Groups and Representations
Contents
Chapter 1: Groups ................................................................................................................................ 3
1.1
Basic Definitions ................................................................................................................... 3
1.2
Groups of Matrices............................................................................................................... 3
1.3
Cyclic Groups ....................................................................................................................... 4
1.4
Symmetric Groups and Alternating Groups .................................................................... 4
1.5
Dihedral Groups................................................................................................................... 4
1.6
Homomorphism and isomorphisms. ................................................................................ 5
Chapter 2: Representations ................................................................................................................. 6
2.1 Representations of Cyclic Groups ........................................................................................... 7
Chapter 3: Generators and Relations............................................................................................... 10
3.1 Generating Sets ......................................................................................................................... 10
3.2 Normal Subgroups................................................................................................................... 10
3.3 Quotient Groups ...................................................................................................................... 11
3.4 Free Groups............................................................................................................................... 13
3.5 Presentations of Groups .......................................................................................................... 17
Chapter 4: Modules............................................................................................................................ 20
4.1 Group Actions .......................................................................................................................... 20
4.2 Representations afforded by
-modules ............................................................................ 21
4.3 Submodules .............................................................................................................................. 22
4.4 Inner Products .......................................................................................................................... 24
Chapter 5: Characters ........................................................................................................................ 27
5.1 Reminder on Conjugacy Classes ............................................................................................ 28
5.2 Schur’s Lemma and Orthogonality ....................................................................................... 30
Chapter 6: Regular Representations ................................................................................................ 33
Chapter 7: Character Tables ............................................................................................................. 36
7.1 Definition and first Examples................................................................................................. 36
7.2 Properties of Character Tables ............................................................................................... 37
7.3 Characters and Normal Subgroups ....................................................................................... 38
7.4 Linear Characters and the Derived Subgroup ..................................................................... 40
7.5 More Examples ......................................................................................................................... 41
Chapter 8: Restriction and Induction .............................................................................................. 44
8.1 Notation and basic properties ................................................................................................ 44
8.2 How to compute an induced character in practice ............................................................. 45
8.3 Induction and Restriction for modules ................................................................................. 48
Chapter 9: Algebraic Integers and Burnside’s Theorem .............................................................. 50
9.1 Algebraic Integers .................................................................................................................... 50
9.2 Burnside’s Theorem ................................................................................................................. 51
Chapter 1: Groups
1.1 Basic Definitions
Group
A Group is a set together with a binary operation
such that the following conditions hold:
mapping
(
)
( )
( )
1. For any
2. There exists
so that for all
,
3. For
there exists
so that
Notation

Let

The number of elements in , denoted
order of .
be a group. For
Order of element
Let
. Then the order of
no such integer exists.
Abelian Group
Two elements
all
,
write
instead of ⏟
or | | is a number in
is the least positive integer
in a group
.
Subgroup
A subgroup of a group
conditions hold:
and
is a non-empty subset
. A group
or
. Note that
such that the two following
is a group in its own right.
1.2 Groups of Matrices
Notation
)
Let (
multiplied e.g.
( )
. Two such matrices can be
(
(
)(
)
(
)
) are called inverses of each other if
)
Let (
multiplication.
( ) be the set of invertible elements of
the identity matrix.
(
if
is Abelian if for
1.
2.
If these conditions hold, write
) .
and called the
such that
are said to commute if
(
and
). This is a group under
Fact:
(
.
)
( )
(
. For
) is not abelian while for
(
)
1.3 Cyclic Groups
Cyclic Group
A Group is said to be cyclic if there exists
called a generator of (it may not be unique).
such that
. In this case
is
Proposition 8 (Classification of Cyclic Groups)
Let be a cyclic group with generator . Then
a. If
b. If
then
. Then
are all the elements of
with no repetitions.
In particular, two cyclic groups are isomorphic if and only if they have equal order.
Notation
Let the Cyclic group of order
(up to isomorphism) be denoted by
,
.
1.4 Symmetric Groups and Alternating Groups
Permutation/Symmetric Group
A permutation of a set is a bijective map from to itself. The Symmetric Group
set of permutations of
with composition the binary operation.
is the
) or simply (
) is defined to be the
If
are distinct then (
element
known as a -cycle such that ( )
for all (indices modulo ) and
( )
whenever
.
Alternating Group
For
the subgroup
alternating group and written
is the subgroup containing all 3-cycles. It is known as the
.
1.5 Dihedral Groups
The set of integers modulo
is denoted
. It is a commutative ring.
Lemma 11
We define
to be the set of mappings from
to itself of the form
where
and
. Then
is a subgroup of the symmetric group on
called
the dihedral group.
Proof
It is clear that
is nonempty and
is a map from
to itself with inverse
hence every element of
is a permutation and thus
. Let
(
)
( )
(
) which
say ( )
and ( )
then ( )
is again of the required form so
. Observe that
and ( )
(
)
so
. Therefore
.
Generating Elements of
We define
to be the unique elements ( )
and ( )
.
Lemma 13
We have
Proof
Since
and
Then observe that ( )
all
hence
and
( )
.
it is enough to prove only one inclusion.
which are both elements of
for
We may refer to
being a rotation and
being a reflection. Therefore every element of
is either a rotation or a reflection but not both.
1.6 Homomorphism and isomorphisms.
Homomorphism
Let
be groups. Then a function
called a homomorphism.
Isomorphism
Let
be groups. An isomorphism
least one isomorphism
we write
such that (
)
( ) ( ) for all
is
is a bijective homomorphism. If there exists at
and say that
are isomorphic.
Direct product of Groups
Let
be groups. The direct product
is the set of pairs (
)( ) (
and multiplication is defined component-wise: (
this makes
into a group.
) where
and
). It is easy to show that
Chapter 2: Representations
Representation
Let
, a group. A representation of
(
). Note that (
)
of dimension
is a homomorphism
Examples
(
) and set ( )
a. Let
dimensional trivial representation.
b. Let
(
and
for all
. This is a representation; the -
) be defined by ( )
(
). This is a
representation.
Lemma 20
Let
(
( )
( )
Proof
Let
(
be a representation and take
. Then is also a representation.
. Then (
)
(
)
) and define
( ) ( )
( )
(
( )
Equivalent Representations
(
) are equivalent if there exists
Two representations
( )
( )
for all
. Denote this as
) by
( ) ( ).
(
) such that
Lemma 22
Equivalence of Representations is an equivalence relation.
Proof
Clearly
If
If
as ( )
then ( )
and
( )( )
( )
( )
then ( )
for all .
so
( )
( )
( )
and ( )
( )
hence ( )
( )
Reminder on Set Theory
Let be an equivalence relation on a set . An equivalence class with respect to or a class is a subset of of the form
where
and is the disjoint union
of all equivalence classes. Then
{ -
}.
Notation
Let
( )
{ -
The main aim of this course is to analyse
| |
.
}
and
( )
∐
( ) and later we will see that
( )
( )
if
2.1 Representations of Cyclic Groups
Lemma 23
Let and be two diagonal
matrices with elements on the diagonal denoted by
and
respectively. Then the following are equivalent:
(
) so that
1. There exists
) ∏ (
)
2. ∏ (
3. There exists a permutation
so that
Proof
[
] If ∏ (
)
the same roots; i.e.
that
( ) for all
∏
()
for all
.
(
) then the two polynomials are equal and hence have
up to reordering. Then simply choose a permutation
so
.
[
] If
then and have the same characteristic polynomial. As and
) and ∏ (
)
both diagonal, their characteristic polynomials are ∏ (
respectively and so these are equal.
[
] Let
be the standard basis of
is diagonal. Similarly, as is diagonal,
invertible matrix representing the linear map
()
Let
()
()
for
()
and define
for
. Then
for
for all
. Then define
( ) . Then observe that
hence
are
because
to be the
( )
.
( )
. We aim to list
Lemma 24
( ) with
a. Let
. Then there exists a unique representation
that ( )
and it satisfies ( )
b. Every representation of is of the form
c.
if and only if
are conjugate.
of
such
for all .
for some
Proof
( ) is uniquely determined by its action on the
a. Since a homomorphism
generator , the representation
sending ( )
is unique and ( )
for
all .
