Overgroups of Irreducible Quasisimple Subgroups in Finite Classical Groups

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1 Overgroups of Irreducible Quasisimple Subgroups in Finite Classical Groups K. Magaard University of Birmingham January 14, 2017

2 Theorem [Aschbacher-Scott 1985] To settle the maximal subgroup problem for general finite groups it suffices to 1. Determine the conjugacy classes of maximal subgroups of the almost simple groups. 2. For every quasisimple group G and all prime divisors l of G determine H 1 (G, V ) for every irreducible F l G-module V. The choices for G are sporadic (for example M ), alternating, of exceptional Lie type, classical (for example PSL n (q)). The possibilities for V are not completely classified.we need all decomposition numbers and minimal fields of definition.

3 Aschbacher s Theorem Let X be a classical group with natural module V = k m, Char(k) = l. Aschbacher defines 8 families C i (X) of geometric subgroups of X. Theorem [Aschbacher 1984] If H X is maximal, then either H C i (X), or H S(X) meaning 1. The generalized Fitting subgroup F (H) is quasisimple; 2. F (H) acts absolutely irreducibly on V ; 3. the action of F (H) on V can not be defined over a smaller field; 4. any bilinear, quadratic or sesquilinear form on V that is stabilized by F (H) is also stabilized by X. Aschbacher calls the collection of subgroups of X which satisfy conditions 1 through 4 above S(X).

4 Some remarks The geometric families C 1 (X) are the stabilizers of subspaces of V. dim(v ) 12 Recent work of Bray, Holt and Roney-Dougal achieves a full classification of conjugacy classes of maximal subgroups. dim(v ) 13 Kleidman and Liebeck classify those situations when a maximal member of C i (X) is not maximal in G. Generally such occurrences are rare.

5 Open Question When are members of S(X) maximal in X? General Observation If H is quasisimple and ϕ IBr l, then there exists a quasisimple classical group X with natural module V affording ϕ and a corresponding homomorphism Φ : H X GL(V ) such that HΦ S(X). Possible obstructions to the maximality of N X (HΦ) in X The definition of S(X) implies that if N X (HΦ) < G < X, then G C 2 (X) C 4 (X) C 6 (X) C 7 (X) S(X). S(X) = Spor Alt CL d CL c EX d EX c

6 Problem 1 For each HΦ, name the possible types of G; i.e. identify the members of C 2 (X) C 4 (X) C 6 (X) C 7 (X) S(X) which may contain HΦ. Problem 2 Name X; i.e., determine the minimal field k of definition of Φ, and the H invariant form(s) on V = k m. Choose X minimal subject to containing HΦ and also determine N GL(V ) (X). Problem 3 For each X N GL(V ) (X) with X X determine N X (HΦ) and N X (HΦ) G. More generally determine the action of N X (HΦ) on the set of G s lying over HΦ. Conclusion: G obstructs the maximality of N X (H) iff N X (H) < X G.

7 Problem 1: Necessary conditions for the existence of obstructions G to the maximality of N X (HΦ) in X. If G C 2 (X), then Φ is an induced representation. If G C 4 (X), then Φ is a tensor (Kronecker) product of two representations of unequal degrees > 1. If G C 7 (X), then Φ is a tensor product of two or more representations of equal degrees; i.e., Φ is tensor induced. If G C 6 (X),then H N X (E) where E is extraspecial resp. symplectic type of order r 1+2s respectively s and dim(v ) = r s resp. 2 s. If G S(X), then (F (G), H, V ) is an irreducible triple; i.e., the irreducible F (G)-module V restricts irreducibly to H.This is a special case of the branching problem.

8 Potential obstructions G to the maximality of H S(X) C 2 HHM DM/NN Seitz HHM Seitz HHM induced C 4 MT BK Stei MT Stei MT tensor prod C 6 MT and Bray inde pend ently r 1+2n Sp 2n (r) C 7 MT tensor ind Spor LM-wip Hu LSS LSS Alt S/KW S/KW S/KW S/KW S/KW S/KW BK/KS BK/KS JS/BK BK/KS BK/KS BK/KS KTS KTS KS/KTS KTS KTS KTS CL d LM-wip Hu D/S/T MRT D/S/T MRT branching CL c LM-wip Hu LSS Seitz LSS Se/S/N rules EX d LM-wip Hu D/S/T MRT D/S/T MRT EX c LM-wip Hu LSS Seitz LSS Seitz G / H Spor Alt CL d CL c EX d EX c H rep n HHM = Hiss Husen Magaard, MT = Magaard Tiep, MRT = Magaard Röhrle Testerman, BK = Bessenrodt Klechshev, DM/NN = Djokovic Malzan / Nett Noeske, D/S/T = Dynkin/Seitz/Testerman, Hu = Husen, JS = Jantzen Seitz, S/KW= Saxl/Kleidman Wales, BK/KS/KST = Brundan Kleshchev/Kleshchev Sheth/Kleshchev Sin Tiep, LSS = Liebeck Saxl Seitz, Se/S/N = Seitz/ Schaefer-Frey / H.N. Nguyen, LM = Le Magaard

9 Irreducible Tensor Decomposable H-modules Recall: If H S(X) is C 4 obstructed, then V is tensor decomposable. Theorem [Seitz, M.-Tiep ]If H S(X) is of Lie type defined over F q, with q = p a and V is a tensor decomposable irreducible H-module, then one of the following is true: p = l and the H-module V is not l-restricted, (Steinberg s Thm.) V is l-restricted, and either p = l = 2 and H is of Lie type B n, C n, F 4 or G 2, or p = l = 3 and H is of type G 2, p l and q 3, H = Sp 2n (5), or Sp 2n (2 a ) H/Z (H) = PSL 3 (4), H = F 4 (2 a ), or 2 F 4 (2 2b+1 ).

