International Scholarly Research Notices

International Scholarly Research Notices / 2011 / Article

Research Article | Open Access

Volume 2011 |Article ID 868096 | 20 pages |

On Irreducible Linear Groups of Nonprimary Degree

Academic Editor: A. Vourdas
Received06 May 2011
Accepted05 Jun 2011
Published17 Aug 2011


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œ‹-solvable irreducible complex linear group 𝐺 with a πœ‹-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 Ξ“=𝐺𝐴 of the group 𝐺 and the group 𝐴 is a group. If 𝐢𝐺(π‘Ž)=𝐢𝐺(𝐴) for each element π‘Žβˆˆπ΄#, 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|βˆ’1,|𝐴|+1,2(|𝐴|βˆ’1) or 2|𝐴|βˆ’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œ‹-solvable irreducible complex linear group 𝐺 of degree 𝑛=2|𝐻|+1, which is not a prime power, the πœ‹-Hall 𝑇𝐼-subgroup 𝐻 is Abelian and normal in 𝐺 (Theorem 1.2).

Condition B
Let us say that the group Ξ“=𝐴𝐺 satisfies B, if πΊβŠ²Ξ“,(|𝐴|,|𝐺|)=1,|𝐴| is uneven, 𝐢𝐺(π‘Ž)=𝐢𝐺(𝐴)=𝐢 for each element π‘Žβˆˆπ΄#, and the group 𝐺 has an extract irreducible complex character of degree 𝑛=2|𝐴|+1, which is π‘Ž-invariant for at least one element π‘Žβˆˆπ΄#.

Theorem 1.1. If the group Ξ“ satisfies Condition B, then 𝑛 is a prime power degree.

Condition D
𝐺 is a πœ‹-solvable irreducible complex linear group of degree 𝑛=2|𝐻|+1 with a nonprimary πœ‹-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,𝑑 stands for 𝑖=1,2,…,𝑑; if πœ“ is the character of a certain group, then Irr(πœ“) is the set of all irreducible components of the character πœ“; 𝑍(πœ“)={π‘”βˆˆπΊβˆ£|πœ“(𝑔)|=πœ“(1)} is the center of the character πœ“;πœ‹=πœ‹(𝐴); if π‘€βŠ†πΊ and πœ‹ is the set of primes, then πœ‹ξ…ž=πœ‹(𝑀)β§΅πœ‹;π‘€πœ‹ is a Hall πœ‹-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 Ξ“=𝐴𝐡 is the group, where π΅βŠ²Ξ“,(|𝐴|,|𝐡|)=1 and |𝐴| is uneven. Then, it satisfies Theorem  13.1 [5]. According to this theorem, there exists one-to-one correspondence πœ‹(𝐡,𝐴)∢Irr𝐴(𝐡)β†’Irr(𝐢𝐡(𝐴)) between the set of all 𝐴-invariant characters of the group 𝐡 and the set of all irreducible characters of the subgroup 𝐢𝐡(𝐴). The set has a range of properties, depending, in particular, on the properties of the subgroup 𝐴. Suppose that πœ’βˆˆIrr𝐴(𝐡). Then, by Lemma  13.3 [5], there is a unique irreducible character ξπœ’ of the group Ξ“ such that ξπœ’π΅=πœ’ and π΄βŠ†ker(detξπœ’). It is called a canonical extension of the character πœ’ to the group Ξ“.

Lemma 2.1 (see [6, Lemma  2]). If πœ’ is an irreducible character of the group 𝐺 stated in Condition B, then πœ’βˆˆIrr𝐴(𝐺).

Lemma 2.2 (see [1, Lemma  10]). If πœ’ is an irreducible character of the group 𝐺 stated in Condition B, then ξπœ’π΄=π‘˜πœŒπ΄+πœ€π›½(1)1𝐴, where π‘˜βˆˆπ™+,𝜌𝐴 is a regular character of the subgroup 𝐴,πœ€=Β±1 and 𝛽 is an irreducible character of the subgroup 𝐢 such that 𝛽=(πœ’)πœ‹(𝐺,𝐴).

Lemma 2.3. If πœ’ is an irreducible character of the group 𝐺 stated in Condition B, then πœ’πΆ=|𝐴|πœƒ+πœ€π›½, where πœƒ is a character of the subgroup 𝐢 or 0 and 𝛽=(πœ’)πœ‹(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 πœ’ 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)πœ’πΆ=𝛽+|𝐴|𝛽1,|𝐴|<𝛽(1)<2|𝐴|,𝛽1(1)=1;(2)πœ’πΆ=𝛽+|𝐴|𝛽1,𝛽(1)<|𝐴|,𝛽1(1)=2;(3)πœ’πΆ=𝛽+2|𝐴|𝛽1,𝛽(1)<|𝐴|,𝛽1(1)=1;(4)πœ’πΆ=𝛽+|𝐴|(𝛽1+𝛽2),𝛽(1)<|𝐴|,𝛽1(1)=𝛽2(1)=1;(5)πœ’πΆ=(|𝐴|+1)𝛽+|𝐴|𝛽1,𝛽(1)=𝛽1(1)=1;(6)πœ’πΆ=(|𝐴|+1)𝛽,𝛽(1)=2;(7)πœ’πΆ=(|𝐴|βˆ’1)𝛽+2|𝐴|𝛽1,𝛽(1)=𝛽1(1)=1;(8)πœ’πΆ=(|𝐴|βˆ’1)𝛽+|𝐴|𝛽1,𝛽(1)=1,𝛽1(1)=2;(9)πœ’πΆ=(|𝐴|βˆ’1)𝛽+|𝐴|(𝛽1+𝛽2),𝛽(1)=𝛽1(1)=𝛽2(1)=1;(10)πœ’πΆ=(|𝐴|βˆ’1)𝛽+|𝐴|𝛽1,𝛽(1)=2,𝛽1(1)=1;(11)πœ’πΆ=(|𝐴|βˆ’1)𝛽,𝛽(1)=3;(12)πœ’πΆ=(2|𝐴|+1)𝛽,𝛽(1)=1;(13)πœ’πΆ=(2|𝐴|βˆ’1)𝛽+|𝐴|𝛽1,𝛽(1)=𝛽1(1)=1;(14)πœ’πΆ=(3|𝐴|βˆ’1)𝛽,𝛽(1)=1;(15)πœ’πΆ=𝛽.
Here, 𝛽=(πœ’)πœ‹(𝐺,𝐴), 𝛽1 and 𝛽2 are distinct irreducible characters of the subgroup 𝐢.