( ) is a homomorphism. Then
b. Let be a representation of ; that is
( )
( ). As a homomorphism is uniquely determined by its
for some
( )
action on , we see that ( )
and ( )
for all . Hence
( )
.
c. Suppose
. Then for all
, ( )
. In particular, ( )
( )
( )
and so and are conjugate.
( ).
Conversely, suppose and are conjugate, that is
for some
Then for all
(
)
(
,
for all
so
)
.
hence
(
)
Lemma 25
(
Let
) be a matrix of finite order. Then
Proof
Say that
for
not an eigenvalue of
(
is diagonalisable.
. We may suppose that is in Jordan Normal form. Note that
as
. If is not diagonalisable, consider a Jordan Block
). This can be done because the eigenvalues
easy to prove that for all
)
Now
so the rows above the diagonal are
contradiction so is diagonalisable.
(
. As
) so
. For
[
{
and
(
this is a
)
denote
]
[
{
. By induction it is
is of the form
(
Put
is
]
( )
[
{
{
Note that
]
(
[
]
)
( )
Proposition 25 (Classification of Representations of )
( ) be the set of -tuples (
)
Let
such that
( ) mapping
there exists a bijection
.
( )
. Then
Proof
1.
is well defined:
(
) (
)
( ) so ( )
Let
. Then ( )
( ).
2.
is injective:
( ) so ( )
( ) and ( ) are conjugate by Lemma 24
Suppose ( )
( ) . Then
and so ( ) and ( ) have equal characteristic polynomials by Lemma 23. I.e.
∏ (
∏ (
for all
)
) . It therefore follows that
and so
.
3.
is surjective:
By Lemma 24 every -dimensional representation of is of the form
where
(
). To prove
and
is in the image of , by Lemma 25, is
diagonalisable so is conjugate to a diagonal matrix . Also
so all diagonal
entries of are powers of . Say
appears times for
so is
∑
conjugate to ( ) where
so
( ) by Lemma 24. Hence ( )
.
( )
Example
Let
and define
(
exists a unique representation
what is
(
)?
) by
with indices in
of
. Then
(namely:
(
)
so there
for all
)
Solution:
(
Put
∑
∑
) and
∑
(
then
)
∑
has distinct eigenvalues
with
because
( )
is
we have
is an eigenvector of
because
and so the eigenvalue is
so all eigenvalues of
(
)
with
( ).
. So
are not repeated and
Chapter 3: Generators and Relations
3.1 Generating Sets
Claim
For a group , let
where
. Then ⋂
Group generated by a subset
Let be a subset of a group then ⟨ ⟩
(this makes sense because
⋂
and
).
Remarks:
1. By the claim ⟨ ⟩
. We say is a generating set for ⟨ ⟩.
⟨ ⟩
2. Every Group has a generating set as
Finitely Generated Group
A Group is finitely generated if there exists a finite set
⟨
⟩
Remark
⟨ ⟩ is the smallest subgroup of
such that
containing .
Proposition 29
Let
be a subset of a group . Then ⟨ ⟩
{
}.
Proof
Let
⟨ ⟩: Let
. Observe that
so
then
.
Also observe that ⟨ ⟩
by definition of ⟨ ⟩.
Conversely, let
say
then
⟨
⟩.
⟨ ⟩.
group operations so
Therefore
Example
⟨ ⟩( ( )
( )
⟨ ⟩ and ⟨ ⟩
so it is closed under
) by an earlier example.
3.2 Normal Subgroups
Notation
Let
be subsets of a group . Then
Cosets
Let
. A left coset (respectively a right coset) of
the form
(Respectively
(Respectively
. For
with respect to is a subset of
) and
of
Claim
Let
a.
b.
. Then:
∐
(
(
)
)
) will be denoted [
From now on, (
Normal Subgroup
Let
. is a normal subgroup of
] and called the index of
if
for all
in .
and denoted
Lemma
The following are equivalent:
1.
2.
3.
4.
is normal in
for all
Example
( )
but not a normal subgroup because (
( ) ( )
b. In an abelian group, every subgroup is normal.
a.
Claim
Let
where
)(
)(
)
(
)(
)(
)
. Then ⋂
Normal Closure
⟨⟨ ⟩⟩ ⋂
Let be a subset of a group . Then ⟨⟨ ⟩⟩
. It is a normal subgroup
⟨⟨
⟩⟩
because of the Claim above. So
is the normal subgroup of generated by or the
normal closure of . In particular, a group is simple if it has no other normal subgroups but
and itself.
3.3 Quotient Groups
Let
. Then
Claim
If
Proof
Say
then
and
It is true that
group.
Examples:
a.
.
.
for some
( )( )
. Then as is normal,
( )
( )
, together with multiplication of cosets (defined above) makes
into a
b.
c. Let
⟨ ⟩ and
Claim
It is true that ⟨ ⟩
⟨
⟩. Then
and
.
.
Proof
We consider the sets of left cosets and right cosets. ⟨ ⟩ ⟨ ⟩ and ⟨ ⟩ ⟨ ⟩ . As
is the
disjoint union of each of the elements of the sets, we observe that since ⟨ ⟩ ⟨ ⟩, we must
⟨ ⟩. Therefore ⟨ ⟩ is normal.
have ⟨ ⟩
Kernel and Image
Let
be a homomorphism. Then the kernel of denoted
( )
and the image of is ( )
. It is true that
Proposition 38
Let
. Then the natural map
homomorphism with kernel .
Proof
is surjective: take a coset
is a homomorphism:
Take
then ( )
:
By definition
taking ( )
. Then clearly ( )
( )
and
( )
.
is a surjective
.
( ) ( )
now
so
.
Theorem 39 (First Isomorphism Theorem)
Let
be a homomorphism with kernel . Then there exists a unique isomorphism
( ) so that
. That is (
)
Proof
Standard proof.
Commutative Diagram
The following diagram is said to commute if
( )
3.4 Free Groups
⟨ ⏟
The expression
|⏟
(
) ⟩ determines
uniquely; this is a presentation for
.
Alphabets and Letters
Let be a set, called an alphabet. Elements of
are called letters (or generators).
Word of Length , Length, Empty Word
A word of
is a -tuple of the form (( )
of length
(
)) where
and
for all . We denote by 1 the “empty word” which is the unique word of length 0. Denote by
( ) the length of the word .
An alternative notation is
. Most people use this despite it causing confusion.
Multiplication of words
We multiply by a process called concatenation; the product of
(( )
(
)) ((
that (
)
( )
((
)
(
)
( )) is simply (( )
( ). The inverse of (( )
(
(
) (
)
( )). Note in particular
)) is defined to be
)).
Remark
The set
is not a group because
multiplication of words is associative.
in general, however
One-Step Reduction
a. Let
and
. We say that
is a 1-step reduction of both
and
. We denote this by
and
b. Let be the reflexive transitive closure of ; equivalently
if and only if there
exists a sequence of 1-step reductions
.
c. Similar notions are
in the usual way.
Ordering of a set
An ordering on a set
is a relation
such that:
1. Reflexivity:
for all
2. Transitivity:
and
implies
3. Antisymmetry: If
and
then
.
( ) so
( )
( ) then
( )
Note that
implies ( )
( )
( ) so ( )
( ) and no letters are removed so
. Therefore the relation
defined by 1-step reduction is an ordering on .
Total Ordering
A total-ordering on a set
4. For any
is an ordering
either
or
on
which additionally satisfies:
.
Reduced Word
We call
a reduced word if
with
(
).
denoted by
( )
Equivalently,
is neither of the form
. The set of reduced words in
nor
for
is
and
.
Lower Bound/ Reduced Lower Bound
) if
In an ordered set (
then we say that is a lower bound of . So a reduced lower
( ) such that
bound (RLB) of
is any reduced
.
Example
Let
Lemma
Let
be distinct letters. Then
so
is an RLB of .
. Then
has a unique RLB denoted ( ).
Proof
Existence:
Take
then consider the set
as
,
. Choose an
element
of least length ( ). This is possible because ( )
by the Well-Ordering
( )
Principle. Then is an RLB, because otherwise there exists
with
so ( )
contradicting the minimality of ( ).
Uniqueness:
We apply induction on ( ). It is clear if ( )
(since then every word is a RLB of itself).
Let ( )
. Assume all shorter words have a unique lower bound. Let
be RLBs of .
We will prove that
.
