Topics in Groups Theory

Prof. SaxlTypeset by Aaron Chan(akyc2@cam.ac.uk)

Last update: March 10, 2010

Books:Alperin and Bell (Springer), John S. Rose (Dover), M. SuzukiRobert A. Wilson, The Finite Simple Groups, Springer 2009This course will follow book of Wilson’s, which is relatively harder to read.

1 Introduction

G is a group, finite; |G| its order.

Theorem 1.1 (Lagrange)G finite, H ≤ G ⇒ |H|

∣∣|G|

Sketch proofDistinct right cosets Hg = hg|h ∈ H cover G and all have size |H|(G : H) the set of all right cosets of H in G. Then |G| = |H| · |G : H|

The map θ : G → H is a homomorphism if (g1 · g2)θ = g1θ · g2θIt is an isomorphism!of groups if also bijectiveker(θ) = g ∈ G|gθ = eHθ is injective ⇔ ker θ = eGker θPG

K ≤ G is a normal subgroup of G, write K PG, if Kg = gK ∀g ∈ GG is simple if there are no normal subgroups other than 1 and G, equivalently, G simple ⇔ anynon-trivial homom from G is injectiveK PG, can form G/K, a quotient group (or factor group) on the set (G : K), by Kg1Kg2 = Kg1g2|G/K| = |G|/|K|

Theorem 1.2 (Isomorphism Theorem)If θ : G → H homomorphism, then ker θPG, Im θ = gθ|g ∈ G ≤ H, and there is an isomorphismG/ ker θ∼=Im θ

Sketch proofWriting K = ker θ, check K PG, Im θ ≤ H and

θ : G/K → Im θ

Kg 7→ gθ

is a well-defined isomorphism

1

So, homomorphic images of G are just quotients of GNote: If K PG, then

π : G G := G/K

g 7→ g := Kg

i.e. quotients of G are homomorphic images of G

θ : G //

π ##HHHH

HHHH

H Im θ

G/Kθ

;;ww

ww

Theorem 1.3 (Second Isomorphism Theorem - Correspondence Theorem)Let K PG. Then every subgroup H of G/K is of the form H/K for some unique H with K ≤ H ≤ GWe get a lattice isomorphism between all subgroups of G/K and all subgroups of G containing K

G/K G

H/K = H H

1G/K K

Moreover, H/K PG/K ⇔ H PG and if so G/KH/K

∼=G/H

ProofIf H PG, define a homomorphism

G/K → G/H

Kg 7→ Hg

This is a homomorphism surjective onto G/H, kernel H/K

Theorem 1.4 (Third Isomorphism Theorem)Let K PG, H ≤ G. Then HK ≤ G (HK := hk|h ∈ H, k ∈ K)H ∩K PH and H/H ∩K ∼=HK/K

ProofThe homomorphism

π|H : H → G/K

h 7→ Kh

has image HK/K, kernel H ∩K

The group of automorphisms is

Aut(G) = θ : G → G isomorphisms

with group operation being composition, we write gx for the image of g ∈ G under x ∈ Aut(G)

Inner automorphisms (conjugation autos) are, for g ∈ G, θg : x 7→ g−1xg for x ∈ G

Then θg is an auto of G and the mapθ : G → Aut(G)

g 7→ θgis a homomorphism.

2

Define the followings:

Group of inner automorphisms Inn(G) := Im θ P Aut(G)

Outer automorphism group Out(G) := Aut(G)/ Inn(G)

Centre of G Z(G) = ker θ = z ∈ G|zx = xz ∀x ∈ G

(Note that for G abelian, Z(G) = G, Inn(G) = e, Aut(G) = Out(G))

Exercise: Aut(D8)∼=D8,Aut(Q8)∼=S4,Aut(A5)∼=S5

Exercise: G = pd-elementary abelian of order pd (i.e. an abelian group of order pd with xp = 1 ∀x ∈ G),then Aut(G) = GLd(p)

G finite.If Z(G) = 1, then G∼=Inn(G)PAut(G)If G is simple, we have Schreier “conjecture ”

Out(G) = Aut(G)/ Inn(G) is “small” and solvable

This is a theorem now, as a consequence of the classification of finite simple groups.E.g. Aut(An) = Sn for n = 5 or n > 6Aut(A6) = Σ6, group of order 2 · 6!

Important definition: G∗ is almost simple if there is a simple group G with G∼=Inn(G)PG∗ ≤ Aut(G)(G is the unique minimal normal subgroup of G∗, and G∗/G ≤ Out(G))

Example: H PK PG ; H PG (Exercise: give an example)

DefinitionK char G (K is characteristic subgroup of G) if Kα(:= α(K)) = K ∀α ∈ Aut(G)i.e. K is α-invariant ∀α ∈ Aut(G)G is characteristically simple group if it has no proper non-trivial characteristic subgroup

K char G ⇒ K PG, since Inn(G) ≤ Aut(G)

Exercise:(1) H char K PG ⇒ H PG(2) H char K char G ⇒ H char G

Example: Z(G) char G, G′ char GG′ = ⟨[x, y]|x, y ∈ G⟩ – commutator subgroup

Exercise: K PG,G/K abelian ⇔ G′ ≤ K

G is perfect if G = G′

3

2 Series, Jordan-Holder Theorem

G is finite (or, if infinite, need some DCC)

DefinitionA series is a chain of subgroups Gi:

1 = G0 PG1 P · · ·PGr = G

A series is normal if also Gi PG ∀i

A proper series for G has all quotients Gi/Gi−1 nontrivial.

A proper series is a composition series if all quotients Gi/Gi−1 are simple (so, a composition series isone that cannot be properly refined)

Theorem 2.1Any finite group has a composition series

ProofIf G simple, stop.Otherwise, let K be a maximal normal subgroup of G; then G/K simple and |K| < |G| so containedin K

Remark. This is constructive; in fact 1PH PG is a series for any H PG

E.g.: Z has no (finite) composition series

Theorem 2.2 (Jordan-Holder)Let G be finite. Then any two composition series have the same quotients, counted with multiplicity

1 = G0 P G1 P · · · P Ga = Gand 1 = H0 P H1 P · · · P Hb = G

two composition series, then a = b and the collectors Gi/Gi−1 and Hi/Hi−1 are the same up toisomorphism

Example:

S4

A4

V4

pppppp

pNNN

NNNN

2MMMMMM 2 2

qqqqqq

1

C6

CCCC

CCCC

C3

BBBB

BBBB

C2

||||

||||

1

ProofPut H = Hb−1, K = Ga−1

Case: H = KSince |H| < |G|, induction applies:We have two composition series of H, so they are “isomorphic”

Case: H = K

4

Note that HK PG ⇒ G = HK by maximality

G = HKhhhhhhh

VVVVVVV

HVVVVVVVV K

hhhhhhhh

H ∩K

Hb−2 Ga−2

Lc−2

H1

UUUUUUUUU G1

iiiiiiiiii

1

Find a composition series for H ∩KThen we have four composition series for G. They are isomorphic series “naturally defined”

Firstly, by induction, the two composition series for H are isomorphic (induction hypothesis), and soare those for K

Hence, the two left series for G are isomorphic, and so are the two on the right.But also the two in the middle are isomorphic:This is because, G/H = HK/H ∼=K/H ∩K, and G/K ∼=H/H ∩K

Remark. All finite groups can be “broken up” into simples, i.e. have series with all quotients simple.On the other hand, building all finite groups out of simple groups is complicated.

E.g. There are, up to isomorphism, 26 different groups of order 26, all have six composition factor C2

E.g. Composition factors C2, A5: C2 ×A5, S5 = Aut(A5), SL2(5)(Z(SL2(5)) = ±I, SL2(5)/±I∼=A5)

A chief series is a proper normal series which cannot be properly refined to a proper normal series.i.e. ∀i @APG s.t. Gi PAPGi+1. The chief factors are characteristically simple

E.g. S4

C2

− A4

C3

− V4

C22

− 1

Any finite group has a chief series.Any normal series can be refined to a chief series.Any two chief series are “isomorphic”

Theorem 2.3Let N be minimal normal subgroup of G. Then N is direct product of isomorphic simple groups, allconjugate in G

ProofLet K be a minimal normal subgroup of N .If K = N , stopIf K N , then K RG⇒ ∃g1 ∈ G with Kg1 = K ⇒ K ∩Kg1 = 1Let K2 := KKg1 (which is PN)Since K ∩Kg1 PN ⇒ K2

∼=K ×Kg1

If K2 = N , stopOtherwise, K2 N ⇒ K2 RG ⇒ ∃h ∈ G with Kh

2 = K2

i.e. ∃k ∈ K s.t. (kkg1)h /∈ K2

Notice (kkg1)h = kh · kg1h (Aim: ∃g2 ∈ G s.t. K2 ∩Kg2 = 1)

⇒

either kh /∈ KK1 ⇒ Kh K2 g2 := h

or kgh1 /∈ KK1 ⇒ Kg1h K2 g2 := g1h

5

⇒ K3 := K2Kg2 PN , and K2 ∩Kg2 = 1, so K3

∼=K ×K2.

Continue this and we get N ∼=K ×Kg1 ×Kg2 × · · · ×Kgk .Finally, K is simple, since, if X PK, then X PN (as N ∼=K × Kg2 × · · · × Kgk), so X = K byminimality of K as normal in N

Remark.

(1) The proof applies to characteristically simple groups N i.e. Characteristically simple group Nare direct products of isomorphic simple groupsIn particular, all chief factors of a finite group are also direct products of isomorphic simplegroups.

(2) If N is characteristically simple group, then N = T1 × · · · × Tk, with each Ti∼=T some simple

group T . Either T = Cp prime p, or T is non-abelian simple

Exercise:

(1) If T = Cp thenN = Vk(p), a vector space over Fp of dimension k. There are pk−1p−1 + pk−1

p2−1· p

k−pp2−p

+· · ·normal subgroup

(2) If T is non-abelian simple, there are precisely 2k normal subgroup. Namely, Ti1 × Ti2 × · · · × Til

2.1 Groups with operators, X-groups

(see J.S. Rose book for more details)

DefinitionX group, G is an X-group if ϕ : X → Aut(G) a homomorphism

Define operation of X on G by gx := g(xϕ)

H ≤ G is an X-subgroup of G, write H ≤X G, if H is X-invariant (Note: H ≤X G ⇒ H is anX-group)If H PG,H ≤X G ⇒ G/H is an X-group via (Hg)x := Hgx

If G1, G2 are X-groups (X fixed) a homomorphism α : G1 → G2 is an X-homomorphism if(gx)α = (gα)x for x ∈ X, g ∈ G1

An X-group is X-simple if it has no non-trivial normal X-invariant subgroupsIf G is an X-group, an X-composition series is a series

1 = G0 PG1 P · · ·PGa = G

with each Gi ≤X Gi+1, Gi PGi+1, which cannot be properly refined.

E.g.

G X X-subgroup X-series X-comp. series

G 1 subgroup series comp. seriesG Inn(G) normal subgroup normal series chief seriesG Aut(G)

Another example: G-modulesG any group, ρ : G → GL(V ) a finite dimensional representation over some field. Set X = Im ρ, thenV is an X-group

In this setup, one proves existence of X-composition series, X-isomorphism theorems and X-Jordan-Holder theorem

6

3 (Finite) Nilpotent Groups

DefinitionThe normal series 1 = G0 PG1 P · · ·PGa = G is a central series if Gi/Gi−1 ≤ Z(G/Gi−1) ∀i

G is nilpotent (of class a) if it has a central series (the shortest such of length a)

Upper central series is the series of groups Zi(G) s.t. Z0(G) = 1, Z1(G) = Z(G), Zi+1(G)PG withZi+1(G)/Zi(G) = Z(G/Zi(G))so, G nilpotent (of class a) ⇔ G/Z(G) nilpotent of class a− 1G nilpotent of class 1 ⇔ G abelian

Theorem 3.1Finite p-groups are nilpotent. If |G| = pe, then class of G < e

Proof|G| = pe. If e = 2 then G is abelian, so class 1.Now Z(G) = 1⇒ |G/Z(G)| < pe ⇒ nilpotent of class < e− 1⇒ G nilpotent of class < e

1 Z(G) Z2(G) Z3(G) ____ G

1 Z(G/Z(G)) _________ G/Z(G)

E.g. G = D2e is nilpotent of class e− 1, (e ≥ 2)If e = 2, D2e = ⟨α, β|α2 = 1, βαβ = α⟩ = V4, so clearIf e > 2, Z(D2e) = 2 and D2e/Z(D2e)∼=D2e−1

Theorem 3.2For finite groups, TFAE:

(1) G is nilpotent

(2) H G ⇒ H NG(H)

(3) Each Sylow subgroup of G is normal in G

(4) G is a direct product of its Sylow subgroups, one for each prime dividing |G|

Exercise:• G nilpotent, H < G ⇒ H nilpotent• Any homomorphic image of G is nilpotent• Direct product of nilpotent groups is nilpotent group

Proof

(1)⇒(2) Let H G. Take i maximal s.t. Zi(G) ≤ HMay assume i = 0 (factor out Zi(G))⇒ Z(G) H. But Z(G) ≤ NG(H) ⇒ H NG(H)

