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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 πΊΜΈβŠ†Ξ“ξ…ž. Then, πΊξ…žβ‰ πΊ. By Lemma 3.3, the subgroup πΊξ…ž is either Abelian or πΊξ…žβŠ†πΆ. If πΊξ…ž is Abelian, 𝐺 is solvable. By Theorem  1 [6], ξπœ’(1) is a prime power. This contradicts the minimality of the group Ξ“. Now let πΊξ…žβŠ†πΆ. Then, we have 𝐢⊲𝐺. Since πœ’(1)=2|𝐴|+1, it is readily seen from Lemma 2.4 that the character πœ’πΆ contains a linear component 𝛽 or a linear component 𝛽1. From the Clifford Theorem, it follows that 𝐢 is Abelian. Thus, πΊξ…ž is Abelian. The case is examined.
Thus, πΊβŠ†Ξ“ξ…ž. Then, we have πΊβŠ†kerπœ“. By Corollary  6.7 [5], πΊβŠ†kerξπœ’. Since the character ξπœ’ 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 πœ“(1)β‰₯2. Then, 𝜏(1)≀2|𝐴|βˆ’1. From Lemma 3.5, it follows that the character 𝜏 is irreducible. Since ker𝜏∩kerπœ“=1, we see that the character 𝜏kerπœ“ is exact. Suppose that the character 𝜏𝐴kerπœ“ is not exact. Then, we have ker𝜏𝐴kerπœ“βŠ†π΄. Since 𝐴 is a 𝑇𝐼-subgroup, it follows that 𝐴⊲𝐴kerπœ“. Thus we have (kerπœ“)πœ‹β€²βŠ†πΆπ‘€(𝐴). From formula (3.8), it follows that [𝑄,𝐴]βŠ†C. A contradiction with the choice of the subgroup 𝑀. Thus, the character 𝜏𝐴kerπœ“ is exact.
Since (kerπœ“)πœ‹β€²ΜΈβŠ†πΆπ‘€(𝐴), we have 𝐴kerπœ‡/kerπœ‡β‹ͺ𝐴kerπœ“/kerπœ‡ for a certain character πœ‡βŠ†Irr(𝜏𝐴kerπœ“). By Theorem [1], we have πœ‡ξ€½||𝐴||||𝐴||,||𝐴||||𝐴||||𝐴||ξ€Ύ(1)βˆˆβˆ’1,+1,2βˆ’2,2βˆ’1(3.31) and πœ‡(1) is a prime power, probably, with the exception of the case when πœ‡(1)=|𝐴|. Besides, it is readily seen that πœ‡(1)=𝜏(1) when πœ‡(1)β‰ |𝐴|βˆ’1.
Let us consider this case separately. Let πœ‡(1)=|𝐴|βˆ’1. Since 𝜏(1)<2|𝐴|, subject to Lemma 3.5, we have 𝜏(1)=2|𝐴|βˆ’2 and by the Clifford Theorem, we obtain 𝜏𝐴kerπœ“=πœ‡+πœ‡π‘”, where π‘”βˆˆπ΄π‘€β§΅π΄kerπœ“. Hence, by Lemma  7 [10], we have ||ξ€·kerπœ“)πœ‹β€²||=||ξ€·βˆΆπ‘(πœ‡)kerπœ“)πœ‹β€²βˆΆπ‘(πœ‡π‘”)||=22𝛼,(3.32) where πœ‡(1)=2𝛼, π›Όβˆˆπ‘. Since [𝑄,𝐴]βŠ†(kerπœ“)πœ‹β€², we see that [𝑄,𝐴]βŠ†π‘(πœ‡)βˆ©π‘(πœ‡π‘”). Since kerπœ‡βˆ©kerπœ‡π‘”=1, by Lemma  5 [1], we have [𝑄,𝐴]βŠ†π‘(𝐴kerπœ“). Then [𝑄,𝐴]βŠ†πΆ. A contradiction with Lemma 3.2. Thus, πœ‡ξ€½||𝐴||,||𝐴||||𝐴||||𝐴||ξ€Ύ(1)∈+1,2βˆ’2,2βˆ’1(3.33) and πœ‡(1) is a prime power, probably, with the exception of the case, when πœ‡(1)=|𝐴|. Since πœ‡(1)=𝜏(1) under the value indicated, it follows that πœ‡(1) and the character 𝜏𝐴kerπœ“ is exact, the lemma is proved.