By definition, there exist
such that
and
. We may assume
and
since otherwise we are done.
Say there is overlap if some letter in
is removed in both arrows
Assume there is no overlap; then there exist
then
and
diagram
with
. Denote
and
.
and
then we have the
Let be a RLB of
. Now,
are both RLBs of ; hence by induction hypothesis
( )
( ). Similarly
are both RLBs of
hence by induction hypothesis
( )
( ). Therefore
.
If there is overlap, assume
and
by the induction hypothesis
Exercise
Let
. Then ( ( ) )
. Then there exists
and
so
. As
( ). As required.
as ( )
(
)
(
as
as
such that
are RLBs of
,
( )) where ( ) is the unique RLB of .
Reduced concatenation
( ). Then define the binary operation
Let
( )
( )
( ) by
(
)
Example
(
)
Theorem 49
The set of reduced words on an alphabet
(
)
equipped with , ( ( ) ) is a group.
Proof
( )
for all
(in the middle there are one-step reductions until everything cancels and you get
1. Likewise for
.
( ). Then
(
)
))
For associativity, let
( (
( ( ))
(
)
( (
) )
((
)
)
(
)
Example 51
( )
a.
b. Claim
(
)
Proof
( ) consists purely of powers of hence generated by
( ) are all distinct so
( )
⏟
so cyclic. Moreover for all
( )
More precisely define a map
by
. Then (
( )
( ) so is a homomorphism. We can define
is an isomorphism.
)
(
as
)
so
Claim
( )
( )
| |
| |
Proof
[ ] If | | | | then there exists a bijection
. We then extend this bijection
( ) ( ).
into an isomorphism of the Free Groups by defining ( )
[ ] Since ( )
( ) there exists an isomorphism which in particular is a bijection.
( ) defines a bijection from to some subset
Restricting this isomorphism to the set
of ( ). As is an isomorphism on the free groups, the set is a generating set for ( )
so | | | | | | as required.
In conclusion the free group (
denote by the free group with
) depends up to isomorphism only on , so we can
generators, or free group of rank .
Unproved Theorem
Every subgroup of a free group is again isomorphic to a free group.
Exercise
⟨ ⟩ and
Let
be groups,
extending ; that is ( )
Solution
Define ( )
Take
( )
( ) for all
. Then
( ) ( )
a map. Then there is at most one homomorphism
( ) for all
.
.
(
for some
and
. Then define
clearly is a homomorphism so is determined uniquely.
)
Theorem 52 (Universal Property of Free Groups)
( ), the free group of . In particular,
Let be a set. Define
. Let
and
’ be a function. Then there exists a unique homomorphism
( )
( ) for all
.
be a group
such that
Proof
Uniqueness:
This follows from the Previous Exercise.
Existence:
Define
by mapping
to
(
Clearly ( )
( ) for all
and
and
( ) for all
( )
)
( )
. To prove
. Then
; that is
( )
is a homomorphism, let
say
So ( )
(
)
( )
(
( ) whenever
( (
))
(
)
( ) ( ) ( ) ( )
( ) ( )
( )
. By induction, ( )
( ( )) so for all
)
( )
( ),
( ) ( ).
Remark
( ) can be characterised by the universal property.
We never need reduced words again.
3.5 Presentations of Groups
Group Presentation
) where is a set and
( ).
a. A group presentation is a pair (
) we associate a group ⟨ | ⟩
( ) ⟨⟨ ⟩⟩
b. For any group presentation (
We say that this group (or any isomorphic group) is presented by this group
)
presentation (
c. We allow various ways of writing down relations. For example
are all equivalent.
⟨ | ⟩ taking ( )
⟨⟨ ⟩⟩. Often we identify
There is a natural map
identity map. Watch out for confusion if is not injective.
Also one simply writes down a word in
with the
and considers it an element of ⟨ | ⟩.
Corollary 55
Let be a group then:
a. There exists a surjective homomorphism from some free group to .
b. There is a presentation for .
Proof
a. Let be a generating set of (if necessary we can take
)
Then let ( ) be the free group generated by . By the Universal Property of Free
( )
groups there exists a unique homomorphism
such that ( )
for all
. Then the image of contains hence contains ⟨ ⟩
because ( )
.
b. Put
. Then ⟨⟨ ⟩⟩
because the kernel of a homomorphism is a normal
subgroup. Then using the first isomorphism theorem on from part a,
( )
( )
( )
⟨ | ⟩
( )
⟨⟨ ⟩⟩
Finitely Presented
A group is finitely presented if it has a presentation (
finite subset of ( ).
Example 56
For
we have
⟨ |
⟩
) where
is a finite set and
is a
Proof
( )
because
as we saw before. Take
is surjective so ⟨ |
is abelian, and
. Then ⟨⟨
map
⟩
( )
⟨
⟩
⟩⟩
⟨
⟩
by the first isomorphism
theorem.
Satisfying Relations
) be a presentation,
Let (
( )
(
)
a group and
whenever
a map. We say that
satisfies
if
.
Theorem 60 (Universal Property for group presentations)
) be a presentation, a group
Let (
a map. The following are equivalent:
1. There exists a homomorphism
2.
satisfies
⟨ |
⟩
such that ( )
Moreover, if these conditions hold, then the homomorphism
( ) for all
is unique.
Proof
Non-examinable.
Example 62
⟨
Prove
(
|
) ⟩
Proof
⟨
( ) ⟩. Then recall that
Let
|
is (by definition) generated by the functions
where ( )
, ( )
. Then observe that
satisfy the relations
( ) by taking
and
. For example
( )
( )
( )
(
)
(
)
so ( )
.
( )
Similarly
.
By Theorem 60, there is a unique homomorphism
mapping ( )
and
( )
. Note that is surjective because ( )
and ( ) contains
and
is
generated by
by an earlier example.
Let
can be written as a word in and . Say
. We may assume
without loss of generality that no exponents in the middle are 0. Using the fact that
and ( )
we find
Using this we can “push” all the ’s to the right in an expression for
for
. Using that
and
|
|.
Example
Prove there is a unique presentation
(
)
(
(
so
) such that ( )
is of the form
so | |
and ( )
where
)
Solution
We apply Theorem 60. From the above example we know
⟨
|
(
) ⟩. We need
to prove that
(
) so
1.
(
)
⟨
|
(
)
(
) ⟩. It can be shown that:
) is finite
[ (
2.
Examples:
(
)
and (
)
satisfy the relations.
the triangle group is (
For
(
satisfy the relations. Observe
(
)
)
]
this is an exercise on sheet 2,
(
)
)
Chapter 4: Modules
We assume now on that all vector spaces are over .
4.1 Group Actions
Action of a Group on a Vector Space
Let be a vector space and a group. An action of
(
)
such that:
1. For all
the map
is a map
mapping
mapping
is linear; that is
)
( )
( )
( )
( )
( )
( ) for all
and
,
(
Equivalently, (
( )
( ) for all
2.
( )
3.
for all
on
)
Example 65
Let
,
⟨ |
⟩. Let
be a vector space with basis {
mapping (
Define
linear maps.
)
} so
on basis vectors and extend
by
Claim
This is an action
Proof
Condition 3 is trivial. Condition 1 is true by hypothesis.
(
)
( )
( ) holds for the basis vectors. But
As to 2: (
)
( ) are linear functions of these basis elements; if they agree on
both sides of ( ( ))
the basis elements then they agree on the entire vector space.
-module
Let be a group. A
-module is a vector space together with a -action on it.
Remark
The ring
is called the group algebra such that
-modules as we define them are the
“same” as modules over
in the sense of ring theory. Hence
has no meaning for this
course, but
-modules do have meaning.
-Homomorphism
Let
be
-modules. A
and
( ( )) for all
Example 69
Let
and ( )
(
)
homomorphism is a linear map
such that ( ( ))
take
and
. Make
into
-modules by (
. It is clear that this makes and into
-modules. Define
. This is a
-homomorphism because
))
(
)
(
)
))
( (
( (
)
(
)
by
4.2 Representations afforded by
-modules
Reminder From Linear Algebra
(
) and a vector space with basis
Let be a vector space with basis
(
). To a linear map
⟩ ( ) by ( )
as associate a matrix ⟨
∑
⟨
⟩ is a bijection from
(
)
. Then
to
and:
1.