(2)⇒(3) Let P be a Sylow p-subgroup of G

Claim: For any G, NG(NG(P )) = NG(P ) for P Sylow in G

7

Proof of Claim:P ∈ Sylp(NG(P )) and P PNG(P ) ⇒ P char NG(P )This is because P is the only subgroup of NG(P ) with order |P |, this implies all automor-phism of NG(P ) sends P to itself

Also, since NG(P )PNG(NG(P ))⇒ P PNG(NG(P ))⇒ NG(NG(P )) = NG(P )

By hypothesis of (2), have NG(P ) = G

(3)⇒(4) Let P1, . . . , Pk be the Sylow p-subgroups of G, one for each prime p∣∣|G|. Each Pi PG

by hypothesis of (3).⇒ P1P2 · · ·Pk PG s.t. |P1P2 · · ·Pk| = |G|⇒ G = P1P2 · · ·Pk, and Pi ∩

∏j =i Pj = 1 by considering their orders

⇒ G∼=P1 × P2 × · · · × Pk

(4)⇒(1) Follows from Exercise above and Theorem 3.1

Corollary 3.3G nilpotent, H < G maximal ⇒ H PG index p, some prime p

ProofNG(H) H ⇒ H PGG/H has no proper subgroups ⇒ cyclic of prime order for some prime p

NoteIf K PG and K ≤ H ≤ G, then H/K ≤ Z(G/K) ⇔ ⟨[h, g]|h ∈ H, g ∈ G⟩ = [H,G] ≤ K

DefinitionLower central series for G nilpotent is the chain Γi(G) s.t. Γ1(G) = G,Γi+1(G) = [Γi(G), G]If 1 = G0 P · · ·PGa = G is a central series, then Γa−i+1(G) ≤ Gi ≤ Zi(G)In particular, ∀c, Γc+1(G) = 1 ⇔ Zc(G) = G

3.1 Two nilpotent characteristic subgroups of a finite group G

G any finite groups

DefinitionThe Fitting subgroup (nilpotent radical of G), F (G), is the maximal subgroup amongst all normalnilpotent subgroups of G

Proposition 3.4In a finite group, there is a unique maximal normal nilpotent subgroup:If H,K are nilpotent normal subgroups of G, so is HK(i.e. F (G) well-defined)

ProofThe last claim is clear if H,K are both p-groups⇒ Op(G), the unique maximal normal p-subgroup of G exist

Claim: F (G) = Op1(G)× · · · ×Opk(G), where the pi are the prime divisors of |G|

8

Proof of Claim:RHS is nilpotent and normalIfK PG, nilpotent, and P is a Sylow p-subgroup ofK, then P charK PG ⇒ P PG ⇒ P ≤ Op(G)⇒ K ≤ RHS

DefinitionG any finite group, the Frattini subgroup of G, Φ(G), is the intersection of all maximal subgroups ofG

Note: Φ char G

Definitiong ∈ G is a non-generator of G if whenever G = ⟨X, g⟩ we have G = ⟨X⟩

Lemma 3.5Φ(G) = g ∈ G|g non-generator ProofIf g /∈ Φ, then ∃M < G maximal with g /∈ M⇒ G = ⟨M, g⟩, but ⟨M⟩ = M < G⇒ g is not a non-generator

Conversely, assume ∃X < G with G = ⟨X, g⟩ but G > ⟨X⟩Let M < G maximal with ⟨X⟩ ≤ MThen g /∈ M ⇒ g /∈ Φ(G)

Denote Sylp(G) as the set of Sylow p-subgroup of G

Proposition 3.6For any G finite, Φ(G) is nilpotent (⇒ Φ(G) ≤ F (G))

ProofUses an important general lemma (Lemma 3.7) known as Frattini argument

Let P ∈ Sylp(Φ(G))Then G = NG(P )Φ(G) so by Lemma 3.5, G = NG(P ),so P PΦ(G). Thus Φ(G) nilpotent.

Lemma 3.7 (Frattini argument)G finite group, K PG, P ∈ Sylp(K)Then G = NG(P )K. Hence G/K ∼=NG(P )/NK(P )

ProofLet g ∈ G. By normality, ∃k′ ∈ K s.t. P g = P k′

⇒ (k′ =: k−1) P gk = (P g)k = P⇒ gk ∈ NG(P ), so g ∈ NG(P )K

G/K = NG(P )K/K ∼=NG(P )/K ∩NG(P ) = NG(P )/NK(P )

Lemma 3.8If G is a p-group, then G/Φ(G) is an elementary abelian group, and hence a vector space over Fp. Infact, Φ(G) = G′Gp

ProofIf M is maximal subgroup of G, then M PG of index p so G′ ≤ Φ(G) (as G/M abelian), andGp = ⟨gp|g ∈ G⟩ ≤ Φ(G) so G′Gp ≤ Φ(G)

9

Equality: If g ∈ G \G′Gp, consider its image in G = G/G′Gp

Then g = 0, a non-zero vector, so let g, g2, . . . , gk be a basis.Then g is not a non-generator (G = ⟨g, g2, . . . , gk,Φ⟩)

DefinitionMinimal generating set for G: no element of the set can be deleted and still generate G

Theorem 3.9 (Burnside’s Basis Theorem)If G is a finite p-group, any two minimal generating sets for G have the same size, dimFp G/Φ(G).

Proofexercise: if g1, . . . gk is a minimal generating set for G, then g1, . . . , gk is a basis for G/Φ(G)

Remark. S5 = ⟨(12), (12345)⟩ = ⟨(12), (23), (34), (45)⟩both are minimal generating set, with different sizeQuestion What is the maximal size of a minimal generating set? What do minimal generating sets ofmaximal size look like?

10

4 Soluble (Solvable) groups

DefinitionA derived series for G is a series G(i) s.t. G(0) = G, G(i+1) = [G(i), G(i)] (so G′ = G(1) = Γ2(G))

A group is soluble if G(v) = 1 for some v

Remark. • Nilpotent ⇒ soluble

• S3 is soluble but not nilpotent

• simple soluble ⇒ Cp for some prime p

Lemma 4.1TFAE:

(1) G is soluble

(2) The chief factors are elementary abelian

(3) The composition factors are cyclic of prime order

Proof(1) ⇒ (2) ⇒ (3) is clear

(3) ⇒ (1):If 1P · · ·PH1 PH0 = G with abelian factors, then G(i) ≤ Hi, by indicator:G(0) = H(0), and G(i+1) = (G(i))′ ≤ H ′

i ≤ Hi+1

Exercise: Subgroups, quotients, direct products of soluble groups are solubleExercise: H,K PG, both H,K soluble ⇒ HK is a soluble normal subgroup of G. Hence, using thefact that if N , G/N soluble then G soluble, we can define the soluble radical of G, i.e. the maximalnormal soluble subgroup of G

Theorem 4.2 (Galois)G finite soluble group, M < G maximal subgroup of G⇒ |G : M | = pa for some prime power pa

ProofLet K PG minimal normal. Then K is elementary abelian p-group, for some prime p. (The proofhere is much easier than before: K is soluble, and K ′ char K so K ′ = 1, so K is abelian. Op(K) charK and Kp char K, so K elementary abelian p-group for some p)

If K ≤ M , induction applies to M/K < G/K, so assume not. Then G = KM (since M KM ≤ G)And K ∩ M = 1: K ∩ M PM , K ∩ M PK (as K is abelian) so K ∩ M PG, so K ∩ M = 1 byminimality of K PG

So |G : M | = |K| = pa, for some a

Remark. Here K is a regular normal subgroup of G on (G : M) (regular = transitive and only identityfixes all point)

11

4.1 Hall’s Theorem on Finite Soluble Groups

π = Set of primesn ∈ N, write n = nπnπ′ where

(pl11 · · · plkk )π =∏pi∈π

plii (π-part)

(pl11 · · · plkk )π′ /∈ π (π′-part)

A π-group H is a finite group with |H| = |H|πA subgroup H of G is a Hall π-subgroup of G if |H| = |G|π, so subgroup of G of maximal possibleπ-order

Remark. π = pHall π-subgroup = Sylow p-subgroupHall π′ subgroup = Hall p-complement subgroup

Theorem 4.3If G is a finite soluble group, π is a set of primes

(1) G has a Hall π-subgroup

(2) Any two are conjugate in G

(3) Any π-subgroup of G is in some Hall π-subgroup

Example: GL3(2) has no Hall 3-complement, and has two non-conjugate Hall 7-complement1 ∗ ∗00

,

1 0 0∗∗

ProofLet |G| = mn, with m = |G|π, n = |G|π′

Induction on n. If n = 1, then trivial. So assume n > 1

Case 1:Assume ∃1 = K PG with |K| = m1n1 with n1 < n

(1) Now |G/K| = mm1

nn1, so by induction, G/K has a subgroup S/K of order m

m1with K ≤ S < G

Then |S| = mn1 < |G|. By induction, S (and hence G) has a subgroup of order m

(2) If H1, H2 are subgroups of G, order m, then in G = G/K, then images H1, H2 are Hall π-subgroups. By induction, ∃x = Kx ∈ G with x−1H2x = H1.⇒ x−1H2Kx = H1K ⇒ x−1H2x,H1 are Hall π-subgroup in H1K, so conjugate. Applyinduction.

(3) Let P be any π-subgroup of G. Then P = PK/K is a π-subgroup in G ⇒ P ⊆ some Hallπ-subgroup S = S/K of G/K (K ≤ S < G)S order mn1. By induction in S, we see that P ⊆ some Hall π-subgroup in G

Case 2:Any non-identity normal subgroup of G has order divisible by n (∀1 = N PG,n

∣∣|N |)Let K be maximal normal in G. Then |K| = pa for some prime power pa, K elementary abelian.Then n|pa, whence n = pa as pa|nm and (n,m) = 1

Let LPG containing K s.t. L = L/K minimal normal in G = G/K. Then |L| = qb for some primepower qb with p = q

12

Let Q be a Sylow p-subgroup of L. |L| = paqb, L = KQ (by Frattini argument)

Gllllllll

NG(Q)

1111

1111

1

L = KQ

xxxx

Q

5555

5555

5

K

1

(1) Claim: NG(Q) is a Hall π-subgroup

Proof of Claim:Firstly, Frattini argument: G = LNG(Q) = KQNG(Q) = KNG(Q)⇒ m

∣∣|NG(Q)K ∩NG(Q) = 1, as

• K ∩NG(Q)PK (since K abelian)• K ∩NG(Q)PNG(Q) (since K PG)• ⇒ K ∩NG(Q)PG, but K minimal normal, and K NG(Q) as Q R G

⇒ |NG(Q)| = m ⇒ H = NG(Q) is a Hall subgroup

(2) Let H2 be any other Hall π-subgroup of G

Since LH2 ≤ G, with |G|∣∣∣|LH2|

⇒ G = LH2

Now |L ∩H2| = qb (as LH2/L∼=H2/L ∩H2)⇒ L ∩H2 is a Sylow q-subgroup of L⇒ Qx = L ∩H2 for some x ∈ L

H2 ≤ NG(L ∩H2) = NG(Q)x = NG(Qx) = Hx ⇒ H2 = Hx (since they are of same order)

(3) Let P be a π-subgroup of G, with |P | = m′ < mNow |G : H| = n which is π′-number|G : PK| = m/m′ which is π-number⇒ P,H ∩ PK have the same order: |G : H ∩ PK| = |G : H||G : PK| (by Lemma 4.4, to beshown) = nm

m′

Gn

ttttttttttm/m′

KKKKKKKKKK

H

m/m′ JJJJJJJJJ PK

nsssssssss

H ∩ PK

So P,H ∩ PK are both Hall π-subgroups in PK⇒ conjugate there by (2)⇒ P is conjugate to a subgroup of H

Lemma 4.4G any finite group let A1, A2 be subgroups of coprime indices n1, n2.Then G = A1A2 (a factorization of G) and |G : A1 ∩A2| = n1n2

13

Proof|G : A1 ∩A2| = |G : A1|︸ ︷︷ ︸

n1

|A1 : A1 ∩A2| = |G : A2|︸ ︷︷ ︸n2

|A2 : A1 ∩A2|

⇒ n1n2

∣∣|G : A1 ∩A2|

|A1A2| =|A1||A2||A1 ∩A2|

=

((|G|/|G : A1|)(|G|/|G : A2|)

|G|/|G : A1 ∩A2|

)=

|G||G : A1 ∩A2|n1n2

⇒ G = A1A2

A1, A2 ≤ G. G is factorisable as G = A1A2 if every g ∈ G is g = a1a2 with ai ∈ Ai

Note: G = A1A2 ⇔ G = A2A1 ⇔ G = Ag1A2;∀g ∈ G

A1, A2 conjugate ⇒ G = A1A2

Exercise: G = A1A2 ⇔ A2 is transitive on (G : A1) ⇔ A1 is transitive on (G : A2)

(this result immediately implies:) G = A1A2 ⇔ |G : A1| = |A2 : A1∩A2| (by using |A1A2| = |A1||A2||A1∩A2|)

Theorem 4.5 (Theorem of Ore)If G is finite soluble, A1, A2 maximal subgroup of G, not conjugate then G = A1A2

Proof as Exercise (hard)

Factorisation in almost simple groups are interesting, important and rare. See later.

