Simple exceptional groups of Lie type are determined by their character degrees more

SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE ARE DETERMINED BY THEIR CHARACTER DEGREES HUNG P. TONG-VIET Abstract. Let G be a finite group. Denote by Irr(G) the set of all irreducible complex characters of G. Let cd(G) = {χ(1) | χ ∈ Irr(G)} be the set of all irreducible complex character degrees of G forgetting multiplicities, and let X1 (G) be the set of all irreducible complex character degrees of G counting multiplicities. Let H be any non-abelian simple exceptional group of Lie type. In this paper, we will show that if S is a non-abelian simple group and cd(S) ⊆ cd(H) then S must be isomorphic to H. As a consequence, we show that if G is a finite group with X1 (G) ⊆ X1 (H) then G is isomorphic to H. In particular, this implies that the simple exceptional groups of Lie type are uniquely determined by the structure of their complex group algebras. 1. Introduction and Notation All groups considered are finite and all characters are complex characters. Let G be a group. Denote by Irr(G) the set of all irreducible characters of G. Let cd(G) be the set of all irreducible character degrees of G forgetting multiplicities, that is, cd(G) = {χ(1) | χ ∈ Irr(G)}, and let X1 (G) be the set of all irreducible character degrees of G counting multiplicities. Observe that X1 (G) is just the first column of the ordinary character table of G. We follow [4] for notation of non-abelian simple groups. In this paper, we will prove the following results. Theorem 1.1. Let H be a simple exceptional group of Lie type. If S is a nonabelian simple group and cd(S) ⊆ cd(H), then S ∼ H. = Corollary 1.2. Let H be a simple exceptional group of Lie type. If G is a perfect group, cd(G) ⊆ cd(H) and |G| ≤ |H|, then G ∼ H. = Corollary 1.3. Let H be a simple exceptional group of Lie type. If G is a group and X1 (G) ⊆ X1 (H), then G ∼ H. = Corollary 1.3 gives a positive answer to [19, Question 11.8(a)] for simple exceptional groups of Lie type. For alternating groups of degree at least 5, sporadic simple groups, and the Tits group, similar results were obtained in [24]. Corollary 1.2 is related to the Huppert’s Conjecture [11] which says that the non-abelian simple groups are determined by the set of their character degrees. Let C be the complex number field and let G be a group. Denote by CG the group algebra of G over C. Date: April 26, 2011. 1991 Mathematics Subject Classification. Primary 20C15 . Key words and phrases. character degree, simple exceptional group. The author is supported by a post-doctoral fellowship from the University of KwaZulu-Natal. 1 2 HUNG P. TONG-VIET Corollary 1.4. Let H be a simple exceptional group of Lie type. If G is a group such that CG ∼ CH, then G ∼ H. = = Basically, this corollary says that if H is a simple exceptional group of Lie type then H is uniquely determined by the structure of its complex group algebra CH. This result is related to the Brauer’s Problem 2 which asks the following question: Which groups can be determined by the structure of their complex group algebras? In [24] and [26], we have shown that the alternating groups of degree at least 5, the sporadic simple groups and the symmetric groups are uniquely determined by the structure of their complex group algebras. In [25], we will establish a similar result for the remaining simple classical groups of Lie type. We fix some notation. If cd(G) = {s0 , s1 , · · · , st }, where 1 = s0 < s1 < · · · < st , then we define di (G) = si for all 1 ≤ i ≤ t. Then di (G) is the ith smallest degree of the non-trivial character degrees of G. We also define t(G) = t + 1, the number of distinct character degrees of G. We put b(G) = st . Then b(G) is the largest character degree of G. 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. 2. Preliminaries The following result is a well-known theorem due to Zsigmondy. Lemma 2.1. ([13, Theorems 5.2.14, 5.2.15]). Let q and n be integers with q ≥ 2 and n ≥ 3. Then q n − 1 has a prime divisor l such that l does not divide q m − 1 for m < n, unless (q, n) = (2, 6). Moreover if l | q k − 1 then n | k. Such an l is called a primitive prime divisor. We denote by ln (q) the smallest primitive prime divisor of q n − 1 for fixed q and n. When n is odd and (q, n) = (2, 3) then there is a primitive prime divisor of q 2n − 1 and we denote this primitive prime divisor by l−n (q). The next result gives a lower bound for the largest character degree of the alternating groups. Recall that k(G) is the number of conjugacy classes of G. Lemma 2.2. Assume that n ≥ 10. Then b(An ) ≥ 2n−1 . Proof. Using [9], we can check that b(An ) ≥ 2n−1 for all 10 ≤ n ≤ 17. Thus we can assume n ≥ 18. Observe that as An Sn , we obtain b(Sn ) ≤ |Sn : An |b(An ) = 2b(An ) (see [12, Exercise 5.4, p. 74]). Thus it suffices to show that b(Sn ) ≥ 2n , for any n ≥ 18. We have |Sn | = n! = χ∈Irr(Sn ) χ(1)2 ≤ k(Sn )b(Sn )2 . It follows from [18, Theorem 1.1] that k(Sn ) ≤ 3(n−1)/2 and hence b(Sn )2 ≥ n!/k(Sn ) ≥ n!/3(n−1)/2 . Thus we only need to verify that n! ≥ 3(n−1)/2 · 22n , for all n ≥ 18. We will prove this inequality by induction on n ≥ 18. Obviously this is true for n = 18. Assume that m! ≥ 3(m−1)/2 · 22m , for some m ≥ 18. We need to show that (m + 1)! ≥ 3m/2 · 22m+2 . By induction hypothesis, we have that (m + 1)! = (m + 1) · m! ≥ (m + 1) · 3(m−1)/2 · 4m . Thus, we need to prove that m + 1 ≥ 4 · 31/2 . As 4 · 31/2 < 4 · 2 = 8 < 18 < m + 1, we obtain (m + 1)! ≥ 3m/2 · 22m+2 , and so n! ≥ 3(n−1)/2 · 22n , for any n ≥ 18. We have shown that b(Sn )2 ≥ 22n , which implies b(Sn ) ≥ 2n . This finishes the proof. The following lemma gives some basic properties of simple groups which satisfy the hypotheses of Theorem 1.1. These results will be used frequently in the next SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 3 Table 1. The orders of simple exceptional groups S B2 (q 2 ), q 2 = 22m+1 2 G2 (q 2 ), q 2 = 32m+1 2 F4 (q 2 ), q 2 = 22m+1 3 D4 (q) 2 E6 (q) G2 (q) F4 (q) E6 (q) E7 (q) E8 (q) 2 d |S| 1 q 4 (q 4 + 1)(q 2 − 1) 1 q 6 (q 6 + 1)(q 2 − 1) 24 12 1 q (q + 1)(q 8 − 1)(q 6 + 1)(q 2 − 1) 1 q 12 (q 8 + q 4 + 1)(q 6 − 1)(q 2 − 1) (3, q + 1) q 36 i∈{2,5,6,8,9,12} (q i − (−1)i )/d 1 q 6 (q 6 − 1)(q 2 − 1) 24 i 1 q i∈{2,6,8,12} (q − 1) 36 i (3, q − 1) q i∈{2,5,6,8,9,12} (q − 1)/d (2, q − 1) q 63 i∈{2,6,8,10,12,14,18} (q i − 1)/d 1 q 120 i∈{2,8,12,14,18,20,24,30} (q i − 1) section. The conclusions (ii) and (iii) are obvious, while (i) and (iv) can be found in [24, Lemma 11]. Lemma 2.3. Let S and H be non-abelian simple groups. If cd(S) ⊆ cd(H) then the following holds: (i) di (S) ≥ di (H), for all i; (ii) b(S) ≤ b(H); (iii) t(S) ≤ t(H); (iv) π(S) ⊆ π(H). The next two results are well-known (see [12, Theorems 3.10, 3.11]), we will use them freely without further reference. Lemma 2.4. Let G be a group. Then the following holds. (a) If |π(G)| ≤ 2, then G is solvable. (b) If χ ∈ Irr(G) then χ(1) | |G|. In Table 1 we list the orders of the simple exceptional groups of Lie type. Table 2 is taken from [13, Table 5.3.A] which gives the Landazuri-Seitz-Zalesskii bounds for non-trivial minimal degrees of the cross-characteristic representations of finite simple groups of Lie type. In Table 3, we give the p-part of the degree of a unipotent character ψ of the simple groups of Lie type, where ψ is not the Steinberg character. We refer to [2] for the classification of unipotent characters and the notion of symbols. In Table 4, we have b(F i24 ) = 336033532800, b(B) = 16547812226400000 and b(M ) = 258823477531055064045234375. The results in this table are taken from [4]. Finally, in Table 5 we list the upper bounds for the largest character degree of the simple exceptional group of Lie type. These upper bounds were obtained in [20, Theorem 2.1]. 3. Proof of the main results We assume the following set up. Let G be a simple simply connected algebraic ¯ group of exceptional type defined over Fp , where p is prime. Let L be the fixed point group of G under a suitable Frobenius morphism such that L/Z(L) is isomorphic to the simple exceptional group of Lie type H. Generically L is the universal covering group of H. We remark that if cd(S) ⊆ cd(H) then cd(S) ⊆ cd(L) for any group S. 4 HUNG P. TONG-VIET Table 2. The Landazuri-Seitz-Zalesskii bounds S B2 (q 2 ), q 2 = 22m+1 2 G2 (q 2 ), q 2 = 32m+1 2 F4 (q 2 ), q 2 = 22m+1 3 D4 (q) 2 E6 (q) G2 (q) F4 (q) 2 E6 (q) E7 (q) E8 (q) e(S) √ Exceptions 2 B2 (8) q(q 2 − 1)/ 2 q 2 (q 2 − 1) √ q 9 (q 2 − 1)/ 2 q 3 (q 2 − 1) q 9 (q 2 − 1) q(q 2 − 1) G2 (3), G2 (4) 6 2 q (q − 1), q odd q 7 (q 3 − 1)(q − 1)/2, q even F4 (2) q 9 (q 2 − 1) q 15 (q 2 − 1) q 27 (q 2 − 1) Table 3. Some unipotent characters of simple groups of Lie type S = S(pb ) Ln (pb ) S2n (pb ), p = 2 S2n (pb ), p > 2 O2n+1 (pb ), p > 2 + O2n (pb ) − O2n (pb ) 3 D4 (pb ) F4 (pb ) 2 F4 (q 2 ) E6 (pb ) 2 E6 (pb ) E7 (pb ) E8 (pb ) Symbol (1n−2 , 2) 0 1 2 ··· n−2 n−1 n 1 2··· n−2 0 1 2 ··· n−2 n−1 n 1 2··· n−2 0 1 2 ··· n−3 n−1 1 2 3··· n−2 n−1 0 1 2 ··· n−2 n−1 1 2··· n−2 φ1,3 φ9,10 2 B2 [a], φ6,25 φ2,16 φ7,46 φ8,91 p-part of degree pb(n−1)(n−2)/2 2 2b(n−1) −1 2 pb(n−1) 2 pb(n−1) 2 pb(n −3n+3) 2 pb(n −3n+2) p7b p10b 1 13 √ q 2 p25b p25b p46b p91b Using the classification of finite simple groups, S is a sporadic group, an alternating group of degree at least 5 or a simple group of Lie type. We will treat the Tits group as a sporadic group. We will prove Theorem 1.1 by a series of propositions. Using [16, Theorem C], the result for Suzuki groups follows easily. Proposition 3.1. If S is a non-abelian simple group with cd(S) ⊆ cd(2 B2 (q 2 )), where q 2 = 22m+1 , m ≥ 1, then S ∼ 2 B2 (q 2 ). = Proof. Let H ∼ 2 B2 (22m+1 ). Assume that S is a non-abelian simple group with = cd(S) ⊆ cd(H), we will show that S ∼ H. By Lemma 2.3(iii) and [16, Theorem = C], we have 4 ≤ t(S) ≤ t(H) = 6 and S ∈ {L2 (r), L3 (4), 2 B2 (22n+1 )}, where r is a prime power. If S = L2 (r) or S = L3 (4) then 3 ∈ π(S). However it is well-known that 3 ∈ π(H), which contradicts Lemma 2.3(iv). Hence S = 2 B2 (22n+1 ). By [17, Theorem 1.1], the only nontrivial character degree of H which is a power of 2 is the degree of the Steinberg character of H of degree |H|2 . As S also possesses a SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 5 Table 4. Sporadic simple groups S M11 M12 J1 M22 J2 M23 HS J3 M24 M cL He Ru Suz ON Co3 Co2 F i22 HN Ly Th F i23 Co1 J4 F i24 B M 2 F4 (2) t(S) d1 (S) d2 (S) d3 (S) b(S) 7 10 11 16 55 11 11 16 45 176 7 56 76 77 209 10 21 45 55 385 16 14 21 36 336 11 22 45 230 2024 18 22 77 154 3200 14 85 323 324 3078 20 23 45 231 10395 17 22 231 252 10395 23 51 153 680 23324 28 378 406 783 118784 36 143 364 780 248832 19 10944 13376 25916 234080 34 23 253 275 255024 51 23 253 275 2095875 52 78 429 1001 2729376 42 133 760 3344 5878125 36 2480 45694 48174 71008476 39 248 4123 27000 190373976 84 782 3588 5083 559458900 98 276 299 1771 551675124 42 1333 299367 887778 3054840657 91 8671 57477 249458 b(F i24 ) 165 4371 96255 1139374 b(B) 170 196883 21296876 842609326 b(M ) 14 26 27 78 2048 Table 5. Upper bounds of b(S) for simple groups of Lie type S F4 (q) G2 (q) ˜ q 28 q8 b(S) S E6 (q) E7 (q) ˜ b(S) q 42 q 70 2 B2 (q 2 ) q6 E8 (q) q 128 2 F4 (q 2 ) q 28 2 E6 (q) q 42 2 G2 (q 2 ) q8 3 D4 (q) q 17 character of degree |S|2 , it follows that |S|2 = |H|2 so that 2(2n + 1) = 2(2m + 1), which implies that m = n. Thus S ∼ H as required. = Therefore we can assume that H is a simple exceptional group of Lie type defined over a field of size q in characteristic p with H = 2 B2 (q 2 ), 2 F4 (2) . Proposition 3.2. If S is a sporadic simple group or the Tits group, then cd(S) cd(H). Proof. Let S be a simple sporadic group or the Tits group. By way of contradiction, assume that cd(S) ⊆ cd(H). We outline our general argument here. By Lemma 6 HUNG P. TONG-VIET 2.3(i), we have d1 (S) ≥ d1 (H) and hence by Table 2, we obtain, apart from some exceptions, d1 (S) ≥ e(H), where e(H) is the Landazuri-Seitz-Zalesskii bound for H. Solving this inequality, we get an upper bound for q. For these values of q, we will show that π(S) ⊆ π(H) and hence cd(S) ⊆ cd(H) by Lemma 2.3(iv). Case 1. H = 2 G2 (q 2 ), q 2 = 32m+1 , m ≥ 1. By Lemma 2.3(iii) and [27], we have t(S) ≤ 11. By Table 4, we only need to consider the following cases {M11 , M12 , J1 , M22 , M23 }. By Lemma 2.3(i), we have d1 (S) ≥ d1 (H) so that by Table 2 we obtain d1 (S) ≥ 32m+1 (32m+1 − 1). As m ≥ 1, we have 32m+1 (32m+1 − 1) ≥ 702 > 56 ≥ d1 (S) for any sporadic simple groups considered above. Case 2. H = G2 (q). By Lemma 2.3(iii) and [3, 7, 8], we have t(S) ≤ 23. By Table 4, we only need to consider the following cases {M11 , M12 , J1 , M22 , J2 , M23 , HS, J3 , M24 , M cL, He, O N, 2 F4 (2) }. Using [4], we can assume that q ≥ 7. Thus by Lemma 2.3(i), we have d1 (S) ≥ d1 (H) so that by Table 2 we obtain d1 (S) ≥ q(q 2 −1), where q ≥ 7. Since q(q 2 − 1) ≥ 336, we see that d1 (S) < 336 ≤ d1 (H) unless S = O N. We have π(O N ) = {2, 3, 5, 7, 11, 19, 31}. As d1 (S) = 10944 ≥ q(q 2 − 