Alternating and Sporadic simple groups are determined by their character degrees more

ALTERNATING AND SPORADIC SIMPLE GROUPS 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) be the set of all irreducible complex character degrees of G forgetting multiplicities, that is, cd(G) = {χ(1) : χ ∈ Irr(G)} and let cd∗ (G) be the set of all irreducible complex character degrees of G counting multiplicities. Let H be an alternating group of degree at least 5, a sporadic simple group or the Tits group. In this paper, we will show that if G is a non-abelian simple group and cd(G) ⊆ cd(H) then G must be isomorphic to H. As a consequence, we show that if G is a finite group with cd∗ (G) ⊆ cd∗ (H) then G is isomorphic to H. This gives a positive answer to Question 11.8(a) in (Unsolved problems in group theory: the Kourovka notebook, 16th edn) for alternating groups, sporadic simple groups or the Tits group. 1. Introduction and Notations 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 cd∗ (G) be the set of all irreducible character degrees of G counting multiplicities. Observe that cd∗ (G) is just the first column of the character table of G. Since |G| = χ∈Irr(G) χ(1)2 , if we know cd∗ (G) then the order of G is known. For solvable groups, cd∗ (G) does not determined the structure of the group, for example, the dihedral group D8 and the quaternion group Q8 have the same character table and hence cd∗ (D8 ) = cd∗ (Q8 ) but they are not isomorphic. For simple groups, the situation is much different as pointed out by Huppert in [8]. In [8], Huppert proposed the following conjecture. Conjecture 1. (Huppert’s Conjecture). Let H be any non-abelian simple group and G be a group such that cd(G) = cd(H). Then G ∼ H × A, where A is abelian. = This conjecture was verified for several families of simple groups, for instance, Huppert [8, 7] verified this conjecture for Sz(q) and P SL2 (q), while Thomas Wakefield [20, 19] verified for P SL3 (q), P SU3 (q 2 ) and 2 G2 (q 2 ). However, this conjecture is still widely open. In this paper, we consider a weaker version of Huppert’s Conjecture which is stated as Question 11.8(a) in [17]. Question 1. Let H and G be groups with cd∗ (G) = cd∗ (H). Is G simple if H is a non-abelian simple group? Date: November 20, 2010. 2000 Mathematics Subject Classification. Primary 20C15. Key words and phrases. character degree, simple group. 1 2 HUNG P. TONG-VIET Assume that H is a non-abelian simple group and G is a finite group satisfying cd∗ (G) = cd∗ (H). The latter condition implies that cd(G) = cd(H) and |G| = |H|. Thus Question 1 follows from Huppert’s Conjecture. By the classification of finite simple groups, every non-abelian simple group is an alternating group of degree at least 5, a simple group of Lie type or one of the 26 sporadic simple groups. In this paper, we consider the case when H is an alternating group, a sporadic simple group or the Tits group. In fact, we will prove the following result. Theorem 1. Let H be an alternating group of degree