On Huppert's Conjecture for the Conway and Fischer Families of Sporadic Simple Groups

ON HUPPERT’S CONJECTURE FOR THE CONWAY AND FISCHER FAMILIES OF SPORADIC SIMPLE GROUPS

S. H. ALAVI, A. DANESHKAH, H. P. TONG-VIET and T. P. WAKEFIELD

(April 2, 2012)

Abstract

Let G denote a finite group and cd(G) the set of irreducible character degrees of G. Huppert conjectured that if H is a finite nonabelian simple group such that cd(G) = cd(H), then G =∼ H × A, where A is an abelian group. He verified the conjecture for many of the sporadic simple groups and we complete its verification for the remainder.

Keywords and phrases: Character degrees; Sporadic simple groups; Huppert’s Conjecture.

1. Introduction

Let G be a ﬁnite group, Irr(G) the set of irreducible characters of G, and denote the set of character degrees of G by cd(G) = {χ(1) : χ ∈ Irr(G)}. This set can be used to gain information about the structure of G. For example, it can be used to determine if G is abelian. Unfortunately, it does not completely determine the structure of G. It is possible for non- isomorphic groups to have the same set of character degrees. For example, the non-isomorphic groups D8 and Q8 not only have the same set of character degrees, but also share the same character table. The character degree set cannot be used to distinguish between solvable and nilpotent groups. For example, if G is either Q8 or S3, then cd(G) = {1, 2}. However, in the late 1990s, Bertram Huppert [6] posed a conjecture which, if true, would sharpen the connection between the character degree set of a nonabelian simple group and the structure of the group.

Huppert’s Conjecture. Let G be a ﬁnite group and H a ﬁnite nonabelian simple group such that the sets of character degrees of G and H are the same. Then G =∼ H × A, where A is an abelian group.

c 0 Australian Mathematical Society 0263-6115/2000 $A2.00 + 0.00

1 The conjecture asserts that the nonabelian simple groups are essentially characterized by the set of their character degrees. It does not extend to solvable groups. Unfortunately, it is not possible to establish the conjecture for all nonabelian simple groups with a single argument. In addition to verifying this conjecture for many of the simple groups of Lie type, Huppert also verified it for many of the sporadic simple groups. The sporadic simple groups not considered by him are the Conway groups, Fischer groups, Mon- ster, and Baby Monster. The conjecture for the Baby Monster, Monster and the Fischer group F i23 was established in [1, 8]. Huppert’s proofs rely upon establishing five steps to achieve the result. Steps 2 and 3 are the most delicate and require the most specific and technical arguments.

1. Show G0 = G00. Hence if G0/M is a chief factor of G, then G0/M =∼ Sk, where S is a nonabelian simple group.

2. Identify H as a chief factor G0/M of G.

3. Show that if θ ∈ Irr(M) and θ(1) = 1, then θg = θ for all g ∈ G0, (i.e. θ is stable under G0), which implies [M,G0] = M 0.

4. Show that M = 1.

0 0 0 ∼ 0 5. Show that G = G × CG(G ). As G/G = CG(G ) is abelian and G0 =∼ H, Huppert’s Conjecture is veriﬁed. We utilize these steps to verify the conjecture for the remaining sporadic simple groups.

2. Preliminaries

In this section, we present some necessary results to establish the conjecture for the Conway and Fischer families of simple groups. If n is an integer, let π(n) denote the set of all prime divisors of n. If G is a group, we write π(G) instead of π(|G|) to denote the set of all prime divisors of the order of G. If N E G and θ ∈ Irr(N), then the inertia group g Pk of θ in G is IG(θ) = {g ∈ G | θ = θ}. If the character χ = i=1 eiχi, where each χi is an irreducible character of G and ei is a nonnegative integer, then those χi with ei > 0 are called the irreducible constituents of χ. The set of all irreducible constituents of θG is denoted by Irr(G|θ). Other notation is standard.

Lemma 2.1. [5, Theorems 19.5 and 21.3] Suppose N E G and χ ∈ Irr(G).

(a) If χN = θ1 + θ2 + ··· + θk with θi ∈ Irr(N), then k divides |G/N|. In particular, if χ(1) is prime to |G/N|, then χN ∈ Irr(N).

2 (b) (Gallagher’s Theorem) If χN ∈ Irr(N), then χψ ∈ Irr(G) for all ψ ∈ Irr(G/N).

Lemma 2.2. [5, Theorems 19.6 and 21.2] Suppose N E G and θ ∈ Irr(N). Let I = IG(θ). I Pk G (a) If θ = i=1 ϕi with ϕi ∈ Irr(I), then ϕi ∈ Irr(G). In particular, ϕi(1)|G : I| ∈ cd(G). (b) If θ extends to ψ ∈ Irr(I), then (ψτ)G ∈ Irr(G) for all τ ∈ Irr(I/N). In particular, θ(1)τ(1)|G : I| ∈ cd(G).

(c) If ρ ∈ Irr(I) such that ρN = eθ, then ρ = θ0τ0, where θ0 is a character of an irreducible projective representation of I of degree θ(1) and τ0 is a character of an irreducible projective representation of I/N of degree e. The following lemma [8, Lemma 3] will be used to verify Step 1.

Lemma 2.3. Let G/N be a solvable factor group of G minimal with respect to being nonabelian. Then two cases can occur. (a) G/N is an r-group for some prime r. Hence there exists ψ ∈ Irr(G/N) b such that ψ(1) = r > 1. If χ ∈ Irr(G) and r - χ(1), then χτ ∈ Irr(G) for all τ ∈ Irr(G/N). (b) G/N is a Frobenius group with an elementary abelian Frobenius kernel F/N. Then f = |G : F | ∈ cd(G) and |F/N| = ra for some prime r, and a is the smallest integer such that ra ≡ 1 (mod f). If ψ ∈ Irr(F ) then either fψ(1) ∈ cd(G) or ra divides ψ(1)2. In the latter case, r divides ψ(1). (1) If no proper multiple of f is in cd(G), then χ(1) divides f for all χ ∈ Irr(G) such that r - χ(1), and if χ ∈ Irr(G) such that a 2 χ(1) - f, then r | χ(1) . (2) If χ ∈ Irr(G) such that no proper multiple of χ(1) is in cd(G), then either f divides χ(1) or ra divides χ(1)2. Moreover if χ(1) is divisible by no nontrivial proper character degree in G, then f = χ(1) or ra | χ(1)2. Let χ ∈ Irr(G). We say that χ is isolated in G if χ(1) is divisible by no proper nontrivial character degree of G and no proper multiple of χ(1) is a character degree of G. In this situation, we also say that χ(1) is an isolated degree of G. We deﬁne a proper power degree of G to be a character degree of G of the form f a for integers f, a > 1. Throughout this paper we use GAP [4] to determine the isolated degrees of the groups under discussion. The next two lemmas will be used to verify Steps 2 and 4. The ﬁrst lemma appears in [2, Theorems 2, 3, 4].

