The Small World Theorem

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1 The Small World Theorem Ulrich Meierfrankenfeld, Bernd Stellmacher June 10, 2002 Assume M(S) 2 and Q!. We investigate the Structure of E/O p (E). For L L define L = L O p (L). In this section we assume Hypothesis 0.1 [hypothesis e structure theorem] (ES1) M(S) 2 and Q!. (ES2) P P(S) and P C. (ES3) P /O p (P ) = SL 2 (q), q a power of 2. (ES4) Y P is a natural module for P/O p (P ). (ES5) N P (S P ) C. (ES6) Y E P is abelian. (ES7) 0 p ( P, E ) = 1. Some remarks on this assumptions. (ES7) follows from E! but not from Q! (example (L n (q). (ES2) to (ES5) follow from P! theorem. But P! have currently been treated only for Y M Q. Let L = N G (P ) and H = E(L C). By (ES7) O p ( H, L ) = 1. By part (a) of the preceeding lemma we can apply the amalgam method to (H, L). Let Γ 0 = Γ(G; L, H) and Γ the connected component of Γ containing L and H. For α Γ. If α = Lg define E α = P g, if α = Hg define E α = (EQ) g, C α = C g and Q α = Q g. Lemma 0.2 [basic es] Let (α, β) be adjacent vertices with α H (a) G α = E α G αβ and G β = E β G αβ. (c) C U (Y β ) = Q α Q β = Q a Q β = Q αβ Syl p (L α ), where U Syl p (G αβ ). (d) G αβ = N Gα (Q αβ ) (e) Y αβ = Y β = [Y α, x] for all x Q αβ \ Q α. 1

2 (f) Let Ṽβ = V β /Y β. Then C Gβ (Ṽβ) C Gβ (Y β ) = Q β In particular G αβ contains a point stablizer for G β on Ṽβ. Proof: By edge transitivity we may assume that α = L and β = H. By (ES5), and the Frattini Argument, L = P (L C) = P (L H). By definition of H, H = (H L)E. Thus (a) holds. Note that Y H L Y H and so the definition of E implies [Y H, E] = 1. Since L = (L H)E we get Y H L = Y H. Let T = S P. Since N P (T ) is a maximal subgroup of L we get N P (T ) = H L and O p (H L) = T. Since Y H L Y L and Y P is a natutal module for P we see that Y H = Y H L = C YL (T ) = Ω 1 Z(T ) and C S (Y H ) = T. Also since Q O p (P ), O p (H)O p (L)/O p (L) is a non-trivial subgroup of T/O p (L) normalized by H L. Thus O p (H)O p (L) = T. For (f) let D = C H (Ṽ ) C H(Y H ). Since [Y L, O p (H)] = Y H, O p (H) D. Note that O p (D) centralizes Y P and so [P, O p (D) O p (P ). Since O p (D) = O p (O p (D)O p (P )] we get P N G (O p (D)). Since also H normalizes O p (D) we conclude that O p (D) = 1, D is a p-group and D = O p (H) G αβ = N G (Q αβ ) Suppose that N G (T ) L. Put M = N G (T ), P T. If O p (M) = 1 then pushing up SL 2 (q) and Ω 1 Z(P ) = 1 gives [O p (P ), O p (P )] Y P. By ES6, V = ZP H O p(l). Thus V is normal in H and L, a contradiction. The last equality in (e) follows since Y P is a the natural module. The following is not needed in the F F -module argumnet: C Gβ ( x) G αβ for all x Y α \ Y β. For (g) let D = C H ( x) and D = C H (x), T acts transitively on the coset Y H x, D = DT. Let x Y l H for some l L. Then D C G(x) and by Q!, D N G (Q l ). Thus D N G ( Q, Q l = N G (P ) = L. Let (α, α ) be a critical pair. Let β = α + 1 and α 1 (a) with α 1 β. Lemma 0.3 [b 2] (a) b > 2. (b) α H. (b) b is odd. b > 2 follows from (ES6) and α H follows from Y H Y L. Suppose that b is even. The by 0.2 Y β = [Y α, Y α ] = Y α 1. Hence by??(c), E β = E α 1. Since b > 3, V α 1 Q β and V α 1 N G (Q β ) = N G(Q α 1 ). Thus V α 1 N Gα 2 (Q α 2Q α 1 ) = N G α 2 (Q α 2α 1 ) = G α 2α 1 2