2.
⟨
⟩⟨
⟩ ⟨
⟩ whenever this makes sense
If is the standard basis of
and
then ( )
represented matrix multiplication).
Notation
If is a
-module and
( )
( ) for all
.
then
⟨
⟩
( )
(the here
is the linear map defined by
Summary:
-modules and representations are the same thing.
Lemma 71
Let be a
( ) ⟨
-module with basis . Define
⟩. Then is a representation of .
(
) (where
) by
Proof
Firstly
for all
because:
( )
(
)
(
)
( ) for all
Also ( ) ( )
because:
( ) ( ) ⟨
⟩⟨
⟩ ⟨
( )
⟩
Next, we use that
for all
. Then
)
so ( )
so for all
, ( ) (
( ) is invertible with inverse ( )
(
).
(
⟨
⟩
)
( )
)
Representation afforded by (
Given a (finite dimensional) vector space with basis and
(
) defined by mapping ( ) ⟨
in Lemma 71
).
afforded by (
Lemma 73
Every representation is afforded by some pair (
basis.
Proof
(
Let
where ( )
) a representation. Put
(
)
Claim
This is an action of
on
.
) where
. For
(
)
so for all
,
. The representation
⟩ is the representation
is a
and
-module and
, define
is a
( )
Proof
For all
,
(
)
( ( )
)
( ) ( )
(
)
So
(
is now a
-module. Let be the standard basis of
) gives rise to some representation where ( ) ⟨
⟩
( ). So is afforded by (
).
so ( ) ⟨
Lemma 74
The representations afforded by (
) and (
(
)
( )
and
. Then by Lemma 71
⟩. Then ( )
( )
) are equivalent.
Proof
) and (
) respectively. Put
Let and be the representations afforded by (
⟨
⟩, the matrix of the identity with respect to and . For all
we have
( )
Hence
and
⟨
⟩⟨
⟩⟨
⟩
⟨
) and (
) are equivalent if and only if
Proof
[ ] Let be the representation afforded by (
) and by (
(
) so that ( )
( )
equivalent so there exists
⟨
⟩. Note that
exists a linear map
such that
invertible. Then for all
⟩
( )
( )
⟨
⟩⟨
⟩⟨
[ ] Let
so
.
). We know they are
for all
. Hence there
is bijective since is
⟩
( ) for all
That is
. Thus ( )
. This shows that
of
-modules. It is an isomorphism because is bijective.
⟨
⟩
is a homomorphism
( ) for all
be the isomorphism of
-modules. Then ( )
and
so let be the invertible matrix corresponding to its linear map. Then
( )
Hence
( )
are equivalent.
Lemma 75
The representations afforded by (
⟨
⟩
and
⟨
⟩
⟨
⟩
⟨
⟩⟨
⟩⟨
⟩
( )
are equivalent.
Intertwiner
is an intertwiner from
to
if ( )
( ) for all
.
4.3 Submodules
Submodules
A submodule of a
-module
is a subspace
Example
⟨ ⟩ act on
Let
by the formula (
1-dimensional submodule of .
)
such that
(
for all
). Then we claim
(
) is the only
Observe ( ) ( ) so it is indeed a submodule. Consider any 1-dimensional submodule
) which is not in
(
) so we get a
of the form ( ). Then ( ) (
contradiction.
Simple Submodule
A
-module is said to be simple if it has no submodules other than 0 and itself.
Example
In particular, every 1-dimensional
-module is simple.
Internal Direct sums
Let be a vector space and
subspaces. We say
if the following equivalent properties hold:
is an internal direct sum
1. Every element of can be written uniquely as
where
2.
and
3. There exists a basis of and a basis of such that
of
Furthermore, if
4.
5.
and
and
is a basis
then the following two properties are also equivalent:
and
and
External Direct Sums
The External Direct Sum of
and
is the set
(
⏟
In practice we assume any two vector spaces
)
and
(
⏟
)
have a direct sum
.
Diagonal Sum/ Irreducible Representations
(
) and
(
) be representations. The diagonal sum
) dimensional representation defined by
( )
(
)( ) (
)
( )
(
) is said to be irreducible if it is not equivalent to any
2. A representation
representation of the form:
1. Let
is the (
( )
(
)
̆
̆
Claim
) is a diagonal sum of two representations of dimension
The representation afforded by (
>0 if and only if there exist non-zero submodules
such that
Corollary
The representation afforded by (
) is irreducible if and only if
is simple.
Dictionary
-module
-homomorphism
Simple
Direct Sum
Representation
Intertwiner
Irreducible
Diagonal Sum
Theorem 81 (Maschke)
Let be a finite group, a finite dimensional
exists a submodule
such that
-module. For every submodule
there
In order to prove Maschke’s Theorem we need to digress into Inner Products. The proof will
be given later.
4.4 Inner Products
Inner Product
An Inner Product or IP on a vector space
that:
1. Bilinearity: ⟨
⟩
⟨
mapping (
is a map
⟩
⟨
⟩ for all
and
)
⟨
⟩ such
.
⟩ ⟨
⟩
2. Hermitian: ⟨
⟩
3. Positive Definite: ⟨
for all
Remarks:


By 2, ⟨
⟨
⟩
⟨
⟩
⟩ so ⟨
⟨
⟩
⟩
⟨
so the third condition makes sense.
⟩ is a consequence of and
Claim
Every finite dimensional vector space has an IP.
Proof
Let
be a basis and put
⟨∑
∑
⟩
∑
It is then clear that this defines an inner product.
Conversely, every IP on a finite dimensional vector space is of this form; we won’t need this
or prove it.
Orthogonal Complement
Let be a vector space with an inner product ⟨ ⟩ and a subspace
⟨ ⟩
complement of is
Note that
is a subspace.
. The orthogonal
Lemma 85
Let be a finite dimensional vector space with an inner product ⟨ ⟩ and let
subspace. Then
.
Proof
Let
.
be a basis of
. Then ⟨
. Let
All we need to prove is that
⟨
⟩
⟨
⟩
by ( ) (
).
Then
so
. Hence
. Let
⟨
⟩
is a linear map and
be a basis of
. Define
so
-invariant Inner Products
Let be a
-module. An inner product ⟨ ⟩ on
and
.
Proposition 87
Let be a finite group and
inner product on .
⟩
be any
is -invariant if ⟨
a finite dimensional
Proof
Let ⟨ ⟩ be any inner product on . For
⟨
⟩
⟨
⟩ for all
-module. Then there exists a -invariant
define
⟩
⟩
∑⟨
We claim that this is an inner product on :
For any
:
⟨
⟩
∑⟨ (
)
⟩
∑ ⟨ ( )
For any
⟩
⟨ ( )
( )
⟩
⟨
⟩
⟩
⟨
⟩
:
⟨
For any
⟩
⟩
∑⟨
⟩
∑⟨
⟨
⟩
:
⟨
⟩
⟩
∑⟨ (
) (
)⟩
⟩
∑⟨
We now show that ⟨ ⟩ is -invariant. Let
⟨
( )
∑⟨
∑⟨(
,
⟨
⟩
. Then:
) (
) ⟩
∑⟨
⟩
⟨
⟩
We are now able to prove Maschke’s Theorem:
Theorem 81 (Maschke)
Let be a finite group, a finite dimensional
exists a submodule
such that
-module. For every submodule
there
Proof
Choose a -invariant inner product ⟨ ⟩ on , which exists by Proposition 87. Put
Lemma 85
. It is left to prove that
is a submodule.
. By
Let
⟨
and
⟩
⟨
so
⟩
. Then ⟨
because
⟩ because ⟨ ⟩ is -invariant. So
because is a submodule. Hence
.
Corollary 88
Let be a finite group and
simple submodules.
Proof
Let
⟩ ⟨
and
a finite dimensional
-module. Then
is a finite sum of
. We proceed by induction on .
For
, we are done. Let
; if is simple we are done so suppose not. Then there are
non-zero submodules
by Maschke’s Theorem so that
. Then in
particular
and
so by induction hypothesis they are direct sums of
simple submodules:
and
then
as required.
Chapter 5: Characters
( ) with
For a matrix
Claim
Let
(
) then the trace is
( )
∑
( ). Then:
( )
( )
1.
