Group Theory
Contents
Chapter 1: Review ................................................................................................................................ 2
Chapter 2: Permutation Groups and Group Actions ...................................................................... 3
Orbits and Transitivity .................................................................................................................... 6
Specific Actions – The Right regular and coset actions .............................................................. 8
The Conjugation Action .................................................................................................................. 8
Chapter 3: Sylow’s Theorems ........................................................................................................... 11
Direct Products ............................................................................................................................... 14
Group Presentations (From Algebra II) ...................................................................................... 15
Semi Direct Products ..................................................................................................................... 15
Groups of Small Order .................................................................................................................. 16
Classification of Groups
with
...................................................................................... 17
............................................................................................................................................ 17
.............................................................................................................................. 18
.......................................................................................................................................... 19
.......................................................................................................................................... 19
.......................................................................................................................................... 20
.......................................................................................................................................... 21
.......................................................................................................................................... 21
Chapter 4: Nilpotent and Soluble (Solvable) Groups ................................................................... 22
Commutators and Commutator Subgroup ................................................................................ 24
Characteristic Subgroups .............................................................................................................. 25
Soluble Groups and Derived Series............................................................................................. 26
Composition Series and the Jordan-Holder Theorem .............................................................. 28
Chapter 5: Permutation Groups and Simplicity Proofs ................................................................ 31
Normal Subgroups of Permutation Groups ............................................................................... 33
The Finite Simple Groups (Classified ~1981/2004) .................................................................. 35
Properties of Finite Fields ......................................................................................................... 36
A Closer Study of
(
) for small ................................................................................... 41
Chapter 6: The Transfer Homomorphism ...................................................................................... 43
Chapter 7: Classification of Simple Groups of order
500 ......................................................... 46
Chapter 1: Review
Notations


or
means that is a subset of (not necessarily subgroup), while
or
means that is a subgroup of .
{
} is a right coset while
{
} is a left
If
and
,
coset. We know from Algebra II that
and if is
finite, | | | | and the distinct right cosets partition .
Theorem 1.1 (Lagrange’s Theorem)
If is finite and
then | ||| |.
Index of a subgroup
| |
|
The value |
is called the index of
Normal Subgroup
is called a normal subgroup of
for all
in
and is denoted |
and is denoted
for all
|
if
.
Quotient Group
If
we can define the quotient group
operations
{
} the set of cosets with group
and
{ } and
All groups have normal subgroups { } and . A group is called simple if
are the only normal subgroups (this is equivalent to having exactly two normal
subgroups).
and
Abelian finite simple groups are exactly those groups which are cyclic of prime order. The
classification of finite simple groups was completed in 1981. The aim of this module is to
classify all finite non-abelian simple groups of order up to 500 with proofs. It turns out there
are only three such examples which have orders 60, 168 and 360.
First we recall some more statements proved in Algebra II.
Proposition 1.2
If
then subgroups of
are of the form
Homomorphism
A map
is a homomorphism if (




)
with
(
) (
and
) for all
.
. In addition:
( )
is a monomorphism if ( )
is an epimorphism if ( )
is an isomorphism if it is both a monomorphism and an epimorphism
is an automorphism if is an isomorphism with
Kernel
If
is a homomorphism then the kernel of
( ).
Note that
Claim
is a monomorphism iff
Claim
For
given by ( )
Theorem 1.3 (First Isomorphism Theorem)
Let
be a homomorphism and denote
( )
iii.
The map
( )
{
}.
( )
( )
, the “quotient” map
i.
ii.
denoted
is an epimorphism.
( )
(not necessarily normal)
( ) defines an isomorphism
( ). Hence
( )
Order of an element
For
, the order of denoted | | is the least
such that
or if there is no
}
such . For
, the set ⟨ ⟩ {
is the cyclic subgroup generated by .
Remark
If | |
, finite then ⟨ ⟩
{
}
while if | |
then ⟨ ⟩
}
⟩ is the subgroup generated by
More generally for {
,⟨
and is
defined to be the intersection of all subgroups containing
which is equivalent to the
set of all products of any length of and
Cyclic Group
A group is called cyclic if it is generated by one element for example
| |
| |
{
}
| | | |
{| | | | | | | |
Chapter 2: Permutation Groups and Group Actions
Let
be a set.
Permutation
A permutation of is a bijection
( ) where | |
and
. Define
Notation
( ) and
If
write the image of
so
means first apply then .
under
( ) be the group of all permutations
as
not ( ), so we get
(
)
Example
{
Take
} and denote
( ) as follows:
(
)(
)
(
)(
)
In cyclic notation
(
)(
)
(
)(
)
So
and
have a cycle of length 2 and a cycle of length 3 hence
( )
cycle type? Why?
is a conjugate of :
For
above, | |
( ) denote | |
| |
| |
o len ths o
| |
Transposition
A permutation of the form (
and
have the same
is oint y les. Therefore in the example
) is a transposition .
Lemma 2.1
If is a finite set, any permutation of
generated by transpositions.
is a composition of transpositions. I.e.
is
Proof
( ) and express it as a product of cycles, so it is enough to express cycles as a
Take
composition of transpositions:
(
) (
)(
) (
)
Even and Odd Permutations
( ) is called even or odd if
transpositions respectively.
is a product of an even or odd number of
Theorem 2.2
No permutation is both even and odd.
Proof
Not instructive, but in the lecture notes.
Clearly, e en e en
homomorphism
odd.
o
( )
o
e en and o
e en e en o
o , so there is a
{
} where ( )
if is even and ( )
if is
( )
( )
Observe that
the First Isomorphism Theorem.
{
( )
Also by the First Isomorphism Theorem if | |
,
|
( )|
( )
( )|
hence |
We will later prove that
( )|
is simple for
|
( ) by
is e en} is a subgroup of
( )
( )
( )
( )
( )
so
Group Actions
We did these in Algebra II, but will use right actions not left actions and use different
notation:
An Action of a Group
on a set
(
definition of an action, we have
the map
and
(since (
is a permutation of
mapping (
)
and
(not
). By
for all
)
. For
, we have
as inverse maps so every permutation is a bijection).
( ) is a homomorphism since
Then
( ) by
Denote
( )
is a map
)
( ) for all
(the action of
{
(
) .
on ). By the First isomorphism,
and
( )
}
Faithful Action
The action on
is called faithful if
( )
or equivalently if
Dihedral Group
Recall that
is the dihedral group of order
(which is often confusingly denoted by
Example
: Consider the regular hexagon below
Define three actions of
On
e
es
on subsets of : actions on
(
)
( )( )
{
ia onals
Observe that
and
{
}
}
(
(
On
erti es
{
)
)(
)(
(
(
)
)
}
are faithful but
)
is not. It has kernel {
} and in fact,
).
Equivalent Actions
Actions of on sets
( )
( )
are called equivalent if there is a bijection
and
.
By examining the cycle type of the three actions on
are inequivalent.
such that
, we conclude that all actions of
Orbits and Transitivity
Let act on . Define
for
if there exists
such that
be an equivalence relation. The equivalence classes are called the orbits of
of is written as .
Example
{
} and take
Let
}{ }{
The orbits of ⟨ ⟩ are {
}{
⟩ are {
The orbits of ⟨
so
hence
. Similarly,
orbit but applying
to
{
}.
. Similar for
Transitive
is called transitive if
there exists
Observe that
⟨
. This is clear to
on . The orbit
(
)( )( ) and
( )( )( ) where
.
} { } while the orbits of ⟨ ⟩ are { } { } { } { } { }.
}. This is deduced because
so
also
because
all give results
so
are all in the same
so {
is an orbit; that is if
so that
.
} is
for all
. Equivalently, for all
⟩ in the previous example is not transitive.
-transitive
For
,
is called -transitive if:
1. | |
2. Given , define pints
that
for
and distinct
. Then there exists
so
.
Observe that 1-transitive is the same as transitive and (
)-transitive implies -transitive.
Examples
1.
2.
}
is -transitive on {
}
)-transitive on {
, the alternating group is (
Proof
Take distinct
and
. Then
{
where
and
are the other two points. Also
Certainly there exists a unique
with
for
are done. Otherwise let
be the transposition (
and
Stabiliser
Let act on
and
{
}
}.
. If
) then
then we
and so
for
. Define the stabiliser of
in
as
( )
{
}
and
( ) is a subgroup of :
( ) then
Let
subgroup.
(
so
)
Note that the kernel of the action is {
action.
} so ⋂
Theorem 2.3 (Orbit-Stabiliser Theorem)
If acts on with finite and
then | |
Proof
Suppose
?
so there exists
|
with
||
.
( ) is a
so
is the kernel of the
| or equivalently |
|
|
may not be unique so for which
So is in the same right coset of
as . So this defines a bijection between
| | | as claimed.
cosets of
in . So |
Notation
For
, denote
If
denote
is
and the right
the two point stabiliser. Similarly,
{
( )
}
{
Example
{
Take
{
} then
|.
}
} and
(
( )
but
.
In general, it is true that
( )
)(
pointwise sta iliser
setwise sta iliser
) and
{
} then
( )
and
, but if
Corollary 2.4
If acts on , then
i.
ii.
iii.
|
| is constant for
in an orbit.
is transitive implies | | | || |
If
is -transitive and | |
then | |
(
)
(
)|
| for any
Proof
i.
ii.
iii.
