Symmetric groups are determined by their character degrees more

SYMMETRIC GROUPS ARE DETERMINED BY THEIR CHARACTER DEGREES HUNG P. TONG-VIET Abstract. Let G be a finite group. Let X1 (G) be the first column of the ordinary character table of G. In this paper, we will show that if X1 (G) = X1 (Sn ), then G ∼ Sn . As a consequence, we show that Sn is uniquely determined by = the structure of the complex group algebra CSn . 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. Put ni = χi (1). We say that (n1 , n2 , · · · , nk ) is the degree pattern of G. Let cd(G) = {χ(1) | χ ∈ Irr(G)} be the set of all irreducible character degrees of G. Following [2], let X1 (G) be the first column of the ordinary character table of G. By a suitable re-ordering of the rows in the character table of G, we can see that X1 (G) coincides with the degree pattern (n1 , n2 , · · · , nk ) of G. We also consider X1 (G) as a multiset consisting of character degrees of G counting multiplicities. 2 Since |G| = χ∈Irr(G) χ(1) , the order of G is known given X1 (G). There are examples showing that non-isomorphic groups may have the same character table and so the first column of their character tables coincide. Using the classification of finite simple groups, it is easy to see that non-abelian simple groups are uniquely determined by their character table. It was shown by Nagao [14] that the symmetric groups Sn are also uniquely determined by their character tables. In [16], we know that the alternating group An of degree at least 5, and the sporadic simple groups are uniquely determined by the first column of their character tables. In this paper, we will prove a similar result for the symmetric groups. ∼ Theorem 1.1. Let G be a finite group. If X1 (G) = X1 (Sn ), then G = Sn . This gives a positive answer to [2, Question 126]. Let C be the complex number field and let G be a group. Denote by CG the group algebra of G over C. Let Gi , i = 1, 2, be groups. By Molien’s Theorem ([2, Theorem 2.13]) we know that CG1 ∼ CG2 if and only if X1 (G1 ) = X1 (G2 ). Therefore, knowing the first column = of the character table of a group G is equivalent to knowing the structure of the group algebra CG. It is known that CG allows us to recognize the Frobenius groups or the p-nilpotent groups ([2, Corollaries 10.11, and 10.27]). Now Theorem 1.1 yields. Corollary 1.2. Let G be a group. If CG ∼ CSn , then G ∼ Sn . = = Date: April 26, 2011. 2000 Mathematics Subject Classification. Primary 20C15. Key words and phrases. character degrees, symmetric groups. Support from the University of KwaZulu-Natal is acknowledged. 1 2 HUNG P. TONG-VIET We should mention that Brauer’s Problem 1 (see [3]) which asks the following: What are the possible degree patterns of finite groups? Little is known about this problem. Now Corollary 1.2 says that there is exactly one isomorphism type of the group algebra with a degree pattern as that of the symmetric groups. We now outline our argument for the proof of Theorem 1.1. Assume that X1 (G) = X1 (Sn ). We first observe that |G : G | = 2, |G| = n!, k(G) = k(Sn ) and cd(G) = cd(Sn ), where k(G) denotes the number of conjugacy classes of G. The result is trivial when n ≤ 4. Hence we will assume that n ≥ 5. Next we will show that G is perfect, that is G = G , by applying [8, Lemma 12.3]. Choose M ≤ G be a normal subgroup of G so that G /M is a chief factor of G. As |G : G | = 2 and G /M is non-abelian, we deduce that G /M ∼ S k , where S is a non-abelian = simple group and k is at most 2. We proceed to show that G /M must be a simple group that is k = 1. This is done by applying Theorem 3.3. We now deduce that either G/M is an almost simple group with socle G /M or G/M ∼ G /M × Z2 . = We now apply Theorem 3.1 which asserts that if H is an almost simple group and cd(H) ⊆ cd(Sn ), n ≥ 5, then the socle of H must be isomorphic