Topics in Groups Theory Prof. Saxl Typeset by Aaron Chan([email protected]) Last update: March 10, 2010

Books: Alperin and Bell (Springer), John S. Rose (Dover), M. Suzuki Robert A. Wilson, The Finite Simple Groups, Springer 2009 This 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 proof Distinct 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 = g1 θ · g2 θ It is an isomorphism!of groups if also bijective ker(θ) = {g ∈ G|gθ = eH } θ is injective ⇔ ker θ = {eG } ker θ P G K ≤ G is a normal subgroup of G, write K P G, if Kg = gK ∀g ∈ G G is simple if there are no normal subgroups other than {1} and G, equivalently, G simple ⇔ any non-trivial homom from G is injective K P G, can form G/K, a quotient group (or factor group) on the set (G : K), by Kg1 Kg2 = Kg1 g2 |G/K| = |G|/|K| Theorem 1.2 (Isomorphism Theorem) If θ : G → H homomorphism, then ker θ P G, Im θ = {gθ|g ∈ G} ≤ H, and there is an isomorphism G/ ker θ ∼ = Im θ Sketch proof Writing K = ker θ, check K P G, 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 G Note: If K P G, then π : G  G := G/K g 7→ g := Kg i.e. quotients of G are homomorphic images of G θ : GH

HH HH H π HHH #

w

w

w

/ Im θ w;

θ

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

H/K = H

H

{1G/K } Moreover, H/K P G/K ⇔ H P G and if so Proof If H P G, define a homomorphism

G/K H/K

G

K

∼ = G/H

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 P G, H ≤ G. Then HK ≤ G (HK := {hk|h ∈ H, k ∈ K}) H ∩ K P H and H/H ∩ K ∼ = HK/K Proof The 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 g x for the image of g ∈ G under x ∈ Aut(G) Inner automorphisms (conjugation autos) are, for g ∈ G, θg : x 7→ g −1 xg for x ∈ G Then θg is an auto of G and the map

θ : G → Aut(G) is a homomorphism. g 7→ θg 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)) ∼ D8 , Aut(Q8 ) ∼ Exercise: Aut(D8 ) = = S4 , Aut(A5 ) ∼ = S5 d d Exercise: G = p -elementary abelian of order p (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) P Aut(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 > 6 Aut(A6 ) = Σ6 , group of order 2 · 6! Important definition: G∗ is almost simple if there is a simple group G with G ∼ = Inn(G) P G∗ ≤ Aut(G) ∗ ∗ (G is the unique minimal normal subgroup of G , and G /G ≤ Out(G)) Example: H P K P G ; H P G (Exercise: give an example) Definition K char G (K is characteristic subgroup of G) if K α (:= 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 P G, since Inn(G) ≤ Aut(G)

Exercise: (1) H char K P G ⇒ H P G (2) H char K char G ⇒ H char G Example: Z(G) char G, G′ char G G′ = ⟨[x, y]|x, y ∈ G⟩ – commutator subgroup Exercise: K P G, G/K abelian ⇔ G′ ≤ K G is perfect if G = G′

3

2

Series, Jordan-H¨ older Theorem

G is finite (or, if infinite, need some DCC) Definition A series is a chain of subgroups Gi : 1 = G0 P G1 P · · · P Gr = G A series is normal if also Gi P G ∀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 is one that cannot be properly refined) Theorem 2.1 Any finite group has a composition series Proof If G simple, stop. Otherwise, let K be a maximal normal subgroup of G; then G/K simple and |K| < |G| so contained in K Remark. This is constructive; in fact 1 P H P G is a series for any H P G E.g.: Z has no (finite) composition series Theorem 2.2 (Jordan-H¨ older) Let G be finite. Then any two composition series have the same quotients, counted with multiplicity 1 = G0 P G1 P · · · and 1 = H0 P H1 P · · ·

P Ga = G P Hb = G

two composition series, then a = b and the collectors {Gi /Gi−1 } and {Hi /Hi−1 } are the same up to isomorphism Example: S4

{{ {{ { { {{

A4

C6 C

C3 B

BB BB BB B

p V4 NNNN NNN ppp p p p 2 2 MMM 2 MMM qqq q q q

{1}

{1}

Proof Put H = Hb−1 , K = Ga−1 Case: H = K Since |H| < |G|, induction applies: We have two composition series of H, so they are “isomorphic” Case: H ̸= K

4

CC CC CC C

|| || | | ||

C2

Note that HK P G

⇒

G = HK by maximality G = HK VVV hh VVVV hVhhhh H VVVVV hhh K VV hhhhh H ∩K

Hb−2

Ga−2 Lc−2

H1 UUUU UUUU U

{1}

ii G1 iiii i i i i

Find a composition series for H ∩ K Then 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 so are 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 @A P G s.t. Gi P A P Gi+1 . The chief factors are characteristically simple C2

C3

C22

E.g. S4 − A4 − V4 − 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 2.3 Let N be minimal normal subgroup of G. Then N is direct product of isomorphic simple groups, all conjugate in G Proof Let K be a minimal normal subgroup of N . If K = N , stop If K  N , then K R G ⇒ ∃g1 ∈ G with K g1 ̸= K ⇒ K ∩ K g1 = {1} Let K2 := KK g1 (which is P N ) Since K ∩ K g1 P N ⇒ K2 ∼ = K × K g1 If K2 = N , stop Otherwise, K2  N ⇒ K2 R G ⇒ ∃h ∈ G with K2h ̸= K2 i.e. ∃k ∈ K s.t. (kk g1 )h ∈ / K2 g1 )h = k h · k g1 h (Aim: ∃g ∈ G s.t. K ∩ K g2 = {1}) Notice (kk 2 2 { either k h ∈ / KK1 ⇒ K h  K2 g2 := h ⇒ h or k g1 ∈ / KK1 ⇒ K g1 h  K2 g2 := g1 h 5

⇒

K3 := K2 K g2 P N , and K2 ∩ K g2 = 1, so K3 ∼ = K × K2 .

Continue this and we get N ∼ = K × K g1 × K g2 × · · · × K gk . Finally, K is simple, since, if X P K, then X P N (as N ∼ = K × K g2 × · · · × K gk ), so X = K by minimality of K as normal in N Remark. (1) The proof applies to characteristically simple groups N i.e. Characteristically simple group N are direct products of isomorphic simple groups In particular, all chief factors of a finite group are also direct products of isomorphic simple groups. (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 then N = Vk (p), a vector space over Fp of dimension k. There are normal subgroup

pk −1 pk −1 pk −p p−1 + p2 −1 · p2 −p +· · ·

(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) Definition X group, G is an X-group if ϕ : X → Aut(G) a homomorphism ϕ Define operation of X on G by g x := g (x ) H ≤ G is an X-subgroup of G, write H ≤X G, if H is X-invariant (Note: H ≤X G ⇒ H is an X-group) If H P G, H ≤X G ⇒ G/H is an X-group via (Hg)x := Hg x If G1 , G2 are X-groups (X fixed) a homomorphism α : G1 → G2 is an X-homomorphism if (g x )α = (g α )x for x ∈ X, g ∈ G1 An X-group is X-simple if it has no non-trivial normal X-invariant subgroups If G is an X-group, an X-composition series is a series 1 = G0 P G1 P · · · P Ga = G with each Gi ≤X Gi+1 , Gi P Gi+1 , which cannot be properly refined. E.g. G X G 1 G Inn(G) G Aut(G)

