On Huppert s Conjecture for 3 D 4 (q), q 3

On Huppert s Conjecture for 3 D 4 (q), q 3 Hung P. Tong-Viet School of Mathematical Sciences, University of KwaZulu-Natal Pietermaritzburg 3209, South Africa Tong-Viet@ukzn.ac.za and Thomas P. Wakefield Department of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, Ohio, U.S. 44555 tpwakefield@ysu.edu May 16, 2011 Abstract Let G be a finite group and cd(g) be the set of all complex irreducible character degrees of G. Bertram Huppert conjectured that if H is a finite nonabelian simple group such that cd(g) = cd(h), then G = H A, where A is an abelian group. In this paper, we verify the conjecture for the family of simple exceptional groups of Lie type 3 D 4(q), when q 3. Keywords: Character degrees, Simple groups, Huppert s Conjecture AMS Classification: 20C15 and 20D05 1 Introduction and notation Let G be a finite group, Irr(G) be the set of all complex irreducible characters of G, and denote the set of character degrees of G by cd(g) = {χ(1) : χ Irr(G)}. In instances where the context is clear, we will refer to character degrees as degrees. Also, context permitting, the greatest common divisor of two integers a and b will be denoted by (a, b). In the late 1990s, Bertram Huppert conjectured that the nonabelian simple groups are essentially determined by the set of their degrees. In (Huppert 2000), he posed the following conjecture. Huppert s Conjecture 1. Let G be a finite group and H be a finite nonabelian simple group such that the sets of character degrees of G and H are the same. Then G = H A, where A is an abelian group. 1

Huppert verified the conjecture on a case-by-case basis for many nonabelian simple groups, including the Suzuki groups, many of the sporadic simple groups, and a few of the simple groups of Lie type (Huppert 2000). In this paper, we verify the conjecture for the family of simple groups 3 D 4 (q). Theorem 1.1. Suppose q 3 and the sets of character degrees of G and 3 D 4 (q) are the same. Then G = 3 D 4 (q) A, where A is an abelian group. In (Huppert 2000), Huppert outlines the following five steps which can be used to verify the conjecture. 1. Show G = G. Then if G /M is a chief factor of G, G /M = S k, where S is a nonabelian simple group. 2. Show G /M = H. 3. Show that if θ Irr(M) and θ(1) = 1, then θ is stable under G, which implies [M, G ] = M. 4. Show that M = 1. 5. Show that G = G C G (G ). As G/G = C G (G ) is abelian and G = H, Huppert s Conjecture is verified. 2 Preliminaries We will require several lemmas to carry out the proof of Huppert s Conjecture. We begin with the following results from Clifford Theory. These are presented as stated in Lemmas 2 and 3 in (Huppert 2000). Lemma 2.1(b) is often referred to as Gallagher s Theorem. Lemma 2.1. Suppose N G and χ Irr(G). (a) If χ N = θ 1 + θ 2 + + θ k with θ j Irr(N), then k divides G : N. In particular, if χ(1) is relatively prime to G : N, then χ N Irr(N). (b) If χ N Irr(N), then χθ Irr(G) for every θ Irr(G/N). Lemma 2.2. Suppose N G and θ Irr(N). Let I = I G (θ) denote the inertia subgroup of θ in G. (a) If θ I = k i=1 φ i with φ i Irr(I), then φ G i Irr(G) for all i. In particular, φ i (1) G : I cd(g). (b) If θ allows an extension θ 0 to I, then (θ 0 τ) G Irr(G) for all τ Irr(I/N). In particular, θ(1)τ(1) G : I cd(g). (c) If ρ Irr(I) such that ρ N = eθ, then ρ = θ 0 τ 0, where θ 0 is a character of an irreducible projective representation of I of degree θ(1) while τ 0 is the character of an irreducible projective representation of I/N of degree e. We will also need the following result, stated as Lemma 4 in (Huppert 2000) and Lemma 12.3 in (Isaacs 1994). 2

Lemma 2.3. Let G/N be a solvable factor group of G, minimal with respect to being nonabelian. Then two cases can occur. (a) G/N is an r-group for some prime r. Hence there exists ψ Irr(G/N) such that ψ(1) = r b > 1. If χ Irr(G) and r χ(1), then χτ Irr(G) for all τ Irr(G/N). (b) G/N is a Frobenius group with an elementary abelian Frobenius kernel F/N. Thus f = G : F cd(g) while F : N = r c for some prime r and F/N is an irreducible module for the cyclic group G/F, hence c is the smallest integer such that r c 1 0 (mod G/F ). If ψ Irr(F ), then either fψ(1) cd(g) or r c divides ψ(1) 2. In the latter case, r divides ψ(1). Moreover the following holds. (i) If no proper multiple of f is in cd(g), then χ(1) divides f for all χ Irr(G) such that r χ(1). (ii) If there exists χ Irr(G) such that no proper multiple of χ(1) is in cd(g), then either f divides χ(1) or r c divides χ(1) 2. Moreover if χ(1) is divisible by no nontrivial proper character degree in G, then f = χ(1) or r c χ(1) 2. Proof. All these statements but the last one appear in Lemma 4 of (Huppert 2000). Suppose χ Irr(G) such that no proper multiple of χ(1) is in cd(g). Let ψ Irr(F ) be an irreducible constituent of χ F. By Lemma 6.8 of (Isaacs 1994), we have that χ(1) = kψ(1) and by Corollary 11.29 of (Isaacs 1994) we obtain k f = G : F. By (b), we have that either fψ(1) cd(g) or r a ψ(1) 2. If the latter case holds then we are done since ψ(1) χ(1). Now assume fψ(1) cd(g). Observe that ψ(1) = χ(1)/k so that ψ(1)f = fχ(1)/k cd(g), where fχ(1)/k is a multiple of χ(1) since k f. As no proper multiple of χ(1) belongs to cd(g), it follows that fχ(1)/k = χ(1), which implies that f = k. Since k divides χ(1), we deduce that f χ(1). The remaining statement follows. We will use the following lemma to establish Step 2. Using tensor induction, the following lemma is proved in (Bianchi 2007). Lemma 2.4. (Lemma 5 (Bianchi 2007)). Let N be a minimal normal subgroup of G so that N = S k, where S is a nonabelian simple group. Let A be the automorphism group of S. If σ Irr(S) extends to A, then σ k Irr(N) extends to G. If G is a group, then we denote by M(G) the Schur multiplier of G. A group H is called a covering group of G if there exists a subgroup A of H such that A H Z(H) and H/A = G. A covering group H of G is called a universal covering group if Z(H) = M(G). The following lemmas will be used in proving Step 3. Lemma 2.5. If q is a prime power with q 5 and q 9, then the Schur multiplier of SL 2 (q) is trivial. The Schur multipliers of SL 2 (4) and SL 2 (9) are cyclic of order 2 and 3, respectively. Proof. Assume first that q 5 and q 9. By Theorem 5.1.4]of (Kleidman 1988), SL 2 (q) is the universal covering group of PSL 2 (q) so that the Schur multiplier 3