1), we deduce that 7 ≤ q ≤ 23. By Lemma 2.3(iv), we have {2, 3, 5, 7, 11, 19, 31} ⊆ π(G2 (q)). However we can check that π(O N ) ⊆ π(G2 (q)) for any 7 ≤ q ≤ 23. Case 3. H = 3 D4 (q). By Lemma 2.3(iii) and [5], we have t(S) ≤ 33. By Table 4, we only need to consider the following cases {M11 , M12 , J1 , M22 , J2 , M23 , HS, J3 , M24 , M cL, He, Ru, O N, 2 F4 (2) }. Using [4], we can assume that q ≥ 3. If q = 3 then by [14], we have d2 (H) ≥ 3942 > d2 (S) unless S = O N. But then π(O N ) ⊆ π(3 D4 (3)), which contradicts Lemma 2.3(iv). Thus q ≥ 4. By Lemma 2.3(i), we have d1 (S) ≥ d1 (H) so that by Table 2 we obtain d1 (S) ≥ q 3 (q 2 − 1). As q ≥ 4, we have q 3 (q 2 − 1) ≥ 960, and so d1 (S) < 960 ≤ d1 (H) for all sporadic simple groups above, unless S = O N. As d1 (O N ) ≥ q 3 (q 2 − 1), we deduce that q ≤ 5 so that 4 ≤ q ≤ 5. However we can check that π(O N ) ⊆ π(3 D4 (q)) for any 4 ≤ q ≤ 5, which contradicts Lemma 2.3(iv). Case 4. H = 2 F4 (q 2 ), q 2 = 22m+1 , m ≥ 1. By Lemma 2.3(iv) and Table 2, we have d1 (S) ≥ d1 (H) ≥ 29m+4 (22m+1 − 1). As m ≥ 1, we have d1 (H) ≥ 57344 > d1 (S) unless S = M. If m ≥ 2 then d1 (H) ≥ 130023424 > d1 (M ). Thus m = 1 and so H = 2 F4 (8). However we can check that π(M ) ⊆ π(2 F4 (8)). Thus cd(S) ⊆ cd(H). Case 5. H = 2 E6 (q) or E6 (q). We have d1 (S) ≥ d1 (H) ≥ q 9 (q 2 − 1) by Table 2. Using [4], we can assume that q ≥ 3. If q ≥ 4, then d1 (H) ≥ 3932160 > d1 (S) for all sporadic groups. Hence q = 3. By [14], we have d2 (H) ≥ 32690203 > 21296876 ≥ d2 (S) for all sporadic groups. Thus cd(S) ⊆ cd(H). Case 6. H = F4 (q). By [4], we can assume that q ≥ 3. Assume first that q is even. Then q ≥ 4 and by Table 2, d1 (H) ≥ q 7 (q 3 − 1)(q − 1)/2 ≥ 1548288 > d1 (S) for all sporadic groups. Thus q is odd and q ≥ 3. By Table 2, we have d1 (H) ≥ q 6 (q 2 − 1). If q ≥ 5, then d1 (H) ≥ 375000 > d1 (S) for all S. Hence q = 3. By [14], we have either d1 (H) ≥ 6643 > d1 (S) or d2 (H) ≥ 83148 > d2 (S) unless S = M. But then π(M ) ⊆ π(F4 (3)). Case 7. H = E7 (q) or E8 (q). Using [14], one can assume that q ≥ 3 since d2 (E8 (2)) ≥ d2 (E7 (2)) > d2 (S) for all S. By Table 2, we have d1 (H) ≥ q 15 (q 2 −1) ≥ 114791256 > d1 (S) for all S. Thus cd(S) ⊆ cd(H). The proof is now complete. Proposition 3.3. If m ≥ 5, then cd(Am ) cd(H). Proof. By way of contradiction, suppose that cd(Am ) ⊆ cd(H). Using [4] and Table 2, we see that d1 (H) ≥ 10 so that by Lemma 2.3(i), we can assume that m ≥ 11. SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 7 Assume that H = G2 (3). Then d1 (H) = 14 and 11 ∈ π(H). As m ≥ 11, we have d1 (Am ) = m − 1 ≥ d1 (H) = 14 so that m ≥ 15. But then 11 ∈ π(Am ), which contradicts Lemma 2.3(iv). Using the same argument, we can assume H = G2 (3), G2 (4), F4 (2). By Lemma 2.3(i, ii) and Table 2, we have d1 (Am ) = m − 1 ≥ d1 (H) ≥ e(H) and b(Am ) ≤ b(H). Combining with Lemma 2.2, we obtain b(H) ≥ 2e(H) . Now by Table 5, we have ˜ b(H) ≥ b(H) ≥ 2e(H) . However it is routine to check that this inequality cannot happen so that cd(Am ) ⊆ cd(H). Proposition 3.4. If S is a simple group of Lie type in cross-characteristic then cd(S) ⊆ cd(H). Proof. Suppose that cd(S) ⊆ cd(H) and that S is a simple group of Lie type in characteristic s = p. Then S possesses a character of degree |S|s which is a nontrivial power of s. As cd(S) ⊆ cd(H), we deduce that H has two different nontrivial prime power degrees. By [17, Theorem 1.1], we deduce that H = G2 (3) and the only nontrivial prime power degrees in H are 26 and 36 . As π(H) = {2, 3, 7, 13} and the characteristic of H is 3, we deduce that S is a simple group of Lie type in characteristic 2 with 3 ≤ |π(S)| ≤ 4, π(S) ⊆ {2, 3, 7, 13} and |S|2 = 26 . However, by checking the list of simple groups with 3 or 4 prime divisors in [10, Lemmas 2, 3], there is no simple groups satisfying these conditions. (Note that in [10, Lemma 3] the author missed the group U4 (3)). This contradiction proves the proposition. Proposition 3.5. If S is a simple group of Lie type of characteristic p as that of H, and cd(S) ⊆ cd(H), then S ∼ H. = Proof. Assume that S is a simple group of Lie type in characteristic p, defined over a field of size r = pb and cd(S) ⊆ cd(H). Observe first that by [17, Theorem 1.1], the only nontrivial character degree of H which is a power of p is the degree of the Steinberg character StH of H of degree |H|p . As S is also a simple group of Lie type in characteristic p, it follows that |S|p = |H|p . We note that the upper bounds for the p-part of the non-Steinberg characters of F4 (q), 2 E6 (q), E6 (q), E7 (q) and E8 (q) are taken from [15]. Case 1. H ∼ 2 G2 (q 2 ), q 2 = 32m+1 , m ≥ 1. We have p = 3 and |H|p = 33(2m+1) . = The character degrees of H are available in [27]. Observe that if χ ∈ Irr(H) with 3 | χ(1) and χ = StH , then χ(1)3 ∈ {3m , 32m+1 }. (a) S = Ln (r), r = pb . We have p = 3 and bn(n − 1) = 6(2m + 1). If n = 2 then r = q 6 so that r − 1 = q 6 − 1 ∈ cd(S) ⊆ cd(2 G2 (q 2 )), which is impossible. If n = 3 then r = q 2 . If S = L3 (q 2 ), then r3 − 1 = q 6 − 1 ∈ cd(S) ⊆ cd(2 G2 (q 2 )), a contradiction. If S = U3 (q 2 ), then it possesses an irreducible character of degree r(r − 1) = q 2 (q 2 − 1), hence q 2 (q 2 − 1) ∈ cd(2 G2 (q 2 )), a contradiction. Hence n ≥ 4. By [2, 13.8], S possesses unipotent characters of degrees (rn − n−1 r)/(r − ) and r2 (rn − n )(rn−3 − n−3 )/((r − )(r2 − 1)) labeled by the symbols (1, n − 1) and (2, n − 2), respectively. Observe that these degrees are not p-powers, so that r = 3m and r2 = 32m+1 . As r2 = 32m+1 = 3 · (3m )2 = 3r2 , we have 1 = 3, a contradiction. (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have p = 3 and bn2 = 3(2m + 1). It follows that n must be odd and so n ≥ 3. By [2, 1 13.8], S possesses a unipotent character labeled by the symbol 0 −n with degree r(rn − 1)(rn−1 − 1)/2(r + 1). By Table 3, S also possesses a unipotent character ψ 2 2 with ψ(1)p = r(n−1) . As r is odd, we have r = 3m and since n ≥ 3, r(n−1) = 32m+1 . 8 HUNG P. TONG-VIET As r = 3m , we have b = m and so mn2 = 3(2m + 1). It follows that m | 3 so that 2 m = 1 or m = 3. If m = 1 then n = 3 but then r(n−1) = 34 = 33 = 32m+1 . Thus 2 m = 3 and so n = 7, which is impossible. (c) S = O2n (r), n ≥ 4. We have p = 3 and bn(n − 1) = 3(2m + 1). As n ≥ 4, bn(n − 1) is even but 3(2m + 1) is always odd. Hence this case cannot happen. (d) S = 2 B2 (22n+1 ) or S = 2 F4 (22n+1 ). These cases cannot happen since S is of characteristic 2. (e) S = 2 G2 (32n+1 ). Since 3(2n + 1) = 3(2m + 1), we have m = n, hence S ∼ H. = (f ) S ∈ {3 D4 (r), G2 (r), F4 (r), E6 (r), 2 E6 (r), E8 (r)}. In these cases, we see that |S|3 is an even power of 3 while |H|3 = 33(2m+1) . Thus these cases cannot happen. (g) S = E7 (r). We have 63b = 3(2m + 1) hence 21b = 2m + 1 and so q 2 = r21 . By [2, 13.9], S possesses unipotent characters of degrees φ7,1 (1) = rΦ7 Φ12 Φ14 and φ27,2 (1) = r2 Φ2 Φ2 Φ9 Φ12 Φ18 . Hence r = 3m and r2 = 32m+1 , which is impossible. 