# On Irreducible Linear Groups of Nonprimary Degree

Authors: A. A. Yadchenko
Journal: ISRN Algebra
 (2011)

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## Abstract

It is established that degree2|A|+1 of irreducible complex linear
group with the group A of cosimple automorphisms of odd order is a prime number
and proved that if degree 2|H|+1 of π-solvable irreducible complex linear group G with a π-Hall TI-subgroup H is not a prime power, then H is Abelian and normal
in G.


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## Body

## 1. Introduction

Suppose thatG is a finite group and A is such a group of its nontrivial automorphisms that (|G|,|A|)=1. Then, A is called a group of cosimple automorphisms of the group G and the semidirect product Γ=GA of the group G and the group A is a group. If CG(a)=CG(A) for each element a∈A#, then A is said to be a strong-centralized group of cosimple automorphisms of the group G.The theorem, proved in the series of papers [1–3], implies that if G is a finite irreducible complex linear group of degree n<2|A| with a nontrivial strong-centralized odd-order group A of cosimple automorphisms, then n=|A|-1,|A|+1,2(|A|-1) or 2|A|-1 and n is a degree of a certain prime number.In this work, it is shown that the degreen=2|A|+1 of the irreducible complex linear group with a nontrivial strongly centralized group of cosimple automorphisms is a degree of a certain prime number (Theorem 1.1), and, on this basis, it is proved that in the π-solvable irreducible complex linear group G of degree n=2|H|+1, which is not a prime power, the π-Hall TI-subgroup H is Abelian and normal in G (Theorem 1.2).Condition B
Let us say that the groupΓ=AG satisfies B, if G⊲Γ,(|A|,|G|)=1,|A| is uneven, CG(a)=CG(A)=C for each element a∈A#, and the group G has an extract irreducible complex character of degree n=2|A|+1, which is a-invariant for at least one element a∈A#.Theorem 1.1.
If the groupΓ satisfies Condition B, then n is a prime power degree.Condition D
G is a π-solvable irreducible complex linear group of degree n=2|H|+1 with a nonprimary π-Hall TI-subgroup H of odd order.Theorem 1.2.
If the groupG satisfies Condition D and n is not a prime power, the subgroup H is Abelian and normal in G.


## 2. Definitions, Notation, and Preliminaries

N is the set of natural numbers; Z+ is the set of nonnegative integers; notation i=1,t¯ stands for i=1,2,…,t; if ψ is the character of a certain group, then Irr(ψ) is the set of all irreducible components of the character ψ; Z(ψ)={g∈G∣|ψ(g)|=ψ(1)} is the center of the character ψ;π=π(A); if M⊆G and π is the set of primes, then π′=π(M)∖π;Mπ is a Hall π-subgroup of the group M. The rest of the notation and definitions are conventional; they can be found, for example, in [4] or [5]. In what follows, by a group character we always mean a complex character, by a group a finite group.Suppose thatΓ=AB is the group, where B⊲Γ,(|A|,|B|)=1 and |A| is uneven. Then, it satisfies Theorem  13.1 [5]. According to this theorem, there exists one-to-one correspondence π(B,A):IrrA(B)→Irr(CB(A)) between the set of all A-invariant characters of the group B and the set of all irreducible characters of the subgroup CB(A). The set has a range of properties, depending, in particular, on the properties of the subgroup A. Suppose that χ∈IrrA(B). Then, by Lemma  13.3 [5], there is a unique irreducible character χ̂ of the group Γ such that χ̂B=χ and A⊆ker(detχ̂). It is called a canonical extension of the characterχto the groupΓ.Lemma 2.1 (see [6, Lemma  2]).
Ifχ is an irreducible character of the group  G stated in Condition B, then  χ∈IrrA(G).Lemma 2.2 (see [1, Lemma  10]).
Ifχ is an irreducible character of the group  G stated in Condition B, then  χ̂A=kρA+εβ(1)1A, where  k∈Z+,ρA is a regular character of the subgroup  A,ε=±1 and  β is an irreducible character of the subgroup  C such that  β=(χ)π(G,A).Lemma 2.3.
Ifχ is an irreducible character of the group  G stated in Condition B, then  χC=|A|θ+ɛβ, where  θ is a character of the subgroup  C or 0 and  β=(χ)π(G,A).Proof.
The proof of the lemma is a literal repeat of the proof of equality (3) of Lemma  11 to [1].Lemma 2.4.
Ifχ is an irreducible character of the group G of degree 2|A|<n<3|A|, stated in Condition B, then one of the following assertions holds:(1)
χC=β+|A|β1,|A|<β(1)<2|A|,β1(1)=1;(2)
χC=β+|A|β1,β(1)<|A|,β1(1)=2;(3)
χC=β+2|A|β1,β(1)<|A|,β1(1)=1;(4)
χC=β+|A|(β1+β2),β(1)<|A|,β1(1)=β2(1)=1;(5)
χC=(|A|+1)β+|A|β1,β(1)=β1(1)=1;(6)
χC=(|A|+1)β,β(1)=2;(7)
χC=(|A|-1)β+2|A|β1,β(1)=β1(1)=1;(8)
χC=(|A|-1)β+|A|β1,β(1)=1,β1(1)=2;(9)
χC=(|A|-1)β+|A|(β1+β2),β(1)=β1(1)=β2(1)=1;(10)
χC=(|A|-1)β+|A|β1,β(1)=2,β1(1)=1;(11)
χC=(|A|-1)β,β(1)=3;(12)
χC=(2|A|+1)β,β(1)=1;(13)
χC=(2|A|-1)β+|A|β1,β(1)=β1(1)=1;(14)
χC=(3|A|-1)β,β(1)=1;(15)
χC=β.
Here,β=(χ)π(G,A), β1 and β2 are distinct irreducible characters of the subgroup C.Proof.
From Lemma2.3, it follows that
(2.1)(χC,β)C=kβ|A|+ε,
where kβ=(θ,β)C. Suppose that β1∈Irr(χC),β1≠β,k1=(χC,β1)C and θ1 is the character C or 0 such that
(2.2)θ=kββ+k1β1+θ1.
Hence, by Lemma 2.3, it follows that
(2.3)χC=|A|(kββ+k1β1+θ1)+εβ.
Hence, by the assumption of the lemma, it follows that
(2.4)|A|kββ(1)+εβ(1)≤χ(1)<3|A|.
Then, |A|kββ(1)<3|A|-εβ(1), that is,
(2.5)kβ<3β(1)-ε|A|.
Hence, kβ≤3.
Assume thatθ≠0 and examine all the possible cases separately.
(1)kβ=0.
Since, by Corollary  2.17 [5], (χC,β)C∈Z+, by equality (2.1), it follows that ε=1. Since χ(1)<3|A|, from Lemma 2.3, it follows that θ(1)≤2, that is, k1β1(1)+θ1(1)≤2.
Suppose thatθ(1)=1. Then, we can assume that θ1=0 and k1β1(1)=1. From equality (2.3), we obtain χC=β+|A|β1,|A|<β(1)<2|A|,β1(1)=1 is assertion (1) of the lemma being proved.
Suppose thatθ(1)=2. Then, k1β1(1)+θ1(1)=2.
Ifθ1=0, then either k1=1 and β1(1)=2, which implies assertion (2) of the lemma, or k1=2 and β1(1)=1, which implies assertion (3) of the lemma.
But ifθ1≠0, then it is clear that k1β1(1)=1 and θ1=β2∈Irr(θ),β1≠β2,β2(1)=1,β(1)<|A|. We obtain assertion (4) of the lemma.
(2)kβ=1.
