The Simple Ree groups are determined by the set of their character degrees more

THE SIMPLE REE GROUPS 2 F4 (q 2 ) ARE DETERMINED BY THE SET OF THEIR CHARACTER DEGREES HUNG P. TONG-VIET Abstract. Let G be a finite group. Let cd(G) be the set of all complex irreducible character degrees of G. In this paper, we will show that if cd(G) = cd(H), where H is the simple Ree group 2 F4 (q 2 ), q 2 ≥ 8, then G ∼ H × A, = where A is an abelian group. This verifies Huppert’s Conjecture for the simple Ree groups 2 F4 (q 2 ) when q 2 ≥ 8. 1. Introduction and Notation All groups considered are finite and all characters are complex characters. For a group G we denote by Irr(G) the set of all irreducible characters of G and let cd(G) = {χ(1) | χ ∈ Irr(G)} be the set of all character degrees of G. Huppert proposed the following conjecture in the late 1990s. Huppert’s Conjecture. Let G be a finite group and let H be a nonabelian simple group. If cd(G) = cd(H), then G ∼ H × A, where A is abelian. = Huppert verified this conjecture for L2 (q) and Sz(q 2 ) in [10] and several small groups. Recently, T. Wakefield verified this conjecture for some families of simple groups of Lie type of Lie rank 2 (see [18]). The proof is based on verifying the following 5 steps outlined in [10], which we call Huppert’s Method. Step 1. Show G = G . It follows that if G /M is a chief factor of G, then G /M ∼ S k , where S is a nonabelian simple group and k ≥ 1. = Step 2. Show G /M ∼ H. = Step 3. If θ ∈ Irr(M ) and θ(1) = 1, then θ is G -invariant, which implies [M, G ] = M . Step 4. Show M = 1, which implies G ∼ H. = Step 5. Show G = G × CG (G ). As G/G ∼ CG (G ) is abelian and G ∼ H, = = Huppert’s Conjecture follows. In this paper, we will verify this conjecture for the simple exceptional group of Lie type 2 F4 (q 2 ), where q 2 = 22m+1 , m ≥ 1. This family of nonabelian simple groups was discovered by Rimhak Ree in 1961 and so called the simple Ree groups. We note that when m = 0, the group 2 F4 (2) is not simple but its derived subgroup 2 F4 (2) is simple. This group is called the Tits group. In his preprint, Huppert already verified the conjecture for this group and so we only need to consider the case when m ≥ 1. The irreducible characters of 2 F4 (q 2 ) were computed by G. Malle [15] and CHEVIE [7] and their maximal subgroups were classified by Malle in [14]. Date: May 28, 2011. 2000 Mathematics Subject Classification. Primary 20C15, secondary 20D05. Key words and phrases. character degrees; simple Ree groups; Huppert’s Conjecture. Support from the University of KwaZulu-Natal is acknowledged. 1 2 HUNG P. TONG-VIET Theorem 1.1. Let G be a finite group and let H be the simple Ree group of type 2 F4 (q 2 ), q 2 = 22m+1 , m ≥ 1. If cd(G) = cd(H), then G ∼ H ×A, where A is abelian. = Huppert’s Method described above was improved by T. Wakefield in [18], especially for Step 2. In this paper, we introduce the notion of an isolated character and use it to simplify the proof of Step 1. For the definition of isolated characters, see the discussion right after the proof of Lemma 2.3. The isolated character behaves like the Steinberg character of the simple groups of Lie type and in fact this is an example of an isolated character (see Lemma 2.4). Now by Lemma 2.3 we can verify Step 1 provided that we know several isolated character degrees instead of all character degrees. This could be used to verify Step 1 for all simple groups of Lie type. In order to verify Step 3, we rely heavily on the criterion for the character extension using Schur multiplier (see [11, Theorem 11.7]) and a result of R. Higgs on the fixed prime power projective character degrees (see [9, Theorem B]). Using the same method, one can verify Step 3 for other simple groups of Lie type. In general, we need to know all maximal subgroups of the simple group H whose indices divide some character degrees of H and also the character degrees and the Schur multipliers of the nonabelian composition factors involved in those maximal subgroups. This is in fact the most difficult step of Huppert’s Method. Finally, in order to verify Step 5, we need to show that the character degree sets of a simple group H and any almost simple group with socle H are different. If n is an integer then we denote by π(n) the set of all prime divisors of n. If G is a group, we will write π(G) instead of π(|G|) to denote the set of all prime divisors of the order of G. Let ρ(G) = ∪χ∈Irr(G) π(χ(1)) be the set of all primes which divide some irreducible character degrees of G. If N G and θ ∈ Irr(N ), then the inertia group of θ in G is denoted by IG (θ). Finally, the set of all irreducible constituents of θG is denoted by Irr(G|θ). Other notation is standard. 