For a character $\chi $ of a finite group $G$, the co-degree of $\chi $ is $\chi ^c(1)=\frac{[G:\ker \chi ]}{\chi (1)}$. Let $p$ be a prime and let $e$ be a positive integer. In this paper, we first show that if $G$ is a $p$-solvable group such that $p^{e+1}\nmid \chi ^c(1)$, for every irreducible character $\chi $ of $G$, then the $p$-length of $G$ is not greater than $e$. Next, we study the finite groups satisfying the condition that $p^2$ does not divide the co-degrees of their irreducible characters

Neda Ahanjideh

Roya Bahramian

Comptes Rendus Mathématique

Bahramian, Roya

Ahanjideh, Neda

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Comptes RendusMathématiqueRoya Bahramian and Neda Ahanjidehp-parts of co-degrees of irreducible charactersVolume 359, issue 1 (2021), p. 79-83Published online: 15 February 2021https://doi.org/10.5802/crmath.158This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/Les Comptes Rendus. Mathématique sont membres duCentre Mersenne pour l’édition scientifique ouvertewww.centre-mersenne.orge-ISSN : 1778-3569Comptes RendusMathématique2021, 359, n 1, p. 79-83https://doi.org/10.5802/crmath.158Group theory / Théorie des groupesp-parts of co-degrees of irreducible charactersRoya Bahramiana and Neda Ahanjideh∗, aa Department of Pure Mathematics, Faculty of Mathematical Sciences, ShahrekordUniversity, P. O. Box 115, Shahrekord, IranE-mails: roya.bahramian98@gmail.com, ahanjideh.neda@sku.ac.irAbstract. For a character χ of a finite group G , the co-degree of χ is χc (1) = [G :kerχ]χ(1) . Let p be a prime andlet e be a positive integer. In this paper, we first show that if G is a p-solvable group such that pe+1 - χc (1),for every irreducible character χ of G , then the p-length of G is not greater than e. Next, we study the finitegroups satisfying the condition that p2 does not divide the co-degrees of their irreducible characters.Mathematical subject classification (2010). 20C15, 20D10, 20D05.Manuscript received 22nd April 2020, revised and accepted 24th November 2020.1. Introduction and preliminariesIn this paper, G is a finite group, p is a prime number and e is a positive integer. Let Z (G) be thecenter of G and let Op (G) and Op ′ (G) be the largest normal p-subgroup and the largest normalp ′-subgroup of G , respectively. Also, Op′(G) is the largest normal subgroup of G whose index in Gis co-prime to p. For a p-solvable group G , the p-length of G , denoted by `p (G), is the minimumpossible number of factors that are p-groups in any normal series of G which every factor is eithera p-group or a p ′-group. Let Irr(G) denote the set of (complex) irreducible characters of G . For anormal subgroup N of G and a character θ of N , let IG (θ) denote the inertia group of θ in G andlet Irr(G|θ) be the set of the irreducible constituents of the induced character θG . Also, we use epto show the p-part of e. For a character χ of G , the number χc (1) = [G :kerχ]χ(1) is called the co-degreeof χ (see [11]). Set Codeg(G) = {χc (1) : χ ∈ Irr(G)}. In [1–3, 11], some properties of the co-degreesof irreducible characters of finite groups have been studied.In [1], it has been proved that the p-length of a finite p-solvable group is not greater than thenumber of the distinct co-degrees of its irreducible characters which are divisible by p. In thispaper, we prove that:Theorem 1. If G is a p-solvable group and pe+1 -χc (1), for every χ ∈ Irr(G), then `p (G) ≤ e.In [8–10], it has been shown that if p2 -χ(1), for every χ ∈ Irr(G), then [G : Op (G)]p ≤ p3. In thispaper, we also prove that:∗Corresponding author.ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/80 Roya Bahramian and Neda AhanjidehTheorem 2. Let G be a non-p-solvable group. If χc (1)p ≤ p, for every χ ∈ Irr(G), then |G|p = p.Corollary 3. If χc (1)p ≤ p, for every χ ∈ Irr(G), then the Sylow p-subgroups of G are elementaryabelian p-groups.In Examples 9, 10 and 11, we show that in Theorem 2, “non-p-solvability” cannot be substi-tuted with “non-solvability” and in Corollary 3, there is not necessarily an upper bound for |G|por |G/Op (G)|p .2. Proofs of the main resultsWe first state a lemma that will be used frequently in this paper without explicit reference.Lemma 4 (cf. [11, Lemma 2.1]). Let N be a normal subgroup of G. Then, Codeg(G/N ) ⊆Codeg(G). Also, if ψ ∈ Irr(N ), then ψc (1) |χc (1), for every χ ∈ Irr(G|ψ).Lemma 5. Let S be a non-abelian simple group.