On Irreducible Linear Groups of Nonprimary Degree

Abstract

It is established that degree 2|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.

1. Introduction

Suppose that 𝐺 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 𝐶_𝐺(𝑎)=𝐶_𝐺(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 degree 𝑛=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 group Γ=𝐴𝐺 satisfies B, if 𝐺⊲Γ,(|𝐴|,|𝐺|)=1,|𝐴| is uneven, 𝐶_𝐺(𝑎)=𝐶_𝐺(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 group Γ 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 group 𝐺 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;𝑀_𝜋 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 that Γ=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)∶Irr_𝐴(𝐵)→Irr(𝐶_𝐵(𝐴)) between the set of all 𝐴-invariant characters of the group 𝐵 and the set of all irreducible characters of the subgroup 𝐶_𝐵(𝐴). The set has a range of properties, depending, in particular, on the properties of the subgroup 𝐴. Suppose that 𝜒∈Irr_𝐴(𝐵). Then, by Lemma  13.3 [5], there is a unique irreducible character 𝜒 of the group Γ such that 𝜒_𝐵=𝜒 and 𝐴⊆ker(det𝜒). It is called a canonical extension of the character 𝜒 to the group Γ.

Lemma 2.1 (see [6, Lemma  2]). If 𝜒 is an irreducible character of the group 𝐺 stated in Condition B, then 𝜒∈Irr_𝐴(𝐺).

Lemma 2.2 (see [1, Lemma  10]). If 𝜒 is an irreducible character of the group 𝐺 stated in Condition B, then 𝜒_𝐴=𝑘𝜌_𝐴+𝜀𝛽(1)1_𝐴, where 𝑘∈𝐙₊,𝜌_𝐴 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. If 𝜒 is an irreducible character of the group 𝐺 stated in Condition B, then 𝜒_𝐶=|𝐴|𝜃+𝜀𝛽, 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. If 𝜒 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)𝜒_𝐶=𝛽+|𝐴|𝛽₁,|𝐴|<𝛽(1)<2|𝐴|,𝛽₁(1)=1;(2)𝜒_𝐶=𝛽+|𝐴|𝛽₁,𝛽(1)<|𝐴|,𝛽₁(1)=2;(3)𝜒_𝐶=𝛽+2|𝐴|𝛽₁,𝛽(1)<|𝐴|,𝛽₁(1)=1;(4)𝜒_𝐶=𝛽+|𝐴|(𝛽₁+𝛽₂),𝛽(1)<|𝐴|,𝛽₁(1)=𝛽₂(1)=1;(5)𝜒_𝐶=(|𝐴|+1)𝛽+|𝐴|𝛽₁,𝛽(1)=𝛽₁(1)=1;(6)𝜒_𝐶=(|𝐴|+1)𝛽,𝛽(1)=2;(7)𝜒_𝐶=(|𝐴|−1)𝛽+2|𝐴|𝛽₁,𝛽(1)=𝛽₁(1)=1;(8)𝜒_𝐶=(|𝐴|−1)𝛽+|𝐴|𝛽₁,𝛽(1)=1,𝛽₁(1)=2;(9)𝜒_𝐶=(|𝐴|−1)𝛽+|𝐴|(𝛽₁+𝛽₂),𝛽(1)=𝛽₁(1)=𝛽₂(1)=1;(10)𝜒_𝐶=(|𝐴|−1)𝛽+|𝐴|𝛽₁,𝛽(1)=2,𝛽₁(1)=1;(11)𝜒_𝐶=(|𝐴|−1)𝛽,𝛽(1)=3;(12)𝜒_𝐶=(2|𝐴|+1)𝛽,𝛽(1)=1;(13)𝜒_𝐶=(2|𝐴|−1)𝛽+|𝐴|𝛽₁,𝛽(1)=𝛽₁(1)=1;(14)𝜒_𝐶=(3|𝐴|−1)𝛽,𝛽(1)=1;(15)𝜒_𝐶=𝛽.
Here, 𝛽=(𝜒)𝜋(𝐺,𝐴), 𝛽₁ and 𝛽₂ are distinct irreducible characters of the subgroup 𝐶.

