# Nonsolvable Groups Whose Degrees of All Proper Subgroups Are the Direct Products of at Most Two Prime Numbers

Authors: Shitian Liu; Xingzheng Tang
Journal: Journal of Mathematics
 (2022)

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## 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.


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## Body

## 1. Introduction

Let(1)n=∏i=1kpiai,where thepis are different prime divisors of n, and define
(2)ωn=∑i=1nai,the number of prime divisors of n. Assume that all groups are finite in this paper. Let IrrG denote the set of all complex irreducible characters of a group G, and let LinG be the set of the linear characters of G. Denote by cdG the set of irreducible character degrees of a group G, i.e., cdG=χ1:χ∈IrrG. Usually, a degree means a complex irreducible character degree in this paper. Let
(3)ωG=maxd∈cdGωd.The structure of a finite groupG with ωG=1 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 groupsG with ωG=2 are determined (see [6, 7]). In particular, if G is nonabelian simple, then G is isomorphic to A5 or A7 where An is an alternating group of degree n. 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 “ωG=2 for a group G” to “ωH≤2 for each proper subgroup H of a group G” 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.
LetG be a finite group, and let ∑G be the set of all proper subgroups of G. A group G is called a T-group if ωG≤2.By Definition1, we have the following definition.Definition 2.
A groupG is named a TS-group if each H∈∑G is a T-group and a non-TS-group otherwise. We call an irreducible character χ∈IrrG a T-character if ωχ1≤2.In generality, aT-group does not mean that it is a TS-group.Example 1.
LetG=A7, where An is an alternating group of degree n. By pp. 10 of [9], we can have cdA7=1,2·3,2·5,2·7,3·5,3·7,5·7, so A7 is a T-group. On the other hand, A7 has a subgroup isomorphic to PSL27. We see that 8∈cdPSL27, so there is an irreducible character χ∈IrrPSL27 with ωχ1=3. Now, PSL27 is not a T-group and so A7 is a non-TS-group.In this paper, we prove the following result.Theorem 3.
LetG be a nonsolvable TS-group. Then, one of the following holds:
(1)
G is isomorphic to PSL2q where q=2rs+1 is a prime for some primes r,s (possibly equal)(2)
G is isomorphic to PSL2pm where p∈2,3,5, and m is a prime such that pm−1/gcd2,q−1=rsa for primes r,s, and a∈0,1(3)
G is isomorphic to 2B28, PSL224(4)
G is isomorphic to S5×A with A abelian(5)
G is isomorphic to A5×A with cdA⊆1,p for some prime p(6)
G has a normal abelian subgroup M such that G/M≅SL25The structure of this paper is formed as follows. In Section2, 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 TS-groups and then that of nonsolvable TS-groups.In a groupG, we will use the notation max G to denote the set of the maximal proper subgroups with respect to subgroup-order divisibility from ∑G. Let Eq be an elementary abelian group of order q, denoting the extraspecial group of order q1+2m by ESq1+2m or q1+2m. Let Cn be the cyclic group of order n. Let MultG be the Schur multiplier of a group G. 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 equationpm−qn=1 with p and q primes and m,n>1 is 32−23=1.Lemma 5 [11, 12].
With the exceptions of the relations2392−2·134=−1 and 35−2·112=1, every solution of the equation pm−2qn=±1 with p,q prime, m,n>1, has exponents m=n=2; i.e., it comes from a unit p−q·21/2 of the quadratic field ℚ21/2 for which the coefficients p and q 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]).
Letq be a prime power and let n be a positive integer.
(1)
Letn≥8. Then, An has a subgroup An−1(2)
Letn≥4, ε=±. Then, PSLnεq has a subgroup isomorphic to SLn−1±q or PSLn−1±q, and SLnεq has a subgroup of the form SLn−1εq(3)
Letn≥2. Then, PSp2nq has a subgroup PSp2n−1q(4)
Letn≥3 and q odd. Then, Ω2n+1q contains a subgroup Ω2n−1q(5)
Letn≥4, ε=±. Then, PΩ2nεq has a subgroup PΩ2n−2εq with q odd or PSp2n−2q with q evenThe following result will be used frequently without reference.Lemma 7.