(i) If p is a prime divisor of the order of the Schur multiplier of S, then |S|p ≥ p2.(ii) If p is a prime divisor of |Out(S)| such that p divides |S|, then |S|p ≥ p2.Proof. Since S is a non-abelian simple group, |S|2 ≥ 4. So, the lemma follows when p = 2. Next,assume that p ≥ 3. If p is a prime divisor of the order of the Schur multiplier of S, then sincep ≥ 3, [7, Section 5.1] shows that S ∼= PSLn(q), p | q −1 and p | n, S ∼= PSUn(q), p | q +1 and p | n,S ∈ {PSL2(9), Al t7,PSU4(3),G2(3), J3, M22,F i22, Mcl ,Suz,B3(3), 2E6(4),F i ′24,O′N }(under isomorphism) and p = 3, S ∼= E6(q) and p = 3 | q −1 or S ∼= 2E6(q) and p = 3 | q +1. Thus,we can check at once that |S|p ≥ p2, as desired in (i). Next, let p be a prime divisor of |Out(S)| and|S|. Then, [8, Lemma 3.1] shows that |S|p > |Out(S)|p ≥ p. Thus, |S|p ≥ p2, as wanted. In order to prove the main results, we need to prove the following propositions:Proposition 6. Let N be a minimal normal subgroup of G.(i) If N is abelian and χ ∈ Irr(G) such that N 6≤ kerχ, then |N | divides χc (1).(ii) If p | |N | and χc (1)p ≤ p, for every χ ∈ Irr(G), then |N |p = p and N is a simple group.Proof. (i). Since N is a minimal normal subgroup of G and N 6= N ∩ kerχEG , N ∩ kerχ ={1}. So, N ∼= N kerχ/kerχ is an abelian normal subgroup of G/kerχ. By Ito’s theorem (see [6,Theorem 6.15]), χ(1) | [ Gkerχ : N kerχkerχ ]= [G : N kerχ]. Thus, |N | |χc (1), as desired in (i).(ii). First suppose that N ≤ Op (G), θ ∈ Irr(N ) − {1N } and χ ∈ Irr(G|θ). Then, N 6≤ kerχ. So,|N | | χc (1), by (i). Thus, |N |p ≤ χc (1)p ≤ p. However, N is a p-group. Hence, |N | = p, as desired.Now, let N be non-abelian. Then, N = S1 × ·· ·×St , where S1, . . . ,St are isomorphic non-abeliansimple groups. For every i ∈ {1, . . . , t }, p | |Si | and there exists θi ∈ Irr(Si )− {1Si } such that p - θi (1),by [6, Corollary 12.2]. Set θ = θ1 × ·· · × θt and let χ ∈ Irr(G|θ). Then, θ ∈ Irr(N ), kerθ = {1} andp - θ(1). Thus, |N |p | θc (1), hence |N |p | χc (1). So, |N |p ≤ p. Consequently, t = 1 and N is a non-abelian simple group. Proposition 7. Let N ≤ Z (G) and G/N be a non-abelian simple group. If p divides |N | and |G/N |,then there exists χ ∈ Irr(G) such that χc (1)p ≥ p2.Proof. Since N ≤ Z (G), N is abelian. Hence, N has a maximal normal subgroup M such that|N /M | = p. However, M ≤ N ≤ Z (G). Thus, M EG and N /M ≤ Z (G/M). If we can show thatthere exists χ ∈ Irr(G/M) such that χc (1)p ≥ p2, then Lemma 4 completes the proof. So, withoutloss of generality, assume that |N | = p. Then, G ′ ∩ N = {1} or N , where G ′ denotes the derivedC. R. Mathématique, 2021, 359, n 1, 79-83Roya Bahramian and Neda Ahanjideh 81subgroup of G . Since G/N is a non-abelian simple group, G ′N = G . Thus, either G ′×N = G andG ′ ∼=G/N or N ≤G ′ =G . In the former case, there exists θ ∈ Irr(G ′)− {1G ′ } such that p - θ(1), by [6,Corollary 12.2]. Set χ= θ×ϕ, for some ϕ ∈ Irr(N )− {1N }. Then, χ ∈ Irr(G), p - χ(1) and kerχ= {1}.Thus, |G|p divides χc (1)p . Hence, χc (1)p ≥ p2, as desired. In the latter case, G is a quasi-simplegroup with Z (G) = N . Consequently, |N | divides |M(G/N )|, the order of the Schur multiplierof G/N . Therefore, p | |M(G/N )|. It follows from Lemma 5(i) that |G/N |p ≥ p2. Since G/N is anon-abelian simple group, there exists ψ ∈ Irr(G/N ) such that p - ψ(1) and kerψ = {1}, by [6,Corollary 12.2]. So, |G/N |p divides ψc (1). Consequently, ψc (1)p ≥ p2. Hence, the propositionfollows because Codeg(G/N ) ⊆ Codeg(G). Proof of Theorem 1. Let G be a minimal counterexample. Then, since the hypothesis is inher-ited by quotients and normal subgroups and `p (G/Op ′ (G)) = `p (G) = `p (Op ′ (G)), we can as-sume that Op ′ (G) = {1} and Op ′ (G) = G . Thus, every minimal normal subgroup M of G is a p-group and `p (G/M) ≤ e. Suppose that M and W are two distinct minimal normal subgroupsof G . Then, since M ∩ W = {1}, `p (G/W ) ≤ e and `p (G/M) ≤ e, `p (G) = `p (G/(M ∩ W )) ≤max{`p (G/M),`p (G/W )} ≤ e, by [5, VI. 6.4]. This is a contradiction. Now let M be the uniqueminimal normal subgroup of G . Let l = `p (G/M) and define a normal series {1} = P0(G/M)EM0(G/M)EP1(G/M)EM1(G/M)E · · ·EPl (G/M)EMl (G/M) =G/M of G/M such that Mi (G/M)Pi (G/M) =Op ′( G/MPi (G/M))and Pi (G/M)Mi−1(G/M) = Op( G/MMi−1(G/M)). Set Pi /M = Pi (G/M) and Mi /M = Mi (G/M). Weclaim that Op ′ (G/M) 6= {1}. If not, M0 = P0. Thus, P1 =Op (G) and {1}EP1EM1E · · ·EPlEMl =Gis a normal series of G such that MiPi =Op ′( GPi)and PiMi−1 =Op( GMi−1), for every 1 ≤ i ≤ l . Therefore,`p (G) = l = `p (G/M) ≤ e. This is a contradiction. Thus, Op ′ (G/M) 6= {1}. Set N /M =Op ′ (G/M). BySchur–Zassenhaus theorem, N has a p-complement L. Then, G = N NG (L) = M NG (L). Since Mis abelian, M ∩ NG (L)EG . However, M is a minimal normal subgroup of G and Op ′ (G) = {1}.Thus, we can check that M ∩ NG (L) = {1}. So, every λ ∈ Irr(M) extends to IG (λ), by [6, Exer-cise 6.18]. Let 1M = λ1, . . . ,λt be the representatives of the action of G on Irr(M). If Oi is the G-orbit of λi , then 1 +Σti=2|Oi |λi (1)2 = Σλ∈Irr(M)λ(1)2 = |M | ≡p 0. Hence, there exists i > 1 suchthat p - |Oi | = [G : IG (λi )]. So, IG (λi ) contains a Sylow p-subgroup P of G . Since λi extends toIG (λi ), there exists λ̂i ∈ Irr(IG (λi )) such that λ̂i M =λi . Set χ= λ̂Gi . By Clifford theory (see [6, The-orem 6.4]), χ ∈ Irr(G) and χ(1) = [G : IG (λi )]. Also, kerχ∩ M is a normal subgroup of G and Mis a minimal normal subgroup of G . Thus, either kerχ∩M = M or kerχ∩M = {1}. In the formercase, M ≤ kerχ, so χM is trivial and λi = λ1, which is a contradiction. Therefore, kerχ∩M = {1}.Consequently, kerχ = {1}, because M is the unique minimal normal subgroup of G . Hence,χc (1) = |IG (λi )|, which is divisible by |P |. So, |G|p ≤ pe . Therefore, `p (G) ≤ e, which is a con-tradiction. Now, the proof is complete. Note that in some parts of the proof of Theorem 1, we follow the ideas in the proof of [4,Theorem 2.3].Proof of Theorem 2. First, let G be a non-p-solvable group of the minimal order such that|G|p ≥ p2. Since the hypothesis is inherited by quotients and normal subgroups, we may assumethat Op ′ (G) = {1} and Op ′ (G) =G . We continue the proof in the following cases:Case a. Assume that Op (G) 6= {1}. Let N ≤ Op (G) be a minimal normal subgroup of G . Then,|N | = p, by Proposition 6(ii). However, G/N is not p-solvable and Codeg(G/N ) ⊆ Codeg(G). So,|G/N |p = p, by minimality of G . Hence, |G|p = p2. Since |N | = p and G/CG (N ) is isomorphic to asubgroup of Aut(N ), we have Op′(G) ≤CG (N ). However, Op ′ (G) =G . So, CG (N ) =G . Consequently,N ≤ Z (G). Set G =G/N and let M/N = M be a minimal normal subgroup of G . If M ≤Op ′ (G), then|N | and |M/N | are co-prime. By Schur–Zassenhaus theorem, M has a p-complement H . SinceN ≤ Z (G), M = H ×N . Thus, M = H N /N ∼= H/(H ∩N ) = H = Op ′ (M) ≤ Op ′ (G) = {1}, which is acontradiction. Now let Op ′ (G) = {1}. Then, since G is not p-solvable and |G|p = p, we get that MC. R. Mathématique, 2021, 359, n 1, 79-8382 Roya Bahramian and Neda Ahanjidehis the unique minimal normal subgroup of G and |M |p = p. So, M is a non-abelian simple group.By Proposition 7, there exists θ ∈ Irr(M) such that θc (1)p ≥ p2. Therefore, χc (1)p ≥ p2, for everyχ ∈ Irr(G|θ). This is a contradiction.Case b. Let Op (G) = {1}. Since Op ′ (G) = {1}, every minimal normal subgroup of G is a non-abeliansimple group of order divisible by p, by Proposition 6(ii). If G has two distinct minimal normalsubgroups M1 and M2, then p | |M1|, |M2|. However, |G|p = p2 and Op ′ (G) =G . Thus, G = M1×M2.So, there exists θ = θ1 ×θ2 ∈ Irr(M1)× Irr(M2) = Irr(G) such that p - θ(1) and kerθ = {1}. Therefore,p2 = |G|p | θc (1), which is a contradiction. Next let G have the unique minimal normal