Proof. From Lemma 2.3, it follows that 𝜒_𝐶,𝛽_𝐶=𝑘_𝛽||𝐴||+𝜀,(2.1) where 𝑘_𝛽=(𝜃,𝛽)_𝐶. Suppose that 𝛽₁∈Irr(𝜒_𝐶),𝛽₁≠𝛽,𝑘₁=(𝜒_𝐶,𝛽₁)_𝐶 and 𝜃₁ is the character 𝐶 or 0 such that 𝜃=𝑘_𝛽𝛽+𝑘₁𝛽₁+𝜃₁.(2.2) Hence, by Lemma 2.3, it follows that 𝜒_𝐶=||𝐴||𝑘_𝛽𝛽+𝑘₁𝛽₁+𝜃₁+𝜀𝛽.(2.3) Hence, by the assumption of the lemma, it follows that ||𝐴||𝑘_𝛽𝛽(1)+𝜀𝛽(1)≤𝜒(1)<3||𝐴||.(2.4) Then, |𝐴|𝑘_𝛽𝛽(1)<3|𝐴|−𝜀𝛽(1), that is, 𝑘_𝛽<3𝛽(1)−||𝐴𝜀||.(2.5) Hence, 𝑘_𝛽≤3.
Assume that 𝜃≠0 and examine all the possible cases separately.
(1) 𝑘_𝛽=0.
Since, by Corollary  2.17 [5], (𝜒_𝐶,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 that 𝜃(1)=1. Then, we can assume that 𝜃₁=0 and 𝑘₁𝛽₁(1)=1. From equality (2.3), we obtain 𝜒_𝐶=𝛽+|𝐴|𝛽₁,|𝐴|<𝛽(1)<2|𝐴|,𝛽₁(1)=1 is assertion (1) of the lemma being proved.
Suppose that 𝜃(1)=2. Then, 𝑘₁𝛽₁(1)+𝜃₁(1)=2.
If 𝜃₁=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 if 𝜃₁≠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) 𝑘_𝛽=1.
Suppose that 𝜀=1. From (2.3), we obtain 𝜒_𝐶=|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 that 𝛽(1)=1. Then, 𝑘₁𝛽₁(1)=1 and 𝜒_𝐶=(|𝐴|+1)𝛽+|𝐴|𝛽₁,𝛽(1)=𝛽₁(1)=1 is assertion (5).
Now, suppose that 𝛽(1)=2. Since 2|𝐴|<𝜒(1)<3|𝐴|, we have 𝑘₁𝛽₁(1)+𝜃₁(1)=0 and, therefore, 𝜒_𝐶=(|𝐴|+1)𝛽 is assertion (6).
Further, suppose that 𝜀=−1. By (2.3), it follows that 𝜒_𝐶(1)=||𝐴||−1𝛽(1)+||𝐴||𝑘₁𝛽₁(1)+𝜃₁(1).(2.6) Since 𝜒(1)<3|𝐴|, we see that 𝛽(1)≤3.
Let 𝛽(1)=1. Then, 𝑘₁𝛽₁(1)+𝜃₁(1)=2.
Let 𝜃₁=0. Then, 𝑘₁𝛽₁(1)=2. If 𝑘₁=2, then 𝛽₁(1)=1. Hence, 𝜒_𝐶=(|𝐴|−1)𝛽+2|𝐴|𝛽₁ is assertion (7). If 𝑘₁=1, then 𝛽₁(1)=2. Then 𝜒_𝐶=(|𝐴|−1)𝛽+|𝐴|𝛽₁ is assertion (8).
Now suppose that 𝜃₁(1)=1. Then, 𝑘₁𝛽₁(1)=1. We obtain 𝜒_𝐶=(|𝐴|−1)𝛽+|𝐴|(𝛽₁+𝛽₂), where 𝛽₁(1)=𝛽₂(1)=1 is assertion (9).
If 𝜃₁(1)=2, then 𝑘₁𝛽₁(1)=0 and it is readily seen that (7), (8) or (9) follow.
Now suppose that 𝛽(1)=2. Hence, 𝜒_𝐶=(|𝐴|−1)𝛽+|𝐴|𝛽₁,𝛽₁(1)=1 is assertion (10).
Suppose that 𝛽(1)=3. Then, 𝑘₁𝛽₁(1)+𝜃₁(1)=0. Assertion (11).(3) 𝑘_𝛽=2.
Let 𝜀=1. Expression (2.3) implies that 𝑘₁𝛽₁(1)+𝜃₁(1)=0. We obtain 𝜒_𝐶=(2|𝐴|+1)𝛽. We see that 𝛽(1)=1 is assertion (12).
Let 𝜀=−1. Then, 𝜒_𝐶=(2|𝐴|−1)𝛽+|𝐴|𝛽₁, where 𝛽(1)=𝛽₁(1)=1 is assertion (13).
(4) 𝑘_𝛽=3.
Expression (2.3) implies that 𝜀=−1,𝜒_𝐶=(3|𝐴|−1)𝛽, where 𝛽(1)=1 is assertion (14).
Now suppose that in Lemma 2.3  𝜃=0. Then, 𝜀=1 and 𝜒_𝐶=𝛽 is assertion (15).
The lemma is proved.

3. Proof of Theorem 1.1

Suppose that 𝜒 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 that Γ 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 character 𝜒 is exact.

