters as follows. Suppose dimC(V) = n and let
B = fv1;:::;vng be an ordered basis of V. For v 2 V, we have v = Pi flivi for uniquely
determined fli 2 K. Put
[v]B =
2
66
64
fl1
?
?
?
fln
3
77
75
(the coordinate vector of v relative to B). If f : V ! V is a linear transformation, the
matrix of f relative to B is given by [f]B = [fiij], where f(vj) = Pi fiijvi(1 ? j ? n).
That is, the jth column of [f]B is the coordinate vector of f(vj). The matrix of f satisfles
the equation [f(v)]B = [f]B[v]B, for all v 2 V. If a second basis, B0, is chosen for V, then
there is an invertible n?n matrix A such that [f]B0 = A[f]BA?1. Recall that the trace of a
matrix is the sum of the diagonal elements. Since the trace in invariant under conjugation,
we see that Tr[f]B = Tr[f]B0. This fact will be called upon shortly.
Now suppose ? : G ! GL(V) is a representation of G. We deflne the character
afiorded by ? to be the function ? : G ! K given by ?(g) = Tr[?(g)]B (g 2 G), where
B is some chosen ordered basis. We have noted that the trace of the matrix of ?(g) is
independent of the basis used when forming that matrix. Therefore, the character ? is also
independent of the choice of basis. Suppose W is a K-subspace of V such that ?(g)(W) W
for all g 2 G. Then the map ?W : G ! GL(W) given by ?W(g) = ?(g)jW is a well-deflned
representation of G. It is called the sub-representation of ? afiorded by W. If ? has no
9
proper, non-trivial sub-representations, then ? is said to be irreducible and the character
it afiords is called an irreducible character. We will see that the irreducible characters
play a critical role in character theory. First, however, let us develop these ideas in the
alternative context of KG-modules.
As mentioned, the notion of a group representation is interchangeable with a second
notion which we now introduce, that of a KG-module. Denote by KG the free K-module
with basis G (or more simply, the K-vector space having the elements of G as basis). That
is, KG consists of all formal sums of the form Pg2G figg, where fig 2 K (Since G is flnite,
the sums are flnite). Note that, as a vector space over K, KG has dimension jGj. The
K-space KG can be made into a ring by deflning multiplication as follows:
?X
x2G
fixx??
X
y2G
flyy? =
X
x2G
?
fixx
X
y2G
flyy
?
=
X
x;y2G
(fixfly)xy:
As a ring, KG has a multiplicative identity: the formal sum 1e, where e denotes the identity
element of G. We may imbed G into KG via g 7! 1g and identify G with its image. Note
that 1a?1b = 1(ab), for a;b 2 G.
The K-vector space structure on KG combines with the ring structure to make KG
a K-algebra. A K-algebra is a ring A with identity that is also a vector space over
K such that the \scalar multiplication" interacts with the ring multiplication as follows:
fi(ab) = (fia)b = a(fib) for all fi 2 K and all a;b 2 A. It is not hard to verify that
KG is indeed a K-algebra and we refer to KG with this structure as the group algebra.
We are now ready to discuss modules over the group algebra and their connections with
representations and characters.
10
Let A be a K-algebra and let V be a K-vector space. Suppose for every v 2 V and
x 2 A that a unique xv 2 V is deflned. Also assume for all x;y 2 A, v;w 2 V, and k 2 K
that
(i) x(v +w) = xv +xw
(ii) (x+y)v = xv +yv
(iii) x(yv) = (xy)v
(iv) x(kv) = k(xv) = (kx)v
(v) 1v = v
Then V is called an A-module. Some authors omit (v) and call an A-module with this
additional property a unitary A-module. Also, an A-module is usually not assumed to be
flnite-dimensional as a K-space.
Suppose again that ? : G ! GL(V) is a representation of a group G. One makes V
into a KG-module by deflning gv = ?(g)(v)(g 2 G;v 2 V), and extending linearly to KG.
Conversely, let V be a KG-module. Then V can be viewed as a (flnite-dimensional)
vector space over K. Here we use the fact that the map K ! KG given by fi 7! fi1, where
fi 2 K, is a ring monomorphism. We may therefore identify K with its image in KG under
this map. Deflne ? : G ! GL(V) by ?(g)(v) = gv. One uses the above properties of a
K-algebra to show that ? is a well-deflned homomorphism, and hence a representation of
G. We call ? the representation of G afiorded by V. If ? is the character afiorded by
?, we say that ? is afiorded by the KG-module V.
If W V is a KG-submodule of V (meaning that sW W for all s 2 KG), then
W afiords a representation of G and this representation is a sub-representation of ? (see
our discussion of representations). A non-zero KG-module is simple if it has no non-zero,
proper submodules. If V is a simple KG-module, then the representation it afiords is
11
irreducible in the sense deflned previously and the character it afiords is an irreducible
character. We may, in this way, pass from representations to KG-modules and back
and develop the theory in either context. Let us take this opportunity to point out that
sums of characters are characters. Indeed, let V1 and V2 be KG-modules afiording the
characters ?1 and ?2 respectively. The direct sum V1 ' V2 becomes a KG-module by
deflning s(v1;v2) = (sv1;sv2) (s 2 KG;vi 2 Vi). In this case, V1 'V2 afiords the character
?1 + ?2 deflned as usual by (?1 + ?2)(g) = ?1(g) + ?2(g). As will be seen, any character
can be written as a sum of irreducible characters.
If one assumes that K =C, the fleld of complex numbers, a simplifled theory results.
This is due to the fact that the complex numbers form an algebraically closed fleld of
characteristic zero. In fact, the simplifled theory continues to hold, with minor adjustments,
for any algebraically closed fleld whose characteristic does not divide jGj. The utility of
these properties for character theory lies in part with two results which we state below:
Maschke?s Theorem and Schur?s Lemma. This theory is covered in detail (and in more
generality) in Chapter 1 and in the beginning of Chapter 2 of [Is].
Deflnition 1.3.1 Let R be a ring. If there is a least positive integer n such that na = 0
for all a 2 R, then R is said to have characteristic n. If no such n exists, R is said to
have characteristic zero.
Note that the fleld of complex numbers has characteristic zero and is algebraically closed.
Theorem 1.3.1.1 (Maschke?s Theorem) Let K be a fleld and let G be a flnite group.
If char(K) does not divide jGj, then every KG-module is a direct sum of simple modules.
12
Thus, given a group G, to know all simple KG-modules is to know all KG-modules. As
an interesting aside, we mention that it is pointed out in Isaacs? book that a KG-module
is external to KG. Therefore, it is not clear how one might determine from KG all simple
KG-modules. One must produce a set of KG-modules large enough to contain copies of
all simple KG-modules. Let us note that KG can be considered a module over itself by
left multiplication. Denote this KG-module by KG?. It can be shown that every simple
KG-module is a submodule of KG?.
Theorem 1.3.1.2 (Schur?s Lemma) Let V and W be simple KG-modules and let
f : V ! W be a KG-module homomorphism.
(i) If V W, then f = 0.
(ii) Assume that K is algebraically closed. If V = W, then f = fi1V for some fi 2 K.
We assume henceforth that K =C. As an illustration of how Maschke?s Theorem and
Schur?s Lemma are used, we include a sketch of their role in a portion of the theory that
will be of interest to us. We need to deflne the notion of a class function.
A function f : G ! C is said to be a class function on G if for each g 2 G,
f(xgx?1) = f(g) for all x 2 G. That is, a class function is constant on conjugacy classes
of G. The class functions of G form a complex vector space, Cl(G). Let fHigki=1 be the
conjugacy classes of G and deflne, for each i, ?i : G ! C by ?i(g) =
8
><
>:
1 if g 2 Hi
0 if g =2 Hi
.
Then the set f?i : 1 ? i ? kg forms an obvious basis of Cl(G) so that the dimension of
Cl(G) over C is equal to the number of conjugacy classes of G.
Characters are class functions and this fact will be used directly at least once in this
work. We have already mentioned the fact that, if K is taken to be C, the irreducible
characters of G play a particularly important role in character theory. It turns out, in this
13
case, that Irr(G) forms a basis for Cl(G) that is orthonormal relative to a certain \inner
product". The proof involves both Maschke?s Theorem and Schur?s Lemma.
Deflnition 1.3.1.3 Let ?, # be class functions on a flnite group G. Deflne the inner
product of ? and # by
(?;#) = (?;#)G = 1jGj
X
g2G
?(g)#(g):
Our \inner product" has the following properties for all x;y;z 2 V and every fi;fl 2C:
(i) (fix+fly;z) = fi(x;z)+fl(y;z)
(ii) (y;x) = (x;y)
(iii) (x;x) ? 0
(iv) (x;x) = 0 ifi x = 0.
From this deflnition, it follows that the inner product is conjugate linear in the second
variable. That is (x;fiy+flz) = fi(x;y)+fl(x;z), where the notation is as in the deflnition.
We have the following.
Theorem 1.3.1.4 Orthogonality Relation ([Is], p.20) Let ?;? 2 Irr(G).
Then (?;?)G = 1jGj
X
g2G
?(g)?(g?1) = ???, where ??? is the Kronecker delta.
The orthogonality relation says that the irreducible characters of G form an orthonormal
set with respect to the inner product deflned above. We note that there is, in fact, a "flrst"
and "second" orthogonality relation. What we have stated is the "flrst". Let us now turn
to the proof that Irr(G) forms a basis for Cl(G). We need one further lemma.
Lemma 1.3.1.5 Let S be a simple CG-module afiording the character ?. Let f 2 Cl(G)
and deflne h = Pg2G f(g)g : V ! V. Then h = jGjn (f;?)1V , where n = dimCV.
14
Sketch of proof for Lemma 1.3.1.5: One shows that h is a CG-module homomorphism. By
Theorem 1.3.1.2, we have h = fi1V for some fi 2C. Thus,
fin = Tr(h) = Tr
X
g2G
f(g)g
?
=
X
g2G
Tr(f(g)g) =
X
g2G
f(g)Tr(g) =
X
g2G
f(g)?(g) = jGj(f;?);
where the last equality follows immediately from Deflnition 1.3.1.3. The result follows.
?
Theorem 1.3.1.6 Let G be a flnite group. Then Irr(G) is a basis for Cl(G).
Sketch of proof: We have seen that Irr(G) forms an orthonormal set in Cl(G). Thus it
su?ces to show that Irr(G) spans Cl(G). For this, it is enough to show that the orthogonal
complement of the subspace generated by Irr(G) is zero (relative to the inner product
deflned in 1.3.1.3). Let f 2 Cl(G) such that (?;f) = 0 for all ? 2 Irr(G). Let V = CG
and set h = Pg2G f(g)g : V ! V. (That is, h(v) =
P
g2G f(g)g
?
v for all v 2 CG.) If
S is a simple submodule of V afiording the character ?, then Lemma 1.3.1.5 gives that the
restriction of h to S equals jGjn (f;?)1S, where n = dimCS. Note that (f;?) = (?;f) (easily
verifled from Deflnition 1.3.1.3). Since the quantity on the right is zero by assumption, we
see that hS is zero. By Theorem 1.3.1.1, V is a direct sum of simple modules. It follows
that h : V ! V is the zero map. Hence Pg2G f(g)g = h(e) = 0. This last quantity is a
linear combination in the vector spaceCG and the elements of G are a basis for that space.
Thus the elements of G form a linearly independent set. This implies that f(g) = 0 for all
g 2 G. That is, f = 0. It follows that f = 0, as desired.
?
15
By Theorem 1.3.1.6, Irr(G) forms a basis over C for Cl(G). The flrst part of the
theorem below follows immediately.
Theorem 1.3.1.7 ([Is], p.16) Every class function ? of G can be uniquely expressed in
the form ? = P?2Irr(G) a??, where a? 2 C. Furthermore, ? is a character if and only if
all of the a? are nonnegative integers and ? 6= 0.
The module viewpoint can be used to give an easy proof of the second statement of the
theorem. Let V be a KG-module afiording the character ?. According to Maschke?s
Theorem, V is a direct sum of simple modules. Therefore, ? is the sum of the irreducible
characters that these simple modules afiord.
More can be said of the fi?. These quantities will play an important role in several of
our arguments. They can be described in terms of the inner product.
Proposition 1.3.1.8 ([Is], p.20) Let ? be a character of G. Then
(i) j?(g)j? ?(e), and
(ii) ?(g?1) = ?(g).
Proposition 1.3.1.9 Let ? 2 Cl(G). Then ? = P 2Irr(G)(?; ) .
Proof: By Theorem 1.3.1.7, ? = P 2Irr(G) fi , for uniquely determined fi 2 C. Fix
? 2 Irr(G) and observe that
(?;?) =
X
2Irr(G)
fi ;?
?
=
X
2Irr(G)
fi ( ;?) = fi?;
where the last equality follows from Theorem 1.3.1.4. The result follows.
?
16
Suppose ? is a character of G. For any ? 2 Irr(G) for which (?;?) 6= 0, we say that ? is
an irreducible constituent of ?. In this case, we see from Proposition 1.3.1.9 that (?;?)
is the multiplicity of ? as an irreducible constituent. Note also that ?(e) = P(?;?)?(e).
Proposition 1.3.1.10 ([Is], p.21) Let ? and ? be (not necessarily irreducible) characters
of a group G. Then (?;?) = (?;?) is a non-negative integer. Also, ? 2 Irr(G) if and only
if (?;?) = 1.
