REPRESENTATIONS OF FINITE GROUPS OF LIE TYPE

Transcription

1 REPRESENTATIONS OF FINITE GROUPS OF LIE TYPE By HUNG NGOC NGUYEN A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA

2 c 2008 Hung Ngoc Nguyen 2

3 ACKNOWLEDGMENTS I am indebted to my advisor, Prof. Pham Huu Tiep, who has given me constant encouragement and devoted guidance over the last four years. I would like to thank Prof. Alexander Dranishnikov, Prof. Peter Sin, Prof. Meera Sitharam and Prof. Alexandre Turull, for serving on my supervisory committee and giving me valuable suggestions. I am also pleased to thank the Department of Mathematics at the University of Florida and the College of Liberal Arts and Sciences for honoring me with the 4-year Alumni Fellowship and the CLAS Dissertation Fellowship. Finally, I thank my parents, my sister, and my wife for all their love and support. 3

4 TABLE OF CONTENTS page ACKNOWLEDGMENTS LIST OF TABLES LIST OF SYMBOLS ABSTRACT CHAPTER 1 INTRODUCTION Overview and Motivation Results IRREDUCIBLE RESTRICTIONS FOR G 2 (Q) Preliminaries Degrees of Irreducible Brauer Characters of G 2 (q) Proofs Small Groups G 2 (3) and G 2 (4) IRREDUCIBLE RESTRICTIONS FOR SUZUKI AND REE GROUPS Suzuki Groups Ree Groups IRREDUCIBLE RESTRICTIONS FOR 3 D 4 (Q) Basic Reduction Restrictions to G 2 (q) and to 3 D 4 ( q) Restriction to Maximal Parabolic Subgroups LOW-DIMENSIONAL CHARACTERS OF THE SYMPLECTIC GROUPS Preliminaries Strategy of the Proofs Centralizers of Semi-simple Elements Unipotent Characters Unipotent Characters of SO 2n+1 (q) and P CSp 2n (q) Unipotent Characters of P (CO 2n(q) 0 ) Unipotent Characters of P (CO + 2n(q) 0 ) Non-unipotent Characters LOW-DIMENSIONAL CHARACTERS OF THE ORTHOGONAL GROUPS Odd Characteristic Orthogonal Groups in Odd Dimension

5 6.2 Even Characteristic Orthogonal Groups in Even Dimension Odd Characteristic Orthogonal Groups in Even Dimension Groups Spin ± 12(3) REFERENCES BIOGRAPHICAL SKETCH

6 Table LIST OF TABLES page 2-1 Degrees of irreducible complex characters of G 2 (q) Degrees of 2-Brauer characters of G 2 (q), q odd Degrees of 3-Brauer characters of G 2 (q), 3 q Degrees of l-brauer characters of G 2 (q), l 5 and l q Degrees of l-brauer characters of G 2 (q), l 5 and l q Fusion of conjugacy classes of P in G 2 (3) Fusion of conjugacy classes of P in G 2 (4) Fusion of conjugacy classes of Q in G 2 (4) Low-dimensional unipotent characters of Sp 2n (q), n 6, q odd, I Low-dimensional irreducible characters of Sp 2n (q), n 6, q odd, II Low-dimensional irreducible characters of Sp 2n (q), n 6, q odd, III

7 LIST OF SYMBOLS, NOMENCLATURE, OR ABBREVIATIONS Let G be a finite group, l be a prime number, and F be an algebraically closed field of characteristic l. We write Z(G) : for the center of G O l (G) : for the maximal normal l-subgroup of G Irr(G) : for the set of irreducible complex characters of G, or the set of irreducible CG-representation IBr l (G) : for the set of irreducible l-brauer characters of G, or the set of absolutely irreducible G-representation in characteristic l d l (G) : for the smallest degree of irreducible FG-modules of dimension > 1 m l (G) : for the largest degree of irreducible FG-modules d 2,l (G) : for the second smallest degree of irreducible FG-modules of dimension > 1 d C (G), m C (G), d 2,C (G) : for d 0 (G), m 0 (G), d 2,0 (G), respectively χ : for the restriction of a character χ Irr(G) to l-regular elements of G Furthermore, we write GL + n (q) : for GL n (q) GL n (q) : for GU n (q) CSp 2n (q) : see the discussion before Lemma P CSp 2n (q) : for the quotient of CSp 2n (q) by its center CO 2n(q), ± CO 2n(q) ± 0 : see the discussion before Lemma P (CO 2n(q) ± 0 ) : for the quotient of CO 2n(q) ± 0 by its center Spin ± n (q) : for the Spin group. Modulo some exceptions, it is the universal cover of the simple orthogonal group P Ω ± n (q) semi-simple element: for an element which is diagonalizable unipotent character: see the discussion at the beginning of 5.1 7

8 Chair: Pham Huu Tiep Major: Mathematics Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy REPRESENTATIONS OF FINITE GROUPS OF LIE TYPE By Hung Ngoc Nguyen August 2008 Let G be a finite quasi-simple group of Lie type. One of the important problems in modular representation theory is to determine when the restriction of an absolutely irreducible representation of G to its proper subgroups is still irreducible. The solution for this problem is a key step towards classifying all maximal subgroups of finite classical groups. We solve this problem for the cases when G is a Lie group of the following types: G 2 (q), 2 B 2 (q), 2 G 2 (q), and 3 D 4 (q). One of the main tools to approach the above problem, and many others, is the classification of low-dimensional representations of finite groups of Lie type. Low-dimensional complex representations were first studied in [70] for finite classical groups and then in [50] for exceptional groups. We extend the results in [70] for symplectic and orthogonal groups to a larger bound. More explicitly, we classify irreducible complex characters of the symplectic groups Sp 2n (q) and orthogonal groups Spin ± n (q) of degrees up to the bound D, where D = (q n 1)q 4n 10 /2 for symplectic groups, D = q 4n 8 for orthogonal groups in odd dimension, and D = q 4n 10 for orthogonal groups in even dimension. 8

9 CHAPTER 1 INTRODUCTION 1.1 Overview and Motivation Finite primitive permutation groups have been studied since the pioneering work of Galois and Jordan on group theory; they have had important applications in many different areas of mathematics. If G is a primitive permutation group with a point stabilizer M then M is a maximal subgroup of G. Thanks to works of Aschbacher, O Nan, Scott [3], and of Liebeck, Praeger, Saxl, and Seitz [47], [48], most problems involving such a G can be reduced to the case where G is a finite classical group. If G is a finite classical group, a fundamental theorem of Aschbacher [2] states that any maximal subgroup M of G is a member of either one of eight families C i, 1 i 8, of naturally defined subgroups of G or the collection S of certain quasi-simple subgroups of G. Conversely, if M 8 i=1c i, then the maximality of M has been determined by Kleidman and Liebeck in [40]. It remains to determine which M S are indeed maximal subgroups of G. This question leads to a number of important problems concerning modular representations of finite quasi-simple groups. One of them is the irreducible restriction problem. Problem A. Let F be an algebraically closed field of characteristic l. Classify all triples (G, V, M) where G is a finite quasi-simple group, V is an irreducible FG-module of dimension greater than one, and M is a proper subgroup of G such that the restriction V M is irreducible. We recall that a finite group G is called quasi-simple if G = [G, G] and G/Z(G) is simple. Moreover, any non-abelian finite simple groups is one of alternating groups, 26 sporadic groups, or finite simple groups of Lie type. The solution for Problem A when G/Z(G) is a sporadic group is computational. When G is a cover of symmetric or alternating group, Problem A is quite complicated and almost done in [6], [41], [42], and [57]. Our main focus is on the case where G is a finite 9