to An , to show that G /M ∼ An . Assume that n = 6. By comparing the orders, G ∼ Sn or G ∼ An ×Z2 . = = = Finally, using the fact that G and Sn have the same number of irreducible characters, we can eliminate the latter case. Thus G must be isomorphic to Sn . In the exceptional case, we have |Out(A6 )| = 4. In this case, G is one of the following groups: A6 × Z2 , P GL2 (9) ∼ A6 .22 , M10 ∼ A6 .23 or S6 . Using [5], we conclude that = = G ∼ S6 . We remark that this argument is based on the Huppert’s method given = in [7]. This method is used to verify the Huppert Conjecture which states that non-abelian simple groups are determined by their sets of character degrees (see [7, 17]). 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)) be the largest prime divisor of the order of G and let ρ(G) = ∪χ∈Irr(G) π(χ(1)) be the set of all primes which divide some 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 λ Let n be a positive integer. We call λ = (λ1 , λ2 , . . . , λr ) a partition of n, written n, provided λi , i = 1, 2, . . . , r are integers, with λ1 ≥ λ2 ≥ · · · ≥ λr > 0 and r ak a1 a2 i=1 λi = n. We collect the same parts together and write λ = ( 1 , 2 , · · · , k ), k with i > i+1 > 0 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. The following result on the minimal character degrees of symmetric groups is due to Rasala [15]. Lemma 2.1. ([15, Result 3]). Let λ be a partition of n. CHARACTER DEGREES OF SYMMETRIC GROUPS 3 (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 )}; (2) d2 (Sn ) = 1 n(n − 3) and λ ∈ {(n − 2, 2), (22 , 1n−4 )}; 2 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 )}; (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 )}; Assume n ≥ 5. Using [5, 6] and Lemma 2.1, we can see that d1 (Sn ) = n − 1 and d2 (Sn ) = n(n − 3)/2 if n = 8 while d2 (S8 ) = 14. Similarly, if n ≥ 6, then d1 (An ) = n − 1 while d1 (A5 ) = 3 (see [16]). The following is well-known. Lemma 2.2. (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 [11, Proposition 5.1]. For 15 ≤ m ≤ 16, the lemma is obvious. The following results on the classification of prime power character degrees of symmetric groups will be used frequently. Lemma 2.3. ([1, Theorem 5.1]). Suppose that Sn possesses a non-trivial irreducible character χ with χ(1) = pd , where p is a prime. Then one of the following holds: (1) n = pd + 1, and χ(1) = pd ; (2) n = 4 and χ(1) = 2; (3) n = 5 and χ(1) = 5; (4) n = 6 and χ(1) ∈ {32 , 24 }; (5) n = 8 and χ(1) = 26 ; (5) n = 9 and χ(1) = 33 ; We refer to [4, 13.8, 13.9] for the classification of unipotent characters and the notion of symbols. Lemma 2.4. Let S be a simple group of Lie type in characteristic p defined over a field of size q. Assume that S = L2 (q), 2 F4 (2) . Then there exist two irreducible characters χi , i = 1, 2, of S such that both χi extend to Aut(S) with 1 < χ1 (1) < χ2 (1) and χ2 (1) = |S|p . In particular, if G is an almost simple group with socle S, where S = L2 (q), 2 F4 (2) , then |S|p > d1 (G). Proof. By the results of Lusztig [9], any unipotent character θ of S has an extension ˜ ˜ θ to the group G1 of inner-diagonal automorphisms of S such that θ and θ have the same inertia group in Aut(S) (see [10, Proposition 2.1]). Moreover, the unipotent characters of G1 remain irreducible upon restriction to S, and these restrictions are all the unipotent characters of S. By [10, Theorem 2.4], all unipotent characters of S extend to their inertia groups in Aut(S). By results of Lusztig, the inertia group of a unipotent character of S is exactly Aut(S) except for several cases explicitly 4 HUNG P. TONG-VIET listed in [10, Theorem 2.5]. Thus we can choose χ2 to be the Steinberg character of S and χ1 to be any unipotent character of S such that χ1 does not appear in [10, Theorem 2.5] and 1 < χ1 (1) < χ2 (1) = |S|p . Assume S is of type An−1 with n ≥ 3. We have G1 = (An−1 )ad (q) = P GLn (q). By [4, 13.8], the unipotent characters of G1 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 StS (1) = |S|p = q n(n−1)/2 , and n ≥ 3, we have StS (1) > χα (1) > 1. Assume S is of type 2 An−1 , where n ≥ 3. Then G1 = (2 An−1 )ad (q 2 ) = P Un (q). By [4, 13.8], the unipotent characters of G1 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 StS (1) = |S|p = q n(n−1)/2 , and n ≥ 3, we have StS (1) > χα (1) > 1. Assume next that S is of type Bn , or Cn where n ≥ 2 and S = S4 (2). Then G1 = (Bn )ad (q) = SO2n+1 (q) or G1 = (Cn )ad (q) = P CSp2n (q). By [4, 13.8], G1 has a unipotent characters χα labeled by the symbol α= 0 1 n − 2 with χα (1) = (q n − 1)(q n − q)/(2(q + 1)). Since |S|p = q n and (n, q) = (2, 2), we see that |S|p > χα (1) > 1. Assume S is of type Dn (q), n ≥ 4. Then G1 = (Dn )ad (q) = P (CO2n (q)0 ). By [4, 13.8], G1 has a unipotent character χα labeled by the symbol α= n−1 1 with χα (1) = (q n − 1)(q n−1 + q)/(q 2 − 1). Since |S|p = q n(n−1) , we see that |S|p > χα (1) > 1. − Assume S is of type 2 Dn (q 2 ), n ≥ 4. Then G1 = (2 Dn )ad (q 2 ) = P (CO2n (q)0 ) and G1 has a unipotent character χα labeled by the symbol α= 1 − n−1 with χα (1) = (q n + 1)(q n−1 − q)/(q 2 − 1). Since |S|p = q n(n−1) , |S|p > χα (1) > 1. For the simple groups of exceptional type, we will use the explicit list of unipotent characters in [4, 13.9]. Assume S is of type G2 (q). Then S has a unipotent character labeled by θ2,1 with degree qΦ2 Φ3 /6. As G2 (2) ∼ U3 (3).2 is not simple, we can assume that q ≥ 3. = 2 Since |S|p = q 6 , we have qΦ2 Φ3 /6 < q 6 so that 1 < θ2,1 (1) < |S|p . 2 Assume S is of type 3 D4 (q 3 ). Then S has a unipotent character labeled by θ1,3 with degree qΦ12 . Since |S|p = q 12 , we have 1 < θ1,3 (1) < |S|p . Assume S is of type F4 (q). Then S has a unipotent character labeled by θ9,2 with degree q 2 Φ2 Φ2 Φ12 . Since |S|p = q 24 , we have 1 < θ9,2 (1) < |S|p . 3 6 Assume S is of type E6 (q). Then S has a unipotent character labeled by θ6,1 with degree qΦ8 Φ9 . Since |S|p = q 36 , we have 1 < θ6,1 (1) < |S|p . Assume S is of type 2 E6 (q 2 ). Then S has a unipotent character labeled by θ2,4 with degree qΦ8 Φ18 . Since |S|p = q 36 , we have 1 < θ2,4 (1) < |S|p . Assume S is of type E7 (q). Then S has a unipotent character labeled by θ7,1 with degree qΦ7 Φ12 Φ14 . Since |S|p = q 63 , we have 1 < θ7,1 (1) < |S|p . CHARACTER DEGREES OF SYMMETRIC GROUPS 5 Assume S is of type E8 (q). Then S has a unipotent character labeled by θ8,1 with degree qΦ2 Φ8 Φ12 Φ20 Φ24 . Since |S|p = q 120 , we have 1 < θ8,1 (1) < |S|p . 4 Assume S is of type 2 B2 (q 2 ), where q 2 = 22m+1 and √ ≥ 1. Then S has a m unipotent character labeled by 2 B2 [a] with degree qΦ1 Φ2 / 2. Since |S|p = q 4 , we have 1 < 2 B2 [a](1) < |S|p . Assume S is of type 2 G2 (q 2 ), where q 2 = √ 2m+1 and m ≥ 1. Then S has a 3 unipotent character θ with degree qΦ1 Φ2 Φ4 / 3. Since |S|p = q 6 , we have 1 < θ(1) < |S|p . Assume S is of type 2 F4 (q 2 ), where q 2 = 22m+1 and m ≥ √ Then S has a 1. unipotent character labeled by 2 B2 [a] with degree qΦ1 Φ2 Φ2 Φ6 / 2. Since |S|p = 4 q 24 , we have 1 < 2 B2 [a](1) < |S|p . This finishes the proof of the first assertion. Now assume G is an almost simple group with socle S, where S = L2 (q), 2 F4 (2) . Let χi ∈ Irr(S), i = 1, 2, be irreducible characters of S obtained above. As both χi extend to Aut(S) and S G ≤ Aut(S), we deduce that each χi extends to ψi ∈ Irr(G) with ψi (1) = χi (1) for i = 1, 2. Thus ψ2 (1) = |S|p > ψ1 (1) = χ1 (1) > 1 so that |S|p > d1 (G) as required. The proof is now complete. Lemma 2.5. Let G be an almost simple group with socle S = L2 (q), where q = pf ≥ 7. If p = 3, then G contains an irreducible character of degree q + δ where q ≡ δ (mod 3). If p = 3, then G contains an irreducible character of degree (q + )/2 or q + , where q ≡ (mod 4). Proof. Assume first that p = 3. It follows from the proof of [13, Proposition 3.7] that the irreducible character of S corresponding to a semisimple element of order 3 in the dual group SL2 (q), of degree q + δ, where q ≡ δ (mod 3), is extendible to Aut(S), and hence G contains an irreducible character of degree q + δ as required. Note that this result fails for L2 (3f ). Now we assume that q = 3f . Observe that L2 (q) always contains irreducible characters χa , χb of degree q − 1 and q + 1, respectively, which are extendible to P GL2 (q). Thus if L2 (q) G ≤ P GL2 (q) then G possesses characters of degree q±1. Now assume that G ≤ P GL2 (q). Recall that the only outer automorphisms of L2 (q) are the diagonal automorphisms and the field automorphisms. It is well-known that S contains two irreducible characters χ± of degree (q + )/2, where q ≡ (mod 4). Now the diagonal automorphisms of S fuse these two irreducible characters while the field automorphisms fix those two. Let θ ∈ {χ+ , χ− }. Then θ ∈ Irr(S) and IAut(S) (θ) = P ΓL2 (q). Thus if S G ≤ P ΓL2 (q), then θ is G-invariant and so θ extends to G as G/S is cyclic. Hence G has an irreducible character of degree (q + )/2. Finally, assume P GL2 (q) G ≤ Aut(L2 (q)). Then the irreducible character µ of P GL2 (q) lying over θ is of degree q + . We see that µ is G-invariant and hence it extends to G, as G/P GL2 (q) is cyclic. Therefore G contains an irreducible character of degree q + . The proof is now complete. Corollary 2.6. If G is an almost simple group then ρ(G) = π(G). Proof. Observe first that for any χ ∈ Irr(G), we have χ(1) divides |G| by [8, Theorem 3.11]. Hence ρ(G) ⊆ π(G). As G is almost simple, it has no normal abelian Sylow p-subgroup, so that by the Ito-Michler Theorem [12], every prime divisor of G must divide χ(1) for some χ ∈ Irr(G), and thus π(G) ⊆ ρ(G). Hence ρ(G) = π(G) as required. Lemma 2.7. Let G and H be groups. Suppose that cd(G) ⊆ cd(H). Then 6 HUNG P. TONG-VIET Table 1. Sporadic simple groups and their automorphism groups S p(S) M11 11 M12 11 M12 .2 11 J1 19 M22 11 M22 .2 11 J2 7 J2 .2 7 M23 23 HS 11 HS.2 11 J3 19 J3 .2 19 M24 23 M cL 11 M cL.2 11 He 17 He.2 17 Ru 29 Suz 13 Suz.2 13 ON 31 O N.2 31 Co3 23 Co2 23 F i22 13 F i22 .2 13 HN 19 HN.2 19 Ly 67 Th 31 F i23 23 Co1 23 J4 43 F i24 29 F i24 .2 29 B 47 M 71 2 F4 (2) 13 2 F4 (2) .2 13 d1 (S) 10 11 22 56 21 21 14 28 22 22 22 85 170 23 22 22 51 102 378 143 143 10944 10944 23 23 78 78 133 266 2480 248 782 276 1333 8671 8671 4371 196883 26 27 d2 (S) 11 16 32 76 45 45 21 36 45 77 77 323 324 45 231 231 153 306 406 364 364 13376 26752 253 253 429 429 760 760 45694 4123 3588 299 299367 57477 57477 96255 21296876 27 52 d3 (S) 16 45 45 77 55 55 36 42 230 154 154 324 646 231 252 252 680 680 783 780 780 25916 37696 275 275 1001 1001 3344 3344 48174 27000 5083 1771 887778 249458 249458 1139374 842609326 78 78 (i) di (G) ≥ di (H), for all i ≥ 1; (ii) If G is almost simple then π(G) ⊆ π(H). Proof. (i) is obvious. (ii) follows from Corollary 2.6 as ρ(G) ⊆ ρ(H) ⊆ π(H). CHARACTER DEGREES OF SYMMETRIC GROUPS 7 3. Proofs of the main results Theorem 3.1. Let G be an almost simple group with socle S and let n ≥ 5 be an integer. If cd(G) ⊆ cd(Sn ) then S ∼ An . = Proof. Using the classification of finite simple groups, S is an alternating group of degree at least 5, a finite simple group of Lie type or one of the 26 sporadic groups. We will treat the Tits group as a sporadic group rather than a group of Lie type. Step 1. Eliminate simple groups of Lie type. By way of contradiction, we assume that S is a simple group of Lie type in characteristic p and cd(G) ⊆ cd(Sn ) with n ≥ 5. By the isomorphisms L2 (4) ∼ L2 (5) ∼ A5 , L2 (9) ∼ A6 and L4 (2) ∼ A8 , we = = = = can assume that S is not one of the groups listed above nor the Tits group. It is well known that the Steinberg