Proof. Suppose that ker𝜒≠1. Since the character 𝜒_𝐺=𝜒 is exact, we have ker𝜒⊆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 that 𝑞∈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]𝐶_𝑄(𝐴), we have 𝑄⊆𝐶. By Exercise 13.2 [5], 𝛽(1)||𝐺∶𝐶||𝜒(1)(3.1) 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|=𝜒(1)−𝜀𝛽(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 that 𝜀=1 in Lemma 2.2. Since 𝑛=2|𝐴|+1, we have 𝑘=2. This case does not satisfy.
Suppose that 𝜀=−1. Since 𝛽(1)≤𝜒(1)<3|𝐴|, we obtain 𝑘=𝜒(1)+𝛽(1)||𝐴||<6,(3.2) 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 that 𝑘=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 that 𝑘=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 that 𝑅 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 𝜒_𝐴𝑅=^𝑓_𝑖=1𝛼_𝑖𝜒_𝑖,(3.3) where 𝛼_𝑖∈𝐍,𝜒_𝑖∈Irr(𝐴𝑅), 𝑖=1,𝑓.
Let us note that by the minimality of the group Γ either 𝑓≠1, or 𝛼₁≠1, that is, 𝜒_𝑖(1)≠𝜒(1) for all 𝑖=1,𝑓.
By the Clifford Theorem 𝜒_𝑖_𝑅=𝑒^𝑖𝑡𝑖_𝑗=1𝜒_𝑖𝑗,(3.4) where 𝜒_𝑖𝑗∈Irr(𝑅), 𝑒_𝑖 are numbers, which divide |𝐴|, 𝑖=1,𝑓, 𝑗=1,𝑡_𝑖.
Suppose that 𝜒_𝑖0𝑗0∈Irr((𝜒)_𝑅). Since 𝑅⊲Γ, by the Clifford Theorem it follows that 𝜒_𝑖𝑗(1) divides 𝑛 for each irreducible component 𝜒_𝑖𝑗 of the character 𝜒_𝑅 and for corresponding elements 𝑥∈Γ⧵𝐼_Γ(𝜒_𝑖0𝑗0) the equality 𝜒_𝑖𝑗=(𝜒_𝑖0𝑗0)^𝑥 is true.
Suppose that 𝜒_𝑖0𝑗0(1)=1 for a certain character 𝜒_𝑖0𝑗0∈Irr((𝜒)_𝑅). Since 𝑅′⊆ker𝜒_𝑖0𝑗0 and 𝑅^⊲Γ, we see that 𝑅^⊆(ker𝜒_𝑖0𝑗0)^𝑥 for all 𝑥∈Γ⧵𝐼_Γ(𝜒_𝑖0𝑗0). Therefore, 𝑅^⊆∩_{𝑥∈Γ⧵𝐼Γ(𝜒𝑖0𝑗0)}(ker𝜒_𝑖0𝑗0)^𝑥=ker𝜒=1. Thus, the subgroup 𝑅 is Abelian.
Now suppose that 𝜒_𝑖𝑗(1)≠1 for all 𝑖=1,𝑓 and for all 𝑗=1,𝑡_𝑖. Since 𝑛 is an uneven number, we have 𝜒_𝑖𝑗(1)≥3. Therefore, if 𝐼_𝐴𝑅(𝜒_𝑖0𝑗0)=𝑅 for a certain character 𝜒_𝑖0𝑗0∈Irr((𝜒)_𝑅), then 𝜒_𝑖0(1)=(𝜒_𝑖0𝑗0)^𝐴𝑅(1)=|𝐴|𝜒_𝑖0𝑗0(1)≥3|𝐴|. Since in this case 𝜒(1)≥𝜒_𝑖0(1), we obtain a contradiction with 𝜒(1)=2|𝐴|+1<3|𝐴|. Thus, 𝐼_𝐴𝑅(𝜒_𝑖𝑗)≠𝑅 for all 𝑖=1,𝑓 and for all 𝑗=1,𝑡_𝑖. Then, by Lemma  1 [6], 𝐼_𝐴𝑅(𝜒_𝑖𝑗)=𝐴𝑅, therefore, 𝑡_𝑖=1 and 𝜒_𝑖(1)=𝑒_𝑖𝜒_𝑖𝑗(1) for all 𝑖=1,𝑓. By Lemma  13.3 [5], for every 𝑖=1,𝑓 there exists a canonical extension 𝜒_𝑖 of the character 𝜒_𝑖1 to the group 𝐴𝑅.
Assume that 𝐴ker𝜒_𝑖0/ker𝜒_𝑖0⋪𝐴𝑅/ker𝜒_𝑖0 for a certain 𝑖₀. By Lemma  9  [1], 𝐴∩ker𝜒_𝑖0=1. Then, 𝐴ker𝜒_𝑖0/ker𝜒_𝑖0≅𝐴/𝐴∩ker𝜒_𝑖0=𝐴. Let us apply the theorem [1] to the factor group 𝐴𝑅/ker𝜒_𝑖0 and its exact irreducible character in terms of Lemma  2.22 [5] (𝜒_𝑖0)^∗. By this theorem, (𝜒_𝑖0)^∗(1)∈{|𝐴|−1,|𝐴|,|𝐴|+1,2|𝐴|−2,2|𝐴|−1} and (𝜒_𝑖0)^∗(1) is a prime power, probably, with the exception of the case, when (𝜒_𝑖0)^∗(1)=|𝐴|. Since (𝜒_𝑖0)^∗(1)=𝜒_𝑖0(1) and 𝜒_𝑖(1) divides 𝑛 for all 𝑖=1,𝑓, it follows that 𝜒_𝑖0(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 that 𝐴ker𝜒_𝑖/ker𝜒_𝑖⊲𝐴𝑅/𝜒_𝑖 for all 𝑖=1,𝑓. Using the fact that ⋂^𝑓𝑖=1ker𝜒_𝑖∩𝑅=ker𝜒∩𝑅 and ker𝜒=1 by Lemma 3.1, then 𝐴₀=⋂^𝑓𝑖=1ker𝜒_𝑖⊆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=∏^𝑓𝑖=1(𝐴𝑅/ker𝜒_𝑖). Since the group 𝐴𝑅 contains a normal Hall 𝜋-subgroup, we have 𝐴⊲𝐴𝑅. Thus, 𝑅⊆𝐶.
The lemma is proved.

Suppose that 𝑞∈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 group Γ and Lemma 3.1, it follows that the character 𝜒_𝐴𝑀 is reducible for every such a subgroup 𝑀. Denote 𝑁=𝑁_Γ([𝑄,𝐴]).

Lemma 3.4. 𝑁≠Γ.

Proof. Assume that 𝑁=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 𝜒_𝐴𝑄. It is clear that 𝐴𝑄≠Γ and 𝐴⋪𝐴𝑄. Therefore, there exists at least one irreducible component 𝜉 of the character 𝜒_𝐴𝑄 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 character 𝜉₁∈Irr(𝜒_𝐴𝑄) such that 𝐴ker𝜉₁/ker𝜉₁⋪𝐴𝑄/ker𝜉₁. Since 𝜉₁(1)=|𝐴| and 𝜒_𝐴𝑄=2|𝐴|+1, we see that 𝜒_𝑄 is a sum of linear irreducible characters 𝜆. Thus, 𝑄^⊆_𝜆∈Irr(𝜒_𝑄)ker𝜆=ker𝜒=1.(3.5) This yields that the subgroup 𝑄 is Abelian. Therefore, 𝑄⊆𝐶_Γ([𝑄,𝐴])⊲Γ. Suppose that 𝐶_Γ([𝑄,𝐴])=Γ, that is, [𝑄,𝐴]⊆𝑍(Γ). Then, [𝑄,𝐴]⊆𝐶 is a contradiction with Lemma 3.2. Thus, 𝐶_Γ([𝑄,𝐴])≠Γ and since 𝐶_Γ([𝑄,𝐴])⊲Γ, we see that, by Lemma 3.3, the subgroup 𝐶_Γ([𝑄,𝐴]) is either Abelian or 𝐶 contains it. If 𝐶_Γ([𝑄,𝐴]) is Abelian, then 𝑄⊲Γ, and we obtain a contradiction with Theorem  6.15 [5], but if 𝐶_Γ([𝑄,𝐴])⊆𝐶, then we obtain a contradiction with Lemma 3.2.
Assume that 𝐴ker𝜓/ker𝜓⊲𝐴𝑄/ker𝜓, where 𝜒_𝐴𝑄=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⊆𝐶_Γ([𝑄,𝐴]). This case has been considered above.
The lemma is proved.