This follows immediately from the Orbit-Stabiliser Theorem
This follows immediately from part i.
We use induction on .
When
,| |
| | from part ii.
Assume true for
; that is
is -transitive implies that
transitive straight from the definition since |
{
}|
{
}
is (
)-
. So by induction
|
(
|
)(
)
(
)|
| also | |
|
| by part ii. So the
result follows.
Specific Actions – The Right regular and coset actions
Right Regular Action
Let be any group. Take
. For
action. Given
,
} { } so it is faithful.
and
we define
( )
so it is transitive. The stabiliser
Coset Action
A generalisation is a coset action. Let
{
and take
{ } gives the right regular action. Note that (
} {
( ) {
so
. This is clearly an
}. Define (
)
}
{
)
so it is transitive. Then
}
{
As the stabiliser is a subgroup, this particular subgroup is called a conjugate of
.
In general, this action is not faithful:
Example
Take
i.
ii.
{
⟨ ⟩
}
{
}. We have two cosets
{
}. Then
and
} kernel of the action so it is not faithful.
( )
( ) so {
⟨ ⟩
Observe that
so
kernel so
}. We have three cosets
{
} then
⟨ ⟩ {
(
a 3-cycle on . Notice that
so ( )
. Then
(
). Therefore | | | |
so the action is faithful.
) is
Theorem 2.5
Any transitive action of on is equivalent to some coset action. (That is, a bijection
between , so that the group action is preserved.
Proof
Let act transitively on . Let
on . Define
by ( )
First we check that
{
. Let
. We claim that
and
for
is well defined: i.e.
is injective because (
.
For equivalence of actions;
. Now
.
is transitive.
is clearly surjective because
)
(
((
) )
)
(
)
The Conjugation Action
acts on
by
} we have a coset action
is a bijection:
. This is clearly an action.
(
) .
Remark
The default meaning of
for
is
.
is a conjugate of in . The orbits are called the conjugacy classes of and denoted
{ } so the action is not transitive if | |
( ). Note that
( ) {
.
} {
}
( ) otherwise known as the Centraliser
of .
( )
( ) { }
Note that
As a generalisation we can take
{
( ) {
Then
is the centraliser of
}
in
Remark
It can be proved that
} the subgroups of
( ) called the normaliser of
}
is the set {
( )
and define an action by
( ) and
in . Also
( )
( ).
Summary of Notation:
Take
then:




( ) is the conjugacy class of which is {
( ) is the centraliser of which is {
( )|| ( )|
finite implies that | | |
( ) is the centre of which is {
}
}
}
Conjugacy of permutations
( ) suppose
Let
is (
)(
and
. Then
so if a cyclic decomposition of
)( ).
is (
Example
(
)(
) and
(
)( ) then
(
Let
note that conjugate permutations have the same cycle type.
Theorem 2.7
Two permutations
cycle types.
( ) are conjugate in
(
)(
This is possible because all
and . It then follows that
)( ). Then simply choose
and all
)(
). In particular
( ) if and only if they have the same
Proof
From the above, we conclude that is true.
( ) of the same cycle type: that is
Conversely, take
and
; that is
)(
)( ) then
(
( ) so that
)(
)( )
for all
appear exactly once in the cyclic decomposition of
.
.
Example
{
Let
} and
(
)(
) and
(
) As
(
cycle type take
. Note that
)(
and
) then
have the same
satisfies
is not unique.
-groups:
For prime, a finite group is called a -group if | |
group if all elements have order a power of ).
that ( )
Note for
{
}
)(
for
(for
{
but for
infinite
is a -
} ( )
Observe that
is a -group. We will show soon that -groups have non-trivial centre.
Lemma 2.8
If
then
is a union of conjugacy classes of
Proof
As
and
for all
. Therefore
Theorem 2.9
Let be a finite -group where is prime and
with
(
)
particular if
take
to conclude that
.
( )
.
then
( )
. In
Proof
∐
( ) for some
By Lemma 2.8,
. This is a disjoint union because
( ) { }. We have
conjugacy classes are either equal or disjoint. Choose
so
| | ∑ | ( )|
∑ | ( )|
∑ |
( )| by the Orbit-Stabiliser
|
(
)|
Theorem. Now observe that
for some
. Since
,| |
mo
y La ran e’s Theorem. There ore in or er to a oi a ontra i tion some
with
since otherwise we get
mo .
( )|
( ) and so ( )
That is |
for some
hence ( )
so
.
Chapter 3: Sylow’s Theorems
Product of subsets:
If
then
{
}
Lemma 3.1
Let
i.
ii.
If
If
or
and
then
then
Proof
i.
ii.
Suppose
(
)(
. Take
and
. Then
) (
)(
) and since is normal,
so
)
so (
. Similarly
( )
(
)
since
by normality.
Therefore
.
( )
Let
and
. Then
because
and
by normality of and . Hence
is normal.
Remark:
Note that if
then
so
.
Theorem 3.2 (Second Isomorphism Theorem)
If
and
then
Proof
Let
to . Then
be the canonical epimorphism ( )
(
)
{
}
. Then consider the restriction
. Moreover,
( )
so applying the
First Isomorphism Theorem, we yield the result
The on erse to La ran e’s Theorem says: “ i en finite and | |
and | , does there
|
|
exist a subgroup
so that
”. In eneral this statement is not true:
has no subgroup of order 6 but |
power.
Sylow -subgroup
Let | |
, prime and
subgroup of . Denote
for
( ) {
|
. However the answer is yes if
. A subgroup
}
| |
Theorem 3.3 (Sylow’s Theorem)
For a finite group let | |
, prime and
i.
( )
for
of order
.
is a prime
is called a Sylow -
ii.
iii.
(containment) Any -subgroup of is contained in some Sylow -subgroup of .
( ) then there exists
(conjugacy) If
with
( )|
(number) |
mo
iv.
We will pro e Sylow’s Theorem in two separate Theorems. The irst one implies
Proposition 3.4
Let be a finite group and
(
).
Proof
Let
| |
| |
{
(
). If
|| |. Then the number of subgroups of
}. Let | |
and
where
then
by right multiplication. That is
of order
. Then using elementary combinatorics,
{
}
so we can define an action of
on
. This is clearly an action.
Let be an orbit of
. If
and
then there exists
( ) then by definition
an element
with
). If
( )
. Now we consider two cases:
Case 1:
( )
.
so
. (I.e. there exists
that is
{
( )
is a subgroup since
and the orbit
| | | | by the orbit-stabiliser
is the set of right cosets of . Then | |
theorem. In particular, is the only subgroup in the orbit.
So if is a subgroup of order
then the orbit of has size
.
. Then
}
Case 2:
( )
Suppose
. Then | |
subgroups of . As | |
| |
|
( )|
| |
| |
, so by Case 1
contains no
| | | because | ||| |
then
| |
{
} then we have shown that equals the number of orbits of size
and the remaining orbits in have
|| |.
Then for some , | |
therefore
| |
(
)
Let
As
is prime there exists a unique inverse
| |
that
(
(
)
)
(
(
)
(
{
} with
mod . Therefore
) It then can be proved using elementary number theory
).
we can avoid this by using the following argument due to G. Higmann
The number
(
)
order | |, in particular
is a function of | | and
mod
so it must be the same for all groups of
is the same for the cyclic group of order | |. Cyclic Groups
have a unique subgroup of order
for all groups of order | |.
Proposition 3.5
( ),
Let
for all
a -subgroup of
|| | so
for a cyclic group so
that is | | is a power of . Then
mod
for some
.
Proof
Take | |
that is (
with
)
{ri ht osets o
. Let
. This is clearly an action.
in }.
Consider the restriction of the action to
. Orbits of
where
. This is a power of , possibly
.
Suppose the orbit of
.
(
has size 1. That is
acts on
on
have size | | |
(
)|
) therefore
Proof of Theorem 3.3
Follow immediately from the conclusion of Proposition 3.4
Since
is a subgroup of order | |
we have
( ) then | | |
| so we must have
If
thus
.
Corollary 3.6 (Cauchy’s Theorem)
If is a finite group and || | for prime then
by right multiplication
( )
by Proposition 3.5 and
has an element of order .
Proof
{ }. Then | |
( ). Take
Choose
Then (
| . Then as
Corollary 3.7
( )|
Let |
| | | | |.
)
so (
we must have
. Then
| | |
) |
for some
hence
for
. That is |
( )| where
. Then
|
( )
and
and so
.
{
} In particular
Proof
( ) and let act on by conjugation. By definition, ( )
( ). By
Let
Theorem 3.3, part iii, this action is transitive. By the orbit stabiliser theorem we have
| |
| |
| ( )|
( ) by definition we have | | | ( )| and so
In particular as
| |
| |
| |
|
| |
| |
| |
Corollary 3.8
|
( )|
iff
so
Proof
By Corollary 3.7, |
Corollary 3.9
( ),
For
( )|
iff |
| |
( )|
( )
( ) iff
iff
for all
iff
( )
Proof
Corollary 3.10 (Frattini Lemma)
( ) then
If
and
( )
Proof
Direct Products
{(
be groups. Then define
)(
) (
). In particular | | |
Let
(
}. Then
)
|
|
|
Proposition 3.11
Let
. Let
i.
ii.
