Finite Groups with Coprime Character Degrees and Codegrees

J. Group Theory 19 (2016), 763–776 DOI 10.1515/jgth-2016-0010 © de Gruyter 2016

Finite groups with coprime character degrees and codegrees

Dengfeng Liang and Guohua Qian Communicated by Gunter Malle

G ker Abstract. We investigate the structure of ﬁnite groups G in which gcd. .1/; j .1/W j / 1 for all irreducible characters of G. D

1 Introduction

In this paper, G always denotes a finite group, all characters are complex characters, and we use Isaacs [4] as a source for standard notation and results from character theory. For a character of G, the second author [8, 11] defined its codegree by G ker c.1/ j W j: D .1/ Clearly, if is irreducible, then c.1/ is an integer divisor of G= ker . Gagola j j and Lewis [1] showed that G is nilpotent if and only if .1/ divides c.1/ for every Irr.G/. The second author [10] obtained a p-local version of the above 2 result for an odd prime p, and proved that G has a normal Sylow p-subgroup if c and only if .1/p < .1/p for every Irr.G/ with p dividing .1/, where mp 2 denotes the p-part of an integer m. Opposite to the case when .1/ divides c.1/, in this paper we will study another extreme case, namely, gcd. .1/; c.1// 1 D for all Irr.G/. 2 An integer m is called a Hall number with respect to a finite group G, pro- vided that m divides G and gcd.m; G =m/ 1. If all of the irreducible character j j j j D degrees of G are Hall numbers with respect to G, then G is called an H1-group. A character of G is called an H-character if .1/ is a Hall number with respect to G= ker , i.e., gcd. .1/; c.1// 1; and G is called an H-group if all its irre- D ducible characters are H-characters. Clearly all linear characters are H-characters, and all H1-groups are H-groups. Note that H1-groups have been studied in [5].

Project supported by the Nature Science Foundation of China (No. 11471054 and 11501017). The corresponding author is Guohua Qian. 764 D. Liang and G. Qian

We say that a group A acts Frobeniusly on a group B if A Ë B is a Frobenius group with kernel B and complement A. A group G is said to be a 2-Frobenius group if there are G-invariant subgroups M and N such that G=N and M are Frobenius groups with M=N and N , respectively, as their Frobenius kernels, and such a 2-Frobenius group is denoted by Frob2.G; M; N /. Let F1.G/ F .G/, D and let Fi .G/ be the ith ascending Fitting subgroup of G. The nilpotent length (or Fitting height) of a solvable group G, denoted by nl.G/, is the smallest number l such that F .G/ G. l D Theorem A. Let G be a solvable group. Then G is an H-group if and only if one of the following holds: (1) G is an abelian group. (2) G M Ë F.G/, where M is cyclic, F.G/ is an abelian Hall subgroup of G, D and every Sylow subgroup P of M acts Frobeniusly on ŒF.G/; P .

(3) G L Ë F2.G/ satisﬁes the following conditions: D (3.1) L is cyclic and has square-free order, and L and F2.G/=F.G/ are coprime. j j j j

(3.2)F 2.G/ is of type (2). (3.3) If a prime p divides both L and F.G/ , then j j j j p p O 0 .G/ Frob2.O 0 .G/; M; N /; D where Op0 .G/=M p and, for a positive integer e, N pep and j j D j j D M=N .pep 1/=.pe 1/. j j D By Theorem A, a solvable H-group has nilpotent length at most 3. Note that a solvable H-group does not need to be an H1-group, the symmetric group of degree 4 is an example; however, a solvable H-group with nilpotent length at most 2 must be an H1-group, see Corollary 2.4 below. Let Sol.G/ be the largest solvable normal subgroup of G, and GS the smallest normal subgroup of G such that G=GS is solvable. For a G-invariant subgroup N , we write Irr.G N/ Irr.G/ Irr.G=N /, cd.G N/ .1/ Irr.G N/ . j D j D ¹ W 2 j º Theorem B. Let G be a nonsolvable group. Then G is an H-group if and only if G satisﬁes the following conditions: (1) G D Ë .L N/, where D;L and N are Hall subgroups of G. D (2) L GS , L=Sol.L/ PSL.2; 2f / for some integer f 2, and we have either D Š Sol.L/ 1 or Sol.L/ 22f and cd.L Sol.L// 22f 1 . D j j D j D ¹ º (3) D is cyclic and D is a square-free divisor of f . j j (4) G=L is a solvable H-group. Finite groups with coprime character degrees and codegrees 765