X-subgroup subgroup normal subgroup

X-series series normal series

X-comp. series comp. series chief series

Another example: G-modules G any group, ρ : G → GL(V ) a finite dimensional representation over some field. Set X = Im ρ, then V is an X-group In this setup, one proves existence of X-composition series, X-isomorphism theorems and X-JordanH¨older theorem 6

3

(Finite) Nilpotent Groups

Definition The normal series 1 = G0 P G1 P · · · P Ga = 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) P G with Zi+1 (G)/Zi (G) = Z(G/Zi (G)) so, G nilpotent (of class a) ⇔ G/Z(G) nilpotent of class a − 1 G nilpotent of class 1 ⇔ G abelian Theorem 3.1 Finite 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)

1

Z3 (G) _ _ _ _ G

Z2 (G)

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 clear If e > 2, Z(D2e ) = 2 and D2e /Z(D2e ) ∼ = D2e−1 Theorem 3.2 For 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) ≤ H May 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 P NG (P ) ⇒ P char NG (P ) This is because P is the only subgroup of NG (P ) with order |P |, this implies all automorphism of NG (P ) sends P to itself Also, since NG (P ) P NG (NG (P )) ⇒ P P NG (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 P G by hypothesis of (3). ⇒ P1 P2 · · · Pk P G s.t. |P1 P2 ·∏ · · Pk | = |G| ⇒ G = P1 P2 · · · Pk , and Pi ∩ j̸=i Pj = {1} by considering their orders ⇒ G∼ = P1 × P2 × · · · × Pk (4) ⇒(1)

Follows from Exercise above and Theorem 3.1

Corollary 3.3 G nilpotent, H < G maximal ⇒ H P G index p, some prime p Proof NG (H) H ⇒ H P G G/H has no proper subgroups ⇒ cyclic of prime order for some prime p Note If K P G and K ≤ H ≤ G, then

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

Definition Lower 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 · · · P Ga = 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 Definition The Fitting subgroup (nilpotent radical of G), F (G), is the maximal subgroup amongst all normal nilpotent subgroups of G Proposition 3.4 In 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) Proof The 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 normal If K P G, nilpotent, and P is a Sylow p-subgroup of K, then P char K P G ⇒ P P G ⇒ P ≤ Op (G) ⇒ K ≤ RHS

Definition G any finite group, the Frattini subgroup of G, Φ(G), is the intersection of all maximal subgroups of G Note: Φ char G Definition g ∈ G is a non-generator of G if whenever G = ⟨X, g⟩ we have G = ⟨X⟩ Lemma 3.5 Φ(G) = {g ∈ G|g non-generator } Proof If 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⟩ ≤ M Then g ∈ /M ⇒ g∈ / Φ(G) Denote Sylp (G) as the set of Sylow p-subgroup of G Proposition 3.6 For any G finite, Φ(G) is nilpotent (⇒ Φ(G) ≤ F (G)) Proof Uses 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 P Φ(G). Thus Φ(G) nilpotent. Lemma 3.7 (Frattini argument) G finite group, K P G, P ∈ Sylp (K) Then G = NG (P )K. Hence G/K ∼ = NG (P )/NK (P ) Proof ′ Let 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 3.8 If G is a p-group, then G/Φ(G) is an elementary abelian group, and hence a vector space over Fp . In fact, Φ(G) = G′ Gp Proof If M is maximal subgroup of G, then M P G of index p so G′ ≤ Φ(G) (as G/M abelian), and Gp = ⟨g p |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⟩) Definition Minimal 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). Proof exercise: 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 size Question What is the maximal size of a minimal generating set? What do minimal generating sets of maximal size look like?

10

4

Soluble (Solvable) groups

Definition A 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 4.1 TFAE: (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 1 P · · · P H1 P H0 = G with abelian factors, then G(i) ≤ Hi , by indicator: G(0) = H (0) , and G(i+1) = (G(i) )′ ≤ Hi′ ≤ Hi+1 Exercise: Subgroups, quotients, direct products of soluble groups are soluble Exercise: H, K P G, both H, K soluble ⇒ HK is a soluble normal subgroup of G. Hence, using the fact that if N , G/N soluble then G soluble, we can define the soluble radical of G, i.e. the maximal normal 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 Proof Let K P G minimal normal. Then K is elementary abelian p-group, for some prime p. (The proof here is much easier than before: K is soluble, and K ′ char K so K ′ = 1, so K is abelian. Op (K) char K and K p 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 P M , K ∩ M P K (as K is abelian) so K ∩ M P G, so K ∩ M = 1 by minimality of K P G 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 identity fixes all point)

11

4.1

Hall’s Theorem on Finite Soluble Groups

π = Set of primes n ∈ N, write n = nπ nπ′ where (pl11 · · · plkk )π = (pl11 · · · plkk )π′

∏

plii

(π-part)

pi ∈π

∈ / π

(π ′ -part)

A π-group H is a finite group with |H| = |H|π A subgroup H of G is a Hall π-subgroup of G if |H| = |G|π , so subgroup of G of maximal possible π-order Remark. π = {p} Hall π-subgroup = Sylow p-subgroup Hall π ′ subgroup = Hall p-complement subgroup Theorem 4.3 If 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-complement       1 ∗ ∗   1 0 0  0  , ∗      0 ∗ Proof Let |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 P G with |K| = m1 n1 with n1 < n (1) Now |G/K| = mm1 nn1 , so by induction, G/K has a subgroup S/K of order mm1 with 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 πsubgroups. By induction, ∃x = Kx ∈ G with x−1 H2 x = H1 . ⇒ x−1 H2 Kx = H1 K ⇒ x−1 H2 x, H1 are Hall π-subgroup in H1 K, so conjugate. Apply induction. (3) Let P be any π-subgroup of G. Then P = P K/K is a π-subgroup in G ⇒ P ⊆ some Hall π-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 P G, 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 L P G containing K s.t. L = L/K minimal normal in G = G/K. Then |L| = q b for some prime power q b with p ̸= q 12

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

NG (Q)

ll G lll lll

11 11 11 11 1

L = KQ

xx xx Q5 55 55 55 K 55

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 (since K abelian) • K ∩ NG (Q) P K • K ∩ NG (Q) P NG (Q) (since K P G) • ⇒ K ∩ NG (Q) P G, 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 | = q b (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 ) = H x

⇒

H2 = H x (since they are of same order)

(3) Let P be a π-subgroup of G, with |P | = m′ < m Now |G : H| = n which is π ′ -number |G : P K| = m/m′ which is π-number ⇒ P, H ∩ P K have the same order: |G : H ∩ P K| = |G : H||G : P K| (by Lemma 4.4, to be shown) = nm m′ t G KK t n ttt t t t tt H JJ JJ JJ J m/m′ JJ