at least 5, a sporadic simple group or the Tits group, and let S be a non-abelian simple group. If cd(S) ⊆ cd(H), then S ∼ H. = Corollary 2. Let H be an alternating group of degree at least 5, a sporadic simple group or the Tits group, and let G be a perfect group. If cd(G) ⊆ cd(H) and |G| ≤ |H|, then G ∼ H. = Corollary 3. Let H be an alternating group of degree at least 5, a sporadic simple group or the Tits group, and let G be a group. If cd∗ (G) ⊆ cd∗ (H), then G ∼ H. = This gives a positive answer to Question 1 for alternating groups, sporadic simple groups and the Tits group. Here are some notations. For a group G, 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. 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 p(G) = max(π(G)), and define p(G) to be the smallest prime such that ˜ p(G) ∈ π(G), and called p(G) a non-divisor of G. Let ρ(G) = ∪χ∈Irr(G) π(χ(1)) be ˜ ˜ the set of all primes which divide some irreducible character degrees of G. 2. Preliminaries Let n be a positive integer. We call λ = (λ1 , . . . , λr ) a partition of n, written λ r n, provided λi , i = 1, . . . , r are integers, with λ1 ≥ · · · ≥ λr > 0 and i=1 λi = n. ak a1 We collect the same parts together and write λ = ( 1 , · · · , k ), with i > i+1 > 0 k for i = 1, · · · , k − 1; ai = 0; and i=1 ai i = n. It is well known that the irreducible complex characters of the symmetric group Sn are parametrized by partitions of n. Denote by χλ the irreducible character of Sn corresponding to partition λ. The irreducible characters of the alternating group An are then obtained by restricting χλ to An . In fact, χλ is still irreducible upon restriction to the alternating group An if and only if λ is not self-conjugate. Otherwise, χλ splits into two irreducible characters of An having the same degree. We first state a result of Rasala on the minimal degrees of symmetric groups. Lemma 4. ([18, Result 3], [10]). Let λ = (λ1 , · · · , λr ) be a partition of n. (a) If n ≥ 15, then the first 6 nontrivial minimal character degrees of Sn are: (1) d1 (Sn ) = n − 1 and λ ∈ {(n − 1, 1), (2, 1n−2 )}; 1 (2) d2 (Sn ) = 2 n(n − 3) and λ ∈ {(n − 2, 2), (22 , 1n−4 )}; 1 (3) d3 (Sn ) = d2 (Sn ) + 1 = 2 (n − 1)(n − 2) and λ ∈ {(n − 2, 12 ), (3, 1n−3 )}; 1 (4) d4 (Sn ) = 6 n(n − 1)(n − 5) and λ ∈ {(n − 3, 3), (23 , 1n−6 )}; 1 (5) d5 (Sn ) = 6 (n − 1)(n − 2)(n − 3) and λ ∈ {(n − 3, 13 ), (4, 1n−4 )}; 1 (6) d6 (Sn ) = 3 n(n − 2)(n − 4) and λ ∈ {(n − 3, 2, 1), (3, 2, 1n−5 )}; CHARACTER DEGREES 3 (b) If n ≥ 22, then the next five smallest character degrees are: (7) d7 (Sn ) = n(n − 1)(n − 2)(n − 7)/24 and λ ∈ {(n − 4, 4), (24 , 1n−8 )}; (8) d8 (Sn ) = (n − 1)(n − 2)(n − 3)(n − 4)/24 and λ ∈ {(n − 4, 14 ), (5, 1n−5 )}; (9) d9 (Sn ) = n(n − 1)(n − 4)(n − 5)/12 and λ ∈ {(n − 4, 22 ), (32 , 1n−6 )}; (10) d10 (Sn ) = n(n − 1)(n − 3)(n − 6)/8 and λ ∈ {(n − 4, 3, 1), (3, 22 , 1n−7 )}; (11) d11 (Sn ) = n(n − 2)(n − 3)(n − 5)/8 and λ ∈ {(n − 4, 2, 12 ), (4, 2, 1n−6 )}; Proof. Part (a) is [18, Result 3]. Now for n ≥ 22, by [18, Main Theorem 1], di (Sn ) = χλ (1), where λ = (n − 4, µ), or its conjugate, and µ is a partition of 4. Now (b) follows from [10, Table 1]. Corollary 5. If n ≥ 15 then di (An ) = di (Sn ) for 1 ≤ i ≤ 4. Moreover if n ≥ 22, then di (An ) = di (Sn ) for 1 ≤ i ≤ 7. Proof. Assume that n ≥ 15. Observe that every partition λ of n in Lemma 4 is not self-conjugate, so that χλ remains irreducible upon restriction to An . For the first part of the corollary, let ϕ be any non-trivial irreducible character of An . Thus there exists a partition µ of n such that ϕ is the restriction of χµ to An . If χµ (1) > d6 (Sn ), then χµ (1)/2 > d6 (Sn )/2 > d4 (Sn ) so that ϕ(1) > d4 (Sn ). Otherwise, µ must be one of the partitions in Lemma 4(a), and so ϕ(1) = χµ (1), by the observation above. Hence the result follows. For the last assertion, we can apply the same argument with d9 (Sn ), where d9 (Sn ) > 2d7 (Sn ). Using [3, 4] and Corollary 5 above, we observe that d1 (An ) = n − 1 for any n ≥ 6 while d1 (A5 ) = 3. Lemma 6. (Tschebyschef). If m ≥ 15, then there is at least one prime p with m/2 < p ≤ m. Proof. If m ≥ 17 then the result follows from [13, Proposition 5.1]. For 15 ≤ m ≤ 16, the lemma is obvious. The prime power degree representations of quasi-simple groups have been classified in [14] and [1]. In the next lemma, we will extract the result for alternating groups, sporadic simple groups and the Tits group. Lemma 7. ([14, Theorem 1.1],[1, Theorem 5.1]). Let S be an alternating group, a sporadic simple group or the Tits group. Suppose that S possesses a non-trivial irreducible character χ with χ(1) = pd , where p is a prime. Then one of the following holds: (a) S is an alternating group: (1) S = Apd +1 , pd + 1 ≥ 5; and χ(1) = pd ; (2) S = A5 and χ(1) ∈ {3, 5}; (3) S = A6 and χ(1) ∈ {23 , 32 }; (4) S = A8 and χ(1) = 26 ; (5) S = A9 and χ(1) = 33 ; (b) S is a sporadic simple group or the Tits group: (6) S ∈ {M11 , M12 } and χ(1) ∈ {11, 16}; (7) S ∈ {M24 , Co2 , Co3 } and χ(1) = 23; (8) S = 2 F4 (2) and χ(1) ∈ {27, 211 }; We refer to [2, 13.8, 13.9] for the classification of unipotent characters and the notion of symbols. Let S be a finite simple group of Lie type in characteristic p and 4 HUNG P. TONG-VIET let G be the adjoint group of the same type as that of S. By results of Lusztig [12], the unipotent characters of G remain irreducible upon restriction to S, and these restrictions are just the unipotent characters of S. Note that every simple group of Lie type S in characteristic p (excluding the Tits group) contains an irreducible character of degree |S|p , which is the size of the Sylow p-subgroup of S, and is called the Steinberg character of S and is denoted by StS . Moreover, this character is extendible to G and we also call it the Steingerg character of G and denote it by StG . Observe that StG (1) = StS (1) = |S|p = |G|p . We will show that the Steinberg character of S is not the minimal non-trivial character of S. Lemma 8. Let S be a finite simple group of Lie type defined