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## Classical Problems in Algebra and Number Theory

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Research Article | Open Access

Volume 2021 |Article ID 6345386 | https://doi.org/10.1155/2021/6345386

Shitian Liu, "On Groups Whose Irreducible Character Degrees of All Proper Subgroups are All Prime Powers", Journal of Mathematics, vol. 2021, Article ID 6345386, 7 pages, 2021. https://doi.org/10.1155/2021/6345386

# On Groups Whose Irreducible Character Degrees of All Proper Subgroups are All Prime Powers

Accepted02 Jun 2021
Published16 Jun 2021

#### Abstract

Isaacs, Passman, and Manz have determined the structure of finite groups whose each degree of the irreducible characters is a prime power. In particular, if is a nonsolvable group and every character degree of a group is a prime power, then is isomorphic to , where and is abelian. In this paper, we change the condition, each character degree of a group is a prime power, into the condition, each character degree of the proper subgroups of a group is a prime power, and give the structure of almost simple groups whose character degrees of all proper subgroups are all prime powers.

#### 1. Introduction

In this paper, we assume that all groups are finite. Denote by the set of all complex irreducible characters of a group , write the set of linear characters of by , and denote the set of character degrees of a group by .

Let be a group and let be different primes. If , then either has an abelian normal subgroup of index or for a prime and is the direct product of a -group and an abelian group (see Theorem 12.5 in ); Isaacs and Passman have identified the structure of groups with and (see [2, 3], respectively).

In this paper, the following problem is considered.

Question: what can we say about the structure of a group if irreducible character degrees of all proper subgroups are all of prime powers.

In order to argue in short, we introduce the following definition.

Definition 1. An irreducible character is a -character if is a prime power, and a non--character otherwise. A group is named a -group if, for every , is a -character, and a non--group otherwise. We also assume that an abelian group is a -group.
Manz in [4, 5] has determined the structure of -groups. In particular, nonsolvable -groups are determined.

Theorem 1 (see ). If is a nonsolvable -group, then , where is abelian and is isomorphic to or .
In order to talk conveniently, we give a definition.

Definition 2. Let be the set of all proper subgroups of a group . A group is called an -group if, for each , is a -group, and a non--group otherwise.
We will prove the following results.

Theorem 2. Let be an almost simple group such that , where is a nonabelian simple group. Assume that is an -group, then is isomorphic to one of the groups: , where is a prime with an odd prime or and is a prime, , , , or .

We will introduce the structure of this paper. In Section 2, we give the structure of simple -groups and some properties of -groups; in Section 3, we give the structure of almost simple -groups. For a group , let be the set of all proper subgroups of and be the set of representatives of maximal subgroups of . The notation and notions are standard, see [1, 6], for instance.

#### 2. Some Lemmas

In this section, we will give some needed results that are used to prove main theorem. We first give some basic results of -group, then some subgroup of the nonabelian simple groups is given, and finally the structure of the nonabelian simple -groups is determined. Now, we give some results about -groups.

Lemma 1. (1) is an -group if and only if, for each , is a -group.(2)Let be an -group and let be a nontrivial normal subgroup of . Then, and are -groups.(3)A -group must be an -group.

Proof. (1)We can get the result by Definition 2.(2) is a -group by Definition 2, and . If is not a -group, then there exists a subgroup with , so and , is a -group. Since is not a -group, there is a character which is not a -character. It follows that is not a -group, a contradiction.(3)Let be a -group. We will show that is an -group. If is not an -group, then there is a subgroup and an irreducible character such that is a non--character. Let , the inertia group of in . Let such that . Then, by Theorem 6.5 of , . We know that , so there is a maximal subgroup and an irreducible character such that is not a -character, so is not a -group. It follows that is not a -group, a contradiction.In generality, a group, which is the direct product of two -groups, is not an -group.

