Upper bounds for the number of irreducible character degrees of a group

Journal of Algebra 403 (2014) 201–222

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Journal of Algebra www.elsevier.com/locate/jalgebra

Upper bounds for the number of irreducible character degrees of a group I.M. Isaacs Mathematics Department, University of Wisconsin, 480 Lincoln Dr., Madison, WI 53703, USA

a r t i c l e

i n f o

Article history: Received 6 August 2013 Available online 29 January 2014 Communicated by Ronald Solomon MSC: 20C15

a b s t r a c t Let G be a p-solvable group for some prime p, and suppose that the largest irreducible character degree of G is kp, where k < p is an integer. We obtain an upper bound on the total number of different irreducible character degrees of G, where the bound depends on k but not on p. © 2014 Elsevier Inc. All rights reserved.

Keywords: Irreducible character Character degree p-Solvable group

1. Introduction Given a finite group G, then as usual, we write cd(G) to denote the set {χ(1) | χ ∈ Irr(G)} of irreducible character degrees of G, and we write b(G) to denote the largest member of cd(G). It is obvious that |cd(G)|  b(G), but this upper bound on the number of irreducible character degrees is almost never attained. In particular, a result of B. Huppert asserts that equality cannot occur if b(G) > 6, and for solvable groups, equality never occurs for b(G) > 4. (See Theorem 32.1 of [4].) In fact, a recent result of M. Lewis and A. Moretó shows that if b(G) is sufficiently large, then |cd(G)| < b(G)ε where ε > 0 is arbitrary. (See Theorem A of [8].) E-mail address: [email protected]. 0021-8693/$ – see front matter © 2014 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jalgebra.2013.12.030

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For certain special values of b(G), very much stronger upper bounds on |cd(G)| are known or conjectured. For example, Lewis and Moretó show in [8] that if b(G) = p, a prime number, then |cd(G)|  4, and in fact, they proved that the upper bound 4 can occur only if p is a Fermat prime and p − 1 is one of the four degrees. (In a private communication, Lewis told me that for solvable groups G with b(G) prime, it is not actually possible to have |cd(G)| = 4.) Lewis and Moretó conjectured that more generally, |cd(G)| is bounded in terms of the number of primes (counting multiplicities) that divide b(G). Our goal here is to establish a different generalization of the case where b(G) is prime, at least for p-solvable groups. Theorem A. Let G be p-solvable for some prime p, and suppose that b(G) = kp, where k < p is an integer. Then |cd(G)| < (5 + 2 ln(k))k. Note that in the Lewis–Moretó situation, k = 1, and Theorem A yields their inequality |cd(G)|  4, at least for p-solvable groups. We obtain our upper bound under a hypothesis that is somewhat weaker than that in Theorem A. We do not require that the integer kp actually lies in cd(G); it is enough to suppose that b(G)  kp and that some member of cd(G) is a multiple of p. The upper bound k(5 + 2 ln(k)) in Theorem A is less precise (but also less technical) than what we actually prove. To state our main result, we will say that an integer f is good with respect to a prime p if f is not divisible by any prime power q > 1 such that q is congruent to −1, 0 or 1 mod p. Thus, for example, integers less than p − 1 are always good for p, and p − 1 is good for p except when p is Fermat. (This is because the smallest that a “bad” divisor q can be is p − 1.) Theorem B. Let G be p-solvable, and suppose that cd(G) contains a multiple of the prime p. Let k < p be an integer, and assume that the largest multiple of p in cd(G) is at most kp. Then the number of members f ∈ cd(G) that are good for p is at most 2k. We stress that in Theorem B, it is not necessary to assume any particular upper bound on the members of cd(G); it is enough that the multiples of p in cd(G) are at most kp. It follows from Theorem B that if G is p-solvable, then cd(G) cannot be the set {1, 2, 3, 4, 5, 6}. To see this, we can take p = 5, and k = 1, and so by Theorem B, there can be at most two degrees that are good for 5. We see, however, that 1, 2, 3 and 6 are good for 5, so these four integers cannot all be degrees. We mention, however, that if G = SL(2, 5), then cd(G) = {1, 2, 3, 4, 5, 6}, and it follows that the p-solvability assumption in Theorem B cannot be dropped. Note that the condition in Theorem B that cd(G) contains some multiple of p is equivalent to assuming that G fails to have an abelian normal Sylow p-subgroup. This result of Itô was originally proved for solvable groups, but the proof goes through essentially unchanged for p-solvable groups. (See Theorem 12.33 and Corollary 12.34 of [5].) In fact, by work of G. Michler using the simple group classification, Itô’s theorem holds for all finite groups, but we shall not need that generality here.

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The following consequence of Theorem B includes Theorem A. Corollary C. Let G be p-solvable, and assume that b(G)  kp, where p is prime and k < p is an integer. Assume also that some member of cd(G) is divisible by p. Then |cd(G)| < k(5 + 2 ln(k)). Proof. Theorem B guarantees that G has at most 2k degrees that are good for p, and so it suffices to count the remaining degrees. There are clearly at most k multiples of p in cd(G), so it remains to determine the possibilities for degrees that are divisible by some integer q > 1, where q ≡ ±1 mod p. (We ignore the additional requirement that q should be a prime power.) First, for q ≡ 1 mod p with q > 1, we can write q = rp + 1, where r  1 is an integer. A degree f divisible by q has the form f = s(rp + 1), where s is an integer. Since f  kp, we have rs < k, and in particular, s < k. We argue that the number of possibilities for f is less than (1 + ln(k))k. If s = 1, there are fewer than k possibilities for r, and hence for f . If s = 2, there are fewer than k/2 possibilities; if s = 3 there are fewer than k/3 possibilities, and so on, up to s = k − 1. The total number of possibilities for f , therefore, is less than     1 1 1  1 + ln(k) k, k 1 + + + ··· + 2 3 k−1 as required. For q ≡ −1 mod p, where q divides a degree f , we have f = s(rp − 1), and thus s(rp − 1)  kp. If s = p, we have already counted this degree, and if s > p, we have rp − 1  kp/s < k  p − 1, and this is impossible. We thus need to consider only integers s < p, and since we have sr  k + s/p and k is an integer, we see that sr  k. Also, r  1, and so 1  s  k, and reasoning as before, we deduce that there are at most k(1 + ln(k)) degrees of this type. Then         cd(G) < 2k + k + k 1 + ln(k) + k 1 + ln(k) = k 5 + 2 ln(k) , and the proof is complete. 2 If k = 1, the statement of Corollary C yields |cd(G)|  4, but since we ignored the requirement that q should be a prime power, we actually get |cd(G)|  3 except when p is a Fermat prime and p − 1 is a degree. This agrees with the Lewis–Moretó theorem. 2. Preliminary results We begin with some general character-theoretic results. Some version of each of these is known, but since their proofs are short, we have decided to sketch them here. The following arguments rely on various facts that can be found in [5].