3 As V α 1 Q β, [V α 1, E β ] is a p-group and so V α 1 Q α 1 Q α 1α and hence [V α 1, Y α ] Y α 1 = Y β Y α. Hence Y α normalizes V α 1, a contradiction. Thus b is odd. Lemma 0.4 [offender on Vbeta] One of the following holds 1. There exists 1 A Q αβ /Q β such A is an offender on Ṽβ. 2. b = 3 and there exists an non-trivial G αβ invariant subgroup A of Q αβ /Q β such that A is a quadratic 2F -offender on Ṽβ. Proof: If Y β Y α 1, then??(c) implies E β = E α, a contradiction since [Y β, E β ] is a p-group, but [V β, E α ] is not. Hence Step 1 [zbza] Y β Y α = 1 Hence by 0.2 [V β Q α, V α Q β ] Y β Y α = 1 and we proved Step 2 [vqvq] [V β Q α, V α Q β ] = 1 Suppose now that b = 3. Since Q α acts transitively on (α + 2) \ {α } get G α+2α = G βα+2α Q α. Hence V β Q α is normal in G α+2α. Also [V α, V β, V β ] [V β, V β = 1. Since G α+2 is doubly transitive on (α + 2), V β Q α /Q α = V α /V α Q β ) Let δ (β). Then no subgroup of Q β is an over-offender on Z δ. This together with Step 2 implies V α Q β /C Vα Q α 1 (V β ) V β /C Vβ (V α Q β ) V β /V β Q α = V β Q α /Q α By the lasy two displayed equations, V β is a 2F offender on lemma holds. So we may assume from now on Ṽα. So Case 2. of the Step 3 b > 3. Suppose that V α Q β. Then by Step 2 [V α, Y α Q α = 1. By 0.2f [V α, Z α ] 1 and since Y α is a natural module for E α and since V α E α we get V α /C Vα (Y α ) q = Y α /C Yα V α Thus 1. holds in this case. So we may assume that for all critcial pairs: 3

4 Step 4 [sym] V α Q β and the situation is symmetric in β and α. If [V β, V α Q β ] = 1 = [V α, V β Q α ], then again (1) holds. So we may assume Step 5 [vvqa] Y β = [Y β, Y α Q β ] V α or Y α [Y α, Y β Q α ] Y β By symmetry in α, α we may assume Step 6 [vvq] Y β = [V β, V α Q β ] V α. Pick µ (β) and t V α Q β with [Z µ, t] 1. Then µ α + 2 and by Step 2, Z µ Q α and we may assume that µ = α. Hence Step 7 [vbq] There exists t V α Q β with [Z α, t] 1. In particular, t Q α Note that Step 8 [O2G] O p (E α ) Q α 1, t. By Step 2 and Step 7 we have V β Q α /Q α Y α Q α /Q α = Y α /C Yα (t) q. We record Step 9 [vbqa] V β Q α /Q α q. We next show: Step 10 If [V α 1, V α 2] = 1 then 1. holds. Suppose [V α 1, V α 2] = 1. Then V α 1 Q α 2 Q α 1. Put A = V α 1 (V β Q α ). Then A V β (V β V α 1 Q α ) V β (Q α 1 Q α ). Thus by 0.2 [A, t] [V β, t][q α 1 Q α, t] Y β Y α. Let X be maximal in A with [X, t] Y β. As Y α = q we have A/X q. Since Yβ X, t normalizes X. By 0.2, [XZ a, Q α 1 ] [V α 1, Q α 1 ] = Y α 1 XZ α. So by Step 8, O p (E α ) normalizes XZ α. Since O p (E α ) is transitively on (α) we conclude that XZ α D α := δ (α) V δ. Put a = V α 1 /A. Then V α 1 D a /D a V α 1 /A A/X aq. Hence V β D a /D α aq. Note that V α 1 Q α 2 Q α 1 G α. Since D α 1 V α 2 we conclude from V β D a /Y D α qa and edge-transitivity that V α /C Vα (V α 1 V β ) V α D α 1/D α 1 = V β D a /D α aq. On the otherhand by definition of a, an isomorphism theorem and Step 9 V α 1 V β Q α /Q α = V α 1 V β Q α /V β Q α V β Q α /Q α aq. By the last two equations 1. holds. So we may assume from now on that 4

