Orthogonal bases of certain symmetry classes of tensors associated with Brauer
characters
by
K. A. A. Indika
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
August 03, 2013
Keywords: Brauer character, projective indecomposable module, block, dihedral group,
symmetrizer, symmetry class of tensors, o-basis, generalized orthogonality relation
Copyright 2013 by K. A. A. Indika
Approved by
Randall R. Holmes, Professor of Mathematics
Narendra K. Govil, Associate Chair & Alumni Professor of Mathematics
Huajun Huang, Associate Professor of Mathematics
Abstract
The main focus of this dissertation is on the existence of an orthogonal basis consisting
of standard symmetrized tensors (o-basis for short) of a symmetry class of tensors associated
with a Brauer character of a  nite group. Most of the work is done for the dihedral group and
some results are given for the symmetric group. The existence of an o-basis of a symmetry
class of tensors associated with an (ordinary) character of a  nite group have been studied
by several authors. My study was motivated by the work done on the existence of such a
basis of a symmetry class of tensors associated with an (ordinary) irreducible character of a
dihedral group.
In Chapter 1 we introduce the basic de nitions in character theory. In this a Brauer
characters, character of a projective indecomposable module (PI) and a block of a  nite
group will be introduced. Also in this chapter a generalised orthogonality relation of blocks
of a  nite group is established. In chapter 2 we introduce the symmetrizer and related
notions. Some general results associated with Brauer characters of a  nite group will also
be given in this chapter. Chapter 3 consists of the results associated with Brauer characters,
PIs and blocks of a dihedral group. Finally Chapter 4 lists some result associated with the
Brauer characters of the symmetric group.
ii
Acknowledgments
This dissertation is a partial ful lment of my Ph.D. program. It consists of the work I
did in the last 5 years at Auburn University. It has been a fruitful experience in my academic
life to have worked at Auburn University with a lot of nice people. I would like to take this
opportunity to thank and mention all the people who supported me and encouraged me in
this journey to achieve this special goal of my life.
It is with great honor I mention the support and guidance of Professor Randall R.
Holmes. As my Ph.D. advisor he took all possible e orts to make my work as smooth as
possible. His incomparable intuition of the subject has played a big role in my improvements
as a researcher. The  exibility and the nature of his guidance made my research experience
a pleasant one. His advice helped me not only with my research but also with my teaching
and my student life. His teaching inspired me and he always happily shared his teaching
experience with me to help me improve my teaching. He is very understanding and had an
extraordinary ability to know what to say and when to say it. I will ever be so grateful to
him not just for the advisor he is but also for the nice person he is.
I am specially thankful to the Dr. Narendra K. Govil for his continuous support since
the  rst day I arrived at the Auburn University. He was always there whenever I needed
his advice. He gladly accepted to be in my Ph.D. advisory committee showing his kindness
to see my success in the program. I will always remember him as a mathematician with a
unique character.
I  rst met Dr. Huajun Huang as a teacher. His humble nature and commitment to his
work has helped me motivate myself in my work. He is very approachable because of his
amicable nature. I am so grateful to him for accepting to be in my Ph.D. advisory committee
and for continuing to support me in achieving my goal.
iii
I thank Dr. Rajesh Amin for kindly accepting to be the university reader of my disser-
tation.
I can not  nd enough words to express my gratitude to my mother M. Kusumawathi
Wanigarathna and my (late) father K. A. George Martin. Their loving care has made me
the person I am today. I owe them for all the success I have achieved in my life. The trust
and the con dence my mother has in me made me reach this level of education.
My wife Shilpa needs a special mention for being there for me and encouraging me to
achieve these heights. Her support in my decisions gives me con dence to do my work well.
Our loving daughter Savana makes my life the most enjoyable one. She keeps me relaxed
after a tiresome day.
Finally my sincere thank goes to each individual responsible in some way to my success
in the program.
iv
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Character Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Group representations and Group Algebra . . . . . . . . . . . . . . . . . . . 1
1.2 Character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Brauer character . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 PIs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.5 Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.6 Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.7 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Symmetrized Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 Symmetrizers associated with ordinary and Brauer characters of G . . . . . . 12
2.2.1 Symmetrizers associated with PIs . . . . . . . . . . . . . . . . . . . . 15
2.2.2 Symmetrizers associated with blocks . . . . . . . . . . . . . . . . . . 16
3 Dihedral group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.1 Brauer characters of Dn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2 PIs of Dn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.3 Blocks of Dn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
3.4 Block idempotent symmetrization . . . . . . . . . . . . . . . . . . . . . . . . 26
3.5 Irreducible symmetrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
3.6 Projective symmetrization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 Symmetric group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
v
4.1 Special case S4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
vi
Chapter 1
Character Theory
1.1 Group representations and Group Algebra
Let G be a  nite group and V be a  nite dimensional vector space over the  eld of
complex numbers C. By GL(V) we denote the group of invertible linear transformations
from V to itself. A representation of G is a group homomorphism  : G! GL(V). The
degree of the representation is the dimension of V.
Denote by CG the vector space over C with the basis G. CG is a ring with the multi-
plication de ned by,
(
X
a2G
 aa)(
X
b2G
 bb) =
X
a;b2G
 a bab:
A C-algebra is a ring A that is also a vector space over C such that (ab) = ( a)b = a( b)
for all  2C and a;b2A. Note that CG is a C-algebra and is called the group algebra of
G over C. CG has an identity given by 1e6= 0, where e is the identity of G. De ne a map
from C to CG by  7! 1, where 1 6= 0 is the identity of CG. This is a well de ned ring
monomorphism, hence C is viewed as a subring of CG.
Let V be a  nite dimensional vector space over C and let  : G! GL(V) be a repre-
sentation of G. Then V can be viewed as a (left) CG module by de ning av =  (a)(v) for
a2G, v2V and extending linearly to CG.
On the other hand, let V be a CG module. Then V is a vector space over C by viewing
C as a subring of CG. When we say that V is a CG module, we always assume that V is
 nite dimensional when viewed as a vector space over C in this way. De ne a map  from G
to GL(V) by  (a)(v) = av for a2G, v2V. Then  is a well de ned group homomorphism
and hence a representation of G called the representation a orded by V.
1
An irreducible representation of G is a representation a orded by a simple CG module.
1.2 Character
Let V be a CG module and let  be the representation of G a orded by V. Let  2G,
so that  ( )2GL(V).
Let M ( ) be the matrix representation of  ( ) corresponding to a  xed basis of V. The trace
of  ( ) is given by tr( ( )) = tr(M ( )): Note that the value of tr( ( )) does not depend on
the choice of the basis since similar matrices have the same trace.
The (ordinary) character of G a orded by  or V is the function  : G!C de ned by
 ( ) = tr( ( )) for  2G. We say  (e) the degree of  where e is the identity element of G.
Theorem 1.1 ([11, Lemma 2.15, page 20]). Let  be a character of G. Let  2G and let
m be the order of  . Then
i)  ( ) is a sum of mth roots of unity,
ii)  ( ) =  (  1).
Two representations  and % of the same degree n are said to be similar if there exists
an invertible matrix P of size n n such that M ( ) = P 1M%( )P for all  2G. It is easy
to observe that the following result holds using the property that tr(AB) = tr(BA) for all
square matrices A;B.
Theorem 1.2 ([11, Lemma 2.3, page 14]).
i) Similar representations of G a ord equal characters.
ii) Characters are constant on the conjugacy classes of G.
A class function is a function on G that is constant on conjugacy classes. The theorem states
that the characters of G are class functions.
2
A character of G is called an irreducible character of G if it is a orded by an irreducible
representation of G (or, equivalently, a simple module of CG). Representations of G a orded
by isomorphic CG-modules are similar. There is a one-to-one correspondence between iso-
morphism classes of CG-modules and similarity classes of representations of G (see [11, page
10]). Therefore in light of the theorem above the number of di erent irreducible characters
of a group G is the same as the number of isomorphism classes of simple CG-modules.
Let Irr(G) denote the set of irreducible characters of G. Maschke?s theorem stated below
provides a way to reduce the study of characters of G to the study of irreducible characters
of G.
Theorem 1.3 (Maschke). Let K be a  eld. If charK - jGj, then every KG-module is a
direct sum of simple KG-modules.
Since char C = 0, Maschke?s theorem holds for the  eld C. Now let V and V0 be
two CG-modules and let  and  0 be the representations they a ord respectively. Let  
be the representation a orded by the direct sum V  V0. Then for any  2G the matrix
representation of  ( ) relative to an ordered basis formed by taking an ordered basis for V
and appending an ordered basis for V0 is given by a block diagonal matrix with blocks the
matrix representations of  ( ) and  0( ) as given by
M ( ) =
0
B@ M ( ) 0
0 M 0( )
1
CA:
Then the character a orded by the CG-module V  V0 is  + 0, since trM ( ) = trM ( ) +
trM 0( ).
Because of the above results we see that to study the characters of G it is enough to look
at the irreducible characters of G. Once all the irreducible characters of G are known the
other characters of G are known as well since they are simply sums of irreducible characters.
3
One gets the number of irreducible characters of G from the number of conjugacy classes of
the group G as stated below.
Theorem 1.4 ([11, Corollary 2.7, page 16]). Let G be a group. The number of irreducible
characters of G equals the number of conjugacy classes of G.
1.3 Brauer character
The Brauer characters are the main focus of this entire thesis. These characters are also
known as modular characters. The modular representation theory was founded by Richard
Brauer in the 1930?s. We begin by setting basic de nitions.
Let R be the ring of algebraic integers in C. Fix a prime p, and let M be a maximal
ideal of R such that pR  M. Set K = R=M. Then K is a  eld. Considering the
natural homomorphism  : R!K we have p7!0, so K has characteristic p. The natural
homomorphism is going from characteristic 0 to characteristic p.
Theorem 1.5 ([11, Lemma 15.1, page 263]). Let U =f 2Cj m = 1 for some integer m
with p-mg and let R;K be as above. Then
i) U R,
ii) the natural homomorphism maps U isomorphically onto Knf0g,
iii) K is algebraically closed and algebraic over its prime  eld.
An element of G is called a p-regular element if its order is not divisible by p. Denote
by ^G the set of all p-regular elements of G.
Let V be a KG-module of  nite dimension n and let  be the representation of G
a orded by V. Let  2 ^G and let  1;:::; n 2Knf0g be the eigenvalues of  ( ). Then
by the theorem above there exist unique  1;:::; n 2 U such that  i 7! i via the natural
4
homomorphism. De ne a function ? : ^G!C by
?( ) =
nX
i=1
 i: (1.1)
Then ? is called the Brauer character of G a orded by  .
Let  2 ^G and suppose  2K is an eigenvalue of  ( ). Then   1 is an eigenvalue of
 (  1) since
 (  1)v =  (  1)  1 ( )v =   1 (e)v =   1v:
