On Huppert's Conjecture for the Monster and Baby Monster more

Noname manuscript No. (will be inserted by the editor) On Huppert’s Conjecture for the Monster and Baby Monster Hung P. Tong-Viet · Thomas P. Wakefield April 26, 2011 Abstract Let G be a finite group. Let cd(G) be the set of all complex irreducible character degrees of G. In this paper, we will show that if cd(G) = cd(H), where H is the Monster or the Baby Monster simple sporadic groups, then G ∼ H × A, where A is an abelian group. = Keywords character degrees, sporadic groups PACS 20C15 (primary), 20D05 (secondary) 1 Introduction and notation All groups considered are finite and all characters are complex characters. Let G be a group and let Irr(G) = {χ1 , χ2 , · · · , χk } be the set of all irreducible characters of G and let cd(G) = {χ(1) | χ ∈ Irr(G)} be the set of all character degrees of G. In [6], Huppert proposed the following conjecture. Huppert’s Conjecture Let G be a finite group and let H be a non-abelian simple group. If cd(G) = cd(H), then G ∼ H × A, where A is abelian. = Hung P. Tong-Viet School of Mathematical Sciences, University of KwaZulu-Natal Pietermaritzburg 3209, South Africa E-mail: Tong-Viet@ukzn.ac.za Thomas P. Wakefield Department of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, Ohio, U.S. 44555 E-mail: tpwakefield@ysu.edu 2 Hung P. Tong-Viet, Thomas P. Wakefield Huppert verified this conjecture for 18 out of 26 sporadic simple groups, L2 (q), and 2 B2 (q 2 ) in [6,7] and a series of pre-prints. The remaining sporadic simple groups except the Monster and the Baby Monster were verified in a pre-print by S. H. Alavi, A. Daneshkhah, H.P. Tong-Viet and T. Wakefield entitled “Huppert’s Conjecture and the sporadic simple groups”. In [12–14], T. Wakefield verified this conjecture for L3 (q), U3 (q), 2 G2 (q 2 ) and S4 (q). The proof is based on the following five steps outlined in [6]. Step 1. Show G = G . Step 2. Show G /M ∼ H. = Step 3. If θ ∈ Irr(M ), θ(1) = 1, then IG (θ) = G , hence [M, G ] = M . Step 4. Show M = 1. Step 5. Show G = G × CG (G ). In this note, we will verify this conjecture for the Monster and Baby Monster simple sporadic groups and thus complete verifying Huppert’s Conjecture for the sporadic simple groups. Theorem 1 Let G be a finite group and let H be the Monster or the Baby Monster sporadic simple group. If cd(G) = cd(H), then G ∼ H × A, where A = is abelian. In this paper, we introduce the notion of an isolated character and use it to simplify the proof of Step 1 described above. In verifying Step 3 for the Baby Monster, we use some results of Frank L¨beck on the character tables of some u covering groups of 2 E6 (2) which are available on his website. If n is an integer then we denote by π(n) the set of all prime divisors of n. If G is a group, we will write π(G) instead of π(|G|) to denote the set of all prime divisors of the order of G. Let ρ(G) = ∪χ∈Irr(G) π(χ(1)) be the set of all primes which divide an irreducible character degrees of G. Finally, if N G and θ ∈ Irr(N ), then the inertia group of θ in G is denoted by IG (θ). Other notation is standard. 2 Preliminaries In this section, we present some results that we will need for the proof of Huppert’s Conjecture. Lemma 1 ([6, Lemma 2]). Suppose N G and χ ∈ Irr(G). (a) If χN = θ1 + θ2 + · · · + θk with θi ∈ Irr(N ), then k divides |G/N |. In particular, if χ(1) is prime to |G/N | then χN ∈ Irr(N ). (b) (Gallagher’s Theorem) If χN ∈ Irr(N ), then χψ ∈ Irr(G) for every ψ ∈ Irr(G/N ). Lemma 2 ([6, Lemma 3]). Suppose N G and θ ∈ Irr(N ). Let I = IG (θ). k G (a) If θI = i=1 ϕi with ϕi ∈ Irr(I), then ϕi ∈ Irr(G). In particular, ϕi (1)|G : I| ∈ cd(G). (b) If θ extends to ψ ∈ Irr(I), then (ψτ )G ∈ Irr(G) for all τ ∈ Irr(I/N ). In particular, θ(1)τ (1)|G : I| ∈ cd(G). On Huppert’s Conjecture for the Monster and Baby Monster 3 (c) If ρ ∈ Irr(I) such that ρN = eθ, then ρ = θ0 τ0 , where θ0 is a character of an irreducible projective representation of I of degree θ(1) while τ0 is the character of an irreducible projective representation of I/N of degree e. The following lemma will be used to verify Step 1 in Huppert’s method of proof. Lemma 3 Let G/N be a solvable factor group of G, minimal with respect to being non-abelian. 