character of S of degree |S|p extends to χ ∈ Irr(G) and hence χ(1) = |S|p is a non-trivial power of p. Assume first that |S|p is not the minimal character degree of Sn , that is |S|p > n − 1. It follows from Lemma 2.3 that n ∈ {5, 6, 8, 9} and |S|p = 5, |S|p ∈ {32 , 23 }, |S|p = 26 , |S|p = 33 , respectively. It is routine to check that these cases cannot happen. Thus |S|p = n − 1 = d1 (Sn ). Now Lemma 2.4 will provide a contradiction unless S = L2 (q), where q = pf ≥ 4. Assume that S = L2 (q) and thus q ≥ 7. As |S|p = q = d1 (Sn ), by Lemma 2.5, G must contain an irreducible character of degree q + 1 and hence q + 1 ∈ cd(Sn ). Since S = L4 (2) ∼ A8 , we have d2 (Sn ) = n(n − 3)/2. We have n − 1 = q and hence = as q ≥ 7, we obtain d2 (Sn ) = n(n − 3)/2 = (q + 1)(q − 2)/2 > q + 1 > q = d1 (Sn ), which contradicts Lemma 2.7(i). This finishes the proof of Step 1. Step 2. Eliminate sporadic simple groups and the Tits group. By way of contradiction, we assume that S is a simple sporadic group or the Tits group and that cd(G) ⊆ cd(Sn ) with n ≥ 5. The character degrees of Sn , where 5 ≤ n ≤ 31 can be found in [6]. Moreover the character degrees of sporadic simple groups and the Tits group together with their automorphism groups are also available in [6]. It is routine to check that cd(G) cd(Sn ) for any 5 ≤ n ≤ 31 and any almost simple group G with socle S, where S is a sporadic simple group or the Tits group. Thus we can assume that n ≥ 32. It follows from Lemma 2.1 that d2 (Sn ) = n(n − 3)/2 ≥ d2 (S32 ) = 464. By Lemma 2.7(i) and Table 1, we only need to consider the following cases: S ∈ {O N, HN, Ly, T h, F i23 , J4 , F i24 , B, M }. (1) S = O N. In this case, we have |Out(S)| = 2 so that either G = S or G = S.2. Assume first that G = S = O N. Then d9 (O N ) = 58653 and since n ≥ 32, by Lemma 2.1, d9 (Sn ) ≥ 62496 > d9 (O N ), which contradicts Lemma 2.7(i). Now assume G = O N.2. Then d7 (G) = 58653 < 62496 ≤ d9 (Sn ) so that d7 (G) ∈ {d7 (Sn ), d8 (Sn )}. However we can check that these equations cannot hold for any n ≥ 32. Thus cd(G) cd(Sn ). (2) S = HN. Then |Out(S)| = 2 so that G = S or G = S.2. From [5], we have d7 (S) = 16929 and d7 (S.2) = 17556. Observe that d7 (G) < 62496 ≤ d7 (Sn ) so that cd(G) cd(Sn ) by Lemma 2.7(i). (3) S = Ly. Since |Out(S)| = 1, we have G = S so that p(S) = 67 ∈ π(Sn ) by Lemma 2.7(ii), and hence n ≥ 67. As d5 (Ly) = 381766 < 718575 ≤ d7 (Sn ), we deduce that d5 (Ly) ∈ {d5 (Sn ), d6 (Sn )}. However we can check that these equations cannot hold for any n ≥ 67. Thus cd(G) cd(Sn ). (4) S = T h. As Out(S) is trivial, we have G = S. Since d1 (G) = 248 < 464 ≤ d2 (Sn ), it follows from Lemma 2.7(i) that d1 (G) = d1 (Sn ) = n − 1 and hence n = 249. But then d2 (Sn ) = n(n − 3)/2 ≥ 30627 > d2 (T h). Thus cd(G) cd(Sn ). 8 HUNG P. TONG-VIET (5) S = F i23 . As Out(S) is trivial, we have G = S. Since d2 (G) = 3588 < 4464 ≤ d4 (Sn ), it follows from Lemma 2.7(i) that d2 (G) ∈ {d2 (Sn ), d3 (Sn )}. However we can check that these equations cannot hold for any n ≥ 32. Thus cd(G) cd(Sn ). (6) S = J4 . Since |Out(S)| = 1, we have G = S so that p(S) = 43 ∈ π(Sn ) and hence n ≥ 43. As d1 (J4 ) = 1333 < 11438 ≤ d4 (Sn ), we deduce that d1 (J4 ) ∈ {di (Sn ) | i = 1, 2, 3}. Solving these equations, we obtain n = 1334. But then d2 (Sn ) = 887777 > 299367 = d2 (J4 ). Thus cd(G) cd(Sn ). (7) S = F i24 . We have G = S or G = S.2. In both cases, we have d1 (G) = 8671 and d2 (G) = 57477. Observe that d1 (G) = 8671 < 8960 ≤ d6 (Sn ) so that d1 (G) ∈ {di (Sn ) | i = 1, · · · , 5}. Solving these equations, we obtain n = 8672. But then d2 (Sn ) > d2 (G). Thus cd(G) cd(Sn ). (8) S = B. Since |Out(S)| = 1, we have G = S so that p(S) = 47 ∈ π(Sn ) and hence n ≥ 47. As d1 (B) = 4371 < 15134 ≤ d4 (Sn ), we deduce that d1 (B) ∈ {d1 (Sn ), d2 (Sn ), d3 (Sn )}. Solving these equations, we have n = 95 or n = 