2. If is invertible then
(
)
( )
Proof
(
1. Write
(
)
∑
2. Put
) and
(
∑
) . Then the (
)th entry of
which is symmetric in
(
so
is ∑
)
so
(
)
in part 1.
If
is a basis of a finite dimensional vector space and
a linear map, we define
( )
⟨
⟩ [Note that this does not depend on the choice of basis because ⟨
⟩
⟨
⟩⟨
⟩⟨
⟩ and then use part 2 of the previous claim to see ⟨
⟩
⟨
⟩].
Character
Let be a group.
1. The character of a representation of is
which maps ( )
2. The character of is the character of some representation of
3. The character of a finite-dimensional
-module is
mapping
Note that if (
) affords
then
.
Proposition 93
Let
be representations of . If
Proof
Let
(
( )
(
) such that ( )
)
( ( ))
then
( )
for all
( )
. Then
( ( ))
( ).
Remark
This is why characters are so useful; they are independent of equivalence class of
representations. Later we’ll prove the converse:
In particular, note that
Proposition 95
Let be a character of a group . Then (
Proof
Say
( ( ))
. Then (
( )
)
( ).
(
)
)
( (
( ) for all
))
.
( ( ) ( ) (
))
( ( ))
( )
5.1 Reminder on Conjugacy Classes
Two elements
in a group are conjugate if there exists
with
. The
relation
if and are conjugate is an equivalence relation with the equivalence classes
are called conjugacy classes.
Put
then
so
Conjugacy Classes
Define ( )
is the conjugacy class of
and ( )
in .
| ( )|.
Easy Exercises:
Let
then it is clear that:
( )
; that is there exists an automorphism
mapping
( )
; that is conjugation defines an action of on itself.
( )
Let be a group and
then ( )
( ). Define
( )
Let
. Then
1.
2.
3.
4.
Centre of a Group
The centre of a group is the set ( )
( ) then is called central.
If
Lemma 95.1
Let be a generating set of a group
The following are equivalent:
1.
for all
2.
for all
3. For every
Example 96
Recall that
are
⟨
,
and
( ), either
|
. For
where
, define
.
or
( ) ⟩. Prove that if
(
( )
so
)
is odd then the conjugacy classes of
and
Solution
is always in ( ). Now
so
is contained in a
(
) as soon as
conjugacy class by Lemma 95.1. Then
for all
. Well
)
)
and (
and
and (
so is a conjugacy class. By the Easy
exercise 4 above,
is also a conjugacy class; they are all distinct if
We have left to prove: Any two elements of
(
So
but
.
are conjugate.
)
and
odd so all elements of
are conjugate.
Example
By the Universal Property, it is easy to show that there exists a unique representation
(
) mapping ( )
(
) and ( )
(
). We want to compute
. By
( ) for only one
Proposition 95, it is sufficient to compute
example, (
)
{
from each conjugacy class. For
} hence we construct the table:
( )
( )
For a group
operations:
denote the set of functions
(
)( )
( ), define (
For
Lemma 99
( )
( )
( )
)
( )
| |
∑
. We make it into a vector space by pointwise
( )
( )
( ) ( )
is an inner product.
Proof
It is a simple matter of checking the conditions in the definition.
Class Function
A class function on a group is a function
such that (
; that is is constant on the conjugacy classes. Let ( )
( ) so
( ) by Proposition 95.
)
( ) for all
Remark
( ) is a subspace of ( ) of dimension ( ).
Lemma 101
Let be a representation of a finite group
1.
2.
( )
( ) is a sum of roots of unity.
3.
(
)
and
.
( )
Proof
( )
1.
( ( ))
(
)
2. Let
be the eigenvalues of ( ). Now | |
so | |
say
. Then
for all ; that is
is a root of unity for all . Therefore
is a sum of roots of unity.
) are
3. The eigenvalues of (
which are
because
. So
(
)
Degree of a Character
The degree of a character
( (
))
is ( ). So
( ( ))
by part 1 of Lemma 101.
for
( )
for all
( )
Irreducible Character
An irreducible character is the character of some irreducible representation. Let ( )
.
5.2 Schur’s Lemma and Orthogonality
Claim
Let
Proof
Let
( )
Let
( )
Let
( )
Hence
be a
-homomorphism. Then
we have ( )
Then for
( )
so
. Then
( )
(
)
. Then (
( )
( )
and
and
( ) and
) so
so (
(
are submodules.
). Then (
( ) for some
.
)
( ) so
)
. Then
. Take
. Then
so
are submodules.
Theorem 102 (Schur’s Lemma)
Let
be simple
-modules. Then:
1. Every
-homomorphism
2. Every
-homomorphism
number.
is either 0 or a
-isomorphism.
is scalar multiplication by some complex
Proof
1. Suppose
. Then
is a submodule of but not so
hence is
injective. Similarly,
is a submodule of but not so
hence is
surjective.
2. Note
because is simple. Let
be an eigenvector with eigenvalue
.
Put
by ( )
. It is clear that is a
-homomorphism since it is a
linear combination of
-homomorphisms. But ( )
so is not
injective. Hence by part 1, is constantly 0 so
for all
hence is scalar
multiplication by scalar .
Lemma 103
Let
be representations of a finite group of degrees
respectively. Take
( ) and
∑
(
) ( ). Then is an intertwiner.
Proof
We must prove that ( )
( )
( ) for all
( )∑ (
∑ (
)
( )
)
. Well
∑ (
( ) ( )
)
( )
( )
∑ (
)
(
)
Retranslating this Lemma in terms of
another linear map
( )
by
-modules, it says “for a linear map
∑
) ( ). Then
(
define
is a
-
homomorphism.”
If
denote its entry in position (
is a matrix let
). Let
be the number of rows of
which equals the number of rows of . Then
(
)
(
∑
)
∑
Theorem 105 (Orthogonality for Characters)
Let
with (
be irreducible representations of a finite group
)
| |
∑
( ) ( ).
Then:
(
)
Proof
Let ( ) be the (
Then ( ( ) )
)
1.
2. (
Put (
)
)
∑
∑
(
Moreover, ∑ ( (
∑
∑
∑
(
(
1.
2.
)
)
)
(
)
for
) matrix with a 1 in position (
)
.
(
) (
))
∑
| |(
) and zeroes elsewhere.
) ( ) which is an intertwiner from
∑ ∑
( )
( )
(
(
(∑
( (
) (
) ( ))
(
) )(∑
( ) )
to
∑
∑ ( (
∑
(
by Lemma 103.
) (
)
) ( ))
( )
)
is an intertwiner between non-equivalent irreducible modules then by
Schur’s Lemma ( )
so the above shows that | |(
)
(
)
is an intertwiner so by Schur’s Lemma (
and
. Then
| |(
)
∑( (
∑ (
))
)
∑∑ (
∑ (
Hence (
∑(
)
∑∑ (
) (
) ( )
) ( )
) is a scalar matrix (
∑(
) (
)
)
∑
) ( )
∑ (
)∑ (
) ( )
| |
)
Corollary 108
Let be a finite group. Then
has at most ( ) non-equivalent irreducible representations.
Proof
( ).
Suppose not. Take
to be non-equivalent irreducible representations with
( ) and
( ) for all , so the are linearly dependent, say
But ( )
∑
and not all
(say
). Then
(
)
(∑
)
(
∑
)
Contradiction.
Later we will prove the converse: ( )
representations.
the number of non-equivalent irreducible
Theorem 109
Let
be a maximal set of non-equivalent irreducible representations of a finite group
. Then for each there exists a unique
such that
where
is
the diagonal sum of copies of .
Moreover,
(
) and is called the multiplicity of in .
Proof
Existence:
This follows immediately from Corollary 88.
Uniqueness:
Assume
and write
(
)
(∑
. Then
)
∑
Corollary 110
Let be a representation of a finite group . Then
(
)
is irreducible if and only if (
)
Proof
[ ] This was proved in Theorem 105 (Orthogonality of Characters).
[ ] Write
(∑
with all
∑
one of them, say
)
∑
. Then
(
non-equivalent. Then
)
∑
and hence
and
is irreducible.
(
. So
)
for all except
Chapter 6: Regular Representations
( ) be a vector space with basis
Let be a group. Let
(a copy of ). So
( ) can be written uniquely as a linear combination ∑
every element of
for
, with only finitely many
.
mapping ( ∑
Define
( )
(∑
)
. This is clearly an action: (∑
) ∑
( )(∑
).