KKKm/m′ KKK KK

ss sss s s n ss

PK

H ∩ PK

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

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

13

Proof |G : A1 ∩ A2 | = |G : A1 | |A1 : A1 ∩ A2 | = |G : A2 | |A2 : A1 ∩ A2 | | {z } | {z } n1 n2 ⇒ n1 n2 |G : A1 ∩ A2( | ) |A1 ||A2 | (|G|/|G : A1 |)(|G|/|G : A2 |) |G||G : A1 ∩ A2 | |A1 A2 | = = = |A1 ∩ A2 | |G|/|G : A1 ∩ A2 | n1 n2 ⇒ G = A1 A2 A1 , A2 ≤ G. G is factorisable as G = A1 A2 if every g ∈ G is g = a1 a2 with ai ∈ Ai Note: G = A1 A2 ⇔ G = A2 A1 ⇔ G = Ag1 A2 ; ∀g ∈ G A1 , A2 conjugate ⇒ G ̸= A1 A2 Exercise: G = A1 A2 ⇔ A2 is transitive on (G : A1 ) ⇔ A1 is transitive on (G : A2 ) (this result immediately implies:) G = A1 A2 ⇔ |G : A1 | = |A2 : A1 ∩ A2 | (by using |A1 A2 | =

|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 = A1 A2 Proof as Exercise (hard) Factorisation in almost simple groups are interesting, important and rare. See later. Definition If |G| = pe11 · · · pekk with the pi distinct primes, a Sylow basis of G is {P1 , . . . , Pk } with |Pi | = pei i s.t. Pi Pj = Pj Pi ∀i, j (so that Pi Pj ≤ G ∀i, j) Note: Pi Pj = Pj Pi ⇒ Pi Pj ≤ G ∀i, j and in fact Pi1 Pi2 · · · Pil ≤ G ∀{i1 , . . . , il } ⊆ {1, . . . , k} − we get Hall subgroups this way Theorem 4.6 If G is a finite soluble group, G has Sylow basis, and any two are conjugate Proof Let |G| = pe11 · · · pekk , with the pi distinct primes. Let Hi be ∩ a (Hall) pi -complement. (i.e. has order |G|/pei i ) Put Pj = i̸=j Hi ej ⇒ |Pj | = p∩ j , using Lemma 4.4 And Pi Pj = i ̸= l ̸= jHl = Pj Pi Conjugacy: Let {P1 , . . . , Pk } and {P1∗ , . . . , Pk∗ } be Sylow bases with |Pi | = |Pi∗ |, so Pi , Pi∗ ∈ Sylpi (G) Put Hi = P1 · · · Pi−1 Pi+1 · · · Pk ∗ P∗ ···P and Hi∗ = P1∗ · · · Pi−1 k i+1 g Claim: ∃g ∈ G with Hi = Hi∗ ∀i Proof of Claim: First, ∃g1 with H1g1 = H1∗ (Hall) g Now assume we found gi−1 with Hj i−1 = Hj∗ ∀j ≤ i − 1 Change notation if necessary so that Hj = Hj∗ ∀ ≤ i − 1 Now G = Hi Pi , and let x ∈ G with Hix = Hi∗ (Hall) ⇒ x = hz with h ∈ Hi , z ∈ Pi ⇒ Hiz = Hi∗ and z ∈ Pi ≤ Hi ∀j < i ⇒ Hjz = Hj∗ for j ≤ i After k steps, finished.  Given claim, we have Pig = Pi∗ ∀i 14

Theorem 4.7 (Wielandt) G finite, H1 , H2 , H3 soluble subgroups of G with |G : Hi | = ni pairwise coprime. Then G soluble Proof Looking for N P G soluble with G/N soluble by induction (then G soluble) Note that G = H1 H2 = H1 H3 = H2 H3 , by Lemma 4.4 May assume Hi ̸= 1 ∀i. Let K minimal normal in 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 = K : K ∩ H1 ∩ H2 {z } | divides n2

so K ≤ H1 ∩ H2 Now let N = ⟨K g |g ∈ G⟩ - the G normal closure of K ⇒ N = ⟨K h1 h2 |hi ∈ Hi ⟩ = ⟨K h2 |h2 ∈ H2 ⟩ ≤ H2 ⇒ N P G, 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 · · · pekk If k = 1, G is nilpotent. Done If k = 2, this is Burnside’s pa q b 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 4.9 G finite soluble group ⇒ CG (F (G)) ≤ F (G) (So G/Z(F (G)) ≤ Aut(F (G))) Proof Put F = F (G), Z = Z(F (G)), C = CG (F (G)). So Z = C ∩ F Assume C F . Let H P G with Z < H G

CF C

CC CC CC C

{ {{ {{ { {{ CC CC HCC C

Z

{{ {{ { { {{

F

H minimal normal in C = C/Z: Since H is elementar yabelian, have H ′ ≤ Z Then 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 prime An , n≥5 Ld (q), d ≥ 2, if d = 2 then q ≥ 4 Ud (q), d ≥ 3, if d = 3 then q ≥ 3 Sp2m (q), m ≥ 2, if m = 2 then q ≥ 3 P SLϵd (q), d ≥ 7 if 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), E6+ (q), E6− (q), E7 (q), E8 (q) 3 D (q) 4 2 B (q) with q = 22a+1 2 2 G (q) with q = 32a+1 2 2 F (q) with q = 22a+1 4 26 sporadic simple groups M11 , M12 , M22 , M23 , M24 , · · · , M In applications, have usually a finite groups acting in some way, .e.g as a group of permutations, or a group of matrices (in this case, this is representation theory, mainly used to study the first 5 families mentioned) 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 roots of polynomial Recall about permutation actions: G finite group, Ω a finite set. G acts on if there is a map

Ω×G → Ω (ω, g) 7→ ωg

s.t. ω1 = ω, ωgh = (ωg)h ∀ω ∈ Ω, g, h ∈ G ϕ :Ω → Ω If so, for g ∈ G, the map g is a permutation on Ω ω 7→ ωg ϕ : G → Sym(Ω) and the map is a homomoprhism, called a permutation representation of G g 7→ ϕg The kernel of the action, G(Ω) := ker ϕ. 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} G is transitive on Ω if αG = Ω ∀α ∈ Ω

⇒

Ω splits into G-orbits

Problems about actions are usually easily reduced to problems about transitive actions (consider the actions 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 ∈ G θ 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 obtained by (Gα x)g = Gα (xg) 16

Lemma 5.1 If 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→ Gα x is bijective, commutes with the actions of G Example: (G : H), (G : K) are G-isomorphic ⇔ H, K are G-conjugate Definition A equivalence relation ρ is a G-congruence on Ω if α ρ β ⇒ αg ρ βg ∀g ∈ G Trivial 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 5.2 G transitive on Ω, ρ a non-trivial G-congruence on Ω, α ∈ Ω, ρ(α) the (equiv.) class of α: ρ(α) := {β ∈ Ω|α ρ β} (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)  G Conversely, 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 = αh2 g some h1 , h2 ∈ H ⇒ h2 gh−1 ⇒ g ∈ H ⇒ Γg = Γ 1 ∈ Gα  H

17



Corollary 5.3 Set up as Lemma 5.2, by (4), G primitive on Ω

⇔

Gα maximal subgroup of G

Corollary 5.4 Set 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 5.5 G primitive on Ω, α ̸= β ⇒ G = ⟨Gα , Gβ ⟩, unless G is Cp of degree p In fact, if G transitive, α ∈ Ω, then fix(Gα )={β|βh = β ∀h ∈ Gα } is a block of a congruence. Proof Let G be transitive. Define α ρ β if Gα = Gβ (note: β ∈ fix(Gα ) ⇒ Gα = Gβ ) Claim:

This is a G-congruence

Proof of Claim: g ∈ G. α ρ β ⇒ 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 = 1 and as Gα maximal, then G is Cp for some prime p Lemma 5.6 Let G transitive on Ω, let 1 ̸= N P G. The N -orbits form a G-invariant partition of Ω Proof Exercise Corollary 5.7 If G is primitive, 1 ̸= N P G ⇒ 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 order dividing 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 Definition G 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 αi g = βi k-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-transitive An is (n − 2)-transitive but not (n − 1)-transitive Example: G = P GLd (q) - projective general linear group = GLd (q)/{scalars} acts on Ω = {1-dim. subspaces of 18

