Group Theory. Contents

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 s... 31 rmal 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 tations 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 rmal Subgroup is called a normal subgroup of and is denoted if Quotient Group for all for all. 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 is an automorphism if is an isomorphism with

Kernel If is a homomorphism then the kernel of denoted { }. te 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 tation If and write the image of under as not, so we get so means first apply then.

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 : For denote o len ths o is oint y les. Therefore in the example above, 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. 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 permutation is both even and odd. t 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, hence so We will later prove that is simple for

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 and for all and for all. For the map is a permutation of (since, we have as inverse maps so every permutation is a bijection). Then is a homomorphism since. Denote by (the action of on ). By the First isomorphism, { } Faithful Action The action on is called faithful if or equivalently if and 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,

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 are inequivalent., we conclude that all actions of 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 Transitive. Similar for { }. 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 { } 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 Let act on and. Define the stabiliser of in as { }

and is a subgroup of : Let then so so is a subgroup. te 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. 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. tation 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 iii. If is -transitive and then for any 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

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 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. te 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. tice 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. 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.. w is clearly surjective because is transitive. is injective because. For equivalence of actions; ( ). The Conjugation Action acts on by. 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. te that { } so the action is not transitive if. { } { } otherwise known as the Centraliser of. te 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 tation: 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. 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 and. It then follows that.

Example Let { } and and As and have the same cycle type take then satisfies. te 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 ). te 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 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. 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. w 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. That is for some hence so and so.

Chapter 3: Sylow s Theorems Product of subsets: If then { } Lemma 3.1 Let i. If or then ii. If and then 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: te that if then so. Theorem 3.2 (Second Isomorphism Theorem) If and then Let be the canonical epimorphism. Then consider the restriction to. Then { }. 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. 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.

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. 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. w 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 subgroups of. As then because, so by Case 1 contains no 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 order, in particular mod is the same for the cyclic group of order. Cyclic Groups

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. 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. 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. 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. 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 iff

By Corollary 3.7, iff iff iff for all iff Corollary 3.9 For, Corollary 3.10 (Frattini Lemma) If and then Direct Products Let be groups. Then define { }. Then. In particular Proposition 3.11 Let. Let { }. Then: i. Easy ii. If and with then iii. Every can be written uniquely as where Theorem 3.12 Let where and and. Then. 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 () hypothesis. () (). Therefore is a homomorphism. is surjective by If () then so therefore and so hence and thus. Therefore is injective and so is an isomorphism.

te {} 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. 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 ) We call an (internal) semi-direct product of.

te 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 semidirect product as follows: Given an action on, define { } with multiplication defined by where is determined by the action. Claim is a group. 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. Easy. Proposition 3.14 If with an odd prime then either or Let, an odd prime. Choose and then so and. Observe because. is a subgroup of by Lemma 3.1 and so we must have. Clearly

. So for some ; therefore is determined by. so if then so mo. So assuming we have or. The two possibilities give and. Proposition 3.15 If for some prime then is abelian and or is abelian by Sheet 2 Q1. Then let { }. Then so or. If then. Assume for all { }. Then choose { }, define and choose and let. Observe. Then or. But which is a contradiction. Therefore and so (because is abelian so and we use Lemma 3.1) but so y La ran e s Theorem so. 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. 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: If not, there exists with. Let so hence. Choose

so. has order so so. Also implies that. Claim Given and is defined uniquely. { } { },, { }, { } { } { } 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 both define a homomorphism so we get two possible semi-direct products:

non a elian ( 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.

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 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. w suppose or : { so we must ha e or { In this ase repla e y an et so Therefore the claim is true. There are four possibilities but two are isomorphic by interchanging get three groups: and. Therefore we w 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.

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. te 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 [te that and does not automatically imply that ]. 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. for all prime. 2. for all for all prime. 3. where where are the primes dividing 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

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. te that if is finite and with then for some maximal. t 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. 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. Let be maximal in. implies so.