By Lemma 3.4, 𝑁≠Γ. For each 𝐴-invariant proper subgroup 𝑀 of the group 𝐺 is such that [𝑄,𝐴]⊆𝑀, we have 𝜒_𝐴𝑀=𝜏+𝜓,(3.6) 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 group 𝐴𝑀/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=𝑁_𝐴𝑀(𝐴)ker𝜓. Then, 𝑀=(ker𝜓)_𝜋′𝐶_𝑀(𝐴).(3.7) 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[𝑄,𝐴]⊆(ker𝜓)_𝜋′.(3.8) Assume that 𝜏=0. Then, 𝜒_𝐴𝑀=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], 𝜉(1)∈||𝐴||−1,||𝐴||,||𝐴||+1,2||𝐴||−2,2||𝐴||−1(3.9) 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 character 𝜏 does not contain any irreducible components of degree |𝐴|−1.

Proof. Assume that there exists 𝜉∈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, 𝜉_{𝐶𝑀(𝐴)}=(|𝐴|−1)𝛽_𝜉,𝛽_𝜉(1)=1, where 𝛽_𝜉=(𝜉_𝑀)𝜋(𝑀,𝐴). By Lemma  10 [1], 𝛼_𝜉=𝜉_𝐴,1_𝐴_𝐴=𝜌_𝐴−1_𝐴𝜆⁻¹,1_𝐴_𝐴(3.10) for a certain linear irreducible character 𝜆 of the subgroup 𝐴. We see that 𝛼_𝜉=0 if 𝜆=1_𝐴 and 𝛼_𝜉=1 if not.
In addition, assume that 𝜏(1)≠𝜉(1), that is, the character 𝜏 contains an irreducible component 𝜉₁.
Besides, assume that 𝜉₁=|𝐴|−1. In the same way, we establish that 𝛼_𝜉1=𝜉₁_𝐴,1_𝐴_𝐴=𝜌_𝐴−1_𝐴𝜆¹⁻¹,1_𝐴_𝐴(3.11) for a certain linear irreducible character 𝜆₁ of the subgroup 𝐴, moreover, 𝛼_𝜉1=0, if 𝜆₁=1_𝐴 and 𝛼_𝜉1=1 in the converse case.
Suppose that 𝜒_𝐴𝑀=𝜉+𝜉₁+𝜂 for a certain character 𝜂 of the subgroup 𝐴𝑀. Then, 𝜂(1)=𝜒(1)−𝜉(1)+𝜉₁(1)=2||𝐴||+1−2||𝐴||−2=3.(3.12) 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 𝐶_𝑀(𝐴), that is, 𝛿_{𝐶𝑀(𝐴)}=𝛽_𝛿. By Lemma  10 [1], we see that 𝛿̂_𝐴=𝛽_𝛿(1)1_𝐴. Then, it is obvious that 𝛼_𝜓=(𝜓_𝐴,1_𝐴)_𝐴=∑_{𝛿∈Irr(𝜓)}(𝛽_𝛿(1)𝜆^𝛿−1,1_𝐴)_𝐴. Here, 𝛿̂ is a canonic extension of the character 𝛿_𝑀 to the group 𝐴𝑀 and 𝜆_𝛿 are linear irreducible characters of the subgroup 𝐴. It is readily seen that 𝛼_𝜓≤𝜓(1)=3.
On the other hand, since 𝐶_𝑀(𝐴)⊆𝐶, we have 𝛽_𝜉∈Irr(𝛽_{𝐶𝑀(𝐴)}), 𝛽_𝜉1∈Irr(𝛽_{𝐶𝑀(𝐴)}), and (𝛽_{𝐶𝑀(𝐴)},𝜓_{𝐶𝑀(𝐴)})_{𝐶𝑀(𝐴)}≠0, where 𝛽 is from Lemma 2.4. Since by Theorem 13.1 [5], 𝛽_𝜉≠𝛽_𝜉1, we have 𝛽(1)≥3. Since 𝜒(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 𝜒_𝐶=𝛽. Suppose that assertion (1) of Lemma 2.4 satisfies. Then, by Lemma 2.2, 𝛼=𝜒_𝐴,1_𝐴_𝐴=𝜌_𝐴+||𝐴||+11_𝐴,1_𝐴_𝐴=||𝐴||+2.(3.13) Since 𝛼=𝛼_𝜉+𝛼_𝜉1+𝛼_𝜓≤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  𝜒_𝐴=𝛽(1)1_𝐴, that is, 𝐴⊆ker𝜒. We obtain a contradiction with Lemma 3.1.
Now assume that 𝜉₁=|𝐴|. Then, (𝜉₁)_𝐴=𝜌_𝐴 and 𝛼_𝜉1=1. Since 𝛽_𝜉∈Irr(𝛽_{𝐶𝑀(𝐴)}) and (𝛽_{𝐶𝑀(𝐴)},𝜓_{𝐶𝑀(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, 𝛼_𝜓≤2, we have 𝛼=𝛼_𝜉+𝛼_𝜉1+𝛼_𝜓≤4. Thus, |𝐴|+2≤4, that is, |𝐴|≤2. A contradiction.
There only remains to consider the case, when 𝜏=𝜉. Since ker𝜏∩ker𝜓=1, then the character 𝜉_ker𝜓 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=𝜉_𝐴ker𝜓 is also exact and from Lemma7 [10] and Lemma2.27 [5] it yields that |𝐴ker𝜓∶𝑍(𝜉̃)|=|𝐴ker𝜓∶𝑍(𝐴ker𝜓)|=|𝐴|2^2𝑟, 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. Let 𝑀 be a solvable 𝐴-invariant subgroup of 𝐺. Assume that 𝐴𝑀 has an exact irreducible character 𝛾 of degree |𝐴|+1=2^𝛽,𝛽∈𝐍. Then, |𝑀∶𝐶_𝑀(𝐴)|=2^𝑚, where 𝑚∈𝐙₊.

