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 χ of G,
the second author [8, 11] defined its codegree by
χc(1)=|G:kerχ|χ(1).
Clearly, if χ is irreducible, then χc(1) is an integer
divisor of |G/kerχ|. Gagola and Lewis [1] showed
that G is nilpotent if and only if χ(1) divides
χc(1) for every χ∈Irr(G).
The second author [10] obtained a p-local version of the above result
for an odd prime p,
and proved that G has a normal Sylow p-subgroup if and only if
χ(1)p<χc(1)p for every χ∈Irr(G) with p dividing χ(1),
where mp denotes the p-part of an integer m.
Opposite to the case when χ(1) divides χc(1),
in this paper we will study another extreme case,
namely,
gcd(χ(1),χc(1))=1 for all χ∈Irr(G).
An integer m is called a Hall number with respect to a finite group G,
provided that m divides |G| and gcd(m,|G|/m)=1.
If all of the irreducible character degrees of G
are Hall numbers with respect to G,
then G is called an ℋ1-group.
A character χ of G is called an H-character if χ(1) is a Hall number with respect to G/kerχ,
i.e., gcd(χ(1),χc(1))=1;
and G is called an H-group if all its irreducible 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].
We say that a group A acts Frobeniusly on a group B if A⋉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), 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 Fl(G)=G.
Theorem ALet G be a solvable group.
Then G is an H-group if and only if one of the following holds:
G=M⋉F(G), where M is cyclic,
F(G) is an abelian Hall subgroup of G,
and every Sylow subgroup P of M acts Frobeniusly on [F(G),P].
G=L⋉F2(G) satisfies the following conditions:
L is cyclic and has square-free order, and
|L| and |F2(G)/F(G)| are coprime.
If a prime p divides both |L| and |F(G)|,
then
Op′(G)=Frob2(Op′(G),M,N),
where |Op′(G)/M|=p and, for a positive integer e,
|N|=pep
and |M/N|=(pep-1)/(pe-1).
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)={χ(1):χ∈Irr(G|N)}.
Theorem BLet G be a nonsolvable group.
Then G is an H-group if and only if
G satisfies the following conditions:
G=D⋉(L×N), where D,L and N are Hall subgroups of G.
L=GS,
L/Sol(L)≅PSL(2,2f) for some integer f≥2, and we have
either Sol(L)=1
or |Sol(L)|=22f and cd(L|Sol(L))={22f-1}.
D is cyclic and |D| is a square-free divisor of f.
G/L is a solvable ℋ-group.
We construct examples showing the existence
of nonsolvable H-groups that are not H1-groups.
Let F be a field with 2f elements, let
α be a field automorphism of F such that
o(α) 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 α acts naturally on SL(2,2f).
Note that α also acts on the elementary abelian 2-group V in the following way,
α(k1u+k2v)=α(k1)u+α(k2)v,
where k1,k2∈F, and u,v constitute a base of the vector space V.
Write
L=SL(2,2f)⋉V and G=〈α〉⋉L.
Then G is a group.
It is known that cd(L|V)={22f-1} (see [12]).
Now one can 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⊴G, λ∈Irr(N), and χ
be a character of G.
As usual, we denote by IG(λ) the inertia group of λ in G,
and we denote by Irr(χ) the set of irreducible constituents of χ.
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 2.1Let G be a finite group and χ∈Irr(G).
Let N be a G-invariant subgroup of kerχ.
Then
χ
may be viewed as an irreducible character of G/N,
and the codegrees of
χ
in G and in G/N are the same.
If M is subnormal in G and
ψ
is an
irreducible constituent of χM, then ψc(1) divides
χc(1).
If
χ
is induced by an irreducible character
ψ
of a subgroup of G, then ψc(1) divides χc(1).
If p is a prime divisor of |G|,
then p divides ψc(1) for some ψ∈Irr(G).
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 ψ∈Irr(M/N) and χ∈Irr(ψG).