over a finite field of size q, where q is a power of a prime p. Then |S|p > d1 (S). Proof. Assume first that S is of type An−1 . Then G = (An−1 )ad (q) = P GLn (q). The result is clear when n = 2 so we assume that n ≥ 3. It suffices to show that G contains a unipotent character of degree less than that of the Steinberg character of G. By [2, 13.8], the unipotent characters of G are parametrized by partitions of n. Let α = (1, n − 1). Then the degree of the unipotent character χα corresponding to α is given by χα (1) = (q n − q)/(q − 1). Since StG (1) = |G|p = q n(n−1)/2 , and n ≥ 3, we have StG (1) > χα (1) > 1. Assume S is of type 2 An−1 , where n ≥ 3. Then G = (2 An−1 )ad (q 2 ) = P Un (q). By [2, 13.8], the unipotent characters of G are again parametrized by partitions of n. Let α = (1, n − 1). Then the degree of the unipotent character χα corresponding to α is given by χα (1) = (q n +(−1)n q)/(q+1). Since StG (1) = |G|p = q n(n−1)/2 , and n ≥ 3, we have StG (1) > χα (1) > 1. Assume next that S is of type Bn or Cn , where n ≥ 2 and S = P Sp4 (2). Then G = (Bn )ad (q) = SO2n+1 (q) or G = (Cn )ad (q) = P CSp2n (q). By [2, 13.8], G has a unipotent character χλ labeled by the symbol λ= 0 1 n − 2 with χλ (1) = (q n − 1)(q n − q)/(2(q + 1)). Since |G|p = q n , we see that |G|p > χλ (1) > 1. Assume S is of type Dn (q), n ≥ 4. Then G = (Dn )ad (q) = P (CO2n (q)0 ). By [2, 13.8], G has a unipotent character χλ labeled by the symbol λ= n−1 1 with χλ (1) = (q n − 1)(q n−1 + q)/(q 2 − 1). Since |G|p = q n(n−1) , we see that |G|p > χλ (1) > 1. Assume S is of type 2 Dn (q 2 ), n ≥ 4. Then G = (2 Dn )ad (q 2 ) = − P (CO2n (q)0 ). By [2, 13.8], G has a unipotent character χλ labeled by the symbol λ= 1 − n−1 with χλ (1) = (q n + 1)(q n−1 − q)/(q 2 − 1). Since |G|p = q n(n−1) , we see that |G|p > χλ (1) > 1. For the remaining groups of exceptional type, the lemma follows from [2, 13.9] or [11]. Lemma 9. (Ito-Michler Theorem, [16]). If p χ(1) for all χ ∈ Irr(G), then G has a normal abelian Sylow p-subgroup. CHARACTER DEGREES 5 Combining the Ito-Michler Theorem with the fact that χ(1) | |G| for all χ ∈ Irr(G), we get the following result. Corollary 10. If S is a non-abelian simple group then ρ(S) = π(S). Proof. Observe first that for any χ ∈ Irr(S), we have χ(1) divides |S| by [9, Theorem 3.11]. Hence ρ(S) ⊆ π(S). As S is non-abelian simple, it has no normal abelian Sylow p-subgroup, so that by the Ito-Michler Theorem, every prime divisor of S must divide χ(1) for some χ ∈ Irr(S), and thus π(S) ⊆ ρ(S). Hence ρ(S) = π(S). Lemma 11. Let S and H be simple groups. Suppose that cd(S) ⊆ cd(H). Then: (i) di (S) ≥ di (H), for all i ≥ 1; (ii) π(S) ⊆ π(H). Proof. (i) is obvious. (ii) follows from Corollary 10 as ρ(S) ⊆ ρ(H). In Table 1, we list the largest prime divisor, the non-divisor and the first three smallest non-trivial character degrees of sporadic simple groups and the Tits group. 