1.3.2 Basic Concepts and Related Theorems
In this section, we introduce some basic concepts in and collect a number of results
from character theory that we will need.
Deflnition 1.3.2.1 For a character ? of G, the positive integer ?(e) is called the degree
of ?. We say ? is linear if ?(e) = 1.
Let ? 2 Irr(G) and let ? : G ! GL(V) be the representation afiording ?. Since ? is
a homomorphism, the matrix of ?(e) is the identity matrix. Thus ?(e), the trace of this
matrix, is the dimension over C of V. In particular, it is a positive integer. As noted in
the remarks before Proposition 1.3.1.10, the degree of ? is the sum of the degrees of its
irreducible constituents (counting multiplicities).
Note that the linear characters are precisely those that are afiorded by representations
into 1-dimensional spaces. This means that linear characters are irreducible.
17
Proposition 1.3.2.2 Let ? be a linear character of G. Then ? is irreducible.
Proof: Let ? : G ! GL(V) be a representation afiording ?. We have dimCV = ?(e) = 1.
Thus V has no proper, non-trivial subspaces so that ? has no proper, non-trivial sub-
representations. That is ?, and thus ? is irreducible.
?
Theorem 1.3.2.3 ([Is], p.16) A group G is abelian if and only if every irreducible char-
acter of G is linear.
Deflnition 1.3.2.4 Let ? be a character of G. The kernel of ? is the subgroup of G deflned
by: ker? = fg 2 G : ?(g) = ?(e)g. We say that ? is faithful if ker? = feg.
Suppose that N C G. There is a one-to-one correspondence between the irreducible
characters of G=N and those irreducible characters of G whose kernels contain N. More
precisely, we have the following result taken from Isaacs.
Proposition 1.3.2.5 ([Is], p.24) Let N C G.
(i) If ? is a character of G and N ker?, then ? is constant
on cosets of N in G and the function b? on G=N deflned by
b?(gN) = ?(g) is a character of G=N.
(ii) If b? is a character of G=N, then the function ? deflned by ?(g) =
b?(gN) is a character of G.
(iii) In both (a) and (b), ? 2 Irr(G) ifi b? 2 Irr(G=N).
Suppose ? 2 Irr(G) and ker? ? G0. Then b? 2 Irr(G=G0). Since G=G0 is abelian, Proposi-
tion 1.3.2.3 gives that ?(e) = b?(e) = 1. That is, ? is linear. The converse is true in fact, as
the proposition below states.
18
Propositon 1.3.2.6 G0 = Tfker(?) : ? 2 Irr(G);?(1) = 1g
Deflnition 1.3.2.7 Let ? be a character of G. The center of ? is the subgroup of G given
by: Z(?) = fg 2 G : j?(g)j = ?(e)g.
Note that ker? Z(?) for all characters ?.
Corollary 1.3.2.8 ([Is], p.27) Let ? 2 Irr(G). Then Z(G) = TfZ(?) : ? 2 Irr(G)g.
Lemma 1.3.2.9 ([Is], p.27) Let ? 2 Irr(G). Then Z(?)=ker? = Z(G=ker?).
This follows from Proposition 1.3.2.5 and Theorem 1.3.2.3.
Proposition 1.3.2.10 ([Is], p.28) Let ? 2 Irr(G). Then ?2(e) ? jG : Z(?)j. Equality
occurs if and only if ? ? 0 on G?Z(?).
Theorem 1.3.2.11 ([Is], p.28) Suppose that ? 2 Irr(G) and that G=Z(?) is abelian.
Then jG : Z(?)j = ?2(e).
The following result follows from remarks accompanying exercise 2.12, [Is], p.31.
Proposition 1.3.2.12 Let ? be a character of G and let n = jGj. Let g 2 G with ?(g) = 0.
Then ?(gm) = 0 for all m 2Z with (m;n) = 1.
Proof: Let ? be a primitive nth root of unity in C. Let m 2Z with (m;n) = 1. Referring
to exercise 2.12 ([Is], p. 31), there exists 2 Gal(Q[?];Q) such that ?(gm) = (?(g)). By
assumption, we have (?(g)) = (0) = 0, where the last equality follows since is a fleld
automorphism. The result is established.
?
19
Theorem 1.3.2.13 Ito?s Theorem ([Is], p.84) Let A C G be abelian. Then ?(e) divides
jG : Aj for all ? 2 Irr(G).
1.3.3 New Characters from Old: Products, Induction, Restriction
and Conjugation
Deflnition 1.3.3.1 Let ? and ? be class functions on a group G. The product ?? of ?
and ? is deflned by ??(g) = ?(g)?(g). If ? and ? are characters, then ?? (see immediately
below).
Let V1 and V2 be CG-modules afiording the characters ?1 and ?2 respectively. The
tensor product V1 ?V2 becomes a CG-module by deflning g(v1 ?v2) = gv1 ?gv2 (g 2 G,
vi 2 Vi) and extending linearly in bothCG and in V1?V2. As aCG-module, V1?V2 afiords
the product character ?1?2.
Theorem 1.3.3.2 ([Is], p.59) Let G be the direct product of subgroups H and K. Let ?
be a character of H and ? be a character of K. Extend ?;? to G by putting ?(h;k) = ?(h)
and ?(h;k) = ?(k). These functions are characters of G so that the product ?? is a
character of G. Moreover, the irreducible characters of G are exactly the products of the
extended irreducible characters of H and K: Irr(G) = f?? : ? 2 Irr(H);? 2 Irr(K)g.
Theorem 1.3.3.2 generalizes in the natural way to a direct product with any flnite number
of factors.
Theorem 1.3.3.3 Suppose H ? G and let ? 2 Irr(G). The restriction, ?H, of ? to H is
a character of H.
20
Keeping the above notation, we see that ?H(e) = ?(e). Thus the restriction of a linear
character is linear and so irreducible by Proposition 1.3.2.2. Let ? 2 Irr(G) and ? 2
Irr(H). When referring to the inner product in H we will suppress the subscript on the
restricted character and write (?;?)H rather than (?H;?)H (see Deflnition 1.3.1.3).
Theorem 1.3.3.4 ([Is], p.81) Let N C G and suppose that ? 2 Irr(G)
with ??;1?N 6= 0. Then N ker?.
Deflnition 1.3.3.5 Let H ? G and let ? 2 Cl(H). The induced class function ?G on
G is deflned by
?G(g) = 1jHj
X
x2G
???xgx?1?
where ??(y) = ?(y) if y 2 H and ??(y) = 0 if y =2 H.
It follows immediately from the deflnition that ?G(e) = jG : Hj??(e). Also, if N C G and
? 2 Cl(N), then ?G ? 0 on G?N.
Lemma 1.3.3.6 ([Is], p.67) Let ? be a character of a subgroup of G. Then ker(?G) =
\
x2G
(ker?)x.
Deflnition 1.3.3.7 Let H ? G, let ? be a character of H and let g 2 G.
The conjugate character, g?, of ? is the character of gH deflned by
the equation g?(gh) := ?(h).
Deflnition 1.3.3.8 Let H C G and let ? 2 Irr(H). The set IG(?) = fg 2 G : g? = ?g is
called the inertial subgroup of ?.
Conjugate characters play a central role in two important theorems concerning char-
acters of subgroups: Clifiord?s Theorem and Mackey?s Theorem. Clifiord?s Theorem deals
with restriction to a normal subgroup.
21
Theorem 1.3.3.9 (Clifiord?s Theorem for Characters) Let H C G, let ? 2 Irr(G),
let ? 2 Irr(H) and assume (?;?)H 6= 0. Then ?H = (?;?)H
X
s2?
s?, where fs?gs2? is a
complete set of distinct conjugates of ?.
We will also be interested in inducing up to G a character of some subgroup H of G and
then restricting this induced character to a subgroup K of G to obtain a character of K.
Mackey?s Theorem provides a description of the restricted character. The theorem involves
the notion of conjugate characters and that of double cosets.
Deflnition 1.3.3.10 Let H;K be subgroups of a group G and let g 2 G. The (H,K)-
double coset containing g is the set HgK := fhgk : h 2 H;k 2 Kg.
Theorem 1.3.3.11 Mackey?s Theorem Let H;K be subgroups of a group G and let ? be
a character of H. Then (?G)K =
X
s2?
?(s?)
sH\K
?K, where ? is a complete set of (H;K)-
double coset representatives in G.
Theorem 1.3.3.12 Frobenius Reciprocity([Is], p.62) Let H be a subgroup of a group
G. Suppose ? 2 Cl(H) and ? 2 Cl(G). Then ??;?H?H = ??G;??G.
Deflnition 1.3.3.13 Let ? be a character of G. We say ? is mononial if ? = ?G, where
? is a linear character of some (not necessarily proper) subgroup of G. The group G is an
M-group if every irreducible character of G is monomial.
Let H ? G, let ? 2 Irr(H) be linear, and let ? = ?G. If ? is non-linear, then
jG : Hj = jG : Hj?(e) = ?(e) > 1 (see the comments after Deflnition 1.3.3.5). That is, H is
a proper subgroup in this case.
Theorem 1.3.3.14 ([Is], p.83) Every nilpotent group is an M-group.
22
1.3.4 Semi-Direct Products
In this section we discuss a generalization of the notion of direct product.
Deflnition 1.3.4.1 Let A;T be subgroups of G. We say G is the internal semi-direct
product of A and T (written AoT) if the following hold:
(i) G = AT,
(ii) A\T = hei.
Our interest in the semi-direct product structure stems in large part from the following
discussion taken from [Ser], p.62. We refer the reader to that passage for a full discussion
and proofs. Consider a semi-direct product, G = AoT, where A is abelian. The irreducible
characters of G shall be described in terms of those for A and T. The group T acts on
Irr(A) as follows: ?t(a) = ?(t?1at), where ? 2 Irr(A), t 2 T, and a 2 A. Choose a
representative ? of an orbit in Irr(A) under the action of T and let T? denote the stabilizer
of ? in T. Put H = AT?. The character ? is extended to an irreducible character of H by
putting ?(at) = ?(a) for a 2 A and t 2 T?. Choose ? 2 Irr(T?). Putting ~? = ??, where
? : H ! H=A ?= T? is the canonical map, one obtains a second irreducible character of
H. Now set ?;? = (?~?)G to obtain an irreducible character of G. That ?;? is indeed an
irreducible character of G and that the above construction gives all irreducible characters
of G is the substance of the next theorem.
Theorem 1.3.4.2 Suppose that G = AoT, where A is abelian. Referring to the above
discussion, let f?ig be a complete set of orbit representatives in Irr(A) under the action of
T and put ?i;? = (?i~?)G for each orbit representative ?i and character ? 2 Irr(T?i). Then
the following statements hold:
23
(i) ?i;? 2 Irr(G),
(ii) If ?i;? = ?0i;?0, then i = i0 and ? = ?0,
(iii) For every ? 2 Irr(G), ? = ?i;? for some i and ? 2 Irr(T?i).
The following Theorem will be used to verify that certain groups of interest are semi-direct
products.
Theorem 1.3.4.3 (Schur-Zassenhaus) Let A C G. If jG : Aj is relatively prime to jAj,
then G is the internal semi-direct product AoT, where T is a subgroup of G with T ?= G=A.
1.3.5 Frobenius Groups
We will have occasion to work with a special class of groups known as Frobenius Groups.
We will, when there is need, refer the reader to this section for the deflnition of these groups
and pertinent results.
Deflnition 1.3.5.1 Let H G, with hei 6= H 6= G. Assume that H \Hg = hei whenever
g 2 G?H. Then H is a Frobenius complement in G. A group that contains a Frobenius
complement is called a Frobenius group.
Theorem 1.3.5.2 (Frobenius, [Is], p.99-100) Let G be a Frobenius group with com-
plement H. Then there exists N C G with HN = G and H \ N = hei. In this case,
CG(x) N for all hei6= x 2 N.
The normal subgroup N above is called the Frobenius kernel of G. It is uniquely deter-
mined by H. (see [Is], p.101).
24
Theorem 1.3.5.3 ([Is], p.94) Let N C G and assume that CG(x) N
for every e 6= x 2 N. Then
(i) For ? 2 Irr(N), with ? 6= 1N, we have IG(?) = N and ?G 2 Irr(G).
(ii) For ? 2 Irr(G) with N *ker?, we have ? = ?G for some ? 2 Irr(N).
Lemma 1.3.5.4 ([Is], p.199) Let G be solvable and assume that G0 is the unique minimal
normal subgroup of G. Then all non-linear irreducible characters of G have equal degree f
and for some prime p one of the following holds:
(i) G is a p-group, Z(G) is cyclic and G=Z(G) is elementary abelian of order f2.
(ii) G is a Frobenius group with an abelian Frobenius complement of order f.
Also, G0 is the Frobenius kernel and is an elementary abelian p-group.
25
Chapter 2
O-Basis Groups
In this chapter, we deflne o-basis groups, make some preliminary observations and
discuss connections to a problem in multi-linear algebra.
2.1 Construction and Deflnition
The o-basis groups were deflned by Holmes in [Hlms]. We give here a brief account of
his construction which will serve to motivate our working deflnition. We also state some
results from [Hlms] which we will use later.
Let G be a flnite group and let H be a subgroup of G. Denote by G=H the set of
left cosets of H in G. The natural left action of G on the set G=H extends linearly to the
complex vector space having this set as basis. Denote this vector space by C(G=H). Let
? 2 Irr(G). Deflne a form B?H on C(G=H) by putting
B?H(aH;bH) = ?(e)jHj
X
h2H
?(a?1bh); (2.1)
and extending linearly in the flrst component and anti-linearly in the second component.