Suppose thatε=1. From (2.3), we obtain χC=|A|(β+k1β1+θ1)+β. Since 2|A|<χ(1)<3|A|, from the latter equality, it is readily seen that β(1)≤2.
Suppose thatβ(1)=1. Then, k1β1(1)=1 and χC=(|A|+1)β+|A|β1,β(1)=β1(1)=1 is assertion (5).
Now, suppose thatβ(1)=2. Since 2|A|<χ(1)<3|A|, we have k1β1(1)+θ1(1)=0 and, therefore, χC=(|A|+1)β is assertion (6).
Further, suppose thatɛ=-1. By (2.3), it follows that
(2.6)χC(1)=(|A|-1)β(1)+|A|(k1β1(1)+θ1(1)).
Since χ(1)<3|A|, we see that β(1)≤3.
Letβ(1)=1. Then, k1β1(1)+θ1(1)=2.
Letθ1=0. Then, k1β1(1)=2. If k1=2, then β1(1)=1. Hence, χC=(|A|-1)β+2|A|β1 is assertion (7). If k1=1, then β1(1)=2. Then χC=(|A|-1)β+|A|β1 is assertion (8).
Now suppose thatθ1(1)=1. Then, k1β1(1)=1. We obtain χC=(|A|-1)β+|A|(β1+β2), where β1(1)=β2(1)=1 is assertion (9).
Ifθ1(1)=2, then k1β1(1)=0 and it is readily seen that (7), (8) or (9) follow.
Now suppose thatβ(1)=2. Hence, χC=(|A|-1)β+|A|β1,β1(1)=1 is assertion (10).
Suppose thatβ(1)=3. Then, k1β1(1)+θ1(1)=0. Assertion (11).(3) kβ=2.
Letε=1. Expression (2.3) implies that k1β1(1)+θ1(1)=0. We obtain χC=(2|A|+1)β. We see that β(1)=1 is assertion (12).
Letε=-1. Then, χC=(2|A|-1)β+|A|β1, where β(1)=β1(1)=1 is assertion (13).
(4)kβ=3.
Expression (2.3) implies that ε=-1,χC=(3|A|-1)β, where β(1)=1 is assertion (14).
Now suppose that in Lemma2.3  θ=0. Then, ε=1 and χC=β is assertion (15).
The lemma is proved.


## 3. Proof of Theorem1.1

Suppose thatχ is an exact irreducible character of degree n=2|A|+1 of the group G given in Condition B, a-invariant for at least one element a∈A#.Suppose thatΓ is a group of the least order that satisfies the assumption of the theorem, but does not satisfy its corollary, that is, |π(n)|>1.Lemma 3.1.
The characterχ̂ is exact.Proof.
Suppose thatkerχ̂≠1. Since the character χ̂G=χ is exact, we have kerχ̂⊆A. Since, by Lemma  1 [7], A is a TI-subgroup in Γ, it follows that A⊲Γ. This contradicts the fact that A is the group of nontrivial automorphisms of the group G.
The lemma is proved.Lemma 3.2.
Suppose thatq∈π(n),  Q is an  A-invariant Sylow  q-subgroup  G. Then,  [Q,A]⊈C.Proof.
Suppose that[Q,A]⊆C. Since, by Lemma  4 [1], Q=[Q,A]CQ(A), we have Q⊆C. By Exercise 13.2 [5],
(3.1)β(1)|G:C|χ(1)
is an integer, where β is from Lemma 2.2. Since Q⊆C, we see that q does not divide |G:C|. Then q divides β(1). From Lemma 2.2, it follows that k|A|=χ̂(1)-ɛβ(1). Hence, q divides k|A|. Since q divides |G|, it follows that q does not divide |A|, therefore, q divides k. By using the fact that q is an odd prime, it follows that k≥3.
Suppose thatε=1 in Lemma 2.2. Since n=2|A|+1, we have k=2. This case does not satisfy.
Suppose thatε=-1. Since β(1)≤χ(1)<3|A|, we obtain
(3.2)k=χ̂(1)+β(1)|A|<6,
that is, k≤5. Since q divides k and q is an uneven number, it follows that it is enough to consider the cases when k=3 or k=5.
Suppose thatk=3. Then, 3|A|=(2|A|+1)+β(1). Hence, |A|-1=β(1). Since χ(1)=2|A|+1, by Lemma 2.4, it follows that only such relations satisfy, where |A|-1=β(1)=2. However, in this case |A|=3. Then, n=2|A|+1=7 is a prime number. A contradiction with the minimality of the group Γ.
Suppose thatk=5. Then, 5|A|=(2|A|+1)+β(1). Therefore, β(1)=3|A|-1>n. We have a contradiction.
The lemma is proved.Lemma 3.3.
Suppose thatR is a proper  A-invariant normal subgroup of the group  G. Then either  R is Abelian, or  R⊆C.Proof.
Let us consider the character(3.3)(χ̂)AR=∑i=1fαiχi,
where αi∈N,χi∈Irr(AR), i=1,f¯.
Let us note that by the minimality of the groupΓ either f≠1, or α1≠1, that is, χi(1)≠χ̂(1) for all i=1,f¯.
By the Clifford Theorem(3.4)(χi)R=ei∑j=1tiχij,
where χij∈Irr(R), ei are numbers, which divide |A|, i=1,f¯, j=1,ti¯.
Suppose thatχi0j0∈Irr((χ̂)R). Since R⊲Γ, by the Clifford Theorem it follows that χij(1) divides n for each irreducible component χij of the character χ̂R and for corresponding elements x∈Γ∖IΓ(χi0j0) the equality χij=(χi0j0)x is true.
Suppose thatχi0j0(1)=1 for a certain character χi0j0∈Irr((χ̂)R). Since R′⊆kerχi0j0 and R′⊲Γ, we see that R′⊆(kerχi0j0)x for all x∈Γ∖IΓ(χi0j0). Therefore, R′⊆∩x∈Γ∖IΓ(χi0j0)(kerχi0j0)x=kerχ̂=1. Thus, the subgroup R is Abelian.
Now suppose thatχij(1)≠1 for all i=1,f¯ and for all j=1,ti¯. Since n is an uneven number, we have χij(1)≥3. Therefore, if IAR(χi0j0)=R for a certain character χi0j0∈Irr((χ̂)R), then χi0(1)=(χi0j0)AR(1)=|A|χi0j0(1)≥3|A|. Since in this case χ̂(1)≥χi0(1), we obtain a contradiction with χ̂(1)=2|A|+1<3|A|. Thus, IAR(χij)≠R for all i=1,f¯ and for all j=1,ti¯. Then, by Lemma  1 [6], IAR(χij)=AR, therefore, ti=1 and χi(1)=eiχij(1) for all i=1,f¯. By Lemma  13.3 [5], for every i=1,f¯ there exists a canonical extension χî of the character χi1 to the group AR.
Assume thatAkerχi0̂/kerχi0̂⋪AR/kerχi0̂ for a certain i0. By Lemma  9  [1], A∩kerχi0̂=1. Then, Akerχi0̂/kerχi0̂≅A/A∩kerχi0̂=A. Let us apply the theorem [1] to the factor group AR/kerχi0̂ and its exact irreducible character in terms of Lemma  2.22 [5] (χi0̂)*. By this theorem, (χi0̂)*(1)∈{|A|-1,|A|,|A|+1,2|A|-2,2|A|-1} and (χi0̂)*(1) is a prime power, probably, with the exception of the case, when (χi0̂)*(1)=|A|. Since (χi0̂)*(1)=χi0̂(1) and χî(1) divides n for all i=1,f¯, it follows that χi0̂(1)=2|A|-1. Thus, n=2|A|+1=s(2|A|-1) for a certain odd s∈N. It is easily seen that it is not true.