(1)
A group is aTS-group G if and only if for every H∈∑G, H is a T-subgroup(2)
LetN be a proper subgroup of a TS-group. Then, N is a T-subgroup(3)
LetN be a nontrivial normal subgroup of a TS-group G. Then, G/N is a T-groupProof.
1 and 2 are obvious by Definition 2
AsN is nontrivial, we have that N is a T-group. Assume that G/N is a non-T-group. Then, G/N is a non-TS-group, so G/N has a non-T-group MN/N for certain M∈maxG. If N≤ΦG, then N≤M, and M/N∩M≅MN/N is a non-T-group. It follows that M is a non-T-group, a contradiction. Now, N≰ΦG and let M be a maximal proper subgroup of G with N≰M. Then, M<G, G=MN, and so G/N=MN/N≅M/M∩N is a non-T-group. It means that M is a non-T-group, a contradiction to the fact M<G.


## 3. NonsolvableTS-Groups

In this section, first, we determine the structure of a nonabelian simpleTS-group and then that of a nonsolvable TS-group.It is well-known that a nonabelian simple group is isomorphic to an alternating groupAn, n≥5, a simple group of Lie type, or a simple sporadic group. So we consider these groups from now on.Lemma 8.
LetG be an alternating group An of degree n≥5. Assuming that G is a TS-group, then G is isomorphic to A5 or A6.Proof.
An irreducible character ofSn, the symmetric group of degree n, is determined by the partition λ of n, and denote such an irreducible character by χλ. Observe that the irreducible characters of An are the restrictions of those of Sn to An. If n≥14 and λ=n−3,13, then by Hook’s formula, one has
(4)χλ1=n!n·n−4!·3!=n−1n−2n−36.See pp. 77 of [14]. Note that for n≥14, λ=n−3,13 is not self-conjugate, so the character degree of Sn is the same as that of An with respect to the partition λ. Hence, ωχλ1>3, a contradiction.If7≤n≤13, then by [9] and Lemma 6, we have a subgroup series:
(5)PSL27<A7<A8<⋯<A13.Note that23∈cdPSL27, so PSL27 is not a T-group; thus, for 13≥n≥7, An is a non-TS-group.Ifn=5, then maxA5=A4,D10,S3. Observe that cdA4=1,3, cdD10=1,2, and cdS3=1,2, so A5 is a TS-group.Ifn=6, then maxA6=A5,32:4,S4. As cdA5=1,3,4,5, cd32:4=1,22, and cdS4=1,2,3, we have that A6 is also a TS-group.It follows thatG is isomorphic to A5 or A6, the desired result.Note thatA7 is a T-group but a non-TS-group as shown in Example 1.Lemma 9.
LetG be a nonabelian simple group of classical Lie type. Assuming that G is a TS-group, then G is isomorphic to one of the groups:
(1)
PSL2q where q=2rsa+1 with a∈0,1 is a prime for some primes r,s (possibly equal)(2)
PSL2pm where p∈2,3,5, m is a prime and pm−1/gcd2,pm−1=rsa for primes r,s, and a∈0,1(3)
PSL224Proof.
A simple group of classical Lie type is isomorphic toPSLnq, n≥2, PSUnq, n≥3, Ω2n+1q, n≥3, PSp2nq, or PΩ2n±q with n≥4. So these groups are considered in what follows.
Case 1: ifG is isomorphic to PSLnq, then G is isomorphic to one of the groups: PSL2q where q=2rs+1 is a prime for some primes r,s (possibly equal); PSL2pm where p∈3,5, m is an odd prime and pm−1=2rsa,a∈0,1; PSL224, PSL22r with r a prime such that 2r−1=r1r2a2 for primes r1,r2 and a2∈0,1
Letn=2.
In Lemma8, we have considered the groups PSL24≅PSL25≅A5 and PSL29≅A6. So q=7 or q=8 or q≥11. Two cases are considered now.
(a)
Ifq is odd, then k=gcd2,q−1=2, so by Table 1, Eq:Cq−1/2∈maxPSL2q and q−1/2∈cdEq:Cq−1/2. Hypothesis shows that q−1/2=r2 for a prime r, or q−1/2=rs for different primes r,s(i)
q−1/2=r2
Ifr=2, then q=9, so G is isomorphic to PSL29≅A6 which is considered in Lemma 8. So r≥3.