If
iii.
Every
{(
and
}
)
with
. Then:
then
can be written uniquely as
where
Proof
Easy
Theorem 3.12
Let
where
and
and
. Then
Proof
We first prove that
for all
,
.
Consider
(called the commutator of
). Observe
(
) and
because is normal so
. Similarly,
is normal so
. So
and so
that is
.
by (
Define
((
)) ((
)
. First ((
)). Therefore
)(
))
(
is a homomorphism.
.
(
) and
because
)
is surjective by
hypothesis.
))
If ((
and thus
then
so
( )
. Therefore
therefore
is injective and so
and so
hence
is an isomorphism.
{(
)} as in the definition is an external product of
Note
and
whereas in Theorem 3.12 is called the internal direct product of and . As a
consequence of Theorem 3.12, we usually just call them direct products.
Corollary 3.13
Suppose
except
. For each , let ̂
with
̂
). Assume
for all
∏
. Then
(the product of all
.
Proof
This is a straightforward induction on , following from Theorem 3.12.
Remark
It is insufficient to assume
for all
.
Group Presentations (From Algebra II)
⟨
|equations in the ⟩
For example, ⟨ |
that is generated by the elements
⟩ for some
es ri es the “lar est” roup
which satisfy the equations on the right.
This is an informal definition. We will only use it when is finite or can be shown to be
inite, so “lar est” makes sense. We woul also nee to pro e that any two su h lar est
groups are isomorphic. We need group presentations here because they are so useful for
defining small finite groups.
The example above is
the dihedral group where is the rotation, a reflection.
|
⟩ for
However ⟨
is isomorphic to
, the direct
product of cyclic groups. In specific examples like the ones above, it is not hard to show that
there is a unique largest such group.
Semi Direct Products
We have seen actions of
the action of ( ) on .
If
on
correspond to homomorphisms
( ) and
is
( ). That is
are groups we can define a group action of on in which ( )
( ) is a bijective homomorphism from to itself. We still write
for the action of ( ) on
. Equivalently we define it by the axioms:



(
(
)
)
( ) is a homomorphism
Semi-Direct Product
Suppose is a group which
and
3.12 but not assuming
)
We call an (internal) semi-direct product of
and
.
and
(as in Theorem
Note that conjugation of elements of by elements of defines a group action
(easy to check it satisfies the three axioms above)
(
)(
)
(
. We write
or
)(
)
on .
) so multiplication in is completely determined by the action
denotes the semi-direct product with action .
Conversely, given groups
and an action
direct product as follows:
{(
Given an action on , define
by (
of
(
) where
of
on
we can define the external semi} with multiplication defined
)
is determined by the action .
Claim
is a group.
Proof
Closure is clear, the identity is (
to check.
Examples:
⟨ ⟩
Take
( )
and ( )
[In general

(
) and (
⟨ ⟩. Then
( )
so
)
for
( )
⟨
)
. Associativity is also easy
⟨ ⟩ where ( )
. Then
( )
and
prime]
( )
|

( )

( ),
( )
then

( ),
( )
( )
( )
(
(
.
( ) mapping
Define
)
⟨
then
⟨
then
⟩
⟩
|
⟨
⟩
|
then
⟩
|
[In general,
is the trivial map so
for all
]
Groups of Small Order
We will now classify (up to group isomorphism) all groups
Claim
If | |
prime then
with | |
(except 16).
.
Proof
Easy.
Proposition 3.14
If | |
with an odd prime then either
Proof
Let | |
, an odd prime. Choose
⟨ |
⟩ and
⟨ |
⟩. Observe
is a subgroup of by Lemma 3.1 and
or
( ) and
( ) then
|
because |
.
so we must have
so
. Clearly
. So
(
for some ; therefore
)
( )
so if
we have
or
The two possibilities give ⟨
⟨ |
Proposition 3.15
If | |
for some prime
is determined by .
)
so |(
then
mo
. So assuming
.
|
⟩
⟩
then
and
.
is abelian and
or
Proof
{ }. Then | ||| | so | |
is abelian by Sheet 2 Q1. Then let
or | |
. If | |
⟨ ⟩
then
.
{ }. Then choose
{ }, define
⟨ ⟩ and choose
Assume | |
for all
⟨ ⟩. Observe | |
| |. Then |
|
|
and let
or . But |
which is a contradiction. Therefore
and so
(because is abelian so
| | so
and we use Lemma 3.1) but | | | | so y La ran e’s Theorem | |
. By theorem 3.12, we have
.
Classification of Groups
with | |
Quickly observe that if | |
,
for prime or | |
for an odd prime then we are
done by the previous two Propositions. It remains to classify | |
.I
start off with another quick Lemma:
Lemma
If is any group and | |
Proof
{ }. Then (
Take
| |
Take
as | ||
for all
{ } then
)
. Moreover
is abelian.
so
.
we must have | |
Case 1:
{ }. Then by the Lemma,
Assume | |
for all
cyclic groups; that is either
or
{ } the only possibility is that
.
is abelian, so is a direct sum of
or
. As | |
for every
{ } for prime then writing with the
[In General if is abelian and
for all
operation addition, can be made into a vector space over a finite field
with elements.
So if finite,
then ⏟
. Groups of this property are called elementary
abelian.
Case 2:
If not, there exists
with | |
. Let
⟨ ⟩ so |
|
hence
. Choose
so
.
has order
so
( )
so
. Also
implies that
.
Claim
Given
and
Proof
{ }
{
{
}
}, {
is defined uniquely.
}
{
,
}


then
then | |
or
and so
.
.
.
is abelian.
⟨
|
If
element of order 8.
If
then ( )
by
and get
Case 2b
In this case


)
{ } so the Claim is true.
The possibilities are that
. If
We now consider the possibilities that
or
| | |
Observe that |
so
or
Case 2a
If
(
,
⟩
. Take
.
and
since
since
was arbitrary in
has no
we can replace
is non-Abelian.
⟨ |
⟩
If
we have
⟨ |
⟩
If
then
we can construct
as a group of complex 2x2 matrices. This proves the existence of
and is necessary to show the relations above are not inconsistent.
Observer that
has
or order 2 and the other 6 elements are of order 4 so
Summarising, | |
| |
By Sylow’s Theorem |
or .
Case 1
( )|
If |
then
⟨ |
⟩
( ). | |
Let
or
( )|
( )
. Also by Corollary 3.7, |
{ }
so
Case 1a
⟨ |
⟩. Then
Take
Either
or
( ) ⟨ ⟩ with
More formally,
both define a homomorphism
or
( )||| | so |
by Corollary 3.8. Since | |
( )|
we may take
by Proposition 3.15.
; the group automorphism is determined by
.
and ( )
. Can have ( )
or ( )
( ) so we get two possible semi-direct products:
⟨
|
⟩
⟨
|
⟩ non a elian
(No need to prove the existence of these groups; their existence is guaranteed by the SemiDirect Product.)
Case 1b
Take
If ( )
⟨ ⟩
( )
(
⟨ ⟩ again we want
( )
( )
) then
( ). There are four possibilities:
( )
( )
⟨
|
⟩
The other three cases give isomorphic groups:
( )
The groups obtained by the cases ( )
are clearly the same as the case
( )
( )
since you can simply interchange and .
( )
⟨ ⟩ ⟨ ⟩ we can simply replace by
If ( )
then ( )
. Since
so
this choice is isomorphic to the two above. So we have a single group
⟨
|
⟩
( )
( ) )
(this is in fact isomorphic to
as | |
Case 2
{
( )|
In this case we take |
. Let
they are all conjugate; that is acts transitively on
( ).
homomorphism
Let
⟨ ⟩. By Corollary 3.7, |
( )) hence
we can assume
If
( )
and
for
( )|
|
| |
( )|
}
( ). By Sylow’s Theorem
by conjugation. That is there is a
( ) so |
for any
and so
(
; that is
( ) for
) and fixes
( )|
; (as
. Then
o
.
⟨ ⟩ then
(
) or
(
) so if necessary replace by
to get
(
) so ⟨
⟩
( ) which has order 12.
( )
( ) so | ( )|
| | so must be an automorphism and
( ).
Summarising, we get 5 groups in total; 4 in Case 1, 1 in Case 2.
| |
( )|
( )|| so then
By Sylow’s Theorem, |
and by Corollary 3.7 |
|
( )|
( )|
( )|
Similarly, |
and divides 5 so |
.
Therefore
( )
,
( )
so
,
implies that
.
(
(This same argument works with any | |
with
prime amd
and
)).
| |
There are 14 isomorphism types. This was proved by Burnside in approximately 1900. The
proof is omitted because it is rather long.
| |
|
( )|
.
Take
and |
mo
( ),
⟨ |
( )| |
so |
( )|
⟩. So
( )
. Therefore
| |
and
is a semi-direct product.
Case 1
⟨ ⟩ cyclic.
As
then the usual argument involving the automorphism group shows
or
each giving the cases
or
respectively.
Case 2
⟨ ⟩
⟨ ⟩
with
Claim
For a suitable choice of
or
and
.
we can assume
or
and similarly
Sketch Proof
( )
In general, if
then as
we have
and so
⟨
⟩
⟨
⟩
(since is abelian). Moreover,
so we can replace by and
can assume
or
. Now suppose
or
:
(
{so we must ha e
(
{
)
by . So we
)
or
In this ase
repla e
y
so
an
et
Therefore the claim is true.