Proof. Since |𝜋|=1, we see that |𝐴|=𝑝^𝑓, 𝑓∈𝐍, therefore, 𝑝^𝑓+1=2^𝛽, 𝛽∈𝐍, that is, 𝑝^𝑓=2^𝛽−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 that 𝑟,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, 𝐶_𝑀(𝐻)=𝐶_𝑀(𝐵)=𝐶_𝑀(𝐴). Since 𝑀 is solvable, by Theorem  6 [9], it follows that [𝑀,𝐻]⊆𝐹(𝑀). Since, by Lemma  4 [1], 𝑀=[𝑀,𝐻]𝐶_𝑀(𝐻)=[𝑀,𝐻]𝐶_𝑀(𝐴), we obtain 𝑀=𝐹(𝑀)𝐶_𝑀(𝐴). Assume that 𝐹=𝐹(𝑀). Let us consider the character 𝛾_𝐴𝐹=^𝑡_𝑖=1𝛼_𝑖𝛾_𝑖,(3.14)𝛾_𝑖∈Irr(𝐴𝐹), 𝛼_𝑖>0, 𝑖=1,2,…,𝑡.
Suppose that 𝐴⊲𝐴𝐹. Then, 𝐹⊆𝐶_𝑀(𝐴). Thus, 𝑀⊆𝐶_𝑀(𝐴), therefore, |𝑀∶𝐶_𝑀(𝐴)|=2⁰=1. In this case, the lemma is true.
Suppose that 𝐴⋪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=2^𝛼, |𝐴| or |𝐴|+1=2^𝛽. Here 𝛼,𝛽∈𝐍.
If 𝛾₁(1)=|𝐴|−1=2^𝛼, then |𝐴|=2^𝛼+1=2^𝛽−1. Hence, 2^𝛽=2^𝛼+2=2(2^𝛼−1+1), that is, 𝛼=1. Thus, |𝐴|=3. This case is impossible.
Suppose that 𝛾₁(1)=|𝐴|. Then, 𝛾(1)=𝛾(1)−𝛾₁(1)=||𝐴||+1−||𝐴||=1,(3.15) where 𝛾=𝛾_𝐴𝐹−𝛾₁. Therefore, all irreducible components of the character 𝛾_𝐹 are linear, that is, the subgroup 𝐹 is Abelian. Since 𝐹⊲𝐴𝑀, we have 𝐶_𝐴𝑀(𝐹)⊲𝐴𝑀. Then, it is not hard to see that 𝐶^∗=𝐶_𝑀(𝐹)⊲𝐴𝑀. Since 𝐹⊈𝐶_𝑀(𝐴) and 𝐹⊆𝐶^∗, it follows that 𝐶^∗⊈𝐶_𝑀(𝐴), that is, 𝐴⋪𝐴𝐶^∗.
Arguing as above for the subgroup 𝐴𝐹, we can show that in the contingency 𝛾_𝐴𝐶^∗ there is an irreducible component 𝛾₁ such that 𝐴ker𝛾₁⋪𝐴𝐶^∗ and 𝛾₁(1)=|𝐴| or |𝐴|+1=2^𝛽. Assume that 𝛾₁(1)=|𝐴|+1=2^𝛽. Since the subgroup 𝐴 is a 𝑇𝐼-set in 𝐴𝐶^∗, from Lemma  1 [6], it yields that 𝐴⊆𝐼_𝐴𝐶^∗(𝛾_1∗), where 𝛾_1∗ is an irreducible component of the character (𝛾₁)_𝐶^∗. Thus, (𝛾₁)_𝐶^∗ is an irreducible character of the subgroup 𝐶^∗. Since 𝐹⊆𝑍(𝐶^∗), we have 𝐹⊆𝑍(𝛾₁). However, since the character 𝛾 is exact, the character 𝛾₁ is exact. Therefore, 𝑍(𝛾₁)=𝑍(𝐴𝐶^∗). Then, 𝐹⊆𝑍(𝐴𝐶^∗), that is, 𝐹⊆𝐶_𝑀(𝐴). This case has already been examined. Therefore, we may assume that 𝛾₁(1)=|𝐴|. Again, we see that 𝐶^∗ is Abelian, that is, 𝐶^∗⊆𝐹. Therefore, 𝐹=𝐶^∗.
Suppose that 𝜆∈Irr(𝛾_𝐹) and 𝐼=𝐼_𝐴𝑀(𝜆). Assume that 𝐼=𝐴𝑀. Since the subgroup 𝐹 is Abelian, we obtain 𝜆(1)=1. Therefore, 𝛾_𝐹=𝛾(1)𝜆, that is, 𝐹⊆𝑍(𝛾)=𝑍(𝐴𝑀), that is, 𝐹⊆𝐶_𝑀(𝐴). The lemma is true in this case.