Vα (q)} =: Pd−1 (q) d −1 |Ω| = qq−1 G is 2-transitive on Ω G is 3-transitive ⇔ d = 2 G is 4-transitive ⇔ d = 2, q = 3, G = P GL2 (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 translation G0 = 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 P AGL, since ∀h ∈ G0 , h−1 tv h = tvh ∀x ∈ V (2) K is a regular normal elementary abelian subgroup (3) K ∩ G0 = 1, AGL = G0 K Any g ∈ G is g = hk for some unique h ∈ H, k ∈ K (h1 k1 )(h2 k2 ) = (h1 h2 )(k1h2 k2 ) Definition Semidirect product H n K: H, K groups, θ : H → Aut(K) Elements of H ⊗ K are of form (h, k) s.t. (h1 , k1 )(h2 , k2 ) = (h1 h2 , k1h2 θ 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 5.8 Let G be primitive on Ω, let K P G 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 is isomorphic (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 Ω is isomorphic to a subgroup of AGLd (p) acting as above In 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 ̸= L P G ⇒ L is transitive ⇒ |K| = |Ω| |L| ⇒ if L ≤ K, then L = K

19

(2) α ∈ Ω, any β is αk for unique k αk ↔ k Ω↔K If g ∈ Gα , αkg = αg −1 kg = αk g ↔ k g (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 = Gα K. 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−1 kh This is a linear transformation: h−1 (k1 k2 )h = h−1 k1 h · h−1 k2 h (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 Proof If 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 = Γg Γg 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

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

# 20



(5.1)

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 5.10 this orbits of Nα on Ω \ {α} have all the same size dividing n − 1 (by Corollary 5.4) Nα P Gα , 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 lemma Claim: G is 2-transitive, N P G and N Frobenius ⇒ ∃K ≤ N characteristic regular subgroup of N Proof Let K = {1} ∪ {x ∈ N |x have no fixed points on Ω} Let n = Ω |K| = n, |N | = nc where c = |Nα | ∪ N =K ∪( Nα \ {1}) α∈Ω

|

{z

}

n(c−1)

K is a transitive set: If k ∈ K \ {1} takes α to β, let g ∈ Gα take β to γ then k g ∈ K \ {1} with k g : α 7→ γ K≤N If not, let k1 , k2 ∈ K with k1 k2−1 ∈ / K ⇒ k1 ̸= k2 and k1 k2−1 fixes some α - then αk1 = αk2 ⇒ K cannot be transitive ⇒ K P G (as k g ∈ K ∀k ∈ K, g ∈ G) Definition Action of G on Ω is semi-regular if g ∈ G fixes any point of Ω ⇒ g = 1 Lemma 5.10 If M transitive subgroup of Sym(Ω), then CSym(Ω) (M ) is a semi-regular Proof Exercise

Simply Primitive Groups Definition G 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 Ω Note r = 2 ⇔ G is 2-transitive

21

For r > 2 the subdegree ni = |Γi (α)|

1 = n0 ≤ n1 ≤ · · · ≤ nr−1

∑r

0 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) ∈ Γi } Γ0 = {(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 orbital Γ and Γ∗ are paired We get complete G-invariant r-colouring of the complete graph on the vertices in Ω G1 = Peterson graph ( ) 5 |Ω| = = 10 G =Aut(Peterson)=S5 2 (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 K1 0, i.e. edges with vertex {i, j} and {i, k}, j ̸= k) ( ) n Exercise: Sn acts on (stablizer Sn−1 × S2 ) as a primitive rank 3 group, subdegrees 1 (↔ 12), 2 ( ) n−2 2n − 2, 2 ( ) n n Sn acts on , 2 > l ≤ 1, rank is l + 1 k Remark. If G is an undirected graph, we have the notion of distance G ≤ Aut(G), G preserves distance Say G is distance transitive on G if given α1 , α2 has distance d and β1 , β2 has distance d, then ∃g ∈ G s.t. αi g = βi What are these on regular symmetric graphs? ( ) n Exercise: Sn is distance transitive on the graph Ω = = l-subsets of [1, n], edges A − B if l |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 points Sim’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 Proof Exercise (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 representation. Lemma 5.12 If G ≤ Sym(Ω), permutation character π 1 ∑ Then ⟨π, 1⟩G = |G| π(g) = #(orb(G, Ω)) g So G is transitive Ω ⇔ ⟨π, 1⟩G = 1 Proof #(orb(G, Ω)) =

∑

|Gα | = #{(α, g) ∈ Ω × G|αg = α} =

∑

π(g)

g∈G

α∈Ω

Lemma 5.13 G acts on Ω1 , Ω2 , with characters π1 , π2 Then ⟨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) Proof First part: ⟨π1 , π2 ⟩G = ⟨π1 π2 , 1⟩G Note π1 π2 is the permutation character of G on Ω1 × Ω2 G is 2-transitive on Ω: π = 1 + χ, χ irreducible G is rank 3: π = 1 + χ1 + χ2 , χi distinct irreducible Definition G 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. If G is distance-transitive on an undirected graph than its permutation character is multiplicityfree Exercise: Show that if G has permutation rank ≤ 5 then G is multiplicity-free Exercise: 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-transitive Also 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 act first) Theorem 6.1 An is (non-abelian) simple for n ≥ 5 (Note for n = 3, A3 is abelian simple) Proof I Induction on n: A5 is simple (prove this). Let n > 5, assume true for n − 1 Let G = An , let 1 ̸= K P G ⇒ Kα P Gα (∼ = 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 above it 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 II Start 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 P G 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] ̸= 1 But fix([g, x]) = fix(g) ∪ {2} (check)



Claim ⇒ K contains all 3-cycles (all conjugate) ⇒ K = An Theorem 6.2 Let G ≤ An , G primitive, G containing a 3-cycle. Then G = An Proof Define G-congruence ρ on [1, n]: α ρ β if α = β on 3-cycle (αβγ) ∈ G for some γ Then ρ is reflexive, symmetric, G-invariant And, ρ is transitive: since α ρ β ρ γ ⇒ ∃(αβδ) ∈ G if δ = γ if δ ̸= γ ∃(βγϵ) so ⟨(αβδ), (βγϵ)⟩ ≤ G is A4 or A5 ⇒ (by α ρ β) (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 α ρ β, β ρ γ) ⇒ (αβγ) ∈ G by 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 Γ Γ is a primitive Jordan set if G is primitive on Γ e.g. if G contains a p-cycle (1 · · · p) then Γ = {1, . . . , p} is a primitive Jordan set Note: 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 6.3 G primitive on Γ, |Ω| = n, if Proof Exercise

n 2

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

Exercise*: If G is primitive on Sn , and p is a prime with G ≥ An Hence: if n2 < p < n − 2 a prime, then p - |G|

n 2

< p < n − 2. If G contains a p-cycle then

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: na11 na22 · · · nakk means ai of ni -cycles, n1 > n2 > · · · > nk ≥ 1, n = ai ni Lemma 6.4 The size of Sn -conjugacy class of elements of type na11 · · · nakk is a1 !(n1

)a1

n! · · · ak !(nk )ak

Centralizer of one such element is a direct product of k wreath products: (C a1 o Sa ) ×(Cna22 o Sa2 ) × · · · × (Cnakk o Sak ) | n1 {z 1 } wreath product Proof n! 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 denominator Also there are ai ! ways to permute these ai lot of ni -cycle. Lemma 6.5 ∑ Conjugacy classes of elements in An : elements in An have type na11 · · · nakk with ( ni ai ) 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 -ccls of equal length Proof Exercise

25

6.1

Structure of Aut(An )