Now suppose thatAkerχî/kerχî⊲AR/χî for all i=1,f¯. Using the fact that ⋂i=1fkerχî∩R=kerχ̂∩R and kerχ̂=1 by Lemma 3.1, then A0=⋂i=1fkerχî⊆A. If A0≠1, then A⊲AR, since A is a TI-subgroup. If A0=1, then the subgroup AR can be isomorphically embedded in the direct product AR¯=∏i=1f(AR/kerχî). Since the group AR¯ contains a normal Hall π-subgroup, we have A⊲AR. Thus, R⊆C.
The lemma is proved.Suppose thatq∈π(n) and Q is a A-invariant Sylow q-subgroup of G that exists by Theorem  6.2.2(i) [4]. By Lemma 3.2,  [Q,A]⊈C.Further,M is an A-invariant proper subgroup of G such that [Q,A]⊆M.From the minimality of the groupΓ and Lemma 3.1, it follows that the character χ̂AM is reducible for every such a subgroup M. Denote N=NΓ([Q,A]).Lemma 3.4.
N≠Γ.Proof.
Assume thatN=Γ. By Lemma 3.3, the subgroup [Q,A] is either Abelian or [Q,A]⊆C. Since, by Lemma 3.2, [Q,A]⊈C, it is enough to consider the case, when [Q,A] is Abelian.
Suppose that[Q,A] is Abelian. Let us consider the character χ̂AQ. It is clear that AQ≠Γ and A⋪AQ. Therefore, there exists at least one irreducible component ξ of the character χ̂AQ such that Akerξ/kerξ⋪AQ/kerξ and since |AQ| is an uneven number, by theorem [1], it follows that for every such a character it is true that ξ(1)∈{|A|,2|A|-1} and 2|A|-1 is a prime power. Assume that ξ(1)=2|A|-1. Then, q divides 2|A|-1 and q divides 2|A|+1. Thus, q=2. A contradiction. Thus, ξ(1)=|A|.
Suppose that there exists one more characterξ1∈Irr(χ̂AQ) such that Akerξ1/kerξ1⋪AQ/kerξ1. Since ξ1(1)=|A| and χ̂AQ=2|A|+1, we see that χ̂Q is a sum of linear irreducible characters λ. Thus,
(3.5)Q′⊆⋂λ∈Irr(χ̂Q)ker⁡λ=ker⁡χ̂=1.
This yields that the subgroup Q is Abelian. Therefore, Q⊆CΓ([Q,A])⊲Γ. Suppose that CΓ([Q,A])=Γ, that is, [Q,A]⊆Z(Γ). Then, [Q,A]⊆C is a contradiction with Lemma 3.2. Thus, CΓ([Q,A])≠Γ and since CΓ([Q,A])⊲Γ, we see that, by Lemma 3.3, the subgroup CΓ([Q,A]) is either Abelian or C contains it. If CΓ([Q,A]) is Abelian, then Q⊲Γ, and we obtain a contradiction with Theorem  6.15 [5], but if CΓ([Q,A])⊆C, then we obtain a contradiction with Lemma 3.2.
Assume thatAkerψ/kerψ⊲AQ/kerψ, where χ̂AQ=ξ+ψ. By Lemma  3 [1], [Q,A]⊆(kerψ)π′. Since Q′⊆kerξ and kerξ∩kerψ=1, we have Q=[Q,A]⊆Z(Q). Then, Q⊆CΓ([Q,A]). This case has been considered above.
The lemma is proved.By Lemma3.4, N≠Γ. For each A-invariant proper subgroup M of the group G is such that [Q,A]⊆M, we have (3.6)χ̂AM=τ+ψ,
where τ and ψ are the characters of the subgroup AM such that Akerξ/kerξ⋪AM/kerξ for each ξ∈Irr(τ) and Akerξ/kerξ⊲AM/kerξ for each ξ∈Irr(ψ).Then, the groupAM/kerψ can be isomorphically embedded in the direct product AM¯=∏ξ∈Irr(ψ)AM/kerξ. Since each factor contains a normal Hall π-subgroup Akerξ/kerξ, we see that Akerψ/kerψ⊲AM/kerψ. It follows that Akerψ⊲AM. By Lemma  2.2.1 [8], AM=NAM(A)kerψ. Then, (3.7)M=(ker⁡ψ)π′CM(A).
Since by Lemma  3 [1], [M,A]⊆(kerψ)π′, [Q,A]⊆M and [[Q,A],A]=[Q,A], by Lemma  6.13 [9], we have(3.8)[Q,A]⊆(ker⁡ψ)π′.
Assume that τ=0. Then, χ̂AM=ψ, that is, ⋂ξ∈Irr(ψ)=kerψ=1. Then, A⊲AM, that is, [Q,A]⊆C. This contradicts Lemma 3.2.Thus,τ≠0. By Theorem [1], (3.9)ξ(1)∈{|A|-1,|A|,|A|+1,2|A|-2,2|A|-1}
and ξ(1) is a prime power, probably, with the exception of the case, when ξ(1)=|A| or ξ(1)=2|A|.Lemma 3.5.
The characterτ does not contain any irreducible components of degree  |A|-1.Proof.
Assume that there existsξ∈Irr(τ) and ξ(1)=|A|-1. Then, the group AM and its irreducible character ξ satisfy Lemma  11 [1]. Since Akerξ/kerξ⋪AM/kerξ, by Lemma  9 [1], A∩kerξ=1, and, by Lemma  10 [1], it yields that assertion (7) of Lemma  11 [1] does not satisfy. Since ξ(1)=|A|-1, we see that only assertion (3) of the mentioned Lemma satisfies, that is, ξCM(A)=(|A|-1)βξ,βξ(1)=1, where βξ=(ξM)π(M,A). By Lemma  10 [1],
(3.10)αξ=(ξA,1A)A=((ρA-1A)λ-1,1A)A
for a certain linear irreducible character λ of the subgroup A. We see that αξ=0 if λ=1A and αξ=1 if not.
In addition, assume thatτ(1)≠ξ(1), that is, the character τ contains an irreducible component ξ1.
Besides, assume thatξ1=|A|-1. In the same way, we establish that
(3.11)αξ1=((ξ1)A,1A)A=((ρA-1A)λ1-1,1A)A
for a certain linear irreducible character λ1 of the subgroup A, moreover, αξ1=0, if λ1=1A and αξ1=1 in the converse case.
Suppose thatχ̂AM=ξ+ξ1+η for a certain character η of the subgroup AM. Then,
(3.12)η(1)=χ̂(1)-(ξ(1)+ξ1(1))=2|A|+1-(2|A|-2)=3.
Suppose that |A|=3 or 5. Then, 2|A|+1=7 or 11 is a prime number. This contradicts the minimality of the group Γ. Thus, |A|≥7. Therefore, η(1)=3<|A|-1. By Theorem  1 [10], Akerη/kerη⊲AM/kerη. Clearly, ψ=η and, as follows from Lemma  11 [1] and Lemma  10 [1], each irreducible component δ of the character ψ is restricted irreducibly to CM(A), that is, δCM(A)=βδ. By Lemma  10 [1], we see that δ̂A=βδ(1)1A. Then, it is obvious that αψ=(ψA,1A)A=∑δ∈Irr(ψ)(βδ(1)λδ-1,1A)A. Here, δ̂ is a canonic extension of the character δM to the group AM and λδ are linear irreducible characters of the subgroup A. It is readily seen that αψ≤ψ(1)=3.