Now suppose that 𝐼≠𝐴𝑀. By Theorem  6.11 [5], there exists an irreducible character 𝜆^ of the subgroup 𝐼 such that (𝜆^)^𝐴𝑀=𝛾. Hence, 𝛾(1)=2^𝛽=|𝐴𝑀∶𝐼|𝜆^(1). Thus, |𝐴𝑀∶𝐼| and 𝜆^(1) is degree 2. Therefore, 𝐴^𝑥⊆𝐼 for a certain element 𝑥∈𝐴𝑀. Assume that 𝐴^𝑥⊲𝐼. Then, 𝐹⊆𝐶_𝑀(𝐴^𝑥). Hence, 𝐹=𝐹^𝑥−1⊆𝐶_𝑀(𝐴). The lemma is true.
Therefore, 𝐴^𝑥⋪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 1=𝛾,𝜆^^𝐴𝑀_𝐴𝑀=𝛾_𝐼,𝜆^_𝐼,(3.16) we have 𝛾_𝐼=𝜆^+𝜆^. Since 𝜆^(1)>1, we obtain 𝜆^(1)=𝛾(1)−𝜆^(1)=||𝐴||+1−𝜆^(1)≤||𝐴||−1.(3.17) Since |𝐴|≠3, by Theorems  1 and 2 [10], we see that 𝐴^𝑥ker𝜆^⊲𝐼. By using the fact that 𝐴^𝑥⋪𝐼, it follows that 𝐴^𝑥ker𝜆^⋪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 that 𝜆^(1)=1. Then, 𝜆^(1)=|𝐴|+1−1=|𝐴|. Since 𝐴^𝑥ker𝜆^⊲𝐼, 𝐴^𝑥⋪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 𝛾_𝐼𝜋′ are linear, that is, the subgroup 𝐼_𝜋′ is Abelian. Since 𝐹=𝐶^∗ is Abelian and 𝐹⊆𝐼_𝜋′, we have 𝐹=𝐼_𝜋′. Then, 𝐼=𝐴^𝑥𝐼_𝜋′=𝐴^𝑥𝐹. Since 𝑀=𝑀^𝑥, we have 𝑀=𝐹𝐶_𝑀(𝐴)=𝐹𝐶_𝑀(𝐴)^𝑥=𝐹^𝑥𝐶_𝑀(𝐴)^𝑥=𝐹𝐶_𝑀(𝐴)^𝑥=𝐹𝐶_𝑀(𝐴^𝑥).(3.18) Further, 𝐴𝑀=(𝐴𝑀)^𝑥=𝐴^𝑥𝑀^𝑥=𝐴^𝑥𝑀=𝐴^𝑥𝐹𝐶_𝑀(𝐴^𝑥)=𝐼𝐶_𝑀(𝐴^𝑥).(3.19) Hence, for each 𝑔∈𝐴𝑀, we obtain that 𝑔=𝑎𝑏, where 𝑎∈𝐼 and 𝑏∈𝐶_𝑀(𝐴^𝑥). Since 𝛾=(𝜆^)^𝐴𝑀, for each ℎ∈𝐴^𝑥 the following expression satisfies 𝛾(ℎ)=1||𝐼||_{𝑔∈𝐴𝑀}𝜆^⁰𝑔ℎ𝑔⁻¹=||𝐼1||_{𝑔∈𝐴𝑀}𝜆^⁰𝑎𝑏ℎ𝑏⁻¹=1𝑎⁻¹||𝐼||_{𝑔∈𝐴𝑀}𝜆^⁰𝑎ℎ^𝑏−1𝑎⁻¹=||𝐼1||_{𝑔∈𝐴𝑀}𝜆^(ℎ)=||𝐴𝑀∶𝐼||𝜆^(ℎ).(3.20) 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 ||𝛾(ℎ)||=||||𝐴𝑀∶𝐼||⋅𝜆^(ℎ)||=||𝐴𝑀∶𝐼||||𝜆^(ℎ)||=||𝐴𝑀∶𝐼||.(3.21)
Since 𝛾(1)=|𝐴𝑀∶𝐼|𝜆^(1), we have 𝛾(1)=|𝐴𝑀∶𝐼|. Then, |𝛾(ℎ)|=|𝐴𝑀∶𝐼|=𝛾(1). By Lemma  2.27(e) [5], 𝐴^𝑥⊆𝑍(𝛾)=𝑍(𝐴𝑀). Hence, 𝐴^𝑥=𝐴⊲𝐴𝑀, that is 𝑀⊆𝐶_𝑀(𝐴), therefore, |𝑀∶𝐶_𝑀(𝐴)|=1. The lemma is true.
There only remains to consider the case, when 𝛾₁(1)=|𝐴|+1=2^𝛽. Since the subgroup 𝐴 is a 𝑇𝐼-set in 𝐴𝐹, from Lemma  1 [6], it follows that 𝐴⊆𝐼_𝐴𝐹(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⊆𝐶_𝑀(𝐴). Then, 𝑀=𝐹𝐶_𝑀(𝐴)=𝐹₂×𝑂𝐶_𝑀(𝐴)=𝐹₂𝐶_𝑀(𝐴).(3.22) Therefore, ||𝑀∶𝐶_𝑀(𝐴)||=||𝐹₂𝐶_𝑀(𝐴)∶𝐶_𝑀(𝐴)||=||𝐹₂∶𝐹₂∩𝐶_𝑀(𝐴)||=2^𝑚,(3.23) where 𝑚>0.
The lemma is proved.