Proof. From Lemma 2.3, it follows that ξ€·πœ’πΆξ€Έ,𝛽𝐢=π‘˜π›½||𝐴||+πœ€,(2.1) where π‘˜π›½=(πœƒ,𝛽)𝐢. Suppose that 𝛽1∈Irr(πœ’πΆ),𝛽1≠𝛽,π‘˜1=(πœ’πΆ,𝛽1)𝐢 and πœƒ1 is the character 𝐢 or 0 such that πœƒ=π‘˜π›½π›½+π‘˜1𝛽1+πœƒ1.(2.2) Hence, by Lemma 2.3, it follows that πœ’πΆ=||𝐴||ξ€·π‘˜π›½π›½+π‘˜1𝛽1+πœƒ1ξ€Έ+πœ€π›½.(2.3) Hence, by the assumption of the lemma, it follows that ||𝐴||π‘˜π›½||𝐴||𝛽(1)+πœ€π›½(1)β‰€πœ’(1)<3.(2.4) Then, |𝐴|π‘˜π›½π›½(1)<3|𝐴|βˆ’πœ€π›½(1), that is, π‘˜π›½<3π›½βˆ’πœ€(1)||𝐴||.(2.5) Hence, π‘˜π›½β‰€3.
Assume that πœƒβ‰ 0 and examine all the possible cases separately.
(1) π‘˜π›½=0.
Since, by Corollary  2.17 [5], (πœ’πΆ,𝛽)πΆβˆˆπ™+, by equality (2.1), it follows that πœ€=1. Since πœ’(1)<3|𝐴|, from Lemma 2.3, it follows that πœƒ(1)≀2, that is, π‘˜1𝛽1(1)+πœƒ1(1)≀2.
Suppose that πœƒ(1)=1. Then, we can assume that πœƒ1=0 and π‘˜1𝛽1(1)=1. From equality (2.3), we obtain πœ’πΆ=𝛽+|𝐴|𝛽1,|𝐴|<𝛽(1)<2|𝐴|,𝛽1(1)=1 is assertion (1) of the lemma being proved.
Suppose that πœƒ(1)=2. Then, π‘˜1𝛽1(1)+πœƒ1(1)=2.
If πœƒ1=0, then either π‘˜1=1 and 𝛽1(1)=2, which implies assertion (2) of the lemma, or π‘˜1=2 and 𝛽1(1)=1, which implies assertion (3) of the lemma.
But if πœƒ1β‰ 0, then it is clear that π‘˜1𝛽1(1)=1 and πœƒ1=𝛽2∈Irr(πœƒ),𝛽1≠𝛽2,𝛽2(1)=1,𝛽(1)<|𝐴|. We obtain assertion (4) of the lemma.
(2) π‘˜π›½=1.
Suppose that πœ€=1. From (2.3), we obtain πœ’πΆ=|𝐴|(𝛽+π‘˜1𝛽1+πœƒ1)+𝛽. Since 2|𝐴|<πœ’(1)<3|𝐴|, from the latter equality, it is readily seen that 𝛽(1)≀2.
Suppose that 𝛽(1)=1. Then, π‘˜1𝛽1(1)=1 and πœ’πΆ=(|𝐴|+1)𝛽+|𝐴|𝛽1,𝛽(1)=𝛽1(1)=1 is assertion (5).
Now, suppose that 𝛽(1)=2. Since 2|𝐴|<πœ’(1)<3|𝐴|, we have π‘˜1𝛽1(1)+πœƒ1(1)=0 and, therefore, πœ’πΆ=(|𝐴|+1)𝛽 is assertion (6).
Further, suppose that πœ€=βˆ’1. By (2.3), it follows that πœ’πΆξ€·||𝐴||𝛽||𝐴||ξ€·π‘˜(1)=βˆ’1(1)+1𝛽1(1)+πœƒ1ξ€Έ(1).(2.6) Since πœ’(1)<3|𝐴|, we see that 𝛽(1)≀3.
Let 𝛽(1)=1. Then, π‘˜1𝛽1(1)+πœƒ1(1)=2.
Let πœƒ1=0. Then, π‘˜1𝛽1(1)=2. If π‘˜1=2, then 𝛽1(1)=1. Hence, πœ’πΆ=(|𝐴|βˆ’1)𝛽+2|𝐴|𝛽1 is assertion (7). If π‘˜1=1, then 𝛽1(1)=2. Then πœ’πΆ=(|𝐴|βˆ’1)𝛽+|𝐴|𝛽1 is assertion (8).
Now suppose that πœƒ1(1)=1. Then, π‘˜1𝛽1(1)=1. We obtain πœ’πΆ=(|𝐴|βˆ’1)𝛽+|𝐴|(𝛽1+𝛽2), where 𝛽1(1)=𝛽2(1)=1 is assertion (9).
If πœƒ1(1)=2, then π‘˜1𝛽1(1)=0 and it is readily seen that (7), (8) or (9) follow.
Now suppose that 𝛽(1)=2. Hence, πœ’πΆ=(|𝐴|βˆ’1)𝛽+|𝐴|𝛽1,𝛽1(1)=1 is assertion (10).
Suppose that 𝛽(1)=3. Then, π‘˜1𝛽1(1)+πœƒ1(1)=0. Assertion (11).(3) π‘˜π›½=2.
Let πœ€=1. Expression (2.3) implies that π‘˜1𝛽1(1)+πœƒ1(1)=0. We obtain πœ’πΆ=(2|𝐴|+1)𝛽. We see that 𝛽(1)=1 is assertion (12).
Let πœ€=βˆ’1. Then, πœ’πΆ=(2|𝐴|βˆ’1)𝛽+|𝐴|𝛽1, where 𝛽(1)=𝛽1(1)=1 is assertion (13).
(4) π‘˜π›½=3.
Expression (2.3) implies that πœ€=βˆ’1,πœ’πΆ=(3|𝐴|βˆ’1)𝛽, where 𝛽(1)=1 is assertion (14).
Now suppose that in Lemma 2.3β€‰β€‰πœƒ=0. Then, πœ€=1 and πœ’πΆ=𝛽 is assertion (15).
The lemma is proved.

3. Proof of Theorem 1.1

Suppose that πœ’ is an exact irreducible character of degree 𝑛=2|𝐴|+1 of the group 𝐺 given in Condition B, π‘Ž-invariant for at least one element π‘Žβˆˆπ΄#.

Suppose that Ξ“ is a group of the least order that satisfies the assumption of the theorem, but does not satisfy its corollary, that is, |πœ‹(𝑛)|>1.

Lemma 3.1. The character ξπœ’ is exact.

Proof. Suppose that kerξπœ’β‰ 1. Since the character ξπœ’πΊ=πœ’ is exact, we have kerξπœ’βŠ†π΄. Since, by Lemma  1 [7], 𝐴 is a 𝑇𝐼-subgroup in Ξ“, it follows that π΄βŠ²Ξ“. 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 π‘žβˆˆπœ‹(𝑛), 𝑄 is an 𝐴-invariant Sylow π‘ž-subgroup 𝐺. Then, ΜΈ[𝑄,𝐴]βŠ†πΆ.

Proof. Suppose that [𝑄,𝐴]βŠ†πΆ. Since, by Lemma  4 [1], 𝑄=[𝑄,𝐴]𝐢𝑄(𝐴), we have π‘„βŠ†πΆ. By Exercise 13.2 [5], ||||𝛽(1)πΊβˆΆπΆπœ’(1)(3.1) is an integer, where 𝛽 is from Lemma 2.2. Since π‘„βŠ†πΆ, we see that π‘ž does not divide |𝐺∢𝐢|. Then π‘ž divides 𝛽(1). From Lemma 2.2, it follows that π‘˜|𝐴|=ξπœ’(1)βˆ’πœ€π›½(1). Hence, π‘ž divides π‘˜|𝐴|. Since π‘ž divides |𝐺|, it follows that π‘ž does not divide |𝐴|, therefore, π‘ž divides π‘˜. By using the fact that π‘ž is an odd prime, it follows that π‘˜β‰₯3.
Suppose that πœ€=1 in Lemma 2.2. Since 𝑛=2|𝐴|+1, we have π‘˜=2. This case does not satisfy.
Suppose that πœ€=βˆ’1. Since 𝛽(1)β‰€πœ’(1)<3|𝐴|, we obtain π‘˜=ξπœ’(1)+𝛽(1)||𝐴||<6,(3.2) that is, π‘˜β‰€5. Since π‘ž divides π‘˜ and π‘ž is an uneven number, it follows that it is enough to consider the cases when π‘˜=3 or π‘˜=5.
Suppose that π‘˜=3. Then, 3|𝐴|=(2|𝐴|+1)+𝛽(1). Hence, |𝐴|βˆ’1=𝛽(1). Since πœ’(1)=2|𝐴|+1, by Lemma 2.4, it follows that only such relations satisfy, where |𝐴|βˆ’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 Ξ“.
Suppose that π‘˜=5. Then, 5|𝐴|=(2|𝐴|+1)+𝛽(1). Therefore, 𝛽(1)=3|𝐴|βˆ’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 π‘…βŠ†πΆ.