10 group of Lie type in characteristic different from l. This has been solved recently in [43] when G is of type A. One of the important ingredients needed to solve Problem A is classification of low-dimensional irreducible representations of finite groups of Lie type. This explains why we have been working simultaneously on the irreducible restriction problem and the problem of determining low-dimension representations. Suppose that G = G(q) is a finite group of Lie type defined over a field of q elements, where q = p n is a prime power. Lower bounds for the degrees of nontrivial irreducible representations G in cross characteristic l (i.e. l p) were found by Landazuri, Seitz and Zalesskii in [45], [59] and improved later by many people. These bounds have proved to be very useful in various applications. We are interested in not only the smallest representation, but more importantly, the low-dimensional representations. Low-dimensional complex representations was first studied by Tiep and Zalesskii [70] for finite classical groups and then by Lübeck [50] for exceptional groups. For representations over fields of cross characteristics, this problem has been studied recently in [5], [23] for SL n (q); [25], [32] for SU n (q); [25] for Sp 2n (q) with q odd; and [24] for Sp n (q) with q even. Let H be one of these groups, and let the smallest degree of nontrivial irreducible representations of H in cross characteristic be denoted by d(h). The main purpose of these papers is to classify the irreducible representations of H of degrees close to d(h), and to prove that there is a relatively big gap between the degrees of these representations and the next degree. A common application of low-dimensional representations is as follows. Suppose we want to prove some statement P involving representations ϕ of finite groups G. First, one tries to reduce to the case when G is quasi-simple. Then, with G being quasi-simple, one shows that P holds if ϕ has degree greater than some certain bound d. At this stage, results on low-dimensional representations should be applied to identify the representation ϕ of degree d and to establish P directly for these representations. We refer to 10

11 [68] for a survey of recent progress and more detailed applications of low-dimensional representations of finite groups of Lie type. 1.2 Results In this dissertation, we have completed Problem A for the cases when G is a Lie type group of the following types: G 2 (q), 2 B 2 (q), 2 G 2 (q) and 3 D 4 (q). The smallest degree of non-trivial irreducible characters of G 2 (q) was determined by Hiss. Based on results of Hiss and Shamash about character tables, decomposition numbers, and Brauer trees of G 2 (q), we found the second smallest degree of irreducible characters of G 2 (q) in cross characteristic. This second degree plays a crucial role in solving Problem A when G = G 2 (q). Another ingredient is the modular representation theory of various large subgroups of G 2 (q), such as SL 3 (q) [64] and SU 3 (q) [20]. The following theorem is the focus of chapter 2. Theorem B. Let G = G 2 (q), q = p n, q 5, p a prime number. Let ϕ be an irreducible character of G in cross characteristic l and M a maximal subgroup of G. Assume that ϕ(1) > 1. Then ϕ M is irreducible if and only if one of the following holds: (i) q 1(mod 3), M = SL 3 (q) : 2 and ϕ is the unique character of the smallest degree q 3 1. (ii) q 1(mod 3), M = SU 3 (q) : 2 and ϕ is the unique character of the smallest degree. In this case, ϕ(1) = q 3 when l = 3 and ϕ(1) = q when l 3. The solution of Problem A for Suzuki and Ree groups is less complicated than G 2 (q) and straightforward. The next two theorems are proved in chapter 3. Theorem C. Let G = Sz(q) = 2 B 2 (q) be the Suzuki group where q = 2 n and n 3 is odd. Let ϕ be an irreducible character of G in characteristic l 2 and M a maximal subgroup of G. Assume that ϕ(1) > 1. Then ϕ M is irreducible if and only if M is G-conjugate to the maximal parabolic subgroup of G and ϕ is the reduction modulo l of any of the two irreducible complex characters of degree (q 1) q/2. 11

12 Theorem D. Let G = 2 G 2 (q) be the Ree group where q = 3 n and n 3 is odd. Let ϕ be an irreducible character of G in characteristic l 3 and M a maximal subgroup of G. Assume that ϕ(1) > 1. Then ϕ M is also irreducible if and only if M is G-conjugate to the maximal parabolic subgroup of G and ϕ is the nontrivial constituent (of degree q 2 q) of the reduction modulo l = 2 of the unique irreducible complex character of degree q 2 q + 1. Unlike the cases of G 2 (q), 2 B 2 (q), and 2 G 2 (q), the cross-characteristic representation theory of 3 D 4 (q) is still not very well understood. To solve Problem A for the case 3 D 4 (q), we need to apply fundamental results of Broué and Michel [4] on unions of l-blocks, as well as results of Geck [21] on basic sets of Brauer characters in l-blocks. Recent results of Himstedt [26] on character tables of parabolic subgroups of 3 D 4 (q) have also proved useful. It turns out that there is no irreducible restrictions for 3 D 4 (q) and this will be shown in chapter 4. Theorem E. Let G = 3 D 4 (q) and let ϕ be any irreducible representation of G in characteristic l coprime to q. If M is any proper subgroup of G and deg(ϕ) > 1, then ϕ M is reducible. Now we move on to the problem of determining low-dimensional complex representations, or equivalently, low-dimensional complex characters. Let we temporarily denote by G either the symplectic groups Sp 2n (q) or the orthogonal groups Spin ± n (q). The smallest nontrivial complex character of G was determined by Tiep and Zalesskii in [70]. Furthermore, when G is the odd characteristic symplectic group, they classified all irreducible complex characters of degrees less than (q 2n 1)/2(q + 1). It turns out that, up to this bound, G has four irreducible characters of degrees (q n ± 1)/2, which are so-called Weil characters, and the smallest unipotent character of degree (q n 1)(q n q)/2(q + 1). We want to extend these results to a larger bound. More explicitly, we classify the irreducible complex characters of G of degrees up to the bound D, where D = (q n 1)q 4n 10 /2 for symplectic groups, D = q 4n 8 for orthogonal groups in odd dimension, and D = q 4n 10 for orthogonal groups in even dimension. When q is 12

13 even, the low-dimensional complex characters of Sp 2n (q) Spin 2n+1 (q) were classified up to a good bound enough (q 2n 1)(q n 1 1)(q n 1 q 2 )/2(q 4 1) (generically) in [24] and therefore it is not in our consideration. Our main results on this topic are the following theorems, which will be proved in chapters 5 and 6. We note that ± could be understood as ±1 and vice verse, depending on the context. Theorem F. Let χ be an irreducible complex character of G = Sp 2n (q), where n 6 and q is an odd prime power. Then either χ(1) > (q n 1)q 4n 10 /2 or χ belongs to a list of q q + 36 characters in Tables 5-1, 5-2, 5-3 (at the end of chapter 5). Theorem G. Let χ be an irreducible complex character of G = Spin 2n+1 (q), where n 5 and q is an odd prime power. Then χ(1) = 1 (1 character), χ(1) = (q 2n 1)/(q 2 1) (1 character), χ(1) = q(q 2n 1)/(q 2 1) (1 character), χ(1) = (q n + α 1 )(q n + α 2 q)/2(q + α 1 α 2 ) (4 characters, α 1,2 = ±1), χ(1) = (q 2n 1)/(q + α) ((q + α 2)/2 characters for each α = ±1), or χ(1) > q 4n 8. Theorem H. Let χ be an irreducible complex character of G = Ω α 2n(q), where n 5, α = ±, (n, q, α) (5, 2, +), and q is a power of 2. Let (q 1)(q 2 + 1)(q 3 1)(q 4 + 1), n = 5, α =, D(n, q, α) = q 4n , otherwise. Then χ(1) = 1 (1 character), χ(1) = (q n α)(q n 1 + αq)/(q 2 1) (1 character), χ(1) = (q 2n q 2 )/(q 2 1) (1 character), χ(1) = (q n α)(q n 1 + αβ)/(q β) ((q β 1)/2 characters for each β = ±1), or χ(1) D(n, q, α). Theorem I. Let χ be an irreducible complex character of G = Spin α 2n(q), where α = ±, n 5, and q is an odd prime power. Let (q 1)(q 2 + 1)(q 3 1)(q 4 + 1), n = 5, α =, D(n, q, α) = q 4n , otherwise. 13