3 Lemma 2.4. If S is a nonabelian simple group, then there exists a nontrivial irreducible character θ of S that extends to Aut(S). Moreover, the following holds: (a) if S is an alternating group of degree at least 7, then S has two characters of consecutive degrees n(n − 3)/2 and (n − 1)(n − 2)/2 that both extend to Aut(S). (b) if S is a sporadic simple group or the Tits group, then S has two nontrivial irreducible characters of coprime degrees which both extend to Aut(S).

Lemma 2.5. ([2, Lemma 5]). Let N be a minimal normal subgroup of G so that N =∼ Sk, where S is a nonabelian simple group. If θ ∈ Irr(S) extends to Aut(S), then θk ∈ Irr(N) extends to G. The following lemma will be used to verify Step 4.

0 00 Lemma 2.6. ([6, Lemma 6]). Suppose that M E G = G and for every λ ∈ Irr(M) with λ(1) = 1, λg = λ for all g ∈ G0. Then M 0 = [M,G0] and |M/M 0| divides the order of the Schur multiplier of G0/M.

0 3. Huppert’s Conjecture for the sporadic simple group F i24

0 3.1. The sporadic simple group F i24 Using [3], we provide relevant 0 properties of the sporadic simple group H = F i24. Observe that π(H) = {2, 3, 5, 7, 11, 13, 17, 23, 29}.

0 Lemma 3.1. Let H be the sporadic simple group F i24. (a) The following are isolated degrees of H: 314 · 72 · 23 · 29, 214 · 52 · 7 · 11 · 13 · 17 · 23, 221 · 52 · 73 · 11, 314 · 17 · 23 · 29, 316 · 11 · 13 · 29.

(b) Let 1 6= χ(1) ∈ cd(H). Then gcd(11 · 13 · 23, χ(1)) > 1. (c) The group H has neither proper power degrees nor consecutive degrees. (d) If K is a maximal subgroup of H such that |H : K| divides some ∼ character degree χ(1) of H, then K = F i23 and χ(1)/|H : K| divides one of the numbers in the set A = {24 · 52 · 7 · 17 · 23, 2 · 33 · 7 · 11 · 13 · 17, 22 · 3 · 11 · 13 · 17 · 23, 23 · 3 · 7 · 11 · 13 · 23, 24 · 3 · 13 · 17 · 23, 2 · 7 · 11 · 17 · 23}.

4 Table 1. Some character degrees of sporadic simple groups and Tits group

Group Character Degree Group Character Degree 2 2 2 M11 χ9 3 · 5 J2 χ6 2 · 3 2 7 6 M12 χ6 3 · 5 Co3 χ32 2 · 3 2 5 5 2 M22 χ3 3 · 5 F i22 χ54 2 · 3 · 5 · 7 2 7 2 M23 χ3 3 · 5 HS χ24 2 · 5 2 18 6 M24 χ3 3 · 5 Co1 χ71 2 · 3 2 2 3 2 0 6 3 McL χ12 2 · 3 · 5 F4(2) χ20 2 · 3 7 10 5 He χ12 2 · 3 · 5 Suz χ43 2 · 3 6 3 5 12 Co2 χ22 3 · 5 F i23 χ82 2 · 3 · 17 12 3 Ru χ34 2 · 3

0 ∼ (e) The Schur multiplier M(F i24) = Z3 and outer automorphism group 0 ∼ Out(F i24) = Z2.

0 ∼ 3.2. Verifying Huppert’s Conjecture for F i24 We assume that H = 0 ∼ F i24 and G is a group such that cd(G) = cd(H). We will show that G = H × A, where A is an abelian group, which conﬁrms Huppert’s Conjecture 0 for F i24. We follow the ﬁve steps of Huppert’s method described in the introduction.

3.2.1. Verifying Step 1 By way of contradiction, suppose that G0 6= G00. Then there exists a normal subgroup N E G where N is maximal such that G/N is solvable and nonabelian. By Lemma 2.3, G/N is an r-group for some prime r or G/N is a Frobenius group. Case 1. G/N is an r-group. Then there exists ψ ∈ Irr(G/N) such that ψ(1) = rb > 1. By Lemma 3.1(c), G has no nontrivial prime power degrees. Hence this case cannot happen. Case 2. G/N is a Frobenius group with Frobenius kernel F/N, |F/N| = ra, 1 < f = |G : F | ∈ cd(G) and ra ≡ 1 (mod f). By Lemma 2.3(b), if χ ∈ Irr(G) such that χ(1) is isolated, then either f = χ(1) or r | χ(1). We observe that there is no prime which divides all the isolated degrees listed in Lemma 3.1(a). Thus f must be one of the isolated degrees in Lemma 3.1(a). By Lemma 2.3(b) again, if χ ∈ Irr(G) with r - χ(1), then χ(1) | f. As f is isolated we deduce that r must divide every nontrivial degree χ(1) of G such that χ(1) 6= f. Assume ﬁrst that f = 314 · 72 · 23 · 29. Then r must divide all of the remaining isolated degrees in Lemma 3.1(a). As gcd(221 · 52 · 73 · 11, 314 · 17 · 23 · 29) = 1, we ﬁnd that f 6= 314 · 72 · 23 · 29 and r ∈ {3, 7, 23, 29}. For each value of r, there exist two distinct isolated degrees in Lemma 3.1(a) which are both prime to r, so f must be equal to both by Lemma 2.3(b)(2), which is impossible. Thus G0 = G00.