Also if  ( ) =  ( 2U), then 1 =  (   ) =  ( ) (  ) =   (  ), so  (  ) =   1. Therefore
?( ) = ?(  1). If ? is a Brauer character of G, then  ?, the complex conjugate of ? is also
a Brauer character [11]. A Brauer character corresponding to a simple KG-module is called
an irreducible Brauer character of the group G. We denote the set of irreducible Brauer
characters of G by IBr(G). The irreducible Brauer characters are linearly independent over
C [11, Theorem 15.5, page 265].
Brauer characters are constant on conjugacy classes. The number of irreducible Brauer
characters is equal to the number of conjugacy classes of G containing p-regular elements of
G as stated by the following theorem.
Theorem 1.6 ([14, Corollary 3, page 150]). The number of classes of simple KG-modules
is equal to the number of p-regular conjugacy classes of G.
Let ^ denote the restriction of an ordinary character  of G to the set ^G of p-regular
elements of G. The following result is a well known relationship between the ordinary
characters and the Brauer characters of a group.
Theorem 1.7 ([11, Theorem 15.6, page 265]). Let  be an ordinary character of G. Then
^ is a Brauer character of G:
5
The character  uniquely determines the Brauer character ^ . We note here that if p
does not divide the order of the group G, then ^G = G and the Brauer characters of G
coincide with the ordinary characters G.
The set of complex valued class functions on ^G form a vector space over C
Theorem 1.8 (R. Brauer). The irreducible Brauer characters of a group G form a basis of
the vector space of complex valued class functions on ^G.
1.4 PIs
We are also interested in the characters associated with projective indecomposable mod-
ules of RG. Note that RG is of  nite dimension so satis es the A.C.C. and D.C.C. Then by [2,
Theorem 14.2, page 81] RG can be written as a direct sum of indecomposable RG-modules.
A summand of this direct sum is a principle indecomposable module (a PIM) of RG. Sim-
ilarly KG can be expressed as a direct sum of principle indecomposable KG-modules. As
direct summands of free modules, PIMs of RG and KG are projective. [5, Theorem I.13.7,
page 44] states that there is a one to one correspondence between the isomorphism classes
of PIMs of RG and those of KG.
Theorem 1.9 ([2, Theorem 54.11, page 372]). Let P be a PIM of KG. Then P has a unique
maximal submodule NP. Two PIMs P and Q are isomorphic if and only if the irreducible
modules P=NP and Q=NQ are isomorphic.
Theorem 1.10 ([2, Corollary 54.14, page 374]). There is a one-to-one correspondence be-
tween the isomorphism classes of PIMs and the isomorphism classes of irreducible KG-
modules.
The character a orded by a PIM P of RG (PI for short) is the character a orded by
CNRP . By the theorem above we have that there is a one to one correspondence between
the irreducible Brauer characters of G and the PIs of G. We will denote by  ? the PI
6
corresponding to ?2 IBr(G). A PI  is a complex valued function de ned on G with the
property that  ( ) = 0 for  2Gn ^G [5, Corollary IV.2.5, page 144].
1.5 Relationships
Let  2 Irr(G) and let ^ be the restriction of  to ^G. Recall by Theorem 1.7 ^ is a
Brauer character of G, so
^ =
X
?2IBr(G)
d ?? (1.2)
for some uniquely determined nonnegative integers d ?. The integers d ? ( 2 Irr(G);?2
IBr(G)) are called the decomposition numbers of G for the prime p. The matrix of size
jIrr(G)j jIBr(G)j with the d ??s as entries is called the decomposition matrix of G.
Let ?2IBr(G). By [14, page 151] we have the following relationships
 ? =
X
 2Irr(G)
d ? ; (1.3)
 ? =
X
 2IBr(G)
c ? ;
where the coe cients c ? are the entries of the matrix C = DDT, with DT the transpose of
D. The matrix C is called the Cartan matrix.
1.6 Block
Let e be a centrally primitive idempotent of the group algebra CG. The block B = Be
corresponding to e is the category of CG-modules V such that eV = V. A CG-module V is
said to belong to B if it is an object of B, that is, if eV = V. By [5, Theorem 7.8, page 23]
a  nitely generated indecomposable CG-module V belongs to a unique block. If V belongs
to B, then every submodule and homomorphic image of V belongs to the same block B.
7
A character or a Brauer character of G is said to belong to a block B if the associated
module belongs to B. If  2Irr(G) belongs to the block B, then ?2IBr(G) belongs to B if
the decomposition number d ? is nonzero. Further each irreducible character and irreducible
Brauer character belongs to a unique block. Two irreducible characters  and  are in a
same block B of G if there is ?2 IBr(G) such that d ? and d ? are both nonzero. In this
case the block is the unique block that contains the Brauer character ? [11].
If  is a character or a Brauer character, we write  2B to mean that  belongs to the
block B. More generally, if S is a set of characters or Brauer characters, we write  2B\S
to mean that  belongs to the block B and  2S.
Let B be a block of G. The Osima idempotent of CG corresponding to B is given by
sB =
X
 2B\Irr(G)
s = 1jGj
X
 2B\Irr(G)
X
 2G
 (e) (  1) : (1.4)
Theorem 1.11 ([11, Theorem 15.30, page 277]). For blocks B and B0 of G, we have
sBsB0 =  BB0sB:
The following theorem holds due to the Equations 1.2 and 1.3.
Theorem 1.12 (Osima).
sB = 1jGj
X
?2B\IBr(G)
X
 2G
?(e) ?(  1) 
= 1jGj
X
?2B\IBr(G)
X
 2^G
 ?(e)?(  1) :
8
1.7 Orthogonality
Let Fun(G;C) denote the set of all functions from G to C. Fun(G;C) is a vector space
over C. For f;g2Fun(G;C) set,
(f;g) = 1jGj
X
 2G
f( )g( ):
( ; ) is an inner product on Fun(G;C). By de nition the characters of G are in Fun(G;C).
Now we will state some known orthogonality relations of characters of G. For  a character
of G and  2G, we have  ( ) =  (  1) by Theorem 1.1. It is a known fact that Irr(G)
forms a basis for the set of class functions from G to C. It is indeed an orthonormal basis
due to the following result.
Theorem 1.13 ([11, Corollary 2.14, page 20]). Let  ; 02Irr(G). Then
( ; 0) = 1jGj
X
 2G
 ( ) 0(  1) =    0:
Theorem 1.14 (Generalized Orthogonality Relation). Let  ; 02Irr(G). For any  2G
X
 2G
 (  ) 0(  1) =    0jGj ( ) (e) :
For complex-valued functions f and g on ^G de ne
(f;g)^= 1jGj
X
 2^G
f( )g( ):
Theorem 1.15 ([5, Lemma 3.3, page 145]). For ?; 2IBr(G), we have
( ?; )^= 1jGj
X
 2^G
 ?( ) (  1) =  ? :
9
We establish an orthogonality relation associated with Osima idempotents of blocks of
a group G which we call the generalized orthogonality relation of blocks. Below we discuss
the formulation of this new result.
Theorem 1.16. For  2G,
X
 2^G
X
?2B\IBr(G)
X
 2B0\IBr(G)
 ?(e)?( ) (e)  (  1 ) =  BB0jGj
X
?2B\IBr(G)
?(e) ?( ):
Proof. By using Theorem 1.12 we have
sBsB0 = 1jGj2
X
 2^G
X
?2B\IBr(G)
 ?(e)?( ) 
X
 2G
X
 2B0\IBr(G)
 (e)  ( ) 
= 1jGj2
X
 2^G
X
 2G
X
?2B\IBr(G)
X
 2B0\IBr(G)
 ?(e)?( ) (e)  ( )  
= 1jGj2
X
 2G
X
 2^G
X
?2B\IBr(G)
X
 2B0\IBr(G)
 ?(e)?( ) (e)  (  1 ) 
and
 BB0sB =
X
 2G
 BB0 1jGj
X
?2B\IBr(G)
?(e) ?( ) :
Now, by Theorem 1.11, sBsB0 =  BB0, so by comparing the coe cients on both sides for a
 xed  2G we get,
X
 2^G
X
?2B\IBr(G)
X
 2B0\IBr(G)
 ?(e)?( ) (e)  (  1 ) =  BB0jGj
X
?2B\IBr(G)
?(e) ?( ):
10
Chapter 2
Symmetrized Tensors
In this chapter we will state some basic de nitions and results of di erent symmetrizers
corresponding to characters discussed in Chapter 1.
2.1 Background
For  xed positive integers n;m set
 n;m =f = ( 1; 2;:::; n)2Znj1  i mg:
Let G be a subgroup of the symmetric group Sn. De ne a right action on  n;m by G as
follows. For  2G and  2 n;m
  = (  (e);:::;  (n)): (2.1)
Consider the relation for  ; 2 n;m given by    if there is an element  2G such that
  =  . This is an equivalence relation on  n;m. We  x a set  of representatives of the
equivalence classes of  n;m with respect to  .
Let V be a complex inner product space of dimension m with orthonormal basis
fe1;e2;:::;emg. V n = V  V      V (n factors) is the nth tensor power of V. For
 2  n;m let e = e 1  e 2      e n. The inner product induced on V n is given by
he ;e i = Qni=1(e i;e i) where ( ; ) is the inner product of V. Under this inner product
fe j  2  n;mg is an orthonormal basis for V n. V n is a CG-module with the action
 e = e   1 for  2G extended linearly to CG.
11
2.2 Symmetrizers associated with ordinary and Brauer characters of G
In the following discussion  stands for either an irreducible ordinary character or an
irreducible Brauer character of G. Denote by S a subset of G where S = ^G when 2IBr(G)
and S = G when  2Irr(G).
The symmetrizer corresponding to  is de ned by,
s =  (e)jSj
X
 2S
 ( ) :
The theorems in this section pertaining to  2 Irr(G) are well known. On the other
hand, we generalize some well-known results for 2Irr(G) to handle the case of 2IBr(G).
Theorem 2.1. The elements s for  2Irr(G) are orthogonal idempotents.
Proof. Let  ; 2Irr(G). Then by Theorem 1.14 we get,
s s =  (e) (e)jGj2
X
 2G
X
 2G
 ( ) ( )  =  (e) (e)jGj2
X
 2G
X
 2G
 ( ) (  1 ) 
=     (e) (e)jGj2
X
 2G
jGj
 (e) ( ) =    
 (e)
jGj
X
 2G
 ( ) =    s :
The symmetry class of tensors V corresponding to  is the image of V n under the
symmetrizer s :
V = s V n:
Corollary 2.2. If  ; 2Irr(G) and  6=  , then the vector spaces V and V are orthogonal.
Proof. If s v = s w for some v;w2V n, then
s v = s (s v) = s (s w) = 0:
12
Let  2 n;m. The standard symmetrized tensor e  corresponding to  is the image of
e 2V n under s :
e  = s e =  (e)jSj
X
 2S
 ( ) e =  (e)jSj
X
 2S
 ( )e   1: (2.2)
By o-basis of a subspace W of V n we mean an orthogonal basis of W that consists of
standard symmetrized tensors. An interesting question to ask is \For which W does there
exist an o-basis?" In 1991 Wang and Gong gave an example in [16] of such an o-basis for a
symmetry class of tensors V with  2 Irr(G) when G is the dihedral group of order eight.
Ever since there have been papers [1, 3, 4, 7, 8, 10, 15] answering the question when such an
o-basis exists. All these papers however address the problem in the ordinary character case.
This dissertation is devoted to answering the question of when an o-basis exists for a
symmetry class of tensors symmetrized by a Brauer symmetrizer for particular choices of G.
Let V = he  j 2Gi and let V  = s (V ). Observe that V  = he   j 2Gi. V  is
called the orbital subspace corresponding to  . Using the orbital subspaces we can write the
symmetry class of tensors as an orthogonal direct sum.
Theorem 2.3 ([9, Theorem 1.1]). We have
V = _
X
 2 
V  (orthogonal direct sum):
In particular, V has an o-basis if and only if V  has an o-basis for each  2 .
Proof. Let  2 n;m. Then  =   for some  2 and  2G, so that e  = e   2V  . This
shows that V is contained in (and hence equals) the indicated sum.
The sets E =fe  j 2Gg,  2 , are pairwise disjoint subsets of the orthogonal set
fe j 2  n;mg and are therefore pairwise orthogonal. For each  2  the subspace V  is
contained in the span of E , so the indicated sum is an orthogonal direct sum.
13
Assume that V has an o-basis B. By the  rst paragraph, B is the union of the sets
B = B\V  ,  2 , and these sets are pairwise disjoint by the second paragraph, so B is
an o-basis for V  for each  2 .
Finally, if V  has an o-basis for each  2 , then the union of these bases is an o-basis
for V .
For  2 n;m let G =f 2Gj  =  g the stabilizer subgroup of  in G.
Theorem 2.4. For  2 n;m and a  xed  2G, we have
(e   ;e  ) =  (e)
2
jSj2
X
 2S
X
 2   1S\G 
 ( ) (  1   1):
Proof. Let  2 n;m and  2S be  xed. Now by using the Equation 2.2 we get,
(e   ;e  ) =  (e)
2
jSj2
X
 2S
X
 2S
 ( ) ( )(e    1;e   1)
=  (e)
2
jSj2
X
 2S
X
 2S
   1 2G 
 ( ) (  1)
=  (e)
2
jSj2
X
 2S
X
 2   1S\G 
 ( ) (  1   1):
Corollary 2.5. For  2 n;m,  2Irr(G), and a  xed  2G
(e   ;e  ) =  (e)jGj
X
 2G  
 ( ):
Proof. From the above Theorem 2.4 we get
(e   ;e  ) =  (e)
2
jGj2
X
 2G
X
 2G 
 ( ) (  1   1) =  (e)
2
jGj2
X
 2G 
X
 2G
 ( ) (  1   1):
14
Then by using the generalized orthogonality relation (Theorem 1.14) we get
(e   ;e  ) =  (e)
2
jGj2
X
 2G 
jGj (  1 )
 (e) =
 (e)
jGj
X
 2G 
 (  1 ) =  (e)jGj
X
 2G  
 ( ):
For  2Irr(G) and for  2 Freese gives the dimension of the orbital subspace V  in
[6]:
dimV  =  (e)jG
 j
X
 2G 
 ( ): (2.3)
2.2.1 Symmetrizers associated with PIs
Let ?2IBr(G) and put  =  ?.
The symmetrizer associated with  is de ned by
s = ?(e)jGj
X
 2G
 ( ) : (2.4)
Note that s 2 CG. For a PI  of G the symmetry class of tensors is de ned by
V = s V n. For  2 n;m the standard symmetrized tensor is de ned by e  = s e and the
orbital subspace is de ned by V   =he   j 2Gi. With a similar argument as in Theorem
2.3 we have
V = _
X
 2 V
 