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 [ ] tice 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. 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 so is abelian so [ ] by Theorem 4.5 part 3.

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 Automorphism Group. is the Outer Lemma 4.6 All Characteristic Subgroups are normal. 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 1. This is Lemma 4.6. 2. This follows from 3 and 1 3. Let. Then Claim The restriction of implies. injective and a homomomorphism immediately imply that is injective and a homomorphism. We only need to check surjectivity. Take then so because. Then

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, rmal 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: Define 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. te 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. 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. for all ( 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

1. [ ] is clear since hence is soluble. implies that the derived series for is a normal series [ ] 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 Because abelian so [ ] [ ] is abelian, 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 Let. Then [ ] [ ] [ ] [ ] Corollary 4.8.2 for all 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 1. is soluble iff Then so for some. 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 soluble. 4. Let be a nilpotent group. We now apply induction on. When we get the desired result. so is

For then by Theorem 4.3 1. So by induction and are soluble so is soluble by part 3. w observe we have an ascending chain of sets: elian roups ilpotent roups Solu le roups roups Composition Series and the Jordan-Holder Theorem Maximal rmal 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. t 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 for all.

Theorem 4.11 (Jordan-Holder) Any two composition series of a finite group are equivalent. 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. w 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: w by Case 1, the series and are equivalent so and the series and are also equivalent so and are equivalent using since 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 called the composition factors of. So in the Example above, it is { } and for 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. [ ] 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 s In some sense, Nilpotent Soluble groups can be regarded as a generalisation of Abelian Groups. By the previous chapter all their composition factors are. w 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.. Fix then let then take (so block. Lemma 5.2 ) then since is a block so so is a Let be transitive and a block. Then. In particular, is prime implies is primitive. 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

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. 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 So now we have - Remarks is primitive. 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. 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. [ ] 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 now by definition of, so. w 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. Let. If then { } implies and { } implies so. Similarly, if then this implies that is transitive since was arbitrary. Then by Lemma 5.4,. w 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. rmal 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. 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. 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 Remark is transitive. 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, 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. Lemma 5.7.1 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. w for, if with then so so. was arbitrary so by definition is a regular normal subgroup. 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. 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 has no regular normal subgroup for.

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. 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 { } { } ( { }) so as is normal 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. w 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 2. The kernel of this action is the subgroup { } of scalar matrices.

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 The order of, and so the order of is { } Let then. For which is So is a multiple of ; there are possibilities. So { { } so

Example so. The small are: Theorem 5.11 For is simple for any field, except for and or 3. Step 1 is -transitive. 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 for all so with respect to the basis ( ) the matrix

becomes so is a transvection. Observe which is also a transvection. Recall that matrices of the form are called elementary matrices. Here is a zero matrix but with a 1 in position. These correspond to elementary row and column operations of the first type. Step 3 is generated by transvections. From Linear Algebra this is equivalent to the statement any can be 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 then. w replace column 1 by column so then. w 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. w applying a proof by induction on, the result follows. The case (the product of 0 transvections). is clear Step 4 is perfect if except for, or. We recall perfect if and only if [ ]. By Step 3, it is sufficient to prove that all transvections are contained in [ ] where but [ ] char so [ ] by Theorem 4.7. Furthermore, using the fact that all transvections are conjugate in, the above says it is sufficient to find a single transvection in [ ].