Lemma 3.9. 𝑛1=|𝐴|.

Proof. By Lemma 3.8, it is enough to show that 𝑛1β‰ |𝐴+1|,2|𝐴|βˆ’2 and 2|𝐴|βˆ’1.
First, let 𝑛1=|𝐴|+1. By Theorem [12], the group 𝐴kerπœ“ is solvable. Then by Lemma 3.6, we have |(kerπœ“)πœ‹β€²βˆΆπΆ(kerπœ“)πœ‹β€²(𝐴)|=2π‘š, where π‘šβˆˆπ™+. From equality (3.8), it follows that [𝑄,𝐴]βŠ†πΆ(kerπœ“)πœ‹β€²(𝐴)βŠ†πΆπ‘€(𝐴). We obtain a contradiction with the choice of the subgroup 𝑀.
Now suppose that 𝑛1=2|𝐴|βˆ’2 or 2|𝐴|βˆ’1. By Lemma 3.8, 𝑛1 is a prime power 𝑠.
Let 𝑆 be a 𝐴-invariant Sylow 𝑠-subgroup from (kerπœ“)πœ‹β€² and 𝑍=𝑍(𝑆). Since, by Lemma 3.8, the character πœ‡ is exact and irreducible, by Lemma  2.27 [5], it follows that 𝑍(πœ‡)=𝑍(𝐴kerπœ“) is a cyclic group. Therefore, π‘βˆ©π‘(πœ‡)βŠ†π‘(𝐴kerπœ“), that is, π‘βˆ©π‘(πœ‡)βŠ†πΆπ‘(𝐴). Since 𝑍 is Abelian, it follows that 𝑍=[𝑍,𝐴]×𝐢𝑍(𝐴). It can be easily seen that |[𝑍,𝐴]| divides |𝑍/π‘βˆ©π‘(πœ‡)|.
Assume that π‘ΜΈβŠ†π‘(πœ‡). Then, by Corollary  1 [14], |𝑍/π‘βˆ©π‘(πœ‡)| divides 𝑛1. Thus, |[𝑍,𝐴]| divides 𝑛1. Thus, we have |[𝑍,𝐴]|=π‘ π‘Ÿ, π‘Ÿβˆˆπ. 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, π‘ π‘Ÿβˆ’1=𝑑|𝐴|, that is, π‘ π‘Ÿ=𝑑|𝐴|+1, where π‘‘βˆˆπ. Since π‘ π‘Ÿβ‰€π‘›1, we have 𝑑=1 and π‘ π‘Ÿ=|𝐴|+1. Therefore, 𝑠=2, as stated above, 𝑛1β‰ |𝐴|+1, and we obtain 𝑛1=2|𝐴|βˆ’2. Since 2π‘Ÿ=|𝐴|+1 and 2|𝐴|βˆ’2 is degree 2, we have |𝐴|=3. As stated above, it is not true. Our assumption that π‘ΜΈβŠ†π‘(πœ‡) is false.
Thus, π‘βŠ†π‘(πœ‡). Then, it is clear that π‘βŠ²π΄π‘€. Since π‘βŠ†kerπœ“, ker𝜏∩kerπœ“=kerξπœ’=1, and the character 𝜏 is irreducible, Corollary  6.7 [5] yields that (πœπ‘,𝜍)𝑍≠0 for a certain nonprincipal irreducible character 𝜍 of the subgroup 𝑍. Then, by Theorem  D [13], ||𝑍||||||Ξ“βˆΆπ΄π‘€πœ(1)=||𝑍||||||π‘›ξπœ’(1)𝜍(1)Ξ“βˆΆπ΄π‘€1𝑛(3.34) is an integer. Since π‘ž divides 𝑛 and π‘ž does not divide |𝑍||Ξ“βˆΆπ΄π‘€|𝑛1, we obtain a contradiction.
The lemma is proved.