 (orthogonal direct sum): (2.5)
Theorem 2.6. For  2G and  2 n;m
(e   ;e  ) = ?(e)
2
jGj2
X
 2G
X
 2G 
 (  1  ) ( ):
15
Proof.
 e 
  ;e
 
 
 = ?(e)2
jGj2
X
 2G
X
 2G
 ( ) ( ) (e   ;e  )
= ?(e)
2
jGj2
X
 2G
X
 2G
    12G 
 ( ) ( )
= ?(e)
2
jGj2
X
 2G
X
 2 G  1\G 
 (  1  ) ( )
= ?(e)
2
jGj2
X
 2G
X
 2G 
 (  1  ) ( ):
2.2.2 Symmetrizers associated with blocks
Let B be a block of G. The symmetrizer corresponding to B is the Osima idempotent
sB of B (Equation 1.4). The symmetry class of tensors is de ned by VB = sBV n. For
 2 n;m the standard symmetrized tensor is de ned by eB = sBe and the orbital subspace
is de ned by VB =heB  j 2Gi.
Lemma 2.7. For  2 n;m
eB =
X
 2B\Irr(G)
e  :
Proof. By Equation 1.4 we get,
eB = sB(e ) =
X
 2B\Irr(G)
s (e ) =
X
 2B\Irr(G)
e  :
Theorem 2.8.
VB = _
X
 2 
VB (orthogonal direct sum):
16
Proof. The argument in the proof is similar to that of the proof of Theorem 2.3, and we
omit the details.
Theorem 2.9. For  2 
dimVB = 1jG
 j
X
 2^G 
X
?2B\IBr(G)
?(e) ?(  1)
= 1jG
 j
X
 2^G 
X
 2B\Irr(G)
 (e) (  1):
Proof. Since sB is an idempotent we have rank (sB) = tr (sB). Then by Equation 1.4 we get,
dimVB = rank (sB) = tr (sB) = tr 1jGj
X
 2G
X
 2B\Irr(G)
 (e) (  1) 
= 1jGj
X
 2G
X
 2B\Irr(G)
 (e) (  1)tr ( ):
Note here that it makes sense to write tr ( ) by viewing  as a linear transformation on V n.
In [6, Equation 13] Freese shows that 1jGjP 2G (e) (  1)tr ( ) =  (e)jG jP 2G  (  1). So we
get,
dimVB =
X
 2B\Irr(G)
 (e)
jG j
X
 2G 
 (  1) = 1jG
 j
X
 2G 
X
 2B\Irr(G)
 (e) (  1):
Now by Equations 1.2 and 1.3 we get, for any  2G,
X
 2B\Irr(G)
 (e) (  1) =
X
?2B\IBr(G)
X
 2B\Irr(G)
?(e)d ? (  1) =
X
?2B\IBr(G)
?(e) ?(  1):
17
So it gives
dimVB = 1jG
 j
X
 2G 
X
?2B\IBr(G)
?(e) ?(  1) = 1jG
 j
X
 2^G 
X
?2B\IBr(G)
?(e) ?(  1)
= 1jG
 j
X
 2^G 
X
 2B\Irr(G)
 (e) (  1):
Theorem 2.10. For  2G,
(eB  ;eB ) = 1jGj
X
 2 G 
X
?2B\IBr(G)
?(e) ?( ): (2.6)
Proof. Recall that V and V are orthogonal for  ; 2Irr(G) with  6=  by Corollary 2.2.
Then using Lemma 2.7 and Corollary 2.5 we get
(eB  ;eB ) = (
X
 2B\Irr(G)
e   ;
X
 2B\Irr(G)
e  ) =
X
 ; 2B\Irr(G)
(e   ;e  )   =
X
 2B\Irr(G)
(e   ;e  )
=
X
 2B\Irr(G)
 (e)
jGj
X
 2G  
 ( ) =
X
 2B\Irr(G)
 (e)
jGj
X
 2 G 
 ( )
= 1jGj
X
 2 G 
X
 2B\Irr(G)
 (e) ( ) = 1jGj
X
 2 G 
X
?2B\IBr(G)
?(e) ?( ):
The following is a useful lemma which we call the translation principle of the orthogo-
nality of symmetrized tensors.
Lemma 2.11. If the standard symmetrized tensors eB  and eB  0 are orthogonal, then eB   
and eB  0 are orthogonal for every  2G.
18
Proof. By Theorem 2.10 we get, for each  2G,
(eB  0 ;eB   ) = 1jGj
X
 2 0 (  ) 1G 
X
?2B\IBr(G)
?(e) ?( )
= 1jGj
X
 2 0  1G 
X
?2B\IBr(G)
?(e) ?( ) = (eB  0;eB  ) = 0:
19
Chapter 3
Dihedral group
In this chapter we focus on the existence of an o-basis of a symmetry class of tensors
associated with a Brauer character, a PI and an Osima idempotent of a block of the dihedral
group.
For an integer n( 3), the dihedral group of degree n is the subgroup Dn of the sym-
metric group Sn, generated by the elements
r =
0
B@ 1 2 ::: n 1 n
2 3 ::: n 1
1
CA s =
0
B@ 1 2 ::: n 1 n
1 n ::: 3 2
1
CA;
That is Dn =frk;srkj0 k n 1g and jDnj= 2n.
Dn with n even has 4 degree one irreducible characters. Let  1; 2; 3; 4 denote these
characters. Dn with n odd has only 2 degree one irreducible characters; we denote them
with  1; 2. For all n the degree two irreducible characters of Dn are given by  h where
1  h < n2 . For each integer k we get  h(rk) = !hk + ! hk = 2 cos 2 hkn where !n = 1 [14,
page 37]. The character table for Dn is given by
rk srk
 1 1 1
 2 1  1
 3 ( 1)k ( 1)k (n even)
 4 ( 1)k ( 1)k+1 (n even)
 h 2 cos(2 hkn ) 0 1 h< n2
20
We observe from the table that if  a character of degree one or degree two of Dn, then for
 2Dn we have
 ( ) =  (  1): (3.1)
Let G = Dn. For a  xed prime p write n = pq? with p - ?. The set ^G of p-regular
elements of G is given by,
^G =
8
><
>:
frapq;srkj0 a<?;0 k<ng; p6= 2;
frapq j0 a<?g; p = 2.
The set of p-regular conjugacy classes of G is,
frapq;r apqg; 0 a ?=2; fsr2kj0 k<n=2g; fsr2k+1j0 k<n=2g; ? even, p6= 2;
frapq;r apqg; 0 a (? 1)=2; fsrkj0 k<ng; ? odd, p6= 2;
frapq;r apqg; 0 a<?=2; p = 2:
So the number of p-regular conjugacy classes is
" =
8
>>>>
<
>>>>
:
?
2 + 3; if ? even, p6= 2;
? 1
2 + 2; if ? odd, p6= 2;
? 1
2 + 1; if p = 2.
(3.2)
3.1 Brauer characters of Dn
Our e ort in this section is to  nd conditions for the existence of an o-basis for the
symmetry class of tensors corresponding to a Brauer character of the dihedral group G = Dn.
We begin by listing the distinct irreducible Brauer characters of G.
Recall for an ordinary character  of G the restriction of  to ^G is denoted by ^ . By 1.7
^ is a Brauer character of G.
21
For each 1  h  n2 , the Brauer character ^ h is of degree 2. The next lemma gives
conditions for when two Brauer characters of degree two are the same.
Lemma 3.1. For 1 i;j < n2 , ^ i = ^ j if and only if either i + j 0 mod ? or i j 0
mod ?.
Proof. Since any degree two character of G is zero on srk for all k it is enough to check
when two characters are the same on the elements rapq 2G. Suppose ^ i = ^ j. Then for any
0 a<? we have,
0 = ^ i(rapq) ^ j(rapq) = 2 cos 2 ap
qi
n  2 cos
2 apqj
n = 4 sin
 apq(i+j)
n sin
 apq(i j)
n :
So  apq(i+j)n = k or  apq(i j)n = k for some integer k. This gives a(i+j)? = k or a(i j)? = k, so
the result follows since ?-a.
Conversely suppose either i + j  0 mod ? or i j  0 mod ?. With out loss of
generality we may assume i+j = k? for some integer k. Then
^ i(rapq) ^ j(rapq) = 4 sin  ap
q(i+j)
n sin
 apq(i j)
n = 4 sin
 apqk?
n sin
 apq(i j)
n
= 4 sin aksin  ap
q(i j)
n = 0:
So ^ i = ^ j as desired.
Some of the restricted degree two characters are not irreducible as given by the following
lemma.
Lemma 3.2. For all n we have ^ k? = ^ 1 + ^ 2 for 1 k < pq2 . When n is even with p6= 2
we also have ^ ?
2 +k?
= ^ 3 + ^ 4 for 0 k< pq 12 .
Proof. For 0 a<?
^ k?(rapq) = 2 cos(2 k?ap
q
n ) = 2 =
^ 1(rapq) + ^ 2(rapq) = ( ^ 1 + ^ 2)(rapq);
22
and for 0 b<n
^ k?(srb) = 0 = ^ 1(srb) + ^ 2(srb) = ( ^ 1 + ^ 2)(srb):
So we have ^ k? = ^ 1 + ^ 2.
Now assume n is even and p6= 2. Then it makes sense to consider the characters of G
given by ^ ?
2 +k?
. Now for 0 a<?, we have
^ ?
2 +k?
(rapq) = 2 cos
 2 ?(1 + 2k)apq
2n
 
