aa r X i v : . [ m a t h . G R ] F e b On p -parts of character degrees ∗ Guohua Qian
Dept. Mathematics, Changshu Institute of Technology, Changshu, Jiangsu, 215500, ChinaE-mail: [email protected]
Yong Yang
Dept. Mathematics, Texas State University, San Marcos, TX 78666, USAE-mail: [email protected]
February 19, 2021
Abstract
Let G be a finite group and p be a prime. In this paper, we get the sharpbound for | G/O p ( G ) | p under the assumption that either p ∤ χ (1) for all χ ∈ Irr( G ) or p ∤ φ (1) for all φ ∈ IBr p ( G ). This would settle two conjectures raised by Lewis, Navarro,Tiep, and Tong-Viet in [7]. Keywords p -parts of degree, character, Brauer character, Sylow p -subgroup. ∗ Project supported by the Nature Science Foundation of China (No. 11871011) and a grant from theSimons Foundation (No. 499532). Introduction
In this paper, G always denotes a finite group, and Irr( G ) is the set of irreducible complexcharacters of G . For a given prime p , let m p be the p -part of an integer m , and let e p ( G ) = max { log p ( χ (1) p ) | χ ∈ Irr( G ) } . It is natural to ask how does this quantity e p ( G ) affect the structure of the Sylow p -subgroups of G . For example, the famous Ito-Michler Theorem [11] asserts that e p ( G ) = 0if and only if G has a normal abelian Sylow p -subgroup. In particular, when this happens,we have | G/O p ( G ) | p = 1.Let b ( P ) denote the largest degree of an irreducible character of P . Moret´o [12,Conjecture 4] conjectured that log p b ( P ) is bounded by some function of e p ( G ). Theconjecture was first proved for solvable groups by Moret´o and Wolf using an orbit theoremof solvable linear groups in [13], and later settled by the authors for arbitrary finite groupsin [21]. In fact, we showed that log p | G/O p ( G ) | p is bounded by a linear function of e p ( G ).The research on the structure of groups under the condition e p ( G ) ≤ e p ( G ) ≤
1, then | G/O p ( G ) | p ≤ p . For arbitrary finite groups, Lewis, Navarro, Tiep,and Tong-Viet proved in [7] that if e p ( G ) ≤
1, then | G/O p ( G ) | p ≤ p . This result wasimproved to | G/O p ( G ) | p ≤ p in [16] by the first author.The alternating group Alt shows that when e ( G ) = 1, the sharp bound for | G/O ( G ) | is 2 . Also, [6, Example 3.3] shows that for a solvable group G with e p ( G ) = 1, the sharpbound for | G/O p ( G ) | p is p . So it is natural to conjecture, as has been asked in [7, Page485], If e p ( G ) = 1 for an odd prime p , whether the sharp bound for | G/O p ( G ) | p is p ? The following result gives a positive answer to the question along with some extra struc-tural information.
Theorem 1.1
Let G be a finite group and p be a prime. If e p ( G ) ≤ , then | G/O p ( G ) | p ≤ p , unless p = 2 and G ∼ = Alt × L , where L has a normal abelian Sylow -subgroup, andin this case, it is clear that | G/O ( G ) | = 2 . Now we turn to consider the similar question for p -Brauer characters, which has alsobeen considered in [7]. As usual, let IBr p ( G ) be the set of irreducible p -Brauer charactersof G . Set e ′ p ( G ) = max { log p ( φ (1) p ) | φ ∈ IBr p ( G ) } . A Brauer character version of the Ito-Michler theorem only holds for the given prime p ,which asserts that e ′ p ( G ) = 0 if and only if G has a normal Sylow p -subgroup ([11, Theorem5.5]). For a finite group G with e ′ p ( G ) ≤
1, it is proved in [7, Theorem 1.5] that | G/O p ( G ) | p ≤ p , if p = 2 ,p , if p ≥ , if p = 3 and Alt is not involved in G,p , if p = 3 and Alt is involved in G.
2t is also conjectured [7, Page 485] that the correct bounds in the above inequalities are p , p and p respectively. We note that the bounds for the odd primes in the aboveinequalities ( p and p ) have been slightly improved to p and p respectively in [8].Generally speaking, “ e ′ p ( G ) ≤
1” does not imply “ e p ( G ) ≤ G has an abelian Sylow p -subgroup with O p ( G ) = 1, then“ e ′ p ( G ) ≤
1” does imply “ e p ( G ) ≤ G is p -solvable with O p ( G ) = 1 and e ′ p ( G ) ≤
1, [17, Corollary 2.6] shows that G has an elementary abelian Sylow p -subgroup.It follows that if G is p -solvable with O p ( G ) = 1, then e ′ p ( G ) ≤ e p ( G ) ≤ | G/O p ( G ) | p when G is p -solvable with e ′ p ( G ) ≤ Corollary 1.2
Let G be a p -solvable group for a prime p . If e ′ p ( G ) ≤ , then | G/O p ( G ) | p ≤ p . Assume that G is non- p -solvable with e ′ p ( G ) ≤