5 3.2.2. Verifying Step 2 Let M ≤ G0 be a normal subgroup of G such that G0/M is a chief factor of G. As G0 is perfect, G0/M is nonabelian. Hence G0/M =∼ Sk for some nonabelian simple group S and some integer k ≥ 1. Claim 1. k = 1. By way of contradiction, assume that k ≥ 2. By Lemma 2.4, S possesses a nontrivial irreducible character θ extendible to Aut(S). By Lemma 2.5, θk ∈ Irr(G0/M) extends to G/M, hence θ(1)k ∈ cd(G), which contradicts Lemma 3.1(c). Hence k = 1 and G0/M =∼ S. Claim 2. S is not an alternating group of degree at least 7. By way of contradiction, assume that S = An, n ≥ 7. By Lemma 2.4, S has nontrivial irreducible characters θ1, θ2 with θ1(1) = n(n − 3)/2, θ2(1) = θ1(1) + 1 = (n − 1)(n − 2)/2 and both θi extend to Aut(S). Thus G possesses two consecutive nontrivial character degrees, contradicting Lemma 3.1(c). Claim 3. S is not a simple group of Lie type. If S is a simple group of 2 0 Lie type in characteristic p, and S 6= F4(2) , then the Steinberg character of S of degree |S|p extends to Aut(S) so that G possesses a nontrivial prime power degree, which contradicts Lemma 3.1(c). ∼ 0 Claim 4. S = F i24. By Claims 1, 2, and 3, S is a sporadic simple group or the Tits group. We will eliminate all other possibilities for S and hence the claim will follow. By the Ito-Michler Theorem (see [6, Lemma 1]) we deduce that every prime divisor of S must divide some character degree of S and, as every character degree of S =∼ G0/M divides some character degree of H, every prime divisor of S is also a prime divisor of H. Thus π(S) ⊆ π(H) = {2, 3, 5, 7, 11, 13, 17, 23, 29}. Hence we only need to consider the simple groups in Table 1. For each group S in Table 1, we exhibit a nontrivial irreducible character θ of S such that θ extends to Aut(S) and gcd(11 · 13 · 23, θ(1)) = 1, which contradicts Lemma 3.1(b). This ﬁnishes the proof of Step 2.

3.2.3. Verifying Step 3 Let θ ∈ Irr(M) with θ(1) = 1 and let I = IG0 (θ). 0 I Ps Assume that I < G and θ = i=1 eiφi, where φi ∈ Irr(I) for i = 1, 2, . . . , s. Let U/M be a maximal subgroup of G0/M containing I/M and let t = |U : 0 0 I|. Then φi(1)|G : I| is a character degree of G by Lemma 2.2(a), so 0 it divides some character degree of G. Thus tφi(1)|G : U| divides some character degree of G and so the index |G0 : U| must divide some character ∼ degree of H. By Lemma 3.1(d), U/M = F i23 and, for each i, tφi(1) divides one of the numbers in A. By inspecting the list of maximal subgroups of F i23 in [3], the index of a maximal subgroup of U/M divides no number in ∼ A so that t = 1 and hence I = U. As the Schur multiplier of I/M = F i23 is trivial and θ is I-invariant, we deduce from [7, Theorem 11.7] that θ extends to θ0 ∈ Irr(I). By Gallagher’s Theorem, τθ0 is an irreducible constituent I of θ for every τ ∈ Irr(I/M), and so τ(1)θ0(1) = τ(1) divides one of the numbers in A. Choose τ ∈ Irr(I/M) = Irr(F i23) with τ(1) = 559458900.

6 This degree divides none of the numbers in A, which is a contradiction. Thus we conclude that θ is G0-invariant. 0 ∼ 0 3.2.4. Verifying Step 4 We have shown that G /M = F i24 and for every θ ∈ Irr(M), if θ(1) = 1, then θ is G0-invariant, so that by Lemma 0 0 2.6, |M : M | divides the order of the Schur multiplier of F i24. Hence |M : M 0| divides 3. Assume ﬁrst that M = M 0. If M is abelian, then we are done. Assume that M is nonabelian. Let N ≤ M be a normal subgroup of G0 such that M/N is a chief factor of G0. It follows that M/N =∼ Sk for some nonabelian simple group S. By Lemma 2.4, S possesses a nontrivial irreducible character ϕ such that ϕk ∈ Irr(M/N) extends to G0/N. Gallagher’s Theorem yields ϕ(1)kτ(1) ∈ cd(G0/N) ⊆ cd(G0) for all τ ∈ Irr(G0/M) ⊆ Irr(G0/N). Since cd(G0/M) = cd(G) and ϕ(1) > 1, if we choose τ ∈ Irr(G0/M) such that τ(1) is the largest degree of H, then ϕ(1)kτ(1) divides no degree of G, a contradiction. If |M : M 0| = 3, then 0 0 ∼ 0 0 G /M = 3·F i24. By [3], 3·F i24 possesses an irreducible character of degree χ182(1) = 405445459419 which divides none of the character degrees of H. Thus M = 1.

0 0 3.2.5. Verifying Step 5 Suppose that G × CG(G ) G and let A = 0 0 ∼ 0 CG(G ). It follows from Step 4 that G = F i24. Then G/A embeds into Aut(G0) and hence G induces on G0 some outer automorphism σ. We have 0 ∼ 0 that | Out(F i24)| = 2, and hence G/A = F i24 · 2. By [3], G/A possesses an irreducible character of degree 410707651200 which is not a degree of 0 0 ∼ 0 F i24. Thus G = G × A. Since A = G/G , we obtain that A is abelian and ∼ 0 ∼ 0 G = G × A = F i24 × A.

4. Huppert’s Conjecture for the sporadic simple group F i22

4.1. Some properties of F i22 We list relevant properties of the character degree set of the sporadic simple group H = F i22 by examining [3].

Lemma 4.1. Let H be the sporadic simple group F i22. (a) The following are isolated degrees of H:

217 · 11, 25 · 35 · 52 · 7, 23 · 38 · 11, 38 · 5 · 11, 39 · 7 · 13, 28 · 52 · 7 · 13.