Then ψ(1)|χ(1) by [4, Lemma 6.8],
and ψc(1)|χc(1) by (2),
so M/N is also an ℋ-group.∎
Lemma 2.2Lemma 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:
G=M⋉F, where M is a cyclic Hall subgroup of G,
F is an abelian Hall subgroup of G,
and if [F,P]>1, then P acts
Frobeniusly on [F,P] for every Sylow subgroup P of M.
G=L⋉D, where L is a cyclic Hall subgroup of G and of square-free order,
D is of type (2).
Lemma 2.3Let G be a solvable H-group.
If G is nilpotent, then G is abelian.
If nl(G)=2, then G=A⋉F(G), where A is cyclic,
F(G) is an abelian Hall subgroup of G.
Furthermore,
every Sylow subgroup P of A acts Frobeniusly on [F(G),P].
Proof.
(1) Observe that χ(1) divides χc(1),
that is, χ(1)2 divides |G/kerχ| for every
irreducible character χ 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 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∈Sylp(G) for some prime p.
Now we need to show that P∩F=1,
or equivalently F is a p′-group, and that P acts Frobeniusly on [F,P].
Assume P∩F>1 and write F=(P∩F)×U, where U=Op′(F).
Let λ be a nonprincipal linear character of P∩F,
let ψ∈Irr(U) and set δ=λ×ψ. Then δ∈Irr(F) and IG(δ)=IG(λ)∩IG(ψ).
Observe that p|λc(1),
it follows by Lemma 2.1
that p|χc(1) for every χ∈Irr(δG).
Since G is an ℋ-group,
all irreducible constituents of δG have p′-degrees.
This implies by [4, Theorem 6.11] that IG(δ)=G.
Then IG(ψ)=G for every ψ∈Irr(U).
Applying [4, Theorem 6.32],
we get that all conjugacy classes of U are G-invariant.
Thus G=P×U is nilpotent, a contradiction.
Hence P∩F=1.
Observe that [F,P]>1 and thus P[F,P] is nonabelian
because P acts faithfully on F.
Since F=[F,P]×CF(P) by [2, Chapter 5, Theorem 2.3],
we get by Lemma 2.1 that
P[F,P]≅G/CF(P) is also a nonabelian ℋ-group.
Note that [[F,P],P]=[F,P].
To see that P acts Frobeniusly on [F,P],
we may assume by induction that F=[F,P] and CF(P)=1.
Let λ∈Irr(F|F), i.e.,
λ is a nonprincipal (linear) character of F.
If P fixes λ,
then by [4, Theorem 13.24], P centralizes a nonidentity element of F.
This implies that CF(P)>1, a contradiction.
Hence IG(λ)<G,
and thus p divides χ(1) for every χ∈Irr(λG).
Assume IG(λ)>F.
Then kerλ is IG(λ)-invariant, and
IG(λ)/kerλ=P1×F/kerλ for some nontrivial p-group P1.
Take a character μ0∈Irr(P1|P1) and set λ0=μ0×λ.
Then p divides the codegree of λ0.
Since λ0∈Irr(λIG(λ)),
χ0:=λ0G is irreducible, and Lemma 2.1 (3) implies that p|χ0c(1).
Now p divides both the degree and the codegree of χ0, a contradiction.
Hence IG(λ)=F for every λ∈Irr(F|F).
This yields by [4, Theorem 13.24] that 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 2.4Let G be a solvable group with nl(G)≤2.
Then G is an H-group 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 Δ
of Irr(N), we put
CΔ(A/N)=CΔ(A)={ψ∈Δ:ψa=ψ for all a∈A},
that is, the subset of
A-invariant members of Δ.
Lemma 2.5Let G=Frob2(G,M,N), where N is elementary abelian.
Set |G/M|=a,
|M/N|=b. Then
cd(G)∪{ab}={1,a}∪{ib:i|a}.
Assume |G/M|=p,
|M/N|=(pep-1)/(pe-1), and |N|=pep,
where p is a prime, e is a positive integer.
Then cd(G|N)={|M/N|}.
Proof.
