# Almost simple groups with no product of two primes dividing three character degrees

From the journal Journal of Group Theory

## Abstract

Let Irr(G) denote the set of complex irreducible characters of a finite group G, and let cd(G) be the set of degrees of the members of Irr(G). For positive integers k and l, we say that the finite group G has the property 𝒫kl if, for any distinct degrees a1,a2,,akcd(G), the total number of (not necessarily different) prime divisors of the greatest common divisor gcd(a1,a2,,ak) is at most l-1. In this paper, we classify all finite almost simple groups satisfying the property 𝒫32. As a consequence of our classification, we show that if G is an almost simple group satisfying 𝒫32, then |cd(G)|8.

### 1 Introduction

Let Mn() be the algebra of n×n matrices over the field of complex numbers . Also, let G be a finite group. Denote by Irr(G)={χ1,,χk} the set of all complex irreducible characters of G. Let cd(G) be the set of all irreducible character degrees of G forgetting multiplicities, that is, cd(G)={χ(1)χIrr(G)}. It follows from Wedderburn’s theorem that the complex group algebra [G] of G admits a decomposition

[G]=Mn1()Mn2()Mnk(),

where ni:=χi(1) for 1ik. Therefore, the complex group algebra [G] determines the character degrees of G and their multiplicities. An important question in character theory is whether one can recover a group or its properties from its character degrees with or without multiplicity. In other words, how much does [G] or cd(G) know about the structure of G?

The other interesting problem in the character theory of finite groups is to determine which set of positive integers containing 1 can occur as character degree sets of some finite groups. Some of the early results about the sets of integers that can occur as the sets of character degrees can be found in [7]. The common way presented in that paper is trying to find the relationship between the character degrees and their prime divisors. Isaacs and Passman [9] studied the groups where the set of degrees consists of 1 and primes, and then Manz [16, 17] studied the groups where the set of degrees consists of only prime powers. Huppert’s σ-ρ conjecture, for which there has been much research, claims that if G is a solvable group, then there is some character degree that is divisible by at least half of the primes dividing character degrees of the group. For general groups, Huppert conjectures that some degree must be divisible by at least a third of the primes dividing character degrees of the group.

In recent papers, the dual problem has been considered. That is, it has been considered what fraction of the character degrees must be divisible by some prime. To aid in investigating of this question, let k and l be positive integers; then we say that a group G has the property 𝒫kl if no product of l primes (not necessarily distinct) divides any of k distinct elements of cd(G). Therefore, a group G has the property 𝒫kl if and only if, for any distinct degrees a1,a2,,akcd(G), the total number of (not necessarily different) prime divisors of the greatest common divisor gcd(a1,a2,,ak) is at most l-1. Clearly, if a finite group G satisfies the property 𝒫kl, then it has the property 𝒫k+1l as well as 𝒫kl+1.

This property for character degrees of finite groups has been studied in some papers. Benjamin [1] proved that if G is a solvable group satisfying the property 𝒫21, 𝒫31 or 𝒫41, then |cd(G)|3, 6 or 9, respectively, and McVey [19] proved that if G is solvable group satisfying 𝒫51, then |cd(G)|12. They present examples showing that this bound is attained in each of these cases. Also, Benjamin proved that if G is a solvable group that satisfies 𝒫k1 and k5, then |cd(G)|k2-3k+4, but this is not the best bound. Lewis [10] could show that if G is a solvable group satisfying 𝒫22, then |cd(G)|9. Hamblin and Lewis [6] proved that if G is a solvable group having 𝒫23, then |cd(G)|462, 515. We note that 27 is the largest known value for |cd(G)|, where G is a solvable group satisfying 𝒫23.

For nonsolvable groups, it follows from [8, Corollary 12.6 and Theorem 12.15] that if G is a nonsolvable group, then |cd(G)|4. Theorems 6.4 and 7.1 of [11] imply that a nonsolvable group G has the property 𝒫21 if and only if GS×A, where S=PSL2(2f) for some integer f2 and A is abelian; hence |cd(G)|=4. Nonsolvable groups satisfying the property 𝒫31 have been studied in [14, 17, 20], in which the bound |cd(G)|6 is attained for such groups. Also, nonsolvable groups with the property 𝒫41 have been studied in [5], in which the bound |cd(G)|8 is attained for such groups. Lewis and White [13] were able to prove |cd(G)|8 if G is a nonsolvable group having the property 𝒫22. Finally, if G is a nonsolvable group satisfying 𝒫23 and G has a composition factor isomorphic to PSL2(q) for some prime power q7, then |cd(G)|19, while, if G has no nonabelian composition factor isomorphic to A5, then |cd(G)|21 (see [12]).