1. For the p = 2 case, we show that G/O ( G ) is a direct product of M and an odd order group (Proposition 3.4 below), andthis gives an answer to the open questions proposed at the end of [7]. For the case when p ≥
3, we show that
G/O p ( G ) has an elementary abelian Sylow p -subgroup (Proposition3.5), whence e p ( G/O p ( G )) ≤
1. Using these facts together with Theorem 1.1, we also getthe upper bound for | G/O p ( G ) | p along with some extra structural information. Theorem 1.3
Let G be a finite group and p be a prime. If e ′ p ( G ) ≤ , then | G/O p ( G ) | p ≤ p , unless p = 2 and G/O ( G ) is a direct product of M and an odd order group, in whichcase | G/O ( G ) | = 2 . We remark that the bounds for | G/O p ( G ) | obtained in Theorems 1.1 and 1.3 are sharp.We now fix some notation that will be used in the paper.Let G ♯ be the set of nonidentity elements of G , let Irr ♯ ( G ) and IBr ♯p ( G ) be the setof nonprincipal irreducible characters and nonprincipal irreducible p -Brauer characters,respectively, of G .Let E ( p n ) be an elementary abelian p -group of order p n . Let C n denote a cyclic groupof order n .Let Sol( G ) be the largest solvable normal subgroup of G ; let Soc( G ) be the socle of G , that is, the product of all minimal normal subgroups of G .For a given prime p , we denote by G Sol p the smallest normal subgroup of G such that G/G
Sol p is p -solvable, and denote by Sol p ( G ) the largest normal p -solvable subgroup of G .The paper is organized as follows. We will prove Theorem 1.1 in section 2 and Theorem1.3 in section 3. p -parts of character degrees We will prove Theorem 1.1 at the end of this section. We begin with listing some knownresults about finite groups G with e p ( G ) ≤ emma 2.1 Let G be a finite group and assume that e p ( G ) ≤ for a prime p . Then (1) If G is solvable, then | G/O p ( G ) | p ≤ p ; (2) If p = 2 and G is nonsolvable, then G is a direct product of Alt and a finite groupwith a normal abelian Sylow -subgroup, in particular | G/O ( G ) | = 2 ; (3) If p = 3 and Alt is involved in G , then O ′ ( G ) /O ( G ) ∼ = Alt , in particular | G/O ( G ) | = 3 ; (4) If p ≥ or, if p = 3 but Alt is not involved in G , then | G/ Sol( G ) | p ≤ p . Proof
These are [6, Theorem 1], [5, Main Theorem], [7, Corollary 3.7] and [16, Lemma3.3], respectively. ✷ The results in the above lemma show that, in order to prove Theorem 1.1, we onlyneed to consider the case when p ≥ G is nonsolvable. Lemma 2.2
Suppose that W is a faithful and irreducible G -module for a solvable group G . Assume that G = O ′ ( G ) , | G | = 3 and | W | ∈ { , , , } . Then G ∼ = C and W ∼ = E (2 ) . Proof
This is a special case of [6, Lemma 2.2]. ✷ Let V be an irreducible G -module. Recall that V is called imprimitive if V can bewritten as V = V ⊕ · · · ⊕ V n for n ≥ V i that are permuted transitively by G ,and that V is primitive if V is not imprimitive.If U ⋉ V is a Frobenius group with complement U and kernel V , then we say that U acts fixed-point-freely on V . Lemma 2.3
Let p, r be different primes with p ≥ , let G be a finite group with G = O p ′ ( G ) that acts faithfully and completely reducibly on an elementary abelian r -group V .Assume that | G | p = p c ≥ p , C p ∼ = O p ( G ) acts fixed-point-freely on V , and that everyelement of V ♯ is fixed by a unique subgroup of order p c − of G . Then G acts primitivelyon V . Proof
Let X be the set of subgroups D of order p c − of G such that C V ( D ) >
1. Thehypothesis implies that V = [ D ∈ X C V ( D ) ,C V ( D ) ∩ C V ( D ) = { } ( . )whenever D and D are different members of X , and that O p ( G ) ∩ D = 1 ( . )for all D ∈ X . Assume that G acts reducibly on V . Write V = W × U , where W and U are nontrivial G -invariant subgroups of V . Let w ∈ W ♯ , u ∈ U ♯ . Let D ∗ ∈ X be suchthat D ∗ centralizes ∗ where ∗ ∈ { w, u, wu } . Since W and U are D wu -invariant, D wu mustcentralize both w and u . Now ( . ) yields D w = D wu = D u .
4y the arbitrariness of u , D w centralizes U . By the arbitrariness of w , D u centralizes W .This implies that D w = D u centralizes V for all w ∈ W ♯ and all u ∈ U ♯ . However G actsfaithfully on V , we get a contradiction. Therefore, G acts irreducibly on V .Assume that V is imprimitive. There exists subgroups V , . . . , V d , d ≥
2, of V suchthat V = V × · · · × V d and G acts transitively on { V , . . . , V d } . Let K be the kernel ofthe action. Note that O p ( G ) ∩ K = 1 would imply C V ( O p ( G )) >