Lemma 3.10. Let 𝑀 be a maximal proper 𝐴-invariant subgroup 𝐺 such that [𝑄,𝐴]βŠ†π‘€. Then, one has 𝑀=[𝑄,𝐴]×𝐢,[𝑄,𝐴]βŠ†π‘(𝑀).

Proof. By assumption of the lemma and by Lemma 3.4, we have π‘βŠ†π΄π‘€.
By Lemma 3.9, 𝑛1=|𝐴|. Since the character πœ‡=𝜏𝐴/kerπœ“ is irreducible, the character 𝜏 is also irreducible.
Since 𝜏(1)=|𝐴|, all irreducible components of the character πœπ‘€ are linear. This means that π‘€ξ…žβŠ†ker𝜏. Since kerπœβŠ²π΄π‘€, 𝐴kerπœ“βŠ²π΄π‘€ and ker𝜏∩kerπœ“=1, we have 𝐴0=ker𝜏∩𝐴kerπœ“βŠ†π΄. But 𝐴 is a 𝑇𝐼-subgroup and 𝐴β‹ͺ𝐴kerπœ“ because in the converse case, by equality (3.8), we have [𝑄,𝐴]βŠ†πΆ, which contradicts Lemma 3.2. Then, we obtain 𝐴0=1. Thus, kerπœβŠ†πΆπ΄π‘€(𝐴kerπœ“), that is, kerπœβŠ†πΆπ‘€(𝐴). Then, π‘€ξ…žβŠ†πΆπ‘€(𝐴), that is, 𝐢𝑀(𝐴)βŠ²π‘€.
By Lemma  3 [1], we have [𝑀,𝐴]βŠ†π‘€βˆ©π΄kerπœ“. By using the fact that π‘€ξ…žβŠ†ker𝜏 and ker𝜏∩kerπœ“=1, it follows that kerπœ“βˆ©π‘€ is Abelian. Thus, the subgroup [𝑀,𝐴] is Abelian. Since, []=]]𝑀,𝐴[[𝑀,𝐴,𝐴×𝐢[𝑀,𝐴](𝐴)(3.35) and [[𝑀,𝐴],𝐴]=[𝑀,𝐴], we conclude that 𝐢[𝑀,𝐴](𝐴)=1. Since []𝑀,π΄βˆ©πΆπ‘€(𝐴)βŠ†πΆ[𝑀,𝐴](𝐴),(3.36) it follows that [𝑀,𝐴]βˆ©πΆπ‘€(A)=1. Since by Lemma  6 [9], [𝑀,𝐴]βŠ²π‘€, 𝐢𝑀(𝐴)βŠ²π‘€ and 𝑀=[𝑀,𝐴]𝐢𝑀(𝐴) by Lemma  4 [1], we have 𝑀=[𝑀,𝐴]×𝐢𝑀(𝐴). By using the fact that [𝑀,𝐴] is Abelian, it follows that [𝑀,𝐴]βŠ†π‘(𝑀).
Since [𝑄,𝐴]βŠ†π‘€, we conclude that [𝑄,𝐴]=[[𝑄,𝐴],𝐴]βŠ†π‘(𝑀). Then, we have π‘€βŠ†π‘. Thus, π΄π‘€βŠ†π‘. Since the reverse inclusion holds as well, we have 𝐴𝑀=𝑁. In view of the fact that π‘„βŠ†π‘ since [𝑄,𝐴]βŠ²π‘„π΄, we obtain 𝐿=βŸ¨π΄π‘€,π‘„βŸ©βŠ†π‘=𝐴𝑀. From the maximality of the subgroup 𝑀, it follows that π‘„βŠ†π‘€.
Further, [𝑀,𝐴]=[𝑀,𝐴]π‘žΓ—π‘‹, where 𝑋 is a Hall π‘žξ…ž-subgroup from [𝑀,𝐴].
Assume that 𝑋≠1.
Let π›ΎβˆˆIrr((𝜏)𝑋). Since 𝑋 is Abelian, it follows that 𝛾(1)=1. Since [𝑀,𝐴]βŠ†kerπœ“, we obtain π‘‹βŠ†kerπœ“. Thus, πœ“π‘‹=πœ“(1)1𝑋.