2. Preliminaries In this section, we present some results that we will need for the proof of Huppert’s Conjecture. Lemma 2.1. ([10, Lemma 2]). Suppose N G and χ ∈ Irr(G). (a) If χN = θ1 +θ2 +· · ·+θk with θi ∈ Irr(N ), then k divides |G/N |. In particular, if χ(1) is prime to |G/N | then χN ∈ Irr(N ). (b) (Gallagher’s Theorem) If χN ∈ Irr(N ), then χψ ∈ Irr(G) for every ψ ∈ Irr(G/N ). Lemma 2.2. ([10, Lemma 3]). Suppose N G and θ ∈ Irr(N ). Let I = IG (θ). k (a) If θI = i=1 ϕi with ϕi ∈ Irr(I), then ϕG ∈ Irr(G). In particular, ϕi (1)|G : i 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 = θ0 τ0 , where θ0 is a character of an irreducible projective representation of I of degree θ(1) while τ0 is the character of an irreducible projective representation of I/N of degree e. The following lemma will be used to verify Step 1. All these statements but the last one appear in [10, Lemma 4]. We will give a proof for completeness. SIMPLE REE GROUPS 3 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 ) such that ψ(1) = rb > 1. If χ ∈ Irr(G) and r χ(1), then χτ ∈ Irr(G) for all τ ∈ Irr(G/N ). (b) G/N is a Frobenius group with an elementary abelian Frobenius kernel F/N. Then f = |G : F | ∈ cd(G) and |F/N | = ra for some prime r, and F/N is an irreducible module for the cyclic group G/F, hence a is the smallest integer such that ra ≡ 1(mod f ). If ψ ∈ Irr(F ) then either f ψ(1) ∈ cd(G) or ra divides ψ(1)2 . In the latter case, r divides ψ(1). (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 χ(1) f, then ra | χ(1)2 . (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 . Proof. Statements (a) and (b) follow from [11, Lemma 2.3] and [11, Theorem 12.4]. Suppose that G/N is a Frobenius group. Now assume that no proper multiple of f is in cd(G), and let χ ∈ Irr(G). Let ψ be an irreducible constituent of χF . By [11, Lemma 6.8], we have that χ(1) = kψ(1) and by [11, Corollary 11.29] we obtain k | f = |G : F |. By (b), we have that either f ψ(1) ∈ cd(G) or ra | ψ(1)2 . Suppose r χ(1). Then r ψ(1) so that f ψ(1) = f χ(1)/k ∈ cd(G). As no proper multiple of f is a character degree of G, we deduce that f χ(1)/k = f so that χ(1) = k | f. Now assume χ(1) f. Then r | χ(1). Since r f, we deduce that r k, hence r | ψ(1) so that f ψ(1) > f. Thus f ψ(1) is not a character degree of G and so ra | ψ(1)2 . As ψ(1) | χ(1), (1) follows. The proof of (2) is exactly the same. Suppose that χ ∈ Irr(G) such that no proper multiple of χ(1) is in cd(G). Let ψ ∈ Irr(F ) be an irreducible constituent of χF . As above, we have that χ(1) = kψ(1), k | f and either f ψ(1) ∈ cd(G) or ra | ψ(1)2 . If the latter case holds then we are done since ψ(1) | χ(1). Now assume f ψ(1) ∈ cd(G). Observe that ψ(1) = χ(1)/k so that ψ(1)f = f χ(1)/k ∈ cd(G), where f χ(1)/k is a multiple of χ(1) since k | f. As no proper multiple of χ(1) belongs to cd(G), it follows that f χ(1)/k = χ(1), which implies that f = k. Since k divides χ(1), we deduce that f | χ(1). The remaining statement is obvious. The proof is now complete. 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. Recall that for χ ∈ Irr(G), χ is said to be of p-defect zero for some prime p if |G|/χ(1) is coprime to p. Lemma 2.4. If S is a simple group of Lie type in characteristic p with S = 2 F4 (2) , then the Steinberg character of S of degree |S|p is an isolated character of S. Proof. The existence of the Steinberg character of S, denoted by StS , is well known. From [13, Theorem 1.1], no proper nontrivial divisor of StS (1) is in cd(S). As StS is the only character of p-defect zero (see [4, Theorem 4]), no proper multiple of StS (1) is in cd(S). This completes the proof. 4 HUNG P. TONG-VIET The next two lemmas will be used to verify Steps 2 and 4. The first lemma appears in [1, Theorems 2, 3, 4]. Lemma 2.5. 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 hold: (i) If S is an alternating group of degree at least 7, then S has two consecutive characters of degrees n(n − 3)/2 and (n − 1)(n − 2)/2 that both extend to Aut(S). (ii) If S is a sporadic simple group or the Tits group, then S has two nontrivial irreducible characters of coprime degrees which both extend to Aut(S). (iii) If S is a simple group of Lie type then the Steinberg character StS of S of degree |S|p extends to Aut(S). Lemma 2.6. ([1, Lemma 5]). Let N be a minimal normal subgroup of G so that N ∼ S k , where S is a nonabelian simple group. If θ ∈ Irr(S) extends to Aut(S), = then θk ∈ Irr(N ) extends to G. The following result due to R. Higgs will be used to verify Step 3. The statement given below can be found in [16, Theorem 2.3]. Lemma 2.7. ([9, Theorem B]). Let N be a normal subgroup of a group G and let θ ∈ Irr(N ) be G-invariant. Assume that χ(1)/θ(1) is a power of a fixed prime p for every χ ∈ Irr(G|θ). Then G/N is solvable. The following lemma will be used to verify Step 4. Lemma 2.8. ([10, Lemma 6]). Suppose that M G = G and that for any λ ∈ Irr(M ) with λ(1) = 1, λg = λ for all g ∈ G . Then M = [M, G ] and |M/M | divides the order of the Schur multiplier of G /M. 3. The simple Ree groups The Ree group F4 (q ), where q 2 = 22m+1 with m ≥ 0, is an exceptional group of Lie type of rank 2 discovered by Ree in [17]. The order of this group is q 24 (q 12 + 1)(q 8 − 1)(q 6 + 1)(q 2 − 1). This group is nonabelian simple unless m = 0. In this case, the group 2 F4 (2) is simple and is called the Tits group. In his preprint, Huppert verified the conjecture for the Tits group so that we can assume m ≥ 1. The character table of this family of simple groups is available in [7] and is reproduced in Table 1. The