Hindawi Journal of Mathematics Volume 2021, Article ID 6345386, 7 pages https://doi.org/10.1155/2021/6345386

Research Article On Groups Whose Irreducible Character Degrees of All Proper are All Prime Powers

Shitian Liu

School of Mathematics, Sichuan University of Arts and Science, Dazhou, Sichuan, 635000, China

Correspondence should be addressed to Shitian Liu; [email protected]

Received 13 May 2021; Accepted 2 June 2021; Published 16 June 2021

Academic Editor: Jie Wu

Copyright © 2021 Shitian Liu. 'is is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 G 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, PSL2(8) � 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.

1. Introduction DP-group otherwise. We also assume that an is a DP-group. In this paper, we assume that all groups are finite. Denote by Manz in [4, 5] has determined the structure of Irr(G) the set of all complex irreducible characters of a DP-groups. In particular, nonsolvable DP-groups are group G, write the set of linear characters of G by Lin(G), determined. and denote the set of character degrees of a group G by cd(G). Theorem 1 (see [5]). If G is a nonsolvable DP-group, then G p, q Let be a group and let be different primes. If G � S × A, where A is abelian and S is isomorphic to A5 or (G) � {}, m G cd 1 , then either has an abelian normal sub- PSL2(8). group of index m or m � pe for a prime p and G is the direct In to talk conveniently, we give a definition. product of a p-group and an abelian group (see 'eorem 12.5 in [1]); Isaacs and Passman have identified the structure Definition 2. Let � G be the set of all proper subgroups of a of groups with cd(G) ⊆ �1, p, q � and cd(G) ⊆ �1, p, p2 � (see group G. A group G is called an SDP-group if, for each [2, 3], respectively). M ∈ � G, M is a DP-group, and a non-SDP-group In this paper, the following problem is considered. otherwise. Question: what can we say about the structure of a group We will prove the following results. if irreducible character degrees of all proper subgroups are all of prime powers. Theorem 2. Let G be an almost such that In order to argue in short, we introduce the following S ≤ G ≤ Aut(S), where S is a nonabelian simple group. Assume definition. that G is an SDP-group, then G is isomorphic to one of the m groups: PSL2(q), where q � 2r + 1 ≥ 5 is a prime with r an m m odd prime or q � 2 and 2 − 1 ≥ 3 is a prime, PSL2(9), (G) 5 Definition 1. An irreducible character χ ∈ Irr is a PSL2(3 ), S5, or PSL3(3). DP-character if χ(1) is a prime power, and a non- DP-character otherwise. A group G is named a DP-group if, We will introduce the structure of this paper. In Section for every χ ∈ Irr(G), χ is a DP-character, and a non- 2, we give the structure of simple SDP-groups and some 2 Journal of Mathematics

properties of SDP-groups; in Section 3, we give the structure not an SDP-group. In fact, A5 and P are DP-groups. Let of almost simple SDP-groups. For a group G, let � G be the G � A5 × P; then, there are a maximal proper P1 of set of all proper subgroups of G and max G be the set of P and an irreducible character χ ∈ Irr(P1) − Lin(P1), so representatives of maximal subgroups of G. 'e notation A5 × P1 has a nonlinear character θχ for some θ ∈ Irr(A5) − and notions are standard, see [1, 6], for instance. Lin(A5) and θχ is a non-DP-character. 2. Some Lemmas 'ere are simple groups whose certain subgroup is a DP-group, but itself not. In this section, we will give some needed results that are used to prove main theorem. We first give some basic results of Example 3. Let J1 be the simple , and let P be SDP-group, then some subgroup of the nonabelian simple the Sylow 2-subgroup of J1. Say N � NG(P), the normalizer ∧ groups is given, and finally the structure of the nonabelian of P in G. 'en, 2 3: 7: 3 is a DP-group since cd(N) � simple SDP-groups is determined. Now, we give some re- {}1, 3, 7 by [7]. sults about SDP-groups. Example 4. Let M11 be the Mathieu group of degree 11. By Lemma 1 pp.18 in [6], M11 has a maximal subgroup H isomorphic to 2: S4. By [7], cd(2: S4) � {}1, 2, 3, 4 and so H is a DP-group. (1) G is an SDP-group if and only if, for each M ∈ � G, We need some subgroup structure of finite simple M is a DP-group. groups to shorten the proof of 'eorem 3. (2) Let G be an SDP-group and let N be a nontrivial q of G. /en, N and G/N are Lemma 2. Let be a prime power.

DP-groups. (1) Let n ≥ 8. /en, An has a subgroup An−1. ε (3) A DP-group must be an SDP-group. (2) Let n ≥ 4 and ε � ± . /en, PSLn(q) has a subgroup ± ± ε isomorphic to SLn−1(q) or PSLn−1(q), and SLn(q) has ε Proof a subgroup of the form SLn−1(q).