First assume that π›ΎβˆˆIrr(πœ“π‘‹). Then, 𝛾=1𝑋. Further, 𝑋char[𝑀,𝐴]⊲𝐿, that is, π‘‹βŠ²πΏ. Then, π‘‹βŠ†ker𝜏. Thus, π‘‹βŠ†kerπœ“βˆ©ker𝜏=kerξπœ’=1. Hence, we obtain that 𝑋=1. We get a contradiction with the assumption.
Now, let π›Ύβˆ‰Irr(πœ“π‘‹), that is, (πœ“π‘‹,𝛾)𝑋=0. Since π‘‹βŠ²πΏ, we apply Theorem  D [13] to the subgroup 𝐴𝑀 and the character ξπœ’π΄π‘€=𝜏+πœ“. By this Theorem, ||𝑋||||||Ξ“βˆΆπ΄π‘€πœ(1)=||𝑋||||||||𝐴||ξπœ’(1)𝛾(1)Ξ“βˆΆπ΄π‘€=||𝑋||||||ξπœ’(1)πΊβˆΆπ‘€ξπœ’(1)(3.37) is an integer. On the other hand, we have π‘„βŠ†π‘€, that is, π‘ž does not divide |πΊβˆΆπ‘€|. Therefore, (π‘ž,|𝑋||πΊβˆΆπ‘€|)=1. But π‘ž divides ξπœ’(1). We obtain a contradiction.
Thus, 𝑋=1, that is, [𝑀,𝐴] is a π‘ž-group.
Since [𝑀,𝐴] is a normal π‘ž-subgroup in 𝑀, it follows that [𝑀,𝐴]βŠ†π‘„. It is readily seen that [𝑀,𝐴]=[[𝑀,𝐴],𝐴]βŠ†[𝑄,𝐴]. Therefore, [𝑀,𝐴]=[𝑄,𝐴].
Thus, we have []𝑀=𝑄,𝐴×𝐢𝑀[](𝐴),𝑄,π΄βŠ†π‘(𝑀).(3.38)
Let us ascertain that assertion (1) of Lemma 2.4 holds and 𝐴[𝑄,𝐴]βŠ†kerπœ“.
Hence, 𝑛=2|𝐴|+1 and, by Lemma 2.4, it follows that only (1)–(5), (12) or (15) satisfy.
If assertion (15) of Lemma 2.4 satisfies, Lemma 2.3 and Lemma 2.2 imply that π΄βŠ†kerξπœ’. This contradicts Lemma 3.1. If assertion (12) of Lemma 2.4 satisfies it follows that πΆβŠ†π‘(𝐺). Since π‘€ξ…žβŠ†ker𝜏, we have (ξπœ’π‘€β€²,1𝑀′)𝑀′≠0. By Corollary  6.7 [5], π‘€ξ…žβŠ†kerξπœ’=1. Thus, the subgroup 𝑀 is Abelian. Then, π‘€βŠ†π‘βˆ—=𝑁𝐺(𝑄). Since 𝑄 is Abelian, by Theorem  6.15 [5], we see that π‘βˆ—β‰ πΊ. Since π‘βˆ—βŠ†π‘€, we obtain π‘βˆ—=𝑀. By the Burnside Theorem, the group 𝐺 contains a normal π‘ž-complement, which is, by Lemma 3.3, either an Abelian group, or is contained in πΆβŠ†π‘(𝐺). Thus, 𝐺 is solvable. In this case, the theorem holds [6].