Lemma 3.7. 𝜓≠0.

Proof. Let us assume the converse, that is, 𝜓=0. Then, (𝜒)_𝐴𝑀=𝜏. Since, by Lemma 3.5, 𝜉(1)≠|𝐴|−1 for 𝜉∈Irr(𝜏), we have 𝜉(1)∈||𝐴||,||𝐴||+1,2||𝐴||−2,2||𝐴||−1,2||𝐴||.(3.24) Suppose that 𝜉(1)∈{2|𝐴|−2,2|𝐴|−1,2|𝐴|}. Then, 𝜒(1)−𝜉(1)=2|𝐴|+1−𝜉(1)≤3. As mentioned above |𝐴|≥7. By Theorem  1 [10], 𝐴ker(𝜒−𝜉)/ker(𝜒−𝜉)⊲𝐴𝑀/ker(𝜒−𝜉). Thus, Irr(𝜒−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 𝜏=𝜉+𝜉₁,(3.25) where 𝜉(1)=|𝐴| and 𝜉₁(1)=|𝐴|+1 is degree 2.
From the Clifford Theorem, it follows that 𝑀^⊆ker𝜉. Since ker𝜉∩ker𝜉₁=ker𝜏=ker𝜒_𝐴𝑀=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 Lemma 3.6, ||𝑀∶𝐶_𝑀(𝐴)||=2^𝑚,𝑚∈𝐙₊,(3.26) where 𝑀=𝑀/𝐾. Here, 𝐾=(ker𝜉₁)_𝜋′. Since (|𝐾|,|𝐴|)=1 and 𝐶=𝐶_𝑀(𝑎)=𝐶_𝑀(𝐴) for all elements 𝑎∈𝐴^#, we have |𝐶_𝑀/𝐾(𝐴)|=|𝐶_𝑀(𝐴)∶𝐶_𝐾(𝐴)|. Thus, ||𝑀∶𝐶_𝑀(𝐴)||=||𝑀||||𝐶^𝐾(𝐴)||||𝐶_𝑀(𝐴)||||𝐾||=2^𝑚,𝑚∈𝐙₊.(3.27)
Since [𝑄,𝐴]⊆[𝑀,𝐴] and [𝑄,𝐴]⊈𝐶, we have 𝐹_𝑞≠1. Since 𝐹_𝑠char𝐹⊲𝐴𝑀, it follows that 𝐹_𝑠⊲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 𝐹_𝑞 is Abelian. Since [𝑄,𝐴]⊆𝐹_𝑞, 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]∩𝐶_𝑀(𝐴)=1. Hence and by the equality (3.27), it follows that [𝑄,𝐴]⊆𝐾.
Further, we see that 𝜏_𝐾=𝜏_𝑀_𝐾=𝜉_𝑀+𝜉₁_𝑀_𝐾=_{𝜆∈Irr(𝜉𝑀)}𝜆_𝐾+𝜉₁(1)1_𝐾,𝜆(1)=1.(3.28) Since 𝐾⊆𝑍(𝜉₁), it follows that 𝐾⊆(⋂_{𝜆∈Irr(𝜉𝑀)}𝑍(𝜆))∩𝑍(𝜉₁). By the fact that the character 𝜏_𝑀 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 character 𝜒_𝐿. Clearly, it is enough to consider the case, when 𝐴ker𝜉^/ker𝜉^⋪𝐿/ker𝜉^ for a certain character 𝜉^∈Irr(𝜒_𝐿). Then, 𝜉∈Irr(𝜉^𝐴𝑀) or 𝜉₁∈Irr(𝜉^𝐴𝑀). Thus, we have 𝜉^(1)≥|𝐴|.
Assume that there exists a character 𝜓^ of the group 𝐿 such that Irr(𝜓^)⊆Irr(𝜒_𝐿) and 𝐴ker𝜓^/ker𝜓^⊲𝐿/ker𝜓^. Then, 𝐴ker𝜓^⊲𝐿, therefore, (𝐴ker𝜓^)∩𝐴𝑀⊲𝐴𝑀. Hence, 𝐴ker𝜓^𝐴𝑀⊲𝐴𝑀. Thus, we have 𝜓^𝐴𝑀=𝜓≠0. A contradiction with the assumption.
Thus, we have 𝜒_𝐿=𝜉^+𝜉^1 and can say that 𝜉^𝐴𝑀=𝜉 and (𝜉^1)_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 𝜉^1. We obtain [𝐿_𝜋′,𝐴]⊆𝐹(𝐿_𝜋′) and [𝑄,𝐴]⊆𝐾_𝐿.(3.29) Here, we have 𝐾_𝐿=(ker𝜉^1)_𝜋′.
Suppose that 𝑞₁∈𝜋([𝐿_𝜋′,𝐴]),𝑞₁≠𝑞 and 𝑄₁ is an invariant Sylow 𝑞₁-subgroup [𝐿_𝜋′,𝐴]. Then, 𝑄₁⊲𝐿. Let 𝜉_𝑞1∈Irr((𝜉^1)_𝑄1). By Theorem  6.11 [5], we have 𝜉′₁=𝜉̃^𝐿𝑞₁ for a certain irreducible character 𝜉̃_𝑞1 of the group 𝐼_𝐿(𝜉_𝑞1). Thus, 𝜉^1(1)=𝜉̃^𝐿𝑞₁(1)=𝜉̃_𝑞1(1)|𝐿∶𝐼_𝐿(𝜉_𝑞1)|. Since 𝜉^1(1) is degree 2, we see that the numbers 𝜉̃_𝑞1(1) and |𝐿∶𝐼_𝐿(𝜉_𝑞1)| are also degrees 2.
Assume that (𝜉^𝑄₁,𝜉_𝑞1)_𝑄1≠0. By the same theorem, we have 𝜉^=(𝜉^𝑞₁)^𝐿 for a certain irreducible character 𝜉^𝑞₁ of the group 𝐼_𝐿(𝜉_𝑞1). Then we have |𝐴|=𝜉^(1)=𝜉^𝑞₁(1)|𝐿∶𝐼_𝐿(𝜉_𝑞1)|. Since |𝐿∶𝐼_𝐿(𝜉_𝑞1)| is degree 2, we obtain 𝐿=𝐼_𝐿(𝜉_𝑞1). Then, 𝜉^𝑄₁=𝑒^𝜉_𝑞1(1) for a certain natural number 𝑒^. Thus, we have |𝐴|=𝑒^𝜉_𝑞1(1). Since 𝜉_𝑞1(1) is a 𝜋^-number, we have 𝜉_𝑞1(1)=1. Then, 𝑄₁⊆𝑍(𝜉^) and, obviously, 𝑄₁⊆𝑍(𝜉^1). By Lemma  5 [1], 𝑄₁⊆𝑍(𝐿). This means that 𝑄₁⊆𝐶_𝐿𝜋′(𝐴).
Now assume that (𝜉^𝑄₁,𝜉_𝑞1)_𝑄1=0. By Theorem  D [13], 𝑡=||𝑄¹||||Γ∶𝐿||𝜉₁(1)𝜒(1)𝜉_𝑞1(1)(3.30) 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 𝜒(1), it follows that 𝑡 is not an integer. We obtain a contradiction. Thus, we have 𝑄₁=1, therefore, [𝐿_𝜋′,𝐴] is a 𝑞-group.
It is obvious that [𝑄,𝐴]⊆[𝐿_𝜋′,𝐴]. Since [𝐿_𝜋′,𝐴]⊲𝐿, we obtain [𝐿_𝜋′,𝐴]⊆𝑄. Then, we have [[𝐿_𝜋′,𝐴],𝐴]⊆[𝑄,𝐴]. Since [[𝐿_𝜋′,𝐴],𝐴]=[𝐿_𝜋′,𝐴], it follows that [𝐿_𝜋′,𝐴]⊆[𝑄,𝐴]. Thus, we have [𝐿_𝜋′,𝐴]=[𝑄,𝐴]. Then, [𝐿_𝜋′,𝐴]⊆𝐾_𝐿. Hence, 𝐴𝐾_𝐿/𝐾_𝐿⊲𝐿/𝐾_𝐿. Therefore, we have 𝐴ker𝜉^1/ker𝜉^1⊲𝐿/ker𝜉^1. Thus, 𝜉^1⊆Irr(𝜓), that is, 𝜓≠0. This contradicts the original assumption.
The lemma is proved.

