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Regular unipotent elements of finite maximal subgroups of classical groups

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Regular unipotent elements of finite maximal subgroups of classical groups

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Content REGULAR UNIPOTENT ELEMENTS OF FINITE MAXIMAL SUBGROUPS OF CLASSICAL GROUPS by Matthew Donner A Dissertation Presented to the FACULTY OF THE USC GRADUATE SCHOOL UNIVERSITY OF SOUTHERN CALIFORNIA In Partial Fulfillment of the Requirements for the Degree DOCTOR OF PHILOSOPHY (MATHEMATICS) August 2018 Copyright 2018 Matthew Donner 1 Acknowledgements I would like to thank my dissertation advisor, Professor Robert Guralnick, for all the help he has provided me. His patience and generosity with his time made it possible for me to complete this dissertation. I would also like to thank my dissertation committee for generously volunteering their time to serve on the committee. i Contents 1 Acknowledgements i List of Tables iii 2 Introduction 1 2.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 Main Results 2 3.1 Results for problem 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 3.2 Results for problem 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 3.3 Results for problem 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 4 Problem 1 18 4.1 Alternating groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 4.2 Classical groups of Lie type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 4.2.1 Special linear groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 4.2.2 Unitary groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 4.2.3 Symplectic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.2.4 Orthogonal groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 4.3 Exceptional groups of Lie type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 4.4 Sporadic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 5 Problem 2 54 6 Problem 3 54 6.1 Jordan normal forms when S =A n . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 6.2 Jordan normal forms when S =PSL(n,q) with n> 2 . . . . . . . . . . . . . . . . . 60 6.3 Jordan normal forms when S =PSL 2 (q) . . . . . . . . . . . . . . . . . . . . . . . . . 61 6.4 Jordan normal forms when S =PSU n (q) . . . . . . . . . . . . . . . . . . . . . . . . 65 6.5 Jordan normal forms when S =PSp 2n (q) . . . . . . . . . . . . . . . . . . . . . . . . 66 6.6 Frobenius-Schur indicators when S ∼ = A n and for Weil representations of classical groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 6.6.1 Frobenius-Schur indicators when S ∼ =A n . . . . . . . . . . . . . . . . . . . . 68 6.6.2 Frobenius-Schur indicators when S ∼ =PSp 2n (q) . . . . . . . . . . . . . . . . . 69 6.6.3 Frobenius-Schur indicators when S ∼ =PSU(n,q) . . . . . . . . . . . . . . . . 69 6.6.4 Frobenius-Schur indicators when S ∼ =PSL n (q) and n> 2 . . . . . . . . . . . 70 7 References 70 ii List of Tables 1 Generic cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 2 Exceptional cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 3 Generic cases with `-|x| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 4 Exceptional cases with `-|x| . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 5 Regular unipotent element when S =A n , standard representation . . . . . . . . . . 14 6 Regular unipotent element when S =PSL 2 (q), generic cases . . . . . . . . . . . . . 15 7 Generic cases with regular unipotent element . . . . . . . . . . . . . . . . . . . . . . 15 8 G with regular unipotent element, exceptional cases . . . . . . . . . . . . . . . . . . 16 9 Alternating groups, generic cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 10 Alternating groups, small cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 11 PSL(2,q) with q6= 9, q Fermat prime . . . . . . . . . . . . . . . . . . . . . . . . . . 23 12 PSL(2,q) with q6= 9, q Marsenne prime . . . . . . . . . . . . . . . . . . . . . . . . 23 13 PSL(2,q), q− 1 = 3(2 i ) for some i . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 14 PSL(2,q) with q6= 9, q + 1 = 3(2 i ) for some i . . . . . . . . . . . . . . . . . . . . . 24 15 PSL(2,q) where x contains a field automorphism . . . . . . . . . . . . . . . . . . . 25 16 SL 2 (q), q≡ 1 mod 4 and q prime . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 17 SL 2 (q), q≡ 3 mod 4 and q prime . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 18 PSL(2,q), q≡ 1 mod 4,|x| odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 19 SL 2 (q), q≡ 3 mod 4,|x| odd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 20 Exceptional case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 21 PSL(2,q) with q− 1 Marsenne prime,|x| odd . . . . . . . . . . . . . . . . . . . . . 28 22 PSL(2,q) with q + 1 Fermat prime,|x| odd . . . . . . . . . . . . . . . . . . . . . . . 29 23 PSL(2,q), q even and x with field automorphism . . . . . . . . . . . . . . . . . . . 29 24 PSL(n,q), n≥ 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 25 PSL 3 (4) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 26 PSU(4, 3) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 27 PSU n (q) with n odd, exceptional cases . . . . . . . . . . . . . . . . . . . . . . . . . 39 28 PSU n (q), generic case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 29 PSp 2n (q) with q even, exceptional cases . . . . . . . . . . . . . . . . . . . . . . . . . 44 30 PSp 2n (q), generic cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 31 PSp 2n (q) with q odd, exceptional cases . . . . . . . . . . . . . . . . . . . . . . . . . 47 32 Small cases from orthogonal groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 33 2 B 2 (8) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 34 G 2 (q), exceptional cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 35 Sporadic groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 iii 2 Introduction In this dissertation we address the issue of representations of almost quasi-simple groupsG where G contains certain prime power order elements. By definition, a given almost quasi-simple group G satisfies S≤G/Z(G)≤Aut(S) for a simple group S. For the following problems we assume that F ∗ (G), the generalized fitting subgroup ofG, is quasisimple and acts absolutely irreducibly. We let q =p a , where p is a prime. Following Tiep [25], we let ∂ 1 ` (T ) denote the smallest degree of an irreducible representation of a finite quasisimple groupT over an algebraically closed field of characteristic `, assuming that the representation is cross-characteristic when T is of Lie type. As described in Tiep [25], the typical situation for a finite quasisimple groupT is to have a few irreducible representations of degree about ∂ 1 ` (T ), followed by a significant gap until the degree of the next irreducible representation. We let ∂ 2 ` (T ) denote the degree of the smallest irreducible representation with degree bigger than 2∂ 1 ` (T ). We let k n,` = 1 if `|n but k n,` = 0 if not. 2.1 Problems 1. Suppose that G≤ GL m (` b ) is an almost quasisimple subgroup where F ∗ (G) is quasisimple and acts absolutely irreducibly. Moreover, suppose that if G is of Lie type the representation is cross-characteristic. For which such G does G contain an element x of prime power order |x| =u t such that|x|≥m− 1 for ` odd and|x|≥m− 2 for ` even? 2. Which element-group pairs from Problem 1 have u6=`? 3. Which elements from Problem 1 correspond to regular unipotent elements of classical algebraic groups? The general strategy of the proof is to exploit the tension between large lower limits on irreducible representations of quasi-simple groups and small upper limits on prime power element orders. For many almost quasi-simple groups, these limits rule out the possibility of a prime power element contained in an almost quasi-simple group as described in Problem 1. Throughout this work, we 1 assume thatG is an almost quasi-simple group containing an elementx of prime power order greater than or equal to m− 2. This will often be assumed without comment. These assumptions lead to a contradiction in most cases, but for the other cases we characterize the elements meeting the assumed conditions. 3 Main Results 3.1 Results for problem 1 We list the possibilities according to the nature ofF ∗ (G). Concerning the notation, we note that u is the prime base of|x|, F. prime refers to a Fermat prime and M. prime refers to a Marsenne prime. Table 1: Generic cases F ∗ (G) Degree Char Ind |x| Conditions A n n− 2 2 =`|n + n−4≤|x|≤n n≥ 17, n6≡ 2 mod 4 A n n− 2 2 =`|n - n−4≤|x|≤n n≥ 17, n≡ 2 mod 4 A n n− 2 26=`|n + n−3≤|x|≤n n≥ 17 A n n− 1 2 =`-n + n−3≤|x|≤n n≥ 17 A n n− 1 26=`-n + n−2≤|x|≤n n≥ 17 PSL(2,q) (q− 1)/2 2 - 1/2(q− 1) q6= 9 F. prime PSL(2,q) (q− 1)/2 2 - (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1)/2 `6= 2 - 1/2(q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1)/2 `6= 2 - (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1)/2 `6= 2 - 2(q− 1) q6= 9 F. prime PSL(2,q) (q + 1)/2 `6= 2 + 1/2(q− 1) q6= 9 F. prime PSL(2,q) (q + 1)/2 `6= 2 + (q− 1) q6= 9 F. prime PSL(2,q) (q− 1) + (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1) q+1 2 l 0 6= 1 - (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1) q+1 2 l 0 6= 1 - 2(q− 1) q6= 9 F. prime PSL(2,q) q `- (q + 1) + (q− 1) q6= 9 F. prime PSL(2,q) (q + 1) q−1 4 l 0 6= 1 + (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q + 1) `6= 2 - (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q + 1) `6= 2 - 2(q− 1) q6= 9 F. prime PSL(2,q) (q− 1)/2 0 1/2(q + 1) q6= 9 M. prime PSL(2,q) (q− 1)/2 0 (q + 1) q6= 9 M. prime 2 Continuation of Table 1 F ∗ (G) Degree Char Ind |x| Conditions 2.PSL 2 (q) (q + 1)/2 `6= 2 0 1/2(q + 1) q6= 9 M. prime 2.PSL 2 (q) (q + 1)/2 `6= 2 0 (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q + 1)/2 `6= 2 0 2(q + 1) q6= 9 M. prime PSL(2,q) (q− 1) + (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q− 1) `6= 2 - (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q− 1) `6= 2 - 2(q + 1) q6= 9 M. prime PSL(2,q) q `- (q + 1) + (q + 1) q6= 9 M. prime PSL(2,q) (q + 1) q−1 2 l 0 6= 1 + (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q + 1) q−1 2 l 0 6= 1, - (q + 1) q6= 9 M. prime `6= 2 2.PSL 2 (q) (q + 1) q−1 2 l 0 6= 1, - 2(q + 1) q6= 9 M. prime `6= 2 2.PSL(2,q) (q− 1)/2 6= 2 - 2/3(q− 1) q− 1 = 3(2 i ), for some i 2.PSL(2,q) (q + 1)/2 `6= 2 0 2/3(q + 1) q + 1 = 3(2 i ), for some i PSL(2,q) (q− 1)/2 2 - q q prime, q≡ 1 mod 4 2.PSL(2,q) (q− 1)/2 `6= 2 - q q prime, q≡ 1 mod 4 PSL(2,q) (q + 1)/2 `6= 2 + q q prime, q≡ 1 mod 4 PSL(2,q) (q− 1) + q q prime, q≡ 1 mod 4 2.PSL(2,q) (q− 1) q+1 2 l 0 6= 1 - q q prime, q≡ 1 mod 4 PSL(2,q) q `- (q + 1) + q q prime, q≡ 1 mod 4 PSL(2,q) (q + 1) q−1 4 l 0 6= 1 + q q prime, q≡ 1 mod 4 2.PSL(2,q) (q + 1) `6= 2 - q q prime, q≡ 1 mod 4 PSL(2,q) (q− 1)/2 0 q q prime, q≡ 3 mod 4 2.PSL(2,q) (q + 1)/2 `6= 2 0 q q prime q≡ 3 mod 4 PSL(2,q) q− 1 + q q prime, q≡ 3 mod 4 2.PSL(2,q) (q− 1) `6= 2 - q q prime, q≡ 3 mod 4 2.PSL(2,q) (q− 1) q+1 2 l 0 6= 1 - q q prime, q≡ 1 mod 4 PSL(2,q) q `- (q + 1) + q q prime, q≡ 3 mod 4 3 Continuation of Table 1 F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) (q + 1) q−1 2 l 0 6= 1 + q q prime, q≡ 3 mod 4 2.PSL(2,q) (q + 1) q−1 2 l 0 6= 1 - q q prime, `6= 2 q≡ 1 mod 4 PSL(2,q) (q− 1)/2 2 - 1/2(q + 1) q + 1 = 2u i , u6= 2 2.PSL(2,q) (q− 1)/2 6= 2 - 1/2(q + 1) q + 1 = 2u i , u6= 2 PSL(2,q) (q + 1)/2 6= 2 + 1/2(q + 1) q + 1 = 2u i , u6= 2 PSL(2,q) (q− 1)/2 0 1/2(q− 1) q− 1 = 2(u i ), u6= 2 2.PSL(2,q) (q + 1)/2 `6= 2 0 1/2(q− 1) q− 1 = 2(u i ), u6= 2 PSL(2,q) q− 1 + q− 1 q− 1 M. prime PSL(2,q) q `-q + 1 + q− 1 q− 1 M. prime PSL(2,q) q + 1 (q−1) ` 0 6= 1 + q− 1 q− 1 M. prime PSL(2,q) q− 1 + q + 1 q + 1 F. prime PSL(2,q) q `-q + 1 + q + 1 q + 1 F. prime PSL(2,q) q + 1 (q−1) ` 0 6= 1 + q + 1 q + 1 F. prime PSL n (q) q n −q q−1 −k n,q,` q n −1 q−1 , n≥ 3, prime u6= 2 PSL n (q) q n −1 q−1 q n −1 q−1 , n≥ 3, prime u6= 2 PSU(n,q) q n −q q+1 q n +1 q+1 , n odd, u6= 2 prime PSU(n,q) q n +1 q+1 q n +1 q+1 , n odd, u6= 2 prime PSp 2n (3) (3 n − 1)/2 (3 n − 1)/2 n odd prime, u6= 2 2.PSp 2n (3) (3 n + 1)/2 (3 n − 1)/2 n odd prime, u6= 2 2.PSp 2n (q) (q n − 1)/2 (q n + 1)/2 n = 2 a , u6= 2, q≡ 3 mod 4 PSp 2n (q) (q n + 1)/2 (q n + 1)/2 n = 2 a , u6= 2, q≡ 3 mod 4 2.PSp 2n (q) (q n − 1)/2 (q n + 1)/2 n = 2 a , u6= 2, q≡ 1 mod 4 PSp 2n (q) (q n + 1)/2 (q n + 1)/2 n = 2 a , u6= 2, 4 Continuation of Table 1 F ∗ (G) Degree Char Ind |x| Conditions q≡ 1 mod 4 Table 2: Exceptional cases F ∗ (G) Degree Char Ind |x| Conditions 3.A 6 3 2 0 2, 3, 4, 5 3.A 6 3 5 0 2, 3, 4, 5 A 6 4 2 - 2, 3, 4, 5 2.A 6 4 5 - 3, 4, 5, 8 A 6 5 5 + 4, 5 3.A 6 6 5 0 5 6.A 6 6 5 0 5, 8 3.A 7 3 5 0 2, 3, 4, 5, 7 A 7 4 2 0 2, 3, 4, 5, 7 2.A 7 4 7 - 3, 4, 5, 7, 8 2.A 7 4 6= 2, 7 0 3, 4, 5, 7, 8 A 7 5 7 + 4, 5, 7 A 7 6 2 + 4, 5, 7 A 7 6 6= 2, 7 + 5, 7 2.A 7 6 3 - 5, 7, 8 3.A 7 6 2 0 4, 5, 7 3.A 7 6 6= 3 0 5, 7 6.A 7 6 6= 2, 3 0 5, 7, 8 A 7 8 5 + 7 A 8 7 6= 2 + 7, 8 2.A 8 8 6= 2 + 7, 8 A 9 7 3 + 7, 8, 9 A 9 8 2 + 7, 8, 9 A 9 8 6= 2, 3 + 8, 9 2.A 9 8 6= 2 + 7, 8, 9 A 10 8 2 - 7, 8, 9 A 10 8 5 + 7, 8, 9 2.A 10 8 5 + 7, 8, 9 A 10 9 6= 2, 5 + 8, 9 A 11 9 11 + 8, 9, 11 A 11 10 2 + 8, 9, 11 A 11 10 6= 2, 11 + 9, 11 A 12 10 2 + 8, 9, 11 A 12 10 3 + 9, 11 A 12 11 6= 2, 3 + 11 A 13 11 13 + 11, 13 5 Continuation of Table 2 F ∗ (G) Degree Char Ind |x| Conditions A 13 12 6= 13 + 11, 13 A 14 12 2, 7 + 11, 13 A 14 13 - 14 + 13 A 15 13 3, 5 + 13 A 15 14 `- 15 + 13 A 16 14 2 + 13, 16 A 16 15 `6= 2 + 16 2.PSL(2, 25) 12 `6= 2 - 16 r f =r d = 2 2.PSL(2, 49) 24 `6= 2 - 32 r f =r d = 2 PSL(2, 5) 2 2 - 3 2.PSL(2, 5) 2 6= 2 - 3 PSL(2, 5) 3 6= 2 + 3 PSL(2, 5) 4 + 3 2.PSL(2, 5) 4 q+1 2 ` 0 6= 1 - 3 PSL 2 (8) 7 + 9 r f = 3 PSL 2 (8) 8 `- 9 + 9 r f = 3 PSL 2 (8) 9 (7) ` 06= 1 + 9 r f = 3 4 2 .PSL 3 (4) 4 3 0 3,5,7,8 2.PSL 3 (4) 6 3 0 5,7,8 6.PSL 3 (4) 6 `6= 2, 3 0 5,7,8 3 1 .PSU(4, 3) 6 2 0 5,7,8,9 6 1 .PSU(4, 3) 6 6= 2, 3 0 5,7,8,9 PSU(3, 3) 6 `6= 3 - 7, 8 PSU(3, 3) 7 7 + 7, 8 PSU(3, 3) 7 7 - 7, 8 PSU(3, 4) 13 `6= 2, 5 0 13, 16 PSU(5, 2) 10 `6= 2 - 9, 11, 16 PSU(5, 2) 11 `6= 2, 3 0 11, 16 PSp 4 (4) 18 `6= 2 + 17 PSp 6 (2) 7 `6= 2 + 7, 8, 9 2.PSp 6 (2) 8 `6= 2 + 7, 8, 9 2.PSp 4 (3) 4 ` = 5 0 4, 5, 8, 9 PSp 4 (3) 5 ` = 5 0 4, 5, 8, 9 PSp 4 (3) 6 ` = 5 + 5, 8, 9 PSp 4 (3) 10 ` = 5 0 9 PSp 6 (3) 13 `6= 3 0 13 2.PSp 6 (3) 14 `6= 2, 3 0 13 2.Ω + 8 (2) 8 `6= 2 0 7, 8 2. 2 B 2 (8) 8 5 + 7,13 2 B 2 (8) 14 6= 2 0 13 G 2 (3) 14 `6= 3 + 13 2.G 2 (4) 12 `6= 2 - 13, 16 M 11 5 3 0 4,5,8,11 M 11 9 11 + 8,11 6 Continuation of Table 2 F ∗ (G) Degree Char Ind |x| Conditions M 11 10 6= 11 + 8,11 M 11 10 6= 2 0 11 M 11 11 6= 2, 3 + 11 2.M 12 6 3 0 5,8,11 M 12 10 2 + 8,11 M 12 10 3 + 11 2.M 12 10 6= 2 0 11 M 12 11 6= 2, 3 + 11 2.M 12 12 6= 2, 3 + 11 3.M 22 6 2 0 4,5,7,8,11 M 22 10 2 0 8,11 2.M 22 10 7 + 11 2.M 22 10 6= 2, 7 0 11 M 23 11 2 0 11,23 M 23 21 23 + 23 M 23 22 6= 2, 23 + 23 M 24 11 2 0 11,23 M 24 22 3 + 23 M 24 23 6= 2,3 + 23 J 1 7 11 + 7,11,19 J 1 14 11 + 19 J 1 20 2 + 19 J 2 6 2 - 4,5,7,8 2.J 2 6 5 - 5,7,8 2.J 2 6 6= 2,5 - 5,7,8 3.J 3 9 2 0 8,9,17,19 J 3 18 3 + 17,19 3.J 3 18 5 0 17,19 3.J 3 18 6= 3,5 0 17,19 Co 1 24 2 + 23 2.Co 1 24 6= 2 + 23 Co 2 22 2 + 23 Co 2 23 6= 2 + 23 Co 3 22 2 - 23 Co 3 22 3 + 23 Co 3 23 6= 2,3 + 23 Ru 28 2 + 29 2.Ru 28 6= 2 0 29 2.Suz 12 3 - 11,13,16 3.Suz 12 2 0 11,13 6.Suz 12 6= 2, 3 0 11,13 7 3.2 Results for problem 2 Table 3: Generic cases with `-|x| F ∗ (G) Degree Char Ind |x| Conditions A n n− 2 2 =`|n + n−4≤|x|≤n n≥ 17, n6≡ 2 mod 4 A n n− 2 2 =`|n - n−4≤|x|≤n n≥ 17, n≡ 2 mod 4 A n n− 2 26=`|n + n−3≤|x|≤n n≥ 17 A n n− 1 2 =`-n + n−3≤|x|≤n n≥ 17 A n n− 1 26=`-n + n−2≤|x|≤n n≥ 17 2.PSL 2 (q) q−1 2 `6= 2 - 1/2(q− 1) q F. prime 2.PSL 2 (q) q−1 2 `6= 2 - (q− 1) q F. prime 2.PSL 2 (q) q−1 2 `6= 2 - 2(q− 1) q F. prime PSL 2 (q) q+1 2 `6= 2 + q−1 2 q F. prime PSL 2 (q) q+1 2 `6= 2 + (q− 1) q F. prime PSL 2 (q) (q− 1) `6=q + (q− 1) q F. prime 2.PSL 2 (q) (q− 1) q+1 2 l 0 6= 1, - (q− 1) q F. prime `6= 2 2.PSL 2 (q) (q− 1) q+1 2 l 0 6= 1 - 2(q− 1) q F. prime `6= 2 PSL 2 (q) q `- (q + 1) + (q− 1) q F. prime PSL 2 (q) (q + 1) q−1 4 l 0 6= 1 + (q− 1) q F. prime `6= 2 2.PSL 2 (q) (q + 1) `6= 2 - (q− 1) q F. prime 2.PSL 2 (q) (q + 1) `6= 2 - 2(q− 1) q F. prime PSL 2 (q) q−1 2 `6= 2 0 q+1 2 q M. prime PSL 2 (q) q−1 2 `6= 2 0 (q + 1) q M. prime 2.PSL 2 (q) q+1 2 `6= 2 0 q+1 2 q M. prime 2.PSL 2 (q) q+1 2 `6= 2 0 (q + 1) q M. prime 2.PSL 2 (q) q+1 2 `6= 2 0 2(q + 1) q M. prime PSL 2 (q) (q− 1) `6= 2 + (q + 1) q M. prime 2.PSL 2 (q) (q− 1) `6= 2 - (q + 1) q M. prime 2.PSL 2 (q) (q− 1) `6= 2 - 2(q + 1) q M. prime PSL 2 (q) q `- (q + 1) + (q + 1) q M. prime PSL 2 (q) (q + 1) q−1 2 l 0 6= 1, + (q + 1) q M. prime `6= 2 2.PSL 2 (q) (q + 1) q−1 2 l 0 6= 1, - (q + 1) q M. prime `6= 2 2.PSL 2 (q) (q + 1) q−1 2 l 0 6= 1, - 2(q + 1) q M. prime `6= 2 2.PSL 2 (q) (q− 1)/2 `6= 2 - 2/3(q− 1) q− 1 = 3(2 i ), for some i 8 Continuation of Table 3 F ∗ (G) Degree Char Ind |x| Conditions 2.PSL 2 (q) (q + 1)/2 `6= 2 0 2/3(q + 1) q + 1 = 3(2 i ), for some i PSL 2 (q) (q− 1)/2 2 - q q prime, q≡ 1 mod 4 2.PSL 2 (q) (q− 1)/2 `6= 2,q - q q prime, q≡ 1 mod 4 PSL 2 (q) (q + 1)/2 `6= 2,q + q q prime, q≡ 1 mod 4 PSL 2 (q) (q− 1) `6=q + q q prime, q≡ 1 mod 4 2.PSL 2 (q) (q− 1) q+1 2 l 0 6= 1 - q q prime, `6=q q≡ 1 mod 4 PSL 2 (q) q `- (q + 1) + q q prime, `6=q q≡ 1 mod 4 PSL 2 (q) (q + 1) q−1 4 l 0 6= 1 + q q prime, `6=q q≡ 1 mod 4 2.PSL 2 (q) (q + 1) `6= 2,q - q q prime, q≡ 1 mod 4 PSL(2,q) (q− 1)/2 0 q q prime, q≡ 3 mod 4 2.PSL(2,q) (q + 1)/2 `6= 2 0 q q prime q≡ 3 mod 4 PSL(2,q) q− 1 + q q prime, q≡ 3 mod 4 2.PSL(2,q) (q− 1) `6= 2 - q q prime, q≡ 3 mod 4 2.PSL(2,q) (q− 1) q+1 2 l 0 6= 1 - q q prime, q≡ 1 mod 4 PSL(2,q) q `- (q + 1) + q q prime, q≡ 3 mod 4 PSL(2,q) (q + 1) q−1 2 l 0 6= 1 + q q prime, q≡ 3 mod 4 2.PSL(2,q) (q + 1) q−1 2 l 0 6= 1 - q q prime, `6= 2 q≡ 1 mod 4 PSL 2 (q) (q− 1)/2 2 - 1/2(q + 1) q + 1 = 2u i , for prime u6= 2 2.PSL 2 (q) (q− 1)/2 6= 2,u - 1/2(q + 1) q + 1 = 2u i , for prime u PSL 2 (q) (q + 1)/2 6= 2,u + 1/2(q + 1) q + 1 = 2u i , for prime u PSL 2 (q) (q− 1)/2 `6=u 0 1/2(q− 1) q− 1 = 2(u i ), for prime u 2.PSL 2 (q) (q + 1)/2 `6= 2,u 0 1/2(q− 1) q− 1 = 2(u i ), for prime u 9 Continuation of Table 3 F ∗ (G) Degree Char Ind |x| Conditions PSL 2 (q) q− 1 `6= 2 + q− 1 q− 1 M. prime PSL 2 (q) q `-q + 1 + q− 1 q− 1 M. prime PSL 2 (q) q + 1 (q−1) ` 0 6= 1, + q− 1 q− 1 M. prime `6= 2 PSL 2 (q) q− 1 `6= 2 + q + 1 q + 1 F. prime PSL 2 (q) q `-q + 1 + q + 1 q + 1 F. prime PSL 2 (q) q + 1 (q−1) ` 0 6= 1, + q + 1 q + 1 F. prime `6= 2 PSL n (q) q n −q q−1 `- q n −1 q−1 q n −1 q−1 , n≥ 3, prime u6= 2 PSL n (q) q n −1 q−1 `- q n −1 q−1 q n −1 q−1 , n≥ 3, prime u6= 2 PSU(n,q) q n −q q+1 `- q n +1 q+1 q n +1 q+1 , n odd, u6= 2 prime PSU(n,q) q n +1 q+1 `- q n +1 q+1 q n +1 q+1 , n odd, u6= 2 prime PSp 2n (3) (3 n − 1)/2 `-|x| (3 n − 1)/2 n odd prime, u6= 2 2.PSp 2n (3) (3 n + 1)/2 `-|x| (3 n − 1)/2 n odd prime, u6= 2 2.PSp 2n (q) (q n − 1)/2 `-|x| (q n + 1)/2 n = 2 a , u6= 2, q≡ 3 mod 4 PSp 2n (q) (q n + 1)/2 `-|x| (q n + 1)/2 n = 2 a , u6= 2, q≡ 3 mod 4 2.PSp 2n (q) (q n − 1)/2 `-|x| (q n + 1)/2 n = 2 a , u6= 2, q≡ 1 mod 4 PSp 2n (q) (q n + 1)/2 `-|x| (q n + 1)/2 n = 2 a , u6= 2, q≡ 1 mod 4 Table 4: Exceptional cases with `-|x| F ∗ (G) Degree Char Ind |x| Conditions 3.A 6 3 2 0 3, 5 3.A 6 3 5 0 2, 3, 4 A 6 4 2 - 3, 5 2.A 6 4 5 - 3, 4, 8 10 Continuation of Table 4 F ∗ (G) Degree Char Ind |x| Conditions A 6 5 5 + 4 6.A 6 6 5 0 8 3.A 7 3 5 0 2, 3, 4, 7 A 7 4 2 0 3, 5, 7 2.A 7 4 7 - 3, 4, 5, 8 2.A 7 4 6= 2, 7 0 3, 4, 5, 7, 8 `-|x| A 7 5 7 + 4, 5 A 7 6 2 + 5, 7 A 7 6 6= 2, 7 + 5, 7 `-|x| 2.A 7 6 3 - 5, 7, 8 3.A 7 6 2 0 5, 7 3.A 7 6 6= 3 0 5, 7 `-|x| 6.A 7 6 6= 2, 3 0 5, 7, 8 `-|x| A 7 8 5 + 7 A 8 7 6= 2 + 7, 8 `-|x| 2.A 8 8 6= 2 + 7, 8 `-|x| A 9 7 3 + 7, 8 A 9 8 2 + 7, 9 A 9 8 6= 2, 3 + 8, 9 `-|x| 2.A 9 8 6= 2 + 7, 8, 9 `-|x| A 10 8 2 - 7, 9 A 10 8 5 + 7, 8, 9 2.A 10 8 5 + 7, 8, 9 A 10 9 6= 2, 5 + 8, 9 `-|x| A 11 9 11 + 8, 9 A 11 10 2 + 9, 11 A 11 10 6= 2, 11 + 9, 11 `-|x| A 12 10 2 + 9, 11 A 12 10 3 + 11 A 12 11 6= 2, 3 + 11 `-|x| A 13 11 13 + 13 A 13 12 6= 13 + 11, 13 `-|x| A 14 12 2, 7 + 11, 13 A 14 13 - 14 + 13 `-|x| A 15 13 3, 5 + 13 A 15 14 `- 15 + 13 `-|x| A 16 14 2 + 13 A 16 15 `6= 2 + 16 2.PSL 2 (25) 12 `6= 2 - 16 r f =r d = 2 2.PSL 2 (49) 24 `6= 2 - 32 r f =r d = 2 PSL 2 (5) 2 2 - 3 2.PSL 2 (5) 2 6= 2 - 3 `-|x| PSL 2 (5) 3 6= 2 + 3 `-|x| PSL 2 (5) 4 + 3 `-|x| 11 Continuation of Table 4 F ∗ (G) Degree Char Ind |x| Conditions 2.PSL 2 (5) 4 1/2(q + 1) ` 06= 1 - 3 `-|x| PSL 2 (8) 7 `6= 3 + 9 r f = 3 PSL 2 (8) 8 `- 9 + 9 r f = 3 PSL 2 (8) 9 (7) ` 06= 1, `6= 3 + 9 r f = 3 4 2 .PSL 3 (4) 4 3 0 5,7,8 2.PSL 3 (4) 6 3 0 5,7,8 6.PSL 3 (4) 6 `6= 2, 3 0 5,7,8 `-|x| 3 1 .PSU(4, 3) 6 2 0 5,7,8,9 `-|x| 6 1 .PSU(4, 3) 6 6= 2, 3 0 5,7,8,9 `-|x| PSU(3, 3) 6 `6= 3 - 7, 8 `-|x| PSU(3, 3) 7 7 + 8 PSU(3, 3) 7 7 - 8 PSU(3, 4) 13 `6= 2, 5 0 13, 16 `-|x| PSU(5, 2) 10 `6= 2 - 9, 11, 16 `-|x| PSU(5, 2) 11 `6= 2, 3 0 11, 16 `-|x| PSp 4 (4) 18 `6= 2, 17 + 17 PSp 6 (2) 7 `6= 2 + 7, 8, 9 `-|x| 2.PSp 6 (2) 8 `6= 2 + 7, 8, 9 `-|x| 2.PSp 4 (3) 4 ` = 5 0 4, 8, 9 PSp 4 (3) 5 ` = 5 0 4, 8, 9 PSp 4 (3) 6 ` = 5 + 8, 9 PSp 4 (3) 10 ` = 5 0 9 PSp 6 (3) 13 `6= 3 0 13 `-|x| 2.PSp 6 (3) 14 `6= 2, 3 0 13 `-|x| 2.