Group Theory. Contents

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Group Theory

Chapter 1: Review

Notations

 or means that is a subset of (not necessarily subgroup), while

or means that is a subgroup of .

 If and , { } is a right coset while { } is a left 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 in and is denoted | |

Normal Subgroup

is called a normal subgroup of and is denoted if

for all _{for all} _.

Quotient Group

If we can define the quotient group { } the set of cosets with group operations and

All groups have normal subgroups { } and . A group is called simple if { } and and are the only normal subgroups (this is equivalent to having exactly two normal

subgroups).

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 with and .

Homomorphism

A map is a homomorphism if ( ) ( ) ( ) for all . In addition:

 is a monomorphism if ( ) ( )

 is an epimorphism if ( )

 is an isomorphism if it is both a monomorphism and an epimorphism

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Kernel

If is a homomorphism then the kernel of denoted ( ) { ( ) }. Note that ( ).

Claim

is a monomorphism iff ( )

Claim

For , the “quotient” map given by ( ) is an epimorphism.

Theorem 1.3 (First Isomorphism Theorem)

Let be a homomorphism and denote ( )

i.

ii. ( ) (not necessarily normal)

iii. The map ( ) 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_{〈 〉}

More generally for { } , 〈 〉 is the subgroup generated by 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 . Define ( ) be the group of all permutations and ( ) where | |

Notation

If ( ) and write the image of under as not ( ), so we get ₍ ₎

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Example

Take { } and denote ( ) as follows:

_{( )( )} _{( )( )}

In cyclic notation

( )( ) ( )( )

So and have a cycle of length 2 and a cycle of length 3 hence and have the same cycle type? Why? is a conjugate of : ( )

Transposition

A permutation of the form ( ) is a transposition .

Lemma 2.1

If is a finite set, any permutation of is a composition of transpositions. I.e. is generated by transpositions.

Proof

Take ( ) and express it as a product of cycles, so it is enough to express cycles as a composition of transpositions:

( ) ( )( ) ( )

Even and Odd Permutations

( ) is called even or odd if is a product of an even or odd number of transpositions respectively.

Theorem 2.2

No permutation is both even and odd.

Proof

Not instructive, but in the lecture notes.

Clearly, e en e en o o e en and o e en e en o o , so there is a homomorphism ( ) { } where ( ) if is even and ( ) if is odd.

Observe that ( ) ( ) { ( ) is e en} is a subgroup of ( ) by the First Isomorphism Theorem.

Also by the First Isomorphism Theorem if | | , ( )_{( )} ( )_{( )} ( ) so

| ( ) ( )| hence | ( )| | ( )|

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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 is a map mapping ( ) (not ). By definition of an action, we have ( ) ( ) for all and for all . For

the map is a permutation of (since ( ) _{, we have}

and as inverse maps so every permutation is a bijection). Then ( ) is a homomorphism since ₍ ₎ _.

Denote ( ) by (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 subsets of : actions on erti es { } ( )

_{( )( )}

On e es { }

( ) ( )( )( )

On ia onals { }

( ) ( )

Observe that _and _{are faithful but} _{is not. It has kernel}_{ _}_{and in fact,}

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Equivalent Actions

Actions of on sets are called equivalent if there is a bijection such that

( ) ( ) and .

By examining the cycle type of the three actions on , we conclude that all actions of

are inequivalent.

Orbits and Transitivity

Let act on . Define for if there exists such that . This is clear to be an equivalence relation. The equivalence classes are called the orbits of on . The orbit of is written as .

Example

Let { } and take ( )( )( ) and ( )( )( ) where . The orbits of 〈 〉 are { } { } { } { } while the orbits of 〈 〉 are { } { } { } { } { }. The orbits of 〈 〉 are { } { }. This is deduced because so also

so hence . Similarly, because so are all in the same orbit but applying to all give results so { } is

. Similar for { }.

Transitive

is called transitive if is an orbit; that is if for all . Equivalently, for all

there exists so that .

Observe that 〈 〉 in the previous example is not transitive.

