character theory and the davenport constant

8/13/2019 Character Theory and the Davenport Constant

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Character Theory and the Davenport Constant

Jesse Ireland

MATH 4F90 Project

April 25, 2012

Brock University

Department of Mathematics

Professor: Dr. Yualin Li

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Character Theory and the Davenport Constant

Jesse Ireland

April 25, 2012

1 Abstract

Representation theory is a means by which abstract groups can be treated as concreteobjects. A well-known theorem due to Cayley tells us that every group can be imbeddedas a subgroup of Sn where n is the order of G. Furthermore, Sn is ismorphic to asubgroup ofGL(n, K), the general linear group, so we can represent any abstract groupas a group of matrices. This in essence, is representation theory. Character theoryis a branch of modern algebra that comes directly from representation theory. Thispaper is an introduction to representations and characters of groups with the goal ofapplying the theory to finding the Davenport constant. It begins with a very quickreview of some concepts from algebra that may be unfamiliar, followed by an introductionto representation theory. We continue with the development of character theory andan introduction to character tables. Finally, we take a look at some novel methodsof applying character theory to the Davenport constant problem, an open problem in

modern algebra. This paper should be accessible to anyone familiar with basic, tointermediate group theory, elementary ring theory and a background in linear algebra.

Key Words: Representation Theory, Character Theory, Group Algebras, The DavenportConstant

2 Introduction

In the very early stages of the development of group theory only a small numberof classes of groups were known. The permutation group was almost synonymous with

the term group prior to the formulation of an abstract definition by the mathematicianArthur Cayley in 1854. The idea of a group didnt catch on right away and it took about30 years before our modern definition was formulated. Another 25 years later the ideaof representing abstract groups in a more concrete sense was put forth by F. Klein. Byrepresenting an abstract group as something concrete such as a group of permutationsor a linear group, these groups can be explored further than the abstract representationallows. Once representation theory was developed it paved the way for characters to bedefined. The character of a representation is a deceptively simple yet powerful idea and

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will be explored in this paper. First, we need some elementary concepts before gettingto representation theory..

2.1 Some Needed Algebra

We begin this section by defining R-modules and then showing how they are related tovector spaces.

Definition 2.1.1 R-ModulesLetRbe a ring. An abelian groupM(written additively) is called an (left)R-Module

(or a left module over R) if for each element a R and each m M there is a productam M such that:

(i) (a + b)m= am + bm,

(ii) a(m1+ m2) =am1+ am2,

(iii) a(bm) = (ab)m,

(iv) 1m= m,

for all a, b R and m, m1, m2 M.

Definition 2.1.2 Generating SetA set S={si}iIof elements of an R-module Mis called a set of generators ofM

ifM=RS=

ni=1 xisi, where n N and xi R.

Definition 2.1.3 Linear IndependenceA setS= {si}iIof elements of an R-module M is calledlinearly independentif, for

any (finite) linear combination of elements ofSwith coefficients in R:

r1s1+ r2s2+ + rtst= 0

implies r1= r2= = rt= 0.

Definition 2.1.4 BasisA set S= {si}iI of elements of an R-module M is called a basis ofM over R (or

R-basis) if it is a linearly indepentent set of generators.

Definition 2.1.5 Free ModuleAn R-module is called freeif it has a basis.

If we impose a restriction on our ring in the definition of R-module, namely that thering be a field, then the module is, by definition, a vector space. Vector spaces havesome very nice properties that are not shared by modules. An example of this is given

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any linearly independent set of vectors, it can always be extended to a basis. This isnot the case with modules. Another example is that given any set of generators of avector space, a subset of the generating set is a basis. Again, this is not true in the more

general module.If we impose a further condition on the Abelian group and a certain condition on thering we can construct an R-algebra. We do this now.

Definition 2.1.6 R-AlgebraLetRbe a commutative ring. AnR-moduleAis called anR-algebraif there is a mul-

tiplication, defined inA, such that, with the addition given in A and this multiplication,Ais a ring and such that the following condition holds:

r(ab) = (ra)b= a(rb),

for all r R, and all a, b A.

We now take a look at a special type of ring that is constructed using a group and aring together.

Definition 2.1.7 Group RingLetR be a ring and G a group (written multiplicitively) then the set of finite sums

RG=

gG

agg: ag R

,

is a ring under the following operatins:

(i) (

gG agg) + (

gG bgg) =

gG(ag+ bg)g

(ii) (

gG agg)(

hG bhh) =

gG,hG(agbh)gh

This ring is called the group ring ofG over R.

We can also very easily make RG into an R-module and in turn, an R-algebra. To dothis simply define a product of elements in RG by elements in R by

(gG agg) = gG(ag)g,and this makes RG into an R-module. In the case whereR is a commutative ring, wehave that RG is an R-algebra.We now have the basic theory that we need in order to look at some representationtheory.

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3 Representation Theory

In this section we shall look at representations of groups. First, we see two equivalent

definitions of group representations, one concrete and one abstract, and then we continueusing the concrete one as it is better suited to our purpose.