(∑
)
)
∑
∑
. In particular
and ( (∑
))
So
is a
-module called the regular module. From now on, assume is finite.
) is
Choosing a total ordering on
, the representation afforded by (
called the regular representation
. Its character
is called the regular character.
Example 112
Take
( )
let
(
) similarly,
Lemma 114
Let be a finite group. Then
( )
(
( )
) then
(
)
(
| | and
(
) and
)
( )
for all
.
Proof
( )
| |.
Firstly,
( ) is in position (
). But this is not on
Let
. Then any nonzero entry of
( )
( )
the diagonal as
so the diagonal consists of zeroes; hence
.
Lemma 115
Let
be the irreducible characters of a finite group . Put
.
⨁
Proof
(
) . By Theorem 109, we have
Let
( )
Also using Lemma 114 above,
unless
(
)
| |
∑
( ) (
⨁
, so:
)
| |
( ). Then
.
( ) ( )
Corollary 117
Let
be the irreducible characters of a finite group , and
| |
.
Proof
By Lemmas 114 and 115, we have | |
( )
∑
( )
( )
( ). Then
∑
.
Theorem 118
Let ( )
. Then | ( )|
Proof
| ( )| and
Let
.
( ). We know that
by Theorem 108. We need to prove that
( ) and
( )
Recall ( )
and
from
). It is sufficient to prove: for any
( )) implies
.
Let
be a representation of
( ) has an inner product (the one restricted
( ) and
( ) (that is (
)
∑
and define
Claim
is an intertwiner; that is
( )
( )
( ) (
( )
).
for all
.
Proof
We use the fact that
( )
is a bijection from
( )
( ) (∑ ( ) (
∑ (
) (
)) (
)
( ):
and
)
∑ ( ) (
∑ ( ) (
)
)
Let be an irreducible representation, so
is an intertwiner between irreducible
itself. By Schur’s Lemma,
is a scalar multiple
for some
and
find
(
)
(
| |(
Now,
(∑ ( ) (
)
))
∑ ( ) ( (
( ). Hence
by assumption, because
| |(
)
( )
(
( ) for
)
∑ ( )
and so
.
for some irreducible
for all representations . In particular it is true for the case
Now, the matrices
.
∑ ( )
)
Observe that if is not irreducible, we can write
and
is a linear map so by the above argument;
Hence
))
and
. To
(
. Then
)
are linearly independent (look at the first row). Hence
Linear Character:
A Character is linear if it has degree 1.
Theorem 119
A finite group
is abelian if and only if all its irreducible characters are linear.
Proof
( ) | ( )|. Let
Let
Corollary 117 we have | |
| |
be the irreducible characters of
and
. Therefore:
and
( ). By
Chapter 7: Character Tables
7.1 Definition and first Examples
( ) and
( ), we often write ( )
( ).
Let be a finite group. If
, for
Let
be the conjugacy classes of , and suppose
. Let
be the
irreducible characters and
the trivial character; ( )
for all .
The matrix ( ( )) is the character table of .
There are non-isomorphic groups with the same character table;
examples.
Example 122
⟨ | ⟩. By Theorem 119,
Let
putting
√
and
has four irreducible linear characters
are such
. Then
we have:
Example 123
Lets compute the character table for
. We know ( )
. First we look for
⟨ |
( ) ⟩ a linear character is just a
linear characters of . Recall that
homomorphism from
. By the Universal Property we are looking for
(the
( ) . Solving for
images of
) satisfying the relations in the presentation;
gives
. We get two rows:
∑
Let
be the last row. We compute ( ) by
∑ ( ) so
find the last row by
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
. We
7.2 Properties of Character Tables
Proposition 124
Let
be irreducible characters of a finite group
(
1.
2.
and
) is again an irreducible character of called the dual of .
( ) ( ) is again an irreducible character of called a twist.
Proof
Let be a representation so that
.
( )
( ( ) ) then is a representation because ( )
1. Define ( )
( )
( ( )
( ) )
( )
( )
( ) ( ) and ( )
( )
.
( ( ) )
( ( ))
( (
is a character of . To prove it is irreducible:
Moreover,
so
( )
(
)
| |
∑ ( ) ( )
| |
))
∑ ( ) ( )
(
)
(
)
(
)
Hence
2. Define
( ) (
( )
( )
is irreducible.
( )
( ) ( ). Then is a representation because ( )
( ) ( )
) ( ) ( )
( ) ( ) because ( )
(it is a linear character). Also
( ) ( )
( ) so is a representation of .
( ) ( ( ))
( ) ( ) ( )( )
Therefore ( )
( ( ))
( ( ) ( ))
so
is a character of . It is irreducible because
(
| |
)
∑
| |
∑
( ) ( )
( ) ( ) ( ) ( )
(
)
so
| |
∑
( ) ( ) ( ) ( )
is irreducible.
Remark
The product of two characters is always a character. The proof of this is out of scope.
Lemma 125:
( ). The following are equivalent:
Let
1. The rows of
are orthonormal.
2.
3.
4. The columns of
are orthonormal.
Proof
If
and
are clear from linear algebra.
, then
so there exists
(
)
(
. Therefore
)
Theorem 126 (Row and Column Orthogonality)
Let be a finite group and
be its irreducible characters. Let
conjugacy classes and
| |. Then for
be its
:
1. (Row Orthogonality):
∑
( ) ( )
| |
2. (Column Orthogonality):
∑
( ) ( )
| |
Proof
1. By Theorem 105 (Orthogonality of characters)
| |
∑
( ) where ( )
( ) (
) is a class function of . Taking terms
from the same conjugacy class together gives the result.
(| |)
2. Write
(
). The rows of
are orthonormal by part 1. By Lemma
125, so are the columns; so we get the expression in part 2.
Example 127
Let be a group, | |
Where
(
and ( )
such that one of its characters is
). We will find the full character table for .
Solution
The trivial character is always there.
is also a character, so we have
∑ ( )
( ) so
Let
be the last row. We find ( ) by:
To find the remaining entries we use orthogonality of columns, getting ( )
( )
( )
( )
.
and
7.3 Characters and Normal Subgroups
Kernel of characters:
Let be a representation of . We define
defined because a representation is determined by
( )
(this is well
up to equivalence and equivalent
characters have the same kernel.)
Lemma 129
Let be a character of . Then
( )
( )
Proof
Let be a representation of
[ ] Take an alement
( )
( ).
[ ] Let
so that
Then ( )
.
so ( )
( ( ))
( ), so
( ). By Lemma 101, there are roots of unity
so that ( )
( )
such that ( )
but ( )
so
. Then as ( ) is diagonalisable, ( )
.
Theorem 130 (Lifting)
Let
be a homomorphism of finite groups with kernel
1. There exists a map
by ( )
2. If is surjective then
where
hence
.
defined
is bijective.
Proof
1. Let
be a character of , say
where is a representation of . Then
is a representation because the composition of two group homomorphisms is a
( ). Also, the kernel of
homomorphism, and its character is
contains
( ).
the kernel of , so
2.
is surjective:
Let be a character of with
say
. Define a representation by
( ) which is well defined because ( )
( )
( )
( ) and is
( ( ))
surjective. Then ( )
.
is injective:
( ). Then
Let ( )
so
.
Theorem 131
Let be a finite group.
1. For irreducible characters
of ,
2. All normal subgroups of arise this way.
Proof
( )
1. Clearly
since is a homomorphism. The intersection of normal
subgroups is again a normal subgroup, so the result follows.
2. Let
and
be the natural epimorphism. Let be any injective
representation of
(for example take
). Then
is a representation of
with kernel by Theorem 130. Write
with irreducible. Put
(
)
(
)
then
.
7.4 Linear Characters and the Derived Subgroup
Commutator
The commutator of and in a group
commutator subgroup of a group is
.
is [
]
[
]
⟨[
. The derived subgroup or
]
]
⟩. Note [
Example 133
1.
2.
is abelian if and only if
and
if
Proposition 134
Let be a group. Then
1.
2. Let
then
is abelian if and only if
Proof
1. Let
For
2.
]
. We prove [
as follows:
is an automorphism of so
[
]
(
)
(
[
]
is abelian
for all
[
]
.