(1) Note that if G=Frob2(G,M,N),
then G is solvable and both G/M 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) is a subset of {p|M/N|,|M/N|} (see (1)),
and all we need to show is
that every nonprincipal linear character of N is invariant under some Sylow
p-subgroup of G. Let Ω be the set of Sylow p-subgroups of G/N
and let α∈Ω.
Clearly |Ω|=|G/N:NG/N(α)|=|M/N|.
Investigating the action of G/N on the elementary abelian group Irr(N),
we get by [4, Theorem 15.16]
that
|CIrr(N)(α)|=|Irr(N)|1/p=|N|1/p=pe.
Since M is a Frobenius group with N as its kernel,
IG(λ)∩M=N for every λ∈Irr(N|N),
and it follows that
CIrr(N)(α)∩CIrr(N)(β)=1N
whenever α,β are different members of Ω.
This implies that
|⋃α∈ΩCIrr(N)(α)|=1+|Ω|(|CIrr(N)(α)|-1)
=1+|M/N|(pe-1)=|N|=|Irr(N)|,
and hence
⋃α∈ΩCIrr(N)(α)=Irr(N).
Now every λ∈Irr(N|N) is fixed
exactly by a Sylow p-subgroup of G/N. Observe that λ is extendible to IG(λ)
because IG(λ)/N≅G/M is cyclic.
By [4, Theorem 6.11, Corollary 6.17],
we see that all irreducible constituents of λG have the same degree |M/N|,
and (2) holds.∎
Lemma 2.6Let G be a solvable H-group with nl(G)=3.
Then:
G/F2(G) has a square-free order.
If a prime p divides |G/F2(G)| and |F(G)|,
then
Op′(G)=Frob2(Op′(G),M,N),
where |Op′(G)/M|=p and, for a positive integer e, we have |N|=pep and
|M/N|=(pep-1)/(pe-1).
Proof.
Write F=F(G) and F2=F2(G).
(1) By induction and Lemma 2.1,
we may assume that G=PF2>F2 for 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.
Suppose Op′(G)<G and write T=Op′(G).
Then G=TF2 because |G/T| and |G/F2| are coprime.
Since nl(G)=nl(TF2)=max{nl(T),nl(F2)},
we have nl(T)=3.
Clearly T=T∩PF2=PF2(T).
Since T/F(T) is an ℋ-group and T/F2(T) is a p-group,
by Lemma 2.3 (2) we see that
F2(T)/F(T) is a p′-group.
Hence
P∩F2(T)=P∩F(T)≤P∩F=(P∩T)∩F
≤P∩F(T)≤P∩F2(T).
This implies that
G/F2=PF2/F2≅P/(P∩F2)
=P/(P∩F)=P/(P∩F2(T))≅T/F2(T),
and then the required result follows by induction.
Suppose that G is equal to Op′(G). Then we observe that Op′(G/F)=G/F and G/F=(PF/F)⋉(F2/F),
it follows that F2/F=[F2/F,PF/F].
Since 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/Φ(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.
Assume V>1. Since |G/F2|=|(G/V)/F2(G/V)|, by induction we get the desired result,
and we are done.
Assume V=1, that is, F is the unique minimal normal subgroup of G.
Let Q be a Sylow subgroup of F2 with Q≠F.
Since [Q,F] is G-invariant,
the unique minimal normality of F implies that [Q,F]=F.
It follows by Lemma 2.3 (2) that F2 is a Frobenius group.
So
G=Frob2(G,F2,F).
Suppose |G/F2|≥p2. By Lemma 2.5 (1),
there exists χ0∈Irr(G) of degree χ0(1)=p|F2/F|.
Clearly χ0∈Irr(G|F) is faithful in G,
and it follows that p divides both χ(1) and χc(1),
a contradiction.
Hence |G/F2|=p, and we are done.
(2) Arguing as in (1),
we may assume G=Op′(G).
Then G=PF2>F2, where P∈Sylp(G).
Using the same arguments as in (1),
we also get that G/F is a Frobenius group with F2/F as its kernel.
Set F=D×V, where D=P∩F>1 and V=Op′(F).
Note that F is abelian and |G/F2|=|G/F|p=p.