14 Then χ(1) = 1 (1 character), χ(1) = (q n α)(q n 1 + αq)/(q 2 1) (1 character), χ(1) = (q 2n q 2 )/(q 2 1) (1 character), χ(1) = (q n α)(q n 1 α)/2(q + 1) (2 characters), χ(1) = (q n α)(q n 1 + α)/2(q 1) (2 characters), χ(1) = (q n α)(q n 1 + αβ)/(q β) ((q β 2)/2 characters for each β = ±1), or χ(1) D(n, q, α). Furthermore, when (n, α) = (5, ), G has exactly q characters of degree D(5, q, ), 1 character of degree q 2 (q 4 + 1)(q 5 + 1)/(q + 1), and no more characters of degrees up to q

15 CHAPTER 2 IRREDUCIBLE RESTRICTIONS FOR G 2 (Q) Let G be a finite group of Lie type G 2 (q) defined over a finite field with q = p n elements, where p is a prime number and n is a positive integer. Let M be a maximal subgroup of G and let ϕ be an absolutely irreducible representation of G in cross characteristic l. The purpose of this chapter is to find all possibilities of ϕ and M such that ϕ M is also irreducible. In order to do that, we use the results about maximal subgroups, character tables, blocks, and Brauer trees of G obtained by many authors. The list of maximal subgroups of G is determined by Cooperstein in [13] for p = 2 and Kleidman in [39] for p odd. The complex character table of G is determined by Chang and Ree ([10]) for p 5, by Enomoto ([16]) for p = 3, and by Enomoto and Yamada ([17]) for p = 2. In a series of papers [29], [33], [34], [61], [62], and [63], Hiss and Shamash have determined the blocks, Brauer trees and (almost completely) the decomposition numbers for G. In order to solve our problem, it turns out to be useful to know the degrees of low-dimensional irreducible Brauer characters of G. The smallest (or the first) degree of non-trivial characters of G was determined by Hiss and will be used here. For the second smallest degree, we compare directly the degrees of irreducible Brauer characters and get the exact formula for it which is given in Theorem Using the first degree of G, we can exclude many maximal subgroups of G by Reduction Theorem The remaining maximal subgroups are treated individually by various tools, including the second degree and modular representation theory of large subgroups of G such as SL 3 (q) [64] and SU 3 (q) [20]. This chapter is organized as follows. In 1, we state some lemmas which will be used later. In 2, we collect the degrees of irreducible Brauer characters and give formulas for the first and second degrees of G 2 (q). 3 is devoted to prove Theorem B. Small groups G 2 (3) and G 2 (4) and their covers will be treated in 4. 15

16 2.1 Preliminaries We record a few statements which will be used throughout until chapter 4. Some of these statements are well-known but we include their proofs for completeness. Hereafter, F is an algebraically closed field of characteristic l. Lemma Let G be a finite group. Suppose V is an irreducible FG-module of dimension greater than one and H is a proper subgroup of G such that the restriction V H is irreducible. Then H/Z(H) mc (H) m l (H) dim(v ) d l (G). Proof. Suppose ϕ IBr l (H) of degree m l (H). Then there exists χ Irr(H) such that d χϕ 0, where d χϕ is the decomposition number associated with χ and ϕ. Since χ = ϕ IBr l (H) d χϕ ϕ, it follows that m C (H) χ(1) = χ(1) ϕ(1) = m l (H). Other inequalities are obvious. Lemma Let G be a simple group and V an irreducible FG-module of dimension greater than one. Then Z G (V ) := {g G g V = λ Id V for some λ F} = 1. Proof. Let X be the FG-representation associated with V. Then Ker(X) = {g G X(g) = X(1)} is a normal subgroup of G. Since G is simple, Ker(X) is either trivial or the whole group G. The case Ker(X) = G cannot happen since X is non-trivial and irreducible. So, Ker(X) = 1. Assume g 0 is any element in Z G (V ). Then X(g 0 ) is a scalar matrix and therefore it commutes with X(g) for every g G. Hence g 0 commutes with g for every g G since Ker(X) = 1. In other words, Z G (V ) Z(G), which implies the lemma. Lemma Let G be a simple group and V an irreducible FG-module of dimension greater than one. Suppose H is a subgroup of G such that V H is irreducible. Then Z(H) = C G (H) = 1. Proof. This is a corollary of Lemma

17 Lemma Let G be a finite group and 1 A H G. Let V be a faithful irreducible FG-module. (i) Suppose that C V (A) := {v V a(v) = v for every a A} = 0. Then V H is reducible. (ii) Suppose that O l (H) 1. Then V H is reducible. Proof. (i) We will argue by contradiction. Suppose that V H is irreducible. By Clifford s theorem, V A = e t i=1 V i where e is the multiplicity of V 1 in V and {V 1,..., V t } is the H-orbit of V 1 under the action of H on the set of all irreducible FA-modules. Since A acts trivially on C V (A) 0, at least one of the V i is the trivial A-module. It follows that all V 1,..., V t are trivial A-modules and therefore A acts trivially on V. This contradicts the faithfulness of V. (ii) Assume the contrary: V H is irreducible. Then it is well-known that O l (H) acts trivially on V. It follows that C V (O l (H)) = V. This and the hypothesis O l (H) 1 lead to a contradiction by (i). Lemma Let G be a finite group and χ Irr(G). Let H be a normal l -subgroup of G and suppose that χ H = t i=1 θ i, where θ 1 is irreducible and θ 1, θ 2,..., θ t are the distinct G-conjugates of θ 1. Then χ is also irreducible. Proof. We have χ H = t i=1 θ i as H is an l -group. Let ψ be an irreducible constituent of χ. Then there is an i {1, 2,..., t} such that θ i is an irreducible constituent of ψ H. Therefore, by Clifford s theory, all θ 1, θ 2,..., θ t are contained in ψ H. This implies that χ H = ψ H. So χ = ψ as desired. Lemma Let τ be an automorphism of a finite group G which fuses two conjugacy classes of subgroups of G with representatives M 1, M 2. Let A and B be subsets of IBr l (G) such that τ(a) = B, τ(b) = A. (i) Assume ϕ M1 is irreducible (resp. reducible) for all ϕ A B. Then ϕ M2 is irreducible (resp. reducible) for all ϕ A B. 17