-transitive

For , is called -transitive if: 1. | |

2. Given , define pints and distinct . Then there exists so that for .

Observe that 1-transitive is the same as transitive and ( )-transitive implies -transitive.

Examples

1. is -transitive on { }

2. , the alternating group is ( )-transitive on { }

Proof

Take distinct and . Then { }

where and are the other two points. Also { }.

Certainly there exists a unique with for . If then we are done. Otherwise let be the transposition ( ) then and so

and for

Stabiliser

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and ( ) is a subgroup of :

Let ( ) then so ₍ ₎ _so _{( )}_{is a}

subgroup.

Note that the kernel of the action is { } so ⋂ is the kernel of the

action.

Theorem 2.3 (Orbit-Stabiliser Theorem)

If acts on with finite and then | | | || | or equivalently | | | |.

Proof

Suppose so there exists with . may not be unique so for which is

?

So is in the same right coset of as . So this defines a bijection between and the right cosets of in . So | | | | as claimed.

Notation

For , denote the two point stabiliser. Similarly,

If denote

( ) { } pointwise sta iliser { } setwise sta iliser

Example

Take { } and ( )( ) and { } then _{( )}and , but if

{ } then _{( )} but . In general, it is true that ( )

Corollary 2.4

If acts on , then

i. | | is constant for in an orbit. ii. is transitive implies | | | || |

Proof

i. This follows immediately from the Orbit-Stabiliser Theorem ii. This follows immediately from part i.

iii. We use induction on .

When , | | | | from part ii.

Assume true for ; that is is -transitive implies that { } is ( ) -transitive straight from the definition since | { }| . So by induction

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Specific Actions – The Right regular and coset actions

Right Regular Action

Let be any group. Take . For and we define . This is clearly an action. Given , so it is transitive. The stabiliser ( ) { } { } so it is faithful.

Coset Action

A generalisation is a coset action. Let and take { }. Define ( )

so { } gives the right regular action. Note that ( ) 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. 〈 〉 { }. We have two cosets { }. Then and

( ) ( ) so { } kernel of the action so it is not faithful. Observe that so kernel so 〈 〉

ii. 〈 〉 { }. We have three cosets { } then ( ) is a 3-cycle on . Notice that so ( ) . Then

( ). Therefore | | | | so the action is faithful.

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 and . Let { } we have a coset action on . Define by ( ) for . We claim that is a bijection:

First we check that is well defined: i.e. . Now

. is clearly surjective because is transitive.

is injective because ( ) ( ) .

For equivalence of actions; (( ) ) ₍ ₎ _{( )} _.

The Conjugation Action

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Remark

The default meaning of for is .

_{is a conjugate of} _in _{. The orbits are called the conjugacy classes of} _{and denoted} ( ). Note that { } so the action is not transitive if | | . ( ) {

_{} { } ( )}_{otherwise known as the Centraliser}

of .

Note that ( ) ( ) { }

As a generalisation we can take { } the subgroups of and define an action by

Then ( ) { _} _{( )}_{called the normaliser of} _in _{. Also} _{( )}

is the centraliser of in is the set { }

Remark

It can be proved that ( ) ( ) and ( ).

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

Let ( ) suppose and . Then _{; that is}

_{so if a cyclic decomposition of} _is₍ ₎₍ _{)( )}_then _is₍ ₎₍ _{)( )}_.

Example

Let ( )( ) and ( )( ) then ( )( ). In particular note that conjugate permutations have the same cycle type.

Theorem 2.7

Two permutations ( ) are conjugate in ( ) if and only if they have the same cycle types.

Proof

From the above, we conclude that is true.

Conversely, take ( ) of the same cycle type: that is ( )( )( )

and ( )( )( ). Then simply choose ( ) so that for all . This is possible because all and all appear exactly once in the cyclic decomposition of

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Example

Let { } and ( )( ) and ( )( ) As and have the same cycle type take

( )( ) then satisfies _{. Note that} _{is not unique.}

-groups:

For prime, a finite group is called a -group if | | for (for infinite is a -group if all elements have order a power of ).