This can be shown to be a well-deflned G-invariant Hermitian form. The term G-invariant
means that B?H(gaH;gbH) = B?H(aH;bH) for all g;a;b 2 G. Put C?H :=C(G=H)=kerB?H,
where kerB?H := fx 2 C(G=H) : B?H(x;y) = 0 for all y 2 C(G=H)g. Then B?H induces
26
a well-deflned form B?H on C?H given by B?H(x;y) = B?H(x;y) (x;y 2 C(G=H)), where x
denotes the coset x+kerB?H. We have the following.
Theorem 2.1.1 ([Hlms] p.135)
(i) dimCC?H = ?(e)(?;1)H, where (?;1)H = 1jHj
X
h2H
?(h).
(ii) The form B?H is positive deflnite.
Holmes deflnes a group G to be an o-basis group if for every H ? G and ? 2 Irr(G)
the vector space C?H has a basis that is orthogonal relative to B?H and consists entirely
of elements of the form aH. Such a basis he calls an o-basis of C?H. He then gives a
characterization of o-basis groups entirely in terms of subgroups and characters without
reference to the linear algebra. This result is given below after one further deflnition.
Theorem 2.1.2 ([Hlms], p.139) The following are equivalent.
(i) G is an o-basis group.
(ii) For each H ? G and each ? 2 Irr(G), there exists at least
?(e)??;1?H cosets of H in G that are mutually orthogonal
relative to B?H.
(iii) For each H ? G and each non-linear ? 2 Irr(G) with
??;1?
H 6= 0, there exist at least ?(e)
??;1?
H cosets of H in
G that are mutually orthogonal relative to B?H.
We will take (iii) of Theorem 2.1.2 as our deflnition of o-basis group. It is obvious in the
theorem that (ii) implies (iii). It seems desirable to provide a brief sketch of the proof for
the remainder of the theorem since the reader will likely be curious about this and some of
the details are not di?cult.
27
Sketch of proof of Theorem 2.1.2: Let H ? G, let ? 2 Irr(G) and assume that (iii) holds.
We prove (i). By assumption, we may assume either (?;1)H = 0 or that ? is linear. Suppose
that (?;1)H = 0. Then by Theorem 2.1.1 part(i), dimCC?H = 0. In this case, the basis of
C?H is empty and satisfles Holmes? original deflnition vacuously. We assume therefore that
(?;1)H 6= 0. Let us suppose that ? is linear. Then the restriction of ?H is linear (since
1 = ?(e) = ?H(e)) and, by Proposition 1.3.2.2, ?H 2 Irr(H). By Proposition 1.3.1.9,
?H = 1H and (?;1)H = 1. Theorem 2.1.1 now gives that dimCC?H = 1. Thus, C?H has
as basis the set fHg, where H := H + ker(B?H). This single set serves as an o-basis, the
orthogonality condition again being satisfled vacuously. This shows that (iii) implies (i).
For (i) implies (ii), observe flrst that for every a;b 2 G, B?H(aH;bH) = B?H(aH;bH)
so that aH and bH are orthogonal relative to B?H if and only if aH and bH are orthogonal
relative to B?H. Assume that G is an o-basis group, let H ? G and ? 2 Irr(G). There
exists an o-basis fa1H;:::;atHg of C?H, (possibly empty with t = 0). By Theorem 2.1.1,
t = ?(e)(?;1)H and, by the above observations, a1H;:::;atH are mutually orthogonal
relative to B?H. This shows that (i) implies (ii) and the proof is complete.
?
As we have said, we take Theorem 2.1.2 part (iii) for our deflnition of o-basis group.
Deflnition 2.1.3 A flnite group G is called an o-basis group if for all H ? G and all
non-linear ? 2 Irr(G) with (?;1)H 6= 0, there exist at least ?(e)(?;1)H cosets of H in G
which are mutually orthogonal relative to B?H.
The next result follows almost immediately from the deflnition.
28
Theorem 2.1.4 Let G be abelian. Then G is o-basis.
Proof: By Theorem 1.3.2.3, every irreducible character of G is linear. Therefore, G sat-
isfles Deflnition 2.1.3 vacuously and is o-basis.
?
Throughout this work, we will often shift focus from a given group to a quotient of
the group. The success of this technique depends on the discussion below and the two
subsequent results. For the proofs, the reader is referred to [Hlms].
Let N C G, let ? 2 Irr(G) and assume that N ker?. For a subgroup H ? G, denote
by bH the image of H under the canonical map G ! G=N. The function b? : bG !C given
by b?(gN) = ?(g) is a well-deflned irreducible character of bG (see Theorem 1.3.2.5, (i) and
(iii)). Let H ? G.
Proposition 2.1.5 ([Hlms], p.137) Let the notation be as in the above paragraph. The
linear map ` : C?H ! Cx62?x62H given by `(gH) = (gN)bH is a well-deflned linear isometry. In
particular, C?H has an o-basis if and only if Cx62?x62H has an o-basis.
Theorem 2.1.6 ([Hlms], p.137) The class of o-basis groups is closed under taking ho-
momorphic images.
The following theorem, also found in [Hlms], has proven useful for inspiration and as a
direct tool in this study.
Theorem 2.1.7 ([Hlms], p.139) Let G be a flnite p-group (p, prime) and assume that
G has an abelian normal subgroup A and a cyclic normal subgroup C with C A satisfying
jG : Aj? p and jA : Cj? p. Then G is an o-basis group.
29
2.2 Connections with Linear Algebra
O-basis groups arose in connection with a problem from multi-linear algebra. We give
a brief description below. For a more in-depth discussion including proofs see [Hlms] and
[Hlms,Tam].
Fix positive integers m and n and put ?m;n = f 2 Zn : 1 ? i ? mg. Let G be
a subgroup of the symmetric group Sn. There is a right action of G on ?m;n given by
= ( (1) :::; (n)) ( 2 ?m;n; 2 G).
Let V be a complex inner product space of dimension m and let fe1;:::emg be an
orthonormal basis of V. To avoid trivialities, one assumes that m ? 2. Denote by V?n
the n-fold tensor power of V. For 2 ?m;n, put e := e 1 ? ??? ? e n 2 V?n. Then
fe : 2 ?m;ng is a basis for V?n.
Let ? 2 Irr(G). The symmetrizer relative to ? is the element of the group algebra
CG of G (see the discussion on KG in section 1.3.1) given by s? := (?(e)=jGj)P 2G ?( ) .
For 2 ?m;n, put e? := s?e , where we view V?n as a left CG-module via e = e ?1
( 2 G). The quantity e? is referred to as a standard symmetrized tensor.
The inner product on V induces an inner product on V?n. If W is a subspace of V?n,
then we call an orthogonal basis of W consisting entirely of standard symmetrized tensors
an o-basis of W relative to G and ?. One may ask about conditions on G which will
guarantee the existence of an o-basis for V?n relative to G and ? for all ? 2 Irr(G). When
discussing this situation, we will suppress reference to ? and talk about the existence on
o-basis relative to G.
Now let G be an arbitrary group. One may ask if there are homomorphisms ? : G ! Sn
such that V?n has an o-basis relative to ?(G). Having flxed G, one might also wonder if
30
some embeddings work while other do not. In [Hlms], Holmes has shown that, for G an
o-basis group, V?n has an o-basis regardless of the homomorphism.
Theorem 2.2.1 ([Hlms], p.138) If G is an o-basis group and ? : G ! Sn(n 2 N) is a
homomorphism, then V?n has an o-basis relative to ?(G).
The reader may recall Cayley?s Theorem which states that for any group G there is a
homomorphic injection of G onto a subgroup of Sn. For each g 2 G, one deflnes ?(g) to
be the permutation of G given by ?(g)(h) = gh (h 2 G). In this case, ?(G) can be viewed
as a subgroup of Sn where n = jGj. This map is called the Cayley embedding. With
the next result, Holmes provides, as he says, a characterization of o-basis groups in terms
of symmertized tensors.
Theorem 2.2.2 ([Hlms], p.139) Let G be a flnite group, let n = jGj, and let ? : G ! Sn
be the Cayley embedding. Then G is an o-basis group if and only if V?n has an o-basis
relative to ?(G).
In this work, we are interested in studying the o-basis property as a tool for distinguish-
ing between abstract groups. However, as the above discussion indicates, those working with
symmetrized tensor spaces may flnd the class of o-basis groups interesting as well.
31
Chapter 3
New Work
3.1 A Generalized Deflnition and Some Preliminary Results
In this section, we deflne a generalization of o-basis group and make some elementary
observations. We will begin to use the terminology immediately. Later, in section 3.3,
we will further explore the generalized notion. Also in this section, we obtain several
preliminary results that will be key to some of the techniques used in later sections. Let us
flx, for the remainder of this work, a flnite group G.
Deflnition 3.1.1 For H ? G and ? 2 Irr(G), say that G is ?H;??-o-basis if there
are at least ?(e)??;1?H cosets of H in G which are mutually orthogonal relative to B?H.
Fix K ? G. If G is ?H;??-o-basis for all subgroups H with K H and all non-linear
? 2 Irr(G) for which (?;1)H 6= 0, we say G is K-o-basis.
Note that G is o-basis (see Deflnition 2.1.3) precisely when G is hei-o-basis so that the
generalized deflnition includes the original. Also, for any two subgroups H;K of G with
H K, whenever G is H-o-basis, G is also K-o-basis.
Since we do not want to refer to the linear algebra involved in Holmes? original deflnition
of o-basis, we have introduced notation in the above deflnition which avoids reference to
that material. Let us restate, for use later, the last part of Theorem 2.1.5 with our new
notation.
Let N C G, let ? 2 Irr(G) and assume that N ker?. For a subgroup H ? G, denote
by bH the image of H under the canonical map G ! G=N. The function b? : bG !C given
32
by b?(gN) = ?(g) is a well-deflned irreducible character of bG (see Theorem 1.3.2.5, (i) and
(iii)).
Theorem 3.1.2 Keeping the above notation, let N C G and let ? 2 Irr(G) such that
N ker(?). Let H ? G. Then G is (H;?)-o-basis if and only if bG is (bH;b?)-o-basis.
In the following lemma and its corollary, we obtain an upper bound on the number of
orthogonal cosets. This will be critical in proving that certain groups are not o-basis.
Lemma 3.1.3 Let ? 2 Irr(G). Suppose that H;K ? G with H K. Assume that no
two cosets of H in K are orthogonal relative to B?H. Then the number of cosets of H in G
which are mutually orthogonal relative to B?H is no greater than jG : Kj.
Proof: By assumtion, for a;b 2 K we have 0 6= B?H?aH;bH? = ?(e)jHj
X
h2H
??a?1bh?. Sup-
pose there are more than jG : Kj cosets of H in G which are orthogonal relative to B?H.
Then at least one coset of K in G contains two cosets of H which are orthogonal. More
precisely, there exists g 2 G and a;b 2 K such that
0 = B?H?gaH;gbH? = ?(e)jHj
X
h2H
??a?1g?1gbh? = ?(e)jHj
X
h2H
??a?1bh?, a contradiction.
?
Keeping the above notation, let aH and bH be two (not necessarily distinct) cosets of H in
K which are orthogonal relative to B?H. Then 0 = B?H(aH;bH) = ?(e)jHj Ph2H ?(a?1bh) so
that Ph2H ?(kh) = 0 for some k 2 K. Conversely, let k 2 K such that Ph2H ?(kh) = 0.
There exist a;b 2 K (a;b not necessarily distinct) such that a?1b = k. Note that aH and
bH are orthogonal relative to B?H. In short, there are two (not necessarily distinct) cosets
of H in K which are orthogonal relative to B?H if and only if for some k 2 K
33
(possibly k = e),Ph2H ?(kh) = 0. Now suppose H = hei. The cosets of H are the singleton
sets fgg (g 2 G). In this case, the above statement becomes: there are two (not necessarily
distinct) \orthogonal elements" in K relative to B?H if and only if ?(k) = 0 for some k 2 K.
We restate the special case of Lemma 3.1.3 when H = hei as a corollary.
Corollary 3.1.4 Let ? 2 Irr(G) and K ? G. Suppose ?(k) 6= 0 for all k 2 K. Then the
number of cosets of hei which are mutually orthogonal relative to B?hei is no greater than
jG : Kj.
We will often flnd it helpful to deal with quotients of G. The utility of this derives from
the result below which follows from Proposition 1.3.2.5 and Theorem 3.1.2.
Theorem 3.1.5 Let N ?G. Then G is N-o-basis if and only if G=N is o-basis.
Proof: For L ? G, let bL denote the image of L under the canonical map G ! G=N, and,
for ? 2 Irr(G) deflne b? : bG !C by b?(gN) = ?(g).
AssumethatGisN-o-basis. LetK ? G=N and? 2 Irr(G=N). BytheCorrespondance
Theorem, K = bH for some H ? G with N H. By Proposition 1.3.2.5 (ii) and (iii), ? = b?
for some ? 2 Irr(G). As G is N-o-basis, G is (H;?)-o-basis. By Theorem 3.1.2, G=N is
(K;?)-o-basis. Since K;? were chosen arbitrarily, this shows that G=N is o-basis.