[
]
) (
)
( )( )( ) ( )
[
]
for all
Proposition 135
Let be a group.
1. A linear character of
is the same thing as the lift to
2. The number of linear characters of
of a linear character of
is the same as | |
Proof
1. [ ] is clear.
[ ] Let be a linear character of . Then is a homomorphism
and is
abelian so ( ) is abelian. By Proposition 134 part 2,
and so by Theorem
130 is a lift (that is
for a character of
and
the natural
map). Clear that
since ( )
so is linear.
2. We know that
is abelian so by Theorem 119, all of its irreducible characters are
linear, and it has | | conjugacy classes so
has precisely | | linear characters. By
Part 1, there is a bijection between the linear characters of and the linear characters
of
so in particular they have the same number of linear characters.
Example
If ( )
then ( ( )
( ))
(
).
Suppose it does. By Proposition 135 | |
divides | |. But | |
so
which is a contradiction.
7.5 More Examples
Example 136
Some generalities on . There exists an index 2 subgroup
, so there exists a linear
character
called the sign character or sign representation.
Let
be the permutation representation of
( )( )
()
|
( )
|
The character of
is the map
Let
. First we compute ( ) and the sizes of the conjugacy classes:
(
)
(
)
(
)
(
)(
)
| |
Let be the trivial character. Let
be the sign character. Let be the permuation character.
Unfortunately, is not irreducible. Consider
then
(
)
(
)
(
)
(
)(
)
| |
Observe that (
Define
)
(
)
so
is irreducible.
then
is an irreducible character since it is a twist. Observe
by looking at the table.
We have one more character call it . We find ( ) by
We find the remaining entries by Column Orthogonality:
(
)
(
)
(
)
(
)(
| |
∑
( ) so
)
| |
Example 137
Take
( )
{
Denote particular elements of
1. Claim
⟨ ⟩
Solution
by
where ( )
}
and ( )
.
( )
.
( )
form
( )
( )
so
are in . Then the composition
(
and so transformations of the
( ))
( )
⟨
hence
⟩.
2. Claim
Solution
( )
( )
( )
(
)
, ( )
,
( )
3. Claim
Any element of can be written uniquely as
where
and
Solution
From part 1, we actually proved that every element of is of the form
then as
we can reduced
into the intervals
. Uniqueness follows from comparison with | |
.
where
and
Remark
In fact,
4. Define
( ) by (
)
Claim
is a homomorphism and there exists a homomorphism
( )
and ( )
Solution
Let
with ( )
and ( )
. Then
so ( )
and ( ) ( )
But ( )
⟨ | ⟩ with
( )
(
)
which defines .
5. Claim
The Conjugacy classes of
are
Solution
( )
( )
( )
is always a conjugacy class. Observe
and
is certainly invariant under conjugation of . We do not need to check
anything else since is generated by
.
(
)
(
)
Therefore is contained in a conjugacy class. Observe ( )
and ( )
so
as two conjugacy classes are either equal or disjoint, they partition the finite group .
Hence
are conjugacy classes as required.
6. We now find the character table for :
Lifting the four irreducible characters of through the homomorphism yields 4
∑
( ) and then use Column
linear characters of . We can then find
by
Orthogonality:
| |
Chapter 8: Restriction and Induction
8.1 Notation and basic properties
Let
( )⏟⏞
be finite groups. We define maps
( )
( )
( )
( )
(
∑
| |
)
( )
{
( )
A more flexible notation is the following:
For an assertion , denote by [ ]
( )
{
. Then
| |
∑ (
]
)[
With the convention that
Proposition 138
and
are again class functions.
Proof
For
this is trivial, since it is a projection of a class function.
For
, take any
then
(
)
| |
∑
(
)
Proposition 139 (Frobenius Reciprocity)
( ) and
( ). Then (
Let
, let
| |
(
∑
)
(
)
( )
)
Proof
(
)
| |
∑
( )̅̅̅̅̅̅
( )
| || |
| |
∑
| || |
(
( )
∑ ( )̅̅̅̅̅̅
Exercise
Let be a finite group,
Then is a character of
∑
(
]̅̅̅̅̅̅̅̅̅̅̅̅̅
(
)
)[
| |
]̅̅̅̅̅̅
( )
)[
( )
∑ ( )̅̅̅̅̅̅̅̅
( ).
if and only if for all
( )(
(
| || |
∑
( )[
]̅̅̅̅̅̅
( )
)
)
Proof
The irreducible characters
are an orthonormal basis of
) is a character if and only if (
)
if and only if ∑ (
.
( ). Now
.
is a character
Corollary 140
Let
. Let
be a character of . Then
is a character of .
Proof
( ). Then (
)
(
) by Frobenius Reciprocity. Since
Let
are character of
, the inner product is an element of
. Then by the exercise,
is a character of .
8.2 How to compute an induced character in practice
It is very easy to compute the induced character of the trivial conjugacy class:
Proposition 141
[
( )
] ( )
Proof
We proceed by direct calculation:
( )
| |
∑ (
]
)[
Proposition 142
Let
be finite groups,
contained in . Then
| |
( ). Let
we say that
( ). Let
[
( )
Moreover, if
∑ ( )[
]
| |
]
| |
( )
| |
[
] ( )
be the conjugacy classes in
∑| | ( )
splits.
Proof
Claim
For
,
}|
|{
| |
| |
| |
Proof
Define
(
. Then
( )
. So |
by
)
( )| is independent of conjugate
elements. Therefore
|
So |
|
∑
|
| |
| |
( )|
( )|
| |
|
| |
|
We now complete the proof:
( )
| |
∑ (
| |
By the claim:
)[
∑|
]
| |
∑ ∑[
| ( )
] (
)
( )
| |
∑
| || |
( )
| |
[
]
| |
∑| | ( )
Example 143
Computing induced characters using 142 and “induction tables”. Namely, we induce to
the irreducible characters of and . The character tables for these groups are the
following:
(
)
(
) (
)
First, for each conjugacy class of we write
, or rather, one element of . For each
( )
conjugacy class of we write | | in the same column as the conjugacy class
such that
.
Then ( ) ( )
(
(
(
(
)
)
(
]
| |
∑
| | ( )
)
)
)
Next for : now there is splitting; (
only splitting occurring).
(
(
(
(
[
( )
, and for the other elements use
)
(
(
(
)(
)
)
)
)
)
)
Example 145
We will compute the full character table for
⟨ ⟩
(
).
) are conjugate in
but not in
(
(
) ( ) [
) (( ))
(
) ((
))
using induction from
. (It’s the
] ( )
(
)
(
)
and
. Write
a. Let be a conjugacy class in . Prove that at most two conjugacy classes of
are in
.
Solution
Let
and assume that
are not conjugate in
and
are not
either;
for
so
hence ( )
so
are
conjugate in .
b. For each conjugacy class in
or , find its cardinality and give one element.
Solution
(
Similarly, for
)
(
)
)(
)
(
(
)(
)
,
(
(
Need to prove
)
(
)
(
)
)
(
)
. Suppose they are; then there exists
with
. By replacing with
( )
Then
.
c. Let be the following linear character of
(
)
we may suppose ( )
for
. Compute (
(
)
(
)(
) and
.
.
)
( )
d. Prove
is a character for
e. Prove that
and
are irreducible.
)
)
f. Find (
and (
(
g. Let
). Let
be the linear character of
h. Prove that
i. Finish the character table.
such that ( )
.
is an irreducible character.
Solution
( )( )
( )( )
(
(
(
(
)
)
)
)
)
(
(
)
)
)
We need to prove that ((
(( )
)
(
(
)
(
(
)(
)
:
)
)
( )( )
So
is conjugate to
(
)
that
(
)(
)(
)
)
so (
(
)
(
)
Since (
is irreducible.
Claim
We claim that
Proof
so ((
(
and (
( )( )
in . So
. Since
(
)
)
)
(
)
)
(
so
is irreducible.
)
)
so (
)
and so
)
( )( )(
)( )(
is conjugate to ( )
(
),
(
)
so
)
(
)
. It remains to prove
(
)(
To prove that
(
)
Define
)
(
(
)
(
)
)(
)
)
)
is an irreducible character, we need to prove (
. This is straightforward and I do not include it here.