3. Proofs of the main results Recall that a group G is said to be an almost simple group if there exists a nonabelian simple group S such that S G ≤ Aut(S). The simple group S is called the socle of G and is denoted by soc(G). Theorem 12. Let S be a simple group and let n ≥ 5 be an integer. If cd(S) ⊆ cd(An ), then S ∼ An . = The following example shows that the conclusion of the theorem may fail if we replace a simple group by an almost simple group. Example 1. Let G = P GL2 (9) and H = A6 . Then G ∼ A6 .22 is an almost simple = group with socle A6 . We have cd(G) = {1, 8, 9, 10} and cd(H) = {1, 5, 8, 9, 10}. Thus cd(G) ⊆ cd(H) but G ∼ H. = Proof. Suppose that cd(S) ⊆ cd(An ). We will show that S ∼ An , where n ≥ 5. We = use the classification of finite simple groups. We consider the following cases: (i) S = Am , m ≥ 5. If n = 5, then cd(A5 ) = {1, 3, 4, 5}. Since cd(Am ) ⊆ cd(A5 ), it follows that every character degree of Am is a prime power, hence by [15, Corollary], S = A5 . Thus we can assume that n ≥ 6, and so d1 (An ) = n − 1 ≥ 5. This implies that m ≥ 6 since d1 (S) ≥ d1 (An ) ≥ 5, by Lemma 11(i). We conclude that d1 (Am ) = m − 1 and d1 (An ) = n − 1. Applying Lemma 11(i) again, we have d1 (Am ) ≥ d1 (An ), and then m − 1 ≥ n − 1. Thus m ≥ n. If m ≤ 17, then 6 ≤ n ≤ m ≤ 17. Using [3] and [4], we can check that the theorem holds in any of these cases. Therefore, we assume from now on that m ≥ 18. It follows that 17 ∈ π(Am ), and hence by Lemma 11(ii), we have 17 ∈ π(An ). As |An | = n!/2, and 17 ∈ π(An ), we conclude that n ≥ 17 and thus m ≥ n ≥ 17. Suppose that d1 (Am ) ≥ d2 (An ). By Corollary 5, we have d1 (Am ) = m − 1 and d2 (An ) = n(n − 3)/2. As d1 (Am ) ≥ d2 (An ), it follows that m − 1 ≥ n(n − 3)/2, hence m ≥ (n−1)(n−2)/2. Since n ≥ 17, (n−1)(n−2)/2−2n = n(n−7)/2+1 > 0, so that m ≥ (n − 1)(n − 2)/2 > 2n, and thus n < m/2 < m. By Lemma 6, there exists a prime p such that m/2 < p ≤ m. Then p ∈ π(Am ) but p ∈ π(An ), as p > n, 6 HUNG P. TONG-VIET Table 1. Sporadic 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) p(S) p(S) d1 (S) ˜ d2 (S) d3 (S) 11 7 10 11 16 11 7 11 16 45 19 13 56 76 77 11 13 21 45 55 7 11 14 21 36 23 13 22 45 230 11 13 22 77 154 19 7 85 323 324 23 13 23 45 231 11 13 22 231 252 17 11 51 153 680 29 11 378 406 783 13 17 143 364 780 31 13 10944 13376 25916 23 13 23 253 275 23 13 23 253 275 13 17 78 429 1001 19 13 133 760 3344 67 13 2480 45694 48174 31 11 248 4123 27000 23 19 782 3588 5083 23 17 276 299 1771 43 13 1333 299367 887778 29 19 8671 57477 249458 47 29 4371 96255 1139374 71 37 196883 21296876 842609326 13 7 26 27 78 which contradicts Lemma 11(ii). Thus d1 (Am ) < d2 (An ), and so d1 (Am ) = d1 (An ), or equivalently, m = n. Therefore S ∼ An as required. = (ii) S is a sporadic simple group or the Tits group. The character degrees of An , where 5 ≤ n ≤ 17 can be found in [3] and [4]. The character degrees of sporadic simple groups are also available in [3]. It is routine to check that the theorem holds in any of these cases. Therefore, we can assume that n ≥ 18. (1) S ∈ {M11 , M12 , J1 , M22 , J2 , M23 , HS, M24 , 2 F4 (2) }. As n ≥ 