of SL 2 (q) is trivial by Theorem 3 of (Harris 1977). Finally if q = 4 or q = 9, then the result follows by using the computer program GAP. Lemma 2.6. Assume that q is a prime power with q 3. If K is a maximal subgroup of SL 2 (q) whose index divides q ± 1 or q, then one of the following cases holds. (1) If q 13, q odd or q 4, q even then K is the Borel subgroup of index q + 1. Moreover q + 1 is the smallest index of a maximal subgroup of SL 2 (q). (2) If q {3, 5, 7, 11} then either K is the Borel subgroup of index q + 1 or K is nonabelian of index q. (3) If q = 9, then K is the Borel subgroup of index 10 and 6 is the smallest index of a maximal subgroup of SL 2 (9). Proof. Note that if q 4 and K is a maximal subgroup of SL 2 (q) then the center Z of SL 2 (q) of order (2, q 1) must lie in K as SL 2 (q) is perfect, so that K/Z is a maximal subgroup of SL 2 (q)/z = PSL 2 (q). The result follows by inspecting the list of maximal subgroup of PSL 2 (q) in (King 2005). The result for SL 2 (3) can be found by using GAP. Finally SL 2 (9) = 2 A 6 possesses a maximal subgroup isomorphic to SL 2 (5) = 2 A 5 with index 6. However 6 does not divide 8, 10 or 9. The only maximal subgroup of SL 2 (9) whose index divides 8, 10 or 9 is the Borel subgroup of index 10. The following lemma will be useful in proving Step 4. It also appears as Lemma 6 in (Huppert 2000). Lemma 2.7. Suppose M G = G and λ g = λ for all g G and all λ Irr(M) such that λ(1) = 1. Then M = [M, G ] and M : M divides the order of the Schur multiplier of G /M. 2.1 Results Concerning the Character Degrees of 3 D 4 (q) We first consider the character degrees of 3 D 4 (q). Denote by Φ n := Φ n (q), the cyclotomic polynomial in variable q. As shown in (Deriziotis 1987), the nontrivial character degrees of 3 D 4 (q) are: qφ 12, q 7 Φ 12, 1 2 q3 Φ 2 2Φ 2 6, Φ 1 Φ 2 3Φ 6 Φ 12, Φ 2 Φ 3 Φ 2 6Φ 12, 1 q3 Φ 2 2Φ 12, q 3 Φ 2 Φ 2 6Φ 12, 1 2 q3 Φ 2 1Φ 2 3, 1 2 q3 Φ 2 1Φ 12, q 12, q 3 Φ 2 Φ 3 Φ 6 Φ 12, q 3 Φ 1 Φ 2 3Φ 12, qφ 2 1Φ 2 3Φ 12, qφ 2 2Φ 2 6Φ 12, qφ 1 Φ 2 3Φ 6 Φ 12, Φ 2 Φ 2 6Φ 12, Φ 2 Φ 3 Φ 6 Φ 12, qφ 2 Φ 3 Φ 2 6Φ 12, q 3 Φ 1 Φ 3 Φ 6 Φ 12, Φ 2 2Φ 3 Φ 2 6Φ 12, Φ 1 Φ 3 Φ 6 Φ 12, Φ 2 1Φ 2 2Φ 2 6Φ 12, Φ 1 Φ 2 Φ 3 Φ 2 6Φ 12, Φ 1 Φ 2 3Φ 12, Φ 2 1Φ 2 3Φ 6 Φ 12, Φ 1 Φ 2 Φ 2 3Φ 6 Φ 12, Φ 2 1Φ 2 2Φ 2 3Φ 12, Φ 2 1Φ 2 2Φ 2 3Φ 2 6. For odd q, we have the additional degrees: Φ 3 Φ 6 Φ 12, qφ 3 Φ 6 Φ 12 q 3 Φ 3 Φ 6 Φ 12 and q 4 Φ 3 Φ 6 Φ 12. We are first interested in determining which degrees of 3 D 4 (q) are nontrivial powers. This reduces to the question of which degrees can be written as y p, where p is a prime. If q is even, (q 1, q + 1) = 1, (q + 1, q 2 + q + 1) = 1, and (q 1, q 2 q + 1) = 1. In addition, (q + 1, q 2 q + 1) = 1 or 3 and (q 1, q 2 + q + 1) = 1 or 3. As Nagell showed in (Nagell 1921), q 2 + q + 1 = y p has no solutions unless p is 3. In that case, as proved in (Ljunggren 1943), 4

the only solutions are (q, y, p) = (18, 7, 3) and (q, y, p) = ( 19, 7, 3). As q is prime or a power of a prime, we see that neither of these cases is possible. Thus q 2 + q + 1 cannot be expressed as y n for n > 1 and q 2 q + 1 can be expressed as y n for n > 1 only when q = 19 and n = 3. Hence q 4 q 2 + 1 and q 4 + q 2 + 1 are never nontrivial powers. By a result in (Nagell 1921), q 2 + q + 1 = 3y n has only trivial solutions. Thus, (q 1)(q 2 + q + 1) and (q + 1)(q 2 q + 1) are not nontrivial powers. As q 8 + q 4 + 1 = (q 2 + q + 1)(q 2 q + 1)(q 4 q 2 + 1), we see that q 8 + q 4 + 1 is never a nontrivial power. For the character degrees of G with relatively prime factors to be a power with prime exponent, each of its factors must be a power with the same prime exponent. Lemma 2.8. For even q, the only possible nontrivial powers among the degrees of 3 D 4 (q) are q 12, 1 2 q3 (q 3 + 1) 2, 1 2 q3 (q 3 1) 2 and (q 6 1) 2. Proof. Assume q > 7 is even. Most of the character degrees of 3 D 4 (q) contain pairwise relatively prime factors. Hence, these degrees can be nontrivial powers only if each of their factors is a nontrivial power. This is not the case for the remaining degrees of 3 D 4 (q). We will also need to know which pairs of character degrees of 3 D 4 (q) are consecutive integers. By examining the degrees of 3 D 4 (q), it is possible to prove the following lemma. Lemma 2.9. The group 3 D 4 (q) has no consecutive degrees. We will also exploit the parity of the degrees of 3 D 4 (q). If q is odd, then G has seven odd degrees. They are: qφ 12, q 7 Φ 12, q 12, Φ 3 Φ 6 Φ 12, qφ 3 Φ 6 Φ 12, q 3 Φ 3 Φ 6 Φ 12 and q 4 Φ 3 Φ 6 Φ 12. Notice that if q is odd, the odd degrees of G are the prime power q 12 or are multiples of Φ 12. If q > 2 is even, 3 D 4 (q) has thirteen nontrivial odd degrees. They are Φ 2 Φ 3 Φ 6 Φ 12, Φ 2 Φ 2 6Φ 12, Φ 2 Φ 3 Φ 2 6Φ 12, Φ 2 2Φ 3 Φ 2 6Φ 12, Φ 1 Φ 3 Φ 6 Φ 12, Φ 2 1Φ 2 2Φ 2 3Φ 2 6, Φ 1 Φ 2 3Φ 12, Φ 1 Φ 2 3Φ 6 Φ 12, Φ 1 Φ 2 Φ 2 3Φ 6 Φ 12, Φ 2 1Φ 2 2Φ 2 6Φ 12, Φ 2 1Φ 2 2Φ 2 3Φ 12, Φ 1 Φ 2 Φ 3 Φ 2 6Φ 12, Φ 2 1Φ 2 3Φ 6 Φ 12. Notice that if q is even, the odd degrees of 3 D 4 (q) again are all multiples of Φ 12 except for the perfect square (q 6 1) 2. We define a mixed degree of G to be a character degree of 3 D 4 (q) which is divisible by q but is not a power of q. The mixed degrees of 3 D 4 (q) are qφ 12, q 7 Φ 12, 1 2 q3 Φ 2 1φ 2 3, 1 2 q3 φ 2 2Φ 2 6, qφ 2 1Φ 2 3Φ 12, 1 2 q3 Φ 2 2Φ 12, qφ 2 Φ 3 Φ 2 6Φ 12, qφ 2 2Φ 2 6Φ 12, q 3 Φ 2 Φ 2 6Φ 12, q 3 Φ 1 Φ 2 3Φ 12, 1 2 q3 Φ 2 1Φ 12, q 3 Φ 2 Φ 3 Φ 6 Φ 12, q 3 Φ 1 Φ 3 Φ 6 Φ 12, qφ 1 Φ 2 3Φ 6 Φ 12. For odd q, we have the additional mixed degrees q 3 Φ 3 Φ 6 Φ 12, qφ 3 Φ 6 Φ 12 and q 4 Φ 3 Φ 6 Φ 12. Notice that the highest power of q on the mixed degrees of 3 D 4 (q) is 7. 3 Step 1: Establishing G = G Assume from now on that H = 3 D 4 (q), where q 3 is power of a prime p. Suppose that G G. Then there exists a solvable factor group G/N of G 5