subgroup,say M . Then, CG (M) = {1}. Consequently, G . Aut(M). By Proposition 6(ii), |M |p = p. Thus,p | |G/M |, hence p | |Out(M)|. Lemma 5(ii) shows that |M |p ≥ p2. This is a contradiction.So, |G|p = p, as desired. Remark 8. If χc (1)p ≤ p, for every irreducible character χ of G , then by Theorems 1 and 2, either|G|p = p or G is a p-solvable group of p-length one.Proof of Corollary 3. If G is non-p-solvable, then |G|p = p, by Theorem 2. Thus, the corollaryfollows. Now, let G be p-solvable. By Theorem 1, G has p-length one. Let L = Op ′ (G) and K /L =Op (G/L). Then, K /L is isomorphic to a Sylow p-subgroup of G . By Lemma 4, Codeg(K /L) = {1, p}.Thus, [3, Lemma 2.4] forces K /L to be elementary abelian, as desired. Example 9. Assume that G = L1 × ·· · ×Lt , where L1, . . . ,Lt are Symmetric groups of degrees 3.Let χ ∈ Irr(G)− {1G }. Then, there exist θ1 ∈ Irr(L1), . . . ,θt ∈ Irr(Lt ) such that χ = θ1 × ·· · × θt . SetΩ1 = {1 ≤ i ≤ t : θi (1) = 2} andΩ2 = {1 ≤ i ≤ t : i 6∈Ω1}. Let χ1 =Πi∈Ω1θi and χ2 =Πi∈Ω2θi . If χ=χ1,then χ(1) = |G|2. Hence, χc (1)2 = 1. Otherwise, fix H = Πi∈Ω2 Li . Then, H/H ′ is an elementaryabelian 2-group of order |H |2 and χ2 ∈ Irr(H/H ′). Therefore, |kerχ2|2 = |H |2/2, by [3, Lemma 2.4].Since, χ = χ1 ×χ2, (Πi∈Ω1 1Li )×kerχ2 ≤ kerχ. Thus, 2|Ω2|−1 ≤ |kerχ|. Also, χ(1) = 2|Ω1|. Therefore,χc (1)2 ≤ 2. This example shows that in Corollary 3, |G/Op (G)|p is not necessarily bounded.Example 10. Let K be an elementary abelian 3-group of order 3n . Then, the cyclic group P = 〈z〉of order 2 acts on K by xz = x2, for every x ∈ K . Let G be a semi-direct product K o P andlet χ ∈ Irr(G) − {1G }. If K ≤ kerχ, then χc (1) = 2. Otherwise, there exists θ ∈ Irr(K ) − {1K } suchthat 〈χK ,θ〉 6= 0. By [3, Lemma 2.4], |kerθ| = 3n−1. It is easy to check that kerθEG . Therefore,kerθ ≤ kerχ. Thus, χc (1)3 | |G/kerθ|3 = 3. Consequently, χc (1)3 ≤ 3. This example shows that inCorollary 3, |Op (G)| and |G/Z (G)|p are not necessarily bounded.Example 11. Let p 6= 2 and S be a non-abelian simple group such that p - |S|. Suppose that P isan elementary abelian p-group of order pn and G = P ×S. For every χ ∈ Irr(G), we can see thatpn−1 | |kerχ|, so χc (1)p | pn/pn−1 = p. This example shows that in Theorem 2, “non-p-solvability”cannot be substituted with “non-solvability”.AcknowledgmentsThe authors would like to thank the referees for careful reading and many constructive com-ments, which substantially helped to improve the quality of the paper.References[1] N. Ahanjideh, “The Fitting subgroup, p-length, derived length and character table”, to appear in Math. Nachr.,https://www.doi.org/10.1002/mana.202000057, 2020.[2] R. Bahramian, N. Ahanjideh, “p-divisibility of co-degrees of irreducible characters”, to appear in Bull. Aust. Math.Soc., https://www.doi.org/10.1017/S0004972720000295, 2020.C. R. Mathématique, 2021, 359, n 1, 79-83Roya Bahramian and Neda Ahanjideh 83[3] N. Du, M. L. Lewis, “Codegrees and nilpotence class of p-groups”, J. Group Theory 19 (2016), no. 4, p. 561-568.[4] E. Giannelli, N. Rizo, A. A. Schaeffer Fry, “Groups with few p ′-character degrees”, J. Pure Appl. Algebra 224 (2020),no. 8, article no. 106338 (15 pages).[5] B. Huppert, Endliche Gruppen I, Grundlehren der Mathematischen Wissenschaften, vol. 134, Springer, 1967.[6] I. M. Isaacs, Character theory of finite groups, Dover Publications, 1994.[7] P. Kleidman, M. Liebeck, The subgroup structure of the finite classical groups, London Mathematical Society LectureNote Series, vol. 129, London Mathematical Society, 1990.[8] M. L. Lewis, G. Navarro, P. H. Tiep, H. P. Tong-Viet, “p-parts of character degrees”, J. Lond. Math. Soc. 92 (2015), no. 2,p. 483-497.[9] M. L. Lewis, G. Navarro, T. R. Wolf, “p-parts of character degrees and the index of the Fitting subgroup”, J. Algebra411 (2014), p. 182-190.[10] G. Qian, “A note on p-parts of character degrees”, Bull. Lond. Math. Soc. 50 (2018), no. 4, p. 663-666.[11] G. Qian, Y. Wang, H. Wei, “Co-degrees of irreducible characters in finite groups”, J. Algebra 312 (2007), no. 2, p. 946-955.C. R. Mathématique, 2021, 359, n 1, 79-83