We should mention that if l2 and k3, then there is no known bound (in terms of k and l) on the number of irreducible character degrees of solvable or nonsolvable groups satisfying the property 𝒫kl.

We consider nonsolvable groups with the property 𝒫32. In this paper, we classify almost simple groups having the property 𝒫32. We should mention that if S is a nonabelian simple group, then a finite group G is said to be an almost simple group with socle S if there is a normal subgroup N of G such that NS and NGAut(N). Then, in later papers, we will use this classification to obtain an upper bound for the number of irreducible character degrees of nonsolvable groups satisfying 𝒫32. Recall that a group G has property 𝒫32 if, for any distinct degrees a1,a2,a3cd(G), the total number of (not necessarily different) prime divisors of the greatest common divisor gcd(a1,a2,a3) is at most 1. Hence a group G satisfies 𝒫32 if no product of any two primes (not necessarily distinct) divides three distinct irreducible character degrees of G.

### Theorem A.

Let S be a nonabelian simple group, and let G be a group such that SGAut(S). Then G has the property P32 if and only if one of the following holds:

1. G=SM11;

2. SA7, and GA7 or S7;

3. SPSL2(pf), pf4, and G is one of the groups listed in Table 1;

4. G=SPSL3(3);

5. G=SPSU3(3);

6. SPSL3(4), GS, Sφτ, Sδ, Sδφτ or Sφ, and

cd(G)={1,20,35,45,63,64},{1,20,35,64,90,126},{1,20,45,63,64,105},{1,20,64,90,105,126}𝑜𝑟{1,20,35,45,64,70,126},

respectively;

7. SSz(q2), q2=22f+1, f1, and G=S, where q2-1=22f+1-1 and 2f+1 are prime numbers, and

cd(G)={1,q4,2f(q2-1),q4+1,(q2-1)(q2-2q+1),(q2-1)(q2+2q+1)};
8. SSz(q2), q2=22f+1, f1, and G=Aut(S), where

q2-1=22f+1-1,2f+1𝑎𝑛𝑑22f+1+2f+1+1

are prime numbers, f1 or 2(mod4), and if f>1, then

cd(G)={1,q4,2f(q2-1),(q4+1)(2f+1),(q2-1)(q2-2q+1)(2f+1),(q2-1)(q2+2q+1),(q2-1)(q2+2q+1)(2f+1)},

and if f=1, then

cd(G)={1,2×7,3×5×7,26,3×5×13,7×13};
9. SSz(q2), q2=22f+1, f1, and G=Aut(S), where

q2-1=22f+1-1,2f+1𝑎𝑛𝑑22f+1-2f+1+1

are prime numbers, f0 or 3(mod4), and

cd(G)={1,q4,2f(q2-1),(q4+1)(2f+1),(q2-1)(q2-2q+1),(q2-1)(q2-2q+1)(2f+1),(q2-1)(q2+2q+1)}.

### Table 1

Almost simple groups with the property 𝒫32 and socle PSL2(q).