18 (ii) Assume ϕ M1 is irreducible for all ϕ A and ϕ M1 is reducible for all ϕ B. Then ϕ M2 is reducible for all ϕ A and ϕ M2 is irreducible for all ϕ B. Proof. It is enough to show that ϕ M1 is irreducible (resp. reducible) for all ϕ A if and only if ϕ M2 is irreducible (resp. reducible) for all ϕ B. This is true because for every ϕ A, we have ϕ M1 = ψ τ τ(m2 ) = (ψ M2 ) τ for some ψ B. Lemma Let G be a finite group. Suppose that the universal cover of G is M.G where M is the Schur multiplier of G of prime order. Then every maximal subgroup of M.G is the pre-image of a maximal subgroup of G under the natural projection π : M.G G. Proof. Let H be a maximal subgroup of M.G. Then π(h) is a subgroup of G. We will show that π(h) is a proper subgroup of G. Assume the contrary π(h) = G. First suppose that there exists an element g G such that π 1 (g) H consists of at least two different elements, say g 1 and g 2. Then g 2 = z.g 1 where 1 z M. Since M is prime, < z >= M and therefore M H. It follows that H = M.G and we get a contradiction. Next we suppose that for each g G, π 1 (g) H consists of exactly one element. Then π H is an isomorphism. In other words, H G and M.G = M : H becomes a split extension. Without loss we may identify H with G. Now every projective complex irreducible representation Φ of G lifts to a linear representation Ψ of M : G. Then Φ also lifts to the linear representation Ψ G of G. Thus the Schur multiplier M of G is trivial, a contradiction again. We have shown that π(h) is a proper subgroup of G. Then π(h) is contained in a maximal subgroup of G, say K. We have H π 1 (K). Since H is maximal, H = π 1 (K) and we are done. Lemma [18] Let G be a finite group and H G. Suppose G : H = p is prime and χ IBr l (G). Then either (i) χ H is irreducible or (ii) χ H = p i=1 θ i, where the θ i are distinct and irreducible. 18

19 Lemma [35, p. 190] Let G be a finite group and H be a normal subgroup of G. Let χ Irr(G) and θ Irr(H) be a constituent of χ H. Then χ(1)/θ(1) divides G/H. Lemma [56] Let B be an l-block of group G. Assume that all χ B Irr(G) are of the same degree. Then B IBr l (G) = {φ} and χ = φ for every χ B Irr(G). Lemma [69, Theorem 1.6] Let G be a finte group of Lie type, of simply connected type. Assume that G is not of type A 1, 2 A 2, 2 B 2, 2 G 2, and B 2. If Z is a long-root subgroup and V is a nontrivial irreducible representation of G, then Z must have nonzero fixed points on V. 2.2 Degrees of Irreducible Brauer Characters of G 2 (q) In this section, we collect the degrees of irreducible characters of G 2 (q) (both complex and Brauer characters) obtained by many people. Then we will recall the value of d l (G) and determine the value of d 2,l (G) when l is coprime to q. The degrees of irreducible complex characters of G 2 (q) can be read off from [10], [16], [17] and are listed in Table 2-1. From this table, we get d C (G) = q 3 + 1, q 1(mod 3), q 3 1, q 2(mod 3), q 4 + q 2 + 1, q 0(mod 3), (2.1) and d 2,C (G) = 1 q(q 6 1)2 (q 2 q + 1), p = 2, 3 or q = 5, 7, q 4 + q 2 + 1, p 5, q > 7, (2.2) for every q 5. Moreover, Irr(G) contains a unique character of degree larger than 1 but less than d 2,C (G) and this character has degree d C (G). The degrees of irreducible l-brauer characters of G 2 (q) when l G and l q can be read off from [29], [33], [34], [61], [62], and [63]. They are listed in Tables 2-2, 2-3, 2-4, and 2-5. Comparing these degrees directly, we easily get the value of d l (G) when l G, l q. Combining this value with formula 2.1, we get that if q 5 and l q then 19

20 d l (G) = q 3 + 1, q 1(mod 3), l 3, q 3, q 1(mod 3), l = 3, q 3 1, q 2(mod 3), l, q 4 + q 2, q 0(mod 3), l = 2, q 4 + q 2 + 1, q 0(mod 3), l 2. (2.3) When l G, we know that IBr l (G) = Irr(G) and the value of d 2,C (G) is given in formula (2.2). Formula (2.2) and direct comparison of the degrees of irreducible l-brauer characters of G when l G yield the following theorem. We omit the details of this direct computation. Theorem Let q 5. We have 1 d 2,2 (G) = q(q 6 1)2 (q 2 q + 1), p = 3 or q = 5, 7, q 4 + q 2, p 5, q 11, d 2,3 (G) = 1 6 q(q 1)2 (q 2 q + 1), q = 5, 7, or p = 2, q 1(mod 3), = q 4 + q 2 + 1, p 5, q 1(mod 3) and q 11, q 4 q 3 + q 2, q 13 and q 1(mod 3), and if l = 0 or l 5, l q then d 2,l (G) = d 2,C (G) = 1 q(q 6 1)2 (q 2 q + 1), p = 2, 3 or q = 5, 7, q 4 + q 2 + 1, p 5, q 11. Moreover, IBr l (G) contains a unique character denoted by ψ of degree larger than 1 but less than d 2,l (G) and this character has degree d l (G). Also from the results about blocks and Brauer trees of G 2 (q), we see that if 3 q then ψ = X 32 in all cases except when l = 3 and q 1(mod 3) where ψ = X 32 1 G. If 3 q then ψ = X 22 except when l = 2 where ψ = X 22 1 G. We also notice that ϕ 18 = X 18 is irreducible in all cases. 20

21 2.3 Proofs In the next Lemmas, we use the notation in [17], [20], [30], and [64]. Lemma equal to χ ( q 2 1 ) 3 q 3 1. (i) When q 1(mod 3), the character X 32 SL3 (q) is irreducible and (ii) When q 1(mod 3), the character X 32 SU3 (q) is irreducible and equal to χ ( q 2 1 ) 3 q Proof. (i) First we assume q is odd. We need to find the fusion of conjugacy classes of SL 3 (q) in G. By [64, p. 487], SL 3 (q) has the following conjugacy classes: C (0) 1, C (0) 2, C (0,0) 3, C (k) 4, C (k) 5, C (k,l,m) 6, C (k) 7 and C (k) 8. For any x G, we denote by K(x) the conjugacy class containing x. Then we have C (0) 1 K(1), C (0) 2 K(u 1 ), C (0,0) 3 K(u 6 ), C (k) 4 K(k 2 ) K(h 1b ), C (k) 5 K(k 2,1 ) K(h 1b,1 ), C (k,l,m) 6 K(h 1 ), C (k) 7 K(h b ) K(h 2b ) and C (k) 8 K(h 3 ). Taking the values of X 32 and χ ( q χ ( q 2 1 ) 3 q 3 1 χ ( q 2 1 ) 3 q 3 1 χ ( q 2 1 ) 3 q 3 1 χ ( q 2 1 ) 3 q 3 1 χ ( q 2 1 ) 3 q 3 1 χ ( q 2 1 ) 3 q 3 1 χ ( q 2 1 ) 3 q 3 1 χ ( q 2 1 ) 3 q 3 1 (C(0) (C(0) 1 ) = X 32 (1) = q 3 1, 2 ) = X 32 (u 1 ) = 1, (C(0,0) (C(k) (C(k) 3 ) = X 32 (u 6 ) = 1, 2 1 ) 3 q ) = X 32 (k 2 ) = X 32 (h 1b ) = q 1, 5 ) = X 32 (k 2,1 ) = X 32 (h 1b,1 ) = 1, (C(k,l,m) (C(k) 6 ) = X 32 (h 1 ) = 0, on these classes, we get 7 ) = X 32 (h b ) = X 32 (h 2b ) = (θ k q θ qk q2 1 3 ) = (ω k + w k ), (C(k) 8 ) = X 32 (h 3 ) = 0, where θ is a primitive (q 2 1)-root of unity and ω = θ q We have shown that X 32 SL3 (q) = χ ( q 2 1 ) 3 q 3 1 if q 1(mod 3) and q odd. The case q is even is proved similarly. (ii) Now we show that X 32 SU3 (q) is irreducible when q 1(mod 3). More precisely, X 32 SU3 (q) = χ ( q 2 1 ) 3 q Let us consider the case q is odd. We need to find the fusion of conjugacy classes of SU 3 (q) in G. From [20, p. 565], SU 3 (q) has the following conjugacy classes: C (0) 1, C (0) 2, C (0,0) 3, C (k) 4, C (k) 5, C (k,l,m) 6, C (k) 7 and C (k) 8. We have C (0) 1 K(1), C (0) 2 K(u 1 ), C (0,0) 3 K(u 4 ), C (k) 4 K(k 2 ) K(h 2a ), C (k) 5 K(k 2,1 ) K(h 2a,1 ), 21

