Main Problems in the Representation Theory of

Finite Groups

Gabriel Navarro

University of Valencia

Bilbao, October 8, 2011

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 1 / 67

This talk is addressed to the non-specialists

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 2 / 67

I. Overview

The Global/Local Conjectures

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 3 / 67

I. Overview

Richard Brauer’s Conjectures (1960’s):

k(B)-Conjecture.

Height Zero Conjecture.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 4 / 67

I. Overview

Richard Brauer’s Conjectures (1960’s):

k(B)-Conjecture.

Height Zero Conjecture.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 4 / 67

I. Overview

Richard Brauer’s Conjectures (1960’s):

k(B)-Conjecture.

Height Zero Conjecture.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 4 / 67

I. Overview

Counting Conjectures (70’s, 80’s, 90’s):

The McKay Conjecture.

Alperin Weight Conjecture.

Dade’s Conjectures.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 5 / 67

I. Overview

Counting Conjectures (70’s, 80’s, 90’s):

The McKay Conjecture.

Alperin Weight Conjecture.

Dade’s Conjectures.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 5 / 67

I. Overview

Counting Conjectures (70’s, 80’s, 90’s):

The McKay Conjecture.

Alperin Weight Conjecture.

Dade’s Conjectures.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 5 / 67

I. Overview

Counting Conjectures (70’s, 80’s, 90’s):

The McKay Conjecture.

Alperin Weight Conjecture.

Dade’s Conjectures.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 5 / 67

I. Overview

A structural conjecture: Broue’s Conjecture.

(Two certain algebras are derived equivalent.)

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 6 / 67

I. Overview

A structural conjecture: Broue’s Conjecture.

(Two certain algebras are derived equivalent.)

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 6 / 67

I. Overview

A structural conjecture: Broue’s Conjecture.

(Two certain algebras are derived equivalent.)

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 6 / 67

I. Overview

We all gather that there should be a hidden theory explaining all these

phenomena, but yet such a theory still remains to be discovered.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 7 / 67

I. Overview

My aim here today is to tell you where we are and what we expect in

the future.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 8 / 67

2. Groups

This talk is about finite groups.

If G is a finite group, then the order of G is jG j, the number of

elements in G .

If you wish to have two good examples of groups in mind:

The symmetric group Sn of bijections f : Ω ! Ω, where

Ω = f1; 2; : : : ; ng (or any set).

The general linear group GL(n;F ) of invertible n n matrices over F .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 9 / 67

2. Groups

This talk is about finite groups.

If G is a finite group, then the order of G is jG j, the number of

elements in G .

If you wish to have two good examples of groups in mind:

The symmetric group Sn of bijections f : Ω ! Ω, where

Ω = f1; 2; : : : ; ng (or any set).

The general linear group GL(n;F ) of invertible n n matrices over F .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 9 / 67

2. Groups

This talk is about finite groups.

If G is a finite group, then the order of G is jG j, the number of

elements in G .

If you wish to have two good examples of groups in mind:

The symmetric group Sn of bijections f : Ω ! Ω, where

Ω = f1; 2; : : : ; ng (or any set).

The general linear group GL(n;F ) of invertible n n matrices over F .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 9 / 67

2. Groups

This talk is about finite groups.

If G is a finite group, then the order of G is jG j, the number of

elements in G .

If you wish to have two good examples of groups in mind:

The symmetric group Sn of bijections f : Ω ! Ω, where

Ω = f1; 2; : : : ; ng (or any set).

The general linear group GL(n;F ) of invertible n n matrices over F .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 9 / 67

2. Groups

This talk is about finite groups.

If G is a finite group, then the order of G is jG j, the number of

elements in G .

If you wish to have two good examples of groups in mind:

The symmetric group Sn of bijections f : Ω ! Ω, where

Ω = f1; 2; : : : ; ng (or any set).

The general linear group GL(n;F ) of invertible n n matrices over F .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 9 / 67

2. Groups

(If there is space for two more groups:

An, the subgroup of even permutations of Sn, and

SL(n;F ), the subgroup of GL(n;F ) of matrices with determinant 1.)

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 10 / 67

2. Groups

(If there is space for two more groups:

An, the subgroup of even permutations of Sn, and

SL(n;F ), the subgroup of GL(n;F ) of matrices with determinant 1.)

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 10 / 67

2. Groups

(If there is space for two more groups:

An, the subgroup of even permutations of Sn, and

SL(n;F ), the subgroup of GL(n;F ) of matrices with determinant 1.)