Ω + 8 (2) 8 `6= 2 0 7, 8 2. 2 B 2 (8) 8 5 + 7,13 2 B 2 (8) 14 6= 2 0 13 `- 13 G 2 (3) 14 `6= 3, 13 + 13 2.G 2 (4) 12 `6= 2 - 13, 16 M 11 5 3 0 4,5,8,11 M 11 9 11 + 8 M 11 10 6= 11 + 8,11 `-|x| M 11 10 6= 2, 11 0 11 M 11 11 6= 2, 3, 11 + 11 2.M 12 6 3 0 5,8,11 M 12 10 2 + 11 M 12 10 3 + 11 2.M 12 10 6= 2, 11 0 11 M 12 11 6= 2, 3, 11 + 11 2.M 12 12 6= 2, 3, 11 + 11 3.M 22 6 2 0 5,7,11 M 22 10 2 0 11 2.M 22 10 7 + 11 2.M 22 10 6= 2, 7, 11 0 11 12 Continuation of Table 4 F ∗ (G) Degree Char Ind |x| Conditions M 23 11 2 0 11,23 M 23 22 6= 2, 23 + 23 M 24 11 2 0 11,23 M 24 22 3 + 23 M 24 23 6= 2,3, 23 + 23 J 1 7 11 + 7,19 J 1 14 11 + 19 J 1 20 2 + 19 J 2 6 2 - 5,7 2.J 2 6 5 - 7,8 2.J 2 6 `6= 2,5 - 5,7,8 `-|x| 3.J 3 9 2 0 9,17,19 J 3 18 3 + 17,19 3.J 3 18 5 0 17,19 3.J 3 18 6= 3,5 0 17,19 `-|x| Co 1 24 2 + 23 2.Co 1 24 6= 2, 23 + 23 Co 2 22 2 + 23 Co 2 23 6= 2, 23 + 23 Co 3 22 2 - 23 Co 3 22 3 + 23 Co 3 23 6= 2,3, 23 + 23 Ru 28 2 + 29 2.Ru 28 6= 2, 29 0 29 2.Suz 12 3 - 11,13,16 3.Suz 12 2 0 11,13 6.Suz 12 6= 2, 3 0 11,13 `-|x| 3.3 Results for problem 3 Here we list the possibilities for the maximal Lie primitive subgroups of the classical algebraic groups containing a regular unipotent element. As with the previous two tables, we list the possi- bilities according to the nature of the quasisimple group F ∗ (G). We note that there are two cases where the Jordan normal form of a possible regular unipotent element is left unresolved: one generic case, where F ∗ (G) =PSL 2 (q) with q a Marsenne prime, ` = 2 and|x| = 1/2(q + 1) and the other the only exceptional case where F ∗ (G) = 3 1 .PSU(4, 3). Concerning the format of the following tables, we list under the column x the Jordan normal form of x by giving a multiset including the sizes of the Jordan blocks, which is sufficient to completely characterize it in these cases since x is 13 unipotent and thus has eigenvalues all equal to one. Also, for the alternating groups we list the cycle decomposition of x. Table 5: Regular unipotent element when S = A n , standard rep- resentation F ∗ (G) Degree Char Ind x Conditions A n n− 2 2 =`|n + |x| =n− 4, (1,...,n−4)(n−3,n−2)(n−1)(n), {n− 4, 2} A n n− 2 2 =`|n + |x| =n− 4, (1,...,n−4)(n−3,n−2,n−1,n), {n− 4, 2} A n n− 1 3 =`-n + |x| =n− 3, (1,...,n−3)(n−2)(n−1)(n), {n− 3, 1} A n n− 1 2 =`-n + |x| =n− 3, (1,...,n−3)(n−2,n−1)(n), {n− 3, 2} A n n− 2 3 =`|n + |x| =n− 3, (1,...,n−3)(n−2,n−1,n), {n− 3, 1} A n n− 2 2 =`|n - |x| =n− 2, (1,...,n−2)(n−1,n), {n− 2} A n n− 2 2 =`|n - |x| =n− 2, (1,...,n−2)(n−1)(n), {n− 2} A n n− 1 26=`-n + |x| =n− 2, (1,...,n−2)(n−1)(n), {n− 2, 1} A n n− 1 `-n + |x| =n− 1, (1,...,n−1)(n), {n− 1} A n n− 1 26=`|n + |x| =n, (1,...,n−1,n), {n− 1} A n n− 2 2 =`|n - |x| =n, n≡ 2 mod 4 (1,...,n−1,n), {n− 1} A n n− 2 2 =`|n + |x| =n, n6≡ 2 mod 4 (1,...,n−1,n), {n− 1} 14 Table 6: Regular unipotent element when S =PSL 2 (q), generic cases F ∗ (G) Degree Char Ind |x|, JNF Conditions PSL 2 (q) 1/2(q− 1) 2 - |x| = 1/2(q− 1), q F. prime {1/2(q− 1)} PSL 2 (q) (q− 1) 2 + |x| = (q− 1), q F. prime {q− 1} 2.PSL 2 (q) (q− 1) 2 - |x| = (q− 1), q F. prime {q− 1} PSL 2 (q) 1/2(q− 1) 2 0 |x| = 1/2(q + 1) q6= 9 M. prime PSL 2 (q) (q− 1) 2 + |x| = (q + 1), q6= 9 M. prime {q + 1} PSL 2 (q) (q + 1) 2 + |x| = (q + 1), q6= 9 M. prime {q + 1} 2.PSL 2 (q) 1/2(q− 1) `6= 2 - |x| = 1/2(q + 1), q + 1 = 2` i {1/2(q− 1)} PSL 2 (q) 1/2(q + 1) `6= 2 + |x| = 1/2(q + 1), q + 1 = 2` i {1/2(q + 1)} PSL 2 (q) 1/2(q− 1) `6= 2 0 |x| = 1/2(q− 1), q− 1 = 2(` i ) {q− 1} 2.PSL 2 (q) (q + 1)/2 `6= 2 0 |x| = 1/2(q− 1), q− 1 = 2(` i ) {q− 1, 1} PSL 2 (q) q− 1 q− 1 + |x| =q− 1, q− 1 M. prime {q− 1} PSL 2 (q) q q− 1 + |x| =q− 1, q− 1 M. prime {q− 1, 1} PSL 2 (q) q− 1 q + 1 + |x| =q + 1, q + 1 F. prime {q + 1} PSL 2 (q) q + 1 q + 1 + |x| =q + 1, q + 1 F. prime {q + 1} Table 7: Generic cases with regular unipotent element F ∗ (G) Degree Char Ind |x|, JNF Conditions PSL n (q) q n −q q−1 − 1 6= 2 + |x| = q n −1 q−1 , n≥ 3, prime { q n −q q−1 − 1} PSL n (q) q n −1 q−1 6= 2 + |x| = q n −1 q−1 , n≥ 3, prime, { q n −1 q−1 } real-valued character PSL n (q) q n −1 q−1 6= 2 0 |x| = q n −1 q−1 , n≥ 3, prime, { q n −1 q−1 } character not 15 Continuation of Table 7 F ∗ (G) Degree Char Ind |x|, JNF Conditions real-valued PSU(n,q) q n −q q+1 6= 2 - |x| = q n +1 q+1 , n odd prime { q n −q q+1 } PSU(n,q) q n +1 q+1 6= 2 + |x| = q n +1 q+1 , n odd prime, { q n +1 q+1 } real-valued character PSU(n,q) q n +1 q+1 6= 2 0 |x| = q n +1 q+1 , n odd prime, { q n +1 q+1 } character not real-valued PSp 2n (3) (3 n − 1)/2 `||x|, 0 |x| = (3 n − 1)/2, n odd prime 6= 2 {(3 n − 1)/2} 2.PSp 2n (3) (3 n + 1)/2 `||x|, 0 |x| = (3 n − 1)/2, n odd prime 6= 2 {(3 n − 1)/2, 1} 2.PSp 2n (q) (q n − 1)/2 `||x|, - |x| = (q n + 1)/2, n = 2 a , 6= 2 {(q n − 1)/2} q≡ 1 mod 4 PSp 2n (q) (q n − 1)/2 `||x|, 0 |x| = (q n + 1)/2, n = 2 a , 6= 2 {(q n − 1)/2} q≡ 3 mod 4 2.PSp 2n (q) (q n + 1)/2 `||x|, + |x| = (q n + 1)/2, n = 2 a , 6= 2 {(q n + 1)/2} q≡ 1 mod 4 PSp 2n (q) (q n + 1)/2 `||x|, 0 |x| = (q n + 1)/2, n = 2 a , 6= 2 {(q n + 1)/2} q≡ 3 mod 4 Table 8: G with regular unipotent element, exceptional cases F ∗ (G) Degree Char Ind |x|, JNF Conditions 3.A 6 3 2 0 4,{3} 3.A 6 3 5 0 5,{3} A 6 4 2 - 4,{4} 2.A 6 4 5 - 5,{4} 3.A 7 3 5 0 5,{3} A 7 4 2 0 4,{4} 2.A 7 4 7 - 7,{4} 2.A 7 4 5 0 5,{4} 3.A 7 6 7 0 7,{6} 6.A 7 6 7 0 7,{7} 2.A 8 8 7 + 7,{7, 1} 2.A 9 8 7 + 7,{7, 1} 2.A 9 8 3 + 9,{7, 1} A 10 8 2 - 8,{8} 2.PSL 2 (5) 2 3 - 3,{2} PSL 2 (5) 3 3 + 3,{2} 16 Continuation of Table 8 F ∗ (G) Degree Char Ind |x|, JNF Conditions PSL 2 (5) 4 3 + 3,{3, 1} PSL 2 (8) 7 3 + 9,{7} r f = 3 6.PSL 3 (4) 6 ` = 7 0 7,{6} 3 1 .PSU(4, 3) 6 2 0 8 6 1 .PSU(4, 3) 6 7 0 7,{6} PSU(3, 3) 6 7 - 7,{6} PSU(3, 3) 6 2 - 8,{6} PSU(3, 3) 7 7 + 7,{7} PSU(3, 3) 7 7 0 7,{7} PSU(3, 4) 12 13 - 13,{12} PSU(5, 2) 10 11 - 11,{10} PSU(5, 2) 11 11 0 11,{11} PSp 4 (4) 18 17 + 17,{16, 1} PSp 6 (2) 7 7 + 7,{7} PSp 6 (2) 7 3 + 9,{7} 2.PSp 6 (2) 8 7 + 7,{7, 1} 2.PSp 6 (2) 8 3 + 9,{7, 1} 2.PSp 4 (3) 4 ` = 5 0 5,{4} PSp 4 (3) 5 ` = 5 0 5{5} PSp 4 (3) 6 ` = 5 + 5{5, 1} PSp 6 (3) 13 13 0 13,{13} 2 B 2 (8) 14 13 + 13,{13, 1} 2.G 2 (4) 12 13 - 13,{12} M 11 9 11 + 11,{12} M 11 10 2 + 8,{8, 2} M 11 10 11 0 11,{10} M 11 11 11 + 11,{11} M 12 10 2 + 8,{8, 2} 2.M 12 10 11 0 11,{10} M 12 11 11 + 11,{10} 2.M 12 12 11 + 11,{11, 1} 3.M 22 6 2 0 8,{6} 2.M 22 10 11 0 11,{10} M 23 21 23 + 23,{21} M 24 23 23 + 23,{23} J 1 7 11 + 11,{7} J 2 6 2 - 8,{7} 2.J 2 6 7 - 7,{6} 2.Co 1 24 23 + 23,{23, 1} Co 2 23 23 + 23,{23} Co 3 23 23 + 23,{23} 2.Ru 28 29 0 29,{28} 6.Suz 12 13 0 13,{12} 6.Suz 12 11 0 11,{12} 17 Continuation of Table 8 F ∗ (G) Degree Char Ind |x|, JNF Conditions 4 Problem 1 We start with a preliminary lemma, found in Guralnick and Tiep [11] as Lemma 2.1, that will be used throughout the paper but not explicitly mentioned. Lemma 4.1. A finite group G is almost quasi-simple exactly when Z(G) = Z(F ∗ (G)) and F ∗ (G)/Z(F ∗ (G)) is simple and non-abelian. In particular, this implies that for any almost quasi-simple group G and for any elementx∈G, we have|x||SM|xZ(G)|, where SM is the order of the Schur multiplier of F ∗ (G)/Z(F ∗ (G)). We use this fact repeatedly throughout the paper. 4.1 Alternating groups We first assume that S ∼ =A n . Then F ∗ (G) ∼ =A n or a covering group of A n . We assume first that F ∗ (G) ∼ = A n , n≥ 17, and that the representation under consideration is of degree at least ∂ 2 ` (F ∗ (G))≥ n 2 −5n 2 , a bound due to James, recorded as Theorem 2.2 of Tiep [25]. For a contradiction, we assume that G contains a prime power order element of order at least m− 2 = ∂ 2 ` (F ∗ (G))− 2≥ n 2 −5n 2 − 2. But n 2 −5n 2 − 2 > n, and the largest possible prime power element of G≤ Aut(A n ) ∼ = S n is n, a contradiction. Thus the only representations of G that allow G to contain a prime power element x with the assumed characteristics are those of degree < ∂ 2 ` (F ∗ (G)). By Lemma 6.1 of Guralnick and Tiep [11], the only such representations are the hearts of the natural permutation modules (also referred to below as the standard representations), of degreen− 2 if`|n and of degreen− 1 if not. We list these representations below in the generic cases table, along with the elements x meeting our required specifications. The remaining cases, namely those where n < 17, we analyze individually by inspection of Hiss and Malle [14] and by 18 considering the fact that any prime power order element of Aut(A n ) ∼ = S n must have order≤ n. We list these cases below. Next we assume that F ∗ (G) ∼ = 2A n and n≥ 13. Then the representation under consideration is of degree at least ∂ 1 ` (F ∗ (G)) = 2 n−3 2 , a result due to Kleshchev and Tiep, recorded as Theorem 2.3 in Tiep [25]. For a contradiction, we assume that G contains a prime power order element of order at least m− 2 = ∂ 1 ` (F ∗ (G))− 2 = 2 n−3 2 − 2. Now, since xZ(G) is a prime power order element of S n , we have|xZ(G)|≤n, and thus we have|x|≤ 2|xZ(G)|≤ 2n. But 2 n−3 2 − 2> 2n, so|x|≤ 2n< 2 n−3 2 − 2≤|x|, a contradiction. The remaining cases we analyze individually by inspection of the faithful, absolutely irreducible representations of F ∗ (G), listed in Hiss and Malle [14], that are of degree small enough so that G contains a prime power order satisfying the assumed conditions. We list these cases below. Table 9: Alternating groups, generic cases F ∗ (G) Degree Char Ind |x| Conditions A n n− 2 2 =`|n + n− 4≤|x|≤n n≥ 17 A n n− 2 26=`|n + n− 3≤|x|≤n n≥ 17 A n n− 1 2 =`-n + n− 3≤|x|≤n n≥ 17 A n n− 1 26=`-n + n− 2≤|x|≤n n≥ 17 Table 10: Alternating groups, small cases F ∗ (G) Degree Char Ind |x| Conditions 3.A 6 3 2 0 2, 3, 4, 5 3.A 6 3 5 0 2, 3, 4, 5 A 6 4 2 - 2, 3, 4, 5 2.A 6 4 5 - 3, 4, 5, 8 A 6 5 5 + 4, 5 3.A 6 6 5 0 5 6.A 6 6 5 0 5, 8 3.A 7 3 5 0 2, 3, 4, 5, 7 A 7 4 2 0 2, 3, 4, 5, 7 2.A 7 4 7 - 3, 4, 5, 7, 8 2.A 7 4 6= 2, 7 0 3, 4, 5, 7, 8 A 7 5 7 + 4, 5, 7 19 Continuation of Table 10 F ∗ (G) Degree Char Ind |x| Conditions A 7 6 2 + 4, 5, 7 A 7 6 6= 2, 7 + 5, 7 2.A 7 6 3 - 5, 7, 8 3.A 7 6 2 0 4, 5, 7 3.A 7 6 6= 2, 3 0 5, 7 6.A 7 6 6= 2, 3 0 5, 7, 8 A 7 8 5 + 7 A 8 7 6= 2 + 7, 8 2.A 8 8 6= 2 + 7, 8 A 9 7 3 + 7, 8, 9 A 9 8 2 + 7, 8, 9 A 9 8 6= 2, 3 + 8, 9 2.A 9 8 6= 2 + 7, 8, 9 A 10 8 2 - 7, 8, 9 A 10 8 5 + 7, 8, 9 2.A 10 8 5 + 7, 8, 9 A 10 9 6= 2, 5 + 8, 9 A 11 9 11 + 8, 9, 11 A 11 10 2 + 8, 9, 11 A 11 10 6= 2, 11 + 9, 11 A 12 10 2 + 8, 9, 11 A 12 10 3 + 9, 11 A 12 11 6= 2, 3 + 11 A 13 11 13 + 11, 13 A 13 12 6= 13 + 11, 13 A 14 12 2, 7 + 11, 13 A 14 13 `- 14 + 13 A 15 13 3, 5 + 13 A 15 14 `- 15 + 13 A 16 14 2 + 13, 16 A 16 15 `6= 2 + 16 4.2 Classical groups of Lie type For this section, we assume thatS is a classical group of Lie type. For eachg∈Aut(S), we have g = h i h d h f h g , where h i is an inner automorphism, h d is a diagonal automorphism, h f is a field automorphism, and h g is a graph automorphism of S by Theorem 2.5.1 of Gorenstein, Lyons and Solomon [5]. We let r f =|h f | andr g =|h g |. Now, h f andh g commute and h i ,h d ∈Inndiag(S)E Aut(S), so the coset Inndiag(S)g = Inndiag(S)h f h g has order equal to lcm(r f ,r g ), which we 20 denote by r fg . Thus g r fg is equal to an innerdiagonal automorphism h 0 i h 0 d where h 0 i is an inner automorphism and h 0 d is a diagonal automorphism. Then if r i is the order of h 0 i we have g r fg ri equal to an inner automorphism h 00 i . Finally, we let|h 00 i | =r i . Thus|g| =r i r d r fg . We collect a few lemmas from Dimuro [4] related to such automorphisms. The notation p b ||y means that p b is the p-part of y. Lemma 4.2. Suppose that p is a prime, that s and n are natural numbers with s≥ 2 and where p| (s− 1). Then, with one exception, p α || (s n − 1) iff p α || (n(s− 1)). The exception occurs when p = 2, n is even, and s≡ 3 (mod 4), in which case p α || (s n − 1) iff p α || (n(s + 1)). Lemma 4.3. Suppose thatp is a prime, and thats andn are natural numbers withp|(s+1). Then if n is odd, p α ||(s n + 1) iff p α ||n(s + 1). If n is even, then p| s n + 1 iff p = 2, in which case 2 1 ||(s n + 1). Lemma 4.4. If x∈GL n (p b ) where x is of p power order, then|x|≤p(n− 1). Lemma 4.5. Suppose S ∼ = X m (s) is an untwisted group and suppose x is a prime power order element of Aut(S) with notation as above and suppose that r f > 1. If r g = 1, then r i r d is equal to the order of an element of Inndiag(X m (s 1/r f )), the group of inner diagonal automorphisms of X m (s 1/r f ). If r g = 2, then r i r d is equal to the order of an element of Inndiag( 2 X m (s 1/r f )). 4.2.1 Special linear groups Now, for the first Lie type case, we assume that S≤ G/Z(G)≤ Aut(S) with S ∼ = PSL(n,q) and q = p a , that n≥ 5, and that the representation under consideration is of degree at least ∂ 2 ` (F ∗ (G))≥ (q n−1 − 1)( q n−2 −q q−1 − 1). (The bound on F ∗ (G) is due to Guralnick and Tiep [11], recorded in Tiep [25].) For a contradiction, we assume thatG contains a prime power order element x with|x|≥m−2 =∂ 2 ` (F ∗ (G))−2≥ (q n−1 −1)( q n−2 −q q−1 −1)−2. But (q n−1 −1)( q n−2 −q q−1 −1)−2> q n −1 q−1 , and, as analyzed below, the largest possible prime power element of a covering group of Aut(PSL(n,q)) is q n −1 q−1 (with a few exceptions we consider separately), a contradiction. Similarly, for n = 4, disregarding the special cases where (n,q) = (4, 2) or (n,q) = (4, 3), the bound on ∂ 2 ` (F ∗ (G)) = (q− 1)(q 3 − 1)/(2,q− 1) rules out the possibility of G containing a prime 21 power element of order at leastm− 2 = (q− 1)(q 3 − 1)/(2,q− 1)− 2 but less than or equal to q n −1 q−1 , as (q− 1)(q 3 − 1)/(2,q− 1)− 2> q n −1 q−1 for n = 4 and q≥ 4. Now, concerning the case where (n,q) = (4, 2), we have S ∼ = A 8 , so we already considered this. And in the case where S ∼ = PSL(4, 3), we can rule this out by inspection of the Atlas [3] and by inspection of Hiss and Malle [14]: from the Atlas,|x|≤ 16, but from Hiss and Malle, the absolutely irreducible faithful representation of smallest degree of any covering group of S is 26, so |x|≥m− 2≥ 26− 2 = 24. As to then = 3 case, a similar constraint on∂ 2 ` (F ∗ (G)) for (n,q)6= (3, 2) and (n,q)6= (3, 4) and using the same upper bound on|x| as above rule this out. As PSL(3, 2) ∼ =PSL(2, 7), we consider this case below. And we consider the S ∼ =PSL(3, 4) case below individually. Concerning then = 2 case we assume in what follows thatq6= 4 andq6= 9, sincePSL(2, 4) ∼ =A 5 and PSL(2, 9) ∼ = A 6 and were thus already considered. We first assume that q is odd and that |x| is even. We note that in this case we have absolutely irreducible, faithful representations of degree 1/2(q± 1), q and q± 1. We have xZ(G)∈Aut(PSL(2,q)) ∼ =PGL(2,q).y, where y acts on PGL(2,q) by mapping each matrix entry to its p th power. Assuming first that x does not contain a field automorphism, we have that|xZ(G)| is equal to the order of an element of PGL(2,q), so |xZ(G)|| 1/2p(q 2 − 1). We note that p-|x| in this case, as p is odd and|x| was assumed to be even. Thus|xZ(G)|| 1/2(q−1)(q +1) (the exponent ofPGL(2,q) without the factor ofp). Now, (q−1)(q +1) 2 | 2(q±1) (since it is impossible for 4 to divide bothq + 1 andq− 1), with equality only whenq is a Marsenne prime or a Fermat prime or the special case where q = 9 by Catalan’s conjecture. However, we disregarded theq = 9 case from the beginning, so we have equality only in the Marsenne prime or a Fermat prime cases. Thus,|xZ(G)|| 1/2(q−1)(q+1)| 1/2(q−1)(q+1) 2 | 1/2(2(q±1)) =q±1. Thus, since the Schur multiplier is of order 2 for all cases considered now, we have|x|| 2|xZ(G)|| 2(q±1), again with equality only possible when q is a Marsenne prime or a Fermat prime. We consider the possibilities stemming from these cases. Now, concerning the absolutely irreducible representations of S ∼ =PSL(2,q) or S ∼ = 2.PSL(2,q), by results from Guralnick and Tiep, recorded in Table 2 of Hiss and Malle [14], we have absolutely irreducible representations of degree (q± 1)/2, q and of 22 order q± 1. This gives a number of possibilities for x meeting the desired conditions that we list. We list below the possibilities that stem from the cases where q is a Fermat prime and then the cases where q is a Marsenne prime. Then we further analyze the cases where q is neither a Fermat prime nor a Marsenne prime below. Table 11: PSL(2,q) with q6= 9, q Fermat prime F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) (q− 