We construct examples showing the existence of nonsolvable H-groups that are f not H1-groups. Let F be a ﬁeld with 2 elements, let ˛ be a ﬁeld automorphism of F such that o.˛/ is square-free and coprime to 2f .22f 1/, and let V be a vector space of dimension 2 over F . Then SL.2; 2f / acts naturally on V and ˛ acts naturally on SL.2; 2f /. Note that ˛ also acts on the elementary abelian 2- group V in the following way,

˛.k1u k2v/ ˛.k1/u ˛.k2/v; C D C where k1; k2 F , and u; v constitute a base of the vector space V . Write 2 L SL.2; 2f / Ë V and G ˛ Ë L: D D h i Then G is a group. It is known that cd.L V/ 22f 1 (see [12]). Now one can j D ¹ º check directly that both L and G are H-groups (see also the proof of Theorem B for details), but neither is an H1-group. Let N E G, Irr.N /, and be a character of G. As usual, we denote by 2 IG./ the inertia group of in G, and we denote by Irr. / the set of irreducible constituents of .

2 Theorem A

We begin with a list of some fundamental results about character codegree, which will be often used freely in the sequel.

Lemma 2.1. Let G be a ﬁnite group and Irr.G/. 2 (1) Let N be a G-invariant subgroup of ker . Then may be viewed as an irreducible character of G=N , and the codegrees of in G and in G=N are the same. c (2) If M is subnormal in G and is an irreducible constituent of M , then .1/ divides c.1/. (3) If is induced by an irreducible character of a subgroup of G, then c.1/ divides c.1/. (4) If p is a prime divisor of G , then p divides c.1/ for some Irr.G/. j j 2 (5) If G is an H-group, then M=N is also an H-group whenever M is subnormal in G and N is normal in M . Proof. Statements (1) and (2) follow from [11, Lemma 2.1]. Statement (3) is [9, Lemma 2.2], and (4) is [11, Theorem A]. (5) Let Irr.M=N / and Irr. G/. Then .1/ .1/ by [4, Lemma 6.8], 2 2 j and c.1/ c.1/ by (2), so M=N is also an H-group. j 766 D. Liang and G. Qian

Lemma 2.2 ([5, Theorem 1.1]). Let G be a solvable group. Then G is an H1-group if and only if one of the following holds: (1) G is abelian. (2) G M Ë F , where M is a cyclic Hall subgroup of G, F is an abelian Hall D subgroup of G, and if ŒF; P > 1, then P acts Frobeniusly on ŒF; P for every Sylow subgroup P of M . (3) G L Ë D, where L is a cyclic Hall subgroup of G and of square-free order, D D is of type (2).

Lemma 2.3. Let G be a solvable H-group. (1) If G is nilpotent, then G is abelian. (2) If nl.G/ 2, then G A Ë F.G/, where A is cyclic, F.G/ is an abelian Hall D D subgroup of G. Furthermore, every Sylow subgroup P of A acts Frobeniusly on ŒF.G/; P .

Proof. (1) Observe that .1/ divides c.1/, that is, .1/2 divides G=ker for j j every irreducible character of a nilpotent group G. We get (1). (2) Write F.G/ F . Note that G=F is nilpotent, by (1) and Lemma 2.1, F and D G=F are abelian. To see the rest of the statement, we may assume by induction and Lemma 2.1 that G PF > F , where P Syl .G/ for some prime p. Now D 2 p we need to show that P F 1, or equivalently F is a p -group, and that P acts \ D 0 Frobeniusly on ŒF; P . Assume P F > 1 and write F .P F/ U , where U Op .F /. Let \ D \ D 0 be a nonprincipal linear character of P F , let Irr.U / and set ı . \ 2 c D Then ı Irr.F / and IG.ı/ IG./ IG. /. Observe that p .1/, it follows by 2 D \ j Lemma 2.1 that p c.1/ for every Irr.ıG/. Since G is an H-group, all irre- j G 2 ducible constituents of ı have p0-degrees. This implies by [4, Theorem 6.11] that IG.ı/ G. Then IG. / G for every Irr.U /. Applying [4, Theorem 6.32], D D 2 we get that all conjugacy classes of U are G-invariant. Thus G P U is nilpo- D tent, a contradiction. Hence P F 1. \ D Observe that ŒF; P > 1 and thus P ŒF; P is nonabelian because P acts faithfully on F . Since F ŒF; P CF .P / by [2, Chapter 5, Theorem 2.3], we get D by Lemma 2.1 that P ŒF; P G=CF .P / is also a nonabelian H-group. Note that Š ŒŒF; P ; P ŒF; P . To see that P acts Frobeniusly on ŒF; P , we may assume D by induction that F ŒF; P and CF .P / 1. D D Let Irr.F F/, i.e., is a nonprincipal (linear) character of F . If P ﬁxes 2 j , then by [4, Theorem 13.24], P centralizes a nonidentity element of F . This implies that CF .P / > 1, a contradiction. Hence IG./ < G, and thus p divides G .1/ for every Irr. /. Assume IG./ > F . Then ker is IG./-invariant, 2 Finite groups with coprime character degrees and codegrees 767