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 10 / 67

2. Groups

In fact, an abstract group G is usually studied via two kind of

homomorphisms:

: G ! Sn (permutation groups)

or

: G ! GL(n;F ) (representation theory).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 11 / 67

2. Groups

In fact, an abstract group G is usually studied via two kind of

homomorphisms:

: G ! Sn (permutation groups)

or

: G ! GL(n;F ) (representation theory).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 11 / 67

2. Groups

In fact, an abstract group G is usually studied via two kind of

homomorphisms:

: G ! Sn (permutation groups)

or

: G ! GL(n;F ) (representation theory).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 11 / 67

2. Groups

Conjugation is a basic operation in group theory:

xg = g1xg :

It partitions G into conjugacy classes.

If H G and g 2 G , then Hg = fhg j h 2 Hg G .

The subgroup H is normal in G if Hg = H for all g 2 G .

We write H / G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 12 / 67

2. Groups

Conjugation is a basic operation in group theory:

xg = g1xg :

It partitions G into conjugacy classes.

If H G and g 2 G , then Hg = fhg j h 2 Hg G .

The subgroup H is normal in G if Hg = H for all g 2 G .

We write H / G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 12 / 67

2. Groups

Conjugation is a basic operation in group theory:

xg = g1xg :

It partitions G into conjugacy classes.

If H G and g 2 G , then Hg = fhg j h 2 Hg G .

The subgroup H is normal in G if Hg = H for all g 2 G .

We write H / G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 12 / 67

2. Groups

Conjugation is a basic operation in group theory:

xg = g1xg :

It partitions G into conjugacy classes.

If H G and g 2 G , then Hg = fhg j h 2 Hg G .

The subgroup H is normal in G if Hg = H for all g 2 G .

We write H / G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 12 / 67

2. Groups

Conjugation is a basic operation in group theory:

xg = g1xg :

It partitions G into conjugacy classes.

If H G and g 2 G , then Hg = fhg j h 2 Hg G .

The subgroup H is normal in G if Hg = H for all g 2 G .

We write H / G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 12 / 67

2. Groups

What is the significance of normal subgroups?

If H / G , then

G=H = fHg j g 2 Gg

is a new group (where Hg = fhg j h 2 Hg G ).

Its multiplication is

(Hg)(Hk) = H(gk)

(multiplication of subsets of G ).

If G is finite, then jG=H j = jG j=jH j.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 13 / 67

2. Groups

What is the significance of normal subgroups?

If H / G , then

G=H = fHg j g 2 Gg

is a new group (where Hg = fhg j h 2 Hg G ).

Its multiplication is

(Hg)(Hk) = H(gk)

(multiplication of subsets of G ).

If G is finite, then jG=H j = jG j=jH j.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 13 / 67

2. Groups

What is the significance of normal subgroups?

If H / G , then

G=H = fHg j g 2 Gg

is a new group (where Hg = fhg j h 2 Hg G ).

Its multiplication is

(Hg)(Hk) = H(gk)

(multiplication of subsets of G ).

If G is finite, then jG=H j = jG j=jH j.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 13 / 67

2. Groups

What is the significance of normal subgroups?

If H / G , then

G=H = fHg j g 2 Gg

is a new group (where Hg = fhg j h 2 Hg G ).

Its multiplication is

(Hg)(Hk) = H(gk)

(multiplication of subsets of G ).

If G is finite, then jG=H j = jG j=jH j.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 13 / 67

2. Groups

If the purpose is to know all finite groups, and you have a normal

subgroup H / G , then by induction you know H and G=H , and then

you have good information about G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 14 / 67

2. Groups

In the remaining case, G has no proper normal subgroups, and we say

that G is simple.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 15 / 67

2. Groups

By choosing a chain of successive maximal normal subgroups

1 = G0 / G1 / G2 / / Gn1 / Gn = G ;

we have that Gi+1=Gi = Ki are simple groups.

These are the composition factors of G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 16 / 67

2. Groups

By choosing a chain of successive maximal normal subgroups

1 = G0 / G1 / G2 / / Gn1 / Gn = G ;

we have that Gi+1=Gi = Ki are simple groups.

These are the composition factors of G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 16 / 67

2. Groups

So, as numbers are built up from prime numbers, finite groups are

built up from simple groups.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 17 / 67

2. Groups

Theorem (The Classification of Finite Simple Groups,

2004)

If G is a finite simple group, then

G = Cp for some prime p.

G = An for n 5.

G is a simple group of Lie type.

G is a sporadic simple group.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 18 / 67

2. Groups

Theorem (The Classification of Finite Simple Groups,

2004)

If G is a finite simple group, then

G = Cp for some prime p.

G = An for n 5.

G is a simple group of Lie type.

G is a sporadic simple group.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 18 / 67

2. Groups

Theorem (The Classification of Finite Simple Groups,

2004)

If G is a finite simple group, then

G = Cp for some prime p.

G = An for n 5.

G is a simple group of Lie type.