DefinitionIf |G| = pe11 · · · pekk with the pi distinct primes, a Sylow basis of G is P1, . . . , Pk with |Pi| = peii s.t.PiPj = PjPi ∀i, j (so that PiPj ≤ G ∀i, j)

Note: PiPj = PjPi ⇒ PiPj ≤ G ∀i, j and in factPi1Pi2 · · ·Pil ≤ G ∀i1, . . . , il ⊆ 1, . . . , k − we get Hall subgroups this way

Theorem 4.6If G is a finite soluble group, G has Sylow basis, and any two are conjugate

ProofLet |G| = pe11 · · · pekk , with the pi distinct primes.Let Hi be a (Hall) pi-complement. (i.e. has order |G|/peii )Put Pj =

∩i=j Hi

⇒ |Pj | = pejj , using Lemma 4.4

And PiPj =∩

i = l = jHl = PjPi

Conjugacy: Let P1, . . . , Pk and P ∗1 , . . . , P

∗k be Sylow bases with |Pi| = |P ∗

i |, so Pi, P∗i ∈ Sylpi(G)

Put Hi = P1 · · ·Pi−1Pi+1 · · ·Pk

and H∗i = P ∗

1 · · ·P ∗i−1P

∗i+1 · · ·Pk

Claim: ∃g ∈ G with Hgi = H∗

i ∀iProof of Claim:First, ∃g1 with Hg1

1 = H∗1 (Hall)

Now assume we found gi−1 with Hgi−1

j = H∗j ∀j ≤ i− 1

Change notation if necessary so that Hj = H∗j ∀ ≤ i− 1

Now G = HiPi, and let x ∈ G with Hxi = H∗

i (Hall)⇒ x = hz with h ∈ Hi, z ∈ Pi

⇒ Hzi = H∗

i and z ∈ Pi ≤ Hi ∀j < i⇒ Hz

j = H∗j for j ≤ i

After k steps, finished.

Given claim, we have P gi = P ∗

i ∀i

14

Theorem 4.7 (Wielandt)G finite, H1,H2,H3 soluble subgroups of G with |G : Hi| = ni pairwise coprime. Then G soluble

ProofLooking for N PG soluble with G/N soluble by induction (then G soluble)Note that G = H1H2 = H1H3 = H2H3, by Lemma 4.4 May assume Hi = 1 ∀i. Let K minimal normalin H1

⇒ K elementary abelian p-group.

WLOG, p - n2

Claim: This implies K ≤ H2

Proof of Claim:By Lemma 4.4 ∣∣K(H1 ∩H2) : H1 ∩H2

∣∣︸ ︷︷ ︸divides n2

=∣∣K : K ∩H1 ∩H2

∣∣so K ≤ H1 ∩H2

Now let N = ⟨Kg|g ∈ G⟩ - the G normal closure of K⇒ N = ⟨Kh1h2 |hi ∈ Hi⟩ = ⟨Kh2 |h2 ∈ H2⟩ ≤ H2

⇒ N PG,N ≤ H2

It follows that N is soluble (as H2 is), and G/N is soluble by induction hypothesis.

Theorem 4.8 (P.Hall)If the finite group G has a Hall p-complement for each prime p

∣∣|G|, then G is soluble

Proof|G| = pe11 · · · pekkIf k = 1, G is nilpotent. DoneIf k = 2, this is Burnside’s paqb Theorem, this uses representation theory

Assume k ≥ 3, let Hi be a Hall pi-complement.By induction, Hi is soluble ∀i (Lemma 4.4)Now use Theorem 4.7

Lemma 4.9G finite soluble group ⇒ CG(F (G)) ≤ F (G)(So G/Z(F (G)) ≤ Aut(F (G)))

ProofPut F = F (G), Z = Z(F (G)), C = CG(F (G)). So Z = C ∩ FAssume C F . Let H PG with Z < H

G

CF

CCCC

CCCC

C

HCC

C

CCC

F

Z

H minimal normal in C = C/Z: Since H is elementar yabelian, have H ′ ≤ ZThen H is nilpotent (so H ≤ F - not so): minimal normal in H,Γ3(H) = [H ′,H] ≤ [Z,C] = 1

15

5 Interlude

In finite simple groups here been classified:(q denote a power of prime)Cp, p primeAn, n ≥ 5Ld(q), d ≥ 2, if d = 2 then q ≥ 4Ud(q), d ≥ 3, if d = 3 then q ≥ 3Sp2m(q), m ≥ 2, if m = 2 then q ≥ 3PSLϵ

d(q), d ≥ 7if d even then G doubly infinite families, denoted by ϵ = ±;if d odd then ϵ is empty

10 families, exceptional groups of Lie type (q denote prime power):G2(q), F4(q), E

+6 (q), E

−6 (q), E7(q), E8(q)

3D4(q)2B2(q) with q = 22a+1

2G2(q) with q = 32a+1

2F4(q) with q = 22a+1

26 sporadic simple groupsM11,M12,M22,M23,M24, · · · ,M

In applications, have usually a finite groups acting in some way, .e.g as a group of permutations, or agroup of matrices (in this case, this is representation theory, mainly used to study the first 5 familiesmentioned)

We will be consider the first case mainly (i.e. acting as groups of permutations)e.g.1: as a group of automorphisms of a graph, e.g. Peterson graph e.g.2: as a group acting on rootsof polynomial

Recall about permutation actions:

G finite group, Ω a finite set. G acts on if there is a mapΩ×G → Ω(ω, g) 7→ ωg

s.t. ω1 = ω, ωgh = (ωg)h ∀ω ∈ Ω, g, h ∈ G

If so, for g ∈ G, the mapϕg : Ω → Ω

ω 7→ ωgis a permutation on Ω

and the mapϕ : G → Sym(Ω)

g 7→ ϕgis a homomoprhism, called a permutation representation of G

The kernel of the action, G(Ω) := kerϕ. Then Gϕ = GΩ := G/G(Ω) ≤ Sym(Ω)The action is faithful if G(Ω) = 1

If G acts on Ω, the orbit of G containing α is αG = αg|g ∈ G ⇒ Ω splits into G-orbitsG is transitive on Ω if αG = Ω ∀α ∈ Ω

Problems about actions are usually easily reduced to problems about transitive actions (consider theactions on orbits). So the assumption of G transitive is common.

If G acts on Ω1,Ω2, then θ : Ω1 → Ω2 is a G-homomorphism if (αg)θ = (αθ)g ∀α ∈ Ω1, g ∈ Gθ is a G-isomorphism if it is also bijective.

Also note that (G : Gα) is the G-space of all right cosets of Gα = g ∈ G|αg = α, with action obtainedby (Gαx)g = Gα(xg)

16

Lemma 5.1If G is transitive on Ω, and α ∈ Ω, then actions of G on Ω and on (G : Gα) are G-isomorphic

Proof

θ : Ω → (G : Gα)

αx 7→ Gαx

is bijective, commutes with the actions of G

Example:(G : H), (G : K) are G-isomorphic ⇔ H,K are G-conjugate

DefinitionA equivalence relation ρ is a G-congruence on Ω if αρβ ⇒ αg ρ βg ∀g ∈ GTrivial means either equality or universal.If G is transitive on Ω, then G is primitive if @ non-trivial G-congruence on Ω

Remark. Kernels of G-homomorphism are “blocks” of G-congruences

Lemma 5.2G transitive on Ω, ρ a non-trivial G-congruence on Ω, α ∈ Ω, ρ(α) the (equiv.) class of α: ρ(α) :=β ∈ Ω|αρβ

(1) The (setwise) stabilizer of ρ(α) in G acts transitively on ρ(α)

(2) ρ(α) is the union of some Gα-orbits, including α

(3) G is transitive on the set Ω/ρ of ρ-classes (blocks), so all the ρ-classes have the same size

(4) Gα Gρ(α) GConversely, if Gα H G, can define a non-trivial G-congruence on Ω by:

(a) γ ρ β ⇔ β = γh, some h ∈ H

(b) γ ρ β ⇒ γg ρ βg ∀g

Proof

(1) Let β ∈ ρ(α), G transitive on Ω⇒ ∃g ∈ G with β = α ⇒ ρ(α) must be kept invariant by g ⇒ g ∈ Gρ(α)

(2) Gα keeps ρ(α) invariant

(3) α ∈ ρ(α) and if ρ(β) another block, ∃g ∈ G s.t. αg = β ⇒ ρ(α)g = ρ(β)

(4) First claim is clear. Proof of converse:If Gα H G, the ρ defined is clearly non-trivial G-congruence.⇒ Ω/ρ = Γg|g ∈ G where Γ = αH

Claim: The blocks form a partition of Ω

Proof of Claim:They cover Ω. So we want: (Γ ∩ Γg = 0 ⇒ Γ = Γg)

β ∈ Γ ∩ Γg ⇒ β = αh1 = αh2g some h1, h2 ∈ H⇒ h2gh

−11 ∈ Gα H ⇒ g ∈ H ⇒ Γg = Γ

17

Corollary 5.3Set up as Lemma 5.2, by (4), G primitive on Ω ⇔ Gα maximal subgroup of G

Corollary 5.4Set up as Lemma 5.2, by (3), all blocks have the same size, didviding |Ω|.If G transitive of prime degree (degree means size of Ω) ⇒ G primitive

Lemma 5.5G primitive on Ω, α = β ⇒ G = ⟨Gα, Gβ⟩, unless G is Cp of degree pIn fact, if G transitive, α ∈ Ω, then fix(Gα)=β|βh = β ∀h ∈ Gα is a block of a congruence.

ProofLet G be transitive. Define αρβ if Gα = Gβ (note: β ∈ fix(Gα) ⇒ Gα = Gβ)

Claim: This is a G-congruence

Proof of Claim:g ∈ G. αρβ ⇒ Gα = Gβ ⇒ Gαg = Gβg ⇒ αg ρ βg

fix(Gα) = ρ(α). So, if G is primitive on Ω, the congruence above is trivial (equlity or universal).If equality, then Gα = Gβ , btoh maximal ⇒ G = ⟨Gα, Gβ⟩If universal, then Gα fixes Ω pointwise ⇒ Gα = 1and as Gα maximal, then G is Cp for some prime p

Lemma 5.6Let G transitive on Ω, let 1 = N PG. The N -orbits form a G-invariant partition of Ω

ProofExercise

Corollary 5.7If G is primitive, 1 = N PG ⇒ N transitive on Ω

Exercise: Let G be transitive of degree prime p, (i.e. |Ω| = p and hence G ≤ Sp)Let N be minimal normal subgroup of G. Then N is simple (possibly Cp), and G/N is abelian of orderdividing p − 1 (use Frattini argument and G ≤ Sp. Note if P is order p, NSp(P ) has order p(p − 1),so G′ = N is simple)

Dichotomy for primitive groups: 2-transitive groups, simply-primitive groups

2-transitive groups

DefinitionG on Ω is 2-transitive on Ω if G is transitive on ordered pairs of distinct points of Ω, i.e. if α1 =α2, β1 = β2, then ∃g ∈ G with αig = βik-transitive is defined similarly on k-tuples

Remark. G is k-transitive on Ω of degree n, then n(n− 1)(n− k + 1)∣∣|G|

Example:Sn on [1, n] is n-transitiveAn is (n− 2)-transitive but not (n− 1)-transitive

Example:G = PGLd(q) - projective general linear group = GLd(q)/scalars acts on Ω = 1-dim. subspaces of

18

Vα(q) =: Pd−1(q)

|Ω| = qd−1q−1

G is 2-transitive on ΩG is 3-transitive ⇔ d = 2G is 4-transitive ⇔ d = 2, q = 3, G = PGL2(3)∼=S4

Example: G = AGLd(p), (d ≥ 1) the group of “symmetries” of Ω = Vd(p)i.e. ⟨translations, linear transformations⟩ ≤ Sym(V )K = tv|v ∈ V where tv : x 7→ x+ v, subgroup of translationG0 = GLd(p) is acting on Ω as linear transformation⇒ transitive on Ω \ 0⇒ 2-transitive on Ω

(Exercise) G is 3-transitive ⇔ d = 2, or d = 1, p = 3(i.e. G∼=S3)(Exercise) G is 4-transitive ⇔ d = 2 = p,AGL2(2)∼=S4

In fact:

(1) K PAGL, since ∀h ∈ G0, h−1tvh = tvh ∀x ∈ V

(2) K is a regular normal elementary abelian subgroup

(3) K ∩G0 = 1, AGL = G0KAny g ∈ G is g = hk for some unique h ∈ H, k ∈ K(h1k1)(h2k2) = (h1h2)(k

h21 k2)

DefinitionSemidirect product H nK:H,K groups, θ : H → Aut(K)Elements of H ⊗K are of form (h, k) s.t.