minimal with respect to being nonabelian. By Lemma 2.3, G/N is an r-group or a Frobenius group. Case 1: G/N is an r-group for some prime r. By Lemma 2.3(a), G/N has a character degree r b > 1. By Theorem 1.1 of (Malle 2001), the only nontrivial prime power degree of G is q 12, hence we deduce that r b = q 12 cd(g/n). Let χ Irr(G) with χ(1) = Φ 2 Φ 3 Φ 6 Φ 12. Then r χ(1). By Lemma 2.1, χ N Irr(N). Let τ Irr(G/N) with τ(1) = q 12. Gallagher s Theorem implies that G has a character degree q 12 χ(1), which is impossible. Case 2: G/N is a Frobenius group with elementary abelian Frobenius kernel F/N, where F : N = r c for some prime r. In addition, f = G : F cd(g) and f divides r c 1. Subcase 2(a) : r p. Let χ Irr(G), χ(1) = q 12. As r χ(1) and no proper multiples and no proper divisors of this degree are in cd(g), we deduce from Lemma 2.3 that f = q 12. Hence r χ(1) for any χ Irr(G) with p χ(1). Now let ψ Irr(G) with ψ(1) = qφ 12 and let γ Irr(F ) be an irreducible constituent of ψ F. As G/F = q 12, by Corollary 11.29 of (Isaacs 1994), we have ψ(1)/γ(1) q 12, which implies that Φ 12 γ(1) and so γ(1)f cd(g). Thus by Lemma 2.3 again, we have r c γ(1) 2 p since r p. As f rc 1, we deduce that q 12 = f r c 1 γ(1) 2 p 1 < Φ2 12 = (q 4 q 2 + 1) 2 < q 8, which is impossible. Subcase 2(b) : p = r. Let ϕ Irr(G), with ϕ(1) = Φ 2 1Φ 2 2Φ 2 3Φ 2 6. As r ϕ(1) and no proper multiples and no proper divisors of this degree are in cd(g), we deduce that f = ϕ(1). Hence no proper multiple of f is in cd(g) and thus χ(1) f for any χ Irr(G) with r χ(1). By choosing χ Irr(G) with χ(1) = Φ 1 Φ 3 Φ 6 Φ 12, we obtain a contradiction. 4 Step 2: Establishing G /M = 3 D 4 (q) Suppose G /M is a chief factor of G. As G = G by Step 1, G /M = S k, where S is a nonabelian simple group. We need to show that G /M = 3 D 4 (q). By the classification of the finite nonabelian simple groups, the possibilities for S include the Tits group, the 26 sporadic simple groups, the alternating groups A n for n 7, the ten families of simple groups of exceptional Lie type, and the six families of simple groups of classical Lie type. We must show that k = 1 and eliminate all possibilities for S except 3 D 4 (q). 5 Eliminating the Tits, Sporadic, and Alternating Groups for all k when q is Even By Lemma 2.8, we have that the character degree set of 3 D 4 (q) contains at most four nontrivial powers when q is even. For even q, this result allows for a simple argument to eliminate the Tits, sporadic, and alternating groups from consideration. To eliminate the alternating groups when k > 1, we need the following result, obtained by the construction of χ r,s outlined in the proof of Lemma 2.3 of (Lewis 2007). 6

Lemma 5.1. If n 7, then Irr(A n ) contains at least four nonlinear irreducible characters of different degrees which extend to Aut(A n ). None of these degrees is a prime or composite power of a prime. To eliminate the sporadic groups, we need the following result, found by checking the Atlas (Conway 1984). Since the Tits group is in the Atlas (Conway 1984), we will include it with the sporadic groups. Lemma 5.2. Let S be a sporadic simple group or the Tits group, and let A be the automorphism group of S. Then there exist at least five nonlinear irreducible characters of S of different degrees which extend to A. With the previous two results, we can prove the following proposition. Proposition 5.3. If q is even and S is isomorphic to an alternating group A n with n 7, a sporadic simple group, or the Tits group, then k = 1. Proof. Suppose that k > 1 and S = A n for some n 7. Lemma 5.1 implies S has nonlinear irreducible characters of distinct degrees, φ 1, φ 2, φ 3, and φ 4, say, which extend to Aut(S). None of these characters has degree that is a power of a prime. Recall that G /M is a chief factor of G. As G = G, G /M = S k, where S is a nonabelian simple group. By Lemma 2.4, φ 1 k, φ 2 k, φ 3 k, and φ 4 k extend to G/M. But as seen in Lemma 2.8, there are at most four nontrivial powers among the character degrees of G/M and one of these is a power of a prime. Thus if S = A n with n 7, then k = 1. The same reasoning applies if S is a sporadic simple group or the Tits group. In that case, Lemma 5.2 shows the existence of at least five irreducible characters of S of different degrees which extend to Aut(S). Again invoking Lemma 2.4 and the fact that cd(g) contains at most four nontrivial powers proves the result. Note that A 5 = PSL2 (5) and A 6 = PSL2 (9) will be considered with the simple groups of classical Lie type. We proved that if S is a sporadic simple group or the Tits group, then k = 1. We will now show that S cannot be one of these groups. Proposition 5.4. If q is even, then the simple group S is not one of the sporadic simple groups or the Tits group. Proof. Now G has at most 32 distinct, nontrivial character degrees. As k = 1, if an irreducible character of S extends to Aut(S), then it must extend to G by Lemma 2.4. Consulting the Atlas (Conway 1984), many of the sporadic groups have more than 32 irreducible characters of distinct nontrivial degrees which extend to Aut(S). This implies that G has more than 32 distinct character degrees, a contradiction. We only need to consider the following cases of sporadic simple groups with 32 or less extendible character degrees. Case 1: S = M 11, S = M 12, S = M 23, S = M 24, or S = J 1 For each of these sporadic simple groups, irreducible characters of consecutive degrees extend. As shown in Lemma 2.9, there are no consecutive integers among the degrees of G. Hence this is not possible. 7