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Lemma 2.1. Suppose that N  G, where G/N is p-solvable. Let θ ∈ Irr(N ) be invariant in G, and assume that the degree θ(1) and the determinantal order o(θ) of θ are powers of p. Then there exists χ ∈ Irr(G|θ) such that χ(1) and o(χ) are powers of p. Proof. We weaken the assumption that θ is invariant in G, and we assume only that θ is invariant in H, where H/N is some p-complement in G/N . There is nothing to prove if N = G, so we assume that N < G, and we proceed by induction on |G : N |. Let M/N be a chief factor of G. If M/N is a p -group, then M ⊆ H, and thus θ is invariant in M . Also, θ(1) and o(θ) are coprime to |M : N |, so θ extends to M , and in this situation, there is a unique extension ϕ such that o(ϕ) = o(θ). Because of the uniqueness, ϕ is invariant in H, and we observe that H/M is a p-complement in G/M . Also, since o(ϕ) = o(θ) and ϕ(1) = θ(1) are powers of p, the inductive hypothesis applies, so there exists χ ∈ Irr(G|ϕ), with the required properties. We can now suppose that M/N is a p-group. The p -group H/M acts on the p-group Irr(M/N ), and it also acts on the set Ω = Irr(M |θ). Furthermore, Irr(M/N ) is a group that acts transitively on Ω by multiplication. An argument using Glauberman’s lemma shows that some member ϕ ∈ Ω is H-invariant, and thus it is invariant in HM , and we observe that HM/M is a p-complement in G/M . Now ϕ(1)/θ(1) divides |M : N |, so ϕ(1) is a p-power. Also, if λ = det(ϕ) and μ = det(θ), then λN is a power of μ, and thus ker(μ) ⊆ ker(λ). Since μ has p-power order, we see that |N : ker(μ)| is a power of p, and thus M/ker(μ) is a p-group. But λ can be viewed as a character of this p-group, so it follows that o(ϕ) = o(λ) is a power of p. The result now follows by the inductive hypothesis. 2 Lemma 2.2. Let N  G, and let λ be a linear character of N , where o(λ) is a power of some prime p. Suppose that there exists χ ∈ Irr(G|λ) with degree not divisible by p. Then λ extends to its stabilizer in G. Proof. Let T be the stabilizer of λ, and let ψ ∈ Irr(T |λ) be the Clifford correspondent of χ with respect to λ. Then ψ G = χ, so ψ(1) divides χ(1), and in particular, p does not divide ψ(1). To show that λ extends to T , it suffices to show that λ extends to S, where S/N is an arbitrary Sylow subgroup of T /N . If S/N is a q-group, with q = p, then since λ has p-power order and degree 1, and it is invariant in S, it follows that λ extends to S, as required. We can now assume that S/N is a p-group. Since p does not divide ψ(1), it also does not divide η(1) for some irreducible constituent η of ψS , and thus ηN is irreducible. But ψN is a multiple of λ, and thus ηN is a multiple of λ, and we deduce that ηN = λ. Thus λ extends to S in this case too. 2 For our purposes, the significance of the extendibility of a character of a normal subgroup to its stabilizer is given by the following well-known fact.

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Lemma 2.3. Let N  G, and suppose that θ ∈ Irr(N ) extends to its stabilizer T in G. Then cd(G|θ) = {|G : T |eθ(1) | e ∈ cd(T /N )}. Proof. By the Clifford correspondence, the members of Irr(G|θ) are exactly the characters η G for η ∈ Irr(T |θ). Also, if ϕ is an extension of θ to T , then by Gallagher’s theorem, the members of Irr(T |θ) are exactly the characters ϕβ for β ∈ Irr(T /N ). Since the degree of (ϕβ)G is |G : T |β(1)ϕ(1) and ϕ(1) = θ(1), the result follows. 2 3. Small orbits In this section we study the situation where a group G acts faithfully on a p-group P in such a way that all of the G-orbits in P are small. Although we will not use the full strength of all results in this section, it is interesting to work as generally as we can, while keeping our arguments elementary. Our key result is the following. Theorem 3.1. Let G act faithfully on a p-group P , and assume that all G-orbits in P have size at most k. (a) If k  (p + 1)/2, then there is a regular G-orbit in P , and thus |G|  k. (b) If k  p − 1, there is a regular G-orbit on P × P , and thus |G|  k2 . We mention that since k < p in Theorem 3.1, it is easy to see that p cannot divide |G|. This is because the action of G is faithful, so a subgroup of order p in G would have an orbit of size p, and this is impossible. We shall not use this observation in the proof of Theorem 3.1, however. It is not clear if Theorem 3.1(a) would remain true if we weaken the restriction on k and assume only that k  p − 1. If we allow orbit sizes as large as p, however, there definitely need not be regular orbits. For an easy counterexample, take P to be a nonabelian group of order p3 , and let G be the group of inner automorphisms of P . The G-orbits in P in this case are the conjugacy classes of P , and these have sizes 1 and p, so there is no orbit with size |G| = p2 . This example may seem unsatisfactory because we have seen that p does not divide |G| under the hypotheses of Theorem 3.1. Perhaps a better question is this: Assuming that p does not divide |G|, what is the largest the integer k can be so that if every G-orbit on P has size at most k, then a regular orbit is guaranteed to exist. In fact, k cannot be as large as p + 1. To see this, observe that a dihedral group G of order 2(p + 1) acts faithfully on an elementary abelian group P of order p2 , and in this situation, all orbits have size p + 1, so there is no regular orbit. (The key observation here is that each of the p + 1 involutions in G outside of the cyclic subgroup of order p + 1 centralizes one of the p + 1 subgroups of order p in P .) For solvable groups G acting faithfully on a group P of coprime order, a result of S. Dolfi [1] asserts that there is always a regular orbit on P × P . In the case where G

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is solvable, therefore, our Theorem 3.1(b) is immediate from Dolfi’s result. Our proof, however, is far easier than Dolfi’s argument, and we do not require solvability, but of course, we rely on the very strong small-orbit assumption, which Dolfi does not need. We mention that Dolfi’s result has recently been generalized by Z. Halasi and K. Podosky [3], who showed that Dolfi’s solvability assumption can be dropped. This requires an appeal to the classification of simple groups. The following easy lemma provides the key to the proofs of both parts of Theorem 3.1. In this result, and throughout the rest of this section, we write Gx to denote the stabilizer in G of a point x ∈ P , where G acts on P . Lemma 3.2. Let G act on a p-group P , and suppose x, y ∈ P . Then there exist point stabilizers Hi in G for 0  i  p, where these subgroups have the property that Hi ∩ Hj = Gx ∩ Gy whenever 0  i < j  p. Proof. Let zi = xi y for 0  i  p − 1, and set zp = x. Let Hi be the stabilizer of zi in G, and let D = Gx ∩ Gy . Since D fixes each of the points zi , we have D ⊆ Hi for 0  i  p. Now let 0  i < j  p. We observe first that x, y ∈ zi , zj . This is clear if j = p since zp = x and zi = xi y. If, on the other hand, j < p, then xj−i = (xj y)(xi y)−1 ∈ zi , zj , and since P is a p-group and j − i is prime to p, it follows that x ∈ xj−i  ⊆ zi , zj , and thus also y ∈ zi , zj , as required. Since Hi ∩ Hj fixes all points in zi , zj , this intersection fixes both x and y, and thus Hi ∩ Hj ⊆ Gx ∩ Gy = D. We have seen, however, that D is contained in both Hi and Hj , and the result follows. 2 Proof of Theorem 3.1. Conclusion (a) is obvious if G is trivial, so we assume that G > 1, and we prove (a) by induction on |G|. Since the action of G on P is faithful, there exists a point x ∈ P that is moved by G, and thus Gx < G. Now Gx acts faithfully on P , and since all Gx -orbits are contained in G-orbits and hence have size at most (p + 1)/2, we can apply the inductive hypothesis to Gx to deduce that there exists y ∈ P such that Gx ∩ Gy = (Gx )y = 1. By Lemma 3.2, there exist point stabilizers Hi ⊆ G such that Hi ∩ Hj = Gx ∩ Gy = 1 for 0  i < j  p. Let m be the minimum of the orders of the subgroups Hi , and note that |G|/m is the size of some G-orbit. Then |G|/m  (p + 1)/2, and we have |G|  m(p + 1)/2. Since the Hi have trivial pairwise intersections, each of these subgroups contains at least m − 1 elements that are not contained in any of the others, and this accounts for a total of at least (p + 1)(m − 1) distinct nonidentity elements of G. We thus have (p + 1)(m − 1) < |G| 

m(p + 1) . 2

Canceling p + 1, we get m − 1 < m/2, or equivalently m < 2. Thus m = 1, and since m is the order of a point stabilizer in G, the proof of (a) is complete.