(h1, k1)(h2, k2) = (h1h2, kh2θ1 k2) ∀hi ∈ H, ki ∈ K

This is a group, with normal subgroup ∼=K and a subgroup ∼=H, s.t. H ∩K = 1, G = HK(Direct product is a special case)

Lemma 5.8Let G be primitive on Ω, let K PG a regular normal subgroup (so K is transitive with Kα = 1 ∀α)

(1) K is minimal normal in G (so it is a direct product of isomorphic simple groups, possibly Cp)

(2) Let α ∈ Ω, we can identify Ω and K in such a way that the actions of Gα on Ω and on K isisomorphic

(3) If G is 2-transitive, then K is elementary abelian

(4) If K is elementary abelian, then K is in fact a vector space Vd(p) over Fp and action of G on Ωis isomorphic to a subgroup of AGLd(p) acting as aboveIn fact, Gα (↔ G0,stabliser of vector 0) is an irreducible subgroup of GLd(p)(This is because if some proper subspace L of K is Gα-invariant, then Gα LGα G)

Proof

(1) If 1 = LPG ⇒ L is transitive ⇒ |K| = |Ω|∣∣|L|

⇒ if L ≤ K, then L = K

19

(2) α ∈ Ω, any β is αk for unique kαk ↔ kΩ ↔ KIf g ∈ Gα, αkg = αg−1kg = αkg ↔ kg

(3) Let G be 2-transitive. By (2), Gα is transitive on Ω \ α⇒ Gα transitive on K \ 1 by conjugations⇒ ∀k ∈ K \ 1, they have the same order p, some prime p⇒ K is a p-group ⇒ elementary abelian

(4) Notation: K is a vector space written multiplicatively.G = GαK.

Action of K:Take k′ ∈ K, we have the correspondence

(k)k′ ↔ (αk)k′ = α(kk′) ↔ kk′

⇒ action by k′ is just by translation

Action of Gα:Take h ∈ Gα, h acts on K by k 7→ h−1khThis is a linear transformation: h−1(k1k2)h = h−1k1h · h−1k2h (and scalar multiple)

Theorem 5.9 (Burnside)Let G be a 2-transitive primitive group on Ω, |Ω| = n. Let N be minimal normal subgroup of G.Then

either (a) N elementary abelian p-group (some p), G ≤ AGLd(p) (pd − 1

∣∣|G0|)or (b) N is non-abelian simple

ProofIf N is regular, then get (a) by Lemma 5.8. So we assume N is not regular (and aim to get (b))

Claim: N is primitive on Ω

Proof of Claim:Assume not. Let Γ be a minimal non-trivial block of N on Ω⇒ |Γ ∩ Γg| ≤ 1, g ∈ G, unless Γ = ΓgΓg is another N -block (possibly in a different system of imprimitivity)⇒ Γ ∩ Γg is also an N -block, so either Γ = Γg or |Γ ∩ Γg| ≤ 1

Write B = Γg|g ∈ G - “lines”. Any 2 points in Ω are on a unique line⇒ all pairs of points are (by G 2-transitivity)

Let α ∈ Ω. Now Nα fixes setwise each line on α

Claim: Nαβ = 1 for α = β

Proof of Claim:Nαβ fixes all points not on line(α, β)Repeat with α, γ and get Nαβ = 1

Hence N is a Frobenius groups (i.e. transitive, Nα = 1 ∀α, and Nαβ = 1 ∀α, β)

⇒ ∃ characteristic regular subgroup K ≤ N (5.1)

(see Remark for proof of this in our case)⇒ K PG ⇒ N = K ⇒ N is reugalr #

20

Recall Theorem 2.3: N minimal normal ⇒ N = T1 × · · · × Tk with Ti∼=T ∀i and T simple.

Also, by Corollary 5.7, each Ti is transitive on Ω

Claim: k = 1, N is simple, non-abelian

Proof of Claim:If k > 1, then Ti is regular.⇒ |N | = |T |k, n = |T | (note n is size of Ω), by Lemma 5.10this orbits of Nα on Ω \ α have all the same size dividing n− 1 (by Corollary 5.4)Nα PGα, Gα transitive on Ω \ α|Nα| = |T |k−1

On the other hand, (n− 1, |Nα|) = 1⇒ Nα = 1 ⇒ N regular⇒ k = 1 ⇒ N simple, non-abelian (if abelian, then regular by the next Lemma 5.10)

We have now completed the proof.

Remark. We give a proof of the statement (5.1) under the set up of the lemmaClaim: G is 2-transitive, N PG and N Frobenius ⇒ ∃K ≤ N characteristic regular subgroup of N

ProofLet K = 1 ∪ x ∈ N |x have no fixed points on ΩLet n = Ω|K| = n, |N | = nc where c = |Nα|

N = K ∪ (∪α∈Ω

Nα \ 1︸ ︷︷ ︸n(c−1)

)

K is a transitive set:If k ∈ K \ 1 takes α to β, let g ∈ Gα take β to γthen kg ∈ K \ 1 with kg : α 7→ γK ≤ NIf not, let k1, k2 ∈ K with k1k

−12 /∈ K ⇒ k1 = k2 and k1k

−12 fixes some α - then αk1 = αk2

⇒ K cannot be transitive⇒ K PG (as kg ∈ K ∀k ∈ K, g ∈ G)

DefinitionAction of G on Ω is semi-regular if g ∈ G fixes any point of Ω ⇒ g = 1

Lemma 5.10If M transitive subgroup of Sym(Ω), then CSym(Ω)(M) is a semi-regular

ProofExercise

Simply Primitive Groups

DefinitionG is simply primitive on Ω if it is primitive but not 2-transitive

Orbits of Gα : Γ0(α) = α,Γ1(α), . . . ,Γr−1(α)where r is the number of these for Gα, called the rank of G on ΩNote r = 2 ⇔G is 2-transitive

21

For r > 2the subdegree ni = |Γi(α)| 1 = n0 ≤ n1 ≤ · · · ≤ nr−1

∑r0 ni = n

We can consider the induced action of G on Ω× Ω:

(α, β)g = (αg, βg) g ∈ G;α, β ∈ Ω

Get orbits Γ0,Γ1, . . . ,Γr−1, these are called orbitals of G⇒ Γi(α) = β ∈ Ω|(α, β) ∈ ΓiΓ0 = (α, α)|α ∈ Ω is called the diagonal orbital|Γi| = n · ni

The orbitals give orbital graphs on Ω (a directed graph):If Γ is an orbital, then Γ∗ = (β, α)|(α, β) ∈ Γ is also an orbitalΓ and Γ∗ are pairedWe get complete G-invariant r-colouring of the complete graph on the vertices in ΩG1 = Peterson graph

|Ω| =(52

)= 10 G =Aut(Peterson)=S5

(Automorphism group of graph G, Aut(G), acts on the set of vertices preserves edge)

Rank=3: n0 = 1,Γ0(1, 2) (trivial edges from i, j to i, j)n1 = 3,Γ1(12) (edges of a Petersen graph, i.e. edges with vertex i, j and k, l with i, j∩k, l = ∅)n2 = 6,Γ2(12) (all other edges of K10, i.e. edges with vertex i, j and i, k, j = k)

Exercise: Sn acts on

(n2

)(stablizer Sn−1 × S2) as a primitive rank 3 group, subdegrees 1 (↔ 12),

2n− 2,

(n− 22

)Sn acts on

(nk

), n

2 > l ≤ 1, rank is l + 1

Remark. If G is an undirected graph, we have the notion of distanceG ≤ Aut(G), G preserves distanceSay G is distance transitive on G if given α1, α2 has distance d and β1, β2 has distance d, then ∃g ∈ Gs.t. αig = βiWhat are these on regular symmetric graphs?

Exercise: Sn is distance transitive on the graph Ω =

(nl

)= l-subsets of [1, n], edges A − B if

|A ∩B| = l − 1

If G is primitive, n1 = 1 ⇒ ni = 1 ∀i ⇒ G regular (see 5.4 or 5 or 6)Exercise: n1 = 2 ⇒ ni = 2 ∀i ⇒ G = D2p on p pointsSim’s Conjecture: n1 fixed ⇒ |Gα bounded

Proposition 5.11 (D.G. Higman)Let G be transitive on Ω. Then G is primitive ⇔ all the non-diagonal orbital graphs are connected

ProofExercise (very very hard)

22

Permutation Character

G acts on Ω, a permutation group, π(g) = |fixΩ(g)| is a character of G of the permutation represen-tation.

Lemma 5.12If G ≤ Sym(Ω), permutation character πThen ⟨π,1⟩G = 1

|G|∑

g π(g) = #(orb(G,Ω))

So G is transitive Ω ⇔ ⟨π,1⟩G = 1

Proof

#(orb(G,Ω)) =∑α∈Ω

|Gα| = #(α, g) ∈ Ω×G|αg = α =∑g∈G

π(g)

Lemma 5.13G acts on Ω1,Ω2, with characters π1, π2Then ⟨π1, π2⟩G = # orbits of G on Ω1 × Ω2

In particular, if G acts transitively on Ω, with character π, then ⟨π, π⟩ = rankΩ(G)

ProofFirst part:

⟨π1, π2⟩G = ⟨π1π2,1⟩GNote π1π2 is the permutation character of G on Ω1 × Ω2

G is 2-transitive on Ω: π = 1+ χ, χ irreducibleG is rank 3: π = 1+ χ1 + χ2, χi distinct irreducible

DefinitionG acts on Ω is multiplicity-free if its permutation character is π = 1 + χ1 + · · · + χv−1 with the χi

distinct

Question: Classify such permutation groups

Remark. IfG is distance-transitive on an undirected graph than its permutation character is multiplicity-free

Exercise: Show that if G has permutation rank ≤ 5 then G is multiplicity-freeExercise: If G is transitive on Ω, ∃ g ∈ G with π(g) = 0

23

6 Alternating Groups

|Sn| = n! , |An| = n!/2, An consist of all the even permutations.Sn, An acts naturally on Ω = [1, n]An is (n− 2)-transitive on [1, n], Sn is n-transitiveAlso recall: An is generated by the 3-cycles (ijk) on [1, n](e.g., (12)(34)=(124)(134), (12)(13)=(123), also note multiplication is defined such that left one actfirst)

Theorem 6.1An is (non-abelian) simple for n ≥ 5 (Note for n = 3, A3 is abelian simple)

Proof IInduction on n:A5 is simple (prove this). Let n > 5, assume true for n− 1

Let G = An, let 1 = K PG ⇒ Kα PGα(∼=An−1) for α ∈ [1, n]Now Gα is simple by induction hypothesis ⇒ Kα = 1 or Kα = Gα

Case Kα = 1: Impossible. Because K is transitive regular, so by Lemma 5.8 and the statement aboveit about AGL being at most 3-transitive (except if n = 4) gives a contradiction

Case Kα = Gα: This implies that K = G (as K would contains a 3-cycles, and hence all 3-cycles, on[1, n], as these conjugate in An)

Proof IIStart as before, get K regular on [1, n] ⇒ |K| = n⇒ K contains a whole An-ccls of elements.But, in fact, the An-ccls are larger than n− 1 (see later)

Proof III(This proof is elementary) Suppose 1 = K PG

Claim: K contains a 3-cycle

Proof of Claim:Let 1 = g ∈ K, fixing as many points of [1, n] as possible.We claim g is a 3-cycle, suppose this is not true, there are 2 possibilities:

(1) all cycles of g have size 2: g = (12)(34) · · · , let x = (345) Then [g, x] = 1, but fixes 1, 2 ∪fix(g) \ 5 #

(2) g = (123 . . . 45 . . .) ⇒ [g, x] = 1But fix([g, x]) = fix(g) ∪ 2 (check)

Claim ⇒ K contains all 3-cycles (all conjugate) ⇒ K = An

Theorem 6.2Let G ≤ An, G primitive, G containing a 3-cycle. Then G = An

ProofDefine G-congruence ρ on [1, n]:

αρβ if α = β on 3-cycle (αβγ) ∈ G for some γ

Then ρ is reflexive, symmetric, G-invariantAnd, ρ is transitive: since αρβ ρ γ ⇒ ∃(αβδ) ∈ G if δ = γif δ = γ ∃(βγϵ) so ⟨(αβδ), (βγϵ)⟩ ≤ G is A4 or A5 ⇒ (by αρβ) (αβγ) ∈ G

24

⇒ α, β lie in a 3-cycle in G for any α, β (as ρ universal)⇒ Let α, β, γ ∈ [1, n] distinct, let (αβγ), (βγϵ) be suitable 3-cycles (using αρβ, β ρ γ) ⇒ (αβγ) ∈ Gby above⇒ have all 3-cycles of An in G ⇒ G = An

G primitive on Ω, say Γ is Jordan set if the pointwise stabiliser in G of Γ = Ω \ Γ is transitive on ΓΓ is a primitive Jordan set if G is primitive on Γe.g. if G contains a p-cycle (1 · · · p) then Γ = 1, . . . , p is a primitive Jordan setNote: If k-transitive on Ω, then any subset of size n− k + 1 is a Jordan set

Exercise: If G is a primitive group with primitive Jordan set (size m), then show that G is (n−m+1)-transitive(Hint: Consider Γ, Γg, Γ \ (Γg ∩ Γ) is a block of non-primitive of GΓ on Γ, so just a singleton)

Corollary 6.3G primitive on Γ, |Ω| = n, if n

2 < p < n− 2, if p∣∣|G| ⇒ G ≥ An (such prime exists for n ≥ 8)

ProofExercise

Exercise*: If G is primitive on Sn, and p is a prime with n2 < p < n− 2. If G contains a p-cycle then

G ≥ An

Hence: if n2 < p < n− 2 a prime, then p - |G|

Recall, conjugacy classes of elements of Sn corresponds to partitions of n:

Two elements of Sn are conjugate in Sn ⇔ they have same cycle-type (in disjoint cycle notation)Notation: na1

1 na22 · · ·nak

k means ai of ni-cycles, n1 > n2 > · · · > nk ≥ 1, n =∑

aini

Lemma 6.4The size of Sn-conjugacy class of elements of type na1

1 · · ·nakk is

n!

a1!(n1)a1 · · · ak!(nk)ak

Centralizer of one such element is a direct product of k wreath products:

(Ca1n1

o Sa1)︸ ︷︷ ︸wreath product

×(Ca2n2

o Sa2)× · · · × (Caknk

o Sak)

Proofn! permutation of n numbers, hence explain the numerator.We need to quotient out those which determine the same cycle:There are ni ways to write the same ni-cycle, and we have ai of those, so have nai

i in the denominatorAlso there are ai! ways to permute these ai lot of ni-cycle.