Proof. Let us consider the character ξ€·ξ€Έξπœ’π΄π‘…=𝑓𝑖=1π›Όπ‘–πœ’π‘–,(3.3) where π›Όπ‘–βˆˆπ,πœ’π‘–βˆˆIrr(𝐴𝑅), 𝑖=1,𝑓.
Let us note that by the minimality of the group Ξ“ either 𝑓≠1, or 𝛼1β‰ 1, that is, πœ’π‘–(1)β‰ ξπœ’(1) for all 𝑖=1,𝑓.
By the Clifford Theorem ξ€·πœ’π‘–ξ€Έπ‘…=𝑒𝑖𝑑𝑖𝑗=1πœ’π‘–π‘—,(3.4) where πœ’π‘–π‘—βˆˆIrr(𝑅), 𝑒𝑖 are numbers, which divide |𝐴|, 𝑖=1,𝑓, 𝑗=1,𝑑𝑖.
Suppose that πœ’π‘–0𝑗0∈Irr((ξπœ’)𝑅). Since π‘…βŠ²Ξ“, by the Clifford Theorem it follows that πœ’π‘–π‘—(1) divides 𝑛 for each irreducible component πœ’π‘–π‘— of the character ξπœ’π‘… and for corresponding elements π‘₯βˆˆΞ“β§΅πΌΞ“(πœ’π‘–0𝑗0) the equality πœ’π‘–π‘—=(πœ’π‘–0𝑗0)π‘₯ is true.
Suppose that πœ’π‘–0𝑗0(1)=1 for a certain character πœ’π‘–0𝑗0∈Irr((ξπœ’)𝑅). Since π‘…β€²βŠ†kerπœ’π‘–0𝑗0 and π‘…ξ…žβŠ²Ξ“, we see that π‘…ξ…žβŠ†(kerπœ’π‘–0𝑗0)π‘₯ for all π‘₯βˆˆΞ“β§΅πΌΞ“(πœ’π‘–0𝑗0). Therefore, π‘…ξ…žβŠ†βˆ©π‘₯βˆˆΞ“β§΅πΌΞ“(πœ’π‘–0𝑗0)(kerπœ’π‘–0𝑗0)π‘₯=kerξπœ’=1. Thus, the subgroup 𝑅 is Abelian.
Now suppose that πœ’π‘–π‘—(1)β‰ 1 for all 𝑖=1,𝑓 and for all 𝑗=1,𝑑𝑖. Since 𝑛 is an uneven number, we have πœ’π‘–π‘—(1)β‰₯3. Therefore, if 𝐼𝐴𝑅(πœ’π‘–0𝑗0)=𝑅 for a certain character πœ’π‘–0𝑗0∈Irr((ξπœ’)𝑅), then πœ’π‘–0(1)=(πœ’π‘–0𝑗0)𝐴𝑅(1)=|𝐴|πœ’π‘–0𝑗0(1)β‰₯3|𝐴|. Since in this case ξπœ’(1)β‰₯πœ’π‘–0(1), we obtain a contradiction with ξπœ’(1)=2|𝐴|+1<3|𝐴|. Thus, 𝐼𝐴𝑅(πœ’π‘–π‘—)≠𝑅 for all 𝑖=1,𝑓 and for all 𝑗=1,𝑑𝑖. Then, by Lemma  1 [6], 𝐼𝐴𝑅(πœ’π‘–π‘—)=𝐴𝑅, therefore, 𝑑𝑖=1 and πœ’π‘–(1)=π‘’π‘–πœ’π‘–π‘—(1) for all 𝑖=1,𝑓. By Lemma  13.3 [5], for every 𝑖=1,𝑓 there exists a canonical extension ξπœ’π‘– of the character πœ’π‘–1 to the group 𝐴𝑅.
Assume that 𝐴kerξ‚Šπœ’π‘–0/kerξ‚Šπœ’π‘–0β‹ͺ𝐴𝑅/kerξ‚Šπœ’π‘–0 for a certain 𝑖0. By Lemma  9  [1], 𝐴∩kerξ‚Šπœ’π‘–0=1. Then, 𝐴kerξ‚Šπœ’π‘–0/kerξ‚Šπœ’π‘–0≅𝐴/𝐴∩kerξ‚Šπœ’π‘–0=𝐴. Let us apply the theorem [1] to the factor group 𝐴𝑅/kerξ‚Šπœ’π‘–0 and its exact irreducible character in terms of Lemma  2.22 [5] (ξ‚Šπœ’π‘–0)βˆ—. By this theorem, (ξ‚Šπœ’π‘–0)βˆ—(1)∈{|𝐴|βˆ’1,|𝐴|,|𝐴|+1,2|𝐴|βˆ’2,2|𝐴|βˆ’1} and (ξ‚Šπœ’π‘–0)βˆ—(1) is a prime power, probably, with the exception of the case, when (ξ‚Šπœ’π‘–0)βˆ—(1)=|𝐴|. Since (ξ‚Šπœ’π‘–0)βˆ—(1)=ξ‚Šπœ’π‘–0(1) and ξπœ’π‘–(1) divides 𝑛 for all 𝑖=1,𝑓, it follows that ξ‚Šπœ’π‘–0(1)=2|𝐴|βˆ’1. Thus, 𝑛=2|𝐴|+1=𝑠(2|𝐴|βˆ’1) for a certain odd π‘ βˆˆπ. It is easily seen that it is not true.
Now suppose that 𝐴kerξπœ’π‘–/kerξπœ’π‘–βŠ²π΄π‘…/ξπœ’π‘– for all 𝑖=1,𝑓. Using the fact that ⋂𝑓𝑖=1kerξπœ’π‘–βˆ©π‘…=kerξπœ’βˆ©π‘… and kerξπœ’=1 by Lemma 3.1, then 𝐴0=⋂𝑓𝑖=1kerξπœ’π‘–βŠ†π΄. If 𝐴0β‰ 1, then π΄βŠ²π΄π‘…, since 𝐴 is a 𝑇𝐼-subgroup. If 𝐴0=1, then the subgroup 𝐴𝑅 can be isomorphically embedded in the direct product βˆπ΄π‘…=𝑓𝑖=1(𝐴𝑅/kerξπœ’π‘–). Since the group 𝐴𝑅 contains a normal Hall πœ‹-subgroup, we have π΄βŠ²π΄π‘…. Thus, π‘…βŠ†πΆ.
The lemma is proved.

Suppose that π‘žβˆˆπœ‹(𝑛) and 𝑄 is a 𝐴-invariant Sylow π‘ž-subgroup of 𝐺 that exists by Theorem  6.2.2(i) [4]. By Lemma 3.2,  ̸[𝑄,𝐴]βŠ†πΆ.

Further, 𝑀 is an 𝐴-invariant proper subgroup of 𝐺 such that [𝑄,𝐴]βŠ†π‘€.

From the minimality of the group Ξ“ and Lemma 3.1, it follows that the character ξπœ’π΄π‘€ is reducible for every such a subgroup 𝑀. Denote 𝑁=𝑁Γ([𝑄,𝐴]).

Lemma 3.4. 𝑁≠Γ.