On the other hand, sinceCM(A)⊆C, we have βξ∈Irr(βCM(A)), βξ1∈Irr(βCM(A)), and (βCM(A),ψCM(A))CM(A)≠0, where β is from Lemma 2.4. Since by Theorem 13.1 [5], βξ≠βξ1, we have β(1)≥3. Since χ̂(1)=2|A|+1, we see that only assertion (1) or assertion (15) of Lemma 2.4 satisfy, that is, β(1)=|A|+1 or χC=β. Suppose that assertion (1) of Lemma 2.4 satisfies. Then, by Lemma 2.2,
(3.13)α=(χ̂A,1A)A=(ρA+(|A|+1)1A,1A)A=|A|+2.
Since α=αξ+αξ1+αψ≤5, we obtain |A|+2≤5. Hence, |A|≤3. As shown above |A|≥7. A contradiction. Suppose that assertion (15) of Lemma 2.4 satisfies. Then, by Lemma 2.2  χ̂A=β(1)1A, that is, A⊆kerχ̂. We obtain a contradiction with Lemma 3.1.
Now assume thatξ1=|A|. Then, (ξ1)A=ρA and αξ1=1. Since βξ∈Irr(βCM(A)) and (βCM(A),ψCM(A))CM(A)≠0, it follows that β(1)≥2. Again, we see that only assertion (1) of Lemma 2.4 satisfies. Since in this case ψ(1)=2, that is, αψ≤2, we have α=αξ+αξ1+αψ≤4. Thus, |A|+2≤4, that is, |A|≤2. A contradiction.
There only remains to consider the case, whenτ=ξ. Since kerτ∩kerψ=1, then the character ξkerψ is exact. Further, A⋪Akerψ, because if not we have [Q,A]⊆C by relation (3.8), and we get a contradiction with Lemma 3.2. Then, the character ξ̃=ξAkerψ is also exact and from Lemma7 [10] and Lemma2.27 [5] it yields that |Akerψ:Z(ξ̃)|=|Akerψ:Z(Akerψ)|=|A|22r, r∈N. Hence, [Q,A]⊆Z(Akerψ), that is, [Q,A]⊆C. A contradiction with Lemma 3.2.
The lemma is proved.Lemma 3.6.
LetM be a solvable  A-invariant subgroup of  G. Assume that  AM has an exact irreducible character  γ of degree  |A|+1=2β,β∈N. Then,  |M:CM(A)|=2m, where  m∈Z+.Proof.
Since|π|=1, we see that |A|=pf, f∈N, therefore, pf+1=2β, β∈N, that is, pf=2β-1. From Lemma  1 [1], it follows that f=1 and p is a Mersenne prime. Thus, |H|=p. By Theorem [11], the assertion of the theorem holds. We obtain a contradiction with the minimality of the group Γ. Therefore, |π|>1.
Suppose thatr,s∈π,r≠s. Since, by Lemma  7  [1], the subgroup A is a twiddle factor of a certain Frobenius group, it follows that by Corollary  1 [10], the subgroup A contains a subgroup H×B of order rs. By assumption B, CM(H)=CM(B)=CM(A). Since M is solvable, by Theorem  6 [9], it follows that [M,H]⊆F(M). Since, by Lemma  4 [1], M=[M,H]CM(H)=[M,H]CM(A), we obtain M=F(M)CM(A). Assume that F=F(M). Let us consider the character
(3.14)γAF=∑i=1tαiγi,γi∈Irr(AF), αi>0, i=1,2,…,t.
Suppose thatA⊲AF. Then, F⊆CM(A). Thus, M⊆CM(A), therefore, |M:CM(A)|=20=1. In this case, the lemma is true.
Suppose thatA⋪AF. Then for a certain i, let us say that i=1, Akerγ1⋪AF. From Theorems  1 and 2 [10] and the assumption of the lemma, we obtain γ1(1)=|A|-1=2α, |A| or |A|+1=2β. Here α,β∈N.
Ifγ1(1)=|A|-1=2α, then |A|=2α+1=2β-1. Hence, 2β=2α+2=2(2α-1+1), that is, α=1. Thus, |A|=3. This case is impossible.
Suppose thatγ1(1)=|A|. Then,
(3.15)γ¯(1)=γ(1)-γ1(1)=|A|+1-|A|=1,
where γ¯=γAF-γ1. Therefore, all irreducible components of the character γF are linear, that is, the subgroup F is Abelian. Since F⊲AM, we have CAM(F)⊲AM. Then, it is not hard to see that C*=CM(F)⊲AM. Since F⊈CM(A) and F⊆C*, it follows that C*⊈CM(A), that is, A⋪AC*.
Arguing as above for the subgroupAF, we can show that in the contingency γAC* there is an irreducible component γ1̃ such that Akerγ1̃⋪AC* and γ1̃(1)=|A| or |A|+1=2β. Assume that γ1̃(1)=|A|+1=2β. Since the subgroup A is a TI-set in AC*, from Lemma  1 [6], it yields that A⊆IAC*(γ1̃*), where γ1̃* is an irreducible component of the character (γ1̃)C*. Thus, (γ1̃)C* is an irreducible character of the subgroup C*. Since F⊆Z(C*), we have F⊆Z(γ1̃). However, since the character γ is exact, the character γ1̃ is exact. Therefore, Z(γ1̃)=Z(AC*). Then, F⊆Z(AC*), that is, F⊆CM(A). This case has already been examined. Therefore, we may assume that γ1̃(1)=|A|. Again, we see that C* is Abelian, that is, C*⊆F. Therefore, F=C*.
Suppose thatλ∈Irr(γF) and I=IAM(λ). Assume that I=AM. Since the subgroup F is Abelian, we obtain λ(1)=1. Therefore, γF=γ(1)λ, that is, F⊆Z(γ)=Z(AM), that is, F⊆CM(A). The lemma is true in this case.
Now suppose thatI≠AM. By Theorem  6.11 [5], there exists an irreducible character λ′ of the subgroup I such that (λ′)AM=γ. Hence, γ(1)=2β=|AM:I|λ′(1). Thus, |AM:I| and λ′(1) is degree 2. Therefore, Ax⊆I for a certain element x∈AM. Assume that Ax⊲I. Then, F⊆CM(Ax). Hence, F=Fx-1⊆CM(A). The lemma is true.
Therefore,Ax⋪I. Let us show that λ′(1)=1.
Let us assume the converse, that is,λ′(1)>1. Since by Frobenius reciprocity law for characters
(3.16)1=(γ,(λ′)AM)AM=(γI,λ′)I,
we have γI=λ′+λ′′. Since λ′(1)>1, we obtain
(3.17)λ′′(1)=γ(1)-λ′(1)=|A|+1-λ′(1)≤|A|-1.
Since |A|≠3, by Theorems  1 and 2 [10], we see that Axkerλ′′⊲I. By using the fact that Ax⋪I, it follows that Axkerλ′⋪I. Then, λ′(1)≥|A|-1. Since the numbers λ′(1) and |A|+1 are of degree 2 and |A|≠3, we obtain λ′(1)=|A|+1, that is, λ′(1)=γ(1). Since γ(1)=|AM:I|λ′(1), we have |AM:I|=1, that is, AM=I. In this case, as shown above, the lemma is true.
We may assume thatλ′(1)=1. Then, λ′′(1)=|A|+1-1=|A|. Since Axkerλ′⊲I, Ax⋪I and |A|≠3, from Theorem  1 and 2 [10], it easily follows that the character λ′′ is irreducible. Therefore, all irreducible components of the character γIπ′ are linear, that is, the subgroup Iπ′ is Abelian. Since F=C* is Abelian and F⊆Iπ′, we have F=Iπ′. Then, I=AxIπ′=AxF. Since M=Mx, we have
(3.18)M=FCM(A)=(FCM(A))x=Fx(CM(A))x=F(CM(A))x=FCM(Ax).