Lemma 6.5Conjugacy classes of elements in An: elements in An have type na1

1 · · ·nakk with (

∑niai) even

For such elements, the conjugacy classes under An is the whole of its Sn-ccl, unless all the ni are odd,all the ai = 1; this is the only case where CSn(x) = CAn(x) so then the Sn-ccl splits into two An-cclsof equal length

ProofExercise

25

6.1 Structure of Aut(An)

Note, Am →Ak ⇒ k ≥ mm ≥ 5 ⇒ any action of Am is on ≥ m points

Lemma 6.6If n > 6, then ∀H ≤ An s.t. H ∼=An−1 ⇒ H = (An)α of some α ∈ [1, n]

Proofn = 4, 5 Use Sylow. So assume n > 6 now.Now An−1 is simple, so H has no permutation of degree k with 1 < k < n− 1Assume H is not a stabilizer of some point.⇒ H is transitive on [1, n] an in fact primitive there.

Now n > 7 (as 7 - |A6|). Let x ∈ H which goes to 3-cycles in An−1 under our isomorphism ϕ : H∼−→

An−1

Now CAn−1(xϕ) = ⟨xϕ⟩ ×A(n−1)−3

⇒ CH(x) > An−4

⇒ CAn(x) ≥ CH(x) > An−4

⇒ x is a 3-cycle on [1, n] (other elements of order 3 have smaller centralizer)⇒ H = An by Corollary 6.3 #

Remark. This fails for n = 6 because(1) A5 acts on transitively on the set of size 6 of its Sylow 5-subgroups(2) CA6(1

331) = CA6(32)

| cclA6(123)| = | cclA6(123)(456)|| cclS6(123)(45)| = | cclS6(123456)|| cclS6(12)| = | cclS6(12)(34)(56)|

Theorem 6.7Aut(An) = Sn, for n > 3, unless n = 6

ProofAssume n ≥ 4, n = 6. Any automorphism permutes the subgroups of An isomorphic to An−1. Thereare precisely n of these, one corresponding to each point of [1, n], so any automorphism induces apermutation of [1, n].⇒ have the injection Aut(An) →Sn

This is obviously surjective because conjugation by element of Sn is an automorphism. ⇒ Aut(A6)∼=S6

Theorem 6.8Aut(A6) = Σ6, a group of order 1440, containing S6 as a subgroup of index 2.

ProofExistence:Let G e a simple group of order 60 (e.g. A5)⇒ G has 6 Sylow 5-subgroup, on which it acts.⇒ ∃ faithful permutation representation ϕ : G → A6

Gϕ ≤ A6 of index 6, this gives a permutation representation A6 → A6 which must be an isomorphism.This automorphism takes Gϕ (now a transitive subgroup of A6) to the stabiliser of a point of A6.Now the elements of order 3 on the left are type 32 (as 3 - 60

6 ), but those on the right are of the type1331

⇒ this automorphism is not induced by conjugations⇒ Aut(A6) > S6

Index 2:Any automorphism fixing the ccl of 3-cycles is induced by conjugation from S6

26

There are only two ccls of elements order 3If θ1, θ2 are two automorphism swapping these two classes (i.e. θ1, θ2 ∈ Σ6 \S6), then θ−1

1 θ2 fixes eachof these ccls, so is in S6

⇒ |Σ6 : S6| = 2

Corollary 6.9G simple order 60 ⇒ G∼=A5

∼=PSL2(5)∼=PSL2(4)

ProofThe proof of Theorem 6.8 uses an arbitrary simple group of order 60.

Example: (Bochart)H a primitive subgroup of Sn, not containing An

Then |H| ≤ n!/[n+12 ]! (e.g. n = 6, S5 on Syl5)

Let k be maximal s.t. H ∩ Sk = 1;⇒ |H||Sn/2| = |HSn/2| ≤ n! and |Γ| = k. Take Γ s.t. |Γ| = k andH ∩ Sym(Γ) = 1

Claim: k ≥ n2

Proof of Claim:∃g(= 1) ∈ H ∩ Sym(Γ), αg = α, α ∈ Γ[g, h] is 3-cycle (check)∃h(= 1) ∈ H ∩ Sym(Γ ∪ α)⇒ H ≥ An #

Sylow p-subgroups of Sn:n = n1p

e1 + n2pe2 + · · ·+ ne1p+ ne1+1 0 ≤ ni < p

Concentrating on the case pe1 , e1 = 1 :p2, e1 = 2: (Cp)

p o Cp

...pnpn+1 (Pn)

p o Cp

Any Sylow p-subgroup is ∏1≤i≤e1

Pei × · · · × Pei︸ ︷︷ ︸ni times

Thus has the right order

How about maximal subgroups H of An, Sn (and other subgroups G with A6 < G ≤ Σ6)?What if H is intransitive on [1, n]? Then H has an orbit of size 1 ≤ k ≤ n

2 and H ≤ Sk × Sn−k

So the maximal subgroups intransitive on [1, n] are precisely Sk × Sn−k for 1 ≤ k ≤ n2

Lemma 6.10The maximal intransitive subgroups of Sn are Sk × Sn−k for 1 ≤ k ≤ n

2These are maximal in Sn unless k = n

2 (n = 2k, Sk × Sk < (Sk × Sk)o S2 < S2k)

ProofLet Sk × Sn−k < X ≤ Sn

Then X on [1, n] is transitive, in fact primitive, (unless n = 2k), since X is then 2-transitive. if not,take α from k, Xα has suborbits of size 1, k − 1, n − k. Take β from n − k, Xβ has suborbits size1, k, n− k − 1 Impossible unless n = 2kBut X contains a transposition and 3-cycle.⇒ X ≥ An

27

Addition to previous lectures:An PSn ≤ Aut(An)θ : Aut(An) → Sn

K = ker θ, K ∩An = 1 ⇒ [K,An] = 1 ⇒ K = 1 so θ injectiveExercise: Sylvester’s construction of an outer automorphism of S6 (Wilson Ex 2.19)

6.2 Wreath Products

Let C be a group, let D be a permutation group on ∆The wreath product C wr∆D is the semidirect product of the base group C∆ (the direct product of|∆| copies of C) by D, with D acting by permuting the components:

(cδ1 , cδ2 , . . . , cδl)d = (cδ1d−1 , cδ2d−1 , . . . , cδld−1)

i.e. C wrD = C∆ oD

Suppose now C acts on Γ as a permutation group. The imprimitive action of C wrD on Γ×∆:Action of base group: (µ, δ)(cδ1 , . . . , cδl) = (γcδ, δ)Action of D: (µ, δ)d = (µ, δd)For δ ∈ Γ, let Γδ = (µ, δ)|µ ∈ Γ (block of impriimitivity), then B = Γδ|δ ∈ ∆ is the set of blocks⇒ C wrD is imprimitive on Γ×∆In particular, have Sk wrSl imprimitive on Γ×∆, with |Γ| = k, |∆| = l

Lemma 6.11Let G be (transitive but) imprimitive on Γ, |Γ| = n so G ≤ Sn, with l blocks of size k. ThenG ≤ Sk wrSk (the full stabilizer of this partition of Γ into blocks). Moreover, these groups Sk wrSl

are maximal in Sn (n = kl)

Proof(Proof of moreover part): Sk wrSl < H ≤ Sn. Then H on [1, n] is transitive, in fact primitive.Hα contains Sk−1 × Sk wrSl−1

But H contains transpositions ⇒ H = Sn (c.f. Theorem 6.2)

Remark. If we are in An rather than Sn, use the 3-cycles in H, unless k = 2. The case k = 2 is harder

and have to use elements of type 22, in fact it is false if n = 8: S2wrS4index 7< AGL(3, 2)

index 15< A8

How about G primitive on [1, n]?Let N1, N2 be minimal normal in G and N1 = N2

Then [N1, N2] = 1, and the N1, N2 are both transitive on [1, n], so regular on [1, n] and CG(N1) =N2, CG(N2) = N1 (c.f. Lemma 5.10)Also, the Ni are non-abelian.It follows that G has at most two minimal normal subgroups.

If two, these are non-abelian, isomorphic to each other:Put N = N1 ×N2 (the socle of G)Put H = Nα ⇒ N = HN1 = HN2 and H ∩Ni = 1⇒ H ∼=H/H ∩N1

∼=HN1/N1 = N1N2/N1∼=N2 and by symmetry H ∼=N1

In fact, by Lemma 5.8, can identify Ω and N2 so that N2 acts by right regular action: n ∗R n2 = nn2

for n ∈ Ω, n2 ∈ N2

⇒ Nacts by left regular action on Ω as above sinceN1 = CSn(N2): n∗Ln1 = n−11 n for n ∈ Ω, n1 ∈ N1

These actions commutes: n ∗L n1 ∗R n2 = n−11 nn2 = n ∗R n2 ∗L n1

So N acts on Ω by n ∗ (n1, n2) = n−11 nn2

This is so-called diagonal action, due to Burnside

28

E.g. A5 ×A5 acting on A5: x(g, h) = g−1xh, G1 = (g, g)|g ∈ A5The G1 orbits are ccls in A5

Now G ≤ NSn(N), but also ∃g ∈ NSn(N) \G:g acts as n 7→ n−1

⇒ G is never maximal in Sn (or An)⇒ N = T 2k where T is non-abelian simple and n = |T |k

Summarizing:

Lemma 6.12If G is primitive subgroup of Sn, then G has at most two minimal normal subgroup.If have two, they have to be regular, non-abelian, isomorphic to (each other) T k with T non-abeliansimple, k ≥ 1, with n = |T |kAnd G is not maximal in Sn (or An)

Exercise: G×G on G by g ∗ (g1, g2) = g−11 gg2 is transitive; it is primitive ⇔ G simple

Lemma 6.13(c.f. Lemma 5.8) If G has a regular elementary abelian normal subgroup, then G ≤ AGLd(p) ≤ Sn,n = pd

In addition, if G transitive, G is primitive ⇔ G0 is an irreducible linear group on Vd(p)

Let G primitive with a unique minimal normal subgroup NIf that is abelian, then |N | = pd and have G0 being an irreducible linear group on Vd(p)

6.3 The primitive (product) action of wreath products

C acts on Γ, D acts on ∆ |Γ| = k, |∆| = lC wrD acts on Γ∆, degree kl (not Γ×∆ as C wrD = C∆ oD)Action by base group C∆:

(γδ1 , . . . , γδl)(cδ1 , . . .) = (γδ1cδ1 , . . . , γδlcδl) (coordinate-wise)

(γδ1 , . . . , γδl)d = (γδ1d−1 , . . . , γδld−1)

Lemma 6.14If D transitive on ∆, C primitive on Γ but not regular Cp, then C wrD is primitive on Γ∆, degree kl

ProofLet γ ∈ Γ. The stailizer of the constant point (γ, γ, . . . , γ) is H = Cγ wrD

Claim: H is maximal in C wrD

Proof of Claim:Exercise

Example:

W (k, l) = Sk wrSl, degree n = kl, with k ≥ 3 is primitiveIf k = 3 or 5, we have an elementary abelian regular normal subgroup⇒ previous case ⇒ assume k ≥ 5This is usually maximal in Sk, n = kl

29

Exercise:

W (k, l) with k ≥ 2 has rank l + 1, subdegrees are 1, l(k − 1),

(l2

)(k − 1)2, . . . ,

(ll

)(k − 1)l

6.4 Diagonal Actions

N = Tm, T non-abelian simple, n = |T |m−1, m ≥ 2

D := (t, . . . , t) ∈ Tm|t ∈ T < N

diagonal subgroup of N , Ω = (N : D)D(T,m) := NSn(T

m). In fact, D(T,m)/Tm∼=Out(T )× Sm (Note that Out(T ) = Aut(T )/T here)Action of N : D(x1, . . . , xm)(t1, . . . , tm) = D(x1t1, . . . , xmtm)Action of Aut(T ): D(x1, . . . , xm)α = D(xα1 , . . . , x

αm)

Action of Sm: D(x1, . . . , xm)π = D(x1π−1 , . . . , xmπ−1)

Example:m = 2, D(T, 2) = (T × T ) · (Aut(T )× C2)may identify Ω with T :

(t1, t2) : t 7→ t−11 tt2

α : t 7→ tα

π(∈ S2) : t 7→ t−1

G < Sn, n = |T |m−1 is of diagonal type if Tm PG ≤ D(T,m). Such G is primitive ⇔ the action ofthe subgroup of Sm on coordinates is primitive.The groups D(T,m) is usually maximal in Sn or An