Proof. Assume that 𝑁=Ξ“. By Lemma 3.3, the subgroup [𝑄,𝐴] is either Abelian or [𝑄,𝐴]βŠ†πΆ. Since, by Lemma 3.2, ΜΈ[𝑄,𝐴]βŠ†πΆ, it is enough to consider the case, when [𝑄,𝐴] is Abelian.
Suppose that [𝑄,𝐴] is Abelian. Let us consider the character ξπœ’π΄π‘„. It is clear that 𝐴𝑄≠Γ and 𝐴β‹ͺ𝐴𝑄. Therefore, there exists at least one irreducible component πœ‰ of the character ξπœ’π΄π‘„ such that 𝐴kerπœ‰/kerπœ‰β‹ͺ𝐴𝑄/kerπœ‰ and since |𝐴𝑄| is an uneven number, by theorem [1], it follows that for every such a character it is true that πœ‰(1)∈{|𝐴|,2|𝐴|βˆ’1} and 2|𝐴|βˆ’1 is a prime power. Assume that πœ‰(1)=2|𝐴|βˆ’1. Then, π‘ž divides 2|𝐴|βˆ’1 and π‘ž divides 2|𝐴|+1. Thus, π‘ž=2. A contradiction. Thus, πœ‰(1)=|𝐴|.
Suppose that there exists one more character πœ‰1∈Irr(ξπœ’π΄π‘„) such that 𝐴kerπœ‰1/kerπœ‰1β‹ͺ𝐴𝑄/kerπœ‰1. Since πœ‰1(1)=|𝐴| and ξπœ’π΄π‘„=2|𝐴|+1, we see that ξπœ’π‘„ is a sum of linear irreducible characters πœ†. Thus, π‘„ξ…žβŠ†ξ™πœ†βˆˆIrr(ξπœ’π‘„)kerπœ†=kerξπœ’=1.(3.5) This yields that the subgroup 𝑄 is Abelian. Therefore, π‘„βŠ†πΆΞ“([𝑄,𝐴])βŠ²Ξ“. Suppose that 𝐢Γ([𝑄,𝐴])=Ξ“, that is, [𝑄,𝐴]βŠ†π‘(Ξ“). Then, [𝑄,𝐴]βŠ†πΆ is a contradiction with Lemma 3.2. Thus, 𝐢Γ([𝑄,𝐴])β‰ Ξ“ and since 𝐢Γ([𝑄,𝐴])βŠ²Ξ“, we see that, by Lemma 3.3, the subgroup 𝐢Γ([𝑄,𝐴]) is either Abelian or 𝐢 contains it. If 𝐢Γ([𝑄,𝐴]) is Abelian, then π‘„βŠ²Ξ“, and we obtain a contradiction with Theorem  6.15 [5], but if 𝐢Γ([𝑄,𝐴])βŠ†πΆ, then we obtain a contradiction with Lemma 3.2.
Assume that 𝐴kerπœ“/kerπœ“βŠ²π΄π‘„/kerπœ“, where ξπœ’π΄π‘„=πœ‰+πœ“. By Lemma  3 [1], [𝑄,𝐴]βŠ†(kerπœ“)πœ‹β€². Since π‘„ξ…žβŠ†kerπœ‰ and kerπœ‰βˆ©kerπœ“=1, we have 𝑄=[𝑄,𝐴]βŠ†π‘(𝑄). Then, π‘„βŠ†πΆΞ“([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]βŠ†π‘€, we have ξπœ’π΄π‘€=𝜏+πœ“,(3.6) where 𝜏 and πœ“ are the characters of the subgroup 𝐴𝑀 such that 𝐴kerπœ‰/kerπœ‰β‹ͺ𝐴𝑀/kerπœ‰ for each πœ‰βˆˆIrr(𝜏) and 𝐴kerπœ‰/kerπœ‰βŠ²π΄π‘€/kerπœ‰ for each πœ‰βˆˆIrr(πœ“).

Then, the group 𝐴𝑀/kerπœ“ can be isomorphically embedded in the direct product βˆπ΄π‘€=πœ‰βˆˆIrr(πœ“)𝐴𝑀/kerπœ‰. Since each factor contains a normal Hall πœ‹-subgroup 𝐴kerπœ‰/kerπœ‰, we see that 𝐴kerπœ“/kerπœ“βŠ²π΄π‘€/kerπœ“. It follows that 𝐴kerπœ“βŠ²π΄π‘€. By Lemma  2.2.1 [8], 𝐴𝑀=𝑁𝐴𝑀(𝐴)kerπœ“. Then, 𝑀=(kerπœ“)πœ‹β€²πΆπ‘€(𝐴).(3.7) Since by Lemma  3 [1], [𝑀,𝐴]βŠ†(kerπœ“)πœ‹β€², [𝑄,𝐴]βŠ†π‘€ and [[𝑄,𝐴],𝐴]=[𝑄,𝐴], by Lemma  6.13 [9], we have[]𝑄,π΄βŠ†(kerπœ“)πœ‹β€².(3.8) Assume that 𝜏=0. Then, ξπœ’π΄π‘€=πœ“, that is, β‹‚πœ‰βˆˆIrr(πœ“)=kerπœ“=1. Then, π΄βŠ²π΄π‘€, that is, [𝑄,𝐴]βŠ†πΆ. This contradicts Lemma 3.2.

Thus, πœβ‰ 0. By Theorem [1], πœ‰ξ€½||𝐴||||𝐴||,||𝐴||||𝐴||||𝐴||ξ€Ύ(1)βˆˆβˆ’1,+1,2βˆ’2,2βˆ’1(3.9) and πœ‰(1) is a prime power, probably, with the exception of the case, when πœ‰(1)=|𝐴| or πœ‰(1)=2|𝐴|.

Lemma 3.5. The character 𝜏 does not contain any irreducible components of degree |𝐴|βˆ’1.