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) = rb > 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. Then f = |G : F | ∈ cd(G) and |F/N | = ra for some prime r, and F/N is an irreducible module for the cyclic group G/F, hence a is the smallest integer such that ra ≡ 1(mod f ). If ψ ∈ Irr(F ) then either f ψ(1) ∈ cd(G) or ra divides ψ(1)2 . In the latter case, r divides ψ(1). 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). Furthermore, 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 divides χ(1). In particular, if χ(1) is divisible by no nontrivial proper character degree in G, then f = χ(1) or r | χ(1). Proof All these statements but the last one appear in [6, Lemma 4]. Suppose that χ ∈ Irr(G) such that no proper multiple of χ(1) is in cd(G). Assume that r χ(1). We will show that f divides χ(1). By Lemma 1, χF = θ1 + · · · + θk , where θi ∈ Irr(F ) and k divides f. As χ(1) = kθ1 (1) and r χ(1), we have r θ1 (1), hence θ1 (1)f ∈ cd(G). Observe that θ1 (1) = χ(1)/k so that θ1 (1)f = f χ(1)/k ∈ cd(G). As no proper multiple of χ(1) belongs to cd(G), it follows that f χ(1)/k = χ(1), which implies that f = k. Since k divides χ(1), so does f. The proof is now complete. Let G be a finite group. We define cd (G) to be a subset of cd(G) consisting of maximal elements of cd(G) ordered by divisibility. Obviously, for any a ∈ cd(G), a ∈ cd (G) if and only if no proper multiple of a is in cd(G). Note that the largest character degree of G is in cd (G), and hence cd (G) is non-empty. Let χ ∈ Irr(G). We say that χ is isolated in G if χ(1) ∈ cd (G) and χ(1) is divisible by no proper non-trivial character degrees of G. In this situation, we also say that χ(1) is an isolated degree of G. Let G be a group. 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)|. Lemma 4 Let G be a group. (a) If (|G : G |, M (G)) = 1, then G has a unique universal covering group and any covering group of G has trivial Schur multiplier. 4 Hung P. Tong-Viet, Thomas P. Wakefield (b) If M (G) = 1 and Z ≤ G ∩ Z(G), then G is a universal covering group of G/Z. Proof (a) is [5, Theorem 3] and (b) is Corollary 1.2 in [5]. The next two lemmas will be used to verify Steps 2 and 4. The first lemma appears in [1, Theorems 2, 3, 4]. Lemma 5 If S is a non-abelian simple group, then there exists a nontrivial irreducible character θ of S that extends to Aut(S). Moreover the following hold: (i) if S is an alternating group of degree at least 7, then S possesses two consecutive characters of degrees n(n − 3)/2 and (n − 1)(n − 2)/2 that both extend to Aut(S). (ii) if S is a sporadic simple group or the Tits group, then S possesses two nontrivial irreducible characters of coprime degrees which both extend to Aut(S). (iii) if S is a simple group of Lie type then the Steinberg character St of S of degree |S|p extends to Aut(S). Lemma 6 ([1, Lemma 5]). Let N be a minimal normal subgroup of G so that N ∼ S k , where S is a non-abelian simple group. If θ ∈ Irr(S) extends to = Aut(S), then θk ∈ Irr(N ) extends to G. The following lemma will be used to verify Step 4. Lemma 7 ([6, Lemma 6]). Suppose that M G = G and that for any λ ∈ Irr(M ) with λ(1) = 1, λg = λ for all g ∈ G . Then M = [M, G ] and |M/M | divides the order of the Schur multiplier of G /M. 3 The Monster The Monster group M is the largest sporadic simple group. Its existence was predicted by Fischer and it was first constructed by Griess in 1980. We collect some information regarding the Monster group in the following lemma. Lemma 8 Let M be the Monster sporadic simple group. (i) M has no nontrivial power degrees nor consecutive degrees. (ii) If 1 = χ ∈ Irr(M) and {17, 41, 47, 59, 71} ∩ π(χ(1)) = ∅, then χ(1) = 1. (iii) The following degrees are isolated degrees: 244 .76 .17.23.41.59.71; 242 .57 .11.23.41.59.71; 320 .76 .112 .17.29.31.41.59.71; 317 .57 .73 .112 .132 .31.59.71; 59 .73 .112 .133 .17.19.23.31.41.47.59; 312 .76 .133 .17.19.23.29.31.41.47.71. (iv) The following degrees are co-prime: 47.59.71, 59 .76 .112 .17.19. (v) The outer automorphism group and the Schur multiplier of M are trivial. (vi) The indices of maximal subgroups of M do not divide any degrees of M. On Huppert’s Conjecture for the Monster and Baby Monster 5 Proof Using [3], (i) − (v) are obvious. For (vi), M has 43 known conjugacy classes of maximal subgroups and any other maximal subgroup of M is almost simple with socle isomorphic to one of the following simple groups L2 (13), U3 (4), U3 (8), Sz(8) (see [2]). In Table 1, we list the structures of the 43 known maximal subgroups of M and their indices. This table is taken from [2, Table 1]. Using [3] we can check that no index in Table 1 divides any degree of the Monster. For the remaining possible maximal subgroups of M, the orders of the full outer automorphism groups of the simple groups listed above are 2, 4, 18, and 3, respectively. Let U be a possible maximal subgroup of M. Then U is almost simple with socle S which belongs to the list above. It follows that the order of U is bounded above by the order of the full automorphism group of S and then the index of U in the Monster is at least |M|/| Aut(S)|, which is larger than the largest character degree of M and so this index cannot divide any character degree of the Monster. Step 1. G = G . By way of contradiction, suppose that G = G . Then there exists a normal subgroup N G such that G/N is solvable and minimal with respect to being non-abelian. By Lemma 3, G/N is an r-group for some prime r or G/N is a Frobenius group. Case 1. G/N is an r-group. Then there exists ψ ∈ Irr(G/N ) such that ψ(1) = rb > 1. However by Lemma 8(i), G has no non-trivial prime power degrees. Hence this case cannot happen. Case 2. G/N is a Frobenius group with Frobenius kernel F/N, |F/N | = ra , 1 < f = |G : F | ∈ cd(G) and ra ≡ 1(mod f ). By Lemma 3(b), if χ ∈ Irr(G) such that χ(1) is isolated then either f = χ(1) or r | χ(1). We observe that there is no prime which divides all the isolated degrees listed in Lemma 8(iii). Thus f must be one of the degrees listed there and hence f is isolated in G. By Lemma 3(b) again, if χ ∈ Irr(G) with r χ(1) then χ(1) | f. As f is isolated we deduce that r must divide every nontrivial degree χ(1) of G, where χ ∈ Irr(G) with χ(1) = f. By Lemma 8(iv), G has two nontrivial co-prime degrees which are not one of the isolated degrees in Lemma 8(iii), and hence these degrees are different from f, so that r must divide both of these degrees, which is impossible. Thus this case cannot happen. Therefore G = G . Step 2. Let M ≤ G be a normal subgroup of G such that G /M is a chief factor of G. As G is perfect, G /M is non-abelian so that G /M ∼ S k for = some non-abelian simple group S and some integer k ≥ 1. Claim 1. k = 1. By way of contradiction, assume that k ≥ 2. By Lemma 5, S possesses a nontrivial irreducible character θ which is extendible to Aut(S) and so by Lemma 6, θk ∈ Irr(G /M ) extends to G/M, hence θ(1)k ∈ cd(G), which contradicts Lemma 8(i). This shows that k = 1. Hence G /M ∼ S. = Claim 2. S is not an alternating group. By way of contradiction, assume that S = An , n ≥ 5. As A5 ∼ L2 (5) and A6 ∼ L2 (9), we can assume that = = n ≥ 7. By Lemma 5, S possesses two nontrivial irreducible characters θ1 , θ2 with θ1 (1) = n(n − 3)/2, θ2 (1) = θ1 (1) + 1 = (n − 1)(n − 2)/2 and both θi extend to Aut(S), so that M possesses two consecutive non-trivial character degrees. This contradicts Lemma 8(i). 6 Hung P. Tong-Viet, Thomas P. Wakefield Table 1 The known maximal subgroups of the Monster Group 2. B 22.2 E6 (2) : S3 21+24 . Co1 + 22 .211 .222. (S3 × M24 ) 23 .26 .212 .218. (L3 (2) × 3S6 ) 25 .210 .220. (S3 × L5 (2)) + 210 .216. O10 (2) 8. O − (3). 2 3 3 8 31+12 . 2. Suz : 2 + + 32 .35 .310. (M11 × 2S4 ) 3 .32 .36 .36 : (L (3) × SD ) 3 3 16 3. F i24 51+6 : 2J2 .4 + − 71+4 : (3 × 2S7 ) + 1+2 13+ : (3 × 4S4 ) 54 : (3 × 2. L2 (25)) : 22 72 : 2. L2 (7) 112 : (5 × 2A5 ) 132 : 2. L2 (13) : 4 41 : 40 + (32 : 2 × O8 (3)). S4 S3 × T h (D10 × HN ). 