Theorem 6.15 (O’Nan Scott)If G < Sn primitive, then G is either almost simple or G is a subgroup of one of:

(1) Sk × Sn−k

(2) Sk wrSn/k

(3) AGLd(p) (affine action), n = pd, a prime power

(4) W (k, l) (product action), n = kl, k ≥ 5

(5) D(T,m) (diagonal action), n = |T |m−1, m ≥ 2, T non-abelian simple

(Proof omitted, see Wilson’s book, also, a more precise revision for primitive permutation groupsgiving a structure)

Corollary 6.16Maximal subgroups of Sn (or An):

(1) Sk × Sn−k, k < n2 , intransitive

(2) Sk wrSl, n = k × l, imprimitive

(3) AGLd(p), n = pd

(4) D(T,m), n = |T |m−1

(5) G primitive almost simple

30

ProofMay appear later (or not)In fact, with a stronger version for G primitive subgroups of Sn:G is almost simple except in cases we have seen (plus one other)

But which are maximal in Sn (or An)?Example:

S5 < S10 on

(52

), (S5 : S3 × S2), maximal? Do not know

S6 < S10 on 3|3 (partition into three 3s), (S6 : S3wrS2)S5 < S6 < S10

Let G = S6, H = S5, have a factorisation of S6 :

G = S6

10

nnnnnnnnnnnn6

LLLLLLLLLL

Gα = S3wrS2

6 PPPPPPPPPPPP S5 = H

10rrrrrrrrrr

S3 × S2

H is transitive, on S6: S3wrS2

So need to know factorisation

G = ABmaximal

qqqqqqqqqqcan take maximal,inductively

IIII

IIII

II

AwrS2

6 MMMMMMMMMM B

uuuu

uuuu

uu

A ∩B

these are now known for G almost simple. The question is answered

Upshot: G almost simple and primitive of degree n ⇒ G maximal in An or Sn unless in a list

Example:G = Sm |A||B| largeLet m

2 < p < m− 2, a prime (exists for m ≥ 8, by Chebychev)⇒ p|m! ⇒ p

∣∣|A| or p∣∣|B|

m! = |AB| = |A||B||A∩B|

By remarks earlier, if A is primitive on [1,m] then A ≥ Am #⇒ A is intransitive on [1,m] (cannot be transitive but imprimitive as p

∣∣|A|)⇒ A = Sk × Sm−k for some k with 1 ≤ k < m

2

Case k = 1 is not very interesting.

Assume k > 1. Then B is transitive on

(mk

)(k-subsets of [1,m]). We say B is k-homogoeneous on

[1,m]

k = 2 : B is 2-transitive on |B| odd (in fact, it is solvable)k > 2 : G is k-homogeneous ⇒ G is (k − 1)-homogeneous

Exercise: If πk is the permutation character of Sm on

(mk

), then πk = πk−1+χk where χk is irreducible

(use ⟨πk, πl⟩ = 1 +min(l, k))⇒ B a is known group

31

So, to find maximal subgroup of Sm (or Am), it is necessary and sufficient to find the maximal subgroupof all smaller almost simple groups

6.5 Doubly transitive representations of Sm or Am

Theorem 6.17 (Maillet)If Sm (or Am) is 2-transitive degree n, then n = m, or one of the below (m ≤ 8)

(1) m = 5;n = 6

(2) m = 6;n = 6 or 10

(3) m = 7, 8;n = 15 and this only happens in Am

Lemma 6.18If G is a transitive permutation group on Ω, and G has a ccl C = 1, then G has a non-trivialsubdegree at most |C|ProofFix c ∈ C. Let α ∈ Ω with β = αc = α. Consider βGα:If ω ∈ βGα, have ω = βh some h ∈ Gα

⇒ ω = αh−1ch with h−1ch ∈ CSketch Proof of Theorem 6.17In Sm, take C = transpositionIn Am, take C = 3-cycles⇒ n− 1 ≤ 1

2m(m− 1)Let H be the stabiliser of a point in [1, n]⇒ |Sm : H| = n and H maximal on Sm

If H is primitive on [1,m], then “small” by Bochert (see Example after Corollary 6.9):[m+12

]! ≤ 1 + 1

2m(m− 1) ⇒ m ≤ 6(If in Am, get m ≤ 8)If H is transitive but not primitive on [1,m], then H = Sk wrSm/k, the action is on k|k| · · · |k, not2-transitive on m > 6

If H is intransitive on [1,m], the action of Sm is on

(nk

), not 2-transitive unless k = 1

7 Linear Groups

Let F be a field, here usually finite F = Fq, q = pl

GLd(F ), group of invertible linear transformation on V = Vd(F )For F = Fq, write GLd(q), group of all non-singular d× d-matrices over F|GLd(q)| = (qd − 1)(qd − q) · · · (qd − qd−1)

SLd(q)PGLd(q) matrices of determinant 1|SLd(q)| = |GLd(q)|/(q − 1)

PGLd(q) := GLd(q)/Z, where Z = scalar matrices in GLLd(q) = PSLd(q) := SLd(q)/Z ∩ SLd(q), has order |SLd(q)|/e, where e = (d, q − 1)

PGLd(q) acts naturally on the set of 1-subspaces of Vd(q) (i.e. the projective space of dimension d−1,

Pd−1(q), of sizeqd−1q−1 )

Recall: PGLd(q) is 2-transitive, and is 3-transitive ⇔ d = 2PSL2(q) is 2-transitive, and is 3-transitive ⇔ q = 2l

32

More on d = 2: “Mobius action” of PGL2(q) of degree q + 1, Fq ∪∞

Lemma 7.1

⟨(x, y)⟩ ↔

x/y y = 0

∞ y = 0

⇒(a bc d

): z 7→ az+b

cz+d ??????

PSL2(q) = z 7→ az+bcz+d |ad− bc is a non-zero square

AGL1(q) = G∞z 7→ az + b|a = 0Gαβγ = 1, any α, β, γ distinctG0∞ = z 7→ αz|a = 0G is 3-transitive, with Gαβγ = 1 ∀αβγ

Generators of PGL2(q) :z 7→ z + 1z 7→ λz (⟨λ⟩ = F×

q )z 7→ −1/z

Lemma 7.2q Group order degree

2 L2(2)∼=S3 6 33 L2(3)∼=A4 12 44 L2(4)∼=A5 60 55 L2(5)∼=A5 60 6

Also: L2(9)∼=A6∼=Ω−

4 (3), L2(7)∼=L3(2), L4(2)∼=A8∼=Ω+

6 (2), L4(2) L3(4)

ProofExercise

Theorem 7.3Ld(q) is simple for d ≥ 2, unless d = 2, q ≤ 3

To prove this, we need Iwasawa’s Lemma:

Lemma 7.4 (Iwasawa’s Lemma)Let G be a perfect group (i.e. G′ = G) acting primitively on a set Ω, let α ∈ Ω, and assume Gα has anormal subgroup K which is abelian (or just solvable) with G = ⟨Kg|g ∈ G⟩If N PG, then N ≤ G(Ω), the kernel of G on Ω, or N = GSo G/G(Ω) is simple

Example:

An is simple: Consider G = An acting on

(n3

), the 3-subsets, (for n = 6, 3|3)

Gα = (Sn−3 × S3) ∩An,K = ⟨(123)⟩ProofAssume N G(Ω), so N transitive on Ω⇒ G = NGα. Hence NK PNGα = GNow, ⟨Kg|g ∈ G⟩ = G⇒ NK = G⇒ G/N ∼=NK/N ∼=K/K ∩N is abelian⇒ G′ ≤ N , but G = G′

⇒ N = G

33

Definitiont ∈ SLd(q) is a transvection if t− 1 has rank 1 and (t− 1)2 = 0(JNF has 1 block size 2, all other size 1) Note: If t is a transvection on V , then wrt some basis B ofV , t has matrix of form 1 0

0. . .

1 0 1

This is because, let W = ker(t− 1), let vd ∈ V \W , let v1 = vd(t− 1)extend the basis v1, . . . , vd − 1 of W , get B = v1, . . . , vd−1, vdAlso, the elementary matrices E(∗) are transvections Xij(λ), Xij(λ)

−1 = Xij(−λ)Transvection subgroup: Xij = Xij(λ)|λ ∈ Fq

Exercise: Al transvections are conjugate in SLd(q) if l = 2. If d = 2, q odd then two ccls

Proposition 7.5SLd(q) is generated by the transvections

ProofLinear algebra: any matrix of det 1 can be reduced to IA by applying elementary row operationsvi := vi + λvj , i = jEach of these operations is just multiplication on the left by a matrix E(∗)

E(n) · · ·E(1)A = I ⇒ A = (E(1))−1 · · · (E(n))−1

Proof of Theorem 7.3Let G = SLd(q). Then G is 2-transitive, hence primitive, on Ω = 1-subspaces of Vd(q)Kernel G(Ω) = scalar matricesLet α = ⟨(1, 0 · · · , 0)⟩

⇒ Gα =

λ 0 · · · 0λ2

... Hd−1

λd

λi ∈ Fq

, K =

1 0 · · · 0λ2

... In−1

λd

∣∣∣λi ∈ Fq

elementary abelian order qd−1 normal in Gα

The elements of K \ 1 are transvections.We shall check they generate G, and each is a commutator in G (shown in next proposition)

Proposition 7.6(d > 1) All transvections are commutators, except when d = 2, q ≤ 3

ProofIf d ≥ 3 1

α 11

11β 1

1−α 1

1

11−β 1

=

11

−(αβ) 1

so if α = 0, take β = α−1γ

d = 2: (α−1

α

)(1β 1

)(α

α−1

)(1β 1

)=

(1 0

α2β 1

)=

(1 0

β(α2 − 1) 0

)So given γ, take α = 0 with α2 = 1 (q > 3)

and β = γ(α2 − 1)−1, to get

(1γ 1

)as a commutator

34

7.1 Some subgroup of GLd(q) or SLd(q)

Parabolic:

(1) B =

∗ 0

. . .

# ∗

∣∣∣∣∣∗ ∈ F ∗,# ∈ F

- Borel subgroup

(2) lower traingular

(3) U =

1 0

. . .

# 1

∣∣∣∣∣# ∈ F

- lower unitriangular matrix |U | = q12d(d−1) a Sylow p-subgroup

of GL, q = pt

B = U o T , with T =

λ1 0. . .

0 λd

|λi ∈ F× diagoonal matrices or split torus

V = Vd(q) B is the stabilizer of a complete flag: 0 = V0 < V1 < V2 · · · < V = Vd

Vi = ⟨v1, . . . , vi⟩

Other flags: 0 < Vi < Vj < · · · < V = Vd

Stabilisers of flags are parabolic subgroupsmaximal parabolic is just a stabilizer of a subspace of W of VPk is the stabilizer of a k-subgroup W of V :

Pk = (

Ak 0

C Bd−k

)|(A,B) ∈ GLk ×GLd−k, C any of k × k matrix

(choose basis e1, . . . , ek of W extend to a basis e1, . . . , ek, . . . , eh for V )

The subgroup Uk = (

Ik∗ Id−k

) is a normal subgroup of Pk;

Lk =

(Ak 0

0 Bd−k

)- complement to Uk in Pk, the Levi subgroup!of linear group of Pk

Then Pk = Uk o Lk

By the way, N= all monomial matrices, then N = NG(T )T = B ∩N , W = N/T - Weyl groupExample: GLd(q), W = Symd

Remark. The subgroups of GLd(q) containing B are precisely the parabolics obtained by stabilizingthe flags obtained by deleting some members of the complete flag for B

Exercise:G = GLd(q) has permutation rank k + 1 for (G : Pk) (Pk stabiliser of W , dim k), 1 ≤ k ≤ d

2And, Pk is a maximal subgroup of GLd(q)

Some other subgroups:some families of geometric subgroupThen theorem of Aschbache tells you that any subgroup H of SL lies in one of these or H/Z(H) isalmost simple, irreducible, etc....