Further,
(3.19)AM=(AM)x=AxMx=AxM=AxFCM(Ax)=ICM(Ax).
Hence, for each g∈AM, we obtain that g=ab, where a∈I and b∈CM(Ax). Since γ=(λ′)AM, for each h∈Ax the following expression satisfies
(3.20)γ(h)=1|I|∑g∈AM(λ′)0(ghg-1)=1|I|∑g∈AM(λ′)0(abhb-1a-1)=1|I|∑g∈AM(λ′)0(a(hb-1)a-1)=1|I|∑g∈AMλ′(h)=|AM:I|λ′(h).
Here, (λ′)0(x)=λ′(x), if x∈I and (λ′)0(x)=0, if x∉I. Since λ′(1)=1, we obtain |λ′(h)|=1. The latter chain of equalities implies that
(3.21)|γ(h)|=||AM:I|⋅λ′(h)|=|AM:I||λ′(h)|=|AM:I|.
Sinceγ(1)=|AM:I|λ′(1), we have γ(1)=|AM:I|. Then, |γ(h)|=|AM:I|=γ(1). By Lemma  2.27(e) [5], Ax⊆Z(γ)=Z(AM). Hence, Ax=A⊲AM, that is M⊆CM(A), therefore, |M:CM(A)|=1. The lemma is true.
There only remains to consider the case, whenγ1(1)=|A|+1=2β. Since the subgroup A is a TI-set in AF, from Lemma  1 [6], it follows that A⊆IAF(γ1)*, where (γ1)* is an irreducible component of the character (γ1)F. Thus, (γ1)F is an irreducible character of the subgroup F. Since the subgroup F is nilpotent, by Theorem  4.21 [5], we have (γ1)F=(γ1)F2×μ, where (γ1)F2 is an irreducible character of the Sylow 2-subgroup F2 from F and μ is an irreducible character of the Hall 2′-subgroup O from F. Since μ(1)=1, we have O⊆Z(γ1)=Z(AF), that is, O⊆CM(A). Then,
(3.22)M=FCM(A)=(F2×O)CM(A)=F2CM(A).
Therefore,
(3.23)|M:CM(A)|=|F2CM(A):CM(A)|=|F2:F2∩CM(A)|=2m,
where m>0.
The lemma is proved.Lemma 3.7.
ψ≠0.Proof.
Let us assume the converse, that is,ψ=0. Then, (χ̂)AM=τ. Since, by Lemma 3.5, ξ(1)≠|A|-1 for ξ∈Irr(τ), we have
(3.24)ξ(1)∈{|A|,|A|+1,2|A|-2,2|A|-1,2|A|}.
Suppose that ξ(1)∈{2|A|-2,2|A|-1,2|A|}. Then, χ̂(1)-ξ(1)=2|A|+1-ξ(1)≤3. As mentioned above |A|≥7. By Theorem  1 [10], Aker(χ̂-ξ)/ker(χ̂-ξ)⊲AM/ker(χ̂-ξ). Thus, Irr(χ̂-ξ)⊆Irr(ψ). This contradicts the assumption. Hence, ξ(1)∈{|A|,|A|+1} and since by the assumption τ(1)=2|A|+1, we obtain
(3.25)τ=ξ+ξ1,
where ξ(1)=|A| and ξ1(1)=|A|+1 is degree 2.
From the Clifford Theorem, it follows thatM′⊆kerξ. Since kerξ∩kerξ1=kerτ=kerχ̂AM=1, it follows that the subgroup kerξ1 is Abelian. By Theorem [12], the factor group AM/kerξ1 is solvable. Then, AM is also solvable. In the course of proving Lemma 3.6, it was established that |π|>1. Likewise, it is shown that [M,A]⊆F=F(M).
On the other hand, by Lemma3.6,
(3.26)|M¯:CM¯(A)|=2m,m∈Z+,
where M¯=M/K. Here, K=(kerξ1)π′. Since (|K|,|A|)=1 and C=CM(a)=CM(A) for all elements a∈A#, we have |CM/K(A)|=|CM(A):CK(A)|. Thus,
(3.27)|M¯:CM¯(A)|=|M||CK(A)||CM(A)||K|=2m,m∈Z+.
Since[Q,A]⊆[M,A] and [Q,A]⊈C, we have Fq≠1. Since FscharF⊲AM, it follows that Fs⊲AM. By using the fact that ξ(1) and ξ1(1) are q′-numbers and from the Clifford Theorem, it is readily seen that the subgroup Fq is Abelian. Since [Q,A]⊆Fq, it yields that [Q,A] is also Abelian. Therefore, [Q,A]=[[Q,A],A]×C[Q,A](A). Since [[Q,A],A]=[Q,A], we have C[Q,A](A)=1. Then, [Q,A]∩CM(A)=1. Hence and by the equality (3.27), it follows that [Q,A]⊆K.
Further, we see that(3.28)τK=(τM)K=(ξM+(ξ1)M)K=∑λ∈Irr(ξM)λK+ξ1(1)1K,λ(1)=1.
Since K⊆Z(ξ1), it follows that K⊆(⋂λ∈Irr(ξM)Z(λ))∩Z(ξ1). By the fact that the character τM is exact and Lemma 3.1 [1], it follows that K⊆Z(M). Then, M⊆N. Since A⊆N and Q⊆N, we obtain L=〈AM,Q〉⊆N. By Lemma 3.4, we have L≠Γ.
Let us consider the characterχ̂L. Clearly, it is enough to consider the case, when Akerξ′/kerξ′⋪L/kerξ′ for a certain character ξ′∈Irr(χ̂L). Then, ξ∈Irr(ξAM′) or ξ1∈Irr(ξAM′). Thus, we have ξ′(1)≥|A|.
Assume that there exists a characterψ′ of the group L such that Irr(ψ′)⊆Irr(χ̂L) and Akerψ′/kerψ′⊲L/kerψ′. Then, Akerψ′⊲L, therefore, (Akerψ′)∩AM⊲AM. Hence, AkerψAM′⊲AM. Thus, we have ψAM′=ψ≠0. A contradiction with the assumption.
Thus, we haveχ̂L=ξ′+ξ1′ and can say that ξAM′=ξ and (ξ1′)AM=ξ1. Let us apply the reasoning analogous to the reasoning applied to the group AM and its characters ξ and ξ1 to the group L and characters ξ′ and ξ1′. We obtain [Lπ′,A]⊆F(Lπ′) and
(3.29)[Q,A]⊆KL.
Here, we have KL=(kerξ1′)π′.
Suppose thatq1∈π([Lπ′,A]),q1≠q and Q1 is an invariant Sylow q1-subgroup [Lπ′,A]. Then, Q1⊲L. Let ξq1∈Irr((ξ1′)Q1). By Theorem  6.11 [5], we have ξ′1=ξ̃q1L for a certain irreducible character ξ̃q1 of the group IL(ξq1). Thus, ξ1′(1)=ξ̃q1L(1)=ξ̃q1(1)|L:IL(ξq1)|. Since ξ1′(1) is degree 2, we see that the numbers ξ̃q1(1) and |L:IL(ξq1)| are also degrees 2.