(b) If 1 6= χ(1) ∈ cd(H), then gcd(7 · 11 · 13, χ(1)) > 1.

(d) If K is a maximal subgroup of H such that |H : K| divides some character degree χ(1) of H then one of the following cases holds.

7 ∼ . (a) K = 2 U6(2) and χ(1)/|H : K| divides one of the numbers in the set A = {22 · 3 · 5 · 11, 23 · 5 · 7, 35}. ∼ + (b) K = O8 (2) : S3 and χ(1)/|H : K| divides 6. ∼ 10 (c) K = 2 : M22 and χ(1)/|H : K| divides 6. ∼ (e) The Schur multiplier M(F i22) = Z6 and outer automorphism group ∼ Out(F i22) = Z2.

4.2. Verifying Huppert’s Conjecture for F i22 Steps 1 and 2 follow in the same way as Steps 1 and 2 of Section 3. In this case we use Lemma 4.1 and apply the argument to the isolated degree 38 ·5·11 and use the fact that gcd(27 · 11, 39 · 7 · 13) = 1. Hence the chief factor G0/M of G is isomorphic to F i22. 4.2.1. Verifying Step 3 Let θ ∈ Irr(M) with θ(1) = 1 and let I = 0 I Ps IG0 (θ). Assume that I < G and θ = i=1 eiφi, where φi ∈ Irr(I) for i = 1, 2, . . . , s. Let U/M be a maximal subgroup of G0/M containing I/M 0 0 and let t = |U : I|. Then φi(1)|G : I| is a character degree of G by Lemma 0 2.2(a). Thus tφi(1)|G : U| divides some character degree of G and hence the index |G0 : U| must divide some character degree of H. By Lemma 4.1(d), one of the following cases holds. ∼ . (a) U/M = 2 U6(2). For each i, tφi(1) divides one of the numbers in A. As U/M is perfect, the center of U/M lies in every maximal subgroup of U/M and so the indices of maximal subgroups of U/M and those of U6(2) are the same. By inspecting the list of maximal subgroups of U6(2) in [3], the index of a maximal subgroup of U6(2) divides no number in A. Thus t = 1 and hence I = U. Let M ≤ L ≤ I such that L/M is isomorphic to the center of I/M and let λ ∈ Irr(L|θ). As LEI, for any ϕ ∈ Irr(I|λ) we have that ϕ(1) divides one of the numbers in A. As above, we deduce that λ is I-invariant. ∼ Let L ≤ T ≤ I such that T/L = U5(2). It follows that λ is T -invariant and ∼ since the Schur multiplier of T/L = U5(2) is trivial, we have that λ extends to λ0 ∈ Irr(T ). By Gallagher’s Theorem, τλ0 is an irreducible constituent of λT for every τ ∈ Irr(T/L). Choose τ ∈ Irr(T/L) with τ(1) = 210 and let I γ = τλ0 ∈ Irr(T |λ). If χ ∈ Irr(I) is any irreducible constituent of γ , then χ(1) ≥ γ(1) by Frobenius reciprocity [7, Lemma 5.2] and χ(1) divides one of the numbers in A, which implies that γ(1) = 210λ(1) ≤ χ(1) ≤ 660, which is impossible. ∼ + (b) U/M = O8 (2) : S3. For each i, tφi(1) divides 6. Let M E W E U ∼ + such that W/M = O8 (2). We have that M EI ∩W EI and M EI ∩W ≤ W . Assume W I. Then I WI ≤ U and t = |U : I| = |U : WI| · |WI : I|. Now |WI : I| = |W : W ∩ I| > 1, and hence t is divisible by |W : W ∩ I|. ∼ + As W/M = O8 (2), t is divisible by the index of some maximal subgroup of + + O8 (2). Thus some index of a maximal subgroup of O8 (2) divides 6, which

8 W Pl is impossible by [3]. Thus W ≤ I ≤ U. Write θ = i=1 fiµi, where µi ∈ Irr(W |θ). As W E I, µi(1) divides 6 for every i. If fj = 1 for some j, then θ extends to θ0 ∈ Irr(W ). By Gallagher’s Theorem, τθ0 is an irreducible W constituent of θ for every τ ∈ Irr(W/M), and so τ(1)θ0(1) = τ(1) divides 6. However we can choose τ ∈ Irr(W/M) with τ(1) = 28 and this degree does not divide 6. Therefore fi > 1, for all i. We deduce that, for each i, fi is the degree of a nontrivial proper irreducible projective representation of + O8 (2). As µi(1) = fiθ(1) = fi, each fi divides 6. This is impossible as the + smallest nontrivial proper projective degree of O8 (2) is 8. ∼ 10 (c) U/M = 2 : M22. For each i, tφi(1) divides 6. Let M E L E U ∼ 10 ∼ such that L/M = 2 . We have that L E U and U/L = M22. The same argument as in part (b) shows that U = IL since the minimal index of a ∼ ∼ maximal subgroup of M22 is 22 by [3]. Hence U/L = I/L1 = M22, where L1 = L ∩ I E I. Let λ ∈ Irr(L1|θ). Then for any ϕ ∈ Irr(I|λ), we have that ϕ(1) divides 6. We conclude that λ is I-invariant as the index of a ∼ I Pl maximal subgroup of I/L1 = M22 is at least 22. Write λ = i=1 fiµi, where µi ∈ Irr(I|λ). Then µi(1) divides 6, for each i. If fj = 1 for some j, then λ extends to λ0 ∈ Irr(I). By Gallagher’s Theorem τλ0 is an irreducible I constituent of λ for every τ ∈ Irr(I/L1), and so τ(1)λ0(1) = τ(1) divides 6. However we can choose τ ∈ Irr(I/L1) with τ(1) = 21 and this degree does not divide 6. Therefore fi > 1 for all i. We deduce that, for each i, fi is the degree of a nontrivial proper irreducible projective representation of M22. Moreover as µi(1) = fiλ(1) = fi, each fi divides 6. However this is impossible as the smallest nontrivial proper projective degree of M22 is 10. Therefore θ is G0-invariant. 0 ∼ 0 4.2.2. Verifying Step 4 Since G /M = F i22 and IG0 (θ) = G , for every linear character θ ∈ Irr(M), Lemma 2.6 implies that |M : M 0| divides six, the 0 order of the Schur multiplier of F i22. Suppose that M = M . If M is abelian, then we are done. If M is not abelian, then by the same argument as in Step 4 of Section 3, a proper multiple of τ(1) = 2729376 ∈ cd(G0/M) divides some degree of G, a contradiction. Suppose that |M : M 0| = 2, 3 or 6. We arrive at a contradiction by considering χ110(1) = 2358720 ∈ cd(2 · F i22), χ164(1) = 3088800 ∈ cd(3 · F i22), and χ198(1) = 3088800 ∈ cd(6 · F i22), respectively. Thus M = 1. 4.2.3. Verifying Step 5 This step follows exactly in the same manner as Step 5 of Section 3 by considering the degree 1965600 ∈ cd(F i22 · 2). ∼ 0 ∼ Therefore G = G × A = F i22 × A, where A is abelian.