and IG./=ker P1 F=ker for some nontrivial p-group P1. Take a char- D acter 0 Irr.P1 P1/ and set 0 0 . Then p divides the codegree of 0. 2 IGj ./ G D Since 0 Irr. /, 0 0 is irreducible, and Lemma 2.1 (3) implies that c 2 WD p .1/. Now p divides both the degree and the codegree of 0, a contradiction. j 0 Hence IG./ F for every Irr.F F/. This yields by [4, Theorem 13.24] that D 2 j P acts Frobeniusly on F . Note that as an abelian complement of a Frobenius group, P is necessarily cyclic, and the proof is complete.

By Lemma 2.2 and Lemma 2.3, we get the following corollary.

Corollary 2.4. Let G be a solvable group with nl.G/ 2. Then G is an H-group Ä if and only if G is an H1-group. Let N be a normal subgroup of G. Then G acts naturally on Irr.N /, and it induces an action of G=N on Irr.N /. For a subgroup A=N of G=N and a subset of Irr.N /, we put

a C.A=N / C.A/ for all a A ; D D ¹ 2 W D 2 º that is, the subset of A-invariant members of .

Lemma 2.5. Let G Frob2.G; M; N /, where N is elementary abelian. D (1) Set G=M a, M=N b. Then cd.G/ ab 1; a ib i a . j j D j j D [ ¹ º D ¹ º [ ¹ W j º (2) Assume G=M p, M=N .pep 1/=.pe 1/, and N pep, where j j D j j D j j D p is a prime, e is a positive integer. Then cd.G N/ M=N . j D ¹j jº Proof. (1) Note that if G Frob2.G; M; N /, then G is solvable and both G=M D and M=N are necessarily cyclic. Now we get (1) by [7, Lemma 1.10]. (2) Since M is a Frobenius group, cd.M N/ M=N . Therefore cd.G N/ j D ¹j jº j is a subset of p M=N ; M=N (see (1)), and all we need to show is that every ¹ j j j jº nonprincipal linear character of N is invariant under some Sylow p-subgroup of G. Let be the set of Sylow p-subgroups of G=N and let ˛ . Clearly 2 G=N N .˛/ M=N . Investigating the action of G=N on the ele- j j D j W G=N j D j j mentary abelian group Irr.N /, we get by [4, Theorem 15.16] that

C .˛/ Irr.N / 1=p N 1=p pe: j Irr.N / j D j j D j j D

Since M is a Frobenius group with N as its kernel, IG./ M N for every \ D Irr.N N/, and it follows that 2 j

C .˛/ C .ˇ/ 1N Irr.N / \ Irr.N / D 768 D. Liang and G. Qian whenever ˛; ˇ are different members of . This implies that ˇ ˇ ˇ [ ˇ ˇ C .N /.˛/ˇ 1 . C .N /.˛/ 1/ ˇ Irr ˇ D C j j j Irr j ˛ 2 1 M=N .pe 1/ N Irr.N / ; D C j j D j j D j j and hence [ C .˛/ Irr.N /: Irr.N / D ˛ 2 Now every Irr.N N/ is ﬁxed exactly by a Sylow p-subgroup of G=N . 2 j Observe that is extendible to IG./ because IG./=N G=M is cyclic. By Š [4, Theorem 6.11, Corollary 6.17], we see that all irreducible constituents of G have the same degree M=N , and (2) holds. j j Lemma 2.6. Let G be a solvable H-group with nl.G/ 3. Then: D (1) G=F2.G/ has a square-free order.