22 C (k,l,m) 6 K(h 2 ), C (k) 7 K(h a ) K(h 1a ) and C (k) 8 K(h 6 ). Taking the values of X 32 and χ ( q 2 1 ) 3 q 3 +1 on these classes, we get χ ( q 2 1 ) 3 q 3 +1 χ ( q 2 1 ) 3 q 3 +1 χ ( q 2 1 ) 3 q 3 +1 χ ( q 2 1 ) 3 q 3 +1 χ ( q 2 1 ) 3 q 3 +1 χ ( q 2 1 ) 3 q 3 +1 χ ( q 2 1 ) 3 q 3 +1 χ ( q 2 1 ) 3 q 3 +1 (C(0) (C(0) 1 ) = X 32 (1) = q 3 + 1, 2 ) = X 32 (u 1 ) = 1, (C(0,0) (C(k) (C(k) 3 ) = X 32 (u 4 ) = 1, 4 ) = X 32 (k 2 ) = X 32 (h 2a ) = q + 1, 5 ) = X 32 (k 2,1 ) = X 32 (h 2a,1 ) = 1, (C(k,l,m) (C(k) 6 ) = X 32 (h 2 ) = 0, 7 ) = X 32 (h a ) = X 32 (h 1a ) = (θ k q θ qk q2 1 3 ) = (ω k + w k ), (C(k) 8 ) = X 32 (h 6 ) = 0, where θ is a primitive (q 2 1)-root of unity and ω = θ q We have shown that, when q 1(mod 3) and q is odd, X 32 SU3 (q) Irr(SU 3 (q)) and X 32 SU3 (q) = χ ( q 2 1 even is proved similarly. We omit the details. ) 3 q The case q Lemma Suppose that q 1(mod 3). Then χ ( q2 1 3 ) q 3 1 IBr l (SL 3 (q)) for l q. Proof. Since q 1(mod 3), GL 3 (q) = SL 3 (q) Z(GL 3 (q)) with Z(GL 3 (q)) Z q 1. Let V be a CGL 3 (q)-module affording the irreducible character χ ( q 2 1 ) 3 q Zq 1. We have dim V = q 3 1. According to [23, 4], V has the form (S C (s, (1)) S C (t, (1))) G where s F q and t has degree 2 over F q. Using Corollary 2.7 of [23], since 1 and 2 are coprime, we see that V is irreducible in any cross characteristic. This implies that V SL3 (q) is also irreducible in any cross characteristic and therefore χ ( q2 1 3 ) q 3 1 l q. IBr l (SL 3 (q)) for every Note: 2-Brauer characters of SU 3 (q) were not considered in [20]. Lemma Suppose that q 1(mod 3). Then χ ( q2 1 3 ) q 3 +1 IBr 2 (SU 3 (q)) when l = 2. Proof. We denote the character χ ( q 2 1 ) 3 q 3 +1 by ρ for short. Since q 1(mod 3), GU 3 (q) = SU 3 (q) Z(GU 3 (q)) with Z(GU 3 (q)) Z q+1. Hence, SU 3 (q) has the same degrees 22

23 of irreducible Brauer characters as GU 3 (q) does. It is shown in [72] that every ϕ IBr 2 (GU 3 (q)) either lifts to characteristic 0 or ϕ(1) = q(q 2 q + 1) 1. This and the character table of SU 3 (q) gives the possible values for degrees of irreducible 2-Brauer characters of SU 3 (q): 1, q 2 q, q 2 q + 1, q(q 2 q + 1) 1, q(q 2 q + 1), (q 1)(q 2 q + 1), q 3, q 3 + 1, and (q + 1) 2 (q 1). Assume ρ is reducible when l = 2. Then ρ is the sum of more than one irreducible 2-Brauer characters of SU 3 (q). In other words, ρ(1) = q is the sum of more than one values listed above. This implies that ρ must include a character of degree either 1, or q 2 q, or q 2 q + 1. Once again, by [72], this character lifts to a complex character that we denote by α. Clearly, ρ and α belong to the same 2-block of SU 3 (q). We will use central characters to show that this can not happen. Let R be the full ring of algebraic integers in C and π a maximal ideal of R containing 2R. It is known that α and ρ are in the same 2-block if and only if ω ρ (K) ω α (K) π, (3.4) where K is any class sum and ω χ is the central character associated with χ. The value of ω χ on a class sum is given below: ω χ (K) = χ(g) K, χ(1) where K is the conjugacy class with class sum K and g is an element in K. Therefore, (3.4) implies that ρ(g) ρ(1) gg α(g) α(1) gg π, (3.5) where g G and g G denotes the length of the conjugacy class containing g. Consider the first case when α(1) = 1. It means that α is the trivial character. In (3.5), take g to be any element in the conjugacy class C (1) 7 (notation in [20]), we have ρ(g) ρ(1) gg α(g) α(1) gg = 1 q 3 +1 q3 (q 3 + 1) 1 1 q3 (q 3 + 1) = q 3 q 3 (q 3 + 1) π. Note that π Z = 2Z. Since q 3 q 3 (q 3 + 1) is an odd number, we get a contradiction. 23