5 Step 11 [va-1va-2] [V α 1, V α 2] 1 Suppose that V α 2 Q a 1. Then by Step 4, V α 1 Q α 2. Note that by Step 8, C Yα 1 (t) = 1. Thus a contradiction to Step 11. Thus 1 [V α 1, V α 2] Y α 1 Y α 2 C Yα 1 (t) = 1 Step 12 [va-1qa-1] V α 2 Q α 1 By Step 4 we get Step 13 [va-1qa-2] V α 1 Q α 2 Since b > 3, t centralizes [V α 2 Q α 1, V α 1 ]. and so [V α 2 Q α 1, V α 1 ] = C Yα 1 (t) = 1 Thus by Step 5 and?? that Y α 2 = [V α 1 Q α 2V α 1] V α 1. Hence there exists 1 x Y α 2 V α 1. Note that t centralizes x and [x, Q α 1 Y α 1 Y α. So by Step 8, O p (E α ) normalizes the coset xy α. Suppose that [x, Q α ] 1. Let R = O p (E α ) and D = [Q α, R]. Since C Yα (R) = 1, the Three Subgroup Lemma implies [x, D] 1. Since R normalizes [x, D] we get [x, D] = Y α. Thus D acts transitively on Y α x and so by the Frattini argument, R = C R (x)d. Since x Y α 2, Q! implies C R (x) C α 2. Also since [E α 2, Q α 2] Q α 2 and E α 2 acts transitively on α 2 we have t V (2) α 1 Q α 2 (V (2) α 1 Q α 2)Q (2) α 2 = (V α 3 Q α 2)Q α 2 (Q α C α 2)Q α 2 The right hand side of this equation is p-group normalized by C R (x) and so t C R(x) Q α /Q α is a p-group. But this contradicts t Q αβ \ Q α and O p (E α ) C R (x)q α. Thus [x, Q α ] = 1, and so x Ω 1 Z(Q α ) = Y α. Since [x, t] = 1 we conclude x Y β. Since also x Y α 2 we conclude that E α 2 C β Since b > 3, V α 1 V (2) α Q β = V (2) α+2 Q β (Q α 2 C β )Q β The right hand side is a p-group normalized by E α 2 and we obtain a contradiction to Step 13. Theorem 0.5 (The abelian E-Structure Theorem) [abelian es] Assume Hypothesis 0.1 ( and maybe that there exists a unique P P(ES) with P N G (P )).) Let V = Y E P and Ṽ = V/[V, O p(e)]. Then 5

6 Proof: G αβ = N G (Q αβ ) Suppose that N G (T ) L. Put M = N G (T ), P T. If O p (M) = 1 then pushing up SL 2 (q) and Ω 1 Z(P ) = 1 gives [O p (P ), O p (P )] Y P. By ES6, V = ZP H O p(l). Thus V is normal in H and L, a contradiction. The last equality in (e) follows since Y P is a the natural module. C Gβ ( x) G αβ for all x Y α \ Y β. For (g) let D = C H ( x) and D = C H (x), T acts transitively on the coset Y H x, D = DT. Let x YH l for some l L. Then D C G(x) and by Q!, D N G (Q l ). Thus D N G ( Q, Q l = N G (P ) = L. Let G be a group of local characteristic p. We say that G has rank 2, provided that there exists P, P P(S) such that P ES and P, P / L. We say that H has rank at least three if G has neither rank 1 nor rank 2. The next lemma shows how P! can be used to obtain information about E/O p (E). Lemma 0.6 [unique component in e] Suppose E!, P!, P uniqueness and that G has rank at least three. Let L = N G (P ) and H = (L C)E. (a) N G (T ) L for all O p (H L) T S. (b) There exists a unique P P H (S) with P L. Moreover, P ES. (c) P /O p ( P ) SL 2 (q).p k. (d) H = K(L H), L H is a maximal subgroup of H and O p (H L) O p (H). (e) H has a unique p-component K. (f) Let Z 0 = C YP (S P ) and V = Y H P. Then Z 0 V and V Q O p (H). (g) Let D = C H (K/O p (K). Then D is the largest normal subgroup of H contained in L and D/O p (H) is isomorphic to a section of the Borel subgroup of Aut(SL 2 (q) (h) Let V = V/Z 0. Then (ha) [V, O p (H)] = 1 (hb) C H (V ) D and C H (V ) C H (Z 0 ) = O p (H). (hc) Let 1 X Y P /Z 0. Then N H (X) H L. (hd) H L contains a point-stabilizer for H on V. Proof: By E!, O p ( L, H ) = 1 and so E L. Since {P } = {P (S) and N G (S) C, N G (S) N G (P ) L. Since E = P ES, N E (S) > there exists P P ES (S) with P L. Since rank G is at least three, P, P L and so by P!, P is uniquely determined. 6