We claim that
every λ∈Irr(D|D) is fixed by a unique Sylow p-subgroup of G.
Clearly p|λc(1), and thus p|χc(1) for every χ∈Irr(λG).
This implies that every irreducible constituent of λG has a p′-degree.
By [4, Theorem 6.11], IG(λ)/D contains a Sylow p-subgroup,
say R/D, of G/D.
Since R/D≅ℤp, by
[4, Theorem 6.26] we conclude that
λ is extendible to IG(λ).
By [4, Theorem 6.11 and Corollary 6.17],
all irreducible characters of IG(λ)/D have p′-degrees.
This implies that R/D is normal in IG(λ)/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.
Let g∈G.
Since D is a normal p-subgroup of G,
CIrr(D)(Pg) must contain a nonprincipal member, say λ0.
The preceding claim tells us that
Pg is the unique Sylow p-subgroup of IG(λ0).
Let σ∈Irr(V) and set ψ=λ0×σ.
Then ψ∈Irr(F) and p|ψc(1).
Since G is an ℋ-group,
all irreducible constituents of ψG have p′-degrees.
It follows that IG(ψ) contains a Sylow p-subgroup,
say P1, of G.
Observe that IG(ψ)=IG(λ0)∩IG(σ),
and that Pg is the unique Sylow p-subgroup of IG(λ0),
we have that Pg=P1∈Sylp(IG(σ)),
and so
IG(σ)≥〈Pg:g∈G〉=Op′(G)=G.
By the arbitrariness of σ,
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 that G=Op′(G), we get that V=1.
Hence F=D is minimal normal in G.
Arguing as in (1),
we get that F2 is a Frobenius group with the kernel F,
and thus G=Frob2(G,F2,F).
Let Δ be the set of Sylow p-subgroups of G/F and let α∈Δ.
Observe that CIrr(F)(α) is a subgroup of Irr(F), and that
Irr(F)=⋃α∈ΔCIrr(F)(α),CIrr(F)(α)∩CIrr(F)(β)=1F
whenever α,β are different members of Δ.
Set |CIrr(F)(α)|=pe.
By [4, Theorem 15.16], we have
|F|=|Irr(F)|=|CIrr(F)(α)|p=ppe.
Then
ppe=|Irr(F)|=|⋃α∈ΔCIrr(F)(α)|=1+|Δ|(pe-1),
and therefore we have (ppe-1)/(pe-1)=|Δ|=|G/F:NG/F(α)|=|F2/F|, we are done.∎
Lemma 2.7Let G be a solvable H-group with nl(G)=3.
Then G=L⋉F2(G), where L is cyclic.
Proof.
Let π be the set of prime divisors of |F2(G)/F(G)|
and let K be a Hall π-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 on the abelian group F(G), we obtain that
F(G)=[F(G),K]×CF(G)(K).
Since
NG(K)∩F(G)=CF(G)(K),
it follows by the Frattini argument that
G=F2(G)NG(K)=F(G)NG(K)=[F(G),K]NG(K)
and that
[F(G),K]∩NG(K)=[F(G),K]∩F(G)∩NG(K)=[F(G),K]∩CF(G)(K)=1.
Therefore NG(K)≅G/[F(G),K] is an ℋ-group.
Observe that F(NG(K))=K×CF(G)(K).
Thus NG(K) has nilpotent length 2.
By Lemma 2.3 (2), K×CF(G)(K)
has a cyclic complement, say L, in NG(K).
Then
G=[F(G),K]NG(K)=[F(G),K](K×CF(G)(K))L=F(G)KL=L⋉F2(G),
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 let χ∈Irr(G).
To see that χ is an ℋ-character, we may assume that
a prime p divides χ(1) and we need only show that χ has a p′-codegree.
Let P∈Sylp(G)
and let π be the set of common prime divisors of |G/F2| and |F|.
Write F=F(G) and F2=F2(G).
Assume p∉π.
If P∩F>1,
then P≤F and thus P is a normal and abelian Sylow p-subgroup of G,
hence χ(1) is a p′-number, we are done.