4372. If the latter case holds then d2 (Sn ) > d2 (B), a contradiction. Thus n = 95. But then d3 (S95 ) = 4371 < d2 (B) < d4 (S95 ) = 133950. Thus cd(G) cd(Sn ). (9) S = M. Since |Out(S)| = 1, we have G = S so that p(S) = 71 ∈ π(Sn ) and hence n ≥ 71. As d1 (M ) = 196883 < 914480 ≤ d7 (Sn ), we deduce that d1 (M ) ∈ {di (Sn ) | i = 1, · · · , 6}. Solving these equations, we obtain n = 196884. But then d2 (Sn ) > 21296876 = d2 (M ). Thus cd(M ) cd(Sn ). Step 3. If S ∼ Am , with m ≥ 5, then m = n. Let λ = (m − 1, 1) be a partition = of m. Since m ≥ 5, λ is not self-conjugate so that the irreducible character χλ of Sm corresponding to λ is still irreducible upon restriction to Am . Note that Aut(Am ) = Sm whenever m = 6 while |Out(A6 )| = 4. Assume first that m = 6. Then G ∈ {Am , Sm } and G contains an irreducible character of degree m − 1. Since cd(G) ⊆ cd(Sn ), we have m − 1 ≥ d1 (Sn ) = n − 1 so that m ≥ n. If m = n then we are done. Hence we assume that m > n ≥ 5. It follows that m ≥ 6 and hence d1 (Am ) = d1 (Sm ) = m − 1 and thus d1 (G) = m − 1 > n − 1 = d1 (Sn ). If m ≤ 17 then 5 ≤ n < m ≤ 17. Using [6], we can check that m = n. So we can assume that m ≥ 18. It follows that 17 ∈ π(G) and so by Lemma 2.7(ii) we have 17 ∈ π(Sn ), which implies that n ≥ 17. Thus 17 ≤ n < m. It follows from Lemma 2.1 that d2 (Sn ) = n(n − 3)/2. Since cd(G) ⊆ cd(Sn ) and d1 (G) > d1 (Sn ), it follows that d1 (G) ≥ d2 (Sn ). Then m − 1 ≥ n(n − 3)/2. Since n ≥ 17, we have m − 2n ≥ n(n − 3)/2 + 1 − 2n = n(n − 7)/2 + 1 > 0 so that m > 2n. Therefore n < m/2 < m. By Lemma 2.2, there exists a prime p such that m/2 ≤ p < m. It follows that p ∈ π(G) but p ∈ π(Sn ) since p > n, which contradicts Lemma 2.7(ii). Thus S ∼ An whenever m ≥ 5, m = 6. Now assume that m = 6, and A6 G ≤ Aut(A6 ). = It follows that G ∈ {A6 , A6 .21 ∼ S6 , A6 .22 ∼ P GL2 (9), A6 .23 ∼ M10 , A6 .22 }. We = = = need to show that n = 6. If G ∈ {A6 , S6 }, then G contains a character of degree 5 so that 5 ≥ n − 1 and hence n ≤ 6. As 10 ∈ cd(G) but 10 ∈ cd(S5 ), we conclude that n = 6. If G ∼ P GL2 (9) then {8, 9} ⊆ cd(Sn ). But this cannot happen by = Lemma 2.3. Assume that one of the last two cases holds. Then {9, 16} ⊆ cd(Sn ) so that by Lemma 2.3, n = 6. The proof is now complete. Remark 3.2. Let λ = (k + 1, 1k ) when n = 2k + 1 and λ = (k, 2, 1k−2 ) when n = 2k. Then λ is a self-conjugate partition of n. We conjecture that χλ (1)/2 ∈ cd(An ) − cd(Sn ) and χλ (1) ∈ cd(Sn ) − cd(An ). If this conjecture is true then, in the situation of Theorem 3.1, we deduce that G ∼ Sn or G ∼ M10 and n = 6. This = = result will be useful in studying the Huppert Conjecture for alternating groups. CHARACTER DEGREES OF SYMMETRIC GROUPS 9 Theorem 3.3. Let G be a group. Assume that |G : G | = 2 and that G ∼ S 2 is = a unique minimal normal subgroup of G, where S is a non-abelian simple group. Then cd(G) cd(Sn ) for any n ≥ 5. Proof. By way of contradiction, assume that cd(G) ⊆ cd(Sn ). Let α ∈ Irr(S) with α(1) > 1 and put θ = α × 1 ∈ Irr(G ). Observe that θ is not G-invariant so that IG (θ) = G hence θG ∈ Irr(G) and so θG (1) = 2α(1) ∈ cd(Sn ). On the other hand, if ϕ = α × α ∈ Irr(G ) then ϕ is G-invariant and since G/G is cyclic, we deduce that ϕ extends to ψ ∈ Irr(G), so that ψ(1) = α(1)2 ∈ cd(Sn ). We conclude that if a ∈ cd(S) − {1} then 2a, a2 ∈ cd(Sn ). Let r ∈ π(S). By Corollary 2.6, r|a for some a ∈ cd(S) − {1}. Since a2 ∈ cd(Sn ), by [8, Theorem 3.11] we have a2 |n!. Thus r2 |n! so that n ≥ 2r. Using the classification of finite simple groups, we consider the following cases. Case S = Am , with m ≥ 5. As m ≥ 5, it follows from the first paragraph that n ≥ 10. Assume first that m ∈ {5, 6, 8, 9, 10}. Observe that m − 1 ∈ cd(S) and hence 2(m − 1), (m − 1)2 ∈ cd(Sn ). For these values of m, we see that (m − 1)2 is a prime power and so by Lemma 2.3, as n ≥ 10, we have d1 (Sn ) = n − 1 = (m − 1)2 . As m ≥ 5, we obtain d1 (Sn ) = (m − 1)2 > 2(m − 1) > 1, which is a contradiction since 2(m − 1) ∈ cd(Sn ). Now assume that m = 7. As above, we have n ≥ 14. As 6 ∈ cd(S), we obtain 6.2 = 12 ∈ cd(Sn ) and so 12 ≥ d1 (Sn ) = n − 1 which implies n ≤ 13, a contradiction. Thus we can assume that m ≥ 11 and hence n ≥ 22. We have m − 1 ∈ cd(S) so that 2(m − 1) and (m − 1)2 are both in cd(Sn ). Similarly, by Lemma 2.1, we have m(m − 3)/2, (m − 1)(m − 2)/2 ∈ cd(S) and so m(m − 3), (m − 1)(m − 2) ∈ cd(Sn ). We will show that m < n. By way of contradiction, assume that m ≥ n. As n ≥ 22, by Lemma 2.2, there exists a prime r such that n/2 < r ≤ n. It follows that the largest power of r dividing n! is r. Since r ≤ n ≤ m, we deduce that r ∈ π(Am ) and so r2 |n!, which is a contradiction. Thus m < n. Observer that 1 < 2(m − 1) < m(m − 3) < (m − 1)(m − 2) < (m − 1)2 , since m ≥ 11. Thus (m − 1)2 ≥ d4 (Sn ) = n(n − 1)(n − 5)/6 by Lemma 2.1. Combining with the fact that m < n, we obtain n(n − 1)(n − 5)/6 ≤ (n − 1)2 so that n(n − 5) ≤ 6(n − 1) or equivalently n(n − 11) + 6 ≤ 0, which is impossible as n ≥ 22. Case S is a finite simple group of Lie type in characteristic p, with S = 2 F4 (2) . Since L2 (4) ∼ L2 (5) ∼ A5 , we can assume that S ∼ L2 (4). Let St be the Steinberg = = = character of S. We can check that St(1) = |S|p ≥ 5. Since St(1) ∈ cd(S), we obtain 2St(1) ∈ cd(Sn ) and St(1)2 ∈ cd(Sn ). By Lemma 2.3, assume first that n − 1 = St(1)2 . Then d1 (Sn ) = St(1)2 > 2St(1), which is a contradiction. Now assume that St(1)2 = n − 1. By Lemma 2.3, n ∈ {5, 6, 8, 9}. Since St(1)2 is an even prime power, one of the following cases holds: n = 6, St(1)2 ∈ {32 , 24 } or n = 8, St(1)2 = 26 . Assume first that n = 6. Then St(1) ∈ {3, 22 } which implies that St(1) ≤ 4, a contradiction as S = L2 (4). Finally, assume that n = 8. Then St(1) = 23 = 8. However 2St(1) = 24 ∈ cd(S8 ), a contradiction. Case S is a sporadic simple group or the Tits group. Recall that p(S) is the largest prime divisor of S. We have n ≥ 2p(S). (1) S ∈ {M11 , M12 , J1 , M22 , J2 , M23 , HS, J3 , M24 , He, Ru, Co3 , Co2 , Co1 , 2 F4 (2) }. These cases can be eliminated as follows: Since n ≥ 2p(S) ≥ 14, we have d2 (Sn ) = n(n − 3)/2 ≥ p(S)(2p(S) − 3). Next observe that 2di (S) ∈ cd(G) ⊆ cd(Sn ), for i = 1, 2 and that 1 < 2d1 (S) < 2d2 (S). In each case, we have p(S)(2p(S) − 3) > 2d2 (S) and hence d2 (Sn ) > 2d2 (S) > 2d1 (S) > 1, which contradicts Lemma 2.7(i). 10 HUNG P. TONG-VIET (2) S ∈ {M cL, Suz, F i22 , HN, Ly, T h, J4 , B}. Since n ≥ 2p(S) ≥ 14, we have d2 (Sn ) = n(n−3)/2 ≥ p(S)(2p(S)−3). We have d2 (Sn ) ≥ p(S)(2p(S)−3) > 2d1 (S) and so 2d1 (S) = d1 (Sn ) = n − 1 so that n = 2d1 (S) + 1. But then d2 (Sn ) = n(n − 3)/2 = (d1 (S) − 1)(2d1 (S) + 1) > 2d2 (S) > 2d1 (S) > 1, which contradicts Lemma 2.7(i) as 2di (S) ∈ cd(G) ⊆ cd(Sn ), where i = 1, 2. (3) S = O N. Then p(S) = 31. We have n ≥ 2p(S) = 62 and so by Lemma 2.1, d7 (Sn ) ≥ 520025. As d8 (S) = 58311, we have 2d8 (S) = 116622 ∈ cd(Sn ). Note that 2di (S) ∈ cd(Sn ) for i = 1, 2, · · · , 8 and so 2di (S) ≥ di (Sn ) for all 1 ≤ i ≤ 8. As d7 (Sn ) ≥ 520025 > 116622 = 2d8 (S), we get a contradiction. (4) If S = F i23 then p(S) = 23. We have n ≥ 2p(S) = 46 and so by Lemma 2.1, d4 (Sn ) ≥ 14145. As d2 (S) = 3588, we obtain 2d2 (S) = 7176 ∈ cd(Sn ). Since d4 (Sn ) > 7176 > 2d1 (S), we must have 7176 ∈ {d2 (Sn ), d3 (Sn )}. However, we can check that these cases cannot happen. (5) S = F i24 . Then p(S) = 29 and n ≥ 2p(S) = 58 so that by Lemma 2.1, d4 (Sn ) ≥ 29203. As {8671, 57477} ⊆ cd(S), we obtain {17342, 114954} ⊆ cd(Sn ). Since d4 (Sn ) > 17342, we have 17342 ∈ {d1 (Sn ), d2 (Sn ), d3 (Sn )}. It follows that 17342 ≤ (n − 1)(n − 2)/2 and hence n ≥ 188. But then d2 (Sn ) ≥ 17390 > 17342. Thus d1 (Sn ) = n − 1 = 17342 hence n = 17343 and so d2 (Sn ) ≥ 150363810 > 114954, a