G is a sporadic simple group.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 18 / 67

2. Groups

Theorem (The Classification of Finite Simple Groups,

2004)

If G is a finite simple group, then

G = Cp for some prime p.

G = An for n 5.

G is a simple group of Lie type.

G is a sporadic simple group.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 18 / 67

2. Groups

Theorem (The Classification of Finite Simple Groups,

2004)

If G is a finite simple group, then

G = Cp for some prime p.

G = An for n 5.

G is a simple group of Lie type.

G is a sporadic simple group.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 18 / 67

2. Groups

A few remarks:

An for n 5 is the reason that polynomial equations of degree

greater than 5 are not solvable by radicals.

PSL(n; q) = SL(n; q)=Z(SL(n; q)) is an example of a simple

group of Lie type.

There are 26 sporadic groups:

M11;M12;M22;M23;M24; J1; J2; J3; J4;Co1;Co2;Co3;Fi22;Fi23;Fi24;

HS ;McL;He;Ru; Suz ;O 0N ;HN ; Ly ;Th;B ;M :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 19 / 67

2. Groups

A few remarks:

An for n 5 is the reason that polynomial equations of degree

greater than 5 are not solvable by radicals.

PSL(n; q) = SL(n; q)=Z(SL(n; q)) is an example of a simple

group of Lie type.

There are 26 sporadic groups:

M11;M12;M22;M23;M24; J1; J2; J3; J4;Co1;Co2;Co3;Fi22;Fi23;Fi24;

HS ;McL;He;Ru; Suz ;O 0N ;HN ; Ly ;Th;B ;M :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 19 / 67

2. Groups

A few remarks:

An for n 5 is the reason that polynomial equations of degree

greater than 5 are not solvable by radicals.

PSL(n; q) = SL(n; q)=Z(SL(n; q)) is an example of a simple

group of Lie type.

There are 26 sporadic groups:

M11;M12;M22;M23;M24; J1; J2; J3; J4;Co1;Co2;Co3;Fi22;Fi23;Fi24;

HS ;McL;He;Ru; Suz ;O 0N ;HN ; Ly ;Th;B ;M :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 19 / 67

2. Groups

A few remarks:

An for n 5 is the reason that polynomial equations of degree

greater than 5 are not solvable by radicals.

PSL(n; q) = SL(n; q)=Z(SL(n; q)) is an example of a simple

group of Lie type.

There are 26 sporadic groups:

M11;M12;M22;M23;M24; J1; J2; J3; J4;Co1;Co2;Co3;Fi22;Fi23;Fi24;

HS ;McL;He;Ru; Suz ;O 0N ;HN ; Ly ;Th;B ;M :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 19 / 67

2. Groups

jM j =

808017424794512875886459904961710757005754368000000000

= 246 320 59 76 112 17 19 23 29 31 41 47 59 71

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 20 / 67

2. Groups

jM j =

808017424794512875886459904961710757005754368000000000

= 246 320 59 76 112 17 19 23 29 31 41 47 59 71

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 20 / 67

2. Groups

We write the order of the groups as jG j = pam, where p does not

divide m.

G has subgroups of order pa (the Sylow p-subgroups of G ).

Also, every p-subgroup Q of G (i.e., jQj = pb) is contained in some

Sylow p-subgroup of G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 21 / 67

2. Groups

We write the order of the groups as jG j = pam, where p does not

divide m.

G has subgroups of order pa (the Sylow p-subgroups of G ).

Also, every p-subgroup Q of G (i.e., jQj = pb) is contained in some

Sylow p-subgroup of G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 21 / 67

2. Groups

We write the order of the groups as jG j = pam, where p does not

divide m.

G has subgroups of order pa (the Sylow p-subgroups of G ).

Also, every p-subgroup Q of G (i.e., jQj = pb) is contained in some

Sylow p-subgroup of G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 21 / 67

2. Groups

As in many other parts of Mathematics, there is a global/local

analysis in Group Theory, which turns out to be fundamental for the

theory.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 22 / 67

2. Groups

The local subgroups are

NG (Q) = fx 2 G jQx = Qg ;

where 1 < Q is any p-subgroup of G .

What do the local subgroups know about G and vice-versa?

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 23 / 67

2. Groups

The local subgroups are

NG (Q) = fx 2 G jQx = Qg ;

where 1 < Q is any p-subgroup of G .

What do the local subgroups know about G and vice-versa?

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 23 / 67

II. Characters

Suppose that G is a group, and let

X : G ! GL(n;C)

be a group homomorphism. This is called a complex representation of

degree n of G .

The character of the representation is the trace function:

= X : G ! C given by

(g) = Trace(X (g)) :

Note that (1) = n is the degree and that is a class function

(constant on conjugacy classes).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 24 / 67

II. Characters

Suppose that G is a group, and let

X : G ! GL(n;C)

be a group homomorphism. This is called a complex representation of

degree n of G .