maximal subgroups of 2 F4 (q 2 ) were determined by G. Malle in [14]. In Table 3, we list the maximal subgroups of 2 F4 (q 2 ) together with their indices. We denote by Φn := Φn (q), the cyclotomic polynomial in variable q. We have Φ1 Φ2 = q 2 − 1, Φ4 = q 2 + 1, Φ8 = q 4 + 1, Φ12 = q 4 − q 2 + 1, Φ24 = q 8 − q 4 + 1. In Table 3, we use the following notation. √ √ u1 := q 2 − 2q + 1, u2 := q 2 + 2q + 1, √ √ √ √ w1 := q 4 − 2q 3 + q 2 − 2q + 1, w2 := q 4 + 2q 3 + q 2 + 2q + 1. Then u1 u2 = Φ8 and w1 w2 = Φ24 . Recall that if n is a positive integer and p is a prime then np and np are the largest p-part and p -part of n, respectively. That is n = np np , where (np , np ) = 1 and np is a p-power. Let i = 3, i = 1, 2, 3, be prime divisors of w1 , w2 and Φ12 , respectively. In the next lemma, we collect some properties of the character degree set of the simple Ree group 2 F4 (q 2 ), q 2 = 22m+1 , where m ≥ 1. 2 2 SIMPLE REE GROUPS 5 Lemma 3.1. Let H be the simple Ree group 2 F4 (q 2 ), q 2 = 22m+1 , m ≥ 1 and let a be a nontrivial character degree of H. Then the following hold. (i) If a = q 24 and ( 1 2 , a) = 1, then a is one of the following degrees: √ √ q 2Φ1 Φ2 Φ2 Φ12 /2, q 4 Φ2 Φ2 Φ2 Φ2 /3, q 13 2Φ1 Φ2 Φ2 Φ12 /2. 4 1 2 4 8 4 (ii) If a = q 24 and ( 3 , a) = 1, then a is one of the following degrees: Φ1 Φ2 Φ2 Φ24 , q 4 Φ2 Φ2 Φ2 Φ24 /6, q 4 Φ2 Φ24 /2, 8 1 2 4 8 q 2 Φ2 Φ2 Φ2 Φ24 , q 6 Φ1 Φ2 Φ2 Φ24 , Φ2 Φ2 Φ2 Φ2 Φ24 , q 4 Φ2 Φ2 Φ2 Φ2 /3. 1 2 8 8 1 2 4 8 1 2 4 8 (iii) We have that (2Φ1 Φ2 Φ4 , a) > 1. (iv) If ( 1 2 3 , a) = 1 then a ∈ {q 24 , q 4 Φ2 Φ2 Φ2 Φ2 /3}. 1 2 4 8 (v) We have that q 4 Φ2 Φ2 Φ2 Φ2 /3 is an isolated degree of H. 1 2 4 8 (vi) If x, y ∈ cd(H) − {1, q 24 }, then (x, y) > 1. (vii) H has no consecutive degrees. √ (viii) If a = q 24 , then a2 ≤ q 13 2/2 = 213m+6 . (ix) If a, b ∈ cd(H) such that b = za, where z > 1 is odd, then z ≥ q 2 − 1. √ (x) The smallest nontrivial character degree of H is q 2Φ1 Φ2 Φ2 Φ12 /2. 4 Proof. Statements (i) − (v) and (viii) − (x) are obvious by checking Table 1. For (vi), if 3 divides both x and y, then we are done. Hence we assume that ( 3 , x) = 1 or ( 3 , y) = 1. Without loss of generality, assume ( 3 , x) = 1. Then x is one of the degrees appearing in (ii). It follows that either 1 2 divides x or 1 2 is prime to x. Assume first that 1 2 | x. If ( 1 2 , y) > 1 then we are done. So assume ( 1 2 , y) = 1 so that y is one of the degrees in (i). In this case, we can see that 2Φ1 Φ2 Φ4 divides y and so we have that (x, y) is divisible by (x, 2Φ1 Φ2 Φ4 ). Applying (iii) we obtain (x, 2Φ1 Φ2 Φ4 ) > 1 so that (x, y) > 1. Now assume that ( 1 2 , x) = 1. It follows that x = q 4 Φ2 Φ2 Φ2 Φ2 /3 and so 2Φ1 Φ2 Φ4 divides x. Using the same argument, we have 1 2 4 8 that (x, y) is divisible by (2Φ1 Φ2 Φ4 , y) which is nontrivial by (iii) so that (x, y) > 1. This proves (vi). Next we will show that H has no two consecutive degrees. By way of contradiction, assume that there exist x, y ∈ cd(H) such that x = y + 1. Since H is nonabelian simple, it has no character of degree 2 so that we can assume y > 1 and then x > y > 1. As x = y + 1, we deduce that (x, y) = 1 and since x > y > 1, by (vi), we have x = q 24 or y = q 24 . It follows that q 24 − 1 ∈ cd(H) or q 24 + 1 ∈ cd(H). However we can check that H has no such degrees. This contradiction proves (vii). Lemma 3.2. Let H be the Ree group 2 F4 (q 2 ), q 2 = 22m+1 , m ≥ 1. If K is a maximal subgroup of H such that the index |H : K| divides some character degree χ(1) of H, then one of the following cases holds: (i) K ∼ Pa , |H : Pa | = Φ4 Φ2 Φ12 Φ24 and = 8 χ(1)/|H : Pa | ∈ {1, q 2 − 1, q 2 , q 2 + 1}. (ii) K ∼ Pb , |H : Pb | = Φ2 Φ8 Φ12 Φ24 and = 4 √ χ(1)/|H : Pb | ∈ {q 2/2(q 2 − 1), u1 (q 2 − 1), u2 (q 2 − 1), (q 2 − 1)2 , q 4 , q 4 + 1}. Proof. If K is one of the parabolic subgroups Pa or Pb , then the result is obvious. 2 For the remaining maximal subgroups of H except 2 F4 (q0 ), we can see that the √ 13 2-part of the indices are larger than q 2/2 so that these indices cannot divide 2α 2 any degrees of H. Finally, assume K ∼ 2 F4 (q0 ), where q 2 = q0 , α prime. Assume = 6 HUNG P. TONG-VIET 2 q0 = 22l+1 . Then α = 2m+1 is odd, so that α ≥ 3 is an odd prime. The 22l+1 24α−24 2 part of the index of 2 F4 (q0 ) in 2 F4 (q 2 ) is q0 . Moreover as this index is not a 2-power, it √ cannot divide the degree of the Steinberg character of H so that 24α−24 14α q0 ≤ q 13 2/2 < q 14 = q0 . It follows that the 24α − 24 < 14α and hence 10α < 24. Thus α ≤ 2, a contradiction. The following results are well known, see for example [10]. We note that the inclusion of the value (q 2 − 1)2 in Lemma 3.4(b) causes no difference as it is less than the smallest index of the maximal subgroups of the Suzuki groups Sz(q 2 ). Lemma 3.3. Let q ≥ 8 be an even prime power. Then the following hold: (a) The Schur multiplier of L2 (q) is trivial and cd(L2 (q)) = {1, q − 1, q, q + 1}. (b) If K is a maximal subgroup of L2 (q) whose index divides some nontrivial character degree of L2 (q), then K is a Frobenius group of index q + 1. Moreover q + 1 is the smallest index of maximal subgroups of L2 (q). Lemma 3.4. Let q 2 = 22m+1 , where m ≥ 1. Then the following hold.: (a) The Schur multiplier of Sz(q 2 ) is trivial when q 2 > 8 while the Schur multiplier of Sz(8) is elementary abelian of order 4, and √ cd(Sz(q 2 )) = {1, q 4 , q 4 + 1, (q 2 − 1)u1 , (q 2 − 1)u2 , q 2(q 2 − 1)/2}. (b) If K is a maximal subgroup of Sz(q 2 ) whose index divides some nontrivial character degree of Sz(q 2 ) or (q 2 − 1)2 , then K is a Frobenius group of index q 4 + 1. Moreover q 4 + 1 is the smallest index of maximal subgroups of Sz(q 2 ). 