Proof. Assume that there exists πœ‰βˆˆIrr(𝜏) and πœ‰(1)=|𝐴|βˆ’1. Then, the group 𝐴𝑀 and its irreducible character πœ‰ satisfy Lemma  11 [1]. Since 𝐴kerπœ‰/kerπœ‰β‹ͺ𝐴𝑀/kerπœ‰, by Lemma  9 [1], 𝐴∩kerπœ‰=1, and, by Lemma  10 [1], it yields that assertion (7) of Lemma  11 [1] does not satisfy. Since πœ‰(1)=|𝐴|βˆ’1, we see that only assertion (3) of the mentioned Lemma satisfies, that is, πœ‰πΆπ‘€(𝐴)=(|𝐴|βˆ’1)π›½πœ‰,π›½πœ‰(1)=1, where π›½πœ‰=(πœ‰π‘€)πœ‹(𝑀,𝐴). By Lemma  10 [1], π›Όπœ‰=ξ€·πœ‰π΄,1𝐴𝐴=πœŒξ€·ξ€·π΄βˆ’1π΄ξ€Έπœ†βˆ’1,1𝐴𝐴(3.10) for a certain linear irreducible character πœ† of the subgroup 𝐴. We see that π›Όπœ‰=0 if πœ†=1𝐴 and π›Όπœ‰=1 if not.
In addition, assume that 𝜏(1)β‰ πœ‰(1), that is, the character 𝜏 contains an irreducible component πœ‰1.
Besides, assume that πœ‰1=|𝐴|βˆ’1. In the same way, we establish that π›Όπœ‰1=πœ‰ξ€·ξ€·1𝐴,1𝐴𝐴=πœŒξ€·ξ€·π΄βˆ’1π΄ξ€Έπœ†1βˆ’1,1𝐴𝐴(3.11) for a certain linear irreducible character πœ†1 of the subgroup 𝐴, moreover, π›Όπœ‰1=0, if πœ†1=1𝐴 and π›Όπœ‰1=1 in the converse case.
Suppose that ξπœ’π΄π‘€=πœ‰+πœ‰1+πœ‚ for a certain character πœ‚ of the subgroup 𝐴𝑀. Then, πœ‚ξ€·πœ‰(1)=ξπœ’(1)βˆ’(1)+πœ‰1ξ€Έ||𝐴||ξ€·2||𝐴||ξ€Έ(1)=2+1βˆ’βˆ’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 Ξ“. Thus, |𝐴|β‰₯7. Therefore, πœ‚(1)=3<|𝐴|βˆ’1. By Theorem  1 [10], 𝐴kerπœ‚/kerπœ‚βŠ²π΄π‘€/kerπœ‚. Clearly, πœ“=πœ‚ and, as follows from Lemma  11 [1] and Lemma  10 [1], each irreducible component 𝛿 of the character πœ“ is restricted irreducibly to 𝐢𝑀(𝐴), that is, 𝛿𝐢𝑀(𝐴)=𝛽𝛿. By Lemma  10 [1], we see that ̂𝛿𝐴=𝛽𝛿(1)1𝐴. Then, it is obvious that π›Όπœ“=(πœ“π΄,1𝐴)𝐴=βˆ‘π›ΏβˆˆIrr(πœ“)(𝛽𝛿(1)πœ†π›Ώβˆ’1,1𝐴)𝐴. Here, ̂𝛿 is a canonic extension of the character 𝛿𝑀 to the group 𝐴𝑀 and πœ†π›Ώ are linear irreducible characters of the subgroup 𝐴. It is readily seen that π›Όπœ“β‰€πœ“(1)=3.
On the other hand, since 𝐢𝑀(𝐴)βŠ†πΆ, we have π›½πœ‰βˆˆIrr(𝛽𝐢𝑀(𝐴)), π›½πœ‰1∈Irr(𝛽𝐢𝑀(𝐴)), and (𝛽𝐢𝑀(𝐴),πœ“πΆπ‘€(𝐴))𝐢𝑀(𝐴)β‰ 0, where 𝛽 is from Lemma 2.4. Since by Theorem 13.1 [5], π›½πœ‰β‰ π›½πœ‰1, we have 𝛽(1)β‰₯3. Since ξπœ’(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 πœ’πΆ=𝛽. Suppose that assertion (1) of Lemma 2.4 satisfies. Then, by Lemma 2.2, 𝛼=ξπœ’π΄,1𝐴𝐴=ξ€·πœŒπ΄+ξ€·||𝐴||ξ€Έ1+1𝐴,1𝐴𝐴=||𝐴||+2.(3.13) Since 𝛼=π›Όπœ‰+π›Όπœ‰1+π›Όπœ“β‰€5, we obtain |𝐴|+2≀5. Hence, |𝐴|≀3. As shown above |𝐴|β‰₯7. A contradiction. Suppose that assertion (15) of Lemma 2.4 satisfies. Then, by Lemma 2.2β€‰β€‰ξπœ’π΄=𝛽(1)1𝐴, that is, π΄βŠ†kerξπœ’. We obtain a contradiction with Lemma 3.1.
Now assume that πœ‰1=|𝐴|. Then, (πœ‰1)𝐴=𝜌𝐴 and π›Όπœ‰1=1. Since π›½πœ‰βˆˆIrr(𝛽𝐢𝑀(𝐴)) and (𝛽𝐢𝑀(𝐴),πœ“πΆπ‘€(𝐴))𝐢𝑀(𝐴)β‰ 0, it follows that 𝛽(1)β‰₯2. Again, we see that only assertion (1) of Lemma 2.4 satisfies. Since in this case πœ“(1)=2, that is, π›Όπœ“β‰€2, we have 𝛼=π›Όπœ‰+π›Όπœ‰1+π›Όπœ“β‰€4. Thus, |𝐴|+2≀4, that is, |𝐴|≀2. A contradiction.
There only remains to consider the case, when 𝜏=πœ‰. Since ker𝜏∩kerπœ“=1, then the character πœ‰kerπœ“ is exact. Further, 𝐴β‹ͺ𝐴kerπœ“, because if not we have [𝑄,𝐴]βŠ†πΆ by relation (3.8), and we get a contradiction with Lemma 3.2. Then, the character Μƒπœ‰=πœ‰π΄kerπœ“ is also exact and from Lemma7 [10] and Lemma2.27 [5] it yields that Μƒ|𝐴kerπœ“βˆΆπ‘(πœ‰)|=|𝐴kerπœ“βˆΆπ‘(𝐴kerπœ“)|=|𝐴|22π‘Ÿ, π‘Ÿβˆˆπ. Hence, [𝑄,𝐴]βŠ†π‘(𝐴kerπœ“), that is, [𝑄,𝐴]βŠ†πΆ. 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=2𝛽,π›½βˆˆπ. Then, |π‘€βˆΆπΆπ‘€(𝐴)|=2π‘š, where π‘šβˆˆπ™+.

