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

Authors: Shitian Liu
Journal: Journal of Mathematics
 (2021)

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## 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, ifG is a nonsolvable group and every character degree of a group G is a prime power, then G is isomorphic to S×A, where S∈A5,PSL28 and A is abelian. In this paper, we change the condition, each character degree of a group G 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.


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## Body

## 1. Introduction

In this paper, we assume that all groups are finite. Denote byIrrG the set of all complex irreducible characters of a group G, write the set of linear characters of G by LinG, and denote the set of character degrees of a group G by cdG.LetG be a group and let p,q be different primes. If cdG=1,m, then either G has an abelian normal subgroup of index m or m=pe for a prime p and G is the direct product of a p-group and an abelian group (see Theorem 12.5 in [1]); Isaacs and Passman have identified the structure of groups with cdG⊆1,p,q and cdG⊆1,p,p2 (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 characterχ∈IrrG is a DP-character if χ1 is a prime power, and a non-DP-character otherwise. A group G is named a DP-group if, for every χ∈IrrG, χ is a DP-character, and a non-DP-group otherwise. We also assume that an abelian group is a DP-group.
Manz in [4, 5] has determined the structure of DP-groups. In particular, nonsolvable DP-groups are determined.Theorem 1 (see [5]).
IfG is a nonsolvable DP-group, then G=S×A, where A is abelian and S is isomorphic to A5 or PSL28.
In order to talk conveniently, we give a definition.Definition 2.
Let∑G be the set of all proper subgroups of a group G. A group G is called an SDP-group if, for each M∈∑G, M is a DP-group, and a non-SDP-group otherwise.
We will prove the following results.Theorem 2.
LetG be an almost simple group such that S≤G≤AutS, where S is a nonabelian simple group. Assume that G is an SDP-group, then G is isomorphic to one of the groups: PSL2q, where q=2rm+1≥5 is a prime with r an odd prime or q=2m and 2m−1≥3 is a prime, PSL29, PSL235, S5, or PSL33.We will introduce the structure of this paper. In Section2, we give the structure of simple SDP-groups and some properties of SDP-groups; in Section 3, we give the structure of almost simple SDP-groups. For a group G, let ∑G be the set of all proper subgroups of G and maxG be the set of representatives of maximal subgroups of G. 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 ofSDP-group, then some subgroup of the nonabelian simple groups is given, and finally the structure of the nonabelian simple SDP-groups is determined. Now, we give some results about SDP-groups.Lemma 1.
(1)
G is an SDP-group if and only if, for each M∈∑G, M is a DP-group.(2)
LetG be an SDP-group and let N be a nontrivial normal subgroup of G. Then, N and G/N are DP-groups.(3)
ADP-group must be an SDP-group.Proof.
(1)
We can get the result by Definition2.(2)
N is a DP-group by Definition 2, and 1<N<G. If G/N is not a DP-group, then there exists a subgroup M∈maxG with N≰M, so G=MN and 1<M<G, M is a DP-group. Since G/N≅M/M∩N is not a DP-group, there is a character χ∈IrrM/N∩M which is not a DP-character. It follows that M is not a DP-group, a contradiction.(3)
LetG be a DP-group. We will show that G is an SDP-group. If G is not an SDP-group, then there is a subgroup M∈∑G and an irreducible character χ∈IrrM such that χ is a non-DP-character. Let I=IGχ, the inertia group of χ in G. Let η∈IrrI such that ηM,χ≠0. Then, by Theorem 6.5 of [1], ηG∈IrrG. We know that χ1η1ηG1, so there is a maximal subgroup K∈∑G and an irreducible character ξ∈IrrK such that ξ is not a DP-character, so K is not a DP-group. It follows that G is not a DP-group, a contradiction.
In generality, a group, which is the direct product of twoDP-groups, is not an SDP-group.Example 1.
LetP,Q be nonabelian p- and q-groups for different primes p,q, respectively. Also, assume that cdP=1,p,p2 and cdQ=1,q. Then, P×Q is not an SDP-group. In fact, let P1∈maxP such that χ∈IrrP1−LinP1. Obviously P1×Q is a subgroup of P×Q. However, by Theorem 4.21 of [1], there is a nonlinear character χ∈IrrP1×Q such that χ is not a DP-character.Example 2.
LetP be nonabelian p-group with cdP=1,d1,d2,…,ds, s≥2, and p>7. Then, A5×P is not an SDP-group. In fact, A5 and P are DP-groups. Let G=A5×P; then, there are a maximal proper subgroup P1 of P and an irreducible character χ∈IrrP1−LinP1, so A5×P1 has a nonlinear character θχ for some θ∈IrrA5−LinA5 and θχ is a non-DP-character.
There are simple groups whose certain subgroup is aDP-group, but itself not.Example 3.
LetJ1 be the simple sporadic group, and let P be the Sylow 2-subgroup of J1. Say N=NGP, the normalizer of P in G. Then, 2∧3:7:3 is a DP-group since cdN=1,3,7 by [7].Example 4.
LetM11 be the Mathieu group of degree 11. By pp.18 in [6], M11 has a maximal subgroup H isomorphic to 2:S4. By [7], cd2:S4=1,2,3,4 and so H is a DP-group.
We need some subgroup structure of finite simple groups to shorten the proof of Theorem3.Lemma 2.
Letq be a prime power.(1)
Letn≥8. Then, An has a subgroup An−1.(2)
Letn≥4 and 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 even.
LetPSLn+q=PSLnq and PSLn−q=PSUnq. Let SLn+q=SLnq and SLn−q=SUnq. Let ε=±.Proof.