4. Verifying Huppert’s Conjecture for the simple Ree groups We are now ready to verify Huppert’s Conjecture for the simple Ree groups. 4.1. Verifying Step 1. Show G = G . By way of contradiction, suppose that G = G . Then there exists a normal subgroup N G of G such that G/N is solvable minimal with respect to being 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 [13, Theorem 1.1], we deduce that ψ(1) = StH (1) = rb and so r = 2. By Thompson’s Theorem [11, Corollary 12.2], G has a nonlinear character χ ∈ Irr(G) such that χ(1) is odd. As (χ(1), |G : N |) = 1, by Lemma 2.1(a), we have χN ∈ Irr(N ) and hence by Gallagher’s Theorem, we obtain that χτ ∈ Irr(G) for all τ ∈ Irr(G/N ). Thus StH (1) < StH (1)χ(1) ∈ cd(G), which contradicts Lemma 2.4. 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)(2), if χ ∈ Irr(G) such that χ(1) is isolated then either f = χ(1) or ra | χ(1)2 . By Lemma 2.4, the Steinberg character of H is isolated in H and hence by Lemma 2.3(b), either f = q 24 or r = 2. Assume first that f = q 24 . As r f, r must be odd. Let ϕ ∈ Irr(G) with √ ϕ(1) = q 2Φ1 Φ2 Φ2 Φ12 /2. As no proper multiple of f is a character of G and 4 ϕ(1) f, we deduce from Lemma 2.3(b)(1) that ra | ϕ(1)2 . As r is odd, we obtain ra | ϕ(1)2 = Φ2 Φ2 Φ4 Φ2 . We have 1 2 4 12 2 Φ1 Φ2 Φ2 Φ12 = (q 4 − 1)(q 2 + 1)(q 4 − q 2 + 1) = (q 4 − 1)(q 6 + 1) < q 10 . 4 As ra | ϕ(1)2 , we deduce that ra ≤ ϕ(1)2 < q 20 . But then as f | ra − 1, we obtain 2 2 f = q 24 ≤ ra − 1 < q 20 − 1 < q 20 , which is impossible. SIMPLE REE GROUPS 7 Thus r = 2. Then f ∈ cd(G) is odd and so f = q 4 Φ2 Φ2 Φ2 Φ2 /3, which is an 1 2 4 8 isolated degree of G by Lemma 3.1(v). Hence ra | (q 4 Φ2 Φ2 Φ2 Φ2 /3)2 by Lemma 1 2 4 8 2.3(b)(2). It follows that ra | q 8 as ra is even. As in the previous case, we have that f divides ra − 1 and since ra ≤ q 8 , we deduce that f ≤ q 8 − 1. However √ as q 8 − 1 < q 10 ≤ q 2Φ1 Φ2 Φ2 Φ12 /2, where the latter is the smallest nontrivial 4 character degree of H by Lemma 3.1(x), we see that f cannot be a character degree of G. This contradiction shows that G = G . 4.2. Verifying Step 2. Let M ≤ G be a normal subgroup of G such that G /M is a chief factor of G. As G is perfect, G /M is nonabelian so that G /M ∼ S k for = some nonabelian simple group S and some integer k ≥ 1. (i) Eliminating the alternating groups. Assume that S = An , n ≥ 7. Let θi , i = 1, 2 be irreducible characters of S obtained from Lemma 2.5(i). Then θ1 (1) = n(n − 3)/2, θ2 (1) = (n − 1)(n − 2)/2 = θ1 (1) + 1 and both θi extend to Aut(An ) ∼ = k Sn . By Lemma 2.6, θi ∈ Irr(G /M ) extend to G/M, hence θi (1)k ∈ cd(G) and θi (1)k , i = 1, 2, are coprime. By Lemma 3.1(vi), one of the degrees θi (1)k , i = 1, 2, must be q 24 . However we have that (n − 1, n − 2) = 1 and (n, n − 3) = (n, 3) so that θi (1)k can never be a power of 2. This shows that S is not an alternating group of degree at least 7. (ii) Eliminating the sporadic simple groups and the Tits group. It follows from [1, Table 1]) that there exist two nontrivial irreducible characters θi , i = 1, 2, such that θi extend to Aut(S), θi (1), i = 1, 2, are coprime and θi (1) are not 2-power. Now argue as in case (i), we obtain a contradiction. (iii) If S is a simple group of Lie type in characteristic p, with S = 2 F4 (2) , then k = 1 and p = 2. By way of contradiction, assume that k ≥ 2. Let θ be the Steinberg character of S. Then θ(1) = |S|p and θ extends to Aut(S). By Lemma 2.6, θk ∈ Irr(G /M ) extends to G/M, hence θ(1)k = |S|k ∈ cd(G). Since q 24 is p the unique nontrivial prime power character degree of G by [13, Theorem 1.1], we s deduce that θ(1)k = q 24 . In particular, we have p = 2. Write θ(1) = q1 . Then sk 24 k q1 = q . Let ψ = τ × θ × · · · × θ ∈ Irr(S ), where τ ∈ Irr(S) with 1 < τ (1) = s(k−1) θ(1). Then ψ(1) = τ (1)q1 q 24 and is nontrivial, so that it must divide some character degree of G, which is different from q 24 . By Lemma 3.1(viii), we have s(k−1) that q1 = q 24(k−1)/k < q 14 and hence 24(k − 1) < 14k, which implies that k ≤ 2. Therefore k = 2. Let C be a normal subgroup of G such that C/M = CG/M (G /M ). Then G C/C ∼ S 2 is a unique minimal normal subgroup of G/C = so that G/C embeds into Aut(S) Z2 , where Z2 is a cyclic group of order 2. Let B = Aut(S)2 ∩ G/C. Then |G/C : B| = 2. As above, let ψ = 1 × θ ∈ Irr(G C/C). Then ψ extends to B and so B is the inertia group of ψ in G/C so that by Lemma s 2.2(a), |G/C : B|ψ(1) = 2ψ(1) ∈ cd(G). Hence 2θ(1) = 2q1 = 2q 12 ∈ cd(G). 