If P∩F=1 and p divides |G/F2|,
then we have |G|p=p by Lemma 2.6, and the required result is obvious.
If P∩F=1 and p divides |F2/F|,
then P∈Sylp(F2).
Let ψ∈Irr(χF2). Then p|ψ(1).
Observe that F2 is an ℋ1-group by Corollary 2.4,
it follows that ψ(1)p=|P|,
and so that χ has a p′-codegree.
Assume p∈π.
Set T=Op′(G) and let ψ∈Irr(χT).
By Lemma 2.6, we have T=Frob2(T,M,N), where |T/M|=p,
|M/N|=(pep-1)/(pe-1), |N|=pep.
If ψ∈Irr(T|N),
then Lemma 2.5 (2) implies that ψ(1)=|M/N| is a p′-number,
so χ(1) is a p′-number, we are done.
If ψ∈Irr(T/N),
then χ∈Irr(G/N), and the required result follows by the fact
that |G/N|p=p.
(⇒) 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)≤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 3.1Lemma 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:
N is a solvable ℋ1-group.
Lemma 3.2Let G=A×B be an H-group, and let
p be a prime divisor of |A|. Then B has a
normal abelian Sylow p-subgroup.
In particular, if B is a nonabelian simple group,
then |A| and |B| are coprime.
Proof.
By Lemma 2.1 (4),
there exists α∈Irr(A) whose codegree is divisible by p.
Let β∈Irr(B) and set χ=α×β.
Then p divides χc(1),
and thus χ(1) is a p′-number.
This implies that every irreducible character β of B has p′-degree,
and it follows by [6] that B has a
normal abelian Sylow p-subgroup. ∎
Lemma 3.3Let G be a nonsolvable H-group
with Sol(G)=1.
Then G=D⋉L, where L≅PSL(2,2f),f≥2,
D is a cyclic Hall subgroup of G whose order is a square-free divisor of f.
Proof.
Assume that G has two subnormal subgroups A and B with AB=A×B.
Since Sol(G)=1, A and B are nonabelian simple groups.
In particular, |A| and |B| are even.
Since A×B is also an ℋ-group,
Lemma 3.2 yields a contradiction.
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.
Since L is the unique minimal normal subgroup of G,
G/L is a subgroup of Out(L)≅ℤf.
Clearly Irr(G|L) is exactly
the set of nonlinear irreducible characters 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 3.4Lemma 3.4 ([12, Lemma 2.5]))
Let N be a normal abelian subgroup of G
with G/N≅PSL(2,2f), f≥2.
Suppose that N is of odd order, and a Sylow
2
-subgroup
R of G acts faithfully on N. Then there exist
λi∈Irr(N|N) such that |IG(λi)|2=|IR(λi)|=2i,
i=1,2,…,f-1.
Suppose that every λ∈Irr(N|N) is fixed by a
unique Sylow
2
-subgroup of G/N. Then N is minimal normal in
G, and if in addition N is a
2
-group, then |N|=22f.
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 λ∈Irr(N|N) such that IG(λ)/N
contains a Frobenius subgroup of order 2k.
Lemma 3.5Let G be an H-group
with G/Sol(G)≅PSL(2,2f), where f≥2.
If Sol(G) has an odd order,
then G=GS×Sol(G)=PSL(2,2f)×Sol(G).
Proof.
Write Sol(G)=N and GS∩N=V.
Then we have G=GSN and thus G/V=N/V×GS/V,
where GS/V≅G/N≅PSL(2,2f).
Suppose GS<G.
Observing that GS is an ℋ-group,
that V=Sol(GS) has an odd order,
and that GS/Sol(GS)≅PSL(2,2f),
we obtain by induction that GS=(GS)S×Sol(GS)=GS×V.
Then V=1, and the result follows.
Suppose GS=G. We need only show N=1.
Assume N>1.
To see a contradiction, 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. Then G=CG(N) is a Schur representation group for
PSL(2,2f) since G=GS.