1, this contradicts theassumption on O p ( G ). Therefore O p ( G ) ≤ K . Clearly G > K , and thus | G/K | p ≥ p because of G = O p ′ ( G ). Combining this together with ( . ), we conclude that K does notcontain any subgroup as a member of X .Let v ∈ V ♯ and D ∈ X be such that D centralizes v . Since D K , we mayassume that D does not normalizes V . Let v ∈ V ♯ and D ∈ X so that D centralizes v v . This implies that D acts on { v , v } . Since D is a p group for an odd prime p , D must centralize both v and v . It follows by ( . ) that D = D . Now D centralizes v and hence normalizes V , a contradiction. Consequently V is a primitive G -module, aswanted. ✷ Let V be an irreducible G -module. Recall that V is quasi-primitive if V N is homoge-neous for all N ✂ G . By Clifford’s Theorem ([10, Theorem 0.1]), a primitive G -module isnecessarily quasi-primitive.Assume that a finite group G acts on a set Ω. For α ∈ Ω, α G := { α g , g ∈ G } is calleda G -orbit of α . If | α G | = | G | , then the orbit α G is called regular; and if | α G | p = | G | p for aprime p , then the orbit α G is called p -regular. Lemma 2.4
Let p, r be different primes with p ≥ , let G be a solvable group with G = O p ′ ( G ) , E ( p ) ∼ = P ∈ Syl p ( G ) and O p ( G ) ∼ = C p . Assume that G acts faithfullyand quasi-primitively on an elementary r -group V . If F ( G ) is not cyclic and G has no p -regular orbit on V , then p = 3 , G ∼ = C × SL(2 , , and V ∼ = E (7 ) . Proof
Clearly O p ( G ) ∼ = C p acts fixed-point-freely on V . Let X be the set of subgroups,different from O p ( G ), of order p of G . Since G has no p -regular orbit on V , every elementof V ♯ is centralized by a member of X . This implies that V ♯ = [ D ∈ X C V ( D ) ♯ . ( . )Since V is a quasi-primitive G -module, every normal abelian subgroup of G is cyclic.Furthermore, since O p ′ ( G ) = G and O p ( G ) ∼ = C p , it is easily verified that every normalcyclic subgroup of G is also central in F ( G ). Let F = F ( G ) and let Z be the socle ofcyclic group Z ( F ). By [10, Corollary 1.10], there exist E ✂ G , U ✂ G such that F = EU, E ∩ U = Z, U = C F ( E ) = Z ( F ) , | F/U | = e ; ( . ) F/U is a completely reducible
G/F -module and a faithful C G ( Z ) /F -module, possibly ofmixed characteristic; furthermore F/U is a direct product of irreducible
G/F -modules ofeven dimension.Note that “ e = 1” would imply F ( G ) := F = U = Z ( F ) is cyclic. Hence e >
1. Since G has no regular orbit on V , it follows by [20, Theorem 3.1] that e = 2 , , , , p ≥
5. The proof of [20, Theorem 3.2] already show that G has a p -regularorbit on V , a contradiction. Therefore p = 3. Clearly C ∼ = O ( G ) ≤ Z ≤ U, and it follows that 3 ∤ e . Consequently p = 3 , e ∈ { , , , } , | | Z | , | G/F | = 3 . Let K = C G ( F/U ) and
F/U = V ×· · ·× V k , where V , . . . , V k are irreducible G/F -modules.Since every V i has even dimension, | V i | ∈ { , , , } . Obviously | G/C G ( V i ) | = 3because O ( G ) ≤ U ≤ C G ( V i ). Now Lemma 2.2 yields G/C G ( V i ) ∼ = C . This also impliesthat G/K = G/ ( k \ i =1 C G ( V i )) ≤ G/C G ( V ) × · · · × G/C G ( V k )is a 3-group. Clearly K < G , and since F ≤ K and | G/F | = 3, we conclude that | G/K | = 3 , K/F = O ′ ( G/F ) . Since
F/U is a faithful C G ( Z ) /F -module, K ∩ C G ( Z ) = F . By G = O ′ ( G ) and theabelianlity of G/C G ( Z ), G/C G ( Z ) is also a 3-group. Consequently C G ( Z ) ≥ K . Now F = K ∩ C G ( Z ) = K and thus G/F ∼ = C . For every D ∈ X , since D ∩ F = D ∩ O ( G ) = 1, we have G = D ⋉ F. ( . )Let W be an irreducible constituent of V U where U = Z ( F ). By [19, Theorem 2.2(7)], | V | = | W | eb ( . )for a positive integer b . Note that 6 | | Z | | | U | and that U acts fixed-point-freely on W . Itfollows that | U | | ( | W | − , | W | ≥ . ( . )Applying (3) and (5) of [18, Lemma 2.4], we conclude that | C V ( D ) | ≤ | W | eb ( . )for all D ∈ X . Now ( . ),( . ) and ( . ) yield | W | eb − | V | − ≤ | X | ( | W | eb − . ( . )Clearly | N G ( P | ≥ | C G ( P ) | ≥ | U | · | P | ≥
18. Consequently | G : N G ( P ) | ≤ | G : F || F |