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For (b), we want to find x, y ∈ P such that Gx ∩ Gy = 1. Choose x and y such that Gx ∩ Gy has the smallest possible order, and write D = Gx ∩ Gy . Our goal is to show that D = 1, so we assume that D > 1, and we work to obtain a contradiction. First, we argue that D cannot be a point stabilizer. To see this, suppose that D = Gz , and note that since D is nontrivial and the action of G is faithful, D must move some point w ∈ P , and thus Gz ∩ Gw = Dw < D. This contradicts the choice of D. By Lemma 3.2, there exist point stabilizers Hi such that Hi ∩ Hj = D whenever 0  i < j  p. Let m be the minimum of the indices |Hi : D|, and note that m > 1 since D is not a point stabilizer. Also, m|D| is the order of one of these point stabilizers, and thus |G|/m|D|  p − 1, and we have |G|  m(p − 1)|D|. Each of the subgroups Hi has order at least m|D|, and their pairwise intersections are D. The sets Hi − D are thus disjoint, and each of these p + 1 sets has cardinality at least m|D| − |D|. This yields   (p + 1) m|D| − |D| < |G|  m(p − 1)|D|. Canceling |D|, and simplifying, we get m < (p + 1)/2. Now let H be one of the point stabilizers Hi with |H| = m|D|, and write H = Gz . If w ∈ P , then by the choice of D, we have |Hw | = |Gz ∩ Gw |  |D|, and thus |H : Hw | =

m|D| p+1 m|D|  =m< . |Hw | |D| 2

We can thus apply part (a) to the action of H on P to deduce that there exists w ∈ P such that |Hw | = 1, and this is a contradiction, since we know that |Hw |  |D| > 1. Finally, if O is a G-orbit in P × P , then the set of first coordinates of points in O forms a G-orbit in P , and similarly for the set of second coordinates. It follows that |O|  k2 , and thus if O is a regular orbit, we have |G|  k2 . 2 There are circumstances where we can produce a regular orbit even if we allow orbit sizes as large as p + 1. (We have seen that in general there need not be such orbits, however.) Theorem 3.3. Let G act faithfully on a p-group P , and suppose that P = X × Y , where X and Y are G-invariant subgroups, and the actions of G on X and on Y are nontrivial. Assume that all G-orbits on the set P − Y have size at most p + 1. Then G has a regular orbit in P − Y . Proof. Let K be the kernel of the action of G on X, and note that K acts faithfully on Y . We work to show that all K-orbits in Y have size at most (p + 1)/2. To see this, note first that because G acts nontrivially on X, we can choose x ∈ X moved by G, and / Y. thus K ⊆ Gx < G. In particular, x = 1, and so x ∈

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Now let y ∈ Y , and note that xy ∈ / Y , so by hypothesis, |G : Gxy |  p + 1. Also, Gxy = Gx ∩ Gy because P = X × Y , and thus we have Ky = K ∩ Gy = K ∩ Gx ∩ Gy = K ∩ Gxy . Also, observe that KGxy ⊆ Gx < G, and thus |G : KGxy |  2. It follows that |K : Ky | = |K : K ∩ Gxy | = |KGxy : Gxy | 

p+1 |G : Gxy |  , 2 2

so all orbits in the faithful action of K on Y have size at most (p + 1)/2, as wanted. We conclude by Theorem 3.1(a) that K has a regular orbit in Y , and thus we can choose y ∈ Y such that Ky = 1, or equivalently, K ∩ Gy = 1. Furthermore, we can assume that Gy < G since otherwise, K = 1, and in that case, we can replace y by an arbitrary element of Y not fixed by G. (We can be sure that such an element exists since the action of G on Y is nontrivial.) Since K ∩ Gy = 1, the action of Gy on X is faithful, and we work next to show that each Gy -orbit in X has size at most (p + 1)/2. This is certainly true for the orbit of the identity of X, so let 1 = x ∈ X and observe that (Gy )x = Gy ∩ Gx = Gxy , and also, since Gy < G, we have |G : Gy |  2. Then   Gy : (Gy )x  = |Gy : Gxy |  |G : Gxy |  p + 1 , 2 2 where the final equality holds by hypothesis because x = 1, and so xy ∈ / Y . We have now shown that all Gy -orbits on X have size at most (p + 1)/2, and we deduce by Theorem 3.1(a) that Gy has a regular orbit on X, and thus there exists x ∈ X such that / Y , and we are done. Finally, if Gxy = (Gy )x = 1. If Gy > 1, then x = 1, and thus xy ∈ Gy = 1, then Gxy = 1 for all x ∈ X, and we are done in this case too. 2 One could attempt to generalize Theorem 3.3 by dropping the assumption that X exists. In other words, we assume that G acts faithfully on a p-group P leaving the subgroup Y ⊆ P invariant, and we suppose that the actions of G on Y and on P/Y are nontrivial. Assuming that all G-orbits in P − Y have size at most p + 1, one might try to show that there is a regular G-orbit in P − Y . We have been unable to accomplish this without assuming that P is abelian, and unfortunately, we need an additional assumption: that p does not divide |G|. (Note that we get this “for free” if we strengthen the upper bound on the orbit sizes from p + 1 to p − 1, and in fact, that is the only case we really need.)

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Theorem 3.4. Let G be a p -group that acts faithfully on an abelian p-group P , and suppose that Y ⊆ P is invariant under G and that the actions of G on Y and on P/Y are nontrivial. Assume that the G-orbits in P − Y all have size at most p + 1. Then G has a regular orbit in P − Y . Proof. We proceed by induction on |P |. Let C be a coset of Y in P for which the stabilizer GC has the least possible order. Since the action of G on P/Y is nontrivial, we have C = Y , and thus if x ∈ C, we have x ∈ P − Y and Gx ⊆ GC . We are done if GC = 1, so we can assume that there is no regular G-orbit in P/Y . Suppose that the action of G on P/Y is faithful. It follows by Theorem 3.1(a) that the G-orbit of C has size exceeding (p + 1)/2, and we have p + 1  |G : Gx | = |G : GC ||GC : Gx | >