Note for that ( ) but for { } ( ) { }

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 then ( ) . In particular if take to conclude that ( ) .

Proof

By Lemma 2.8, ∐ ( )for some . This is a disjoint union because conjugacy classes are either equal or disjoint. Choose so ( ) { }. We have

| | ∑| ( )| ∑| ( )| ∑| ( )| 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 .

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Chapter 3: Sylow’s Theorems

Product of subsets:

If then { }

Lemma 3.1

Let

i. If or then

ii. If and then

Proof

i. Suppose . Take and . Then

( )( ) ( ₎₍ ₎_{and since} _{is normal,} _so _so₍ ₎_{. Similarly}

( ) ₍ ₎ _since _{by normality.}

Therefore .

ii. 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 be the canonical epimorphism ( ) . Then consider the restriction to . Then ( ) { } . Moreover, ( ) so applying the First Isomorphism Theorem, we yield the result

has no subgroup of order 6 but | | . However the answer is yes if is a prime power.

Sylow -subgroup

Let | | , prime and for . A subgroup of order is called a Sylow -subgroup of . Denote ( ) { | | }

Theorem 3.3 (Sylow’s Theorem)

For a finite group let | | , prime and for . i. ( )

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ii. (containment) Any -subgroup of is contained in some Sylow -subgroup of . iii. (conjugacy) If ( ) then there exists with

iv. (number) | ( )| mo

We will pro e Sylow’s Theorem in two separate Theorems. The irst one implies .

Proposition 3.4

Let be a finite group and || |. Then the number of subgroups of of order

( ).

Proof

Let { | | }. Let | | where . Then using elementary combinatorics,

| | ( ). If and then { } so we can define an action of on by right multiplication. That is . This is clearly an action.

Let be an orbit of . If and then there exists _{. (I.e. there exists}

an element with ). If ( ) then by definition so that is

( ) . Now we consider two cases:

Case 1:

( ) . Then is a subgroup since ( ) and the orbit { }

is the set of right cosets of . Then | | _{| | | |}_{by the orbit-stabiliser}

theorem. In particular, is the only subgroup in the orbit.

So if is a subgroup of order then the orbit of has size _.

Case 2:

Suppose ( ) . Then | | _|| |_{( )|} | |_{| |} _{, so by Case 1} _{contains no}

subgroups of . As | | _then _{| | |}_because_{| ||| |}

Let { | | } then we have shown that equals the number of orbits of size

_{and the remaining orbits in} _have _{|| |}_.

Then for some , | | _therefore | |

( )

As is prime there exists a unique inverse { } with mod . Therefore

| | ( ) (

)

( ) It then can be proved using elementary number theory

that (

)

( ).

we can avoid this by using the following argument due to G. Higmann

The number (

)

is a function of | | and so it must be the same for all groups of

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have a unique subgroup of order for all || | so for a cyclic group so mod for all groups of order | |.

Proposition 3.5

Let ( ), a -subgroup of that is | | is a power of . Then _{for some} .

Proof

Take | | with . Let {ri ht osets o in }. acts on by right multiplication that is ( ) . This is clearly an action.

Consider the restriction of the action to . Orbits of on have size | | | ( )|

where . This is a power of , possibly .

Suppose the orbit of has size 1. That is ( ) therefore _.

Proof of Theorem 3.3

Follow immediately from the conclusion of Proposition 3.4

Since _{is a subgroup of order}_{| |} _{we have} _{( )}

If ( ) then | | | _|_{so we must have} _{by Proposition 3.5 and}

thus _.

Corollary 3.6 (Cauchy’s Theorem)

If is a finite group and || | for prime then has an element of order .

Proof

Choose ( ). Take { }. Then | | for some . Then and

Then ( ) so ( _{) |} _hence _for_{and so}_so | . Then as we must have . That is | | .

Corollary 3.7

Let | ( )| . Then | | | ( )| where ( ) { _}_{In particular} | | | | |.