18, by Corollary 5, we have d2 (An ) = n(n−3)/2 ≥ 135. From Table 1 we see that d2 (S) < 135. Thus d2 (S) < d2 (An ) for all n ≥ 18. It follows from Lemma 11(i) that cd(S) cd(An ). (2) S = M cL. As n ≥ 18, by Corollary 5, we have d4 (An ) = n(n − 1)(n − 5)/6 ≥ 663. From Table 1, d3 (S) = 252 < d4 (An ). It follows from Lemma 11(i) that d3 (S) = d3 (An ), and hence di (S) = di (An ) for all 1 ≤ i ≤ 3. However, this cannot happen since d2 (S) + 1 = d3 (S), while d2 (An ) + 1 = d3 (An ) for all n ≥ 15. Thus cd(S) cd(An ). CHARACTER DEGREES 7 (3) S ∈ {J3 , He, F i22 , HN }. We have d1 (S) < 135 ≤ d2 (An ). Thus d1 (S) = d1 (An ) = n − 1, hence n = d1 (S) + 1. However, we can check that for these values of n, d2 (An ) > d2 (S) > d1 (An ). Therefore, cd(S) cd(An ). (4) S = Ru. As 29 ∈ π(S), by Lemma 11(ii), we deduce that n ≥ 29. By Corollary 5, we have d4 (An ) = n(n − 1)(n − 5)/6 ≥ 3248 > d3 (S). It follows that d3 (S) = d3 (An ) and hence di (S) = di (An ), for 1 ≤ i ≤ 3. However, this cannot happen since d2 (S) + 1 = d3 (S). Thus cd(S) cd(An ). (5) S = Suz. We have d2 (S) = 364 < 663 ≤ d4 (An ) as n ≥ 18. Hence d2 (S) = d3 (An ) or d2 (S) = d2 (An ). If the first case holds then (n − 1)(n − 2)/2 = 364. However, this equation has no integer solutions. Thus d2 (S) = d2 (An ), and hence d1 (S) = d1 (An ). Then n − 1 = 143 or n = 144. But then d2 (An ) = 10152 > 364 = d2 (S), which is a contradiction. Thus cd(S) cd(An ). (6) S = O N. As 31 ∈ π(S), by Lemma 11(ii), we have n ≥ 31. By Corollary 5, we have d7 (An ) = n(n−1)(n−2)(n−7)/24 ≥ 26970. Then d4 (S) = 26752 < d7 (An ), so that d4 (S) ∈ {d4 (An ), d5 (An ), d6 (An )}. However, we can check that these equations have no integer solutions. Hence cd(S) cd(An ). (7) S ∈ {Co3 , Co2 }. As 23 ∈ π(S), by Lemma 11(ii), we deduce that n ≥ 23. Now by Corollary 5, d4 (An ) ≥ 1518 > d3 (S), hence d3 (S) = d3 (An ). It follows that di (S) = di (An ) for all 1 ≤ i ≤ 3. However, this cannot happen since d2 (S) + 1 = d3 (S). Thus cd(S) cd(An ). (8) S ∈ {Ly, F i23 , J4 }. These cases can be eliminated in the same way. We will outline the argument here. By Lemma 11(ii), we deduce that n ≥ p(S) > 22. By Corollary 5, d4 (An ) = n(n − 1)(n − 5)/6. Since n ≥ p(S), we can check that d4 (An ) > d1 (S) and so d1 (S) ∈ {d1 (An ), d2 (An ), d3 (An )}. Solving these equations, we get d1 (S) = d1 (An ) or n = d1 (S) + 1. But then d2 (An ) = n(n − 3)/2 > d2 (S). Thus cd(S) cd(An ). (9) S = T h. As 31 ∈ π(S), by Lemma 11(ii), we deduce that n ≥ 31. By Corollary 5, d2 (An ) ≥ 434 > d1 (S), and so d1 (S) = d1 (An ). Solving this equation, we get n = 249. But d2 (An ) = 30627 > d2 (S). Thus cd(S) cd(An ). (10) S = Co1 . As 23 ∈ π(S), by Lemma 11(ii), we deduce that n ≥ 23. By Corollary 5, d4 (An ) ≥ 1518 > d2 (S), and so d2 (S) = d3 (An ) or d2 (S) = d2 (An ). As the first equation has no integer solutions, we must have d2 (S) = d2 (An ) and hence d1 (S) = d1 (An ). The latter equation implies n = 277. Then d2 (An ) = n(n − 3)/2 = 37949 > 299 = d2 (S), a