24 Secondly, if α(1) = q 2 q then α = χ q 2 q. In (3.5), take g to be any element in the conjugacy class C (1) 7, we have ρ(g) ρ(1) gg α(g) α(1) gg = 1 q 3 +1 q3 (q 3 +1) 0 q 2 q q3 (q 3 +1) = q 3 π. Since q 3 is an odd number, we again get a contradiction. Finally, if α(1) = q 2 q + 1 then α = χ (u) q 2 q+1 for some 1 u q. In (3.5), take g to be any element in the conjugacy class C (q+1) 7. Note that since q 1(mod 3), we have ρ(g) ρ(1) gg α(g) α(1) gg = 1 q 3 +1 q3 (q 3 + 1) 1 q 2 q+1 q3 (q 3 + 1) = q 3 q 3 (q + 1) π, which leads to a contradiction since q 3 q 3 (q + 1) is odd. Proof is completed. Theorem (Reduction Theorem). Let G = G 2 (q), q = p n, q 5, and p a prime number. Let ϕ be an irreducible character of G in cross characteristic and M a maximal subgroup of G. Assume that ϕ(1) > 1 and ϕ M is also irreducible. Then M is G-conjugate to one of the following groups: (i) maximal parabolic subgroups P a, P b, (ii) SL 3 (q) : 2, SU 3 (q) : 2, (iii) G 2 (q 0 ) with q = q 2 0, p 3. Proof. By Lemma 2.1.1, we have d l (G) M. Moreover, from formulas (2.1) and (2.3), we have d l (G) q 3 1 if 3 q and d l (G) q 4 + q 2 if 3 q for every l q. Therefore, M q 3 1 if 3 q, q 4 + q 2 if 3 q. (3.6) Here, we will only give the proof for the case p 5. The proofs for p = 2 and p = 3 are similar. According to [39], if M is a maximal subgroup of G = G 2 (q), q = p n, p 5, then M is G-conjugate to one of the following groups: 1. P a, P b, maximal parabolic subgroups, 2. (SL 2 (q) SL 2 (q)) 2, involution centralizer, L 3 (2), only when p = q, 4. SL 3 (q) : 2, SU 3 (q) : 2, 5. G 2 (q 0 ), q = qo α, α prime, 24

25 6. P GL 2 (q), p 7, q 11, 7. L 2 (8), p 5, 8. L 2 (13), p 13, 9. G 2 (2), q = p 5, 10. J 1, q = 11. Consider for instance the case 5) with α 3. Then M = q0(q )(q0 2 1) < q0 7 q 7/3 < q 3 1 for every q 5. This contradicts (3.6). The cases 2), 3), 6) - 10) are excluded similarly. Proof of Theorem B. By Reduction Theorem, M must be G-conjugate to one of the following subgroups: (i) maximal parabolic subgroups P a, P b, (ii) SL 3 (q) : 2, SU 3 (q) : 2, (iii) G 2 (q 0 ) with q = q0, 2 p 3. Case 1: M = P a. Let Z := Z(P a), the center of the derived subgroup of P a. We know that P a is the normalizer of Z in G and therefore Z is nontrivial. In fact, Z is a long-root subgroup of G. Let V be an irreducible FG-module affording the character ϕ. By Lemma , Z must have nonzero fixed points on V. In other words, C V (Z) = {v V a(v) = v for every a Z} = 0. Therefore V Pa is reducible by Lemma Equivalently, ϕ Pa is reducible. Case 2: M = P b. Using the results about character tables of P b in [1], [16], and [17], we have m C (P b ) = q(q 1)(q 2 1) for q 5. Suppose that ϕ Pb is irreducible, then ϕ(1) q(q 1)(q 2 1) by Lemma If 3 q then d l (G) q 4 + q 2 because of formula (2.3). Then we have d l (G) q 4 + q 2 > q(q 1)(q 2 1) ϕ(1) and this cannot happen. So q must be coprime to 3. It is easy to check that m C (P b ) < d 2,l (G) for every q 8. Therefore, when q 8, the inequality ϕ(1) q(q 1)(q 2 1) can hold only if ϕ is the nontrivial character of smallest degree. When q = 5 or 7, checking directly, we see that besides the nontrivial 25

26 character of smallest degree, ϕ can be ϕ 18 = X 18 of degree 1 6 q(q 1)2 (q 2 q + 1). Recall that P b = q 6 (q 2 1)(q 1), which is not divisible by X 18 (1) for q = 5 or 7. It follows that X 18 Pb is reducible and so is ϕ 18 Pb. We have shown that the unique possibility for ϕ is the nontrivial character ψ of smallest degree when 3 q. Recall that X 32 (1) = q 3 + ɛ, which does not divide P b and therefore X 32 Pb is reducible. If l = 3 and q 1(mod 3) then ψ = X 32 1 G. Assume that ψ Pb = X 32 Pb 1 Pb is irreducible. The reducibility of X 32 Pb and the irreducibility of X 32 Pb 1 Pb implies that X 32 Pb = λ + µ where λ = 1 Pb, µ Irr(P b ) and µ IBr 3 (P b ). We then have µ(1) = X 32 (1) 1 = q 3, which is a contradiction since P b has no irreducible complex character of degree q 3. It remains to consider the case when l 3 or q is not congruent to 1 modulo 3. Then ψ Pb = X 32 Pb, which is reducible as noted above. Case 3: M = SL 3 (q) : 2. From [64], we know that m C (SL 3 (q)) = (q +1)(q 2 +q +1) for every q 5. Therefore, m C (SL 3 (q) : 2) 2(q + 1)(q 2 + q + 1). Since ϕ SL3 (q):2 is irreducible, ϕ(1) 2(q + 1)(q 2 + q + 1). Similar arguments as above show that q is not divisible by 3. By Theorem 2.2.1, the inequality ϕ(1) 2(q + 1)(q 2 + q + 1) can hold only if ϕ is the nontrivial character ψ of smallest degree or ϕ 18 when q = 5. Note that X 18 (1) = 280 and SL 3 (q) : 2 = 744, 000 when q = 5 and therefore X 18 (1) SL 3 (q) : 2. Hence, X 18 SL3 (q):2 is reducible and so is ϕ 18 SL3 (q):2 when q = 5. Again, the unique possibility for ϕ is ψ when 3 q. If l = 3 and q 1(mod 3) then ψ = X 32 1 G. Assume that ψ SL3 (q):2 is irreducible. Let V be an irreducible FG-module, char(f) = 3, affording the character ψ. Then V SL3 (q):2 is an irreducible F(SL 3 (q) : 2)-module. Let σ be a generator for the multiplicative group F q and I be the identity matrix in SL(3, F q ). Consider the matrix T = σ q 1 3 I. We have < T >= Z(SL 3 (q)) and hence < T > SL 3 (q) : 2. Since ord(t ) = 3, < T > O 3 (SL 3 (q) : 2) and therefore O 3 (SL 3 (q) : 2) is nontrivial. It follows that V SL3 (q):2 is reducible by Lemma