5. Huppert’s Conjecture for the sporadic simple group Co1

5.1. Some properties of Co1 We provide relevant properties of the character degree set of the sporadic simple group H = Co1 by examining

9 [3]. Observe that π(H) = {2, 3, 5, 7, 11, 13, 23}.

Lemma 5.1. Let H be the sporadic simple group Co1. (a) The following are isolated degrees of H :

218 · 36, 218 · 52 · 7 · 11, 219 · 7 · 11, 37 · 53 · 7 · 11, 37 · 53 · 7 · 23, 212 · 3 · 53 · 11 · 13.

(b) If 1 6= χ(1) ∈ cd(H) and gcd(11·13·23, χ(1)) = 1, then χ(1) = 218 ·36.

(c) The group H has neither consecutive degrees nor nontrivial prime power degrees and the only proper nontrivial power degree of H is 218 · 36.

(d) If K is a maximal subgroup of H such that |H : K| divides some character degree χ(1) of H, then one of the following cases holds. ∼ (a) K = Co2 and χ(1)/|H : K| divides one of the numbers in the set A = {3 · 7 · 11 · 23, 23 · 33 · 5}. ∼ 1+8. + (b) K = 2+ O8 (2) and χ(1)/|H : K| divides 1. ∼ (e) The Schur multiplier M(Co1) = Z2 and outer automorphism group Out(Co1) is trivial.

5.2. Verifying Huppert’s Conjecture for Co1 Steps 1 and 2 follow exactly in the same manner as Steps 1 and 2 in Section 3, respectively. We use Lemma 5.1 and apply the argument to the isolated degree 218 · 36 and use the fact that gcd(218 · 36, 11 · 13 · 23) = 1. Consequently, one can show 0 that the chief factor G /M of G is isomorphic to Co1. 5.2.1. Verifying Step 3 Let θ ∈ Irr(M) with θ(1) = 1 and let I = 0 I Ps IG0 (θ). Assume that I < G and θ = i=1 eiφi, where φi ∈ Irr(I) for i = 1, 2, . . . , s. Let U/M be a maximal subgroup of G0/M containing I/M 0 0 and let t = |U : I|. Then φi(1)|G : I| is a character degree of G by Lemma 0 2.2(a). Thus tφi(1)|G : U| divides some character degree of G and so the index |G0 : U| must divide some character degree of H. By Lemma 5.1(d), one of the following cases holds. ∼ (a) U/M = Co2. For each i, tφi(1) divides one of the numbers in A. By inspecting the list of maximal subgroups of Co2 in [3], the index of a maximal subgroup of Co2 divides no number in A so that t = 1 and hence I = U. ∼ As the Schur multiplier of I/M = Co2 is trivial, we have that θ extends to I θ0 ∈ Irr(I). By Gallagher’s Theorem, τθ0 is an irreducible constituent of θ for every τ ∈ Irr(I/M). Choose τ ∈ Irr(I/M) with τ(1) = 23 · 11 · 23 and let 3 γ = τθ0 ∈ Irr(I|θ). Then γ(1) = 2 · 11 · 23 divides no number in A, which is a contradiction.

10 ∼ 1+8. + (b) U/M = 2+ O8 (2). For each i, tφi(1) divides 1. It follows that I t = 1 and all φi(1) = 1. Hence I = U and every irreducible constituent of θ is linear so that θ extends to θ0 ∈ Irr(I). By Gallagher’s Theorem, τθ0 are I all irreducible constituents of θ , where τ ∈ Irr(I/M). Hence τ(1)θ0(1) = τ(1) = 1, which implies that all irreducible characters of I/M are linear and ∼ 1+8. + so I/M = 2+ O8 (2) is abelian, which is a contradiction. Therefore θ is G0-invariant.

0 ∼ 0 5.2.2. Verifying Step 4 As G /M = Co1 and IG0 (θ) = G for every linear character θ ∈ Irr(M), Lemma 2.6 implies that |M : M 0| divides two, the order of the Schur multiplier of Co1. Considering the largest degree τ(1) of Co1, the same argument as in Step 4 of Section 3 shows that if |M : M 0| = 1, then M is abelian and thus M = 1. If |M : M 0| = 2, then 0 0 ∼ G /M = 2 · Co1. By [3], 2 · Co1 possesses an irreducible character of degree χ167(1) = 1021620600, which divides no degree of H. Thus M = 1. 0 0 ∼ 5.2.3. Verifying Step 5 Let A = CG(G ). As Out(Co1) = 1 and G = Co1 is simple, we deduce that G = G0A and G0 ∩A = 1. Therefore G = G0 ×A =∼ ∼ 0 Co1 × A, where A = G/G is abelian.

6. Huppert’s Conjecture for the sporadic simple group Co2

6.1. Some properties of Co2 We provide relevant properties of the character degree set of the sporadic simple group H = Co2 by examining [3]. Observe that π(H) = {2, 3, 5, 7, 11, 23}.

Lemma 6.1. Let H be the sporadic simple group Co2.

(a) We have that 218 · 7 is an isolated degree of H. Moreover, the two degrees 22 · 35 · 52 · 23 and 2 · 36 · 53 · 11 of H are maximal with respect to divisibility and no nontrivial divisor of the greatest common divisor of both of these degrees is a degree of H.