7 Let T be as in (a). It follows easily from P! that QO p (L) is a Sylow p subgroup of P O p (L). Since QO p (L) O p (H L) T we conclude that T is a Sylow p-subgroup of T P. Suppose that M := P T, N G (T ) inl. Then by Pushing up?? and Q! P q 2 SL 2 (q). Since P, P L we get P N G (Y P ) and??(dd) gives the contradiction P N G (P ) = L. Thus M L. If P M, then T P and so O p (P ) O p ( P ), a contradiction to ( P 2b). Hence P M. By the uniqueness of P and since S M, M L. Thus (a) holds. Let P P H (S) with P L. Suppose that P P. Then P ES and so S E O p (P ) Since EO p (H L) is normalized by E(L H) = H we get O p (H L) O p (P ). Thus by (a) P L, a contradiction. So (b) holds. (a) is obvious. Let H L < M H. Then M L and so P M. Let R P(H)(S). If R = P, then R M, if R P, then R L and again R M. Since also N H (S) H L M we get M = H. By (a) O p (H L) O p (H). Let N be a normal subgroup of H minimal with respect to N L. By the uniqeness of P, P NS. Hence O p ( P ) N and since O p ( P )L, N = O p ( P ) H E. Next let F be the largest normal subgroup of H contained in L. Then [O p (H L), F ] O p (H L) F O p (F ) O p (H). Note that [O p (H), N] is normal in N(H L) = N and so [O p (H), N] = N and we conclude that [N, F ] O p (H). In particular F N/O p (N) Z(N/O p (N). Suppose that N is solvable. Then the mimimality of N implies that N/O p (N) is a r-group for some prime r p. In particular H N < N N (H N) and the maximality of H N implies H N H. Thus H N F. Suppose that S does not act irreducible on N/H N. The by coprime action there exists an S-invarinat R N with R L and O p ( P )R. Then by (b) RS L, a contradiction. So S acts irreducible on N/H N. Thus N = (H N)O p ( P ) and N = [N, O p (H)] O p ( P ). Note that P, N is normalizes by P, H L and N and so by L, H, it follows that O p P, N = 1 and so also O p ( P, P = 1, a contradiction. Thus N is not solvable and so the product of p-components. Let K 1 be a p-component of N. By minimality of N, N = K1 H L If S does not act tranisitively on the p-components of N, we can choose K 1 such that O p ( P )K1 S. But then K 1 L, a contradiction. Thus N = K1 S. Suppose that K 1 P lies in the unique maximal subgroup of P containing S. Since K 1 P is subnormal in P the structure of P implies [K 1 P, S] is a p-group. Thus K 1 N and K 1 P is a p-group. Hence N S O p ( P ). Since N N (N S)/N S is a p-group we conclude from coprimes action that N L projects onto N K1 (N S)F/F. Thus [K 1 L)F, N K1 (N S)] [K 1 L)F, N L] L. Conjugation with S yields [N L, N N (N S) N L. Thus H = P, L N H (N L) and so N L = F L. Since L contains a Sylow p-subgroup of N, we conclude that NF/F is a p -group. Let r be a prime divividing the order of NF/F and R/F an S-invariant Sylow p-subgroup of NF/F. Then RS is not contained in L and so P RS. Thus P = is a {r, p} group and r is unique. Thus NF/F is a r-group, a contradiction. We proved that K 1 P is not contained in the unique maximal subgroup of P containing S. Since [K 1 P, (K 1 P ) g is a p-group for all g P \ N H (K 1 ) the structure of P implies O p ( P )K 1. Thus S N H (K 1 ) and so N = K 1. Thus (e) is proved. 7

