#### Abstract

It is established that degree of irreducible complex linear group with the group of cosimple automorphisms of odd order is a prime number and proved that if degree of -solvable irreducible complex linear group with a -Hall -subgroup is not a prime power, then is Abelian and normal in .

#### 1. Introduction

Suppose that is a finite group and is such a group of its nontrivial automorphisms that . Then, is called a group of cosimple automorphisms of the group and the semidirect product of the group and the group is a group. If for each element , then is said to be a strong-centralized group of cosimple automorphisms of the group .

The theorem, proved in the series of papers , implies that if is a finite irreducible complex linear group of degree with a nontrivial strong-centralized odd-order group of cosimple automorphisms, then or and is a degree of a certain prime number.

In this work, it is shown that the degree 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 -solvable irreducible complex linear group of degree , which is not a prime power, the -Hall -subgroup is Abelian and normal in (Theorem 1.2).

Condition B
Let us say that the group satisfies B, if is uneven, for each element , and the group has an extract irreducible complex character of degree , which is -invariant for at least one element .

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

Condition D
is a -solvable irreducible complex linear group of degree with a nonprimary -Hall -subgroup of odd order.

Theorem 1.2. If the group satisfies Condition and is not a prime power, the subgroup is Abelian and normal in .

#### 2. Definitions, Notation, and Preliminaries

is the set of natural numbers; is the set of nonnegative integers; notation stands for ; if is the character of a certain group, then is the set of all irreducible components of the character ; is the center of the character ; if and is the set of primes, then is a Hall -subgroup of the group . The rest of the notation and definitions are conventional; they can be found, for example, in  or . In what follows, by a group character we always mean a complex character, by a group a finite group.

Suppose that is the group, where and is uneven. Then, it satisfies Theorem  13.1 . According to this theorem, there exists one-to-one correspondence between the set of all -invariant characters of the group and the set of all irreducible characters of the subgroup . The set has a range of properties, depending, in particular, on the properties of the subgroup . Suppose that . Then, by Lemma  13.3 , there is a unique irreducible character of the group such that and . It is called a canonical extension of the character to the group .

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

Lemma 2.2 (see [1, Lemma  10]). If is an irreducible character of the group stated in Condition B, then , where is a regular character of the subgroup and is an irreducible character of the subgroup such that .

Lemma 2.3. If is an irreducible character of the group stated in Condition B, then , where is a character of the subgroup or 0 and .

Proof. The proof of the lemma is a literal repeat of the proof of equality (3) of Lemma  11 to .

Lemma 2.4. If is an irreducible character of the group of degree , stated in Condition B, then one of the following assertions holds: (1);(2);(3);(4);(5);(6);(7);(8);(9);(10);(11);(12);(13);(14);(15).
Here, , and are distinct irreducible characters of the subgroup .

Proof. From Lemma 2.3, it follows that where . Suppose that and is the character or 0 such that Hence, by Lemma 2.3, it follows that Hence, by the assumption of the lemma, it follows that Then, , that is, Hence, .
Assume that and examine all the possible cases separately.
(1) .
Since, by Corollary  2.17 , , by equality (2.1), it follows that . Since , from Lemma 2.3, it follows that , that is, .
Suppose that . Then, we can assume that and . From equality (2.3), we obtain is assertion (1) of the lemma being proved.
Suppose that . Then, .
If , then either and , which implies assertion (2) of the lemma, or and , which implies assertion (3) of the lemma.
But if , then it is clear that and . We obtain assertion (4) of the lemma.
(2) .
Suppose that . From (2.3), we obtain . Since , from the latter equality, it is readily seen that .
Suppose that . Then, and is assertion (5).
Now, suppose that . Since , we have and, therefore, is assertion (6).
Further, suppose that . By (2.3), it follows that Since , we see that .
Let . Then, .
Let . Then, . If , then . Hence, is assertion (7). If , then . Then is assertion (8).
Now suppose that . Then, . We obtain , where is assertion (9).
If , then and it is readily seen that (7), (8) or (9) follow.
Now suppose that . Hence, is assertion (10).
Suppose that . Then, . Assertion (11).(3) .
Let . Expression (2.3) implies that . We obtain . We see that is assertion (12).
Let . Then, , where is assertion (13).
(4) .
Expression (2.3) implies that ,, where is assertion (14).
Now suppose that in Lemma 2.3  . Then, and is assertion (15).
The lemma is proved.

#### 3. Proof of Theorem 1.1

Suppose that is an exact irreducible character of degree of the group given in Condition B, -invariant for at least one element .

Suppose that is a group of the least order that satisfies the assumption of the theorem, but does not satisfy its corollary, that is, .

Lemma 3.1. The character is exact.