p+1 |GC : Gx |, 2

and thus |GC : Gx | < 2. We deduce that GC fixes every element of C, and thus GC centralizes Y . Writing N = CG (Y ), we see that N ⊇ GC > 1. We are assuming that P is abelian and that p does not divide |G|, so we can write P = [P, N ] × CP (N ) by Fitting’s lemma. Also, N  G, and thus both factors are G-invariant. Since N > 1 and the action of G on P is faithful, it follows that [P, N ] > 1, and so the action of G on the first factor is nontrivial. Also, Y ⊆ CP (N ), and since G acts nontrivially on Y , it follows that the action of G on the second factor is also nontrivial. Furthermore, because Y is contained in CP (N ), we see that every G-orbit of elements outside of the second factor has size at most p + 1. We can thus apply Theorem 3.3 to complete the proof in this case. We can now assume that the action of G on P/Y is not faithful, and we write M = CG (P/Y ), so 1 < M  G. Now [P, M ] ⊆ Y and P = [P, M ] × CP (M ), where both factors are G-invariant. Since the action of G on P is faithful and M > 1, we have [P, M ] > 1, and thus G acts nontrivially on the first factor. Also, since [P, M ] ⊆ Y and the action of G on P/Y is nontrivial, it follows that G acts nontrivially on P/[P, M ], and hence the action of G on the second factor is nontrivial. If [P, M ] = Y , we are done by Theorem 3.3, so we can assume that [P, M ] < Y , and hence we can write Y = [P, M ] × U , where U = Y ∩ CP (M ) > 1. We wish to apply the inductive hypothesis to the action of G on P/U with respect to the subgroup Y /U , so there are a number of items that must be checked. First, to see that the actions of G on Y /U and (P/U )/(Y /U ) are nontrivial, observe that there are G-isomorphisms Y /U ∼ = [P, M [ and (P/U )/(Y /U ) ∼ = P/Y , and we know that the actions of G on [P, M [ and P/Y are nontrivial. Next, to compute G-orbit sizes in (P/U ) − (Y /U ), let B be a coset of U not contained in Y , and let x ∈ B. Then x ∈ / Y , and Gx ⊆ GB , and thus |G : GB |  |G : Gx |  p + 1, as needed. Finally, we must show that the action of G on P/U is faithful, so we let K be the kernel of this action. Since U ⊆ Y , we see that K acts trivially on P/Y , and thus K ⊆ M . Then [P, K] ⊆ U ∩ [P, M ] = 1, and it follows that K = 1, as required. By the inductive

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hypothesis, therefore, there exists a coset B of U with B  Y and GB = 1. If x ∈ B, then x ∈ / Y and Gx ⊆ GB = 1, so x lies in a regular orbit. 2 The following contains the small-orbit facts that we will actually need. Corollary 3.5. Let G act faithfully on an abelian p-group P , and let Y < P be G-invariant. Assume that all G-orbits in P − Y have size at most k. (a) If k  (p + 1)/2, then there is a regular G-orbit in P − Y , and in particular, |G|  k. (b) If k  (p − 1), then |G|  k2 . Proof. We can certainly assume that G is nontrivial. If G acts trivially on Y , then all G-orbits in P have size at most k, and the result is immediate from Theorem 3.1. (In this case, a regular G-orbit in P would necessarily be contained in P − Y .) We can thus assume that the action of G on Y is nontrivial. Because k < p, a subgroup of G of order p fixes all points in Y − P . But Y − P generates P since Y < P , and it follows that G has no subgroup of order p, so p does not divide |G|. If G acts nontrivially on P/Y , therefore, it follows by Theorem 3.4 that G has a regular orbit in P − Y , and we are done in this case. We can now assume that the action of G on P/Y is trivial, and since |G| and |P | are coprime, it follows that P = Y C, where C = CP (G). Also, C  Y since Y < P , and we can choose x ∈ C with x ∈ / Y . For y ∈ Y , we have xy ∈ / Y and Gy = Gxy has index at most k in G. It follows that all G-orbits in P have size at most k, and we are done by Theorem 3.1 since if y ∈ Y lies in a regular orbit, then xy also lies in a regular orbit and xy ∈ P − Y . 2 4. Frobenius actions Recall that an action via automorphisms of a group G on a group P is said to be “Frobenius” if for every nonidentity element x ∈ P , the stabilizer Gx is trivial, or equivalently, every G-orbit in P is regular. In particular, in the cases where G = 1 or P = 1, we consider the (trivial) actions to be Frobenius. We offer the following well-known fact without proof. Lemma 4.1. Let G act on P , and suppose that X  P is G-invariant. Then the action of G on P is Frobenius if and only if the actions of G on X and P/X are Frobenius. Corollary 4.2. Let G act on P , and let V be a collection of normal subgroups of P such that for every member V ∈ V, the action of G on P/V is Frobenius. Then the action of  G on P/D is Frobenius, where D = V. Proof. It suffices to show that if X, Y  P and the actions of G on P/X and P/Y are Frobenius, then the action of G on P/(X ∩ Y ) is Frobenius. Since the action on P/X is

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Frobenius, it suffices by Lemma 4.1 to show that the action of G on X/(X ∩ Y ) is also Frobenius. But X/(X ∩ Y ) is G-isomorphic to XY /Y , and since the action of G on P/Y is Frobenius, the action on XY /Y is Frobenius by Lemma 4.1. The result follows. 2 For our purposes, the main application of Frobenius actions is the following. Lemma 4.3. Let G act on an abelian group P , and let Z ⊆ P . Suppose that G centralizes Z, and assume that the action of G on P/Z is Frobenius. Then every orbit in the action of G on Irr(P ) has size 1 or |G|. Proof. Suppose that λ ∈ Irr(P ) lies in a G-orbit with size smaller than |G|. Then some nonidentity element x ∈ G fixes λ, and our task is to show that the whole group G fixes λ. Write K = ker(λ), and note that since λ is linear and fixed by x, it follows that x acts trivially on P/K, and thus x acts trivially on P/KZ. But P/KZ is a homomorphic image of P/Z, on which the action of G is Frobenius, and thus the action of G on P/KZ is Frobenius. Since x is a nonidentity element of G that acts trivially on P/KZ, it follows that P/KZ is the trivial group, and thus KZ = P . Then P/K = KZ/K ∼ = Z/(Z ∩ K), and this isomorphism respects the action of G. But G acts trivially on Z/(Z ∩ K), and it follows that G acts trivially on P/K, and thus G stabilizes λ, as required. 2 The following result collects some fairly standard facts. Lemma 4.4. Let θ ∈ Irr(P ), where P is a p-group of class 2 and θ(1) = p, and write V = Z(θ). Then P/V is elementary abelian of order p2 . Also, if G is a group that acts on P and stabilizes θ, let K = CG (P/V ), and assume that G/K is a p -group. Then the action of G/K on P/V is Frobenius. Proof. Since P has class 2, we know that θ vanishes on P − V , and because θV is homogeneous, θ is fully ramified with respect to V , and thus |P : V | = θ(1)2 = p2 . Also, since θ is fully ramified with respect to V and |P/V | = p2 , it is well known that P/V must be elementary, and that it carries a symplectic form determined by θ. Since θ is G-invariant, it follows that V is G-invariant and that the associated symplectic form on P/V is G-invariant. We can thus embed G/K into Sp(2, p) = SL(2, p), and it suffices to show that the identity is the only p -element of SL(2, p) that has a nonzero fixed point in the underlying two-dimensional vector space over the field of order p. Assuming that x ∈ SL(2, p) has a nonzero fixed point, it has an eigenvalue equal to 1, and thus both eigenvalues are 1 since det(x) = 1. If also x is a p -element, then it is diagonalizable, and thus x = 1, as required. 2 Theorem 4.5. Let Z = Z(P ), where P is a p-group, and let G be a p -group that acts faithfully on P and centralizes Z. Assume that P/Z is abelian and that cd(P ) = {1, p}. Then the action of G on P/Z is Frobenius.