contradiction. Thus cd(S) cd(An ). (11) S ∈ {F i24 , M }. By Lemma 11(ii), we deduce that n ≥ p(S) > 22. By Corollary 5, d7 (An ) = n(n − 1)(n − 2)(n − 7)/24. As n ≥ p(S), we can check that d7 (An ) > d1 (S) so that d1 (S) ∈ {di (An ) | 1 ≤ i ≤ 6}. Solving these equations, we get n = d1 (S) + 1. However d2 (An ) = n(n − 3)/2 > d2 (S). Thus cd(S) cd(An ). (12) S = B. As 47 ∈ π(S), by Lemma 11(ii), we deduce that n ≥ 47. By Corollary 5, d4 (An ) ≥ 15134 > d1 (S), and so d1 (S) ∈ {di (An ) | 1 ≤ i ≤ 3}. Solving these equations, we get n = 4372 and d1 (S) = d1 (An ) or n = 95 and d1 (S) = d3 (An ). If the first case holds then d2 (An ) = 9550634 > d2 (S). If n = 95, then d4 (An ) = 133950. As d2 (S) = 96255, we see that d3 (An ) = 4371 < d2 (S) < d4 (An ). Thus cd(S) cd(An ). (iii) S is a finite simple group of Lie type in characteristic p, and S = 2 F4 (2) . Recall that S contains a prime power character degree which is the degree of the Steinberg character StS , with StS (1) = |S|p , the size of the Sylow p-subgroup of S. We consider the following cases: 8 HUNG P. TONG-VIET (1) n = 5. Then cd(A5 ) = {1, 3, 4, 5}. Since cd(S) ⊆ cd(A5 ), it follows that every character degree of S is a prime power, hence by [15, Corollary], S ∼ A5 ∼ L2 (4) ∼ = = = L2 (5) or S ∼ L2 (8). If the first case holds then we are done. Otherwise, S ∼ L2 (8). = = However, cd(L2 (8)) = {1, 7, 8, 9} cd(A5 ). Hence the theorem holds in this case. (2) n = 6. We have cd(A6 ) = {1, 5, 8, 9, 10} and π(A6 ) = {2, 3, 5}. It follows from Burnside’s pa q b Theorem ([9, Theorem 3.10]) and Lemma 11(ii) that |π(S)| = 3. By [6, Theorem 1], S is isomorphic to one of the following simple groups: A5 , L3 (2) ∼ L2 (7), A6 , L2 (8), L2 (17), L3 (3), U3 (3), or U4 (2). = Since π(S) = {2, 3, 5}, it follows that S ∈ {A5 , A6 , U4 (2)}. Obviously cd(A5 ) cd(A6 ) and cd(U4 (2)) cd(A6 ), and so S = A6 ∼ L2 (9). = (3) n = 8. We have π(A8 ) = {2, 3, 5, 7}, and the only prime power character degrees in A8 are 7 and 26 . Thus either p = 7 and |S|7 = 7 or p = 2 and |S|2 = 26 . By Burnside’s pa q b Theorem and Lemma 11(ii) again, we have 3 ≤ |π(S)| ≤ 4. If |π(S)| = 3 then [3] and [6, Theorem 1] imply that cd(S) cd(A8 ). Hence |π(S)| = 4. Then S is isomorphic to one of the groups in [5, Lemma 3]. Assume first that S is of characteristic 7, that is p = 7. As |S|7 = 7, the Lie rank of S must be 1. As the Sylow 7-subgroup of S is of size 7, the only possibility for S is L2 (7). However, |π(L2 (7))| = 3, which is a contradiction as |π(S)| = 4. Thus p = 2 and |S|2 = 26 . Also, by Lemma 11(ii), we have π(S) ⊆ π(A8 ) = {2, 3, 5, 7}. Using these properties of S together with [5, Lemma 3], we deduce that either S = L4 (2) or S = L3 (4). If the latter case holds, then S = L3 (4). However we can check that cd(L3 (4)) cd(A8 ). Thus S = L4 (2) ∼ A8 . = (4) n = 9. We have π(A9 ) = {2, 3, 5, 7}, and the only prime power character degrees in A9 are 23 and 33 . Argue as in the previous case to conclude S is of characteristic 3 with |S|3 = 33 or S is of