#### 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 [1]); 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 [5]). *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 [1], . 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 [1], 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 [7].

*Example 4. *Let be the Mathieu group of degree 11. By pp.18 in [6], has a maximal subgroup H isomorphic to . By [7], 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 [8], , the symmetric group of degree , has a subgroup with . Take , , and so has a subgroup as .(2)By Proposition 4.1.4 of [9], has a maximal subgroup isomorphic to Take , 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 [9], has a subgroup .(4)By Table 3.5.D of [9], has a subgroup with form with , . Taking , has a subgroup .(5)By Table 3.5.E of [9], 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 [9], 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 [10]). *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 [12]). *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 [13]). *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 satisfy*

*where is the regular character of .*

The following result is due to White [14].

Lemma 7 (see Theorem A of [14]). *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 [6]) and , a contradiction. It follows that is not an -group and also not a -group. Let . We know from Appendix in [8] 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 [6], . 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 [6] and Theorem 4.21 of [1]. 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 .

##### 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 [6], all maximal subgroups of are of the forms and . Note that and , so is an -group. Now, . If , then by pp.5 in [6], ; 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 3–5, , 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.

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 [6], has maximal subgroups with the forms: , , and , so by GAP, one obtains that , so . If , then by pp.23 in [6], 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.

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 [6], has a subgroup and , a contradiction to the hypothesis.

###### 2.1.3. Let

We know from [17] 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 [6], so , a contradiction. Hence, , so by pp.200 in [16], has a subgroup . Notice that implies . Thus, by Table 2 in [18], . By hypothesis, , so or by Lemmas 3 and 5. If , then has a maximal subgroup , so by [7], , a contradiction. If , then, by pp.66 in [6], has a subgroup , whence by [7], , a contradiction to the assumption.

###### 2.2.2. Let

We see from [17] thatso is not a -group.

By Lemma 2, one hasso for is not a -group.

##### 2.3. with , with , Odd, with , or a Simple Group of Exceptional Lie Type

First, we consider the simple groups of exceptional Lie type, where : has abelian Hall subgroups and of order and , respectively; is a Frobenius group with kernel , , and a cyclic noninvariant factor of order 6. denotes the central product of and . Consider ; then, by Table 4, has a Frobenius subgroup with cyclic complement of order 6 and abelian kernel of order , i.e., is isomorphic to a cyclic group of order 6. It follows that is abelian, so . This means that there is a proper subgroup of and a character such that is divisible by 6, so is not a -character. It follows that is a non--group. Now, we do with other cases. By Table 4, we obtain subgroup series and either or Note that and that and are non--groups, so , , , , , and are non--groups. Now, consider with ; then, by Theorem 9 of [21], has a subgroup , and , where and , for . It follows that is not a -group, so is a non--group. Second, we consider the groups: with , with , odd, with .Notice that , and with , , and are non--groups. It follows from Table 5 and Lemma 2 that the groups with , with , is odd, and with are non--group. So, we need to consider . We know from pp. 89 in [6] that has a as a subgroup. As , contains a non--subgroup, so is a non--group.

#### 3. Proof of Theorem 2

Now, we will prove Theorem 2.

Let for some prime . Note that the order of outer-automorphism group of is

##### 3.1. Proof of Theorem 2

*Proof. *If is simple, then by Theorem 3, is isomorphic to one of the groups. , where , is a prime with an odd prime or and a prime; ; . So, if is nonsimple, then we consider the following cases: with a prime, and , , and is a prime with an odd prime and . Case 1: with a prime. If , then and . As is a not simple group, is possible isomorphic to , where is the symmetric group of degree . By pp. 2 in [6], . Note that , , , and , so is an -group. Thus, . If , then, by pp. 6 in [6], , so is not an -group. If , then we easily get that does divides the order of , so is a maximal subgroup of . We see that for is not a -group, so is a non--group. Case 2: . Let . Now, by pp. 4 in [6], , so is possibly isomorphic to . Note that and that is a non--group, so each is a non--group. Case 3: . Now, . Thus, possibly is isomorphic to , , , or . It is easy to see that is a maximal subgroup of . So, is a non--group as is not a -group and so are and . We know that has a normal subgroup of index 2, but is a non--group. Thus, is a non--group. Case 4: is a prime with an odd prime. In this case, . We know that and that is a non--group. Hence, is not an -group. Case 5: . Now, . From pp. 13 in [6], , so is not a prime power. Thus, is a non--group.

#### Data Availability

No data were used to support the findings of the study.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

The project was supported by the NSF of China (Grant no 11871360) and Opening Project of Sichuan Province University Key Laborstory of Bridge Non-destruction Detecting and Engineering Computing (Grant no 2019QYJ02).