12 24 Obviously 1 < 2q < q , which leads to a contradiction again by using [13, Theorem 1.1]. Thus k = 1. (iv) If S is a simple group of Lie type in characteristic 2 and S = 2 F4 (2) , then ∼ 2 F4 (q 2 ). We will prove this by eliminating other possibilities for S. Assume that S= S is a simple group of Lie type in characteristic 2 and S is not the Tits group. We have shown that G /M ∼ S and |S|2 = q 24 = 212(2m+1) . Observe that if θ ∈ Irr(S) = is extendible to Aut(S), then θ extends to G/C, where C/M = CG/M (G /M ), so that θ(1) ∈ cd(G). In fact, we will choose θ to be a unipotent character of S, so that by results of Lusztig, θ is extendible to Aut(S) apart from some exceptions 8 HUNG P. TONG-VIET (see [12, Theorem 2.5]). We refer to [2, 13.8, 13.9] for the classification of unipotent characters and the notion of symbols. In Table 2, for each simple group of Lie type S in characteristic p, we list the p-part of some unipotent character of S that is extendible to Aut(S). (a) Case S ∼ Ln (2b ), where b ≥ 1 and n ≥ 2. We have bn(n − 1) = 24(2m + 1). = If n = 2 then b = 12(2m + 1) so that S = L2 (q 24 ) and hence S has a character of degree q 24 + 1. Obviously this degree does not divide any degree of G since q 24 +1 |2 F4 (q 2 )|. Next if n = 3 then b = 4(2m+1) and so S = L3 (q 8 ). By [2, (13.8)], S possesses a unipotent character parametrized by the partition (1, 2) of degree q 8 (q 8 + 1). However by checking Table 1, 2 F4 (q 2 ) has no such degree. If n = 4, then b = 2(2m+1) so that S = L4 (q 4 ). In this case, the unipotent character parametrized by the partition (2, 2) has degree q 8 (q 8 + 1). As above, this degree does not belong to cd(G). Thus we can assume that n ≥ 5. By Table 2, S possesses a unipotent character χ different from the Steinberg character with χ(1)2 = 2b(n−1)(n−2)/2 . By Lemma 3.1(viii), we have b(n − 1)(n − 2)/2 < 7(2m + 1). Multiplying both sides by 2n, we obtain bn(n − 1)(n − 2) = 24(2m + 1)(n − 2) < 14n(2m + 1), and so 24(n − 2) < 14n. Thus 5n < 24 so that n < 5, which is a contradiction. (b) Case S ∼ S2n (q1 ), or O2n+1 (q1 ), where q1 = 2b , b ≥ 1, n ≥ 2 and S = S4 (2). = As S2n (2b ) ∼ O2n+1 (2b ), we can assume S = S2n (q1 ) and S = S4 (2). We have = bn2 = 12(2m + 1). If n = 2 then b = 3(2m + 1) and so S = S4 (q 6 ). By [2, 1 (13.8)], S possesses a unipotent character labeled by the symbol 0 − 2 of degree 6 6 2 2 2 q (q − 1) /2. However by checking Table 1, F4 (q ) has no such character degree. 3 If n = 3 then 3b = 4(2m + 1) and so q1 = q 8 . By [2, (13.8)], S possesses a unipotent 12 3 2 2 character labeled by the symbol 1 with degree q1 (q1 − q1 + 1)(q1 + q1 + 1) = 8 2 2 q (q1 − q1 + 1)(q1 + q1 + 1), which leads to a contradiction since G has no degree 4 whose 2-part is q 8 . If n = 4 then 4b = 3(2m+1) and hence q1 = q 6 . By Table 2, there 9b−1 exists a unipotent character χ with χ(1)2 = 2 . As 4b = 3(2m + 1) and m ≥ 1, we have that 9b − 1 = 6(2m + 1) + (6m − 1)/4 = 13m + 6 + (2m − 1)/4 > 13m + 6, which contradicts Lemma 3.1(viii). Hence we can assume that n ≥ 5. By Table 2, S possesses a nontrivial irreducible character χ different from the Steinberg character 2 with χ(1)p = 2b(n−1) −1 . Since b(n − 1)2 − 1 ≥ b(n − 1)2 − b = bn(n − 2), by Lemma 3.1(viii), we have bn(n − 2) < 7(2m + 1). Multiplying both sides by n, we obtain bn2 (n − 2) = 12(n − 2)(2m + 1) < 7n(2m + 1) so that 5n < 24 and hence n < 5, a contradiction. (c) Case S ∼ O2n (q1 ), where q1 = 2b , b ≥ 1, and n ≥ 4. We have bn(n − 1) = = + 12(2m+1). If n = 4, then b = 2m+1 hence q1 = q 2 . Now if S = O8 (q 2 ), then S has − 2(42 −3·4+3) 14 a unipotent character χ with χ(1)2 = q = q and if S = O8 (q 2 ), then S 2 has a unipotent character χ with χ(1)2 = q 2(4 −3·4+2) = q 12 , by Table 2.√ However 2 14 13 2/2 and G has no such degrees by Table 1 since q ≥ 8. Observe that q > q √ q 13 2/2 = q 12 if and only if q 2 = 2. Thus we can assume that n ≥ 5. By Table 2, S possesses a unipotent character χ different from the Steinberg character with χ(1)2 ≥ 2b(n−1)(n−2) . By Lemma 3.1(viii), we have b(n − 1)(n − 2) < 7(2m + 1). Multiplying both sides by n, we obtain bn(n − 1)(n − 2) = 12(n − 2)(2m + 1) < 7n(2m + 1) so that 5n < 24 and hence n < 5, which is a contradiction. (d) Case S ∼ G2 (q1 ), where q1 = 2b , b ≥ 1. We have 6b = 12(2m + 1) and so = b = 2(2m+1). Thus S = G2 (q 4 ), where q 4 > 2 so that S has an irreducible character of degree q 24 − 1 by [5, Table IV-2]. However this degree divides no degrees of G since q 24 − 1 = Φ1 Φ2 Φ3 Φ4 Φ6 Φ8 Φ12 Φ24 |2 F4 (q 2 )|. SIMPLE REE GROUPS 9 2 2 (e) Case S ∼ 2 B2 (q1 ), where q1 = 22n+1 , n ≥ 1. We have 2(2n + 1) = 12(2m + 1) = and so 2n + 1 = 6(2m + 1), which is impossible. 2 2 (f ) Case S ∼ 2 G2 (q1 ), where q1 = 32n+1 , n ≥ 1. This case cannot occur as the = 2 2 characteristic of G2 (q1 ) is 3. 2 2 (g) Case S ∼ 2 F4 (q1 ), where q1 = 22n+1 , n ≥ 1. We have 12(2n+1) = 12(2m+1) = ∼ 2 F4 (q 2 ). so that n = m. Thus S = (h) For the remaining cases, we can argue as follows. We have |S|2 = 212(2m+1) . Lemma 3.1(viii) yields χ(1)2 ≤ 213m+6 , where χ is a unipotent character different from the Steinberg character listed in Table 2. Using these two properties, we will obtain a contradiction. For example, assume S ∼ E8 (q1 ), where q1 = 2b , b ≥ 1. We = have 120b = 12(2m + 1) and so 10b = 2m + 1. By Table 2, S possesses a unipotent character χ with χ(1)2 = 291b . By Lemma 3.1(viii), we have 91b < 7(2m + 1). Thus 91b < 7 · 10b = 70b, a contradiction. This completes the proof of Step 2. 