Definition 3.0.1 (Linear) RepresentationLetG be a group, R be a commutative ring, andVbe a free R-module of finite rank.

A representation ofG over R, with representation space V, is a group homomorphismT :G GL(V), where GL(V) denotes the group ofR-automorphisms ofV. The rankof V is called the degree of the representation T and will be denoted by deg(T). Weshall denote by Tg the image ofg under the mapping T.

Definition 3.0.2 Matrix Representation

Let G be a group, and let R be a commutative ring. A matrix representation ofG over R of degree n is a group homomorphism : G GL(n, R), where GL(n, R)denotes the group ofn n non-singular matrices over R under the operation of matrixmultiplication.

Since we can define an isomorphism : GL(V) GL(n, R), when we fix a basis ofV, Tis a matrix representation and 1 is a representation so these two types ofrepresentations are analogous. Let us now consider an example.

Example 3.0.3 LetCn be the cyclic group of order n given by Cn= a: an = 1. We

can define n representations ofCn of the form : Cn GL(1,C) given by : ai i,

0 i n 1, where denotes an n th root of unity. The degree of this representationis 1 since the group is mapped to a group of 11 matrices.

By changing the type of root of unity in the above example we obtain differenttypes of representations. If we taketo be 1 then we obtain what is called the trivialrepresentationsince every element is mapped the identity ofGL(1,C). On the other handif we taketo be a primitive root of unity then we get a different type of representation,a so-calledfaithful representation. To see what this means we must first define the kernelof a representation.

Definition 3.0.4 Kernel of a RepresentationLet be a representation of a group G over a ring, R. Then the kernel of the

representation is given by

ker() ={g G : (g) =In}.

Now, we call a representation faithful if the kernel is the set consisting of only theidentity element ofG. We next consider the concept of equivalent representations.

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Definition 3.0.5 Equivalent Matrix RepresentationsTwo matrix representations A : G GL(n, K) andB : G GL(n, K) of a groupG

over a fieldKare said to beequivalentif there exists an invertible matrix,Q GL(n, K)

such thatAg =QBgQ1, for all g G.

Definition 3.0.6 Reducibility and IrreducibilityA matrix representationT :G GL(n, K) is calledreducibleif there exists a matrix

UGL(n, K) such that, for all g G, we have that the matrix U TgU1 is of the form

U TgU1 =

A(g) B(g)

0 C(g)

.

Also if the above condition holds then each matrix Tg is similar to a matrix of the form

U TgU1 = D(g) 00 E(g) .Furthermore, the functions defined by g A(g) and g C(g) as well as the functionsdefined by g D(g) and g E(g) are all representations.A representation is called irreducible if the representation space V admits no properT-invariant subspaces. That is if there is no subspace W V such that Tg(W) W.This is just another way of stating that you cant find a basis such that the matrices areof the form described above.

We shall now give a very important technical result about the number of irreduciblerepresentations that will be extremely important in our discussion of character theory.

Theorem 3.0.7 LetG be a finite group and let Kbe a splitting field forG. Then

KGr

i=1

Mni(K),

where r is the number of conjugacy classes of G. Hence, the number of irreducible,non-equivalent, representations ofG over K is also r and we have that

|G|=r

i=1

n2i .

Before continuing on to group characters we shall look at a special representation.

Example 3.0.8 The Regular RepresentationThe regular F G-module is constructed from the group algebra as follows. LetV =

F G, so that V is a vector space of dimension n = |G| over a field F. For all u,v V , F, and g , h G we have

vg V

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v(gh) = (vg)h

v1 =v

(v)g= (vg)

(u + v)g= ug + vg

hence V is an F G-module.

Now we can define the regularF G-module as well as the the regular representation.

Definition 3.0.9 The Regular F G-module and The Regular RepresentationLet G be a finite group and F be R or C. The vector space F G, with the natural

multiplicationvg, v V, g G, is called the regularF G-module.The matrix representation g [g]B (where [g]Bdenotes the coordinate vector ofg in

F G) obtained by using the basis B={g1, g2, . . . gn} ofF G is called the regular repre-sentation ofGover F.

So the regular representation is a faithful representation of degree |G|.

4 Group Characters

4.1 Basic Concepts and Definitions

The notion of a character was introduced by the mathematician G. Frobenius withinspiration from R. Dedekind [4]. We shall now consider characters but we will define

them in terms ofCG-modules and focus on matrix representations. Before defining whata character is however, it would be prudent to recall the trace of a matrix and a coupleof its properties.

Definition 4.1.1 TraceLetM = (aij) be an nn matrix. Then the trace ofMis given by

tr(M) =ni=1

aii

Provided that the entries of two matrices A, and B are from a commutative ring we

have tr(AB) = tr(BA). Also it is good to note that in general tr(AB) = tr(A) tr(B).The first relation is very important because it tells us that given a linear transformationT, the trace ofT is independent of the basis. This is easily seen as

tr(Q1T Q) =tr(Q1(T Q)) =tr(Q1QT) =tr(T).