Proof. Suppose that . Since the character is exact, we have . Since, by Lemma  1 , is a -subgroup in , it follows that . This contradicts the fact that is the group of nontrivial automorphisms of the group .
The lemma is proved.

Lemma 3.2. Suppose that , is an -invariant Sylow -subgroup . Then, .

Proof. Suppose that . Since, by Lemma  4 , , we have . By Exercise 13.2 , is an integer, where is from Lemma 2.2. Since , we see that does not divide . Then divides . From Lemma 2.2, it follows that . Hence, divides . Since divides , it follows that does not divide , therefore, divides . By using the fact that is an odd prime, it follows that .
Suppose that in Lemma 2.2. Since , we have . This case does not satisfy.
Suppose that . Since , we obtain that is, . Since divides and is an uneven number, it follows that it is enough to consider the cases when or .
Suppose that . Then, . Hence, . Since , by Lemma 2.4, it follows that only such relations satisfy, where . However, in this case . Then, is a prime number. A contradiction with the minimality of the group .
Suppose that . Then, . Therefore, . We have a contradiction.
The lemma is proved.

Lemma 3.3. Suppose that is a proper -invariant normal subgroup of the group . Then either is Abelian, or .

Proof. Let us consider the character where , .
Let us note that by the minimality of the group either , or , that is, for all .
By the Clifford Theorem where , are numbers, which divide , , .
Suppose that . Since , by the Clifford Theorem it follows that divides for each irreducible component of the character and for corresponding elements the equality is true.
Suppose that for a certain character . Since and , we see that for all . Therefore, . Thus, the subgroup is Abelian.
Now suppose that for all and for all . Since is an uneven number, we have . Therefore, if for a certain character , then . Since in this case , we obtain a contradiction with . Thus, for all and for all . Then, by Lemma  1 , , therefore, and for all . By Lemma  13.3 , for every there exists a canonical extension of the character to the group .
Assume that for a certain . By Lemma  9  , . Then, . Let us apply the theorem  to the factor group and its exact irreducible character in terms of Lemma  2.22  . By this theorem, and is a prime power, probably, with the exception of the case, when . Since and divides for all , it follows that . Thus, for a certain odd . It is easily seen that it is not true.
Now suppose that for all . Using the fact that and by Lemma 3.1, then . If , then , since is a -subgroup. If , then the subgroup can be isomorphically embedded in the direct product . Since the group contains a normal Hall -subgroup, we have . Thus, .
The lemma is proved.

Suppose that and is a -invariant Sylow -subgroup of that exists by Theorem  6.2.2(i) . By Lemma 3.2,  .

Further, is an -invariant proper subgroup of such that .

From the minimality of the group and Lemma 3.1, it follows that the character is reducible for every such a subgroup . Denote .

Lemma 3.4. .

Proof. Assume that . By Lemma 3.3, the subgroup is either Abelian or . Since, by Lemma 3.2, , it is enough to consider the case, when is Abelian.
Suppose that is Abelian. Let us consider the character . It is clear that and . Therefore, there exists at least one irreducible component of the character such that and since is an uneven number, by theorem , it follows that for every such a character it is true that and is a prime power. Assume that . Then, divides and divides . Thus, . A contradiction. Thus, .
Suppose that there exists one more character such that . Since and , we see that is a sum of linear irreducible characters . Thus, This yields that the subgroup is Abelian. Therefore, . Suppose that , that is, . Then, is a contradiction with Lemma 3.2. Thus, and since , we see that, by Lemma 3.3, the subgroup is either Abelian or contains it. If is Abelian, then , and we obtain a contradiction with Theorem  6.15 , but if , then we obtain a contradiction with Lemma 3.2.
Assume that , where . By Lemma  3 , . Since and , we have . Then, . This case has been considered above.
The lemma is proved.

By Lemma 3.4, . For each -invariant proper subgroup of the group is such that , we have where and are the characters of the subgroup such that for each and for each .

Then, the group can be isomorphically embedded in the direct product . Since each factor contains a normal Hall -subgroup , we see that . It follows that . By Lemma  2.2.1 , . Then, Since by Lemma  3 , , and , by Lemma  6.13 , we have Assume that . Then, , that is, . Then, , that is, . This contradicts Lemma 3.2.

Thus, . By Theorem , and is a prime power, probably, with the exception of the case, when or .

Lemma 3.5. The character does not contain any irreducible components of degree .