The character of the representation is the trace function:

= X : G ! C given by

(g) = Trace(X (g)) :

Note that (1) = n is the degree and that is a class function

(constant on conjugacy classes).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 24 / 67

II. Characters

Suppose that G is a group, and let

X : G ! GL(n;C)

be a group homomorphism. This is called a complex representation of

degree n of G .

The character of the representation is the trace function:

= X : G ! C given by

(g) = Trace(X (g)) :

Note that (1) = n is the degree and that is a class function

(constant on conjugacy classes).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 24 / 67

II. Characters

There is always a character of degree 1:

G ! C

g 7! 1

called the trivial character.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 25 / 67

II. Characters

The complex space cf(G ) of class-functions

G ! C

has dimension k(G ), the number of conjugacy classes of G . Also, it

is a Hermitian space with inner product

[; ] =1

jG j

Xx2G

(x)(x) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 26 / 67

II. Characters

Values of the characters?

Since X (g)m = In, by elementary linear algebra we have that X (g) is

similar to a diagonal matrix with roots of unity. So (g) is an

algebraic integer, in the cyclotomic field QjG j.

In fact, a deep theorem of Brauer, with consequences in number

theory, asserts that complex characters can be afforded by

representations

X : G ! GL(n;QjG j) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 27 / 67

II. Characters

Values of the characters?

Since X (g)m = In, by elementary linear algebra we have that X (g) is

similar to a diagonal matrix with roots of unity. So (g) is an

algebraic integer, in the cyclotomic field QjG j.

In fact, a deep theorem of Brauer, with consequences in number

theory, asserts that complex characters can be afforded by

representations

X : G ! GL(n;QjG j) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 27 / 67

II. Characters

Values of the characters?

Since X (g)m = In, by elementary linear algebra we have that X (g) is

similar to a diagonal matrix with roots of unity. So (g) is an

algebraic integer, in the cyclotomic field QjG j.

In fact, a deep theorem of Brauer, with consequences in number

theory, asserts that complex characters can be afforded by

representations

X : G ! GL(n;QjG j) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 27 / 67

II. Characters

If you have a representation

X : G ! GL(n;C)

and fix M 2 GL(n;C), then you have another representation

Y = M1XM

(which is called similar, and denoted X Y).

Theorem

The representations X and Y are similar if and only if X = Y .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 28 / 67

II. Characters

If you have a representation

X : G ! GL(n;C)

and fix M 2 GL(n;C), then you have another representation

Y = M1XM

(which is called similar, and denoted X Y).

Theorem

The representations X and Y are similar if and only if X = Y .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 28 / 67

II. Characters

If Y and Z are representations, then you can build another one, the

direct sum 0@Y 0

0 Z

1A :

In particular, a sum of two characters is a character.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 29 / 67

II. Characters

Definition

A character of G is irreducible if it is not the sum of two other

characters. In this case, we write 2 Irr(G ).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 30 / 67

II. Characters

Theorem (Fundamental of Character Theory)

If G is a finite group, then Irr(G ) is an orthonormal basis of cf(G ).

A finite group has associated a square complex invertible matrix

X (G ) = (i(gj)) ;

which is called the character table of G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 31 / 67

II. Characters

Theorem (Fundamental of Character Theory)

If G is a finite group, then Irr(G ) is an orthonormal basis of cf(G ).

A finite group has associated a square complex invertible matrix

X (G ) = (i(gj)) ;

which is called the character table of G .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 31 / 67

II. Characters

Character Table of A5:

X.1 1 1 1 1 1X.2 3 -1 0 A *AX.3 3 -1 0 *A AX.4 4 0 1 -1 -1X.5 5 1 -1 0 0

A = -E(5)-E(5)^4 = (1-ER(5))/2 = -b5

1

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 32 / 67

II. Characters

We are interested in the degrees (1) of the irreducible characters,

that is, in the number n in

X : G ! GL(n;C) :

It is possible to prove that (1) divides jG j.

We also haveX

2Irr(G)

(1)2 = jG j ;

equation that follows from:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 33 / 67

II. Characters

We are interested in the degrees (1) of the irreducible characters,

that is, in the number n in

X : G ! GL(n;C) :

It is possible to prove that (1) divides jG j.

We also haveX

2Irr(G)

(1)2 = jG j ;

equation that follows from:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 33 / 67

II. Characters

We are interested in the degrees (1) of the irreducible characters,

that is, in the number n in

X : G ! GL(n;C) :

It is possible to prove that (1) divides jG j.