3 6 Case 2. H = 2 F4 (q 2 ), q 2 = 22m+1 , m ≥ 1. We have p = 2 and |H|2 = 212(2m+1) . Using the list of character degrees of 2 F4 (q 2 ) in [15], we see that if χ ∈ Irr(H) with χ(1) is even, then either χ(1)2 = 2m or 22m+1 | χ(1). Moreover χ(1)2 ≤ 213m+6 < q 13 if χ = StH . (a) S = Ln (r), n ≥ 2. Then bn(n−1) = 24(2m+1). If n = 2, then b = 12(2m+1) so r = q 24 and S = L2 (q 24 ). We have r + 1 = q 24 + 1 ∈ cd(S) but q 24 + 1 ∈ cd(H), a contradiction. Thus n ≥ 3. By Table 3, S possesses a unipotent character ψ with ψ(1)p = r(n−1)(n−2)/2 . It follows that (n−1)(n−2)b/2 < 13(2m+1)/2. Multiplying both sides by 2n, we obtain bn(n−1)(n−2) = 24(2m+1)(n−2) < 13n(2m+1). After simplifying, we have 24(n − 2) < 13n and so n ≤ 4. If n = 3 then b = 4(2m + 1) so that r = q 8 . Hence S = L3 (q 8 ). By [22], S possesses a character of degree q 24 − . However we see that 2 F4 (q 2 ) has no such degrees. Thus n = 4 and so S = L4 (r), where r = q 4 . By [2, 13.8], S possesses a unipotent character of degree r2 (rn − n )(rn−3 − n−3 )/(r − )(r2 − 1) = r2 (r2 + 1) = q 8 (q 8 + 1). However we can check that q 8 (q 8 + 1) ∈ cd(2 F4 (q 2 )). (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. As p = 2, S = S2n (r). Then bn2 = 12(2m + 1). By Table 3, S possesses a unipotent character 2 ψ with ψ(1)2 = 2b(n−1) −1 . It follows that b(n − 1)2 − 1 < 13(2m + 1)/2 and so b(n − 1)2 ≤ 13(2m + 1)/2 = 13bn2 /24. Hence 24(n − 1)2 ≤ 13n2 . Solving this inequality, we obtain 2 ≤ n ≤ 3. If n = 2 then r = q 6 and S possesses a character of degree (r − 1)(r2 + 1) = (q 6 − 1)(q 12 + 1) by [6]. However we can check that H has no such degree. Assume that n = 3. Then 3b = 4(2m + 1) and hence 6b = 8(2m + 1) > 6 so that l8(2m+1) (2) exists and since l8(2m+1) (2) ∈ π(S), by Lemma 2.3(iv), we deduce that l8(2m+1) (2) ∈ π(H), which is impossible. (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 12(2m + 1). By [2, 13.8], S possesses a unipotent character of degree r(rn − )(rn−2 + )/(r2 − 1). If r = 2m then b = m. We have mn(n − 1) = 12(2m + 1) > 24m so that n(n − 1) > 24 and hence n ≥ 6. Thus 12(2m + 1) = mn(n − 1) ≥ 30m. It follows that 1 ≤ m ≤ 2. If m = 1 then n(n − 1) = 36. However this equation has no integer solutions. Thus m = 2. Then r = 22 and n = 6. Hence S = O12 (4) and H = 2 F4 (25 ). But then π(O12 (4)) ⊆ π(2 F4 (32)). Therefore 2m + 1 ≤ b. Thus n(n − 1) ≤ 12 and so as n ≥ 4, we deduce that n = 4. Then b = 2m + 1 and hence r = q 2 > 2. We have S = O8 (22m+1 ) and H = 2 F4 (22m+1 ). However l3 (22m+1 ) ∈ π(S) − π(H), which contradicts Lemma 2.3(iv). SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 9 (d) S = 2 B2 (r2 ), r2 = 22n+1 . Then 2(2n + 1) = 12(2m + 1) and so r2 = q 12 . By [23], S possesses a character of degree r4 + 1 = q 24 + 1. However q 24 + 1 ∈ cd(H). (e) S = 2 F4 (r2 ), r2 = 22n+1 . Then 12(2n + 1) = 12(2m + 1) hence m = n so that ∼ H. S= (f ) S = 2 G2 (32n+1 ). This case cannot happen as S is of characteristic 3. (g) S = 3 D4 (r). Then 12b = 12(2m + 1) and so r = q 2 . By Table 3, S possesses a unipotent character ψ with ψ(1)p = p7b . As b = 2m + 1, we see that 7b = 14(2m + 1) > 13m + 6, a contradiction. (h) S = 2 E6 (r) or E6 (r). Then 36b = 12(2m + 1) and so r3 = q 2 . By Table 3, S possesses a unipotent character ψ with ψ(1)p = r25 . As r3 = q 2 , we have r25 > r24 = (r3 )8 = q 16 > q 13 , a contradiction. (i) S = G2 (r). Then 6b = 12(2m + 1) and so r = q 4 . As r is even, using [8], S possesses an irreducible character ψ ∈ Irr(S) with ψ(1)2 = r3 = q 12 . However this cannot happen by checking the list of character degrees of H. (j) S = F4 (r). Then 24b = 12(2m + 1) and so 2b = 2m + 1, which is impossible. (k) S = E7 (r). Then 63b = 12(2m + 1) and so 21b = 4(2m + 1). By [2, 13.9], S possesses a unipotent character of degree φ7,1 (1) = rΦ7 Φ12 Φ14 . If r = 2m then b = m so that 21m = 4(2m + 1) and hence 13m = 4, a contradiction. Thus 2m + 1 ≤ b. But then 21b = 4(2m + 1) ≤ 4b, which is impossible. (l) S = E8 (r). Then 120b = 12(2m + 1) and so 10b = (2m + 1) or r10 = q 2 . By Table 3, S possesses a unipotent character ψ with ψ(1)p = r91 . As r10 = q 2 , we have r91 > r90 = q 18 > q 13 , a contradiction. Case 3. H = 3 D4 (q), q = pa . Then |H|p = p12a . The character degrees of H are available in [5, Table 4.4]. Observe that if χ ∈ Irr(H) with χ = StH and p | χ(1) then q = χ(1)p or q 3 /(2, q) | χ(1)p . In any cases q | χ(1)p and χ(1)p ≤ q 7 . (a) S = Ln (r), n ≥ 2. Then bn(n−1) = 24a. If n = 2, then b = 12a hence r = q 12 . We have r + 1 = q 12 + 1 ∈ cd(S) but q 12 + 1 ∈ cd(H), a contradiction. Thus n ≥ 3. By Table 3, S possesses a unipotent character ψ with ψ(1)p = pb(n−1)(n−2)/2 . Thus b(n − 1)(n − 2)/2 ≤ 7a. Multiplying both sides by n, we obtain bn(n − 1)(n − 2)/2 = 12a(n − 2) ≤ 7an and so 5n ≤ 24 hence n ≤ 4. We conclude that 3 ≤ n ≤ 4. If n = 4 then r = q 2 . By [2, 13.8], L4 (q 2 ) possesses a unipotent character labeled by the symbol (2, 2) of degree r2 (r2 + 1) = q 4 (q 4 + 1). However we can check that q 4 (q 4 + 1) ∈ cd(H) since q 4 + 1 |H|, and so cd(S) ⊆ cd(H). If n = 3 then r = q 4 and S = L3 (q 4 ). By [22], S possesses a unipotent character of degree r3 − = q 12 − . However we can see that q 12 − |H|. Thus cd(S) ⊆ cd(H). (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have bn2 = 12a. By [2, 13.8], S possesses a unipotent character χ ∈ Irr(S) labeled by 1 the symbol 0 −n with degree χ(1) = r(rn − 1)(rn−1 − 1)/2(r + 1). Assume at first that p is odd. If r = q then b = a. It follows that n2 = 12, a contradiction. Thus q 3 | χ(1)p = r and so b ≥ 3a. Hence 12a = bn2 ≥ 3an2 so that n2 ≤ 4 hence n = 2. Thus r = q 3 and so S = S4 (q 3 ) and H = 3 D4 (q). However using [5, 21], we can check that cd(S4 (q 3 )) cd(3 D4 (q)). Now assume p = 2. Then S = S2n (r). If r = 2 then b = 1 so that n2 = 12a. By 2 Table 3, S has a unipotent character ψ with ψ(1)2 = 2(n−1) −1 . Thus (n − 1)2 − 1 ≤ 7a. Combining this with n2 = 12a, we deduce that 12((n − 1)2 − 1) ≤ 7n2 so that 2 ≤ n ≤ 4. However for these values of n, the equation n2 = 12a is impossible. Thus r > 2 and so χ(1)2 = 2b−1 > 1. If b − 1 = a then (a + 1)n2 = 12a. Hence