Case 2: S = M 22, S = J 2, S = HS, or S = Ru For each of these possibilities for S, it is possible to find three or more odd degrees of S which are relatively prime and extend to Aut(S). This contradicts the fact that all but one odd degrees of G are multiples of q 4 q 2 + 1. Case 3: S = Suz, S = McL, S = He, S = O N, or S = J 3 In each of these cases, S has three characters of odd degree which extend to Aut(S). But the greatest common divisor of these degrees cannot be q 4 q 2 +1, which contradicts the fact that all but one odd degrees of G are multiples of q 4 q 2 + 1. Case 4: S = HN Now S has irreducible characters of exactly 32 distinct degrees which extend to Aut(S). Thus the degrees of S and those of G must be in a one-to-one correspondence. As S has only seven odd degrees, we have a contradiction. Case 5: S = 2 F 4 (2), the Tits group The Tits group has irreducible characters of eight distinct nontrivial degrees which extend. The only power of a prime which extends is 27. The only character degree of G that is a power of a prime is q 12, so we have a contradiction. We have shown that S is neither a sporadic group nor the Tits group. We will now show that S is not an alternating group A n with n 7. Proposition 5.5. If q is even, then the simple group S is not an alternating group A n with n 7. Proof. As shown in (Conway 1984), the simple groups A 7, A 8, and A 10 have irreducible characters of consecutive integer degrees which extend to Aut(A n ), and hence to G by Lemma 2.4. By Theorem 3 of (Bianchi 2007), A n, n 7 possesses two irreducible characters θ i, i = 1, 2, with degrees n(n 3)/2 and n(n 3)/2 + 1 = (n 1)(n 2)/2 and both θ i, i = 1, 2, extend to Aut(A n ) and hence both θ i extend to G by Lemma 2.4. But the degree set of G contains no consecutive degrees. 6 Eliminating the Tits, Sporadic, and Alternating Groups for all k when q is Odd Proposition 6.1. If q is odd, then the simple group S is not an alternating group A n with n 7, a sporadic simple group, or the Tits group. Proof. If S has an irreducible character χ of odd degree which extends to Aut(S), then χ(1) k is an odd degree of G /M, hence an odd degree of G by Lemma 2.4. We will find two or more irreducible characters of S of distinct odd degrees which extend to Aut(S), say χ and φ, with χ(1) > φ(1). If χ(1) k and φ(1) k are not powers of a prime or have Φ 12 as a common factor, then we have a contradiction. This follows from the earlier remark that the odd degrees of G are either a power of a prime or a multiple of Φ 12. Case 1: S = A 7, S = A 8, or S = A 10 8

As shown in the Atlas (Conway 1984), A 7 and A 8 have irreducible characters of degree 21 and 35 which extend to Aut(A 7 ) and Aut(A 8 ), respectively. As (21, 35) = 7, we must have that q 4 q 2 + 1 = 7 k, which is a contradiction. Thus the k th power of these degrees cannot be degrees of G. As shown in the Atlas (Conway 1984), A 10 has seven irreducible characters of odd degree which extend to S 10, hence their k th powers are degrees of G. But the greatest common divisor among the degrees of A 10 which are not prime powers is 1. This is a contradiction. Case 2: S = A 2m, m 6 Adopting the construction and notation used in the proof of Lemma 2.3 of (Lewis 2007), we have the following degree of an irreducible character of S which extends to Aut(S): χ 1,0 (1) = 2m 1. As shown in Theorem 3 of (Bianchi 2007), we also have the irreducible character χ of S which extends to Aut(S) of degree χ(1) = m(2m 3). Now χ 1,0 is odd for any m while χ(1) is odd for odd m. Examining the odd degrees of G, we see that q 12 (2m 1) k since (2m 1) k is a factor of (χ 3,0 (1)) k and q 12 is not a factor of any of the other degrees of G. For odd m, the greatest common divisor of χ 1,0 (1) and χ(1) is 1. This contradicts the fact that, excluding the prime power degree q 12, the odd degrees of G share a common factor of q 4 q 2 +1. For even m, the construction outlined in Lemma 2.3 of (Lewis 2007) provides characters of 35 or more distinct degrees which extend to Aut(S). This is a contradiction as G has only 32 distinct, nontrivial character degrees. Thus, for any k 1, the k th powers of these degrees cannot be degrees of G. Case 3: S = A 2m+1, m 4 Adopting the construction and notation of Lemma 2.3 of (Lewis 2007), we have the following degrees of irreducible characters of S which extend to Aut(S): χ 2,0 (1) = m(2m 1), χ 2,1 (1) = (2m 1)(2m + 1)(2m 3)/3, χ 2,3 (1) = m(2m 1)(2m + 1)(2m 3)(2m 7)/15, and, as constructed in Theorem 3 of (Bianchi 2007), χ(1) = (m 1)(2m + 1). Now χ 2,0 and χ 2,3 (1) are odd for odd m while χ 2,1 is odd for all m. None of these degrees is a power of a prime since (m, 2m 1) = 1 and (2m 1, 2m + 1) = 1 or 2. For odd m, the k th powers of these odd degrees are degrees of G. As none of these degrees is a power of a prime, examining the odd degrees of G, we see that one of the quotients of these three degrees is a power of a prime. The quotients of χ 2,1 (1) and χ 2,0 (1) as well as χ 2,3 (1) yield χ 2,1 (1)/χ 2,0 (1) = (2m + 1)(2m 3)/3m, χ 2,3 (1)/χ 2,0 (1) = (2m + 1)(2m 3)(2m 7)/15, and χ 2,3 (1)/χ 2,1 (1) = m(2m 7)/5. None of these is a power of a prime. So these cannot extend to the irreducible characters of odd degree of G. If m is even, χ(1) is an odd degree. But χ(1) and χ 2,0 (1) are relatively prime and neither χ(1) nor χ 2,0 (1) are powers of a prime. This contradicts the fact that, excluding the prime power degree q 12, the odd degrees of G share a common factor of Φ 12. This eliminates from consideration the alternating groups A n with n 7. 9

Next, consider the possibility that S is a sporadic simple group or the Tits group. Now G has at most 32 distinct, nontrivial character degrees. By the same argument as presented in the proof of Proposition 5.4, we must only consider sporadic simple groups with 32 or less extendible character degrees. We can eliminate S = M 11, S = M 12, S = M 22, S = M 23, S = J 2, S = HS, S = 2 F 4 (2), and S = J 1 as each of these groups has two or more characters of odd degree which extend to Aut(S) and are relatively prime. We can eliminate S = M 24 as it has more than seven odd degrees. Similarly, S = Suz and S = Ru can be eliminated as they both have exactly seven odd degrees, none of which is a power of a prime. We can eliminate S = McL, S = He, S = HN, S = O N, and S = J 3 as these groups have three or more characters of odd degree which extend to Aut(S) and whose common divisor cannot be q 4 q 2 + 1. This eliminates all the sporadic simple groups and 2 F 4 (2) from consideration. 7 Elimination of the Groups of Lie Type To eliminate the groups of Lie type, we will require the Steinberg character of these groups. The properties of the Steinberg character can be found in Chapter 6 of (Carter 1985). We will rely on the fact that the degree of the Steinberg character is a power of a prime, which is proved in Theorem 6.4.7 of (Carter 1985). Lemma 7.1. If S = S(q 1 ) is a simple group of Lie type, and S PSL 2 (q 1 ), then S possesses an irreducible character of mixed degree, i.e., an irreducible character whose degree is divisible by p but is not a power of p. Proof. Examining the list of unipotent characters for the simple groups of exceptional Lie type found in Section 13.9 of (Carter 1985), we see that it is true for these groups. For the simple classical groups of Lie type, the degrees of the unipotent characters labeled by the symbols in Table 1 have mixed degrees. Proposition 7.2. If S = S(q 1 ) is a simple group of Lie type and S PSL 2 (q 1 ), then k 2. Proof. Suppose k 2. The Steinberg character χ of S extends to Aut(S) so χ(1) k = q 12. Let χ(1) = q 1 j. As G /M = S k, there is an irreducible character of G /M found by multiplying k 1 copies of χ with another irreducible character of S of mixed degree, say τ. Then χ k 1 τ is of mixed degree. As the degrees of G /M divide the degrees of G, we must have that the degree of this character divides one of the mixed degrees of G. The highest power of q on any mixed degree of G is 7. Now q 12 = q 1 jk implies q = q 1 jk/12. The exponent on q 1 in (χ k 1 τ)(1) is at least j(k 1) + 1. Now j(k 1) + 1 7jk/12 reduces to 12jk 12j + 12 7jk, which implies j(5k 12) + 12 0. Thus k 2. Since the sporadic, Tits, and alternating groups have been eliminated as possibilities for S, we have that S is a simple group of Lie type and, thus, k 2. 10