= 2 cos
 
 (1 + 2k)a
 
;
so ^ ?
2 +k?
(rapq) equals 2 if a is even and 2 if a is odd, and also we have that ^ 3(rapq)+ ^ 4(rapq)
equals 2 if a is even and  2 if a is odd. For 0  k < n we have ^ ?
2 +k?
(srk) = 0 =
^ 3(srk) + ^ 4(srk). So we get ^ ?
2 +k?
= ^ 3 + ^ 4 as desired.
Let
 =
8
>>>
><
>>>>
:
4; if ? even, p6= 2;
2; if ? odd, p6= 2;
1; if p = 2.
(3.3)
For each 1 j  , the Brauer character ?1j = ^ j is of degree 1.
Theorem 3.3. Let G = Dn. The complete list of distinct irreducible Brauer characters of
G is
?1j = ^ j, for 1 j  , ?2i = ^ i for 1 i< ?2.
Proof. For each 1 j  , the Brauer character ^ j is of degree one and hence is irreducible.
We see the distinctness of these characters by observing the character values for srk in the
character table above. So we have  distinct irreducible Brauer character of degree one of G.
To see the distinctness of the degree two Brauer characters in the given list observe that
for any i;j such that 1 i<j < ?2 we have j i<? and i+j <?, so ?-j i;i+j which
gives that ^ i6= ^ j by Lemma 3.1. Now to show that they are irreducible assume that ?2i is
23
not irreducible for some 1 i< ?2. Then ?2i is a sum of two irreducible Brauer characters of
degree one. So we write ?2i = ?1j +?1k. Now r2pq 2 ^G and ?2i(r2pq) = ?1j(r2pq) +?1k(r2pq) = 2.
Now since 1  i < ?2 we have 0 < 4 pqin = 4 i? < 4 ?2? = 2 . So ?2i(r2pq) = 2 cos 4 pqin 6= 2.
This gives a contradiction. Therefore ?2i is irreducible. The number of irreducible Brauer
characters of G equals the number of p-regular conjugacy classes of G by Theorem 1.6. Using
Equation 3.3 we see that the number of all characters in the given list is
?
2 + 3; if ? even, p6= 2;
? 2
2 + 2; if ? even, p6= 2;
? 2
2 + 1 if p = 2:
This is same as the number of p-regular conjugacy classes as given by Equation 3.2, so the
indicated set is a complete set of irreducible Brauer characters of G.
Assume p6= 2. For 1 i<?=2, put
Ai =fi;k?+i;k? ij1 k p
q 1
2 g
and note that jAij= pq.
Lemma 3.4. Let 1 i<?=2. We have ^ a = ^ i = ?2i for all a2Ai.
Proof. Let 1 i<?=2. No proof is needed when a = i. Suppose a = k? + i or a = k? i.
Then a i = k? or a+i = k?. So by Lemma 3.1 we get ^ a = ^ i = ?2i.
3.2 PIs of Dn
Let G = Dn. Let  1j; 2i be the PIs corresponding to ?1j;?2i 2IBr(G) where j = 1;2;3;4
and 1 i< ?2. Let  2Irr(G). By Equation 1.2 we have
^ =
X
?2IBr(G)
d ??;
where d ? are uniquely de ned nonnegative integers.
24
In the following tables we give the values of the decomposition matrix entries d ? cor-
responding to  2Irr(G) and ?2IBr(G).
For odd n we get
 ? d ?
 j ?1j 1 (j = 1;2)
 k? ?1j 1 (1 k pq 12 , j = 1;2)
 a ?2i 1 (a2Ai; 1 i< ?2)
and for even n we get
 ? d? 
 j ?1j 1 (j = 1;2;3;4)
 k? ?1j 1 (1 k pq 12 , j = 1;2)
 ?
2 +k?
?1j 1 (0 k pq 12  1, j = 3;4)
 a ?2i 1 (a2Ai; 1 i< ?2)
Theorem 3.5. Let G = Dn. Then the complete list of PIs of G is
 1j =  j +
pq 1
2X
k=1
 k? for j = 1;2,
 1j =  j +
pq 1
2  1X
k=0
 ?
2 +k?
for j = 3;4,
 2i =
X
a2Ai
 a; for 1 i< ?2:
Proof. The proof follows from the tables above and the Equation 1.3.
25
3.3 Blocks of Dn
Let G = Dn. The relationships of characters belonging to a block of G can be understood
by means of the decomposition numbers of G for a considered prime p. Recall that both
 ; 2 Irr(G) belong to the block containing ? 2 IBr(G) if both decomposition numbers
d ? and d ? are nonzero. Also recall that each  2 Irr(G) belongs to a unique block. Now
by observing the tables above one notes the consistence of characters in blocks of G as
given below. For the blocks we give the notation Bab, where a gives the lowest degree of
the irreducible Brauer characters it contains and b gives the lowest index of the degree a
irreducible characters it contains.
 ?1j; j; k? in a block B11 where j = 1;2 and 1 k pq 12 .
 ?1j; j; ?
2 +k?
in a block B13 where j = 3;4 and 0 k pq 12  1.
 for 1 i< ?2, ?2i; a in a block B2i where a2Ai.
3.4 Block idempotent symmetrization
In this section we will establish necessary and su cient conditions for the existence of
an o-basis of the symmetry class of tensors VB corresponding to an Osima idempotent sB of
a block B of G = Dn. We will consider the two cases namely B1j the block containing degree
one irreducible characters of G and B2i the block consisting only of degree two irreducible
characters of G separately when  nding the o-basis.
First we state a specialized formula for the inner product given in Theorem 2.10.
Corollary 3.6. Let G = Dn and let B be a block of G. Then
(eB  ;eB ) = 1jGj
X
 2 G \Cn
X
?2B\IBr(G)
?(e) ?( ):
26
Proof. Fix 0 k n 1. We observe from the character table for G that  1(srk)+ 2(srk) =
0 and  3(srk) + 4(srk) = 0. Therefore by using Theorem 3.5 we get,
X
?2B11\IBr(G)
?(e) ?(srk) = ?11(e) 11(srk) +?12(e) 12(srk)
=  1(srk) + 2(srk) + 2
pq 1
2X
k=1
 k?(srk) = 0
and
X
?2B13\IBr(G)
?(e) ?(srk) = ?13(e) 13(srk) +?14(e) 14(srk)
=  3(srk) + 4(srk) + 2
pq 1
2X
k=1
 ?
2 +k?
(srk) = 0:
Also for  xed i with 1 i< ?2 we have
X
?2B2i\IBr(G)
?(e) ?(srk) = ?2i(e) 2i(srk) = ?2i(e)
X
a2Ai
 a(srk) = 0:
Now write B for B11;B13, or B2i . By Theorem 2.10 we have
(eB  ;eB ) = 1jGj
X
 2 G 
X
?2B\IBr(G)
?(e) ?( ) = 1jGj
X
 2 G \Cn
X
?2B\IBr(G)
?(e) ?( ):
Lemma 3.7. Let G = Dn. Let B be a block of G. Fix  2 . Let G \Cn =hrki with kjn
and let t be the largest such that ptjk.
i) If H =fra0;ra1;:::;rapt 1gis a list of coset representatives ofhrptiin Cn, thenfeB  j 2
Hg is an orthogonal set. In particular feB  j 2H0g is an orthogonal set where H0 =
f1;r;:::;rpt 1g.
27
ii) If G  Cn, then feB  ;eB s j 2H0g is an orthogonal set.
iii) If eB and eB  are orthogonal for some  2 ^G\Cn, then feB  ;eB   j 2 H0g is an
orthogonal set.
iv) If G  Cn and if eB and eB  are orthogonal for some  2 ^G\Cn, then
feB  ;eB   ;eB s ;eB s  j 2H0g is an orthogonal set.
Proof. i) Take rax;ray 2 H. Then raxhrpti 6= rayhrpti and pt - ax  ay. Now we will
show that there are no regular elements in the set raxr ayG \Cn. Take rax ay+mk 2
raxr ayG \Cn for some integer m = 0;:::;nk  1. Assume ax ay + mk = m0pq. Then
ax ay = m0pq mk, implying ptjax ay which is a contradiction. Therefore by Corollary
3.6 we get (eB raxr ay;eB ) = 0. Now for any rx;ry 2H0 with x6= y, we have pt - x y, so
rx;ry are distinct coset representatives of hrpti in Cn. Therefore this is a special case of the
argument above so feB  j 2H0g is an orthogonal set.
ii) By part i) we know that (eB  ;eB  ) = 0 for  ; 2H0. To show that the setfeB  eB s j 2
H0g is orthogonal it remains to show that (eB s ;eB  ) = 0 and (eB s ;eB s ) = 0 for  ; 2
H0. First for (eB s ;eB  ) we have, (eB s ;eB  ) = (eB s   1;eB ) = 0 by Corollary 3.6 since
s   1G \Cn =; when G  Cn. Now for the other case
(eB s ;eB s ) = (eB s s ;eB ) = (eB   1 ;eB ) = (eB    1;eB ) = (eB  ;eB  ) = 0;
by part i) and this completes the proof.
iii) Suppose (eB ;eB  ) = 0 for some  2 ^G\Cn. To show that feB  ;eB   j 2H0g is an
orthogonal set we only need to show that (eB   ;eB   ) = 0 and (eB   ;eB  ) = 0 since part i)
takes care of the case for two elements of the form eB  where  2H0. It is easily seen by the
translation principle (Lemma 2.11) and part i) above that (eB   ;eB   ) = (eB  ;eB  ) = 0:
28
Now consider (eB   ;eB  ). We have  = rx; = ry and  = r pq for some integers x;y;
and  with 0 x;y<pt. Then
(eB   ;eB  ) = (eB ryr pq;eB rx) = (eB r pq+y x;eB ):
Now to show that (eB   ;eB  ) = 0 it is enough by Corollary 3.6 to show that r pq+y xG 
has no p-regular elements. Assume to the contrary. Then for some integer m, we have
rmk+ pq+y x 2 r pq+y xG \ ^G, implying rmk+ pq+y x = rm0pq for some integer m0. But
this implies mk +  pq + y x m0pq = jn for some integer j, which is equivalent to
y x = m0pq mk  pq +jn implying ptjy x. This is a contradiction since 0 x;y<pt.
iv) Suppose (eB ;eB  )=0. To show feB  ;eB   ;eB s ;eB s  j 2H0g is an orthogonal set, we
only need to show that (eB s  ;eB  ) = 0;(eB s ;eB   ) = 0;(eB s  ;eB   ) = 0;(eB s ;eB s  ) = 0
and (eB s  ;eB s  ) = 0 since all other combinations of elements in the set are shown to be
orthogonal in the three previous parts. Note that we have (eB s  ;eB  ) = (eB s    1;eB ). Now
since G  Cn we see that s    1G \Cn = ; whence (eB s    1;eB ) = 0 by Corollary 3.6.
The argument for the cases (eB s ;eB   ) and (eB s  ;eB   ) is the same. Next due to part iii)
(eB s ;eB s  ) = (eB s (s  ) 1;eB ) = (eB s s  ;eB ) = (eB   1  ;eB ) = (eB     1;eB ) = (eB   ;eB  )
= 0:
Finally by part ii) we get (eB s  ;eB s  ) = (eB s ;eB s ) = 0 and it completes the proof.
Now we look at the block B11 of G.
Lemma 3.8. Fix  2 . We have G \Cn =hrai with ajn. Let t be the largest such that
ptja. Then
dimVB11 =
8>
<
>:
2pt; if G  Cn;
pt; otherwise.
29
Proof. First observe that for any 0  <?,
 k?(r pq) = 2 cos 2 k? p
q
n = 2:
Write ? = ?1j. Then by Theorem 2.9, Theorem 3.5 and the fact that  1( ) +  2( ) = 0 for
all  62Cn and  1( ) + 2( ) = 2 for all  2Cn we get
dimVB11 = 1jG
 j
X
 2^G 
X
?2B11\IBr(G)
?(e) ?( ) = 1jG
 j
X
 2^G \Cn
 
 1( ) + 2( ) + 2
pq 1
2X
k=1
 k?( )
 
= 1jG
 j
 
2 + (pq 1)2
 
j^G \Cnj= 2p
q
jG j
n
apq t =
2
jG j
n
ap
t:
Now if G  Cn, thenjG j= na, in which case we get dimVB11 = 2pt and otherwisejG j= 2na
which gives dimVB11 = pt.
Let n be even and consider the block B13 of G containing the characters  3 and  4.
Lemma 3.9. Let G = Dn with n even and  x  2 . We have G \Cn =hrai with ajn.
Let t be the largest such that ptja. Then
dimVB13 =
8
>>>
><
>>>>
:
0; if a is odd;
2pt; if G  Cn, a is even;
pt; if G *Cn, a is even.
Proof. By Theorem 2.9, Theorem 3.5 and the fact that  3( ) + 4( ) = 0 for all  62Cn we
get,
dimVB13 = 1jG
 j
X
 2^G 
X
?2B13\IBr(G)
?(e) ?( ) = 1jG
 j
X
 2^G 
 
 3( ) + 4( ) + 2
pq 1
2X
k=1
 ?
2 +k?
( )
 
= 1jG
 j
X
 2^G \Cn
 
 3( ) + 4( ) + 2
pq 1
2X
k=1
 ?
2 +k?
( )
 
:
30
Suppose that a is odd. Let a = a0pt for some odd integer a0. Then ^G \Cn =fr apq t j0 
  ?a0 1g. Note that ?a0 is even.
 ?
2 +k?
(r apq t) = 2 cos 2 (
?
2 +k?) ap
q t
n = 2 cos 2 (
1
2 +k) a
0 = 2 cos  a0;
which is equal to 2 or  2 depending upon whether  is even or odd. Also  3(r apq t) =
 4(r apq t) equals 1 or 1 according as  is even or odd respectively. So the sum  3(r apq t)+
 4(r apq t)+2P
pq 1
2
k=1  ?2 +k?(r
 apq t) on the right side of the formula for dimVB13 above equals
2pq if  is even and  2pq if  is odd. Therefore in the case a is odd we get
dimVB13 = 1jG
 j
 