Assume that assertion (3)–(5) of Lemma 2.4 satisfy. Since 𝛽(1)=𝛽1(1)=1 and 𝛽2(1)=1 (assertion (4)), it follows that πΆξ…ž=1, that is, the subgroup 𝐢 is Abelian. Therefore, 𝐢𝑀(𝐴) is Abelian. Then by equality (3.38), the subgroup 𝑀 is Abelian. This case is examined.
Suppose that assertion (2) of Lemma 2.4 satisfies.
In addition, assume that the character 𝛽1 is imprimitive, that is, 𝛽1=πœ†πΆ for a certain linear irreducible character πœ† of the subgroup 𝐢0 of index 2 in 𝐢. Then, it is readily seen that 𝐢0 is Abelian. Thus, the subgroup 𝑄 is Abelian. Since we have [𝑄,𝐴]βŠ†π‘(𝑀) by equality (3.38), and we can think that π‘βˆ—=𝑀, by Theorem  6.4.4 [4], π‘ž divides |πΊβˆΆπΊξ…ž|. Thus, πΊξ…žβ‰ πΊ. This case is also examined.
Now assume that the character 𝛽1 is primitive. By Theorem  14.23 [5], |πΆβˆΆπ‘(𝐢)|=12,24 or 60, where 𝐢=𝐢/ker𝛽1 and, by Lemma  2.27 [5], 𝑍(𝐢)=𝑍(𝛽1)/ker𝛽1. Thus, |πΆβˆΆπ‘(𝛽1)|=12,24 or 60. Since 𝛽(1)=1, it follows that 𝑍(𝛽1)βŠ†π‘(𝛽). Since ker𝛽1∩ker𝛽=1, by Lemma  5 [1], we have 𝑍(𝛽1)=𝑍(𝐢). Then, |πΆβˆΆπ‘(𝐢)|=12,24 or 60. Let π‘žβ‰ 3,5. Then, 𝐢𝑀(𝐴)βŠ†π‘(𝐢). Thus, 𝑄 is Abelian and as above we obtain that π‘ž divides |πΊβˆΆπΊξ…ž|. This case is examined. Suppose that π‘ž=3,5. Then, the factor group πΆπ‘žπ‘(𝐢)/𝑍(𝐢) is cyclic. Since 𝑍(𝐢)βŠ†π‘(πΆπ‘žπ‘(𝐢)), πΆπ‘žπ‘(𝐢) 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 Lemma 2.4 holds, that is, πœ’πΆ=𝛽+|𝐴|𝛽1,𝛽1(1)=1. Then, 𝛽(1)=|𝐴|+1. From Lemma 2.2, it follows that ξπœ’π΄=𝜌𝐴+𝛽1(1)1𝐴. Thus, ξπœ’π΄=𝜌𝐴+(|𝐴|+1)1𝐴, therefore, we have (ξπœ’π΄,1𝐴)𝐴=|𝐴|+2. On the other hand, ξ€·ξπœ’π΄,1𝐴𝐴=ξ€·ξ€·ξπœ’π΄π‘€ξ€Έπ΄,1𝐴𝐴=ξ€·(𝜏+πœ“)𝐴,1𝐴𝐴=ξ€·πœ‡π΄,1𝐴𝐴+ξ€·πœ“π΄,1π΄ξ€Έπ΄ξ€·πœ“=1+𝐴,1𝐴𝐴.(3.39) Here, we take into account that πœ‡π΄=𝜌𝐴. Thus, |𝐴|+2=1+(πœ“π΄,1𝐴)𝐴, that is, (πœ“π΄,1𝐴)𝐴=|𝐴|+1. Since πœ“(1)=|𝐴|+1, we conlude π΄βŠ†kerπœ“. Since, by equality (3.8), [𝑄,𝐴]βŠ†kerπœ“, it follows that 𝐴[𝑄,𝐴]βŠ†kerπœ“.