Proof. Assume that there exists and . Then, the group and its irreducible character satisfy Lemma  11 . Since , by Lemma  9 , , and, by Lemma  10 , it yields that assertion (7) of Lemma  11  does not satisfy. Since , we see that only assertion (3) of the mentioned Lemma satisfies, that is, , where . By Lemma  10 , for a certain linear irreducible character of the subgroup . We see that if and if not.
In addition, assume that , that is, the character contains an irreducible component .
Besides, assume that . In the same way, we establish that for a certain linear irreducible character of the subgroup , moreover, , if and in the converse case.
Suppose that for a certain character of the subgroup . Then, Suppose that or 5. Then, or 11 is a prime number. This contradicts the minimality of the group . Thus, . Therefore, . By Theorem  1 , . Clearly, and, as follows from Lemma  11  and Lemma  10 , each irreducible component of the character is restricted irreducibly to , that is, . By Lemma  10 , we see that . Then, it is obvious that . Here, is a canonic extension of the character to the group and are linear irreducible characters of the subgroup . It is readily seen that .
On the other hand, since , we have , , and , where is from Lemma 2.4. Since by Theorem 13.1 , , we have . Since , we see that only assertion (1) or assertion (15) of Lemma 2.4 satisfy, that is, or . Suppose that assertion (1) of Lemma 2.4 satisfies. Then, by Lemma 2.2, Since , we obtain . Hence, . As shown above . A contradiction. Suppose that assertion (15) of Lemma 2.4 satisfies. Then, by Lemma 2.2  , that is, . We obtain a contradiction with Lemma 3.1.
Now assume that . Then, and . Since and , it follows that . Again, we see that only assertion (1) of Lemma 2.4 satisfies. Since in this case , that is, , we have . Thus, , that is, . A contradiction.
There only remains to consider the case, when . Since , then the character is exact. Further, , because if not we have by relation (3.8), and we get a contradiction with Lemma 3.2. Then, the character is also exact and from Lemma7  and Lemma2.27  it yields that , . Hence, , that is, . A contradiction with Lemma 3.2.
The lemma is proved.

Lemma 3.6. Let be a solvable -invariant subgroup of . Assume that has an exact irreducible character of degree . Then, , where .

Proof. Since , we see that , , therefore, , , that is, . From Lemma  1 , it follows that and is a Mersenne prime. Thus, . By Theorem , the assertion of the theorem holds. We obtain a contradiction with the minimality of the group . Therefore, .
Suppose that . Since, by Lemma  7  , the subgroup is a twiddle factor of a certain Frobenius group, it follows that by Corollary  1 , the subgroup contains a subgroup of order . By assumption B, . Since is solvable, by Theorem  6 , it follows that . Since, by Lemma  4 , , we obtain . Assume that . Let us consider the character , , .
Suppose that . Then, . Thus, , therefore, . In this case, the lemma is true.
Suppose that . Then for a certain , let us say that , . From Theorems  1 and 2  and the assumption of the lemma, we obtain , or . Here .
If , then . Hence, , that is, . Thus, . This case is impossible.
Suppose that . Then, where . Therefore, all irreducible components of the character are linear, that is, the subgroup is Abelian. Since , we have . Then, it is not hard to see that . Since and , it follows that , that is, .
Arguing as above for the subgroup , we can show that in the contingency there is an irreducible component such that and or . Assume that . Since the subgroup is a -set in , from Lemma  1 , it yields that , where is an irreducible component of the character . Thus, is an irreducible character of the subgroup . Since , we have . However, since the character is exact, the character is exact. Therefore, . Then, , that is, . This case has already been examined. Therefore, we may assume that . Again, we see that is Abelian, that is, . Therefore, .
Suppose that and . Assume that . Since the subgroup is Abelian, we obtain . Therefore, , that is, , that is, . The lemma is true in this case.
Now suppose that . By Theorem  6.11 , there exists an irreducible character of the subgroup such that . Hence, . Thus, and is degree 2. Therefore, for a certain element . Assume that . Then, . Hence, . The lemma is true.
Therefore, . Let us show that .
Let us assume the converse, that is, . Since by Frobenius reciprocity law for characters we have . Since , we obtain Since , by Theorems  1 and 2 , we see that . By using the fact that , it follows that . Then, . Since the numbers and are of degree 2 and , we obtain , that is, . Since , we have , that is, . In this case, as shown above, the lemma is true.
We may assume that . Then, . Since , and , from Theorem  1 and 2 , it easily follows that the character is irreducible. Therefore, all irreducible components of the character are linear, that is, the subgroup is Abelian. Since is Abelian and , we have . Then, . Since , we have Further, Hence, for each , we obtain that , where and . Since , for each the following expression satisfies Here, , if and , if . Since , we obtain . The latter chain of equalities implies that
Since , we have . Then, . By Lemma  2.27(e) , . Hence, , that is , therefore, . The lemma is true.
There only remains to consider the case, when . Since the subgroup is a -set in , from Lemma  1 , it follows that , where is an irreducible component of the character . Thus, is an irreducible character of the subgroup . Since the subgroup is nilpotent, by Theorem  4.21 , we have , where is an irreducible character of the Sylow 2-subgroup from and is an irreducible character of the Hall 2′-subgroup from . Since , we have , that is, . Then, Therefore, where .
The lemma is proved.