Assume that(ξQ1′,ξq1)Q1≠0. By the same theorem, we have ξ′=(ξq1′)L for a certain irreducible character ξq1′ of the group IL(ξq1). Then we have |A|=ξ′(1)=ξq1′(1)|L:IL(ξq1)|. Since |L:IL(ξq1)| is degree 2, we obtain L=IL(ξq1). Then, ξQ1′=e′ξq1(1) for a certain natural number e′. Thus, we have |A|=e′ξq1(1). Since ξq1(1) is a π′-number, we have ξq1(1)=1. Then, Q1⊆Z(ξ′) and, obviously, Q1⊆Z(ξ1′). By Lemma  5 [1], Q1⊆Z(L). This means that Q1⊆CLπ′(A).
Now assume that(ξQ1′,ξq1)Q1=0. By Theorem  D [13],
(3.30)t=|Q1||Γ:L|ξ1(1)χ̂(1)ξq1(1)
is an integer. Since Q⊆L, we see that |Γ:L| is a q′-number. Since ξ1(1) is a q′-number and q divides χ̂(1), it follows that t is not an integer. We obtain a contradiction. Thus, we have Q1=1, therefore, [Lπ′,A] is a q-group.
It is obvious that[Q,A]⊆[Lπ′,A]. Since [Lπ′,A]⊲L, we obtain [Lπ′,A]⊆Q. Then, we have [[Lπ′,A],A]⊆[Q,A]. Since [[Lπ′,A],A]=[Lπ′,A], it follows that [Lπ′,A]⊆[Q,A]. Thus, we have [Lπ′,A]=[Q,A]. Then, [Lπ′,A]⊆KL. Hence, AKL/KL⊲L/KL. Therefore, we have Akerξ1′/kerξ1′⊲L/kerξ1′. Thus, ξ1′⊆Irr(ψ), that is, ψ≠0. This contradicts the original assumption.
The lemma is proved.Lemma 3.8.
The characterμ=τAkerψ is an exact irreducible character of degree  n1=|A|,|A|+1,  2|A|-2 or  2|A|-1 and  n1 is a prime power, probably, with the exception of the case, when  μ(1)=|A|.Proof.
By Lemma3.7, we have ψ≠0.
As stated above, the characterχ̂AM cannot be irreducible.
Letψ(1)=1. Then, we have Γ′⊆kerψ. Suppose that G⊈Γ′. Then, G′≠G. By Lemma 3.3, the subgroup G′ is either Abelian or G′⊆C. If G′ is Abelian, G is solvable. By Theorem  1 [6], χ̂(1) is a prime power. This contradicts the minimality of the group Γ. Now let G′⊆C. Then, we have C⊲G. Since χ(1)=2|A|+1, it is readily seen from Lemma 2.4 that the character χC contains a linear component β or a linear component β1. From the Clifford Theorem, it follows that C is Abelian. Thus, G′ is Abelian. The case is examined.
Thus,G⊆Γ′. Then, we have G⊆kerψ. By Corollary  6.7 [5], G⊆kerχ̂. Since the character χ̂ is exact, we have G=1. Since A is a group of nontrivial automorphisms of the group G, it follows that the latter statement is impossible.
Therefore, we obtainψ(1)≥2. Then, τ(1)≤2|A|-1. From Lemma 3.5, it follows that the character τ is irreducible. Since kerτ∩kerψ=1, we see that the character τkerψ is exact. Suppose that the character τAkerψ is not exact. Then, we have kerτAkerψ⊆A. Since A is a TI-subgroup, it follows that A⊲Akerψ. Thus we have (kerψ)π′⊆CM(A). From formula (3.8), it follows that [Q,A]⊆C. A contradiction with the choice of the subgroup M. Thus, the character τAkerψ is exact.
Since(kerψ)π′⊈CM(A), we have Akerμ/kerμ⋪Akerψ/kerμ for a certain character μ⊆Irr(τAkerψ). By Theorem [1], we have
(3.31)μ(1)∈{|A|-1,|A|,|A|+1,2|A|-2,2|A|-1}
and μ(1) is a prime power, probably, with the exception of the case when μ(1)=|A|. Besides, it is readily seen that μ(1)=τ(1) when μ(1)≠|A|-1.
Let us consider this case separately. Letμ(1)=|A|-1. Since τ(1)<2|A|, subject to Lemma 3.5, we have τ(1)=2|A|-2 and by the Clifford Theorem, we obtain τAkerψ=μ+μg, where g∈AM∖Akerψ. Hence, by Lemma  7 [10], we have
(3.32)|(ker⁡ψ)π′:Z(μ)|=|(ker⁡ψ)π′:Z(μg)|=22α,
where μ(1)=2α, α∈N. Since [Q,A]⊆(kerψ)π′, we see that [Q,A]⊆Z(μ)∩Z(μg). Since kerμ∩kerμg=1, by Lemma  5 [1], we have [Q,A]⊆Z(Akerψ). Then [Q,A]⊆C. A contradiction with Lemma 3.2. Thus,
(3.33)μ(1)∈{|A|,|A|+1,2|A|-2,2|A|-1}
and μ(1) is a prime power, probably, with the exception of the case, when μ(1)=|A|. Since μ(1)=τ(1) under the value indicated, it follows that μ(1) and the character τAkerψ is exact, the lemma is proved.Lemma 3.9.
n1=|A|.Proof.
By Lemma3.8, it is enough to show that n1≠|A+1|,2|A|-2 and 2|A|-1.
First, letn1=|A|+1. By Theorem [12], the group Akerψ is solvable. Then by Lemma 3.6, we have |(kerψ)π′:C(kerψ)π′(A)|=2m, where m∈Z+. From equality (3.8), it follows that [Q,A]⊆C(kerψ)π′(A)⊆CM(A). We obtain a contradiction with the choice of the subgroup M.
Now suppose thatn1=2|A|-2 or 2|A|-1. By Lemma 3.8, n1 is a prime power s.
LetS be a A-invariant Sylow s-subgroup from (kerψ)π′ and Z=Z(S). Since, by Lemma 3.8, the character μ is exact and irreducible, by Lemma  2.27 [5], it follows that Z(μ)=Z(Akerψ) is a cyclic group. Therefore, Z∩Z(μ)⊆Z(Akerψ), that is, Z∩Z(μ)⊆CZ(A). Since Z is Abelian, it follows that Z=[Z,A]×CZ(A). It can be easily seen that |[Z,A]| divides |Z/Z∩Z(μ)|.
Assume thatZ⊈Z(μ). Then, by Corollary  1 [14], |Z/Z∩Z(μ)| divides n1. Thus, |[Z,A]| divides n1. Thus, we have |[Z,A]|=sr, r∈N. Since A[Z,A] is a Frobenius group with a kernel [Z,A], the order of its kernel can be compared with 1 modulo |A|. Therefore, sr-1=d|A|, that is, sr=d|A|+1, where d∈N. Since sr≤n1, we have d=1 and sr=|A|+1. Therefore, s=2, as stated above, n1≠|A|+1, and we obtain n1=2|A|-2. Since 2r=|A|+1 and 2|A|-2 is degree 2, we have |A|=3. As stated above, it is not true. Our assumption that Z⊈Z(μ) is false.
Thus,Z⊆Z(μ). Then, it is clear that Z⊲AM. Since Z⊆kerψ, kerτ∩kerψ=kerχ̂=1, and the character τ is irreducible, Corollary  6.7 [5] yields that (τZ,ς)Z≠0 for a certain nonprincipal irreducible character ς of the subgroup Z. Then, by Theorem  D [13],
(3.34)|Z||Γ:AM|τ(1)χ̂(1)ς(1)=|Z||Γ:AM|n1n
is an integer. Since q divides n and q does not divide |Z||Γ:AM|n1, we obtain a contradiction.