1)/2 2 - 1/2(q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1)/2 `6= 2 - 1/2(q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1)/2 `6= 2 - (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1)/2 `6= 2 - 2(q− 1) q6= 9 F. prime PSL(2,q) (q + 1)/2 `6= 2 + 1/2(q− 1) q6= 9 F. prime PSL(2,q) (q + 1)/2 `6= 2 + (q− 1) q6= 9 F. prime PSL(2,q) (q− 1) + (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1) q+1 2 l 0 6= 1 - (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q− 1) q+1 2 l 0 6= 1 - 2(q− 1) q6= 9 F. prime PSL(2,q) q `- (q + 1) + (q− 1) q6= 9 F. prime PSL(2,q) (q + 1) q−1 4 l 0 6= 1 + (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q + 1) `6= 2 - (q− 1) q6= 9 F. prime 2.PSL 2 (q) (q + 1) `6= 2 - 2(q− 1) q6= 9 F. prime End of Table Table 12: PSL(2,q) with q6= 9, q Marsenne prime F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) (q− 1)/2 0 1/2(q + 1) q6= 9 M. prime PSL(2,q) (q− 1)/2 0 (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q + 1)/2 `6= 2 0 1/2(q + 1) q6= 9 M. prime 2.PSL 2 (q) (q + 1)/2 `6= 2 0 (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q + 1)/2 `6= 2 0 2(q + 1) q6= 9 M. prime PSL(2,q) (q− 1) + (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q− 1) `6= 2 - (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q− 1) `6= 2 - 2(q + 1) q6= 9 M. prime PSL(2,q) q `- (q + 1) + (q + 1) q6= 9 M. prime PSL(2,q) (q + 1) q−1 2 l 0 6= 1 + (q + 1) q6= 9 M. prime 2.PSL 2 (q) (q + 1) q−1 2 l 0 6= 1, - (q + 1) q6= 9 M. prime `6= 2 2.PSL 2 (q) (q + 1) q−1 2 l 0 6= 1, - 2(q + 1) q6= 9 M. prime `6= 2 23 Continuation of Table 12 F ∗ (G) Degree Char Ind |x| Conditions End of Table Next, concerning the case where q is neither a Marsenne prime nor a Fermat prime, by the same reasoning as above, we have that|x|| 2(q± 1). Thus,|x|| 2(q± 1) 2 ≤ 2/t(q± 1) for some t6= 2 where t| (q± 1) and (t, 2) = 1. For the cases where 2(q± 1) = 3(2 j ) for some j, we have the possibility of an x with|x| = 2/3(q± 1)≥m− 2 for the irreducible representations of degree (q± 1)/2. We list these representations below. So we assume that t≥ 5. That is, we assume that 3- 2(q± 1). Of the cases where t = 5, we only have 2/5(q± 1)≥ m− 2≥ (q− 1)/2− 2 when q = 11 or 19. Concerning the cases with q = 11 or 19, we analyze them individually. For q = 11, we have 4 = 2/5(q− 1)≥m− 2 form = (q− 1)/2 = 5. However, the absolutely irreducible, faithful representation of order 5 has F ∗ (G) ∼ =PSL(2, 11), but not F ∗ (G) ∼ = 2.PSL(2, 11), so for such an x we must have|x| =|xZ(G)|. Thus we have|x| =|xZ(G)| = 1/5(q− 1) = 2 < 3 = m− 2, so this is ruled out. For q = 19, we have 8 = 2/5(q + 1)≥ m− 1 for m = (q− 1)/2 = 9. But for the same reason as the q = 11 case, we can rule out this possibility. The cases where t≥ 7 give no possibilities. Table 13: PSL(2,q), q− 1 = 3(2 i ) for some i F ∗ (G) Degree Char Ind |x| Conditions 2.PSL(2,q) (q− 1)/2 6= 2 - 2/3(q− 1) q− 1 = 3(2 i ), for some i End of Table Table 14: PSL(2,q) with q6= 9, q + 1 = 3(2 i ) for some i F ∗ (G) Degree Char Ind |x| Conditions 2.PSL(2,q) (q + 1)/2 `6= 2 0 2/3(q + 1) q + 1 = 3(2 i ), for some i End of Table 24 Next we assume thatx contains a field automorphism. Then by Lemma 4.5, taken from Dimuro [4], r i r d is equal to the order of an element in Inndiag(PSL(2,q 1/r f )) ∼ = (PGL(2,q 1/r f )), hence r i r d |(q 1/r f ± 1) 2 | (q 1/r f ± 1). Thus we have|xZ(G)|| r i r d r f | (q 1/r f ± 1)r f . Thus, as the Schur multiplier in this case has order 2, we have|xZ(G)|| 2r f (q 1/r f ± 1). But in all cases except for q∈{3 4 , 5 2 , 7 2 } where r f = 2, we have 2r f (q 1/r f ± 1) < (q− 1)/2− 2, so|x|≤ 2r f (q 1/r f ± 1) < (q− 1)/2)− 2≤|x|, a contradiction. We analyze the exceptional cases individually: We eliminate the q = 3 4 case by direct computation. For q = 5 2 , we have an element|xZ(G)| involving a field automorphism of order r i r d r f = [1/2(q 1/2 − 1)](2)(2) = 8, and thus|x|| 2|xZ(G)| = 16. And moreover we have an absolutely irreducible representation whenF ∗ (G) ∼ = 2.PSL(2, 25) with degree less than 16. We list the possibilities in the following table. And for the q = 7 2 case, we have an element|x| involving a field automorphism of order r i r d r f = [1/2(q 1/2 + 1)](2)(2) = 16 and thus |x|| 2|xZ(G)| = 2(16) = 32. And moreover we have an absolutely irreducible representation when F ∗ (G) ∼ = 2.PSL(2, 49) with degree less than 32. Table 15: PSL(2,q) where x contains a field automorphism F ∗ (G) Degree Char Ind |x| Conditions 2.PSL(2, 25) 12 `6= 2 - 16 r f = 2 2.PSL(2, 49) 24 `6= 2 - 32 r f = 2 End of Table Next we assume that|x| and q are odd and that x does not contain a field automorphism. We first assume that p||x|. Then by Lemma 4.4,|x|| p, so|x| = p. Now, disregarding the (n,q) = (2, 9), case, ifa> 1 (recall that q =p a ), thenp< (q− 1)/2− 2 (we recall that (q− 1)/2 is the smallest degree of an irreducible representation in this case), so|x| =p< (q− 1)/2− 2≤|x|, a contradiction. Concerning the (n,q) = (2, 9) case, this was already considered, since PSL(2, 9) ∼ = A 6 . Now, concerning the a = 1 case, we have possibilities for every absolutely irreducible faithful representation of G, as|x| = p is greater than or equal to m− 1 in each case. We list these representations in the following table. 25 Table 16: SL 2 (q), q≡ 1 mod 4 and q prime F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) (q− 1)/2 2 - q q prime, q≡ 1 mod 4 2.PSL(2,q) (q− 1)/2 `6= 2 - q q prime q≡ 1 mod 4 PSL(2,q) (q + 1)/2 `6= 2 + q q prime, q≡ 1 mod 4 PSL(2,q) (q− 1) + q q prime, q≡ 1 mod 4 2.PSL(2,q) (q− 1) q+1 2 l 0 6= 1 - q q prime, q≡ 1 mod 4 PSL(2,q) q `- (q + 1) + q q prime, q≡ 1 mod 4 PSL(2,q) (q + 1) q−1 4 l 0 6= 1 + q q prime, q≡ 1 mod 4 2.PSL(2,q) (q + 1) `6= 2 - q q prime, q≡ 1 mod 4 End of Table Table 17: SL 2 (q), q≡ 3 mod 4 and q prime F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) (q− 1)/2 0 q q prime, q≡ 3 mod 4 2.PSL(2,q) (q + 1)/2 `6= 2 0 q q prime q≡ 3 mod 4 PSL(2,q) q− 1 + q q prime, q≡ 3 mod 4 2.PSL(2,q) (q− 1) `6= 2 - q q prime, q≡ 3 mod 4 2.PSL(2,q) (q− 1) q+1 2 l 0 6= 1 - q q prime, q≡ 1 mod 4 PSL(2,q) q `- (q + 1) + q q prime, q≡ 3 mod 4 PSL(2,q) (q + 1) q−1 2 l 0 6= 1 + q q prime, q≡ 3 mod 4 2.PSL(2,q) (q + 1) q−1 2 l 0 6= 1 - q q prime, `6= 2 q≡ 1 mod 4 End of Table 26 Next we assume that p - |x| and that x does not contain a field automorphism. Now, as r d | (n,q− 1) = (2,q− 1) but|x| is odd, we must have r d = 1. Similarly, as the Schur multiplier has order dividing 2, we must have|x| =|xZ(G)|. Thus|x| is equal to the order of an element of PSL(n,q). Thus, since|x| is odd and p -|x|,|x|| 1/2(q± 1) by Section 3.1 of Kantor and Seress [19]. Now, if|x| = 1/2(q± 1), we have|x| = 1/2(q± 1)≥ m− 1≥ 1/2(q± 1)− 2 for m = 1/2(q± 1). We list these possibilities below. But with the exception of (n,q) = (2, 5) or (2, 7), which we analyze individually, for any representation of degree greater than or equal to q− 1, we have|x|| 1/2(q± 1)< (q− 1)− 2≤m− 2≤|x|, so|x|<|x|, a contradiction. Table 18: PSL(2,q), q≡ 1 mod 4,|x| odd F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) (q− 1)/2 2 - 1/2(q + 1) q + 1 = 2u i , for prime u6= 2 2.PSL(2,q) (q− 1)/2 6= 2 - 1/2(q + 1) q + 1 = 2u i , for prime u6= 2 PSL(2,q) (q + 1)/2 6= 2 + 1/2(q + 1) q + 1 = 2u i , for prime u6= 2 End of Table Table 19: SL 2 (q), q≡ 3 mod 4,|x| odd F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) (q− 1)/2 0 1/2(q− 1) q− 1 = 2(u i ), for prime u6= 2 2.PSL(2,q) (q + 1)/2 `6= 2 0 1/2(q− 1) q− 1 = 2(u i ), for prime u6= 2 End of Table Table 20: Exceptional case F ∗ (G) Degree Char Ind |x| Conditions PSL(2, 5) 2 2 - 3 2.PSL(2, 5) 2 6= 2 - 3 PSL(2, 5) 3 6= 2 + 3 27 Continuation of Table 20 F ∗ (G) Degree Char Ind |x| Conditions PSL(2, 5) 4 + 3 2.PSL(2, 5) 4 q+1 2 ` 0 6= 1 - 3 End of Table Next we assume that|x| is odd, q is odd and that x contains a field automorphism. We note that in this case we must have r f ≥ 3, as x is odd. As above, we must have that|x| =|xZ(G)| and thatr d = 1. Thus, as before, we haver i equal to an odd order element ofPSL(2,q 1/r f ). Thus |x||r f (1/2(q 1/r f ± 1)). But|x||r f (1/2(q 1/r f ± 1))< 1/2(q− 1)− 2≤m− 2≤|x| in all cases, so |x|<|x|, a contradiction. For the next case we assume that q is even. We note that in this case we have absolutely irreducible, faithful representations of degreeq andq±1, so in particular we havem≥q−1. We start by assuming that|x| is even. Then by Lemma 4.4,|x| = 2. Thus, we have 2 =|x|≥m− 1≥q− 2, so 4≥q. Now, the only q with q even and q≤ 4 are q = 2 or q = 4. But for q = 2, PSL(2,q) is not simple and for q = 4, we have that PSL(2, 4) ∼ =A 5 and it was thus already considered. Next we assume that|x| is odd. In this case, we must have r d = 1 since|x| is odd but r d | (2,q− 1) and we must also have that|x| =|xZ(G)| since the Schur multiplier of S is even if it is non-trivial. Now, we start by assuming that r f = 1. Then|x| = r i and r i equals the order of an element of PSL(2,q). Thus, since p -|x|, we have|x|| q± 1, and in particular|x|≤ q + 1. But we also have|x|≥ m− 1≥ (q− 1)− 1 = q− 2. So q− 2≤|x|≤ q + 1 and|x|| q± 1. It follows that |x| = q− 1 or|x| = q + 1. Now, by Catalan’s conjecture, we only have that q− 1 or q + 1 is a power of 2 if q = 9 or if q is a Fermat prime or a Marsenne prime. Thus, since we disregarded the q = 9 case, we only have possibilities stemming from the Fermat prime or Marsenne prime cases. We have representations of G of order q− 1, q and q + 1 that satisfy m− 2≤|x|. We list these representations in the following table. Table 21: PSL(2,q) with q− 1 Marsenne prime,|x| odd F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) q− 1 + q− 1 q− 1 M. prime 28 Continuation of Table 21 F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) q `-q + 1 + q− 1 q− 1 M. prime PSL(2,q) q + 1 (q− 1) ` 06= 1 + q− 1 q− 1 M. prime End of Table Table 22: PSL(2,q) with q + 1 Fermat prime,|x| odd F ∗ (G) Degree Char Ind |x| Conditions PSL(2,q) q− 1 + q + 1 q + 1 F. prime PSL(2,q) q `-q + 1 + q + 1 q + 1 F. prime PSL(2,q) q + 1 (q− 1) ` 06= 1 + q + 1 q + 1 F. prime End of Table Finally we assume thatr f > 1 with the other assumptions from above, namely thatq is even and |x| is odd. We note that, in this case, we must have r f |a. Now, by Lemma 4.5, we have r i equal to the order of an element inPSL(2,q 1/r f ). Then|x||r f (q 1/r f ± 1). Also, as before,|x|≥m− 2≥ (q− 1)− 2 =q− 3. But except forq = 8,r f (q 1/r f ± 1)<q− 3, so|x|≤r f (q 1/r f ± 1)<q− 3≤|x|, a contradiction. In the q = 8 case, we have an element x of order 9 with a field automorphism of order three and with r i =q 1/r f + 1 = 8 1/3 + 1 = 2 + 1 = 3. We note that, in this special case, we have|x|≥m− 2 for m =q± 1 and for m =q. We list these possibilities below. Table 23: PSL(2,q), q even and x with field automorphism F ∗ (G) Degree Char Ind |x| Conditions PSL 2 (8) 7 + 9 r f = 3 PSL 2 (8) 8 `- 9 + 9 r f = 3 PSL 2 (8) 9 (7) ` 06= 1 + 9 r f = 3 End of Table Now, returning to the cases where n≥ 3, the only representations not ruled out above are those of degree less than ∂ 2 ` (F ∗ (G)). To list these representations, we let k n,q,` = 1 if `| q n −1 q−1 but k n,q,` = 0 if not. Now, excluding the cases where (n,q) = (3, 2), (3, 4), (4, 2), (4, 3), (6, 2) or (6, 3), by results due to Guralnick and Tiep, recorded as Theorem 4.1 of Tiep [25], such representations 29 are found by reducing the q− 1 Weil representation over C modulo `. Reducing modulo ` gives (q− 1) ` 0 inequivalent irreducible modular representations, (q− 1) ` 0− 1 of degree q n −1 q−1 and one of degree q n −q q−1 −k n,q,` . Thus in any case, our representations under consideration are of degree greater than q n −q q−1 −k n,q,` ≥ q n −1 q−1 − 2. Thus we concern ourselves with prime power order elements of order greater than or equal to ( q n −1 q−1 − 2)− 2 = q n −1 q−1 − 4. We show that the only such possible element has order q n −1 q−1 where n is prime. To that end, we assume that x∈ G such that|x| = u t with u a prime. We first assume that u =p. We note that, as|Z(G)|| (n,q− 1) (this is the order of the Schur multiplier of PSL n (q)), (|Z(G)|,p) = 1. So|x| =|xZ(G)|. Furthermore, diagonal automorphisms have order dividing (n,q−1). But ((n,q−1),p) = 1, sox does not contain a diagonal automorphism. Sox is a product of a inner, field and graph automorphisms, with order r i r f r g . Finally, as u = p, r i ≤ p(n− 1) by Lemma 4.4. Then, apart from the special cases where (n,q) = (3, 2), (3, 3) or (4, 2) (we eliminate (n,q) = (3, 3) by direct computation and the other two cases are considered elsewhere), we have: |x| =|xZ(G)|, =r i r f r g , ≤ (n− 1)p(a)(2), ≤ (n− 1)(q)(2), < q n − 1 q− 1 − 4. Next we assume that u|(q− 1). By Proposition 3.2 of Kantor and Seress [19], r i |(q m − 1) for some m≤n. Thus by Lemma 4.2, either r i |m(q− 1) or r i |m(q + 1). Thus, in either case, we have r i ≤n(q + 1). Also, r d |(n,q− 1) u , r f |a, and r g |2. Then for n≥ 8: |x|||xZ(G)|(n,q− 1) u (as any element in Z(G) divides (n,q− 1)), ≤ (r i r d r f r g )(n,q− 1) u , ≤ (q + 1) u (n)(n,q− 1) u (a)(2)(n,q− 1) u . 30 Thus for n≥ 8, we have: |x|≤ (q + 1) u (n)(n,q− 1) u (a)(2)(n,q− 1) u , ≤n(a)(2)(q + 1)(q− 1)(q− 1), ≤nq 4 , ≤q n−1 (as n≥ 8), < (q n − 1)/(q− 1)− 4. Similarly,|x|< (q n − 1)/(q− 1)− 4 for n< 8. Next we assume that u-q− 1. In this case, (|Z(G)|,|x|) = 1, and similarly, x cannot contain a diagonal automorphism, as the order of any diagonal automorphism divides (n,q−1). Thus the only outer automorphisms of concern are of graph field type. We first assume that r g =r f = 1. In this case, we have|x| =|xZ(G)| =r i , andr i divides either (q n − 1)/(q− 1)(n,q− 1) = (q n − 1)/(q− 1), (q n−1 − 1)/(n,q− 1) = q n−1 − 1 or q i − 1 for some i ≤ n− 2 by Secton 3.1 of Kantor and Seress [19]. But the only one of these possibilities greater than or equal to (q n − 1)/(q− 1)− 4 is |x| = (q n −1)/(q−1) except for the special case where (n,q) = (3, 2). But since this was considered above, we disregard it here. Before we list this possibility where|x| = (q n −1)/(q−1) below, we show that for (q n −1)/(q−1) to be a prime power,n must be prime andu must be odd. To that end, we let (q n −1)/(q−1) =u i and we assume for a contradiction that n is not prime. Then n =ab fora,b> 1. Now, by Zsigmondy’s theorem, q n − 1 has a primitive prime divisor t. Now, since t- q− 1, t| (q n − 1)/(q− 1) = u i . It follows that t = u. But q n − 1 = q ab − 1 = (q a − 1)(a a(b−1) + q a(b−2) + ... + q a + 1) = (q− 1)(q a−1 +q a−2 +... +q + 1)(a a(b−1) +q a(b−2) +... +q a + 1), so u i = (q n − 1)/(q− 1) = (q a−1 +q a−2 +...+q+1)(a a(b−1) +q a(b−2) +...+q a +1). Thust =u| (q a−1 +q a−2 +...+q+1)| (q a −1), so t|q a − 1 where 1<a<n, a contradiction. Next we show that|x| = (q n − 1)/(q− 1) is not a two-power. In other words, we show that u6= 2. If q is even, then q n − 1 is odd and thus (q n − 1)/(q− 1) is odd. Alternatively, if q is odd, then as in the above paragraph,u is a primitive prime divisor ofq n − 1. But 2|q n − 1 and 2|q− 1, 31 so 2 is not a primitive prime divisor of q n − 1, and thus u6= 2. Finally, we note that in the following representations, F ∗ (G) = PSL n (q): for a contradiction, we suppose that F ∗ (G) =i.PSL n (q) for some i> 1. Then i| (n,q− 1), since the Schur multiplier in this case is of order (n,q− 1). Thus i =n since n is a prime. And i| (q− 1), so q≡ 1 (mod i). Then (q n − 1)/(q− 1)≡q n−1 +... +q + 1, ≡ 1 n−1 +... + 1 + 1 (mod i), = 1 +... + 1 | {z } n times , =n, ≡ 0 (mod i). Thus (q n − 1)/(q− 1)≡ 0 (mod i), so i| (q n − 1)/(q− 1) = u j for some j. It follows that i = u, since i =n is prime. Thus i =u is a primitive prime divisor of (q n − 1)/(q− 1), in contradiction to the fact that i|q− 1. Table 24: PSL(n,q), n≥ 3 F ∗ (G) Degree Char Ind |x| Conditions PSL n (q) q n −q q−1 −k n,q,` q n −1 q−1 , n≥ 3, prime u6= 2 PSL n (q) q n −1 q−1 q n −1 q−1 , n≥ 3, prime u6= 2 End of Table Now, before moving to the case where r f > 1, we consider the special cases that arose above where (n,q) = (3, 2), (3, 4), (4, 2), (4, 3), (6, 2) or (6, 3). For (n,q) = (3, 2), this was considered above, as PSL 3 (2) ∼ = PSL 2 (7). For (n,q) = (4, 2), this was also considered above, as PSL 4 (2) ∼ = A 8 . We elminimate (n,q) = (4, 3), (6, 2) and (6, 3) by inspection of Hiss and Malle [14] and by using Magma for element orders. Lastly, we consider the case where (n,q) = (3, 4) by inspection of 32 Hiss and Malle [14] and by using Magma. We list the possibilities stemming from this case below. Table 25: PSL 3 (4) F ∗ (G) Degree Char Ind |x| Conditions 4 2 .PSL 3 (4) 4 3 0 3,5,7,8 2.PSL 3 (4) 6 3 0 5,7,8 6.PSL 3 (4) 6 `6= 2, 3 0 5,7,8 End of Table Next we assume that r f > 1. We note that this implies that q > 3. Then, by Lemma 4.5, r i = r i r d is the order of an element of Inndiag(PSL n (q 1/r f ))≤ Aut(PSL n (q 1/r f )), and so r i = r i r d ≤ (q n/r f − 1)/(q 1/r f − 1) by Table 2.1 of Dimuro [4]. Then, disregarding the exceptional case considered above of (n,q) = (3, 4), we have: |x|≤r g r f r i r d , ≤ 2(r f )(q n/r f − 1)/(q 1/r f − 1), < (q n − 1)/(q− 1)− 4. Thus,|x|< (q n − 1)/(q− 1)− 4≤|x|, a contradiction. Finally we assume that r f = 1 and r g = 2. In this case, we must have that n≥ 3 and that u = 2. Then as 2 =u6=p, p must be odd. Thus, we have that u|(q− 1) and that u| (q + 1). Now, by Proposition 3.2 of Kantor and Seress [19], r i |(q m − 1) for some m≤ n. Thus by Lemma 4.2, eitherr i |m(q− 1) orr i |m(q + 1). Thus, in either case, we have r i ≤n(q + 1). We first assume that 33 n≤ 4 and q≥ 9. Then: |x|≤r g r i , ≤ 2(n(q + 1)), = 2nq + 2n, ≤ 8q + 8 (as n≤ 4), < 8(9) + 8, < 8(9) + 9, = 9 2 , ≤q 2 , ≤q n−1 (as n≥ 3), ≤q n−1 + ((q n−2 +... +q + 1)− 4) (as n≥ 3 and q≥ 9), = (q n − 1)/(q− 1)− 4. Thus,|x|< (q n − 1)/(q− 1)− 4≤|x|, a contradiction. Next we assume that n≥ 5. Then: |x|≤r g r i , ≤ 2(n(q + 1)), ≤ (q− 1)n(q + 1), ≤q 2 n, <q 2 q n−3 (as n≥ 5), =q n−1 , ≤q n−1 + (q n−2 +... +q + 1)− 4, = (q n − 1)/(q− 1)− 4. 