(1)
By Appendix in [8], Sn, the symmetric group of degree n, has a subgroup Sn−k×Sk with 1≤k<n/2. Take k=1, Sn−1≤Sn, and so An has a subgroup An−1 as Sn:An=2.(2)
By Proposition 4.1.4 of [9], PSLn±q has a maximal subgroup H isomorphic to(1)am,n−mεq−ε,n⋅PSLmεq×PSLn−mεq⋅bm,n−mε.Takem=1, and PSLnεq has a subgroup isomorphic to SLn−1εq or PSLn−1εq.We know thatGLnεq has a subgroup GLn−1εq, and SLnεq⊴¯GLnεq. Note that GLnεq/SLnεq is isomorphic to a cyclic group of order q−1, so SLnεq has a subgroup isomorphic to SLn−1εq.(3)
Note thatSp2nq=2⋅PSp2nq for odd q and Sp2nq=PSp2nq for even q. Then, by Table 3.5.C of [9], PSp2nq has a subgroup PSp2n−2q.(4)
By Table 3.5.D of [9], Ω2n+1q has a subgroup with form Ω2m+1q⊥Ω2n−m+1εq with 1≤m<n, ε=±. Taking m=1, Ω2n+1q has a subgroup Ω2n−1q.(5)
By Table 3.5.E of [9], Ω2n+q has a subgroup of the form PΩ2mεq⊥PΩ2n−mεq or Ω2m+1⊥PΩ2n−mεq, with 1≤m<n, ε∈+,−,∘, q odd if m odd, or the form Sp2n−2q with q even.Takingm=1 and Ω2n+q:PΩ2n+q=gcd4,q−1, Wn has a subgroup PΩ2n−1+q or PSp2n−2q.By Table 3.5.F of [9], we also have that Ω2n− has a subgroup of the form either or Ω2n−1−q for q odd or Sp2n−2q with q even.Takingm=1 and Ω2n−q:PΩ2n−q=gcd4,q+1, Wn has PΩ2n−1−q for odd q or PSp2n−1q for even q.The following three lemmas are important to give the structure of the simple groups as Theorem3.Lemma 3 (see [10]).
The only solution of the diophantine equationpm−qn=1 with p and q prime and m,n>1 is 32−23=1.Lemma 4 (see [10, 11]).
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.Lemma 5 (Zsigmondy theorem, see [12]).
Letp be a prime and let n be a positive integer. Then, one of the following holds:(i)
There is a primitive primep′ for pn−1, that is, p′|pn−1, but p′∤pm−1, for every 1≤m<n(ii)
p=2,n=1 or 6(iii)
p is a Mersenne prime and n=2
We say that a groupA acts Frobeniusly on a group B if BA is a Frobenius group with kernel B and complement A. A group G is a 2-Frobenius group if there is a normal series N⊲¯M⊲¯G such that G/N and M are Frobenius groups with M/N, N as their Frobenius kernels, respectively.Lemma 6 (see Theorems 13.3 and 13.8 of [13]).
LetG=K:H be a Frobenius group with kernel K and complement H. Then,(i)
H||K|−1.(ii)
Any subgroup ofH of order p2 or pq is cyclic, where p and q are primes.(iii)
IfH is even, K is abelian.(iv)
In any case,K is nilpotent.(v)
Assume thatK has hK conjugacy classes, H has hH conjugacy classes, and G has characters with forms:(a)
hH is irreducible characters χ1,…,χhH with K in their kernel; if μ1,…,μhH are the irreducible characters of H, then these satisfy χihk=μih, for all h∈H and k∈K.(b)
Wheneverτ≠1K is an irreducible character of K, then τG is an irreducible character of G. This gives hK−1/H irreducible characters of G with K not in their kernel. Such τG satisfy(2)τG|H=τ1ρH,where ρH is the regular character of H.The following result is due to White [14].Lemma 7 (see Theorem A of [14]).
LetS=PSL2q, with q=pf>3, for some prime p, A=AutS, and let S≤H≤A. Given G=PGL2q if δ¯∈H, G=S if δ¯∉H and set H:G=d=2am for m odd. If p is odd, set ε=−1q−1/2. The set of irreducible character degrees of H is(3)cdH=1,q,q+ε2∪q−12ai:i|m∪q+1j:j|d,with the following exceptions:(i)
Ifp is odd with H≠Sφ or if p=2, then q+ε/2 is not a degree of H(ii)
Iff is odd, p=3, and H=Sφ, then i≠1(iii)
Iff is odd, p=3, and H=A, then j≠1(iv)
Iff is odd, p=2,3 or 5, and H=Sφ, then j≠1(v)
Ifp≡2mod4, p=2 or 3, and H=Sφ, then j≠2Theorem 3.
LetG be a nonabelian simple SDP-group. Then, G is isomorphic to one of the groups:(1)
PSL29, PSL235, and PSL2q, where q=2rm+1≥5 is a prime with r prime or q=2m and 2m−1≥3 is a prime(2)
PSL33Proof.
It is well-known that a nonabelian simple groupG 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: ifG is isomorphic to An with n≥5, then G is isomorphic to A5 or A6.Ifn=7, then A7 has a subgroup L27 by (pp.10 in [6]) and 2.3∈cdPSL27, a contradiction. It follows that A7 is not an SDP-group and also not a DP-group.Letn≥8. We know from Appendix in [8] that the symmetric group Sn of degree n contains a subgroup Sn−1, so An−1<An. It follows from Lemma 2 that there exists a subgroup series:(4)A7<A8<⋯<An−1<An,soAn with n≥8 is not an SDP-group.Ifn=6, then by pp.4 in [6], maxA6=A5,32:4,S4. We can see that cdA5=1,3,4,5, cd32:4=1,4, and cdS4=1,2,3, so G≅A6 is an SDP-group by Lemma 1. If n=5, then by Theorem 1 and Lemma 1, A5 is an SDP-group.It follows thatG is isomorphic to A5 or A6.Claim 2: a sporadic simple group or a Tits group is not anSDP-group.LetG be a sporadic simple group or a Tits group and let H be a proper subgroup of G; then, we can get Table 1 by [6] and Theorem 4.21 of [1]. For each χi∈IrrH, χi is not a DP-character. It follows that a sporadic simple group or a Tits group is a non-SDP-group.Claim 3: ifG is isomorphic to a simple group of classical Lie type, then G is isomorphic to PSL29, PSL235, PSL2q, where q=2rm+1≥5 is a prime with r a prime or q=2m and 2m−1≥3 is a prime, and PSL33.Table 1
Simple sporadic groups.
GHCharacterDegreeM11PSL211χ42.5J1PSL211χ42.5J2A4×A5χ203.5HSM11χ22.5M24M23χ22.11HePSL32χ42.3SuzA7χ22.3Co3A4×S5χ32.3Fi22S10χ55.7LyG25χ222⋅31Fi23S12χ75.11J4211:M24χ332⋅5BS4×2F42χ322⋅13F242′M11χ22.5M12M11χ22.5M22PSL211χ42.5M23A7χ22.3J3PSL219χ42⋅32McLM22χ23.7RuA8χ32.7ONA7χ22.3Co2M23χ22.11HNA12χ45.11ThPSL219:2χ32⋅32Co1A9×S3χ73.7Fi24′He:2χ32.3.17M25:S6χ172.5