Ifq is a prime, then q=2r2+1 with gcd2,r=1. We see from Table 1 that PSL2q possibly contains Eq:Cq−1/2,Dq+1,Dq+1,S4,A5,A4 as its maximal subgroups, so every H∈maxPSL2q is a T-group, so PSL2q is a TS-group.
Ifq is a prime power, then by Lemma 5, q=35, r=11, or q=p2 for some prime p≥5. If q=p2 for some prime p≥5, we see that q−1/2=p2−1/2=p−1p+1/2 is divisible by 4=22. It follows from hypothesis and Lemma 4 that q=9 and sop=3≥5, a contradiction. It is easy to check that PSL235 is a TS-group by Table 1.
(ii)
q−1/2=rs for different primes r,s
Without loss of generality, we can assume thats>r. If r=2, then q=4s+1≥13.
Ifq is a prime power, say q=pm, then m≥3 is odd (if m is even, 8q−1 forces 4=rs≥6, a contradiction).
Ifm=r1a1r2a2, then PSL2q has a subgroup of the form PSL2q0.gcd2,r1 with q=q0r1. Note that PSL2q0.2 or PSL2q0 is a subgroup of PSL2q and that
(6)cdPSL2q0=1,q0+−1q−1/22,q0,q0−1,q0+1,cdPSL2q0.2=1,q0,q0−1,q0+1,
by [16]. As q0>9 is odd, one of the numbers q0−1 and q0+1 has the form 4s for some s>1, so PSL2q0.2 and PSL2q0 are not T-groups. It follows that m≥3 is a prime. If p≥7, then PSL2p is a subgroup of PSL2q. Since PSL2p is not a T-group for p≥7, one has that PSL2q is not a TS-group. Thus, p=3,5. Now, Table 1 shows that maxPSL2pm possibly contains Eq:Cq−1/2,Dq+1,Dq+1,S4,A5,A4 as its members. Thus, PSL2pm where m is an odd prime and p∈3,5 with pm−1=2rsa, a∈0,1, is a TS-group.
Ifq is a prime, then q=2rs+1 is a prime. By Table 1, we get that PSL2q is a TS-group.
(b)
Ifq is even, then q=8 or q≥24, say q=2s, s≥4
Letq=8, then by [9] (pp. 6), maxPSL28=E23:C7,D18,D14, so PSL28 is a TS-group.
Lets=r1a1r2a2≥4 with a1,a2≥1 integers and r1,r2 primes. Without loss of generality, we can assume that r1>r2, and then, PSL22s has a subgroup of the form PSL22s/r2a2. We see that 2s/r2a2∈cdPSL22s/r2a2, so ω2s/r2a2>3 because r1>r2≥2 shows s/r2a2≥r1>r2≥2. Thus, s≔r1a1 say.
Ifr1=2 and a1≥2, then let q0=22a1−1; we get that PSL22s has a subgroup PSL2q0. If a1=2, then PSL224 is a TS-group as PSL2q0 is a T-subgroup. If a1>2, then let χ∈IrrPSL2q0 with χ1=q0, ωχ1=2a1−1>22−1=2, so PSL2q0 is not a T-group.
Ifr≔r1≥5 is odd, then a1≥2 or a1=1. Note that
(7)maxPSL22s=E2s:C2s−1,D22s−1,D22s+1,and that PSL22ra1 has a subgroup PSL22ra1−1. Note that 2ra1−1∈PSL22ra1−1, and for a1≥2, ω2ra1−1=ra1−1≥r2−1=r≥3, so PSL22s is not a TS-group. If a1=1, then PSL22s with s an odd prime and 2s−1=rsa for a∈0,1 is a TS-group.
Letn=3. If q=2, then PSL32≅PSL27 is considered as above. So q≥3. If q=q0b with b a prime, then PSL3q has PSL3q0.gcdb,gcd3,q−1 as its subgroup. So PSL3q0∈∑PSL3q. By [17],
(8)q0−1q02+q0+1,q0+1q02+q0+1∈cdPSL3q0.