Proof

Let ( ) and let act on by conjugation. By definition, ( ) ( ). By Theorem 3.3, part iii, this action is transitive. By the orbit stabiliser theorem we have

| | | | | ( )|

In particular as ( ) by definition we have | | | ( )| and so

| |

| | | | | | |

| | | |

Corollary 3.8

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Proof

By Corollary 3.7, | ( )| iff _|_{( )|}| | iff ( ) iff _{for all}_iff

Corollary 3.9

For ( ), ( ) ( )

Proof

Corollary 3.10 (Frattini Lemma)

If and ( ) then ( )

Proof

Direct Products

Let be groups. Then define {( ) }. Then

Proposition 3.11

Let . Let {( ) } . Then:

i.

ii. If and with then

iii. Every 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,}₍ ₎_and _because

is normal so . So and so that is .

Define by ( ) . First (( )( )) ( ) (( )) (( )). Therefore is a homomorphism. is surjective by hypothesis.

If (( )) then so _therefore_{and so}_hence

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Note {( )} as in the definition is an external product of 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 with . For each , let ̂ ∏

(the product of all except ). Assume ̂ for all . Then .

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, ⟨ | _⟩_{for some}_{es ri es the “lar est” roup}

that is generated by the elements 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.

However ⟨ | _⟩_for_{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 on correspond to homomorphisms ( ) and is the action of ( ) on .

If are groups we can define a group action of on in which ( ) ( ). That is

( ) 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 and and (as in Theorem 3.12 but not assuming )

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Note that conjugation of elements of by elements of defines a group action of on .

_{(easy to check it satisfies the three axioms above)}

( )( ) ( ) so multiplication in is completely determined by the action . We write or denotes the semi-direct product with action .

Conversely, given groups and an action of on we can define the external semi-direct product as follows:

Given an action on , define {( ) } with multiplication defined by ( )( ) ( ) where is determined by the action .

Claim

is a group.

Proof

Closure is clear, the identity is ( ) and ( ) ₍ ₍ ₎

. Associativity is also easy to check.

Examples:

Take 〈 〉 〈 〉. Then ( ) 〈 〉 where ( ) . Then ( ) and

( ) and ( ) so ( ) . [In general ( ) for prime]

 Define ( ) mapping ( ) then

⟨ | _⟩

 ( ) ( ) 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 with | | (except 16).

Claim

If | | prime then .

Proof

Easy.

If | | with an odd prime then either or

Proof

Let | | , an odd prime. Choose ( ) and ( ) then so ⟨ | ⟩ and ⟨ | ⟩. Observe because | | .

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. So for some ; therefore is determined by .

( ) ( ) so if then so |( ) mo . So assuming

we have or .

The two possibilities give ⟨ | _⟩ and ⟨ | _⟩

.

If | | for some prime then is abelian and or

Proof

is abelian by Sheet 2 Q1. Then let { }. Then | ||| | so | | or | | . If | |

then 〈 〉 .

Assume | | for all { }. Then choose { }, define 〈 〉 and choose

. 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 | | for all { } then is abelian.

Proof

Take { }. Then ( ) . Moreover so .

| |

Take as | || we must have | |

Case 1:

Assume | | for all { }. Then by the Lemma, is abelian, so is a direct sum of cyclic groups; that is either or or . As | | for every

{ } the only possibility is that .

[In General if is abelian and for all { } for prime then writing with the 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:

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so . has order so ₍ ₎ so . Also implies that

_.

Claim

Given and _{is defined uniquely.}

Proof

{ } { } _, _, ₍ ₎ { }, { } { } _{ _}_{so the Claim is true.}

The possibilities are that . If or then | | and so . We now consider the possibilities that or .

Observe that | _{| | |}_so _or _.

Case 2a

If _then _{is abelian.}

 If ⟨ | ⟩ since has no element of order 8.

 If then ( ) . Take since was arbitrary in we can replace by and get .

Case 2b

In this case _and _{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, | | or

| |

By Sylow’s Theorem | ( )| . Also by Corollary 3.7, | ( )||| | so | ( )|

or .

Case 1

If | ( )| then ( ) { } by Corollary 3.8. Since | | we may take

⟨ | ⟩

Let ( ). | | so or by Proposition 3.15.