Conversely, assume that G=N is o-basis. Let H ? G with N H and let ? 2 Irr(G)
such that (?;1)H 6= 0. Then (?;1)N 6= 0. Since N C G, Theorem 1.3.3.4 gives that
N ker(?). Since G=N is o-basis, G=N is (bH;b?)-o-basis. By Theorem 3.1.2, G is (H;?)-
o-basis and it follows that G is N-o-basis.
?
34
3.2 O-basis Groups and Nilpotency
In this section, we address two questions.
1) Which nilpotent groups are o-basis?
2) Are all o-basis groups nilpotent?
The possibility of a special connection between o-basis groups and nilpotentcy is im-
plied by two facts. First, every group that has been identifled as o-basis is also nilpotent
(see Theorem 1.1.1). Second, the o-basis property picks out from the dihedrals exactly the
nilpotent groups (see Theorem 1.1.3). Not all nilpotent groups are o-basis however. In
[Hlms] (p.143), Holmes constructs an example of order 34 that is not o-basis. What condi-
tions, then, on a nilpotent group are su?cient for the group to be o-basis? Our flrst result
gives such a condition: that G0 Z(G). Note that this condition implies nilpotentcy.
Theorem 3.2.1 Suppose that G0 Z(G). Then G is o-basis.
Proof: Let H ? G and let ? 2 Irr(G). For any subgroup K ? G, let bK denote the
image of K under the canonical map G ! G=ker?. Deflne b? : bG !Cby b?(gker?) = ?(g)
for all g 2 G. By Lemma 1.3.2.5, b? 2 Irr(bG) and, by Theorem 3.1.2, G is (H;?)-o-basis if
and only if bG is (bH;b?)-o-basis. It su?ces, therefore, to show that bG is (bH;b?)-o-basis. Note
that ker(b?) = feg
We flrst show that bG=Z(b?) is abelian. Note that (bG)0 Z(bG). For since G0 Z(G),
we have (bG)0 = cG0 [Z(G) [Z(?) = Z(bG), where the last containment and equality
follow respectively from Corollary 1.3.2.8 and Lemma 1.3.2.9. Also, Z(bG) = Z(b?). Indeed,
Corollary 1.3.2.8 gives that Z(bG) Z(b?). Conversely, suppose that gker? 2 Z(b?). Then
j?(g)j = jb?(gker?)j = b?(eker?) = ?(e) so that gker? 2 [Z(?) = Z(bG). This shows that
(bG)0 Z(b?) so that bG=Z(b?) is abelian as claimed. By Theorem 1.3.2.11, b?(e)2 = jbG : Z(b?)j
35
and so, by Proposition 1.3.2.10, b? ? 0 on bG?Z(b?). Since Z(b?) = Z(bG), we have b? ? 0 on
bG?Z(bG).
Assume that (b?;1)x62H 6= 0. In this case, (b?;1)x62H\Z(x62G) 6= 0. As bH \Z(bG)? bG, Proposi-
tion 1.3.3.4 gives that bH \Z(bG) ker(b?) = feg. Since b? ? 0 on bG?Z(bG) we have
(b?;1)x62H = 1jbHj
X
h2x62H
b?(h) = b?(e)jbHj :
Therefore, b?(e)(b?;1)x62H = b?(e)
2
jbHj =
jbG : Z(bG)j
jbHj =
jbGj
jZ(bG)j?jbHj = j
bG : bHZ(bG)j, where the last
equality holds since bH \Z(bG) = feg. Let fai bHZ(bG) : 1 ? i ? tg be a complete set of coset
representatives of bHZ(bG) in bG. Suppose i 6= j. Then, for all h 2 bH, a?1i ajh =2 Z(bG). Since
b? ? 0 on bG?Z(bG), b?(a?1i ajh) = 0 for each h 2 bH. Therefore,
Bx62?x62H(ai bH;aj bH) = b?(e)jbHj
X
h2x62H
b?(a?1i ajh) = 0:
The b?(e)(b?;1)x62H cosets fai bH : 1 ? i ? tg form a mutually orthogonal collection of cosets of
bH in bG. This shows that G is (bH;b?)-o-basis and the proof is complete.
?
To obtain the following corollary to Theorem 3.2.1, we will use two lemmas.
Lemma 3.2.2 ([Is], p.75, ex. 5.14(a)) Let G be non-abelian and
let f = minf?(e) : ? 2 Irr(G);?(e) > 1g. If jG0j? f, then G0 Z(G).
Lemma 3.2.3 ([Is], p.38) Let ? 2 Irr(G). Then ?(e) divides jGj.
36
Corollary 3.2.4 Suppose G is a p-group with jG0j = p. Then G is o-basis.
Proof: Since G0 6= hei, G is non-abelian. By Theorem 1.3.2.3, G has non-linear ir-
reducible characters. Let ? be an arbitrary non-linear irreducible character of G. By
Theorem 3.2.3, ?(e) divides jGj so that ?(e) ? p. Since ? was chosen arbitrarily and
jG0j = p ? ?(e), Lemma 3.2.2 gives that G0 Z(G). By Theorem 3.2.1, G is o-basis.
?
We note that, in Corollary 3.2.4, one can easily show that G0 Z(G) without using
characters. However, our approach has the advantage of giving the reader further exposure
to elementary character theory.
We will have occasion to call upon Theorem 3.2.1 in several results. We will also give a
slight generalization the theorem (Theorem 3.3.5). In addition to this, Theorem 3.2.1 raises
an interesting question for possible future study. Any group with G0 Z(G) has nilpotence
class no greater than 3 (see the deflnition of the lower central series in the notation). Holmes?
example of order 34 that is not o-basis is easily seen to have nilpotence class 4. We ask if
nilpotence class less than or equal to 3 is a necessary condition for a nilpotent group to be
o-basis. This is an open question, and one the author looks forward to considering in the
future.
In studying nilpotent groups one might wish to narrow the focus by concentrating
on p-groups. In our next result, we flnd that there is some valid grounds for doing so.
More precisely, we show that a nilpotent group is o-basis precisely when each of its Sylow
subgroups is o-basis. Before proving this, we remind the reader of some basic information
about Sylow subgroups.
37
Deflnition 3.2.5 Let p be a prime and let G be a group. A p-subgroup of G is a subgroup
whose order (cardinality as a set) is a power of p. A Sylow p-subgroup is a p-subgroup
that is not properly contained in any other p-subgroup.
Thus a Sylow p-subgroup is a maximal p-subgroup. The reader may recall that, given a
prime p, if a Sylow p-subgroup is normal in G, then it is the unique Sylow p-subgroup of G
for that prime. Our result for nilpotent groups depends heavily on Lemma 3.2.6 below.
Lemma 3.2.6 ([Rob], p.134 ex. 12) For 1 ? i ? n, let pi denote a prime such that
pi 6= pj whenever i 6= j and let Gi denote a pi-group . Let G = Qni=1 Gi, the direct product
of the Gi. Let H ? G. Then H = Qni=1 Hi, where Hi = H \Gi.
Theorem 3.2.7 ([Rob], p.126)Let G be a flnite group. Then G is nilpotent if and only
if G is the direct product of its Sylow subgroups.
Theorem 3.2.8 Assume that G is nilpotent. Then G is o-basis if and only if every Sylow
subgroup of G is o-basis.
Proof: LetfPigmi=1 be the distinct Sylow subgroups of G. By Theorem 3.2.7, G = Qmi=1 Pi.
Suppose that G is o-basis and flx 1 ? j ? m. Put D = Qmi=1;i6=j Pi. Note that D C G and
let ? : G ! G=D denote the canonical map. Then ?(G) = Pj. Theorem 2.1.6 gives that Pj
is o-basis. As j was chosen arbitrarily, the forward direction is proved.
Suppose now that Pi is o-basis for each 1 ? i ? m. Let H ? G and let ? 2 Irr(G)
be non-linear such that (?;1)H 6= 0 (see Deflnition 2.1.3). For each i, put pi = jPij (so
pi is a prime). Since Pi C G, Pi is the unique pi-subgroup of G. It follows that pi 6= pj
whenever i 6= j. Therefore, Lemma 3.2.6 gives that H = Qmi=1 Hi, where Hi = H\Pi. Also,
38
Theorem 1.3.3.2 gives that ? = ?1????m, where ?i 2 Irr(Pi) is extended to G by putting
?i(Qmj=1 gj) = ?i(gi).
For convenience, let g1???gm denote the element (g1;:::;gm) 2 G. Let a = a1???am
and b = b1???bm be elements of G. We claim that
B?H(aH;bH) = B?1H1(a1H1;b1H1)???B?mHm(amHm;bmHm): (3.1)
Proceed byinduction onm. The formulaisobviousforthe casem = 1. Therefore, flxm > 1.
Put ^H = Qm?1i=1 Hi, ^G = Qm?1i=1 Pi, ^? = ?1????m?1, ^a = a1???am?1, and ^b = b1???bm?1.
Note that ^? 2 Irr(^G) and assume
B^?^H(^a ^H;^b ^H) = B?1H1(a1H1;b1;H1)???B?m?1Hm?1(am?1Hm?1;bm?1Hm?1):
We have,
B?H(aH;bH) = ?(e)jHj
X
h2H
?(a?1bh)
= ?1(e)????m(e)jH
1j???jHmj
X
h1???hm2H
?(a?11 b1h1???a?1m bmhm)
= ?1(e)????m(e)jH
1j???jHmj
X
h1???hm2H
?1(a?11 b1h1)????m(a?1m bmhm)
= ?m(e)jH
mj
X
hm2Hm
?
?m(a?1m bmhm)^?(e)j ^Hj
X
h2^H
^?(^a?1^bh)
?
= ?m(e)jH
mj
X
hm2Hm
?m(a?1m bmhm)B?1H1(a1H1;b1H1)???B?m?1Hm?1(am?1Hm?1;bm?1Hm?1)
= B?1H1(a1H1;b1H1)???B?mHm(amHm;bmHm);
where the next to last equality follows from by the induction hypothesis. The claim follows.
39
Also observe that
?(e)(?;1)H = ?(e)jHj
X
h2H
?(h) = ?(e)jHj
X
h2H
?(a?1ah)
= B?H(aH;aH)
= B?1H1(a1H1;a1H1)???B?mHm(amHm;amHm)
= ?1(e)(?1;1)H1 ????m(e)(?m;1)Hm;
where the next to last equality follows from equation 3.1. Thus, we have
?(e)(?;1)H = ?1(e)(?1;1)H1 ????m(e)(?m;1)Hm: (3.2)
For each 1 ? i ? m, let Ai denote a set (non-empty, as will be shown) of distinct cosets
representatives of Hi in Pi such that B?iHi(aHi;bHi) = 0 whenever a;b 2 Ai with a 6= b.
Since (?;1)H 6= 0, equation 3.2 gives that (?i;1)Hi 6= 0 for all i. Since Pi is o-basis, we may
assume that Ai contains at least ?i(e)(?i;1)Hi elements.
Now let a = a1???am and b = b1???bm be elements of G with a 6= b such that ai;bi 2 Ai
for each i. As a 6= b, there is at least one i such that ai 6= bi. Then, by deflnition of Ai,
B?iHi(aiH;biH) = 0. It follows from equation 3.1 that B?H(aH;bH) = 0. We see that
the set cartesian product A1 ????? Am forms a collection of coset representatives of H
in G that are mutually orthogonal relative to B?H. The cardinality of this collection is
?1(e)(?1;1)H1 ????m(e)(?m;1)Hm, and we have already noted that this last quantity is
equal to ?(e)(?;1)H. Thus G has an (H;?)-o-basis. It follows that G is o-basis as desired.
?
Suppose G is a direct product of a flnite number of groups. It follows immediately from
Theorem 2.1.6 that whenever G is o-basis, each of the direct factors is o-basis as well. The
truth or falsity of the converse however remains an open question. Critical to the converse
40
in the above argument is the fact that the primes of the distinct Sylow subgroups are all
distinct. Lemma 3.2.6 then gives that any subgroup is the direct product of its intersection
with the factors. The absence of this property for subgroups of general direct products has
been the primary obstacle to the analogous result for that case. For example, it is suspected
(but not proven) that the "diagonal" subgroup could fail for some non-linear character.
In light of the above result, we see that the question of which o-basis groups are
nilpotent can be "reduced" in some sense to that of which prime power groups are o-basis.
Let us brie y consider this. Suppose p is a prime. Since any group of order p2 is abelian,
Theorem 2.1.4 gives that any such group is o-basis. Holmes? has shown that any group
of order p3 is o-basis (see Theorem 1.1.1). However, in [Hlms], Holmes also provided an
example of order 34 that is not o-basis. We take a closer look at groups of order p4 beginning
with the fact that any such group G is Z(G)-o-basis (see Deflnition 3.1.1).
Theorem 3.2.9 Suppose that jGj = p4. Then G is Z(G)-o-basis. That is, every group of
order p4 is Z-o-basis.
Proof: Since G is a p-group, we have by Theorem 1.2.4 that Z(G) is non-trivial. Thus
jG=Zj ? p3. If jG=Zj ? p2, then G=Z is abelian and so o-basis by Theorem 1.1.1 (i).
Suppose that jG=Zj = p3. Then G=Z is o-basis by Theorem 1.1.1 (iv). By Theorem 3.1.5,
G is Z-o-basis if and only if G=Z is o-basis, and the result follows.
?