Note, Am ,→ Ak ⇒ k ≥ m m ≥ 5 ⇒ any action of Am is on ≥ m points Lemma 6.6 If n > 6, then ∀ H ≤ An s.t. H ∼ = An−1 ⇒ H = (An )α of some α ∈ [1, n] Proof n = 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 − 1 Assume 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 (13 31 ) = CA6 (32 ) | cclA6 (123)| = | cclA6 (123)(456)| | cclS6 (123)(45)| = | cclS6 (123456)| | cclS6 (12)| = | cclS6 (12)(34)(56)| Theorem 6.7 Aut(An ) = Sn , for n > 3, unless n = 6 Proof Assume n ≥ 4, n ̸= 6. Any automorphism permutes the subgroups of An isomorphic to An−1 . There are precisely n of these, one corresponding to each point of [1, n], so any automorphism induces a permutation 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 6.8 Aut(A6 ) = Σ6 , a group of order 1440, containing S6 as a subgroup of index 2. Proof Existence: 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 type 13 31 ⇒ 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 3 If θ1 , θ2 are two automorphism swapping these two classes (i.e. θ1 , θ2 ∈ Σ6 \ S6 ), then θ1−1 θ2 fixes each of these ccls, so is in S6 ⇒ |Σ6 : S6 | = 2 Corollary 6.9 G simple order 60 ⇒ G ∼ = A5 ∼ = P SL2 (5) ∼ = P SL2 (4) Proof The 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+1 2 ]! (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 and H ∩ Sym(Γ) = 1 Claim:

k≥

n 2

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 = n1 pe1 + n2 pe2 + · · · + ne1 p + ne1 +1

0 ≤ ni < p

Concentrating on the case pe1 , e1 = 1 : p2 , e1 = 2: (Cp )p o Cp .. . pn pn+1 Any Sylow p-subgroup is

(Pn )p o Cp ∏ 1≤i≤e1

Pe × · · · × Pei } | i {z 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 ≤ n2 and H ≤ Sk × Sn−k So the maximal subgroups intransitive on [1, n] are precisely Sk × Sn−k for 1 ≤ k ≤ n2 Lemma 6.10 The maximal intransitive subgroups of Sn are Sk × Sn−k for 1 ≤ k ≤ n2 These are maximal in Sn unless k = n2 (n = 2k, Sk × Sk < (Sk × Sk ) o S2 < S2k ) Proof Let 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 size 1, k, n − k − 1 Impossible unless n = 2k But X contains a transposition and 3-cycle. ⇒ X ≥ An

27

Addition to previous lectures: An P Sn ≤ Aut(An ) θ : Aut(An ) → Sn K = ker θ, K ∩ An = {1} ⇒ [K, An ] = 1 ⇒ K = 1 so θ injective Exercise: 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 = (cδ1 d−1 , cδ2 d−1 , . . . , cδl d−1 ) i.e. C wr D = C ∆ o D Suppose now C acts on Γ as a permutation group. The imprimitive action of C wr D 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 wr D is imprimitive on Γ × ∆ In particular, have Sk wr Sl imprimitive on Γ × ∆, with |Γ| = k, |∆| = l Lemma 6.11 Let G be (transitive but) imprimitive on Γ, |Γ| = n so G ≤ Sn , with l blocks of size k. Then G ≤ Sk wr Sk (the full stabilizer of this partition of Γ into blocks). Moreover, these groups Sk wr Sl are maximal in Sn (n = kl) Proof (Proof of moreover part): Sk wr Sl < H ≤ Sn . Then H on [1, n] is transitive, in fact primitive. Hα contains Sk−1 × Sk wr Sl−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: S2 wr S4

index 7


2 : G is k-homogeneous ⇒ G is (k − 1)-homogeneous ( ) m Exercise: If πk is the permutation character of Sm on , then πk = πk−1 +χk where χk is irreducible k (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 subgroup of 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 6.18 If G is a transitive permutation group on Ω, and G has a ccl C = ̸ {1}, then G has a non-trivial subdegree at most |C| Proof Fix c ∈ C. Let α ∈ Ω with β = αc ̸= α. Consider βGα : If ω ∈ βGα , have ω = βh some h ∈ Gα ⇒ ω = αh−1 ch with h−1 ch ∈ C Sketch Proof of Theorem 6.17 In Sm , take C = {transposition} In Am , take C = {3-cycles} ⇒ n − 1 ≤ 21 m(m − 1) Let H be the stabiliser of a point in [1, n] ⇒ |Sm : H| = n and H maximal on Sm If H is on [1, m], then “small” by Bochert (see Example after Corollary 6.9): ] primitive [ m+1 1 ! ≤ 1 + 2 m(m − 1) ⇒ m ≤ 6 2 (If in Am , get m ≤ 8) If H is transitive but not primitive on [1, m], then H = Sk wr Sm/k , the action is on k|k| · · · |k, not 2-transitive on m > 6 ( ) n If H is intransitive on [1, m], the action of Sm is on , not 2-transitive unless k = 1 k

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)| = (q d − 1)(q d − q) · · · (q d − q d−1 ) SLd (q) P GLd (q) matrices of determinant 1 |SLd (q)| = |GLd (q)|/(q − 1) P GLd (q) := GLd (q)/Z, where Z = {scalar matrices in GL} Ld (q) = P SLd (q) := SLd (q)/Z ∩ SLd (q), has order |SLd (q)|/e, where e = (d, q − 1) P GLd (q) acts naturally on the set of 1-subspaces of Vd (q) (i.e. the projective space of dimension d − 1, d −1 ) Pd−1 (q), of size qq−1 Recall: P GLd (q) is 2-transitive, and is 3-transitive ⇔ d = 2 P SL2 (q) is 2-transitive, and is 3-transitive ⇔ q = 2l 32

More on d = 2: “M¨obius action” of P GL2 (q) of degree q + 1, Fq ∪{∞} Lemma 7.1{ x/y ⟨(x, y)⟩ ↔ ∞ ( ⇒

a b c d

y ̸= 0 y=0

) : z 7→

az+b cz+d

??????

P SL2 (q) = {z 7→ az+b cz+d |ad − bc is a non-zero square} AGL1 (q) = G∞ {z 7→ az + b|a ̸= 0} Gαβγ = 1, any α, β, γ distinct G0∞ = {z 7→ αz|a ̸= 0} G is 3-transitive, with Gαβγ = 1 ∀αβγ Generators of P GL2 (q) : z 7→ z + 1 z 7→ λz (⟨λ⟩ = F× q ) z 7→ −1/z Lemma 7.2 q Group order degree ∼ 2 L2 (2) = S3 6 3 3 L2 (3) ∼ 12 4 = A4 4 L2 (4) ∼ 60 5 = A5 5 L2 (5) ∼ 60 6 = A5 − ∼ ∼ Also: L2 (9) = A6 = Ω4 (3), L2 (7) ∼ = L3 (2), L4 (2) ∼ = A8 ∼ = Ω+ 6 (2), L4 (2)  L3 (4) Proof Exercise Theorem 7.3 Ld (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 a normal subgroup K which is abelian (or just solvable) with G = ⟨K g |g ∈ G⟩ If N P G, then N ≤ G(Ω) , the kernel of G on Ω, or N = G So G/G(Ω) is simple Example: An is simple: Consider G = An acting on