By equality (3.38) for a complete proof of the lemma, it is enough to make sure that 𝐢=𝐢𝑀(𝐴).
Since 𝐢𝑀(𝐴)βŠ†πΆ, it remains to show that πΆβŠ†πΆπ‘€(𝐴).
Suppose that π‘βˆˆπΆ. Then, we have ξπœ’π΄π‘€π‘=𝜏(𝑐)+πœ“(𝑐). Since π΄βŠ†kerπœ“, we have π΄βŠ†kerπœ“(𝑐). Hence, [𝑄𝑐,𝐴]βŠ†kerπœ“(𝑐). Therefore, ξπœ’π΄[𝑄,𝐴]=(𝜏)𝐴[𝑄,𝐴]+πœ“π΄[𝑄,𝐴]=(𝜏)𝐴[𝑄,𝐴]+πœ“(1)1𝐴[𝑄,𝐴],ξπœ’π΄[𝑄𝑐,𝐴]=ξ‚€πœ1(𝑐)𝐴[𝑄𝑐,𝐴]+πœ“π΄ξ€Ίπ‘„(𝑐)𝑐,𝐴=ξ€·πœ(𝑐)𝐴[𝑄𝑐,𝐴]+πœ“(𝑐)(1)1𝐴[𝑄,𝐴].(3.40) Thus, ξ€·ξπœ’π΄[𝑄,𝐴],1𝐴[𝑄,𝐴]𝐴[𝑄,𝐴]+ξ€·ξπœ’π΄[𝑄𝑐,𝐴],1𝐴[𝑄𝑐,𝐴]𝐴[𝑄𝑐,𝐴]=ξ‚€ξ€·πœπ΄[𝑄,𝐴],1𝐴[𝑄,𝐴]𝐴[𝑄,𝐴]+ξ‚€ξ€·πœ+πœ“(1)(𝑐)𝐴[𝑄𝑐,𝐴],1𝐴[𝑄𝑐,𝐴]𝐴[𝑄𝑐,𝐴]+πœ“(𝑐)ξ€·(1)=2πœ“(1)=2𝛽(1)>𝛽(1)+1=ξπœ’π΄,1𝐴𝐴=ξ€·πœŒπ΄+𝛽(1)1𝐴,1𝐴𝐴β‰₯ξ€·ξπœ’π΄[𝑄,𝐴]∩𝐴[𝑄𝑐,𝐴],1𝐴[𝑄,𝐴]∩𝐴[𝑄𝑐,𝐴]𝐴[𝑄,𝐴]∩𝐴[𝑄𝑐,𝐴].(3.41) Here we take into account that ||𝐴||||𝐴||πœ“(1)=ξπœ’(1)βˆ’πœ(1)=+𝛽(1)βˆ’=𝛽(1),(3.42) and consider that 𝛽(1)=|𝐴|+1>1. Therefore, by the Brouwer Theorem  5.19 [5], the subgroup 𝐴[]𝑄𝐷=𝑄,𝐴,𝐴𝑐,𝐴(3.43) is proper.
Let π·π‘ž be a 𝐴-invariant Sylow π‘ž-subgroup 𝐷 such that [𝑄,𝐴]βŠ†π·π‘ž. Such a subgroup exists by Theorem  6.2.2(iii) [4]. Let 𝑀𝐷 be a maximal 𝐴-invariant subgroup of the group 𝐺, containing π·πœ‹β€². Let us apply the reasoning concerning the group 𝐴𝑀 to the character ξ„žπœ’π΄π‘€π· and the group 𝐴𝑀𝐷. We get 𝑀𝐷=[𝑀𝐷,𝐴]×𝐢𝑀𝐷(𝐴),[𝑀𝐷,𝐴]βŠ†π‘(𝑀𝐷).
Since [𝑄,𝐴]βŠ†π‘€π·, we have [[𝑄,𝐴],𝐴]βŠ†[𝑀𝐷,𝐴]. Since [[𝑄,𝐴],𝐴]=[𝑄,𝐴], it follows that [𝑄,𝐴]βŠ†[𝑀𝐷,𝐴]. Therefore, [𝑄,𝐴]βŠ†π‘(𝑀𝐷). Thus, π‘€π·βŠ†π‘πœ‹β€². Since [𝑀,𝐴]βŠ†π‘(𝑀) and [𝑀,𝐴]=[𝑄,𝐴], as ascertained above, we have π‘€βŠ†π‘πœ‹β€². From the maximality of the subgroup 𝑀 and Lemma 3.4, it follows that 𝑀=π‘πœ‹β€². Then, π‘€π·βŠ†π‘€. Thus, [𝑄𝑐,𝐴]βŠ†π‘€. Since 𝑀=[𝑄,𝐴]×𝐢𝑀(𝐴), as ascertained above, we have [𝑄,𝐴]=[𝑄𝑐,𝐴]=([𝑄,𝐴])𝑐. Therefore, π‘βˆˆπ‘πœ‹β€². Then, cβˆˆπ‘€, that is, πΆβŠ†πΆπ‘€(𝐴).
The lemma is proved.