contradiction. (6) S = M. Then p(S) = 71 and n ≥ 2p(S) = 142. By Lemma 2.1, d4 (Sn ) ≥ 457169. As {196883, 21296876} ⊆ cd(S), we obtain {393766, 42593752} ⊆ cd(Sn ). Since d4 (Sn ) > 393766, we have 393766 ∈ {d1 (Sn ), d2 (Sn ), d3 (Sn )}. It follows that 393766 ≤ (n − 1)(n − 2)/2 and hence n ≥ 889. As d2 (Sn ) ≥ 393827 > 393766, we have d1 (Sn ) = n − 1 = 393766 hence n = 393767 and so d2 (Sn ) ≥ 77525634494 > 42593752 > 393766, a contradiction. The proof is complete. Proof of Theorem 1.1. Suppose that X1 (G) = X1 (Sn ). It follows that |G| = |Sn | = n!, |G : G | = 2, k(G) = k(Sn ) and cd(G) = cd(Sn ). For n ≤ 3, the result is trivial. If n = 4, then the result follows from [2, Chapter 17, Exercise 2]. Thus from now on, we assume that n ≥ 5. We first show that G = G . By way of contradiction, assume that G < G . Let N ≤ G be a normal subgroup of G maximal such that G/N is solvable and G /N is the unique minimal normal subgroup of G/N. By [8, Lemma 12.3], all non-linear irreducible characters of G/N have equal degree f and either G/N is a p-group, Z(G/N ) is cyclic and G/N/Z(G/N ) is elementary abelian of order f 2 or G/N is a Frobenius group with an abelian Frobenius complement of order f, and G /N is the Frobenius kernel and is an elementary abelian p-group. Assume first that G/N is a p-group. As G/N/Z(G/N ) is abelian, we have G /N ≤ Z(G/N ). Since |G/N : G /N | = 2 and G/N is non-abelian, we deduce that G /N = Z(G/N ) and so G/N/Z(G/N ) is a cyclic group of order 2, which is a contradiction as G/N/Z(G/N ) is elementary abelian of order f 2 . Thus the second situation holds. It follows that f = |G/N : G /N | = |G : G | = 2. Therefore 2 ∈ cd(G) = cd(Sn ), which is impossible as the minimal non-trivial irreducible character degree of Sn is n − 1 ≥ 4 as n ≥ 5. Let M ≤ G be a normal subgroup of G so that G /M is a chief factor of G and so G /M ∼ S k , where S is a non-abelian simple group where k ≥ 1. Let = C/M = CG/M (G /M ). Then M ≤ C G. Assume first that C = M. Then G /M is the unique minimal normal subgroup of G/M. Since |G/M : G /M | = |G : G | = 2, we deduce that k is at most 2. However CHARACTER DEGREES OF SYMMETRIC GROUPS 11 k cannot be 2 by Theorem 3.3. Thus k = 1 and so G/M is an almost simple group with socle G /M and cd(G/M ) ⊆ cd(Sn ). By Theorem 3.1, we have G /M ∼ An . It = follows that |G/M | = 2|G /M | = n! = |Sn | and so M = 1 as |G| = |Sn |. Thus G is an almost simple group with socle An and |G| = n!. If n = 6 then as Aut(An ) = Sn we deduce that G ∼ Sn . Now assume that n = 6. Then G is isomorphic to one of = the following groups S6 ∼ A6 .21 , P GL2 (9) ∼ A6 .22 and M10 ∼ A6 .23 . Using [5], we = = = can see that G must be isomorphic to S6 . Finally assume that C = M. It follows that C/M is a non-trivial normal subgroup of G/M and so C/M ∩ G /M is trivial. Thus G < G C ≤ G. Since |G : G | = 2, we deduce that G = G C and hence G/M = G /M × C/M, where C/M is a cyclic subgroup of order 2. Thus cd(G /M ) = cd(G/M ) ⊆ cd(Sn ). Applying Theorem 3.1 again, we obtain G /M ∼ An , and hence G/M ∼ An ×Z2 . By comparing the orders, = = we deduce as in previous case that M = 1 and so G ∼ An × Z2 . We now show that = this case cannot happen. In fact, we have k(G) = 2k(An ), and hence it suffices to show that k(Sn ) < 2k(An ). Let λ be a partition of n and denote by χλ the irreducible character of Sn associated to λ. If λ is not self-conjugate, that is λ = λ , where λ denotes the conjugate of λ, then (χλ )An = (χλ )An ∈ Irr(An ). Otherwise, (χλ )An = χλ+ + χλ− , where χλ+ , χλ− ∈ Irr(An ) are two non-equivalent irreducible characters of the same degree. Let ps (n) be the number of self-conjugate partitions of n. Then k(An ) = (k(Sn ) − ps (n))/2 + 2ps (n). Hence k(Sn ) = 2k(An ) − 3ps (n). So it suffices to show that ps (n) ≥ 1 for n ≥ 5. 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