(b) If 1 6= χ(1) ∈ cd(H) and gcd(11 · 23, χ(1)) = 1, then χ(1) ∈ {36 · 53, 218 · 7}.

(c) The group H has no consecutive degrees. The only proper nontrivial power degree of H is 36 · 53 and the only prime power degree of H is 23.

(d) If K is a maximal subgroup of H such that |H : K| divides some character degree χ(1) of H, then one of the following cases holds. ∼ (a) K = U6(2) : 2 and χ(1)/|H : K| divides one of the numbers in 4 5 the set A1 = {2 · 5 · 7 · 11, 2 · 5 · 7, 3 }.

11 ∼ 10 (b) K = 2 : M22 : 2 and χ(1)/|H : K| divides one of the numbers 2 2 in the set A2 = {3 · 5, 2 · 3}. ∼ 1+8 (c) K = 2+ : S6(2) and χ(1)/|H : K| divides one of the numbers in the set A3 = {5 · 7, 3 · 5, 2}. (d) K =∼ McL and χ(1)/|H : K| divides 22 · 11.

(e) The Schur multiplier and the outer automorphism group of Co2 are both trivial.

6.2. Verifying Huppert’s Conjecture for Co2 6.2.1. Verifying Step 1 Suppose that G0 6= G00. Then there exists a normal subgroup N E G where N is maximal such that G/N is solvable and nonabelian. By Lemma 2.3, G/N is either an r-group for some prime r or a Frobenius group. Case 1. G/N is an r-group. Then there exists ψ ∈ Irr(G/N) such that ψ(1) = rb > 1. By Lemma 6.1(c), we obtain that rb = 23. Let χ ∈ Irr(G) such that χ(1) = 218 · 7. As gcd(|G/N|, χ(1)) = 1, by Lemma 2.1, we have that χN ∈ Irr(N). By Gallagher’s Theorem, χψ ∈ Irr(G) so that χ(1)ψ(1) ∈ cd(G), which is impossible as χ(1) is an isolated degree of G by Lemma 6.1(a). Case 2. G/N is a Frobenius group with Frobenius kernel F/N, |F/N| = ra, 1 < f = |G : F | ∈ cd(G) and ra ≡ 1 (mod f). By Lemma 2.3(b), if χ ∈ Irr(G) such that χ(1) is isolated, then either f = χ(1) or r | χ(1). By Lemma 6.1(a), 218 · 7 is an isolated degree of G so that either f = 218 · 7 or r ∈ {2, 7}. Assume ﬁrst that f = 218 · 7. By Lemma 2.3(b) again, if χ ∈ Irr(G) with r - χ(1), then χ(1) | f. As f is isolated we deduce that r must divide every nontrivial degree χ(1) of G such that χ(1) 6= f. However G possesses two relatively prime degrees 23 and 52 · 11 diﬀerent from f, which leads to a contradiction. Hence r ∈ {2, 7}. By Lemma 6.1(a), the 2 5 2 6 3 degrees χ1(1) = 2 · 3 · 5 · 23 and χ2(1) = 2 · 3 · 5 · 11 of G are maximal with respect to divisibility and are both prime to 7, so that if r = 7, then by Lemma 2.3(b), f must divide both of these degrees. This is impossible by Lemma 6.1(a) and thus r = 2. By Lemma 2.3(b) again, if ra divides 2 a 4 a 4 χi(1) for i = 1 or 2, then r ≤ 2 and hence f ≤ r − 1 ≤ 2 − 1 = 15, which is impossible as the smallest nontrivial degree of Co2 is 23. Thus we conclude that f divides both χi(1), which again contradicts Lemma 6.1(a). Thus G0 = G00. 6.2.2. Verifying Step 2 Let M ≤ G0 be a normal subgroup of G such that G0/M is a chief factor of G. As G0 is perfect, G0/M is nonabelian. Hence G0/M =∼ Sk for some nonabelian simple group S and some integer k ≥ 1.

Claim 1. S is not an alternating group of degree at least 7. If S = An for n ≥ 7, then by Lemma 2.4, two nontrivial irreducible characters θ1, θ2

12 of consecutive degrees θ1(1) = n(n − 3)/2, θ2(1) = θ1(1) + 1 extend to k Aut(S), so that by Lemma 2.5, θi (1) ∈ cd(G). By Lemma 6.1(c), G has only one nontrivial power degree, so that k = 1. But then G possesses two consecutive nontrivial character degrees, contradicting Lemma 6.1(c). Claim 2. S is not a simple group of Lie type. If S is a simple group of 2 0 Lie type in characteristic p and S 6= F4(2) , then the Steinberg character k of S of degree |S|p extends to Aut(S). By Lemma 2.5, |S|p ∈ cd(G) and hence G possesses a nontrivial prime power degree. By Lemma 6.1(c), we k deduce that |S|p = 23, and so k = 1, p = 23 and |S|p = 23. It follows that the Lie rank of S is 1 and the only possibility for S is L2(23). By [3], L2(23) possesses an irreducible character of degree 22 which is extendible to Aut(S). By Lemma 2.5, G has an irreducible character of degree 22, which is a contradiction. ∼ Claim 3. S = Co2. By Claims 1, 2, and 3, S is a sporadic simple group or the Tits group. If k ≥ 2, Lemmas 2.4(b) and 2.5 imply G would possess two distinct nontrivial proper power degrees, which contradicts Lemma 6.1(c). Thus k = 1. As π(S) ⊆ π(H) = {2, 3, 5, 7, 11, 13, 23}, we only need to consider the simple groups in Table 1 excluding Co2. For each such simple group S, there exists an irreducible character θ ∈ Irr(S) which is extendible to Aut(S). By Lemma 2.5, we obtain that θ(1) ∈ cd(G) with gcd(θ(1), 11 · 13 · 23) = 1. It follows from Lemma 6.1(b) that θ(1) ∈ {36 · 53, 218 · 7}, which is impossible by inspecting Table 1.