n2 < 12 so that n = 2 or n = 3. If n = 2 then 4(a + 1) = 12a or 2a = 1, which is 10 HUNG P. TONG-VIET impossible. Hence n = 3 and 9(a + 1) = 12a, which implies that a = 3. Thus b = 4 and so S = S6 (16) and H = 3 D4 (8). But then π(S) π(H), which contradicts Lemma 2.3(iv). Thus q 3 /2 = 23a−1 | 2b−1 . It follows that b ≥ 3a, hence n2 ≤ 4 so that n = 2, b = 3a and so r = q 3 . This leads to a contradiction as in the case when p is odd. (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 12a. By [2, 13.8], S possesses a unipotent character χ of degree r(rn − )(rn−2 + )/(r2 − 1). If r = q then a = b so that n(n − 1) = 12 or n = 4. We have S = O8 (q) and H = 3 D4 (q). We can check that χ(1) ∈ cd(H). If q 3 /(2, q) | r then b ≥ 3a − 1 ≥ 2a and so 12a = bn(n − 1) ≥ 2an(n − 1) so that n(n − 1) ≤ 6, which is impossible as n ≥ 4. (d) S = 2 B2 (r2 ), r2 = 22n+1 . Then 2(2n + 1) = 12a and so 2n + 1 = 6a, which is absurd. (e) S = 2 F4 (r2 ), r2 = 22n+1 . Then 12(2n + 1) = 12a and so 2n + 1 = a. √ By [2, 13.9], S possesses a unipotent character of degree 2 B2 [a](1) = rΦ1 Φ2 Φ2 Φ6 / 2. 4 √ However this is impossible as r/ 2 = 2n < q = r2 . (f ) S = 2 G2 (r2 ), r2 = 32n+1 . Then 3(2n + 1) = 12a which is impossible. (g) S = 3 D4 (r). Then 12b = 12a and so r = q. Hence S ∼ H. = (h) S = 2 E6 (r) or E6 (r). Then 36b = 12a and so r3 = q. By Table 3, S possesses a unipotent character ψ with ψ(1)p = r25 . As r3 = q, we have r25 > r21 = (r3 )7 = q 7 , a contradiction. (i) S = G2 (r). Then 6b = 12a and so r = q 2 . By [3, 7, 8], S possesses an irreducible character ψ with ψ(1)2 = r3 = q 6 . However this cannot happen by checking the list of character degrees of H. (j) S = F4 (r). Then 24b = 12a and so q = r2 . By [2, 13.9], S possesses a unipotent character of degree φ9,10 (1) = r10 Φ2 Φ2 Φ12 . As r10 = q 5 , we can check 3 6 that this degree does not belong to cd(H). (k) S = E7 (r). Then 63b = 12a and so 21b = 4a. By [2, 13.9], S possesses a unipotent character of degree φ7,1 (1) = rΦ7 Φ12 Φ14 . It follows that b ≥ a and so 4a = 21b ≥ 21a, which is impossible. (l) S = E8 (r). Then 120b = 12a and so 10b = a or r10 = q. By Table 3, S possesses a unipotent character ψ with ψ(1)p = r91 . As r10 = q, we have r91 > r90 = q 9 > q 7 , a contradiction. Case 4. H = 2 E6 (q), q = pa . Then |H|p = p36a . If χ ∈ Irr(H) with χ = StH , then χ(1)p ≤ p25a . (a) S = Ln (r), r = pb , n ≥ 2. Then bn(n − 1) = 72a. If n = 2, then b = 36a hence r = q 36 . We have r + 1 = q 36 + 1 ∈ cd(S) but q 36 + 1 |H|, a contradiction. Thus n ≥ 3. By Table 3, S has a unipotent character ψ ∈ Irr(S) with ψ(1)p = pb(n−1)(n−2)/2 . It follows that b(n − 1)(n − 2) ≤ 50a. Multiplying both sides by n, we obtain bn(n − 1)(n − 2) = 72a(n − 2) ≤ 50an hence 11n ≤ 72 so that n ≤ 6. Assume first that l bn (p) exists. As rn − n | |S|, we deduce that l bn (p) ∈ π(H) by Lemma 2.3(iv). Thus bn ≤ 18a. Multiplying both sides by n − 1, we obtain bn(n − 1) = 72a ≤ 18a(n − 1), hence n ≥ 5. Thus 5 ≤ n ≤ 6. If n = 5 then 5b = 18a. It follows that 4b = 72a/5 > 12a so that l4b (p) ∈ π(S) ⊆ π(H) exists and hence 4b | 18a = 5b, which is impossible. If n = 6 then 5b = 12a. We also have 4b = 48a/5 > 9a and so arguing as above, we have l4b ∈ π(H). It follows that 4b | 18a or 4b | 12a = 5b. Obviously the latter case cannot happen and so 4b | 18a and then 8b | 36a = 15b, which is impossible. Thus l bn (p) does not SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 11 exist. It follows that bn = 6 or 2bn = 6. In both cases, we have bn ≤ 6 so that n − 1 = 72a/bn ≥ 72a/6 = 12a ≥ 12, which is a contradiction as n ≤ 6. (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have bn2 = 36a. If p = 2 and 2bn = 6, then bn = 3 so that n = 3, b = 1 and hence 9 = 36a, which is absurd. Thus l2bn (p) exists and so l2bn (p) ∈ π(H) as l2bn (p) ∈ π(S). We deduce that 2bn ≤ 18a or bn ≤ 9a. Multiplying both sides by n, we obtain bn2 = 36a ≤ 9an. Hence n ≥ 4. By Table 3, S has a unipotent character 2 ψ with ψ(1)p ≥ pb(n−1) −b and so bn(n − 2) ≤ 25a. Multiplying both sides by n, we have bn2 (n − 2) = 36a(n − 2) ≤ 25an. Thus 4 ≤ n ≤ 6. If n = 4 then 4b = 9a. We have 6b = 27a/2 > 13a and so l6b (p) ∈ π(S) ⊆ π(H) exists and hence 6b | 18a = 8b, which is impossible. Similarly, if n = 5 then 25b = 36a. We have 10b = 360a/25 = 72a/5 > 12a and so l10b (p) ∈ π(H). Hence 10b | 18a, and so 20b | 36a = 25b, a contradiction. If n = 6 then b = a so that q = r. However we see that l5 (q) ∈ π(S) but l5 (q) ∈ π(H), which contradicts Lemma 2.3(iv). (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 36a. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1)(n−2) . Thus b(n − 1)(n − 2) ≤ 25a. Multiplying both sides by n and simplifying, we obtain 11n ≤ 72 and so 4 ≤ n ≤ 6. If n = 4 then b = 3a and hence r = q 3 . We have S = O8 (q 3 ) and H = 2 E6 (q). We have l9 (q) ∈ π(S) but l9 (q) ∈ π(H), which contradicts Lemma 2.3(iv). If n = 5 then 5b = 9a. As 8b = 72a/5 > 12a, l8b (p) ∈ π(S) ⊆ π(H) exists and hence 8b | 18a = 10b, which is impossible. Assume that n = 6. Then 5b = 6a. As above, we have l8b (p) ∈ π(H) and since 8b = 48a/5 > 9a, we have 8b | 10a, 8b | 12a = 10b or 8b | 18a = 15b. Obviously the last two cases cannot happen. If 8b | 10a, then 24b | 30a = 25b, which is again impossible. (d) S = 2 B2 (r2 ), r2 = 22n+1 or 2 G2 (r2 ), r2 = 32n+1 . Then equation |S|p = |H|p cannot happen. (e) S = 2 F4 (r2 ), r2 = 22n+1 . Then 12(2n + 1) = 36a and so 2n + 1 = 3a or 2 r = q 3 . We have r12 + 1 = q 18 + 1 | |S| so that l36 (q) ∈ π(H), which is impossible. (f ) S = 3 D4 (r). Then 12b = 36a and so r = q 3 . We observe that q 36 − 1 = 12 r − 1 = (r4 − 1)(r8 + r4 + 1) and r8 + r4 + 1 | |S| so that l36 (q) | |S| but l36 (q) ∈ π(H), a contradiction. (g) S = 2 E6 (r). Then 36b = 36a and so r = q. Thus S ∼ H. = (h) S = G2 (r). Then 6b = 36a and so r = q 6 . We have r6 − 1 = q 36 − 1 | |S| so that l36 (q) ∈ π(H), which is impossible. (i) S = F4 (r). Then 24b = 36a and so q 3 = r2 . We have r6 − 1 = q 9 − 1 | |S| so that l9 (q) ∈ π(H). However l9 (q) |H|, a contradiction. (j) S = E6 (r). Then 36b = 36a and so r = q. We have l9 (q) ∈ π(S) but l9 (q) ∈ π(H), a contradiction. (k) S = E7 (r). Then 63b = 36a and so 7b = 4a. By Table 3, S has a unipotent character ψ with ψ(1)p = r46 . As r7 = q 4 , we have r46 = q 184/7 > q 25 , a contradiction. (l) S = E8 (r). Then 120b = 36a and so 20b = 9a or r20 = q 9 . By Table 3, S has a unipotent character ψ with ψ(1)p = r91 . As r20 = q 9 , we have r91 = q 819/20 > q 25 , a contradiction. Case 5. H = G2 (q), q = pa . Then |H|p = p6a . The character degrees of G2 (q) are available in [3, 7, 8]. Observe that if χ ∈ Irr(H) with χ = StH and p | χ(1) then q/(6, q) | χ(1)p and χ(1)p ≤ p3a . 