characteristic 2 and |S|2 = 23 . Moreover π(S) ⊆ π(A9 ) = {2, 3, 5, 7}. Assume first that S is of characteristic 3. It follows from [6, Theorem 1] and [5, Lemma 3] that S = U3 (3). But then cd(U3 (3)) cd(A9 ) by using [3]. Thus S is of characteristic 2 and |S|2 = 23 . It follows from [6, Theorem 1] and [5, Lemma 3] that S = L2 (8) or S = L3 (2) ∼ L2 (7). However we can check = that cd(S) cd(A9 ) in any of these cases. (5) n = 7 or n ≥ 10. We have StS (1) = |S|p ∈ cd(An ). By Lemma 7, we have n − 1 = |S|p . As n ≥ 7, it follows that d1 (An ) = n − 1 = |S|p . By Lemma 8, d1 (S) < |S|p = d1 (An ) which contradicts Lemma 11(i). Thus cd(S) cd(An ). The proof is now complete. Theorem 13. Let S be a simple group and let H be a sporadic simple group or the Tits group. If cd(S) ⊆ cd(H), then S ∼ H. = Proof. Suppose that cd(S) ⊆ cd(H). We consider the following cases: (i) S is a sporadic simple group or the Tits group. This case can be dealt with easily using [3]. (ii) S = An , n ≥ 5. If n ≤ 17, then we can check that the theorem holds by using [3] and [4]. Thus we can assume that n ≥ 18. It follows from Lemma 11(ii) that π(An ) ⊆ π(H). Thus as {7, 11, 13, 17} ⊆ π(An ), it implies that {7, 11, 13, 17} ⊆ π(H). From Table 1, H ∈ {F i23 , F i24 , B, M }. Now by Lemma 11(i), we have d1 (An ) = n − 1 ≥ d1 (H) ≥ 782. Thus n ≥ 783 and hence 73 ∈ π(An ). As above, 73 ∈ π(H) but this cannot happen since the largest prime divisor of a sporadic simple group is 71. Therefore cd(An ) cd(H), for any sporadic simple group H. CHARACTER DEGREES 9 (iii) S is a simple group of Lie type in characteristic p and S = 2 F4 (2) . In this case, H contains a prime power character degree pa , which is the degree of the Steinberg character of S. Now applying Lemma 7, H is one of the following groups: (a) H ∈ {M11 , M12 } and pa ∈ {11, 16}; (b) H = 2 F4 (2) and pa ∈ {27, 211 }; (c) H ∈ {M24 , Co2 , Co3 } and pa = 23; If H is one of the groups in (a) and (b) then |π(H)| = 4. By Burnside’s pa q b Theorem ([9, Theorem 3.10]) and Lemma 11(ii), we have 3 ≤ |π(S)| ≤ 4. By inspecting the lists in [6, Theorem 1] and [5, Lemma 3], we can see that cd(S) cd(H) in any of these cases. Finally, assume that H is one of the groups in (c). As pa = 23, S must be a finite simple group of Lie type in characteristic 23 and has Lie rank 1 since |S|23 = 23. Notice that every simple group of Lie type with Lie rank 1 is one of the following groups: L2 (pm ), U3 (pm ), 2 B2 (22m+1 ), 2 G2 (32m+1 ), (m ≥ 1). Hence S = L2 (23). We have d1 (L2 (23)) = 11 but d1 (H) = 23 > d1 (L2 (23)), which contradicts Lemma 11(i). Therefore cd(S) cd(H). This completes the proof. Proof of Theorem 1. This is done by combining Theorems 12 and 13. Proof of Corollary 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, 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 complete. = Proof of Corollary 3. Assume that G is a group satisfying cd∗ (G) ⊆ cd∗ (H). It follows that |G| ≤ |H| and cd(G) ⊆ cd(H). 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