6.2.3. Verifying Step 3 Let θ ∈ Irr(M) with θ(1) = 1 and let I = IG0 (θ). 0 I Ps Assume that I < G and θ = i=1 eiφi, where φi ∈ Irr(I) for i = 1, 2, . . . , s. Let U/M be a maximal subgroup of G0/M containing I/M and let t = |U : 0 0 I|. Then φi(1)|G : I| is a character degree of G by Lemma 2.2(a), so 0 it divides some character degree of G. Thus tφi(1)|G : U| divides some character degree of G and the index |G0 : U| must divide some character degree of H. By Lemma 6.1(d), one of the following cases holds. ∼ (a) U/M = U6(2) : 2. For each i, tφi(1) divides one of the numbers in ∼ A1. Let M ≤ W ≤ U such that W/M = U6(2). Then W E U. Assume that W I. Since t = |U : I| = |U : WI| · |WI : I| and |WI : I| = |W : ∼ W ∩ I|, the index of some maximal subgroup of W/M = U6(2) divides t and so divides some number in A1, which is a contradiction by [3]. Thus ∼ W ≤ I ≤ U. Let M ≤ V ≤ W such that V/M = U5(2). We have that θ is V -invariant and, since the Schur multiplier of V/M is trivial, θ extends to θ0 ∈ Irr(V ). By Gallagher’s Theorem, τθ0 is an irreducible constituent of θV for every τ ∈ Irr(V/M). Choose τ ∈ Irr(V/M) with τ(1) = 1215 and let I γ = τθ0 ∈ Irr(V |θ). If χ ∈ Irr(I) is an irreducible constituent of γ , then χ(1) ≥ γ(1) by Frobenius reciprocity [7, Lemma 5.2] and also χ(1) divides one of the numbers in A1, which implies that γ(1) = 1215 ≤ χ(1) ≤ 770, which is impossible.

13 ∼ 10 (b) U/M = 2 : M22 : 2. For each i, tφi(1) divides one of the numbers ∼ 10 in A2. Let M E L E W E U be a subgroup of U such that L/M = 2 and ∼ W/L = M22. Let L1 = L ∩ I and X = W ∩ I. As I ≤ WI ≤ U, we obtain t = |U : I| = |U : WI| · |WI : I| = |U : WI| · |W : X|. Now X ≤ XL ≤ W so that |W : X| = |W : XL| · |XL : X|. Assume that |W : XL| > 1. ∼ Then L ≤ XL W and W/L = M22 so an index of a maximal subgroup ∼ of W/L = M22 must divide a number in A2. This is impossible by checking ∼ ∼ [3]. Thus W = XL and W/L = X/L1 = M22. Let λ ∈ Irr(L1|θ). As X X = W ∩I EI, we deduce that for any irreducible constituent ϕ of λ , ϕ(1) X Pl divides 12 or 45. Write λ = i=1 fiµi, where µi ∈ Irr(X|λ). If fj = 1 for some j then λ extends to λ0 ∈ Irr(X). By Gallagher’s Theorem, τλ0 is an X irreducible constituent of λ for every τ ∈ Irr(X/L1) and τ(1)λ0(1) = τ(1) divides 45 or 12. However we can choose τ ∈ Irr(X/L1) with τ(1) = 385 and this degree does not divide 45 and 12. Therefore fi > 1 for all i. We conclude, for each i, fi is the degree of a nontrivial proper irreducible projective representation of M22. As µi(1) = fiθ(1) = fi, each fi divides 45 Pl 2 or 12. By [3], we obtain that all fi = 45. As |X/L1| = i=1 fi , we conclude that 27 · 32 · 5 · 7 · 11 = 34 · 52l, which is impossible. ∼ 1+8 (c) U/M = 2+ : S6(2) For each i, tφi(1) divides one of the numbers ∼ 1+8 in A3. Let M E L E U be subgroup of U such that L/M = 2+ . Let L1 = L ∩ I and X = U ∩ I. As I ≤ IL ≤ U, we obtain t = |U : I| = ∼ |U : IL| · |IL : I|. Now L ≤ IL ≤ U and U/L = S6(2). Assume that ∼ |U : IL| > 1. The index of some maximal subgroup of U/L = S6(2) must divide some number in A3. This is impossible by checking [3]. Thus U = IL ∼ ∼ I Pl and then U/L = I/L1 = S6(2). Let λ ∈ Irr(L1|θ). Write λ = i=1 fiµi, where µi ∈ Irr(I|λ). If fj = 1 for some j then λ extends to λ0 ∈ Irr(I). By I Gallagher’s Theorem, τλ0 is an irreducible constituent of λ for every τ ∈ Irr(I/L1) and then τ(1)λ0(1) = τ(1) divides some number in A3. However we can choose τ ∈ Irr(I/L1) with τ(1) = 512 and this degree divides no number in A3. Therefore fi > 1 for all i. We deduce that, for each i, fi is the degree of a nontrivial proper irreducible projective representation of S6(2). As µi(1) = fiθ(1) = fi, each fi divides 2, 15 or 35. This is impossible by checking [3]. ∼ (d) U/M = McL. For each i, tφi(1) divides 44. By inspecting the list of maximal subgroups of McL in [3], no index of a maximal subgroup of McL I Ps divides 44 so that t = 1 and hence I = U. Recall that θ = i=1 eiφi, where φi ∈ Irr(I) for i = 1, 2, . . . , s. Assume ﬁrst that ej = 1 for some j. Then θ extends to θ0 ∈ Irr(I). By Gallagher’s Theorem, τθ0 is an irreducible I constituent of θ for every τ ∈ Irr(I/M) and then τ(1)θ0(1) = τ(1) divides 44. We can choose τ ∈ Irr(I/M) = Irr(McL) with τ(1) = 231 and this degree does not divide 44. Therefore ei > 1 for all i. We deduce that, for each i, ei is the degree of a nontrivial proper irreducible projective representation of McL. As φi(1) = eiθ(1) = ei, each ei divides 44. It follows that ei ≤ 44

14 for each i and ei is the degree of a nontrivial proper irreducible projective representation of McL. Using [3], we deduce that there is no such projective degree. Therefore θ is G0-invariant. 0 ∼ 0 6.2.4. Verifying Steps 4 and 5 Since G /M = Co2 and IG0 (θ) = G for linear characters θ ∈ Irr(M), by Lemma 2.6, |M : M 0| divides the order of 0 the Schur multiplier of Co2, which is 1. Thus M = M . By considering the largest degree τ(1) of Co2, the same argument as in Step 4 of Section 3 shows that M is abelian. Therefore M = 1. Since Out(Co2) = 1, Step 5 follows in the same way as Step 5 of Section 5.