Example 1. Let be nonabelian - and -groups for different primes , respectively. Also, assume that and . Then, is not an -group. In fact, let such that . Obviously is a subgroup of . However, by Theorem 4.21 of , there is a nonlinear character such that is not a -character.

Example 2. Let be nonabelian -group with , , and . Then, is not an -group. In fact, and are -groups. Let ; then, there are a maximal proper subgroup of and an irreducible character , so has a nonlinear character for some and is a non--character.
There are simple groups whose certain subgroup is a -group, but itself not.

Example 3. Let be the simple sporadic group, and let be the Sylow 2-subgroup of . Say , the normalizer of in . Then, is a -group since by .

Example 4. Let be the Mathieu group of degree 11. By pp.18 in , has a maximal subgroup H isomorphic to . By , and so is a -group.
We need some subgroup structure of finite simple groups to shorten the proof of Theorem 3.

Lemma 2. Let be a prime power.(1)Let . Then, has a subgroup .(2)Let and . Then, has a subgroup isomorphic to or , and has a subgroup of the form .(3)Let . Then, has a subgroup .(4)Let , and odd. Then, contains a subgroup .(5)Let , . Then, has a subgroup with odd or with even.Let and . Let and . Let .

Proof. (1)By Appendix in , , the symmetric group of degree , has a subgroup with . Take , , and so has a subgroup as .(2)By Proposition 4.1.4 of , has a maximal subgroup isomorphic toTake , and has a subgroup isomorphic to or .We know that has a subgroup , and . Note that is isomorphic to a cyclic group of order , so has a subgroup isomorphic to .(3)Note that for odd and for even . Then, by Table 3.5.C of , has a subgroup .(4)By Table 3.5.D of , has a subgroup with form with , . Taking , has a subgroup .(5)By Table 3.5.E of , has a subgroup of the form or , with , , odd if odd, or the form with even.

Taking and , has a subgroup or .

By Table 3.5.F of , we also have that has a subgroup of the form either or for odd or with even.

Taking and , has for odd or for even .

The following three lemmas are important to give the structure of the simple groups as Theorem 3.

Lemma 3 (see ). The only solution of the diophantine equation with p and q prime and is .

Lemma 4 (see [10, 11]). With the exceptions of the relations and , every solution of the equation with prime, , has exponents , i.e., it comes from a unit of the quadratic field for which the coefficients and are primes.

Lemma 5 (Zsigmondy theorem, see ). Let be a prime and let be a positive integer. Then, one of the following holds:(i)There is a primitive prime for , that is, , but , for every (ii) or 6(iii) is a Mersenne prime and We say that a group acts Frobeniusly on a group if is a Frobenius group with kernel and complement . A group is a 2-Frobenius group if there is a normal series such that and are Frobenius groups with , as their Frobenius kernels, respectively.

Lemma 6 (see Theorems 13.3 and 13.8 of ). Let be a Frobenius group with kernel and complement . Then,(i).(ii)Any subgroup of of order or is cyclic, where and are primes.(iii)If is even, is abelian.(iv)In any case, is nilpotent.(v)Assume that has conjugacy classes, has conjugacy classes, and has characters with forms:(a) is irreducible characters with in their kernel; if are the irreducible characters of , then these satisfy , for all and .(b)Whenever is an irreducible character of , then is an irreducible character of . This gives irreducible characters of with not in their kernel. Such satisfywhere is the regular character of .

The following result is due to White .

Lemma 7 (see Theorem A of ). Let , with , for some prime , , and let . Given if , if and set for odd. If is odd, set . The set of irreducible character degrees of iswith the following exceptions:(i)If is odd with or if , then is not a degree of (ii)If is odd, , and , then (iii)If is odd, , and , then (iv)If is odd, or 5, and , then (v)If , or 3, and , then

Theorem 3. Let be a nonabelian simple -group. Then, is isomorphic to one of the groups:(1), , and , where is a prime with prime or and is a prime(2)