(3) Let n ≥ 2. /en, PSp2n(q) has a subgroup PSp2(n−1)(q). (1) We can get the result by Definition 2. (4) Let n ≥ 3, and q odd. /en, Ω2n+1(q) contains a N N G (2) is a DP-group by Definition 2, and 1 < < . If subgroup Ω2n−1(q). G N / is not a DP-group, then there exists a subgroup (5) Let n ≥ 4, ε � ± . /en, PΩε (q) has a subgroup M G N M G � MN M G 2n ∈ max with ≰ , so and 1 < < , PΩε (q) with q odd or PSp (q) with q even. M is a DP-group. Since (G/N) � (M/M ∩ N) is not 2n−2 2n−2 + − a DP-group, there is a character χ ∈ Irr(M/N ∩ M) Let PSLn (q) � PSLn(q) and PSLn (q) � PSUn(q). Let + − which is not a DP-character. It follows that M is not a SLn (q) � SLn(q) and SLn (q) � SUn(q). Let ε � ± . DP-group, a contradiction. (3) Let G be a DP-group. We will show that G is an Proof SDP-group. If G is not an SDP-group, then there is a subgroup M ∈ � G and an irreducible character (1) By Appendix in [8], Sn, the of χ ∈ Irr(M) such that χ is a non-DP-character. Let degree n, has a subgroup Sn−k × Sk with 1 ≤ k < n/2. Take k � 1, S ≤ S , and so A has a subgroup A I � IG(χ), the inertia group of χ in G. Let η ∈ Irr(I) n−1 n n n−1 as |S : A | � 2. such that [ηM, χ] ≠ 0. 'en, by 'eorem 6.5 of [1], n n G G ± η ∈ Irr(G). We know that χ(1)|η(1)|η (1), so there (2) By Proposition 4.1.4 of [9], PSLn (q) has a maximal is a maximal subgroup K ∈ � G and an irreducible subgroup H isomorphic to (K) character ξ ∈ Irr such that ξ is not a DP-char- aε K G m,n−m ε ε ε acter, so is not a DP-group. It follows that is not � � · PSLm(q) × PSLn−m(q)� · �bm,n−m �. (1) a DP-group, a contradiction. (q − ε, n) In generality, a group, which is the direct product of two ε Take m � 1, and PSLn(q) has a subgroup isomorphic DP-groups, is not an SDP-group. □ ε ε to SLn−1(q) or PSLn−1(q). ε ε We know that GLn(q) has a subgroup GLn−1(q), and Example 1. Let P, Q be nonabelian p- and q-groups for ε ε ε ε SLn(q) ⊴ GLn(q). Note that (GLn(q)/SLn(q)) is different primes p, q, respectively. Also, assume that cd(P) � ε 2 isomorphic to a of order q − 1, so SLn(q) �1, p, p � and cd(Q) � �1, q �. 'en, P × Q is not an ε has a subgroup isomorphic to SLn−1(q). SDP-group. In fact, let P1 ∈ max P such that (3) Note that Sp n(q) � 2 · PSp n(q) for odd q and χ ∈ Irr(P1) − Lin(P1). Obviously P1 × Q is a subgroup of 2 2 P × Q. However, by 'eorem 4.21 of [1], there is a nonlinear Sp2n(q) � PSp2n(q) for even q. 'en, by Table 3.5.C of [9], PSp n(q) has a subgroup PSp n− (q). character χ ∈ Irr(P1 × Q) such that χ is not a DP-character. 2 2 2 (4) By Table 3.5.D of [9], Ω2n+1(q) has a subgroup with ε Example 2. Let P be nonabelian p-group with form Ω2m+1(q) ⊥ Ω2(n−m)+1(q) with 1 ≤ m < n, ε � ±. cd(P) � �1, d1, d2, ... , ds�, s ≥ 2, and p > 7. 'en, A5 × P is Taking m � 1, Ω2n+1(q) has a subgroup Ω2n−1(q). Journal of Mathematics 3