Proof. Since |πœ‹|=1, we see that |𝐴|=𝑝𝑓, π‘“βˆˆπ, therefore, 𝑝𝑓+1=2𝛽, π›½βˆˆπ, that is, 𝑝𝑓=2π›½βˆ’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 Ξ“. Therefore, |πœ‹|>1.
Suppose that π‘Ÿ,π‘ βˆˆπœ‹,π‘Ÿβ‰ π‘ . 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×𝐡 of order π‘Ÿπ‘ . By assumption B, 𝐢𝑀(𝐻)=𝐢𝑀(𝐡)=𝐢𝑀(𝐴). Since 𝑀 is solvable, by Theorem  6 [9], it follows that [𝑀,𝐻]βŠ†πΉ(𝑀). Since, by Lemma  4 [1], 𝑀=[𝑀,𝐻]𝐢𝑀(𝐻)=[𝑀,𝐻]𝐢𝑀(𝐴), we obtain 𝑀=𝐹(𝑀)𝐢𝑀(𝐴). Assume that 𝐹=𝐹(𝑀). Let us consider the character 𝛾𝐴𝐹=𝑑𝑖=1𝛼𝑖𝛾𝑖,(3.14)π›Ύπ‘–βˆˆIrr(𝐴𝐹), 𝛼𝑖>0, 𝑖=1,2,…,𝑑.
Suppose that 𝐴⊲𝐴𝐹. Then, πΉβŠ†πΆπ‘€(𝐴). Thus, π‘€βŠ†πΆπ‘€(𝐴), therefore, |π‘€βˆΆπΆπ‘€(𝐴)|=20=1. In this case, the lemma is true.
Suppose that 𝐴β‹ͺ𝐴𝐹. Then for a certain 𝑖, let us say that 𝑖=1, 𝐴ker𝛾1β‹ͺ𝐴𝐹. From Theorems  1 and 2 [10] and the assumption of the lemma, we obtain 𝛾1(1)=|𝐴|βˆ’1=2𝛼, |𝐴| or |𝐴|+1=2𝛽. Here 𝛼,π›½βˆˆπ.
If 𝛾1(1)=|𝐴|βˆ’1=2𝛼, then |𝐴|=2𝛼+1=2π›½βˆ’1. Hence, 2𝛽=2𝛼+2=2(2π›Όβˆ’1+1), that is, 𝛼=1. Thus, |𝐴|=3. This case is impossible.
Suppose that 𝛾1(1)=|𝐴|. Then, 𝛾(1)=𝛾(1)βˆ’π›Ύ1||𝐴||||𝐴||(1)=+1βˆ’=1,(3.15) where 𝛾=π›Ύπ΄πΉβˆ’π›Ύ1. Therefore, all irreducible components of the character 𝛾𝐹 are linear, that is, the subgroup 𝐹 is Abelian. Since πΉβŠ²π΄π‘€, we have 𝐢𝐴𝑀(𝐹)βŠ²π΄π‘€. Then, it is not hard to see that πΆβˆ—=𝐢𝑀(𝐹)βŠ²π΄π‘€. Since πΉΜΈβŠ†πΆπ‘€(𝐴) and πΉβŠ†πΆβˆ—, it follows that πΆβˆ—ΜΈβŠ†πΆπ‘€(𝐴), that is, 𝐴β‹ͺπ΄πΆβˆ—.
Arguing as above for the subgroup 𝐴𝐹, we can show that in the contingency π›Ύπ΄πΆβˆ— there is an irreducible component 𝛾1 such that 𝐴ker𝛾1β‹ͺπ΄πΆβˆ— and 𝛾1(1)=|𝐴| or |𝐴|+1=2𝛽. Assume that 𝛾1(1)=|𝐴|+1=2𝛽. Since the subgroup 𝐴 is a 𝑇𝐼-set in π΄πΆβˆ—, from Lemma  1 [6], it yields that π΄βŠ†πΌπ΄πΆβˆ—(𝛾1βˆ—), where 𝛾1βˆ— is an irreducible component of the character (𝛾1)πΆβˆ—. Thus, (𝛾1)πΆβˆ— is an irreducible character of the subgroup πΆβˆ—. Since πΉβŠ†π‘(πΆβˆ—), we have πΉβŠ†π‘(𝛾1). However, since the character 𝛾 is exact, the character 𝛾1 is exact. Therefore, 𝑍(𝛾1)=𝑍(π΄πΆβˆ—). Then, πΉβŠ†π‘(π΄πΆβˆ—), that is, πΉβŠ†πΆπ‘€(𝐴). This case has already been examined. Therefore, we may assume that 𝛾1(1)=|𝐴|. Again, we see that πΆβˆ— is Abelian, that is, πΆβˆ—βŠ†πΉ. Therefore, 𝐹=πΆβˆ—.
Suppose that πœ†βˆˆIrr(𝛾𝐹) and 𝐼=𝐼𝐴𝑀(πœ†). Assume that 𝐼=𝐴𝑀. Since the subgroup 𝐹 is Abelian, we obtain πœ†(1)=1. Therefore, 𝛾𝐹=𝛾(1)πœ†, that is, πΉβŠ†π‘(𝛾)=𝑍(𝐴𝑀), that is, πΉβŠ†πΆπ‘€(𝐴). The lemma is true in this case.
Now suppose that 𝐼≠𝐴𝑀. By Theorem  6.11 [5], there exists an irreducible character πœ†ξ…ž of the subgroup 𝐼 such that (πœ†ξ…ž)𝐴𝑀=𝛾. Hence, 𝛾(1)=2𝛽=|π΄π‘€βˆΆπΌ|πœ†ξ…ž(1). Thus, |π΄π‘€βˆΆπΌ| and πœ†ξ…ž(1) is degree 2. Therefore, 𝐴π‘₯βŠ†πΌ for a certain element π‘₯βˆˆπ΄π‘€. Assume that 𝐴π‘₯⊲𝐼. Then, πΉβŠ†πΆπ‘€(𝐴π‘₯). Hence, 𝐹=𝐹π‘₯βˆ’1βŠ†πΆπ‘€(𝐴). The lemma is true.