For the case, we compute directly that [ ] so when there exists with so [ ] contains a transvection. When and we use a similar argument: [ ] For the case and or we observe that: [ ] That is [ ] 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 that is abelian and. Then is simple. [te that this is also used in analogous proofs for simplicity of other classical groups.) Let with. primitive implies transitive by Corollary 5.7. Then for, by Theorem 5.5, is maximal so or. In particular, so. Therefore for all so hence. Then w this is abelian since is abelian so [ ] by Theorem 4.5. But is perfect by hypothesis so. As arbitrary, we have shown simple. Step 6 Completion of the We apply Iwasawa s Theorem with and { - }.. By Step 1 we know that is 2-transitive hence primitive by Theorem 5.3. By Step 4, is perfect except for the cases and or. It remains to find the subgroup. Take. Then { that is for. Therefore

{ } Let be the set { }. Denote by the expression. It is elementary to check that so is an abelian subgroup of. Define a map by. This is a homomorphism: So { }. te that. Hence is the kernel of some homomorphism; is a normal subgroup;. It remains to show that. Let be a transvection. If we show we are done since by Step 3, is generated by all transvections. We know that there exists with. Let. Observe for the case that: And. So every transvection is conjugate in to some element of. For the general case, again let then take so that is. Then so every transvection is conjugate in to an element of. Therefore all transvections are contained in for some. By Step 3, is generated by transvections so. There ore y Iwasawa s Theorem is simple.

A Closer Study of for small. First we recall some facts about the case : { } { } the order of this is Let, and. by the Orbit-Stabiliser Theorem. { }. Let { }; the subgroup in the previous proof. Hence and. Recall for some prime so is a Sylow -subgroup of ; that is. Let { }.,. Define By Theorem 5.5 primitive implies that is maximal is so. We write down the Generators for small ; In this case for and. { } { } So so similarly. Therefore. w { } { }. Then as before where and and Then and so and { } where and. Then { } with. Let and so for We have and and. Clearly, and so.

. We have { } with. We have, and where With and. We shall see later that. In this case and. { } with. We have, and where In the same way as before we determine that and and and is omitted, although it is no more difficult to compute. In this case { } where and. In this case and. We can take { } where, and. We choose { } where. We have, and where We can then calculate that

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, for some giving a permutation of. depends on and. We can write permutation of { }. giving a We can write for some. Define [ ]. We will prove this is a homomorphism order of in the product.. Since is a elian, it oesn t matter a out the [ ] [ ] Lemma 6.1 is a homomorphism. Let. Then and for. Then ( ) But also,. Hence [ ] ([ ] ) ([ ] ) The map is a permutation of { } and so { } { }. Since [ ] is abelian, we get [ ] [ ]. Therefore te is actually independent of the choice of coset representatinves (assuming ). Lemma 6.2 For, [ ] where { } { } and and each Rather than a proof, I will give an example: Example: then

Therefore: So [ ]. Here and. rmal -complement Let be a finite group,. Then is called a normal -complement in if and so 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) Let be a finite group, and. If ( ) (that is in the centre of its normaliser) then has a normal -complement. So unless, then is not simple. te that ( ) and so is abelian. Therefore BTT can only be applied if of BTT is abelian. We need two Lemmas. Lemma 6.4 Let be a finite abelian group with ( ) then mapping is an automorphism. The map is clearly a homomorphism. If then. As ( ) this implies that. Hence is injective. As is finite, is an isomorphism and thus an automorphism. Lemma 6.5 Let be finite, and abelian. If are conjugate in then they are conjugate in. This is not true in general if is not abelian. We have for some. So and. Since are abelian,. Therefore ( ). By Sylow Conjugacy applied to there exists with so. w. That is, they are conjugate in. ( ) therefore is abelian so [ ]. So the transfer. By Lemma 6.2, for then where { } { } and and and so by Lemmas 6.4, 6.5, and hence are conjugate in

that is for some. But ( ) so and thus for, We have so ( ) since so by Lemma 6.4, is an automorphism of. So let (and so is normal). As is an automorphism,. Let, since is surjective, there exists with so 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!)] Take then so { } so for, so therefore hence is not simple by BTT.