It is easy to see that for at least one of the numbersq0−1q02+q0+1,q0+1q02+q0+1, say d, we have ωd>2 if q is a prime power. Thus, q≥3 is a prime. By pp. 191 of [18], SO3q∈maxPSL3q. We know that SO3q≅SL2q and that
(9)cdSL2q=1,q−12,q+12,q−1,q,q+1,
by [16], so q=3,5 because one of the numbers q−1 and q+1 with q≥7 has the form 4s for some s≥2. If q=3, then 32.2.S4∈maxPSL33, so by [19], 24∈cd32.2.S4; if q=5, then 52:4S5∈maxPSL35, so by [19], 25·3∈cd52:4S5. Thus, PSL3q with q=3,5 is not a TS-group. Now, we have shown that PSL3q with q≥3 is not a TS-group and so is SL3q as
(10)PSL3q=SL3qZSL3q,cdPSL3q⊆cdSL3q,
by [20]. In particular, neither SL3q nor PSL3q is a T-group.
Letn≥4. If n=4 and q=2, then PSL42≅A8 is considered in Lemma 8. Now, n≥5, so by Lemma 6, we obtain that
(11)eitherPSL3q<PSLnqorSL3q<PSLnq.
This implies thatPSLnq with n≥4 is not a TS-group since PSL3q and SL3q are not T-groups.
Case 2:PSUnq with n≥3
Letn=3. If q=2, then PSU32 is solvable, so q≥3. By [9], we have that
(12)ES31+2:C8∈maxPSU33,ES41+2:C15∈maxPSU34,ES51+2:C8∈maxPSU35,ES71+2:C48∈maxPSU37,ES81+2:C21∈maxPSU38,
so by [19], we get that PSU3q for q∈3,4,5,7,8 is not a TS-group. Thus, we can assume that q≥9; then, by pp. 200 of [18], ESq1+2:Cq2−1/k∈maxPSU3q where k=gcd3,q+1, and q2−1/k∈cdESq1+2:Cq2−1/k. Observe that ESq1+2:Cq2−1/k is a Frobenius group and that there in G does exist an irreducible character χ∈IrrESq1+2 with χ1=q (note that ZESq1+2=q, and ESq1+2/ZESq1+2 is abelian, so by Theorem 2.31 of [21], for χ∈IrrG, χ12=ESq1+2:ZESq1+2=q2; hence, χ1=q).
Ifq is even, then ωq>4 as q>23. If q is odd, then q≥9 and
(13)q2−1k∈cdESq1+2:Cq2−1/k.
Observe thatq2−1/k is divisible by eight and that k∈1,3, so ωq2−1/k≥3.
It follows thatPSU3q with q≥3 is a non-TS-group.
Letn≥4. If n=4 and q=2, then PSU42 has 24:A5 as a subgroup. As 22·5∈cd24:A5, 24:A5 is not a T-group. Now, assume that n≥5; then, by Lemma 6, we have that
(14)eitherPSU3q<PSUnqorSU3q<PSUnq.
Forq≥3, by [20], we have that SU3q and PSU3q are not T-groups. It follows that PSUnq with n≥4 is not a TS-group.
Case 3:Ω2n+1q with n≥3, q odd
Ifn=3, then q7:1/2GL2q×SO3q by [18] (pp. 213). We know that q+12∈cdGL2q×SO3q, so ωq+12>2 for an odd q. Thus, Ω7q is not a TS-group. In particular, Ω7q is not a T-group.
Letn≥4. Then, by Lemma 6, a subgroup series is obtained:
(15)Ω7q<Ω9q<⋯<Ω2n+1q,so Ω2n+1q is not a TS-group since Ω7q is not a T-group.
Case 4:PSp2nq with n≥2
Letn=2. If q=2, then PSp42≅S6 is not simple, so q≥3, and by pp. 209 of [18] PSp4q has a subgroup PSp2q2.2≅PSL2q2.2. We know that PSL2q2 is a normal subgroup of PSL2q2.2 and that q4−1∈cdPSL2q2, so PSp4q is not a TS-group; in particular, PSp4q is not a T-group.
Letn≥3. Then from Lemma 6, PSp2nq contains a subgroup PSp4q, so PSp2nq is not a TS-group.
Case 5:PΩ2nεq with n≥4 and ε=±
Ifn=4, then PΩ8εq has a subgroup Ω7q, so by Case 3, Ω7q is neither a T-group nor a TS-group.
Ifn≥5, then by Lemma 6, PΩ2nεq is not a TS-group as PΩ2nεq contains a subgroup isomorphic to either Ω7q with q odd or PSp4q with q even. Note that Ω7q with q odd and PSp4q with q are non-T-groups, so we rule out this case.Lemma 10.