Lemma 3.7. .

Proof. Let us assume the converse, that is, . Then, . Since, by Lemma 3.5, for , we have Suppose that . Then, . As mentioned above . By Theorem  1 , . Thus, . This contradicts the assumption. Hence, and since by the assumption , we obtain where and is degree 2.
From the Clifford Theorem, it follows that . Since , it follows that the subgroup is Abelian. By Theorem , the factor group is solvable. Then, is also solvable. In the course of proving Lemma 3.6, it was established that . Likewise, it is shown that .
On the other hand, by Lemma 3.6, where . Here, . Since and for all elements , we have . Thus,
Since and , we have . Since , it follows that . By using the fact that and are -numbers and from the Clifford Theorem, it is readily seen that the subgroup is Abelian. Since , it yields that is also Abelian. Therefore, . Since , we have . Then, . Hence and by the equality (3.27), it follows that .
Further, we see that Since , it follows that . By the fact that the character is exact and Lemma 3.1 , it follows that . Then, . Since and , we obtain . By Lemma 3.4, we have .
Let us consider the character . Clearly, it is enough to consider the case, when for a certain character . Then, or . Thus, we have .
Assume that there exists a character of the group such that and . Then, , therefore, . Hence, . Thus, we have . A contradiction with the assumption.
Thus, we have and can say that and . Let us apply the reasoning analogous to the reasoning applied to the group and its characters and to the group and characters and . We obtain and Here, we have .
Suppose that and is an invariant Sylow -subgroup . Then, . Let . By Theorem  6.11 , we have for a certain irreducible character of the group . Thus, . Since is degree 2, we see that the numbers and are also degrees 2.
Assume that . By the same theorem, we have for a certain irreducible character of the group . Then we have . Since is degree 2, we obtain . Then, for a certain natural number . Thus, we have . Since is a -number, we have . Then, and, obviously, . By Lemma  5 , . This means that .
Now assume that . By Theorem  D , is an integer. Since , we see that is a -number. Since is a -number and divides , it follows that is not an integer. We obtain a contradiction. Thus, we have , therefore, is a -group.
It is obvious that . Since , we obtain . Then, we have . Since , it follows that . Thus, we have . Then, . Hence, . Therefore, we have . Thus, , that is, . This contradicts the original assumption.
The lemma is proved.

Lemma 3.8. The character is an exact irreducible character of degree , or and is a prime power, probably, with the exception of the case, when .

Proof. By Lemma 3.7, we have .
As stated above, the character cannot be irreducible.
Let . Then, we have . Suppose that . Then, . By Lemma 3.3, the subgroup is either Abelian or . If is Abelian, is solvable. By Theorem  1 , is a prime power. This contradicts the minimality of the group . Now let . Then, we have . Since , it is readily seen from Lemma 2.4 that the character contains a linear component or a linear component . From the Clifford Theorem, it follows that is Abelian. Thus, is Abelian. The case is examined.
Thus, . Then, we have . By Corollary  6.7 , . Since the character is exact, we have . Since is a group of nontrivial automorphisms of the group , it follows that the latter statement is impossible.
Therefore, we obtain . Then, . From Lemma 3.5, it follows that the character is irreducible. Since , we see that the character is exact. Suppose that the character is not exact. Then, we have . Since is a -subgroup, it follows that . Thus we have . From formula (3.8), it follows that . A contradiction with the choice of the subgroup . Thus, the character is exact.
Since , we have for a certain character . By Theorem , we have and is a prime power, probably, with the exception of the case when . Besides, it is readily seen that when .
Let us consider this case separately. Let . Since , subject to Lemma 3.5, we have and by the Clifford Theorem, we obtain , where . Hence, by Lemma  7 , we have where , . Since , we see that . Since , by Lemma  5 , we have . Then . A contradiction with Lemma 3.2. Thus, and is a prime power, probably, with the exception of the case, when . Since under the value indicated, it follows that and the character is exact, the lemma is proved.

Lemma 3.9. .

Proof. By Lemma 3.8, it is enough to show that and .
First, let . By Theorem , the group is solvable. Then by Lemma 3.6, we have , where . From equality (3.8), it follows that . We obtain a contradiction with the choice of the subgroup .
Now suppose that or . By Lemma 3.8, is a prime power .
Let be a -invariant Sylow -subgroup from and . Since, by Lemma 3.8, the character is exact and irreducible, by Lemma  2.27 , it follows that is a cyclic group. Therefore, , that is, . Since is Abelian, it follows that . It can be easily seen that divides .
Assume that