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Definition 4.1.2 CharacterLetG be a group, Fa field, and V a free FG-module over F. LetT :GGL(V) be

a representation ofG over F. Then, thecharacter, ofG affordedby the representation

T is the mapping : G F defined by (g) =tr(Tg), g G. If the representationTis irreducible (reducible) then is called an irreducible (reducible) character.The paragraph preceding the definition shows that the character is independent of thebasis, B.

Definition 4.1.3 DegreeIf is the character of the CG-module, V, then the dimension of V is called the

degree of.

From here on we shall mostly restrict ourselves to complex representations. Letsconsider an example of the characters of a finite abelian group of rank 2.

Example 4.1.4 Let G= a, b : al =bm = 1, ab= ba. Then, if we leta =e2il and

b = e2im we can construct the following mappings fora and b:

a0: ai 1 b0 : b

i 1

a1: ai ia b1 : b

i ib...

...

a(l1): ai lia b(m1) : b

i mib

Now we can define lm= |G| representations that are irreducible and non-equivalentfrom these mappings as follows:

00(aibj) =a0(a

i)b0(aj)

01(aibj) =a0(a

i)b1(aj)

...

xy(aibj) =ax(a

i)by(aj)

where 0 x l 1 and 0 y m 1. Since all of these representations are ofdegree one the image of each element under the mapping is identical to the character.Also, the reader should notice that we have constructed |G| representations of degreeone, hence every irreducible representation of an abelian group of rank 2 is of degreeone. This will become important later on when we look at the Davenport constant.

We next give another result relating characters and CG-modules.

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Theorem 4.1.5 Isomorphic CG-modules have the same character.

Proof Suppose V and Ware isomorphic CG-modules. Then there exists a basis B1

ofVand a basis B2 ofW such that

[g]B1 = [g]B2 for all g G

Consequently, tr[g]B1=tr[g]B2 for all g G, and so V and Whave the same character.

Definition 4.1.6 Reducibility and IrreducibilityWe say that is a character ofG if is the character of some CG-module. Further-

more, is anirreducible (reducible) character ofG if is the character of an irreducible(reducible) CG-module.

Let us consider now an example to illustrate these definitions.

Example 4.1.7 Let G= D8 = a, b: a4 = b2 = 1,bab1 =a1 with the representa-

tion: G GL(2,C) defined by : aibj AiBj where

A=

0 11 0

B =

1 00 1

Then we have the following table showing, explicitly, the mapping of elements under and the character of. Notice the entry corresponding to(1) shows the degree of therepresentation is two. This is as it should be since we are considering a mapping into

GL(2,C

).

g 1 a a2 a3

(g)

1 00 1

0 11 0

1 00 1

0 11 0

(g) 2 0 -2 0

g b ab a2b a3b

(g)

1 00 1

0 11 0

1 00 1

0 11 0

(g) 0 0 0 0

We should note here that in general the character is not a homomorphism but ifdeg() = 1 then it is in fact a homomorphism. Characters of degree 1 are called linearcharactersand the character corresponding to the trivial representation is called thetrivial character ofG or theprincipal character ofG. We next show that the characteris constant on each conjugacy class.

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Theorem 4.1.13 Ifreg is the regular character then

reg(1) =|G|, and

reg(g) = 0 ifg = 1

Proof Letg1, g2, . . . , gn be the elements ofG, and let Bbe the basis g1, g2, . . . , gnofCG. By Theorem 4.1.9 reg= dim(CG) =|G|.Now let g G with g = 1. Then for 1 i n, we have gig = gj for some j with j =i.Therefore the ith row of the matrix [g]B has zeros in every place except column j; inparticular, the ii-entry is zero for all i. It follows that

reg(g) = tr[g]B= 0.

We shall finish this section with an example that will illustrate the ideas presentedso far.

Example 4.1.14 Let G = D6 = a, b : a3 = b2 = 1,bab1 = a1. For the sake

of brevity, we shall not derive the irreducible representations but rather list them. Thethree (Theorem 3.1.7 shows us there are only three) irreducible representations are givenby

1: a (1), 1: b (1)

2

: a

(1), 2

: b

(

1)

3: a

00 1

, 3 : b

0 11 0

where = e

2i3 . Tabulating the irreducible characters together with the regular

character gives the table

g 1 a a2 b ab a2b

1 1 1 1 1 1 12 1 1 1 1 1 1

3 2 1 1 0 0 0

reg(g) 6 0 0 0 0 0

The reader should notice a few things about this table. The characters take on thesame values for conjugate group elements as mentioned before. Also note the kernel ofeach representation, hence 3 and the regular representation are faithful while 1 and2 are not. Finally, notice that reg =1+2+ 23. This is no coincidence. We shallexplore this in greater detail in the next section.

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4.2 Character Tables

In this section we shall define the character table and explore different ways to calculate

irreducible characters without having to construct the correspondingC

G-modules.