LetG be a simple group of exceptional Lie type. Assuming that G is a TS-group, then G is isomorphic to 2B28.Proof.
We see thatG is isomorphic to 2B2q with q=22m+1≥8, 2G2q with q=32m+1, m≥1, G2q, 3D4q, F4q, 2F4q2, E6εq, E7q, or E8q. We deal with these case by case.
The following subgroup series are obtained from Table2:
(16)G2q>SU3q2.2>SL3q2,2F4q2>SU3q2.2>SU3q2.
We know thatPSLnq≅SLnq/ZSLnq and PSUnq≅SUnq/ZSUnq, so SU3q2 and SL3q2 are non-TS-groups since Cases 1 and 2 in Lemma 9 show that PSU3q2 and PSL3q2 are non-T-groups. So G2q and 2F2q2 are not TS-groups. Note that 2F42′ is not a TS-group by [9]. From Table 2, 3D4q, F4q, and E6εq are non-TS-groups. Now, if PΩ12+q is not a TS-group, then E7q and E8q are not TS-groups. Now, we will show that PΩ12+q is not a TS-group. In fact, PΩ12+q contains a subgroup Ω7q or PSp4q which are not a T-group, so PΩ12+q is not a TS-group.
Now, we can conclude thatG is possibly isomorphic to 2B2q with q=22m+1≥8 or 2G2q with q=32m+1, m≥1. Thus, two cases are considered.
Case 1:2B2q with q=22m+1≥8
We know from pp. 385 of [10] that 2B222m+1 has a maximal subgroup of the form E22m+1+2m+1:C22m+1−1 which is a Frobenius group. So there is an irreducible character χ∈IrrE22m+1+2m+1 with a maximal degree with respect to divisibity in cdE22m+1+2m+1. Observe that if m≥2, then by Theorems 13.3 and 13.8 of [24],
(17)χ122m+1−1∈cdE22m+1+2m+1,
so by hypothesis,22m+1−1 is a prime. On the other hand, ZE22m+1+2m+1 is of order 22m+1, so E22m+1+2m+1/ZE22m+1+2m+1 is a 2-group, so there is a nonlinear character χ∈IrrE22m+1+2m+1 such that χ12 divides 22m+1, i.e., χ12m. Notice that χ1=maxcdE22m+1+2m+1. If χ1=2 and let P=E22m+1+2m+1, then P/P′=2, and P′=ΦP, so ZP=P′, P is an extraspecial 2-group, so m=1. (In fact, we know from (17) that for m≥2, E22m+1+2m+1:C22m+1−1 is not a T-group.) So by [9] (pp. 28), max2B28=23+3:7,13:4,5:4,D14. By [19], we may get that the groups 23+3:7, 13:4, 5:4, and D14 are T-groups, so 2B28 is a TS-group.
Case 2:2G2q with q=32m+1≥33
From [10] (pp. 398), 2G2q has a maximal subgroup Eq1+1+1:Cq−1 which is a Frobenius group. Let χ1=k≠1 for χ∈IrrEq1+1+1:Cq−1. Then, by Theorems 13.3 and 13.8 of [24], kq−1∈cdEq1+1+1:Cq−1. It follows that 2G2q is not a TS-group.Lemma 11.
A simple group of sporadic group is not aTS-group.Proof.
By [9] (pp. 18), PSL211∈maxM11 and 12∈cdPSL211, so PSL211 is not a T-group. Now, [9] (pp. 238) shows that M11 is a subgroup of these groups: M12, M23, HS, M24, McL, Suz, ON, Co3, Co2, Fi22, HN, Ly, Fi23, Co1, J4, Fi24, B, and M.
By checking [9] J1, M22, J2, J3, He, Ru, and Th are not TS-groups as PSL211 is a subgroup of J1,M22, PSL29 is a subgroup of J2,J3,Th, and PSL32≅PSL27 is a subgroup of He,Ru.Theorem 12.
LetG be a nonabelian simple TS-group. Then, G is isomorphic to the following:
(1)
PSL2q where q=2rsa+1 with a∈0,1 is a prime for some primes r,s (possibly equal)(2)
PSL2pm where p∈2,3,5, m is a prime and pm−1/gcd2,pm−1=rsa for some primes r,s, and a∈0,1(3)
PSL224(4)
2B28Proof.