### 2.1.PSLnq, n≥2

#### 2.1.1. Letn=2

Ifq=4 or 8, then PSL24 and PSL28 are DP-groups by Theorem 1, so Lemma 1 (3) implies that PSL24 and PSL28 are SDP-groups. It follows that G is isomorphic to PSL24 or PSL28. Thus, q≥9 or q=7. If q=7, then by pp.3 in [6], all maximal subgroups of PSL27≅PSL32 are of the forms S4 and 7:3. Note that cdS4=1,2,3 and cd7:3=1,3, so PSL27 is an SDP-group. Now, q≥9. If q=9, then by pp.5 in [6], maxL29=A5,32:4,S4; hence ,PSL29 is an SDP-group since A5 is a DP-group, cd32:4=1,4 and cdS4=1,2,3. Therefore, G is isomorphic to PSL27 or PSL29.By Table2, PSL2q has a subgroup of the form Eq:Zq−1/2. Note from Lemma 6 that cdEq:Zq−1/2=1,q−1/2, so by hypothesis, q−1/2 is a prime power or a prime. Two cases now are considered as follows: q>9 odd and q≥16 even.Case 1:q>9 odd.Letq=pm with m≥2. We first assume that q−1/2 is a prime. If m≥2 is even, then p−1p+1 divides pm−1. Note that p−12p+12=2k, for some k≥3, so assumption shows q−1/2=2s for some s≥3. Now, Lemma 3 shows that q−1/2=2s has no solution in ℕ. If m≥3 is odd, then q−1/2=p−1/2⋅pm−1+pm−2+⋯+1. It is easy to see that pm−1+pm−2+⋯+1 is odd, so q−1/2 cannot be a prime. Assume now that q−1/2 is a prime power and say q−1=2rn for some integer n≥0 and a prime r.(i)
Ifn=0, then q=3, but PSL23 is solvable, a contradiction.(ii)
Ifn≥1 and r=2, then by Lemmas 3–5, q=5,9>9, or q=2n+1+1 is a prime. It follows that G is isomorphic to PSL2q with q=2n+1+1≥17 a prime.(iii)
Ifn≥2 and r≥3, then by Lemma 4, q=35, r=11, and n=2 or q−1=p2−1=p+1p−1=2r2(rule out this case since 8 divides p+12p−12). So, we have G≅PSL235.Letq=p be a prime. Then, the fact that q−1/2 is a prime or a prime power implies that q=p=1+2rm, where m≥1 is an integer and r is a prime, so G≅PSL2q with q=1+2rm, for some integer m≥1 and r a prime.Case 2:q≥24 even.Table 2
PSL2q, q≥5 (chap II, Theorem 8.27 of [15]).
maxL2qConditionC1Eq:Zq−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=Fp5Now, Table2 implies that PSL2q has a maximal subgroup Eq:Zq−1. By Lemma 6, cdEq:Zq−1=1,q−1. By assumption, q−1 is a prime or a prime power. If q−1 is a prime and say q=2m≥24, then G≅PSL2q with q=2m≥24, q−1 a prime. If q−1 is a prime power, then by Lemma 3, it is impossible.


#### 2.1.2. Letn=3

Ifq=2, then since PSL32≅PSL27, one has that PSL32 is an SDP-group. If q=3, then by [6], PSL33 has maximal subgroups with the forms: 32:2S4, 13:3, and S4, so by GAP, one obtains that cd32:2S4=cdM9:S3=1,2,3,4,8,16, so G≅PSL33. If q=4, then by pp.23 in [6], PSL34 has a maximal subgroup PSL27, but 2.3∈cdPSL27, a contradiction to the hypothesis. Now, we assume that q≥5. By Table 3, PSL3q has a subgroup Eq2:1/kGL2q>PGL2q. Note that cdPGL2q=1,q,q−1,q+1. By assumption, q−1 and q+1 are prime powers. Note that q−1,q,andq+1 are consecutive integers, so one of the numbers q−1,q,q+1 is divisible by 3.Table 3
PSL3q(see page 191 of [16]).
maxL3qConditionC1Eq2:1/kGL2qk=gcd3,q−1C2Zq−1/k2⋅S3q≥5C3Zq2+q+1/k⋅3q≠4C5PSL3q0⋅δ,bq=q0b, b primeC632⋅SL23q=p≡1mod932⋅Q8q=p≡4,7mod9C8SO3qq oddPSU3q0q=q02SA6p≡1,2,4,7,8,13,mod15, Fq=Fp5,−3PSL272<q=p≡1,2,4mod7Letq=pm be odd with m≥2. Then, q=3n for some integer n. Since q≥5, one has that n≥2 and that q−1 and q+1 are divisible by 2. It follows that q=3≱5, a contradiction.Letq=pm be even with m≥3. Then, 3 divides q−1 or q+1. If q−1=3 or q−1=3s, for some s≥2, then, by Lemma 3, q=4≱5 a contradiction. If q+1=3 or q+1=3s, for some s≥2, then q=2 or q=8 by Lemma 3. If G≅PSL38, then, by pp.74 in [6], PSL38 has a subgroup PSL27 and 2.3∈cdPSL27, a contradiction to the hypothesis.