Case 1a

Take ⟨ | ⟩. Then ; the group automorphism is determined by

_Either _or _.

More formally, ( ) 〈 〉 with and ( ) _{. Can have}_{( )}_or_{( )}

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⟨ | _⟩ ⟨ | _⟩_{non a elian}

(No need to prove the existence of these groups; their existence is guaranteed by the Semi-Direct Product.)

Case 1b

Take 〈 〉 〈 〉 again we want ( ). There are four possibilities:

( ) ( ) ( ) ( )

If ( ) ( ) _{( )} then

⟨ | _⟩

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 .

If ( ) ( ) then ( ) . Since 〈 〉 〈 〉 we can simply replace by 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 { } ( ). By Sylow’s Theorem they are all conjugate; that is acts transitively on by conjugation. That is there is a homomorphism ( ).

Let 〈 〉. By Corollary 3.7, | ( )| _{| ( )|}| | for any ( ) so | ( )| ; (as

( )) hence ( ) for and so ( ) for . Then o we can assume and ; that is ( ) and fixes .

If 〈 〉 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.

| |

By Sylow’s Theorem, | ( )| and by Corollary 3.7 | ( )|| so then

| ( )|

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.

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| |

| ( )| mo and | ( )| | so | ( )| . Therefore ( ) | | and

.

Take ( ), ⟨ | ⟩. So 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 and we can assume _or _{and similarly} _or _.

Sketch Proof

In general, if _{then as} _{we have} _{and so} _{( )} (since is abelian). Moreover, 〈 〉 〈 〉 so we can replace by and by . So we can assume _or _{. Now suppose} _or _:

{

_{( )}

so we must ha e _or

{

_{( )} In this ase so

repla e y _{an et}

Therefore the claim is true.

There are four possibilities but two are isomorphic by interchanging and . Therefore we get three groups:

⟨ | _⟩ ⟨ | _⟩

Now observe that in the element has order 6 but no element of has order 6 because

| | for all and so .

Summarising we get five groups of order 18; 2 in case 1 and 3 in case 2, two of which are abelian.

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| |

By Sylow’s Theorem | ( )| and | ( )|| so | ( )| and so ( ) and . Take ( ), then . Where ⟨ | ⟩. There are two possibilities for ; either or .

Case 1

Assume . Then | ( )| and the elements of ( ) are the maps . For the possible ( ) ( ) can be any element of ( ) so we get 4 possibilities:

If ( ) then

If ( ) ( ) then ⟨ | _⟩_{. As} _{we have} ₍ ₎ ₍ ₎ _{and so}

So by replacing by _{we get the same semi-direct product as the previous case.}

The final case is that ( ) ( ₎_{. Then} _{⟨ |} _⟩

Observe that in , ₍ ₎ _so _centralises _{; in fact} ₍ ₎_and _has

order 10. But in , _and₍ ₎_and _{has no element of order 10 so} .

Case 2

Suppose 〈 〉 〈 〉 as usual we get _and _{similar to} | | case 1b. This leads to two possible groups

and ⟨ | ⟩

Therefore we get 5 groups of order 20: three in case 1 and two in case 2.

| |

| ( )| and divides 3 so | ( )| hence ( ) and denote

⟨ | ⟩.

Let ⟨ | ⟩ ( ). Then .

Need action ( ) and ( ) so ( ) for some with . Then

so .

If _so_and_{give isomorphic groups (swap} _{). So we}

get two groups corresponding to the cases and respectively:

⟨ | _⟩

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.

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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

2. for all ( ) for all | | | prime.

3. 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 | | ( ) so by hypothesis.

Let ̃ (excluding ). Then | | | | | || | | | | | where | | . Then ̃ by Lemma 3.1. In particular for we have | | ̃| so ̃ | | and thus ̃ since it is divisible by all . In addition, ̃ since orders of all the elements are coprime. Then by Corollary 3.13 .

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 , then ( ) (nontrivial centre) 2. If then is nilpotent

3. If then is nilpotent

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1. Assume . As for and , so by Theorem 2.9 ( ) . Take ( ) { } then we have ( ) ( ).

