II Representation Theory - Integrality in the group algebra

13Integrality in the group algebra

II Representation Theory

13 Integrality in the group algebra

The next big result we are going to have is that the degree of an irreducible

character always divides the group order. This is not at all an obvious fact. To

prove this, we make the cunning observation that character degrees and friends

are not just regular numbers, but algebraic integers.

Definition

(Algebraic integers)

A complex number

a ∈ C

is an algebraic integer

is a root of a monic polynomial with integer coefficients. Equivalently,

such that the subring of C given by

Z[a] = {f(a) : f ∈ Z[X]}

is finitely generated. Equivalently,

is the eigenvalue of a matrix, all of whose

entries lie in Z.

We will quote some results about algebraic integers which we will not prove.

Proposition.

(i) The algebraic integers form a subring of C.

(ii)

a ∈ C

is both an algebraic integer and rational, then

is in fact an

integer.

(iii)

Any subring of

which is a finitely generated

-module consists of algebraic

integers.

The strategy is to show that given a group

and an irreducible character

the value

|G|

χ(1)

is an algebraic integer. Since it is also a rational number, it must

be an integer. So |G| is a multiple of χ(1).

We make the first easy step.

Proposition.

is a character of

and

g ∈ G

, then

(

) is an algebraic

integer.

Proof.

We know

(

) is the sum of roots

th roots of unity (where

is the order

). Each root of unity is an algebraic integer, since it is by definition a root of

−

1. Since algebraic integers are closed under addition, the result follows.

We now have a look at the center of the group algebra

. Recall that the

group algebra

of a finite group

is a complex vector space generated by

elements of G. More precisely,

CG =







g∈G

g : α

∈ C







Borrowing the multiplication of

, this forms a ring, hence an algebra. We now

list the conjugacy classes of G as

{1} = C

, ··· , C

Definition

(Class sum)

The class sum of a conjugacy class

of a group

g∈C

g ∈ CG.

The claim is now

lives in the center of

(note that the center of the

group algebra is different from the group algebra of the center). Moreover, they

form a basis:

Proposition.

The class sums

, ··· , C

form a basis of

(

). There exists

non-negative integers a

ij`

(with 1 ≤ i, j, ` ≤ k) with

`=1

ij`

The content of this proposition is not that

can be written as a sum of

’s. This is true because the

’s form a basis. The content is that these are

actually integers, not arbitrary complex numbers.

Definition

(Class algebra/structure constants)

The constants

ij`

as defined

above are the class algebra constants or structure constants.

Proof.

It is clear from definition that

−1

. So we have

∈ Z

(

Also, since the

’s are produced from disjoint conjugacy classes, they are linearly

independent.

Now suppose z ∈ Z(CG). So we can write

z =

g∈G

By definition, this commutes with all elements of

. So for all

h ∈ G

, we must

have

−1

= α

So the function

g 7→ α

is constant on conjugacy classes of

. So we can write

= α

for g ∈ C

. Then

g =

j=1

Finally, the center Z(CG) is an algebra. So

`=1

ij`

for some complex numbers

ij`

, since the

span. The claim is that

ij`

∈ Z

≥0

for all

i, j`

. To see this, we fix

∈ C

. Then by definition of multiplication, we

know

ij`

= |{(x, y) ∈ C

× C

: xy = g

}|,

which is clearly a non-negative integer.

Let

G → GL

(

) be an irreducible representation of

over

affording

the character

. Extending linearly, we get a map

CG → End V

, an algebra

homomorphism. In general, we have the following definition:

Definition

(Representation of algebra)

Let

be an algebra. A representation

of A is a homomorphism of algebras ρ : A → End V .

Let

z ∈ Z

(

). Then

(

) commutes with all

(

) for

g ∈ G

. Hence, by

Schur’s lemma, we can write

(

) =

for some

∈ C

. We then obtain the

algebra homomorphism

= ω

= ω : Z(CG) → C

z 7→ λ

By definition, we have ρ(C

) = ω(C

)I. Taking traces of both sides, we know

g∈C

χ(g) = χ(1)ω(C

We also know the character is a class function. So we in fact get

χ(1)ω(C

) = |C

|χ(g

where g

∈ C

is a representative of C

. So we get

ω(C

) =

χ(g

)

χ(1)

Why should we care about this? The thing we are looking at is in fact an

algebraic integer.

Lemma. The values of

) =

χ(g)

χ(1)

are algebraic integers.

Note that all the time, we are requiring χ to be an irreducible character.

Proof.

Using the definition of

ij`

∈ Z

≥0

, and the fact that

is an algebra

homomorphism, we get

)ω

) =

`=1

ij`

Thus the span of

{ω

(

) : 1

≤ j ≤ k}

is a subring of

and is finitely generated

as a

-module (by definition). So we know this consists of algebraic integers.

That’s magic.

In fact, we can compute the structure constants

ij`

from the character table,

namely for all i, j, `, we have

ij`

|G|

)||C

s=1

)χ

−1

)

(1)

where we sum over the irreducible characters of

. We will neither prove it nor

use it. But if you want to try, the key idea is to use column orthogonality.

Finally, we get to the main result:

Theorem. The degree of any irreducible character of G divides |G|, i.e.

(1) | |G|

for each irreducible χ

It is highly non-obvious why this should be true.

Proof. Let χ be an irreducible character. By orthogonality, we have

|G|

χ(1)

g∈G

χ(g)χ(g

−1

)

χ(1)

i=1

|χ(g

)χ(g

−1

)

i=1

|χ(g

)

χ(1)

χ(g

)

−1

Now we notice

|χ(g

)

χ(1)

is an algebraic integer, by the previous lemma. Also,

(

−1

) is an algebraic

integer. So the whole mess is an algebraic integer since algebraic integers are

closed under addition and multiplication.

But we also know

|G|

χ(1)

is rational. So it must be an integer!

Example.

Let

be a

-group. Then

(1) is a

-power for each irreducible

If in particular

|G|

, then

(1) = 1

, p

. But the sum of squares of the

degrees is

|G|

. So it cannot be that

(1) =

. So

(1) = 1 for all

irreducible characters. So G is abelian.

Example.

No simple group has an irreducible character of degree 2. Proof is

left as an exercise for the reader.