Definition 4.2.1 Class FunctionA mapping : G C is called a class function if is constant of the conjugacy

classes ofG, that is, ifx = g1yg for x, y ,g G implies (x) =(y).

Now Theorem 4.1.8 tells us that characters, and hence, linear combinations of char-acters, are class functions. The next theorem shows the converse is also the case.

Theorem 4.2.2 Every class function : G C can be uniquely expressed in the form

=

ri=1

aii, ai C, 1 i r,

whereibelong to a set of irreducible, non-equivalent characters. Thus, {1, 2, . . . , r}is a basis of the C-vector space of all class functions ofG over C.

Proof We know from Theorem 3.7 that if the number of irreducible representationsofG over C is r, then the number of conjugacy classes ofG is also r. We shall denotethese classes by C1, C2, . . . C r. Also, denote by Ei : G C the class function such thatEi(a) = 1 ifa Ci and Ei(a) = 0 otherwise, 1 i r. Then, it is easily seen that theset {E1, . . . , E r} is a basis of the vector space of all class functions. Hence, the dimension

of this space is precisely r and the result will follow if we show that {1, . . . , r} is alinearly independent set. Assume, by way of contradiction, that there exist coefficientsi C, 1 i r such that

ri=1 ii = 0. Let {e1, . . . , er} be the primitive central

idempotents ofCG. Then, we have that:

0 =

ri=1

ii

(ej) =

ri=1

ii(ej) =jdeg(Tj),

thus j = 0, 1 j r.

Here it should be noted that given a class function of the form

=r

i=1

aii, ai C, 1 i r,

is a character if and only if = 0, ai Z, ai 0, 1 i r. The characters i suchthat the corresponding coefficientai is non-zero are called the constituents of.

We shall now show this in the case of the regular character.

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Example 4.2.3 LetG be a finite group of order n. We shall denote by : G Ctheregular character, which is afforded by the regular representation ofG over C that wasconstructed earlier. Let

CGr

i=1

Mni(C)

be the decomposition of CG as a direct sum of simple components. If Ii Cni is

the irreducible CG-module corresponding to the ith simple component, we see that, asCG-modules, we have that

Mni(C) Ii Ii nitimes

and henceCG I1 I1

n1times Ir Ir

nrtimesIfTi denotes the irreducible representation ofG over C corresponding to Ii, and i itscorresponding character, 1 i r, then we can write the regular representation T ofGover C as:

T =r

i=1

niTi

Computing the trace of each side we get

=r

i=1

nii.

Since ni =deg(Ti) =i(1), this can be written as

=r

i=1

i(1)i.

Hence the aforementioned sum in the example ofD6.

We are now in a position to describe the character table of a group. Since the numberof irreducible complex representations, and hence the number of irreducible characters,is equal to the number of conjugacy classes of a group, by tabulating the irreduciblecharacters together with a class representative of each conjugacy class we obtain an

n n array or matrix. This gives the character table of a group.

Definition 4.2.4 Let1, . . . , k be the irreducible characters ofG and letg1, . . . gk berepresentatives of the conjugacy classes ofG. The k k matrix whose ij -entry isi(gj)is called the character table ofG.

To illustrate this definition we simply note that the conjugacy classes of D6 areC1= {1}, C2= {a, a

2}, and C3= {b,ab,a2b} and so the character table ofD6 is

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g 1 a b

1 1 1 12 1 1 13 2 1 0

We shall now develop a couple of methods of constructing character tables withoutexplicit construction of the corresponding CG-modules. The first thing we shall do isconstruct an inner product of characters. This will be very helpful for deconstructingreducible characters in terms of known irreducible characters, determining if a givencharacter is irreducible, and to fill in a partially known character table. To this end,we shall start with a more general theorem and construct an inner product using thetheorem.

Theorem 4.2.5 Generalized Orthogonality RelationLeth be an element in a finite group G. For every pair of irreducible characters i,

j , we have that1

|G|

gG

j(gh)i(g1) =ij

i(h)

j(1),

where ij denotes the Kronecker delta; ie. ii = 1 and ij = 0 ifi =j .

Proof See [4] Theorem 5.1.14.

We now state two useful consequences of the Generalized Orthogonality Relation.These two relations are very helpful in the construction of character tables. The firstfollows quickly from the theorem by taking g = 1 and recalling that the character is aclass function. The second can be proved by taking two matrices, A and B say, withA = (aij) where aij = i(aj) and B = (bij) where bij =

Ci|G|j(ai) and ah as below,

then using the fact that the character is a class function and then finally using the firstrelation.

Corollary 4.2.6 Let 1, . . . , k be the irreducible characters of G and let a1, . . . , akbe representatives of the conjugacy classes ofG. Then the following relations hold foranyr, s {1, . . . , k}.

The row orthogonality relations:

ki=1

r(ai)s(ai)

|CG(ai)| =ij .

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The column orthogonality relations:

k

i=1 i(ar)i(as) =rs|CG(ar)|.where |CG(g)| denotes the index of the centralizer in G.