2. Let ( ) for some hence because nilpotent and so . By the second isomorphism theorem which has order coprime to , so ( )so the Sylow -subgroups of are normal in and so is nilpotent. 3. Let ( ) for some and so . Then for any by Lemma 3.1

therefore and so by the third isomorphism theorem which has order coprime to so ( ). Hence the Sylow p-subgroups of are normal in

and so 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 implies or . Note that if is finite and with then for some maximal . Not all infinite groups have maximal subgroups.

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 _{( )} _{( )}. By induction _{( )} nilpotent so _{( )} ( )

(

( )) ( ) for some .

So _{( )} _{( )} implies that so ( ). Case 2:

( ) but ( ) ( ) and also ( ) so ( ) ( ) and ( ) so

( ).

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( ) ( )

Assume ( ). If is not nilpotent, then for some ( ) so ( ) so ( )

for some maximal in . By condition iii, ( ) but ( ) by Corollary 3.9. Contradiction.

Commutators and Commutator Subgroup

Commutator

Let . Define the commutator to be [ ]

Notice that [ ] and [ ] _{[ ]}_.

Commutator Subgroup

The Commutator subgroup of , denoted [ ] is by definition

[ ] ⟨[ ]| ⟩

In general, the set {[ ] } is not a subgroup. The smallest counterexample is of order 32.

Theorem 4.5

1. [ ]

2. _{[ ]} is abelian.

3. If 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).

2. For all , _{[ ]}_therefore_{[ ] [ ] [ ] [ ]} [ ] [ ] so _{[ ]} is abelian.

3. abelian implies for all therefore _so [ ] .

Remark

It should be clear that [ ] is abelian. is called perfect if [ ]. It follows that non abelian simple groups are perfect.

Examples:

is done in the notes, and [ ] on Sheet 5.

Take then [( ) ( )] ( )( )( )( ) ( )( ) [ ] so similarly ( )( ) [ ] and ( )( ) [ ]. Therefore

{ ( )( ) ( )( ) ( )( )} [ ]

In fact because it is a union of conjugacy classes. Moreover

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Characteristic Subgroups

Characteristic

A subgroup is called characteristic in if for all ( ). We abbreviate this to .

For any group and any the map (conjugation by ) mapping

is an automorphism and so ( ).

Inner Automorphisms

The group of inner automorphisms is denoted ( ) { }

Observe that this truly is a group because and ( ) _so_{( ) ( )}

In fact it is a normal subgroup: Let ( ). Then (

) ( ₎ _so _{and thus}_{( ) ( )}_.

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 for all ( ). In particular, for all ( ) ( ) so this means _{for all}_{and so}_.

Theorem 4.7

Let be a group. Then: 1.

2. and

3. and

4. [ ]

5. ( )

6. ( ), and finite

Proof

1. This is Lemma 4.6. 2. This follows from 3 and 1

3. Let ( ). Then

Claim

The restriction of implies ( ).

Proof

injective and a homomomorphism immediately imply that is injective and a homomorphism. We only need to check surjectivity.

Take then ( ) _{( )}_so

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so so is surjective.

Therefore _since_so_.

4. Let ( ) so [ ] ( ₎ _[ _]_so _{permutes the Commutators}

then [ ] ⟨[ ]| ⟩ so [ ] [ ] that is [ ] char .

5. Take ( ) and let ( ) then [ ] for all implies [ ] for all so [ ] for all . As is a bijection from to we have

[ ] for all so ( ). Hence ( ) ( ) that is ( )

6. Take ( ). If | ( )| then there exists a unique ( ). Then

| | | | ( ) 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 is called a

subnormal subgroup of ). It is called a normal series if for all .

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 is 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 ( ) for all by Theorem 4.7 hence this series is a normal series.

The series may stabilise ( ) ( ) for some . (this is always true if is finite) but for infinite groups this may not be the case e.g. free groups.

Theorem 4.8

1. is soluble if and only if ( ) for some .