$p$-parts of co-degrees of irreducible characters

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Comptes RendusMathématiqueRoya Bahramian and Neda Ahanjidehp-parts of co-degrees of irreducible charactersVolume 359, issue 1 (2021), p. 79-83.<https://doi.org/10.5802/crmath.158>© Académie des sciences, Paris and the authors, 2021.Some rights reserved.This article is licensed under theCreative Commons Attribution 4.0 International License.http://creativecommons.org/licenses/by/4.0/Les Comptes Rendus. Mathématique sont membres duCentre Mersenne pour l’édition scientifique ouvertewww.centre-mersenne.orgComptes RendusMathématique2021, 359, n 1, p. 79-83https://doi.org/10.5802/crmath.158Group theory / Théorie des groupesp-parts of co-degrees of irreducible charactersRoya Bahramiana and Neda Ahanjideh∗, aa Department of Pure Mathematics, Faculty of Mathematical Sciences, ShahrekordUniversity, P. O. Box 115, Shahrekord, IranE-mails: roya.bahramian98@gmail.com, ahanjideh.neda@sku.ac.irAbstract. For a character χ of a finite group G , the co-degree of χ is χc (1) = [G :kerχ]χ(1) . Let p be a prime andlet e be a positive integer. In this paper, we first show that if G is a p-solvable group such that pe+1 - χc (1),for every irreducible character χ of G , then the p-length of G is not greater than e. Next, we study the finitegroups satisfying the condition that p2 does not divide the co-degrees of their irreducible characters.Mathematical subject classification (2010). 20C15, 20D10, 20D05.Manuscript received 22nd April 2020, revised and accepted 24th November 2020.1. Introduction and preliminariesIn this paper, G is a finite group, p is a prime number and e is a positive integer. Let Z (G) be thecenter of G and let Op (G) and Op ′ (G) be the largest normal p-subgroup and the largest normalp ′-subgroup of G , respectively. Also, Op′(G) is the largest normal subgroup of G whose index in Gis co-prime to p. For a p-solvable group G , the p-length of G , denoted by `p (G), is the minimumpossible number of factors that are p-groups in any normal series of G which every factor is eithera p-group or a p ′-group. Let Irr(G) denote the set of (complex) irreducible characters of G . For anormal subgroup N of G and a character θ of N , let IG (θ) denote the inertia group of θ in G andlet Irr(G|θ) be the set of the irreducible constituents of the induced character θG . Also, we use epto show the p-part of e. For a character χ of G , the number χc (1) = [G :kerχ]χ(1) is called the co-degreeof χ (see [11]). Set Codeg(G) = {χc (1) : χ ∈ Irr(G)}. In [1–3, 11], some properties of the co-degreesof irreducible characters of finite groups have been studied.In [1], it has been proved that the p-length of a finite p-solvable group is not greater than thenumber of the distinct co-degrees of its irreducible characters which are divisible by p. In thispaper, we prove that:Theorem 1. If G is a p-solvable group and pe+1 -χc (1), for every χ ∈ Irr(G), then `p (G) ≤ e.In [8–10], it has been shown that if p2 -χ(1), for every χ ∈ Irr(G), then [G : Op (G)]p ≤ p3. In thispaper, we also prove that:∗Corresponding author.ISSN (electronic) : 1778-3569 https://comptes-rendus.academie-sciences.fr/mathematique/80 Roya Bahramian and Neda AhanjidehTheorem 2. Let G be a non-p-solvable group. If χc (1)p ≤ p, for every χ ∈ Irr(G), then |G|p = p.Corollary 3. If χc (1)p ≤ p, for every χ ∈ Irr(G), then the Sylow p-subgroups of G are elementaryabelian p-groups.In Examples 9, 10 and 11, we show that in Theorem 2, “non-p-solvability” cannot be substi-tuted with “non-solvability” and in Corollary 3, there is not necessarily an upper bound for |G|por |G/Op (G)|p .2. Proofs of the main resultsWe first state a lemma that will be used frequently in this paper without explicit reference.Lemma 4 (cf. [11, Lemma 2.1]). Let N be a normal subgroup of G. Then, Codeg(G/N ) ⊆Codeg(G). Also, if ψ ∈ Irr(N ), then ψc (1) |χc (1), for every χ ∈ Irr(G|ψ).Lemma 5. Let S be a non-abelian simple group.