+ (5) By Table 3.5.E of [9], Ω2n(q) has a subgroup of the [h(K) − 1]/|H| irreducible characters of G with ε ε ε G form PΩ2m(q)⊥PΩ2(n−m)(q) or Ω2m+1⊥PΩ2(n−m) K not in their . Such τ satisfy (q), with 1 ≤ m < n, ε ∈ {}+, −, ∘ , q odd if m odd, or the form Sp (q) with q even. G 2n−2 □ τ |H � τ(1)ρH, (2) + + Taking m � 1 and |Ω2n(q): PΩ2n(q)| � gcd(4, q − 1), where ρH is the regular character of H. + Wn has a subgroup PΩ2(n−1)(q) or PSp2n−2(q). − By Table 3.5.F of [9], we also have that Ω2n has a sub- 'e following result is due to White [14]. − group of the form either or Ω2(n−1)(q) for q odd or Sp2n−2(q) with q even. Lemma 7 (see Theorem A of [14]). Let S � PSL2(q), with − − f Taking m � 1 and |Ω2n(q): PΩ2n(q)| � gcd(4, q + 1), q � p > 3, for some prime p, A � Aut(S), and let S ≤ H ≤ A. − Wn has PΩ2(n−1)(q) for odd q or PSp2(n−1)(q) for even q. Given G � PGL2(q) if δ ∈ H, G � S if δ ∉ H and set |H: G| � 'e following three lemmas are important to give the d � 2am for m odd. If p is odd, set ε � (−1)(q− 1)/2. /e set of structure of the simple groups as 'eorem 3. irreducible character degrees of H is (q + ε) cd(H) ��1, q, � ∪ �(q − 1)2ai: i|m� ∪ �(q + 1)j: j|d �, (3) Lemma 3 (see [10]). /e only solution of the diophantine 2 equation pm − qn � 1 with p and q prime and m, n > 1 is with the following exceptions: 32 − 23 � 1. (i) If p is odd with H ≠ S〈φ〉 or if p � 2, then (q + ε)/2 is H Lemma 4 (see [10, 11]). With the exceptions of the relations not a degree of 2392 − 2 · 134 � −1 and 35 − 2 · 112 � 1, every solution of the (ii) If f is odd, p � 3, and H � S〈φ〉, then i ≠ 1 m n equation p − 2q � ±1 with p, q prime, m, n > 1, has ex- (iii) If f is odd, p � 3, and H � A, then j ≠ 1 m � n � p − q · (1/2) ponents 2, i.e., it comes from a unit 2 of (iv) If f is odd, p � 2, 3 or 5, and H � S〈φ〉, then j ≠ 1 the quadratic field Q(2(1/2)) for which the coefficients p and q are primes. (v) If p ≡ 2(mod 4), p � 2 or 3, and H � S〈φ〉, then j ≠ 2 Lemma 5 (Zsigmondy theorem, see [12]). Let p be a prime and let n be a positive . /en, one of the following holds: Theorem 3. Let G be a nonabelian simple SDP-group. /en, (i) /ere is a primitive prime p′ for pn − 1, that is, G is isomorphic to one of the groups: n m p′|(p − 1), but p′∤(p − 1), for every 1 ≤ m < n 5 m (1) PSL2(9), PSL2(3 ), and PSL2(q), where q � 2r + (ii) p � 2, n � 1 or 6 1 ≥ 5 is a prime with r prime or q � 2m and 2m − 1 ≥ 3 (iii) p is a Mersenne prime and n � 2 is a prime (2) PSL (3) We say that a group A acts Frobeniusly on a group B if 3 [B]A is a with kernel B and complement A.A group G is a 2-Frobenius group if there is a normal series Proof. It is well-known that a nonabelian simple group G is N ⊲ M ⊲ G such that G/N and M are Frobenius groups with isomorphic to a sporadic simple group, a simple group of M/N, N as their Frobenius kernels, respectively. exceptional Lie type, an , or a simple group of classical Lie type. 'us, in the following, we consider the Lemma 6 (see Theorems 13.3 and 13.8 of [13]). Let G � simple groups case by case. K: H be a Frobenius group with kernel K and complement H. /en, Claim 1: if G is isomorphic to An with n ≥ 5, then G is isomorphic to A5 or A6. (i) |H| | (|K| − 1). If n � 7, then A7 has a subgroup L2(7) by (pp.10 in [6]) H p2 pq p (ii) Any subgroup of of order or is cyclic, where and 2.3 ∈ cd(PSL2(7)), a contradiction. It follows that q and are primes. A7 is not an SDP-group and also not a DP-group. (iii) If |H| is even, K is abelian. Let n ≥ 8. We know from Appendix in [8] that the (iv) In any case, K is nilpotent. symmetric group Sn of degree n contains a subgroup S A A (v) Assume that K has h(K) conjugacy classes, H has n−1, so n−1 < n. It follows from Lemma 2 that there h(H) conjugacy classes, and G has characters with exists a subgroup series: forms: A7 < A8 < ··· < An−1 < An, (4)

so An with n ≥ 8 is not an SDP-group. (a) h(H) is irreducible characters χ1, ... , χh(H) with 2 K in their kernel; if μ1, ... , μh(H) are the irre- If n � 6, then by pp.4 in [6], max A6 � �A5, 3 : 4,S4 �. 2 ducible characters of H, then these satisfy We can see that cd(A5) � {}1, 3, 4, 5 , cd(3 : 4) � {}1, 4 , χi(hk) � μi(h), for all h ∈ H and k ∈ K. and cd(S4) � {}1, 2, 3 , so G � A6 is an SDP-group by (b) Whenever τ ≠ 1K is an irreducible character of K, Lemma 1. If n � 5, then by 'eorem 1 and Lemma 1, A5 then τG is an irreducible character of G. /is gives is an SDP-group. 4 Journal of Mathematics