2. is soluble if and only if has a normal series with abelian factors

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1. [ ] is clear since ( ) implies that the derived series for is a normal series hence is soluble.

[ ] Let be a subnormal series with

abelian. We prove by

induction that ( ) for all . Take then ( ) .

Assume true for then by induction and using the fact that is abelian,

( ) _[( )( )_{] [} _]

Because abelian so ( )

2. [ ] by definition

[ ] If is soluble then ( ) for some by 1. But then the derived series is a normal series with abelian quotients.

Lemma 4.8.1

Let . Then [ ] [ ]

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. is soluble iff ( ) for some .

Then ( ) ( ) so ( ) so is soluble.

2. soluble implies ( ) so ( ) { } then by Corollary 4.8.2, ( )( ) { }

so is soluble.

3. If is soluble then ( ) ( )( ) for some so ( ) .

If is soluble then ( )_{for some} _{and so} ( )( )_so _is

soluble.

4. Let be a nilpotent group. We now apply induction on | |. When | | we get the desired result.

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For | | then ( ) by Theorem 4.3 1. So by induction _{( )} and ( ) are soluble so 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 if and implies or .

Simple Group

A Group is simple if and implies or . So is a maximal normal subgroup is simple.

A subnormal series is a composition series of if each is

simple. Not all groups have them for example ( ).

Example:

Take ⟨ | _⟩

What are the maximal normal subgroups of ?

is simple, hence cyclic of prime order so | | or . Since ( ), since

implies abelian. So | | so | |

So | | so there is a unique subgroup ( ), 〈 〉 . Then 〈 〉 〈 〉

has three subgroups of order 2; 〈 〉 〈 〉 〈 〉 so there are three possible so that | | :

〈 〉 〈 〉 〈 〉

We consider these in turn:

〈 _〉

There is no normal subgroup of order 2 so 〈 〉 so the composition series is of .

Similarly, with composition series

Since _{〈 〉 〈 〉}

which has two normal subgroups 〈 〉

and 〈 〉 so we have two composition series

〈 〉

Therefore has four composition series altogether.

Equivalent Series

Let ( ) and ( ) be two composition series of . Then we say those series are equivalent if and there is a permutation of

{ } such that ( )

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Theorem 4.11 (Jordan-Holder)

Any two composition series of a finite group are equivalent.

Proof

We apply induction on | |. When | | this is clear. Let ( )and( ) be two series as above.

Case 1

If then the result follows by applying induction to and .

Case 2

If so (because and both normal in and applying Lemma 3.1). Moreover simple and simple implies and are both maximal normal subgroups of

, so the previous statement proves . Now by the Second Isomorphism Theorem

Let and let be a composition series for we then construct the following diagram:

We have four composition series in the diagram:

( ) ( ) ( ) ( )

Now by Case 1, the series ( )and( ) are equivalent so and the series ( )and( )are also equivalent so

( )and( )are equivalent using ( ) since and

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Composition Factors

The multiset of isomorphism types of factors in the composition series of are

called the composition factors of . So in the Example above, it is { } and for it is

{ }

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.

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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 is a subset with | | and but that for all ,

or .

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 { } 〈( )〉 then { } and { } are blocks but { } is not. More generally, for ( ) on { }, if | and then { } is a block. e.g. taking , then { } { } is a block.

Lemma 5.1

If is a block then is a block for all .

Proof

| | | | | | | |. Fix then let then take _(so _{) then}

since is a block so _so _{is a}

block.

Lemma 5.2

Let be transitive and a block. Then | ||| |. In particular, | | is prime implies is primitive.

Proof

By Lemma 5.1, the sets { } partition and | | | | for all hence | | | |. So { } form a block system.

Example

Let { } take ( ), where [

]

{ }

Then in the case we have [

]. Then ( )( )( ) and

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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 .

Suppose is a block and so | | . Let . Let { }. By 2-transitivity there exists so that . Therefore so a block implies that . But so . As { } arbitrary, hence . 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 ( ) ( )).

 For not prime then 〈( )〉 is transitive but not primitive.

