Abstract
Huppert and Manz have determined the nonsolvable groups whose character degrees are products of at most two prime numbers. In this paper, we change the condition from “degrees of a group are products of at most two prime divisors” to “degrees of all proper groups of a group are products of at most two prime divisors” and determine the structure of finite groups with such condition.
1. Introduction
Let
where the s are different prime divisors of , and define the number of prime divisors of . Assume that all groups are finite in this paper. Let denote the set of all complex irreducible characters of a group , and let be the set of the linear characters of . Denote by the set of irreducible character degrees of a group , i.e., . Usually, a degree means a complex irreducible character degree in this paper. Let
The structure of a finite group with is determined by Isaacs and Passman and Manz; see [1–3], respectively. The influence of Brauer characters with prime-power degrees on the structure of finite groups is considered in [4, 5].
Finite groups with are determined (see [6, 7]). In particular, if is nonabelian simple, then is isomorphic to or where is an alternating group of degree . Recently, Miraali and Robati furthered Huppert’s results and identified almost simple groups whose degrees are divisible by at most two primes; see Theorem 3.6 of [8].
Inspired by the works of [6, 8], we change the condition from “ for a group ” to “ for each proper subgroup of a group ” and will determine the structure of nonsolvable groups whose degrees of all proper subgroups are the direct products of at most two primes. In order to shorten arguments, we give a definition.
Definition 1. Let be a finite group, and let be the set of all proper subgroups of . A group is called a -group if .
By Definition 1, we have the following definition.
Definition 2. A group is named a -group if each is a -group and a non--group otherwise. We call an irreducible character a -character if .
In generality, a -group does not mean that it is a -group.
Example 1. Let , where is an alternating group of degree . By pp. 10 of [9], we can have , so is a -group. On the other hand, has a subgroup isomorphic to . We see that , so there is an irreducible character with . Now, is not a -group and so is a non--group.
In this paper, we prove the following result.
Theorem 3. Let be a nonsolvable -group. Then, one of the following holds: (1) is isomorphic to where is a prime for some primes (possibly equal)(2) is isomorphic to where , and is a prime such that for primes , and (3) is isomorphic to , (4) is isomorphic to with abelian(5) is isomorphic to with for some prime (6) has a normal abelian subgroup such that
The structure of this paper is formed as follows. In Section 2, some results are given which will be used in the proof of our main theorem. In Section 3, we first give the structure of simple -groups and then that of nonsolvable -groups.
In a group , we will use the notation max to denote the set of the maximal proper subgroups with respect to subgroup-order divisibility from . Let be an elementary abelian group of order , denoting the extraspecial group of order by or . Let be the cyclic group of order . Let be the Schur multiplier of a group . For the other notation and notions, we can refer to [9, 10] for instance.
2. Some Lemmas
In this section, some results about elementary number theory, Frobenius groups, and also subgroup structure of a simple classical Lie group are given.
Lemma 4 [11]. The only solution of the Diophantine equation with and primes and is .
Lemma 5 [11, 12]. 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.
In order to prove our main result, we need some information about certain subgroup structure of a nonabelian simple group.
Lemma 6 (Lemma 2 of [13]). Let be a prime power and let be a positive integer. (1)Let . Then, has a subgroup (2)Let , . 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
The following result will be used frequently without reference.
Lemma 7. (1)A group is a -group if and only if for every , is a -subgroup(2)Let be a proper subgroup of a -group. Then, is a -subgroup(3)Let be a nontrivial normal subgroup of a -group . Then, is a -group
Proof. and are obvious by Definition 2
As is nontrivial, we have that is a -group. Assume that is a non--group. Then, is a non--group, so has a non--group for certain . If , then , and is a non--group. It follows that is a non--group, a contradiction. Now, and let be a maximal proper subgroup of with . Then, , , and so is a non--group. It means that is a non--group, a contradiction to the fact .