Proposition 7.3. The simple group S is not PSL 2 (q 1 ) for any k 1. Proof. Suppose that k = 1. The degree of the Steinberg character of PSL 2 (q 1 ) is q 1. If q 1 is prime, it is impossible for q 1 = q 12. If q 1 is composite, then q 1 + 1 = q 12 + 1 divides a degree of G, which is impossible. Now consider the possibility that k = 2. Here q 1 2 = q 12. Again, if q 1 is prime, it is not possible for q 1 2 = q 12. If q 1 is composite, then (q 6 + 1) 2 is a degree of G /M, hence must divide a degree of G. This is a contradiction. Finally, consider the possibility that k > 2. Then q 1 k = q 12. Hence q = q 1 k/12. Consider the irreducible character of G /M found by multiplying k 1 copies of the Steinberg character with a character of S of degree q 1 1. As the degree of this character must divide a degree of G, we find that k 1 7k/12, which implies k 2, a contradiction. 7.1 Eliminating Groups of Lie Type when k 2 We have proved that G /M = S, where S = S(q 1 ) is a nonabelian simple group of Lie type defined over the field of q 1 elements. We now want to show that S = 3 D 4 (q). We will begin by eliminating the possibility that S is a different simple group of exceptional Lie type. Proposition 7.4. If S is a simple group of exceptional Lie type, then k = 1 and S = 3 D 4 (q). Proof. Case 1: S = G 2 (q 1 ) The Steinberg character of S = G 2 (q 1 ) has degree q 1 6. Suppose first that k = 1. As the Steinberg character of S extends to the character of G of degree q 12, we must have that q 12 = q 1 6, which implies q 1 = q 2. The irreducible character φ 2,2 of S has degree 1 2 q 1Φ 2 (q 1 ) 2 Φ 6 (q 1 ) = 1 2 q2 Φ 4 (q) 2 Φ 12 (q). This degree must divide a mixed degree of G, which is impossible. Now consider the possibility that k = 2. In this case, we have that q 1 12 = q 12, so q = q 1 and the product of the Steinberg character of S and φ 2,2 is an irreducible character of G /M, hence its degree must divide a degree of G. But the degree of this character is 1 2 q7 Φ 2 (q) 2 Φ 6 (q), which divides no mixed degree of G. Case 2: S = F 4 (q 1 ) The Steinberg character of S = F 4 (q 1 ) has degree q 1 24. Suppose first that k = 1. As the Steinberg character of S extends to the character of G of degree q 12, we must have that q 12 = q 1 24 which implies q = q 1 2. Consider the character φ 9,10 of F 4 (q 1 ) of degree q 1 10 Φ 3 (q 1 ) 2 Φ 6 (q 1 ) 2 Φ 12 (q 1 ) = q 5 Φ 3 (q 1 ) 2 Φ 6 (q 1 ) 2 Φ 12 (q 1 ). This degree must divide a mixed degree of G. But the power on q 1 in the mixed degrees of G is less than 5 except for the mixed degree q 7 Φ 12 (q). But this degree of S does not divide this degree. Now consider the possibility that k = 2. In this case, we have that q 1 48 = q 12, so q = q 1 4. The product of the Steinberg character of S and φ 9,10 is an irreducible character of G /M, hence its degree must divide a degree of G. But the degree of this character is q 1 34 Φ 3 (q 1 ) 2 Φ 6 (q 1 ) 2 Φ 12 (q 1 ). This degree divides no degree of G. Case 3: S = 2 B 2 (q 1 2 ) or S = 2 F 4 (q 1 2 ), q 1 2 = 2 2n+1, n 1 11

Suppose first that k = 1. Then the Steinberg character of S is of degree q 4 1 so q 4 1 = q 12, implying q 1 = q 3. But q is an integer while q 1 is not an integer, so we have a contradiction. Now suppose that k = 2. Then we have that q 8 1 = q 12, so q 1 = q 3/2. For S = 2 B 2 (q 2 1 ), q 4 1 + 1 = ((q 3/2 ) 4 + 1) 2 = (q 6 + 1) 2 must divide a degree of G, which is a contradiction. For S = 2 F 4 (q 2 1 ), ɛ (1) = 1 2 q 6 (q 3/2 ) 13 Φ 1 (q 3/2 )Φ 2 (q 3/2 )Φ 4 (q 3/2 ) 2 Φ 6 (q 3/2 ) must divide a degree of G, which results in a contradiction. Case 4: S = 2 G 2 (q 2 1 ), q 2 1 = 3 2m+1, m 1 Suppose first that k = 1. Then q 6 1 = 3 6m+3 is an odd power of 3 while q 12, the degree of the Steinberg character of 3 D 4 (q), is an even power of 3. Hence q 16 q 12. Next suppose that k = 2. In this case, we have that q 12 1 = q 12, so q = q 1. But q is an integer, while q 1 is not. Thus S 2 G 2 (q 2 1 ). Case 5: S = 3 D 4 (q 1 ) Suppose first that k = 2. Then q 24 1 = q 12, so q = q 2 1. Now the product of the Steinberg character of S and the unipotent character labeled by the symbol φ 1,3 is an irreducible character of G /M with degree q1 19 Φ 12 (q 1 ). But then this degree divides no degrees of 3 D 4 (q). Hence k = 1 and so q = q 1 and all divisibility relations are satisfied. Case 6: S is isomorphic to one of the remaining simple groups of exceptional Lie type. For the remaining simple groups of exceptional Lie type, we will use the same general argument. Recall S = S(q 1 ) is a simple group of exceptional Lie type defined over a field of q 1 elements. Suppose the Steinberg character of S has degree q j 1. First assume that k = 1. By Lemma 2.4, q 12 = q j 1, so q = q j/12 1. For each of the remaining possibilities for S, there is a mixed degree of S whose power on q 1 is greater than 7j/12. As the mixed degrees of G have power at most 7j/12 on q 1, we have a contradiction. Table 1 exhibits the degree of the Steinberg character of S and a character of appropriate degree which will result in a contradiction. Now suppose that k = 2. In this case, the mixed degrees of G have power at most 14j/12 on q 1 and q 12 = q 2j 1, so q = q j/6 1. The product of the Steinberg character of S and the given character in Table 1 is an irreducible character of G /M, hence its degree must divide a degree of G. But the degree of this character does not divide a degree of G, which is a contradiction. Proposition 7.5. The simple group S is not a simple group of classical Lie type. Proof. To eliminate most of the simple groups of classical Lie type, we will use the same argument as in Case 6 of Proposition 7.4. As S = S(q 1 ) is a simple group of classical Lie type defined over a field of q 1 elements, the Steinberg character of S has degree q 1 j for some j. Suppose first that k = 1. By Lemma 2.4, q 12 = q 1 j, so q = q 1 j/12. For most of the possibilities for S, we can find a mixed degree of S whose power on q 1 is greater than 7j/12. As the mixed degrees of G have power at most 7j/12 on q 1, we have a contradiction. 12