(2) If a prime p divides G=F2.G/ and F.G/ , then j j j j p p O 0 .G/ Frob2.O 0 .G/; M; N /; D where Op0 .G/=M p and, for a positive integer e, we have N pep j j D j j D and M=N .pep 1/=.pe 1/. j j D Proof. Write F F.G/ and F2 F2.G/. D D (1) By induction and Lemma 2.1, we may assume that G PF2 > F2 for D a Sylow p-subgroup P of G. Since G=F is also an H-group, we get by part (2) of Lemma 2.3 that G=F2 is a cyclic p-group and that P F2 P F . p p \ D \ Suppose O 0 .G/ < G and write T O 0 .G/. Then G TF2 because G=T D D j j and G=F2 are coprime. Since nl.G/ nl.TF2/ max nl.T /; nl.F2/ , we have j j D D ¹ º nl.T / 3. Clearly T T PF2 P F2.T /. Since T=F.T / is an H-group and D D \ D T=F2.T / is a p-group, by Lemma 2.3 (2) we see that F2.T /=F.T / is a p0-group. Hence P F2.T / P F.T / P F .P T/ F \ D \ Ä \ D \ \ P F.T / P F2.T /: Ä \ Ä \ This implies that

G=F2 PF2=F2 P=.P F2/ D Š \ P=.P F/ P=.P F2.T // T=F2.T /; D \ D \ Š and then the required result follows by induction. Finite groups with coprime character degrees and codegrees 769

Suppose that G is equal to Op0 .G/. Then we observe that Op0 .G=F / G=F D and G=F .PF=F / Ë .F2=F /, it follows that F2=F ŒF2=F; PF=F . Since D D G=F is an H-group, Lemma 2.3 (2) implies that G=F is a Frobenius group with F2=F as its kernel. Since F2=F acts nontrivially on F=ˆ.G/, we may take a chief factor F=V of G such that F2=F acts nontrivial on F=V . As G=F is a Frobenius group, we have F2.G=V / F2=V . Hence nl.G=V / 3. D D Assume V > 1. Since G=F2 .G=V /=F2.G=V / , by induction we get the j j D j j desired result, and we are done. Assume V 1, that is, F is the unique minimal normal subgroup of G. Let Q D be a Sylow subgroup of F2 with Q F . Since ŒQ; F is G-invariant, the unique 6D minimal normality of F implies that ŒQ; F F . It follows by Lemma 2.3 (2) D 2 that F2 is a Frobenius group. So G Frob2.G; F2;F/. Suppose G=F2 p . D j j By Lemma 2.5 (1), there exists 0 Irr.G/ of degree 0.1/ p F2=F . Clearly 2 D j j c 0 Irr.G F/ is faithful in G, and it follows that p divides both .1/ and .1/, 2 j a contradiction. Hence G=F2 p, and we are done. j j D p (2) Arguing as in (1), we may assume G O 0 .G/. Then G PF2 > F2, D D where P Syl .G/. Using the same arguments as in (1), we also get that G=F is 2 p a Frobenius group with F2=F as its kernel. Set F D V , where D P F > 1 D D \ and V Op .F /. Note that F is abelian and G=F2 G=F p p. D 0 j j D j j D We claim that every Irr.D D/ is ﬁxed by a unique Sylow p-subgroup 2 j of G. Clearly p c.1/, and thus p c.1/ for every Irr.G/. This implies j j G 2 that every irreducible constituent of has a p0-degree. By [4, Theorem 6.11], IG./=D contains a Sylow p-subgroup, say R=D, of G=D. Since R=D Zp, by Š [4, Theorem 6.26] we conclude that is extendible to IG./. By [4, Theorem 6.11 and Corollary 6.17], all irreducible characters of IG./=D have p0-degrees. This implies that R=D is normal in IG./=D, and the claim follows. Let us investigate the action of G=D on Irr.D/. By [14, Lemma 4], Irr.D/ is G=D-irreducible, and this implies that D is minimal normal in G. g Let g G. Since D is a normal p-subgroup of G, CIrr.D/.P / must con- 2 g tain a nonprincipal member, say 0. The preceding claim tells us that P is the unique Sylow p-subgroup of IG.0/. Let Irr.V / and set 0 . Then 2 D Irr.F / and p c.1/. Since G is an H-group, all irreducible constituents of G2 j have p0-degrees. It follows that IG. / contains a Sylow p-subgroup, say P1, g of G. Observe that IG. / IG.0/ IG./, and that P is the unique Sylow D \g p-subgroup of IG.0/, we have that P P1 Syl .IG.//, and so D 2 p g p IG./ P g G O 0 .G/ G: h W 2 i D D By the arbitrariness of , all the irreducible characters of V are G-invariant. Since gcd . V ; G=V / 1, it follows that V is a direct factor of G. By the assumption j j j j D that G Op0 .G/, we get that V 1. Hence F D is minimal normal in G. D D D 770 D. Liang and G. Qian