3. Nonsolvable -Groups
In this section, first, we determine the structure of a nonabelian simple -group and then that of a nonsolvable -group.
It is well-known that a nonabelian simple group is isomorphic to an alternating group , , a simple group of Lie type, or a simple sporadic group. So we consider these groups from now on.
Lemma 8. Let be an alternating group of degree . Assuming that is a -group, then is isomorphic to or .
Proof. An irreducible character of , the symmetric group of degree , is determined by the partition of , and denote such an irreducible character by . Observe that the irreducible characters of are the restrictions of those of to . If and , then by Hook’s formula, one has
See pp. 77 of [14]. Note that for , is not self-conjugate, so the character degree of is the same as that of with respect to the partition . Hence, , a contradiction.
If , then by [9] and Lemma 6, we have a subgroup series:
Note that , so is not a -group; thus, for , is a non--group.
If , then . Observe that , , and , so is a -group.
If , then . As , , and , we have that is also a -group.
It follows that is isomorphic to or , the desired result.
Note that is a -group but a non--group as shown in Example 1.
Lemma 9. Let be a nonabelian simple group of classical Lie type. Assuming that is a -group, then is isomorphic to one of the groups: (1) where with is a prime for some primes (possibly equal)(2) where , is a prime and for primes , and (3)
Proof. A simple group of classical Lie type is isomorphic to , , , , , , , or with . So these groups are considered in what follows.
Case 1: if is isomorphic to , then is isomorphic to one of the groups: where is a prime for some primes (possibly equal); where , is an odd prime and ; , with a prime such that for primes and
Let .
In Lemma 8, we have considered the groups and . So or or . Two cases are considered now.
(a)If is odd, then , so by Table 1, and . Hypothesis shows that for a prime , or for different primes (i)If , then , so is isomorphic to which is considered in Lemma 8. So .
If is a prime, then with . We see from Table 1 that possibly contains as its maximal subgroups, so every is a -group, so is a -group.
If is a prime power, then by Lemma 5, , , or for some prime . If for some prime , we see that is divisible by . It follows from hypothesis and Lemma 4 that and so, a contradiction. It is easy to check that is a -group by Table 1.
(ii) for different primes Without loss of generality, we can assume that . If , then .
If is a prime power, say , then is odd (if is even, forces , a contradiction).
If , then has a subgroup of the form with . Note that or is a subgroup of and that
by [16]. As is odd, one of the numbers and has the form for some , so and are not -groups. It follows that is a prime. If , then is a subgroup of . Since is not a -group for , one has that is not a -group. Thus, . Now, Table 1 shows that possibly contains as its members. Thus, where is an odd prime and with , , is a -group.
If is a prime, then is a prime. By Table 1, we get that is a -group.
(b)If is even, then or , say , Let , then by [9] (pp. 6), , so is a -group.
Let with integers and primes. Without loss of generality, we can assume that , and then, has a subgroup of the form . We see that , so because shows . Thus, say.
If and , then let ; we get that has a subgroup . If , then is a -group as is a -subgroup. If , then let with , , so is not a -group.
If is odd, then or . Note that
and that has a subgroup . Note that , and for , , so is not a -group. If , then with an odd prime and for is a -group.
Let . If , then is considered as above. So . If with a prime, then has as its subgroup. So . By [17],
It is easy to see that for at least one of the numbers , say , we have if is a prime power. Thus, is a prime. By pp. 191 of [18], . We know that and that
by [16], so because one of the numbers and with has the form for some . If , then , so by [19], ; if , then , so by [19], . Thus, with is not a -group. Now, we have shown that with is not a -group and so is as
by [20]. In particular, neither nor is a -group.
Let . If and , then is considered in Lemma 8. Now, , so by Lemma 6, we obtain that
This implies that with is not a -group since and are not -groups.