Next suppose that k = 2. In this case, we have that q 2j 1 = q 12 j/6, so q = q 1 and the product of the Steinberg character of S and another character of mixed degree is an irreducible character of G /M, hence its degree must divide a degree of G. This can be shown to result in a contradiction. Table 1 exhibits the degree of the Steinberg character of S and the symbol corresponding to an irreducible character of S of appropriate degree which will result in a contradiction in both cases. All notation is adapted from Section 13.8 of (Carter 1985). We will proceed by examining each family of simple groups of classical Lie type separately. For S = PSL l+1 (q 1 ) or S = PSU l+1 (q 2 1 ), if k = 1, we must have that l(l 1)/2 7l(l + 1)/24, which is not satisfied for l 4. Hence, if S is one of these groups, then l 3. Now suppose that k = 2. Then l(l 1) 14l(l + 1)/24, which is not satisfied for l 4. Hence, in either case, if S is one of these groups, then l 3. We will examine each of these possibilities for l separately. Case 1: S = PSL 2 (q 1 ) = PSU 2 (q 2 1 ) for q 1 4 This case was handled in Proposition 7.3. Case 2: S = PSL l+1 (q 1 ) or S = PSU l+1 (q 2 1 ) for l = 2 or l = 3 For 2 l 3, we will use the same general argument. Suppose the Steinberg character of S has degree q j 1. Again, let us first assume that k = 1. By Lemma 2.4, q 12 = q j 1, so q = q j/12 1. For 2 l 3, there is a mixed degree of S = PSL l+1 (q 1 ) or S = PSU l+1 (q 2 1 ) which does not divide a degree of G. Table 2 exhibits the degree of the Steinberg character of S and a character of appropriate degree which will result in a contradiction. All notation is adapted from (Simpson 1973) and Section 13.8 of (Carter 1985). Next suppose that k = 2. Then q 2j 1 = q 12 so q = q j/6 1. Again, the product of the degree of the Steinberg character of S and the character degree of S present in Table 2 will result in a contradiction. Next, we consider the possibility that S = Ω 2l+1 (q 1 ) or S = PSp 2l (q 1 ). First consider k = 1. As shown in Table 1, we must have that l 2 2l 7l 2 /12, which is not satisfied for l 5. Now suppose that k = 2. Then 2l 2 2l 14l 2 /12, which is not satisfied for l 3. Suppose that 2 l 4. When l = 2, for k = 1, we have q 4 1 = q 12, so q 1 = q 3. These groups have unipotent degree 1 2 q3 (q 6 + 1) corresponding to the symbol ( ) 0 1 2. This degree divides no degree of G. For k = 2, we have 8 q1 = q 12 and the product 1 2 q 1 5 (q 2 1 +1) must divide a degree of G. But this is not possible. For the remaining possibilities for l, we need only consider the case when k = 1. For l = 3, the Steinberg character has degree q 9 1, so q = q 3/4 1. From Section 13.8 of (Carter 1985), we see that S has a unipotent character χ( α of degree χ α (1) = q1(q 4 1 2 + 1)(q1 3 + 1)/(2q 1 + 2) corresponding to the symbol 1 2 3 ) 0 1. This is a mixed degree of S, hence must divide a mixed degree of G. But it contains the factor q 4 + 1, which does not divide any of the degrees of G. For l = 4, we have that q1 4 = q 3. The group S has a unipotent character labeled by the symbol ( ) 1 2 l 0 1 for Bl and ( ) 1 l 1 1 for Cl with degrees 1 2 q 1 4 Φ 3 (q 1 )Φ 6 (q 1 )Φ 8 (q 1 ) and q 3 1 Φ 2 2(q 1 )Φ 8 (q 1 ), respectively. But this degree divides no degree of 3 D 4 (q). 13

Next, we consider the possibility that S = PΩ ± 2l (q 1 ), where l 4. Suppose that k = 1. As shown in Table 1, we must have that l 2 3l + 2 7l(l 1)/12, which is not satisfied for l 5. Now suppose that k = 2. Then 2l 2 4l + 2 14l(l 1)/12, which is not satisfied for all l. Thus k = 1 and l = 4. Assume first that S = PΩ + 8 (q 1 ). Then q 1 = q 3. By Section 13.8 of ) (Carter 1985), S possesses a unipotent character χ α labeled by the symbol ( 1 3 0 2 with degree q 3 Φ 4 2(q)Φ 6 (q)/2. However this degree divides no degree of 3 D 4 (q). Assume that S = PΩ 8 (q 1 ). Then q 1 = q 3. By Section 13.8 of (Carter 1985), S possesses a unipotent character χ labeled by the symbol ( 1 3 ) of degree χ(1) = q(q 4 + 1). However this degree divides no degree of 3 D 4 (q). 8 Step 3: Establishing I G (θ) = G We are now ready to prove Step 3 of the Huppert s method. Suppose I G (θ) = I G for some θ Irr(M) with θ(1) = 1. Let U be maximal such that I U G. If θ I = s i=1 e iφ i, for φ i Irr(I), then by Lemma 2.2, φ i (1) G : I is a degree of G and thus divides some degree of G. We will need to find indices of maximal subgroups of 3 D 4 (q) which divide character degrees of 3 D 4 (q). From the list of maximal subgroups of 3 D 4 (q) in (Kleidman 1988), we have Table 3. Let q = p a and recall that the notation is as follows: [q 5 ] denotes an unspecified group of order q 5, A : B denotes a split extension, A B denotes a central product, and A B denotes a non-split extension. Let d = (2, q 1) and recall that if N G and λ Irr(N), then the set of irreducible constituents of λ G is denoted by Irr(G λ). Lemma 8.1. The only maximal subgroups of 3 D 4 (q) whose indices divide a degree of G are the maximal parabolic subgroups with structure [q 9 ] : (SL 2 (q 3 ) Z q 1 ) d and [q 11 ] : (Z q2 1 SL 2 (q)) d. Proof. The indices of the parabolic subgroups with structure [q 9 ] : (SL 2 (q 3 ) Z q 1 ) d or [q 11 ] : (Z q2 1 SL 2 (q)) d divide many degrees of G. The indices of all of the remaining subgroups in Table 3 must divide mixed degrees of 3 D 4 (q). A close examination of the mixed degrees of 3 D 4 (q) shows that these indices have exponents on q too large to divide a mixed degree of G. For the subgroup 3 D 4 (q 0 ) of Table 3, note that q = q 0 α so the power on q 0 in the mixed degrees of 3 D 4 (q) is at most 7α. To divide a degree of 3 D 4 (q), we must have that the power on q 0 in the index of the subgroup must satisfy 12α 12 7α. This implies α 2. Certainly α 1 as we are considering a maximal subgroup. Suppose that α = 2. Then the index of the maximal subgroup is q 0 12 (q 0 2 + 1)(q 0 6 + 1)(q 0 8 q 0 4 + 1). As q = q 0 2, this index can be expressed in terms of q as q 6 (q + 1)(q 3 + 1)(q 4 q 2 + 1). Clearly this index divides no degree of G. Hence the only maximal subgroups of 3 D 4 (q) whose indices divide a degree of G are the parabolic subgroups. The character tables of the parabolic subgroups of 3 D 4 (q) are available in (Himstedt 2004,2007). Suppose U/M is a maximal parabolic subgroup of 14