Table It follows that G is isomorphic to A5 or A6. 1: Simple sporadic groups. Claim 2: a sporadic simple group or a Tits group is not G H Character Degree an SDP-group. M11 PSL2(11) χ4 2.5 Let G be a sporadic simple group or a Tits group and let J1 PSL2(11) χ4 2.5 H be a proper subgroup of G; then, we can get Table 1 J2 A4 × A5 χ20 3.5 M by [6] and 'eorem 4.21 of [1]. For each χ ∈ Irr(H), χ HS 11 χ2 2.5 i i M M is not a DP-character. It follows that a sporadic simple 24 23 χ2 2.11 ( ) group or a Tits group is a non-SDP-group. He PSL3 2 χ4 2.3 Suz A7 χ 2.3 G 2 Claim 3: if is isomorphic to a simple group of classical Co3 A4 × S5 χ3 2.3 5 Lie type, then G is isomorphic to PSL2(9), PSL2(3 ), Fi22 S10 χ5 5.7 m 2 PSL2(q), where q � 2r + 1 ≥ 5 is a prime with r a Ly G2(5) χ2 2 · 31 m m prime or q � 2 and 2 − 1 ≥ 3 is a prime, and Fi23 S12 χ7 5.11 11 2 PSL (3). J4 2 : M24 χ3 3 · 5 3 □ 2 2 B S4× F4(2) χ3 2 · 13 2 F4(2)′ M11 χ2 2.5 2.1. PSLn(q), n ≥ 2 M12 M11 χ2 2.5 M22 PSL2(11) χ4 2.5 n � q � ( ) ( ) 2.1.1. Let 2. If 4 or 8, then PSL2 4 and PSL2 8 are M23 A7 χ2 2.3 2 DP-groups by 'eorem 1, so Lemma 1 (3) implies that J3 PSL2(19) χ4 2 · 3 M PSL2(4) and PSL2(8) are SDP-groups. It follows that G is McL 22 χ2 3.7 Ru A χ 2.7 isomorphic to PSL2(4) or PSL2(8). 'us, q ≥ 9 or q � 7. If 8 3 A q � 7, then by pp.3 in [6], all maximal subgroups of ON 7 χ2 2.3 Co2 M23 χ2 2.11 PSL2(7) � PSL3(2) are of the forms S4 and 7: 3. Note that HN A12 χ4 5.11 cd(S4) � {}1, 2, 3 and cd(7: 3) � {}1, 3 , so PSL2(7) is an 2 Th PSL2(19): 2 χ3 2 · 3 SDP-group. Now, q ≥ 9. If q � 9, then by pp.5 in [6], A × S 2 Co1 9 3 χ7 3.7 max L2(9) � �A5, 3 : 4,S4 �; hence ,PSL2(9) is an SDP-group Fi24′ He: 2 χ3 2.3.17 since A is a DP-group, cd(32: 4) � {}1, 4 and cd(S ) � 5 5 4 M 2 : S6 χ17 2.5 {}1, 2, 3 . 'erefore, G is isomorphic to PSL2(7) or PSL2(9). (q) By Table 2, PSL2 has a subgroup of the form Table 2: PSL2(q), q ≥ 5 (chap II, 'eorem 8.27 of [15]). Eq: Z(q−1)/2. Note from Lemma 6 that cd(Eq: Z(q−1)/2) � �1, (q − 1)/2 �, so by hypothesis, (q − 1)/2 max L2(q) Condition is a prime power or a prime. Two cases now are considered as C E : Z k � gcd(q − 1, 2) follows: q > 9 odd and q ≥ 16 even. 1 q (q−1)/k C2 D2(q−1)/k q ∉ {}5, 7, 9, 11 C3 D2(q−1)/k q ∉ {}7, 9 Case 1: q > 9 odd. b C5 PSL2(q0)·(k, b) q � q , b a prime, q0 ≠ 2 m 0 Let q � p with m ≥ 2. We first assume that (q − 1)/2 C6 S4 q � p ≡ ±1(mod 8) A q � p , , , , ( ) is a prime. If m ≥ 2 is even, then (p − 1)(p + 1) divides 4 ≡ 3 5 13 27 37 mod√ 40� m k (p − 1). Note that (p − 1)2(p + 1)2 � 2 , for some S A5 q ≡ ±1(mod 10),Fq � Fp[ 5 ] 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 Let q � p be a prime. 'en, the fact that (q − 1)/2 is a ((q − 1)/2) � ((p − 1)/2)·(pm− 1 + pm− 2 + · · · + 1). It prime or a prime power implies that q � p � 1 + 2rm, is easy to see that pm− 1 + pm− 2 + · · · + 1 is odd, so (q − where m ≥ 1 is an integer and r is a prime, so G � m 1)/2 cannot be a prime. Assume now that (q − 1)/2 is a PSL2(q) with q � 1 + 2r , for some integer m ≥ 1 and prime power and say q − 1 � 2rn for some integer r a prime. n r ≥ 0 and a prime . Case 2: q ≥ 24 even. n � q � ( ) (i) If 0, then 3, but PSL2 3 is solvable, a Now, Table 2 implies that PSL (q) has a maximal contradiction. 2 subgroup Eq: Zq−1. By Lemma 6, cd(Eq: Zq−1) � �1, q − 1 �. (ii) If n ≥ 1 and r � 2, then by Lemmas 3–5, By assumption, q − 1 is a prime or a prime power. If q − 1 is a q � 5, 9 > 9, or q � 2n+1 + 1 is a prime. It follows m 4 prime and say q � 2 ≥ 2 , then G � PSL2(q) with n+1 m 4 that G is isomorphic to PSL2(q) with q � 2 + q � 2 ≥ 2 , q − 1 a prime. If q − 1 is a prime power, then by 1 ≥ 17 a prime. Lemma 3, it is impossible. (iii) If n ≥ 2 and r ≥ 3, then by Lemma 4, q � 35, r � n � q − � p2 − � (p + 11, and 2 or 1 1 2.1.2. Let n � 3. If q � 2, then since PSL3(2) � PSL2(7), one )(p − ) � r2 1 1 2 (rule out this case since 8 divides has that PSL3(2) is an SDP-group. If q � 3, then by [6], (p + ) (p − ) G � ( 5) 2 1 2 1 2). So, we have PSL2 3 . PSL3(3) has maximal subgroups with the forms: 3 : 2S4, Journal of Mathematics 5