Lemma 3.8. The character 𝜇=𝜏_𝐴ker𝜓 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 Lemma 3.7, we have 𝜓≠0.
As stated above, the character 𝜒_𝐴𝑀 cannot be irreducible.
Let 𝜓(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], 𝜒(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 𝜒_𝐶 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⊆ker𝜒. Since the character 𝜒 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 obtain 𝜓(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 𝜏_ker𝜓 is exact. Suppose that the character 𝜏_𝐴ker𝜓 is not exact. Then, we have ker_{𝜏𝐴ker𝜓}⊆𝐴. Since 𝐴 is a 𝑇𝐼-subgroup, it follows that 𝐴⊲𝐴ker𝜓. Thus we have (ker𝜓)_𝜋′⊆𝐶_𝑀(𝐴). From formula (3.8), it follows that [𝑄,𝐴]⊆C. A contradiction with the choice of the subgroup 𝑀. Thus, the character 𝜏_𝐴ker𝜓 is exact.
Since (ker𝜓)_𝜋′⊈𝐶_𝑀(𝐴), we have 𝐴ker𝜇/ker𝜇⋪𝐴ker𝜓/ker𝜇 for a certain character 𝜇⊆Irr(𝜏_𝐴ker𝜓). By Theorem [1], we have 𝜇(1)∈||𝐴||−1,||𝐴||,||𝐴||+1,2||𝐴||−2,2||𝐴||−1(3.31) 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. Let 𝜇(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 𝜏_𝐴ker𝜓=𝜇+𝜇^𝑔, where 𝑔∈𝐴𝑀⧵𝐴ker𝜓. Hence, by Lemma  7 [10], we have ||ker𝜓)_𝜋′∶𝑍(𝜇)||=||ker𝜓)_𝜋′∶𝑍(𝜇^𝑔)||=2^2𝛼,(3.32) where 𝜇(1)=2^𝛼, 𝛼∈𝑁. Since [𝑄,𝐴]⊆(ker𝜓)_𝜋′, we see that [𝑄,𝐴]⊆𝑍(𝜇)∩𝑍(𝜇^𝑔). Since ker𝜇∩ker𝜇^𝑔=1, by Lemma  5 [1], we have [𝑄,𝐴]⊆𝑍(𝐴ker𝜓). Then [𝑄,𝐴]⊆𝐶. A contradiction with Lemma 3.2. Thus, 𝜇(1)∈||𝐴||,||𝐴||+1,2||𝐴||−2,2||𝐴||−1(3.33) 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 𝜏_𝐴ker𝜓 is exact, the lemma is proved.

Lemma 3.9. 𝑛₁=|𝐴|.

Proof. By Lemma 3.8, it is enough to show that 𝑛₁≠|𝐴+1|,2|𝐴|−2 and 2|𝐴|−1.
First, let 𝑛₁=|𝐴|+1. By Theorem [12], the group 𝐴ker𝜓 is solvable. Then by Lemma 3.6, we have |(ker𝜓)_𝜋′∶𝐶_{(ker𝜓)𝜋′}(𝐴)|=2^𝑚, where 𝑚∈𝐙₊. From equality (3.8), it follows that [𝑄,𝐴]⊆𝐶_{(ker𝜓)𝜋′}(𝐴)⊆𝐶_𝑀(𝐴). We obtain a contradiction with the choice of the subgroup 𝑀.
Now suppose that 𝑛₁=2|𝐴|−2 or 2|𝐴|−1. By Lemma 3.8, 𝑛₁ is a prime power 𝑠.
Let 𝑆 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)⊆𝐶_𝑍(𝐴). Since 𝑍 is Abelian, it follows that 𝑍=[𝑍,𝐴]×𝐶_𝑍(𝐴). It can be easily seen that |[𝑍,𝐴]| divides |𝑍/𝑍∩𝑍(𝜇)|.
Assume that