4.3. Verifying Step 3. Suppose θ ∈ Irr(M ) with θ(1) = 1 and let I = IG (θ). We need to show that I = G . By way of contradiction, suppose I < G . Write s θI = i=1 ei µi , where µi ∈ Irr(I). As M ≤ I < G and G /M ∼ H, there exists a = subgroup U such that I ≤ U and U/M is a maximal subgroup of G /M. By Lemma 2.2(a), we have |G : I|µi (1) = |G : U ||U : I|µi (1) divides some character degree of G. Thus the index |G : U | = |G /M : U/M | of 2 F4 (q 2 ) must divide some character degree of 2 F4 (q 2 ) and hence by Lemma 3.2, U/M ∼ Pa or U/M ∼ Pb , where Pa and = = Pb are maximal parabolic subgroups of 2 F4 (q 2 ). Note that both unipotent radicals [q 22 ] and [q 20 ] of the parabolic subgroups Pa and Pb , respectively, are nonabelian by using the commutator relations (see [8]). Let t = |U : I| and recall that if N G and λ ∈ Irr(N ), then Irr(G|λ) denotes the set of all irreducible constituents of λG . Case 1 : U/M ∼ Pa . Recall that Pa ∼ [q 22 ] : (L2 (q 2 ) × (q 2 − 1)). Let L and V be = = subgroups of U containing M such that L/M ∼ [q 22 ] and V /M ∼ L2 (q 2 ), and let = = W = LV. Then M W U, L ∩ V = M, and W/L ∼ V /M ∼ L2 (q 2 ). By Lemma = = 3.2(i), tµi (1) divides q 2 ± 1 or q 2 . We consider the following cases: Case 1(a) : tµj (1) | q 2 ± 1 for some j. Then t is odd. As L U, we deduce that I ≤ IL ≤ U so that t = |U : IL| · |IL : I|. Now |IL : I| = |L : L ∩ I|. As |L : L ∩ I| divides |L : M | = q 20 , if L ≤ I, then |L : L ∩ I| > 1 is even so that t is even, a contradiction. Thus L ≤ I ≤ U. Assume W ≤ I. Then I WI ≤ U and t = |U : W I| · |W I : I|. As |W I : I| = |W : W ∩ I| and |W I : I| > 1, we deduce that |W : W ∩ I| > 1 and divides q 2 ± 1. Observe that W/L ∼ L2 (q 2 ) and = since L ≤ W ∩ I W, we deduce that |W : W ∩ I| is divisible by some index of a maximal subgroup of L2 (q 2 ). By Lemma 3.3(b), we have |W : W ∩ I| = q 2 + 1 and W ∩ I/L is isomorphic to the Borel subgroup of L2 (q 2 ), in particular W ∩ I/L is nonabelian. Furthermore t = q 2 +1 and hence tµi (1) | q 2 ±1, for all i, which implies that µi (1) = 1 for all i. Thus θ extends to θ0 ∈ Irr(I). By Gallagher’s Theorem, we have θI = τ ∈Irr(I/M ) τ θ0 and so τ (1) = 1 for all τ ∈ Irr(I/M ), which implies that I/M is abelian, which is a contradiction as I/M possesses a nonabelian section W ∩ I/L. Thus W ≤ I. Let λ be an irreducible constituent of (µj )L . Then λM = eθ for some integer e. Now λ(1) = e divides |L/M | = q 22 by [11, Corollary 11.29]. By [11, Lemma 6.8], we have that λ(1) | µj (1), which yields that λ(1) is odd. Thus e = 1 and so λ is an extension of θ to L. Since L/M ∼ [q 22 ] is nonabelian, it possesses = a nonlinear irreducible character τ with even degree. By Gallagher’s Theorem, γ = τ λ ∈ Irr(L|θ) and γ(1) = τ (1)λ(1) is even. If µk is any irreducible constituent 10 HUNG P. TONG-VIET of γ I then as L I, by [11, Lemma 6.8] we have that γ(1) | µk (1) and since tµk (1) divides q 2 ± 1 or q 2 , we deduce that tµk (1) | q 2 and so as t is odd, t = 1 and hence I = U and µk (1) is a 2-power for any µk ∈ Irr(I|γ). Let L ≤ J = IW (γ) and suppose that J < W. Let δ ∈ Irr(J|γ). We have δ W ∈ Irr(W |γ) and δ W (1) = |W : J|δ(1). Since W I, we deduce that |W : J|δ(1) | q 2 and hence |W : J| ≤ q 2 , and |W : J| is divisible by the index of some maximal subgroup of W/L ∼ L2 (q 2 ), so that by = Lemma 3.3(b), we obtain |W : J| ≥ q 2 + 1, a contradiction. Thus γ is W -invariant and every irreducible constituent of Irr(W |γ) is a 2-power. By Lemma 2.7, we deduce that W/L ∼ L2 (q 2 ) is solvable, which is impossible as q 2 ≥ 8. Thus this = case cannot happen. Case 1(b) : tµi (1) | q 2 for all i. Then t | q 2 and all µi (1) are 2-powers. We will show that I/M is nonsolvable. If t = 1, then I = U, hence I/M ∼ Pa is = nonsolvable. Assume t > 1. As W/M ∼ [q 22 ] : L2 (q 2 ) is nonsolvable, if W ≤ I, = then we are done. So assume W ≤ I. Let X = W ∩ I. Then X W. Since |W I : I| = |W : W ∩ I| = |W : X|, we have t = |U : W I| · |W : X|, and hence |W : X| is a nontrivial divisor of q 2 . If L ≤ X, then X/L is a proper subgroup of W/L ∼ L2 (q 2 ) and hence |W : X| ≥ q 2 +1 by Lemma 3.3(b), which is impossible as = 1 < |W : X| ≤ q 2 . Thus L ≤ X. Since L W and X = W ∩ I ≤ W, we deduce that X XL ≤ W and L XL ≤ W. It follows that |W : X| = |W : XL| · |XL : X| is a nontrivial divisor of q 2 . If |W : XL| > 1, then as L ≤ XL W, by Lemma 3.3(b), we obtain |W : XL| ≥ q 2 + 1, which is impossible since |W : XL| ≤ |W : X| ≤ q 2 . Thus W = XL and so L2 (q 2 ) ∼ W/L ∼ X/X ∩ L. We have M X ∩ L X I and = = X/X ∩L ∼ L2 (q 2 ) so that I/M is nonsolvable. Hence µi (1) are 2-power for all i, θ is = I-invariant and I/M is nonsolvable. Now Lemma 2.7 will provide a contradiction. Case 2 : U/M ∼ Pb . Recall that Pb ∼ [q 20 ] : (Sz(q 2 ) × (q 2 − 1)). Let L and V = = be subgroups of U containing M such that L/M ∼ [q 20 ], V /M ∼ Sz(q 2 ) and let = = W = LV. It follows that L U, W U, V ∩ L = M, and W/L ∼ V /M ∼ Sz(q 2 ). Let = = √ B = {q 4 , q 4 + 1, (q 2 − 1)u1 , (q 2 − 1)u2 , q 2(q 2 − 1)/2, (q 