2pq ?2a0 + 2pq ?2a0
 
= 0:
Now suppose a is even. Then any p-regular element inhraican be expressed as r2 pq for
some integer  . We have
 ?
2 +k?
(r2 pq) = 2 cos 2 (
?
2 +k?)2 p
q
n = 2 cos 2 (
1
2 +k)2 = 2;
in which case the proof is similar to the proof of Lemma 3.8 above. So we get dimVB13 = 2pt
if G is contained in Cn and dimVB13 = pt if G is not contained in Cn.
Theorem 3.10. Let G = Dn. Write B = B11. The symmetry class of tensors VB has an
o-basis.
Proof. By Theorem 2.8 we have
VB = _
X
 2 
VB ;
so it su ces to show that VB has an o-basis for each  2 . Let  2 . By Lemma 3.8 if
G * Cn, then dimVB = pt. Therefore by part i) of Lemma 3.7 the set feB  j 2H0g is an
orthogonal basis. If G  Cn, then dimVB = 2pt and in this case feB  ;eB s j 2H0g is an
orthogonal basis by part ii) of Lemma 3.7.
31
Theorem 3.11. Let G = Dn with n even. Write B = B13. The symmetry class of tensors
VB has an o-basis.
Proof. Due to Theorem 2.8 it su ces to show that VB has an o-basis for each  2  . Let
 2  . We have G \Cn = hrai for some integer a. If a is odd, then by Lemma 3.9
dimVB = 0, so VB has an o-basis. Suppose a is even. Then dimVB = pt if G * Cn, so
by part i) of Lemma 3.7 the set feB  j 2H0g is an orthogonal basis and dimVB = 2pt if
G  Cn, so feB  ;eB s j 2H0g is an orthogonal basis by part ii) of Lemma 3.7.
Now we will bring our attention to the blocks consisting only of degree two characters
of G. For each 1 i< ?2 the block B2i contains ?2i 2IBr(G) and this is the only irreducible
Brauer character of G it contains. Below is a statement for conditions when the dimension
of the orbital subspace VB2i corresponding to a  2 is not zero.
Theorem 3.12. Fix  2  n;m. Then for 1  i < ?2 we have dimVB2i 6= 0 if and only if
^G \Cn hrn0i, where n0 = ngcd(n;i).
Proof. Suppose ^G \Cn6 hrn0i. We have ^G \Cn =hrbiwith bjn, so rb62hrn0i. Fix i with
1 i< ?2. Then for  2 ^G we have by Theorem 3.5 and Lemma 3.4
 2i( ) =
X
a2Ai
 a( ) =jAij i( ) = pq i( ): (3.4)
Now by Theorem 2.9
dimVB2i = 1jG
 j
X
 2^G 
?2i(e) 2i( ) = ?
2
i(e)
jG j
X
 2^G 
pq i( ) = p
q?2
i(e)
jG j
X
 2^G \Cn
 i( )
= p
q?2
i(e)
jG j
n
b 1X
j=0
 i(rjb) = p
q?2
i(e)
jG j
n
b 1X
j=0
(!ibj +! ibj);
where !ib is an nbth root of unity. Now Pnb 1j=0 (!ibj + ! ibj) 6= 0 if !ib = 1. But if !ib = 1,
then there is an integer m such that ib = mn, so i00b = mn0 where i00 = igcd(n;i). Now since
32
gcd(n0;i00) = 1 we get n0jb which is a contradiction. Therefore Pnb 1j=0 (!ibj + ! ibj) = 0 and
hence dimVB2i = 0.
Conversely suppose ^G \Cn hrn0i. Note that for any integer c we have
 i(rcn0) = 2 cos 2 icn
0
n = 2 cos 2 i
00c = 2;
and in this case dimVB2i =  2i(e)jG j P 2^G \Cnpq i( )6= 0.
Theorem 3.13. Let G = Dn. Fix i with 1 i< ?2. Let G \Cn =hrki with kjn and let t
be the largest integer such that ptjk. If ^G \Cn hrn0i, then
dimVB2i =
8>
<
>:
4pt; if G  Cn
2pt; otherwise.
Proof. Let  2 ^G \Cn hrn0i. Then  = rcn0 for some integer c and
 i( ) =  i(rcn0) = 2 cos 2 icn
0
n = 2 cos 2 i
00c = 2:
Therefore by Theorem 2.9 and Equation 3.4
dimVB2i = 1jG
 j
X
 2^G 
?2i(e) 2i( ) = ?
2
i(e)
jG j
X
 2^G 
pq i( ) = p
q?2
i(e)
jG j
X
 2^G \Cn
 i( )
= 2p
q
jG j
X
 2^G \Cn
2 = 4p
q
jG jj
^G \Cnj= 4pq
jG j
n
apq t =
4
jG j
n
ap
t:
Now if G  Cn, then jG j = na, so dimVB2i = 4pt, and if G 6 Cn, then jG j = 2na , so
dimVB2i = 2pt.
Theorem 3.14. Let G = Dn and assume dimV  2. For  xed i with 1  i < ?2 write
B = B2i . The space VB has an o-basis if and only if ?0 0 mod 4, where ?0 = ?gcd(?;i).
33
Proof. Suppose VB has an o-basis. Then by Theorem 2.8 it follows that VB has an o-basis
for each  2  . Let  = (1;2;2;:::;2). Then G = f1;sg, so G \Cn = hrni. Now since q
is the largest such that pqjn, by Theorem 3.13 dimVB = 2pq. The space VB has an o-basis,
that is, an orthogonal basis of the form E = feB  x j x 2G;1  x 2pqg. Consider the
subgroup J =hrpqiG of G. The index of J in G is pq, so by the pigeonhole principle there
is at least one right coset of J containing  x and  y for some 1  x;y  2pq with x6= y.
Then we have  x  1y 2 J, so  x  1y = rmpq for some integer m and  2 G . Therefore
 x  1y G = rmpq G = rmpqG =frmpq;sr mpqg. Then by Corollary 3.6 and Equation 3.4
0 = (eB  x;eB  y) = 1jGj
X
 2 x  1y G \Cn
?2i(e) 2i( ) = ?
2
i(e)
jGj  
2
i(r
mpq) = ?
2
i(e)
jGj p
q i(rmpq):
So we get 0 =  i(rmpq) = 2 cos 2 impqn = cos 2 im? which gives 2 im? = (2k + 1) 2 for some
integer k. Let i0 = igcd(?;i). Then
4i0m = 4imgcd(?;i) = (2k + 1)?gcd(?;i) = (2k + 1)?0:
So ?0 is divisible by 4.
Conversely suppose ?0 0 mod 4. Fix  2 such that dim VB 6= 0. Then by Theorem
3.12 we have ^G \Cn  hrn0i where n0 = ngcd(n;i). Let ^G \Cn = hrai where ajn. Note
that a = a0n0 for some integer a0. We will  rst show that there exists  2 ^Cn such that
(eB  ;eB ) = 0. Let  = r pq where  = ?04 2Z. Note here that the set of p-regular elements of
r pqG \Cn is the same as r pq ^G \Cn since for some  2 ^Cn,   2 ^Cn if and only if  2 ^Cn.
Then by Corollary 3.6
(eB r pq;eB ) = 1jGj
X
 2r pqG \Cn
?2i(e) 2i( ) = ?
2
i(e)
jGj
X
 2r pq ^G \Cn
 2i( )
= ?
2
i(e)
jGj
n
a 1X
 =0
 2i(r pq+ a) = ?
2
i(e)
jGj
n
a 1X
 =0
pq i(r pq+ a):
34
Note that since 4j?0, the number i0 = igcd(?;i) is odd. Therefore using i00 = igcd(n;i) and  = ?04
we get
 i(r pq+ a) = 2 cos 2 i( p
q + a0n0)
n = 2 cos
 2 i0gcd(?;i) 
? +
2 i00gcd(n;i) a0n0
n
 
= 2 cos
  i0
2 + 2 i
00 a0
 
= 2 cos  i
0
2 = 0;
so (eB r pq;eB ) = 0. Let t be the largest such that pt ja. Then by Theorem 3.13, we have
dimVB = 2pt if G 6 Cn, in which case feB  ;eB   j 2H0g is an o-basis by Lemma 3.7 part
iii), and dimVB = 4pt if G  Cn, in which case feB  ;eB   ;eB s ;eB s  j 2H0g is an o-basis
by Lemma 3.7 part iv).
3.5 Irreducible symmetrization
In this section we will give necessary and su cient conditions for the existence of an
o-basis of the symmetry class of tensors corresponding to an irreducible Brauer character of
Dn. We will also show the existence of an orthogonal basis for the Brauer symmetry class
of tensors that consists of ordinary standard symmetrized tensors in the case of degree two
irreducible Brauer characters of Dn.
Let G = Dn. Recall that ?2i 2IBr(G) for 1 i< ?2 is of degree two and  i; jl i; jl+i2
Irr(G) for 1 j pq 12 are of degree two. Recall also that  2i denotes the PI corresponding
to ?2i and that ^ 2i denotes the restriction of  2i to ^G.
Lemma 3.15. For each 1 i< ?2 we have
^ 2i = pq?2i:
Proof. By using Theorem 3.5 we write
^ 2i = X
a2Ai
^ a =
X
a2Ai
?2i =jAij?2i = pq?2i
35
Lemma 3.16. For each 1 i< ?2,
s?2i = jGjpqj^Gj
X
a2Ai
s a:
Proof. By the de nition of a symmetrizer s?2i = ?2i(e)j^Gj P 2^G?2i( ) . Now by the Lemma 3.15
above
s?2i = 2j^Gj
X
 2^G
1
pq
^ 2i( ) = 2
j^Gj
X
 2G
1
pq 
2
i( ) ;
where the second equality holds because  2i vanishes o of ^G. Now by Theorem 3.5 we get
s?2i = 2pqj^Gj
X
 2G
X
a2Ai
 a( ) = jGjpqj^Gj
X
a2Ai
2
jGj
X
 2G
 a( ) = jGjpqj^Gj
X
a2Ai
 a(e)
jGj
X
 2G
 a( ) 
= jGjpqj^Gj
X
a2Ai
s a;
which completes the proof.
Lemma 3.17. For each 1 i< ?2 we have
V?2i = _
X
a2Ai
V a (orthogonal direct sum):
Proof. Let v2V n. Then s?2i(v)2V?2i. Let g = jGjpqj^Gj. By using Theorem 3.16 we get
s?2i(v) =
X
a2Ai
s a(gv);
so V?2i  Pa2AiV a. Now to show the other inclusion, note that for some arbitrary va2V n,
P
a2Ais a(va) is in
P
a2AiV a. Therefore there exists an element
P
a2Ais a(
1
gva) in V
 n such
36
that
X
a2Ai
s a(va) = g
X
a2Ai
s a
 X
a2Ai
s a(1gva)
 
= s?2i
 X
a2Ai
s a(1gva)
 