(i) If p is a prime divisor of the order of the Schur multiplier of S, then |S|p ≥ p2.(ii) If p is a prime divisor of |Out(S)| such that p divides |S|, then |S|p ≥ p2.Proof. Since S is a non-abelian simple group, |S|2 ≥ 4. So, the lemma follows when p = 2. Next,assume that p ≥ 3. If p is a prime divisor of the order of the Schur multiplier of S, then sincep ≥ 3, [7, Section 5.1] shows that S ∼= PSLn(q), p | q −1 and p | n, S ∼= PSUn(q), p | q +1 and p | n,S ∈ {PSL2(9), Al t7,PSU4(3),G2(3), J3, M22,F i22, Mcl ,Suz,B3(3), 2E6(4),F i ′24,O′N }(under isomorphism) and p = 3, S ∼= E6(q) and p = 3 | q −1 or S ∼= 2E6(q) and p = 3 | q +1. Thus,we can check at once that |S|p ≥ p2, as desired in (i). Next, let p be a prime divisor of |Out(S)| and|S|. Then, [8, Lemma 3.1] shows that |S|p > |Out(S)|p ≥ p. Thus, |S|p ≥ p2, as wanted. In order to prove the main results, we need to prove the following propositions:Proposition 6. Let N be a minimal normal subgroup of G.(i) If N is abelian and χ ∈ Irr(G) such that N 6≤ kerχ, then |N | divides χc (1).(ii) If p | |N | and χc (1)p ≤ p, for every χ ∈ Irr(G), then |N |p = p and N is a simple group.Proof. (i). Since N is a minimal normal subgroup of G and N 6= N ∩ kerχEG , N ∩ kerχ ={1}. So, N ∼= N kerχ/kerχ is an abelian normal subgroup of G/kerχ. By Ito’s theorem (see [6,Theorem 6.15]), χ(1) | [ Gkerχ : N kerχkerχ ]= [G : N kerχ]. Thus, |N | |χc (1), as desired in (i).(ii). First suppose that N ≤ Op (G), θ ∈ Irr(N ) − {1N } and χ ∈ Irr(G|θ). Then, N 6≤ kerχ. So,|N | | χc (1), by (i). Thus, |N |p ≤ χc (1)p ≤ p. However, N is a p-group. Hence, |N | = p, as desired.Now, let N be non-abelian. Then, N = S1 × ·· ·×St , where S1, . . . ,St are isomorphic non-abeliansimple groups. For every i ∈ {1, . . . , t }, p | |Si | and there exists θi ∈ Irr(Si )− {1Si } such that p - θi (1),by [6, Corollary 12.2]. Set θ = θ1 × ·· · × θt and let χ ∈ Irr(G|θ). Then, θ ∈ Irr(N ), kerθ = {1} andp - θ(1). Thus, |N |p | θc (1), hence |N |p | χc (1). So, |N |p ≤ p. Consequently, t = 1 and N is a non-abelian simple group. Proposition 7. Let N ≤ Z (G) and G/N be a non-abelian simple group. If p divides |N | and |G/N |,then there exists χ ∈ Irr(G) such that χc (1)p ≥ p2.Proof. Since N ≤ Z (G), N is abelian. Hence, N has a maximal normal subgroup M such that|N /M | = p. However, M ≤ N ≤ Z (G). Thus, M EG and N /M ≤ Z (G/M). If we can show thatthere exists χ ∈ Irr(G/M) such that χc (1)p ≥ p2, then Lemma 4 completes the proof. So, withoutloss of generality, assume that |N | = p. Then, G ′ ∩ N = {1} or N , where G ′ denotes the derivedC. R. Mathématique, 2021, 359, n 1, 79-83Roya Bahramian and Neda Ahanjideh 81subgroup of G . Since G/N is a non-abelian simple group, G ′N = G . Thus, either G ′×N = G andG ′ ∼=G/N or N ≤G ′ =G . In the former case, there exists θ ∈ Irr(G ′)− {1G ′ } such that p - θ(1), by [6,Corollary 12.2]. Set χ= θ×ϕ, for some ϕ ∈ Irr(N )− {1N }. Then, χ ∈ Irr(G), p - χ(1) and kerχ= {1}.Thus, |G|p divides χc (1)p . Hence, χc (1)p ≥ p2, as desired. In the latter case, G is a quasi-simplegroup with Z (G) = N . Consequently, |N | divides |M(G/N )|, the order of the Schur multiplierof G/N . Therefore, p | |M(G/N )|. It follows from Lemma 5(i) that |G/N |p ≥ p2. Since G/N is anon-abelian simple group, there exists ψ ∈ Irr(G/N ) such that p - ψ(1) and kerψ = {1}, by [6,Corollary 12.2]. So, |G/N |p divides ψc (1). Consequently, ψc (1)p ≥ p2. Hence, the propositionfollows because Codeg(G/N ) ⊆ Codeg(G). Proof of Theorem 1. Let G be a minimal counterexample. Then, since the hypothesis is inher-ited by quotients and normal subgroups and `p (G/Op ′ (G)) = `p (G) = `p (Op ′ (G)), we can as-sume that Op ′ (G) = {1} and Op ′ (G) = G . Thus, every minimal normal subgroup M of G is a p-group and `p (G/M) ≤ e. Suppose that M and W are two distinct minimal normal subgroupsof G . Then, since M ∩ W = {1}, `p (G/W ) ≤ e and `p (G/M) ≤ e, `p (G) = `p (G/(M ∩ W )) ≤max{`p (G/M),`p (G/W )} ≤ e, by [5, VI. 6.4]. This is a contradiction. Now let M be the uniqueminimal normal subgroup