Which groups of order p4 are o-basis? We explore this question as follows. Suppose G
is a group of p4 that is not o-basis. Then there exists a subgroup H ? G and ? 2 Irr(G)
such that G is not (H;?)-o-basis. We derive some necessary conditions on H, ? and G. We
will have need of the following two lemmas.
41
Lemma 3.2.10 ([Is] p.204) Let A C G with A abelian and G=A cyclic. Then jAj =
jG0j?jA\Z(G)j.
Theorem 3.2.11 ([Karp], p.803) Let ? be an irreducible character of a nilpotent group G.
Then ?(e)2 divides jG : Z(?)j.
Theorem 3.2.12 Let p be prime and suppose jGj = p4 . Let ? 2 Irr(G) and let H ? G
such that G is not (H;?)-o-basis. Then the following hold.
(i) jZ(G)j = p,
(ii) ? is faithful and ?(e) = p,
(iii) There exist A C G with A abelian and jG : Aj = p.
Also, ? ? 0 on G?A.
(iv) Z(G) ? G0 ? A and jA : G0j = jG0 : Z(G)j = p,
(v) If ?(a) = 0 for some a 2 G0, then ? ? 0 on G0 ?Z(G). In this case,
H 6G.
Proof: Throughout, let Z denote the center of G and K the kernel of ?. Suppose flrst
that jZj? p2. Then G=Z is a p-group with order no greater than p2. It follows that G=Z is
abelian so that G0 Z. By Theorem 3.2.1, G is o-basis, a contradiction. By Theorem 1.2.4,
jZj > 1. Thus, jZj = p and (i) is established.
For the flrst condition of (ii), suppose, in view of a contradiction, that K 6= feg. As
K C G, we have by Theorem 1.2.5 that K \ Z 6= feg. Since jZj = p, it follows that
Z K. Deflne b? : G=Z ! C by putting b?(gZ) = ?(g) for all g 2 G. Since Z K,
Proposition 1.3.2.5 gives that b? 2 Irr(G=Z). By Theorem 3.2.9, G is Z-o-basis. Thus G=Z
is o-basis by Theorem 3.1.5 so that G=Z has a (?(H);b?)-o-basis, where ? : G ! G=Z is
42
the canonical map. It follows from Theorem 3.1.2 that G is (H;?)-o-basis, contrary to our
assumption. This shows that ? is faithful.
Recalling that all p-groups are nilpotent, we see from Theorem 3.2.11 that ?(e)2 divides
jG : Z(?)j. By Corollary 1.3.2.8, Z(?) ? Z(G) 6= feg so that jG : Z(?)j ? p3. Since ?
is non-linear, ?(e) = p, and (ii) is fully established. By Theorem 1.3.3.14, G is an M-
group. Choose a subgroup A of G and linear character ? 2 Irr(A) such that ? = ?G.
We have p = ?(e) = ?G(e) = jG : Aj?(e) = jG : Aj (see remark after Deflnition 1.3.3.5).
Note that A is a maximal subgroup of G. For suppose L ? G with A ? L ? G. Then
p = jG : Aj = jG : Lj?jL : Aj. Either jG : Lj = 1 so that L = G or jL : Aj = 1 so that L = A.
By Theorem 1.2.6, A C G. Thus, ? ? 0 on GnA. (see comments after Deflnition 1.3.3.5)
For (iii), it remains to show that A is abelian. Note that j?(z)j = ?(e) 6= 0 for all z 2 Z(?).
Since ? ? 0 on G?A, Z(?) A so that, by Corollary 1.3.2.8, Z(G) A. Suppose that
A is not abelian. As jAj = p3, A is extra-special (see Deflnition 1.2.7 and remarks). We
have that Z(A) = A0 and jZ(A)j = p. Since Z(G) A, if follows that Z(G) Z(A). Thus,
as jZ(G)j = p, we have that Z(A) = Z(G). Since ? is linear, Z(A) = A0 ker?, the last
containment being given by Proposition 1.3.2.6. By Lemma 1.3.3.6, ker? = Tx2G(ker?)x.
If follows that Z ker?, a contradiction since ? is faithful. As claimed therefore, A is
abelian and the proof of (iii) is complete.
By Lemma 3.2.10, p3 = jAj = jZ(G)j?jG0j = pjG0j so that jG0j = p2. By Theorem 1.2.5,
Z(G) G0. Also, since A C G and G=A is abelian, we have that G0 A. This establishes
(iv).
Suppose that G0 is cyclic. Then putting C = G0 in Theorem 2.1.2, we see that G is
o-basis, a contradiction. Thus G0 cannot be cyclic. It follows that G0 ?= Zp ?Zp. Thus,
43
every proper, non-trivial subgroup of G0 has order p. Since no two of these subgroups can
intersect non-trivially, there are p + 1 such subgroups and G0 is their union. Since G is
a p-group, every non-trivial, normal subgroup must intersect the center non-trivially. If
follows that Z(G) is the only normal subgroup of G having order p. Therefore, Z(G) is the
only subgroup of G0 that is normal in G. Since G0 A and A is abelian, G=A acts on the
subgroups of G0 by conjugation. For any given subgroup, the order of the orbit under this
action is either 1 or p. Moreover, any subgroup having orbit of size 1 is normal in G. It
follows that G=A acts transitively on the p non-central subgroups of G0.
For (v), suppose that ?(a) = 0 for some a 2 G0. Then a =2 Z(G) since Z(G) Z(?)
(Theorem 1.3.2.8). Put K = hai. Then ?K is a character of K. Putting ?K = ? in
Proposition 1.3.2.12, it follows, since jKj = p, that ? ? 0 on K ? hei. Since G acts
transitively on the p non-central subgroups of G0, we see that G ? Z(G) is the union of
the conjugates of K. Recalling that character values are invariant under conjugation (see
the discussion on class functions in section 1.3.1) we have that ? ? 0 on G0 ? Z(G), as
claimed in (v). Since G is not (H;?)-o-basis, it follows from the deflnition that (?;1)H 6= 0.
Suppose, to obtain a contradiction, that H C G. Then H ker? by Theorem 1.3.3.4
so that H = hei. Let fai : 1 ? i ? pg and fgi : 1 ? i ? pg be complete sets of coset
representatives of Z in G0 and of A in G respectively. Put S = fgjai : 1 ? i;j ? pg. Choose
gj1aii;gj2ai2 2 S and put y = a?1i1 g?1j1 gj2ai2. We have B?H(gj1aiiH;gj2ai2H) = ?(e)?(y). If
j1 6= j2, then y 2 G?A and ?(y) = 0. If j1 = j2, then y = a?1i1 ai2 2 G0 ?Z and ?(y) = 0.
This gives jSj = p2 cosets of H which are mutually orthogonal relative to B?H. Finally, note
that ?(e)(?;1)H = ?(e)2 = p2. This shows that G is (H;?)-o-basis, a contradiction. It
follows that H 6G and the proof is complete.
44
?
All of the results so far in this section have dealt with the flrst of two questions posed
at its beginning. Namely, which nilpotent groups are o-basis. Let us now turn to the second
question. Are all o-basis groups nilpotent? With the dihedrals in mind, we will consider
our second question for a particular class of \dihedral-like" groups. Every dihedral group
has a cyclic normal subgroup of index 2. We consider those groups G having an abelian
normal subgroup A of index pn, where p is a prime and n ? 1. Let us state our question as
a conjecture.
Conjecture 3.2.13 Let p be prime and A C G be abelian such that jG : Aj = pn for some
positive integer n. Suppose that G is o-basis. Then G is nilpotent.
We are able to prove the conjecture in the case that G=A is abelian (Theorem 3.2.17). When
G=A is non-abelian, we show the answer is still a?rmative when n = 3 in Theorem 3.2.20.
In Theorem 3.2.21, we obtain the conjecture for the case n = 4, but only after adding
some additional conditions. Finally, we show that nilpotency is a necessary and su?cient
condition for o-basisness when A is cyclic and n = 1. The following three lemmas set the
stage.
Lemma 3.2.14 Let p be prime and A C G be abelian with jG : Aj = pn for some n ? 1.
Assume that G is not nilpotent. Then there is an integer m ? 1 such that the following
hold:
(i) G=Zm is not abelian,
(ii) AZm=Zm C G=Zm is abelian,
(iii) jG=Zm : AZm=Zmj = pk for some k ? 1,
(iv) ?jG=Zm : AZm=Zmj;jAZm=Zmj? = 1.
45
Proof: Assume G is not nilpotent. Then there is an integer m ? 1 such that Zm = Zm+i,
for all i ? 0 and G 6= Zm. Fix such an m and note that G=Zm 6= hei and Z(G=Zm) =
hei. Since AZm C G, the Correspondance Theorem gives that AZm=Zm C G=Zm. Also,
AZm=Zm ?= A=(A\Zm). The latter is abelian, being the homomorphic image of an abelian
group. This verifles (i). Observe that Zm + A. Indeed, suppose that Zm ? A. Then
G=Zm ?= (G=A)=(Zm=A). The latter is a p-group since we have assumed G=A is a p-
group. Therefore, Z(G=Zm) 6= hei, a contradiction. It follows that AZm=Zm is a non-trivial
subgroup of G=Zm with jG=Zm : AZm=Zmj = jG : AZmj. Now note that this last quantity
dividesjG : Ajand is therefore equal to pk for some k ? 0. If k = 0, then G=Zm = AZm=Zm.
In this case, G=Zm is abelian by (i). This again contradicts the fact that Z(G=Zm) = hei.
Therefore, k > 0 and (ii) is established.
Let T denote the quotient group G=AZm. There is a well-deflned action of T on
(AZm)=Zm given by conjugation and Z(G=Zm) is precisely the set, (AZm=Zm)T, of flxed
points of AZm=Zm under this action. Theorem 1.2.1 gives that
1 = jZ(G=Zm)j = j(AZm=Zm)Tj?jAZm=Zmj (mod p):
Thus (jAZmj;jG=Zm : AZm=Zmj) = 1 and (iii) is established.
?
46
Lemma 3.2.15 Let p be prime and A C G be abelian such that jG : Aj = pn for some
integer n ? 1. Assume that G is not nilpotent. Then there exists K C G such that the
following hold:
(i) G=K is a non-abelian internal semidirect product of AK=KoT,
where T ?= (G=K)??AK=K?,
(ii) jG=K : AK=Kj = pn for some n ? 1, and
(iii) AK=K is a non-trivial elementary abelian q-group for some
prime q distinct from p.
Proof: Since G is not nilpotent, Lemma 3.2.14 applies. Choose an integer m satisfying
the conclusions of Lemma 3.2.14. Suppose we are able to construct a normal subgroup K
of G=Zm with K AZm=Zm such that K satisfles (i) - (iii) above for G=Zm. Then the
Correspondance Theorem gives a normal subgroup K of G which also satisfles properties
(i) - (iii). If, in addition, G=Zm possessed the properties assumed of G, there would be
no loss of generality in working with G=Zm rather G. Since G=Zm also possesses property
(iii) of Lemma 3.2.14, this would be to our advantage. Let us note that, by (i) and (ii) of
Lemma 3.2.14, G=Zm does indeed possess the properties assumed here of G. Thus, we may,
with loss generality, identify G with G=Zm, A with AZm=Zm and assume, by 3.2.14 (iii),
that (jAj;p) = 1.
Denote by Q the set of prime divisors of jAj. Since A is abelian, every subgroup of
A is normal in A. In particular, any Sylow subgroup of A is normal in A and is therefore
unique. For q 2 Q, denote by Aq the unique Sq subgroup of A and put Dq = Qr2Q
r6=q
Ar if
jQj > 1. If jQj = 1, put Dq = hei. Note that Aq C G for all q 2 Q. Indeed, choose g 2 G.
Since A C G, (Aq)g A. Also, j(Aq)gj = jAqj so that (Aq)g is also a Sylow q-subgroup of A.
By uniqueness, (Aq)g = Aq. Now flx q 2 Q. Since A is nilpotent, it is the direct product of
47
its Sylow subgroups so that Dq ? A. It follows from the above discussion that each direct
summand of Dq C G so that Dq C G. Note that, since (jAj;p) = 1, q 6= p.
For each subset S G, denote by bS the image of S under the canonical map G !
G=Dq. Note that bA is a non-trivial abelian q-group. We are ready to establish (iii). The
Fundamental Theorem of Finitely Generated Abelian Groups gives that bA = Qki=1 Ai, where
Ai ?= Zqmi and mi ? 1 for all i. Put m = maxfmig1?i?k. Deflne subgroups Bi of bA as
follows: let Bi = Ai whenever mi < m, let Bi be the unique subgroup of Ai of order qmi?1
whenever mi = m. Put B = Qki=1 Bi. Observe that B char bA. To see this, flx 1 ? i ? k
and choose bi such that Bi = ?bifi. Let 2 Aut(bA). It su?ces to show that b i 2 B. We
write b i = Qki=1 ai, where ai 2 Ai and calculate
qm > jbij = jb i j =
flfl
flfl
kY
i=1
ai
flfl
flfl = lcmfjaij : 1 ? i ? kg:
Therefore, jaij < qm so that ai 2 Bi and b i 2 B, as desired. By Theorem 1.2.3, B C G. It
follows that there is a subgroup K C G, such that G=K ?= ?G=Dq?=B. Since B ( bA, we
have jG=K : A=Kj = jG : Aj so that (ii) holds. Also A=K ?= bA=B and bA=B is an elementary
abelian q-group so that (iii) holds. Finally, since (jA=Kj;p) = 1, conclusion (i) follows from
Theorem 1.3.4.3. The proof is complete.