#### 2.1.3. Letn≥4

We know from [17] that(5)cdPSL3q=1,qq+1,q2+q+1,q+1q−12,q−1q2+q+1,q3,qq2+q+1,q+1q2+q+1,so PSL3q is not a DP-group.Now, Lemma2 shows that there exists a subgroup series:(6)PSL3q<PSLnqorSL3q<PSLnq,so PSLnq for n≥4 is a non-SDP-group.


### 2.2.PSUnq, n≥3

#### 2.2.1. Letn=3

Ifq=2, then PSU32≅32:Q8 is solvable, so q≥3. If q=3, then PSU33 has a subgroup PSL27 by pp.14 in [6], so 2.3∈cdPSL27, a contradiction. Hence, q≥4, so by pp.200 in [16], PSU3q has a subgroup GU2q. Notice that PSU2q≅PSL2q implies GU2q≅GL2q. Thus, by Table 2 in [18], cdGU2q=1,q−1,q,q+1. By hypothesis, πq−1=1=πq+1, so q=4 or q=8 by Lemmas 3 and 5. If q=4, then PSU34 has a maximal subgroup 2∧2+4:15, so by [7], cd2∧2+4:15=1,22.3,3.5, a contradiction. If q=8, then, by pp.66 in [6], PSU38 has a subgroup 2∧3+6:21, whence by [7], cd2∧3+6:21=1,3.7,23⋅7, a contradiction to the assumption.


#### 2.2.2. Letn≥4

We see from [17] that(7)cdPSU3q=1,qq−1,q2−q+1,q−1q2−q+1,qq2−q+1,q3,q+1q2−q+1,q−1q+12,so PSU3q is not a DP-group.By Lemma2, one has(8)PSU3q<PSUnqorSU3q<PSUnq,so PSUnq for n≥4 is not a SDP-group.Table 4
A proper maximal subgroupH of G.
GHConditionReferencesG22qNAiq=32m+1≥27[19]G2qSL3q⋅23|q(Table1 in [20])SL3q⋅23∤qD34qG2qq square(Table1 in [20])D34qF4qD34q⋅S3(Table1 in [20])E6εF4qε=±1(Table1 in [20])E7qSL2q∘PΩ12+q⋅dZL∩M=d(Table1 in [20])E8qSL2q∘E7q⋅dZL∩M=d(Table1 in [20])


### 2.3.PSp2nq with n≥2, Ω2n+1q with n≥3, q Odd, PΩ2nεq with n≥4, or a Simple Group of Exceptional Lie Type

First, we consider the simple groups of exceptional Lie type, whereNAi: G22q has abelian Hall subgroups A2 and A3 of order q+1−3m+1 and q+1+3m+1, respectively; NAi is a Frobenius group with kernel Ai, i=2,3, and a cyclic noninvariant factor of order 6. A∘B denotes the central product of A and B.ConsiderG22q; then, by Table 4, G has a Frobenius subgroup NAi with cyclic complement of order 6 and abelian kernel Ai of order q±3q+1, i.e., NAi/Ai is isomorphic to a cyclic group of order 6. It follows that NAi/Ai is abelian, so NAi/Ai∈cdNAi. This means that there is a proper subgroup H of G and a character χ∈IrrH such that χ1 is divisible by 6, so χ is not a DP-character. It follows that G22q is a non-SDP-group.Now, we do with other cases. By Table4, we obtain subgroup series(9)E8q>E7q>PΩ12q,and either(10)E6εq>F4q>D34q>G2q>SL3,q⋅2>SL3q,for3|q,or(11)E6εq>F4q>D34q>G2q>SU3,q⋅2>SU3q,for3∤q.Note thatPΩ12+q>PΩ4+q and that PSL3q≅SL3q/ZSL3q and PSU3q2≅SU3q2/ZSU3q2 are non-DP-groups, so E8q, E7q, E6εq, F4q, D34q, and G2q are non-SDP-groups.Now, considerB22q with q=22m+1≥8; then, by Theorem 9 of [21], B22q has a subgroup Eq1+1:Zq−1, and dq−1∈cdEq1+1:Zq−1, where q=22m+1≥8 and χ1=d≠1, for χ∈IrrEq1+1. It follows that Eq1+1:Zq−1 is not a DP-group, so B22q is a non-SDP-group.Second, we consider the groups:PSp2nq with n≥2, Ω2n+1q with n≥3, q odd, PΩ2nεq with n≥4.Notice thatPΩ4+q=PSL2q×PSL2q, and PΩ4−q=PSL2q2 with q>2, G2q, and Ω7q are non-DP-groups. It follows from Table 5 and Lemma 2 that the groups PSp2nq with n≥2, Ω2n+1q with n≥3, q is odd, and PΩ2nεq with n≥4 are non-SDP-group. So, we need to consider PΩ8−2. We know from pp. 89 in [6] that PΩ8−2 has a PSL27 as a subgroup. As 2.3∈cdPSL27, PΩ8−2 contains a non-DP-subgroup, so PΩ8−2 is a non-SDP-group.Table 5
PSp2nq, n≥2, Ω2n+1q, n≥3, q is odd, and PΩ2nεq,n≥4.
GHConditionReferencesPSp4qPΩ4+qq even(pp. 209 in [16])PΩ4−qq odd(pp. 209 in [16])Ω7qG2qq odd(pp. 213 in [16])PΩ8+qΩ7q(pp. 214 in [16])PΩ8−qPΩ4−q(pp. 215 in [16])