Therefore, 𝐴π‘₯β‹ͺ𝐼. Let us show that πœ†ξ…ž(1)=1.
Let us assume the converse, that is, πœ†ξ…ž(1)>1. Since by Frobenius reciprocity law for characters ξ‚€ξ€·πœ†1=𝛾,ξ…žξ€Έπ΄π‘€ξ‚π΄π‘€=𝛾𝐼,πœ†ξ…žξ€ΈπΌ,(3.16) we have 𝛾𝐼=πœ†ξ…ž+πœ†ξ…žξ…ž. Since πœ†ξ…ž(1)>1, we obtain πœ†ξ…žξ…ž(1)=𝛾(1)βˆ’πœ†ξ…ž||𝐴||(1)=+1βˆ’πœ†ξ…ž||𝐴||(1)β‰€βˆ’1.(3.17) Since |𝐴|β‰ 3, by Theorems  1 and 2 [10], we see that 𝐴π‘₯kerπœ†ξ…žξ…žβŠ²πΌ. By using the fact that 𝐴π‘₯β‹ͺ𝐼, it follows that 𝐴π‘₯kerπœ†ξ…žβ‹ͺ𝐼. Then, πœ†ξ…ž(1)β‰₯|𝐴|βˆ’1. Since the numbers πœ†ξ…ž(1) and |𝐴|+1 are of degree 2 and |𝐴|β‰ 3, we obtain πœ†ξ…ž(1)=|𝐴|+1, that is, πœ†ξ…ž(1)=𝛾(1). Since 𝛾(1)=|π΄π‘€βˆΆπΌ|πœ†ξ…ž(1), we have |π΄π‘€βˆΆπΌ|=1, that is, 𝐴𝑀=𝐼. In this case, as shown above, the lemma is true.
We may assume that πœ†ξ…ž(1)=1. Then, πœ†ξ…žξ…ž(1)=|𝐴|+1βˆ’1=|𝐴|. Since 𝐴π‘₯kerπœ†ξ…žβŠ²πΌ, 𝐴π‘₯β‹ͺ𝐼 and |𝐴|β‰ 3, from Theorem  1 and 2 [10], it easily follows that the character πœ†ξ…žξ…ž is irreducible. Therefore, all irreducible components of the character π›ΎπΌπœ‹β€² are linear, that is, the subgroup πΌπœ‹β€² is Abelian. Since 𝐹=πΆβˆ— is Abelian and πΉβŠ†πΌπœ‹β€², we have 𝐹=πΌπœ‹β€². Then, 𝐼=𝐴π‘₯πΌπœ‹β€²=𝐴π‘₯𝐹. Since 𝑀=𝑀π‘₯, we have 𝑀=𝐹𝐢𝑀(𝐴)=𝐹𝐢𝑀(𝐴)π‘₯=𝐹π‘₯𝐢𝑀(𝐴)π‘₯𝐢=𝐹𝑀(𝐴)π‘₯=𝐹𝐢𝑀(𝐴π‘₯).(3.18) Further, 𝐴𝑀=(𝐴𝑀)π‘₯=𝐴π‘₯𝑀π‘₯=𝐴π‘₯𝑀=𝐴π‘₯𝐹𝐢𝑀(𝐴π‘₯)=𝐼𝐢𝑀(𝐴π‘₯).(3.19) Hence, for each π‘”βˆˆπ΄π‘€, we obtain that 𝑔=π‘Žπ‘, where π‘ŽβˆˆπΌ and π‘βˆˆπΆπ‘€(𝐴π‘₯). Since 𝛾=(πœ†ξ…ž)𝐴𝑀, for each β„Žβˆˆπ΄π‘₯ the following expression satisfies 1𝛾(β„Ž)=||𝐼||ξ“π‘”βˆˆπ΄π‘€ξ€·πœ†ξ…žξ€Έ0ξ€·π‘”β„Žπ‘”βˆ’1ξ€Έ=1||𝐼||ξ“π‘”βˆˆπ΄π‘€ξ€·πœ†ξ…žξ€Έ0ξ€·π‘Žπ‘β„Žπ‘βˆ’1π‘Žβˆ’1ξ€Έ=1||𝐼||ξ“π‘”βˆˆπ΄π‘€ξ€·πœ†ξ…žξ€Έ0ξ‚€π‘Žξ‚€β„Žπ‘βˆ’1ξ‚π‘Žβˆ’1=1||𝐼||ξ“π‘”βˆˆπ΄π‘€πœ†ξ…ž||||πœ†(β„Ž)=π΄π‘€βˆΆπΌξ…ž(β„Ž).(3.20) Here, (πœ†ξ…ž)0(π‘₯)=πœ†ξ…ž(π‘₯), if π‘₯∈𝐼 and (πœ†ξ…ž)0(π‘₯)=0, if π‘₯βˆ‰πΌ. Since πœ†ξ…ž(1)=1, we obtain |πœ†ξ…ž(β„Ž)|=1. The latter chain of equalities implies that ||||=||||||𝛾(β„Ž)π΄π‘€βˆΆπΌβ‹…πœ†ξ…ž||=||||||πœ†(β„Ž)π΄π‘€βˆΆπΌξ…ž||=||||(β„Ž)π΄π‘€βˆΆπΌ.(3.21)
Since 𝛾(1)=|π΄π‘€βˆΆπΌ|πœ†ξ…ž(1), we have 𝛾(1)=|π΄π‘€βˆΆπΌ|. Then, |𝛾(β„Ž)|=|π΄π‘€βˆΆπΌ|=𝛾(1). By Lemma  2.27(e) [5], 𝐴π‘₯βŠ†π‘(𝛾)=𝑍(𝐴𝑀). Hence, 𝐴π‘₯=π΄βŠ²π΄π‘€, that is π‘€βŠ†πΆπ‘€(𝐴), therefore, |π‘€βˆΆπΆπ‘€(𝐴)|=1. The lemma is true.
There only remains to consider the case, when 𝛾1(1)=|𝐴|+1=2𝛽. Since the subgroup 𝐴 is a 𝑇𝐼-set in 𝐴𝐹, from Lemma  1 [6], it follows that π΄βŠ†πΌπ΄πΉ(𝛾1)βˆ—, where (𝛾1)βˆ— 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)𝐹2Γ—πœ‡, where (𝛾1)𝐹2 is an irreducible character of the Sylow 2-subgroup 𝐹2 from 𝐹 and πœ‡ is an irreducible character of the Hall 2β€²-subgroup 𝑂 from 𝐹. Since πœ‡(1)=1, we have π‘‚βŠ†π‘(𝛾1)=𝑍(𝐴𝐹), that is, π‘‚βŠ†πΆπ‘€(𝐴). Then, 𝑀=𝐹𝐢𝑀𝐹(𝐴)=2𝐢×𝑂𝑀(𝐴)=𝐹2𝐢𝑀(𝐴).(3.22) Therefore, ||π‘€βˆΆπΆπ‘€||=||𝐹(𝐴)2𝐢𝑀(𝐴)βˆΆπΆπ‘€||=||𝐹(𝐴)2∢𝐹2βˆ©πΆπ‘€||(𝐴)=2π‘š,(3.23) where π‘š>0.
The lemma is proved.