There are four possibilities but two are isomorphic by interchanging
get three groups:
⟨
⟨
Now observe that in
| |
for all
|
|
the element
and so
and . Therefore we
⟩
has order 6 but no element of
.
⟩
has order 6 because
Summarising we get five groups of order 18; 2 in case 1 and 3 in case 2, two of which are
abelian.
| |
( )|
By Sylow’s Theorem |
( ) and
. Take
possibilities for ; either
and |
( ), then
or
( )|| so |
( )|
. Where
⟨ |
and so
⟩. There are two
.
Case 1
Assume
( )|
( ) are the maps
. Then |
and the elements of
( ) ( ) can be any element of
( ) so we
. For the possible
get 4 possibilities:
If ( )
then
) then
If ( ) (
⟨ |
⟩. As
we have
(
)
(
)
and so
So by replacing by
we get the same semi-direct product as the previous case.
). Then
The final case is that ( ) (
⟨
⟩
|
Observe that in
order 10. But in
.
(
,
,
)
so
and ( )
( ) and
centralises ; in fact
and
has no element of order 10 so
Case 2
⟨ ⟩ ⟨ ⟩ as usual we get
Suppose
| |
case 1b. This leads to two possible groups
and
⟨
and
has
similar to
⟩
|
Therefore we get 5 groups of order 20: three in case 1 and two in case 2.
| |
|
( )|
⟨ |
⟨ |
Let
Need action
and divides 3 so |
( )|
( ) and
hence
denote
⟩.
⟩
( ). Then
( ) and (
.
)
so
for some with
. Then
.
If
so
and
get two groups corresponding to the cases
⟨
( )
so
|
give isomorphic groups (swap
and
respectively:
). So we
⟩
Observe that the second group is non-abelian so we get two isomorphism classes.
Note that in all our cases, all our groups were constructed from cyclic groups as direct or
semi-direct products, apart from
which we had to construct as a matrix group.
Chapter 4: Nilpotent and Soluble (Solvable) Groups
Theorem 4.1 (Third Isomorphism Theorem)
Let be a group,
and
with
(hence
). Then
an
[Note that
and
does not automatically imply that
].
Proof
Define
by (
)
. Because
}
{
}
{
,
( )
is well defined and
. So
.
and so by the first isomorphism theorem
Theorem 4.2
Let be a finite group. The following are equivalent:
1. |
2.
3.
( )|
for all | | | prime.
( ) for all | | | prime.
for all
( ) where
where
are the primes dividing | |
Proof
1 2 by Corollary 3.8
3
2 by Proposition 3.11 i.
To prove 2 3:
Let
be the primes dividing | |
Let ̃
(excluding
. Then ̃
| |
( ) so
). Then | |
| |
by hypothesis.
|
||
by Lemma 3.1. In particular for
and thus ̃
since it is divisible by all
. In addition,
elements are coprime. Then by Corollary 3.13
|
we have
̃
| |
| |
where
| | ̃ | so ̃
since orders of all the
.
Nilpotent Group
A finite group satisfying the three properties of Theorem 4.2 is called nilpotent. The
definition for infinite groups is different; see sheet 5.
Theorem 4.3
Let be a finite nilpotent group. Then:
1. If
2. If
3. If
Proof
, then ( )
(nontrivial centre)
then is nilpotent
then
is nilpotent
| |
1. Assume
Take
2. Let
and
, so by Theorem 2.9 ( )
.
)
( ).
because nilpotent and so
. By the
. As
for
( ) { } then we have (
( ) for some hence
second isomorphism theorem
which has order coprime to , so
( )so the Sylow -subgroups of
( ) for some and so
3. Let
therefore
and so
for any
is nilpotent.
by Lemma 3.1
and so by the third isomorphism theorem
order coprime to
and so
are normal in
. Then
which has
( ). Hence the Sylow p-subgroups of
so
are normal in
is nilpotent.
Examples:




Abelian Groups are nilpotent
All groups of order
for prime are nilpotent.
Direct products of nilpotent groups are nilpotent. (Condition 3 of 4.2) .
is not nilpotent since it has 3 Sylow 2-subgroups so they cannot be normal
subgroups.
Maximal Subgroup
A subgroup of a group
is maximal if
but
Note that if is finite and
with
infinite groups have maximal subgroups.
then
implies
or
.
for some maximal . Not all
Theorem 4.4
The following are equivalent for finite groups:
1.
is nilpotent
( )
2.
and
implies that
3. All maximal subgroups are normal.
Proof
( ) ( ) Let
so
. As is nilpotent by Theorem 4.3, ( )
We proceed by induction on | |. If | |
there is nothing to prove.
Case 1:
( )
So
so
( )
Case 2:
( )
( )
( )
( )
implies that
but ( )
( ).
( ) Let
. By induction
( )
so
( ) and also
be maximal in .
( )
nilpotent so
( )
(
( )
( )
)
( )
for some .
( ).
( ) so
( ) implies
( )
( )
( ) and
so
( ) so
.
( ) ( )
Assume ( ). If is not nilpotent, then
for some
for some maximal in . By condition iii, ( )
Contradiction.
but
( ) so
( )
( )
so ( )
by Corollary 3.9.
Commutators and Commutator Subgroup
Commutator
Let
. Define the commutator to be [
Notice that [
]
and [
]
]
[
].
Commutator Subgroup
] is by definition
The Commutator subgroup of , denoted [
[
] ⟨[
]|
⟩
]
} is not a subgroup. The smallest counterexample is of
In general, the set {[
order 32.
Theorem 4.5
1. [
2.
]
[
]
3. If
is abelian.
and
abelian, then [
]
that is [
]
is the ‘lar est’ a elian quotient
group of .
Proof
1. It follows from Theorem 4.7 (i) and (iv).
[
] therefore
2. For all
,
[
3.
[
] [
] so
[
]
abelian implies
]
.
for all
Examples:
is done in the notes, and [
In fact
| |
so
]
[
[
] [
]
therefore
]
[
so
[
is called perfect if
]. It follows
on Sheet 5.
)(
)] (
)(
)(
)(
) (
then [(
] and ( )( ) [
]. Therefore
)( ) [
{ ( )( ) ( )( ) ( )( )} [
because it is a union of conjugacy classes. Moreover
is abelian so
]
is abelian.
Remark
]
It should be clear that [
is abelian.
that non abelian simple groups are perfect.
Take
similarly (
[
] by Theorem 4.5 part 3.
)(
]
)
[
] so
Characteristic Subgroups
Characteristic
A subgroup
this to
.
is called characteristic in
For any group and any
the map
( ).
is an automorphism and so
Inner Automorphisms
The group of inner automorphisms is denoted
Observe that this truly is a group because
In fact it is a normal subgroup: Let
( )
( ).
and thus
if
( ). We abbreviate
for all
(conjugation by ) mapping
( ) {
and ( )
( ). Then
}
( )
so
(
)
(
( )
)
so
Outer Automorphisms and the Outer Automorphism Group
( )
( ) is called an outer automorphism and
( )
( )
( )
is the Outer
Automorphism Group.
Lemma 4.6
All Characteristic Subgroups are normal.
Proof
Let
. Then
( ) so this means
for all
( ). In particular,
for all
and so
.
for all
( )
Theorem 4.7
Let be a group. Then:
1.
2.
3.
]
4. [
5. ( )
6.
and
and
( ),
and
finite
Proof
1. This is Lemma 4.6.
2. This follows from 3 and 1
( ). Then
3. Let
Claim
( ).
The restriction
of implies
Proof
injective and a homomomorphism immediately imply that
homomorphism. We only need to check surjectivity.
Take
then
( )
( ) so
because
is injective and a
. Then
so
so
is surjective.
Therefore
since
so
.
]
[
] so permutes the Commutators
( ) so [
(
)
4. Let
] ⟨[
]|
]
[
] that is [
] char .
⟩ so [
then [
]
]
( ) and let
( ) then [
5. Take
for all
implies [
for
]
all
so [
for all
. As is a bijection from to we have
[
]
( ). Hence ( )
( ) that is ( )
for all
so
( ). If |
( )|
( ). Then
6. Take
then there exists a unique
| | | |
( ) so
that is
.
Soluble Groups and Derived Series
Let
be a group and suppose
is a series of length .
Subnormal series, Normal Series
The above is called a subnormal series if
for all (in this case
subnormal subgroup of ). It is called a normal series if
for all .
is called a
Soluble
For a group , we say it is Soluble if it has a subnormal series with each quotient
abelian.
Examples:


Observe that Abelian Groups are Soluble.
Let
be the set { ( )( ) ( )( ) ( )( )} then has series
]
and as [
,
and
because it is a union of conjugacy
classes. Moreover,
are abelian and
is abelian so this proves
Soluble.
{
} { }
implies
is soluble. Note that this series is a
subnormal series but not a normal series.
All groups up to order 23 are soluble. In fact,
of order 60 is the smallest group
which is not soluble. Moreover
is simple for
so this implies that
are
not soluble for
.