## 2.1.PSLnq, n≥2

### 2.1.1. Letn=2

Ifq=4 or 8, then PSL24 and PSL28 are DP-groups by Theorem 1, so Lemma 1 (3) implies that PSL24 and PSL28 are SDP-groups. It follows that G is isomorphic to PSL24 or PSL28. Thus, q≥9 or q=7. If q=7, then by pp.3 in [6], all maximal subgroups of PSL27≅PSL32 are of the forms S4 and 7:3. Note that cdS4=1,2,3 and cd7:3=1,3, so PSL27 is an SDP-group. Now, q≥9. If q=9, then by pp.5 in [6], maxL29=A5,32:4,S4; hence ,PSL29 is an SDP-group since A5 is a DP-group, cd32:4=1,4 and cdS4=1,2,3. Therefore, G is isomorphic to PSL27 or PSL29.By Table2, PSL2q has a subgroup of the form Eq:Zq−1/2. Note from Lemma 6 that cdEq:Zq−1/2=1,q−1/2, so by hypothesis, q−1/2 is a prime power or a prime. Two cases now are considered as follows: q>9 odd and q≥16 even.Case 1:q>9 odd.Letq=pm with m≥2. We first assume that q−1/2 is a prime. If m≥2 is even, then p−1p+1 divides pm−1. Note that p−12p+12=2k, for some k≥3, so assumption shows q−1/2=2s for some s≥3. Now, Lemma 3 shows that q−1/2=2s has no solution in ℕ. If m≥3 is odd, then q−1/2=p−1/2⋅pm−1+pm−2+⋯+1. It is easy to see that pm−1+pm−2+⋯+1 is odd, so q−1/2 cannot be a prime. Assume now that q−1/2 is a prime power and say q−1=2rn for some integer n≥0 and a prime r.(i)
Ifn=0, then q=3, but PSL23 is solvable, a contradiction.(ii)
Ifn≥1 and r=2, then by Lemmas 3–5, q=5,9>9, or q=2n+1+1 is a prime. It follows that G is isomorphic to PSL2q with q=2n+1+1≥17 a prime.(iii)
Ifn≥2 and r≥3, then by Lemma 4, q=35, r=11, and n=2 or q−1=p2−1=p+1p−1=2r2(rule out this case since 8 divides p+12p−12). So, we have G≅PSL235.Letq=p be a prime. Then, the fact that q−1/2 is a prime or a prime power implies that q=p=1+2rm, where m≥1 is an integer and r is a prime, so G≅PSL2q with q=1+2rm, for some integer m≥1 and r a prime.Case 2:q≥24 even.Table 2
PSL2q, q≥5 (chap II, Theorem 8.27 of [15]).
maxL2qConditionC1Eq:Zq−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=Fp5Now, Table2 implies that PSL2q has a maximal subgroup Eq:Zq−1. By Lemma 6, cdEq:Zq−1=1,q−1. By assumption, q−1 is a prime or a prime power. If q−1 is a prime and say q=2m≥24, then G≅PSL2q with q=2m≥24, q−1 a prime. If q−1 is a prime power, then by Lemma 3, it is impossible.


### 2.1.2. Letn=3

Ifq=2, then since PSL32≅PSL27, one has that PSL32 is an SDP-group. If q=3, then by [6], PSL33 has maximal subgroups with the forms: 32:2S4, 13:3, and S4, so by GAP, one obtains that cd32:2S4=cdM9:S3=1,2,3,4,8,16, so G≅PSL33. If q=4, then by pp.23 in [6], PSL34 has a maximal subgroup PSL27, but 2.3∈cdPSL27, a contradiction to the hypothesis. Now, we assume that q≥5. By Table 3, PSL3q has a subgroup Eq2:1/kGL2q>PGL2q. Note that cdPGL2q=1,q,q−1,q+1. By assumption, q−1 and q+1 are prime powers. Note that q−1,q,andq+1 are consecutive integers, so one of the numbers q−1,q,q+1 is divisible by 3.Table 3
PSL3q(see page 191 of [16]).
maxL3qConditionC1Eq2:1/kGL2qk=gcd3,q−1C2Zq−1/k2⋅S3q≥5C3Zq2+q+1/k⋅3q≠4C5PSL3q0⋅δ,bq=q0b, b primeC632⋅SL23q=p≡1mod932⋅Q8q=p≡4,7mod9C8SO3qq oddPSU3q0q=q02SA6p≡1,2,4,7,8,13,mod15, Fq=Fp5,−3PSL272<q=p≡1,2,4mod7Letq=pm be odd with m≥2. Then, q=3n for some integer n. Since q≥5, one has that n≥2 and that q−1 and q+1 are divisible by 2. It follows that q=3≱5, a contradiction.Letq=pm be even with m≥3. Then, 3 divides q−1 or q+1. If q−1=3 or q−1=3s, for some s≥2, then, by Lemma 3, q=4≱5 a contradiction. If q+1=3 or q+1=3s, for some s≥2, then q=2 or q=8 by Lemma 3. If G≅PSL38, then, by pp.74 in [6], PSL38 has a subgroup PSL27 and 2.3∈cdPSL27, a contradiction to the hypothesis.


### 2.1.3. Letn≥4

We know from [17] that(5)cdPSL3q=1,qq+1,q2+q+1,q+1q−12,q−1q2+q+1,q3,qq2+q+1,q+1q2+q+1,so PSL3q is not a DP-group.Now, Lemma2 shows that there exists a subgroup series:(6)PSL3q<PSLnqorSL3q<PSLnq,so PSLnq for n≥4 is a non-SDP-group.