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Proof. Let Y be the set of G-invariant characters of degree p in Irr(P ). We argue that every nonprincipal character τ ∈ Irr(P  ) is a constituent of θP  for some member θ ∈ Y. One way to see this is to observe that P  ⊆ Z, and thus τ is invariant in the semidirect product Γ = P G because both P and G centralize Z. By Lemma 2.1, there exists a character χ ∈ Irr(Γ |τ ), with p-power degree. Then χP is irreducible since |Γ : P | = |G| is a p -number. Writing θ = χP , we see that θ is G-invariant, and that τ is a constituent of θP  , as wanted. Also, since τ is not principal, θ cannot be linear, and thus θ(1) = p, and so θ ∈ Y.  Let D = Z(θ), where θ runs over Y. Since Z = Z(P ) ⊆ Z(θ) for all θ ∈ Irr(P ), we have Z ⊆ D, and we argue that Z = D by showing that [D, P ] = 1. Since [D, P ] ⊆ P  , it suffices to show that [D, P ] ⊆ ker(τ ) for each character τ ∈ Irr(P  ). We can assume that τ is nonprincipal, so we can choose θ ∈ Y lying over τ . Then [D, P ] ⊆ [Z(θ), P ] ⊆ ker(θ), and since [D, P ] ⊆ P  and τ is a constituent of θP  , we have [D, P ] ⊆ ker(τ ), as wanted. We know now that Z = D is the intersection of subgroups of the form V = Z(θ) for θ ∈ Y. To show that the action of G on P/Z is Frobenius, it suffices by Corollary 4.2 to show that the action of G on P/V is Frobenius for each such subgroup V . We show first that V = Z(θ) is abelian. Since cd(P ) = {1, p}, it follows by a result of D.S. Passman and the author [7] that either P has an abelian subgroup of index p, or else |P : Z| = p3 . (See also Theorem 12.11 of [5].) Supposing first that there exists an abelian subgroup A ⊆ P such that |P : A| = p, we show that V ⊆ A, so V is abelian, as claimed. Otherwise, V A = P , and since V is central modulo ker(θ) and A is abelian, it follows that P/ker(θ) is abelian, which is not the case since θ ∈ Irr(P ) is nonlinear. In the remaining case, where P has no abelian subgroup of index p, we have |P : Z| = p3 . Since Z ⊆ V and |P : V | = p2 , we have |V : Z| = p, and in this case too, V is abelian. Next, we argue that the action of G on P/V is faithful. To see this, let K be the kernel of this action, so that [P, K] ⊆ V , and this yields [P, K, V ] = 1 since V is abelian. Also, [V, P, K] ⊆ [P  , K], and we recall that P  ⊆ Z and that K ⊆ G and G centralizes Z, and thus [P  , K] ⊆ [Z, K] = 1, and we have [V, P, K] = 1. By the three-subgroups lemma, [K, V, P ] = 1, and thus [K, V ] ⊆ Z. Then [P, K] = [P, K, K, K] ⊆ [V, K, K] ⊆ [Z, K] = 1, where the first equality holds because |K| is coprime to |P |. By assumption, however, the action of G on P is faithful, and thus K = 1, and it follows that the action of G on P/V is faithful, as claimed. By Lemma 4.4, this action is Frobenius, and we deduce that the action of G on P/Z is Frobenius, as required. 2 Corollary 4.6. Under the hypotheses of Theorem 4.5, each linear character of P lies in a G-orbit of size 1 or |G|. Proof. In the action of G on the abelian group P/P  , we see that the action on Z/P  is trivial and by Theorem 4.5, the action on P/Z is Frobenius. The result then follows by Lemma 4.3. 2

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5. Normal Sylow subgroups We begin with a preliminary result. Lemma 5.1. Let P  G, where P/P  is a p-group and P  is abelian, and let b = b(G/P ). Suppose that the size of each G-orbit on the set of linear characters of P lies in the set {1, n} for some fixed integer n. If f ∈ cd(G) is not divisible by p, then either f  b, or else f /n is an integer and f /n  b. Proof. Let f ∈ cd(G), where f is not divisible by p, and choose χ ∈ Irr(G) with χ(1) = f . Let λ be an irreducible constituent of χP , and note that p does not divide λ(1). Since |P : P  | is a power of p, it follows that λP  is irreducible. Also, P  is abelian, and hence λP  is a linear character, so λ is linear and P  ⊆ ker(λ). Then λ can be viewed as a character of the p-group P/P  , and in particular, the order of λ is a power of p. Let T be the stabilizer of λ in G, and note that by hypothesis, either |G : T | = 1 or |G : T | = n. Also, since λ lies under the character χ of G having p -degree f , it follows by Lemma 2.2 that λ has an extension to T . By Lemma 2.3, therefore,   f ∈ cd(G|λ) = |G : T |e  e ∈ cd(T /P ) , and thus either f = e or f = ne, where e is an integer such that e  b(T /P )  b(G/P ) = b. Then either f  b or f /n is an integer and f /n  b, as required. 2 Next, we consider the case where G has a normal Sylow p-subgroup. Recall that in this situation, G has an irreducible character with degree divisible by p precisely when the Sylow p-subgroup is nonabelian. Theorem 5.2. Suppose that G has a normal nonabelian Sylow p-subgroup, and suppose that the multiples of p in cd(G) are all at most kp, where k < p is an integer. Then one of the following occurs. (1) There exists an integer n such that for each p -degree f ∈ cd(G), either f  k, or else f /n is an integer and f /n  k. (2) There exists an integer m  k2 such that each p -degree f ∈ cd(G) divides m. Proof. We proceed by induction on |G|. Since k < p, no member of cd(G) can be divisible by a power of p exceeding p, and thus no irreducible character of a normal subgroup of G can have p-power degree exceeding p. Let P be minimal among nonabelian normal p-subgroups of G, and observe that cd(P ) = {1, p}. Also, P ⊆ S, where S is the given normal Sylow p-subgroup of G. Let M ⊆ P  be minimal normal in G. If χ ∈ Irr(G) has p -degree, then the restriction of χ to P is a sum of linear characters of P , and thus M ⊆ P  ⊆ ker(χ). It follows that the p -degrees in cd(G) are exactly the p -degrees in cd(G/M ). If S/M is nonabelian,