We now define a useful notation that also tells us a little about the structure of thespace of class functions from G to C, namely the inner product of characters.

Definition 4.2.7 Let and be characters of a group G. Then the inner product of and is given by

, = 1

|G| gG (g)(g).It is easy to verify that , does in fact satisfy the axioms of an inner product and

that not only does the set of irreducible characters form a basis, but the basis is in factan orthonormal basis as evident from the above corollary. In short we have , = 0and, = 1 for distinct irreducible characters , and . It is good to note that whilethe two relations are different they contain the exact same information. That is whateverinformation we can obtain using one relation we can also obtain the same informationusing the other one. The only difference is that a calculation may be simplified by usingone over the other. We shall now use a couple of examples to illustrate the usefulness of

the orthogonality relations as well as the inner product.

Example 4.2.8 Suppose we have obtained all of the linear characters of some groupG of order 12 with exactly four conjugacy classes with representatives g1, g2, g3, g4 andthe partial character table is as below.

gi g1 g2 g3 g4|CG(gi)| 12 4 3 3

1 1 1 1 12 1 1

2

3 1 1 2

4

where = e2i3 . Since the entries in the first column are the degrees of the irreducible

characters they are positive integers. Hence by the column orthogonality relation we

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have,

12 + 12 + 12 + 4(1)2 = 12

4(1)2 = 9

4(1) = 3

Next we can use the first column to obtain the entries in the rest of the columns using

4i=1

i(g1)i(gr) = 0 with r = 2, 3, 4

These computations give the rest of the character table to be

gi g1

g2

g3

g4|CG(gi)| 12 4 3 3

1 1 1 1 12 1 1

2

3 1 1 2

4 3 -1 0 0

We next consider a method by which we can decompose characters and in turn CG-modules. We will not prove this or even explore this in much detail rather the ideahere is to introduce this method and simply claim that it works in general. We shalldemonstrate this by decomposing the regular character ofD6.

Example 4.2.9 LetG = D6. ThenG has three irreducible characters, which we shalldenote by 1, 2, 3. Since the set of irreducible characters is a basis for the space ofclass functions from D6 to C we can write reg= d11+ d22+ d33. Now we can usethe inner product to find each di.

reg, 1= 1

|G|

gG

reg(g)1(g) =3

i=1

1

|CG(gi)|reg(gi)1(gi) =

1

6reg(1)1(1) = 1

computing d2, and d3 we get

reg, 2= 2(1) = 1

reg, 3= 3(1) = 2

This gives reg = 1+2+ 23 as we had previously. Since we will not prove this wejust note that this works because the inner product acts as a projection since these arejust vectors in a vector space.

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The last thing we will do in our development of character theory is show a method tofind all of the linear characters of a group G. We will first show how to use factor groupsof a group G to help construct the character table ofG. Then, using a particular factor

group, we can construct all linear characters of a group. This method is useful becausein general the character table of a factor group of a group G is easier to construct thanthe character table ofG itself. This portion is a little more advanced so we shall onlyprovide an outline of the main proof and instead direct the reader to [3] for the fullproof.

Theorem 4.2.10 Assume N G, and let be a character ofG/N. Define : G Cby

(g) = (gN) g G.

Then is a character ofG, and and have the same degree.

Proof Let : G/N GL(n,C) be a representation of G/N with character . Thefunction : G GL(n,C) that is given by the composition

g gN (gN) g G

is a homomorphism fromG to GL(n,C). Thus is a representation ofG. The character of satisfies

(g) = tr((g)) = tr((gN)) = (gN)

for all g G. Moreover, (1)(N), so and have the same degree.

Definition 4.2.11 IfN G and is a character ofG/N, then the character ofGthat is given by

(g) = (gN)g G

is called the lift of to G.

Now that we have defined the lift of a character we shall jump right to the mainresult as it is a very technical process to get to it and is beyond the scope of this paper.Before the statement of the main theorem recall the derived subgroup is the smallestnormal subgroup with abelian factor group. It is given by G =[g, h] :g, h Gwhere[g, h] =ghg1h1.

Theorem 4.2.12 The linear characters of G are precisely the lifts to G of the irre-ducible characters ofG/G. In particular, the number of distinct linear characters ofGis equal to |G/G|, and so divides |G|.

The main ideas in the proof rely on the fact that G/G is abelian and that theirreducible characters of an abelian group are all linear hence the lifts ofG/G are alllinear characters. The next part relies on the fact that irreducible characters of a factor

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group G/Ncorrespond to irreducible characters ofG that have N in there kernel andthat if is a linear character ofG, thenG Kerhence the lifts ofG/G give all of thelinear characters ofG.

We end this section with the construction of a character table using this method.