(1) Pk - the stabilizer of a k-subspace W of V

(2) V = V1 ⊕ · · · ⊕ Vk, d = ka, dimVi = aGLa(F )wr Symk

35

(3) V = V1 ⊗ V2, dimVi = di, d = d1d2, d1 = d2GLd1(F )⊗GLd2(F )

(4) V = V1 ⊗ · · · ⊗ Vk, dimVi = a, d = ak

GLa(F )wr Symk

(5) Subfield subgroup Fq′ < Fq

V = W ⊗ Fq, dimFq′ W = d

GLd(q′) < GLd(q)

(6) Extension field subgroups GLd/e(qe) < GLd(q)

(7) “Extraspecial”

(8) Classical

Upslot: Questions about maximal subgroups of linear groups can be reduced to question about modularrepresentation theory of quasi-almost simple groups

7.2 Automorphism group of PSLd(q)

PSLd(q), q = pf , Outer automorphisms come in different flavours:

Ld(q) P PGLd(q) P PΓLd(q) P Aut

(index) (d,q−1)Inn diag

ffield auto

2 if d>21 if d=2

graph auto

ΓLd(q): semilinear transfromation of V = Vd(q)(r + w)θ = rθ + wθ∃σ ∈ Γ = Gal(Fq /Fp)(λv)θ = λσvθ (σ depends on θ)Fix a basis, (

a11 . . . a1d... · · ·

)σ

=

(aσ11 . . . aσ1d... · · ·

)ΓLd(q) = GLd(q)o Γfinally, a graph auto: A 7→ A−t (inverse transpose of A) an auto of GL Exercise: Graph auto is innerif d = 2, outer otherwiseAut := PΓLo ⟨τ⟩

Theorem 7.7 (Steinberg, for groups of Lie type)The Aut. group of L2(q) is PΓL2(q)The Aut. group of Ld(q) with d > 2 in PΓLd(q) : C2 (semicolon denotes semidirect product)

SketchNeed to show no more automorphisms. Let G = PSLd(q), V = Vd(q), ϕ ∈ Aut(Ld(q)), q = pf

Any p-local subgroup (i.e. normaliser of a a p-subgroup) stabilise (set-wsie) a subspace⇒ in Pk for some k:H = NG(Q), Q a p-subgroup, let

W := fixV (Q)

is a subspace kept invariant by NG(Q). Note that if the ccls of stabilisers of hyperplanes is not ϕ-invariant, adjust ϕ by a graph automorphism. (note that it would have been swapped with the cclsof stabilisers of 1-spaces, as the only possibility by structure of Pk).

So assume ϕ keeps the set of hyperplanes invariant, and then it follows that it keeps the set of k-spacesinvariant for all k. In particular, ϕ acts on the set of 1-spaces.

36

G = SLd(q) is transitive on the set of complete flags, so may assume that ϕ stabilises a complete flags,and in fact, may assume that ϕ stabilises ⟨v1⟩, ⟨v2⟩, . . . , ⟨vd⟩, with v1, . . . , vd a basis. Moreover, usinga diagonal automorphism, if necessary, to adjust ϕ, we may assume ϕ fixes v1, . . . , vd.

We now claim that such ϕ is induced by a Galois automorphism of Fq /Fp, i.e. ϕ is semilinear on V1λ 1

. . .

1µ 1

. . .

=

1λ+ µ 1

. . .

ϕ : ↓ ↓ 1

λσ 1. . .

1µσ 1

. . .

=

1λσ + µσ 1

. . .

λσ ∈ Fq, (λ+ µ)σ = λσ + µσ

and

λλ−1

I

µµ−1

I

→ (λµ)σ = λσµσ

⇒ σ ∈ Gal(Fq) A general proof: In Carter’s book, Groups of Lie type

7.3 Some isomorphisms and interesting actions

Exercise: Any simple group of order 168 is isom. to L2(7)

Lemma 7.8L2(2)∼=L2(7)

ProofL3(2) acts on P2(2): V = V3(2)

points: 1-subspace of V

lines: 2-subspace of V

incidence: subspace of V

Fano plane (any 2 points on a unique line (of size 3)) (see picture)

In fact, can ??? mod 7 points so that 013, 124, 235, . . . are lines

F7

g : u 7→ u+ 1 on F7 (points to points, lines to lines), g = (0123456)h : u 7→ 2u in N(⟨g⟩), h = (124)(365)t = (12)(36) ∈ NG(⟨h⟩)These generate L3(2)

Let G = L3(2) (or any simple group of order 168)np(G) = 8, |NG(P )| = 21Let P ∈ Syl7(G), say P = ⟨g⟩Number the Sylow 7-subgroups as P = ∞, 0, 1, . . . , 6Choose one of the Sylow 7-subgroup as 0 and g : z 7→ z + 1, then all the numbering are determined

37

Let h ∈ NG(P ), order 3, h : z 7→ 2z (NG(P ) = NA7(P ))∃t ∈ NG(⟨h⟩), order 2, inverting h (know this is subgroup L2(2))Now the stabiliser of any point in Syl7(G) has odd order, so t is fixed-point-free in this action of degree8:

(0∞)︸ ︷︷ ︸fix h

(1x)(2y)(4z)

(x, y, z ∈ 3, 5, 6 but not known which yet)Conjugating t by h or h−1, we may ssume t : 1 ↔ 6Then 2t = 3, since 2t = 1ht = 1th−1 = 6h−1 = 3, so we get

(0∞)(16)(23)(45)

Thus, t : z 7→ −1/z−1

⇒ L3(2) . L2(7), so ∼= by order

Lemma 7.9A6

∼=L2(9), S6∼=PΣL2(9) (Σ field autos), Σ6

∼=PΓL2(9)

ProofProduce an action of A6 degree 10 (3|3 - stab. is (S3wrS2) ∩A6)Put F9 ∪∞ structure on it to get A6 . L2(9)H = NA6(⟨(123)⟩, ⟨(456)⟩)∼=(S3wrS2) ∩A6

123|456 is fixed by HAn element h of order 4 in H: (14)(2536)Let Fq = 0,±1,±i,±(1± i)Let g1 : z 7→ z + 1, g2 : z 7→ z + iNow h also fixes partition 156|234Rest of notation is now fixed, t : z 7→ −1/z−1

⇒ A6 . L2(9)The above part demonstrates actions of A6 on F9 ∪∞ = P1(9)It yields A6 can be “embedded” into L2(9)Now (56) acts as z 7→ z3 a field automorphism⇒ S6

∼=PΣL2(9)

Note: Gal(Fpl /F) is generated by z 7→ zp and is cyclic of order l

Some 2-transitive actions of Ld(q):

Ld(q)is 2-transitive on qd−1q−1 (2 actions for d > 2, 1 action for d = 2)

plus: L2(7),deg 7 (2 action)L2(9),deg 6L2(11),deg 11PΣL2(8)∼= 2G2(3),deg 28L4(2)∼=A8,deg 8

8 symplectic Groups Spd(q) ≤ GLd(q)

DefinitionIf f a bilinear alternating non-singular form on V = Vd(q), f is called symplectic form (We are takingdefinition f(v, v) = 0 ∀v for alternating because f(v, w) = −f(w, v) is weaker when char=2)Non-degenerate (non-singular): ∀v = 0,∃w s.t. f(v, w) = 0⇒ V ⊥ := w|f(v, w) = 0 = V ∀v = 0

38

Lemma 8.1If f is a symplectic form on V , there exists basis e1, f1, e2, f2, . . . em, fn with f(ei, ej) = 0 = f(fi, fj)and f(ei, fj) = δijProofConstruct:Let e1 = 0. Take f1 ∈ V \ e⊥1 (non-empty)with (after scaling) f(e1, f1) = 1. Continue this process in ⟨e1, f1⟩⊥

DefinitionSuch ei, fi is a hyperbolic pairThen the matrix of f under this basis is

1−1

1−1

. . .

or taking e1, e2, . . . , f1, f2, . . .

1

. ..

1

−1

. ..

−1

Let G = Sp2m(q) preserves this form, i.e. G = g ∈ GL2m(q), f(vg, wg) = f(v, w) ∀v, w ∈ V In terms of matrices A ∈ GL2m(q)|AtJA = J

Sp2m(q) is regular in its action on the symplectic bases. (transitive with trivial stabliser)

Lemma 8.2|Sp2m(q)| = (q2m − 1)(q2m−1)|Sp2m−2(q)| = · · · = qm

2∏mi=1 q

2i − 1

Lemma 8.3Sp2(q)∼=SL2(q),

Proof

At

(1

−1

)A =

(1

−1

)for ad− bc = 1

Z(Sp2m(q)) = ±I, PSp2m(q)∼=Sp2m(q)/±I

Definitionsymplectic transvections are those transvections in Sp

tv,λ : x 7→ x− λf(x, v)v

E.g. v = e1 :f(x, e1) = 0 for other basis

39

f(x, f1) = 1, so under basis e1, e2, . . . , em, fm, . . . , f1:

1. . .

1

1

λ. . .

1

Proposition 8.4Sp2n(q) is generated by these transvections

ProofG = ⟨ these transvections ⟩, will show G is transtive on the set of symplectic basis

(1) G is transitive on the set of all vectors:To go from v to w. If v⊥w, then vtv−w,λ = v + λf(v, w)(v − w)Take λ = −f(v, w)−1 we get vtv−w,λ = wIf v⊥w, choose x ∈ V \ (v⊥ ∪ w⊥)and go v 7→ x 7→ w by the above method

(2) G is transitive on the set of hyperbolic pairs:(v, w1) and (v, w2) are hyperbolic pairIf w1 * w2, take tw1−w2,λ as in is to send w1 7→ w2

If w1⊥w2, go w1 7→ w1 + v 7→ w2 which fixing v

(3) G transitive on the set of symplectic basis by inductionIf B′ = e′1, f ′

1, . . . may assume e′1 = e1, f′1 = f1 by (2)

Then work in ⟨e1, f1⟩⊥ − V2m−2(q) with symplectic form obtained by restriction. Induction dothe rest

Corollary 8.5Sp2m(q) ⊆ SL2m(q)

Proposition 8.6Transvections in Sp2m(q) are commutators, unless (2m, q) is one of (2, 2), (2, 3)(4, 2)

ProofThis is true as Sp2(q) = SL2(q) for q ≥ 3So only have to check cases Sp4(3) and Sp6(2) (exercise)(Sp2m(q) = [SL2m(q), SL2m(q)])

Theorem 8.7PSp2m(q) is simple, except for PSp2(2)∼=S3, PSp2(3)∼=A4, PSp4(2)∼=S6

ProofUse Iwasawa’s Lemma 8.8

Lemma 8.8 (Iwasawa)

The action of Sp2m(q) on the q2m−1q−1 points of P2m−1(q) is rank 3, with subdegrees 1, q(q2m−2−1)/(q−

1), q2m−1

This stabilizer of ⟨v⟩ is P1 =

λ 0 0∗ A 0∗ ∗ λ−1

∣∣∣∣∣λ ∈ F×q , A ∈ Sp2m−2(q)

with normal subgroup

10 Iλ 0 1

=

tv,λ|λ ∈ Fq

40

The action is primitive, kernel is ±IProofv ∈ V , suborbits ⟨v⟩ (size 1), ⟨w⟩|w ∈ v⊥ − ⟨v⟩, ⟨w⟩|w ∈ V \ v⊥ (size q2m−1)

The action is primitive since any non-trivial block would consist of ⟨v⟩ together with one of the

G⟨v⟩-orbits, but the size does not divide qd−1q−1

Finally, G⟨v⟩ = P1 - see below about Pk

Proposition 8.9Sp4(2)∼=S6, so not perfect

ProofS6 on U = V6(2) natural action as permutations of coordinates.W = (x1, . . . , x6)|

∑xi ≡ 0 mod 2

f(x, y) =∑

xiyi, bilinear form on UW⊥ = (1, 1, . . . , 1)⇒ W ⊇ W⊥

V = W/W⊥ a 4-dimensional space over F2 with a symplectic form preserved by S6

⇒ S6 ≤ Sp4(2) on V⇒ S6

∼=Sp4(2) by order.

Exercise: S2m+2 ≤ Sp2m(2)

8.1 Parabolic subgroups in Sp2m(q)

V = V2m(Fq)A complete symplectic flag: 0 < W1 = ⟨e1⟩ < W2 = ⟨e1, e2⟩ < · · · < Wm = W⊥

m = ⟨e1, . . . , em⟩ <W⊥

m−1 = ⟨e1, . . . , em, fm⟩ Wi isotropic, so f |Wi is 0, for each 1 ≤ i ≤ mB Borel stabilises such a complete flagThe parabola are stabilisers of some flagsMaximal parabolic subgroup Pk := Stabilisers of Wk with Wk ⊆ W⊥

k

Pk stabilise W = ⟨e1, . . . , ek⟩ and W⊥ = ⟨e1, . . . , em, fk+1, . . . , fm⟩ (dim 2m− k)W⊥/W = ⟨ek+1, . . . , em, fk+1, . . . , fm⟩ -Symplectic space dim 2(m− k)

Pk contains a Levi subgroup GLk × Sp2(m−k):

A

B

A−t

(on W , W⊥/W , V/W⊥)

Kernel of the homomorphism Pk → Lk is Qk =

I∗ I∗ ∗ I

Qk is the unipotent radical of Pk, order q

k(k+1)/2q2k(m−k), often a “special” p-group:

Q′ = Z(Q) = Φ(Q), order qk(k+1)/2

Example2m = 4, k = 1

Q1 =

1α 1β 0 1γ β −α 1

, e.g. q = 3, can get

1

11

1 1

as a product of commutator

k = 2, Q2 =

(IC I

) ∣∣C = Ct

41

Exercise: P1, Pm in general?

Remark. Sp2m(q) is Cm(q) as group of Lie type

The other interesting stabilisers of subspace W have W ∩W⊥ = 0⇒ V = W ⊕W⊥ (dimW = 2k, dimW⊥ = 2(m− k))

Nk = Sp2k(q)× Sp2(m−k)(q) < Sp2m(q)

(Sp4(2)∼=S6 deg 6, 10)2-transitive actions of Sp2m(q): 2m = 2, deg q + 1 and deg q if q ∈ 5, 7, 11and Sp2m(2) deg 2m−1(2m ± 1) on the two classes of orthogonal forms.