34 Thus,|x|< (q n − 1)/(q− 1)− 4≤|x|, a contradiction. By direct computation, with the exceptions of (n,q) = (3, 2), (3, 3), (3, 4), (3, 5) and (4, 2), we see that for the remaining cases,|x| < (q n − 1)/(q− 1)− 4≤|x|, a contradiction. But these are equivalent to cases considered elsewhere except for (n,q) = (3, 3) and (3, 5). For (n,q) = (3, 3), the greatest element of 2-prime order is 8, but (q n − 1)/(q− 1)− 4 = 9, so we rule this out. Similarly, for (n,q) = (3, 5), the greatest element of 2-prime order is also 8, but (q n − 1)/(q− 1)− 4 = 27 in this case, so we rule this out. 4.2.2 Unitary groups Next we assume that S≤G/Z(G)≤Aut(S) with S ∼ =PSU(n,q) and q 2 =p f , or f = 2a. We first assume that n is even, so n≥ 4. From the Landazuri-Seitz-Zalesskii bound taken from Table 1 of Tiep [25], assuming that (n,q)6= (4, 2) or (4, 3), the representation of G is of degree at least ∂ 1 ` (F ∗ (G))≥ (q n − 1)/(q + 1). For a contradiction, we assume that G contains a prime power order element|x| of order at least m− 2≥ ∂ 1 ` (F ∗ (G))− 2≥ q n −1 q+1 − 2. And we first assume that (n,q + 1)6= q + 1. Then (n,q+1)≤ (q+1)/2. But in this case, with the exceptions of (n,q) = (4, 2), (6, 2), (4, 5), (4, 7), (6, 3) and (4, 9) (we consider these cases individually), we have that|xZ(G)|≤ q n−1 +1 q+1 by Table 2.1 of Dimuro [4]. Moreover, assuming that (n,q)6= (4, 2) or (6, 2), the Schur multiplier of PSU(n,q) is cyclic of order (n,q + 1) so|x|≤ (n,q + 1)|xZ(G)|≤ (q + 1)/2|xZ(G)|. Thus, disregarding the 35 exceptional cases, we have: |x|| (q + 1,n)|xZ(G)|, ≤ (q + 1)/2|xZ(G)|, ≤ (q + 1)/2( q n−1 + 1 q + 1 ), = (q n−1 + 1)/2, < q n − 1 q + 1 − 2, ≤|x|, a contradiction. We now consider individually the exceptional cases of (n,q) = (4, 2), (6, 2), (4, 5), (4, 7), (6, 3) or (4, 9) from above. We note that (n,q) = (4, 3) is not included, since n = q + 1 in this case and is thus considered elsewhere. We take information from Table 2.1 of Dimuro [4] regarding the bounds on prime power order elements. To begin, concerning the case where (n,q) = (4, 2), we consider it elsewhere since PSU(4, 2) ∼ = PSp 4 (3). For (n,q) = (6, 2), we have PSU(6, 2) ∼ = Fi 21 , so we consider this case below. For (n,q) = (4, 5), we have that|xZ(G)|≤ 32. Thus, since the Schur Multiplier is of order 2 in this case, we have that|x|≤ 2|xZ(G)|≤ 64. Then|x|≤ 64 and |x|≥m− 2≥∂ 1 ` (F ∗ (G))− 2≥ q n −1 q+1 − 2 = 104− 2 = 102> 64≥|x|, so|x|>|x|, a contradiction. For (n,q) = (4, 7), we have Schur multiplier of order (n,q +1) = (4, 8) = 4, so|x|≤ 4|xZ(G)|. Now, |xZ(G)|≤ 64, so|x|≤ 4|xZ(G)|≤ 4(64) = 256. But|x|≥m− 2≥∂ 1 ` (F ∗ (G))− 2≥ q n −1 q+1 − 2 = 300−2 = 298> 256≥|x|, so|x|>|x|, a contradiction. For (n,q) = (6, 3), we have Schur multiplier of order (n,q + 1) = (6, 4) = 2, so|x|≤ 2|xZ(G)|. Now,|xZ(G)|≤ 64, so|x|≤ 2(64) = 128. But |x|≥m− 2≥∂ 1 ` (F ∗ (G))− 2≥ q n −1 q+1 − 2 = 182− 2 = 180> 128≥|x|, so|x|>|x|, a contradiction. For (n,q) = (4, 9), we have Schur multiplier of order (n,q + 1) = (4, 10) = 2, so|x|≤ 2|xZ(G)|. But|xZ(G)|≤ 128, so|x|≤ 2|xZ(G)|≤ 2(128) = 256. Then|x|≥ m− 2≥ ∂ 1 ` (F ∗ (G))− 2≥ q n −1 q+1 − 2 = 656− 2 = 654> 256≥|x|, so|x|>|x|, a contradiction. Next we assume that (n,q + 1) = q + 1. Now, if u- q + 1, and thus|xZ(G)| =|x|, then by 36 Table 2.1 of Dimuro [4],|x|≤ q n−1 +1 q+1 < q n −1 q+1 − 2, provided that (n,q)6= (4, 2) (but in this case (q + 1,n)6= (q + 1), so we can disregard it). But q n −1 q+1 − 2≤|x|, so|x|<|x|, a contradiction. To conclude this case, we address the exceptional case not addressed above (as q + 1 =n), namely the case where (n,q) = (4, 3). We list the possibilities from this case by inspection of Hiss and Malle [14] and the Atlas [3]. Table 26: PSU(4, 3) F ∗ (G) Degree Char Ind |x| Conditions 3 1 .PSU(4, 3) 6 2 0 5,7,8,9 6 1 .PSU(4, 3) 6 6= 2, 3 0 5,7,8,9 End of Table Thus we assume thatu|q+1. As before, we only consider elementsx with|x|≥∂ 1 ` (F ∗ (G))−2≥ q n −1 q+1 −2. In this case,u|q+1|q 2 −1. Now, by Section 3.2 of Kantor and Seress [19],r i |q m −(−1) m for somem≤n. Ifm is odd, thenr i |q m + 1. Thus, asu|q + 1 andr i |q m + 1, by Lemma 4.3, we have r i |m(q + 1), hence r i ≤n(q + 1). Alternatively, if m is even, then r i |q m − 1 = (q 2 ) m/2 − 1. Thus, as r|q 2 − 1 and r i | (q 2 ) m/2 − 1, r i | (m/2)(q 2 − 1) by Lemma 4.2. Thus r i | (m/2)(q + 1)2, as the biggest common factor of q + 1 and q− 1 is at most 2. So r i | m(q + 1)≤ n(q + 1). Thus for both m even and m odd, we have r i ≤ n(q + 1). Thus we have an upper bound on |xZ(G)| given by r i r d r f ≤n(q + 1)(n,q + 1)(f). Thus, as the Schur multiplier of SU n (q) is cyclic of order (n,q + 1) (with the exception of (n,q) = (4, 3) that we consider separately), we have |x|≤n(q + 1)(n,q + 1)(f)(n,q + 1). We also have an upper bound on|x| of q n−1 +1 q+1 (n,q + 1) found by multiplying the bound on|xZ(G)| with the order of the Schur multiplier. But the minimum of these two upper bounds is less than the lower bound on|x| of q n −1 q+1 − 2 in all cases. (The inequality does not hold in the case (n,q) = (6, 2), but this case is irrelevant here since (n,q + 1)6=q + 1.) Next we assume that n is odd, so n≥ 3. We start by ruling out the representations of degree 37 m≥∂ 2 ` (F ∗ (G)). To that end, for a contradiction we assume that there is anx∈G with|x|≥m−2 for a representation of degree m≥∂ 2 ` (F ∗ (G)). We first consider the case where n = 3. We first assume that n = 3 and q≥ 7. By Table 2.1 of Dimuro [4], we have|xZ(G)|≤ q 3 +1 (q+1)(3,q+1) so|x|≤ (3,q +1)|xZ(G)| = (q 3 +1)/(q +1) =q 2 −q +1. Also, Hiss and Malle [15] give the following bounds on∂ 2 ` (PSU 3 (q)) that depend on the parameters involved: 1/3(q− 1)(q 2 −q + 1), 1/6(q− 1)(q 2 + 3q + 2, and 1/3(2q 3 −q 2 + 2q− 3). We show that each bound is sufficient to rule out these cases. We note that q− 1≥ 6, as q≥ 7. Now, for the first bound, we have|x|≥ 1/3(q− 1)(q 2 −q + 1)− 2≥ 1/3(6)(q 2 −q + 1)− 2 = 2(q 2 −q + 1)− 2 = (q 2 −q + 1) + (q 2 −q + 1)− 2> (q 2 −q + 1). Thus|x|> (q 2 −q + 1)≥|x|, a contradiction. For the second bound, we have|x|≥ 1/6(q−1)(q 2 +3q+2)−2≥ (1/6)(6)(q 2 +3q+2)−2 = (q 2 +3q+2)−2 = q 2 + 3q > q 2 −q + 1. Thus|x| > q 2 −q + 1≥|x|, a contradiction. For the last bound, we have |x|≥ 1/3(2q 3 −q 2 + 2q− 3)− 2 = 1/3(q 3 + (q 3 −q 2 ) +q + (q− 3))− 2 > 1/3(q 3 +q)− 2 = 1/3(q)(q 2 +q)> 1/3(6)(q 2 +q) = 2(q 2 +q)>q 2 −q +1. Thus|x|>q 2 −q +1≥|x|, a contradiction. Concerning the cases where n = 3 and q< 7, these are either considered elsewhere, or we can rule these cases by direct computation using the bounds on|x| found in Table 2.1 of Dimuro [4] and the bounds on ∂ 2 ` (PSU(3,q)) found in Hiss and Malle [15]. Next we consider the case where n≥ 5 and n is odd. Now, we assume that (n,q)6= (5, 2) (we consider the case where (n,q) = (5, 2) below) and we assume for a contradiction that x is a prime power order element of G where the representation of G is of degree m≥ ∂ 2 ` (PSU n (q)). By Table 2.1 of Dimuro [4], we have|xZ(G)|≤ q n +1 (q+1)(n,q+1) , and thus|x|≤ (n,q + 1)|xZ(G)|≤ (q n + 1)/(q + 1). Now, using the bound ∂ 2 ` (PSU(n,q)) = (q n−2 −1)(q−1)(q n−2 −1) q+1 of Hiss and Malle 38 recorded as Theorem 4.2 in Tiep [25], we have: |x|≥m− 2, ≥ (q n−2 − 1)(q− 1)(q n−2 − 1) q + 1 − 2, ≥ (q n−2 − 1)(q n−2 − 1) q + 1 − 2, ≥ q 2n−4 − 2q n−2 + 1 q + 1 − 2, = q 2n−4 − 2q n−2 + 1 q + 1 − (2q + 2)/(q + 1), = q 2n−4 − 2q n−2 + 1− 2q− 2 q + 1 , = q 2n−4 − (2q n−2 + 2q + 2) + 1 q + 1 , ≥ 2q 2n−5 − (2q n−2 + 2q + 2) + 1 q + 1 , = q 2n−5 + (q 2n−5 − (2q n−2 + 2q + 2)) + 1 q + 1 , > q 2n−5 + 1 q + 1 , ≥ q n + 1 q + 1 (since n≥ 5). Thus|x|> q n +1 q+1 ≥|x|, a contradiction. Now, we address the representations of degree less than ∂ 2 ` (F ∗ (G)). We assume that (n,q)6= (3, 3), (3, 4), (3, 5) or (5, 2) in what follows. We now consider these special cases individually and list the possibilities in the following table. Table 27: PSU n (q) with n odd, exceptional cases F ∗ (G) Degree Char Ind |x| Conditions PSU(3, 3) 6 `6= 3 - 7, 8 PSU(3, 3) 7 7 + 7, 8 PSU(3, 3) 7 7 - 7, 8 PSU(3, 4) 13 `6= 2, 5 0 13, 16 PSU(5, 2) 10 `6= 2 - 9, 11, 16 PSU(5, 2) 11 `6= 2, 3 0 11, 16 39 Continuation of Table 27 F ∗ (G) Degree Char Ind |x| Conditions End of Table Now, we have∂ 1 ` (F ∗ (G)) = (q n −q)/(q +1) = (q n +1)/(q +1)−1. Thus we considerx such that |x|≥m− 2≥ [(q n + 1)/(q + 1)− 1]− 2 = (q n + 1)/(q + 1)− 3. Assuming (n,q)6= (3, 3), (3, 4), (3, 5) or (5, 2) (we considered these exceptional cases above separately), we note that by Table 2.1 of Dimuro [4] we have|xZ(G)|≤ (q n + 1)/(q + 1)(n,q + 1). First we assume that u| (n,q + 1). We note that u6= 2 in this case, as n is odd. Now, by similar reasoning to the n even case, we haver i ≤ (n− 1)(q + 1). Thus, asr d | (n,q + 1) andr f |a (r f cannot be even, as u is odd, so we divide out the factor of two), we have|xZ(G)|≤ r i r d r f ≤ (n− 1)(q + 1)(n,q + 1)(a). Thus, as the Schur multiplier in this case is cyclic of order (n,q + 1), we have|x|≤ (n,q + 1)|xZ(G)|≤ (n,q + 1)(n− 1)(q + 1)(n,q + 1)(a). This upper bound rules out all cases except (n,q) for n = 3 and q∈{2, 3, 4, 5, 8, 11, 17} or (n,q) = (5, 2), (5, 4) or (9, 2). All of these cases are eliminated individually by direct computation. Next we assume thatu- (n,q+1). In this caser d = 1 and|x| =|xZ(G)|, as the Schur multiplier is of order (n,q + 1) and u- (n,q + 1). If (n,q + 1)6= 1, then (n,q + 1)≥ 3, since 2- (n,q + 1) as n is odd. Thus, by Table 2.1 of Dimuro [4], |x| =|xZ(G)|≤ (q n + 1)/(q + 1)(n,q + 1)≤ (q n + 1)/(q + 1)3 < (q n + 1)/(q + 1)− 3≤|x|, a contradiction. Thus we only consider the case where (n,q + 1) = 1. Thus we have|x| =|xZ(G)|≤ (q n + 1)/(q + 1)(n,q + 1) = (q n + 1)/(q + 1). But also|x|≥m− 2≥ (q n + 1)/(q + 1)− 3. We determine all such possible elements of G where S≤G/Z(G)≤Aut(S) with S ∼ =SU n (q) for n odd, given that (n,q + 1) = 1. To that end, we first assume that r f = 1. We also start by assuming that u = p. Then by Lemma 4.4,|x| = r i ≤ (n− 1)p. But except for (n,q) = (3, 3), and (5, 2), we have (n− 1)p < (q n + 1)/(q + 1)− 3≤|x|, a contradiction. Moreover, these exceptions were already considered as special cases. Next, we assume u6= p. Then again x = r i and by section 3.2 of Kantor and Seress [19], r i | (q m − (−1) m ) for some m≤ n, and furthermore r i | (q n−1 − 1)/(n,q + 1) = q n−1 − 1 if m =n− 1 and r i | (q n − 1)/(q + 1)(n,q + 1) = (q n − 1)/(q + 1) if m =n. But if m≤n− 2, then 40 |x| = r i ≤ (q n−2 + 1). But, except for (n,q) = (3, 2), which we already considered separately, we have (q n−2 + 1)≤ (q n + 1)/(q + 1)− 3≤|x|, so|x|<|x|, a contradiction. Next for m = n− 1, we have x| (q n−1 − 1), as (−1) n−1 = 1 since n is odd. But in this case n− 1 is even, so q n−1 − 1 = (q (n−1)/2 − 1)(q (n−1)/2 + 1). But unless (n,q) = (3, 3) (a case we already considered separately), (q (n−1)/2 − 1) and (q (n−1)/2 + 1) cannot both be prime powers. Furthermore, ifq is even, then both factors are odd, so we must be able to factor out an odd number. Alternatively, if q is odd, then both factors are even, and so if|x| is even we could factor out an odd number, but if|x| is odd, we could factor out 2 from both (q (n−1)/2 − 1) and (q (n−1)/2 + 1). In any case, we can factor out at least 3, so|x|≤ (q n−1 − 1)/3< (q n + 1)/(q + 1)− 3≤|x|. Thus, apart from the exceptional cases, we have|x|<|x|, a contradiction. Next form =n, we have|x| =r i | (q n +1)/(q +1). But, excluding (n,q) = (3, 2), a case already considered, a factor of (q n + 1)/(q + 1) is only≥ (q n + 1)/(q + 1)− 3 if the factor is (q n + 1)/(q + 1) itself. Thus for x to satisfy both that|x|≥ (q n + 1)/(q + 1)− 3 and that|x|| (q n + 1)/(q + 1), we must have|x| = (q n + 1)/(q + 1). Before we list this generic possibility in the table below, we add a few constraints. We first show thatn must be prime. For a contradiction, we suppose thatn =st wheres andt are both odd (since n must be odd) and each is not equal to one. Thenq n +1 =q st +1 = (q s +1)(q st−s −q st−2s +q st−3s − ...−q st−(t−1)s +1) = (q +1)(q s−1 −q s−2 +...−q +1)(q st−s −q st−2s +q st−3s −...−q st−(t−1)s +1), so (q n + 1)/(q + 1) = (q s−1 −q s−2 +...−q + 1)(q st−s −q st−2s +q st−3s −...−q st−(t−1)s + 1). But (n,q)6= (3, 2), since in this case PSU(n,q) is not simple, so by Zsigmondy’s theorem q n + 1 has a primitive prime divisor t. Then t-q + 1, so t| (q n + 1)/(q + 1) =u i . It follows that t =u. Then t =u| (q s−1 +q s−2 +... +q+1)|q s +1, since (q s−1 +q s−2 +... +q+1) is au-power, as it is a factor of (q s−1 +q s−2 +... +q + 1)(q st−s −q st−2s +q st−3s −...−q st−(t−1)s + 1) = (q n + 1)/(q + 1) =u i . Thus t =u|q s + 1 with 1<s<n, a contradiction. Next, for|x| = (q n + 1)/(q + 1) to be u-power with u prime, we must have u6= 2. First, if q is even, then q n + 1 is odd, so (q n + 1)/(q + 1) is odd. Second, if q is odd, then as in the above paragraph, we have that u is a primitive prime divisor of q n + 1. But 2|q n + 1 and 2|q + 1 and thus cannot be a primitive prime divisor of q n + 1. 41 The representations we list along with the possibilities for x are the only representations less than ∂ 2 ` (F ∗ (G)), a result due to Hiss and Malle, recorded as Theorem 4.2 of Tiep [25]. Before listing these possibilities, we show that F ∗ (G) =PSU(n,q). To that end, we assume for a contradiction that F ∗ (G) = i.PSU(n,q) where i > 1. First, we have i| (n,q + 1), the order of the Schur multiplier in this case. But n is prime andi> 1, soi =n. Also, sincei| (q + 1),q≡−1 (mod i). Thus (q n + 1)/(q + 1) =q n−1 −q n−2 +q n−3 −... +q 2 −q + 1, ≡ (−1) n−1 − (−1) n−2 + (−1) n−3 −... + (−1) 2 − (−1) + 1 (mod i), = 1 +... + 1 | {z } n times (since n is odd), =n, ≡ 0 (mod i). Thus (q n + 1)/(q + 1)≡ 0 (mod i), so i| (q n + 1)/(q + 1) = u j for some j. Thus i = u, since i =n is prime. Then i =u is a primitive prime divisor of (q n + 1). But i| (q + 1), a contradiction. Table 28: PSU n (q), generic case F ∗ (G) Degree Char Ind |x| Conditions PSU(n,q) (q n −q)/(q + 1) (q n + 1)/(q + 1), n odd, u6= 2 prime PSU(n,q) (q n + 1)/(q + 1) (q n + 1)/(q + 1), n odd, u6= 2 prime End of Table Lastly we assume that r f > 1. First, we assume that r f is even. Then u = 2. We first assume that q is also even. Then r d = 1, since r d |n with n odd. Also, as before, r i ≤ (n− 1)p, so|x|≤ r i r d r f ≤ (n− 1)p(r f )≤ (n− 1)p(2a)< (q n +1)/(q +1)− 3 provided that (n,q)6= (3, 3), (3, 4), (3, 5) or (5, 2), cases already considered above. Then (q n +1)/(q+1)−3≤|x|, so|x|<|x|, a contradiction. Next we assume that q is odd, so q + 1 is even. Thus, we have u|q 2/r f + 1 and u|q 2/r f − 1. 42 Furthermore, we have that r i r d is equal to the order of an element in PSU(n,q 2/r f ) by Lemma 2.2.3 of Dimuro [4], so r i r d | (q 2m/r f − (−1) m ) for some m≤n. Thus by Lemma 4.2 and Lemma 4.3, r i r d | m(q 2/r f − 1) or r i r d | m(q 2/r f + 1). But since n is odd but|x| is even, if m = n, then r i r d ≤ n/3(q 2/r f + 1)≤ (n− 1)(q 2/r f + 1). Thus, in any case, r i r d ≤ (n− 1)(q 2/r f + 1), so|x|≤ r f (n− 1)(q 2/r f + 1)≤ (2a)(n− 1)(q + 1). This is less than (q n + 1)/(q + 1)− 3 unless (n,q) = (3, 9). However, in the (n,q) = (3, 9) case, we haver f (n−1)(q 2/r f +1)< (q n +1)/(q+1)−3, so|x|≤r f (n− 1)(q 2/r f + 1)< (q n + 1)/(q + 1)− 3≤|x|, so|x|<|x|, a contradiction. Next we assume that r f is odd. Then r i r d is equal to the order of an element in PSU(n,q 1/r f ) by Lemma 2.2.3 of Dimuro [4]. Thus by Table 2.1 of Dimuro [4], r i r d ≤ (q n/r f + 1)/(q 1/r f + 1), so |x|≤r f r d r i ≤r f (q n/r f + 1)/(q 1/r f + 1)< (q n + 1)/(q + 1)− 3≤|x|, a contradiction. 4.2.3 Symplectic groups For the next case, we assume that S≤G/Z(G)≤Aut(S) with S ∼ =PSp 2n (q) and q =p f . We first assume that q is even. First, we disregard the case where (n,q) = (2, 2), where PSp 4 (2) 0 is a simple group, as PSp 4 (2) 0∼ =A 6 and was thus already considered. Now, for (n,q)6= (2, 2), we assume for a contradiction that G contains a prime power order element of order greater than or equal to m− 2≥ ∂ 1 ` (F ∗ (G))− 2. But by the Landazuri-Seitz-Zalesskii bound, we have ∂ 1 ` (F ∗ (G))− 2 = (q n −1)(q n −q) 2(q+1) − 2. Moreover, disregarding the case where (n,q) = (3, 2), the Schur multiplier is trivial since it is of order (2,q− 1) and q is even. Thus by Table 2.1 of Dimuro [4], we have a bound on|xZ(G)| =|x| of q n + 1. But with the exceptions of (n,q) = (2, 2), (2, 4), and (3, 2), q n + 1 < (q n −1)(q n −q) 2(q+1) − 2, so|x| =|xZ(G)|≤ q n + 1 < (q n −1)(q n −q) 2(q+1) − 2≤ m− 2≤|x|, so|x|<|x|, a contradiction. We now consider the exceptional cases from above. As noted above, we disregard the case where (n,q) = (2, 2), where PSp 4 (2) 0 is a simple group, as PSp 4 (2) 0∼ = A 6 and was thus already considered. For (2, 4), we have a possibility stemming from the faithful, absolutely irreducible representation of degree 18 and the element x∈ G where|x| = 17. We list this possibility in the following table. For (n,q) = (3, 2), we have possibilities from the faithful, irreducible representations of degree 7 and 8 ofPSp 6 (2) and 2.PSp 6 (2), respectively, sincePSp 6 (2) and 2.PSp 6 (2) have elements of order 7, 8 and 9. We list these possibilities in the following table. 