27 If l 3 and q 1(mod 3) then ψ = X 32. So ψ SL3 (q):2 = X 32 SL3 (q):2. Note that X 32 (1) = q and SL 3 (q) : 2 = 2q 3 (q 3 1)(q 2 1). Therefore X 32 (1) SL 3 (q) : 2 for every q 5. Hence X 32 SL3 (q):2 as well as ψ SL3 (q):2 are reducible in this case. When q 1(mod 3), we have ψ = X 32. By Lemmas and 2.3.2, we get that ψ SL3 (q) = X 32 SL3 (q) = χ ( q2 1 3 ) q 3 1 Theorem B. IBr l (SL 3 (q)) for l q, as we claim in the item (i) of Case 4: M = SU 3 (q) : 2. According to [64], we have m C (SU 3 (q)) = (q + 1) 2 (q 1) for every q 5 and therefore m C (SU 3 (q) : 2) 2(q + 1) 2 (q 1) for q 5. Hence ϕ(1) 2(q + 1) 2 (q 1) by the irreducibility of ϕ SU3 (q):2. Again, q must be coprime to 3. By Theorem 2.2.1, the inequality ϕ(1) 2(q + 1) 2 (q 1) can hold only if ϕ is the nontrivial character ψ of smallest degree or ϕ 18 of degree 1 6 q(q 1)2 (q 2 q + 1) when q = 5. Note that ϕ 18 = X 18. By [20], the degrees of irreducible complex characters of SU 3 (5) are: 1, 20, 125, 21, 105, 84, 126, 144, 28 and 48. When q = 5, X 18 (1) = 280. Therefore, X 18 SU3 (5) is the sum of at least 3 irreducible characters. Since SU 3 (5) is a normal subgroup of index 2 of SU 3 (5) : 2, by Clifford s theorem, X 18 SU3 (5):2 is reducible when q = 5. This implies that ϕ 18 SU3 (5):2 is also reducible when q = 5. In summary, the unique possibility for ϕ is ψ when 3 q. If q 1(mod 3) then ψ = X 32 of degree q 3 1. Recall that SU 3 (q) : 2 = 2q 3 (q 3 + 1)(q 2 1), which is not divisible by q 3 1 for every q 5. Hence X 32 SU3 (q):2 is reducible and so is ψ SU3 (q):2. Now it remains to consider q 1(mod 3). First we assume that l = 2. Then ψ = X 32. By Lemmas and 2.3.3, we have ψ SU3 (q) = X 32 SU3 (q) = χ ( q2 1 3 ) q 3 +1 IBr 2 (SU 3 (q)). Therefore ψ SU3 (q):2 is irreducible when l = 2. Next, if l = 3 then ψ = X 32 1 G. By Lemma 2.3.1, we have ψ SU3 (q) = X 32 SU3 (q) 1 SU3 (q) = χ ( q2 1 ) 3 q SU3 (q) = χ q 3, which is an irreducible 3-Brauer character of SU 3 (q) by [20, p. 573]. Finally, if l 2, 3 then ψ = X 32. By Lemma 2.3.1, we have X 32 SU3 (q) = χ ( q2 1 ) 3 q 3 +1 which is irreducible again by [20]. Therefore, ψ SU3 (q):2 is also irreducible, as we claim in the item (ii) of Theorem B. 27

28 Case 5: M = G 2 (q 0 ) with q = q 2 0, 3 q. Since ϕ M is irreducible, ϕ(1) G 2 (q 0 ) = q 6 0 (q 6 0 1)(q 2 0 1) < q 7 < d 2,l (G) for every q 5. Therefore, by Theorem 2.2.1, the unique possibility for ϕ is ψ. Since q = q 2 0, q 1(mod 3). If l = 3 then ψ = X 32 1 G. Hence, ψ G2 (q 0 ) = X 32 G2 (q 0 ) 1 G2 (q 0 ). Recall that X 32 (1) = q and G 2 (q 0 ) = q0(q )(q0 2 1) = q 3 (q 3 1)(q 1). It is easy to see that (q 3 + 1) q 3 (q 3 1)(q 1) for every q 5. So X 32 G2 (q 0 ) is reducible. Assume that X 32 G2 (q 0 ) 1 G2 (q 0 ) is irreducible. Then X 32 G2 (q 0 ) = λ + µ where λ = 1 G2 (q 0 ), µ Irr(G 2 (q 0 )) and µ IBr 3 (G 2 (q 0 )). We then have µ(1) = X 32 (1) 1 = q 3 = q0. 6 So µ is the Steinberg character. From [33], we know that the reduction modulo 3 of the Steinberg character is reducible, which contradicts µ IBr 3 (G 2 (q 0 )). If l 3 then ψ = X 32. Therefore, ψ G2 (q 0 ) = X 32 G2 (q 0 ). From the reducibility of X 32 G2 (q 0 ) as noted above, ψ G2 (q 0 ) is also reducible. 2.4 Small Groups G 2 (3) and G 2 (4) In this section, we mainly use results and notation of [12] and [36]. We notice that the universal cover of G 2 (3), which is 3 G 2 (3), has two pairs of complex conjugate irreducible characters of degree 27. The purpose of this section is to prove the following theorem. Theorem Let G {G 2 (3), 3 G 2 (3), G 2 (4), 2 G 2 (4)}. Let ϕ be a faithful irreducible character of G in cross characteristic l of degree greater than 1 and M a maximal subgroup of G. Then we have (i) When G = G 2 (3), ϕ M is irreducible if and only if one of the following holds: (a) M = U 3 (3) : 2, ϕ is the unique irreducible character of degree 14 when l = 7 or ϕ is any of the irreducible characters of degree 14, 64 when l 3, 7; (b) M = 2 3 L 3 (2), ϕ is the unique irreducible character of degree 14 when l 2, 3. (ii) When G = 3 G 2 (3), the universal cover of G 2 (3), ϕ M is irreducible if and only if one of the following holds: (a) M = 3.P or 3.Q where P, Q are maximal parabolic subgroups of G 2 (3), ϕ is any of the four irreducible characters of degree 27; 28

29 (b) M = 3.(U 3 (3) : 2), ϕ is any of the two complex conjugate irreducible characters of degree 27 when l 2, 3, 7; (c) M = 3.(L 3 (3) : 2), ϕ is any of the two complex conjugate irreducible characters of degree 27 when l 2, 3, 13; (d) M = 3.(L 2 (8) : 3), ϕ is any of the four irreducible characters of degree 27 when l 2, 3, 7. (iii) When G = G 2 (4), ϕ M is irreducible if and only if one of the following holds: (a) M = U 3 (4) : 2, ϕ is the unique irreducible character of degree 64 when l = 3 or ϕ is the unique irreducible character of degree 65 when l 2, 3; (b) M = J 2, ϕ is any of the two irreducible characters of degree 300 when l 2, 3, 7. (iv) When G = 2 G 2 (4), the universal cover of G 2 (4), ϕ M is irreducible if and only if one of the following holds: (a) M = 2.P or 2.Q where P, Q are maximal parabolic subgroups of G 2 (4), ϕ is the unique irreducible character of degree 12; (b) M = 2.(U 3 (4) : 2), ϕ is the unique irreducible character of degree 12 or ϕ is any of the two irreducible characters of degree 104 when l 2, 5; (c) M = 2.(SL 3 (4) : 2), ϕ is the unique irreducible character of degree 12 when l 2, 3. Lemma Theorem holds in the case G = G 2 (3), M = U 3 (3) : 2. Proof. According to [12, p. 14], we have m C (U 3 (3)) = 32 and m C (U 3 (3) : 2) = 64. Thus, if ϕ M is irreducible then ϕ(1) 64. Inspecting the character tables of G 2 (3) in [12, p. 60] and [36, p. 140, 142, 143], we see that ϕ(1) = 14 or 64. Note that G 2 (3) has a unique irreducible complex character of degree 14 which is denoted by χ 2 and every reduction modulo l 3 of χ 2 is still irreducible. Now we will show that χ 2 U3 (3) = χ 6, which is the unique irreducible character of degree 14 of U 3 (3). Suppose that χ 2 U3 (3) χ 6, then χ 2 U3 (3) is reducible and it is the sum of more than one 29