Case 2: with
Let . If , then is solvable, so . By [9], we have that
so by [19], we get that for is not a -group. Thus, we can assume that ; then, by pp. 200 of [18], where , and . Observe that is a Frobenius group and that there in does exist an irreducible character with (note that , and is abelian, so by Theorem 2.31 of [21], for , ; hence, ).
If is even, then as . If is odd, then and
Observe that is divisible by eight and that , so .
It follows that with is a non--group.
Let . If and , then has as a subgroup. As , is not a -group. Now, assume that ; then, by Lemma 6, we have that
For , by [20], we have that and are not -groups. It follows that with is not a -group.
Case 3: with , odd
If , then by [18] (pp. 213). We know that , so for an odd . Thus, is not a -group. In particular, is not a -group.
Let . Then, by Lemma 6, a subgroup series is obtained:
so is not a -group since is not a -group.
Case 4: with
Let . If , then is not simple, so , and by pp. 209 of [18] has a subgroup . We know that is a normal subgroup of and that , so is not a -group; in particular, is not a -group.
Let . Then from Lemma 6, contains a subgroup , so is not a -group.
Case 5: with and
If , then has a subgroup , so by Case 3, is neither a -group nor a -group.
If , then by Lemma 6, is not a -group as contains a subgroup isomorphic to either with odd or with even. Note that with odd and with are non--groups, so we rule out this case.
Lemma 10. Let be a simple group of exceptional Lie type. Assuming that is a -group, then is isomorphic to .
Proof. We see that is isomorphic to with , with , , , , , , , , or . We deal with these case by case.
The following subgroup series are obtained from Table 2:
We know that and , so and are non--groups since Cases 1 and 2 in Lemma 9 show that and are non-T-groups. So and are not -groups. Note that is not a -group by [9]. From Table 2, , , and are non--groups. Now, if is not a -group, then and are not -groups. Now, we will show that is not a -group. In fact, contains a subgroup or which are not a -group, so is not a -group.
Now, we can conclude that is possibly isomorphic to with or with , . Thus, two cases are considered.
Case 1: with
We know from pp. 385 of [10] that has a maximal subgroup of the form which is a Frobenius group. So there is an irreducible character with a maximal degree with respect to divisibity in . Observe that if , then by Theorems 13.3 and 13.8 of [24],
so by hypothesis, is a prime. On the other hand, is of order , so is a 2-group, so there is a nonlinear character such that divides , i.e., . Notice that . If and let , then , and , so , is an extraspecial 2-group, so . (In fact, we know from (17) that for , is not a -group.) So by [9] (pp. 28), . By [19], we may get that the groups , , , and are -groups, so is a -group.
Case 2: with
From [10] (pp. 398), has a maximal subgroup which is a Frobenius group. Let for . Then, by Theorems 13.3 and 13.8 of [24], . It follows that is not a -group.
Lemma 11. A simple group of sporadic group is not a -group.
Proof. By [9] (pp. 18), and , so is not a -group. Now, [9] (pp. 238) shows that is a subgroup of these groups: , , , , , , , , , , , , , , , , , and .
By checking [9] , , , , , , and are not -groups as is a subgroup of , is a subgroup of , and is a subgroup of .
Theorem 12. Let be a nonabelian simple -group. Then, is isomorphic to the following: (1) where with is a prime for some primes (possibly equal)(2) where , is a prime and for some primes , and (3)(4)
Proof. We conclude the result from Lemmas 8, 10, and 11.
Theorem 13. Let be an almost simple -group with socle , . Then, is isomorphic to one of the groups: (1) where is a prime for some primes (possibly equal)(2) where , and is a prime such that for primes , and (3), (4)
Proof. If is simple, then by Theorem 12, we have (1)-(3). So three cases are considered, and also, we assume that is nonsimple.
Case 1: where is a prime for some primes (possibly equal) and
In this case, is possibly isomorphic to . Note that has a normal subgroup with index and that
by [16], so hypothesis shows that or (in fact, if is odd, then one of the numbers or can be written by for some ). It follows that is isomorphic to .
