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 satisfywhere 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).