2 − 1)2 }. By Lemma 3.2(ii), we deduce that for each i, tµi (1) divides one of the members of B. Case 2(a) : t is odd. As L U, we have I ≤ IL ≤ U so that t = |U : IL|·|IL : I|. As |IL : I| = |L : L ∩ I|, if L ≤ I, then |L : L ∩ I| > 1 is even so is t, a contradiction. Thus L ≤ I ≤ U. Assume that W ≤ I. Then I W I ≤ U and t = |U : W I| · |W I : I|. As |W I : I| = |W : W ∩ I| and |W I : I| > 1, we deduce that |W : W ∩ I| > 1 and divides one of the members in B. Observe that W/L ∼ Sz(q 2 ) = and since L ≤ W ∩ I W, we deduce that |W : W ∩ I| is divisible by some index of a maximal subgroup of Sz(q 2 ). By Lemma 3.4(b), we have |W : W ∩ I| = q 4 + 1 and W ∩ I/L is isomorphic to the Borel subgroup of Sz(q 2 ), in particular W ∩ I/L is nonabelian. It follows that t = q 4 + 1 and hence tµi (1) = q 4 + 1, as q 4 + 1 divides no other members of B, which implies that µi (1) = 1 for all i. Now arguing as in the first paragraph of Case 1(a), we obtain a contradiction. Therefore W ≤ I ≤ U. Let λ be an irreducible constituent of θL . We have that λ(1) = eθ(1) for some integer e. By [11, Corollary 11.29], we deduce that e | q 20 . If e = 1 then θ extends to λ ∈ Irr(L) so that as L/M is nonabelian, L/M has a nontrivial irreducible character τ of even degree and then by Gallagher’s Theorem γ = τ λ ∈ Irr(L|θ) with γ(1) is even. If e > 1, then obviously e is even and so we choose γ = λ ∈ Irr(L|θ) and γ(1) is even. In both cases, we can choose γ ∈ Irr(L|θ) such that γ(1) is even. Let J be the stabilizer in W of γ. Write γ J = δ1 + δ2 + · · · + δk , where δi ∈ Irr(J). Since L W I, the degrees of irreducible constituents of γ W divide some µi (1) and SIMPLE REE GROUPS 11 W since µi (1) divides one of the member of B, we deduce that δi (1) = |W : J|δi (1) √ 2 4 divides either q or q 2(q − 1)/2 for all i, as δi (1) is even since γ(1) | δi (1). By Lemma 3.4(b), we can deduce that γ is W -invariant and if ϕ ∈ Irr(W |γ) then as √ W I, we have that ϕ(1) divides q 4 or q 2(q 2 − 1)/2. Assume first that q 2 > 8. By Lemma 3.4(a), the Schur multiplier of W/L ∼ = Sz(q 2 ) is trivial so that by [11, Theorem 11.7], γ extends to γ0 ∈ Irr(W ). Hence by Gallagher’s Theorem, τ γ0 are all the irreducible constituents of γ W , where τ ∈ √ Irr(W/L), and thus τ (1)γ0 (1) = γ(1)τ (1) divides (q 2 − 1)q 2/2, or q 4 . But this is impossible since W/L ∼ Sz(q 2 ) has an irreducible character of degree q 4 + 1, which = divides none of the degrees above. Now assume q 2 = 8. We have W/L ∼ Sz(8), γ ∈ Irr(L) is W -invariant and all = irreducible constituents of γ W divide 64 or 14. Write γ W = f1 φ1 + f2 φ2 + · · · , where φi ∈ Irr(W ). Then φi (1) = fi γ(1) divides 64 or 14. If fi = 1 for some i, then γ extends to γ0 ∈ Irr(W ), and hence argue as in the previous case to obtain a contradiction. Thus fi > 1 for all i. Also by Lemma 2.2(c), all fi are character degrees of irreducible projective representations of W/L ∼ Sz(8). Thus = all fi > 1 are character degrees of irreducible projective representations of Sz(8) with fi dividing 64 or 14. Using [3], we have cd(Sz(8)) = {1, 14, 35, 64, 65, 91}, and the projective but not ordinary degrees of Sz(8) are 40, 56, 64, 104. It follows that fi is either 14 or 64. Since γ W = f1 φ1 + · · · + ft φt , and φi (1) = fi γ(1), we deduce t that |Sz(8)| = i=1 fi2 . Let a and b be the numbers of fi s which equal 14 and 64, respectively. Then 82 · 5 · 7 · 13 = 142 a + 642 b. Obviously both a and b are nonzero. We have 4 · 16 · 5 · 7 · 13 = 4 · 72 a + 4 · 16 · 82 b. After simplifying, we obtain 16 · 5 · 7 · 13 = 72 a + 16 · 82 b, and then b = 7b1 and a = 16a1 , where a1 , b1 ≥ 1 are integers. Thus 5 · 13 = 7a1 + 82 b1 so that 65 = 7a1 + 64b1 . As b1 ≥ 1, we deduce that b1 = 1 and then 1 = 7a1 , which is impossible. √ Case 2(b) : t > 1 is even. Then tµi (1) divides q 4 or q 2(q 2 − 1)/2 for all i. Let X = W ∩ I I. Assume first that L ≤ X. Then L ≤ I. Suppose that W ≤ I. Then I W I ≤ U and t = |U : W I| · |W I : I|. As |W I √I| = |W : X| and : |W I : I| > 1, we deduce that |W : X| > 1 and divides q 4 or q 2(q 2 − 1)/2. Hence |W : X| ≤ q 4 and since L ≤ X W, we deduce that |W : X| is divisible by the index of some maximal subgroup of Sz(q 2 ), which is impossible by Lemma 3.4(b). Thus W ≤ I. But then t = |U : I| divides |U : W | = q 2 − 1, which is an odd number, a contradiction. Thus L ≤ X. It follows that X XL ≤ W. We have that |W : X| = |W : XL| · |XL : X| > 1 and |W : X| ≤ t ≤ q 4 as t = |U : W I| · |W : X|. Assume XL < W. We have L ≤ XL < W and X ≤ XL < W. Since W/L ∼ = Sz(q 2 ), we deduce that |W : XL| is divisible by the index of some maximal subgroup of Sz(q 2 ), and hence by Lemma 3.4(b), |W : XL| ≥ q 4 + 1, which is impossible since |W : XL| ≤ |W : X| ≤ q 4 . Hence W = XL and so W/L ∼ XL/L ∼ X/X ∩ L. Let L1 = X ∩ L. Then = = M L1 X I and X/L1 ∼ Sz(q 2 ). Let λ ∈ Irr(L1 |θ). Observe that the degree of = every irreducible constituent of λX must divide some degree µi (1) so that it divides √ q 4 or q 2(q 2 − 1)/2. Now Lemma 3.4(b) yields that λ is X-invariant. Applying the same argument as in the last two paragraphs of Case 2(a) for λ ∈ Irr(L1 |θ) and L1 X with X/L1 ∼ Sz(q 2 ), we obtain a contradiction. = This finishes the proof of Step 3. 