Automorphism groups: PSp2m(q)PPGSp2m(q) (diagonal (2, q − 1)), g : f(vg, wg) = λgf(v, w) ∈ Fq

PSp2m(q)PPGSp2m(q)PPΓSp2m(q)PAut (this last P has index 1 except for Sp4(2f) then index2)

9 Unitary Groups

F = Fq2 > Fq a quadratic extension, V = Vd(q2)

σ : α 7→ αq ⟨σ⟩ = Gal(Fq2 /Fq)

α 7→ α N(α) = αq+1, N : Fq2 → Fq2 onto

Let f be a σ-Hermitian, left linear, non-singular form

GUd(q) = g ∈ GLd(q2) | f(vg, wg) = f(v, w)

If dimV ≥ 2 ⇒ ∃ isotropic vectors:Since, let v ∈ V with f(v, v) = 0Let u ∈ v⊥

If u not isotropic, consider f(u+ λv, u+ λv) = f(u, u) + λλf(v, v) - can make it zero by choice of λ

Unitary bases:

(1) d = 2m e1, f1, . . . , em, fm as usual: f(ei, fj) = δij ,etc.d = 2m+ 1 e1, f1, . . . , fm, v v⊥ei, fj , f(v, v) = 1⇒ all forms are equivalent, for each dimension

(2) Also, have an orthonormal bases

Lemma 9.1

|GUd(q)| = q12d(d−1)(qd − (−1)d)(qd−1 − (−1)d−1) · · · (q2 − 1)(q + 1)

ProofWrite

zd = # isotropic vectors in dimension d

yd = # vectors norm 1 ⇒ q2d = 1 + zd + (q − 1)yd

Also, zd+1 = zd + (q2 − 1)yd = −qzd + (q2d − 1)(q + 1)

Since z0 = 0 = z1, can solve recurrence: zd = (qd − (−1)d)(qd−1 − (−1)d−1)⇒ yd = qd−1(qd − (−1)d)The order of GU follows, by induction along an orthonormal basis

42

|SU | = |GU |/(q + 1) , |PSU | = |GU |/(q + 1) gcd(d, q + 1)

In particular, PSL2(q)∼=PSU2(q)

Lemma 9.2PSU3(q) acts 2-transitively on the isotropic points of P2(q

2). There are q3 + 1 of these; for larger d,this leads to a primitive rank 3 action.

Proofe isotopic. Let f1, f2 isotropic in V \ ⟨e⊥⟩ (Note: no isotropic vectors in e⊥ as d = 3)Can swap e, f1, to e, f2, using an element in GU3(q)⇒ (GU3)⟨e⟩ transitive of degree q3, and SU3 has index q + 1 in GU3

⇒ still transitive

For d ≥ 3, PSUd(q) is simple except for PSU3(2) (order 72, 2-transitive on 9), c.f. Iwasawa Lemma

Maximal parabolic subgroups: Pk = stabiliser of a totally isotropic k-vector space = GLk(q2) ×

GUd−2k(q)

10 Orthogonal Groups

Quadratic form Q : V → F = Fq

s.t. Q(λv) = λ2Q(v)Q(u+ v) = Q(u) +Q(v) + f(u, v), with f bilinear symmetric form⇒ 2Q(v) = f(v, v)In char F = 2, have Q ↔ f symmetricIf char F = 2, f does not determine Q, and f is alternating

10.1 Case: Odd characteristic

work with f symmetric, bilinear, non-degenerate

Lemma 10.1Two equivalence classes of such form:

Either ∃ orthonormal basis for V

Or ∃ orthonormal basis withf(vi, vi) = 1 ∀i < df(vd, vd) = α some non-square α

ProofFind v1 s.t. f(v1, v1) = 0. “Normalize” to either 1 or α (α fixed non-square). Now continue in v⊥1

If f(v1, v1) = f(v2, v2) = α, can replace inside ⟨v1, v2⟩Choose λ2 + µ2 s.t. not a square in F ; normalize to α−1

⇒ λv1 + µv2, µv1 − λv2 are orthonormal

d = 2m+ 1 odd: Get the same group both forms GO2m+1(q) d = 2m even: The groups different.More useful distinction:

(1) maximal totally isotropic space has dimension (called Witt index) m, Q+2m

(2) Witt index m− 1, Q−2m

43

WriteGOϵ

2m(q) = g ∈ GL2m(q)|Qϵ2m(vg) = Q(v) ∀v (ϵ = ±)

Lemma 10.2If d = 2, there may or may not exists isotropic vectors: according to type of Q and q ≡ ±1 mod 4

Proofq ≡ 1 mod 4: If f(u, u) = 1 = f(v, v) an u⊥v, let i =

√−1 ∈ F

f(u+ iv, u+ iv) = 0 ⇒ u+ iv isotropicIf f(u, v) = 1, f(v, v) = α, with u⊥v⇒ f(u+ λv) = 1 + λ2α⇒ f(u+ λv) = 0 with λ ∈ F

q ≡ 3 mod 4: Other way round

If d > 2, isotropic vector exist, so the forms are:

(1) Q+2m ↔ f+ : e1, f1, . . . , em, fm

(2) Q−2m ↔ f− : e1, f2, . . . , em−1, fm−1, d, d

′ with ⟨d, d′⟩ = O−2 , d, d

′⊥ei, fj ∀i, j, f(d, d) = 1, (infact have f(d, d′) = 1, see later)

Related groups:

GO+2 (q) = D2(q−1) , GO−

2 (q) = D2(q+1)

d = 2:GO+

2 (q) = D2(q−1):e1, f1, cyclic group order q − 1: e1 7→ λe1, f1 7→ λ−1f1t inverting: e1 ↔ f1No more elements: ⟨e1⟩, ⟨f1⟩ are the only isotropic points

GO−2 (q) = D2(q+1):

Let V = Fq2 - dim 2 over Fq

Q(v) = N(v) = vq+1 This is a quadratic form, no isotropic vectorsCyclic group order q + 1: N(λ) = 1 ⇒ v 7→ λv is an isometryinverted by t : v ↔ vq

No others- e.g. the stabiliser of 1 conssits of i and v ↔ vq (check)

Lemma 10.3

|GO2m−1(q)| = 2qm2(q2m − 1)(q2m−2 − 1) · · · (q2 − 1)

|GOϵ2m(q) = 2qm(m−1)(q2 − 1)(q4 − 1) · · · (q2m−2 − 1)(qm − ϵ1)

ProofBy induction, going up in steps of 2.Let zm = #(non-zero) isotropic vectors in dimension 2m+ 1 or 2m

Claim: zm = q2m − 1 for Q2m+1 and zm = (qm − ϵ)(qm−1 + ϵ) for Qϵ2m

Proof of Claim:Correct for dimension 1, 2 (both ϵ)

Let dimV = n+ 2, V = U ⊕W (i.e. U⊥W ) with U dimension 2 with some type, W dimension nof same type.

44

Isotropic v ∈ V is u+ w - both norm 0 but not both zero vector, OR, norms are λ(= 0),−λ⇒ zm+1 = (2q − 1)(1 + zm) + (q − 1)(qn − 1 − zm) − 1 = qzm + (q − 1)(qn + 1) - now obtainedformula in each case

Having chosen e1, need f1 with e1, f1 hyperbolic:

Claim: This can be done in qn−1 ways (hence formulae follow)

Proof of Claim:Number of isotropic vectors in e⊥1 :e⊥1 /⟨e1⟩ is a space of same type - so zm−1 isotropic vectors ⟨e1⟩+ v⇒ qzm−1 + (q − 1) isotropic vectors in e⊥1⇒ 1

q−1(zm − qzm−1 − q + 1) choices of f1

10.2 Even characteristic

Work with quadratic form Q over F = Fq, char 2Let f be the associated bilinear alternating form - f(v, v) = 0

rad f = w|f(v, w) = 0 ∀vradQ = w ∈ rad(f)|Q(w) = 0

radQ ≤ rad f ≤ V (subspaces), codimension of rad f over radQ is 0 or 1:Q|rad f : rad f → F semilinear Q(v + w) = Q(v) +Q(w), Q(λv) = λ2Q(v)Norm of v=Q(v), v is isotropic if Q(v) = 0Q non-singular: radQ = 0Q non-degenerate: rad f = 0

Let Q non-singular ⇒ rad f has dimension ≤ 1Look at dimV = 2m + 1 odd ⇒ rad f dim1 because V/ rad(V ) has dimension 2m even, with anon-singular alternating form

GO2m+1(q) ≤ Sp2m(q)

Note: in fact, |PΩ2m+1(q)| = |PSp2m(q)| for q odd, NOT isomorphic if 2m > 4

Now look at dimV = 2m evenNow handle things as before, choose a symplectic type basis as much as poss with Q(v) = 0Note: Q(e) = 0, f(e, f) = 1 ⇒ can adjust f to have Q(f) = 0 (as Q(f + λe) = Q(f) + λ, so replacef by f +Q(f)e)

If dimV > 2, ∃ isotropic vectors: there are vectors u⊥vif Q(v) = 0, then Q(v + λu) = Q(v) + λ2Q(u)⇒ Take λ2 = Q(v)/Q(u), then v + λu isotropic

So consider dimV = 2: v, w basisLet Q(v) = 1 = f(v, w), Q(w + λv) = Q(w) + λ2 + λ⇒ at most 2 forms, and in fact, exactly two:Q+

2 : e1, f1 hyperbolic pair (isotropic vectors exists)Q−

2 : v, w s.t. Q(v) = 1, f(v, w) = 1, Q(w) = µ ∈ F s.t. X2 +X + µ irreducible (no isotropic vectors)

Now get:Q+

2m : e1, f1, . . . , em, fm -isotropic, etc.Q−

2m : e1, f1, . . . , em−1, fm−1, v, w where v, w are as in Q−2

45

GOϵ2m(q)

|SO2m| = 12 |GO2m| for q odd

SO = GO for q evenPSOϵ

2m NOT simple, has a subgroup index PΩϵ2m(q) which is simple if dim > 4 (hard to prove the

existence of this, see Wilson)

Some isomorphisms:

(1) PΩ3(q)∼=L2(q), q odd, W = V2, V = S2W (see Wilson page 96, and example sheet)

(2) PΩ4(q)∼=L2(q)× L2(q)

(3) PΩ−4 (q)

∼=L2(q2)

(4) L4(q)∼=PΩ+6 (q) on

∧2 V4

GOϵ2m ≤ Sp2m(q)

q = 2 ⇒ two 2-transitive actions

Sp2m(q)

ooooooooooo

OOOOOOOOOOO

GO+2m

NNNNNNNNNNNGO−

2m

ppppppppppp

Sp2(m−1)(q)× 2

46

Index

acting, 16almost simple group, 3Automorphism group, Aut(G), 2

blocks, 17Borel subgroup, 35Burnside’s paqb Theorem, 15Burnside’s Basis Theorem, 10Burnside’s Theorem, 20

characteristic subgroup, 3characteristically simple group, 3commutator subgroup, 3Correspondence Theorem, 2

degree, 18subdegree, 22

diagonal action, 28diagonal subgroup, 30diagonal type, 30distance, 22distance transitive, 22

elementary abelian p-group, 3

factor group, 1factorisable, 14factorization of group, 13faithful action, 16Fano plane, 37Fitting subgroup, F (G), 8fix(Gα), 18Frattini argument, 9Frattini subgroup, Φ(G), 9Frobenius groups, 20

G-congruencetrivial, 17

G-congruence, 17G-homomorphism, 16G-isomorphism, 16Gp, 9GΩ, 16

Hall π-subgroup, 12Hall p-complement subgroup, 12homomorphism

of groups, 1hyperbolic pair, 39

Inner automorphism, Inn(G), 2Isomorphism Theorem, 1isotropic, 45

Jordan set, 25primitive, 25

Jordan-Holder Theorem, 4

k-homogeneous, 31k-transitive, 18kernel of action G(Ω), 16

Levi subgroup, 35of Sp2m(q), 41

line, 20

minimal generating set, 10multiplicity-free, 23

nilpotent group, 7nilpotent radical of G, F (G), 8non-generator, 9normal closure, 15normal subgroup, 1

Op(G), 8orbit αG, 16orbital, 22

diagonal, 22paired Γ∗, 22

Orthogonal group GOd(q)ϵ, 44

Outer automorphism group, Out(G), 3

p-local subgroup, 36P.Hall Theorem, 15parabolic, 35perfect, 3permutation actions, 16permutation representation, 16π′-part of n, 12π-group, 12π-part of n, 12primitive, 17projective general linear group PGLd(q), 18PSUd(q), 43

quotient group, 1

rankof groups, 21

regular group, 11ρ(α), 17

Second Isomorphism Theorem, 2semi-regular, 21semidirect product, 19series, 4

47

central, 7chief, 5composition, 4derived, 11lower central, 8normal, 4proper, 4upper central, 7

simple group, 1simply primitive, 21socle, 28soluble group, 11soluble radical, 11split torus, 35SUd(q), 43Sylp(G), 9Sylow basis, 14

Third Isomorphism Theorem, 2transitive on Ω, 16transvection, 34

symplectic, 39

unipotent radical, 41Unitary group GUd(q), 42unitriangular matrix, 35

W (k, l) = Sk wrSl, 29Witt index, 43wreath product, 25wreath product C wr∆D, 28

48