G /M. We have I/M U/M G /M. Then G /M : U/M U/M : I/M φ i (1) divides a degree of G. It follows from Lemma 8.1 that either U/M = [q 9 ] : (SL 2 (q 3 ) Z q 1 ) d or U/M = [q 11 ] : (Z q2 1 SL 2 (q)) d, where d = (2, q 1) and q 3. Let t = U : I and recall that θ Irr(M) with θ(1) = 1, I = I G (θ), and θ I = s i=1 e iφ i, where φ i Irr(I), e i 1 for all i. 8.1 The parabolic subgroups P a with structure [q 9 ] : (SL 2 (q 3 ) Z q 1 ) d We have G : U = Φ 2 Φ 3 Φ 6 Φ 12 and hence tφ i (1) divides one of the following numbers: q 3, q 3 1, q 3 + 1, q(q 2 q + 1), (q 1)(q 2 q + 1). Denote by A the set consisting of the numbers above. Observe that the first three numbers are character degrees of SL 2 (q 3 ), and the latter two numbers are less than q 3 + 1. Let L, V be subgroups of U containing M such that L/M = [q 9 ], and V/M = SL 2 (q 3 ), and let W = LV U. It follows that L U, W U, L V = M and W/L = SL 2 (q 3 ). Case 1. p t. As L U, we have I IL U so that t = U : IL IL : I. Now IL : I = L : L I and L : L I divides L/M. Thus if L I, then L : L I > 1 and is divisible by p so that p t, which is a contradiction. Therefore we can assume that L I U. Assume that W I. Then I W I U and t = U : W I W I : I. As W I : I = W : W I and W I : I > 1, we deduce that W : W I > 1 and divides one of the members of A. Observe that W/L = SL 2 (q 3 ) and since L W I W, we deduce that W : W I is divisible by some index of a maximal subgroup of SL 2 (q 3 ). As q 3, we obtain q 3 27, so that by Lemma 2.6, we must have that W : W I = q 3 + 1 and W I/L is isomorphic to the Borel subgroup of SL 2 (q 3 ), in particular W I/L is nonabelian. Moreover t = q 3 +1 and hence since tφ i (1) q 3 +1, we deduce that φ i (1) = 1 for all i. Then θ extends to φ Irr(I). Hence by Lemma 2.2(b) we obtain θ I = τ Irr(I/M) τφ and so τ(1) = 1 for any τ Irr(I/M), which implies that I/M is abelian. This is a contradiction as I/M possesses a nonabelian section namely W I/L. Thus we conclude that W I. Subcase 1(a). p φ j (1) for some j. Let λ Irr(L) be an irreducible constituent of φ j when restricted to L. Since (φ j ) M = e j θ, we deduce that λ M = eθ for some integer e. Now λ(1) = λ(1)/θ(1) = e must divide L/M by Corollary 11.29 of (Isaacs 1994). However by Clifford theory, we know that λ(1) φ j (1) which implies that λ(1) is coprime to p. Thus λ(1) = 1 and so λ is an extension of θ to L. By Lemma 2.1, we have θ L = τ Irr(L/M) λτ and hence τ(1)λ(1) = τ(1) φ i (1), where φ i is an irreducible constituent of λ I. As L/M = q 1+8 is the unipotent radical of the parabolic subgroup P a of 3 D 4 (q), by (Himstedt 2004,2007), L/M possesses an irreducible character τ of degree q 4. Let γ = τλ Irr(L θ). Then γ(1) = q 4 and since L I, we deduce that γ(1) ϕ(1) for any ϕ Irr(I γ) and hence q 4 ϕ(1), which is impossible as ϕ(1) q 3 + 1. Subcase 1(b). p φ i (1) for all i. Then tφ i (1) divides q 3 or q(q 2 q + 1) 15

for all i. As W I U, t U : W = q 1 and (q 1, q 2 q + 1) = 1, we deduce that t = 1 since p t. Hence I = U and (φ i (1), q 1) = 1 for all i. It follows that all φ i are irreducible upon restriction to W. Let λ be any irreducible constituent of θ L. We will show that λ is W -invariant. Suppose not. Let J be the stabilizer in W of λ. Since W I, if ϕ Irr(W λ) then ϕ(1) must divide some φ k so that ϕ(1) divides q 3 or q(q 2 q + 1). Now for δ Irr(J λ), we have that δ W Irr(W γ) so that δ W (1) = W : J δ(1) divides q 3 or q(q 2 q + 1). As L J W, and q 3 27, by Lemma 2.6 we deduce that W : J q 3 + 1, which is a contradiction as W : J q 3. Thus λ is W -invariant. By Lemma 2.5, the Schur multiplier of W/L = SL 2 (q 3 ) is trivial so that by Theorem 11.7 of (Isaacs 1994), λ extends to λ 0 Irr(W λ). By Gallagher s Theorem, τλ 0 are all irreducible constituents of λ W, where τ Irr(W/L) and hence τ(1)λ 0 (1) divides q 3 or q(q 2 q + 1), which is impossible as W/L possesses an irreducible character of degree q 3 + 1. Case 2. p t. Then t > 1 and so tφ i (1) divides q 3 or q(q 2 q + 1) for all i. Let X = W I I. Assume first that L X. Then L I. Suppose W I. Then I W I U and t = U : W I W I : I. As W I : I = W : W I and W I : I > 1, we deduce that W : W I > 1 and divides q 3 or q(q 2 q + 1), and hence W : W I q 3. Observe that W/L = SL 2 (q 3 ) and since L W I W, we deduce that W : W I is divisible by some index of a maximal subgroup of SL 2 (q 3 ). As q 3 27, by Lemma 2.6, we obtain that W : W I q 3 + 1, which is impossible. Thus W I. Let λ be any irreducible constituent of θ L. Arguing as in Subcase 1(b), we obtain that λ is W -invariant. By Lemma 2.5, the Schur multiplier of W/L = SL 2 (q 3 ) is trivial so that by Theorem 11.7 of (Isaacs 1994), λ extends to λ 0 Irr(W λ). By Gallagher s Theorem, τλ 0 are all irreducible constituents of λ W, where τ Irr(W/L) and hence τ(1)λ 0 (1) divides q 3 or q(q 2 q+1), which is impossible as W/L possesses an irreducible character of degree q 3 + 1. Hence L X and so L XL and X XL. As t = U : W I W : X, we have that W : X t and hence W : X q 3. Since X XL W, we obtain W : X = W : XL XL : X, where XL : X = L : L X. Assume XL W. Then W : XL is divisible by the index of some maximal subgroup of W/L = SL 2 (q 3 ), which is impossible as W : XL W : X q 3. Thus W = XL and so W/L = X/X L = SL 2 (q 3 ). Let L 1 = X L X. Then X/L 1 = SL2 (q 3 ). Let λ be any irreducible constituent of θ L1. Since M L 1 X I, we deduce that every irreducible constituent of λ X divides either q 3 or q(q 2 q + 1), so that with the same argument as in the previous paragraph, we deduce that λ is X-invariant and then as the Schur multiplier of X/L 1 is trivial, we obtain a contradiction as in Subcase 1(b). 8.2 The parabolic subgroup P b with structure [q 11 ] : (SL 2 (q) Z q 3 1) d We have G : U = (q 3 + 1)(q 8 + q 4 + 1) and hence tφ i (1) divides one of the following numbers: q, q 1, q + 1. Observe that these numbers are character 16