43 Table 29: PSp 2n (q) with q even, exceptional cases F ∗ (G) Degree Char Ind |x| Conditions PSp 4 (4) 18 `6= 2 + 17 PSp 6 (2) 7 `6= 2 + 7, 8, 9 2.PSp 6 (2) 8 `6= 2 + 7, 8, 9 End of Table Next we assume that q is odd. In this case, with the exception of (n,q) = (2, 3), a case we consider individually below, the bound on∂ 2 ` (F ∗ (G)) is the same as the bound above for theq even case, namely we have ∂ 2 ` (F ∗ (G)) = (q n −1)(q n −q) 2(q+1) . Moreover, we have a bound on|xZ(G)| found in Table 2.1 of Dimuro [4] given by xZ(G)≤ (q n + 1)/(2,q− 1) = (q n + 1)/2. Thus|x|≤ 2|xZ(G)|≤ q n + 1, since the Schur multiplier in this case has order 2. But q n + 1 < (q n −1)(q n −q) 2(q+1) − 2 for (n,q)6= (2, 3), so|x|≤q n + 1< (q n −1)(q n −q) 2(q+1) − 2≤|x|, a contradiction. Thus we concern ourselves with representations of degree less than∂ 2 ` (F ∗ (G)). Thus, the degree of our representation is one of two Weyl representations of degree (q n ± 1)/2, a result due to Guralnick, Magaard, Saxl and Tiep, recorded as Theorem 4.2 in Tiep [25], so m = (q n ± 1)/2. We note that of the two representations of degree (q n ± 1)/2, 2.PSp 2n (q) acts faithfully for the even degree representation and PSp 2n (q) acts faithfully for the odd degree representation. Also, since q is odd, we assume in this case that r g = 1, since r g > 1 only if n = 2 =p f . Now we assume that x is an element of order at least m− 2≥∂ 1 ` (F ∗ (G))− 2≥ (q n − 1)/2− 2 and show that this only occurs when|x| = (q n ± 1)/2 apart from a few exceptions. We first assume that r f = 1 and that u = p. In this case, as p is odd, u is odd. Thus, as the Schur multiplier in this case divides 2, we must have|x| =|xZ(G)|. Now, since u = p, r i < 2n(p) by Lemma 4.4. But (q n − 1)/2− 2 > 2n(p) for all but the following exceptional cases: (n,q) = (2, 3), (3, 3), (2, 5), and (2, 7). Disregarding these exceptions, we have|x|< 2n(p)≤ (q n − 1)/2− 2≤|x|, a contradiction. Now we consider the exceptional cases of (n,q) = (2, 5) and (n,q) = (2, 7) and we consider the other exceptional cases below. We take information on the faithful, absolutely irreducible representations from Hiss and Malle [14] and we take information on x from GAP. For (n,q) = (2, 5), there are absolutely irreducible, faithful representations ofPSp 4 (5) 44 of degree 12 and 13, but then a considerable gap from there. But there is nox∈G with|x| a power ofp = 5 and|x|≥m−2. For (n,q) = (2, 7), there are absolutely irreducible, faithful representations of PSp 4 (7) of degree 24 and 25 and an absolutely irreducible, faithful representation of 2.PSp 4 (7) of degree 24, but then a considerable gap thereafter. But there is no x∈G with|x|≥m− 2 and |x| a power of 7. Next we assume thatr f = 1 andu6=p. We first assume thatu = 2 and then assume thatu6= 2. We note that the Schur multiplier is of order 2 and thus|x|| 2|xZ(G)|. Also, r d | (2,q− 1) = 2, so in particularr d ≤ 2. Now, from Section 3.3 of Kantor and Seress [19],r i | (q m ± 1) for somem≤n. Also, since u = 2 and thus u| q− 1 and u| q + 1, we have by Lemma 4.2 and Lemma 4.3 that r i |m(q±1)≤n(q+1) for somem≤n. Thus,|xZ(G)|≤r i r d r f r g ≤n(q+1)(2)(1)(1) =n(q+1)(2). Thus|x|≤ 2|xZ(G)|≤ 2n(q + 1)(2) = 4n(q + 1). But 4n(q + 1) < (q n − 1)/2− 2 for all cases except (n,q) for n = 2 and q∈{3, 5, 7, 9, 11, 13, 17} and (n,q) = (3, 3), (3, 5) or (4, 3). We consider the cases where (n,q) = (2, 3) or (3, 3) below, and we eliminate the other cases by comparing the Landazuri-Seitz-Zalesskii bound on ∂ 1 ` (F ∗ (G)) with the upper bound on 2 power order elements of G found by using GAP or in the ATLAS [3]. Next we assume that u6= 2. Now, from Section 3.3 of Kantor and Seress [19], r i | (q m ± 1) for some m≤ n and in particular r i | (q m ± 1)/(2,q− 1) if m = n. Now, r d | (2,q− 1) = 2, so r d = 1 or r d = 2. But since u6= 2, we must have r d = 1. Thus, 1 = r g = r f = r d . Furthermore, the Schur multiplier has order dividing 2, so|x| =|xZ(G)|. Thus,|x| = r i . Now, the case where r i = (q n ± 1)/(2,q− 1) gives a possibility that we list in the following table. But when r i | (q m ± 1) for m < n, we have that (q m ± 1) < (q n − 1)/2− 2 except for (n,q) = (2, 3) and (n,q) = (3, 3). Thus for (n,q)6= (2, 3) and (n,q)6= (3, 3), we have|x| = r i | (q m ± 1) < (q n − 1)/2− 2≤|x|, a contradiction. Next we further analyze the constraints on elements x∈PSp 2n (q) with|x| = (q n ± 1)/2 =u i for some i. Now, we first assume that|x| = (q n − 1)/2 with n even. Then n = 2t for some t. Then 1/2(q n − 1) = 1/2(q 2t − 1) = 1/2(q t − 1)(q t + 1), which is only a u-prime power if u = 2, since the greatest common divisor ofq t − 1 andq t + 1 is 2. Moreover, 1/2(q t − 1)(q t + 1) is only a two-power when both q t − 1 and q t + 1 are two-powers. But this only occurs when q t − 1 = 2, for otherwise 45 4 would divide both factors. Thus q t − 1 = 2, and it follows that q = 3 and t = 1. Thus this case only occurs when (n,q) = (2t,q) = (2, 3), a case we consider below. Next we assume that n is odd. Now, by Zsigmondy’s theorem, (q n − 1) has a primitive prime divisort. Sincet| (q n − 1) = (q− 1)(q n−1 +q n−2 +... +q + 1) = (q− 1)u i for somei,t| (q− 1)u i . Butt-q− 1, sot|u i , and it follows thatt =u. Thus|x| = 1/2(q n − 1) = 1/2(q− 1)(q n−1 +q n−2 + ... +q + 1) = 1/2(q− 1)u i , where u-q− 1. Thus|x| is only an u-power when 1/2(q− 1) = 1, or equivalently when q = 3. To add another constraint to this case, n must be prime: for a contradiction, we assume that n = st for s and t odd and not equal to 1. Then q n − 1 = q st − 1 = (q s − 1)(q st−s +q st−2s + ... +q st−(t−1)s + 1) = (q− 1)(q s−1 +q s−2 +... +q + 1)(q st−s +q st−2s +... +q st−(t−1)s + 1), so u i =|x| = 1/2(q n −1) = 1/2(q−1)(q s−1 +q s−2 +... +q +1)(q st−s +q st−2s +... +q st−(t−1)s +1) = 1/2(3− 1)(q s−1 +q s−2 +... +q + 1)(q st−s +q st−2s +... +q st−(t−1)s + 1) = (q s−1 +q s−2 +... + q + 1)(q st−s +q st−2s +... +q st−(t−1)s + 1), sou| (q s−1 +q s−2 +... +q + 1)| (q− 1)(q s−1 +q s−2 + ... +q + 1) =q s − 1 where 1<s<st, contradicting the fact that u is a primative prime divisor of q st − 1. Next, we add a constraint to the case where|x| = (q n +1)/2 =u i for somei: for|x| to be a prime power, n must be a two-power. For a contradiction we suppose that n is not a two-power. The n =ts, wheret> 1 is odd ands is a two-power (possibly trivial). Thenq n +1 =q ts +1 = (q s ) t +1 = (q s + 1)((q s ) t−1 − (q s ) t−2 + (q s ) t−3 −... + (q s ) t−(t−2) − (q s ) t−(t−1) + 1). Butq n + 1 has a primitive prime divisor r by Zsigmondy’s theorem. Now, r6= 2, since 2| q + 1, so r| 1/2(q n + 1) = u i . It follows that r = u. Now, from above, (q s + 1)| (q n + 1), so 1/2(q s + 1)| 1/2(q n + 1). But 1/2(q n + 1) is a u-power, so 1/2(q s + 1) is a u-power, and thus in particular u| 1/2(q s + 1). Thus r =u| 1/2(q s + 1)| (q s + 1), so r| (q s + 1) with 1<s<n, a contradiction to the fact that r is a primitive prime divisor of q n + 1. Finally, we note that in both cases we have that 2-|x| by Catalan’s conjecture: (3 n −1)/2 cannot be a two power unless n = 2, a case we consider below, and (q n + 1)/2 can never be a two-power. We now list the possibilities involving x with|x| = 1/2(q n ± 1) reflecting these number-theoretical constraints that we have imposed. 46 Table 30: PSp 2n (q), generic cases F ∗ (G) Degree Char Ind |x| Conditions PSp 2n (3) (3 n − 1)/2 (3 n − 1)/2 n odd prime, u6= 2 2.PSp 2n (3) (3 n + 1)/2 (3 n − 1)/2 n odd prime, u6= 2 2.PSp 2n (q) (q n − 1)/2 (q n + 1)/2 n = 2 a , u6= 2, q≡ 3 mod 4 PSp 2n (q) (q n + 1)/2 (q n + 1)/2 n = 2 a , u6= 2, q≡ 3 mod 4 2.PSp 2n (q) (q n − 1)/2 (q n + 1)/2 n = 2 a , u6= 2, q≡ 1 mod 4 PSp 2n (q) (q n + 1)/2 (q n + 1)/2 n = 2 a , u6= 2, q≡ 1 mod 4 End of Table We now consider the two exceptional cases from above. For (n,q) = (2, 3), we use Hiss and Malle [14] and the Atlas of Finite Group Representations [32] to list the possibilities next. We note that sincePSp 4 (3) ∼ =PSU(4, 2) and since we only consider cross-characteristic representations, we disregard the representations of characteristic 2 and 3. For (n,q) = (3, 3), we have an an absolutely irreducible, faithful representation ofPSp 6 (3) of degree 13 and one an absolutely irreducible, faithful representation of 2.PSp 6 (3) of degree 14, but then a considerable gap from there. And the only element x with|x|≥m− 2 for either of these representations has order|x| = 13. Table 31: PSp 2n (q) with q odd, exceptional cases F ∗ (G) Degree Char Ind |x| Conditions 2.PSp 4 (3) 4 ` = 5 0 4, 5, 8, 9 PSp 4 (3) 5 ` = 5 0 4, 5, 8, 9 PSp 4 (3) 6 ` = 5 + 5, 8, 9 PSp 4 (3) 10 ` = 5 0 9 PSp 6 (3) 13 `6= 3 0 13 2.PSp 6 (3) 14 `6= 2, 3 0 13 47 Continuation of Table 31 F ∗ (G) Degree Char Ind |x| Conditions End of Table Next we suppose that r f 6= 1. We first assume that u = 2 and then that u6= 2. To that end, we assume that u = 2. Then r d ≤ 2 and since the Schur multiplier is of order 2, we have |x|| 2|xZ(G)|. Now by Lemma 4.5, r i r d is equal to the order of an element in Aut(PSp 2n (q 1/r f )), so r i r d ≤ (q n/r f + 1)/(2,q 1/r f − 1) by Table 2.1 of Dimuro [4]. Thus|x|| 2|xZ(G)| = 2r i r d r f r g ≤ 2(q n/r f + 1)/(2,q 1/r f − 1)(r f )(1) = 2(q n/r f + 1)/(2)(r f ) = r f (q n/r f + 1). But (r f )(q n/r f + 1) < (q n − 1)/2− 2≤|x|, so|x|<|x|, a contradiction. Lastly we assume that r f 6= 1 and that u6= 2. Then r d = 1, since r d | 2 but|x| is odd. Furthermore, as the Schur multiplier is of order 2, we have|x| =|xZ(G)|. Now, as above, we have thatr i r d =r i is equal to an element ofAut(PSp 2n (q 1/r f )), sor i =r i r d ≤ (q n/r f + 1)/(2,q 1/r f − 1) by Table 2.1 of Dimuro [4]. Thus,|x| =|xZ(G)| =r i r d r f r g =r i r f ≤r f (q n/r f + 1)/(2,q 1/r f − 1). But r f (q n/r f + 1)/(2,q 1/r f − 1)< (q n − 1)/2− 2≤|x|, so|x|<|x|, a contradiction. 4.2.4 Orthogonal groups Next we suppose that S ≤ G/Z(G)≤ Aut(S) with S ∼ = P Ω + 2n (q), P Ω − 2n (q), or P Ω 2n+1 (q). We start by assuming that n > 3 and we consider separately the cases where S ∼ = P Ω + 8 (2) and S ∼ = P Ω 7 (3). Then from Hoffman [16], recorded in Table II of Tiep [25], we have ∂ 1 ` (F ∗ (G))≥ (q n −1)(q n−1 −q) q 2 −1 . For a contradiction, we assume thatG contains a prime power order element|x| of order at least m−2 =∂ 1 ` (F ∗ (G))−2≥ (q n −1)(q n−1 −q) q 2 −1 −4. Now, by a bound from Guest, Morris, Praeger and Spiga [13], we have|xZ(G)|≤q n+1 /(q−1). Now, the order of the Schur multiplier for any of these groups is at most 4, so|x|≤ 4|xZ(G)|≤ 4q n+1 /(q− 1)≤ 4q n+1 . However, the Schur multiplier is trivial whenever q is even. In this case, we have|x| =|xZ(G)|≤q n+1 /(q− 1)< (q n −1)(q n−1 −q) q 2 −1 − 4≤|x|, so|x| <|x|, a contradiction. Now we consider the case where q is odd and so a Schur multiplier of order up to 4 is possible. In this case, we have|x| = 4|xZ(G)|≤ 4q n+1 /(q− 1). But except for (n,q) = (4, 3), (4, 5), or (4, 7), we have|x|≥ (q n −1)(q n−1 −q) q 2 −1 − 4> 4q n+1 /(q− 1)≥|x|, so|x|>|x|, 48 a contradiction. Moreover, we rule out the listed exceptional cases using the more specific bounds on|xZ(G)| found in Table 2.1 of Dimuro [4]. Now, we deal with the small cases from above, namely the cases where S ∼ =P Ω + 8 (2) orP Ω 7 (3), by inspection of Hiss and Malle [14], by using bounds from Table 2.1 in Dimuro [4], and by inspection of the Atlas [3]. We now list the possibilities stemming from these two cases. Table 32: Small cases from orthogonal groups F ∗ (G) Degree Char Ind |x| Conditions 2.Ω + 8 (2) 8 `6= 2 0 7, 8 End of Table Next for the n = 3 case, we note that this only occurs when S ∼ = P Ω 2n+1 (q). In this case, disregarding the case whereq = 3 (we considered this case above), again by a bound from Hoffman [16], we have∂ 1 ` (F ∗ (G))≥ (q 2n −1)/(q 2 −1). Also, by the bound from Table 2.1 of Dimuro [4], and given that the Schur multiplier is of order at most 2 in this case, we have|x|≤ 2|xZ(G)|≤ 2(q n + 1)/2 = (q 3 + 1). For a contradiction, we assume thatG contains a prime power order element|x| of order at leastm−2. Then|x|≥m−2≥ (q 2n −1)/(q 2 −1)−2 = (q 6 −1)/(q 2 −1)−2>q 3 +1≥|x|, a contradiction. 4.3 Exceptional groups of Lie type We next consider the case whereS is an exceptional finite simple group of Lie type. We use the Landazuri-Seitz-Zalesskii bounds on the lowest degree irreducible representations of these groups and their covering groups. We use bounds on the prime power element orders of the automorphism groups of the exceptional groups found in Table 2.1 of Dimuro [4]. First we suppose thatS ∼ = 2 B 2 (q). We first assume thatq> 8, orq≥ 32. We evaluate theq = 8 case by inspection of the Atlas of Finite Group Representations [32] and by inspection of Hiss and 49 Malle [14] and we list the possibilities below. Now, for the q > 8 case, from Table 2.1 of Dimuro [4],|xZ(G)|≤ q + p q/2 + 1. But the Schur multiplier is trivial in this case, so|x| =|xZ(G)|≤ q + p q/2 + 1. Now, for a contradiction, we assume that G contains a prime power order elementx with|x|≥m− 2≥∂ 1 ` ( 2 B 2 (q))− 2≥ (q− 1) p q/2− 2. But (q− 1) p q/2− 2>q + p q/2 + 1≥|x|, so|x|>|x|, a contradiction. Table 33: 2 B 2 (8) F ∗ (G) Degree Char Ind |x| Conditions 2. 2 B 2 (8) 8 5 + 7,13 2 B 2 (8) 14 6= 2 0 13 End of Table Next we suppose that S ∼ = 2 G 2 (q). We note that q = 3 b where b > 1 and odd. The special case where b = 1 and thus 2 G 2 (q) = 2 G 2 (3) is not simple, but its derived subgroup 2 G 2 (3) 0 is simple. However, 2 G 2 (3) 0∼ =PSL 2 (8) and is considered elsewhere. Now, assuming thatb> 1 andb odd, from Table I in Tiep [25], we use the Landazuri-Seitz-Zalesskii bound on ∂ 1 ` ( 2 G 2 (q)) given by q(q− 1). From Table 2.1 of Dimuro [4], we have a bound on|xZ(G)| given byq + √ 3q + 1. But the Schur multiplier is trivial in this case, so|x| =|xZ(G)|. Now, for a contradiction, we suppose that G contains an element x of prime power order where|x|≥m− 2≥∂ 1 ` ( 2 G 2 (q))− 2≥q(q− 1)− 2. But given that q≥ 27, we have q(q− 1)− 2 > q + √ 3q + 1≥|xZ(G)| =|x|, so|x| >|x|, a contradiction. Next we assume that S ∼ = G 2 (q). From Table II in Tiep [25], for q > 4 we use the Landazuri- Seitz-Zalesskii bound on∂ 1 ` ( 2 G 2 (q)) given byq 3 −1. For a contradiction, we assume thatG contains an element x of prime power order with|x|≥m− 2≥ (q 3 − 1)− 2 =q 3 − 3. We note that since q> 4, the Schur multiplier is trivial, so|x| =|xZ(G)|. Now, from Table 2.1 of Dimuro [4], we have that|xZ(G)|≤q 2 +q +1. But|x| =|xZ(G)|≤q 2 +q +1< 3q 2 <q 3 −3≤|x|, so|x|<q 3 −3≤|x|, a contradiction. Now G 2 (2) is not simple, but G 2 (2) 0 is simple. However, G 2 (2) 0∼ = PSU 3 (3) and 50 thus was already considered. For q = 3 and q = 4 we list the possibilities in the following table, found by inspection of Hiss and Malle [14] and considering element orders of Aut(G 2 (q)) found in the Atlas of Finite Group Representations [32]. Table 34: G 2 (q), exceptional cases F ∗ (G) Degree Char Ind |x| Conditions G 2 (3) 14 `6= 3 + 13 2.G 2 (4) 12 `6= 2 - 13, 16 End of Table Next we suppose thatS ∼ = 3 D 4 (q). From Table I in Tiep [25], we use the bound on ∂ 1 ` ( 3 D 4 (q)), due to L¨ ubeck, Maagard, Malle and Tiep, given byq 5 −q 3 +q−1. Now for a contradiction we suppose thatG contains a prime power order elementx with|x|≥∂ 1 ` ( 3 D 4 (q))−2≥ (q 5 −q 3 +q−1)−2 =q 5 − q 3 +q−3. From Table 2.1 of Dimuro [4], we have that|xZ(G)|≤q 4 −q 2 +1. But the Schur multiplier in this case is trivial, so|x| =|xZ(G)|. Then|x| =|xZ(G)|≤q 4 −q 2 + 1<q 5 −q 3 +q− 3≤|x|, so|x|<|x|, a contradiction. Next we assume that S ∼ = 2 F 4 (2) 0 or S ∼ = 2 F 4 (q). We first consider the case of 2 F 4 (2) 0 . Given the bound on|x| of 16 found in Table 2.1 of Dimuro [4] and by inspection of the cross-characteristic faithful irreducible representations of 2 F 4 (q) 0 found in the Hiss and Malle [14], we rule this case out. And now we assume that q≥ 8. From Table 2.1 of Dimuro [4], we have a bound on|xZ(G)| given by q 2 + p 2q 3 +q + √ 2q + 1 < 5q 2 . Moreover, the Schur multiplier in this case is trivial, so|x| =|xZ(G)|. And from Table I in Tiep [4], we use the Landazuri-Seitz-Zalesskii bound on ∂ 1 ` ( 2 F 4 (q)) given by (q 5 −q 4 ) p q/2. Now for a contradiction we suppose that G contains a prime power order element x with|x|≥m− 2≥∂ 1 ` ( 2 F 4 (q))− 2≥ (q 5 −q 4 ) p q/2− 2. But since q≥ 8, we have (q 5 −q 4 ) p q/2− 2≥ 5q 2 >|x|, so|x|>|x|, a contradiction. Next we assume that S ∼ = F 4 (q). We first consider the case of F 4 (2). From Hiss and Malle [14], the smallest irreducible representation in this case is of degree 52. From Dimuro [4], we have |xZ(G)|≤ 32. Moreover, F 4 (2) has Schur multiplier of order 2, but 2.F 4 (2) : 2 still has a 2 power element order bound of 32. Thus, we rule this case out. 51 And now we assume that q > 2. From Table 2.1 of Dimuro [4], we have a bound on|xZ(G)| given by q 4 + 1. Furthermore, the Schur multiplier in this case is trivial, so|x| =|xZ(G)|. And from Table I in Tiep [25], we use the Landazuri-Seitz-Zalesskii bound on∂ 1 ` (F 4 (q)) given byq 8 −q 7 . Now for a contradiction we suppose that G contains a prime power order element x with|x|≥ m− 2≥q 8 −q 7 − 2. But q 8 −q 7 − 2>q 4 + 1≥|x|, so|x|>|x|, a contradiction. Next we assume thatS ∼ = 2 E 6 (q). We first consider the case of 2 E 6 (2). We rule this case out by inspection of the cross characteristic irreducible representations of 2 E 6 (2) and its covering groups in Hiss and Malle [14] and by inspection of the Atlas [3]. And now we assume that q > 2. From Table 2.1 of Dimuro [4], we have a bound on|xZ(G)| given by q 6 −q 3 + 1. Moreover, the Schur multiplier in this case is trivial, so|x| =|xZ(G)|. And from Table I in Tiep [25], we use the Landazuri-Seitz-Zalesskii bound on ∂ 1 ` ( 2 E 6 (q)) given by q 11 −q 9 . Now for a contradiction we suppose that G contains a prime power order element x with |x|≥ m− 2≥ q 11 −q 9 − 2. But q 11 −q 9 − 2 > q 6 −q 3 + 1, so|x| =|xZ(G)|≤ q 6 −q 3 + 1 < q 11 −q 9 − 2≤|x|, so|x|<|x|, a contradiction. Next we consider E 6 (q). From Table 2.1 of Dimuro [4], we have a bound on|xZ(G)| given by q 6 +q 3 + 1. Moreover, the Schur multiplier in this case is of order at most 3, so|x|≤ 3|xZ(G)|≤ 3q 6 + 3q 3 + 3. And from Table I in Tiep [25], we use the Landazuri-Seitz-Zalesskii bound on ∂ 1 ` (E 6 (q)) given by q 11 −q 9 . Now, for a contradiction, we suppose that G contains a prime power order elementx with|x|≥m− 2≥q 11 −q 9 − 2. Butq 11 −q 9 − 2> 3q 6 + 3q 3 + 3, so|x|≥m− 2≥ q 11 −q 9 − 2> 3q 6 + 3q 3 + 3≥|x|. Thus|x|>|x|, a contradiction. Next we assume that S ∼ =E 7 (q). From Table 2.1 of Dimuro [25], we have a bound on|xZ(G)| given by q 7 −1 q−1 ≤ q 8 . Moreover, the Schur multiplier in this case is of order at most 2, so|x|≤ 2|xZ(G)|≤ 2q 8 . And from Table I in Tiep [25], we use the Landazuri-Seitz-Zalesskii bound on ∂ 1 ` (E 7 (q)) given by q 17 −q 15 . Now for a contradiction we suppose that G contains an element x of prime power order such that|x|≥ m− 2≥ q 17 −q 15 − 2. But q 17 −q 15 − 2 > 2q 8 , so |x|≥q 17 −q 15 − 2> 2q 8 ≥|x|. Thus|x|>|x|, a contradiction. For the last exceptional group of Lie type, we assume thatS ∼ =E 8 (q). From Table 2.1 of Dimuro [4], we have a bound on|xZ(G)| given byq 8 +q 7 −q 5 −q 4 −q 3 +q+1≤ 7q 8 ≤q 3 q 8 ≤q 11 . Moreover, 52 the Schur multiplier in this case is trivial, so|x| =|xZ(G)|. And from Table I in Tiep [25], we use the Landazuri-Seitz-Zalesskii bound on∂ 1 ` (E 8 (q)) given byq 29 −q 27 . Now for a contradiction we assume that G contains a prime power order element x with|x|≥m− 2≥∂ 1 ` (E 8 (q))− 2 =q 29 −q 27 − 2. Butq 29 −q 27 −2>q 11 , so|x|≥q 29 −q 27 −2>q 11 ≥|xZ(G)| =|x|. Thus|x|>|x|, a contradiction. 4.4 Sporadic groups Lastly we consider the case where F ∗ (G) is a sporadic simple group or a covering group of a sporadic simple group. We consult Hiss and Malle for low dimensional representations of F ∗ (G) and the Atlas of Finite Group Representations [32] for the possible prime power element orders of G and we list the representations and elements not ruled out by our assumed constraints. Table 35: Sporadic groups F ∗ (G) Degree Char Ind |x| Conditions M 11 5 3 0 4,5,8,11 M 11 9 11 + 8,11 M 11 10 6= 11 + 8,11 M 11 10 6= 2 0 11 M 11 11 6= 2, 3 + 11 2.M 12 6 3 0 5,8,11 M 12 10 2 + 8,11 M 12 10 3 + 11 2.M 12 10 6= 2 0 11 M 12 11 6= 2, 3 + 11 2.M 12 12 6= 2, 3 + 11 3.M 22 6 2 0 4,5,7,8,11 M 22 10 2 0 8,11 2.M 22 10 7 + 11 2.M 22 10 6= 2, 7 0 11 M 23 11 2 0 11,23 M 23 21 23 + 23 M 23 22 6= 2, 23 + 23 M 24 11 2 0 11,23 M 24 22 3 + 23 M 24 23 6= 2,3 + 23 J 1 7 11 + 7,11,19 J 1 14 11 + 19 J 1 20 2 + 19 J 2 6 2 - 4,5,7,8 53 Continuation of Table 35 F ∗ (G) Degree Char Ind |x| Conditions 2.J 2 6 5 - 5,7,8 2.J 2 6 6= 2,5 - 5,7,8 3.J 3 9 2 0 8,9,17,19 J 3 18 3 + 17,19 3.J 3 18 5 0 17,19 3.J 3 18 6= 3,5 0 17,19 Co 1 24 2 + 23 2.Co 1 24 6= 2 + 23 Co 2 22 2 + 23 Co 2 23 6= 2 + 23 Co 3 22 2 - 23 Co 3 22 3 + 23 Co 3 23 6= 2,3 + 23 Ru 28 2 + 29 2.Ru 28 6= 2 0 29 2.Suz 12 3 - 11,13,16 3.Suz 12 2 0 11,13 6.Suz 12 6= 2, 3 0 11,13 End of Table 5 Problem 2 For this, we simply run through the tables compiled throughout problem 1, modifying them as appropriate to ensure that `-|x|. The result is listed above in the Main Results section. 