of G . Let l = `p (G/M) and define a normal series {1} = P0(G/M)EM0(G/M)EP1(G/M)EM1(G/M)E · · ·EPl (G/M)EMl (G/M) =G/M of G/M such that Mi (G/M)Pi (G/M) =Op ′( G/MPi (G/M))and Pi (G/M)Mi−1(G/M) = Op( G/MMi−1(G/M)). Set Pi /M = Pi (G/M) and Mi /M = Mi (G/M). Weclaim that Op ′ (G/M) 6= {1}. If not, M0 = P0. Thus, P1 =Op (G) and {1}EP1EM1E · · ·EPlEMl =Gis a normal series of G such that MiPi =Op ′( GPi)and PiMi−1 =Op( GMi−1), for every 1 ≤ i ≤ l . Therefore,`p (G) = l = `p (G/M) ≤ e. This is a contradiction. Thus, Op ′ (G/M) 6= {1}. Set N /M =Op ′ (G/M). BySchur–Zassenhaus theorem, N has a p-complement L. Then, G = N NG (L) = M NG (L). Since Mis abelian, M ∩ NG (L)EG . However, M is a minimal normal subgroup of G and Op ′ (G) = {1}.Thus, we can check that M ∩ NG (L) = {1}. So, every λ ∈ Irr(M) extends to IG (λ), by [6, Exer-cise 6.18]. Let 1M = λ1, . . . ,λt be the representatives of the action of G on Irr(M). If Oi is the G-orbit of λi , then 1 +Σti=2|Oi |λi (1)2 = Σλ∈Irr(M)λ(1)2 = |M | ≡p 0. Hence, there exists i > 1 suchthat p - |Oi | = [G : IG (λi )]. So, IG (λi ) contains a Sylow p-subgroup P of G . Since λi extends toIG (λi ), there exists λ̂i ∈ Irr(IG (λi )) such that λ̂i M =λi . Set χ= λ̂Gi . By Clifford theory (see [6, The-orem 6.4]), χ ∈ Irr(G) and χ(1) = [G : IG (λi )]. Also, kerχ∩ M is a normal subgroup of G and Mis a minimal normal subgroup of G . Thus, either kerχ∩M = M or kerχ∩M = {1}. In the formercase, M ≤ kerχ, so χM is trivial and λi = λ1, which is a contradiction. Therefore, kerχ∩M = {1}.Consequently, kerχ = {1}, because M is the unique minimal normal subgroup of G . Hence,χc (1) = |IG (λi )|, which is divisible by |P |. So, |G|p ≤ pe . Therefore, `p (G) ≤ e, which is a con-tradiction. Now, the proof is complete. Note that in some parts of the proof of Theorem 1, we follow the ideas in the proof of [4,Theorem 2.3].Proof of Theorem 2. First, let G be a non-p-solvable group of the minimal order such that|G|p ≥ p2. Since the hypothesis is inherited by quotients and normal subgroups, we may assumethat Op ′ (G) = {1} and Op ′ (G) =G . We continue the proof in the following cases:Case a. Assume that Op (G) 6= {1}. Let N ≤ Op (G) be a minimal normal subgroup of G . Then,|N | = p, by Proposition 6(ii). However, G/N is not p-solvable and Codeg(G/N ) ⊆ Codeg(G). So,|G/N |p = p, by minimality of G . Hence, |G|p = p2. Since |N | = p and G/CG (N ) is isomorphic to asubgroup of Aut(N ), we have Op′(G) ≤CG (N ). However, Op ′ (G) =G . So, CG (N ) =G . Consequently,N ≤ Z (G). Set G =G/N and let M/N = M be a minimal normal subgroup of G . If M ≤Op ′ (G), then|N | and |M/N | are co-prime. By Schur–Zassenhaus theorem, M has a p-complement H . SinceN ≤ Z (G), M = H ×N . Thus, M = H N /N ∼= H/(H ∩N ) = H = Op ′ (M) ≤ Op ′ (G) = {1}, which is acontradiction. Now let Op ′ (G) = {1}. Then, since G is not p-solvable and |G|p = p, we get that MC. R. Mathématique, 2021, 359, n 1, 79-8382 Roya Bahramian and Neda Ahanjidehis the unique minimal normal subgroup of G and |M |p = p. So, M is a non-abelian simple group.By Proposition 7, there exists θ ∈ Irr(M) such that θc (1)p ≥ p2. Therefore, χc (1)p ≥ p2, for everyχ ∈ Irr(G|θ). This is a contradiction.Case b. Let Op (G) = {1}. Since Op ′ (G) = {1}, every minimal normal subgroup of G is a non-abeliansimple group of order divisible by p, by Proposition 6(ii). If G has two distinct minimal normalsubgroups M1 and M2, then p | |M1|, |M2|. However, |G|p = p2 and Op ′ (G) =G . Thus, G = M1×M2.So, there exists θ = θ1 ×θ2 ∈ Irr(M1)× Irr(M2) = Irr(G) such that p - θ(1) and kerθ = {1}. Therefore,p2 = |G|p | θc (1), which is a contradiction. Next let G have the unique minimal normal subgroup,say M . Then, CG (M) = {1}. Consequently, G . Aut(M). By Proposition 6(ii), |M |p = p. Thus,p | |G/M |, hence p | |Out(M)|. Lemma 5(ii) shows that |M |p ≥ p2. This is a contradiction.So, |G|p = p, as desired. Remark 8. If χc (1)p ≤ p, for every irreducible character χ of G , then by Theorems 1 and 2, either|G|p = p or G is a p-solvable group of p-length