2V?2i;
so Pa2AiV a  V?2i as desired. In the above computation we have used the Lemma 3.16
and the fact that the symmetrizers corresponding to ordinary irreducible characters are
orthogonal projections (Theorem 2.1). Since ordinary symmetrized spaces are orthogonal by
Corollary 2.2 we have the result.
Recall that for  2Irr(G) a standard decomposable symmetrized tensor corresponding
to  is given by e  where  2 n;m.
Theorem 3.18. For 1  i < ?2, V?2i has an orthogonal basis consisting of decomposable
tensors of the form e  ,  2Irr(G), if and only if ?0 0 mod 4, where ?0 = ?gcd(?;i).
Proof. Suppose V?2i has an orthogonal basis of the stated form. Then in particular V i has an
orthogonal basis due to Lemma 3.17. Therefore by [10, Theorem 3.1] we get n 0 mod 4i2
where i2 is the power of 2 such that ii2 is odd. This means that i2 is a factor of ? and further
gcd(?;i)
i2 = gcd(
?
i2;
i
i2 ) is odd. Then since ?
0 = ?
gcd(?;i), for some integer m we get
4m = ni
2
= p
q?
i2 =
pq?0gcd(?;i)
i2 = p
q?0gcd(?;i)
i2 ;
so 4j?0 as desired.
Conversely suppose ?0 0 mod 4. Let ?2 be the largest factor of ? that is expressed
as a power of 2. Now since 4 j ?0 = ?gcd(?;i) it is the case that 4 j ?2gcd(?2;i2), which gives
gcd(?2;i2) = i2, so 4i2 j?2 and hence 4i2 j?. So we have n 0 mod 4i2 and therefore V i
has an o-basis by [10, Theorem 3.1]. Now, since 4i2 j?, we have ? = a4i2 for some integer a,
so for a  xed k where 1 k pq 12 we have
k? i = k(a4i2) i2i20 = i2(ka4 i20);
37
where i20 = ii2 is odd. Then if (k? i)2 is the largest factor of k? i as a power of 2, then
(k? i)2 = i2, so n 0 mod 4(k? i)2. Therefore V k? i has an orthogonal basis by [10,
Theorem 3.1]. This means V a has an o-basis for each a2Ai and hence by Lemma 3.17 we
get the result.
Recall for a character  of G, V  =he   j 2Gi.
Lemma 3.19. For 1 i< ?2 and  2 n;m
V?2i = _
X
a2Ai
V a (orthogonal direct sum):
Proof. Fix 1  i < ?2 and  2  n;m. Take s?2i(w) 2V?2i for some w2V . Let g = jGjpqj^Gj.
Then by Lemma 3.16,
s?2i(w) = g
X
a2Ai
s a(w) =
X
a2Ai
s a(gw);
and this gives the inclusion V?2i  Pa2AiV a : For the other inclusion consider arbitrary
wa2V and set
v :=
X
a2Ai
s a(wa)2
X
a2Ai
V a :
Note that v is also in V , so 1gv2V . Then
v = g
X
a2Ai
s a(1gv) = s?2i(1gv)2V?2i ;
so we have Pa2AiV a  V?2i .
Since orbital subspaces are orthogonal by Theorem 2.3 we have an orthogonal direct
sum as desired.
38
For  2  n;m recall that G is the stabilizer subgroup of  and   = f 2  j
P
 2G  ( )6= 0g. Using Theorem 2.3 and Equation 2.3 we have
V = _
X
 2 
V  = _
X
 2  
V  : (3.5)
For 1 i< ?2 put  i = Sa2Ai   a.
Theorem 3.20. For 1 i< ?2 we have
V?2i = _
X
 2 i
V?2i :
Proof. By using Lemma 3.17 and Equation 3.5 we get,
V?2i = _
X
a2Ai
V a = _
X
a2Ai
_X
 2  a
V a = _
X
 2 i
_X
a2Ai
V a = _
X
 2 i
V?2i ;
where the last equality is due to Lemma 3.19.
Lemma 3.21. For each 1 i< ?2,  2 m;n, and  2G, we have
(e?2i  ;e?2i ) = g2
X
a2Ai
(e a  ;e a );
where g =jGj=(pqj^Gj).
Proof. Fix 1 i< ?2,  2 m;n, and  2G. By Lemma 3.16 we get,
e?2i = s?2ie = g
X
a2Ai
s ae = g
X
a2Ai
e a ;
so
(e?2i  ;e?2i ) = g2(
X
a2Ai
e a  ;
X
a2Ai
e a ) = g2
X
a2Ai
(e a  ;e a ):
39
Lemma 3.22. For a  xed integer a consider the list of numbers of the form k? + a where
k = 0;:::;pq 1. Then for any integer 0   q, there are exactly pq  numbers in the list
that are divisible by p .
Proof. Let 0   q. We  rst show that for any list of p numbers of the form k?+a where
k = 0;:::;p  1 there is exactly one number in the list divisible by p . We will show that
the remainders when divided by p of the numbers k?+a where k = 0;:::;p  1 are distinct.
Let 0 k1 <k2  p  1 and write k1?+a = Q1p +R1 and k2?+a = Q2p +R2 with
Qi;Ri2Z. Assume R1 = R2. Then Q1 <Q2 since if Q1  Q2, then k1  k2. Now we have
k1?+a Q1p = k2?+a Q2p )(Q2 Q1)p = (k2 k1)?:
This is a contradiction since p - (k2 k1)?. So all the remainders of the numbers in the list
when divided by p are distinct and hence are 0;1;:::;p  1. Therefore there is exactly one
number of the form k?+a where k = 0;:::;p  1 that is divisible by p .
Now we consider the list of pq numbers of the form k? +a where k = 0;:::;pq 1 and
a is any integer. By the above result we have that there is exactly one number in the list of
the  rst p numbers that is divisible by p . Say p jk0? + a where 0  k0 p  1. Then
to be divisible by p , a number in the list fk? + ajk = 0;:::;pq 1g must have the form
(k0 +bp )?+a where b = 0;:::;pq   1. Therefore there are exactly pq  numbers in the list
divisible by p .
Theorem 3.23. Fix i where 1 i< ?2. There are exactly pq  elements of the set Ai that
are divisible by p .
Proof. For a = k? i 2 Ai where 1  k  pq 12 note that p j k? i if and only if
p j pq? (k? i) = (pq k)? + i. Then by writing i = 0? + i we see that the number
of elements of Ai that are divisible by p is the same as the number of integers k? + i,
0 k pq 1, that are divisible by p , which number is pq  by Lemma 3.22.
40
Let n0a = ngcd(n;a) for a2Ai.
Lemma 3.24. Fix a2Ai. Then n0a = pq  a?0, where  a is the largest such that p a ja.
Proof. Let  a with 0   a q, be the largest such that p a ja. Observe that  a is also the
largest such that p a j agcd(?;i) since p- gcd(?;i). Then gcd(pq?0; agcd(?;i)) = gcd( pq?gcd(?;i); agcd(?;i)) =
tp a for some integer t with p - t. Now since t j pq?0 we get t j ?0. If a = i we get
a
gcd(?;i) =
i
gcd(?;i) = i
0, so tji0. On the other hand, if a = k? i we get a
gcd(?;i) =
k? i
gcd(?;i) = k?
0 i0
and since tjk?0 i0 and tj?0 we get tji0. Then since gcd(?0;i0) = 1 it should be that t = 1
and this gives gcd(pq?0; agcd(?;i)) = p a. Now
n0a = p
q?
gcd(pq?;a) =
pq?
gcd(?;i) gcd(pq?0; agcd(?;i)) =
pq?
gcd(?;i) p a = p
q  a?0:
Recall for  2Irr(G),   =f 2 jP 2G  ( )6= 0g.
Lemma 3.25. Let 1 h< n2 . We have  2  h if and only if G is of the form H or HT
where H 6hrn0hi and T =hsi
Proof. Suppose  2   h. In [10] it is shown that if  2   h, then G is of the form H or
HT, where H 6hrn0hiand T =htiwith t2 = 1 and t62Cn. So the desired result follows since
s62Cn and s2 = 1. To prove the other direction suppose G is of the form H or HT. First
assume G is of the form H. We have G = hrmn0hi for some integer m such that mn0h jn.
Then for any r mn0h 2G with  2Z we have
 h(r mn0h) = !h mn0h +! h mn0h:
Now since h00 = hgcd(n;h) and n0h = ngcd(n;h) we get,
 h(r mn0h) = !h00 mn +! h00 mn = 2:
41
So P 2G  h( )6= 0, and hence  2  h.
Now suppose G is of the form HT, so that G =fr mn0h;sr mn0h j0   nmn0
h
 1g for
some m. For each  we have  h(sr mn0h) = 0. So P 2G  h( ) = P 2G \Cn h( ), which is
the same as the sum in the case of G = H and hence is nonzero. So  2  h.
Theorem 3.26. Let 1 i< ?2. The space V?2i has an o-basis if and only if either dimV = 1
or ?0 0 mod 4, where ?0 = ?gcd(?;i).
Proof. Put ? = ?2i. First suppose V? has an o-basis and assume dimV 6= 1. Now by
Theorem 3.20 the space V? has an o-basis for all  2 i = Sa2Ai   a. Let  = (1;2;:::;2).
Then G = f1;sg. So for all a2Ai,  2   a since  a(e) = 2 and  a(s) = 0 and hence
 2  i. By Equation 2.3 we have dimV a = 2 for each a2Ai. Then by Lemma 3.19 we
have dimV? = 2pq, so V? has a nonempty o-basis B. We may assume B contains e? .
By Lemma 3.21 and Corollary 2.5,
(e?  ;e? ) =
X
a2Ai
(e a  ;e a ) =
X
a2Ai
 a(e)
jG j
X
 2G  
 a( ) =  a(e)jG
 j
X
 2G  \Cn
X
a2Ai
 a( ):
First note that (e? s;e? )6= 0 because with  = s we get G  \Cn =f1;sgs\Cn =f1gwhence
the above sum is not zero. Now we will show that (e? rmpq;e? ) = 0 or (e? srmpq;e? ) = 0 for some
1 m ? 1. Assume to the contrary, that is for all 1 m ? 1, we have (e? rmpq;e? )6= 0
and (e? srmpq;e? )6= 0. Then for 0 x pq 1 we havejfe?  j 2rxhrpqig\Bj 1, because
for e? rxrm1pq;e? rxrm2pq 2fe?  j 2rxhrpqig with 1 m1 <m2  ? 1 we get
(e? rxrm2pq;e? rxrm1pq) = (e? r(m2 m1)pq;e? )6= 0:
So jfe?  j 2Cng\Bj pq. Now by our observation (e? s;e? )6= 0 and by our assumption
we getfe?  j 2shrpqig\B =;. For 1 x pq 1 we havejfe?  j 2srxhrpqig\Bj 1.
So jfe?  j 2GnCng\Bj< pq. Therefore we get dimV? < 2pq, which contradicts with
the observation above. So (e? rmpq;e? ) = 0 or (e? srmpq;e? ) = 0 for some 1 m ? 1.
42
Note that since G = f1;sg we have G rmpq \Cn = frmpqg and G srmpq \Cn =
frmpqg. Recall since rmpq is p-regular we have  k? i(rmpq) =  i(rmpq), so Pa2Ai a(rmpq) =
pq i(rmpq). So for  2frmpq;srmpq j1 m ? 1g we have
0 = (e?  ;e? ) = 2jG
 j
X
 2G  \Cn
X
a2Ai
 a( ) =
X
a2Ai
 a(rmpq) = pq i(rmpq)
= 2pq cos 2 mp
qi
n :
Therefore, 2 mpqin = (2c+ 1) 2 for some integer c, so
4i0m = 4imgcd(?;i) = (2c+ 1)?gcd(?;i) = (2c+ 1)?0:
So ?0 is divisible by 4.
Conversely suppose that dimV = 1. Then following the same argument in [9, Theorem
2.2], V? =he? iwith  = (1;:::;1), so V? has o-basisfe? gor;accordingly as dimV? is 1 or
0.
Now suppose ?0 0 mod 4. To show that there is an o-basis for V?, it is enough by
Theorem 3.20 to show that there is an o-basis for V? for each  2 i. Fix  2 i. Then there
is a2Ai such that  2  a and by Lemma 3.25, G is of the form H or HT where H 6hrn0ai
and T =hsi. Now let  a2Ai be such that hrn0 ai is the smallest for which H 6hrn0 ai. Let   a
be the largest such that p  a j a. Then by Lemma 3.24 we have n0 a = pq   a?0.
We claim that if p  a j a for some a 2 Ai, then  2   a. Fix a 2 Ai and let  a be
the largest such that p a ja. Then if p  a ja it follows that p  a  p a, so n0a = pq  a?0  
pq   a?0 = n0 a. Therefore H 6hrn0 ai6hrn0ai. Then it follows from Lemma 3.25 that,  2  a
as claimed.
Also we claim that  2  a only when p  a ja. To see this assume there is a2Ai such
that  2   a, but p  a - a. Note that H  hrn0ai by Lemma 3.25. Letting  a be the largest
43
such that p a ja we get p a < p  a, so n0 a = pq   a?0 < pq  a?0 = n0a. This gives hrn0ai6hrn0 ai
which is a contradiction since hrn0 ai is the smallest subgroup to contain H. Therefore the
claim holds and then by Lemma 3.19 we can write
V? = _
X
a2Ai
V a = _
X
a2Ai
p  aja
V a :
By Lemma 3.22 there are pq   a summands in the direct sum above. In the proof of [10,
Theorem 3.1] it is shown that for  2 Irr(G) of degree two, dimV  = 4 if G = H and
dimV  = 2 if G = HT. Then we have dimV? = 4pq   a if G = H and dimV? = 2pq   a if
G = HT.
Suppose G = H. Write G = hrmn0 ai for some integer m with mn0 a jn. Now we will
show thatfe?  ;e? s j 2Xgwhere X =frx?04 j0 x 2pq   a 1gis an orthogonal basis for
V? . To show this we will compare all possible combinations of elements for orthogonality.
Let  ; 2X. Consider the elements e? s and e?  . Then since G s   1\Cn =; we have
(e? s ;e?  ) = (e? s   1;e? ) = 2jG
 j
X
 2G s   1\Cn
X
a2Ai
 a( ) = 0:
For elements e? s and e? s we see that (e? s ;e? s ) = (e?   1 ;e? ) = (e?  ;e?  ), so it is enough
to check the orthogonality of elements of the form e?  . In this case it is su cient to check
that (e?  ;e? ) = 0 for each  2 X with  6= 1. Fix  = rx?04 2 X with  6= 1. Then
G  \Cn =fr mn0 a+x?04 j0   nmn0
 a
 1g, so we get
(e?  ;e? ) = 2jG
 j
X
 2G  \Cn
X
a2Ai
 a( ) = 2jG
 j
n
mn0 a 1X
 =0
X
a2Ai
 a(r mn0 a+x?04 )
= 2jG
 j
n
mn0 a 1X
 =0
X
a2Ai
 a(r mpq   a?0+x?04 ) = 2jG
 j
n
mn0 a 1X
 =0
X
a2Ai
 a(r?04 (4 mpq   a+x)):
44
We observe here that for any rt2Cn
X
a2Ai
 a(rt) =  i(rt) +
pq 1
2X
k=1
 k?+i(rt) +
pq 1
2X
k=1
 k? i(rt)
=  i(rt) +
pq 1
2X
k=1
 k?+i(rt) +
pq 1X
k=pq+12
 pq? (k?+i)(rt)
= !ti +! ti +
pq 1
2X
k=1
(!t(k?+i) +! t(k?+i)) +
pq 1X
k=pq+12
(!t(pq? (k?+i)) +! t(pq? (k?+i)))
=
pq 1X
k=0
(!t(k?+i) +! t(k?+i)):
So with t = ?04 (4 mpq   a +x) we get,
(e?  ;e? ) = 2jG
 j
n
mn0 a 1X
 =0
pq 1X
k=0
(!?04 (4 mpq   a+x)(k?+i) +! ?04 (4 mpq   a+x)(k?+i))
= 2jG
 j
n
mn0 a 1X
 =0
 