. Let be an automorphism of
. It is clear that
is irreducible because
(
(
(
)
(
mapping ( )
is. Then
)
(
)
)
(
)
(
) (
)
). Define
Therefore the full character table is as follows:
(
)(
)
(
)
(
Now we can find all normal subgroups of . Observe that
( )
( )
and
.
Since every normal subgroup of
is a of the form ⋂
(
( ), we see that the only normal subgroups of
are
(
)
)
(
)
(
)
(for some subset
. That is,
is simple.
8.3 Induction and Restriction for modules
Definition 146
Let
be finite groups, a
-module. We define
to be a
character ( ) . This is well defined (up to isomorphism) because:
-module with
1. ( ) is a character of .
2. A
-module is determined by its character.
Definition 147
Let be a finite group, a
-module. An imprimitivity condition or ID is a sequence
(
) of non-zero subspaces (not submodules) such that:
1.
2. Invariance:
3. Transitivity:
It is a proper ID if
We are interested in
to by transitivity.
.
is called primitive if it has no proper ID.
(
)
. Note that
(
) is conjugate
Proposition 148
Let be a
-module with character
(
Then
) ,
Proof
Write
. Let
and with ID
(
. Put
).
is a character of .
be a left-transversal for
. Then as runs through ,
in ; that is
runs through the
∐
and
. Put
, so:
⨁
Let (
( ) (for
) be a basis for . Then
) using this basis.
The contribution to ( ) of
is 0 unless
; that is
and
fixed. Write
From now on, assume
(
) ∑
Now (
of
( )
)
∑
(
∑∑
∑
is
for all . Then
. So the contribution
)
∑ ∑
( )
| |
.
so the contribution to ( ) of
(
). So
is ∑
∑
is a basis for . We calculate
(
)
⏞
| |
∑∑
(
∑
)
⏞
)
( )
.
| |
∑
(
)
( )
Proposition 148.1 (Not in notes)
Let
, a
-module. Then
(
On which
acts by (
)( )
(
) for all
( )
Proof
Non-examinable. See 8.13 for the details.
( ) is meaningful for | |
called the induced module.
and
and
replaced by any commutative ring, and
Chapter 9: Algebraic Integers and Burnside’s Theorem
Easy Exercises:
1. Every abelian finite simple group is of prime order and thus cyclic.
2. Let be a finite simple group of order
for prime. Then
and
The smallest non-abelian finite simple group is
and |
.
|
Theorem (Burnside)
There is no finite simple group of order
where
are distinct primes.
(The proof will use characters, although they are not mentioned in the proof.)
9.1 Algebraic Integers
Lemma 150
∑
a. A polynomial
is said to be monic (of degree ) if
b. A complex number is called an algebraic number if there exists
.
[ ]
such
that ( )
. The set of algebraic numbers is denoted .
c. A complex number is called an algebraic integer (AI) if there exists a monic
[ ] such that ( )
polynomial
. The set of algebraic integers is denoted .
[ ], monic of least degree so that ( )
d. Let
then there exists a unique
It is called the minimum polynomial (MP) of .
e. Two algebraic numbers are said to be conjugate over if they have the same
minimum polynomials.
.
Proof
We only need to prove , but this follows by applying the Well-Ordering Principle to the
[ ] ( )
degrees of the set
. Note that this set is non-empty because is
an algebraic number, and we can multiply through by a constant to make monic without
changing the degree.
Example
√
√ have the same minimum polynomial
Example
(
) then every complex eigenvalue
Let
(
).
so they are conjugate.
of
is in because it is the root of
Theorem 152
a.
b.
c.
is a subfield of
is a subring of
[ ] be such that ( )
d. Let
and
divides in [ ].
e. The MP of an
is in [ ]
. Then the minimum polynomial of
f.
Let
. Then every conjugate to
conjugate and
are conjugate.
is of the form
where
are
Proof
The proof of all these statements belongs to a course in Algebraic Number Theory or Galois
Theory.
Example
Observe that the converse of f is false; take
conjugate and
are conjugate. But
same minimum polynomial.
√ and
√ and
√ . Then
are
which do not have the
Examples
be any complex root of 1, say
. Then
because ( )
where
.
2. Let be conjugate to . Then is also a root of 1. Let be the MP of (hence also of
) then | in [ ] so ( )
because ( )
.
⟨ ⟩ ⟨ ⟩; that is they
3. Conversely, two roots of 1, say and are conjugate
generate the same multiplicative group. We won’t need to prove this.
1. Let
9.2 Burnside’s Theorem
Lemma 154
Let be a character of a finite group
and
. Then
( )
1.
2. If
is irreducible then
( )
|
( )
|
Proof
a. We know that there are roots of unity
such that ( )
follows because
and is a ring.
b. Write
and
. For any representation put ( )
( ). Then
is an intertwiner because for all
,
( )
( )
( )
(∑
Since is irreducible, it follows that
scalar matrix; that is
. Also
|
| ( ( ))
|
( )) ( )
∑
(
)
∑
. The result
∑
( )and
∑
( )
is irreducible and so by Schur’s Lemma
∑
( ) ∑
( ( ))
is a
| ( ). Therefore
( )
|
( )
|
We are left to prove
. By Lemma 115
for some representation . So
(
)
(
). But (
)
(
)
every eigenvalue of
is also an eigenvalue of
| |. So every eigenvalue of (
) is in , hence so is .
where
Theorem 155
Let be an irreducible character of a finite group . Then ( )|| |.
Proof
be the conjugacy classes of . By row orthogonality, ∑
Let
| |.
| | ( ) ( )
Divide by ( ); by 154
| |
( )
( )
∑| |
( )
( )⏟
⏟
Therefore the LHS
Clearly,
| |
( )
so
| |
( )
because is a ring therefore
| |
( )
.
so ( )|| |.
Proposition 156
Let be an irreducible representation of ,
is a scalar matrix where
( |
. If
|)
( )
then
or ( )
.
Proof
( ). Choose
Write
|
such that
|
. Multiply both sides by , then
by Lemma 154
⏟( )
|
|
be the eigenvalues of ( ). Then
Let
|
( )
|
⏟( )
| |
hence
(
) also
are roots of 1 so
. We now consider the three cases:
Case 1
| |
so
Case 2
| |
Case 3
In this case
( )
and so
( )
.
. As ( ) is diagonalisable hence ( ) is a scalar matrix.
| |
. We will deduce a contradiction, so this case never occurs.
[ ] because
∏ (
) and say
Let be the MP for . Then
and
( ) ( )
Therefore
.
Let
then
is conjugate to . By 152 part f,
is of the form
(
.
)
where is some conjugate to . But
is a root of and every conjugate of a root of 1 is
| |
itself a root of 1, so is a root of 1 and hence | |
. As before
because has
MP so
.
|
|
Multiplying over all and using that | |
we have
which is a
contradiction because
.
Proposition 157
Let be a prime number and
with a conjugacy class of size
. Then there is no non-abelian finite simple group of
.
Proof
Let be a non-abelian finite simple group and
Let
be irreducible character of with
simple so
or . If
Claim
( ) is not a scalar matrix unless
an element with | |
. Note
trivial. Write
for all . Now is
then
is not trivial so
.
.
Proof
( ) has a non-trivial central element ( ). (Note ( )
If it were then ( )
( ).
because
and
). As
we have
( ). As is simple, ( ) is simple hence ( ( ))
Therefore
( ( ))
( ) so ( ) is abelian hence is abelian. Contradiction.
By 156 if
and
∑
then
( )̅̅̅̅̅̅̅
( )
∑
( )
( )
. By orthogonality of columns,
∑
( )
∑
( )
|
Therefore ∑
⏟( ) ⏟
. Hence as is a ring,
. Clearly,
so
.
Contradiction.
Theorem 158 (Burnside’s
Theorem)
Let
be distinct prime numbers and
group of order
.
. Then there is no non-abelian finite simple
Proof
Let be such a group. Then ( )
. is simple, so ( ) {
( )
. So there is just one conjugacy class of size 1, namely
conjugacy classes of and
. As | ||| | by 157, | if
After interchanging and if necessary, we may assume | ∑
| (| |
∑| |)
}. is non-abelian, so
. Let
be the
. Therefore | ∑ | |.
| |. Therefore
.