Define
( )
( )
and
( )
( )
[
] and
(
)
[
()
()
] for all
. Then the series
( )
is called the derived series for . Moreover
by Theorem 4.7 hence this series is a normal series.
(
)
The series may stabilise ( )
for some . (this is always true if
infinite groups this may not be the case e.g. free groups.
( )
1.
is soluble if and only if
2.
is soluble if and only if
( )
for all
is finite) but for
Theorem 4.8
Proof
is
for some .
has a normal series with abelian factors
1. [ ] is clear since (
hence is soluble.
)
implies that the derived series for
[ ] Let
be a subnormal series with
induction that
()
for all
is a normal series
abelian. We prove by
.
( )
Take
then
.
Assume true for then by induction and using the fact that
(
)
]
[ ( ) ( )] [
Because
is abelian,
( )
abelian so
2. [ ] by definition
[ ] If is soluble then ( )
for some
normal series with abelian quotients.
by 1. But then the derived series is a
Lemma 4.8.1
. Then [
Let
]
[
]
Proof
]
⟨[
⟩
⟨[
]
⟩
Corollary 4.8.2
( )
( )
( )
for all
Proof
We apply induction. The Lemma gives us the induction step.
Theorem 4.9
1. If
is soluble and
then
is soluble
2. If
and
is soluble then
is soluble
3. If
and
and
are both soluble then
is soluble
4. All nilpotent groups are soluble
Proof
1.
Then
2.
4.
( )
( )
If
for some .
( )
so
soluble implies
so
3.
( )
is soluble iff
( )
so
( )
so
is soluble.
( )
{ } then by Corollary 4.8.2, ( )
is soluble.
is soluble then
( )
( )
( )
for some
so
( )
.
)
(
If is soluble then ( )
for some and so (
soluble.
Let be a nilpotent group. We now apply induction on | |.
When | |
we get the desired result.
)
so
is
{ }
For | |
soluble so
then ( )
by Theorem 4.3 1. So by induction
( )
and ( ) are
is soluble by part 3.
Now observe we have an ascending chain of sets:
elian roups
ilpotent roups
Solu le
roups
roups
Composition Series and the Jordan-Holder Theorem
Maximal Normal Subgroup
is a maximal normal subgroup of
Simple Group
A Group is simple if
So
if
and
and
implies
is a maximal normal subgroup
implies
or
is simple.
What are the maximal normal subgroups
So | |
so | |
So | |
is
of ?
is simple, hence cyclic of prime order so | |
abelian.
if each
⟩
|
implies
.
.
A subnormal series
is a composition series of
).
simple. Not all groups have them for example (
Example:
Take
⟨
or
( ),
so there is a unique subgroup
has three subgroups of order 2; ⟨ ⟩
⟨
⟨ ⟩
⟨
⟩
( ),
or . Since
⟨ ⟩
. Then
⟩ so there are three possible
⟨
⟩
⟨
since
⟨ ⟩
⟨ ⟩
so that | |
⟩
We consider these in turn:
⟨
There is no normal subgroup of order 2 so
of .
Similarly,
with composition series
⟨ ⟩ ⟨ ⟩
Since
and ⟨ ⟩ so we have two composition series
⟨ ⟩
⟩
so the composition series is
which has two normal subgroups
⟨ ⟩
⟨ ⟩
Therefore
has four composition series altogether.
Equivalent Series
( ) and
Let
series of . Then we say those series are equivalent if
{
} such that
()
()
for all
.
( ) be two composition
and there is a permutation of
:
Theorem 4.11 (Jordan-Holder)
Any two composition series of a finite group
Proof
We apply induction on | |. When | |
are equivalent.
this is clear.
Let ( ) and ( ) be two series as above.
Case 1
If
then the result follows by applying induction to
Case 2
If
so
Moreover
(because
simple and
and
simple implies
, so the previous statement proves
and
both normal in
and
.
and applying Lemma 3.1).
are both maximal normal subgroups of
.
Now by the Second Isomorphism Theorem
Let
and let
construct the following diagram:
be a composition series for
we then
We have four composition series in the diagram:
()
( )
( )
( )
Now by Case 1, the series ( ) and (
also equivalent so
(
) and (
) are equivalent using
) are equivalent so
(
) since
and the series ( ) and (
) are
and
As equivalence of series is an equivalence relation, it follows that ( )
( )
.
Composition Factors
The multiset of isomorphism types of factors
in the composition series of are
} and for it is
called the composition factors of . So in the Example above, it is {
{
}
Proposition 4.12
A finite group is soluble if and only if it its composition factors are all
that is cyclic of
prime order.
Proof
[ ] If all the composition factors are cyclic then
is soluble by definition.
[ ] Conversely, if is soluble then it has a subnormal series with abelian factors, so its
composition factors are all abelian. So by sheet 1, they are all cyclic of prime order.
So
soluble and | |
then the composition factors are {
}
Unfortunately a finite group is not determined by its composition factors but knowledge of
the composition factors is useful for studying a finite group.
Chapter 5: Permutation Groups and Simplicity Proofs
In some sense, Nilpotent Soluble groups can be regarded as a generalisation of Abelian
Groups. By the previous chapter all their composition factors are .
Now we will study non-abelian simple groups.
Block
Let act on . A block for
or
.
with | |
is a subset
and
but that for all
,
Primitive Action
An action on is primitive if it is transitive and there are no blocks.
Imprimitive Action
is imprimitive if transitive and there are blocks. (Im)primitivity only applies to transitive
actions.
Example:
{
}
⟨(
on
) on {
generally, for (
e.g. taking
, then {
Lemma 5.1
If is a block then
Proof
| |
)⟩ then { } and { } are blocks but {
}, if | and
then {
} { } is a block.
is a block for all
| |
|
|
| |. Fix
} is not. More
} is a block.
.
then let
) then
since
then take
(so
is a block so
so
is a
block.
Lemma 5.2
Let
be transitive and
primitive.
a block. Then | ||| |. In particular, | | is prime implies
Proof
} partition
By Lemma 5.1, the sets {
} form a block system.
So {
and | |
|
| for all
hence | |
Example
{
Let
} take
( ), where [
]
{
}
we have [
Then in the case
(
)(
)(
)
]. Then (
)(
)(
)
and
is
| |.
We have two block systems if
. Each row is a block and the set of all rows form a
block system. Similarly for the columns. In fact
.
Theorem 5.3
If
is 2-transitive then
is primitive.
Proof
Let
be 2-transitive. Then for any
and
there exists
so that
and
.
{ }. By 2-transitivity there
Suppose is a block and so | |
. Let
. Let
exists
so that
. Therefore
so a block implies that
.
{ } arbitrary, hence
But
so
. As
. Contradiction. Therefore
there are no blocks and so
is primitive.
So now we have Remarks


)⟩
For prime we know ⟨(
. This is primitive by Lemma 5.2. Hence by
Theorem 5.3 it cannot be 2-transitive. (In particular it cannot map the pair ( )
( )).
)⟩ is transitive but not primitive.
For not prime then ⟨(
Lemma 5.4
If
is transitive and
transitive, then
.
Proof
Let
as
Therefore
Theorem 5.5
Let | |
and
of for any
and there exists a subgroup
is transitive then there exists
so
.
transitive. Then
with
with
is primitive if and only if
and
so
.
is a maximal subgroup
.
Proof
[ ] Take
and assume
is maximal in . Let be a block and define
{
}.
} partition replace by
Since the sets {
if necessary and assume
.
Then
so
so
. By maximality,
or
.
If
then let
so there exists
with
(because
is transitive) so
so
so
. But as
arbitrary then this
implies | |
so is not a block. Contradiction.
If
let
so by transitivity of
, there exists
with
. Then
now by definition of ,
so
. Now as
arbitrary,
so is not a block.
Contradiction.
Hence there are no blocks and so
is primitive.
[ ] Assume
is primitive. Let
for some
. We want to prove
or
.
{ } implies
{ } implies
Let
. If | |
then
and
so
. Similarly, if
then this implies that
is transitive since
was arbitrary.
Then by Lemma 5.4,
.
| | | | we will prove is a block, giving a
Now we consider the case when
contradiction to
primitive.
Let
and
so there exists
such that
. Since
there exist
so that
and
(
. But then
so
So we have proved that if
Contradiction to
primitive.
. So
)
then
so
so by definition
so
is a block.
Normal Subgroups of Permutation Groups
Theorem 5.6
Let
be transitive and
1.
2.
3.
with
. Then one of the following holds:
{ } and so
and so
is a block.
is transitive.
Proof
{ }. Then
Suppose
so
for all
If
hence
so
then
. So if
and
transitive and
and
for all
then
implies that
.
is transitive by definition.
In general let
. This is a subgroup of by Lemma 3.1 containing
and
. Assume that both 1. and 2. do not hold. It was proved in Theorem 5.5 that
a block, so we have 3.
Corollary 5.7
If
is primitive and
with
then
is
is transitive.
Proof
We apply Theorem 5.6. As
, condition 1. does not hold. As
is primitive there are
no blocks so condition 3. does not hold. Therefore we must have 2. and so
is transitive.
Remark
By Theorem 5.3, if
is 2-transitive then
is primitive. Hence every 2-transitive action
with a normal subgroup acting non-trivially on implies that
is transitive.
We now consider the case where
.