Table 13: 3, and S4, so by GAP, one obtains that 3: PSL3(q)(see page 191 of [16]). 2 cd(3 : 2S4) � cd(M9: S3) � {1, 2, 3, 4, 8, 16}, so G � PSL3 max L3(q) Condition (3). If q � 4, then by pp.23 in [6], PSL3(4) has a maximal C1 Eq2 : (1/k)GL2(q) k � gcd(3, q − 1) subgroup PSL2(7), but 2.3 ∈ cd(PSL2(7)), a contradiction to 2 the hypothesis. Now, we assume that q ≥ 5. By Table 3, C2 Z(q−1)/k · S3 q ≥ 5

(q) E 2 : ( k) (q) (q) PSL3 has a subgroup q 1/ GL2 > PGL2 . Note C3 Z(q2+q+1)/k · 3 q ≠ 4 that cd(PGL2(q)) � �1, q, q − 1, q + 1�. By assumption, q − 1 C PSL (q )·(δ, b) q � qb, b prime and q + 1 are prime powers. Note that q − 1, q, and q + 1 are 5 3 0 0 32 · SL (3) q � p ≡ 1(mod9) q − , q, q + C 2 consecutive , so one of the numbers 1 1 is 6 32 · Q q � p ≡ 4, 7(mod9) divisible by 3. 8 SO (q) q odd q � pm m q � n C 3 Let be odd with ≥ 2. 'en, 3 for some 8 PSU (q ) q � q2 integer n. Since q ≥ 5, one has that n ≥ 2 and that q − 1 and 3 0 0 p , , , , , , ( ) q + 1 are divisible by 2. It follows that q � 3≱5, a A ≡ 1 2 4 7 8√13� √mod15��� , S 6 F � F [ 5 , −3 ] contradiction. q p PSL (7) 2 < q � p ≡ 1, 2, 4(mod7) Let q � pm be even with m ≥ 3. 'en, 3 divides q − 1 or 2 q + 1. If q − 1 � 3 or q − 1 � 3s, for some s ≥ 2, then, by PSL (7) and 2.3 ∈ cd(PSL (7)), a contradiction to the Lemma 3, q � 4≱5 a contradiction. If q + 1 � 3 or q + 1 � 3s, 2 2 hypothesis. for some s ≥ 2, then q � 2 or q � 8 by Lemma 3. If G � PSL3(8), then, by pp.74 in [6], PSL3(8) has a subgroup 2.1.3. Let n ≥ 4. We know from [17] that