one.Proof of Corollary 3. If G is non-p-solvable, then |G|p = p, by Theorem 2. Thus, the corollaryfollows. Now, let G be p-solvable. By Theorem 1, G has p-length one. Let L = Op ′ (G) and K /L =Op (G/L). Then, K /L is isomorphic to a Sylow p-subgroup of G . By Lemma 4, Codeg(K /L) = {1, p}.Thus, [3, Lemma 2.4] forces K /L to be elementary abelian, as desired. Example 9. Assume that G = L1 × ·· · ×Lt , where L1, . . . ,Lt are Symmetric groups of degrees 3.Let χ ∈ Irr(G)− {1G }. Then, there exist θ1 ∈ Irr(L1), . . . ,θt ∈ Irr(Lt ) such that χ = θ1 × ·· · × θt . SetΩ1 = {1 ≤ i ≤ t : θi (1) = 2} andΩ2 = {1 ≤ i ≤ t : i 6∈Ω1}. Let χ1 =Πi∈Ω1θi and χ2 =Πi∈Ω2θi . If χ=χ1,then χ(1) = |G|2. Hence, χc (1)2 = 1. Otherwise, fix H = Πi∈Ω2 Li . Then, H/H ′ is an elementaryabelian 2-group of order |H |2 and χ2 ∈ Irr(H/H ′). Therefore, |kerχ2|2 = |H |2/2, by [3, Lemma 2.4].Since, χ = χ1 ×χ2, (Πi∈Ω1 1Li )×kerχ2 ≤ kerχ. Thus, 2|Ω2|−1 ≤ |kerχ|. Also, χ(1) = 2|Ω1|. Therefore,χc (1)2 ≤ 2. This example shows that in Corollary 3, |G/Op (G)|p is not necessarily bounded.Example 10. Let K be an elementary abelian 3-group of order 3n . Then, the cyclic group P = 〈z〉of order 2 acts on K by xz = x2, for every x ∈ K . Let G be a semi-direct product K o P andlet χ ∈ Irr(G) − {1G }. If K ≤ kerχ, then χc (1) = 2. Otherwise, there exists θ ∈ Irr(K ) − {1K } suchthat 〈χK ,θ〉 6= 0. By [3, Lemma 2.4], |kerθ| = 3n−1. It is easy to check that kerθEG . Therefore,kerθ ≤ kerχ. Thus, χc (1)3 | |G/kerθ|3 = 3. Consequently, χc (1)3 ≤ 3. This example shows that inCorollary 3, |Op (G)| and |G/Z (G)|p are not necessarily bounded.Example 11. Let p 6= 2 and S be a non-abelian simple group such that p - |S|. Suppose that P isan elementary abelian p-group of order pn and G = P ×S. For every χ ∈ Irr(G), we can see thatpn−1 | |kerχ|, so χc (1)p | pn/pn−1 = p. This example shows that in Theorem 2, “non-p-solvability”cannot be substituted with “non-solvability”.AcknowledgmentsThe authors would like to thank the referees for careful reading and many constructive com-ments, which substantially helped to improve the quality of the paper.References[1] N. Ahanjideh, “The Fitting subgroup, p-length, derived length and character table”, to appear in Math. Nachr.,https://www.doi.org/10.1002/mana.202000057, 2020.[2] R. Bahramian, N. Ahanjideh, “p-divisibility of co-degrees of irreducible characters”, to appear in Bull. Aust. Math.Soc., https://www.doi.org/10.1017/S0004972720000295, 2020.C. R. Mathématique, 2021, 359, n 1, 79-83Roya Bahramian and Neda Ahanjideh 83[3] N. Du, M. L. Lewis, “Codegrees and nilpotence class of p-groups”, J. Group Theory 19 (2016), no. 4, p. 561-568.[4] E. Giannelli, N. Rizo, A. A. Schaeffer Fry, “Groups with few p ′-character degrees”, J. Pure Appl. Algebra 224 (2020),no. 8, article no. 106338 (15 pages).[5] B. Huppert, Endliche Gruppen I, Grundlehren der Mathematischen Wissenschaften, vol. 134, Springer, 1967.[6] I. M. Isaacs, Character theory of finite groups, Dover Publications, 1994.[7] P. Kleidman, M. Liebeck, The subgroup structure of the finite classical groups, London Mathematical Society LectureNote Series, vol. 129, London Mathematical Society, 1990.[8] M. L. Lewis, G. Navarro, P. H. Tiep, H. P. Tong-Viet, “p-parts of character degrees”, J. Lond. Math. Soc. 92 (2015), no. 2,p. 483-497.[9] M. L. Lewis, G. Navarro, T. R. Wolf, “p-parts of character degrees and the index of the Fitting subgroup”, J. Algebra411 (2014), p. 182-190.[10] G. Qian, “A note on p-parts of character degrees”, Bull. Lond. Math. Soc. 50 (2018), no. 4, p. 663-666.[11] G. Qian, Y. Wang, H. Wei, “Co-degrees of irreducible characters in finite groups”, J. Algebra 312 (2007), no. 2, p. 946-955.C. R. Mathématique, 2021, 359, n 1, 79-83

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$p$ -parts of co-degrees of irreducible characters

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ppp-parts of co-degrees of irreducible characters

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$p$ -parts of co-degrees of irreducible characters