18 = 16 | F | = 16 | F : U || U | ≤ e ( | W | − . P ∼ = E (3 ) contains exactly 3 = − − − X , and it follows that | X | ≤ | G : N G ( P ) | ≤ e ( | W | − . ( . )By ( . ) and ( . ), we conclude that | W | eb − ≤ e ( | W | − | W | eb − | W | eb + 1 ≤ e ( | W | − , ( . )in particular, | W | eb − < e . Since | W | ≥ . ) and e ∈ { , , , } , the above inequality implies e ≤ b = 1.Note that if e = 4 and b = 1, then ( . ) yields a contradiction. Consequently, e = 2 and b = 1.Suppose that U = Z ( G ) and let D ∈ X . Since D acts nontrivially on the cyclicgroup U by ( . ), there exists a 3 ′ -subgroup Y of U of prime order such that D actsfixed-point-freely on Y . Considering the action of D ⋉ Y on V and observing that Y actsfixed-point-freely on V , we conclude from [10, Lemma 0.34] that dim C V ( D ) = dim V for all D ∈ X . By ( . ), ( . ) and ( . ), we get that6 | dim V and that | V | − ≤ | X | ( | V | / − ≤ | V | / − | V | / − < | V | / − , which is clearly impossible. Consequently U = Z ( G ).By ( . ), we have F/E ≤ Z ( G/E ). Since E ≥ Z ≥ O ( G ), F/E is a 3 ′ -group , and itfollows that F = E because G = O ′ ( G ). Now Z ( G ) = U = U ∩ F = U ∩ E = Z via ( . ). Recall that F/U ∼ = E (2 ) because e = 2. Obviously G = O ′ ( G ) also implies Z = Z ( G ) ∼ = C . Now it is easy to see that O ( F ) ∼ = Q , G = O ( G ) × H, where H = D ⋉ O ( F ) ∼ = SL(2 , D ∈ X , and that | X | = 3 | G : N G ( P ) | = 12 . By ( . ), we get that | W | − ≤ | W | − | W | ≤
11. Since 6 = | U | divides | W | −
1, we have | W | = 7. Recall that | V | = | W | eb where e = 2 and b = 1. Consequently V ∼ = E (7 ), and we are done. ✷ In order to prove Theorem 2.6, Propositions 3.4 and 3.5, we also need the followinglemma. 7 emma 2.5
Let P be a Sylow p -subgroup of a finite group G , and let Z ≤ P ∩ Z ( G ) .Assume that P is either abelian or split over Z . Then Z ∩ G ′ = 1 . Proof
By the hypothesis, every λ ∈ Irr( Z ) is extendible to P . It follows by [2, Theorem6.26] that λ extends to λ ∈ Irr( G ). Observe that λ is linear and G ′ ≤ ker λ . We haveker λ = Z ∩ ker λ ≥ Z ∩ G ′ . Therefore Z ∩ G ′ ≤ T λ ∈ Irr( Z ) ker λ = 1. ✷ Theorem 2.6
Let H be a finite group and p be an odd prime. Assume that e p ( H ) ≤ .Then | H/O p ( H ) | p ≤ p . Proof
Assume the result is not true. Let H be a counterexample with minimal or-der. Note that the hypothesis e p ( H ) ≤ H . We will deduce a contradiction via several steps.(1) O p ( H ) = Φ( H ) = 1, H = O p ′ ( H ), | H/ Sol( H ) | p = p , | Sol( H ) | p = p and | H | p = p .Suppose that O p ( H ) >
1. Since e p ( H/O p ( H )) ≤ H is a minimal counterexam-ple, we have | ( H/O p ( H )) /O p ( H/O p ( H )) | p ≤ p , that is, | H/O p ( H ) | p ≤ p , a contradiction.Therefore O p ( H ) = 1.Suppose that Φ( H ) >
1. Note that Φ( H ) ≤ O p ′ ( H ) and O p ( H/ Φ( H )) = 1 because O p ( H ) = 1. Since e p ( H/ Φ( H )) ≤ H is a minimal counterexample, we concludethat | ( H/ Φ( H )) /O p ( H/ Φ( H )) | p ≤ p . This implies | H/O p ( H ) | p ≤ p , a contradiction.Therefore Φ( H ) = 1.Since e p ( O p ′ ( H )) ≤
1, we also have H = O p ′ ( H ).By Lemma 2.1(3), we may assume that if p = 3, then Alt is not involved in H . By(1) and (4) of Lemma 2.1, we have that | H/ Sol( H ) | p ≤ p and | Sol( H ) | p ≤ p . Since H isa counterexample, we get that | H/ Sol( H ) | p = p , | Sol( H ) | p = p and | H | p = p .(2) H = G ⋉ V , where G is a maximal subgroup of H with G = O p ′ ( G ), O p ( G ) = h a i ∼ = C p , V ∼ = E ( r n ) is a unique minimal normal subgroup of H where r = p is a prime,furthermore G acts faithfully on V .Let N be a minimal normal subgroup of H . Assume that p divides | N | . Since O p ( H ) = 1, we have N ∩ Sol( H ) = 1. Observe that neither N nor Sol( H ) contains anormal abelian Sylow p -subgroup. By the Ito-Michler Theorem, there exist θ ∈ Irr( N )and θ ∈ Irr(Sol( H )) such that p divides θ (1) and θ (1). Since θ θ ∈ Irr( N × Sol( H )),we get that e p ( H ) ≥ e p ( N × Sol( H )) ≥ log p (( θ θ )(1)) p ≥
2, a contradiction. Hence N isnecessarily a p ′ -group. Note that since H is a minimal counterexample, we also concludethat O p ( H/N ) > N of G .Suppose that H admits different minimal normal subgroups, say N , N . Let D i ✂ H be such that D i /N i = O p ( H/N i ). Since p | | D i | and O p ( D i ) ≤ O p ( H ) = 1 , there exists θ i ∈ Irr( D i ) such that p | θ i (1) for every i ∈ { , } . Note that D ∩ D = 1 and D D = D × D ✂ H . It follows that θ θ ∈ Irr( D × D ) and that e p ( H ) ≥ log p (( θ θ )(1)) p ≥ H admits a unique minimal normal subgroup, say V .Since Sol( H ) is nontrivial by (1), the uniqueness of V implies V ≤ Sol( H ), and hence V ∼ = E ( r n ) for some prime power r n , where r = p . Since Φ( H ) = 1, there exists a maximalsubgroup G of H such that H = G ⋉ V . Since H = O p ′ ( H ), we have G = O p ′ ( G ).8bserve that C G ( V ) is normal in H . The uniqueness of V implies C G ( V ) = 1, thatis, G acts faithfully on V .Observe that 1 < O p ( H/V ) ∼ = O p ( G ) ≤ Sol( H ) and | Sol( H ) | p = p . It follows that O p ( G ) has order p or p . In particular O p ( G ) is abelian. Obviously O p ( G ) acts faithfullyand coprimely on V . By [10, Lemma 18.1], there exists θ ∈ Irr( O p ( G ) ⋉ V ) of degree | O p ( G ) | . Since e p ( O p ( G ) ⋉ V )) ≤ e p ( H ) ≤