!?04 (4 mpq   a+x)i
pq 1X
k=0
!?04 (4 mpq   a+x)k?
+! ?04 (4 mpq   a+x)i
pq 1X
k=0
! ?04 (4 mpq   a+x)k?
 
= 2jG
 j
n
mn0 a 1X
 =0
 
!?04 (4 mpq   a+x)i
pq 1X
k=0
(!?04 (4 mpq   a+x)?)k
+! ?04 (4 mpq   a+x)i
pq 1X
k=0
(! ?04 (4 mpq   a+x)?)k
 
:
In order to proceed, we need the fact that if j is a positive integer and  2C is a jth
root of unity with  6= 1, then
j 1X
k=0
 k =  
j 1
  1 = 0:
45
Note that !?04 (4 mpq   a+x)? is a pqth root of unity for all  . If for some  xed  the expression
?0
4 (4 mp
q   a + x) is not a multiple of pq, then we have Ppq 1
k=0 (!
 ?04 (4 mpq   a+x)?)k = 0 by the
preceding observation.
On the other hand if a  xed  is such that ?04 (4 mpq   a + x) is a multiple of pq, then
since p - ?0 it should be that 4 mpq   a + x = z pq for some integer z . Then for such  , on
the right hand side of the above equation we get,
!?04 (4 mpq   a+x)i
pq 1X
k=0
(!?04 (4 mpq   a+x)?)k +! ?04 (4 mpq   a+x)i
pq 1X
k=0
(! ?04 (4 mpq   a+x)?)k
= !?04 z pqi
pq 1X
k=0
!?04 z pq?k +! ?04 z pqi
pq 1X
k=0
! ?04 z pq?k = pq(!?04 z pqi +! ?04 z pqi)
= pq(!?4z pqi0 +! ?4z pqi0) = pq(!n4z i0 +! n4z i0) = pq! n4z i0(!n2z i0 + 1)
= pq! n4z i0(( 1)z i0 + 1):
Now clearly i0 = igcd(?;i) is odd because when 4 j?0 = ?gcd(?;i) we have 4 j?0 = ?2gcd(?2;i2),
so i2 = gcd(?2;i2) = gcd(?;i)2, that is, the largest powers of 2 dividing i and gcd(?;i),
respectively, are equal. Here we claim that z is also an odd number. Observe that x =
z pq 4 mpq   a = pq   a(z p  a 4 m), but since x is an integer such that 0 x 2pq   1
we get z p  a 4 m = 1, so z p  a = 4 m + 1 and therefore z is odd as claimed. Then the
last expression of the above equation pq! n4z i0(( 1)z i0+1) = 0 and this gives us the desired
result that (e?  ;e? ) = 0.
Now suppose G = HT. Write H = hrmn0 ai with m an integer satisfying mn0 a jn. In
this case we will show that fe?  j 2Xg where X = f = rx?04 j0  x 2pq   a 1g is an
orthogonal basis for V? . Let  2X not be the identity element. It is su cient to check that
(e?  ;e? ) = 0. We have HT \Cn = H \Cn, so the computation is the same as in the case
of G = H, whence we have shown the desired result.
46
We will state the result for the existence of an o-basis in the case of a degree one
irreducible Brauer character as it appears in [9].
Theorem 3.27 ([9, Theorem 2.2]). Let 0  j < ", and put ? = ^ j. The space V? has an
o-basis if and only if at least one of the following holds:
i) dimV = 1,
ii) p = 2,
iii) m is not divisible by p.
3.6 Projective symmetrization
In this section we will discuss the existence of an o-basis associated with a PI of G = Dn.
To prevent the redundancy of some computations to follow we introduce some notation below.
" =
8
><
>:
2; if n is odd;
4; if n is even.
Tj =
8>
><
>>:
fk?j1 k pq 12 g; j = 1;2;
f?2 +k?j0 k pq 12  1g; j = 3;4.
(3.6)
Then using Theorem 3.5 we can write
 1j =  j +
X
t2Tj
 t; for 1 j "; (3.7)
 2i =  i +
pq 1
2X
k=1
( k?+i + k? i); for 1 i< ?2: (3.8)
47
First considering the PIs corresponding to degree one Brauer characters and using the
Equation 2.4 the symmetrizer is given by
s 1i = ?
1
i(e)
jGj
X
 2G
 1i( ) :
Proposition 3.28. Fix j with 1 j ". Then
s 1j = s j + 12
X
t2Tj
s t:
Proof. By Equations 2.4 and 3.7 we get
s 1j = ?
1
j(e)
jGj
X
 2G
 1j( ) = ?
1
j(e)
jGj
X
 2G
( j( ) +
X
t2Tj
 t( )) 
= 1jGj
X
 2G
 j( ) +
X
t2Tj
1
jGj t( ) 
=  
1
j(e)
jGj
X
 2G
 j( ) +
X
t2Tj
 t(e)
2jGj t( ) 
= s j + 12
X
t2Tj
s t:
Recall that  is a set of representatives of the orbits of  n;m under the action given in
Equation 2.1. Then for 1 j " by Equation 2.5 we have
V 1j = _
X
 2 
V  1j :
Theorem 3.29. Fix j with 1 j " and  x  2 . Then
V  1j = V j _+ _
X
t2Tj
V t (orthogonal direct sum)
48
Proof. Let s 1j(v)2V  1j . Then it is clear from Proposition 3.28 above that s 1j(v)2V j +
P
t2Tj V
 t
 . To show the other inclusion consider s j(w) +
P
t2Tj s t(wt)2V
 j
 +
P
t2Tj V
 t
 .
Note that s j(w) +Pt2Tj s t(2wt)2V . Then
s j(w) +
X
t2Tj
s t(wt) = s j(w) + 12
X
t2Tj
s t(2wt) = s j(w) + 12
X
t2Tj
X
b
s bs t(2wt)
=
 
s j + 12
X
b
s b
  
s j(w) +
X
t2Tj
s t(2wt)
 
=
 
s j + 12
X
b
s b
  
s j(w) +
X
t2Tj
s t(2wt)
 
= s 1j
 
s j(w) +
X
t2Tj
s t(2wt)
 
2V  1j ;
where we have used that s s =    s for all  ; 2 Irr(G). This shows that V  1j =
V j + Pt2Tj V t . The orthogonality follows from the argument in the proof of Theorem
2.3.
Fix j with 1 j " and  2 n;m. We have by Theorem 2.6
(e 1j  ;e 1j ) = ?
1
j(e)
2
jGj2
X
 2G
X
 2G 
 1j(  1  ) 1j( ):
Below we state as a corollary a useful form of this inner product.
Corollary 3.30.
(e 1j  ;e 1j ) = 12jGj
X
 2G 
 
2 j(  1 ) +
X
t2Tj
 t(  1 )
 
= 12jGj
X
 2G 
 
 j(  1 ) +  1j(  1 )
 
:
49
Proof. Using Equation 3.7, Theorem 1.13 and Theorem 1.14 we get,
(e 1j  ;e 1j ) = ?
1
j(e)
2
jGj2
X
 2G
X
 2G 
 1j(  1  ) 1j( )
= ?
1
j(e)
2
jGj2
X
 2G
X
 2G 
 
 j(  1  ) +
X
t2Tj
 t(  1  )
  
 j( ) +
X
u2Tj
 u( )
 
= ?
1
j(e)
2
jGj2
X
 2G
X
 2G 
 
 j(  1  ) j(  1) +
X
t2Tj
 t(  1  ) t(  1)
 
= ?
1
j(e)
2
jGj2
X
 2G 
 X
 2G
 j(  1  ) j(  1) +
X
t2Tj
X
 2G
 t(  1  ) t(  1)
 
= ?
1
j(e)
2
jGj2
X
 2G 
 jGj
 j(e) j( 
 1 ) +X
t2Tj
jGj
 t(e) t( 
 1 )
 
:
Now since ?1j(e) =  1j(e) = 1 and  t(e) = 2 for each t2Tj we get,
(e 1j  ;e 1j ) = 12jGj
X
 2G 
 
2 j(  1 ) +
X
t2Tj
 t(  1 )
 
= 12jGj
X
 2G 
 
 j(  1 ) +  1j(  1 )
 
:
Theorem 3.31. For i = 1;2;3;4 the space V 1j has an o-basis if and only if at least one of
the following holds.
i) dimV = 1,
ii) p = 2,
iii) n is not divisible by p.
Proof. If dimV = 1, then V 1j =he 1j i, where  = (1;1;:::;1), so V 1j has o-basisfe 1j gor;
according as dimV 1j is 1 or 0.
Assume p = 2. Then  1j is an ordinary irreducible character of the group ^G =hrpqi 
Cn. Then using Freese?s result for the dimension of an orbital subspace [6], dimV  1j is at
most one and hence has an o-basis. So V 1j has an o-basis by Equation 2.5.
50
Assume n is not divisible by p. Then ^G = G and hence  1j =  j. Since  j is of degree
one, each orbital subspace has dimension at most one and hence V 1j has an o-basis by the
same argument as in the above paragraph.
Now assume that none of the three conditions stated in the theorem holds. Let  =
(1;2;:::;2). which is in  n;m since dimV  2. We show that (e 1j  ;e 1j )6= 0 for every  2G.
Note that G =f1;sg.
First let  2Gn ^G. Since p6= 2 we have  2Cn. Using Corollary 3.30 we get,
2jGj(e 1j  ;e 1j ) = ( j +  1j)(  1) + ( j +  1j)(  1s) =  j( ) + 2 j(s )6= 0:
Now let  2 ^G. Then  2frapq;srb j 0  a < ?;0  b < ng. Assume  = rapq for some
0 a<?. We have
 k?( ) =  k?(rkpq) = 2 cos 2 j?kp
q
n = 2;
and
 ?
2 +k?
( ) =  ?
2 +k?
(rapq) = 2 cos 2 (
?
2 +k?)r
apq
n = 2 cos ap
q = ( 1)a2:
We also note that for a  xed j we have that  j(rapq) and  t(rapq) (t2Tj) are all positive or all
negative at the same time for a given a. Recall by Equation 3.1 we have  j(r apq) =  j(rapq)
and  t(r apq) =  t(rapq). So
2jGj(e 1j  ;e 1j ) = ( j +  1j)(r apq) + ( j +  1j)(r apqs)
= 2 j(rapq) + 2 j(srapq) +
X
t2Tj
 t(rapq)6= 0
51
using that p6= 2 and that n is divisible by p so that Tj is nonempty. Now assume  = srb
for some 0 b<n. By Corollary 3.30 we get
(e 1j  ;e 1j ) = 12jGj
 
( j +  1j)(srb) + ( j +  1j)(srbs)
 