Regular Action
An action
is called regular if it is transitive and
for
Regular normal subgroup
For a group action
, if
is a regular normal subgroup of .
, and
is regular then
.
Examples
{ ( )( ) ( )( ) ( )( )}. is a regular normal subgroup of
1. Take
and
.
2. Affine Groups:
( ) of affine
Take
for a field, we define a group ( )
transformations. We say that for a map
,
if there exists
and a
non-singular linear map
such that
for all
.
[If
and we restrict
to the set of orthogonal maps we get the isometry group
of
.] We must first prove that is a group.
For
if
with (
(
we have
)
and (
)
(
)
)
so
. (Closure).
Given
define
by
and
then
so
.
As the composition of functions is associative and the identity map is in , we have
a group.
{
} be called the translation subgroup. Then for
Let
and
we have
(
so
hence
.
is clearly transitive. Now for
, if
.
was arbitrary so by definition
)
(
)
with
then
so
is a regular normal subgroup.
so
Lemma 5.7.1
Let be a finite set. If
is 2-transitive and a regular normal subgroup of then is an
elementary abelian -group for some dividing | |. Hence
for some and in
particular | |
.
Proof
Assume
. For
{
}. As
,
is regular, for all
so
there exists a unique
. As
so that
is 2-transitive then for all
{ } there exists
{ } are
with
so ( )
. So all elements
conjugate in so all have the same order. Therefore every non-identity element has order
for some prime so | |
for some . By Theorem 2.9 we conclude that ( )
. As
( )
(
)
we must have
by Theorem 4.7. As all elements of are conjugate and
( )
this implies that ( )
and so is abelian.
Therefore
by the Fundamental Theorem of Abelian Groups.
Proposition 5.8
has no regular normal subgroup for
.
Proof
Suppose not and let
be a regular normal subgroup. As in the previous proof there exist
{
}. Let
(
)
with
for all
. Since
}. As
( ) if and only if
. Therefore ( ) {
unique
we have
( )
| so (
we have
and hence
| |
)
. As
we have
so
. But by Lemma 5.7.1, for some prime and integer we have
but this is impossible since is not a prime power.
Theorem 5.9
is not simple for
.
Proof
We apply induction on . As
is
transitive, take a normal subgroup so that
. Then by Corollary 5.7 and Theorem 5.3 this implies
is transitive. By
Proposition 5.8, is not a regular normal subgroup so for some
the stabiliser
{
(
) (
}
)
. Moreover;
{ }
{ }
and as
{ })
(
we have so
so as
. Hence
is normal
is simple.
We must now do
.
Take
. Then by Sheet 2, Q4 part ii, the conjugacy classes of
have representations
( )( ) (
)(
)(
) using combinatorics, the sizes of these conjugacy classes
are
respectively. Then | |
. But any
combination will not yield a factor of 60, so we get
simple.
The Finite Simple Groups (Classified ~1981/2004)
0.
for prime (the abelian ones)
1.
for
2. Groups of Lie type over finite fields.
3. 26 sporadic groups
Here we deal with the linear classical groups,
(
).
) be the group of invertible
Let be any field. Let (
matrices with entries in , or
)
equivalently, invertible non-singular linear maps
. Now (
the
) is
multiplicative subgroup of which is abelian. We usually assume
and so (
non-abelian.
There are two actions of
(
):
1. Right multiplication of row vectors by matrices. This has two orbits { } and
2. Let be the set of 1-dimensional subspaces of
. That is
{⟨ ⟩
⟨ ⟩. The Projective Action.
Then the action is defined ⟨ ⟩
Theorem 5.10
(
) is 2-transitive for
1.
2. The kernel of this action is the subgroup
{
} of scalar matrices.
{ }
{ }}.
Proof
1. Let ⟨ ⟩ ⟨ ⟩ and ⟨ ⟩ ⟨ ⟩
.⟨ ⟩ ⟨ ⟩
are linearly independent so
we can extend to a basis
of . Similarly we have a basis
of
. So there is a linear map mapping
for
(which is invertible since
the domain and the image are linearly independent) hence there exists
(
) with ⟨ ⟩
⟨ ⟩ and ⟨ ⟩
.
⟨ ⟩
{
2. The kernel is the set
{ }} that is the set {
⟨ ⟩
{ }} so in particular applying to basis elements
{
we obtain that
Therefore by the first isomorphism theorem
(
)
(
)
(
(
Moreover
{
)
)
}.
(
)
is a homomorphism and
}
(
) so by the Second Isomorphism
Theorem:
(
(
)
)
(
)
(
)
Properties of Finite Fields
 All finite fields have order
for some prime and
 For each
there is a unique field of order (Galois Theory - the splitting field
of
).
(
). Take
 Also, finite implies
cyclic. Let
be the rows of . Take
so | |
. is any non-zero vector (there are
possibilities). Su[[pse
we have chosen
. They must be linearly independent. Choose ; to get
linearly independent we must not be in the subspace spanned by
⟩ {∑
denoted ⟨
. Hence the number of
} so has size
possibilities for
|
is
(
)|
The order of , | |
(
) is |
(
. Hence
|
)|
(
)(
|
(
(
)|
) then
|
)
|
)
so the order of
)|
)|
|
(
)|
}|
is
(
|
|
{
(
(
|
∏(
|{
(
{
)
|
(
)|
⟨ ⟩. For which
; there are
)
(
|
(
|
)|
So is a multiple of
So
)
and |
|
|
Let
(
)|
possibilities.
(
)
(
)(
)
} so |
(
)|
Example
|
|
(
)|
(
(
)|
(
The small
)(
|
(
)
)
)|
so |
(
)|
(
(
(
(
(
(
(
)|
)|
)|
)|
)|
)
)|
.
) are:
|
|
|
|
|
|
Theorem 5.11
(
For
(
(
)
(
|
|
(
(
(
)
)|
)|
)
|
(
)|
(
)
(
)
(
) is simple for any field , except for
(
)
and | |
)
or 3.
Proof
Step 1
(
) is -transitive.
Proof
Let ⟨ ⟩ ⟨ ⟩ and ⟨ ⟩ ⟨ ⟩
.⟨ ⟩ ⟨ ⟩
are linearly independent so
we can extend to a basis
of . Similarly we have a basis
of
. So there is a linear map mapping
for
(which is invertible since
the domain and the image are linearly independent) hence there exists
(
) with ⟨ ⟩
⟨ ⟩ and ⟨ ⟩
( )
(
) by
. Let
and define
⟨ ⟩ ⟨ ⟩
⟨ ⟩.
and
for
so
and ⟨ ⟩
Step 2 – Basic Properties of Transvections
(
) is a transvection if its matrix with respect to some basis of
(
) that is the identity matrix but with a 1 in the (
From linear algebra, changing the basis replaces by
which shows that all transvections are conjugate.
Consider a matrix
position (
(
for all
) position.
for some
) that is the identity matrix with a
). If this matrix is using the basis
so with respect to the basis (
then
is
(
),
in the
and
(
)) the matrix
becomes (
) so
is a transvection.
Observe (
)
(
Recall that matrices of the form
zero matrix but with a 1 in position (
column operations of the first type.
) which is also a transvection.
are called elementary matrices. Here
is a
). These correspond to elementary row and
Step 3
(
) is generated by transvections.
Proof
(
) can be
From Linear Algebra this is equivalent to the statement “any
reduced to the identity using elementary row and column operations of the first
type.” This is the operation o a in a multiple o a row or olumn to another row
or column.
As
, we know there is a non-zero entry in the first row. Add a multiple (if
necessary) of the relevant column to column 2 to get
Now replace column 1 by column
(
then
so then
)
.
. Now subtract
multiples of the first row and column from other columns to get a matrix of the form
(
) but denote the lower-right hand corner matrix of size (
(
) by . Then
(
).
Now applying a proof by induction on , the result follows. The case
(the product of 0 transvections).
Step 4
(
) is perfect if
]
only if [
.
except for
,| |
or . We recall
)
is clear
perfect if and
Proof
] where
By Step 3, it is sufficient to prove that all transvections are contained in [
] char
(
) but [
(
) so [
(
) by Theorem 4.7.
]
), the above
Furthermore, using the fact that all transvections are conjugate in (
].
says it is sufficient to find a single transvection in [
For the case
, we compute directly that
) (
[(
)]
(
)
so when | |
there exists
with
so [
transvection. When
and | |
we use a similar argument:
) (
[(
and | |
For the case
[(
That is [
] contains a
)]
or | |
) (
we observe that:
)]
(
)
] contains a transvection in all cases so in all cases we have [
]
.
Step 5 – Iwasawa’s Theorem
Let
be primitive with perfect. Suppose there exists
for any
such
⟨
|
⟩. Then
that is abelian and
is simple.
[Note that this is also used in analogous proofs for simplicity of other classical
groups.)
Proof
Let
with
.
primitive implies
transitive by Corollary 5.7.
Then for
, by Theorem 5.5,
is maximal so
or
.
In particular,
⟨
|
so
⟩
hence
is abelian so [
Now this is abelian since
by hypothesis so
.
As
. Therefore
. Then
arbitrary, we have shown
]
for all
by Theorem 4.5. But
so
is perfect
simple.