?
Theorem 3.2.16 Let p;q be distinct primes. Let A C G be an elementary abelian q-group
with jG : Aj = pn for some integer n ? 1. Suppose ? 2 Irr(G) is non-linear. Then ?(a) 6= 0
for all a 2 A.
48
Proof: Let a 2 A and assume that ?(a) = 0. Put L = ?afi. Note that ?L is a character
of L (see Deflnition 1.3.3.3). Since A is elementary abelian, L ?=Zq (see Deflntion 1.2.10).
Therefore, for each b 2 L with b 6= e, we have that b = ai, where (i;q) = 1. Putting ?L = ?
in Proposition 1.3.2.12, gives that ? vanishes on L?feg.
We have (?L;?L)L = 1jLj
X
l2L
j?(l)j2 = ?(e)
2
jLj =
?(e)2
q . By Proposition 1.3.1.10, this
last quantity should be an integer. However, by Theorem 1.3.2.13, ?(e) divides jG : Aj, a
power of p. This is a contradiction. It follows that ?(a) 6= 0 for all a 2 A.
?
Case 1: G=A is abelian.
Theorem 3.2.17 Let p be prime and A C G be abelian such that jG : Aj = pn for some
integer n ? 1. Assume also that G=A is abelian. If G is o-basis then G is nilpotent.
Proof: We prove the contrapositive. Assume that G is not nilpotent. We show that G
is not o-basis. Note that Theorem 3.2.14 applies and choose an integer m such that the
conclusions of that theorem hold. By Theorem 2.1.6, it su?ces to show that G=Zm is not
o-basis. By the Third Isomorphism Theorem, (G=Zm)=(AZm=Zm) ?= G=(AZm). The latter
is a quotient of G=A and so abelian. It follows that G0 AZm so that (G=Zm)0 AZm=Zm.
Note that?G=Zm?0 AZm=Zm. This, along with 3.2.14 (ii) and (iii), shows that G=Zm
has all the properties we have assumed of G. Without loss of generality, we replace G with
G=Zm, A with AZm=Zm and assume, citing 3.2.14 (iii), that (jAj;p) = 1.
Because G is non-abelian (see 3.2.14 (i)), there are normal subgroups of G with non-
abelian quotient, the identity comprising one such subgroup. Since G is flnite, we may
choose from among the set of such subgroups one that is maximal with respect to contain-
ment. Let us make such a choice and call the subgroup K. Observe that every non-trivial,
49
normal subgroup of G=K contains ?G=K?0. For suppose H is a non-trivial, normal sub-
group of G=K that does not contain (G=K)0. Then H = H=K for some normal subgroup
H of G, where H % K. Recalling that (G=K)0 = G0K=K, we have H + G0. It follows
that G=H is non-abelian, contradicting the maximality of K. Therefore, (G=K)0 is the
unique minimal (non-trivial) normal subgroup of G=K. In addition, since G0 A, G0 is
abelian. It follows that G is solvable so that G=K is solvable. These arguments show that
G=K satisfles the hypotheses of Lemma 1.3.5.4. Since G=K is non-abelian and G0 A,
we have that K + A. Thus AK=K is a nontrivial subgroup of G=K. Since (jAj;p) = 1,
G=K is not a p-group. Conclusion (b) of Lemma 1.3.5.4 therefore applies. We have that
G=K is a Frobenius group having Frobenius kernel (G=K)0 AK=K and that (G=K)0 is
an elementary abelian q-group for some prime q. Since (jAj;p) = 1, q 6= p.
By Theorem 2.1.6, it is enough to show that G=K is not o-basis. Recalling that G=K
is non-abelian, flx a non-linear ? 2 Irr(G=K).
Note that ker?+(G=K)0 (see remarks after Proposition 1.3.2.5). Theorem 1.3.5.3 (b) gives
that ? = `G=K for some ` 2 Irr((G=K)0). Now ?(e) = `G=K(e) = jG=K : (G=K)0j?`(e) =
jG=K : (G=K)0j, where the last equality holds since ` is linear ((G=K)0 being abelian).
By Theorem 1.3.2.13, jG=K : (G=K)0j divides jG=K : AK=Kj. But jG=K : AK=Kj divides
jG=K : (G=K)0j since (G=K)0 AK=K. Thus the indices are equal, AK=K = (G=K)0 and
?(e) = jG=K : AK=Kj. Since AK=K is proper and elementary abelian, Theorem 3.2.16
gives that ? never vanishes on AK=K. Put H = hei. For G=K to be (H;?)-o-basis, there
must be as least ?(e)2 = jG=K : AK=Kj2 orthogonal cosets of H in G=K relative to B?H.
However, since ? never vanishes on AK=K, Corollary 3.1.4 gives that there are at most
jG=K : AK=Kj such cosets. This shows that G=K is not o-basis and completes the proof.
50
?
Case II: G=A is not abelian.
We now turn to the situation where G=A is not abelian. Note that, in this case,
jG : Aj ? p3. In Lemma 3.2.18 and Theorem 3.2.19 below, we will assume that G is a
semi-direct product. We refer the reader to section 1.3.4 for a discussion of semi-direct
products and we adopt here the notation of that section.
Lemma 3.2.18 Let p;q be distinct primes. Let G = AoT be an internal semi-direct
product. Assume that A is an elementary abelian q-group and that jTj = pn for some
n ? 1. Suppose that, for some ? 2 Irr(A), jT?j = pk, where 0 ? k < n2. Then G is not
o-basis.
Proof: Let ? 2 Irr(T?), put H = AT? and ? = ?;?. We show that G is not ?hei;??-o-
basis. We have ?(e) = jG : Hj? ?;?(e) ? jG : Hj = pn?k. Since n > k, ? is non-linear. By
Theorem 3.2.16, ? never vanishes on A. Applying Corollary 3.1.4 with K = A gives that
the number of cosets of hei which are orthogonal relative to B?hei is no greater than jG : Aj.
However, the required number of cosets is ?(e)(?;1)hei = ?(e)2 ? p2n?2k > pn = jG : Aj.
?
Theorem 3.2.19 Let p;q be distinct primes. Let G = AoT be an internal semi-direct
product. Assume that A is an elementary abelian q-group and that jTj = pn for some n ? 1.
Suppose there exists ? 2 Irr(A) such that T? is a proper, normal subgroup of T. Then G is
not o-basis.
Proof: Choose ? 2 Irr(A) satisfying the hypotheses. Let ? = 1T?, put H = AT? and
? = ?;? := (?~?)G (for convenience, we will suppress the subscript and write simply ).
51
Note that ?(e) = jG : Hj ? (?~?)(e) ? jG : Hj. We show that ?(h) 6= 0 for all h 2 H.
Since T? C T, the Correspondance Theorem gives that H C G. We may therefore apply
Clifiord?s Theorem (Theorem 1.3.3.9) to ?H. By Frobenius Reciprocity (Theorem 1.3.3.12),
(?H;?~?)H = (?; )G = (?;?)G = 1. Let at 2 H, where a 2 A and t 2 T?. Observe that
?H(at) = (?; )H
X
s2?
s (at) = X
s2?
(sats?1) =
X
s2?
((sas?1)(sts?1))
where ? is a complete set of coset representatives of H in G. Since we are free to choose any
such set, we choose ? to be a complete set of representatives for T? in T. Since s 2 T, we
have sts?1 2 T so that ((sas?1)(sts?1)) = ?(sas?1)~?(sts?1). Now recalling that T? C T,
we have sts?1 2 T? for each s 2 ? so that ~?(sts?1) = ?(sts?1) = 1, where the last equality
holds since ? = 1T?. The above sum then becomes
X
s2?
?s(a). Putting t = e in the above
computation shows that this last sum is actually ?(a). By Theorem 3.2.16, ?(a) 6= 0. We
have, therefore, shown that ? never vanishes on H. By Theorem 3.1.4, there are at most
jG : Hj cosets of hei which are orthogonal relative to B?hei. But the required number of
orthogonal cosets is ?(e)(?;1)hei = ?2(e) ?jG : Hj2. Therefore, G is not o-basis.
?
Theorem 3.2.20 Let A C G be abelian with jG : Aj = p3. If G is o-basis, then G is
nilpotent.
Proof: We prove the contrapositive. Assume that G is not nilpotent. Then Lemma 3.2.15
applies. Choose a normal subgroup K of G satisfying the conclusions of that lemma. By
Theorem 2.1.6, it su?ces to show that G=K is not o-basis.
By 3.2.15 (i) and (ii), G=K is a non-abelian semi-direct product of AK=KoT,where
52
jTj = pn for some n ? 1. Therefore, Theorem 1.3.4.2 (concerning semi-direct products) ap-
plies to G=K and we adopt the notation established there. Observe that if ? 2 Irr(AK=K)
and T? = T, then (AK=K)T? = (AK=K)T = G=K so that ?;?(e) = (?~?)G(e) = 1. As
we shall show, it follows that, for some character ? 2 Irr(AK=K), T? 6= T. For suppose
that T? = T for all ? 2 Irr(AK=K). Since, by Theorem 1.3.4.2 (iii), every character of G
has the form ?;? for some ? 2 Irr(AK=K) and ? 2 Irr(T?), we have that ?(e) = 1 for
all ? 2 Irr(G=K). Theorem 1.3.2.3 then gives that G=K is abelian, contrary to 3.2.15 (i).
Therefore, choose ? 2 Irr(AK=K) such that T? 6= T. By Lemma 3.2.15 (iii), AK=K is
an elementary abelian q-group for prime q 6= p. Applying Lemma 3.2.18 to G=K, we may
assume that jT?j = p2. Since jT : T?j = p, it follows that T? C T. By Theorem 3.2.19, G is
not o-basis.
?
In the next result, we will again take of advantage of semi-direct product structure. As
before, we refer the reader to section 1.3.4 for a discussion of semi-direct products and the
associated notation.
Theorem 3.2.21 Let p be prime and let A C G be abelian with jG : Aj = p4. Suppose that
G is o-basis. Then exactly one of the following holds:
(i) G is nilpotent.
(ii) There exists K C G with K A such that G=K is the in-
ternal semidirect product of A=K and T, where T ?= G=A.
Moreover, there exists ? 2 Irr(A=K) such that T? 6= G=A.
Finally, T? \Z(G=A) = hei for all such characters ?.
53
Proof: As in previous results, we prove the contrapositive. Suppose that G is not nilpo-
tent. By Lemma 3.2.15 (i), there is a normal subgroup K of G with K A such that
G=K is a non-abelian semidirect product (A=K)oT for some subgroup T ? G=K, where
T ?= (G=K)=(A=K) ?= G=A, where the last isomorphism is given by the Third Isomorphism
Theorem. Let us choose such a K. By Lemma 3.2.15 (ii), jG=K : A=Kj = p4, and, by
3.2.15 (iii), A=K is a non-trivial elementary abelian q-group for some prime q 6= p (see
Deflnition 1.2.10). If we can show that G=K is not o-basis, it will follow by Theorem 3.1.5
that G is not o-basis. By Theorem 3.2.17, G=K is not o-basis if T is abelian. We may
assume, therefore, that T is non-abelian.
Let us suppose, to obtain a contradiction, that T? = T for all ? 2 Irr(A=K). Then
(A=K)T? = (A=K)T = G=K and ?;?(e) = 1 for all pairs ?;?, where ? 2 Irr(A=K) and ? 2
Irr(T?). In this case, every irreducible character of G=K is linear by Theorem 1.3.4.2 (iii).
By Theorem 1.3.2.3, G=K is abelian. But we have noted that G=K is non-abelian. Assume
that (ii) does not hold, and choose ? 2 Irr(A=K) with T? 6= T such that T? \Z(T) 6= hei.
Note that jT?j > 1. By Lemma 3.2.18, we may assume that jT?j ? p2. Suppose that
jT?j = p3. Then T? C T by Theorem 1.2.6. It follows from Lemma 3.2.19 that G is not
o-basis. Assume, therefore, that jT?j = p2.
Fix ? 2 Irr(T?). Note that T? is abelian by reason of order and therefore ? is linear
by Theorem 1.3.2.3. Put H = AT?, H = A(T? \Z(T)), and ? = ?;?. Note that ?(a) 6= 0
for all a 2 A=K by Theorem 3.2.16. We show that ?(h) 6= 0 for all h 2 H.
Since H=(A=K) Z(T), H C G by the Correspondance Theorem. Let az 2 H, where
a 2 A and z 2 T? \Z(T). Applying Mackey?s Theorem, Theorem 1.3.3.11, we have
54
?H(az) = ?(?~?)G?H(az) =
X
s2?
?
s(?~?)
sH\H
?H
(az);
where ? is a complete set of H ?H double coset representatives in G=K.
As H C G and H H, we have sH \H = H so that the sum becomes
X
s2?
?
s(?~?)
H
?H
(az) =
X
s2?
s(?~?)(az) = X
s2?
(?~?)(s?1azs) =
X
s2?
?(s?1as)~?(s?1zs):
For g 2 G, HgH = gHH = gH, where the next to last and last equalities hold
since H C G=K and H H respectively. The elements of ? can thus be chosen to be
coset representatives of H in G and therefore we may assume these elements to be coset
representatives of T? in T. Recall that ~? = ??, where ? : H ! H=(A=K) ?= T? is the
canonical map. Since z 2 T? \ Z(T), we have s?1zs = z for all s 2 ? and ?(s?1zs) =
?(z) = z. Thus ~?(s?1zs) = ?(z). We have
X
s2?