12 HUNG P. TONG-VIET (a) S = Ln (r), n ≥ 2. Then bn(n − 1) = 12a. If n = 2, then b = 6a hence r = q 6 . We have r + 1 = q 6 + 1 ∈ cd(S) but q 6 + 1 |H|, a contradiction. Thus n ≥ 3. By Table 3, S possesses a unipotent character ψ ∈ Irr(S) with ψ(1)p = pb(n−1)(n−2)/2 . It follows that b(n − 1)(n − 2) ≤ 6a. Multiplying both sides by n and simplifying, we obtain 3 ≤ n ≤ 4. If n = 4 then b = a so that S = L4 (q) and H = G2 (q). However l4 (q) ∈ π(S) but l4 (q) ∈ π(H). If n = 3 then b = 2a so that r = q 2 . We have S = L3 (q 2 ) and S possesses a character of degree q 2 (q 2 + ). However, we can check that this degree does not belong to cd(H). (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have bn2 = 6a. By Table 3, S possesses a unipotent character ψ ∈ Irr(S) with ψ(1)p ≥ pbn(n−2) . Thus bn(n − 2) ≤ 3a. It follows that 2 ≤ n ≤ 4. If n = 2 then r2 = q 3 and S = S4 (r) possesses a unipotent character labeled by the symbol 102 with degree r(r2 + 1)/2. Assume first that p is odd. Since r = q 3/2 > q, we deduce from the list of character degrees of G2 (q) that r = q 2 or r = q 3 . But then r ≥ q 2 > q 3/2 , a contradiction. 1 Thus p = 2. We have r/2 = 2 q 3/2 ≥ q. Thus r/2 ∈ {q, q 2 , q 3 }. If r = 2q, then q = 4 and so S = S4 (8) and H = G2 (4). However by using [9], cd(S4 (8)) cd(G2 (4)). For the remaining cases, we have r ≥ 2q 2 so that r2 = q 3 ≥ 4q 4 , which is impossible. If n = 3 then 3b = 2a. We have l4b (p) ∈ π(S) so that l4b (p) ∈ π(H). As 4b > 2a, we deduce that 4b | 6a = 9b, which is impossible. If n = 4 then 8b = 3a. If l6b (p) exists then l6b (b) ∈ π(H). As 6b = 18a/8 > 2a, we have 6b | 6a = 16b, which is impossible. Hence 6b = 6 and p = 2. Hence b = 1 and 3a = 8, a contradiction. (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 6a. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1)(n−2) . Thus b(n − 1)(n − 2) ≤ 3a. Multiplying both sides by n and simplifying, we obtain n ≤ 4 and so n = 4 and 2b = a and hence q = r2 . We have S = O8 (r) and H = G2 (r2 ). By [2, 13.8], S possesses a unipotent character ϕ of degree r(r4 − )(r2 + )/(r2 − 1). Thus a − 1 ≤ b so that a = 2b ≥ 2(a − 1) hence a ≤ 2. Since a = 2b ≥ 2, we obtain a = 2 and b = 1. Then S = O8 (2) and H = G2 (4). Using [4] we can see that cd(O8 (2)) ⊆ cd(G2 (4)). (d) S = 2 B2 (r2 ), r2 = 22n+1 . Then 2(2n + 1) = 6a, hence 2n + 1 = 3a. By [23], S possesses a character of degree r4 +1 = q 6 +1. But then q 6 +1 |H|, a contradiction. (e) S = 2 F4 (r2 ), r2 = 22n+1 . Then 12(2n + 1) = 6a and so 2(2n + 1) = a. By Table 3, S has a unipotent character ψ with ψ(1)2 = r13n+6 . Hence 13n + 6 ≤ 3a = 6(2n + 1). It follows that 13n + 6 ≤ 12n + 6, which is impossible. (f ) S = 2 G2 (r2 ), r2 = 32n+1 . Then 3(2n + 1) = 6a, which is impossible. (g) S = 3 D4 (r). Then 12b = 6a and so a = 2b. By [2, 13.9], S possesses a unipotent character of degree φ1,3 (1) = rΦ12 . Thus b ≥ a − 1 and hence a = 2b ≥ 2(a − 1) so that a ≤ 2. As a = 2b ≥ 2, we have a = 2 and b = 1. Then S = 3 D4 (2) and H = G2 (4). Using [4] we can see that cd(3 D4 (2)) ⊆ cd(G2 (4)). (h) S = 2 E6 (r). Then 36b = 6a and so a = 6b. By [2, 13.9], S possesses a unipotent character of degree φ2,4 (1) = rΦ8 Φ12 . Thus b ≥ a − 1 and hence a = 6b ≥ 6(a − 1) so that a = 1, which is impossible as a = 6b ≥ 6. (i) S = G2 (r). Then 6b = 6a and so r = q. Thus S ∼ H. = (j) S = F4 (r). Then 24b = 6a and so a = 4b. By [2, 13.9], S possesses a unipotent character of degree φ9,2 (1) = r2 Φ2 Φ2 Φ12 . Thus 2b ≥ a − 1 and hence 3 6 a = 4b ≥ 2(a − 1) so that a ≤ 2, which is impossible as a = 4b ≥ 4. (k) S = E6 (r). Then 36b = 6a and so a = 6b. By [2, 13.9], S possesses a unipotent character of degree φ6,1 (1) = rΦ8 Φ9 . Thus b ≥ a − 1 and hence a = 6b ≥ 6(a − 1) so that a ≤ 1, which is impossible as a = 6b ≥ 6. SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 13 (l) S = E7 (r). Then 63b = 6a and so 2a = 21b. By [2, 13.9], S possesses a unipotent character of degree φ7,1 (1) = rΦ7 Φ12 Φ(14). Thus b ≥ a − 1 and hence 2a = 21b ≥ 21(a − 1) so that a ≤ 1, which is impossible as a = 21b/2 > 10. (m) S = E8 (r). Then 120b = 6a and so a = 20b. By [2, 13.9], S possesses a unipotent character of degree φ8,1 (1) = rΦ2 Φ8 Φ12 Φ20 Φ24 . Thus b ≥ a − 1 and 4 hence a = 20b ≥ 20(a − 1) so that a ≤ 1, which is impossible as a = 20b ≥ 20. Case 6. H = F4 (q), q = pa . Then |H|p = p24a . If χ ∈ Irr(H) with χ = StH then χ(1)p ≤ p16a . (a) S = Ln (r), n ≥ 2. Then bn(n−1) = 48a. If n = 2, then b = 24a hence r = q 24 . We have r + 1 = q 24 + 1 ∈ cd(S) but q 24 + 1 |H|, a contradiction. Thus n ≥ 3. By Table 3, S possesses a unipotent character ψ with ψ(1)p = pb(n−1)(n−2)/2 . It follows that b(n − 1)(n − 2) ≤ 32a. Multiplying both sides by n and simplifying, we obtain 3 ≤ n ≤ 6. Assume that l bn (p) exists. Then l bn (p) ∈ π(H) by Lemma 2.3(iv). Thus bn ≤ 12a. Multiplying both sides by n − 1, we have bn(n − 1) = 48a ≤ 12a(n − 1), and hence n ≥ 5. Thus 5 ≤ n ≤ 6. If n = 5 then 5b = 12a. It follows that 4b = 48a/5 > 9a so that l4b (p) exists and since l4b (p) ∈ π(S), by Lemma 2.3(iv), l4b (p) ∈ π(H) so that 4b | 12a = 5b, which is impossible. If n = 6 then 5b = 8a. As 6b = 48a/5 > 9a, l6b exists and so as above l6b (p) ∈ π(H) It follows that 6b | 12a and hence 12b | 24a = 15b, which is impossible. Thus l bn (p) does not exist and 3 ≤ n ≤ 6. It follows that bn = 6 or 2bn = 6. In both cases we obtain bn ≤ 6 and so n − 1 = 48a/bn ≥ 48a/6 = 8a ≥ 8, which is a contradiction as n ≤ 6. (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have bn2 = 24a. If 2bn = 6, then bn = 3 so that n = 3, b = 1 and hence 9 = 24a, which is absurd. Thus l2bn (p) ∈ π(S) ⊆ π(H) exists and so 2bn ≤ 12a or equivalently bn ≤ 6a. Multiplying both sides by n, we obtain bn2 = 24a ≤ 6an. Hence n ≥ 4. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pbn(n−2) and so bn(n − 2) ≤ 16a. Multiplying both sides by n, we have bn2 (n − 2) = 24a(n − 2) ≤ 16an. Thus 4 ≤ n ≤ 6. If n = 4 then 2b = 3a. We have 6b = 9a and so l6b (p) exists. Hence 6b | 12a = 8b, which is impossible. If n = 5 then 25b = 24a. We have 10b = 240a/25 = 48a/5 > 9a and so arguing as above, we have 10b | 12a. Hence 40b | 48a = 50b, a contradiction. If n = 6 then 3b = 2a. We have 10b = 20a/3 > 6a so that l10b (a) exists and so 10b | 8a = 12b or 10b | 12a = 18b. However we can see that both cases are impossible. (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 24a. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1)(n−2) . Thus b(n − 1)(n − 2) ≤ 16a. Multiplying both sides by n and simplifying, we obtain 3(n − 2) ≤ 2n and so 4 ≤ n ≤ 6. If n = 4 then b = 2a and hence r = q 2 . We have S = O8 (q 2 ) and H = F4 (q). By [2, 13.8], S possesses a unipotent character ϕ of degree r(r4 − )(r2 + )/(r2 − 1). However we can check that ϕ(1) |H|. If n = 5 then 5b = 6a. As 8b = 48a/5 > 9a, l8b (p) ∈ π(S) so that l8b (p) ∈ π(H). We deduce that 8b | 12a = 10b, which is impossible. If n = 6 then 5b = 4a. As above, we have 8b = 32a/5 > 6a, so that l8b (p) ∈ π(H) and hence 8b | 8a = 10b or 8b | 12a = 15b. Obviously both cases are impossible. (d) S = 2 B2 (r2 ), 2 F4 (r2 ) or 2 G2 (r2 ), where r2 = 22n+1 , 22n+1 or 32n+1 , respectively. In these cases, the equation |S|p = |H|p cannot happen. (e) S = 3 D4 (r). Then 12b = 24a and so r = q 2 . We observe that q 24 − 1 = 12 r − 1 = (r4 − 1)(r8 + r4 + 1) and r8 + r4 + 1 | |S| so that l24 (q) | |S| but l24 (q) ∈ π(H), which contradicts Lemma 2.3(iv). 14 HUNG P. TONG-VIET (f ) S = 2 E6 (r) or E6 (r). Then 36b = 24a and so 3b = 2a. By Table 3, S has a unipotent character ψ with ψ(1)p = p25b . Thus 25b ≤ 16a. Hence 25b ≤ 24b, a contradiction. (g) S = G2 (r). Then 6b = 24a and so r = q 4 . We have r6 − 1 = q 24 − 1 | |S| so that l24 (q) ∈ π(H), which is impossible. (h) S = F4 (r). Then 24b = 24a and so q = r so that S ∼ H. = (i) S = E7 (r). Then 63b = 24a and so 21b = 8a. By Table 3, S possesses a unipotent character ψ with ψ(1)p = r46 . As r21 = q 8 , we have r46 = q 8.46/24 > q 16 , a contradiction. (j) S = E8 (r). Then 120b = 24a and so a = 5b or q = r5 . By Table 3, S possesses a unipotent character ψ with ψ(1)p = r91 . As r5 = q, we have r91 > r90 = q 18 > q 16 , a contradiction. Case 7. H = E6 (q), q = pa . Then |H|p = p36a . If χ ∈ Irr(H) with χ = StH , then χ(1)p ≤ p25a . (a) S = Ln (r), n ≥ 2. Then bn(n−1) = 72a. If n = 2, then b = 36a hence r = q 36 . We have r + 1 = q 36 + 1 ∈ cd(S) but q 36 + 1 |H|, a contradiction. Thus n ≥ 3. By Table 3, S has a unipotent character ψ ∈ Irr(S) with ψ(1)p = pb(n−1)(n−2)/2 . It follows that b(n − 1)(n − 2) ≤ 50a. Multiplying both sides by n and simplifying, we obtain 11n ≤ 72 so that n ≤ 6. Assume first that l bn (p) exists. We deduce that l bn (p) ∈ π(S) ⊆ π(H) by Lemma 2.3(iv). Thus bn ≤ 12a. Multiplying both sides by n − 1, we have bn(n − 1) = 72a ≤ 12a(n − 1). Hence n ≥ 7, a contradiction. Thus l bn (p) does not exist. Arguing as in Case 4(a), we get a contradiction. (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have bn2 = 36a. If p = 2 and 2bn = 6, then bn = 3 so that n = 3, b = 1 and hence 9 = 36a, which is absurd. Thus l2bn (p) ∈ π(S) ⊆ π(H) exists and so 2bn ≤ 12a or equivalently bn ≤ 6a. Multiplying both sides by n and simplifying, we obtain 2 n ≥ 6. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1) −b and so bn(n−2) ≤ 25a. Multiplying both sides by n and simplifying, we obtain n ≤ 6. Thus n = 6 and then b = a so that q = r. However l10 (q) ∈ π(S) − π(H), contradicting Lemma 2.3(iv). (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 36a. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1)(n−2) . Thus b(n−1)(n−2) ≤ 25a. Multiplying both sides by n and simplifying, we obtain 11n ≤ 72 and so 4 ≤ n ≤ 6. If 2(n − 1)b = 6 then b = 1 and n = 4 so that 36a = 12, which is impossible. Thus l2(n−1)b (p) ∈ π(S) ⊆ π(H) exists and hence 2b(n − 1) ≤ 12a. Multiplying both sides by n and simplifying, we obtain n ≥ 6. Therefore n = 6 and so 5b = 6a. As 8b = 48a/5 > 9a, l8b (p) exists and lies in π(S) so that by Lemma 2.3(iv), l8b (p) ∈ π(H). Since 8b > 9a, we must have 8b | 12a = 10b, which is impossible. (d) S = 2 E6 (r). Then 36b = 36a and so r = q. Thus S = 2 E6 (q) and H = E6 (q). However we have l18 (q) ∈ π(S) but l18 (q) ∈ π(H), which contradicts Lemma 2.3(iv). (e) S = E6 (r). Then 36b = 36a and so r = q. Hence S ∼ H. = For the remaining simple exceptional groups of Lie type, using the same argument as in Case 4, we deduce that cd(S) ⊆ cd(H) in any of these cases. Case 8. H = E7 (q), q = pa . Then |H|p = p63a . If χ ∈ Irr(H) with χ = StH , then χ(1)p ≤ p46a . (a) S = Ln (r), r = pb . Then bn(n − 1) = 126a. If n = 2, then b = 63a hence r = q 63 . We have r + 1 = q 63 + 1 ∈ cd(S) but q 63 + 1 |H|, a contradiction. Thus n ≥ 3. By Table 3, S has a unipotent character ψ ∈ Irr(S) with ψ(1)p = SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 15 pb(n−1)(n−2)/2 . It follows that b(n − 1)(n − 2) ≤ 92a. Multiplying both sides by n and simplifying, we obtain n ≤ 7. Assume first that l bn (p) exists. We deduce that l bn (p) ∈ π(H) by Lemma 2.3(iv). Thus bn ≤ 18a. Multiplying both sides by n − 1 and simplifying, we obtain n ≥ 8, a contradiction. Thus l bn (p) does not exist. It follows that bn = 6 or bn = 3. In both cases, we have bn ≤ 6 so that n − 1 = 126a/bn ≥ 126a/6 = 21a ≥ 21, a contradiction as n ≤ 7. (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have bn2 = 63a. If p = 2 and 2bn = 6, then bn = 3 so that n = 3, b = 1 and hence 9 = 63a, which is impossible. Thus l2bn (p) ∈ π(S) ⊆ π(H) exists and hence 2bn ≤ 18a so that bn ≤ 9a. Multiplying both sides by n and simplifying, we 2 obtain n ≥ 7. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1) −b and so bn(n − 2) ≤ 46a. Multiplying both sides by n and simplifying, we obtain n ≤ 7. Thus n = 7 and then 7b = 9a. We see that 12b = 108a/7 > 15a so that l12b (p) ∈ π(S) ⊆ π(H) exists and hence 12b | 18a = 14b, which is impossible. (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 63a. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1)(n−2) . Thus b(n − 1)(n − 2) ≤ 46a. Multiplying both sides by n and simplifying, we obtain 17n ≤ 126 and so 4 ≤ n ≤ 7. If 2(n − 1)b = 6 then b = 1 and n = 4 so that 63a = 12 which is impossible. Thus l2(n−1)b (p) ∈ π(S) ⊆ π(H) exists and hence 2b(n − 1) ≤ 18a. Multiplying both sides by n and simplifying, we obtain n ≥ 7. Therefore n = 7 and so 2b = 3a. We have 10b = 15a so that l10b (p) ∈ π(S) ⊆ π(H) exists so that 10b | 18a = 12b, which is impossible. (d) S = 2 B2 (r2 ). Then r4 = q 63 . By [23], S possesses a character of degree 4 r + 1 = q 63 + 1. But then q 63 + 1 |H|, a contradiction. (e) S = 2 F4 (r2 ), r2 = 22n+1 . Then 12(2n+1) = 63a. We have l12(2n+1) (2) ∈ π(S) so that by Lemma 2.3, l12(2n+1) (2) ∈ π(H). We have 12(2n + 1) = 63a and so l63a (2) ∈ π(H), and then l63 (q) ∈ π(H), which is impossible. (f ) S = 2 G2 (r2 ), r2 = 32n+1 . Then 3(2n + 1) = 63a so that 2n + 1 = 21a. We have l6(2n+1) (3) ∈ π(S) so that l6(2n+1) (3) ∈ π(H). As 6(2n + 1) = 126a, we deduce that l126a (3) ∈ π(H), which is impossible. (g) S = 3 D4 (r). Then 12b = 63a and so 4b = 21a or r4 = q 21 . We observe that 36 q − 1 = r12 − 1 = (r4 − 1)(r8 + r4 + 1) and r8 + r4 + 1 | |S| so that l36 (q) | |S| but l36 (q) ∈ π(H), which contradicts Lemma 2.3(iv). (h) S = 2 E6 (r). Then 36b = 63a and so 4b = 7a. We have 12b = 21a and 12b p − 1 | |S| so that l21 (q) ∈ π(H), which is impossible as 21a > 18a. (i) S = G2 (r). Then 6b = 63a and so 2b = 21a. We have r6 − 1 = q 63 − 1 | |S| so that l63 (q) ∈ π(H), which is impossible. (j) S = F4 (r). Then 24b = 63a and so 8b = 21a. As p8b − 1 = q 21 − 1 | |S|, we deduce that l21 (q) ∈ π(H), which is impossible. (k) S = E6 (r). Then 36b = 63a and so 4b = 7a. We have p12b − 1 = q 21 − 1 | |S| so that l21 (q) ∈ π(H), a contradiction. (l) S = E7 (r). Then 63b = 63a and so b = a or q = r. Hence S ∼ H. = (m) S = E8 (r). Then 120b = 63a and so 40b = 21a. As l30b (p) exists and belongs to π(S), it follows that l30b (p) ∈ π(H). As 30b = 63a/4 > 15a, we deduce that 30b | 18a and hence 120b = 63a | 72a, which is impossible. Case 9. H = E8 (q), q = pa . Then |H|p = p120a . If χ ∈ Irr(H) with χ = StH , then χ(1)p ≤ p91a . 16 HUNG P. TONG-VIET (a) S = Ln (r), r = pb . Then bn(n − 1) = 240a. If n = 2, then b = 120a hence r = q 120 . We have r + 1 = q 120 + 1 ∈ cd(S) but q 120 + 1 |H|, a contradiction. Thus n ≥ 3. By Table 3, S has a unipotent character ψ ∈ Irr(S) with ψ(1)p = pb(n−1)(n−2)/2 . It follows that b(n − 1)(n − 2) ≤ 182a. Multiplying both sides by n and simplifying, we obtain n ≤ 8. Assume first that l bn (p) exists. We deduce that l bn (p) ∈ π(H) by Lemma 2.3(iv). Thus bn ≤ 30a. Multiplying both sides by n − 1, we have bn(n − 1) = 240a ≤ 30a(n − 1). Hence n ≥ 9, which contradicts the assertion n ≤ 8 above. Thus l bn (p) does not exist. It follows that bn = 6 or bn = 3, and hence bn ≤ 6 so that n − 1 = 240a/bn ≥ 240a/6 = 40a ≥ 40, a contradiction. (b) S = S2n (r), n ≥ 2, S = S4 (2) or S = O2n+1 (r), n ≥ 3, r odd. We have bn2 = 120a. If p = 2 and 2bn = 6, then bn = 3 so that n = 3, b = 1 and hence 9 = 120a, which is impossible. Thus l2bn (p) ∈ π(S) ⊆ π(H) exists and hence 2bn ≤ 30a so that bn ≤ 15a. Multiplying both sides by n, we obtain bn2 = 120a ≤ 15an. Hence n ≥ 8. By Table 3, S has a unipotent character ψ with 2 ψ(1)p ≥ pb(n−1) −b and so bn(n − 2) ≤ 91a. Multiplying both sides by n, we 2 have bn (n − 2) = 120a(n − 2) ≤ 91an. Thus 29n ≤ 240 and so n ≤ 8. Hence n = 8 and then 8b = 15a. We see that 14b = 105a/4 > 26a so that l14b (p) exists and l14b (p) ∈ π(H) as l14b (p) ∈ π(S). It follows that 14b | 30a = 16b, which is impossible. (c) S = O2n (r), n ≥ 4. We have bn(n − 1) = 120a. By Table 3, S has a unipotent character ψ with ψ(1)p ≥ pb(n−1)(n−2) . Thus b(n − 1)(n − 2) ≤ 91a. Multiplying both sides by n and simplifying, we obtain 29n ≤ 240 and so 4 ≤ n ≤ 8. If 2(n − 1)b = 6 then b = 1 and n = 4 so that 120a = 12 which is impossible. Thus l2(n−1)b (p) ∈ π(S) ⊆ π(H) exists and hence 2b(n − 1) ≤ 30a. Multiplying both sides by n and simplifying, we obtain n ≥ 8. Therefore n = 8 and so 7b = 15a. We have 12b = 180a/7 > 25a so that l12b (p) ∈ π(S) ⊆ π(H) exists and then 12b | 30a = 14b, which is impossible. (d) S = 2 B2 (r2 ), 2 F4 (r2 ) or 2 G2 (r2 ), where r2 = 22n+1 , 22n+1 or 32n+1 , respectively. In these cases, the equation |S|p = |H|p cannot happen. (e) S = 3 D4 (r). Then 12b = 120a and so b = 10a or r = q 10 . We have r6 − 1 = 60 q − 1 | |S| so that l60 (q) ∈ π(H), which is impossible. (f ) S = 2 E6 (r) or E6 (r). Then 36b = 120a and so 3b = 10a. We have 12b = 40a and p12b − 1 = q 40 − 1 | |S| so that l40 (q) ∈ π(H), which is impossible as 40a > 30a. (g) S = G2 (r). Then 6b = 120a and so b = 20a. We have r6 − 1 = q 120 − 1 | |S| so that l120 (q) ∈ π(H), which is impossible. (h) S = F4 (r). Then 24b = 120a and so b = 5a. As p8b − 1 = q 40 − 1 | |S|, we deduce that l40 (q) ∈ π(H), which is impossible. (i) S = E7 (r). Then 63b = 120a and so 21b = 40a. We have 18b = 240a/7 > 30a so that l18b (p) exists and hence l18b (p) ∈ π(H) so that 18b ≤ 30a, a contradiction. (j) S = E8 (r). Then 120b = 120a and so b = a. Hence S ∼ H. = Proof of Theorem 1.1. This follows from Propositions 3.1-3.5. Proof of Corollary 1.2. Assume that G is a perfect group satisfying cd(G) ⊆ cd(H) and |G| ≤ |H|. Let N be a maximal normal subgroup of G. Then G/N is a non-abelian simple group. As cd(G/N ) ⊆ cd(G) and cd(G) ⊆ cd(H), it follows that cd(G/N ) ⊆ cd(H). By Theorem 1.1, we obtain G/N ∼ H. Hence |G/N | = |H|. = Since |G| ≤ |H|, we conclude that |N | = 1. Hence N is a trivial subgroup of G, so that G ∼ H. The proof is now complete. = SIMPLE EXCEPTIONAL GROUPS OF LIE TYPE 17 Proof of Corollary 1.3. Assume that G is a group satisfying X1 (G) ⊆ X1 (H). It follows that |G| ≤ |H| and cd(G) ⊆ cd(H). Moreover, as H is non-abelian simple, H has a unique irreducible complex character of degree 1, which is the trivial character of H and as X1 (G) ⊆ X1 (H), G also possesses a unique character of degree 1. Thus G must be perfect. Now the result follows from Corollary 1.2. Proof of Corollary 1.4. Assume that CG ∼ CH. By Molien’s Theorem (see = [1, Theorem 2.13]), the first columns of the ordinary character tables of G and H coincide so that X1 (G) = X1 (H). Hence the result follows from Corollary 1.3. Acknowledgment. The author is grateful to the referee for his or her comments. References [1] Berkovich, Y., Zhmud, E.: Characters of finite groups, Part 1. Translations of Mathematical ´ Monographs, 172, 181. AMS, Providence, RI (1997) [2] Carter, R.: Finite Groups of Lie Type. Conjugacy classes and complex characters. John Wiley and Sons, New York (1985) [3] Chang, B., Ree, R.: The characters of G2 (q). 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Ann. of Math.(2) 75, 105–145 (1962) [24] Tong-Viet, H.P.: Alternating and sporadic simple groups are determined by their character degrees. Algebr. Represent. Theory doi:10.1007/s10468-010-9247-1 (2010) [25] Tong-Viet, H.P.: Simple classical groups of Lie type are determined by their character degrees. Preprint. [26] Tong-Viet, H.P.: Symmetric groups are determined by their character degrees. J. Algebra doi:10.1016/j.jalgebra.2010.11.018 (2011) [27] Ward, H.N.: On Ree’s series of simple groups. Trans. Amer. Math. Soc. 121, 62–89 (1966) E-mail address: Tong-Viet@ukzn.ac.za School of Mathematical Sciences, University of KwaZulu-Natal, Pietermaritzburg 3209, South Africa