= 12jGj
 
( j +  1j)(srb) + ( j +  1j)(r b)
 
= (e 1j rb;e 1j ):
Now we get (e 1j  ;e 1j ) 6= 0 since (e 1j rb;e 1j ) 6= 0 by the two previous cases. By Equation
2.3 we have dimV t =  t(e)jG j P 2G  t( ) = 2 for each t 2 Tj. As observed earlier, Tj is
nonempty, so dimV  1j > 1. So we conclude that V 1j does not have an o-basis and the proof
is complete.
Now we will consider the PIs corresponding to degree two Brauer characters of G = Dn.
Corollary 3.32. For 1 i< ?2
s 2i = p
qj^Gj
jGj s?2i:
Proof. Let 1 i< ?2. By Equation 2.4 and Lemma 3.15,
s 2i = ?
2
i(e)
jGj
X
 2G
 2i( ) = ?
2
i(e)
jGj
X
 2^G
^ 2i( ) = pq?2i(e)
jGj
X
 2^G
?2i( ) = p
qj^Gj
jGj s?2i:
Recall that V 2i = s 2i(VNn) and V?2i = s?2i(VNn).
Theorem 3.33. For 1 i< ?2 we have
V 2i = V?2i:
Proof. Take s 2i(w)2V 2i. Then using Corollary 3.32 we get
s 2i(w) = p
qj^Gj
jGj s?2i(w) = s?2i(
pqj^Gj
jGj w)2V?2i;
52
which shows the inclusion V 2i  V?2i. Then take s?2i(u) 2V?2i. Again using Corollary 3.32
we see that
s?2i(u) = jGjpqj^Gjs 2i(u) = s 2i( jGjpqj^Gju)2V 2i;
which gives the other inclusion. So we get V 2i = V?2i.
Theorem 3.34. Fix i where 1 i< ?2 and let  2 n;m. Then
e 2i = p
qj^Gj
jGj e
?2i
 :
Proof. By Corollary 3.32 we get that,
e 2i = s 2i(e ) = p
qj^Gj
jGj s?2i(e ) =
pqj^Gj
jGj e
?2i
 :
Lemma 3.35. Fix i with 1  i < ?2. Then V 2i has an o-basis if and only if V?2i has an
o-basis.
Proof. We observe that,
(e 2i ;e 2i ) = (p
qj^Gj
jGj e
?2i
 ;
pqj^Gj
jGj e
?2i
 ) = (
pqj^Gj
jGj )
2(e?2i ;e?2i ):
So (e 2i ;e 2i ) = 0 if and only if (e?2i ;e?2i ) = 0.
Theorem 3.36. Fix i with 1  i < ?2. Then V 2i has an o-basis if and only if either
dimV = 1 or ?0 is divisible by 4, where ?0 = ?=gcd(?;i).
Proof. The result follows from Lemma 3.35 and Theorem 3.26.
Theorem 3.37. Fix i with 1  i < ?2. Then V 2i has an orthogonal basis consisting of
decomposable tensors of the form e  if and only if ?0 0 mod 4. Where ?0 = ?gcd(?;i).
Proof. The result follows from Theorem 3.33 and Theorem 3.18.
53
Chapter 4
Symmetric group
In this chapter we will discuss some results associated with Brauer characters of a
symmetric group.
For some positive integer n, the symmetric group Sn of degree n is the group of permu-
tations of the set f1;2;:::;ng with the binary operation de ned by function composition.
The number of elements of Sn is n!. A permutation  of Sn is given by
 =
0
B@ 1 2    n
 (e)  (2)     (n)
1
CA
or
 = (i11;:::;i1r1)(i21;:::;i2r2)   (is1;:::;isrs);
where 1 iab n, iab = icd implies a = c and b = d, and  (iab) = ia(b+1) (b<ra);  (iara) =
ia1. The latter is in a factored form, where the factors are disjoint cycles. The length of a
cycle is the number of numbers that appear in the cycle. The lengths of the cycles r1;:::;rs
of  , when arranged in non-increasing order is called the cycle type of  . Two permutations
are conjugate in Sn if and only if they have the same cycle type ([13, Theorem 2.4, page
292]). In particular, the number of conjugacy classes of Sn is equal to the number of di erent
cycle types of the elements of Sn. So the number of irreducible characters of Sn is the same
as the number of di erent cycle types of the elements of Sn.
The order of a cycle equals the length of the cycle. Let  be a permutation with the
cycle type (r1;r2;:::;rk). Then it can easily be observed that the order of  equals the least
54
common multiple of the numbers r1;r2;:::;rk. In light of this a p-regular element of Sn is a
permutation with cycle type consisting of numbers not divisible by p.
In the following study we will just consider the principal Braur character of G = Sn,
which is the character  given by  ( ) = 1 for each  2G. As the Brauer character a orded
by the trivial KG-module  is irreducible.
Theorem 4.1. Let G = Sn with n  3. Assume that dimV  2 and p 6= 2. Then
(e   ;e  ) 6= 0 for all  2G, where  = (1;:::;1;2). In particular, if dimV  > 1, then V 
does not have an o-basis.
Proof. Let  = (1;:::;1;2). We can assume  to be the representative of the orbit containing
it, so  2 . Observe that G =f 2Snj (n) = ng = Sn 1:
By Theorem 2.4, for any  2G we have
(e   ;e  ) =  (e)
2
j^Gj2
X
 2^G
X
 2   1 ^G\G 
 ( ) (  1   1) =  (e)
2
j^Gj2
X
 2^G
X
 2   1G \^G
 ( ) (  1)
= 1j^Gj2
X
 2^G
j   1G \ ^Gj:
So (e   ;e  ) = 0 only when    1G \ ^G =;for some  2 ^G. We will show that for all  2G
there is some  2 ^G such that    1G \ ^G6=;, which implies that (e   ;e  )6= 0.
We claim here that the cyclic group H =h(1;2;:::;n)i is a set of right coset represen-
tatives of G in G. For h1;h2 2H with h1 6= h2 we have h1h 12 (n) 6= n, so G h1 6= G h2.
AlsojG : G j= n =jHj. Now since G H = G we havefe   j 2Gg=fe  hjh2Hg. So it is
enough to show (e  h;e  )6= 0 for all h2H. Let h2H. If h = e, then letting  = e2 ^G we
get  h 1G \ ^G = G \ ^G, and this latter set contains the transposition (1;2), since n 3
and p6= 2, so it is nonempty as desired. Now assume that h6= e. Then h(n)6= n and there is
1 m n 1 such that h(m) = n. Now let  = (m;n) and observe that we get  h 1(n) = n,
so  h 1 2Sn 1 = G . Therefore,  2 ^G, since p6= 2, and  h 1G \ ^G = G \ ^G6= ; as
desired.
55
If dimV  > 1, then V  does not have an o-basis, so by Theorem 2.3 the space V does
not have an o-basis.
The following example is a case, where we do not have an o-basis for V .
Example 4.2. Let G = S3. Assume dimV  2 and p = 3. Let  be the principal Brauer
character of G and let  = (1;1;2). Then dimV  > 1.
Proof. Note since p = 3 we have ^G = f1;(a;b);(a;c);(b;c)g. Write e  = e (112). Then for
 = (a;b;c)2G we get e   = e (211). Now by Equation 2.2 we get
e (112) = 14(2e(112) +e(211) +e(121));
e (211) = 14(2e(211) +e(112) +e(121)):
By inspection we see that e (112) and e (211) are linearly independent, so dimV  > 1.
The alternating group G = An is the subgroup of Sn consisting of all the even per-
mutations of Sn. Let  be the irreducible Brauer character of G with  ( ) = 1 for all
 2G.
Theorem 4.3. Let G = An. Assume that dimV  2, n( 3) is odd, and p 6= 2. Then
(e   ;e  ) 6= 0 for all  2G, where  = (1;:::;1;2). In particular, if dimV  > 1, then V 
does not have an o-basis.
Proof. Let  = (1;:::;1;2). We can assume  to be the representative of the orbit containing
it, so  2  . Observe that G = f 2 Anj (n) = ng = An 1. Since n is odd H =
h(1;2;:::;n)i G. Following the same argument as in the proof of Theorem 4.1 we can
show that H is a set of right coset representatives of G in G. The argument to show
(e   ;e  )6= 0 for all  2G is the same as in the proof of Theorem 4.1.
56
4.1 Special case S4
The symmetric group G = S4 of degree 4 is the group of permutations of a setfa;b;c;dg.
Let p = 2. Then there are two, 2-regular conjugacy classes and
^G =f1;(abc);(acb);(abd);(adb);(acd);(adc);(bcd);(bdc)g:
The Brauer character table of G in this case is (see [12, page 431])
( ) (   )
?1 1 1
?2 2  1
Theorem 4.4. Assume that dimV > 1. The space V?i does not have an o-basis for i 2
f1;2g.
Proof. Fix i2f1;2g and put ? = ?i. To show V? does not have an o-basis it is enough by
Theorem 2.3 to show that V? does not have an o-basis for some  2 .
Let  = (1;1;1;2). Then G =f1;(ab);(ac);(bc);(abc);(acb)g.
Let H = f1;(ab)(cd);(ac)(bd);(ad)(bc)g and observe that G = G H. So fe?  j  2
Gg=fe?  j 2Hg. We will compute the value of (e?  ;e? ) using the formula
(e?  ;e? ) = ?(e)
2
j^Gj2
X
 2^G
X
 2   1G \^G
?( )?(  1) (4.1)
57
(see proof of Theorem 4.1). The following table lists the products    1, with  2 ^G and
 2H.
 n 1 (abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
1 1 (abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
(ab)(cd) (ab)(cd) (acd) (bcd) (adc) (bdc) (abc) (abd) (acb) (adb)
(ac)(bd) (ac)(bd) (bdc) (abd) (acb) (acd) (adb) (bcd) (adc) (abc)
(ad)(bc) (ad)(bc) (adb) (adc) (bcd) (abc) (bdc) (acb) (abd) (acd)
Now we will look at the partP 2^GP 2   1G \^G?( )?(  1) on the right side of the Equation
4.1. Note from the table above that    1 is even for all  and  . Now since ^G does not
contain odd cycles we can neglect the products of    1 with the elements of the form (  ) in
G when considering    1G \ ^G.
When  = 1, for all cases of  6= 1 we get:
   1( ) = (  )(  )( ) = (  )(  ) and    1(   ) = (  )(  )(   ) = (   ) (two times):
From the table we see if  = (   ), then for each  2H we have    1 = (   ). The
table below lists the products    1 (columns) of the form (   ) with the even permutations
in G (rows).
(abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
1 (abc) (acb) (abd) (adb) (acd) (adc) (bcd) (bdc)
(abc) (acb) 1 (ad)(bc) (bcd) (abd) (ab)(cd) (ac)(bd) (adc)
(acb) 1 (abc) (acd) (ac)(bd) (ad)(bc) (bdc) (adb) (ab)(cd)
58
Assume that ? = ?1. Since ?( ) = 1 for all  2 ^G and    1G \ ^G6=;we get (e?  ;e? )6= 0
for all  2H. Now write e? = e?(1112). Then e? (ad) = e?(2111). Now by Equation 2.2 we get
e?(1112) = 19(3e(1112) + 2e(2111) + 2e(1211) + 2e(1121));
e?(2111) = 19(3e(2111) + 2e(1112) + 2e(1211) + 2e(1121)):
By inspection we note that e?(1112) and e?(2111) are linearly independent, so dimV?  2. So
we conclude that V? does not have an o-basis and hence V? does not have an o-basis.
Now assume that ? = ?2. Then to evaluate (e?  ;e? ) for each  6= 1 in H we observe
from the computations above in the cases of  = 1 = ( ) and  of the form (   ) that
X
 2^G
X
 2   1G \^G
?( )?(  1) = 4?
 
( )
 
?
 
(   )
 
+ 16?
 
(   )
 
?
 
(   )
 
:
So by the Brauer character table given above and Equation 4.1 we get
(e?  ;e? ) =  (e)
2
j^Gj2
 
4(2)( 1) + 16( 1)( 1)
 
6= 0:
Now note that by Equation 2.2 we get
e?(1112) = 29( 2e(2111) 2e(1211) 2e(1121));
e?(2111) = 29( 2e(1112) 2e(1211) 2e(1121)):
By inspection e?(1112) and e?(2111) are linearly independent implying dimV?  2. So we
conclude that V? has no o-basis and therefore V? does not have an o-basis.
59
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