Since PSL(2,2f) has trivial multiplier whenever
2f>4, while PSL(2,4)≅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 Lemma 3.4 (1),
there exists
λ∈Irr(N|N) such that |IG(λ)|2=2.
Let χ be an irreducible constituent of λG.
We have 2≤χ(1)2<|G|2.
Observe that χ is faithful in G,
it follows that 2 also divides the codegree of χ,
a contradiction. Hence N=1 as wanted. ∎
Lemma 3.6Let G be an H-group
with G/Sol(G)≅PSL(2,2f), f≥2.
Assume Sol(G)>1 and GS=G.
Then |Sol(G)|=22f and cd(G|Sol(G))={22f-1}.
Proof.
Write N=Sol(G).
Since G/N′ is an ℋ-group and GS=G,
we get by Lemma 3.5 that N/N′ is a 2-group.
Let λ be a nonprincipal linear character of N.
Then 2 divides λc(1).
This implies that all irreducible constituents of λG
have even codegrees and thus have odd degrees.
By [4, Theroem 6.11],
IG(λ)/N contains a Sylow 2-subgroup, say P/N, of G/N.
Assume that λ is G-invariant and λ is extendible to G.
Since PSL(2,2f) has an irreducible character of degree 2f,
we get by [4, Corollary 6.17] that
λG has an irreducible constituent of degree 2f,
a contradiction.
Assume that λ is G-invariant and λ does not extend to G.
Then G/N is isomorphic to PSL(2,4) because
PSL(2,2f) has a trivial multiplier
whenever f≥3,
and then λG has an irreducible constituent of even degree 2,
a contradiction.
Hence λ is not G-invariant.
By [3, Theorem 8.27], we get
that
P/N≤IG(λ)/N≤NG/N(P/N),
where NG/N(P/N) is a Frobenius group of order 2f(2f-1).
In particular, P/N is the unique Sylow 2-subgroup of IG(λ)/N.
By Lemma 3.4 (2), we get
that N/N′ is a chief factor of G and of order 22f.
Assume N′>1 and let N′/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) tells us that N/N′ is a cyclic 2-group,
which is clearly impossible.
Hence N′=1 and thus |N|=22f.
Suppose that a
subgroup of odd order k>1 fixes a member in Irr(N|N).
By Lemma 3.4 (3),
there exists a character λ0∈Irr(N|N)
such that IG(λ0)/N
contains a Frobenius subgroup of order 2k.
This contradicts the fact that
IG(λ0)/N contains a unique Sylow 2-subgroup of G/N.
Therefore for every λ∈Irr(N|N),
IG(λ) is exactly a Sylow 2-subgroup of G.
Since all irreducible constituents of λG have odd degrees,
we get that cd(G|N)={22f-1}. ∎
Lemma 3.7Lemma 3.7 ([13])
Let S=PSL(2,2f),
where f≥2,
and S≤G≤Aut(S).
Let |G:S|=d=2am, where m is odd,
and let φ be a field automorphism of S of order f.
Then
cd(G)={1,2f}∪{(2f-1)2al:l|m}∪{(2f+1)j:j|d},
with the following exceptions:
If f is odd, and G=S〈φ〉, then j≠1.
If f≡2(mod 4),
and G=S〈φ〉,
then j≠2.
Proof of Theorem B.
(⇒)
Suppose that G is a nonsolvable ℋ-group.
Set L=GS, K=Sol(G), M=LK and E=L∩K.
Clearly G/K is a nonsolvable ℋ-group with Sol(G/K)=1,
and M/K≅L/E=(L/E)S.
By Lemma 3.3, we have that
L/E≅M/K≅PSL(2,2f)
and G/M is a cyclic group of order d,
where d is a square-free divisor of f with gcd(d,2f(22f-1))=1.
As M/E=L/E×K/E is again an ℋ-group, by Lemma 3.2 we see that |L/E| and |K/E| are coprime.
In particular, |K/E| is odd.
We claim that d is coprime to |M/E|.
To see this, we may assume by induction that d=|G/M| is a prime p and E=1.
By Lemma 3.7,
there exists a character ψ∈Irr(G/K)⊂Irr(G) such that ψ(1)=p(2f-1).