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

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

33

Definition t ∈ 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 of V , 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 , vd } Also, 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 7.5 SLd (q) is generated by the transvections Proof Linear algebra: any matrix of det 1 can be reduced to IA by applying elementary row operations vi := vi + λvj , i ̸= j Each 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 7.3 Let G = SLd (q). Then G is 2-transitive, hence primitive, on Ω = {1-subspaces of Vd (q)} Kernel G(Ω) = {scalar matrices} Let α = ⟨(1,0· · · , 0)⟩      1 0 ··· 0 λ 0 ··· 0             λ2      λ 2     , K= . λi ∈ Fq λi ∈ F q ⇒ Gα =  .        ..  .. In−1  Hd−1              λd λd elementary abelian order q d−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 Proof If d ≥ 3        1 1 1 1 1 α 1   1  −α 1   = 1 1  1 β 1 1 −β 1 −(αβ) 1 so if α ̸= 0, take β = α−1 γ d = 2:

(

)(

α−1 α

)( )( ) ( ) ( ) 1 α 1 1 0 1 0 = = β 1 α−1 β 1 α2 β 1 β(α2 − 1) 0

So given γ, take α ̸= 0 with α(2 ̸= 1)(q > 3) 1 and β = γ(α2 − 1)−1 , to get as a commutator γ 1

34

7.1

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

Parabolic:

   0    ∗    . ∗ . (1) B =  .  ∗ ∈ F , # ∈ F  - Borel subgroup    # ∗ (2) lower traingular    1 0     1   . . .  # ∈ F - lower unitriangular matrix |U | = q 2 d(d−1) a Sylow p-subgroup (3) U =      # 1 of GL, q = pt

  λ1 0   .. B = U o T , with T = {  |λi ∈ F × } diagoonal matrices or split torus . 0 λd 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 subgroups maximal parabolic is just a stabilizer of a subspace of W of V Pk is the ( stabilizer of)a k-subgroup W of V : Ak 0 Pk = { |(A, B) ∈ GLk × GLd−k , C any of k × k matrix} C Bd−k (choose basis e1 , . . . , ek of W extend to a basis e1 , . . . , ek , . . . , eh for V ) ) ( Ik } is a normal subgroup of Pk ; The subgroup Uk = { ∗ Id−k ) ( Ak 0 - complement to Uk in Pk , the Levi subgroup!of linear group of Pk Lk = 0 Bd−k Then Pk = Uk o Lk By the way, N = all monomial matrices, then N = NG (T ) T = B ∩ N , W = N/T - Weyl group Example: GLd (q), W = Symd Remark. The subgroups of GLd (q) containing B are precisely the parabolics obtained by stabilizing the 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 ≤ And, Pk is a maximal subgroup of GLd (q)

d 2

Some other subgroups: some families of geometric subgroup Then theorem of Aschbache tells you that any subgroup H of SL lies in one of these or H/Z(H) is almost simple, irreducible, etc.... (1) Pk - the stabilizer of a k-subspace W of V (2) V = V1 ⊕ · · · ⊕ Vk , d = ka, dim Vi = a GLa (F ) wr Symk 35

(3) V = V1 ⊗ V2 , dim Vi = di , d = d1 d2 , d1 ̸= d2 GLd1 (F ) ⊗ GLd2 (F ) (4) V = V1 ⊗ · · · ⊗ Vk , dim Vi = 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 (q e ) < GLd (q) (7) “Extraspecial” (8) Classical Upslot: Questions about maximal subgroups of linear groups can be reduced to question about modular representation theory of quasi-almost simple groups

7.2

Automorphism group of P SLd (q)

P SLd (q), q = pf , Outer automorphisms come in different flavours: Ld (q) (index)

P (d,q−1) Inn diag

P

P GLd (q)

P ΓLd (q)

f field auto

P

Aut

2 if d>2 1 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 = λσ vθ (σ depends on θ) Fix a basis, ( )σ ( σ ) a11 . . . a1d a11 . . . aσ1d = .. .. . ··· . ··· ΓLd (q) = GLd (q) o Γ finally, a graph auto: A 7→ A−t (inverse transpose of A) an auto of GL Exercise: Graph auto is inner if d = 2, outer otherwise Aut := P ΓL o ⟨τ ⟩ Theorem 7.7 (Steinberg, for groups of Lie type) The Aut. group of L2 (q) is P ΓL2 (q) The Aut. group of Ld (q) with d > 2 in P ΓLd (q) : C2 (semicolon denotes semidirect product) Sketch Need to show no more automorphisms. Let G = P SLd (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 ϕinvariant, adjust ϕ by a graph automorphism. (note that it would have been swapped with the ccls of 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-spaces invariant 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, using a 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 V 



1 λ 1  ϕ:



↓

1  λσ 1 

and

..

 λ  λ−1

1  µ 1  .



..

 µ  µ−1

I ⇒

7.3

..

. 

..



1  λ + µ 1  = 

↓

1  µσ 1  .



 ..

  .



1  λσ + µσ 1  =  .

 ..

 

 

λσ ∈ Fq , (λ + µ)σ = λσ + µσ

.

→ (λµ)σ = λσ µσ

I

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

Some isomorphisms and interesting actions

Exercise: Any simple group of order 168 is isom. to L2 (7) Lemma 7.8 L2 (2) ∼ = L2 (7) Proof L3 (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 )| = 21 Let P ∈ Syl7 (G), say P = ⟨g⟩ Number the Sylow 7-subgroups as P = ∞, 0, 1, . . . , 6 Choose 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 degree 8: (0∞)(1x)(2y)(4z) | {z } fix h

(x, y, z ∈ {3, 5, 6} but not known which yet) Conjugating t by h or h−1 , we may ssume t : 1 ↔ 6 Then 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 7.9 A6 ∼ = L2 (9), S6 ∼ = P ΣL2 (9) (Σ field autos), Σ6 ∼ = P ΓL2 (9) Proof Produce an action of A6 degree 10 (3|3 - stab. is (S3 wr S2 ) ∩ A6 ) Put F9 ∪{∞} structure on it to get A6 . L2 (9) H = NA6 (⟨(123)⟩, ⟨(456)⟩) ∼ =(S3 wr S2 ) ∩ A6 123|456 is fixed by H An 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 + i Now h also fixes partition 156|234 Rest 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→ z 3 a field automorphism ⇒ S6 ∼ = P ΣL2 (9) Note: Gal(Fpl / F) is generated by z 7→ z p and is cyclic of order l Some 2-transitive actions of Ld (q): d −1 Ld (q)is 2-transitive on qq−1 (2 actions for d > 2, 1 action for d = 2) plus: L2 (7), deg 7 (2 action) L2 (9), deg 6 L2 (11), deg 11 P ΣL2 (8) ∼ = 2 G2 (3), deg 28 ∼ L4 (2) = A8 , deg 8

8

symplectic Groups Spd (q) ≤ GLd (q)

Definition If f a bilinear alternating non-singular form on V = Vd (q), f is called symplectic form (We are taking definition 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 8.1 If 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 ) = δij Proof Construct: Let e1 ̸= 0. Take f1 ∈ V \ e⊥ 1 (non-empty) with (after scaling) f (e1 , f1 ) = 1. Continue this process in ⟨e1 , f1 ⟩⊥ Definition Such ei , fi is a hyperbolic pair Then the matrix of f under this basis is   −1     

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



1

     

1 −1 ..



.

