Transitive permutation groups of prime-squared degree. Dave Witte

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1 1 Transitive permutation groups of prime-squared degree Dave Witte Department of Mathematics Oklahoma State University Stillwater, OK Main author: Edward Dobson Department of Mathematics and Statistics Mississippi State University Mississippi State, MS June 30, 2000

2 2 Thm (Burnside). Let G be a transitive perm grp of degree p; and P be a Sylow p-subgroup of G. Then either 1) P G;or 2) G is doubly transitive. This leads a complete classification of permutation groups of degree p. 1) G p = p P regular repof Z p, so G N Sp (Z p )=Aff(1,p) = Z p Z p. Choice of G divisor d of p 1 Aff(1,p):G = d. 2) Classification of Finite Simple Groups, all doubly transitive groups are known. Want to do the same for groups of degree p 2.

3 Thm. Let H S p doubly transitive; S be a min l normal subgroup of H; and N = N Sp (S). Then S is simple, S H N, N/S is cyclic, and, after replacing H by a conjugate, either: S = Z p, N = A(1,p), and N/S = Z p 1 ;or S = A p, N = S p, and N/S = Z 2 ;or p =11and S = H = N = PSL(2, 11); or p =11and S = H = N = M 11 ;or p =23and S = H = N = M 23 ;or p = qd 1 q 1 for some q = rm and d, and S = PSL(d, q), N = PΓL(d, q), N/S = Z m. Conj. primes of the form (q d 1)/(q 1). E.g. any Mersenne prime p =2 d 1. 3

4 4 Thm (Wielandt). Let G S p 2 transitive. Then 1) P G;or 2) G is doubly transitive; or 3) G is imprimitive; or 4) G has an imprimitive subgroup of index 2, and P = Z p Z p, for any P Syl p (G). Defn. Let G act transitively on Ω. B Ωisablock if 1 < B < Ω and g G, either Bg = B or Bg B =. I.e., B = { Bg g G } partitions Ω. G is imprimitive if block B. 2) Use the classification of dbly transitive grps. a) Z 2 p <H Aff(2,p) b) A p 2, S p 2 c) PSL(d, q) H PΓL(d, q) a) C. Hering [Huppert III, Rmk. XII.7.5, p. 386]

5 5 Analogue of Burnside s Theorem. Main Thm. G S p 2 transitive, P Syl p (G) 1) P G;or 2) G is doubly transitive; or 3) exp(p )=p and P {p 2,p p };or 4) P = p p+1. Prop. G S p 2 imprim, P elem abel of order p 2 G is equivalent to a subgroup of S p S p.

6 6 Defn. Let H, L S p. Then H L = H L p S p 2, where (l 1,...,l p ) h =(l 1 h 1,...,l ph 1 ). H L acts faithfully on Z p Z p by (i, j) h =(i h,j) (i, j) (l 1,...,l p ) =(i, j l i ) {i} Z p is a block, so H L is imprimitive. Note that e L fixes each block. is equiva- Exer. Any imprimitive subgroup of S p 2 lent to a subgroupof S p S p. Exer. Z p Z p is a Sylow p-subgrpof S Zp Z p = S p 2. For 0 i p,! K i e p Z p = (Z p ) p, s.t. K i Z p Z p and K i = p i. E.g., K p = e p Z p and K i 1 =[Z p Z p,k i ]. Exer. If P Z p Z p, P K p, and P = p i, then P K p = K i 1.

7 Fact. Z p K p 1 is the unique transitive subgroup of Z p Z p with exponent p and order p p. K p 1 = {(z 1,...,z p ) (Z p ) p p i=1 z i =0}. Fact. For 0 i<p, N e Sp (K i )=Z p K p. Prop. G S p 2 imprim, P = Z p K p 1 H S p Z p, s.t. G is equiv to H K p 1. Eg. Let H, L S p transitive. G = H L S p 2 is imprim, and G p = p p+1. Prop. G S p 2 imprim, G p = p p+1. 1) H, L S p transitive, such that L is simple; 2) K/L p ( N Sp (L)/L ) p H-invariant; 3) φ: H N Sp (L) p /K crossed homomorphism; such that G is equivalent to { (h, v) H N Sp (L) p φ(h) =vk } S p S p. 7

8 8 Remaining problems. For PSL(d, q) H PΓL(d, q): 1) Find H-invariant subgroups K of (Z r k) p (where q = r m ). (k = 1 done [Bardoe-Sin]) 2) For each subgrp K, calculate H 1( H, (Z r k) p /K ) and find an explicit rep of each coho class. For H = PSL(2, 11), M 11, M 23 : Problem 2.

9 9 Main Thm. Let G S p 2, P Syl p (G). Then either P G, or G is doubly transitive, or P = p p+1, or exp(p )=p and P {p 2,p p }. Proof. Assume G is not dbly trans, P p p+1, and either exp(p )=p 2 or P / {p 2,p p }. P = Z p Z p, so Wielandt s Thm G imprim. Let B be a block system for G. Defn. Fix G (B) ={ g G B B,Bg= B }. Strategy: show Fix G (B) and G/B are solvable. Lem. G solvable, imprimitive, P p p+1 P G.