Example 4.2.13 Let G = Q8 =a, b : a4 = 1, b2 = a2, b1ab = a1, the quaternion

group of order 8. It is easy to see that the conjugacy classes are C1 ={1}, C2 ={a2},

C2 = {a, a3}, C2 = {b, a

2b}, C2 = {ab,a3b} and with a bit of work we can find that

G = {1, a2}. This gives G/G = {G, aG, bG,abG} C2 C2. Using the fact thatabelian groups admit only linear characters, it is easy to see that the character table ofG/G is

gi G aG bG abG

1 1 1 1 12 1 1 -1 -13 1 -1 1 -14 1 -1 -1 1

Lifting the characters ofG/G effectively means that we insert |G| columns andCCrows where C and Cdenote the number of conjugacy classes ofG and G respectively.Hence we get for the character table ofG

gi 1 a2 a b ab

|CG(gi)| 8 8 4 4 4

1 1 1 1 12 1 1 -1 -13 1 -1 1 -14 1 -1 -1 15

Since a2 G the column corresponding to a2 is the same as the column corre-sponding to 1 for the linear characters. Then the orthogonality relations can be used toconstruct the rest of the table and we get

gi 1 a2 a b ab|CG(gi)| 8 8 4 4 4

1 1 1 1 1 12 1 1 1 -1 -13 1 1 -1 1 -14 1 1 -1 -1 15 2 -2 0 0 0

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5 The Davenport Constant

5.1 Finding the Davenport Constant

The Davenport constant was proposed by H. Davenport in 1966. It is an open problemin modern algebra that has proved quite difficult to solve. We shall not discuss theapplications of the Davenport constant but rather look at methods to determine thevalue of the constant for different groups. The Davenport constant is only defined forfinite Abelian groups but there is a generalization of the Davenport constant for arbitrarygroups. Before getting into this section, however, there is a bit of terminology that weneed. It should be noted that the rank of a finite Abelian group is the number of cycliccomponents in the decomposition under the Fundmental Theorem of Finite AbelianGroups. Also, a set consisting of one generator of each of the cyclic components of afinite Abelian group will be refered to as a basis. Now, we can move on to the Davenportconstant.

Definition 5.1.1 The Davenport ConstantLetG be a finite Abelian group. Define s = D(G) to be the smallest positive integer

such that, for any sequence g1, g2, . . . , gs (repetition allowed) of group elements, thereexist indices

1 i1< < it s

or which gi1gi2 git = 1. D(G) is called the Davenport constant.

The Fundamental Theorem of Finite Abelian Groups shows us that we can alwaysdefine a finite Abelian group to be a group under the operation of addition. This means

that the Davenport constant problem can be written in the form of a zero-sum sequenceproblem. The Davenport constant is related as such to the zero-sum constant proposedby Erdos, Ginzburg, and Ziv.There are not many classes of groups for which the Davenport constant is know but wenow take a look at some of the groups for which the constant is know.

Theorem 5.1.2 The Davenport constant ofZn, D(Zn), is n.

Proof If G = g : gn = 1 then the sequence g1, g2, . . . , gn1, where gi = g has nosubsequence of product 1 since |g|> n 1. Hence D(Zn) n.Letg i1 , g i2 , . . . , g in be a sequence of length n. Now consider the n products given by:

pj =n

j=1

pij

Thus we have pj g for all j and hence n elements ofg. If the pj are distinct thenone of them is 1. Otherwise, pl = pk for some k > l. Thus, multiplying by g

l yields1 =gil+1 gik , a subsequence of product 1. Hence, D(Zn) n. Thus D(Zn) =n.

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So, the Davenport constant problem is solved for cyclic groups. Another very simpleclass of group is the finite p-groups. It turns out that the Davenport constant problemis rather simple to solve in this case as well. We next look at a lemma that we shall

need in order to prove what the value of the Davenport constant is for finite Abelianp-groups.

Lemma 5.1.3 Let G= Zpe1Zpe2Zper . Ifg1, . . . , gk G and k 1+

r(pei1),

then(1 g1)(1 g2) . . . (1 gk) 0 (mod p).

Proof Letx1, x2, . . . , xr be a basis for G so that each xi has order pei. If, for some i,

gi = uv, we may, in a sense, decompose the product

J= (1 g1) (1 gk)

to the form

J= (1g1) (1gi1)(1u)(1gi+1) (1gk)+u(1g1) (1gi1)(1v)(1gi+1) (1gk)

using the identity1 uv= (1 u) + u(1 v).

Since each gi is the product of the basis elements xj we may, be repeated applying thisdecomposition, arrive at the following expression for J.

J= gJwhere each g G and each J is a product of the form

J = (1 x1)f1(1 x2)

f2 (1 xr)fr .

In the product we also fave that each fi is a nonnegative integer and

rfi =k. Now,sincek 1 +

r(p

ei 1) we have

rfi 1 +

r(pei 1), hence we must have at least

one fi pei . But

(1 xi)pe1 0.

Thus, J 0 (mod p) for each , and this completes the proof.

We are now ready to prove the theorem due to Olsen on the Davenport constant forfinite p-groups.

Theorem 5.1.4 (Olsen) LetG= Zpe1 Zpe2 Zper . ThenD(G) = 1+r

i=1(pei

1).