We conclude the result from Lemmas8, 10, and 11.Theorem 13.
LetG be an almost simple TS-group with socle S, S≤G≤AutS. Then, G is isomorphic to one of the groups:
(1)
PSL2q where q=2rs+1 is a prime for some primes r,s (possibly equal)(2)
PSL2pm where p∈2,3,5, and m is a prime such that pm−1/gcd2,q−1=rsa for primes r,s, and a∈0,1(3)
2B28, PSL224(4)
S5Proof.
IfG is simple, then by Theorem 12, we have (1)-(3). So three cases are considered, and also, we assume that G is nonsimple.
Case 1:PSL2q where q=2rsa+1 is a prime for some primes r,s (possibly equal) and a∈0,1
In this case,G is possibly isomorphic to PGL2q. Note that PGL2q has a normal subgroup PSL2q with index gcd2,q−1 and that
(18)cdPSL2q=1,q+−1q−1/22,q−1,q,q+1,by [16], so hypothesis shows that q+1=4 or q−1=4 (in fact, if q is odd, then one of the numbers q−1 or q−1 can be written by 4r for some r≥1). It follows that G is isomorphic to S5.
Case 2:PSL2pm where p∈2,3,5, m is a prime and pm−1/gcd2,q−1=rsa for primes r,s, and a∈0,1(i)
Letp=2
Ifm=2, then PSL24≅PSL25 is done in Case 1. If m≥3, then G is possibly isomorphic to PSL22m.m. We see that PSL22m.m has PSL22m as its subgroup and that 2m∈cdPSL22m, so hypothesis forces m≤2, a contradiction. It means that PSL22m.m is not a TS-group.
(ii)
Letp=3
Ifm=2, then the order of the outer-automorphism group of PSL29 is 4, so by [9], G is possibly isomorphic to PSL29.21, PSL29.22, PSL29.23, or PSL29.22. Notice that PSL29 is a normal subgroup of these groups (PSL29.21, PSL29.22, PSL29.23, and PSL29.22) and that cdPSL29=1,5,8,9,10, so PSL29 is not a T-group. It follows that the groups PSL29.21, PSL29.22, PSL29.23, and PSL29.22 are non-TS-groups.
Letm≥3. Then, the order of the outer-automorphism group of PSL23m is 2m, and so by Corollary 6.5 of [25], G possibly has one of the structures: PGL23m, PSL23m.m, PGL23m.m, and PSL23m.2m. Note that PGL23m, PSL23m.m, PGL23m.m, and PSL23m.2m have a subgroup of the form PSL23m. We know that 3m∈cdPSL23m, so hypothesis shows that m≤2. If m=1, then PSL23 is solvable; if m=2, then 8∈cdPSL232. Thus, PGL23m, PSL23m.m, PGL23m.m, and PSL23m.2m are not TS-groups.
(iii)
Letp=5
Then, we can get from Corollary 6.5 of [25] that the possible groups G are isomorphic to PSL23m.m, PGL23m, PSL23m.2m, and PGL23m.m. Note that these groups always have a subgroup PSL23m which is not a T-group. Thus, PSL23m.m, PGL23m, PSL23m.2m, and PGL23m.m are not TS-groups
Case 3:2B28 and PSL224
In this case, the outer-automorphism group of2B28 is of order 3, so G is possibly isomorphic to 2B28.3. Now, by [9] (pp. 28), 13:12∈max2B28.3, so 22·3∈cd13:12. It follows that 13:12 is not a T-group and so 2B28.3 is not a TS-group.
We can get from [9] (pp. 12) that PSL224.2 and PSL224.4 are not TS-groups as PSL224 is a subgroup of both PSL224.2 and PSL224.4.Table 1
PSL2q, q≥5 (Chap II Theorem 8.27 of [15]).
maxHConditionC1Eq:Cq−1/kk=gcdq−1,2C2D2q−1/kq∈5,7,9,11C3D2q+1/kq∈7,9C5PSL2q0.k,bq=q0b, b a prime, q0≠2C6S4q=p≡±1mod8A4q=p≡3,5,13,27,37mod40SA5q≡±1mod10,Fq=Fp5Table 2
Exceptional groups of Lie type.