10 Not all tensor product factorizations lead to C 4 (X) or C 7 (X) obstructions If H = 4 1 PSL 3 (4), then 8 a 8 d = 64 a. The Frobenius-Schur indicator of 8 a and 8 d is 0, whereas it is + for 64 a. Thus H S(Ω + 64 (l)) for l 2, 3, 5, 7, yet the maximal C 4(Ω + 64 (l)) subgroups are N X (Sp 8 (l) Sp 8 (l)) and N X (Ω ± 8 (l) Ω± 8 (l)). Another example: H = M 24 S(Ω (l)), via it s largest irreducible character χ 26, whenever (l, M 24 ) = 1. Here χ 26 = χ 3 χ 5 = χ 4 χ 5 = χ 3 χ 6 = χ 4 χ 6 The indicator of χ 26 is +, whereas those of χ 4, χ 5, χ 6 χ 7 are 0. If H = M, then 11 of the 194 ordinary irreducible characters are tensor decomposable.of these, 8 are C 4 -obstructed, and 3 are not. For example χ 55 = χ 2 χ 16 is C 4 obstructed, whereas χ 185 = χ 6 χ 17 is not.

11 S(X) obstructions: The case H = M 11 How often does H S(X), imply H maximal in X? Morally this should happen most of the time, but... Theorem If H M 11 and H S(X), then N X (H) is maximal in X iff F (X) = SL 5 (3). ϕ Type of X C 2 C 4,7 C 6 S Lie S altspor ϕ 2 Ω 9 (11) A 11 ϕ 3 SL 10 (11) 2.M 12 ϕ 4 SL 10 (11) 2.M 12 ϕ 5 Ω 11 (11) 2 10 M 11 < 2 10 A 11 M 12 < A 12 ϕ 6 Ω + 16 (11) Spin+ 9 (11) 2.A 11, M 12 ϕ 7 Ω + 44 (11) Ω 9(11) A 11 ϕ 8 Ω 55 (11) 2 54 M 11 < 2 54 A 55 PΩ 11 (11) M 12 < A 12 Table: Obstructions to the maximality of M 11 embeddings; the case l = 11

12 Irreducible Imprimitive Representations Theorem [Hiß, M. 2016] If H is quasisimple and l is not a divisor of H, then all imprimitive elements of IBr l (H) are known. Theorem[Hiß, Husen, M. 2015]If F (H) is quasisimple and V is irreducible and imprimitive, then one of the following is true: 1. H is of Lie type and characteristic p l and V is Harish-Chandra induced or H has an exceptional Schur multiplier, 2. H Lie(p), p = l, F (H) {SL 2 (5), SL 2 (7) = SL 3 (2), Sp 4 (3)}, 3. H = A n and V is from one of three families, or l n, H = 2 A n and the block stabilizer is intransitive, or n F (H) Z (F (H)) {M 11, M 12, M 22, M 24, HS, McL, ON, Co 2, Fi 22, Co 1, Fi 24 }

13 Theorem 7.3 [Hiß, Husen, M.] due to Lusztig Let s G be semisimple such that C G (s) is contained in a proper split Levi subgroup L of G. Let L be a split Levi subgroup of G dual to L. Then every ordinary irreducible character of G contained in E(G, [s]) is Harish-Chandra induced from a character of E(L, [s]). If s is l-regular for some prime l not dividing q, then every irreducible l-modular character of G contained in E l (G, [s]) is Harish-Chandra induced from a Brauer character lying in E l (L, [s]). Corollary: The proportion of primitive irreducible characters in Irr(SL n (q)) approaches 1/n for large values of q.

14 V irreducible and imprimitve does not necessarily imply the existence of a C 2 (X) obstruction. If H = SL 2 (q), l is odd, and V affords R T,Θ with T = q 1, then V is irreducible and imprimitive of dimension q + 1 = H : T p. In fact V = Ind H B (L) is irreducible and imprimitive. Here B is a Borel subgroup of H,and L affords Infl B T (Θ) IBr l (B). Then H k S m GL m (k), and H projects to a transitive subgroup of S m. The module V is a self-dual H-module which implies that H must be in Z 2 S m O m (k); which is impossible if Θ > 2. If Θ = 2, then q is odd and R T,Θ and is not irreducible. Even though roughly half of the irreducible H representations are induced, none are C 2 (X) obstructed.

15 Examples of C 2 (X) type obstructions Let H = SL 3 (5), 5 < l 31, and V = Ind G P (Θ), where P is one of the two classes of the maximal parabolics and 1 Θ is a linear character of P. If Θ = 4, then H SL ɛ 31(l) where l ɛ mod 4. If X = SL ɛ 31(l), then N X (H) is C 2 ( X) obstructed in X iff no element of X induces a graph automorphism on F (H). If Θ = 2, then H X = Ω 31 (l) and Aut(H) SO 31 (l). Also Aut(H) Ω 31 (l) iff l 1 mod 4. So N Ω31 (l)(h) is C 2 (Ω 31 (l)) obstructed iff l 3 mod 4. If l 3 mod 4, then N X (H) is not C 2 ( X) obstructed if SO 31 (l) X.

16 THANK YOU!