We also haveX

2Irr(G)

(1)2 = jG j ;

equation that follows from:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 33 / 67

II. Characters

Theorem (Wedderburn)

CG = Mat(n1;C) Mat(nk ;C) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 34 / 67

II. Characters

A digression: Not all the problems are global/local, of course.

Brauer’s Problem 1 is: what algebras are the group algebras?

For instance,

C C C C CMat(5;C)

is not a group algebra.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 35 / 67

II. Characters

A digression: Not all the problems are global/local, of course.

Brauer’s Problem 1 is: what algebras are the group algebras?

For instance,

C C C C CMat(5;C)

is not a group algebra.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 35 / 67

II. Characters

A digression: Not all the problems are global/local, of course.

Brauer’s Problem 1 is: what algebras are the group algebras?

For instance,

C C C C CMat(5;C)

is not a group algebra.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 35 / 67

III. The Conjectures

Now, that you know the environment in which we live, I can tell you

about some of the problems that we face.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 36 / 67

III. The Conjectures

Counting Conjectures

If G is a finite group and p is a prime, let

Irrp0(G ) = f 2 Irr(G ) of degree (1) not divisible by pg.

CONJECTURE (McKay, 1971)

If G is a finite group and P is a Sylow p-subgroup of G , then

jIrrp0(G )j = jIrrp0(NG (P))j :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 37 / 67

III. The Conjectures

Counting Conjectures

If G is a finite group and p is a prime, let

Irrp0(G ) = f 2 Irr(G ) of degree (1) not divisible by pg.

CONJECTURE (McKay, 1971)

If G is a finite group and P is a Sylow p-subgroup of G , then

jIrrp0(G )j = jIrrp0(NG (P))j :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 37 / 67

III. The Conjectures

Counting Conjectures

If G is a finite group and p is a prime, let

Irrp0(G ) = f 2 Irr(G ) of degree (1) not divisible by pg.

CONJECTURE (McKay, 1971)

If G is a finite group and P is a Sylow p-subgroup of G , then

jIrrp0(G )j = jIrrp0(NG (P))j :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 37 / 67

III. The Conjectures

Character Table of G = A5. Set p = 5.

X.1 1 1 1 1 1X.2 3 -1 0 A *AX.3 3 -1 0 *A AX.4 4 0 1 -1 -1X.5 5 1 -1 0 0

A = -E(5)-E(5)^4 = (1-ER(5))/2 = -b5

1

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 38 / 67

III. The Conjectures

Character Table of NG (P) = D10.

X.1 1 1 1 1X.2 1 -1 1 1X.3 2 0 A *AX.4 2 0 *A A

A = E(5)^2+E(5)^3 = (-1-ER(5))/2 = -1-b5

1

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 39 / 67

III. The Conjectures

There is no doubt that the McKay conjecture is true. If instead of

Mathematics this were Physics...

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 40 / 67

III. The Conjectures

There is no doubt that the McKay conjecture is true. If instead of

Mathematics this were Physics...

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 40 / 67

III. The Conjectures

To make it even more intriguing:

CONJECTURE (Isaacs, N, 2001)

For every integer k , then

jf 2 Irrp0(G ) j(1) k mod pgj =

jf 2 Irrp0(NG (P)) j(1) k mod pgj :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 41 / 67

III. The Conjectures

To make it even more intriguing:

CONJECTURE (Isaacs, N, 2001)

For every integer k , then

jf 2 Irrp0(G ) j(1) k mod pgj =

jf 2 Irrp0(NG (P)) j(1) k mod pgj :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 41 / 67

III. The Conjectures

Character Table of A5. Set p = 5.

X.1 1 1 1 1 1X.2 3 -1 0 A *AX.3 3 -1 0 *A AX.4 4 0 1 -1 -1X.5 5 1 -1 0 0

A = -E(5)-E(5)^4 = (1-ER(5))/2 = -b5

1

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 42 / 67

III. The Conjectures

Character Table of D10.

X.1 1 1 1 1X.2 1 -1 1 1X.3 2 0 A *AX.4 2 0 *A A

A = E(5)^2+E(5)^3 = (-1-ER(5))/2 = -1-b5

1

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 43 / 67

III. The Conjectures

This suggests bijections

: Irrp0(G ) ! Irrp0(NG (P))

satisfying (1) (1) mod p.

There is evidence that these bijections should be compatible with

automorphisms and with certain Galois actions. However...

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 44 / 67

III. The Conjectures

This suggests bijections

: Irrp0(G ) ! Irrp0(NG (P))

satisfying (1) (1) mod p.

There is evidence that these bijections should be compatible with

automorphisms and with certain Galois actions. However...