GH∈maxGReferenceG2qSL3q.2, SU3q2.2Table 1 of [22]3D4qG2qTable 1 of [22]F4q3D4q3Table 1 of [22]E6εqF4qTable 1 of [22]E7qPΩ12+qTable 1 of [22]E8qE7qTable 1 of [22]2F4q2SU3q2.2Main theorem of [23]Now, we will prove Theorem3.Proof.
LetN be a minimal nonabelian normal subgroup of G. Then, we get that
(19)N=S1×S2×⋯×Sn,where Si is isomorphic to a nonabelian simple group S for each 1≤i≤n. If n≥2, we obtain by Theorem 4.21 of [21] that for τ1∈IrrS1\LinS1, τ2∈IrrS2\LinS2 and λ∈LinSi, for i≥3,
(20)τ1×τ2×λ⋯×λ∈IrrN.
Thus,N is a non-T-group, a contradiction. Now, we get that N=S is a nonabelian simple group which is isomorphic to one of the groups satisfying Theorem 12.
We know thatN is isomorphic to a subgroup of G/CGN and that G/CGN is isomorphic to a subgroup of AutN, the automorphism group of N; then, G/CGN is an almost simple group satisfying Theorem 13.
LetS be the set of the groups:
(i)
PSL2q where q=2rs+1 is a prime for some primes r,s (possibly equal)(ii)
PSL2pm where p∈2,3,5, and m is a prime such that pm−1/gcd2,q−1=rsa for primes r,s, and a∈0,1(iii)
2B28, PSL224(iv)
S5
IfM∈S\S5,A5, we have that G/CGN is isomorphic to S and that NCGN⊲G. From N∩CGN=1, we conclude that G is isomorphic to N×CGN (if G>NCGN, NCGN is a T-group, a contradiction). If CGN≠1, one has that N is a T-group, a contradiction. It follows that G is isomorphic to S.
IfM=S5, we may get that G is isomorphic to S5×CGN as S5∩CGN=1. Notice that in this case, N is isomorphic to PSL25. If CGN is nonabelian, we have from the fact, PSL25≅A5<S5, that PSL25×CGN is a T-group, a contradiction. So CGN is abelian.
IfM=A5≅PSL25, one has that N≅PSL25 and that G′∩CGPSL25≤C2 as the order of the Schur multiplier of PSL25 is two. If G′∩CGPSL25=1, we have
(21)G′≅G′G′∩CGPSL25≅G′CGPSL25/CGPSL25≅PSL25,and so G is isomorphic to PSL25×CGN. By hypothesis, CGN is a T-group. Let A=CGN. Then, we easily get that G is a TS-group if cdA⊆1,p for some prime p and a non-TS-group if for different primes p,r, pr∈cdA, or p,r∈cdA.
IfG′∩CGPSL25≅C2, we have G/K≅SL25 for some normal subgroup K of G. If K is nonabelian, we assume that cdK=1,p for some prime p by above arguments. Let θ∈IrrK with θ1=p and I=IGθ be the inertia subgroup of θ in G. Then, I is isomorphic to K, S4K, Q12K, E5:C4K, or G by [10] (pp. 377). If I=K, we have that IS4Kθ=I∩S4K=I, θS4K∈IrrS4K by Theorem 6.16 of [21], and so θS4K1=S4K:Iθ1=23·3·p. Now, by Corollary 11.29 of [21], for χ∈IrrS4Kθ, 23·3=χ1/θ1S4K/K. It follows that there is an irreducible character δ∈IrrS4K with 12δ1. Now, S4K is a non-T-group, and so we also rule out the case I=S4K. If I=Q12K, we have that I has a subgroup isomorphic to C6K, so hypothesis shows that C6K is a T-group. We see that C6≅C6K/K is cyclic and so 6∈cdC6K. Note that I∩C6K=IC6Kθ where θ∈IrrK with θ1=p. It follows from Corollary 6.17 of [21] that 6p∈cdS4K, a contradiction. Similarly, we also rule out when I=E5:C4K. If I=G, we can get a contradiction by Corollary 6.17 of [21] too. It follows that K 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 to2B28 or PSL2q for certain q. Note that A7 is a T-group but not a TS-group.



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*Source: 1455299-2022-10-08.xml*