This completes the proof of the theorem.

Let us show that 𝑛 is a primary number.

Let us assume the converse. Let π‘ž1 and π‘ž2 be distinct odd primes from πœ‹(𝑛).

Let 𝑄1 and 𝑄2 be 𝐴-invariant Sylow π‘žπ‘–-subgroups 𝐺, 𝑁𝑖=𝑁Γ(𝑄𝑖). Assume that 𝑁1=Ξ“. From Lemma 3.3, it follows that 𝑄1 is Abelian or 𝑄1βŠ†πΆ. 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 π‘π‘–βŠ†π΄π‘€π‘–,𝑖=1,2. By Lemma 3.10, 𝑀𝑖=[𝑄𝑖,𝐴]×𝐢,𝑖=1,2, that is, π‘€π‘–βŠ†π‘πΊ(𝐢),𝑖=1,2. Therefore, 𝐢⊲𝐿=βŸ¨π‘€1,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 π‘ž1β‰ π‘ž2, we obtain |[𝑄1,𝐴]|β‰ |[𝑄2,𝐴]|. A contradiction.

Now suppose that 𝐿=𝐺, that is, 𝐢⊲𝐺. Since, by Lemma 3.10, 𝑀=𝐢×[𝑄,𝐴], it follows that π‘€βŠ†πΆπΊ(𝐢)⊲𝐺. By Lemma 3.3, 𝐢𝐺(𝐢) is either Abelian or 𝐢𝐺(𝐢)βŠ†πΆ. In the first case, 𝑀 is Abelian. It was considered above. In the second case, we have π‘€βŠ†πΆ. A contradiction with Lemma 3.2.

Thus, our assumption on the fact that 𝑛 will divide by two distinct odd primes does not satisfy. The Theorem 1.1 is proved.

4. Proof of Theorem 1.2

Let us carry out induction over the order of the group 𝐺.

Let us examine the subgroup π‘‚πœ‹β€²,πœ‹(𝐺). By definition, put π‘‚πœ‹β€²,πœ‹(𝐺)π‘‚πœ‹β€²(𝐺)=π‘‚πœ‹ξ‚΅πΊπ‘‚πœ‹β€²ξ‚Ά(𝐺).(4.1) Since the factor group 𝐺/π‘‚πœ‹β€²(𝐺) is πœ‹-solvable and π‘‚πœ‹β€²(𝐺/π‘‚πœ‹β€²(𝐺)=1, we have π‘‚πœ‹(𝐺/π‘‚πœ‹β€²(𝐺))β‰ 1. Since π»π‘‚πœ‹β€²(𝐺)/π‘‚πœ‹β€²(𝐺) is a πœ‹-Hall 𝑇𝐼-subgroup in 𝐺/π‘‚πœ‹β€²(𝐺), we obtain π‘‚πœ‹β€²,πœ‹(𝐺)=π»π‘‚πœ‹β€²(𝐺). Denote 𝑋=π‘‚πœ‹β€²,πœ‹(𝐺).

Identity substitution of the group 𝐺 defines an exact irreducible character πœ‘ of degree 𝑛.

First let us assume that |πΊβˆΆπ‘‹|β‰ 1. By the Clifford Theorem πœ‘π‘‹=𝑒𝑑𝑖=1πœ‘π‘–,(4.2) where πœ‘π‘–=1,π‘‘βˆˆIrr(𝑋) and 𝑒, 𝑑 are numbers, dividing |πΊβˆΆπ‘‹|.