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)|At JA = J} Sp2m (q) is regular in its action on the symplectic bases. (transitive with trivial stabliser) Lemma 8.2 2 ∏m 2i |Sp2m (q)| = (q 2m − 1)(q 2m−1 )|Sp2m−2 (q)| = · · · = q m i=1 q − 1 Lemma 8.3 Sp2 (q) ∼ = SL2 (q), Proof ( ) ( ) 1 1 At A= for ad − bc = 1 −1 −1 Z(Sp2m (q)) = {±I}, P Sp2m (q) ∼ = Sp2m (q)/{±I} Definition symplectic 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 8.4 Sp2n (q) is generated by these transvections Proof G = ⟨ 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 = w If 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 pair If 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 induction If B ′ = {e′1 , f1′ , . . .} may assume e′1 = e1 , f1′ = f1 by (2) Then work in ⟨e1 , f1 ⟩⊥ − V2m−2 (q) with symplectic form obtained by restriction. Induction do the rest

Corollary 8.5 Sp2m (q) ⊆ SL2m (q) Proposition 8.6 Transvections in Sp2m (q) are commutators, unless (2m, q) is one of (2, 2), (2, 3)(4, 2) Proof This is true as Sp2 (q) = SL2 (q) for q ≥ 3 So only have to check cases Sp4 (3) and Sp6 (2) (exercise) (Sp2m (q) = [SL2m (q), SL2m (q)]) Theorem 8.7 P Sp2m (q) is simple, except for P Sp2 (2) ∼ = S3 , P Sp2 (3) ∼ = A4 , P Sp4 (2) ∼ = S6 Proof Use Iwasawa’s Lemma 8.8 Lemma 8.8 (Iwasawa) 2m −1 The action of Sp2m (q) on the q q−1 points of P2m−1 (q) is rank 3, with subdegrees 1, q(q 2m−2 − 1)/(q − 2m−1 1), q      0    1  λ 0   0 I = , A ∈ Sp (q) with normal subgroup This stabilizer of ⟨v⟩ is P1 =  ∗ A 0  λ ∈ F× 2m−2 q     λ 0 1 ∗ ∗ λ−1 {tv,λ |λ ∈ Fq } 40

The action is primitive, kernel is {±I} Proof v ∈ V , suborbits {⟨v⟩} (size 1), {⟨w⟩|w ∈ v ⊥ − ⟨v⟩}, {⟨w⟩|w ∈ V \ v ⊥ } (size q 2m−1 ) The action is primitive since any non-trivial block would consist of {⟨v⟩} together with one of the d −1 G⟨v⟩ -orbits, but the size does not divide qq−1 Finally, G⟨v⟩ = P1 - see below about Pk Proposition 8.9 Sp4 (2) ∼ = S6 , so not perfect Proof S6 on U = V6 (2) natural action as permutations of coordinates. ∑ W = {(x1∑ , . . . , x6 )| xi ≡ 0 mod 2} f (x, y) = xi yi , bilinear form on U W ⊥ = {(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 ) ⊥ = ⟨e , . . . , e ⟩ < A complete symplectic flag: 0 < W1 = ⟨e1 ⟩ < W2 = ⟨e1 , e2 ⟩ < · · · < Wm = Wm 1 m ⊥ Wm−1 = ⟨e1 , . . . , em , fm ⟩ Wi isotropic, so f |Wi is 0, for each 1 ≤ i ≤ m B Borel stabilises such a complete flag The parabola are stabilisers of some flags Maximal parabolic subgroup Pk := Stabilisers of Wk with Wk ⊆ Wk⊥ Pk stabilise W = ⟨e1 , . . . , ek ⟩ and W ⊥ = ⟨e1 , . . . , em , fk+1 , . . . , fm ⟩ (dim 2m − k) W ⊥ /W = ⟨ek+1 , . . . , em , f k+1 , . . . , f m ⟩ -Symplectic space dim 2(m − k)   A  (on W , W ⊥ /W , V /W ⊥ ) B Pk contains a Levi subgroup GLk × Sp2(m−k) :  A−t   I  Kernel of the homomorphism Pk → Lk is Qk =  ∗ I ∗ ∗ I k(k+1)/2 2k(m−k) q , often a “special” p-group: Qk is the unipotent radical of Pk , order q Q′ = Z(Q) = Φ(Q), order q k(k+1)/2 Example 2m =  4, k=1    1 1             1 α 1  as a product of commutator  , e.g. q = 3, can get  Q1 =    β 0 1 1            1 1 γ β −α 1 {( ) } I C = C t k = 2, Q2 = C I

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 ⊥ (dim W = 2k, dim W ⊥ = 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, 11} and Sp2m (2) deg 2m−1 (2m ± 1) on the two classes of orthogonal forms. Automorphism groups: P Sp2m (q) P P GSp2m (q) (diagonal (2, q − 1)), g : f (vg, wg) = λg f (v, w) ∈ Fq P Sp2m (q) P P GSp2m (q) P P ΓSp2m (q) P Aut (this last P has index 1 except for Sp4 (2f ) then index 2)

9

Unitary Groups

F = Fq2 > Fq a quadratic extension, V = Vd (q 2 ) σ : α 7→ αq α 7→ α

⟨σ⟩ = Gal(Fq2 / Fq ) N (α) = αq+1 , N : Fq2 → Fq2 onto

Let f be a σ-Hermitian, left linear, non-singular form GUd (q) = {g ∈ GLd (q 2 ) | f (vg, wg) = f (v, w) If dim V ≥ 2 ⇒ ∃ isotropic vectors: Since, let v ∈ V with f (v, v) ̸= 0 Let 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 1

|GUd (q)| = q 2 d(d−1) (q d − (−1)d )(q d−1 − (−1)d−1 ) · · · (q 2 − 1)(q + 1) Proof Write zd = # isotropic vectors in dimension d yd = # vectors norm 1

⇒

q 2d = 1 + zd + (q − 1)yd

Also, zd+1 = zd + (q 2 − 1)yd = −qzd + (q 2d − 1)(q + 1) Since z0 = 0 = z1 , can solve recurrence: zd = (q d − (−1)d )(q d−1 − (−1)d−1 ) ⇒ yd = q d−1 (q d − (−1)d ) The order of GU follows, by induction along an orthonormal basis 42

|SU | = |GU |/(q + 1) ,

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

In particular, P SL2 (q) ∼ = P SU2 (q) Lemma 9.2 P SU3 (q) acts 2-transitively on the isotropic points of P2 (q 2 ). There are q 3 + 1 of these; for larger d, this leads to a primitive rank 3 action. Proof e 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 q 3 , and SU3 has index q + 1 in GU3 ⇒ still transitive For d ≥ 3, P SUd (q) is simple except for P SU3 (2) (order 72, 2-transitive on 9), c.f. Iwasawa Lemma Maximal parabolic subgroups: Pk = stabiliser of a totally isotropic k-vector space = GLk (q 2 ) × GUd−2k (q)

10

Orthogonal Groups

Quadratic form Q : V → F = Fq s.t. Q(λv) = λ2 Q(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 symmetric If 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 10.1 Two equivalence classes of such form: Either

∃ orthonormal basis for V

Or

∃ orthonormal basis with

f (vi , vi ) = 1 ∀i < d f (vd , vd ) = α some non-square α

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

Write ϵ GO2m (q) = {g ∈ GL2m (q)|Qϵ2m (vg) = Q(v) ∀v}

(ϵ = ±)

Lemma 10.2 If d = 2, there may or may not exists isotropic vectors: according to type of Q and q ≡ ±1 mod 4 Proof √ q ≡ 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 isotropic If f (u, v) = 1, f (v, v) = α, with u⊥v ⇒ f (u + λv) = 1 + λ2 α ⇒ 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− e1 , f2 , . . . , em−1 , fm−1 , d, d′ with ⟨d, d′ ⟩ = O2− , d, d′ ⊥ei , fj ∀i, j, f (d, d) = 1, (in 2m ↔ f : fact have f (d, d′ ) = 1, see later)