8 By P! uniqueness Z 0 is normal in C and [Z 0, Q] = 1. Since Y P is a natural module for P we get [Y P, Q] [V, O p (H) [V, O P (L H)] Z 0. So Y P acts trivial an all factors of 1 Z 0 Q and since C is of characteristic p, Y P Q. This proves (f) and (ha). To prove (g) note that N D and so by uniqueness of N, D L and so D F. But as seen above F D and so D = F. Let D 0 be maximal in D with [P, D 0 ] O p (P )]. Then H and P normalize O p (D 0 ) and so O p (D 0 ) = 1. Thus D 0 is a p-group and (g) holds. Note that C H (V ) N H (Y P ) H L and So C H (V ) D. Let R = O p (C H (V ) C H (Z 0 )). Then R) centralizes Y P and [R, P ] C P (Y P ) = O p (P ) O p (H L). But R is normal in H L and so R = O p (R) = O p (RO p (H L). Thus H and P both normalizes R and so R = 1. Hence (hb) holds. Let e Y P \Z 0 with ez 0 X. Let g N H (X). Since H L acts transitively on Y P \Z 0, there exists h H L with e gh = e. Let t P with e Z0 t. Then [e, Qt ] = 1 and so gh C t. Thus gh N G ( Q, Q t = N G (P ) = L. Hence g L and (hc) holds. (hd) follows from (hc). 1 The Small World Theorem Given Q! and P P (S). We say that b = 2 for P if b > 1 for P and Y E P If neither b = 1 nor b = 2 for P we say that b is at least three for P. is not abelian. Theorem 1.1 (The Small World Theorem) [the small world theorem] Suppose E! and let P P (S). Then one of the following holds: 1. G has rank 1 or b = 1 or b = 2 for P. 3. A rank three sitiation described below. Proof: Assume that G has rank at least three and that b is at least three. In the exceptional case of the P!-theorems (?? and?? one easily sees that b = 2 for P. Thus P! holds. Also in the exceptionell case of the P! Theorem?? one gets b = 2 for P. Thus (strong) P! holds. We proved Step 1 [ P! and wp!] P! and P! hold. 0.6 gives us a good amount of information about E. We use the notation introduced in 0.6. Since H, L / L, we can apply the amalgam method to the pair (H, L). A non-trivial argument shows Step 2 [offender on V] One of the following holds: 8

9 1. O p (H L)/O p (H) contains a non-trivial quadratic offender on Ṽ. 2. There exists a non-trivial normal subgroup A of H L/O p (H L) and normal subgroups Y P Z 1 Z 2 Z 3 V of H L such that: (a) A and V/Z 3 are isomorphic as F p C H L (Y P )-modules. (b) Z 3 /Z 2 A. (c) [V, A] Z 2 C V (A). In particular, A is a quadratic 2F -offender. (d) [x, A] = Y P for all x Z 3 \ Z 2. (e) Z 1 is a natural SL 2 (q)-module for P C H (Z 0 ). Using?? and?? (and the Z -theorem to deal with the case A = 2) it is not too difficult to derive Step 3 [e-structure] K/O p (K) = SL n (q), (n 3), Sp 2n (q),(n 2) or G 2 (q),( p = 2). Moreover, if W is a maximal submodule of V, then V/W is the natural module for K/O p (K) and H L contains a point-stabilizer on V/W. Let M M( P, P ) with M maximal. Suppose first that M P, P. Then by the Structure Theorem?? M /O p (M ) = SL n (q), n 4 or Sp 2n (q), n 3. Moreover, R := O p (M C O p ( P ) M C K In the case of Sp 2n (q) we have R/O p (R) = Sp 2n 2 (q). So Step 3 implies K/O p (K) = Sp 2m (q) and R H L, a contradiction. Thus M /O p (M ) = SL n (q) and R/O p (R) = SL n 1 (q). Let R be a parabolic subgroup of KS minimal with RS < R. Then R K/O p (R K) = SL n (q) or Sp 2n 2 (q). Let M = M S, R. Since R O p ( P ) R S (M ) S the maximality of M implies that M / L. Now M has a geometry of type A n or C 2n with all its residues classical we conclude that M has a normal subgroup SL n+1 (q) or Sp 2n (q). But this contradicts the assumption that V is abelian. (Remark: One does not have to identify M first to obtain this contradiction. Indeed an easy geometric argument shows that M has rank at most three on M /M C. But then V abelian gives the contradiction Z M 0 abelian. ) We conclude that M P, P. Let P be the unique elemenent of PKS(S) with P N G (O p ( P )). By maximality of M we obtain P, P, P / L. This is rank three situation eluded to in (c). 9

10 Lemma 1.2 [quadratic normal point stabilizer theorem] Let H be a finite group and V a faithful irreducible F p H-module. Let P be a point stabilizer for H on V and A P. Suppose that (i) F (H) is quasi simple and H = A H (ii) A P and A > 2. (iii) A acts quadratically on V. Then one of the following holds: 1. H = SL n (q), Sp 2n (q),or G 2 (q) and V is the natural module. Moreover, 2. p = 2, H is a group of Lie Typ in char p, and H is contained in a long root subgroup of H. 3. Who knows References 10