4.4. Verifying Step 4. Show M = 1. We have shown that G /M ∼ 2 F4 (q 2 ) = and for any θ ∈ Irr(M ), if θ(1) = 1, then θ is G -invariant so that by Lemma 12 HUNG P. TONG-VIET Table 1. Character degrees of 2 F4 (q 2 ) Degree 1√ q 2Φ1 Φ2 Φ2 Φ12 /2 4 q 2 Φ12 Φ24 Φ1 Φ2 Φ2 Φ24 8 q 4 · u2 · w1 · Φ2 Φ2 Φ12 /12 1 1 2 q 4 · u2 · w2 · Φ2 Φ2 Φ12 /12 2 1 2 q 4 Φ2 Φ2 Φ2 Φ24 /6 1 2 4 q 4 · w1 · Φ2 Φ2 Φ2 Φ12 /4 1 2 4 q 4 · u2 · w2 · Φ2 Φ12 /4 1 4 q 4 · w2 · Φ2 Φ2 Φ2 Φ12 /4 1 2 4 q 4 · u2 · w1 · Φ2 Φ12 /4 2 4 q 4 Φ2 Φ2 Φ12 Φ24 /3 1 2 q 4 Φ2 Φ2 Φ2 Φ2 /3 1 2 4 8 q 4 Φ2 Φ24 /2 8 u1 · Φ1 Φ2 Φ2 Φ12 Φ24 4 Φ2 Φ8 Φ12 Φ24 4 u2 · Φ1 Φ2 Φ2 Φ12 Φ24 4 q 2 Φ2 Φ2 Φ2 Φ24 1 2 8 Φ1 Φ2 Φ2 Φ12 Φ24 8 q 10 Φ12 Φ24 Φ4 Φ2 Φ12 Φ24 √ 8 q 2 · u1 · Φ2 Φ2 Φ2 Φ12 Φ24 /2 1 2 4 √ q 13 2Φ1 Φ2 Φ2 Φ12 /2 4 √ q √2Φ1 Φ2 Φ2 Φ8 Φ12 Φ24 /2 4 q 2 · u2 · Φ2 Φ2 Φ2 Φ12 Φ24 /2 1 2 4 u2 · Φ2 Φ2 Φ2 Φ12 Φ24 1 1 2 4 w1 · Φ2 Φ2 Φ2 Φ2 Φ12 1 2 4 8 q 4 · u1 · Φ1 Φ2 Φ2 Φ12 Φ24 4 u1 · Φ1 Φ2 Φ2 Φ8 Φ12 Φ24 4 Φ2 Φ2 Φ2 Φ12 Φ24 1 2 8 q 2 Φ1 Φ2 Φ2 Φ12 Φ24 8 Φ2 Φ2 Φ2 Φ8 Φ12 Φ24 1 2 4 q 6 Φ1 Φ2 Φ2 Φ24 8 Φ1 Φ2 Φ4 Φ2 Φ12 Φ24 8 Φ2 Φ2 Φ2 Φ2 Φ24 1 2 4 8 q 24 q 2 Φ4 Φ2 Φ12 Φ24 8 q 4 Φ2 Φ8 Φ12 Φ24 4 Φ2 Φ2 Φ12 Φ24 4 8 w2 · Φ2 Φ2 Φ2 Φ2 Φ12 1 2 4 8 q 4 · u2 · Φ1 Φ2 Φ2 Φ12 Φ24 4 u2 · Φ1 Φ2 Φ2 Φ8 Φ12 Φ24 4 u2 · Φ2 Φ2 Φ2 Φ12 Φ24 2 1 2 4 Multiplicity 1 2 1 1 1 1 1 2 1 2 1 1 2 √ 1 q(q + 2)/4 (q 2√ 2)/2 − (q − 2)q/4 1 (q 2 − 2)/2 1 (q 2√ 2)/2 − (q + 2)q/2 2 q2 − 2 √ √ (q2− 2)q/2 (q + 2√2)(q − 2)q/96 (q + 2)(q 2 +√ 1)q/12 √ (q + √2)q/4 (q − 2)q(q + 2)2 /8 (q 2 − 8)(q 2 − 2)/48 (q 2 − 2)/2 2 (q − 2)q 2 /16 1 (q 2 − 2)q 2 /4 (q 2 − 2)(q 2 + 1)/6 1 (q 2 − 2)/2 (q 2 − 2)/2 2 (q √ 8)(q 2 − 2)/16 − (q − 2)(q 2 +√ 1)q/12 (q − √2)q/4 √ q+ √ 2)q(q − 2)2 /8 (q − 2 2)(q 2 − 2)q/96 SIMPLE REE GROUPS 13 Table 2. Some unipotent characters of simple groups of Lie type S = S(pb ) Ln (pb ), n ≥ 3 S2n (pb ), p = 2 S2n (pb ), p > 2 O2n+1 (pb ), p > 2 + O2n (pb ) − O2n (pb ) 3 D4 (pb ) F4 (pb ) 2 F4 (q 2 ) E6 (pb ) 2 E6 (pb ) E7 (pb ) E8 (pb ) Symbol (1n−2 , 2) 0 1 2 ··· n−2 n−1 n 1 2··· n−2 0 1 2 ··· n−2 n−1 n 1 2··· n−2 0 1 2 ··· n−3 n−1 1 2 3··· n−2 n−1 0 1 2 ··· n−2 n−1 1 2··· n−2 φ1,3 φ9,10 2 B2 [a], φ6,25 φ2,16 φ7,46 φ8,91 p-part of degree pb(n−1)(n−2)/2 2 2b(n−1) −1 2 pb(n−1) 2 pb(n−1) 2 pb(n −3n+3) 2 pb(n −3n+2) p7b p10b 1 √ q 13 2 p25b p25b p46b p91b Table 3. The maximal subgroups of 2 F4 (q 2 ) Group Index Pa = [q 22 ] : (L2 (q 2 ) × (q 2 − 1)) (q 12 + 1)(q 6 + 1)(q 4 + 1) Pb = [q 20 ] : (Sz(q 2 ) × (q 2 − 1)) (q 12 + 1)(q 6 + 1)(q 2 + 1) 2 18 12 3.U3 (q ) : 2 q (q + 1)(q 4 + 1)(q 2 − 1)/2 24 4 (Zq2 +1 × Zq2 +1 ) : GL2 (3) q (q + 1)2 (q 2 − 1)2 .Φ12 Φ24 /(3.24 ) √ √ (Zq2 − 2q+1 × Zq2 − 2q+1 ) : [96] q 24 (q 4 − 1)2 .u2 .Φ12 Φ24 /(3.25 ) 2 √ √ (Zq2 + 2q+1 × Zq2 + 2q+1 ) : [96] q 24 (q 4 − 1)2 .u2 .Φ12 Φ24 /(3.25 ) 1 Zq4 −√2q3 +q2 −√2q+1 : 12 q 24 (q 8 − 1)2 w2 .Φ12 /(3.22 ) q 24 (q 8 − 1)2 w1 .Φ12 /(3.22 ) Zq4 +√2q3 +q2 +√2q+1 : 12 2 P GU3 (q ) : 2 q 18 (q 4 + 1)(q 2 − 1)Φ24 /2 2 Sz(q ) 2 q 16 (q 6 + 1)(q 2 + 1)Φ24 /2 2 Sz(q ) : 2 q 20 (q 8 − 1)(q 6 + 1)Φ24 /2 24α 12α 8α 6α 2α q0 (q0 +1)(q0 −1)(q0 +1)(q0 −1) 2 2 2α F4 (q0 ), q 2 = q0 , α prime q 24 (q 12 +1)(q 8 −1)(q 6 +1)(q 2 −1) 0 0 0 0 0 2.8, |M : M | divides the order of the Schur multiplier of 2 F4 (q 2 ). As the Schur multiplier of 2 F4 (q 2 ), q 2 ≥ 8, is trivial, we deduce that M = M . If M is abelian then we are done. Assume that M is nonabelian. Let N ≤ M be a normal subgroup of G such that M/N is a chief factor of G . It follows that M/N ∼ = S k , for some nonabelian simple group S. By Lemmas 2.5 and 2.6, S possesses a nontrivial irreducible character ϕ such that ϕk ∈ Irr(M/N ) which extends to G /N. Gallagher’s Theorem yields ϕ(1)k τ (1) ∈ cd(G /N ) ⊆ cd(G ) for any τ ∈ Irr(G /M ) ⊆ Irr(G /N ). Since cd(G /M ) = cd(G) and ϕ(1) > 1, if we choose τ to be the Steinberg character of G /M, then ϕ(1)k τ (1) cannot divide any degree of G, a contradiction. Thus M = 1. 4.5. Verifying Step 5. G = G ×CG (G ). It follows from Step 4 that G ∼ 2 F4 (q 2 ) = is a nonabelian simple group. Let C = CG (G ). Then G/C is almost simple with 14 HUNG P. TONG-VIET socle 2 F4 (q 2 ). Assume that G × C < G. Then G induces some outer automorphism on G . Note that the only nontrivial outer automorphisms of 2 F4 (q 2 ) are field automorphisms. Let σ be a nontrivial outer automorphism of G . By [6, Theorem C], σ does not fix some conjugacy class of G , and so by [11, Theorem 6.32], the action of σ on the conjugacy classes of G is permutation isomorphic to the action of σ on Irr(G ), so that σ does not fix some nontrivial irreducible character ψ ∈ Irr(G ). Let γ ∈ Irr(G) be an irreducible constituent of ψ G . As ψ is not σ-invariant, we deduce that γ(1) = zψ(1), where z > 1 and z | |Out(G )| = 2m + 1. We have ψ(1) > 1, zψ(1) ∈ cd(G) and ψ(1) ∈ cd(G ) = cd(G). Thus ψ(1) and zψ(1) are in cd(G) with z > 1 being odd, so that by Lemma 3.1(ix), we have that z ≥ q 2 − 1. But then as z | 2m + 1, we have 2m + 1 ≥ z ≥ 22m+1 − 1, which is impossible as m ≥ 1. Thus G = G × C. It follows that C ∼ G/G is abelian. The proof is now complete. = Acknowledgment. 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