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 44 / 67

III. The Conjectures

THEOREM (Isaacs, Malle, N, 2007)

If all simple groups are “good”, then the McKay conjecture is true.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 45 / 67

III. The Conjectures

Alternating groups are “good”. Sporadic groups are “good”. Most

groups of Lie type have been proved to be “good” (O. Brunat, G.

Malle, B. Spath, etc.)

In order to complete the proof of the McKay conjecture, what is

needed is a deeper knowledge of the character theory of certain

quasisimple groups of Lie type. Some of this depends on ongoing

work by G. Lusztig.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 46 / 67

III. The Conjectures

Alternating groups are “good”. Sporadic groups are “good”. Most

groups of Lie type have been proved to be “good” (O. Brunat, G.

Malle, B. Spath, etc.)

In order to complete the proof of the McKay conjecture, what is

needed is a deeper knowledge of the character theory of certain

quasisimple groups of Lie type. Some of this depends on ongoing

work by G. Lusztig.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 46 / 67

III. The Conjectures

If the McKay conjecture deals with characters of degree

(1)p = 1 ;

then the Alperin Weight Conjecture (AWC) deals with characters

2 Irr(G ) such that

(1)p = jG jp :

Set k0(G ) = jf 2 Irr(G ) j(1)p = jG jpgj:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 47 / 67

III. The Conjectures

If the McKay conjecture deals with characters of degree

(1)p = 1 ;

then the Alperin Weight Conjecture (AWC) deals with characters

2 Irr(G ) such that

(1)p = jG jp :

Set k0(G ) = jf 2 Irr(G ) j(1)p = jG jpgj:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 47 / 67

III. The Conjectures

If the McKay conjecture deals with characters of degree

(1)p = 1 ;

then the Alperin Weight Conjecture (AWC) deals with characters

2 Irr(G ) such that

(1)p = jG jp :

Set k0(G ) = jf 2 Irr(G ) j(1)p = jG jpgj:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 47 / 67

III. The Conjectures

The following is AWC (in the Knorr-Robinson form).

Alperin Weight Conjecture (Alperin, 1987)

If G is a finite group, then

k0(G ) =XC

(1)jC jk(NG (C )) ;

where C runs over G -representatives of chains C of p-subgroups

1 = P0 < P1 < < Pn of length n = jC j, and NG (C ) =T

NG (Pj).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 48 / 67

III. The Conjectures

The following is AWC (in the Knorr-Robinson form).

Alperin Weight Conjecture (Alperin, 1987)

If G is a finite group, then

k0(G ) =XC

(1)jC jk(NG (C )) ;

where C runs over G -representatives of chains C of p-subgroups

1 = P0 < P1 < < Pn of length n = jC j, and NG (C ) =T

NG (Pj).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 48 / 67

III. The Conjectures

In 2011 two independent reductions of AWC to simple groups have

been published. One is due to L. Puig, and the other is due to

N-Tiep.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 49 / 67

III. The Conjectures

To finish with the counting conjectures, we also have Dade’s

Conjectures (1992-1994) which locally count 2 Irr(G ) with a fixed

(1)p = pd .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 50 / 67

III. The Conjectures

Brauer’s Conjectures

I would not be telling the whole truth without mentioning

characteristic p-theory and Brauer blocks. These are the tools to go

as deep as possible. Part of the richness of this theory is that we

work at three levels:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 51 / 67

III. The Conjectures

Brauer’s Conjectures

I would not be telling the whole truth without mentioning

characteristic p-theory and Brauer blocks. These are the tools to go

as deep as possible. Part of the richness of this theory is that we

work at three levels:

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 51 / 67

III. The Conjectures

At the characteristic zero level: The cyclotomic field Qn,

n = jG j.

At the integers level: R the algebraic integers in Qn.

At the characteristic p level: F = R=P , where P is a prime ideal

containing p.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 52 / 67

III. The Conjectures

At the characteristic zero level: The cyclotomic field Qn,

n = jG j.

At the integers level: R the algebraic integers in Qn.

At the characteristic p level: F = R=P , where P is a prime ideal

containing p.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 52 / 67

III. The Conjectures

At the characteristic zero level: The cyclotomic field Qn,

n = jG j.

At the integers level: R the algebraic integers in Qn.

At the characteristic p level: F = R=P , where P is a prime ideal

containing p.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 52 / 67

III. The Conjectures

We can decompose

FG = B1 Bs ;

where the Bi ’s are indecomposable two sided ideals (the blocks of G ).

These are algebras (not necessarily matrix algebras).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 53 / 67

III. The Conjectures

Now, if 2 Irr(G ), we can prove that there is

X : G ! GL(n; S)

affording , where S = f= j 2 R ; 2 R Pg Qn.