Chapter 7: Classification of Simple Groups of order 500 Proposition 7.1 Let be a finite non abelian simple group. 1. If acts on with then is faithful (that is ) and and 2. If with then, and (where means isomorphi to a subgroup of) 3. If, then for. 1. Let be the kernel of on. Then (since ) so simple implies that and hence. Suppose then using the second isomorphism theorem Hence. so and hence simple implies. Therefore so is abelian which is a contradiction. Hence. If then is soluble so it has a subnormal series with abelian 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. For, apply part 1 to the special case of the action of on by Theorem 7.2 conjugation. 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 1. Let be a simple, By Sylow s Theorem, and divides 15 so It cannot be 1, since then has by Corollary 3.8 which would give a contradiction. By Proposition 7.1 part 3, 3. If then for then 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 { }. Let. We may assume. w, there are only two groups of order 21; and. If is abelian then is not simple by BTT.. Recall that 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. so. Since, we have Hence. Let. and implies (as ). Lemma Let, let { }. If. Then { } Let and. We must prove that is for all then since. w using this notation, { } so { } that is fixes and or interchanges them. Since { }. w 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 since they are all of order 2. Recall and so. We now determine : So or. So and so by replacing with or. We can therefore assume wlog that. w but so w is transitive. so so is generated by. We have proved that. 3. n Examinable, but structured similar to the above with further complications. Theorem 7.3 The only non-abelian simple groups of order have order 60, 168, 360. If prime then is not non-abelian simple as by Theorem 2.9. and

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. Requires Character Theory. See Groups and Representations. During the rest of the proof, will denote a Sylow -subgroup of.. and We can now omit all possibilities except 6 cases, which we postpone until later: Order Reason t 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 possible : contradiction. possible contradiction. [ ] [ ] [ ] 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. te that if we have some prime with then we cannot have and and. To speed things up, from now on we exclude such numbers 2 times odd no 66 2 times odd 67 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 contradicting simple. contradicting BTT Contradiction to 7.1 part 3. so but

, 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. POSTPONED hence. By the Remark above, is abelian contradicting BTT. POSTPONED so but abelian. Contradiction to BTT. so ontra i tion to La ran e s Theorem.. POSTPONED. contradicts BTT no contradicts BTT so abelian contradicting BTT. abelian by remark, so contradicts BTT. so abelian so contradicts BTT. Remember that we are omitting prime powers, 2 times odd and with for some. POSTPONED POSTONED

but 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. Since { } so is a single 11-cycle. We may assume WLOG that. so must fix some element other than, say. Therefore so as maps we must have but then which is a contradiction. Lemma Let be a finite group. Then and By the Second Isomorphism Theorem, implies is a -group. As maximal -groups and, this implies that and so has order 1, so is a -group. Moreover, we cannot have since and. Order We have. contradiction, so. Take,, { }. Since, fixes a unique point, since if fixes some other by definition of the conjugation action, hence contradicting the Lemma. { }. The orbits of on { } have or ; since there exists an orbit of length 3. Let be in such an orbit.. Let then by Orbit Stabiliser

Theorem. By the Lemma, and and are both abelian. Let as so. As so. Therefore, so If then and so contradiction. If then contradiction to 7.1. Therefore. Since by BTT, has a normal 3-complement with. but also and by Theorem 4.4 so so and. Contradiction. Order. Let { } and let so.. Let with. so assume. Then or. If then so action is not faithful. Contradiction to 7.1 Therefore so. Contradiction to 7.1. Order. Then { } Take. Then so { } consists of two 11-cycles. We may assume that. have conjugation. so there exists with. Case 1 { } say then Case 2 { } so permutes the set { } and { } so fixes at least one point in both of these orbits so. Contradiction. Order. { }. As in the case, has a orbit of length 3 say { } on { }. Let and. By the Lemma, and. But contradicting 7.1., then by Theorem 4.4 ii so. so ( ) so hence Order so { }. { } the orbits of on { } have size 2,4,8. So there exists an orbit { } of size 2. Let so by the Orbit-Stabiliser Theorem and so and hence ( ) then ; contradiction since. This completes the of Theorem 7.3