Related groups: GO2+ (q) = D2(q−1)

,

GO2− (q) = D2(q+1)

d = 2: GO2+ (q) = D2(q−1) : e1 , f1 , cyclic group order q − 1: e1 7→ λe1 , f1 7→ λ−1 f1 t inverting: e1 ↔ f1 No more elements: ⟨e1 ⟩, ⟨f1 ⟩ are the only isotropic points GO2− (q) = D2(q+1) : Let V = Fq2 - dim 2 over Fq Q(v) = N (v) = v q+1 This is a quadratic form, no isotropic vectors Cyclic group order q + 1: N (λ) = 1 ⇒ v 7→ λv is an isometry inverted by t : v ↔ v q No others- e.g. the stabiliser of 1 conssits of i and v ↔ v q (check) Lemma 10.3 2

|GO2m−1 (q)| = 2q m (q 2m − 1)(q 2m−2 − 1) · · · (q 2 − 1) ϵ |GO2m (q) = 2q m(m−1) (q 2 − 1)(q 4 − 1) · · · (q 2m−2 − 1)(q m − ϵ1)

Proof By induction, going up in steps of 2. Let zm = #(non-zero) isotropic vectors in dimension 2m + 1 or 2m Claim: zm = q 2m − 1 for Q2m+1 and zm = (q m − ϵ)(q m−1 + ϵ) for Qϵ2m Proof of Claim: Correct for dimension 1, 2 (both ϵ) Let dim V = n + 2, V = U ⊕ W (i.e. U ⊥W ) with U dimension 2 with some type, W dimension n of 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)(q n − 1 − zm ) − 1 = qzm + (q − 1)(q n + 1) - now obtained formula in each case  Having chosen e1 , need f1 with e1 , f1 hyperbolic: Claim:

This can be done in q n−1 ways (hence formulae follow)

Proof of Claim: Number of isotropic vectors in e⊥ 1: /⟨e ⟩ is a space of same type so zm−1 isotropic vectors ⟨e1 ⟩ + v e⊥ 1 1 ⇒ 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 2 Let f be the associated bilinear alternating form - f (v, v) = 0 rad f

= {w|f (v, w) = 0 ∀v}

rad Q = {w ∈ rad(f )|Q(w) = 0} rad Q ≤ rad f ≤ V (subspaces), codimension of rad f over rad Q is 0 or 1: Q|rad f : rad f → F semilinear Q(v + w) = Q(v) + Q(w), Q(λv) = λ2 Q(v) Norm of v=Q(v), v is isotropic if Q(v) = 0 Q non-singular: rad Q = 0 Q non-degenerate: rad f = 0 Let Q non-singular ⇒ rad f has dimension ≤ 1 Look at dim V = 2m + 1 odd ⇒ rad f dim 1 because V / rad(V ) has dimension 2m even, with a non-singular alternating form GO2m+1 (q) ≤ Sp2m (q) Note: in fact, |P Ω2m+1 (q)| = |P Sp2m (q)| for q odd, NOT isomorphic if 2m > 4 Now look at dim V = 2m even Now handle things as before, choose a symplectic type basis as much as poss with Q(v) = 0 Note: Q(e) = 0, f (e, f ) = 1 ⇒ can adjust f to have Q(f ) = 0 (as Q(f + λe) = Q(f ) + λ, so replace f by f + Q(f )e) If dim V > 2, ∃ isotropic vectors: there are vectors u⊥v if Q(v) ̸= 0, then Q(v + λu) = Q(v) + λ2 Q(u) ⇒ Take λ2 = Q(v)/Q(u), then v + λu isotropic So consider dim V = 2: v, w basis Let 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) 2 Q− 2 : v, w s.t. Q(v) = 1, f (v, w) = 1, Q(w) = µ ∈ F s.t. X + 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 Q2 45

ϵ (q) GO2m |SO2m | = 12 |GO2m | for q odd SO = GO for q even ϵ P SO2m NOT simple, has a subgroup index P Ωϵ2m (q) which is simple if dim > 4 (hard to prove the existence of this, see Wilson)

Some isomorphisms: (1) P Ω3 (q) ∼ = L2 (q), q odd, W = V2 , V = S 2 W (see Wilson page 96, and example sheet) (2) P Ω4 (q) ∼ = L2 (q) × L2 (q) 2 ∼ (3) P Ω− 4 (q) = L2 (q )

(4) L4 (q) ∼ = P Ω+ 6 (q) on

∧2

V4

ϵ ≤ Sp GO2m 2m (q) q = 2 ⇒ two 2-transitive actions

Sp2m (q)

+ GO2m

o ooo o o oo ooo

OOO OOO OOO OO

NNN NNN NNN NN

pp ppp p p p ppp

Sp2(m−1) (q) × 2

46

− GO2m

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

Jordan set, 25 primitive, 25 Jordan-H¨older Theorem, 4

blocks, 17 Borel subgroup, 35 Burnside’s pa q b Theorem, 15 Burnside’s Basis Theorem, 10 Burnside’s Theorem, 20

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

characteristic subgroup, 3 characteristically simple group, 3 commutator subgroup, 3 Correspondence Theorem, 2 degree, 18 subdegree, 22 diagonal action, 28 diagonal subgroup, 30 diagonal type, 30 distance, 22 distance transitive, 22 elementary abelian p-group, 3 factor group, 1 factorisable, 14 factorization of group, 13 faithful action, 16 Fano plane, 37 Fitting subgroup, F (G), 8 fix(Gα ), 18 Frattini argument, 9 Frattini subgroup, Φ(G), 9 Frobenius groups, 20 G-congruence trivial, 17 G-congruence, 17 G-homomorphism, 16 G-isomorphism, 16 Gp , 9 GΩ , 16 Hall π-subgroup, 12 Hall p-complement subgroup, 12 homomorphism of groups, 1 hyperbolic pair, 39 Inner automorphism, Inn(G), 2 Isomorphism Theorem, 1 isotropic, 45

Levi subgroup, 35 of Sp2m (q), 41 line, 20 minimal generating set, 10 multiplicity-free, 23 nilpotent group, 7 nilpotent radical of G, F (G), 8 non-generator, 9 normal closure, 15 normal subgroup, 1 Op (G), 8 orbit αG, 16 orbital, 22 diagonal, 22 paired Γ∗ , 22 Orthogonal group GOd (q)ϵ , 44 Outer automorphism group, Out(G), 3 p-local subgroup, 36 P.Hall Theorem, 15 parabolic, 35 perfect, 3 permutation actions, 16 permutation representation, 16 π ′ -part of n, 12 π-group, 12 π-part of n, 12 primitive, 17 projective general linear group P GLd (q), 18 P SUd (q), 43 quotient group, 1 rank of groups, 21 regular group, 11 ρ(α), 17 Second Isomorphism Theorem, 2 semi-regular, 21 semidirect product, 19 series, 4

47

central, 7 chief, 5 composition, 4 derived, 11 lower central, 8 normal, 4 proper, 4 upper central, 7 simple group, 1 simply primitive, 21 socle, 28 soluble group, 11 soluble radical, 11 split torus, 35 SUd (q), 43 Sylp (G), 9 Sylow basis, 14 Third Isomorphism Theorem, 2 transitive on Ω, 16 transvection, 34 symplectic, 39 unipotent radical, 41 Unitary group GUd (q), 42 unitriangular matrix, 35 W (k, l) = Sk wr Sl , 29 Witt index, 43 wreath product, 25 wreath product C wr∆ D, 28

48