6 Problem 3 We now determine which of the elements from Problem 1 correspond to regular unipotent elements of classical algebraic groups. This question is motivated by the more general question of which closed maximal subgroups of the classical algebraic groups contain regular unipotent elements. Now, by a reduction theorem due to Liebeck and Seitz, found for example as Theorem 18.6 in Malle and Testerman [21], such maximal subgroups occur either as ‘geometric’ subgroups stabilizing some structure on the underlying vector space, or as almost quasi-simple groups whose quasi- 54 simple normal subgroups act irreducibly. Concerning the geometric-type subgroups containing regular unipotent elements, these have been addressed in a paper by Saxl and Seitz [23] where they classify the maximal closed reductive subgroups of positive dimension that contain regular unipotent elements. Moreover, in general, almost quasi-simple subgroups of Lie type with same- characteristic representations extend to subgroups of positive dimension, and thus are also treated by Saxl and Seitz. So concerning Lie type subgroups we limit ourselves to Lie type subgroups with cross-characteristic representations. Thus we address the aspect of the problem concerning which finite almost quasi-simple subgroups (cross-characteristic for Lie type subgroups) contain regular unipotent elements. Such subgroups give the possibilities for the finite, closed subgroups containing regular unipotent elements that are maximal as Lie primitive subgroups. (A subgroup is Lieprimitive if it is not contained in a proper closed subgroup of positive dimension and if it does not contain a subgroup of the form O ` 0 (H F ), where H is the classical group and F is a Frobenius endomorphism of H). Now we describe the regular unipotent elements of classical groups: unipotent elements have Jordan normal form with 1’s along the main diagonal, i.e., all the eigenvalues are 1. And a regular elementg hasC H (g) as small as possible for the elements ofH. For the purpose of our computations, we use the characterization of the regular unipotent elements of the classical groups found in Saxl and Seitz [23]: For classical groups of root structure typeA n ,B n , orC n , regular unipotent elements in their Jordan normal form consist of a single Jordan block. For classical groups of root structure typeD n , regular unipotent elements in their Jordan normal form contain two Jordan blocks, of size 2n-1 and 1 in the case where `6= 2, and of size 2n− 2 and 2 if ` = 2. We note that by the next theorem this implies that a regular unipotent element x in a classical group has order greater than or equal tom− 2 in all cases, and has order greater than or equal to m− 1 if`6= 2, wherem is the degree of the classical group. Next, we record the following fact about the orders of regular unipotent elements. Theorem 6.1. A unipotent element M∈GL n (k) has order ` b , where ` is the characteristic of the underlying field andb is the smallest positive integer such that ` b is greater than or equal to the size of the biggest Jordan block of the matrix. 55 Proof. Let M be a regular unipotent element in Jordan normal form. Then M =I +N, where I is the identity matrix and N is a nilpotent matrix with ones or zeroes on the superdiagonal and zeroes everywhere else. Now, M ` b = (I +N) ` b = I ` b +N ` b = I +N ` b , as the cross terms vanish since the ground field has characteristic `. ButN ` b will equal zero when` b is greater than or equal to the largest Jordan block. We note that, by this theorem, regular unipotent elements have `-power order, where ` is the characteristic of the underlying vector space. Now, for an almost quasi-simple group G, we have S≤ G/Z(G)≤ Aut(S) for a finite non- abelian simple group S. Moreover, as in the reduction theorem from Liebeck and Seitz, we assume that F ∗ (G), the quasi-simple normal subgroup of G and a covering group of S, acts irreducibly. We note that for a regular unipotent elementx∈G≤H, whereH is a classical group, we have |x| =|xZ(H)|, as (`,Z(H)) = 1. ButZ(G) =Z(F ∗ (G))≤Z(H), sinceF ∗ (G) acts irreducibly and thus its central elements are also central in H by Schur’s Lemma. Thus (`,Z(G))|(`,Z(H)) = 1, so (`,Z(G)) = 1. Thus for a regular unipotent element x ∈ G, we have|x| = |xZ(G)|. In our computations we will assume this without comment. We also point out now that since we are dealing exclusively with cross-characteristic representations, and since `||x| for all regular unipotent elements x, we can disregard all possibilities from above where p||x|. The results of Problem 1 give possibilities for regular unipotent elements. But we can narrow the list down considerably by eliminating elements that are not of `-power order. Moreover, we have sharper bounds on|x|, since|x| =|xZ(G)| in this case. For the sake of organization, we list the possible orders of x assuming that it must be of `-power without explicitly specifying this. We list the possibilities by reproducing the tables from from Problem 1 with adjustments to ensure that `||x|, and also by analyzing the Jordan normal form of x. We list the order of x, and the Jordan normal form of x (abbreviated in the tables as “JNF”) by listing a set containing the sizes of the Jordan blocks. This suffices to completely characterize the Jordan normal form since we are dealing with unipotent elements whose eigenvalues are all 1. For a single Jordan block of size n, we list this as{n}. For Jordan blocks of sizen−1 and 1, we list this as{n−1, 1}. Lastly, for Jordan 56 blocks of sizen− 2 and 2, we list this as{n− 2, 2}. We eliminate from consideration anyx that has Jordan normal form outside of these possibilities, as such anx cannot be regular unipotent. Lastly, we list the Frobenius-Schur indicators of the representations under consideration, which specify to which classical groups the given almost quasi-simple group belongs. 6.1 Jordan normal forms when S =A n We now determine the Jordan normal form of the smallest degree representations of A n . Such representations are called the standard representations, or the deleted permutation representations, and are the only representations of degree small enough to be relevant in the generic case. (We deal with the other relevant representations in the small cases individually). Now, for the standard representation, we have two cases: either `| n or `- n. When `| n, the standard representation is constructed by restricting the natural permutation representation to the subspace of vectors whose coordinates sum to zero, and then by quotienting out by the subspace < (1, 1,..., 1) >. The representation in this case has degree n− 2. When ` - n, the standard representation is constructed simply by restricting the natural permutation representation to the subspace of vectors whose coordinates sum to zero, and in this case the degree is n− 1. Thus when `|n, we concern ourselves with elements of order greater than or equal ton−4 in the case where` = 2 and elements of order greater than or equal to n− 3 in the case where `> 2. When `- n we concern ourselves with elements of order greater than or equal ton− 2 when`6= 2 and greater than or equal ton− 3 when ` = 2. We first assume that x is of ` power order and has order n− 4. We note that in this case we must have` = 2, so`|n, and thus the standard representation has degree n− 2 and is constructed as above. Now, elements of order n− 4 in Aut(A n ) ∼ = S n are conjugate to elements with cycle decomposition with one of the following four forms: (1, 2,...,n− 4)(n− 3)(n− 2)(n− 1)(n), (1, 2,...,n− 4)(n− 3,n− 2)(n− 1)(n), (1, 2,...,n− 4)(n− 3,n− 2)(n− 1,n), or (1, 2,...,n− 4)(n−3,n−2,n−1,n). First, an element with cycle decomposition of the form (1, 2,...,n−4)(n− 3)(n− 2)(n− 1)(n) fixes the three-space generated by (1, 1,..., 1, 0, 0, 0, 0), (0, 0,..., 0, 1, 1, 0, 0), and (0, 0,..., 0, 1, 0, 1, 0), and thus cannot be a regular unipotent element, which must have a fixed 57 space of dimension less than three. An element with cycle decomposition of the form (1, 2,...,n− 4)(n− 3,n− 2)(n− 1)(n) fixes two one-spaces, one of them generated by the vector (1, 1,..., 1, 0, 0, 0, 0), contained in the n− 4 dimensional x-cyclic subspace generated by (1, 0, 0,..., 0, 0, 0, 1, 0). This n− 4 space has an x- invariant complementary two-space that includes the other one-dimensional fixed space (the one- space generated by (0, 0,..., 0, 1, 1, 0, 0)) that is generated by the following linearly independent set of two vectors: {(0, 0,...., 0, 1, 1, 0, 0), (0, 0,...., 0, 1, 0, 1, 0)}. Thus, the Jordan normal form of x consists of two Jordan blocks, one of size n− 4 and the other of size 2. An element with cycle decomposition (1, 2,...,n − 4)(n − 3,n − 2)(n − 1,n) fixes the three-space generated by the following linearly independent set of vectors: {(1,..., 1, 0, 0, 0, 0), (0,..., 0, 1, 1, 0, 0), (1, 0, 1, 0,..., 1, 0, 1, 0, 1, 0)}. Thus, x cannot be regu- lar unipotent, since regular unipotent elements have a fixed space of dimensional at most two. The last element of order n− 4 to consider has cycle decomposition of the form (1, 2,...,n− 4)(n− 3,n− 2,n− 1,n). This has a two dimensional fixed space. One of the one-dimensional fixed spaces, generated by (1,..., 1, 0, 0, 0, 0), is contained the (n− 4)-dimensional x-cyclic sub- space generated by (1, 0, 0,..., 0, 1, 0, 0, 0). The other one-dimensional fixed space, generated by (1, 0, 1, 0,..., 1, 0, 1, 0, 1, 0), is contained in a two-dimensional x-invariant subspace, com- plementary to the n− 4 space above, generated by the following linearly independent set: {(1, 0, 1, 0, 1, 0,...., 1, 0, 1, 0, 1, 0), (1, 1, 0, 0, 1, 1, 0, 0,..., 1, 1, 0, 0, 1, 1, 0, 0)}. It follows that the Jor- dan normal form of x consists of two Jordan blocks, one of size n− 4 and the other of size 2. Next, we consider the `-power order elements of order (n− 3). Such elements are conjugate to elements that have one of the following cycle decompositions: (1, 2,...,n− 3)(n− 2)(n− 1)(n), (1, 2,...,n− 3)(n− 2,n− 1)(n) or (1, 2,...,n− 3)(n− 2,n− 1,n). We start with the first case, assuming thatx has cycle decomposition (1, 2,...,n− 3)(n− 2)(n− 1)(n). Now, for`> 3, we have ` - n and thus a representation of degree n− 1, so an element of order n− 3 cannot be regular unipotent as `6= 2. Thus we have only two subcases to consider, where ` = 2 or ` = 3. First we assume that` = 2. Then` = 2-n, asn is odd, so the representation is of degree n− 1. But in this 58 case, x fixes a three-dimensional space (and thus cannot be regular unipotent), generated by the following linearly independent set of vectors:{(1, 1,..., 1, 0, 0, 0), (0,..., 0, 1, 1, 0), (0,..., 0, 1, 0, 1)}. Next we assume that ` = 3. Then n− 3 is a three-power, so ` = 3| (n− 3) + 3 =n, and thus the representation has degreen−2. Now, in this case,|x| has a two dimensional fixed space. One of the fixed one-spaces, generated by the vector (1, 1,..., 1, 0, 0, 0), is contained in thex-cyclic (n−3)-space generated by (1, 0,..., 0, 2, 0, 0). The other fixed one-space, generated by the vector (0, 0,..., 0, 1, 2), is complementary to this (n− 3)-space. It follows that the Jordan normal form ofx has two Jordan blocks, one of size n− 3 and the other of size one. Next, for an element of cycle decomposition (1, 2,...,n− 3)(n− 2,n− 1)(n), we must have that ` = 2, since an element of this cycle decomposition has order a multiple of|(n− 2,n− 1)| = 2. In this case, ` = 2| n− 3, so n = (n− 3) + 3 must be odd. Thus ` - n, so our representation has degree n− 1. Now, x fixes two one-spaces, one generated by (1, 1,..., 1, 0, 0, 0), contained in the x-cyclic (n− 3)-space generated by (1, 0, 0,..., 0, 1, 0, 0). The other fixed one-space, generated by (0, 0,..., 0, 1, 1, 0), is contained in a two dimensional x-invariant space, complementary to the above (n− 3)-subspaces, and generated by the following linearly independent set of two vectors: {(0, 0,..., 0, 1, 1, 0), (0, 0,... 0, 1, 0, 1)}. Thus the Jordan normal form of x consists of two Jordan blocks, one of size n− 3 and the other of size 2. The last type of element of order n− 3 has cycle decomposition (1, 2,...,n− 3)(n− 2,n− 1,n). We note that x must be of three-power order, since|x| must be a multiple of|(n− 2,n− 1,n)|. Thus ` = 3, and it follows that ` | (n− 3) + 3 = n, so the standard representation has degree n− 2 and is constructed as explained above. Now, x fixes two one-spaces, one of them generated by (1, 1,..., 1, 0, 0, 0) and contained in the x-cyclic (n− 3)-space generated by (1, 0, 0,..., 0, 0, 2, 0, 0). The other fixed one-space is generated by (1, 2, 0, 1, 2, 0,..., 1, 2, 0, 1, 2, 0), a subspace complementary to the above n− 3 space. It follows that the Jordan normal form of x has two Jordan blocks, one of size n− 3 and another of size one. Next we consider`-power order elements of ordern−2. Such elements are conjugate to elements that have cycle decomposition of the form (1, 2,...,n−2)(n−1,n) or (1, 2,...,n−2)(n−1)(n). We start with the first case, assuming that x is an element of cycle decomposition (1, 2,...,n− 2)(n− 59 1,n). We note that x must be of two-power order, since it must be a multiple of|(n− 1,n)| = 2. Then ` = 2| (n− 2) + 2 = n, so the representation is of degree n− 2. Now, since x has only a one-dimensional fixed space generated by (1, 1,..., 1, 0, 0), the Jordan normal form of x must consist of a single Jordan block. For the second case, we assume that x has cycle decomposition (1, 2,...,n− 2)(n− 1)(n). We also first assume that ` = 2, so n− 2 is a two-power and thus ` = 2| (n− 2) + 2 = n. Thus our representation is of degree n− 2. Now, the fixed space of x is one-dimensional, generated by the vector (1, 1,..., 1, 0, 0), so x consists of a single Jordan block. For the case `6= 2, since n− 2 is an `-power and 26= `, we must have ` - (n− 2) + 2 = n, so our representation is of degree n−1. Nowx has a two-dimensional fixed space, where one fixed one-space, generated by the vector (1, 1,..., 1, 0, 0), is contained in the x-cyclic (n− 2)-space generated by (1, 0, 0,..., 0,−1, 0). The other one-space, generated by the vector (0, 0,... 0, 1,−1), is complementary to the above (n− 2)- space and thus the Jordan normal form of x consists of a two Jordan blocks, one of size n− 2 and the other of size one. Next we consider`-power order elements of ordern−1. Such elements are conjugate to elements that have cycle decomposition (1, 2,...,n−1)(n). Since`|n−1, we have`-n, so our representation is of degreen−1. But in this casex has only a one-dimensional fixed space, the one-space generated by the vector (1, 1,..., 1, 0), so x consists of a single Jordan block. Finally, we consider `-power order elements of order n. Such elements are conjugate to the element with cycle decomposition (1, 2, 3,...,n− 1,n), and since n is an`-power, we have `|n, so the representation is of degreen−2. Now, an elementx with cycle decomposition (1, 2, 3,...,n−1,n) has only a one-dimensional fixed space, the one-space generated by the vector (0, 1, 2,...,`− 2,`− 1, 0, 1, 2,...,`− 2,`− 1,..., 0, 1, 2,...,`− 2,`− 1), so the Jordan normal form of x consists of a single Jordan block. 6.2 Jordan normal forms when S =PSL(n,q) with n> 2 Next we determine the Jordan normal form of x when x∈ PSL(n,q) with n > 2 and|x| = (q n − 1)/(q− 1). 60 Now, we have two types of relevant representations to consider, one of degree (q n −q)/(q−1)−1 and the others of degree (q n − 1)/(q− 1). The representation of degree (q n −q)/(q− 1)− 1 is a composition factor of k G P , where P is the stabilizer of a one space, and k refers to the trivial representation. Now, the action of G can be viewed as the permutation action of G on the coset space of P . But no non-trivial power of x belongs to P since every non-trivial power of x acts irreducibly, so x is conjugate to a (q n − 1)/(q− 1)-cycle on the coset space, and thus its fixed space is one-dimensional. But the fixed space of x in a composition factor is also one-dimensional, so the Jordan normal form of x consists of a single Jordan block. The representations of degree (q n −1)/(q−1) are of the formV G P , the induced representation from a one-dimensional representation ofP . But again, since no non-trivial power of x belongs toP ,x is conjugate to a (q n − 1)/(q− 1)- cycle, so its fixed space is also one-dimensional in this case. So the Jordan normal form of x in this case is also a single Jordan block. 