## 2.1.1. Letn=2

Ifq=4 or 8, then PSL24 and PSL28 are DP-groups by Theorem 1, so Lemma 1 (3) implies that PSL24 and PSL28 are SDP-groups. It follows that G is isomorphic to PSL24 or PSL28. Thus, q≥9 or q=7. If q=7, then by pp.3 in [6], all maximal subgroups of PSL27≅PSL32 are of the forms S4 and 7:3. Note that cdS4=1,2,3 and cd7:3=1,3, so PSL27 is an SDP-group. Now, q≥9. If q=9, then by pp.5 in [6], maxL29=A5,32:4,S4; hence ,PSL29 is an SDP-group since A5 is a DP-group, cd32:4=1,4 and cdS4=1,2,3. Therefore, G is isomorphic to PSL27 or PSL29.By Table2, PSL2q has a subgroup of the form Eq:Zq−1/2. Note from Lemma 6 that cdEq:Zq−1/2=1,q−1/2, so by hypothesis, q−1/2 is a prime power or a prime. Two cases now are considered as follows: q>9 odd and q≥16 even.Case 1:q>9 odd.Letq=pm with m≥2. We first assume that q−1/2 is a prime. If m≥2 is even, then p−1p+1 divides pm−1. Note that p−12p+12=2k, for some k≥3, so assumption shows q−1/2=2s for some s≥3. Now, Lemma 3 shows that q−1/2=2s has no solution in ℕ. If m≥3 is odd, then q−1/2=p−1/2⋅pm−1+pm−2+⋯+1. It is easy to see that pm−1+pm−2+⋯+1 is odd, so q−1/2 cannot be a prime. Assume now that q−1/2 is a prime power and say q−1=2rn for some integer n≥0 and a prime r.(i)
Ifn=0, then q=3, but PSL23 is solvable, a contradiction.(ii)
Ifn≥1 and r=2, then by Lemmas 3–5, q=5,9>9, or q=2n+1+1 is a prime. It follows that G is isomorphic to PSL2q with q=2n+1+1≥17 a prime.(iii)
Ifn≥2 and r≥3, then by Lemma 4, q=35, r=11, and n=2 or q−1=p2−1=p+1p−1=2r2(rule out this case since 8 divides p+12p−12). So, we have G≅PSL235.Letq=p be a prime. Then, the fact that q−1/2 is a prime or a prime power implies that q=p=1+2rm, where m≥1 is an integer and r is a prime, so G≅PSL2q with q=1+2rm, for some integer m≥1 and r a prime.Case 2:q≥24 even.Table 2
PSL2q, q≥5 (chap II, Theorem 8.27 of [15]).
maxL2qConditionC1Eq:Zq−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=Fp5Now, Table2 implies that PSL2q has a maximal subgroup Eq:Zq−1. By Lemma 6, cdEq:Zq−1=1,q−1. By assumption, q−1 is a prime or a prime power. If q−1 is a prime and say q=2m≥24, then G≅PSL2q with q=2m≥24, q−1 a prime. If q−1 is a prime power, then by Lemma 3, it is impossible.


## 2.1.2. Letn=3

Ifq=2, then since PSL32≅PSL27, one has that PSL32 is an SDP-group. If q=3, then by [6], PSL33 has maximal subgroups with the forms: 32:2S4, 13:3, and S4, so by GAP, one obtains that cd32:2S4=cdM9:S3=1,2,3,4,8,16, so G≅PSL33. If q=4, then by pp.23 in [6], PSL34 has a maximal subgroup PSL27, but 2.3∈cdPSL27, a contradiction to the hypothesis. Now, we assume that q≥5. By Table 3, PSL3q has a subgroup Eq2:1/kGL2q>PGL2q. Note that cdPGL2q=1,q,q−1,q+1. By assumption, q−1 and q+1 are prime powers. Note that q−1,q,andq+1 are consecutive integers, so one of the numbers q−1,q,q+1 is divisible by 3.Table 3
PSL3q(see page 191 of [16]).
maxL3qConditionC1Eq2:1/kGL2qk=gcd3,q−1C2Zq−1/k2⋅S3q≥5C3Zq2+q+1/k⋅3q≠4C5PSL3q0⋅δ,bq=q0b, b primeC632⋅SL23q=p≡1mod932⋅Q8q=p≡4,7mod9C8SO3qq oddPSU3q0q=q02SA6p≡1,2,4,7,8,13,mod15, Fq=Fp5,−3PSL272<q=p≡1,2,4mod7Letq=pm be odd with m≥2. Then, q=3n for some integer n. Since q≥5, one has that n≥2 and that q−1 and q+1 are divisible by 2. It follows that q=3≱5, a contradiction.Letq=pm be even with m≥3. Then, 3 divides q−1 or q+1. If q−1=3 or q−1=3s, for some s≥2, then, by Lemma 3, q=4≱5 a contradiction. If q+1=3 or q+1=3s, for some s≥2, then q=2 or q=8 by Lemma 3. If G≅PSL38, then, by pp.74 in [6], PSL38 has a subgroup PSL27 and 2.3∈cdPSL27, a contradiction to the hypothesis.


## 2.1.3. Letn≥4

We know from [17] that(5)cdPSL3q=1,qq+1,q2+q+1,q+1q−12,q−1q2+q+1,q3,qq2+q+1,q+1q2+q+1,so PSL3q is not a DP-group.Now, Lemma2 shows that there exists a subgroup series:(6)PSL3q<PSLnqorSL3q<PSLnq,so PSLnq for n≥4 is a non-SDP-group.


## 2.2.PSUnq, n≥3

### 2.2.1. Letn=3

Ifq=2, then PSU32≅32:Q8 is solvable, so q≥3. If q=3, then PSU33 has a subgroup PSL27 by pp.14 in [6], so 2.3∈cdPSL27, a contradiction. Hence, q≥4, so by pp.200 in [16], PSU3q has a subgroup GU2q. Notice that PSU2q≅PSL2q implies GU2q≅GL2q. Thus, by Table 2 in [18], cdGU2q=1,q−1,q,q+1. By hypothesis, πq−1=1=πq+1, so q=4 or q=8 by Lemmas 3 and 5. If q=4, then PSU34 has a maximal subgroup 2∧2+4:15, so by [7], cd2∧2+4:15=1,22.3,3.5, a contradiction. If q=8, then, by pp.66 in [6], PSU38 has a subgroup 2∧3+6:21, whence by [7], cd2∧3+6:21=1,3.7,23⋅7, a contradiction to the assumption.