1, we have O p ( G ) = h a i ∼ = C p .(3) Let P ∈ Syl p ( G ), J = O p ′ (Sol( G )) and P J = P ∩ J . Let J ∈ { J, G } and let | J | p = p c . Then(3a) For every λ ∈ Irr ♯ ( V ), we have that I J ( λ ) ∩ h a i = 1 and that I J ( λ ) admits anormal abelian Sylow p -subgroup of order p c − ;(3b) P ∼ = E ( p ), P J ∼ = E ( p );(3c) J acts faithfully and primitively on V and Irr( V ).Note that | P | = p and | P J | = p by (1).Clearly h a i = O p ( G ) ≤ J , J ✂ G and thus e p ( J ⋉ V ) ≤
1. Let λ ∈ Irr ♯ ( V ) and let T = I J ( λ ). Since G acts faithfully and irreducibly on V , h a i acts fixed-point-freely on V .This implies that T ∩ h a i = 1 and hence | T | p ≤ p c − . Assume that | T | p ≤ p c − . By [2,Theorem 6.11], there exists χ ∈ Irr( J ⋉ V | λ ) of degree divisible by p , a contradiction.Assume that | T | p = p c − and that a Sylow p -subgroup of T is nonnormal or nonabelian.Note that λ is extendible to T ⋉ V and that T has an irreducible character of degreedivisible by p . By [2, Corollary 6.17], there exists θ ∈ Irr( T ⋉ V | λ ) of degree divisible by p ; and by [2, Theorem 6.11], there exists χ ∈ Irr( J ⋉ V | θ ) of degree divisible by p , acontradiction. Consequently, T admits a normal abelian Sylow p -subgroup of order p c − .Assume that P/ h a i is cyclic. Since O p ( G/ h a i ) = 1, it is well-known [22] that G/ h a i admits an irreducible character of degree divisible by | P/ h a i| = p , a contradiction. Hence P/ h a i ∼ = E ( p ). Let λ ∈ Irr ♯ ( V ) and P ∈ Syl p (I G ( λ )). Without loss of generality, we mayassume P ≤ P . Note that h a i ∩ P = 1 by (3a), P ∼ = P/ h a i and h a i ✂ P . We get that P = h a i × P ∼ = E ( p ) and P J ∼ = E ( p ).Clearly O p ′ ( J ) = J and J acts faithfully and completely irreducibly on V andIrr( V ). It follows by (3a) and Lemma 2.3 that Irr( V ) is a faithful primitive J -module,and so is V . Note that we only use the primitivity of J -module Irr( V ) in the sequel.(4) F ( J ) is cyclic.Observe that Irr( V ) is a faithful primitive J -module by (3), and that J has no p -regular orbit on Irr( V ) because e p ( J ⋉ V ) ≤ | P J | = p . Assume that F ( J ) is notcyclic. Applying Lemma 2.4, we get that p = 3 , J ∼ = C × SL(2 , , V ∼ = Irr( V ) ∼ = E (7 ) . Since V is the unique minimal normal subgroup of H , we have C G ( V ) = 1 and G ≤ GL(2 , J in GL(2 , G/J is a { , } -group.By Burnside’s p a q b Theorem,
G/J is solvable, and so is H , a contradiction. Consequently F ( J ) is cyclic.(5) There exist elements b, c ∈ P ♯ and normal subgroups B, C of G such that95a) P = h a i × h b i × h c i , G = h a i × ( BC ), where h a i = O p ( G );(5b) h a i × B = J , B = h b i ⋉ O p ′ ( J ) with B = O p ′ ( B );(5c) C is nonsolvable with h c i ∈ Syl p ( C ) and C = O p ′ ( C ), B ∩ C = O p ′ ( B ) = O p ′ ( J ).Since C p ∼ = h a i < P J and P J ∼ = E ( p ) by (3b), we may write P J = h a i × h b i , where o ( a ) = o ( b ) = p . Since P J ∈ Syl p ( J ) is abelian, the p -solvable subgroup J with J = O p ′ ( J )has a normal p -complement, that is, J = P J ⋉ O p ′ ( J ) . Let M ✁ G be maximal such that P J ∩ M = 1 and O p ′ ( J ) ≤ M . We claim that G = P J ⋉ M .Observe that E ( p ) ∼ = P J M/M = P J O p ′ ( J ) M/M = J M/M ✂ G/M.
Since
G/M admits an abelian Sylow p -subgroup, G/C G ( P J M/M ) is a p ′ -group. Since G = O p ′ ( G ), P J M/M lies in the center of
G/M . By Lemma 2.5, we have ( P J M/M ) ∩ ( G/M ) ′ = 1, that is, P J M ∩ G ′ M = M . This also implies that P J ∩ G ′ ≤ P J ∩ G ′ M = P J ∩ ( P J M ∩ G ′ M ) = P J ∩ M = 1 . By the maximality of M , we have G ′ M = M and thus G ′ ≤ M . Note that G/M is anelementary abelian p -group because G = O p ′ ( G ) and E ( p ) ∼ = P ∈ Syl p ( G ). Consequently P J M/M is split over
G/M . Let