The lemma is proved.Lemma 3.10.
LetM be a maximal proper  A-invariant subgroup  G such that  [Q,A]⊆M. Then, one has  M=[Q,A]×C,[Q,A]⊆Z(M).Proof.
By assumption of the lemma and by Lemma3.4, we have N⊆AM.
By Lemma3.9, n1=|A|. Since the character μ=τA/kerψ is irreducible, the character τ is also irreducible.
Sinceτ(1)=|A|, all irreducible components of the character τM are linear. This means that M′⊆kerτ. Since kerτ⊲AM, Akerψ⊲AM and kerτ∩kerψ=1, we have A0=kerτ∩Akerψ⊆A. But A is a TI-subgroup and A⋪Akerψ because in the converse case, by equality (3.8), we have [Q,A]⊆C, which contradicts Lemma 3.2. Then, we obtain A0=1. Thus, kerτ⊆CAM(Akerψ), that is, kerτ⊆CM(A). Then, M′⊆CM(A), that is, CM(A)⊲M.
By Lemma  3 [1], we have [M,A]⊆M∩Akerψ. By using the fact that M′⊆kerτ and kerτ∩kerψ=1, it follows that kerψ∩M is Abelian. Thus, the subgroup [M,A] is Abelian. Since,
(3.35)[M,A]=[[M,A],A]×C[M,A](A)
and [[M,A],A]=[M,A], we conclude that C[M,A](A)=1. Since
(3.36)[M,A]∩CM(A)⊆C[M,A](A),
it follows that [M,A]∩CM(A)=1. Since by Lemma  6 [9], [M,A]⊲M, CM(A)⊲M and M=[M,A]CM(A) by Lemma  4 [1], we have M=[M,A]×CM(A). By using the fact that [M,A] is Abelian, it follows that [M,A]⊆Z(M).
Since[Q,A]⊆M, we conclude that [Q,A]=[[Q,A],A]⊆Z(M). Then, we have M⊆N. Thus, AM⊆N. Since the reverse inclusion holds as well, we have AM=N. In view of the fact that Q⊆N since [Q,A]⊲QA, we obtain L=〈AM,Q〉⊆N=AM. From the maximality of the subgroup M, it follows that Q⊆M.
Further,[M,A]=[M,A]q×X, where X is a Hall q′-subgroup from [M,A].
Assume thatX≠1.
Letγ∈Irr((τ)X). Since X is Abelian, it follows that γ(1)=1. Since [M,A]⊆kerψ, we obtain X⊆kerψ. Thus, ψX=ψ(1)1X.
First assume thatγ∈Irr(ψX). Then, γ=1X. Further, Xchar[M,A]⊲L, that is, X⊲L. Then, X⊆kerτ. Thus, X⊆kerψ∩kerτ=kerχ̂=1. Hence, we obtain that X=1. We get a contradiction with the assumption.
Now, letγ∉Irr(ψX), that is, (ψX,γ)X=0. Since X⊲L, we apply Theorem  D [13] to the subgroup AM and the character χ̂AM=τ+ψ. By this Theorem,
(3.37)|X||Γ:AM|τ(1)χ̂(1)γ(1)=|X||Γ:AM||A|χ̂(1)=|X||G:M|χ̂(1)
is an integer. On the other hand, we have Q⊆M, that is, q does not divide |G:M|. Therefore, (q,|X||G:M|)=1. But q divides χ̂(1). We obtain a contradiction.
Thus,X=1, that is, [M,A] is a q-group.
Since[M,A] is a normal q-subgroup in M, it follows that [M,A]⊆Q. It is readily seen that [M,A]=[[M,A],A]⊆[Q,A]. Therefore, [M,A]=[Q,A].
Thus, we have(3.38)M=[Q,A]×CM(A),[Q,A]⊆Z(M).
Let us ascertain that assertion (1) of Lemma2.4 holds and A[Q,A]⊆kerψ.
Hence,n=2|A|+1 and, by Lemma 2.4, it follows that only (1)–(5), (12) or (15) satisfy.
If assertion (15) of Lemma2.4 satisfies, Lemma 2.3 and Lemma 2.2 imply that A⊆kerχ̂. This contradicts Lemma 3.1. If assertion (12) of Lemma 2.4 satisfies it follows that C⊆Z(G). Since M′⊆kerτ, we have (χ̂M′,1M′)M′≠0. By Corollary  6.7 [5], M′⊆kerχ̂=1. Thus, the subgroup M is Abelian. Then, M⊆N*=NG(Q). Since Q is Abelian, by Theorem  6.15 [5], we see that N*≠G. Since N*⊆M, we obtain N*=M. By the Burnside Theorem, the group G contains a normal q-complement, which is, by Lemma 3.3, either an Abelian group, or is contained in C⊆Z(G). Thus, G is solvable. In this case, the theorem holds [6].
Assume that assertion (3)–(5) of Lemma2.4 satisfy. Since β(1)=β1(1)=1 and β2(1)=1 (assertion (4)), it follows that C′=1, that is, the subgroup C is Abelian. Therefore, CM(A) is Abelian. Then by equality (3.38), the subgroup M is Abelian. This case is examined.
Suppose that assertion (2) of Lemma2.4 satisfies.
In addition, assume that the characterβ1 is imprimitive, that is, β1=λC for a certain linear irreducible character λ of the subgroup C0 of index 2 in C. Then, it is readily seen that C0 is Abelian. Thus, the subgroup Q is Abelian. Since we have [Q,A]⊆Z(M) by equality (3.38), and we can think that N*=M, by Theorem  6.4.4 [4], q divides |G:G′|. Thus, G′≠G. This case is also examined.
Now assume that the characterβ1 is primitive. By Theorem  14.23 [5], |C¯:Z(C¯)|=12,24 or 60, where C¯=C/kerβ1 and, by Lemma  2.27 [5], Z(C¯)=Z(β1)/kerβ1. Thus, |C:Z(β1)|=12,24 or 60. Since β(1)=1, it follows that Z(β1)⊆Z(β). Since kerβ1∩kerβ=1, by Lemma  5 [1], we have Z(β1)=Z(C). Then, |C:Z(C)|=12,24 or 60. Let q≠3,5. Then, CM(A)⊆Z(C). Thus, Q is Abelian and as above we obtain that q divides |G:G′|. This case is examined. Suppose that q=3,5. Then, the factor group CqZ(C)/Z(C) is cyclic. Since Z(C)⊆Z(CqZ(C)), CqZ(C) is Abelian. We obtain that the subgroup Q is Abelian as well. The case is examined.
It remains to consider the case when assertion (1) of Lemma2.4 holds, that is, χC=β+|A|β1,β1(1)=1. Then, β(1)=|A|+1. From Lemma 2.2, it follows that χ̂A=ρA+β1(1)1A. Thus, χ̂A=ρA+(|A|+1)1A, therefore, we have (χ̂A,1A)A=|A|+2. On the other hand,
(3.39)(χ̂A,1A)A=((χ̂AM)A,1A)A=((τ+ψ)A,1A)A=(μA,1A)A+(ψA,1A)A=1+(ψA,1A)A.
Here, we take into account that μA=ρA. Thus, |A|+2=1+(ψA,1A)A, that is, (ψA,1A)A=|A|+1. Since ψ(1)=|A|+1, we conlude A⊆kerψ. Since, by equality (3.8), [Q,A]⊆kerψ, it follows that A[Q,A]⊆kerψ.
By equality (3.38) for a complete proof of the lemma, it is enough to make sure that C=CM(A).
SinceCM(A)⊆C, it remains to show that C⊆CM(A).