7. Huppert’s Conjecture for the sporadic simple group Co3

7.1. Some properties of Co3 We provide relevant properties of the character degree set of the sporadic simple group H = Co3 by examining [3]. Observe that π(H) = {2, 3, 5, 7, 11, 23}.

Lemma 7.1. Let H be the sporadic simple group Co3. (a) We have that 27 · 36 and 36 · 53 are isolated degrees of H. The degrees 27 ·52 ·23, 27 ·52 ·7·11, and 53 ·7·11·23 of H are maximal with respect to divisibility. No nontrivial divisor of both 27 · 52 · 23 and 27 · 52 · 7 · 11 is a degree of H.

(b) If 1 6= χ(1) ∈ cd(H) and gcd(11·23, χ(1)) = 1 then χ(1) ∈ {27 ·36, 36 · 53, 27 · 7}.

(c) The group H has no consecutive degrees. The only proper nontrivial power degree of H is 36 · 53 and the only prime power degree of H is 23.

(d) If K is a maximal subgroup of H such that |H : K| divides some character degree χ(1) of H then K =∼ McL : 2 and χ(1)/|H : K| divides one of the numbers in the set A = {22 · 3 · 7 · 11, 2 · 3 · 5 · 7}.

(e) The Schur multiplier and the outer automorphism group of Co3 are both trivial.

7.2. Verifying Huppert’s Conjecture for Co3 7.2.1. Verifying Step 1 This step follows similarly to Step 1 of Section 6. We use Lemmas 2.3 and 7.1. If G/N is an r-group, we make the same argument by taking χ(1) = 27 · 36. If G/N is a Frobenius group with Frobenius kernel F/N, |F/N| = ra and 1 < f = |G : F | ∈ cd(G), then applying Lemma 2.3(b) to χ, we deduce that either f = 27 · 36 or r ∈ {2, 3}. If f = 27 · 36, then again by Lemma 2.3(b), two relatively prime degrees

15 23 and 52 · 11 lead us to a contradiction. If r ∈ {2, 3}, by considering 7 2 7 2 χ1(1) = 2 · 5 · 23 and χ2(1) = 2 · 5 · 7 · 11, we deduce that r = 2. In this case, the degrees 36 · 53 and 53 · 7 · 11 · 23 lead us to a contradiction. Thus G0 = G00.

7.2.2. Verifying Step 2 The same argument as in Step 2 of Section 6 0 shows that the chief factor G /M of G is isomorphic to Co3. To eliminate sporadic simple groups except for Co3, use Lemma 7.1(b) and the fact that Co2 possesses an irreducible character of degree χ60(1) = 2095875 which divides no degree of Co3.

7.2.3. Verifying Step 3 Let θ ∈ Irr(M) with θ(1) = 1 and let I = IG0 (θ). 0 I Ps Assume that I < G and θ = i=1 eiφi, where φi ∈ Irr(I) for i = 1, 2, . . . , s. Let U/M be a maximal subgroup of G0/M containing I/M and let t = |U : 0 0 I|. Then φi(1)|G : I| is a character degree of G by Lemma 2.2(a), so 0 it divides some character degree of G. Thus tφi(1)|G : U| divides some character degree of G and the index |G0 : U| must divide some character ∼ degree of H. By Lemma 7.1(d), U/M = McL : 2. Then for each i, tφi(1) divides one of the numbers in A. Let M ≤ W ≤ U such that W/M =∼ McL. Suppose that W I. Since I ≤ WI ≤ U and |WI : I| = |W : W ∩ I|, we deduce that the index of some maximal subgroup of W/M =∼ McL divides t, and so divides some number in A, which is a contradiction by [3]. Thus W Pl W ≤ I ≤ U. Write θ = i=1 fiµi, where µi ∈ Irr(W ) for i = 1, 2, . . . , s. As W E I, we deduce that, for each i, µi(1) divides one of the numbers in A. Assume ﬁrst that fj = 1 for some j. Then θ extends to θ0 ∈ Irr(W ). W By Gallagher’s Theorem, τθ0 is an irreducible constituent of θ for every τ ∈ Irr(W/M) and τ(1)θ0(1) = τ(1) divides one of the numbers in A. However we can choose τ ∈ Irr(I/M) = Irr(McL) with τ(1) = 10395 and this degree divides none of the numbers in A. Therefore fi > 1 for all i. We deduce that, for each i, fi is the degree of a nontrivial proper irreducible projective representation of McL. As µi(1) = fiθ(1) = fi, each fi divides one of the numbers in A. It follows that, for each i, fi ≤ 924, fi is the degree of a nontrivial proper irreducible projective representation of McL, and fi divides 210 or 924. Using [3], no such projective degree exists. Thus θ is G0-invariant.

7.2.4. Verifying Steps 4 and 5 Note that the order of the Schur multiplier of Co3 is 1. Applying Lemma 2.6 and considering the largest degree τ(1) of Co3, the same argument as in Step 4 of Section 3 shows that M = 1. Since Out(Co3) = 1, Step 5 follows exactly in the same manner as Step 5 in Section 5.

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S. H. Alavi A. Daneshkah Centre for the Mathematics of Department of Mathematics Symmetry and Computation Faculty of Science School of Mathematics and Statistics Bu-Ali Sina University The University of Western Australia Hamedan, Iran 35 Stirling Hw, Crawley WA 6009 e-mail: [email protected] Australia e-mail: [email protected]

H. P. Tong-Viet T. P. Wakefield School of Mathematical Sciences Department of Mathematics and Statistics North-West University (Mafikeng) Youngstown State University Mmabatho 2735 Youngstown, Ohio South Africa United States 44555 e-mail: [email protected] e-mail: tpwakefi[email protected]