30 irreducible characters of degree less than 14. Note that U 3 (3) has exactly one conjugacy class of elements of order 6, which is denoted by 6A. If χ 2 U3 (3) is sum of two irreducible characters, then the degree of these characters is 7. So χ 2 U3 (3)(6A) is 0, 2 or 4. This cannot happen since the value of χ 2 on any class of elements of order 6 of G 2 (3) is 1 or 2. If χ 2 U3 (3) is sum of more than two irreducible characters, then χ 2 U3 (3)(6A) 2 which cannot happen, neither. In summary, we have χ 2 U3 (3) = χ 6. We also see that every reduction modulo l 3 of χ 6 is still irreducible. Hence, if l 3 and ϕ is the l-brauer character of G 2 (3) of degree 14, then ϕ U3 (3) is irreducible and so is ϕ U3 (3):2. By [12, p. 14], U 3 (3) : 2 has one complex characters of degree 64 which we denote by χ. Also, G 2 (3) has two irreducible complex characters of degree 64 that are χ 3 and χ 4 as denoted in [12, p. 60]. We will show that χ 3,4 U3 (3):2 = χ. Note that the two conjugacy classes of (U 3 (3) : 2)-subgroups of G 2 (3) are fused under an outer automorphism τ of G 2 (3), which stabilizes each of χ 3 and χ 4. Hence without loss we may assume that U 3 (3) : 2 is the one considered in [16, p. 237]. Checking directly, it is easy to see that the values of χ 3 and χ 4 coincide with those of χ at every conjugacy classes except the class of elements of order 3 at which we need to check more. U 3 (3) : 2 has two classes of elements of order 3, 3A and 3B. Using [16, p. 237] to find the fusion of conjugacy classes of U 3 (3) in G 2 (3) and the values of χ 3 and χ 4 (which are θ 12 (k) in [16]), we see that the classes 3A, 3B of U 3 (3) are contained in the classes 3A, 3E of G 2 (3), respectively. We also have χ 3 (3A) = χ 4 (3A) = χ(3a) = 8 and χ 3 (3E) = χ 4 (3E) = χ(3b) = 2. Thus, χ 3 U3 (3):2 = χ 4 U3 (3):2 = χ. By [36, p. 140, 142, 143], any reduction modulo l 3 of χ 3 as well as χ 4 is irreducible. Also, the reduction modulo l of χ is irreducible for every l 3, 7. Therefore, χ 3 U3 (3):2 and χ 4 U3 (3):2 are the same and irreducible for every l 3, 7. When l = 7, m 7 (U 3 (3)) = 28 and m 7 (U 3 (3) : 2) = 56. Since χ 3 (1) = χ 4 (1) = 64, χ 3 U3 (3):2 and χ 4 U3 (3):2 are reducible when l = 7. Lemma Theorem holds in the case G = G 2 (3), M = 2 3 L 3 (2). 30

31 Proof. We have 2 3 L 3 (2) = So m C (2 3 L 3 (2)) 1344 < 37 and therefore the unique possibility for ϕ is the reduction modulo l 3 of the character χ 2 of degree 14. When l G 2 (3) (l 2, 3, 7, and 13), from [44], we know that χ 2 23 L 3 (2) is irreducible. When l = 2, we have m 2 (2 3 L 3 (2)) = m 2 (L 3 (2)) 168 < 13. So χ 2 23 L 3 (2) is reducible when l = 2. When l = 13, since L 3 (2), χ 2 23 L 3 (2) is irreducible. The last case we need to consider is l = 7. Let E = L 3 (2) which is an elementary abelian group of order 2 3. Since χ 2 E L3 (2) is irreducible and L 3 (2) acts transitively on E\{1} and Irr(E)\{1 E }, χ 2 E = 2 α Irr(E)\{1 E } α. Let I be the inertia group of α in E L 3 (2). By Clifford s theory, we have χ 2 E L3 (2) = Ind E L 3(2) I (ρ) for some ρ Irr(I) and ρ E = 2α. We also have I = E L 3(2) Irr(E)\{1 E } modulo 7 of ρ is itself and therefore it is irreducible. Hence when l = 7. = Thus, the reduction χ2 E L3 (2) is also irreducible Lemma Theorem holds in the case G = 3 G 2 (3), M = 3.P or 3.Q where P, Q are maximal parabolic subgroups of G 2 (3). Proof. First, we consider M = 3.P where P is one of two maximal parabolic subgroups which is specified in [16, p. 217]. If ϕ 3.P is irreducible then ϕ(1) m C (3.P ) (3.P )/Z(3.P ) P = = 108. Inspecting the character tables (both complex and Brauer) of 3 G 2 (3) in [12] and [36] (note that we only consider faithful characters), we have ϕ(1) = 27 and ϕ is actually the reduction modulo l 3 of one of four irreducible complex characters of degree 27 of 3 G 2 (3). From now on, we denote these characters by χ 24, χ 24 (corresponding to the line χ 24 in [12, p. 60]) and χ 25, χ 25 (corresponding to the line χ 25 in [12, p. 60]). Now we will show that χ 24 3.P is irreducible. Note that if g 1 and g 2 are the pre-images of an element g G 2 (3) under the natural projection π : 3 G 2 (3) G 2 (3), then χ 24 (g 1 ) = ωχ 24 (g 2 ) where ω is a cubic root of unity. Therefore we have [χ 24 3.P, χ 24 3.P ] 3.P = 1 3 P x 3.P χ 24(x)χ 24 (x) = 1 P g P χ 24(g)χ 24 (g), where g is a fixed pre-image of g under π. We choose g so that the value of χ 24 at g is printed in [12, p. 60]. The fusion 31

32 of conjugacy classes of P in G 2 (3) is given in [16, p. 217]. By comparing the orders of centralizers of conjugacy classes of G 2 (3) in [16, p. 239] with those in [12, p. 60], we can find a correspondence between conjugacy classes of G 2 (3) in these two papers. The length of each conjugacy class of P can be computed from [16, p. 217, 218]. All the above information is collected in Table 2-6. From this table, we see that the value of χ 24 is zero at any element g for which the order of g is 3, 6 or 9. We denote by χ 24 (X) the value χ 24 (g) for some g X. Then we have g P χ 24(g)χ 24 (g) = A 1 (χ 24 (A 1 )) 2 + B 11 (χ 24 (B 11 )) 2 + B 21 (χ 24 (B 21 )) 2 + D 1 (χ 24 (D 1 )) 2 (4.7) + D 2 (χ 24 (D 2 )) E 2 (1) (χ 24 (E 2 (1))) 2. By [16, p. 217], the class D 1 of P is contained in the class D 11 of G 2 (3). This class is the class 4A or 4B according to the notation in [12, p. 60]. First we assume D 11 is 4A. Then by looking at the values of one character of degree 273 of G 2 (3) both in [16] and [12], it is easy to see that D 2 D 12 = 12A and E 2 (i) E 2 = 8A. Therefore, Equation (4.7) becomes g P χ 24(g)χ 24 (g) = ( 1) ( 1) 2 = 11, 664. Next, we assume D 11 is 4B. Then D 12 = 12B and E 2 = 8B and Equation (4.7) becomes g P χ 24(g)χ 24 (g) = = 11, 664. So in any case, we have [χ 24 3.P, χ 24 3.P ] 3.P = 1 P χ 24 3.P is irreducible. g P χ 24(g)χ 24 (g) = 1. That means Note that χ 24 is irreducible for every l 3 by [36, p. 140, 142, 143]. We will show that χ 24 3.P is also irreducible for every l 3. The structure of P is [3 5 ] : 2S 4. We denote by O 3 the maximal normal 3-subgroup (of order 3 5 ) of P. Then the order of any element 32