2 (52 : (4 ◦ Q8 ) × U3 (5)) : S3 (7 : 3 × He) : 2 (72 : (3 × 2A4 ) × L2 (7)).2 (13 : 6 × L3 (3)). 2 72 .7.72 : GL2 (7) 53. 53. (2 × L3 (5)) 52 .52 .54 : (S3 × GL2 (5)) (L2 (11) × M12 ) : 2 (A5 × A12 ) : 2 M11 × A6 . 22 (L3 (2) × S4 (4) : 2). 2 (A5 × U3 (3) : 31 ) : 2 A3 . (2 × S4 ) 6 (A7 × (A5 × A5 ) : 22 ) : 2 3 :S S5 3 L2 (11)2 : 4 L2 (71) L2 (59) L2 (29) : 2 L2 (19) : 2 Index 24 .37 .74 .11.132 .29.41.59.71 27 .310 .57 .11.132 .23.29.31.41.47.59.71 11 .55 .74 .11.132 .17.19.29.31.41.47.59.71 3 316 .58 .75 .11.133 .17.19.29.31.41.47.59.71 316 .58 .75 .112 .133 .17.19.23.29.31.41.47.59.71 316 .58 .75 .112 .133 .17.19.23.29.31.41.47.59.71 315 .57 .75 .112 .133 .19.23.29.31.47.59.71 235 .58 .75 .112 .132 .17.19.23.29.31.47.59.71 231 .57 .75 .11.132 .17.19.23.29.31.41.47.59.71 238 .58 .76 .11.133 .17.19.23.29.31.41.47.59.71 238 .59 .76 .112 .132 .17.19.23.29.31.41.47.59.71 224 .57 .73 .11.132 .19.31.41.47.59.71 236 .317 .75 .112 .133 .17.19.23.29.31.41.47.59.71 241 .317 .58 .112 .133 .17.19.23.29.31.41.47.59.71 241 .318 .59 .76 .112 .17.19.23.29.31.41.47.59.71 241 .318 .53 .76 .112 .132 .17.19.23.31.41.47.59.71 242 .319 .59 .73 .112 .133 .17.19.23.29.31.41.47.59.71 243 .319 .57 .76 .133 .17.19.23.29.31.41.47.59.71 241 .319 .59 .75 .112 .17.19.23.29.31.41.47.59.71 243 .320 .58 .76 .112 .133 .17.19.23.29.31.47.59.71 230 .35 .57 .75 .112 .132 .17.19.23.29.31.47.59.71 230 .39 .56 .74 .112 .132 .17.23.29.41.47.59.71 230 .314 .52 .75 .11.133 .17.19.23.29.31.47.59.71 237 .317 .54 .75 .112 .133 .17.19.23.29.31.47.59.71 235 .316 .57 .72 .112 .133 .19.23.29.31.41.47.59.71 239 .317 .59 .73 .112 .133 .17.19.23.29.31.41.47.59.71 240 .316 .59 .76 .112 .13.17.19.23.29.31.41.47.59.71 241 .318 .59 .112 .133 .17.19.23.29.31.41.47.59.71 240 .319 .76 .112 .133 .17.19.23.29.41.47.59.71 240 .318 .76 .112 .133 .17.19.23.29.31.41.47.59.71 237 .316 .57 .76 .133 .17.19.23.29.31.41.47.59.71 34 .314 .56 .75 .11.133 .17.19.23.29.31.41.47.59.71 2 238 .316 .57 .76 .11.133 .17.19.23.29.31.41.47.59.71 233 .317 .57 .75 .112 .133 .19.23.29.31.41.47.59.71 238 .315 .58 .75 .112 .133 .17.19.23.29.31.41.47.59.71 233 .313 .56 .76 .112 .133 .17.19.23.29.31.41.47.59.71 236 .316 .56 .75 .112 .133 .17.19.23.29.31.41.47.59.71 236 .316 .56 .76 .112 .133 .17.19.23.29.31.41.47.59.71 240 .318 .57 .76 .133 .17.19.23.29.31.41.47.59.71 243 .318 .58 .75 .112 .133 .17.19.23.29.31.41.47.59 244 .319 .58 .76 .112 .133 .17.19.23.29.31.41.47.71 243 .319 .58 .75 .112 .133 .17.19.23.31.41.47.59.71 245 .318 .58 .76 .112 .133 .17.23.29.31.41.47.59.71 Claim 3. S is not a simple group of Lie type. If S is a simple group of Lie type in characteristic p, and S = 2 F4 (2) , then the Steinberg character of S of degree |S|p extends to Aut(S) so that G possesses a character of degree |S|p , which contradicts Lemma 8(i). Claim 4. S is isomorphic to the Monster. By Claims 1, 2, and 3, S is a sporadic simple group or the Tits group. We will eliminate all other possibilities On Huppert’s Conjecture for the Monster and Baby Monster 7 for S and hence the claim will follow. Let = {17, 41, 47, 59, 71}. If ∩π(S) = ∅, then S possesses a non-trivial character θ that extends to χ ∈ Aut(S) so that G possesses a non-trivial character of degree χ(1) = θ(1) and π(χ(1)) ∩ = ∅, which contradicts Lemma 8(ii). Therefore we only need to consider sporadic simple group S with ∩ π(S) = ∅. Thus S ∈ {J3 , He, F i23 , F i24 , B}. However in these cases we can see that there exist a non-trivial character θ such that θ extends to Aut(S) and π(θ(1)) ∩ = ∅, which leads to a contradiction as above. The pairs (S, θ(1)) are listed below: (J3 , 22 .34 ), (He, 27 .3.5), (F i23 , 22 .3.13.23), (F i24 , 13.23.29), (B, 33 .5.23.31). This finishes the proof of Step 2. Step 3. If θ ∈ Irr(M ), then IG (θ) = G . Let θ ∈ Irr(M ) and let I = IG (θ). Assume that I < G . Let U/M be a maximal subgroup of G /M containing I/M. If µ is an irreducible constituent of θI then µ(1)|G : I| is a character degree of G by Lemma 2(a), so it divides some character degree of G. Thus µ(1)|G : U ||U : I| divides some character degree of G and so the index of some maximal subgroup of