U/M be a complement of P J M/M in G/M . We havethat P J ∩ U = P J ∩ ( P J M ∩ U ) = P J ∩ M = 1and that G = P J M U = P J U = P J ⋉ U . Obviously U = M and the claim follows.Let B = h b i ⋉ O p ′ ( J ). Clearly J = P J ⋉ O p ′ ( J ) = ( h a i × h b )) ⋉ O p ′ ( J ) = h a i ⋉ B = h a i× B . Since J = O p ′ ( J ), we have B = O p ′ ( B ). Note that h b i M ✂ G because G = P J ⋉ M by the claim, and that( P J ∩ h b i M ) O p ′ ( J ) = P J O p ′ ( J ) ∩ h b i M = J ∩ h b i M where the first equality follows from O p ′ ( J ) ≤ M ≤ h b i M . This implies that B = h b i ⋉ O p ′ ( J ) = ( P J ∩ h b i M ) O p ′ ( J ) = J ∩ h b i M ✂ G. Let h c i = P ∩ M . Clearly P = h a i × h b i × h c i . Let C = (cid:26) h c i ⋉ [ O p ′ ( G ) , h c i ] , if G is p − solvable ,G Sol p , otherwise . Assume that G is p -solvable. Since G admits an abelian Sylow p -subgroup with G = O p ′ ( G ), we have G = P ⋉ O p ′ ( G ). Thus C = h c i ⋉ [ O p ′ ( G ) , h c i ] is normal in G , and C = O p ′ ( C ). Now let us investigate the case when G is non- p -solvable. Clearly C = G Sol p is also normal in G , and also C = O p ′ ( C ). Observe that | C | p = | G Sol p | p ≥ p because G isnot p -solvable, and that C = G Sol p ≤ M because G/M ∼ = E ( p ) is p -solvable. It followsthat | C | p = | M | p = p and P ∩ C = P ∩ M = h c i . Now we always have h a i , B, C ✂ G, P ∩ B = h b i , P ∩ C = h c i , B = O p ′ ( B ) and C = O p ′ ( C ). Now h a i BC is normal in G with p ′ -index. Since G = O p ′ ( G ), we have G = h a i BC = h a i × ( BC ) . Since J = h a i × B is solvable but G is not solvable, C is necessarily nonsolvable.Now it suffices to show that B ∩ C = O p ′ ( B ). Let D = B ∩ C . Note that D is a p ′ -group because | BC | p = p = | B | p | C | p . Consequently | B/D | p = | B | p = p = | C | p = | C/D | p . Assume that
C/D has a normal Sylow p -subgroup. Since O p ′ ( C ) = C and | C/D | p = p ,we have | C/D | = p . However D ≤ B < J ≤ Sol( G ), we get that C is solvable, acontradiction. Consequently C/D admits a nonnormal Sylow p -subgroup. Assume that B/D also admits a nonnormal Sylow p -subgroup. By the Ito-Michiler Theorem, we maytake θ b ∈ Irr(
B/D ) and θ c ∈ Irr(
C/D ) such that p divides θ b (1) and θ c (1). It follows that e p ( H ) ≥ e p ( G ) ≥ e p ( B/D × C/D ) ≥ log p (( θ b θ c )(1)) p ≥
2, a contradiction. Therefore,
B/D must admit a normal Sylow p -subgroup. Since B = O p ′ ( B ) and | B/D | p = p , wehave B/D ∼ = C p . Consequently D = O p ′ ( B ), as required.(6) Final contradiction.Since J = O p ′ ( J ) and F ( J ) is cyclic by (4), J/F ( J ) = J/C J ( F ( J )) is an abelian p -group. Since | P J | = p and | F ( J ) | p = | O p ( G ) | = p , we have J/F ( J ) ∼ = C p . We keep all notations in (5). Let D = O p ′ ( B ) and let us investigate BC ✂ G . By (5), itis easy to see that B = h b i ⋉ D, D = B ∩ C = O p ′ ( B ) = O p ′ ( J ) = O p ′ ( F ( J )) . Since O p ( G ) ∼ = C p , we have D >
D/S be a G -chief factor. Since D = O p ′ ( F ( J )) is a cyclic p ′ -group, D/S ∼ = C q for some prime q = p . Since G = O p ′ ( G ) and G/C G ( D/S ) is abelian, all p ′ -elements of G are contained in C G ( D/S ), that is, O p ( G ) ≤ C G ( D/S ). Consequently O p ( BC ) ≤ C G ( D/S ) ∩ BC = C BC ( D/S ) . ( . )Suppose that D/S ≤ Φ( BC/S ). Since
B/S ✂ BC/S and h b i S/S ∈ Syl p ( B/S ), weconclude by the Frattini Argument that
BC/S = (
B/S )( N BC/S ( h b i S/S )) = (
D/S )( N BC/S ( h b i S/S )) = N BC/S ( h b i S/S ) . It follows that h b i S/S ✂ B/S . Consequently
B/S = h b i S/S × D/S and O p ′ ( B ) < B , acontradiction.Suppose that D/S ∩ Φ( BC/S ) = 1. Let λ ∈ Irr ♯ ( D/S ). We also view λ as a characterof D and write T = I BC ( λ ). Since BC/S is split over
D/S , λ is extendible to T . Since11 /S ∼ = C q , we know that an element g ∈ BC fixes λ if and only if g centralizes D/S .Consequently T = C BC ( D/S ) . ( . )Assume that T ∩ B = D . Since T ∩ B ≥ D and B/D ∼ = C p , we have T ∩ B = B ,that is, B ≤ T . Now B centralizes D/S by ( . ), and this also yields O p ′ ( B ) < B , acontradiction. Therefore T ∩ B = D , and in particular | T | p ≤ p . Note that e p ( BC/S ) ≤ p does not divide χ (1) for any irreducible constituent χ of λ BC . By [2,Theorem 6.11, Corollary 6.17], T /D admits a normal abelian Sylow p -subgroup of order p . Observe that BC/O p ( BC ) ≤ E ( p ) , T ∩ B = D, and BC > T = C BC ( D/S ) ≥ O p ( BC )by ( . ) and ( . ). It follows that T ✂ BC, BT ✂ BC and that | BT | p = | B | p | T | p | B ∩ T | p = | B | p | T | p | D | p = p , that is, BT is a normal subgroup of BC with p ′ -index. Note that O p ′ ( BC ) = BC because G = h a i × BC and G = O p ′ ( G ). Consequently, BC = BT and BC/D = (
B/D )( T /D ) =
B/D × T /D.
Since O p ′ ( BC ) = BC , we also have O p ′ ( T /D ) =
T /D . Recall that
T /D admits a normalabelian Sylow p -subgroup of order p , and it follows that T /D ∼ = C p . Now BC/D ∼ = E ( p ).Since D is cyclic, BC and G, H are necessarily solvable, a contradiction. ✷ Proof of Theorem 1.1.