Suppose thatc∈C. Then, we have χ̂AMc=τ(c)+ψ(c). Since A⊆kerψ, we have A⊆kerψ(c). Hence, [Qc,A]⊆kerψ(c). Therefore,
(3.40)χ̂A[Q,A]=(τ)A[Q,A]+ψA[Q,A]=(τ)A[Q,A]+ψ(1)1A[Q,A],χ̂A[Qc,A]=(τ1(c))A[Qc,A]+ψA[Qc,A](c)=(τ(c))A[Qc,A]+ψ(c)(1)1A[Q,A].
Thus,
(3.41)(χ̂A[Q,A],1A[Q,A])A[Q,A]+(χ̂A[Qc,A],1A[Qc,A])A[Qc,A]=((τA[Q,A],1A[Q,A])A[Q,A]+ψ(1))+((τ(c))A[Qc,A],1A[Qc,A])A[Qc,A]+ψ(c)(1)=2ψ(1)=2β(1)>β(1)+1=(χ̂A,1A)A=(ρA+β(1)1A,1A)A≥(χ̂A[Q,A]∩A[Qc,A],1A[Q,A]∩A[Qc,A])A[Q,A]∩A[Qc,A].
Here we take into account that
(3.42)ψ(1)=χ̂(1)-τ(1)=|A|+β(1)-|A|=β(1),
and consider that β(1)=|A|+1>1. Therefore, by the Brouwer Theorem  5.19 [5], the subgroup
(3.43)D=〈A[Q,A],A[Qc,A]〉
is proper.
LetDq be a A-invariant Sylow q-subgroup D such that [Q,A]⊆Dq. Such a subgroup exists by Theorem  6.2.2(iii) [4]. Let MD be a maximal A-invariant subgroup of the group G, containing Dπ′. Let us apply the reasoning concerning the group AM to the character χAMD̂ and the group AMD. We get MD=[MD,A]×CMD(A),[MD,A]⊆Z(MD).
Since[Q,A]⊆MD, we have [[Q,A],A]⊆[MD,A]. Since [[Q,A],A]=[Q,A], it follows that [Q,A]⊆[MD,A]. Therefore, [Q,A]⊆Z(MD). Thus, MD⊆Nπ′. Since [M,A]⊆Z(M) and [M,A]=[Q,A], as ascertained above, we have M⊆Nπ′. From the maximality of the subgroup M and Lemma 3.4, it follows that M=Nπ′. Then, MD⊆M. Thus, [Qc,A]⊆M. Since M=[Q,A]×CM(A), as ascertained above, we have [Q,A]=[Qc,A]=([Q,A])c. Therefore, c∈Nπ′. Then, c∈M, that is, C⊆CM(A).
The lemma is proved.This completes the proof of the theorem.Let us show thatn is a primary number.Let us assume the converse. Letq1 and q2 be distinct odd primes from π(n).LetQ1 and Q2 be A-invariant Sylow qi-subgroups G, Ni=NΓ(Qi). Assume that N1=Γ. From Lemma 3.3, it follows that Q1 is Abelian or Q1⊆C. In the first case we obtain a contradiction with Theorem  6.15 [5]; in the second case we obtain a contradiction with Lemma 3.2. Thus, we have Ni≠Γ,i=1,2. Let Mi be maximal proper A-invariant subgroups G such that Ni⊆AMi,i=1,2. By Lemma 3.10, Mi=[Qi,A]×C,i=1,2, that is, Mi⊆NG(C),i=1,2. Therefore, C⊲L=〈M1,M2〉. Since Mi are maximal A-invariant subgroups G, we see that either L=M1=M2, or L=G. Suppose that L≠G. Then, |M1|=|M2|. Thus, |[Q1,A]|=|[Q2,A]|. Since q1≠q2, we obtain |[Q1,A]|≠|[Q2,A]|. A contradiction.Now suppose thatL=G, that is, C⊲G. Since, by Lemma 3.10, M=C×[Q,A], it follows that M⊆CG(C)⊲G. By Lemma 3.3, CG(C) is either Abelian or CG(C)⊆C. In the first case, M is Abelian. It was considered above. In the second case, we have M⊆C. A contradiction with Lemma 3.2.Thus, our assumption on the fact thatn will divide by two distinct odd primes does not satisfy. The Theorem 1.1 is proved.


## 4. Proof of Theorem1.2

Let us carry out induction over the order of the groupG.Let us examine the subgroupOπ′,π(G). By definition, put (4.1)Oπ′,π(G)Oπ′(G)=Oπ(GOπ′(G)).
Since the factor group G/Oπ′(G) is π-solvable and Oπ′(G/Oπ′(G)=1, we have Oπ(G/Oπ′(G))≠1. Since HOπ′(G)/Oπ′(G) is a π-Hall TI-subgroup in G/Oπ′(G), we obtain Oπ′,π(G)=HOπ′(G). Denote X=Oπ′,π(G).Identity substitution of the groupG defines an exact irreducible character φ of degree n.First let us assume that|G:X|≠1. By the Clifford Theorem (4.2)φX=e∑i=1tφi,
where φi=1,t¯∈Irr(X) and e, t are numbers, dividing |G:X|.In addition, assume thate≠1 or t≠1. Then, the numbers φi=1,t¯(1)<2|H|. Therefore, the factor groups X/kerφi,i=1,t¯ satisfy the theorem proved in [1–3]. If (4.3)Hker⁡φiker⁡φi⋬Xker⁡φi,i=1,t¯,
from this theorem, it follows that φi=1,t¯(1)=|H|-1,|H|,|H|+1,2|H|-2 or 2|H|-1. Since the numbers φi=1,t¯(1) divide φ(1), it is readily seen that the numbers φi=1,t¯(1) cannot possess the stated values. Therefore, (4.4)Hker⁡φiker⁡φi⊴Xker⁡φi,i=1,t¯,
that is, X/kerφi contain normal π-Hall subgroups Hkerφi/kerφi,i=1,t¯. Since ⋂i=1t(kerφi)=1, we see that X can be isomorphically embedded in the direct product (4.5)X¯=Xker⁡φ1×⋯×Xker⁡φt.
Since X¯ contains a normal π-Hall subgroup, it follows that X also contains a normal π-Hall subgroup H, that is, H⊴G. Commutativily of the subgroup H results from the faithfulness of the character φ and the fact that φ(1) is a π′-number.Now suppose thate=t=1. Then, the character φX is irreducible. Since |X|<|G|, by induction we see that either φX(1) is a prime power or H⊴G. Since φX(1)=n, in both cases the theorem holds.Now suppose that|G:X|=1, that is, G=X=HOπ′(X). Let us apply the notation to Condition B and Theorem 1.1. For this, suppose that X=Γ, H=A, Oπ′(X)=G. Then, φ is an irreducible character of the group Γ=AG. Since (|A|,φ(1))=1, it is readily seen that the character φG contains an A-invariant faithful irreducible component φ*. It is obvious that φG=φ*∈IrrA(G).Suppose thatkerφ*̂≠1. Thus, we have kerφ*̂⊆A. Since A is a TI-subgroup in Γ, it follows that A⊲Γ. This contradicts the fact that A is a group of nontrivial automorphisms of the group G.Therefore,kerφ*̂=1. Since φ*(1)=φ(1) and, by assumption of the theorem being proved, φ(1)=n is not a prime power, it follows that φ*̂(1) is not a prime power either. Taking into account that by Lemma  6 [1] CG(a)=CG(A)=C for each element a∈A#, by Theorem 1.1 we again see that A⊲Γ. A contradiction.The theorem is proved.



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*Source: 868096-2011-08-17.xml*