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therefore, we can apply the inductive hypothesis to the group G/M , and the result will follow. We can thus assume that S  ⊆ M , and since M ⊆ P  ⊆ S  , we have M = P  = S  . In particular, P  is minimal normal in G, and thus P  ⊆ Z, where Z = Z(P ). Also, since S/P  is abelian and P ⊆ S, it follows that the action of S on P/P  is trivial Write L = {σ ∈ Irr(Z) | P   ker(σ)}, and note that no member of L can extend to P because P  is contained in the kernel of every linear character of P . We have L = Irr(Z) − Irr(Z/P  ), and since P  is nontrivial, Irr(Z/P  ) is a proper subgroup of Irr(Z). It follows that L generates the whole group Irr(Z), and in particular, L is nonempty. Also, L is invariant under the action of G on Irr(Z). Let C = CG (Z), and note that P ⊆ C. Since C centralizes Z, it acts trivially on Irr(Z), and in fact it is the kernel of the action of G on Irr(Z), and thus G/C acts faithfully on Irr(Z). Let σ ∈ L, and let T be the stabilizer of σ in G. Then C ⊆ T and |G : T | is the size of the (G/C)-orbit containing σ. Let θ ∈ Irr(T |σ), and note that θZ is a multiple of σ. Since σ does not extend to p, it follows that no constituent of θP is linear, and thus all irreducible constituents of θP have degree p. Then θ(1) is a multiple of p, and we can write θ(1) = ap for some integer a depending on θ. Write χ = θG , so that χ is irreducible by the Clifford correspondence. We have χ(1) = |G : T |θ(1) = a|G : T |p, and thus a|G : T |  k. In particular, |G : T |  k, and we see that in the faithful action of G/C on the group Irr(Z), all orbits contained in L have size at most k  p − 1. Recall that S ∈ Sylp (G). Every S-orbit in L has p-power size and is contained in some G-orbit, which has size less than p. It follows that S acts trivially on L, and since L generates Irr(Z), we conclude that S acts trivially on Irr(Z). Then S ⊆ C, so G/C is a p -group. We propose to apply Lemma 5.1 to the group C with respect to the normal subgroup P , and so we argue that there exists an integer n such that the size of each C-orbit of linear characters of P lies in the set {1, n}. To see this, choose a Hall p -subgroup X of C. Since C = SX and S acts trivially on P/P  , the C-orbits of linear characters of P are exactly the X-orbits. Now let Y = CX (P ), so that X/Y acts faithfully on P , and we check that the hypotheses of Theorem 4.5 are satisfied for the action of X/Y on P . First, X/Y centralizes Z since X ⊆ C = CG (Z). Also, P  ⊆ Z, so P/Z is abelian, and finally, we recall that cd(P ) = {1, p}. By Theorem 4.5, therefore, the action of X/Y on P/Z is Frobenius, and thus by Corollary 4.6, each linear character of P lies in an (X/Y )-orbit of size 1 or n, where n = |X/Y |. Since the C-orbits of linear characters are the X-orbits, which are the (X/Y )-orbits, this establishes our claim. To apply Lemma 5.1, we need information about b = b(C/P ), and we argue that if there exists a regular (G/C)-orbit in L, then b  k/m, where m = |G : C|. To see this, suppose that σ ∈ L is in a regular (G/C)-orbit, so C is the subgroup we previously called T , the stabilizer of σ in G. By Lemma 2.1 we can choose θ ∈ Irr(C|σ)

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having p-power degree, and thus θ(1)  p. We have seen, however, that θP has no linear constituents, and thus θ(1) = p and θP is irreducible. Now C is the full stabilizer of θP in G since θZ is a multiple of σ and C is full the stabilizer of σ. Since |G : C| = m and θ(1) = p, Lemma 2.3 yields  mep  e ∈ cd(C/P ) ⊆ cd(G),



and thus for all e ∈ cd(C/P ), we have me  k, and so e  k/m, and it follows that b = b(C/P )  k/m, as wanted. Continuing to assume that there is a regular (G/C)-orbit in L, let f be a p -degree in cd(G). We can write f = de, where d is some divisor of |G : C| = m and e is a p -degree in cd(C). By Lemma 5.1 applied to the group C with respect to the normal subgroup P , we see that either e  b, or e/n is an integer such that e/n  b. Since b  k/m and d  m, it follows that db  k, and we have f = de  db  k,

or else

de f =  db  k, n n

and in the latter case, f /n = d(e/n) is an integer. We have now shown that conclusion (1) of the theorem holds if there exists a regular (G/C)-orbit in L. We can now assume that there is no such regular orbit. We can apply Corollary 3.5(a) to the action of G/C on Irr(Z), where the proper subgroup Irr(Z/P  ) plays the role of Y . Since there is no regular (G/C)-orbit in L, there must exists σ ∈ L in a (G/C)-orbit with size exceeding (p + 1)/2, and by Corollary 3.5(b), we have m = |G/C|  k 2 . As before, we let T be the stabilizer of σ in G, so that |G : T | > (p + 1)/2. If θ ∈ Irr(T |σ), is arbitrary, we saw previously that we can write θ(1) = ap for some integer a, and that a|G : T |  k  p − 1. Since |G : T | > (p + 1)/2, it follows that a = 1, and we conclude that every member of Irr(T |σ) has degree p, and thus every such character restricts irreducibly to P . (Recall that the restrictions to P of characters in Irr(T |σ) have no linear constituents.) It follows by Gallagher’s theorem that cd(T /P ) = {1}, and thus T /P is abelian, and it follows that C/P is abelian, so C  ⊆ P . We suppose now that C centralizes P/P  , and we show that in this case, C/P  is abelian. To see this, recall that we can write C = SX, where X is a p-complement in C, and thus it suffices to show that each of [S, S], [X, X] and [S, X] is contained in P  . That the first of these commutators is contained in P  is clear since we know that S  = P  . The second is contained in P  because [X, X] ⊆ X ∩ C  ⊆ X ∩ P = 1. Finally, [S, X] = [S, X, X] ⊆ [P, X] ⊆ [P, C] ⊆ P  ,

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where the equality holds because the action of X on S is coprime; the first containment holds because [S, X] ⊆ C  ⊆ P , and the final containment holds because we are assuming that C centralizes P/P  . Continuing to assume that C centralizes P/P  , suppose that χ ∈ Irr(G) has p -degree f . We have seen that P  ⊆ ker(χ), and thus χ ∈ Irr(G/P  ). Since C/P  is an abelian normal subgroup of G/P  , we conclude by a result of Itô that f divides |G : C| = m, and since m  k2 , conclusion (2) of the theorem holds in this case. We can now assume that C acts nontrivially on P/P  , and we derive a contradiction, thereby completing the proof. Recall that the action of X/Y on P/Z is Frobenius, where X is a p-complement in C and Y = CX (P ). Now C = SX and we are assuming that C acts nontrivially on P/P  . Since P  = S  , we see that S acts trivially on P/P  , and it follows that X acts nontrivially, and thus Y < X. As before, let σ ∈ L lie in a G-orbit of size exceeding (p + 1)/2, let T be the stabilizer of σ in G and recall that if θ ∈ Irr(T |σ), then θ has degree p and θ restricts irreducibly to P . It follows that each member of Irr(P |σ) has degree p and is invariant in T , and in particular, every such character is invariant in C. Let ϕ ∈ Irr(P |σ), and write V = Z(ϕ). We know by Lemma 4.4 that |P : V | = p2 , and we argue that C centralizes V /Z. To see this, observe that ϕV is a multiple of some linear character λ, and note that since ϕ is C-invariant, λ must also be C-invariant. Now let α ∈ Irr(V /Z). Since P/Z is abelian, α extends to a linear character β of P , and we see that βϕ ∈ Irr(P |σ), and thus βϕ is C-invariant. But (βϕ)V is a multiple of αλ, and so αλ is invariant in C. Since λ is linear we can write α = (αλ)λ−1 , and thus α is C-invariant. Thus C acts trivially on Irr(V /Z), and thus the action of C on V /Z is also trivial, as claimed, and in particular, X/Y acts trivially on V /Z. We know, however, that the action of the nontrivial group X/Y on P/Z is Frobenius, so its action on V /Z is also Frobenius, and it follows that V /Z is trivial, and thus Z = V has index p2 in P . Now P is a p-group and |P : Z(P )| = p2 , and thus P has an abelian subgroup A with index p. By Lemma 4.6 of [6], we have |P  | = |A : Z(P )| = p. Also, X/Y acts trivially on Z/P  , and since its action on P/Z is Frobenius, it follows that Z/P  = CP/P  (X/Y ). By Fitting’s lemma, therefore, we can write P/P  = (Z/P  ) × (W/P  ), where W/P  = [(P/P  ), X]. Then W/P  = [(P/P  ), C] since C = SX and S centralizes P/P  , and it follows that W G. Also, |W/P  | = |P : Z| = p2 , and so |W | = p3 . Furthermore, ZW = P is nonabelian and Z = Z(P ), and it follows that W is nonabelian. Recall that P was chosen to be minimal among nonabelian normal p-subgroups of G. Thus W = P , and we have Z = P  has order p. It follows that the stabilizer in G of an arbitrary nonidentity element of Z centralizes Z, and so this stabilizer is C. The action of G/C on Z is thus Frobenius, and in particular, all orbits of G/C on L are regular. This is a contradiction, and the proof is now complete. 2 In the situation of Theorem 5.2, it is clear that if (1) holds, there are at most 2k degrees in cd(G) that are not divisible by p. Using a bit of elementary number theory, we will obtain the same conclusion if (2) holds.