Arguing as in (1), we get that F2 is a Frobenius group with the kernel F , and thus G Frob2.G; F2;F/. D Let be the set of Sylow p-subgroups of G=F and let ˛ . Observe that 2 CIrr.F /.˛/ is a subgroup of Irr.F /, and that [ Irr.F / C .˛/; C .˛/ C .ˇ/ 1F D Irr.F / Irr.F / \ Irr.F / D ˛ 2 whenever ˛; ˇ are different members of . Set C .˛/ pe. By [4, Theo- j Irr.F / j D rem 15.16], we have F Irr.F / C .˛/ p ppe: j j D j j D j Irr.F / j D Then ˇ ˇ pe ˇ [ ˇ e p Irr.F / ˇ C .F /.˛/ˇ 1 .p 1/; D j j D ˇ Irr ˇ D C j j ˛ pe 2 e and therefore we have .p 1/=.p 1/ G=F N .˛/ F2=F , D j j D j W G=F j D j j we are done.

Lemma 2.7. Let G be a solvable H-group with nl.G/ 3. Then G L Ë F2.G/, D D where L is cyclic.

Proof. Let be the set of prime divisors of F2.G/=F.G/ and let K be a Hall j j -subgroup of G. Since G=F.G/ and F2.G/ are H-groups, by Lemma 2.3 (2) we get that K F2.G/=F.G/. In particular, K is nilpotent. Since K acts coprimely Š on the abelian group F.G/, we obtain that F.G/ ŒF.G/; K C .K/. Since D F.G/ NG.K/ F.G/ C .K/, it follows by the Frattini argument that \ D F.G/ G F2.G/NG.K/ F.G/NG.K/ ŒF.G/; KNG.K/ D D D and that

ŒF.G/; K NG.K/ ŒF.G/; K F.G/ NG.K/ \ D \ \ ŒF.G/; K C .K/ 1: D \ F.G/ D Therefore NG.K/ G=ŒF.G/; K is an H-group. Š Observe that F.NG.K// K C .K/. Thus NG.K/ has nilpotent length 2. D F.G/ By Lemma 2.3 (2), K C .K/ has a cyclic complement, say L, in NG.K/. F.G/ Then

G ŒF.G/; KNG.K/ ŒF.G/; K.K C .K//L D D F.G/ F.G/KL L Ë F2.G/; D D we are done. Proof of Theorem A. ( ) Suppose that G is of type (1) or (2). By Lemma 2.2, ( G is an H1-group, and so is an H-group. Suppose that G is of type (3) and Finite groups with coprime character degrees and codegrees 771 let Irr.G/. To see that is an H-character, we may assume that a prime p 2 divides .1/ and we need only show that has a p -codegree. Let P Syl .G/ 0 2 p and let be the set of common prime divisors of G=F2 and F . Write F F.G/ j j j j D and F2 F2.G/. D Assume p . If P F > 1, then P F and thus P is a normal and abelian 62 \ Ä Sylow p-subgroup of G, hence .1/ is a p -number, we are done. If P F 1 0 \ D and p divides G=F2 , then we have G p p by Lemma 2.6, and the required j j j j D result is obvious. If P F 1 and p divides F2=F , then P Syl .F2/. Let \ D j j 2 p Irr. F /. Then p .1/. Observe that F2 is an H1-group by Corollary 2.4, it 2 2 j follows that .1/p P , and so that has a p0-codegree. D j j p Assume p . Set T O 0 .G/ and let Irr. T /. By Lemma 2.6, we 2 D 2 ep e have T Frob2.T; M; N /, where T=M p, M=N .p 1/=.p 1/, D j j D j j D N pep. If Irr.T N/, then Lemma 2.5 (2) implies that .1/ M=N j j D 2 j D j j is a p -number, so .1/ is a p -number, we are done. If Irr.T=N /, then 0 0 2 Irr.G=N /, and the required result follows by the fact that G=N p p. 2 j j D . / Let G be a solvable H-group. If G is nilpotent, then G is abelian by ) Lemma 2.3 (1). If G has nilpotent length 2, then G is of type (2) by Lemma 2.3 (2). Assume that G has nilpotent length at least 3. Since F2.G/ is an H-group, it follows that F2.G/=F.G/ is cyclic by Lemma 2.3 (2). This implies that G=F2.G/ Ä Aut.F2.G/=F.G// is abelian, and hence G has nilpotent length 3. By Lemma 2.6 and Lemma 2.7, G is of type (3). The proof now is complete.