Let τ be an irreducible constituent of ψL.
Clearly IG(τ)=M.
Assume that the claim is false.
Then p divides |K|,
and by Lemma 2.1 (4) we may take σ∈Irr(K) such that p|σc(1).
Set η=σ×τ and let χ be an irreducible constituent of ηG.
Clearly we have IG(η)=IG(σ)∩IG(τ)=M.
Then p|χ(1) by [4, Theorem 6.11], and p|χc(1) 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 conditions stated in Theorem B,
where N=K and D is a complement of M in G.
Assume E>1. We have E=L∩K=Sol(L).
Since L is an ℋ-group with L=LS,
we get by Lemma 3.6 that E has order 22f and cd(L|E)={22f-1}. Since d is coprime to |M/E|=1, M is a normal Hall subgroup of G,
and thus M has a complement, say D, in G.
Let λ∈Irr(E|E).
As shown in the proof of Lemma 3.6,
IL(λ) is exactly a Sylow 2-subgroup of L.
Therefore the L-orbit of λ has size
22f-1. Since |Irr(E|E)|=22f-1,
Irr(E|E) forms an L-orbit and so an M-orbit.
As |K/E| is coprime to |L|,
counting the orbit size we get that every λ∈Irr(E|E) is K-invariant.
This implies that E is a direct factor of K.
Write K=N×E, where N=O2′(K).
Since both L and N are G-invariant Hall subgroups of G,
we get that M=LK=LN=L×N. Now G=D⋉(L×N) satisfies all the conditions in Theorem B.
(⇐)
Let χ∈Irr(G) and λ∈Irr(χL).
To see that χ is an ℋ-character,
we may assume that a prime p divides χ(1); we need only show that χ has a p′-codegree.
Since G/L is a solvable ℋ-group,
we may assume that λ is nonprincipal.
Assume that p divides |D|.
Since D is a Hall subgroup of G and has a square-free order,
the required result is obvious.
Assume that p divides |L|. Then p divides λ(1).
If λ∈Irr(L/Sol(L)), then λ(1)∈{2f,2f+1,2f-1},
thus λ(1)p=|G/Sol(L)|p,
and so χ∈Irr(G/Sol(L)) has a p′-codegree.
If λ∈Irr(L|Sol(L)), then λ(1)=22f-1, thus λ(1)p=|G|p,
and so χ has a p′-codegree.
Assume that p divides |N|. Then λ(1)p=1.
Clearly N×L≤T:=IG(λ).
Let σ0=1N×λ. Since |N| and |L| are coprime,
σ0 is the unique extension of λ to N×L with o(σ0)=o(λ).
Thus σ0 is T-invariant,
and therefore extends to λ0∈Irr(T) because T/(L×N) is cyclic.
Clearly N≤kerλ0.
By [4, Theorem 6.11, Corollary 6.17], we have that
χ=(λ0τ)G
for some τ∈Irr(T/L).
Since |G/T|λ0(1) is a p′-number,
we see that
χ(1)p=τ(1)p.
To see that χ has a p′-codegree, we may assume that
p≤χ(1)p=τ(1)p<|G|p.
Write K/L=Op′(T/L) and let ψ∈Irr(τK).
Observe that T,K are normal in G,
ψ(1)p=τ(1)p<|G|p=|K/L|p,
that K/L is an ℋ-group and ψ∈Irr(K/L).
By Theorem A, we get that
K/L=Frob2(K/L,A/L,B/L),
where B/L is a p-group with B/L=kerψ, ψ(1)=p=|K/B|p.
Observe that 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(λ0).
Since B≤kerψ, we get that Op(B)≤B≤kerτ.
Therefore Op(B)≤ker(λ0)∩ker(τ), and so
Op(B)≤ker((λ0τ)G)=kerχ.
This implies that
χ(1)p=τ(1)p=ψ(1)p=p=|K/B|p=|G/Op(B)|p≥|G/kerχ|p,
so χ has a p′-codegree,
we are done. ∎
and
Guohua Qian