It follows directly from Theorem 2.6 and Lemma 2.1(2). ✷ Remark 2.7
Let H be a finite group with | H/O p ( H ) | p = p and e p ( H ) = 1 for an oddprime p , assume that | U/O p ( U ) | p ≤ p for all proper quotient groups U of H . Let usinvestigate the structure of H . Suppose that p = 3 and Alt is involved in H . By Lemma 2.1(3), we have O ′ ( H ) /O ( H ) ∼ = Alt . Suppose that p ≥ p = 3 and Alt is not involved in H . Using the samearguments as in the proof of Theorem 2.6, we obtain that(1) O p ( H ) = Φ( H ) = 1, | H | p = p , | Sol( H ) | p ≥ p ;(2) H admits a unique minimal normal subgroup, say V ; V is an elementary r -groupfor a prime r = p ; H = G ⋉ V for a maximal subgroup G of H ; O p ( G ) ∼ = C p ;(3) Every λ ∈ Irr ♯ ( V ) is fixed by a unique subgroup of G of order p . It follows fromLemma 2.3 that Irr( V ) and V are faithful primitive O p ′ ( G )-module. ✷ We do not find an example of non- p -solvable group H with e p ( H ) = 1 and | H/O p ( H ) | p = p for a prime p ≥
5. Can one construct such an example?It would be also desirable for one to obtain a detailed structure of finite solvable groups G with e p ( G ) = 1 (or e ′ p ( G ) = 1) and | G/O p ( G ) | p = p .12 On p -parts of Brauer character degrees We will prove Theorem 1.3 at the end of this section. We begin with listing some knownresults about finite groups G with e ′ p ( G ) ≤ Lemma 3.1
Let p be a prime and G be a finite p -solvable group with O p ( G ) = 1 . Then e ′ p ( G ) ≤ if and only if e p ( G ) ≤ . Proof
This is [7, Corollary 1.4].
Proof of Corollary 1.2.
It follows directly from Theorem 1.1 and Lemma 3.1.
Lemma 3.2
Let p be a prime, let G be a finite group with O p ( G ) = 1 , and assume that G has an abelian Sylow p-subgroup. If e ′ p ( G ) ≤ , then e p ( G ) ≤ . Proof
This is [7, Theorem 1.3]. ✷ Lemma 3.3
Let p be a prime and let G be a non- p -solvable group with G = O p ′ ( G ) and Sol p ( G ) = 1 . If e ′ p ( G ) ≤ , then G is a nonabelian simple group, furthermore the followinghold: (1) If p = 2 , then G ∼ = M ; (2) If p ≥ , then either | G | p = p or, p = 3 and G ∼ = Alt . Proof
It follows directly from Lemmas 4.5 and 4.3 of [7]. ✷ Let p be a prime and G be a finite group. Recall that χ ∈ Irr( G ) is said to have p -defectzero if χ (1) p = | G | p . By [14, Theorem 3.8], if χ is of p -defect zero, then χ ∈ IBr p ( G ),where χ is the restriction of χ to the set of p -regular elements of G . Proposition 3.4
Let G be a nonsolvable group. Then e ′ ( G ) ≤ if and only if G/O ( G ) is a direct product of M and an odd order group. Proof
Since IBr ( G ) = IBr ( G/O ( G ), we may assume by induction that O ( G ) = 1.Since e ′ ( M ) = 1 by [3], we only need to prove the necessity. Note that all odd ordergroups are solvable.Suppose that e ′ ( G ) ≤ W = O ′ ( G/ Sol( G )). Then W is nonsolvable with e ′ ( W ) ≤
1. Clearly W = O ′ ( W ) and Sol ( W ) = Sol( W ) = 1. Hence W ∼ = M byLemma 3.3. Since Out( M ) ∼ = C by [1], G/ Sol( G ) = W × ( K/ Sol( G )), where K/ Sol( G )is of odd order. This implies that K = Sol( G ) and G/ Sol( G ) ∼ = M . ( . )Assume that G > G ′ and let M ✂ G be such that | G : M | is a prime. Note that M isnonsolvable with e ′ ( M ) ≤
1. Since O ( M ) ≤ O ( G ) = 1, we conclude by induction that13 = S × U where S ∼ = M and U = O ′ ( M ) = Sol( M ). Note that if M = M Sol( G ),then Sol( G ) ≤ M , and this contradicts ( . ). Therefore G = M Sol( G ) = SU Sol( G ) = S Sol( G ) = S × Sol( G ) . Now e ′ ( G ) ≤ e ′ (Sol( G )) = 0. Therefore Sol( G ) admits a normal Sylow p -subgroup. Since O ( G ) = 1, Sol( G ) has odd order, and we are done. Hence we mayassume that O ( G ) = 1 , G = G ′ , G/ Sol( G ) ∼ = M . ( . )Now it suffices to show that G ∼ = M . Assume that this is not true. By ( . ), wemay take N ✂ G maximal so that G/N has an odd order chief factor. By ( . ) andthe maximality of N , we get that N ≤ Sol( G ), G/N has a minimal normal subgroup
V /N ∼ = E ( r n ) for an odd prime r , V /N is the unique minimal normal subgroup of
G/N ,and that Sol( G ) /V = O ( G/V ). In order to see a contradiction, we may assume that N = 1 . Since e ′ (Sol( G )) ≤
1, we have e (Sol( G )) ≤ G ) /V is an abelian 2-group of order at most 4. Note that e (Sol( G )) =log | Sol( G ) /V | by [10, Lemma 18.1]. It follows that | Sol( G ) /V | ≤
2. Since G = G ′ , weconclude that G/V ∼ = M or 2 .M . Assume that