Lemma 3.7. πœ“β‰ 0.

Proof. Let us assume the converse, that is, πœ“=0. Then, (ξπœ’)𝐴𝑀=𝜏. Since, by Lemma 3.5, πœ‰(1)β‰ |𝐴|βˆ’1 for πœ‰βˆˆIrr(𝜏), we have πœ‰ξ€½||𝐴||,||𝐴||||𝐴||||𝐴||||𝐴||ξ€Ύ(1)∈+1,2βˆ’2,2βˆ’1,2.(3.24) Suppose that πœ‰(1)∈{2|𝐴|βˆ’2,2|𝐴|βˆ’1,2|𝐴|}. Then, ξπœ’(1)βˆ’πœ‰(1)=2|𝐴|+1βˆ’πœ‰(1)≀3. As mentioned above |𝐴|β‰₯7. By Theorem  1 [10], 𝐴ker(ξπœ’βˆ’πœ‰)/ker(ξπœ’βˆ’πœ‰)βŠ²π΄π‘€/ker(ξπœ’βˆ’πœ‰). Thus, Irr(ξπœ’βˆ’πœ‰)βŠ†Irr(πœ“). This contradicts the assumption. Hence, πœ‰(1)∈{|𝐴|,|𝐴|+1} and since by the assumption 𝜏(1)=2|𝐴|+1, we obtain 𝜏=πœ‰+πœ‰1,(3.25) where πœ‰(1)=|𝐴| and πœ‰1(1)=|𝐴|+1 is degree 2.
From the Clifford Theorem, it follows that π‘€ξ…žβŠ†kerπœ‰. Since kerπœ‰βˆ©kerπœ‰1=ker𝜏=kerξπœ’π΄π‘€=1, it follows that the subgroup kerπœ‰1 is Abelian. By Theorem [12], the factor group 𝐴𝑀/kerπœ‰1 is solvable. Then, 𝐴𝑀 is also solvable. In the course of proving Lemma 3.6, it was established that |πœ‹|>1. Likewise, it is shown that [𝑀,𝐴]βŠ†πΉ=𝐹(𝑀).
On the other hand, by Lemma 3.6, ||π‘€βˆΆπΆπ‘€(||𝐴)=2π‘š,π‘šβˆˆπ™+,(3.26) where 𝑀=𝑀/𝐾. Here, 𝐾=(kerπœ‰1)πœ‹β€². Since (|𝐾|,|𝐴|)=1 and 𝐢=𝐢𝑀(π‘Ž)=𝐢𝑀(𝐴) for all elements π‘Žβˆˆπ΄#, we have |𝐢𝑀/𝐾(𝐴)|=|𝐢𝑀(𝐴)∢𝐢𝐾(𝐴)|. Thus, ||π‘€βˆΆπΆπ‘€||=||𝑀||||𝐢(𝐴)𝐾||(𝐴)||𝐢𝑀||||𝐾||(𝐴)=2π‘š,π‘šβˆˆπ™+.(3.27)
Since [𝑄,𝐴]βŠ†[𝑀,𝐴] and ΜΈ[𝑄,𝐴]βŠ†πΆ, we have πΉπ‘žβ‰ 1. Since 𝐹𝑠charπΉβŠ²π΄π‘€, it follows that πΉπ‘ βŠ²π΄π‘€. By using the fact that πœ‰(1) and πœ‰1(1) are π‘žξ…ž-numbers and from the Clifford Theorem, it is readily seen that the subgroup πΉπ‘ž is Abelian. Since [𝑄,𝐴]βŠ†πΉπ‘ž, it yields that [𝑄,𝐴] is also Abelian. Therefore, [𝑄,𝐴]=[[𝑄,𝐴],𝐴]×𝐢[𝑄,𝐴](𝐴). Since [[𝑄,𝐴],𝐴]=[𝑄,𝐴], we have 𝐢[𝑄,𝐴](𝐴)=1. Then, [𝑄,𝐴]βˆ©πΆπ‘€(𝐴)=1. Hence and by the equality (3.27), it follows that [𝑄,𝐴]βŠ†πΎ.
Further, we see that 𝜏𝐾=ξ€·πœπ‘€ξ€ΈπΎ=ξ€·πœ‰π‘€+ξ€·πœ‰1𝑀𝐾=ξ“πœ†βˆˆIrr(πœ‰π‘€)πœ†πΎ+πœ‰1(1)1𝐾,πœ†(1)=1.(3.28) Since πΎβŠ†π‘(πœ‰1), it follows that β‹‚πΎβŠ†(πœ†βˆˆIrr(πœ‰π‘€)𝑍(πœ†))βˆ©π‘(πœ‰1). By the fact that the character πœπ‘€ is exact and Lemma 3.1 [1], it follows that πΎβŠ†π‘(𝑀). Then, π‘€βŠ†π‘. Since π΄βŠ†π‘ and π‘„βŠ†π‘, we obtain 𝐿=βŸ¨π΄π‘€,π‘„βŸ©βŠ†π‘. By Lemma 3.4, we have 𝐿≠Γ.
Let us consider the character ξπœ’πΏ. Clearly, it is enough to consider the case, when 𝐴kerπœ‰ξ…ž/kerπœ‰ξ…žβ‹ͺ𝐿/kerπœ‰ξ…ž for a certain character πœ‰ξ…žβˆˆIrr(ξπœ’πΏ). Then, πœ‰βˆˆIrr(πœ‰ξ…žπ΄π‘€) or πœ‰1∈Irr(πœ‰ξ…žπ΄π‘€). Thus, we have πœ‰ξ…ž(1)β‰₯|𝐴|.
Assume that there exists a character πœ“ξ…ž of the group 𝐿 such that Irr(πœ“ξ…ž)βŠ†Irr(ξπœ’πΏ) and 𝐴kerπœ“ξ…ž/kerπœ“ξ…žβŠ²πΏ/kerπœ“ξ…ž. Then, 𝐴kerπœ“ξ…žβŠ²πΏ, therefore, (𝐴kerπœ“ξ…ž)βˆ©π΄π‘€βŠ²π΄π‘€. Hence, 𝐴kerπœ“ξ…žπ΄π‘€βŠ²π΄π‘€. Thus, we have πœ“ξ…žπ΄π‘€=πœ“β‰ 0. A contradiction with the assumption.
Thus, we have ξπœ’πΏ=πœ‰ξ…ž+πœ‰ξ…ž1 and can say that πœ‰ξ…žπ΄π‘€=πœ‰ and (πœ‰ξ…ž1)𝐴𝑀=πœ‰1. Let us apply the reasoning analogous to the reasoning applied to the group 𝐴𝑀 and its characters πœ‰ and πœ‰1 to the group 𝐿 and characters πœ‰ξ…ž and πœ‰ξ…ž1. We obtain [πΏπœ‹β€²,𝐴]βŠ†πΉ(πΏπœ‹β€²) and []𝑄,π΄βŠ†πΎπΏ.(3.29) Here, we have 𝐾𝐿=(kerπœ‰ξ…ž1)πœ‹β€².
Suppose that π‘ž1βˆˆπœ‹([πΏπœ‹β€²,𝐴]),π‘ž1β‰ π‘ž and 𝑄1 is an invariant Sylow π‘ž1-subgroup [πΏπœ‹β€²,𝐴]. Then, 𝑄1⊲𝐿. Let πœ‰π‘ž1∈Irr((πœ‰ξ…ž1)𝑄1). By Theorem  6.11 [5], we have πœ‰β€²1=Μƒπœ‰πΏπ‘ž1 for a certain irreducible character Μƒπœ‰π‘ž1 of the group 𝐼𝐿(πœ‰π‘ž1). Thus, πœ‰ξ…ž1Μƒπœ‰(1)=πΏπ‘ž1Μƒπœ‰(1)=π‘ž1(1)|𝐿∢𝐼𝐿(πœ‰π‘ž1)|. Since πœ‰ξ…ž1(1) is degree 2, we see that the numbers Μƒπœ‰π‘ž1(1) and |𝐿∢𝐼𝐿(πœ‰π‘ž1)| are also degrees 2.
Assume that (πœ‰ξ…žπ‘„1,πœ‰π‘ž1)𝑄1β‰ 0. By the same theorem, we have πœ‰ξ…ž=(πœ‰ξ…žπ‘ž1)𝐿 for a certain irreducible character πœ‰ξ…žπ‘ž1 of the group 𝐼𝐿(πœ‰π‘ž1). Then we have |𝐴|=πœ‰ξ…ž(1)=πœ‰ξ…žπ‘ž1(1)|𝐿∢𝐼𝐿(πœ‰π‘ž1)|. Since |𝐿∢𝐼𝐿(πœ‰π‘ž1)| is degree 2, we obtain 𝐿=𝐼𝐿(πœ‰π‘ž1). Then, πœ‰ξ…žπ‘„1=π‘’ξ…žπœ‰π‘ž1(1) for a certain natural number π‘’ξ…ž. Thus, we have |𝐴|=π‘’ξ…žπœ‰π‘ž1(1). Since πœ‰π‘ž1(1) is a πœ‹ξ…ž-number, we have πœ‰π‘ž1(1)=1. Then, 𝑄1βŠ†π‘(πœ‰ξ…ž) and, obviously, 𝑄1βŠ†π‘(πœ‰ξ…ž1). By Lemma  5 [1], 𝑄1βŠ†π‘(𝐿). This means that 𝑄1βŠ†πΆπΏπœ‹β€²(𝐴).
Now assume that (πœ‰ξ…žπ‘„1,πœ‰π‘ž1)𝑄1=0. By Theorem  D [13], ||𝑄𝑑=1||||||πœ‰Ξ“βˆΆπΏ1(1)ξπœ’(1)πœ‰π‘ž1(1)(3.30) is an integer. Since π‘„βŠ†πΏ, we see that |Ξ“βˆΆπΏ| is a π‘žξ…ž-number. Since πœ‰1(1) is a π‘žξ…ž-number and π‘ž divides ξπœ’(1), it follows that 𝑑 is not an integer. We obtain a contradiction. Thus, we have 𝑄1=1, therefore, [πΏπœ‹β€²,𝐴] is a π‘ž-group.
It is obvious that [𝑄,𝐴]βŠ†[πΏπœ‹β€²,𝐴]. Since [πΏπœ‹β€²,𝐴]⊲𝐿, we obtain [πΏπœ‹β€²,𝐴]βŠ†π‘„. Then, we have [[πΏπœ‹β€²,𝐴],𝐴]βŠ†[𝑄,𝐴]. Since [[πΏπœ‹β€²,𝐴],𝐴]=[πΏπœ‹β€²,𝐴], it follows that [πΏπœ‹β€²,𝐴]βŠ†[𝑄,𝐴]. Thus, we have [πΏπœ‹β€²,𝐴]=[𝑄,𝐴]. Then, [πΏπœ‹β€²,𝐴]βŠ†πΎπΏ. Hence, 𝐴𝐾𝐿/𝐾𝐿⊲𝐿/𝐾𝐿. Therefore, we have 𝐴kerπœ‰ξ…ž1/kerπœ‰ξ…ž1⊲𝐿/kerπœ‰ξ…ž1. Thus, πœ‰ξ…ž1βŠ†Irr(πœ“), that is, πœ“β‰ 0. This contradicts the original assumption.
The lemma is proved.

Lemma 3.8. The character πœ‡=𝜏𝐴kerπœ“ is an exact irreducible character of degree 𝑛1=|𝐴|,|𝐴|+1, 2|𝐴|βˆ’2 or 2|𝐴|βˆ’1 and 𝑛1 is a prime power, probably, with the exception of the case, when πœ‡(1)=|𝐴|.

Proof. By Lemma 3.7, we have πœ“β‰ 0.
As stated above, the character ξπœ’π΄π‘€ cannot be irreducible.
Let πœ“(1)=1. Then, we have Ξ“ξ…žβŠ†kerπœ“. Suppose that