### 2.2.2. Letn≥4

We see from [17] that(7)cdPSU3q=1,qq−1,q2−q+1,q−1q2−q+1,qq2−q+1,q3,q+1q2−q+1,q−1q+12,so PSU3q is not a DP-group.By Lemma2, one has(8)PSU3q<PSUnqorSU3q<PSUnq,so PSUnq for n≥4 is not a SDP-group.Table 4
A proper maximal subgroupH of G.
GHConditionReferencesG22qNAiq=32m+1≥27[19]G2qSL3q⋅23|q(Table1 in [20])SL3q⋅23∤qD34qG2qq square(Table1 in [20])D34qF4qD34q⋅S3(Table1 in [20])E6εF4qε=±1(Table1 in [20])E7qSL2q∘PΩ12+q⋅dZL∩M=d(Table1 in [20])E8qSL2q∘E7q⋅dZL∩M=d(Table1 in [20])


## 2.2.1. Letn=3

Ifq=2, then PSU32≅32:Q8 is solvable, so q≥3. If q=3, then PSU33 has a subgroup PSL27 by pp.14 in [6], so 2.3∈cdPSL27, a contradiction. Hence, q≥4, so by pp.200 in [16], PSU3q has a subgroup GU2q. Notice that PSU2q≅PSL2q implies GU2q≅GL2q. Thus, by Table 2 in [18], cdGU2q=1,q−1,q,q+1. By hypothesis, πq−1=1=πq+1, so q=4 or q=8 by Lemmas 3 and 5. If q=4, then PSU34 has a maximal subgroup 2∧2+4:15, so by [7], cd2∧2+4:15=1,22.3,3.5, a contradiction. If q=8, then, by pp.66 in [6], PSU38 has a subgroup 2∧3+6:21, whence by [7], cd2∧3+6:21=1,3.7,23⋅7, a contradiction to the assumption.


## 2.2.2. Letn≥4

We see from [17] that(7)cdPSU3q=1,qq−1,q2−q+1,q−1q2−q+1,qq2−q+1,q3,q+1q2−q+1,q−1q+12,so PSU3q is not a DP-group.By Lemma2, one has(8)PSU3q<PSUnqorSU3q<PSUnq,so PSUnq for n≥4 is not a SDP-group.Table 4
A proper maximal subgroupH of G.
GHConditionReferencesG22qNAiq=32m+1≥27[19]G2qSL3q⋅23|q(Table1 in [20])SL3q⋅23∤qD34qG2qq square(Table1 in [20])D34qF4qD34q⋅S3(Table1 in [20])E6εF4qε=±1(Table1 in [20])E7qSL2q∘PΩ12+q⋅dZL∩M=d(Table1 in [20])E8qSL2q∘E7q⋅dZL∩M=d(Table1 in [20])


## 2.3.PSp2nq with n≥2, Ω2n+1q with n≥3, q Odd, PΩ2nεq with n≥4, or a Simple Group of Exceptional Lie Type

First, we consider the simple groups of exceptional Lie type, whereNAi: G22q has abelian Hall subgroups A2 and A3 of order q+1−3m+1 and q+1+3m+1, respectively; NAi is a Frobenius group with kernel Ai, i=2,3, and a cyclic noninvariant factor of order 6. A∘B denotes the central product of A and B.ConsiderG22q; then, by Table 4, G has a Frobenius subgroup NAi with cyclic complement of order 6 and abelian kernel Ai of order q±3q+1, i.e., NAi/Ai is isomorphic to a cyclic group of order 6. It follows that NAi/Ai is abelian, so NAi/Ai∈cdNAi. This means that there is a proper subgroup H of G and a character χ∈IrrH such that χ1 is divisible by 6, so χ is not a DP-character. It follows that G22q is a non-SDP-group.Now, we do with other cases. By Table4, we obtain subgroup series(9)E8q>E7q>PΩ12q,and either(10)E6εq>F4q>D34q>G2q>SL3,q⋅2>SL3q,for3|q,or(11)E6εq>F4q>D34q>G2q>SU3,q⋅2>SU3q,for3∤q.Note thatPΩ12+q>PΩ4+q and that PSL3q≅SL3q/ZSL3q and PSU3q2≅SU3q2/ZSU3q2 are non-DP-groups, so E8q, E7q, E6εq, F4q, D34q, and G2q are non-SDP-groups.Now, considerB22q with q=22m+1≥8; then, by Theorem 9 of [21], B22q has a subgroup Eq1+1:Zq−1, and dq−1∈cdEq1+1:Zq−1, where q=22m+1≥8 and χ1=d≠1, for χ∈IrrEq1+1. It follows that Eq1+1:Zq−1 is not a DP-group, so B22q is a non-SDP-group.Second, we consider the groups:PSp2nq with n≥2, Ω2n+1q with n≥3, q odd, PΩ2nεq with n≥4.Notice thatPΩ4+q=PSL2q×PSL2q, and PΩ4−q=PSL2q2 with q>2, G2q, and Ω7q are non-DP-groups. It follows from Table 5 and Lemma 2 that the groups PSp2nq with n≥2, Ω2n+1q with n≥3, q is odd, and PΩ2nεq with n≥4 are non-SDP-group. So, we need to consider PΩ8−2. We know from pp. 89 in [6] that PΩ8−2 has a PSL27 as a subgroup. As 2.3∈cdPSL27, PΩ8−2 contains a non-DP-subgroup, so PΩ8−2 is a non-SDP-group.Table 5
PSp2nq, n≥2, Ω2n+1q, n≥3, q is odd, and PΩ2nεq,n≥4.
GHConditionReferencesPSp4qPΩ4+qq even(pp. 209 in [16])PΩ4−qq odd(pp. 209 in [16])Ω7qG2qq odd(pp. 213 in [16])PΩ8+qΩ7q(pp. 214 in [16])PΩ8−qPΩ4−q(pp. 215 in [16])


## 3. Proof of Theorem 2

Now, we will prove Theorem2.Letq=pf for some prime p. Note that the order of outer-automorphism group of PSL2q is(12)OutPSL2q=gcd2,q−1⋅f.