3 Theorem B

Lemma 3.1 ([5, Theorem 1.2]). Let G be a nonsolvable group. Then G is an H1-group if and only if G has normal Hall subgroups M and L that satisfy: G M is square-free, j W j L PSL.2; 2f /, f 2, Š M N L, where N CG.L/, D D

N is a solvable H1-group.

Lemma 3.2. Let G A B be an H-group, and let p be a prime divisor of A . D j j Then B has a normal abelian Sylow p-subgroup. In particular, if B is a nonabelian simple group, then A and B are coprime. j j j j Proof. By Lemma 2.1 (4), there exists ˛ Irr.A/ whose codegree is divisible 2 by p. Let ˇ Irr.B/ and set ˛ ˇ. Then p divides c.1/, and thus .1/ 2 D is a p0-number. This implies that every irreducible character ˇ of B has p0-degree, and it follows by [6] that B has a normal abelian Sylow p-subgroup. 772 D. Liang and G. Qian

Lemma 3.3. Let G be a nonsolvable H-group with Sol.G/ 1. Then G DËL, D D where L PSL.2; 2f /; f 2, D is a cyclic Hall subgroup of G whose order is Š a square-free divisor of f .

Proof. Assume that G has two subnormal subgroups A and B with AB A B. D Since Sol.G/ 1, A and B are nonabelian simple groups. In particular, A and D j j B are even. Since A B is also an H-group, Lemma 3.2 yields a contradiction. j j Hence G possesses a unique minimal normal subgroup, say L, and L is a nonabelian simple group. Observe that L is an H-group and thus is an H1-group, it follows by Lemma 3.1 that L PSL.2; 2f / for some integer f 2. Š Since L is the unique minimal normal subgroup of G, G=L is a subgroup of Out.L/ Z . Clearly Irr.G L/ is exactly the set of nonlinear irreducible charac- Š f j ters of G, and each of them is faithful in G. This implies that all the irreducible character degrees of G are Hall numbers with respect to G. So G is an H1-group, and Lemma 3.1 implies the required result. Note that by the Schur–Zassenhaus theorem the G-invariant Hall subgroup L has a complement in G.

Lemma 3.4 ([12, Lemma 2.5])). Let N be a normal abelian subgroup of G with G=N PSL.2; 2f /, f 2. Š (1) Suppose that N is of odd order, and a Sylow 2-subgroup R of G acts faithfully i on N . Then there exist i Irr.N N/ such that IG.i / 2 IR.i / 2 , 2 j j j D j j D i 1; 2; : : : ; f 1. D (2) Suppose that every Irr.N N/ is ﬁxed by a unique Sylow 2-subgroup 2 j of G=N . Then N is minimal normal in G, and if in addition N is a 2-group, then N 22f . j j D (3) Suppose that N is a 2-group and that a subgroup of odd order k > 1 ﬁxes a member in Irr.N N/. Then there exists Irr.N N/ such that IG./=N j 2 j contains a Frobenius subgroup of order 2k.

Lemma 3.5. Let G be an H-group with G=Sol.G/ PSL.2; 2f /, where f 2. Š If Sol.G/ has an odd order, then G GS Sol.G/ PSL.2; 2f / Sol.G/. D D Proof. Write Sol.G/ N and GS N V . Then we have G GS N and thus D \ D D G=V N=V GS =V , where GS =V G=N PSL.2; 2f /. D Š Š Suppose GS < G. Observing that GS is an H-group, that V Sol.GS / has D an odd order, and that GS =Sol.GS / PSL.2; 2f /, we obtain by induction that Š GS .GS /S Sol.GS / GS V . Then V 1, and the result follows. D D D Suppose GS G. We need only show N 1. Assume N > 1. To see a con- D D tradiction, by induction we may assume that N is minimal normal in G. Clearly N is an elementary abelian p-group for an odd prime p. Assume CG.N / > N . Finite groups with coprime character degrees and codegrees 773