M must divide some character degree of M, which is impossible by Lemma 8(vi). Thus IG (θ) = G for all θ ∈ Irr(M ). Step 4. M = 1. As the Schur multiplier of M is trivial, it follows from Lemma 7 and Step 3 that |M/M | = 1 and hence M = M . If M is abelian then we are done. Assume that M is non-abelian. Let N be a normal subgroup of G such that M/N is a chief factor of G . As M = M , we deduce that M/N ∼ S k , where S is a non-abelian simple group. By Lemma 5, S possesses = a nontrivial irreducible character θ that extends to Aut(S). By Lemma 6, θk ∈ Irr(M/N ) extends to ϕ ∈ Irr(G ), and so ϕ(1) = θ(1)k > 1. By Lemma 1(b), ϕτ ∈ Irr(G ) for any τ ∈ Irr(G /M ). Take τ ∈ Irr(G /M ) such that τ (1) is the largest character degree of G /M. Then τ (1) < τ (1)θ(1) ∈ cd(G ) so that τ (1)θ(1) must divide some character degree of G, which is impossible. Thus M = 1 as required. Step 5. G = G × CG (G ). It follows from Step 4 that G ∼ M is a non= abelian simple group. Let C = CG (G ). Then G/C is an almost simple group with socle G . As Out(M) = 1, we deduce that G = G C. As G is simple, we have G ∩ C = 1 so that G = G × C = G × CG (G ). It follows that CG (G ) ∼ G/G is abelian. The proof is now complete. = 4 The Baby Monster The Baby Monster B was also discovered by B. Fischer. It was constructed by Leon and Sims in 1977. The outer automorphism group of the Baby Monster is trivial but the Schur multiplier is cyclic of order 2. The universal covering group of B is 2. B, which is an involution centralizer in the Monster. The character table of the Baby Monster is available in [3] and the list of its maximal subgroups can be found in [15]. Lemma 9 Let B be the Baby Monster sporadic simple group. 8 Hung P. Tong-Viet, Thomas P. Wakefield (i) Two degrees 238 .23.31.47 and 26 .313 .56 .11.13.19 are isolated degrees of B. (ii) The following two odd degrees φi (1), i = 1, 2, are in cd (B), which are divisible by a unique nontrivial proper degree ϕi (1), i = 1, 2, respectively, in B. φ1 (1) = 33 .56 .72 .11.23.31 and ϕ1 (1) = 33 .5.23.31; φ2 (1) = 313 .72 .11.13.17.19.31.47 and ϕ2 (1) = 3.31.47. (iii) If 1 = χ(1) ∈ cd(B) and (23.31.47, χ(1)) = 1 then χ(1) = 26 .313 .56 .11.13.19. (iv) B has no proper power degrees nor consecutive degrees. (v) If U is a maximal subgroup of B such that |B : U | divides some character degree χ(1) of B then U ∼ 2. (2 E6 (2)) : 2 and χ(1) = 25 .34 .56 .7.17.19.23.31.47. = (vi) Two degrees 3.31.47 and 239 .11.19.23 of B are coprime. (vii) The group 2. B has a character degree 211 .33 .5.7.13.31.47, which divides no character degree of B. Proof The list of maximal subgroups of the Baby Monster and their indices is given in Table 2. This table is taken from [15, Table I]. The other information can be found in [3]. Lemma 10 Let L = 2 E6 (2) be the simple group of exceptional Lie type. (a) L has no non-trivial projective irreducible representation whose degree divides 22 .52 .7.17.19. (b) The smallest index of a maximal subgroup of L is 3968055. ˜ Proof By [3], the Schur Multiplier of L is of order 22 × 3. Let L be the full cov˜ = ˜ ˜ ering group of L. Then Z(L) ∼ Z2 × Z3 . Hence Z(L) is non-cyclic and hence L 2 ˜ has no faithful irreducible characters. Thus if χ ∈ Irr(L), then χ must be an irreducible character of one of the following groups: 2 E6 (2), 2.2 E6 (2), 2 .2 E6 (2), 2 .2 E6 (2), 22 .2 E6 (2), 3.2 E6 (2), 6.2 E6 (2), 6 .2 E6 (2), 6 .2 E6 (2). The 3-fold covering group of 2 E6 (2) is the simply connected group of Lie type and the 6-fold covering groups of type 6.2 E6 (2) are isomorphic via an outer automorphism (see [9, 3.3]). The character tables of 3.2 E6 (2) and 6.2 E6 (2) are available in [10] while the character tables of the remaining groups are stored in [4] and [3]. As the complex projective irreducible representations of 2 E6 (2) are ordinary irreducible characters of one of the covering groups of 2 E6 (2), it follows that the degrees of non-trivial projective irreducible representations of 2 E6 (2) which are smaller than 22 .52 .7.17.19 = 226100 are among the following degrees: 1938, 2432, 45696, 