Step 6 – Completion of the Proof
(
) and
We apply Iwasawa’s Theorem with
{ (
). By Step 1 we know that
is 2-transitive hence
5.3. By Step 4, is perfect except for the cases
and
or
It remains to find the subgroup .
Take
⟨(
)⟩. Then
. Therefore
{
(
)
that is (
}.
primitive by Theorem
.
)
(
) for
Let
be the set
(
expression
(
{(
)
{(
)
(
}
(
Let
. Observe for the
(
And
)
(
(
)
)
. Then (
conjugate in
)
.
is a normal subgroup;
⟩.
⟨
.
⟩ we are done since by Step 3,
) with
(
is
).
case that:
(
)(
)(
)
(
then take
)
(
)
(
) to an element of
(
)(
(
. So every transvection is conjugate in
For the general case, again let
is
by the
)
. Note that
is the kernel of some homomorphism;
⟨
It remains to show that
Let
be a transvection. If we show
generated by all transvections.
) (
(
)
)
We know that there exists
(
.
)
. This is a homomorphism:
)(
{(
Hence
}. Denote (
) by ( )
(
}
)
). It is elementary to check that
) so is an abelian subgroup of
Define a map
So
(
(
)
)
) to some element of
) so
)
(
.
that
so every transvection is
.
Therefore all transvections are contained in
for some
. By Step 3, is generated
⟨
|
⟩. There ore y Iwasawa’s Theorem
by transvections so
is simple.
(
A Closer Study of
|
(
)|
(
) for small .
)(
|
)
(
)|
|
|
(
(
|
)
Let
{(
| |
. Recall | |
( ).
Let
{(
)
}; the subgroup
⟨ ⟩ for
)(
(
so
(
Then
(
| |
|
⟨
(
(
)
{(
).
)
) similarly
⟩⟨
⟩⟨
⟩⟨
⟩}
⟨ ⟩
⟨(
) and
| and
⟩
⟨
(
{
⟩
⟨
{
(
(
⟨
⟩
)(
)(
)(
⟨
⟩. We
{
)}
(
)
). Therefore
}. Then as before
)⟩
)(
⟩
⟨(
} where
⟩
⟨
⟩
and | |
(
}
)
)
(
Clearly,
⟨
)(
) and
(
⟩
)
)⟩
{
. Then
}
)
⟨ ⟩ for
so
(
)(
and
⟩. Let
)
⟩ where
(
)(
⟨
(
We have
⟨
and
) so
(
⟨
(
so
)
.
⟨ ⟩ and
)
is maximal is
)(
(
. Define
and
)
and | |
is a Sylow -subgroup of ; that is
,
)
(
)
)⟩
{⟨
with
)
)
in the previous proof. Hence
.| |
}
so
(
) (
:
} the order of this is
⟩
| |
) by the Orbit-Stabiliser Theorem.
for some prime
)
(
Now
(
}.
In this case
)(
)
)
(
By Theorem 5.5
primitive implies that
write down the Generators for small ;
So
⟨(
)(
) (
First we recall some facts about the case
{⟨ ⟩} {⟨
| ( )| |
( )|
⟨( )⟩, and
⟨( )⟩. | |
(
Let
{(
(
)|
)(
( ) and |
)
)
(
|
|
) and
|
so
(
)(
(
).
)
.
| |
⟨ ⟩
|
⟨
{
. We have
⟨ ⟩. We have
|
⟩
(
(
With
(
)
)
} with
⟨ ⟩ and
⟨ ⟩,
)
(
(
)(
)(
⟩
⟨
⟩
⟩ where
(
(
.
⟨
{
⟩
)
). We shall see later that
} with
⟨
)
(
)|
⟩. We have
)
(
(
In the same way as before we determine that
(
)( )( )(
)
and
⟨
⟨ ⟩,
⟩
⟨ ⟩
⟨ ⟩ and
)
) and
(
)(
)
and is omitted, although it is no more difficult to compute. In this case
} where
and
.
In this case | |
and |
|
{
⟩
We choose
⟨
⟩
⟨
⟨
⟨
(
} where
⟨ ⟩
⟨
⟩
⟨
⟩
⟨ ⟩ and
)
We can then calculate that
(
)(
)
(
} where
,
.
⟩
⟩,
{
. We can take
and
We have
⟨
.
(
(
{
⟩
⟨
)
) and
In this case | |
and | |
⟨ ⟩
⟨ ⟩
⟨ ⟩
⟨ ⟩
⟨
⟩ where
|
⟨
⟨
(
(
)(
)(
⟩
⟩
⟨
⟨
⟨
⟩
⟨
⟩
⟩.
⟩ where
)
)(
)(
⟨
)(
)
(
)
)
(
(
)(
)
)(
)
Chapter 6: The Transfer Homomorphism
Let
{
[
be a finite index subgroup of a group ( itself need not be finite). Let
} where |
|
and
. Then we can define a homomorphism
, the abelianisation of .
]
acts on the coset space by right multiplication. That is, for any
some giving a permutation of . depends on and . We can write
}.
permutation of {
( )
We can write
( )
for some
prove this is a homomorphism
. Define ( )
. Since [
[
]
]
[
,
for
giving a
]∏
( ). We will
is a elian, it oesn’t matter a out the
( ) in the product.
order of
Lemma 6.1
is a homomorphism.
Proof
Let
(
But also,
(
]
[
)
( )
]∏
is abelian, we get [
Therefore (
Note
)
and
( )
)
( )
( ( )
( )
]∏
for
( )
)
(
( ) ( )
is a permutation of {
The map
Since [
( )
. Then
). Hence
]∏
([
} and so {
( )
. Then
( )
[
( )) ([
( )
]∏
}
]∏
}.
( ) ( )
[
]∏
where {
}
{
each
Proof
Rather than a proof, I will give an example:
Example:
(
{ ( )
( ).
is actually independent of the choice of coset representatinves
).
Lemma 6.2
For
, ( )
( ))
)(
) then
( )
( )
( )
( )
( )
( )
} and
(assuming
and
Therefore:
( )
( )
( )
[
So ( )
](
( )
)(
( )
)(
). Here
and
.
Normal -complement
Let be a finite group,
and
so
( ). Then
is called a normal -complement in if
is a semi- ire t pro u t. (They on’t ne essarily exist.)
Example
For
, it has a normal -complement but no normal -complement.
Theorem 6.3 (Burnside’s Transfer Theorem) (BTT)
( ). If
Let be a finite group, | | | and
( ( )) (that is in the centre of its
normaliser) then has a normal -complement. So unless
,| |
then is not
simple.
Note that
applied if
( )
( ) and so
( ( ))
is abelian. We need two Lemmas.
Lemma 6.4
Let be a finite abelian group with
is an automorphism.
is abelian. Therefore BTT can only be
( | |)
Proof
The map is clearly a homomorphism. If
this implies that
. Hence is injective. As
isomorphism and thus an automorphism.
Lemma 6.5
Let be finite,
conjugate in
( ) and
then
is finite,
abelian. If
( ). This is not true in general if
mapping ( )
then
( | |)
. As
is an
are conjugate in
then they are
is not abelian.
Proof
We have
for some
. So
and
. Since
are
( ). Therefore
abelian,
( ( )). By Sylow Conjugacy
( ) with
( )
applied to ( ) there exists
so
( ). Now ( )
. That is, they are conjugate in ( ).
Proof of BTT
( ( )) therefore
then ( )
∏
is abelian so [
where {
and so by Lemmas 6.4, 6.5,
]
. So the transfer
} {
} and
and hence
. By Lemma 6.2, for
and
are conjugate in
( ) that is
and thus for
for some
( ). But
(
( )) so
,
( )
∏
∏
|
| so ( | |)
( ) so by Lemma 6.4, | is an
We have
since
automorphism of . So let
(and so is normal). As | is an automorphism,
( ) so
. Let
, since | is surjective, there exists
with ( )
and so
. Therefore
.
Corollary 6.6
If is finite, | |
for odd and
then is not simple.
[Also, | | odd then is not simple by Feit-Thompson (VERY HARD!)]
Proof
Take
therefore
( ) then | |
(
) hence
{
} so for
so
is not simple by BTT.
( ),
so
Chapter 7: Classification of Simple Groups of order
500
Proposition 7.1
Let be a finite non abelian simple group.
( ) and
1. If acts on with
then
is faithful (that is
) and
| |
|
|
2. If
with
then
, and
(where means “isomorphi to
a subgroup of)
( )| then
3. If || |,
for
.
|
Proof
1. Let be the kernel of on . Then
(since
) so simple implies
that
and hence
.
( ) then using the second isomorphism theorem
Suppose
( )
( )
|
| |
| |
|
( )
( )
( )
( )| |
( )
( )|
( )
Hence |
.
so
( )|
|
| |
Hence
If | |
and hence
simple implies
( )
. Therefore
so
is abelian which is a contradiction.
( ).
( ) is soluble so it has a subnormal series with abelian
then
quotients. Therefore
is soluble which is a contradiction. Therefore | |
.
2. We apply part 1 to the special case of the action of on the set of cosets of .
3. If
, has a normal Sylow -subgroup so cannot be simple.
( ) by
For
, apply part 1 to the special case of the action of on
conjugation.