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Recall that τ (n) is the number of divisors of n, and that τ is a multiplicative function. Lemma 5.3. For all integers n  1, we have τ (n) 

√

3n.

Proof. Since τ is multiplicative, it suffices to show the following:

(1) τ (n)  (9/4)n if n is a power of 2. (2) τ (n)  (4/3)n if n is a power of 3. √ (3) τ (n)  n if n is a power of a prime p > 3. Suppose now that p is prime and n = pe , so τ (n) = e + 1. We must show that (e + 1)2  cpe , where ⎧ ⎨ 9/4 if p = 2, c = 4/3 if p = 3, ⎩ 1 if p  5. Writing f = e + 1, we want f 2  dpf for f  1, where d = c/p, and so ⎧ ⎨ 9/8 d = 4/9 ⎩ 1/p

if p = 2, if p = 3, if p  5.

It is easy to check the relevant inequalities for 1  f  3, so working by induction on f , it suffices to check that (f + 1)2 p f2 for f  3. In fact, for f  3, we have (f + 1)2  2, f2 since this inequality simplifies to f 2 − 2f − 1  0, or equivalently, (f − 1)2  2, and this holds for f  3. 2 In particular, if m is an integer that is at most k2 , then τ (m)  We have now established the following.

√

3m 

√

3k < 2k.

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Corollary 5.4. Suppose that G has a normal nonabelian Sylow p-subgroup, and assume that the multiples of p in cd(G) are all at most kp, where k < p is an integer. Then the number of p -degrees in cd(G) is at most 2k, and so |cd(G)|  3k. 6. A technical result In this section, we prove the following. Theorem 6.1. Let S  G, and assume that G/S is a p -group, S/S  is a p-group and S  is an abelian p -group. Let U = Op (S), and B = CG (S/S  U ), and write n = |G : B|. Suppose that cd(G) contains a multiple of p and that all multiples of p in cd(G) are at most kp, where k < p is an integer. Then for each p -degree f ∈ cd(G), either f  k or else f /n is an integer and f /n  k. In particular, there are at most 2k degrees in cd(G) that are prime to p. Proof. First, we consider the case where U = 1. Let P ∈ Sylp (S), and note that S = S  P and that P is an abelian Sylow p-subgroup of G. Now P acts faithfully on S  since Op (S) = 1, and hence P also acts faithfully on the p -group Irr(S  ). Since P is abelian, we conclude that there is a regular orbit in this action. (See Corollary 3.4 of [6] for this consequence of Brodkey’s theorem.) It follows that there exists θ ∈ Irr(S) with θ(1) = |P |, and in particular, θ(1) is a power of p. Since θ(1) divides some member of cd(G) and k < p, we deduce that |P | = θ(1)  p. On the other hand, we know by hypothesis that cd(G) contains a multiple of p, so p divides |G|, and thus |P | = θ(1) = p. We argue next that b(G/S)  k. To see this, let T be the stabilizer of θ in G, and note that p does not divide |T : S| because G/S is a p -group. Since S/S  is a p-group, every linear character of S has p-power order, and since θ has degree p, it follows that θ extends to T , and thus by Lemma 2.3, we have  tep  e ∈ cd(T /S) ⊆ cd(G),



where t = |G : T |. Then te  k for all e ∈ cd(T /S), and it follows that b(T /S)  k/t. To compute b(G/S), let ψ ∈ Irr(G/S) and let β be an irreducible constituent of ψT . Then β ∈ Irr(T /S), and ψ is a constituent of β G . This yields ψ(1)  β G (1) = tβ(1)  tb(T /S)  t

k = k, t

and thus b(G/S)  k, as wanted. Recall that |S/S  | = |P | = p, and that in this case, B = CG (S/S  ). Since S/S  has prime order, the stabilizer in G of each nonidentity element of S/S  is B, and thus the action of G/B on S/S  is Frobenius. Since |G/B| = n, it follows that the size of each G-orbit of linear characters of S lies in the set {1, n}. By Lemma 5.1 with S in place of P , we see that if f is a p -degree in cd(G), then either f  b(G/S), or else f /n is

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an integer such that f /n  b(G/S). We have seen that b(G/S)  k, however, and this establishes the desired conclusion in the special case where U = 1. We proceed now to the general case. Let f be a p -degree in cd(G), and choose χ ∈ Irr(G) with χ(1) = f . Now U is abelian since it is a p-subgroup of S, and we let λ be a linear constituent of χU . Let H be the stabilizer of λ in G, and write h = |G : H|. Since λ extends to H by Lemma 2.2, we can apply Lemma 2.3 to deduce that  he  e ∈ cd(H/U ) = cd(G|λ) ⊆ cd(G),



and thus the largest multiple of p in cd(H/U ) is at most kp/h. Also, we see that f = he for some p -degree e ∈ cd(H/U ). Using the standard bar convention, write G = G/U . We will show that H satisfies the hypotheses of the theorem with S in place of S. First, observe that U ⊆ Z(S), and thus S stabilizes λ, so S ⊆ H. Then S ⊆ H, as needed, and in fact, S  H. Since U ⊆ S, we have H/S ⊆ G/S ∼ = G/S, which is a p -group. Also (S) = S  = S  U , so S/(S) ∼ = S/S  U ,  which is a homomorphic image of S/S , and so is a p-group, and furthermore, (S) = S  is an abelian p -group since it is a homomorphic image of the abelian p -group S  . Now U = Op (S), so Op (S) = 1. It follows that the section of H that plays the role of S/S  U in G is S/(S) ∼ = S/S  U , and thus the centralizer of this section in H is B ∩ H, and its centralizer in H is B ∩ H. Next, we consider character degrees of H. First, observe that S  U < S since otherwise, U is a full Sylow p-subgroup of G, and this is not the case since U is abelian, and yet we know that cd(G) contains a multiple of p. Then p divides |H|, and since Op (H) = Op (S) is trivial, it follows that H does not have a normal Sylow p-subgroup, and thus cd(H) contains a multiple of p. Since Op (S) is trivial, the first part of the proof applies to H, with k replaced by [k/h], the greatest integer in k/h. Writing r = |H : B ∩ H|, we deduce that for each p -member e ∈ cd(H) either e  k/h, or else e/r  k/h, where e/r is an integer. Recall now that f was an arbitrary p -degree in cd(G), and that f = eh for some p -degree e ∈ cd(H). If e  k/h, then f = eh  k, and otherwise, e/r  k/h, and thus f /r = eh/r  k. Also, observe that r = |H : (B ∩ H)|  |G : B| = n, so in the second case, we have f /n  f /r  k, as wanted. To complete the proof, we must show that in the case where e/r is an integer, it is also true that f /n is an integer. Now       hr = |G : H|H : (B ∩ H) = |G : B ∩ H| = |G : B|B : (B ∩ H) = nB : (B ∩ H), and thus hr/n = |B : (B ∩ H)| is an integer. Then f /n = e(h/n) = (e/r)(hr/n) is an integer, as required. This completes the proof. 2 7. The general case Recall our definition: an integer f is good for a prime p if f is not divisible by any prime power q > 1 such that q is congruent to −1, 0 or 1 modulo p. The relevance of