Now, by using the canonical : S ! F = R=P , we have a way to

connect representations in characteristic 0 with characteristic p:

X : G ! GL(n;F ) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 54 / 67

III. The Conjectures

Now, if 2 Irr(G ), we can prove that there is

X : G ! GL(n; S)

affording , where S = f= j 2 R ; 2 R Pg Qn.

Now, by using the canonical : S ! F = R=P , we have a way to

connect representations in characteristic 0 with characteristic p:

X : G ! GL(n;F ) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 54 / 67

III. The Conjectures

Now, if 2 Irr(G ), we can prove that there is

X : G ! GL(n; S)

affording , where S = f= j 2 R ; 2 R Pg Qn.

Now, by using the canonical : S ! F = R=P , we have a way to

connect representations in characteristic 0 with characteristic p:

X : G ! GL(n;F ) :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 54 / 67

III. The Conjectures

It turns out that we can partition

Irr(G ) =[

B2Bl(G)

Irr(B) :

Every complex 2 Irr(G ) can be associated to a p-block.

The trivial character belongs to the principal block, the most

important block.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 55 / 67

III. The Conjectures

It turns out that we can partition

Irr(G ) =[

B2Bl(G)

Irr(B) :

Every complex 2 Irr(G ) can be associated to a p-block.

The trivial character belongs to the principal block, the most

important block.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 55 / 67

III. The Conjectures

It turns out that we can partition

Irr(G ) =[

B2Bl(G)

Irr(B) :

Every complex 2 Irr(G ) can be associated to a p-block.

The trivial character belongs to the principal block, the most

important block.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 55 / 67

III. The Conjectures

All Brauer’s conjectures deal with defect groups.

Every block B of G has canonically associated a conjugacy class of

p-subgroups D of G , the defect groups of B .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 56 / 67

III. The Conjectures

All Brauer’s conjectures deal with defect groups.

Every block B of G has canonically associated a conjugacy class of

p-subgroups D of G , the defect groups of B .

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 56 / 67

III. The Conjectures

For instance, the defect groups D of the principal block are the Sylow

p-subgroups of G . In other extreme case, B has defect group D = 1

if and only if Irr(B) = fg which happens if and only if

(1)p = jG jp.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 57 / 67

III. The Conjectures

For instance, the defect groups D of the principal block are the Sylow

p-subgroups of G . In other extreme case, B has defect group D = 1

if and only if Irr(B) = fg which happens if and only if

(1)p = jG jp.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 57 / 67

III. The Conjectures

The following is the Height Zero Conjecture.

Brauer’s Height Zero Conjecture for principal blocks

If B is the principal block of G and D is a Sylow p-subgroup of G ,

then all 2 Irr(B) have degree not divisible by p if and only if D is

abelian.

Both implications are quite deep.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 58 / 67

III. The Conjectures

The following is the Height Zero Conjecture.

Brauer’s Height Zero Conjecture for principal blocks

If B is the principal block of G and D is a Sylow p-subgroup of G ,

then all 2 Irr(B) have degree not divisible by p if and only if D is

abelian.

Both implications are quite deep.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 58 / 67

III. The Conjectures

The following is the Height Zero Conjecture.

Brauer’s Height Zero Conjecture for principal blocks

If B is the principal block of G and D is a Sylow p-subgroup of G ,

then all 2 Irr(B) have degree not divisible by p if and only if D is

abelian.

Both implications are quite deep.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 58 / 67

III. The Conjectures

Using a reduction to simple groups by T. Berger and R. Knorr, and

previous work by Cabanes-Enguehard, Blau-Ellers, it has been recently

announced the solution of half of the Height Zero Conjecture:

THEOREM (R. Kessar, G. Malle, 2011)

Suppose that B has defect group D. If D is abelian, then all

2 Irr(B) have height zero.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 59 / 67

III. The Conjectures

Using a reduction to simple groups by T. Berger and R. Knorr, and

previous work by Cabanes-Enguehard, Blau-Ellers, it has been recently

announced the solution of half of the Height Zero Conjecture:

THEOREM (R. Kessar, G. Malle, 2011)

Suppose that B has defect group D. If D is abelian, then all

2 Irr(B) have height zero.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 59 / 67

III. The Conjectures

The other half of the conjecture is even more difficult. It was proven

for p-solvable groups by D. Gluck and T. Wolf in the 1980’s. A

reduction to simple groups is expected to be published very soon.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 60 / 67

III. The Conjectures

Next is Brauer’s k(B)-conjecture:

Brauer’s k(B)-conjecture

If B is a block of G with defect group D, then

k(B) = jIrr(B)j jDj :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 61 / 67

III. The Conjectures

Next is Brauer’s k(B)-conjecture:

Brauer’s k(B)-conjecture

If B is a block of G with defect group D, then

k(B) = jIrr(B)j jDj :