Theorem 7.2
1. All finite simple groups of order 60 are isomorphic to
2. All finite simple groups of order 168 are isomorphic to
3. All finite simple groups of order 360 are isomorphic to
(
(
)
)
(
(
(
)
)
)
Proof
1. Let be a simple, | |
( )|
( )|
By Sylow’s Theorem, |
and divides 15 so |
( ) has
It cannot be 1, since then
by Corollary 3.8 which would give a
contradiction.
( )| 3. If |
( )|
( ) then
By Proposition 7.1 part 3, |
then for
|
( )|
so ( )
a group of order 4. So ( ) is abelian and hence
( ( )) then by BTT has a normal 2-complement so not simple.
( )|
Therefore |
. By Proposition 7.1,
but | | | |
so
.
2. Let be simple, | |
( ). |
( )|
( )|
Let
and divides 24 so |
or 8. It cannot
be 1 by Corollary 3.8 so |
( )|
. Take
( ). We may assume
Let
{
}.
⟨ ⟩. Now, |
( )|
| |
| |
. Recall that
⟩. If
there are only two groups of order 21;
and ⟨
|
( ) is abelian then is not simple by BTT.
⟩.
Therefore ( ) ⟨ |
Consider the action of on as conjugation so ( )
.| |
and
so is
a 7-cycle. We might as well assume (after reordering the elements of ) that
(
). | |
(so is a product of 3-cycles) and
so as
, must
fix some other point in . We may assume that fixes , so
.
(
)
(
). Since
so
, we have
Hence
(
)(
). Let
⟨ ⟩. | |
and
implies
).
Lemma
( ), let
( ) {
Let
{
}
Then
Proof
( ). We must prove
Let
and
then
(
Now using this notation,
)
since
( )
{
(as
}. If
that is
( ).
for all
.
} so
( )
{
}
that is
( ) fixes
|
and or interchanges them. Since |
.
| { }|
Now ( ) cannot be abelian since then it contradicts BTT. Therefore ( )
⟨ |
⟩ so
(
)
|
As |
we have |
for any
(
is 2-transitive) and so cannot
|
contain . Therefore is a product of 4 transpositions. The same applies to
and
since they are all of order 2.
(
)(
) and
(
)(
). We now
Recall
so
determine :
( )
( )
(
)
So
or . So
and
so by replacing with
or
. We can therefore assume wlog that
.
|
|
(
)( )( )(
)
Now
but
so
⟩ is transitive. |⟨ ⟩|
⟩|
| | so is generated
Now ⟨
so |⟨
⟨
⟩.
⟨
⟩
(
).
by
We have proved that
3. Non Examinable, but structured similar to the above with further complications.
Theorem 7.3
The only non-abelian simple groups of order
Proof
If | |
prime then
have order 60, 168, 360.
is not non-abelian simple as ( )
by Theorem 2.9.
Also, if | |
for
odd then
is not simple by Corollary 6.6
During this proof we will not use:
Theorem (Burnside)
Any group of order
for
prime is soluble.
Proof
Requires Character Theory. See Groups and Representations.
During the rest of the proof,
|
( )|.
and
will denote a Sylow -subgroup of .
|
|
| |
We can now omit all possibilities except 6 cases, which we postpone until later:
Order
Reason
Not Simple by Definition
Prime Powers
2 times odd
Prime powers
2 times odd
|
|
|
|
contradiction to Theorem 7.1 ii.
2 times odd
|
|
2 times odd
|
|
|
|
2 times odd
|
|
From now on, to speed things up, we omit prime powers and 2 times odd:
|
|
|
|
and |
no possible
|
|
|
|
No possible :
|
contradiction.
|
|
No possible
|
contradiction.
[
]
[
]
[
]
No possible
and
|
contradiction.
|
|
Must have
so
is abelian and
( ( )) contradicting BTT.
( )|
[
]
By Theorem 7.2 all non-abelian simple groups of order 60 are
isomorphic to .
Note that if we have some prime
and
numbers
and
with | |
then we cannot have
|| |
. To speed things up, from now on we exclude such
2 times odd
|
66
67
|
|
2 times odd
|
|
no
|
|
|
contradicts 7.1.
| ( )|
|
so ( ) abelian
contradicting BTT.
|
no
|
|
contradicts 7.1
| ( )|
contradicts BTT
|
|
|
( )|
so
by Proposition 7.1 but this
| |
ontra i ts La ran e’s Theorem e ause
|
by Proposition 7.1 but | |
so |
and since
is simple we apply 7.1 part 2 (to ) so
which
is a contradiction.
so | ( )|
contradicting BTT
|
| ,
so no
|
Contradiction to 7.1
| ( )|
( ) abelian Contradiction to BTT
No
No
No
|
|
|
( )|
|
contradicting
contradicting BTT
no
|
|
No
No
No
Contradiction to 7.1 part 3.
No
|
No
|
so
but
|
|
simple.
No
, no
| |
but
Remark
Recall that if a group with | |
and
prime and
and
then by Sylow,
so
abelian.
| ( )|
so ( ) abelian therefore
(
))
contradicting BTT.
(
|
so | ( )|
contradicting BTT.
No
POSTPONED
No
No
|
hence | ( )|
. By the Remark
(
)
above,
is abelian contradicting BTT.
POSTPONED
No
|
so
but | | | |
No
No
| ( )|
( ) abelian.
Contradiction to BTT.
|
|
so
ontra i tion to La ran e’s Theorem.
|
|
No
.
POSTPONED
No
.
contradicts BTT
no
No
No
contradicts BTT
|
|
No
| ( )|
so ( ) abelian contradicting BTT.
| ( )|
( ) abelian by
remark, so contradicts BTT.
No
so | ( )|
abelian so contradicts BTT.
Remember that we are omitting prime powers, 2 times odd and | |
with
for some | .
|
|
|
|
No
No
POSTPONED
No
No
POSTONED
No
No
|
No
|
but | | |
No
No
No
No
No
POSTPONED
| (
)|
so
|
(
) abelian, contradicting BTT.
All the above orders were covered by Theorems in the course and arithmetic. The postponed
orders need more Group Theory. In all cases, we consider the conjugation action of on
( ) for some suitable .
During all cases we will use the following notation:
( )
( )
{
}
( )
( )
Order
| |
We have
. So | ( )|
so ( )
by Proposition 3.14. As
( )
abelian, this contradicts BTT, so assume | ( )| ⟨
|
{ }| so
⟩
|
. Since | |
is a single 11-cycle. We may assume WLOG that
(
). | |
so must fix some element other than , say . Therefore
(
)
(
) as maps
so
we must have
( )(
)( )( )(
) but then
( ) which is a contradiction.
Lemma
Let be a finite group. Then || | and
( )
( )
Proof
By the Second Isomorphism Theorem,
implies
As
( )
( )
( )
( )
( )
( )
and
( )
, this implies that
( )
( )
so
( )
is a -group
is a -group.
maximal -groups and
( )
( )
. Moreover, we cannot have
( )
has order 1, so
since
and | |
| |.
Order
We have
.
contradiction, so
.
{
}. Since | |
( ),
( ),
Take
, fixes a unique point
,
( )
( )
since if fixes some other
by definition of the conjugation action,
{
}|
(
)
|
hence
contradicting the Lemma.
.
{ } have | |
The orbits of on
or ; since
there exists an orbit of length 3. Let
be in such an orbit. |
|
. Let
( ) then |
( )|
| |
|
|
by Orbit Stabiliser
Theorem. By the Lemma,
so
.
( )|
( )|
As |
so |
| ||| | || | | |
|
If | |
then |
Therefore | |
. Since |
| |
( ) but also
.
| ( )|
( )|
and |
and
and
( ) as
are both abelian. Let
. Therefore | |
, so
If | |
then
and so
contradiction.
contradiction to 7.1.
( )|
by BTT, has a normal 3-complement with
( ) and
( ) by Theorem 4.4 so | ( ) so
. Contradiction.
Order
⟨ ⟩
{
( )
. Let
(
} and let
). Let
or
. If
Therefore
so
so |
|
|
( )|
| |
.
with | |
. | ( )|
so assume
. Then
then
so action is not faithful. Contradiction to 7.1
( )(
)( )
( ). Contradiction to 7.1.
Order
} Take
( ) {
. Then ( ) | |
( ) { } consists of two 11-cycles. We may assume that
)(
). |
( )|
( )
( ) conjugation
have
( ) with
so there exists
.
. Then
⟨ ⟩| |
(
| ( )|
Case 1
{
} say
( )(
then
Case 2
{
} so permutes the set {
} and {
of these orbits so
. Contradiction.
)
(
)
so
( ).
( )
} so fixes at least one point in both
Order
.
} on
{
| |
| |
| |
{
( )
{ }. Let
,| |
( ) so |
}. As in the case | |
, has a orbit of length 3 say
( ). By the Lemma,
and
. But
and
( ) so
then by Theorem 4.4 ii
( ⏟ ( ))|
⏞
so |
( )|
( )
hence |
.
⏟ ( )|
contradicting 7.1.
Order
}. |
{ }|
{ } have size 2,4,8.
( ) {
so
the orbits of on
So there exists an orbit { } of size 2. Let
so | |
by the Orbit-Stabiliser
( ) so
( ) and hence |
( )|
Theorem and
then |
;
( ( ))|
contradiction since
.
This completes the Proof of Theorem 7.3