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Proof First, let x1, x2, . . . , xr be a basis for G so that each xi has order pei. Now if

we form the sequence in which each xi occurs pei 1 times, then it is clear that this

sequence has no subsequence with product 1. Hence, D(G) 1 + (pei 1).

For the other direction, look at all subsequences ofg1, g2, . . . , gk that have product g.LetEg be the number of subsequences with product g of even length and let Og be thenumber of subsequences with product g of odd length. Using the lemma we get:

Eg Og

0 (modp) ifg = 11 (mod p) ifg = 1

To see this is the case expand the expression in the lemma to getgG

1 + (Eg Og) g 0 mod p.

Were interested in when g = 1 and we notice that we cannot have E(1) =O(1) = 0and the theorem is proved.

Another one of the few classes for which the Davenport constant is known is thegroups of rank two. To prove the theorem concerning the Davenport constant of groupsof rank two we require a lemma. But first, a definition.

Definition 5.1.5 The elementary Abelianp-group if order pk is the group

Zp Zp Zp

k timesunder the usual operation.

Lemma 5.1.6 Let E be the elementary abelian p-group of order p2 (p prime). Ifg1, g2, . . . , gs E and s 3p 2, then there exists indices 1 i1 < i2 < < it swith 1 t p such that gi1gi2 git = 1.

Proof The previous theorem givesD(P) = 1 + k(p 1) ifP is the elementary Abelianp-group of order pk. Now, embed E into the elementary Abelian group F of order p3.Letx F, but x /E. Since D(F) = 3p 2, some subsequence ofxg1, xg2, . . . , x gs hasproduct 1. Sincex has order p (otherwise x E) we must have that the subsequence

with product 1 must be of length p or 2p. That is (after a rearrangement of subscripts)g1g2 gep= 1 where e = 1 or e = 2. Ife = 1 then we are finished so, let e = 2. But thesequence g1, g2, . . . , g2p has a proper subsequence with product 1 since D(E) = 2p 1.Thus, the sequence contains a subsequence of length not exceeding p with product 1.

We now present the theorem due to Olsen giving the value of the Davenport constantof finite abelian groups of rank two.

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Theorem 5.1.7 Let G = H Kbe the direct product of Abelian groups H and Kof orders h and k, respectively, where h|k. Ifg1, g2, . . . , gs G and s h + k 1, thengi1gi2 git = 1 for some 1 i1< < it s.

Proof We proceed by induction on the order ofH,h. We have already proved the casein which h = 1 in Theorem 4.1.3. So, assume thath > 1 and let p be a prime divisorofh. Then we also have that p|k. Let H1 be a subgroup ofH and K1 be a subgroupofKwith [H : H1] = [K : K1] =p. Let h = ph1 and k = pk1. Also let Q = H1 K1.We assume that the theorem is true for Q= H1 K1. Note here that the factor groupG/Q Eand by our assumption on the value ofs we have

s h + k 1

=p(h1+ k1 2) + 2p 1

p(2 + 1 2) + 2p 1

= 3p 1

3p 2

Considering the image of the sequence g1, g2, . . . , gs under the natural homomorphismfromG to G/Q we get a sequence of length greater than or equal to 3p 2 inG/Q Eand thus there is a subsequence of length at most p with product 1 by Lemma 4.1This means that there is a subsequence ofg1, g2, . . . , gs in Gwith product q1 Q sincethe kernel of the natural homomorphism is Q. Now continuing this process with otherelements ofQ we can construct pair-wise disjoint subsets S1, S2, . . . , S u1 of the indexset{1,2,...,s} of sizes 1 |Sj | p, such that

iSj

gi = qj ,

so that qj Q and where u 1 = h1 +k1 2. Since the length of each sequenceS1, S2, . . . , S u1 is at most p we have at least 2p 1 indices left since s p(h1+ k12 ) + 2p 1. Now since D(E) = 2p 1 there is a subsetSu of the remaining indices suchthat

iSu

gi = qu Q.

Sinceu = h1+ k1 1, some subsequence ofq1, q2, . . . , q u has product 1. This completesthe proof.

We have proved what the value of the Davenport constant is for a few classes ofAbelian groups. These are actually most of the groups for which the Davenport constantis known. It is in general, very difficult to determine the Davenport constant for a group.To deal with this issue mathematicians study the problem from a slightly different angleto help develop insight into the problem and make progress toward determining the valueof the constant. There is a good deal of research dedicated to bounding the Davenportconstant from above and this is the focus of the last section of this paper.

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Hence,kg = 0 for everyg Gsinceg was chosen arbitrarily. Now, it follows immediatelythat = 0, so the theorem has been proven.

We now introduce a new concept but we will not explore it in detail as it is outsideof the scope of this paper. We introduce it simply to show how character theory canbe used to bound the Davenport constant. In what followsv w denotes the usual dotproduct. Also we denote by F an algebraically closed field of characteristic 0.