f S Then G CG.N / is a Schur representation group for PSL.2; 2 / since G G . D D Since PSL.2; 2f / has trivial multiplier whenever 2f > 4, while PSL.2; 4/ Š PSL.2; 5/ has multiplier of even order 2, we get a contradiction. So now we have CG.N / N . In particular, a Sylow 2-subgroup R of G acts faithfully on N . By D Lemma 3.4 (1), there exists Irr.N N/ such that IG./ 2 2. Let be an G 2 j j j D irreducible constituent of . We have 2 .1/2 < G 2. Observe that is faith- Ä j j ful in G, it follows that 2 also divides the codegree of , a contradiction. Hence N 1 as wanted. D Lemma 3.6. Let G be an H-group with G=Sol.G/ PSL.2; 2f /, f 2. Assume Š Sol.G/ > 1 and GS G. Then Sol.G/ 22f and cd.G Sol.G// 22f 1 . D j j D j D ¹ º Proof. Write N Sol.G/. Since G=N is an H-group and GS G, we get by D 0 D Lemma 3.5 that N=N 0 is a 2-group. Let be a nonprincipal linear character of N . Then 2 divides c.1/. This implies that all irreducible constituents of G have even codegrees and thus have odd degrees. By [4, Theroem 6.11], IG./=N contains a Sylow 2-subgroup, say P=N , of G=N . Assume that is G-invariant and is extendible to G. Since PSL.2; 2f / has an irreducible character of degree 2f , we get by [4, Corollary 6.17] that G has an irreducible constituent of degree 2f , a contradiction. Assume that is G-invariant and does not extend to G. Then G=N is isomor- phic to PSL.2; 4/ because PSL.2; 2f / has a trivial multiplier whenever f 3, and then G has an irreducible constituent of even degree 2, a contradiction. Hence is not G-invariant. By [3, Theorem 8.27], we get that

P=N IG./=N N .P=N /; Ä Ä G=N where N .P=N / is a Frobenius group of order 2f .2f 1/. In particular, P=N G=N is the unique Sylow 2-subgroup of IG./=N . By Lemma 3.4 (2), we get that N=N 0 is a chief factor of G and of order 22f . Assume N 0 > 1 and let N 0=E be a chief factor of G. Observe that N=E is a nonabelian but solvable H-group. By Lemma 2.3 (1), N=E is not nilpotent, and thus F.N=E/ N =E because N=N is a chief factor of G. Now Lemma 2.3 (2) D 0 0 tells us that N=N is a cyclic 2-group, which is clearly impossible. Hence N 1 0 0 D and thus N 22f . j j D Suppose that a subgroup of odd order k > 1 ﬁxes a member in Irr.N N/. By j Lemma 3.4 (3), there exists a character 0 Irr.N N/ such that IG.0/=N con- 2 j tains a Frobenius subgroup of order 2k. This contradicts the fact that IG.0/=N contains a unique Sylow 2-subgroup of G=N . Therefore for every Irr.N N/, 2 j G IG./ is exactly a Sylow 2-subgroup of G. Since all irreducible constituents of have odd degrees, we get that cd.G N/ 22f 1 . j D ¹ º 774 D. Liang and G. Qian