### 3.1. Proof of Theorem 2

Proof.
IfG is simple, then by Theorem 3, G is isomorphic to one of the groups. PSL2q, where q=2rm+1≥5, is a prime with r an odd prime or q=2m and 2m−1≥3 a prime; PSL29; PSL235. So, if G is nonsimple, then we consider the following cases: q=2f with 2f−1 a prime, and q=9, q=35, and q=2rm+1≥5 is a prime with r an odd prime and PSL33.Case 1:q=2f with 2f−1 a prime.Iff=2, then PSL24≅PSL25≅A5 and OutA5=2. As G is a not simple group, G is possible isomorphic to S5, where Sn is the symmetric group of degree n. By pp. 2 in [6], maxS5=A5,S4,5:4,2×S3. Note that cdA5=1,3,4,5, cdS4=1,2,3, cd5:4=1,4, and cdS3=1,2, so S5 is an SDP-group. Thus, G≅S5.Iff=3, then, by pp. 6 in [6], 7:6∈maxPSL28⋅3, so PSL28⋅3 is not an SDP-group.Iff≥5, then we easily get that f does divides the order of PSL22f, so PSL22f is a maximal subgroup of PSL22f⋅f. We see that PSL22f for f≥5 is not a DP-group, so PSL22f⋅f is a non-SDP-group.Case 2:q=9.LetS=PSL29⋅21,PSL29⋅22,PSL29⋅23,PSL29⋅22.Now, by pp. 4 in [6], OutPSL29=4, so G is possibly isomorphic to A∈S. Note that PSL29∈∑A and that PSL29 is a non-DP-group, so each A∈S is a non-SDP-group.Case 3:q=35.Now,OutPSL235=2⋅5. Thus, G possibly is isomorphic to PGL235, PGL235⋅5, PSL235⋅5, or PSL235⋅10. It is easy to see that PSL235 is a maximal subgroup of PSL235⋅2. So, PSL235⋅5 is a non-SDP-group as PSL235 is not a DP-group and so are PSL235⋅10 and PGL235⋅5. We know that PGL235 has a normal subgroup PSL235 of index 2, but PSL235 is a non-DP-group. Thus, PGL235 is a non-SDP-group.Case 4:q=2rm+1 is a prime with r an odd prime.In this case,OutPSL2q=2. We know that PSL2q⊴¯PGL2q and that PSL2q is a non-DP-group. Hence, PGL2q is not an SDP-group.Case 5:PSL33.Now,OutPSL33=2. From pp. 13 in [6], 13:6∈maxPSL33⋅2, so 6∈cd13:6 is not a prime power. Thus, PSL33⋅2 is a non-SDP-group.


## 3.1. Proof of Theorem 2

Proof.
IfG is simple, then by Theorem 3, G is isomorphic to one of the groups. PSL2q, where q=2rm+1≥5, is a prime with r an odd prime or q=2m and 2m−1≥3 a prime; PSL29; PSL235. So, if G is nonsimple, then we consider the following cases: q=2f with 2f−1 a prime, and q=9, q=35, and q=2rm+1≥5 is a prime with r an odd prime and PSL33.Case 1:q=2f with 2f−1 a prime.Iff=2, then PSL24≅PSL25≅A5 and OutA5=2. As G is a not simple group, G is possible isomorphic to S5, where Sn is the symmetric group of degree n. By pp. 2 in [6], maxS5=A5,S4,5:4,2×S3. Note that cdA5=1,3,4,5, cdS4=1,2,3, cd5:4=1,4, and cdS3=1,2, so S5 is an SDP-group. Thus, G≅S5.Iff=3, then, by pp. 6 in [6], 7:6∈maxPSL28⋅3, so PSL28⋅3 is not an SDP-group.Iff≥5, then we easily get that f does divides the order of PSL22f, so PSL22f is a maximal subgroup of PSL22f⋅f. We see that PSL22f for f≥5 is not a DP-group, so PSL22f⋅f is a non-SDP-group.Case 2:q=9.LetS=PSL29⋅21,PSL29⋅22,PSL29⋅23,PSL29⋅22.Now, by pp. 4 in [6], OutPSL29=4, so G is possibly isomorphic to A∈S. Note that PSL29∈∑A and that PSL29 is a non-DP-group, so each A∈S is a non-SDP-group.Case 3:q=35.Now,OutPSL235=2⋅5. Thus, G possibly is isomorphic to PGL235, PGL235⋅5, PSL235⋅5, or PSL235⋅10. It is easy to see that PSL235 is a maximal subgroup of PSL235⋅2. So, PSL235⋅5 is a non-SDP-group as PSL235 is not a DP-group and so are PSL235⋅10 and PGL235⋅5. We know that PGL235 has a normal subgroup PSL235 of index 2, but PSL235 is a non-DP-group. Thus, PGL235 is a non-SDP-group.Case 4:q=2rm+1 is a prime with r an odd prime.In this case,OutPSL2q=2. We know that PSL2q⊴¯PGL2q and that PSL2q is a non-DP-group. Hence, PGL2q is not an SDP-group.Case 5:PSL33.Now,OutPSL33=2. From pp. 13 in [6], 13:6∈maxPSL33⋅2, so 6∈cd13:6 is not a prime power. Thus, PSL33⋅2 is a non-SDP-group.



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*Source: 6345386-2021-06-16.xml*