6.3 Jordan normal forms when S =PSL 2 (q) Next we determine the Jordan normal form of x when x ∈ Aut(PSL(2,q)). We start by assuming that|x| = (q− 1)/2. Then x belongs to the normalizer of a Sylow p-subgroup Q. But Q is abelian and the elements of Q are each diagonalizable, so the elements of Q are simultaneously diagonalizable and thus there is a basis of common eigenvectors of Q. Now, since x∈ N(Q), x permutes these common eigenspaces: for an eigenvectorv from one of the common eigenspaces and for an element y∈Q, we haveyxv =xy 0 v =xλv =λxv for somey 0 ∈Q (since y∈N(Q)), and for some λ (since v is also an eigenvector of y 0 ∈Q). Now, the representations of concern are of degree (q− 1)/2 or (q + 1)/2. We first show that one of the orbits of the action of x on the eigenspaces must be of size (q− 1)/2: for a contradiction, suppose that none of the orbits has size (q− 1)/2. Then the sizes of the orbits divide (q− 1)/4. Then x (q−1)/4 is diagonal but non-trivial. Now, since x (q−1)/4 is diagonal, it centralizes Q. But C(Q) = Q, and x (q−1)/4 / ∈ Q, a contradiction. It follows that in the case of (q− 1)/2 degree representation, x permutes the (q− 1)/2 one-dimensional eigenspaces in a single orbit. Thus the fixed space of x is one-dimensional, so the Jordan normal form of x consists of a single Jordan 61 block. For the case where the degree is (q + 1)/2, x has two orbits, one of size (q− 1)/2 and the other trivial. It follows that the Jordan normal form of x consists of two Jordan blocks, one of size (q− 1)/2 and the other of size one. Before starting with the next case, we give a few introductory definitions. We first define that a trivial intersection set (aT.I. set) is a set that intersects its conjugates either fully or trivially. Also, an indecomposable kG-module is Q-projective if it is a summand of a module of the form V G Q for some kQ-module V . The vertex of a kG-module, defined up to conjugacy, is a minimal subgroup relative to which the module is projective. We note that over a fieldk of characteristic`, the vertex of every indecomposable kG-module has vertex an `-group. Thus a vertex of an indecomposable kG-module W is a minimal `-group Q∈G such that W is a summand of V G Q for some kQ-module V . A projective module is a 1-projective module, a summand of the module induced from the trivial group. We also note that an indecomposable projective module over a field of characteristic ` has degree a multiple of the order of an `-Sylow subgroup. Next, we suppose that|x| = (q + 1)/2 where ` is odd. We note that by Theorem 1 of Blau [2], cyclic Sylow subgroups of finite simple groups are T.I. sets. Thus, since A =< x > is an `-Sylow subgroup in this case and is cyclic, it is a T.I. set. Moreover, its normalizer is dihedral of order 2(q +1), and the complementH ofA inN(A) is the Klein four-group. Now, letW be akG-module corresponding to one of the representations either of degree (q + 1)/2 or (q− 1)/2. By Green correspondence, W| N(A) =Y⊕X, where Y is a non-projective indecomposable N(A)-module and each summand of X is projective relative to a subgroup of the form N(A)∩Q y for an element y / ∈N(A), where Q is the vertex of W contained in A. But since A is a T.I. set, A∩A y = 1 and thus N(A)∩A y = 1 for y6∈ N(A): for if not, then the product (N(A)∩A y )A would be a Sylow `-subgroup of N(A) of greater order than|A|, a contradiction. Now, since N(A)∩A y = 1 and Q≤A, we have N(A)∩Q y = 1. Thus X is a direct sum of projective indecomposable modules. Next, we analyze the Jordan normal form of x in both the projective indecomposables and the non-projective indecomposables. We start with the projective indecomposables. Now, N(A) = AoH where (|A|,|H|) = 1, so by Proposition 8.3.2 of Webb [29], the projective indecomposable kN(A) modules in this case, where k is the underlying field of the representation, are of the form 62 S N(A) H , where S is a simple kH module. But the simple kH modules are one-dimensional, so S N(A) H acts on a (q + 1)/2 dimensional vector space over k, and x permutes a basis for this space in an orbit of length (q + 1)/2. Thus, x has a one-dimensional fixed space, and it follows that the Jordan normal form of x consists of a single Jordan block. Next, to analyze the non-projective indecomposables, we note that by Proposition 11.2.1 of Webb, every non-projective indecomposable occurs as a quotient module of a projective indecomposable. But a single Jordan block remains a single Jordan block under quotients, so x has Jordan normal form a single Jordan block in the projective indecomposable kN(A)-modules also. Now we analyze how x acts in W , the kG module from above of degree 1/2(q± 1) where W| N(A) =Y⊕X. We start by assuming that the degree is (q− 1)/2. To figure out the structure of Y⊕X, we note that the degree of Y and X must sum to the total degree of the N(A)-module, Y consists of a single non-projective indecomposable module of degree less than|A|, andX consists of projective indecomposable summands all of dimension|A|. Thus, in this case, Y⊕X must haveX trivial. ThenW| N(A) consists of an indecomposable non-projectivekN(A) module, and thus by the above reasoning x consists of a single Jordan block. For the case where the dimension is (q + 1)/2, the Y -part of Y ⊕X must be trivial. Thus W| N(A) in this case consists of an indecomposable projective, so again by the above reasoning the Jordan normal form of x consists of a single Jordan block. Next we assume that|x| =q±1 with` odd. Then the relevant representations have degreeq±1 orq. These cases are similar to the (q +1)/2 case above: as above, the Sylow`-subgroupA is cyclic and aT.I. set, andH ∼ =N(A)/A, the cyclic complement of A inN(A), has simplekH-modules all of dimension one. And by the same reasoning as above, the Jordan normal form of x has a single Jordan block for any indecomposable N(A)-module. Now, by Green correspondence, since A is a T.I. set, W| N(A) = Y ⊕X where Y consists of at most one non-projective indecomposable and each summand ofX is projective indecomposable. Next we analyze the breakdown of Y⊕X again using the facts that the degree of Y and X must sum to the total degree of the N(A)-module, Y consists of a single non-projective indecomposable summand of degree less than|A|, andX consists of projective indecomposable summands all of dimension|A|. 63 We first assume that|x| =q−1 with` odd. Then the module must have degreeq−1 orq. In the q− 1 case,W consists of a projective indecomposable, and thus the Jordan normal form of x has a single Jordan block. In the case where the dimension is q, the direct sum decomposition ofW| N(A) consists of a one-dimensional non-projective indecomposable and one projective indecomposable of dimension q− 1, so the Jordan normal form of x consists of two blocks, one of size one and the other of size q− 1. Next, we assume that|x| = q + 1. Then the kG-module W has dimension q− 1,q, or q + 1. In the q− 1 or q case, the kN(A)-module W| N(A) consists of a non-projective indecomposable and thus the Jordan normal form of x consists of a single Jordan block. For the q + 1 case, the module consists of a single projective indecomposable and thus the Jordan normal form of x consists of a single Jordan block. Next we assume that|x| =q± 1 with ` even. We start by assuming that|x| =q− 1. We note that there are no relevant (q + 1)/2 representations in this case. And we also note that an element of `-power order equal to (q− 1) is a diagonal automorphism of PSL 2 (q), and the representations with characteristic ` = 2 of degree (q− 1)/2 do not extend to include such an automorphism. Thus we concern ourselves only with representations of degree q± 1 or q. Now, as with the case of two-power elements of order (q− 1)/2, x normalizes a p-Sylow subgroup and thus permutes its common eigenspaces. For a degreeq−1 representation,x permutes the eigenspaces in a single orbit of length q− 1, so the Jordan normal form of x consists of a single Jordan block. For the degree q representations, x permutes the eigenspaces in two orbits, one of length q− 1 and the other of length one. Thus the Jordan normal form of x consists of two blocks, one of size q− 1 and the other of size one. Finally, we disregard the degree q + 1 representation since the representations of degree q + 1 including x are not irreducible. Lastly we assume that|x| =q + 1 with ` even. Then the relevant representations are of degree q− 1 andq + 1 (there is no relevant degree q representation in this case). The q− 1 degree module is the non-trivial composition factor of k G P , where P is the stabilizer of a one-space. Now, x is a (q− 1)-cycle in the permutation action on the coset space of P . Thus the fixed space of x is one- dimensional, so its Jordan normal form has a single Jordan block, and x remains a single Jordan block in our composition factor ofk G P . The degreeq +1 representations are induced representations 64 of the formW G P , whereW is a one-dimensional representation ofP . Butx has no non-trivial power in P , so it acts as a (q + 1)-cycle. Thus its fixed space is one-dimensional, and the Jordan normal form of x consists of a single Jordan block. 6.4 Jordan normal forms when S =PSU n (q) Next, to determine the Jordan normal form of elementsx∈PSU(n,q) with|x| = (q n +1)/(q+1) an `-power, we first show that < x > is a Sylow `-subgroup, implying that < x > is a T.I. set. First, we note that|PSU(n,q)| = q 1/2(n−1)(n) (n+1)(q+1) Q n−1 i=1 (q i+1 − (−1) i+1 ). We show that the only factor of|PSU(n,q)| that` divides is (q n + 1). This follows from Zsigmondy’s Theorem: By Zsigmondy’s Theorem, q 2n − 1 has a primitive prime power divisor t. Thus t| (q n − 1)(q n + 1). But since t does not divide q n − 1, it must divide q n + 1. Also, since t - q 2 − 1, t - q + 1| (q 2 − 1). Thus t| (q n + 1)/(q + 1) and therefore t = `. Now, since t- (q 2(i+1) − 1) for any i with 1≤ i≤ n− 2, t- (q i+1 − (−1) i+1 )| (q 2(i+1) − 1), so ` =t- (q i+1 − (−1) i+1 ) for any i with 1≤i≤n− 2. Thus the `-part of|PSU(n,q)| is (q n + 1)/(q + 1). Thus A =< x > is an `-Sylow cyclic subgroup of PSU(n,q) and hence a T.I. set by Theorem 1 of Blau [2]. Next we determine the Jordan normal form of x∈ PSU(n,q) for the relevant representations ofPSU(n,q), those of degree (q n −q)/(q− 1) = (q n + 1)/(q + 1)− 1 or (q n + 1)/(q + 1). LetW be the kG-module corresponding to one of these representations. Then, since A =<x> is a T.I. set, we have W| N(A) = Y⊕X by Green correspondence, where Y is a non-projective kN(A)-module and X is a direct sum of projective indecomposable kN(A) modules. To begin to analyze the structure of Y⊕X, we first let H be the complement of A in N(A), which is cyclic in this case. Then it follows from Theorem 8.3.2 of Webb [29] that the projective indecomposable kN(G)-modules are induced from simple kH-modules (which are one-dimensional since H is cyclic), since N(A) ∼ = AoH with|H| relatively prime to|A| and A cyclic. Thus the projective indecomposable modules have dimension|A|, and the indecomposable kN(A) modules are exactly the homomorphic images of the projective indecomposables by Proposition 11.2.1 of Webb [29]. Now, in a projective indecomposable, which has the form S N(A) H for a simple one- dimensionalkH-moduleS,x is conjugate to a (q n +1)/(q +1)-cycle. Thusx has a one-dimensional 65 fixed space, and so has Jordan normal form a single Jordan block. Moreover, the non-projective indecomposable modules are quotient modules of projective indecomposables, and a single Jordan block remains a single Jordan block under quotients. Thus, the Jordan normal form of x has a single Jordan block also for the non-projective indecomposable kN(A)-modules. Finally, we analyze the two cases above, where W is either a (q n + 1)/(q + 1)− 1 degree or a (q n + 1)/(q + 1)− 1 degree kG-module. If the degree is (q n + 1)/(q + 1)− 1, then W| N(A) is a non-projective indecomposable module, and thus the Jordan normal form of x consists of a single Jordan block. If the degree is (q n + 1)/(q + 1), the module is projective indecomposable, and thus the Jordan normal form of x also consists of a single Jordan block in this case. 6.5 Jordan normal forms when S =PSp 2n (q) To analyze the Jordan normal form of elements x∈ PSp 2n (q) with|x| = 1/2(q± 1), we first assume that|x| = 1/2(q n − 1). We note that, as shown above, q = 3 in this case. We start by showing that < x > is a Sylow `-subgroup of PSp 2n (3). We first note that |PSp 2n (q)| = q n 2 (2,q−1) Q n i=1 (q 2i − 1). We show that the only factor of|PSp 2n (q)| that ` divides is q n − 1. First, disregarding the (n,q) = (2, 3) case considered separately, we have by Zsigmondy’s theorem that t| (q n − 1) for some prime t with t- (q i − 1) for i<n. Then t6= 2 since 2| (q− 1). Thus t| 1/2(q n − 1) = ` j for some j, so t = `. Thus, since `| (q n − 1) but `- (q i − 1) for i < n, the order of q (mod `) is n. Thus ` divides (q j − 1) if and only if n| j. Thus for any factor q 2i − 1 of|PSp 2n (q)|, `| q 2i − 1 if and only if n| 2i, and n| 2i if and only if n| i since n is odd. But for the factors q 2i − 1 of|PSp 2n (q)|, i≤ n, so if i| n, i = n. Thus the only factor of |PSp 2n (q)| that` divides isq 2n −1. Also, the only prime` that dividesq n −1 andq n +1 is two, but 1/2(q n − 1) = 1/2(3 n − 1) is a non-trivial two-power only whenn = 2 by Catalan’s conjecture. But we considered that case where (n,q) = (2, 3) separately, so we disregard it now. Thus `- q n + 1, so the `-part of|PSp 2n (3)| equals 1/2(3 n − 1). Thus, disregarding the case where (n,q) = (2, 3), <x> is a cyclic Sylow `-subgroup of PSp 2n (3), and hence a T.I. set by Theorem 1 of Blau [2]. Next we determine the Jordan normal form of x∈ PSp 2n (q) for the relevant representations, those of degree 1/2(q n ±1). LetW be thekG module corresponding to one of these representations. 66 Then, since A =< x > is a T.I. set, it follows from Green correpondence that W| N(A) = Y⊕X, where Y is a non-projective kN(A)-module and X is a direct sum of projective indecomposable kN(A) modules. To anaylize Y⊕X, we first let H be the complement of A in N(A), which is cyclic in this case. Now, it follows from Theorem 8.3.2 of Webb [29] that the projective indecom- posable kN(A)-modules are induced from simple kH-modules (which are one-dimensional since H is cyclic), since N(A) ∼ =AoH with|H| relatively prime to|A| and A cyclic. Thus the projective indecomposable modules have dimension|A|, and the indecomposable kN(A) modules are exactly the homomorphic images of the projective indecomposables by Proposition 11.2.1 of Webb [29]. Next we analyze the Jordan normal form ofx in the indecomposablekN(A) modules. Now, in a projective indecomposable, which has the formS N(A) H for a simple one-dimensionalkH-moduleS,x is conjugate to a 1/2(q n −1)-cycle, and thus has a one-dimensional fixed space. Thus,x has Jordan normal form a single Jordan block. Moreover, the non-projective indecomposable modules occur as quotients of non-projective indecomposables, and single Jordan blocks remain single Jordan blocks under quotients. Thusx also consists of a single Jordan block in the non-projective indecomposable kN(A) modules. Finally, we analyze the Jordan normal form of x for a kG-module W by analyzing W| N(A) = Y⊕X for the two cases whereW has degree 1/2(q n − 1) or 1/2(q n + 1). We do this using the fact that the degree ofY plus the degree ofX must sum to the degree ofW , the fact thatY has degree less than 1/2(q n − 1), and the fact that each summand of X has degree|A| = (q n − 1)/2. First, if the degree ofW is (q n −1)/2, then the only way for the degree ofX andY to sum to this, given the above constraints, is for Y to be trivial and for X to consist of a single projective indecomposable. So W| N(A) is a projective indecomposable module, and thus x consists of a single Jordan block. If the degree is (q n + 1)/2, Y must consist of a one degree non-projective indecomposable and X must consist of a single projective indecomposable, and thus the Jordan normal form of x consists of two Jordan blocks, one of size one and the other of size (q n − 1)/2. For the next case, we assume that|x| = 1/2(q n + 1). We start by showing that < x > is a Sylow `-subgroup of PSp 2n (q). We note again that|PSp 2n (q)| = q n 2 (2,q−1) Q n i=1 (q 2i − 1). Now, by Zsigmondy’s theorem, q 2n − 1 has a primitive prime divisor t. Then t- (q n − 1), so t| (q n + 1) 67 (since t| (q 2n − 1) = (q n − 1)(q n + 1) but t- (q n − 1)). But t6= 2 since 2| (q + 1)| (q 2 − 1), so t| 1/2(q n + 1). It follows that t = `. Thus ` = t- q 2i − 1 for any i < n. Thus the only factor of |PSp 2n (q)| that ` divides is q 2n − 1. But again `-q n − 1, so the `-part of|PSp 2n (q)| is a factor of q n + 1. But 1/2(q n + 1) is an `-power, so the `-part of|PSp 2n (q)| is 1/2(q n + 1). Thus <x> is a Sylow `-subgroup of PSp 2n (q). Now that we have established that <x> is a Sylow`-subgroup ofPSp 2n (q), we determine the Jordan normal form of x. By the same argument as the above case where|x| = 1/2(q− 1), we see that for a degree 1/2(q− 1) module, the Jordan normal form of x consists of a single Jordan block and for modules of degree 1/2(q + 1) the Jordan normal form of x also consists of a single Jordan block. 6.6 Frobenius-SchurindicatorswhenS ∼ =A n andforWeilrepresentations of classical groups Next we discuss the Frobenius-Schur indicators of the generic cases of the alternating groups and the classical groups. The relevant representations forS ∼ =A n are the standard representations. And when S is a classical group, all the representations under consideration are `-modular Weil representations, which are constituents of the complex Weil representations reduced modulo `. We make use of the following result concerning`-reduced Weil representations. Assuming`6= 2, for V a complex irreducible Weil module whose reduction modulo ` contains M, the Frobenius- Schur indicators of M and V are equal by Theorem 2.8 of Willems [30], a result due to Thompson and Willems. We note that for the classical groups considered below, we have `6= 2, so this result applies. 6.6.1 Frobenius-Schur indicators when S ∼ =A n We start with the case where F ∗ (G) = A n with n≥ 14 where the representation under con- sideration is the standard representation. When `6= 2, all such representations are self-dual and orthogonal by Theorem 1.1 of Turull [27]. And for the case where ` = 2, we have the following result from Lemma 6.2 of Benson [1]: if n≡ 2 (mod 4), the representation preserves a symplectic 68 form but not an orthogonal form; if n≡ 0 or±1 (mod 8), then the representation preserves an orthogonal form of ‘plus’ type; and if n≡ 3, 4, or 5 (mod 8), then the representation preserves an orthogonal form of ‘minus’ type. Using this information in combination with the above information determined above on the Jordan normal forms of the possible regular unipotent elements, we list the regular unipotent elements in this case in Table 5 above. 6.6.2 Frobenius-Schur indicators when S ∼ =PSp 2n (q) Next we assume that S ∼ = PSp 2n (q). Here the `-modular Weil representations have degree 1/2(q± 1). The following results come from Gow [6] and [7]: if q = 3 (mod 4), the Weil characters are not real, so the Frobenius-Schur indicator is 0. Ifq = 1 (mod 4), the Weil characters are real, so we have Frobenius-Schur indicators of either− or +. The Weil representations of degree 1/2(q n −1) in this case have Schur indicator−. Alternatively, the Weil representations of degree 1/2(q n + 1) in this case have Frobenius-Schur indicator +. 6.6.3 Frobenius-Schur indicators when S ∼ =PSU(n,q) Next we assume that S ∼ = PSU(n,q). First, we have one irreducible Weil representations of degree (q n −q)/(q + 1) (up to equivalence), which has a real-valued character, and a Frobenius Schur indicator of− by Corollary 4.5 of Tiep [26] whenq is odd. The Frobenius-Schur indicator in the general case, including when q is even, is−, a result found as Corollary 3 in Szechtman [24]. The rest of the Weil representations (there are (q +1) ` 0−1 more, up to equivalence) have degree (q n + 1)/(q + 1). Of these representations, the ones with real-valued characters are given by a combinatorial criterion found in Gow and Vinroot [9] as Lemma 3.1. By Theorem 4.2 of Gow [8], these real-valued characters have Schur index equal to 1 over Q, and thus have Frobenius-Schur indicator of +. We list a representative for both cases of modular representations, one where the corresponding complex representation has a real-valued character and another whose corresponding complex representation has a character that is not real-valued. 69 6.6.4 Frobenius-Schur indicators when S ∼ =PSL n (q) and n> 2 Lastly we assume that S ∼ =PSL n (q) with n> 2. First, we have one irreducible Weil represen- tation of Frobenius-Schur indicator + of degree (q n −q)/(q− 1)− 1 (up to equivalence). The rest of the Weil representations (there are (q− 1) ` 0− 1 more, up to equivalence) have degree (q n − 1)/(q− 1). Of these representations, the ones with real-valued characters are given by a combinatorial criterion expressed in Proposition 6.2 of Turull [28]. 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Abstract (if available)