Lemma 5.4

If is transitive and and there exists a subgroup with and transitive, then .

Proof

Let as is transitive then there exists with so _.

Therefore so .

Theorem 5.5

Let | | and transitive. Then is primitive if and only if is a maximal subgroup of for any .

Proof

[ ] Take and assume is maximal in . Let be a block and define { }.

Since the sets { } partition replace by 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

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Contradiction.

Hence there are no blocks and so is primitive.

[ ] Assume is primitive. Let for some . We want to prove or

.

Let . If | | then { } implies and { } implies so

. Similarly, if then this implies that is transitive since was arbitrary. Then by Lemma 5.4, .

Now we consider the case when | | | | we will prove is a block, giving a contradiction to primitive.

Let and so there exists such that . Since there exist

so that and . So

so _so . But then so ( )

So we have proved that if then so by definition is a block. Contradiction to primitive.

Normal Subgroups of Permutation Groups

Theorem 5.6

Let be transitive and with . Then one of the following holds: 1. { } and so

2. and so is transitive. 3. is a block.

Proof

Suppose { }. Then . So if and and then so _hence _{transitive and} _{for all}_{implies that}

for all so .

If then 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} _is

a block, so we have 3.

Corollary 5.7

If is primitive and with then 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.

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Regular Action

An action is called regular if it is transitive and for .

Regular normal subgroup

For a group action , if , and is regular then is a regular normal subgroup of .

Examples

1. Take { ( )( ) ( )( ) ( )( )}. is a regular normal subgroup of and .

2. Affine Groups:

Take for a field, we define a group ( ) ( ) of affine

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 we have ₍ ₎ ₍ ₎ _so with ( ) and ( ) . (Closure).

Given define by _and _then_so _.

As the composition of functions is associative and the identity map is in , we have a group.

Let { } be called the translation subgroup. Then for and

we have

( ₎ ₍ ₎

so _hence_.

is clearly transitive. Now for , if with then so so

. was arbitrary so by definition is a regular normal subgroup.

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 { }. As is regular, for all there exists a unique so that

. For , so _{. As} _{is 2-transitive then for all} { } there exists with so ( ) . So all elements { } are 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

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Proof

Suppose not and let be a regular normal subgroup. As in the previous proof there exist unique with for all { }. Let ( ) . Since

we have ( ) if and only if . Therefore ( ) { }. As

( ) we have | so ( ) . As we have so

and hence . 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; { } ( { }) so as is normal

( ) ( ) and as { } we have so . Hence 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, ( ).

Let be any field. Let ( ) be the group of invertible matrices with entries in , or equivalently, invertible non-singular linear maps . Now ( ) the

multiplicative subgroup of which is abelian. We usually assume and so ( ) is 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 {〈 〉 { }}. Then the action is defined 〈 〉 〈 〉. The Projective Action.

Theorem 5.10

1. ( ) is 2-transitive for

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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 ).

 Also, finite implies cyclic. Let ( ). Take 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 〈 〉 {∑

} so has size . Hence the number of

possibilities for is _{. Hence}

| ( )| ( )( ) ( _{) ∏(} ₎

( ) is | ( )| | ( )| | ( )| | ( )| | ( )|

| ( )|

| ( )| |{ }|

Let ( ) then 〈 〉. For which is _{( ) |} _| _|

So is a multiple of ; there are possibilities.

So ( ) { { ( ) ( )( )} so | ( )| ( )

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Example

| ( )| ( )( ) | ( )| | ( )|

( ) so | ( )| . The small ( ) 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 〈 〉 . Let ( ) and define ( ) by

_and _for_so_and_〈 _〉 _〈 _{〉 〈} _〉 _〈 _〉_.

Step 2 – Basic Properties of Transvections

( ) is a transvection if its matrix with respect to some basis of is

(

) that is the identity matrix but with a 1 in the ( ) position.

From linear algebra, changing the basis replaces by _{for some}_{( )}_,

which shows that all transvections are conjugate. Consider a matrix (

) that is the identity matrix with a in the

position ( ). If this matrix is using the basis then and