degrees of SL 2 (q). Let L, V be subgroups of U containing M such that L/M = [q 11 ], and V/M = SL 2 (q), and let W = LV U. It follows that L U, W U, L V = M and W/L = SL 2 (q). Case 1 : tφ j (1) q±1 for some j. Then t is relatively prime to p. As L U, we deduce that I IL U so that t = U : IL IL : I. Now IL : I = L : L I. Thus if L I, then L : L I > 1 and is divisible by p so that p t, which is a contradiction. Therefore we can assume that L I U. Let λ Irr(L) be an irreducible constituent of φ j when restricted to L. Since (φ j ) M = e j θ, we deduce that λ M = eθ for some integer e. Now λ(1) = λ(1)/θ(1) = e must divide L/M by Corollary 11.29 of (Isaacs 1994). However by Clifford theory, we know that λ(1) φ j (1) which implies that λ(1) is coprime to p. Thus λ(1) = 1 and so λ is an extension of θ to L. As the unipotent radical L/M = [q 11 ] of the parabolic subgroup P b is nonabelian, we can choose τ Irr(L/M) such that τ(1) > 1 is divisible by p. By Gallagher s Theorem, we have that γ = λτ Irr(L θ) and p γ(1). Since L I, we have that if φ Irr(I γ) then φ(1) q. Now let φ k be an irreducible constituent of γ I. As γ(1) φ k (1), we have that tφ k (1) q and since t is coprime to p, we must have t = 1 and so I = U. Assume first that q 4 so that W/L = SL 2 (q) is nonsolvable. We next show that γ is W -invariant. Note that if φ Irr(W γ) then φ(1) q. Suppose that γ is not W -invariant and let J = I W (γ). Then L J < W. Let δ be an irreducible constituent of γ J. Then δ J (1) = W : J δ(1) q, and W : J is divisible by the index of a maximal subgroup of W/L = SL 2 (q), q 4. By Lemma 2.6, we deduce that q = 5, 7 or q = 11 and also J/L is isomorphic to a nonabelian subgroup of SL 2 (q) of index q. Thus W : J = q and hence δ(1) = 1, which is impossible as p γ(1) and γ(1) δ(1) so that δ(1) 1. Thus γ is W -invariant and every irreducible constituent of γ W is a power of a fixed prime p, so that by Theorem 2.3 of (Moreto 2006), W/L is solvable, which is a contradiction. Now let q = 3. Then the unipotent radical L/M = 3 2+3+6 of the parabolic subgroup P b possesses an irreducible character τ of degree 3 3 by Section 6 of (Himstedt 2004). In this case, we choose γ = τλ Irr(L θ) and then γ(1) = 3 3 λ(1). However this is a contradiction as 3 3 > 4 = q +1 and γ(1) divides q = 3. Case 2 : tφ i (1) q for all i. Then t q and all φ i (1) are p-powers. We claim that I/M is nonsolvable when q 4 and nonabelian when q = 3. If t = 1, then I = U and hence I/M = [q 11 ] : (SL 2 (q) Z q 3 1) d is obviously nonsolvable when q 4 and nonabelian when q = 3 as SL 2 (3) is nonabelian. Thus we assume that t > 1. If W I, then as W/L = SL 2 (q), we deduce that I/M is nonsolvable if q 4 and nonabelian if q = 3. Hence we assume W I and t > 1. Then I W I U and so t = U : W I W I : I = U : W I W : W I. As W I, we have that W I W so that W : W I is a nontrivial divisor of q. Let X = W I. Assume that L X. Then X/L is a proper subgroup of W/L = SL 2 (q) and hence there exists a subgroup K of W such that X K W and K/L is isomorphic to a maximal subgroup of SL 2 (q) whose index W : K is a divisor of q. It follows from Lemma 2.6 that q = 3, 5, 7 or q = 11, W : K = q and K/L is nonabelian. Thus q {3, 5, 7, 11}, W : X = q and X/L is nonabelian. Hence t = q and then all φ i (1) = 1. 17

Gallagher s Theorem now yields that I/M is abelian, which is impossible as I/M possesses a nonabelian section X/L. Thus L X. Since L W and X = W I W, we deduce that X XL W and L XL W. It follows that W : X = W : XL XL : X is a nontrivial divisor of q. If W = XL, then SL 2 (q) = W/L = XL/L = X/X L and hence since M X L X I, we deduce that I/M is nonsolvable if q 4 and nonabelian if q = 3. Therefore we assume that W : XL = 1 and hence W : XL is a nontrivial divisor of q, XL/L = X/X L and L XL W. Using the same argument, we complete the proof of our claim. Now assume that q 4. Then θ Irr(M), θ(1) = 1, θ is I-invariant and for any χ Irr(I θ), we have that χ(1) is a power of a fixed prime p, so that by Theorem 2.3 of(moreto 2006), I/M is solvable, which is a contradiction. Now let q = 3. As tφ i (1) 3 for all i, and t > 1, we deduce that t = 3 and hence φ i (1) = 1 for all i. Gallagher s Theorem now yields that I/M is abelian, which contradicts our claim above. This finishes the proof of Step 3. 9 Step 4: Establishing M = 1 By Step 2, we know that G /M = 3 D 4 (q). Hence, when paired with this step, we have that G = 3 D 4 (q). For all q, the Schur multiplier of 3 D 4 (q) is trivial. Thus M = M by Step 3 and Lemma 2.7. If M is abelian, then we are done. Suppose M = M 1. Let M/N = T k be a chief factor of G, where T is a nonabelian simple group. Then T possesses a nonprincipal irreducible character λ which is extendible to Aut(T ). By Lemma 2.4, λ k Irr(M/N) extends to ψ Irr(G /N). Let χ Irr(G /M) be such that χ(1) = q 12. By Gallagher s Theorem, we have ψχ Irr(G ) so that ψ(1)χ(1) = λ(1) k q 12 cd(g ), and hence λ(1) k q 12 must divide some character degree of G, which is impossible. Hence M = 1. 10 Step 5: Establishing G = G C G (G ) We will show that G = G C G (G ). As G/G = C G (G ) and G/G is abelian, this will prove Huppert s Conjecture for this family of simple groups. Let q = p a. Suppose G C G (G ) G. Then G induces on G some outer automorphism σ. By Theorem C of (Feit 1989), some conjugacy class of G is not fixed by σ. As the irreducible characters of G separate the conjugacy classes of G, there exists some χ Irr(G ) such that χ is not fixed by σ. Let ψ Irr(G) such that [ψ G, χ] > 0. Then ψ(1) = eχ(1) where e > 1 and e divides Out(G ) = a. Now χ(1) > 1 and eχ(1) cd(g). As χ Irr(G ) and G = 3 D 4 (q), we have that χ(1) and eχ(1) are both character degrees of G. Examining the character degrees of G to find degrees with proper multiples which are also degrees of G shows that e q. As e a, we have a e q = p a, a contradiction. Thus G = G C G (G ). This concludes the verification of the five steps of Huppert s method and proves Theorem 1.1. 18