?(s?1as)?(z) = ?(z)
X
s2?
?s(a) = ?(t)?(a);
where the last equality is obtained by carrying out the above computation with t = e. Since
?(a) 6= 0, this shows that ? never vanishes on H.
By Corollary 3.1.4 with N = H, there are at most p3 cosets of hei in G=K which are
orthogonal with respect to B?hei.
But ?(e) = jG : Hj(?~?)(1) = jG : Hj = p2. Thus, in order for G=K to be (hei;?)-o-basis, it
is required that there be at least ?(e)(?;1)hei = ?(e)2 = p4 mutually orthogonal cosets. It
follows that G=K, and so G, is not o-basis and the proof is complete.
55
?
Lemma 3.2.22 Let K ? G and let N C G with K N. Let ? 2 Irr(G) and assume that
G is (K;?)-o-basis. Then G is (N;?)-o-basis.
Proof: Let H ? G with N H. Assume that (?;1)H 6= 0. Then (?;1)N 6= 0 and
N ker? by Lemma 1.3.3.4. Deflne b? : G=ker? ! C by b?(gker?) = ?(g) for all g 2 G.
Since G is (K;?)-o-basis, Theorem 3.1.2 gives that G=ker? is (hei;b?)-o-basis. A second
application of Theorem 3.1.2, gives that G is (N;?)-o-basis, as desired.
?
Theorem 3.2.23 Suppose A C G, that A is cylic and jG : Aj = p. Then G is o-basis if
and only if G is nilpotent.
Proof: If G is o-basis, it follows immediately from Theorem 3.2.17 that G is nilpotent.
Turning, therefore, to the other direction, let us assume that G is nilpotent. We may also
assume that G is non-abelian (see Proposition 1.3.2.3). Note that A is a maximal subgroup.
For suppose K ? G with A K. We have K=A ? G=A ?= Zp so that either K=A = A or
K=A = G=A. Since A K, it follows that either K = A or K = G. Let ? 2 Irr(G) be non-
linear. Deflne ^? : G=ker? !Cby ^?(g?ker?) = ?(g). By Lemma 1.3.2.5, ^? 2 Irr(G=ker?).
Let us note that ker(^?) = feg. Moreover, since j^?(g ? ker?)j = j?(g)j, we have Z(^?) =
Z(?)=ker? = Z(G=ker?), where the last equality is given by Lemma 1.3.2.9. Next observe
that ker? is a proper subgroup of A. For suppose flrst, to obtain a contradiction, that
A ker?. Then G0 ker? so that G=ker? is abelian. It follows from Lemma 1.3.2.3 that
? is linear, a contradiction. Assume then that neither of the groups A and ker? is contained
in the other. In this case, A?ker? properly contains A. Since A is maximal, A?ker? = G.
56
Thus G=ker? = (A?ker?)=ker? ?= A=(A\ker?). This last group is abelian, and we again
have G0 ker?, a contradiction. It follows then that ker? is properly contained in A. From
this we deduce that feg 6= A=ker? C G=ker?, that jG=ker? : A=ker?j = jG : Aj = p
and that A=ker? is cyclic since A is. Note also that G=ker? is nilpotent since G is. We
have verifled therefore that G=ker? has all of the properties assumed of G. Finally, by
Proposition 2.1.5, G is ?-o-basis if and only if G=ker? is ^?-o-basis. We lose no generality,
then, in identifying G with G=ker?, and ? with ^?. With this identiflcation, we may assume
that ? is faithful, ? is non-linear and that Z(?) = Z(G).
By Theorem 1.3.2.13, ?(e) divides jG : Aj. Thus, ?(e) = p. Let ? 2 Irr(A) such that
(?;?)A 6= 0 (that this is possible follows from Proposition 1.3.1.9). Then 1 ? (?;?)A =
(?;?G), where the last equality is by Theorem 1.3.3.12. Now ?G(e) = jG : Aj?(e). Since
?(e) = 1 by Lemma 1.3.2.3, we have ?G(e) = p so that (?;?G) ? 1. This shows that
(?;?G) = 1 so that ? = ?G. Thus ? ? 0 on G?A (see note after Deflnition 1.3.3.5)and
it follows that ker? Z(?) A. We note here, for later use, that ? is faithful. Indeed,
since A is cyclic, every subgroup of A is characteristic in A. Characteristic subgroups of
normal subgroups are themselves normal. Therefore, every subgroup of A is normal in G.
In particular, ker(?) C G. Therefore, ker(?) = \x2G[ker(?)]x = ker? = feg, where the next
to last equality is given by Lemma 1.3.3.6 since ? = ?G.
Let H ? G and assume that (?;1)H 6= 0. Note that G0 A. If A H, then G0 H
and G is H-o-basis by Theorem 3.3.1. Let us now assume that A*H and H *A. We will
show that if G is (H \A;?)-o-basis, then G is (H;?)-o-basis. From this, we will conclude
that it su?ces to assume H A. Since A is maximal, we have jGj = jAHj = jAj?jHjjH \Aj.
Dividing by jHj, we have jG : Hj = jA : H \ Aj. Also, dividing by jAj, we have that
57
jH : H \Aj = jG : Aj = p. Let S denote a complete set of coset representatives for H \A
in A. Let a;b 2 S with a 6= b so that a(A \ H) 6= b(A \ H). Suppose aH = bH. Then
a?1b 2 H. However, a?1b 2 A so that a(H\A) = b(H\A), a contradiction. If follows that
the set faH : a 2 Sg consists of distinct cosets of H in G. Since jSj = jA : H\Aj = jG : Hj,
we see that S comprises a complete set of coset representatives for H in G. Now let a 2 S.
We claim that a(H \A) = (aH)\A. Indeed, let g 2 a(H \A). Then g = ah, h 2 H \A.
Since h 2 H, g 2 aH. Since h 2 A, g 2 A. Thus g 2 (aH) \ A. Conversely, suppose
g 2 (aH) \ A. Then g = ah and g = a for some h 2 H and a 2 A. We have ah = a
so that h = a?1a 2 A. Thus g 2 a(H \ A) and the claim follows. Let a;b 2 S. Then
B?H\A?a(H \A);b(H \A)? = ?(e)jH \Aj
X
h2H\A
?
?
a?1bh
?
= jH : H \Aj? ?(e)jH \Aj?jH : H \Aj
X
h2H\A
?
?
a?1bh
?
= jH : H \Aj? ?(e)jHj
X
h2H
?
?
a?1bh
?
= jH : H \Aj?B?H?aH;bH?
where, in obtaining the next to last equality, we have used the facts that a?1b(H \A) =
(a?1b)H\A and that ? ? 0 on G?A. Thus, whenever a;b 2 S and a(H\A) and b(H\A)
are orthogonal relative to B?H\A, aH and bH are orthogonal relative to B?H.
Assume that G is (?;H \ A)-o-basis. Then there are ?(e)(?;1)H\A cosets of H \ A
which are mutually orthogonal relative to B?H\A. Since jG : Aj = ?(e), some coset of
A must contain (?;1)H\A of these cosets. By G-invariance of B?H\A, every coset of A,
and in particular A itself, must contain (?;1)H\A such cosets. Choose S = fai : 1 ?
i ? (?;1)H\Ag S such that ai(A\H) and aj(A\H) are orthogonal relative to B?H\A
whenever i 6= j. Then faiH : ai 2 Sg consists of (?;1)H\A mutually orthogonal cosets of
H in G. We recall that ? ? 0 on G?A and calculate
58
(?;1)H\A = 1jH \Aj
X
h2H\A
?(h) = jH : H \Aj 1jH \Aj?jH : H \Aj
X
h2H
?(h)
= jH : H \Aj 1jHj
X
h2H
?(h)
= p?(?;1)H
= ?(e)(?;1)H
It follows that G is (H;?)-o-basis whenever G is (H \A;?)-o-basis. Therefore, we assume,
without loss of generality, that H A. Since A is cyclic, every subgroup of A is character-
istic in A. Since A C G, Lemma 1.2.3 gives that H C G. By Lemma 3.2.22, it su?ces to
show that G is (E;?)-o-basis, where E denotes the identity subgroup.
Recall that Z(?) A. Since G is nilpotent, Theorem 3.2.11 gives that ?(e)2 divides
jG : Z(?)j. It follows that p divides jA : Z(?)j. There is, therefore, a subgroup of A=Z(?)
of order p.This subgroup is of the form D=Z(?), where Z(?) D A and jD : Z(?)j =
p. We claim that ?D = Pp?1i=0 ?i, where the ?i are distinct, irreducible characters of D.
By Theorem 1.3.3.11, ?D = (?G)D = P 2?? ? A\D?D, where ? is a complete set of
(D;A)-double coset representatives in G. We remind the reader that ? 2 Irr( A), where
A = A ?1. Since A C G and D A, we have that A = A and A \ D = D. Thus
?D = P 2??( ?)D?D = P 2?( ?)D. For each 2 ?, we have D A = DA = A since
D C G and D A. It follows that ? may be taken to be the set fxi : 0 ? i ? p?1g, where
x 2 G?A is chosen so that G=A = hxAi. That is, ?D = Pp?1i=0(xi?)D. Fix 0 ? i ? p?1
and let ? 2 Irr(D) such that ?xi?;??D ? 1. Since ? is linear, we have ?xi?;??D ? 1 so
that xi?D = ?. That is, xi?D is irreducible. It remains to show that the xi? are all distinct.
Suppose that xi? = xj?for somei 6= j. Choosed 2 D?Z(G) such thatD=Z(G) = hd?Z(G)i.
Then ?(xid) = xi?(d) = xj?(d) = ?(xjd). Recalling that ? is a faithful homomorphism, we
59
have that xid = xjd so that xj?id = d. Since i 6= j, G = hxj?i;Ai. If follows that d 2 Z(G),
a contradiction. We put ?i = xi? and the claim is established.
Put Z = Z(G). We now show that ? ? 0 on D ? Z. By Theorem 1.3.3.9, ?Z =
(?;?)Z Pg g? for some ? 2 Irr(Z), where (?;?)Z 6= 0 and the g? are the distinct conjugates
of ? under G. For all g 2 G and z 2 Z, we have that g? 2 Irr(Z) and g?(z) = ?(gz) = ?(z)
so that g? = ?. Thus, ?Z = (?;?)Z ?? so that p = ?Z(e) = (?;?)Z ??(e) = (?;?)Z and we
have ?Z = p?. Now observe that p = (?;?)Z = (?D;?)Z = ?Pp?1i=0 ?i;??Z = Pp?1i=0(?i;?)Z.
For each i, ?i is linear so that (?i;?) is either 1 or 0. It follows that (?i;?)Z = 1 for all i.
By Theorem 1.3.3.12, (?i;?D)D = (?i;?)Z = 1 for each i. Also, ?D(e) = jD : Zj?(e) = p.
It follows that ?D = Pp?1i=0 ?i = ?D. Since Z C D, it follows that ?D ? 0 on D?Z.
Note that B?E(aE;bE) = ?(e)?(a?1b) for all a;b 2 G. If aA 6= bA, then a?1b =2 A
and the above quantity is zero since ? ? 0 on G?A. That is, any two elements of G in
distinct cosets of A are orthogonal relative to B?E. By the same reasoning, a is orthogonal
to b relative to B?E whenever a;b 2 D and aZ 6= bZ. Let fai : 1 ? i ? pg be a complete set
of distinct cosets of Z in D. Then the set Sp?1j=0fxjai : 1 ? i ? pg, where x0 is chosen to be
the identity of G, comprises a set of p2 cosets of E which are mutually orthogonal relative
to B?E. Since ?(e)(?;1)E = p2, we have shown that G is (E;?)-o-basis. As noted, it follows
from this that G is (H;?)-o-basis. As H and ? were chosen arbitrarily, G is o-basis and the
proof is complete.
?
60
3.3 The Upper and Lower Central Series
In section 3.1, we deflned the notion of K-o-basis. One reason for doing so is to use
the generalized notion as a kind of fllter for distinguishing between groups that is flner than
that provided by the original notion of o-basis. To apply the generalized notion to all groups
in a given class, the subgroup K must be chosen so that it is deflned for all groups in that
class. For example, the notion of Z-o-basis makes sense for all flnite groups. In this section,
we consider two series of \universal subgroups", the upper and lower central series. These
series are deflned in the notation section at the end of this work. It is expected that fewer
groups G will be n(G)-o-basis than n?1-o-basis. One discovers quickly that every group
is 2-o-basis (Theorem 3.3.1). Our main result in this section, Theorem 3.3.4, is that every
group is in fact 3-o-basis. Finally, in Theorem 3.3.5, we also present a slight generalization
of Theorem 3.2.1.
Theorem 3.3.1 The group G is G0-o-basis. That is, every group is G0-o-basis.
proof: By Theorem 3.1.5, G is G0-o-basis if and only if G=G0 is o-basis. The latter holds
by Theorem 2.1.4 since G=G0 is abelian.
?
Lemma 3.3.2 Let N?G and for H ? G let bH denote the image of H under the canonical
map G ! G=N. Let M ? G with N M. Then bH cM ) H M for all H ? G.
Proof: Let H ? G, assume that bH cM and let x 2 H. Then xN = yN for some y 2 M
so that y?1x = n for some n 2 N. Thus x = yn 2 M by closure since N M. It follows
that H M.