G/V ∼ = M . By [9, Lemma 2.3], there exists χ ∈ Irr( G ) of 2-defect zero.This leads to χ ∈ IBr ( G ). Now e ′ ( G ) ≥ log ( χ (1)) = 7, and we get a contradiction.Assume that G/V ∼ = 2 .M . Note that “ V ≤ Φ( G )” would imply Sol( G ) is nilpotent,and this contradicts the unique minimal normality of V . Consequently Φ( G ) = 1, whence G = H ⋉ V, where H ∼ = G/V ∼ = 2 .M . Let λ ∈ IBr ♯ ( V ) and write T = I H ( λ ). Clearly IBr ( V ) = Irr( V ) because V has odd order.Note that Sol( G ) = Z ( H ) ⋉ V is a Frobenius group with V as its kernel. It follows that T ∩ Z ( H ) = 1 . ( . )Since 4 does not divide φ (1) for any φ ∈ IBr ( G | λ ), we conclude by ( . ) and CliffordCorrespondence ([14, Theorem 8.9]) that a Sylow 2-subgroup P of T must be isomorphicto a Sylow 2-subgroup of M . Now ( . ) also implies that P × Z ( H ) ∈ Syl ( H ) . Applying Lemma 2.5, we have Z ( H ) ∩ H ′ = 1. However H = H ′ because G = G ′ , we geta contradiction. ✷ Proposition 3.5
Let G be a finite group and p be an odd prime. Assume that e ′ p ( G ) ≤ .Then G/O p ( G ) admits an elementary abelian Sylow p -subgroup. roof By induction, we may assume that O p ( G ) = 1 and G = O p ′ ( G ). Let P ∈ Syl p ( G )and let N = Sol p ( G ). Since N is p -solvable and e ′ p ( N ) ≤
1, we have e p ( N ) ≤ | P ∩ N | ≤ p by Theorem 1.1. Assume that P ∩ N is a cyclic groupof order p . It follows by [22] that N has an irreducible character of degree divisibleby p , a contradiction. Hence P ∩ N ≤ E ( p ). Now we may assume P > P ∩ N .Observe that G/N is non- p -solvable with e ′ p ( G/N ) ≤
1, while
G/N = O p ′ ( G/N ) andSol p ( G/N ) = 1. It follows from Lemma 3.3 that
G/N is a nonabelian simple group, andthat either | G/N | p = p , or p = 3 and G/N ∼ = Alt . In particular, G/N always has anelementary abelian Sylow p -subgroup. Now the required result holds obviously for thecase when P ∩ N = 1. Consequently, we may assume that P > P ∩ N ∼ = C p or E ( p ) . Let V be a minimal normal subgroup of G . Assume that V (cid:2) N . Since G/N is anonabeian simple group, we have G = N × V where V ∼ = G/N . Since both V and N have elementary abelian Sylow p -subgroups, P is elementary abelian, and we are done.We therefore may assume that V ≤ N = Sol p ( G ). Now V ≤ O p ′ ( G ) because O p ( G ) = 1.Note that if O p ( G/V ) = 1, then the induction implies the required result. Hence we mayalso assume that
D/V := O p ( G/V ) > D of G . Clearly D ≤ N and O p ( D ) = 1. Since P ∩ N ≤ E ( p ), we have C p ≤ D/V ≤ E ( p ). By [15, Corollary 2.8] when V is nonabelianor [10, Lemma 18.1] when V is abelian, we may take λ ∈ Irr ♯ ( V ) = IBr ♯p ( V ) such thatI D ( λ ) = V , that is, I G ( λ ) ∩ D = V . Since p does not divide φ (1) for any φ ∈ IBr p ( G | λ ),we conclude by [14, Theorem 8.9] that D/V ∼ = C p and that I G ( λ ) contains a Sylow p -subgroup Y of order | P | /p . Note that D/V is centralin every Sylow p -subgroup of G/V . This implies that(
Y V /V ) × ( D/V ) ∈ Syl p ( G/V ) . Applying Lemma 2.5, we get that (
D/V ) ∩ ( G/V ) ′ = 1, that is, D ∩ G ′ V = V. ( . )Since G/N is a nonabelian simple group, we have G = N G ′ . Observe that G/G ′ isa p -group because G = O p ′ ( G ) and that P ∩ N is elementary abelian. It follows that G/G ′ ∼ = N/ ( N ∩ G ′ ) is an elementary abelian p -group. Now all subgroups of G/G ′ arecomplemented in G/G ′ . Let R/G ′ be a complement of DG ′ /G ′ in G/G ′ . We have G = RD, R ∩ DG ′ = G ′ . ( . )Using ( . ), ( . ) and the Dedekind’s Modular Law, we get that D ∩ RV = V ( D ∩ R ) = V ( D ∩ DG ′ ∩ R ) = V ( D ∩ G ′ ) = V ( D ∩ G ′ V ∩ G ′ ) = V ( V ∩ G ′ ) = V, and that G/V = RD/V = (
D/V ) × ( RV /V ) . ( . )15ow RV is a proper normal subgroup of G with O p ( RV ) ≤ O p ( G ) = 1. Applying theinductive hypothesis to RV , we conclude that RV has an elementary abelian Sylow p -group. This implies by ( . ) that G/V has an elementary abelian Sylow p -subgroup.Therefore P ∼ = P V /V is elementary abelian. ✷ Suppose that e p ( G ) = 1 for an odd prime p . By Theorem 1.1 and a result in [22], wealso conclude that G/O p ( G ) admits an elementary abelian Sylow p -subgroup of order atmost p . Corollary 3.6
Let G be a finite group with O p ( G ) = 1 for an odd prime p . If e ′ p ( G ) ≤ ,then e p ( G ) ≤ . Proof
It follows from Proposition 3.5 and Lemma 3.2. ✷ Proof of Theorem 1.3. If p = 2, the required result follows from Corollary 1.2 andProposition 3.4. For the p ≥ e p ( G/O p ( G )) ≤ ✷ ACKNOWLEDGMENT
The authors would like to thank Dr. Yanjun Liu for his helpful discussions.
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