46683, 46683, 48620. However we can check that none of these degrees divides 22 .52 .7.17.19. This proves (a). Finally (b) follows from [11, Theorem 4]. Step 1. G = G . By way of contradiction, suppose that G = G . Then there exists a normal subgroup N G such that G/N is solvable and minimal with respect to being non-abelian. By Lemma 3, G/N is an r-group for some prime r or G/N is a Frobenius group. Case 1. G/N is an r-group. Then there exists ψ ∈ Irr(G/N ) such that ψ(1) = rb > 1. However by Lemma 9(iv), G has no nontrivial prime power degrees. Hence this case cannot happen. On Huppert’s Conjecture for the Monster and Baby Monster Table 2 The maximal subgroups of the Baby Monster Group 2. (2 E6 (2)) : 2 21+22. Co2 F i23 29+16 S8 (2) Th (22 × F4 (2)) : 2 22+10+20 (M22 : 2 × S3 ) 25+5+10+10 L5 (2) S3 × F i22 : 2 2[35] (S5 × L3 (2)) HN : 2 + O8 (3) : S4 31+8 : 21+6. U4 (2).2 (32 : D8 × U4 (3).22 ).2 5 : 4 × HS : 2 S4 × 2 F4 (2) 32+3+6 (S4 × 2S4 ) S5 × M22 : 2 (S6 × L3 (4).2).2 53. L3 (5) 51+4 : 21+4 A5 .4 (S6 × S6 ).4 52 : 4S4 × S5 L2 (49).2 L2 (31) M11 L3 (3) L2 (17) : 2 L2 (11) : 2 47 : 23 Index 23 .34 .54 .23.31.47 37 .53 .7.13.17.19.31.47 223 .54 .7.19.31.47 38 .54 .7.11.13.19.23.31.47 2.33 .53 .11.17.23.47 214 .37 .54 .11.19.23.31.47 310 .55 .7.13.17.19.23.31.47 2.311 .55 .7.11.13.17.19.23.47 222 .33 .54 .7.17.19.23.31.47 311 .55 .7.11.13.17.19.23.31.47 226 .37 .7.13.17.23.31.47 226 .54 .7.11.17.19.23.31.47 227 .55 .72 .11.13.17.19.23.31.47 228 .35 .55 .7.11.13.17.19.23.31.47 229 .311 .52 .7.13.17.19.23.31.47 226 .39 .54 .72 .11.17.19.23.31.47 234 .56 .72 .11.13.17.19.23.31.47 230 .310 .54 .7.13.17.19.23.31.47 229 .39 .54 .7.11.13.17.19.23.31.47 236 .312 .72 .11.13.17.19.23.47 232 .312 .72 .11.13.17.19.23.31.47 231 .39 .54 .72 .11.13.17.19.23.31.47 233 .311 .53 .72 .11.13.17.19.23.31.47 236 .312 .54 .11.13.17.19.23.31.47 236 .312 .55 .72 .11.13.17.19.23.47 237 .311 .55 .72 .11.13.17.19.23.31.47 237 .310 .56 .72 .11.17.19.23.31.47 236 .311 .53 .72 .11.13.19.23.31.47 238 .312 .55 .72 .13.17.19.23.31.47 239 .313 .56 .72 .11.13.17.19.23 9 Case 2. G/N is a Frobenius group with Frobenius kernel F/N, |F/N | = ra , 1 < f = |G : F | ∈ cd(G) and ra ≡ 1(mod f ). By Lemma 3(b), if χ ∈ Irr(G) such that χ(1) is isolated then either f = χ(1) or r | χ(1). We observe that there is no prime which divides all the isolated degrees listed in Lemma 9(i) except 2. Assume that r = 2. By Lemma 9(ii) and Lemma 3(b), we deduce that f divides both φi (1), i = 1, 2 as r = 2 φi (i) for i = 1, 2. Hence f divides the greatest common divisor of φi (1), i = 1, 2, which is 33 .72 .11.31 so that 1 < f < φ2 (1). Thus f = ϕ2 (1) = 3.31.47 by Lemma 9(ii). But then f 33 .72 .11.31, a contradiction. Thus r > 2 and f must be one of the degrees in Lemma 9(i). Hence f is isolated in G. By Lemma 3(b) again, if χ ∈ Irr(G) with r χ(1) then χ(1) | f. As f is isolated we deduce that r must divide every nontrivial degree χ(1) of G such that χ(1) = f. By Lemma 9(vi), B has two nontrivial coprime degrees which are not one of the isolated degrees in Lemma 9(i), and hence these degrees are different from f, so that r must divide both of these degrees, which is impossible. Thus this case cannot happen. 10 Hung P. Tong-Viet, Thomas P. Wakefield Step 2. Let M ≤ G be a normal subgroup of G such that G /M is a chief factor of G. As G is perfect, G /M is non-abelian so that G /M ∼ S k for = some non-abelian simple group S and some integer k ≥ 1. Claim 1. k = 1. By way of contradiction, assume that k ≥ 2. By Lemma 5, S possesses a nontrivial irreducible character θ which is extendible to Aut(S) and so by Lemma 6, θk ∈ Irr(G /M ) extends to G/M, hence θ(1)k ∈ cd(G), which contradicts Lemma 9(iv). This shows that k = 1. Hence G /M ∼ S. = Claim 2. S is not an alternating group. By way of contradiction, assume that S = An , n ≥ 5. As A5 ∼ L2 (5) and A6 ∼ L2 (9), we can assume that = = n ≥ 7. By Lemma 5, S possesses two nontrivial irreducible characters θ1 , θ2 with θ1 (1) = n(n − 3)/2, θ2 (1) = θ1 (1) + 1 = (n − 1)(n − 2)/2 and both θi extend to Aut(S), so that G possesses two consecutive