2 2 2 3 2 2 cd PSL3(q) � ��1, q(q + 1), q + q + 1, (q + 1)(q − 1) , (q − 1)�q + q + 1 �, q , q� q + q + 1 �, (q + 1)�q + q + 1 ��, (5)

so PSL3(q) is not a DP-group. q ≥ 4, so by pp.200 in [16], PSU3(q) has a subgroup GU2(q). Now, Lemma 2 shows that there exists a subgroup series: Notice that PSU2(q) � PSL2(q) implies GU2(q) � GL2(q). 'us, by Table 2 in [18], cd(GU2(q)) � �1, q − 1, q, q + 1�. By PSL3(q) < PSLn(q) (6) hypothesis, |π(q − 1)| � 1 � |π(q + 1)|, so q � 4 or q � 8 by or SL3(q) < PSLn(q), Lemmas 3 and 5. If q � 4, then PSU3(4) has a maximal subgroup 2∧(2 + 4): 15, so by [7], cd(2∧(2 +4): 15) so PSLn(q) for n ≥ 4 is a non-SDP-group. � �1, 22.3, 3.5 �, a contradiction. If q � 8, then, by pp.66 in [6], PSU3(8) has a subgroup 2∧(3 + 6): 21, whence by [7], ( ( + ): ) � �, . , 3 · � 2.2. PSUn(q), n ≥ 3 cd 2∧ 3 6 21 1 3 7 2 7 , a contradiction to the assumption. 2 2.2.1. Let n � 3. If q � 2, then PSU3(2) � 3 : Q8 is solvable, so q ≥ 3. If q � 3, then PSU3(3) has a subgroup PSL2(7) by pp.14 in [6], so 2.3 ∈ cd(PSL2(7)), a contradiction. Hence, 2.2.2. Let n ≥ 4. We see from [17] that

2 2 2 3 2 2 cd PSU3(q) � ��1, q(q − 1), q − q + 1, (q − 1)�q − q + 1 �, q� q − q + 1 �, q , (q + 1)�q − q + 1 �, (q − 1)(q + 1) �, (7)

so PSU3(q) is not a DP-group. respectively; N(Ai) is a Frobenius group with kernel By Lemma 2, one has Ai, i � 2, 3, and a cyclic noninvariant factor of order 6. A ∘ B denotes the central product of A and B. PSU (q) < PSU (q)or SU (q) < PSU (q), (8) 3 n 3 n 2 Consider G2(q); then, by Table 4, G has a Frobenius so PSUn(q) for n ≥ 4 is not a SDP-group. subgroup N(Ai) with cyclic complement��� of order 6 and abelian kernel Ai of order q ± 3q + 1, i.e., N(Ai)/Ai is isomorphic to a cyclic group of order 6. It follows that ε 2.3. PSp2n(q) with n ≥ 2, Ω2n+1(q) with n ≥ 3, q Odd, PΩ2n(q) N(Ai)/Ai is abelian, so |N(Ai)/Ai| ∈ cd(N(Ai)). 'is with n ≥ 4, or a Simple Group of Exceptional Lie Type. means that there is a proper subgroup H of G and a First, we consider the simple groups of exceptional Lie character χ ∈ Irr(H) such that χ(1) is divisible by 6, so 2 2 type, where N(Ai): G2(q) has abelian Hall subgroups χ is not a DP-character. It follows that G2(q) is a non- m+1 m+1 A2 and A3 of order q + 1 − 3 and q + 1 + 3 , SDP-group. 6 Journal of Mathematics