Lemma 3.7 ([13]). Let S PSL.2; 2f /, where f 2, and S G Aut.S/. Let D Ä Ä G S d 2am, where m is odd, and let ' be a field automorphism of S of j W j D D order f . Then cd.G/ 1; 2f .2f 1/2al l m .2f 1/j j d ; D ¹ º [ ¹ W j º [ ¹ C W j º with the following exceptions: (i) If f is odd, and G S ' , then j 1. D h i 6D (ii) If f 2 .mod 4/, and G S ' , then j 2. Á D h i 6D Proof of Theorem B. . / Suppose that G is a nonsolvable H-group. Set L GS , ) D K Sol.G/, M LK and E L K. Clearly G=K is a nonsolvable H-group D D D \ with Sol.G=K/ 1, and M=K L=E .L=E/S . By Lemma 3.3, we have that D Š D L=E M=K PSL.2; 2f / and G=M is a cyclic group of order d, where d is Š Š a square-free divisor of f with gcd.d; 2f .22f 1// 1. As M=E L=E K=E D D is again an H-group, by Lemma 3.2 we see that L=E and K=E are coprime. j j j j In particular, K=E is odd. j j We claim that d is coprime to M=E . To see this, we may assume by induction j j that d G=M is a prime p and E 1. By Lemma 3.7, there exists a character D j j D Irr.G=K/ Irr.G/ such that .1/ p.2f 1/. Let be an irreducible 2 D constituent of L. Clearly IG./ M . Assume that the claim is false. Then p D divides K , and by Lemma 2.1 (4) we may take Irr.K/ such that p c.1/. j j 2 j Set and let be an irreducible constituent of G. Clearly we have D c IG./ IG./ IG./ M . Then p .1/ by [4, Theorem 6.11], and p .1/ D \ D j j by Lemma 2.1, a contradiction. Now the claim follows. Clearly G=L is a solvable H-group. Assume E 1. Then G satisfies all the D conditions stated in Theorem B, where N K and D is a complement of M in G. D Assume E > 1. We have E L K Sol.L/. Since L is an H-group with D \ D L LS , we get by Lemma 3.6 that E has order 22f and cd.L E/ 22f 1 . D j D ¹ º Since d is coprime to M=E 1, M is a normal Hall subgroup of G, and thus j j D M has a complement, say D, in G. Let Irr.E E/. As shown in the proof 2 j of Lemma 3.6, IL./ is exactly a Sylow 2-subgroup of L. Therefore the L-orbit of has size 22f 1. Since Irr.E E/ 22f 1, Irr.E E/ forms an L-orbit j j j D j and so an M -orbit. As K=E is coprime to L , counting the orbit size we get j j j j that every Irr.E E/ is K-invariant. This implies that E is a direct factor of K. 2 j Write K N E, where N O2 .K/. Since both L and N are G-invariant Hall D D 0 subgroups of G, we get that M LK LN L N . Now G D Ë .L N/ D D D D satisfies all the conditions in Theorem B. . / Let Irr.G/ and Irr. L/. To see that is an H-character, we may ( 2 2 assume that a prime p divides .1/; we need only show that has a p0-codegree. Since G=L is a solvable H-group, we may assume that is nonprincipal. Finite groups with coprime character degrees and codegrees 775

Assume that p divides D . Since D is a Hall subgroup of G and has a square- j j free order, the required result is obvious. Assume that p divides L . Then p divides .1/. If Irr.L=Sol.L//, then f f f j j 2 .1/ 2 ; 2 1; 2 1 , thus .1/p G=Sol.L/ p, and so Irr.G=Sol.L// 2 ¹ C º D j j 2f 2 has a p -codegree. If Irr.L Sol.L//, then .1/ 2 1, thus .1/p G p, 0 2 j D D j j and so has a p0-codegree. Assume that p divides N . Then .1/p 1. Clearly N L T IG./. j j D Ä WD Let 0 1N . Since N and L are coprime, 0 is the unique extension D j j j j of to N L with o.0/ o./. Thus 0 is T -invariant, and therefore extends D to 0 Irr.T / because T=.L N/ is cyclic. Clearly N ker 0. By [4, Theo- 2 Ä rem 6.11, Corollary 6.17], we have that

G .0/ D for some Irr.T=L/. Since G=T 0.1/ is a p -number, we see that 2 j j 0 .1/p .1/p: D To see that has a p0-codegree, we may assume that

p .1/p .1/p < G p: Ä D j j p Write K=L O 0 .T=L/ and let Irr.K /. Observe that T;K are normal in G, D 2 .1/p .1/p < G p K=L p, that K=L is an H-group and Irr.K=L/. D j j D j j 2 By Theorem A, we get that

K=L Frob2.K=L; A=L; B=L/; D where B=L is a p-group with B=L ker , .1/ p K=B p. Observe that D D D j j T and L are normal Hall subgroups of G, and B=L is a characteristic subgroup of T=L; it follows that B is characteristic in G and therefore Op.B/ is normal in G. Since L < B N L, we have that B=L Op.B/ N ker.0/. Since Ä Š Ä Ä B ker , we get that Op.B/ B ker . Therefore Op.B/ ker.0/ ker./, Ä Ä Ä Ä \ and so G Op.B/ ker..0/ / ker : Ä D This implies that

.1/p .1/p .1/p p K=B p G=Op.B/ p G=ker p; D D D D j j D j j j j so has a p0-codegree, we are done.

Acknowledgments. The authors are grateful to the referee for his/her valuable comments. 776 D. Liang and G. Qian

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Received October 7, 2015; revised January 8, 2016.

Author information Dengfeng Liang, Department of Mathematics, Beijing Technology and Business University, Beijing, 100048, P.R. China. E-mail: [email protected] Guohua Qian, Department of Mathematics, Changshu Institute of Technology, Changshu, Jiangsu, 215500, P.R. China. E-mail: [email protected]