nontrivial character degrees which contradicts Lemma 9(iv). Claim 3. S is not a simple group of Lie type. If S is a simple group of Lie type in characteristic p, and S = 2 F4 (2) , then the Steinberg character of S of degree |S|p extends to Aut(S) so that G possesses a nontrivial prime power degree, which contradicts Lemma 9(iv). Claim 4. S is isomorphic to the Baby Monster. By Claims 1, 2, and 3, S is a sporadic simple group or the Tits group. We will eliminate all other possibilities for S and hence the claim will follow. It follows from Lemma 9(iii) that if χ ∈ Irr(G) and (19.23.31.47, χ(1)) = 1, then χ must be the trivial character of G. Let = {19, 23, 31, 47}. Argue as in the proof of Claim 4 for the Monster group. We need to show that for any sporadic simple group or the Tits group with S = B, and π(S) ∩ = ∅, then S possesses a nontrivial irreducible character θ such that π(θ(1)) ∩ = ∅ and θ extends to Aut(S). Thus we only need to consider the following sporadic simple groups: M23 , J1 , M24 , J3 , O N, Co3 , Co2 , HN, Ly, T h, F i23 , Co1 J4 , F i24 , M. Now by using [3], for each simple sporadic group S above, except for the Monster, we list a character degree that satisfies the required conditions in Table 3. For the Monster group, there is a character of degree 23 .58 .7.132 .17.19.29.41.59.71 which is extendible to Aut(B) and coprime to 11.23.31.47 and so by Lemma 9(iii), it must be trivial, a contradiction. This finishes the proof of Step 2. Step 3. If θ ∈ Irr(M ), θ(1) = 1, then IG (θ) = G . Let θ ∈ Irr(M ) and s I = IG (θ). Assume that I < G and θI = i=1 ei µi , where µi ∈ Irr(I), i = 1, 2, · · · , s. Let U/M be a maximal subgroup of G /M containing I/M. Then µi (1)|G : I| is a character degree of G by Lemma 2(a), so it divides some character degree of G. Thus µi (1)|G : U ||U : I| divides some character degree of G and so the index |G : U | must divide some character degree of B. By Lemma 9(v), we have U/M ∼ 2. (2 E6 (2)) : 2, |G : U | = |B : 2. (2 E6 (2)) : 2| = = 23 .34 .54 .23.31.47, and the character whose degree is divisible by this index is χ(1) = 25 .34 .56 .7.17.19.23.31.47. Thus µi (1)|U : I| | 22 .52 .7.17.19. Let L be normal subgroups of U such that M ≤ L and L/M ∼ 2.2 E6 (2). Then L U = and |U : L| = 2. Assume first that L ≤ I. Let φj be an irreducible constituent of θL . Then φj is an irreducible constituent of (µi )L for some i. Since L ≤ I ≤ U and |U : L| = 2, we deduce that |I : L| ≤ 2 so that φj (1) = µi (1) or µi (1)/2 by [8, On Huppert’s Conjecture for the Monster and Baby Monster 11 Corollary 6.19]. As µi (1) | 22 .52 .7.17.19, we deduce that φj (1) | 22 .52 .7.17.19. As θ(1) = 1, we have M/Ker(θ) ≤ Z(L/Ker(θ)) and so φj is an irreducible projective character of L/M ∼ 2.2 E6 (2) whose degree divides 22 .52 .7.17.19. = ˜ ˜ Let L be the universal covering group of 2 E6 (2). As L is perfect, by Lemma ˜ ˜ 4(a), the Schur multiplier of L is trivial so that by Lemma 4(b), L is a universal covering group of 2.2 E6 (2). It follows that the irreducible projective character of 2.2 E6 (2) is also an irreducible projective character of 2 E6 (2). Now by Lemma 10(a), φj (1) = 1 so that φj is an extension of θ to L. As φj is chosen arbitrarily, we deduce that all irreducible constituents of θL are linear. Now let φ be any irreducible constituent of θL . Then φ is an extension of θ to L. By Lemma 1(b), the characters φ τ, where τ ∈ Irr(L/M ), are all the irreducible constituents of θL . Since L/M ∼ 2.2 E6 (2) is non-abelian, there exists τ ∈ Irr(L/M ) such that = τ (1) > 1 so that φ (1)τ (1) > 1, which contradicts the claim above. Thus L ≤ I so that I ∩ L < L U. Let I1 = I ∩ L. Then I1 = IL (θ) < L. If I ≤ L then I = I1 < L, and |U : I| = |U : L||L : I| so that |L : I1 | = |L : I| divides |U : I|. If I ≤ L then U = IL, as L is maximal in U, so that |U : I| = |L : I1 |. Thus |L : I1 | | |U : I|, so |L : I1 | | 22 .52 .7.17.19, since µi |U : I| | 22 .52 .7.17.19. As I1 < L, there exists a subgroup T of L such that M ≤ I1 ≤ T < L and T /M is a maximal subgroup of L/M ∼ 2.2 E6 (2). = We have |L : I1 | = |L : T ||T : I1 | so that |L : T | divides 22 .52 .7.17.19 = 226100, where |L : T | is the index of some maximal subgroup of 2.2 E6 (2). Since 2.2 E6 (2) is quasi-simple, its center must lie in any maximal subgroups so that the index of a maximal subgroup of 2.2 E6 (2) is also an index of a maximal subgroup of 2 E6 (2). By Lemma 10(b), the smallest index of a maximal subgroup of 2 E6 (2) is 3968055. Thus |L : T | ≥ 3968055, a contradiction as |L : T | ≤ 226100. Therefore IG (θ) = G for any θ ∈ Irr(M ) with θ(1) = 1. Step 4. M = 1. By Step 3 and Lemma 7, we have M = [G , M ] and |M : M | divides M (B). As |M (B)| = 2, either |M : M | = 2 or |M : M | = 1. Assume that |M : M | = 2. As M = [G , M ] we deduce that G /M is a universal covering group of G /M ∼ B and so G /M ∼ 2. B. However = = by Lemma 9(vii), 2. B has an irreducible character whose degree divides no character degree of B. Thus M = M . Let N ≤ M be a normal subgroup of G such that M/N is a chief factor of G . Then M/N ∼ S k for some non-abelian = simple group S. By Lemma 5, there exists an irreducible character τ ∈ Irr(S) such that τ (1) > 1 and τ extends to Aut(S). By Lemma 6, τ k extends to G and so by Lemma 1(b), we have τ k β ∈ Irr(G ) for all β ∈ Irr(G /M ). Let χ ∈ Irr(G /M ) = Irr(B) such that χ(1) is the largest character degree of B. Then τ (1)k χ(1) divides some character degree of G, which is impossible. Hence M = 1. Step 5. G = G × CG (G ). It follows from Step 4 that G ∼ B is a non= abelian simple group. Let C = CG (G ). Then G/C is an almost simple group with socle G . As Out(B) = 1, we deduce that G = G C. As G is simple, we have G ∩ C = 1 so that G = G × C = G × CG (G ). It follows that CG (G ) ∼ G/G is abelian. The proof is now complete. = 12 Hung P. Tong-Viet, Thomas P. Wakefield Table 3 Character degrees of sporadic simple groups Group M23 J1 M24 J3 ON Co3 Co2 Degree 2.11 7.11 32 .5 22 .33 29 .73 52 .11 52 .11 Group HN Ly Th F i23 Co1 J4 F i24 Degree 2.34 .5.11 28 .7.67 23 .33 .53 23 .33 .11.13 22 .3.52 .7.13 221 .33 .5.7 52 .73 .11.17 Acknowledgements: The first author is financially supported from the University of KwaZulu-Natal. References 1. Bianchi, M., Chillag, D., Lewis, M., and Pacific, E.: Character degree graphs that are complete graphs. Proc. Amer. Math. Soc. 135 no. 3, 671–676 (2007) 2. Bray, J.N. and Wilson, R.A.: Explicit representations of maximal subgroups of the Monster. J. Algebra 300 no. 2, 834–857 (2006) 3. Conway, J.H., Curtis, R.T., Norton, S.P., Parker, R.A., and Wilson, R.A.: Atlas of Finite Groups. Oxford University Press, Eynsham 1985 4. The GAP Group: GAP - Groups, Algorithms, and Programming, Version 4.4.10, http://www.gap-system.org (2007) 5. Harris, M.E.: A universal mapping problem, covering groups and automorphism groups of finite groups. Rocky Mountain J. Math. 7 no. 2, 289–295 (1977) 6. Huppert, B.: Some simple groups which are determined by the set of their character degrees I. Illinois J. Math. 44 no. 4, 828–842 (2000) 7. Huppert, B.: Some simple groups which are determined by the set of their character degrees. II. Rend. Sem. Mat. Univ. Padova 115, 1-13 (2006) 8. Isaacs, I.M.: Character theory of finite groups, AMS Chelsea Publishing, Providence, RI 2006 9. L¨beck, F.: Smallest degrees of representations of exceptional groups of Lie type. Comm. u Algebra 29 no. 5, 2147–2169 (2001) 10. L¨beck, F.: http://www.math.rwth-aachen.de/ ∼Frank.Luebeck/chev/2E62.html. u 11. Vasil ev, A.V.: Minimal permutation representations of finite simple exceptional groups of twisted type. Algebra and Logic 37 no. 1, 9–20 (1998) 12. Wakefield, T.: Verifying Huppert’s Conjecture for P SL3 (q) and P SU3 (q 2 ). Comm. Algebra 37 no. 8, 2887–2906 (2009) 13. Wakefield, T.: Verifying Huppert’s Conjecture for 2 G2 (q 2 ). Algebras and Representation Theory (2010). doi 10.1007/s10468-009-9206-x 14. Wakefield, T.: Verifying Huppert’s Conjecture for P Sp4 (q) when q > 7. Algebras and Representation Theory (2010). DOI: 10.1007/s10468-010-9246-2 15. Wilson, R. A.: The maximal subgroups of the Baby Monster I. J. Algebra 211 no. 1, 1–14 (1999)