Table Table 4: A proper maximal subgroup H of G. 5: PSp2n(q), n ≥ 2, Ω2n+1(q), n ≥ 3, q is odd, and PΩε (q), n ≥ 4. G H Condition References 2n 2 2m+1 G H Condition References G2(q) N(Ai) q � 3 ≥ 27 [19] + SL (q)· 2 3|q (Table 1 in PΩ4 (q) q even (pp. 209 in [16]) 3 PSp (q) − G2(q) 4 PΩ (q) q odd (pp. 209 in [16]) SL3(q)· 2 3∤q [20]) 4 G (q) (Table 1 in Ω7(q) G2(q) q odd (pp. 213 in [16]) 3D (q) 2 √� q square 4 3D ( q ) [20]) + 4 PΩ8 (q) Ω7(q) (pp. 214 in [16]) P −(q) P −(q) F (q) 3D (q)· S (Table 1 in Ω8 Ω4 (pp. 215 in [16]) 4 4 3 [20]) (Table 1 in Eε F (q) ε � ± 1 q � pf p 6 4 [20]) Let for some prime . Note that the order of outer-automorphism group of PSL2(q) is E (q)( (q) P + (q)) · d |Z(L M)| � d (Table 1 in � � 7 SL2 ∘ Ω12 ∩ � � [20]) �Out PSL2(q) �� � gcd(2, q − 1)· f. (12) (Table 1 in E (q)(SL (q) ∘ E (q)) · d |Z(L ∩ M)| � d 8 2 7 [20]) 3.1. Proof of /eorem 2 Now, we do with other cases. By Table 4, we obtain subgroup series Proof. If G is simple, then by 'eorem 3, G is isomorphic to one of the groups. PSL (q), where q � 2rm + 1 ≥ 5, is a prime E (q) E (q) P (q), ( ) 2 8 > 7 > Ω12 9 with r an odd prime or q � 2m and 2m − 1 ≥ 3 a prime; 5 PSL2(9); PSL2(3 ). So, if G is nonsimple, then we consider and either the following cases: q � 2f with 2f − 1 a prime, and q � 9, ε 3 q � 35, and q � 2rm + 1 ≥ 5 is a prime with r an odd prime E6(q) > F4(q) > D4(q) > G2(q) > SL(3, q)· 2 > SL3(q), for 3|q, and PSL (3). (10) 3 Case 1: q � 2f with 2f − 1 a prime. or If f � 2, then PSL2(4) � PSL2(5) � A5 and ε 3 |Out(A )| � 2. As G is a not simple group, G is possible E6(q) > F4(q) > D4(q) > G2(q) > SU(3, q)· 2 > SU3(q), for 3∤ q. 5 isomorphic to S , where S is the symmetric group of (11) 5 n degree n. By pp. 2 in [6], max S5 � �A5,S4, 5: 4, 2 × S3�. + + Note that cd(A5) � {}1, 3, 4, 5 , cd(S4) � {}1, 2, 3 , Note that PΩ12(q) > PΩ4 (q) and that 2 cd(5: 4) � {}1, 4 , and cd(S3) � {}1, 2 , so S5 is an PSL3(q) � (SL3(q)/Z(SL3(q))) and PSU3(q ) � 2 2 SDP-group. 'us, G � S5. (SU3(q )/Z(SU3(q ))) are non-DP-groups, so E8(q), ε 3 If f � 3, then, by pp. 6 in [6], 7: 6 ∈ max PSL (8)· 3, so E7(q), E6(q), F4(q), D4(q), and G2(q) are non- 2 SDP-groups. PSL2(8)· 3 is not an SDP-group. 2 2m+1 If f ≥ 5, then we easily get that f does divides the order Now, consider B2(q) with q � 2 ≥ 8; then, by 2 1+1 of PSL (2f), so PSL (2f) is a maximal subgroup of 'eorem 9 of [21], B2(q) has a subgroup Eq : Zq−1, 2 2 1+1 2m+1 ( f)· f ( f) f and d(q − 1) ∈ cd(Eq : Zq−1), where q � 2 ≥ 8 PSL2 2 . We see that PSL2 2 for ≥ 5 is not a 1+1 ( f)· f and χ(1) � d ≠ 1, for χ ∈ Irr(Eq ). It follows that DP-group, so PSL2 2 is a non-SDP-group. 1+1 2 Eq : Zq−1 is not a DP-group, so B2(q) is a non- Case 2: q � 9. SDP-group. Let S � �PSL2(9)· 21, PSL2(9)· 22, PSL2(9)· 23, (q) n 2 Second, we consider the groups: PSp2n with ≥ 2, PSL2(9)· 2 }. ε Ω2n+1(q) with n ≥ 3, q odd, PΩ2n(q) with n ≥ 4. Now, by pp. 4 in [6], |Out(PSL2(9))| � 4, so G is + A S Notice that PΩ4 (q) � PSL2(q) × PSL2(q), and possibly isomorphic to ∈ . Note that − 2 ( ) � A ( ) PΩ4 (q) � PSL2(q ) with q > 2, G2(q), and Ω7(q) are non- PSL2 9 ∈ and that PSL2 9 is a non-DP-group, so DP-groups. It follows from Table 5 and Lemma 2 that the each A ∈ S is a non-SDP-group. 5 groups PSp2n(q) with n ≥ 2, Ω2n+1(q) with n ≥ 3, q is odd, and Case 3: q � 3 . ε PΩ2n(q) with n ≥ 4 are non-SDP-group. So, we need to 5 − − Now, |Out(PSL2(3 ))| � 2 · 5. 'us, G possibly is iso- consider PΩ8 (2). We know from pp. 89 in [6] that PΩ8 (2) 5 5 5 − morphic to PGL2(3 ), PGL2(3 )· 5, PSL2(3 )· 5, or has a PSL2(7) as a subgroup. As 2.3 ∈ cd(PSL2(7)), PΩ8 (2) 5 5 − PSL2(3 )· 10. It is easy to see that PSL2(3 ) is a contains a non-DP-subgroup, so PΩ8 (2) is a non- 5 5 maximal subgroup of PSL2(3 )· 2. So, PSL2(3 )· 5 is a SDP-group. 5 non-SDP-group as PSL2(3 ) is not a DP-group and so 5 5 are PSL2(3 )· 10 and PGL2(3 )· 5. We know that 5 5 PGL2(3 ) has a normal subgroup PSL2(3 ) of index 2, 3. Proof of Theorem 2 5 5 but PSL2(3 ) is a non-DP-group. 'us, PGL2(3 ) is a Now, we will prove 'eorem 2. non-SDP-group. Journal of Mathematics 7