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this definition is its role in the following theorem of D.L. Winter and B. Newton. (The solvable case of this result is in Winter’s paper [11], and it also appears as Theorem 14.5 of [5]. Using the classification of simple groups, Newton extended Winter’s result to p-solvable groups in [9].) Theorem 7.1 (Winter–Newton). Let G be p-solvable, and suppose that Irr(G) contains a faithful character of degree f , where f is good for the prime p. Then G has a normal Sylow p-subgroup. Corollary 7.2. Let G be p-solvable, and let K be the intersection of the kernels of all characters χ ∈ Irr(G) such that χ(1) is good for p. Then G/K has a normal Sylow p-subgroup. Proof. It suffices to observe that G/K is a subdirect product of groups satisfying the hypotheses of Theorem 7.1. 2 In both Theorem 7.1 and Corollary 7.2, it is easy to see that the normal Sylow subgroup of G must be abelian, but we will not need that additional information here. We thank R. Solomon for his help with the proof of the following result, which we need to establish Theorem B in the case where the given p-solvable group is not solvable. Theorem 7.3. Let M  S where S/M is a p-group and M is a p -group. Assume that M is a direct product of nonabelian simple groups and that a Sylow p-subgroup of S is not normal. Then there exists θ ∈ Irr(S) such that θ(1) is divisible by p and θ(1) > p2 . Proof. It suffices to show that M has an irreducible character α that is not invariant in S and such that α(1) > p. By hypothesis, M is a direct product of nonabelian simple groups, and it follows that these subgroups Xi are permuted by S. Also, every irreducible character of M can be  uniquely written as a product αi , where ker(αi ) contains the product of all Xj with j = i. Suppose first that at least one of the simple factors is not normal in S. Then we  can choose the characters αi so that the set {αi } is not S-invariant, and thus α = αi is not S-invariant. Also, we can suppose that all of the αi are nonlinear, and since there are at least p factors in this case, we have α(1)  2p > p, as wanted. We can now assume that each simple direct factor of M is normal in S, and thus by passing to a homomorphic image of S, we can assume that M is simple and that there is a faithful action of a nontrivial p-group P on M . To complete the proof in this case, we show that every nonprincipal irreducible character of M has degree exceeding p, and that at least one such character is not P -invariant. Since the simple p -group M has an automorphism of order p, it follows from the classification of simple groups that M is of Lie type, defined over a field of order q = rp , for some prime power r not divisible by p. (See Theorem 7.1.2 of [2].) Then M has a subgroup isomorphic to one of SL(2, q), P SL(2, q) or Sz(2, q) by Lemma 2.4 of [10], and

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it follows that M has a subgroup B = EC, where E is elementary abelian of order q and C is cyclic. Also, C either acts transitively on the set of nonprincipal characters of E, or else there are two orbits of equal size. If χ ∈ Irr(M ) is nonprincipal, therefore, then χB has an irreducible constituent ψ whose kernel does not contain E, and thus   χ(1)  ψ(1)  (q − 1)/2 = rp − 1 /2 > p, where the final inequality holds because r and p cannot both be 2. Finally, we show that some irreducible character of M is not P -invariant. Assuming that this is false, Brauer’s permutation lemma (Theorem 6.32 of [5]) implies that every class of M is P -invariant. Since the class sizes are not divisible by p, each class of M contains an element of CM (P ), and thus M is the union of the conjugates of CM (P ). As is well known, the conjugates of a proper subgroup can never cover the whole group. (This follows since if H < G, the conjugates overlap, and their number is at most |G : H|.) It follows that CM (P ) = M , and this is a contradiction. 2 We can now prove Theorem B, which we restate here. Theorem 7.4. Let G be p-solvable, and suppose that cd(G) contains a multiple of the prime p. Let k < p be an integer, and assume that the largest multiple of p in cd(G) is at most kp. Then the number of members f ∈ cd(G) that are good for p is at most 2k. Proof. We proceed by induction on |G|. If G has a normal Sylow p-subgroup, then since cd(G) contains a multiple of p, this Sylow subgroup must be nonabelian. We can thus apply Corollary 5.4 to deduce that the number of p -degrees in cd(G) is at most 2k. Since integers that are good for p are not divisible by p, the result follows in this case, so we can assume that a Sylow p-subgroup of G is not normal. Let K be the intersection of the kernels of all characters χ ∈ Irr(G) such that χ(1) is good for p. Then Corollary 7.2 guarantees that G/K has a normal Sylow p-subgroup, and we conclude that K > 1. Now choose a minimal normal subgroup M of G with M ⊆ K. The degrees in cd(G) that are good for p are exactly the degrees in cd(G/M ) that are good for p, so the result will follow by the inductive hypothesis if cd(G/M ) contains some multiple of p. We can assume, therefore, that cd(G/M ) contains no multiple of p, and thus G/M has a normal abelian Sylow p-subgroup S/M , and S does not have a normal Sylow p-subgroup. In particular, S is not a p-group, and thus M is not a p-group, and since G is p-solvable, it follows that M is a p -group. If the minimal normal subgroup M is nonabelian, then it is a direct product of nonabelian simple groups. We conclude by Theorem 7.3 that Irr(S) contains a character θ such that θ(1) is a multiple of p exceeding p2 . If χ ∈ Irr(G) lies over θ, then since k < p, it follows that χ(1) is a multiple of p exceeding kp, and this is a contradiction. We conclude that M is abelian. We know that S/M is an abelian normal Sylow p-subgroup of G/M , and in particular, S  ⊆ M . But S is not abelian since G does not have a normal Sylow p-subgroup, and

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since S   G and M is minimal normal, we have S  = M , and we see that the hypotheses of Theorem 6.1 are satisfied with respect to the integer k. We conclude that there are at most 2k degrees in cd(G) that are not divisible by p, and this is an upper bound for the number of degrees in cd(G) that are good for p. 2 References [1] S. Dolfi, Large orbits in coprime actions of solvable groups, Trans. Amer. Math. Soc. 360 (2008) 135–152. [2] D. Gorenstein, R. Lyons, R. Solomon, The Classification of the Finite Simple Groups, Number 3, AMS Math. Surveys Monogr., vol. 40.3, 1998. [3] Z. Halasi, K. Podoski, Every coprime linear group admits a base of size two, arXiv:1212.0199, 2012. [4] B. Huppert, Character Theory of Finite Groups, Walter de Gruyter & Co., Berlin, 1998. [5] I.M. Isaacs, Character Theory of Finite Groups, AMS Chelsea Publishing, Providence, 2006, Corrected reprint of the 1976 original. [6] I.M. Isaacs, Finite Group Theory, Grad. Stud. Math., AMS, Providence, 2008. [7] I.M. Isaacs, D.S. Passman, Groups whose irreducible representations have degrees dividing pe , Illinois J. Math. 8 (1964) 446–457. [8] M.L. Lewis, A. Moretó, Bounding the number of character degrees in terms of the largest one, J. Algebra Appl. 13 (2) (2014), http://dx.doi.org/10.1142/S0219498813500965. [9] B. Newton, On the degrees of complex p-solvable linear groups, J. Algebra 288 (2005) 384–391. [10] I.M.I. Riera, A. Turull, Boosting an analogue of Jordan’s theorem for finite groups, arXiv: 1310.6518v2. [11] D.L. Winter, On finite solvable linear groups, Illinois J. Math. 15 (1971) 425–428.