10 Lem. G solvable, imprimitive, P p p+1 P G. 10 Proof. We always assume P Z p Z p. We have Fix P (B) =K p P = K i for some i. Fix G (B) solvable G N Sp S p (K i ). Because N Sp S p (K i ) normalizes B B K i B = K p, then G N Sp S p (K p )=S p Aff(1,p). G/B solvable G Aff(1,p) Aff(1,p). Because p 1 < P <p p+1, we have 0 <i<p, so N e Sp (K i )=Z p K p. Therefore G ( Aff(1,p) Z p) Kp. This has a unique Sylow p-subgroup.

11 11 Lem. If P p p+1 and P = Z p Z p, then Fix G (B) is solvable. Proof. Let N Fix G (B) minimal normal in G. Suppose N is not a p-group, so N = L 1 L m is a direct product of nonabelian simple groups. Defn. supp(l i )={ B B L i B e }. [L i,l j ]=e supp(l i ) supp(l j )=. G-action on B is primitive supp(l i ) {1,p}; m {1,p}. m = p (Z p ) p N P,so G p = p p+1. m =1 Fix G (B) N e Sp (L 1 ) = N SB (L 1 B ) (because C SB (L 1 B )=e) so Fix G (B) p = p. Therefore P = p 2. We now know P = Z p 2 (and m = 1).

12 12 Suppose P is cyclic ( = Z p 2), and m =1. Let g be any p -element of N L1 (P p ). Because P p = K 1, we know that g Z pk p, so g normalizes Z p Z p. Therefore, [P, g] isap-subgroupof Fix G (B). Each of P and g normalizes P p, and P p is a Sylow p-subgroupof Fix G (B), so [P, g] P p. Therefore g normalizes P, and centralizes P/P p. Because g is a p -element, then g centralizes P. Thus, P p is in the center of its normalizer in L 1, so L 1 has a normal p-complement. Because L 1 is simple, this is nonsense.

13 13 Lem. Assume Fix G (B) is solvable; and G/B is not solvable. Then P {p 2,p p,p p+1 }. Proof. Let K Syl p ( FixG (B) ),sok G. Then G N Sp S p (K), and K p N Sp S p (K), so G N Sp S p (K p )=S p Aff(1,p). Because K is invariant under Z p e, and the Z p -invariant subgroupof order p i is unique, we know that K is normalized by Aff(1,p) e. From the list of doubly transitive groups, we see that Aff(1,p) is maximal in S p. Therefore N Sp Aff(1,p)(K)/B = S B. Then the following result of modular representation theory implies that K {1,p,p p 1,p p }.

14 14 Prop. Let χ: S p 1 Z p be a homomorphism. Nontrivial invariant subspaces of Ind S p S p 1 χ have either dimension 1 or codimension 1. More generally, coding theorists have calculated the automorphism group of any code admitting Aff(d, q).

15 15 Lem. Assume Fix G (B) has a unique Sylow p-subgroup Q; exp(p )=p 2 ; and P p p+1 Then P G. Proof. We have G Q (S p Z p) (Z p ) p K p 1 Let φ: G/Q Aff(1,p) = S p Aff(1,p) (the projection). WMA φ not faithful; else G/B solvable, so done. G/B primitive ker φ transitive ker φ p = p; therefore, the image of φ is a p -group. Therefore, every p-element of G is in the kernel of φ, sop Z p K p 1. So every element of P has order p.

16 M. Bardoe and P. Sin, The permutation modules for GL(n +1, F q ) acting on P n (F q ) and F n+1 q, London Math. Soc. (to appear). P.J. Cameron, Finite Permutation groups and finite simple groups, Bull. London Math. Soc , W. Feit, Some consequences of the classfication of finite simple groups, Proc. Symp. Pure Math. 37 (1980) B. Cooperstein and G. Mason, eds, The Santa Cruz Conference on Finite Groups, Amer. Math. Soc., W. C. Huffman, Codes and Groups, in V.S. Pless and W.C. Huffman, eds., Handbookof Coding Theory, vol. 2, Elsevier, 1998, pp A. S. Kleshchev and A. A. Premet, On second degree cohomology of symmetric and alternating groups, Comm. Alg. 21(2) (1993) B. Mortimer, The modular permutation representations of the known doubly transitive groups, Proc. London Math. Soc. (3) 41 (1980) H. Wielandt, Finite Permutation Groups, Academic Press, New York, H. Wielandt, Permutation groups through invariant relations and invariant functions, in: B. Huppert and H. Schneider, eds., Mathematische Werke = Mathematical Works / Helmut Wielandt, vol. 1, de Gruyter, Berlin, 1994, pp QA3.W V1 [Thm. 16.2, 16.3] 16