Case 4: q � 2rm + 1 is a prime with r an odd prime. [13] L. Dornhoff, Group Representation /eory. Part A: Ordinary In this case, |Out(PSL (q))| � 2. We know that Representation /eory, Marcel Dekker, Inc., New York, NY, 2 USA, 1971. PSL (q) ⊴ PGL (q) and that PSL (q) is a non- 2 2 2 [14] D. L. White, “Character degrees of extensions of PSL (q) and DP-group. Hence, PGL (q) is not an SDP-group. 2 2 SL2 (q),” Journal of Group /eory, vol. 16, no. 1, pp. 1–33, 2013. Case 5: PSL3(3). [15] B. Huppert, Endliche Gruppen. I. Die Grundlehren der Now, |Out(PSL (3))| � 2. From pp. 13 in [6], Mathematischen Wissenschaften, Band 134, Springer-Verlag, 3 Berlin, Germany, 1967. 13: 6 ∈ max PSL3(3)· 2, so 6 ∈ cd(13: 6) is not a prime ( )· [16] P. B. Kleidman, /e subgroup structure of some finite simple power. 'us, PSL3 3 2 is a non-SDP-group. □ groups, (Ph.D. 'esis), Trinity College, Cambridge, CA, USA, 1987. Data Availability [17] F. L¨ubeck,“Character degrees and their multiplicities for some groups of lie type of rank <9,” 2007, https://www.math. No data were used to support the findings of the study. rwth-aachen.de/Frank.Luebeck/chev/DegMult/index.html? LANG�en. [18] W. A. Simpson and J. S. Frame, “'e character tables for SL (3, Conflicts of Interest q), SU (3, q2), PSL (3, q), PSU (3, q2),” Canadian Journal of 'e authors declare that they have no conflicts of interest. Mathematics, vol. 25, pp. 486–494, 1973. [19] V. M. Levchuk and Y. N. Nuzhin, “'e structure of ree groups,” Algebra I Logika, vol. 24, no. 1, pp. 26–41, 1985. Acknowledgments [20] M. W. Liebeck and J. Saxl, “On the orders of maximal sub- groups of the finite exceptional groups of lie type,” Proceedings 'e project was supported by the NSF of China (Grant no of the London Mathematical Society, vol. 55, no. 2, pp. 299– 11871360) and Opening Project of Sichuan Province Uni- 330, 1987. versity Key Laborstory of Bridge Non-destruction Detecting [21] M. Suzuki, “On a class of doubly transitive groups,” Annals of and Engineering Computing (Grant no 2019QYJ02). Mathematics, vol. 79, no. 2, pp. 105–145, 1962.

References [1] I. M. Isaacs, Character /eory of Finite Groups, Dover Pub- lications, Inc., New York, NY, USA, 1994. [2] I. M. Isaacs and D. S. Passman, “A characterization of groups in terms of the degrees of their characters. II,” Pacific Journal of Mathematics, vol. 24, pp. 467–510, 1968. [3] D. S. Passman, “Groups whose irreducible representations have degrees dividing p2,” Pacific Journal of Mathematics, vol. 17, pp. 475–496, 1966. [4] O. Manz, “Endliche auflosbare¨ gruppen, deren samtliche¨ charaktergrade primzahlpotenzen sind,” Journal of Algebra, vol. 94, no. 1, pp. 211–255, 1985. [5] O. Manz, “Endliche nicht-aufl¨osbare gruppen, deren samtliche¨ charaktergrade primzahlpotenzen sind,” Journal of Algebra, vol. 96, no. 1, pp. 114–119, 1985. [6] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson, Atlas of Finite Groups: Maximal Subgroups and Ordinary Characters For Simple Groups, Oxford University Press, Oxford, UK, 1985. [7] T. Breuer, “'e GAP character table library,” 2012, http:// www.math.rwth-aachen.de/Thomas.Breuer/ctbllib.̃ [8] M. Aschbacher and L. Scott, “Maximal subgroups of finite groups,” Journal of Algebra, vol. 92, no. 1, pp. 44–80, 1985. [9] P. Kleidman and M. Liebeck, /e Subgroup Structure of the Finite Classical Groups, Cambridge University Press, Cam- bridge, CA, USA, volume 129 of London Mathematical So- ciety Lecture Note Series, 1990. [10] A. Khosravi and B. Khosravi, “A new characterization of some alternating and symmetric groups. II,” International Journal of Mathematics and Mathematical Sciences, vol. 30, no. 4, pp. 953–967, 2004. [11] P. Crescenzo, “A Diophantine equation which arises in the theory of finite groups,” Advances in Mathematics, vol. 17, no. 1, pp. 25–29, 1975. [12] K. Zsigmondy, “Zur theorie der potenzreste,” Monatshefte fur¨ Mathematik und Physik, vol. 3, no. 1, pp. 265–284, 1892.