61
?
Lemma 3.3.3 Let m ? 1, n ? 0 and put m = m(G) and Zn = Zn(G).
Then m Zn , m+1 Zn?1.
Proof: For a subset S G, let bS denote the image of S under the canonical map G !
G=Zn?1. Assume flrst that m Zn. We recall that m+1 = ?f[a;b] : a 2 m;b 2 Ggfi.
Therefore, let a 2 m and b 2 G, and note that it is enough to show that [a;b] 2 Zn?1.
We show [a;b]Zn?1 = Zn?1. Indeed, a 2 Zn by assumption and cZn = Z(bG). Thus
[a;b]Zn?1 = [aZn?1;bZn?1] = Zn?1. It follows that [a;b] 2 Zn?1, and thus m+1 Zn?1,
as desired.
Assume now that m+1 Zn?1. Applying Lemma 3.3.2 with N = Zn?1, we see that
it su?ces to show c m bZn. Now c m = ?f[a;b]Zn?1 : a 2 m?1;b 2 Ggfi. Therefore, let
a 2 m?1 and let b;g 2 G. Then [a;b] 2 m so that ?[a;b];g? 2 m+1 Zn?1, where this
last containment is by assumption. We have
?[a;b]Z
n?1;gZn?1
? = ?[a;b];g?Z
n?1 = Zn?1:
It follows that b m Z(bG). Since Z(bG) = bZn, the desired containment is established.
?
Theorem 3.3.4 The group G is 3-o-basis. That is, every group is 3-o-basis.
Proof: For a subset S G, let bS denote the image of S under the canonical map
G ! G= 2. By Theorem 3.1.5, it su?ces to show that bG is o-basis. Note that 3(bG) =
\ 3(G) = hei = Z0(bG). By Lemma 3.3.3, (bG)0 = 2(bG) Z1(bG) = Z(bG). Theorem 3.2.1
gives that bG is o-basis, as desired.
62
?
Our last result is a slight generalization of Theorem 3.2.1.
Theorem 3.3.5 Let m ? 1, n ? 0 and assume that m Zn. Then G is Zm+n?2-o-basis.
Proof: By Theorem 3.3.4, it su?ces to show that 3 Zm+n?2. We proceed by induction
on m. Assume flrst that m = 0. Then G = 1 Zn so that G = Zn. If n = 0, then
G = Z0 = hei so that G is abelian and o-basis. The result holds since Zm+n?2 = Z?1 = hei.
Assume that n > 0. Since G = Zn, we have G=Zn?1 = Zn=Zn?1 = Z(G=Zn?1), where the
last equality holds by deflnition of Zn for n > 0. Thus G=Zn?1 is abelian and 2 Zn?1.
It follows from Lemma 3.3.3 that 3 Zn?2, establishing the case m = 0.
Now let m ? 0 and assume that m Zl ) 3 Zm+l?2 for all non-negative integers
l. Suppose that m+1 Zn. We wish to show that 3 Zm+n?1. Applying the induction
hypothesis with l = n+1, it su?ces to show that m Zjn+1. We obtain this by applying
the backwards implication in the conclusion of Lemma 3.3.3 to our assumption
that m+1 Zn.
?
63
Chapter 4
Conclusions
Our work may be divided into main subdivisions. First, we have attempted to explore
the connections between the o-basis property and nilpotency. We have posed two questions.
First, which nilpotent groups are o-basis? In [Hlms], Holmes has given an example of
a group of order 34 that is not o-basis. Thus, it is known that not all nilpotent groups are
o-basis. We have shown that whenever G0 Z(G) (a condition implying nilpotency), G is
o-basis. This result has been used in several subsequent arguments. In addition, it raises
a question for possible future study. If G0 Z(G), then G has nilpotence class no greater
than 3 (see the notation section). Also, Holmes? example of order 34 has class 4. We ask,
therefore, if nilpotence class 3 is a necessary condition for a (nilpotent) group to be o-basis,
leaving this question, for the moment, open.
We have also shown that a nilpotent group is o-basis if and only if each of its Sylow
subgroups are. Therefore, the question can be "reduced" in some sense to which p-groups
are o-basis. Suppose G is a group of order pn, where p is prime. We have taken the approach
of considering the question for increasing values of n. Since all abelian groups are o-basis,
we immediately flnd that G is o-basis if n ? 2. That G is o-basis for n = 3 was established
by Holmes in [Hlms] (see Theorem 1.1.1). We have attempted to better understand groups
of order p4 by working from the assumption that such a group is not o-basis. In this case,
there exists a subgroup H ? G and ? 2 Irr(G) such that G is not (H;?)-o-basis. We derive
from these assumptions some conditions on H, ? and G. We note that G begins to look
very much like Holmes? example of order 34.
64
Our second question concerning the o-basis property and nilpotentcy is, "Are all o-
basis groups nilpotent?". In this, there are two motivating facts. First, all examples so
far of o-basis groups have been nilpotent. Second, the dihedral groups that are o-basis
are precisely those that are nilpotent. With the dihedrals in mind, we have narrowed the
question to those groups G having an abelian, normal subgroup A with non-trivial prime
power index pn. We have asked if nilpotentcy is a necessary condition for these groups to
be o-basis. We have shown that if G=A is abelian, the answer is a?rmative. When G=A is
non-abelian, in which case n ? 3, we have some limited results. In this case, nilpotentcy is
necessary for o-basisness if n = 3. In the case n = 4, we obtain the result only after adding
certain technical conditions. Finally, we have shown that nilpotentcy is a necessary and
su?cient condition for o-basisness if A is cyclic and n = 1.
Our second major objective has been to explore the idea of generalizing the o-basis
property. To this end, we deflned the notion of a K-o-basis group, where K is a subgroup
of G. In order for this notion to be used to distinguish between groups in a given class,
the subgroup K must be chosen so that it makes sense for all of the groups in that class.
For example, we may choose K to be the central subgroup. Since the center is deflned for
all groups, it makes sense to ask which groups are Z-o-basis. The upper and lower central
series are two series of subgroups deflned for all groups. In section 3.3, we have obtained
some results with K coming from these series. One obtains quickly that every flnite group
is 2-o-basis. Our main result along these lines is that every group is in fact 3-o-basis.
Theorem 3.2.1, mentioned above, was important to obtaining this result. In this same
section, we also extend Theorem 3.2.1 somewhat.
65
Having summarized what has been done, we are positioned to look ahead. There seems
to be quite a bit of room for further study with o-basis groups. Neither of our two questions
concerning the o-basis property and nilpotentcy have been answered fully. The question of
whether or not all o-basis groups are nilpotent is still open. As a more manageable goal, on
may attempt to prove Conjecture 3.2.13. In our attempts at this conjecture, we only made it
to n = 4. Even here, we were obliged in Theorem 3.2.21 to add certain technical conditions
to make the argument work. It is not known whether these conditions are actually needed.
One may therefore attempt to remove them. We have already mentioned the possibility
that all o-basis groups are in fact not only nilpotent but of nilpotence class 3.
There are also open questions about how the o-basis property behaves with regard to
basic group-theoretic operations. For example, it is unknown if a subgroup of an o-basis
group o-basis. It is also unknown whether or not the direct product of a number of o-basis
groups is o-basis (see the discussion following Theorem 3.2.8). Along these lines, it may
be worth noting that, since the deflnition of o-basis involves characters, it may be that
\o-basis" is not a purely group-theoretic property.
Room for further study also exists in experimenting with alternative means of general-
izing the notion of o-basis group. There are at least two alternative generalizations. First,
we have chosen, in the character theory, the complex numbers for the base fleld (or at least
a fleld that is algebraically closed and whose characteristic does not divide jGj). However,
theory exists for the case in which the fleld has prime characteristic (see [Is], Ch. 15). This
involves the notion of Brauer characters. One might attempt to derive a notion of o-basis
that in terms of Brauer characters rather than ordinary C-characters.
66
Another possible generalization is to allow the group G to be inflnite. In this case, one
deals with arbitrary compact groups as opposed to flnite groups. In this case, the group G is
endowed with a certain topology, and a linear representation of G in V is a homomorphism
? : G ! GL(V) which is continuous with respect to this topology. A portion of the theory
of theC-characters of flnite groups has an analog in the setting of compact groups, and one
might try to deflne the notion of o-basis in this setting.
Future researchers may do well to familiarize themselves with current research on p-
groups, zeros of characters and number theory. The study of o-basis groups may well hold
the potential for furthering progress in these on-going flelds of research. Also, considering
their connection with tensor spaces, the author feels he has reason to believe that the study
of o-basis groups holds promise as a challenging and signiflcant endeavor.
67
Bibliography
[Greub] Werner Greub, Linear Algebra, fourth edition, Sringer-Verlag, New-York Heidel-
berg Berlin, 1975.
[Hlms] R. Holmes, Orthogonality of Cosets Relative to Irreducible Characters of Finite
Groups, Linear and Multilinear Algebra, March-April 2004 Vol.52, No.2, pp. 133-143.
[Hlms,Tam] R. Holmes and Tin-Yau Tam, Symmetry Classes of Tensors Associated with
Certain Groups, Linear and Multilinear Algebra, 1992 Vol.32, pp. 21-31.
[Hun] T.W. Hungerford, Algebra, Springer New-York, 1974.
[Is] I.M. Isaacs, Character Theory of Finite Groups, Academic Press, New-York San Fran-
cisco London, 1976.
[Karp] G. Karpilovsky, Group Representations, volume 1, part B: Introduction to Group
Representations and Characters, North-Holland, Amsterdam London New-York Tokyo,
1992.
[Rob] D.J.S. Robinson, A Course in the Theory of Groups, Springer-Verlag, New York
Heidelberg Berlin, 1982.
[Ser] Jean-Pierre Serre, Linear Representations of Finite Groups, Springer-Verlag, New-
York Heidelberg Berlin, 1977.
[Suz] M. Suzuki, Group Theory I, Springer-Verlag, New-York Heidelberg Berlin, 1982.
68
Notation
Let G denote a flnite group. We use the following notational conventions. Entries are
arranged according to category and, after that, roughly in the order that they appear in
the text. References to relevant sections of the text are at the far right. Also, the reader
will want to take note of comments that accompany some of the entries.
jAj, the cardinality of the set A
e, the identity element of G
G?S, fg 2 G : g =2 Sg
H ? G, H is a subgroup of G
hSi, the subgroup generated by the subset S G
G=N, the quotient group of left cosets of the normal subgroup N
of G
gx, gx = x?1gx, where x;g 2 G
xg, xg = xgx?1, where x;g 2 G
Hg, Hg = fhg : h 2 Hg, where g 2 G and H ? G
gH, gH = fgh : h 2 Hg, where g 2 G and H ? G
CG(x), CG(x) := fg 2 G : g?1xg = xg, where x 2 G. This is called
the centralizer of x in G. It is a subgroup of G.
Aut(G), the automorphism group of G
K char G, K is a characteristic subgroup of G Def. 1.2.2
Th. 1.2.3
lcm(S), the least common multiple of a flnite set, S, of integers
69
Y X, the set of flxed points of a set Y under the action of a set
X. That is Y X = fy 2 Y : yx = y for all x 2 Xg, where
y 7! yx is the image of y 2 Y under the action of x 2 X.
[a;b], the commutator aba?1b?1, where a;b 2 G
G0, the commutator subgroup of G
n(G), the nth term of the lower central series of G (see below)
The lower central series of G is the series of subgroups
deflned recursively as follows: 1(G) = G and, for n > 1,
n(G) := ?f[a;b] : a 2 n?1(G);b 2 Ggfi.
Note that n(G) C G for all n ? 1 and that G0 = 2(G).
A group G is nilpotent , n(G) = hei for some n.
In this case, the smallest such n is called the nilpotence
class of G. This can also be shown to be the smallest n such
that Zn(G) = G (see below).
Z(G), the center of the group G
Zn(G), the nth term of the upper central series of G
The upper central series is the series of subgroups of G
deflned recursively as follows. For n ? 0, put Zn(G) = hei
and let Z1(G), or simply Z(G), denote the center of G.
For n > 1, Zn(G) is the unique subgroup of G containing
Zn?1 such that Zn(G)=Zn?1(G) = Z(G=Zn?1).
z, the conjugate of the complex number z
jzj, the modulus of the complex number z
70
1G, the principal character of G, 1G(g) = 1 for all g 2 G.
Note that 1G is linear and so irreducible. (see Def 1.3.2.1 and
Propostion 1.3.2.2.)
ker?, the kernel of the character ?, ker? := fg 2 G : ?(g) = ?(e)g
Z(?), the center of the character ?, Z(?) := fg 2 G : j?(g)j = ?(e)g
For all characters ?, ker? Z(?) and both subgroups are
normal in G
?K, the restriction of the character ? to the subgroup K sec. 1.3.3
?G, the character of G induced from ?, where ? is a character
of some subgroup of G
sec. 1.3.3
IG(?), the inertial subgroup of ?, where ? is a character of some
subgroup of G
Def. 1.3.3.8
Cl(G), the set of class functions on G Sec. 1.3.1
??;??
G
??;??
G =
1
jGj
X
g2G
?(g)?(g), where ?;? are class functions on
G.
sec. 1.3.1
T?, the stabilizer of ? in T sec 1.3.4
Gal(Q;?), the galois group of Q[?] over Q
B?H, eqn. 2.1
Qn
i=1 Gi, the direct product of the groups Gi
71