Abstract We investigate certain almost quasi-simple subgroups containing certain prime power order elements with an application to the question of which maximal subgroups of classical groups contain a regular unipotent element.

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Creator Donner, Matthew James (author)

Core Title Regular unipotent elements of finite maximal subgroups of classical groups

School College of Letters, Arts and Sciences

Degree Doctor of Philosophy

Degree Program Mathematics

Publication Date 07/16/2018

Defense Date 04/27/2018

Publisher University of Southern California (original), University of Southern California. Libraries (digital)

Tag algebra,group theory,maximal subgroups,OAI-PMH Harvest,regular unipotent elements

Format application/pdf (imt)

Language English

Contributor Electronically uploaded by the author (provenance)

Advisor Guralnick, Robert (committee chair), Montgomery, Susan (committee member), Pilch, Krzysztof (committee member)

Creator Email donner.matthew@gmail.com,mdonner@usc.edu

Permanent Link (DOI) https://doi.org/10.25549/usctheses-c89-16587

Unique identifier UC11671716

Identifier etd-DonnerMatt-6400.pdf (filename),usctheses-c89-16587 (legacy record id)

Legacy Identifier etd-DonnerMatt-6400.pdf

Dmrecord 16587

Document Type Dissertation

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Rights Donner, Matthew James

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Source University of Southern California (contributing entity), University of Southern California Dissertations and Theses (collection)

Access Conditions The author retains rights to his/her dissertation, thesis or other graduate work according to U.S. copyright law. Electronic access is being provided by the USC Libraries in agreement with the a...

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