Case 2: where , is a prime and for primes , and (i)Let If , then is done in Case 1. If , then is possibly isomorphic to . We see that has as its subgroup and that , so hypothesis forces , a contradiction. It means that is not a -group.
(ii)Let If , then the order of the outer-automorphism group of is , so by [9], is possibly isomorphic to , , , or . Notice that is a normal subgroup of these groups (, , , and ) and that , so is not a -group. It follows that the groups , , , and are non--groups.
Let . Then, the order of the outer-automorphism group of is , and so by Corollary 6.5 of [25], possibly has one of the structures: , , , and . Note that , , , and have a subgroup of the form . We know that , so hypothesis shows that . If , then is solvable; if , then . Thus, , , , and are not -groups.
(iii)Let Then, we can get from Corollary 6.5 of [25] that the possible groups are isomorphic to , , , and . Note that these groups always have a subgroup which is not a -group. Thus, , , , and are not -groups
Case 3: and
In this case, the outer-automorphism group of is of order 3, so is possibly isomorphic to . Now, by [9] (pp. 28), , so . It follows that is not a -group and so is not a -group.
We can get from [9] (pp. 12) that and are not -groups as is a subgroup of both and .
Now, we will prove Theorem 3.
Proof. Let be a minimal nonabelian normal subgroup of . Then, we get that
where is isomorphic to a nonabelian simple group for each . If , we obtain by Theorem 4.21 of [21] that for , and , for ,
Thus, is a non--group, a contradiction. Now, we get that is a nonabelian simple group which is isomorphic to one of the groups satisfying Theorem 12.
We know that is isomorphic to a subgroup of and that is isomorphic to a subgroup of , the automorphism group of ; then, is an almost simple group satisfying Theorem 13.
Let be the set of the groups:
(i) where is a prime for some primes (possibly equal)(ii) where , and is a prime such that for primes , and (iii), (iv)If , we have that is isomorphic to and that . From , we conclude that is isomorphic to (if , is a -group, a contradiction). If , one has that is a -group, a contradiction. It follows that is isomorphic to .
If , we may get that is isomorphic to as . Notice that in this case, is isomorphic to . If is nonabelian, we have from the fact, , that is a -group, a contradiction. So is abelian.
If , one has that and that as the order of the Schur multiplier of is two. If , we have
and so is isomorphic to . By hypothesis, is a -group. Let . Then, we easily get that is a -group if for some prime and a non--group if for different primes , , or .
If , we have for some normal subgroup of . If is nonabelian, we assume that for some prime by above arguments. Let with and be the inertia subgroup of in . Then, is isomorphic to , , , , or by [10] (pp. 377). If , we have that , by Theorem 6.16 of [21], and so . Now, by Corollary 11.29 of [21], for , . It follows that there is an irreducible character with . Now, is a non--group, and so we also rule out the case . If , we have that has a subgroup isomorphic to , so hypothesis shows that is a -group. We see that is cyclic and so . Note that where with . It follows from Corollary 6.17 of [21] that , a contradiction. Similarly, we also rule out when . If , we can get a contradiction by Corollary 6.17 of [21] too. It follows that is abelian, the desired result.
4. Conclusion
In this paper, we change the condition from “the degrees of a group are direct products of two prime numbers” to “the degrees of all proper subgroups of a group are direct products of two prime numbers” and get that if a nonsolvable group that all proper subgroups have degrees which are the direct products of at most two prime numbers, then it has a section isomorphic to or for certain . Note that is a -group but not a -group.
Data Availability
No data were used to support this 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), the High-Level Talent of Sichuan University of Arts and Science (Grant No. 2021RC001Z), and the Opening Project of Sichuan Province University Key Laboratory of Bridge Nondestruction Detecting and Engineering Computing (Grant No. 2022QYJ04).