In addition, assume that 𝑒≠1 or 𝑑≠1. Then, the numbers πœ‘π‘–=1,𝑑(1)<2|𝐻|. Therefore, the factor groups 𝑋/kerπœ‘π‘–,𝑖=1,𝑑 satisfy the theorem proved in [1–3]. If 𝐻kerπœ‘π‘–kerπœ‘π‘–β‹¬π‘‹kerπœ‘π‘–,𝑖=1,𝑑,(4.3) from this theorem, it follows that πœ‘π‘–=1,𝑑(1)=|𝐻|βˆ’1,|𝐻|,|𝐻|+1,2|𝐻|βˆ’2 or 2|𝐻|βˆ’1. Since the numbers πœ‘π‘–=1,𝑑(1) divide πœ‘(1), it is readily seen that the numbers πœ‘π‘–=1,𝑑(1) cannot possess the stated values. Therefore, 𝐻kerπœ‘π‘–kerπœ‘π‘–βŠ΄π‘‹kerπœ‘π‘–,𝑖=1,𝑑,(4.4) that is, 𝑋/kerπœ‘π‘– contain normal πœ‹-Hall subgroups 𝐻kerπœ‘π‘–/kerπœ‘π‘–,𝑖=1,𝑑. Since ⋂𝑑𝑖=1(kerπœ‘π‘–)=1, we see that 𝑋 can be isomorphically embedded in the direct product 𝑋𝑋=kerπœ‘1𝑋×⋯×kerπœ‘π‘‘.(4.5) Since 𝑋 contains a normal πœ‹-Hall subgroup, it follows that 𝑋 also contains a normal πœ‹-Hall subgroup 𝐻, that is, 𝐻⊴𝐺. Commutativily of the subgroup 𝐻 results from the faithfulness of the character πœ‘ and the fact that πœ‘(1) is a πœ‹ξ…ž-number.

Now suppose that 𝑒=𝑑=1. Then, the character πœ‘π‘‹ is irreducible. Since |𝑋|<|𝐺|, by induction we see that either πœ‘π‘‹(1) is a prime power or 𝐻⊴𝐺. Since πœ‘π‘‹(1)=𝑛, in both cases the theorem holds.

Now suppose that |πΊβˆΆπ‘‹|=1, that is, 𝐺=𝑋=π»π‘‚πœ‹β€²(𝑋). Let us apply the notation to Condition B and Theorem 1.1. For this, suppose that 𝑋=Ξ“, 𝐻=𝐴, π‘‚πœ‹β€²(𝑋)=𝐺. Then, πœ‘ is an irreducible character of the group Ξ“=𝐴𝐺. Since (|𝐴|,πœ‘(1))=1, it is readily seen that the character πœ‘πΊ contains an 𝐴-invariant faithful irreducible component πœ‘βˆ—. It is obvious that πœ‘πΊ=πœ‘βˆ—βˆˆIrr𝐴(𝐺).

Suppose that ξ‚Šπœ‘kerβˆ—β‰ 1. Thus, we have ξ‚Šπœ‘kerβˆ—βŠ†π΄. Since 𝐴 is a 𝑇𝐼-subgroup in Ξ“, it follows that π΄βŠ²Ξ“. This contradicts the fact that 𝐴 is a group of nontrivial automorphisms of the group 𝐺.

Therefore, ξ‚Šπœ‘kerβˆ—=1. Since πœ‘βˆ—(1)=πœ‘(1) and, by assumption of the theorem being proved, πœ‘(1)=𝑛 is not a prime power, it follows that ξ‚Šπœ‘βˆ—(1) is not a prime power either. Taking into account that by Lemma  6 [1] 𝐢𝐺(π‘Ž)=𝐢𝐺(𝐴)=𝐢 for each element π‘Žβˆˆπ΄#, by Theorem 1.1 we again see that π΄βŠ²Ξ“. A contradiction.

The theorem is proved.

Acknowledgment

The research is carried out within the Project no. Ph09URB-001 of June 2, 2009.