Proof. It is well-known that a nonabelian simple group is isomorphic to a sporadic simple group, a simple group of exceptional Lie type, an alternating group, or a simple group of classical Lie type. Thus, in the following, we consider the simple groups case by case.Claim 1: if is isomorphic to with , then is isomorphic to or .If , then has a subgroup by (pp.10 in ) and , a contradiction. It follows that is not an -group and also not a -group.Let . We know from Appendix in  that the symmetric group of degree contains a subgroup , so . It follows from Lemma 2 that there exists a subgroup series:so with is not an -group.If , then by pp.4 in , . We can see that , , and , so is an -group by Lemma 1. If , then by Theorem 1 and Lemma 1, is an -group.It follows that is isomorphic to or .Claim 2: a sporadic simple group or a Tits group is not an -group.Let be a sporadic simple group or a Tits group and let be a proper subgroup of ; then, we can get Table 1 by  and Theorem 4.21 of . For each , is not a -character. It follows that a sporadic simple group or a Tits group is a non--group.Claim 3: if is isomorphic to a simple group of classical Lie type, then is isomorphic to , , , where is a prime with a prime or and is a prime, and .

 Character Degree 2.5 2.5 3.5 2.5 2.11 2.3 2.3 2.3 5.7 5.11 2.5 2.5 2.5 2.3 3.7 2.7 2.3 2.11 5.11 3.7 2.3.17 2.5
##### 2.1. ,
###### 2.1.1. Let

If or 8, then and are -groups by Theorem 1, so Lemma 1 (3) implies that and are -groups. It follows that is isomorphic to or . Thus, or . If , then by pp.3 in , all maximal subgroups of are of the forms and . Note that and , so is an -group. Now, . If , then by pp.5 in , ; hence , is an -group since is a -group, and . Therefore, is isomorphic to or .

By Table 2, has a subgroup of the form . Note from Lemma 6 that , so by hypothesis, is a prime power or a prime. Two cases now are considered as follows: odd and even.Case 1: odd.Let with . We first assume that is a prime. If is even, then divides . Note that , for some , so assumption shows for some . Now, Lemma 3 shows that has no solution in . If is odd, then . It is easy to see that is odd, so cannot be a prime. Assume now that is a prime power and say for some integer and a prime .(i)If , then , but is solvable, a contradiction.(ii)If and , then by Lemmas 35, , or is a prime. It follows that is isomorphic to with a prime.(iii)If and , then by Lemma 4, , , and or (rule out this case since 8 divides ). So, we have .Let be a prime. Then, the fact that is a prime or a prime power implies that , where is an integer and is a prime, so with , for some integer and a prime.Case 2: even.

 Condition , a prime,

Now, Table 2 implies that has a maximal subgroup . By Lemma 6, . By assumption, is a prime or a prime power. If is a prime and say , then with , a prime. If is a prime power, then by Lemma 3, it is impossible.

###### 2.1.2. Let

If , then since , one has that is an -group. If , then by , has maximal subgroups with the forms: , , and , so by GAP, one obtains that , so . If , then by pp.23 in , has a maximal subgroup , but , a contradiction to the hypothesis. Now, we assume that . By Table 3, has a subgroup . Note that . By assumption, and are prime powers. Note that are consecutive integers, so one of the numbers is divisible by 3.

 Condition , prime odd ,

Let be odd with . Then, for some integer . Since , one has that and that and are divisible by 2. It follows that , a contradiction.

Let be even with . Then, 3 divides or . If or , for some , then, by Lemma 3, a contradiction. If or , for some , then or by Lemma 3. If , then, by pp.74 in , has a subgroup and , a contradiction to the hypothesis.

###### 2.1.3. Let

We know from  thatso is not a -group.

Now, Lemma 2 shows that there exists a subgroup series:so for is a non--group.

##### 2.2. ,
###### 2.2.1. Let

If , then is solvable, so . If , then has a subgroup by pp.14 in , so , a contradiction. Hence, , so by pp.200 in , has a subgroup . Notice that <