Definition 5.2.2 CoverLet M =

vi,j

, (1 i m; 1 j n) be an m n matrix with entries in Fk. Acover ofM is a set P={w1, . . . , wn} of vectors in F

k such that for each i {1, . . . , m}there exists a j {1, . . . , n} with vi,j wj {0, 1}. We say that P is a proper cover ifvi,j wj = 1 for all i and the chosen j .

We illustrate the definition with an example.

Example 5.2.3 Let F = C and Mbe the following matrix with entries in C2.

M=

(1, 0) (1, 1) (1 i, i)(1, 2) (2, 1) (2 3i, 1 + 3i)

Then, P = {(1, 0), (0, 1), (1, 1)} is a proper cover of M. To see this is the case we

compute the dot products vi,j wj to get the following:

v1,1 w1= (1, 0) (1, 0) = 1

v2,1 w1= (1, 2) (1, 0) = 1

v1,2 w2= (1, 1) (0, 1) = 1

v2,2 w2= (2, 1) (0, 1) = 1

v1,3 w3= (1 i, i) (1, 1) = 1

v2,3 w3= (2 3i, 1 + 3i) (1, 1) = 1

Definition 5.2.4 LetS Fk be a finite set of elements ofFk. We say that S admits

an (m,n,k)-cover if all m n matrices Mwith entries in Shave a proper cover.

We are now ready for the final theorem. This is the main theorem of [1] and it isincluded for the sole purpose of giving an example of how one can apply character theoryto the Davenport constant problem.

Theorem5.2.5 Let Un be the set of all nth roots of unity. IfUn admits an (n

r, l , k)-cover, then lk D(l) (Zrn).

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Proof Since G = Zrn is Abelian, we have that, for each character G and each

g G, (g) Un by Therem 4.1.10. We can therefore view the character table of Gas an nr nr matrix with elements in Un. We need to prove that given kl elements

g1, g2, . . . , gkl G = Zrn, there exist indices i1, i2, . . . , is with gi1 gis = 1 (s l).Consider the columns of the character table of G that correspond to the elementsg1, . . . , gkl. They form an n

r kl matrix T. Construct an nr l matrixT with el-ements in Ukn , obtained from T by replacing, in each row, the elemenets with indices

jl + 1, . . . , j l+ k (0 j k 1) by a k-tuple that will be regarded as an entry ofT.Since Un admits an (n

r, l , k)-cover,T admits a cover, say P = {v1, . . . , vl}, and letvi = (v

1i , v

2i , . . . , v

ki) U

kn .

We now define

wi =

(i1)l+k

j=(i1)l+1

vj

i gj .

Since P is a proper cover ofT, we have that, for each character G, there existsan index i = i() such that (wi) = 1.

Now, consider the element

J= (1 w1)(1 w2) (1 wl)

of the group algebra A. We claim thatJ= 0. Indeed, by Theorem 5.2.1 it suffices toshow that (J) = 0 for all irreducible characters . But, as shown above, there always

exists an index i = i() with 1 (wi) = 0. Hence, the claim will follow at once if weshow that

(J) =l

i=1

(1 (wi)),

which holds since is a group homomorphism. Now were left with

0 = (1 w1)(1 w2) (1 wl),

from which it follows that there exist indices i1, i2, . . . , is with gi1 gis = 1.

6 ConclusionWe shall conclude this paper with some final remarks. First, the reader should nowunderstand the basic ideas and motivation behind representation theory as well as theconnection to characters theory. The reader should also have an understanding of simplemethods of constructing character tables. The character table was not explored verythoroughly and the interested reader could continue on to see how the character tableof a group gives a lot of immediate information. Examples of such information include

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immediate classification of a group as simple or not simple and, moreover, determinationof all normal subgroups. A special case of this is the centre of the group. We canalso determine whether or not a group is solvable, nilpotent or abelian. More advanced

thoery that can be explored in character theory includes induced characters and a tensorproduct of characters. Finally, the reader may be interested in exploring the applicationsof the Davenport constant. In order to do this, however, a stronger background incombinatorics and number theory would be very beneficial as the level of difficulty israther higher than the difficulty of this paper.

References

[1] Dimitrov, Vesselin. Zero-sum Problems in Finite Groups.Research Science Institues,Massachusetts Institute of Technology, 2007.

[2] Dummit, D. S. & Foote, R. M. Abstract Algebra, 3rd ed. New Jersey: John Wileyand Sons Inc., 2004.t

[3] James, G. & Liebeck, M. Representations and Characters of Groups, 2nd ed. Cam-bridge: Cambridge University Press, 2001.

[4] Milies, C. P. & Sehgal, S. K., An Introduction to Group Rings, Dordrecht, TheNetherlands: Kluwer Academic Publishers, 2002.

[5] Olson, J. E. A combinatorial problem on finite Abelian groups, I. J. Number Theory1 (1969), 8-10.

[6] Olson, J. E. A combinatorial problem on finite Abelian groups, II.J. Number Theory1 (1969), 195-199.

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character theory and the davenport constant

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