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 61 / 67

III. The Conjectures

The k(B)-conjecture for p-solvable groups has remained a challenge

for many years. This is an incredibly difficult theorem, only possible

because of the work of R. Knorr, then G. Robinson and J. Thompson,

and finally D. Gluck, K. Magaard, U. Riese and P. Schmid. There is

no reduction to simple groups (so far).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 62 / 67

III. The Conjectures

The k(B)-conjecture for p-solvable groups has remained a challenge

for many years. This is an incredibly difficult theorem, only possible

because of the work of R. Knorr, then G. Robinson and J. Thompson,

and finally D. Gluck, K. Magaard, U. Riese and P. Schmid. There is

no reduction to simple groups (so far).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 62 / 67

III. The Conjectures

The k(B)-conjecture for p-solvable groups has remained a challenge

for many years. This is an incredibly difficult theorem, only possible

because of the work of R. Knorr, then G. Robinson and J. Thompson,

and finally D. Gluck, K. Magaard, U. Riese and P. Schmid. There is

no reduction to simple groups (so far).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 62 / 67

III. The Conjectures

By Brauer’s First Main Theorem:

A block B of G with defect group D uniquely determines a block b

of NG (D) (with defect group again D).

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 63 / 67

III. The Conjectures

Conjecture (Broue)

If D is abelian, then the algebras B and b are derived equivalent.

This explains the McKay conjecture (with congruences), but only

when D is abelian. Many cases of this deep conjecture have been

proven, among them J. Chuang and R. Rouquier have done the

symmetric group case.

A big challenge: what is going on if D is not abelian?

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 64 / 67

III. The Conjectures

Conjecture (Broue)

If D is abelian, then the algebras B and b are derived equivalent.

This explains the McKay conjecture (with congruences), but only

when D is abelian. Many cases of this deep conjecture have been

proven, among them J. Chuang and R. Rouquier have done the

symmetric group case.

A big challenge: what is going on if D is not abelian?

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 64 / 67

III. The Conjectures

Conjecture (Broue)

If D is abelian, then the algebras B and b are derived equivalent.

This explains the McKay conjecture (with congruences), but only

when D is abelian. Many cases of this deep conjecture have been

proven, among them J. Chuang and R. Rouquier have done the

symmetric group case.

A big challenge: what is going on if D is not abelian?

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 64 / 67

III. The Conjectures

Conjecture (Broue)

If D is abelian, then the algebras B and b are derived equivalent.

This explains the McKay conjecture (with congruences), but only

when D is abelian. Many cases of this deep conjecture have been

proven, among them J. Chuang and R. Rouquier have done the

symmetric group case.

A big challenge: what is going on if D is not abelian?

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 64 / 67

IV. Rational Groups

Rational Groups.

A group G is rational if (g) 2 Q for all g 2 G , 2 Irr(G ). Or

equivalently: if whenever hxi = hyi, then x and y are G -conjugate.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 65 / 67

IV. Rational Groups

Rational Groups.

A group G is rational if (g) 2 Q for all g 2 G , 2 Irr(G ). Or

equivalently: if whenever hxi = hyi, then x and y are G -conjugate.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 65 / 67

IV. Rational Groups

Rational Groups.

A group G is rational if (g) 2 Q for all g 2 G , 2 Irr(G ). Or

equivalently: if whenever hxi = hyi, then x and y are G -conjugate.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 65 / 67

IV. Rational Groups

Two are the main problems on rational groups.

CONJECTURE

If Cp is a composition factor of a rational group, then p 5

J. Thompson has proved that p 13 (2008).

CONJECTURE

If G is a rational group and P 2 Syl2(G ), then P is rational.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 66 / 67

IV. Rational Groups

Two are the main problems on rational groups.

CONJECTURE

If Cp is a composition factor of a rational group, then p 5

J. Thompson has proved that p 13 (2008).

CONJECTURE

If G is a rational group and P 2 Syl2(G ), then P is rational.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 66 / 67

IV. Rational Groups

Two are the main problems on rational groups.

CONJECTURE

If Cp is a composition factor of a rational group, then p 5

J. Thompson has proved that p 13 (2008).

CONJECTURE

If G is a rational group and P 2 Syl2(G ), then P is rational.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 66 / 67

IV. Rational Groups

Two are the main problems on rational groups.

CONJECTURE

If Cp is a composition factor of a rational group, then p 5

J. Thompson has proved that p 13 (2008).

CONJECTURE

If G is a rational group and P 2 Syl2(G ), then P is rational.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 66 / 67

IV. Rational Groups

THEOREM (Isaacs, N)

There is a rational group G of order 29 3 such that P 2 Syl2(G ) is

not rational.

Gabriel Navarro (University of Valencia) Problems in Representation Theory of Groups Bilbao, October 8, 2011 67 / 67