 S |G:H|, H=PGL2⁢(q) or PSL2⁢(q) cd⁡(G) 1 PSL2⁢(q) 1 {1,2f-1,2f,2f+1}, q=2f {1,q+ε2,q±1,q}, q odd, ε=(-1)q-12 {1,q-1,q,q+1}, q⩾5 odd 2 PSL2⁢(2f), q=2f, 2f+1 is prime 2⁢r, r is an odd prime {1,q,2⁢(q±1),2⁢r⁢(q±1),q+1,r⁢(q+1)}, ∤⁡rq2-1 3 PSL2⁢(2f), q=2f, 2f+1 is prime 2⁢r, r is an odd prime, ∤⁡rq-1 and r≠q+1 {1,q,2⁢(q±1),2⁢r⁢(q±1),r⁢(q+1)} 4 PSL2⁢(2f), q=2f, 2f+1 is prime 2⁢r, r is an odd prime, ∤⁡rq-1 {1,q,2⁢(q-1),2⁢r⁢(q±1),(q+1),r⁢(q+1)} 5 PSL2⁢(2f), 2f-1 is prime r2, r is an odd prime, ∤⁡rq2-1 {1,q,q-1,r⁢(q±1),r2⁢(q±1)} 6 PSL2⁢(2f) r, r is an odd prime {1,2f,2f-1,f⁢(2f±1)}, f=r {1,2f,2f±1,r⁢(2f±1)}, f>r 7 PSL2⁢(2f) 2 {1,2f,2⁢(2f±1),2f+1} 8 PSL2⁢(pf), p>2 2 {1,q,2⁢(q±1),q+1} 9 PSL2⁢(3f), G=Aut⁡(S) r, r is an odd prime {1,3f,f⁢(3f±1),3f-1}, f=r 10 PSL2⁢(pf), p>3 or p=3 and G≠Aut⁡(S) r, r is an odd prime, ∤⁡rq2-1 {1,q,q-1,r⁢(q-1),q+1,r⁢(q+1)} 11 PSL2⁢(9) 2 {1,5,9,10,16} 12 PSL2⁢(q), q≠9 and q+12 is an odd prime 2 {1,(q+1)/2,q,q+1,2⁢(q±1)} 13 PSL2⁢(3f) r, r is an odd prime {1,3f,3f-12,f⁢(3f±1)}, f=r 14 PSL2⁢(5f) r, r is an odd prime {1,5f,5f+12,5f-1,f⁢(5f±1)}, f=r 15 PSL2⁢(q), q=pf, p>5 r, r is an odd prime, ∤⁡rq2-1 {1,q,q+ε2,q±1,r⁢(q±1)}, f=r, ε=(-1)q-12 16 PSL2⁢(q) r, r is an odd prime, ∤⁡rq2-1 {1,q,q+ε2,q±1,r⁢(q±1)}, f>r, ε=(-1)q-12 17 PSL2⁢(9) 2, G≅M10 {1,9,10,16} 18 PSL2⁢(q), q≠9 2 {1,q,q+1,2⁢(q±1)}

Since, in all cases of Theorem A, the almost simple group G satisfying the property 𝒫32 has at most eight irreducible character degrees, we obtain the following corollary.

### Corollary 1.1.

If G is an almost simple group with the property P32, then

|cd(G)|8.

#### Notation

For a positive integer n, we denote the set of all prime divisors of n by π(n). If G is a finite group, then we write π(G) instead of π(|G|) for the set of all prime divisors of the order of G. If NG and θIrr(N), then the inertia group of θ in G is denoted by IG(θ). We write Irr(Gθ) for the set of all irreducible constituents of θG. The greatest common divisor of three integers a, b and c is gcd(a,b,c). Denote by Φk:=Φk(q) the value of the k-th cyclotomic polynomial evaluated at q. Other notation is standard.

### 2 Sporadic and alternating groups

In this section, we determine the finite simple groups satisfying the property 𝒫32. Lemmas 2.1 and 2.2 and the lemmas in Section 3 imply that a simple group S has the property 𝒫32 if and only if SM11, A7, PSL2(q) (for some prime power q4), PSL3(3), PSU3(3), PSL3(4) or Sz(q2) (q2-1 is a prime).

### Lemma 2.1.

If S is one of the sporadic simple groups, then S has the property P32 if and only if SM11.

### Proof.

Since cd(M11)={1,2×5,11,24,22×11,32×5,5×11}, M11 has the property 𝒫32. For other sporadic simple groups, Table 3 lists three distinct degrees, which are divisible by a product of two primes, and so other sporadic simple groups do not satisfy 𝒫32. Notation for the characters is as in the Atlas [3]. ∎

### Lemma 2.2.

Let S be the simple alternating group An for n5. Then SAn has the property P32 if and only if n=5, n=6 or n=7.

### Proof.

We have

cd(A5)={1,3,4,5},cd(A6)={1,5,8,9,10},cd(A7)={1,6,10,14,15,21,35};

hence the simple groups A5,A6 and A7 have the property 𝒫32. The group A8 has character degrees 20, 28 and 56, the group A9 has character degrees 28, 48 and 56, and the group A10 has character degrees 36, 84 and 160, which are all divisible by 4. Hence, for 8n10, An does not satisfy 𝒫32.

We now assume SAn, n11, and consider the irreducible character χr,s of the symmetric group Sn corresponding to the partition (n-s-r,s+1,1r-1). As shown in [18], this partition exists, provided r1, s0 and r+2s+1n, in which case

χr,s(1)=(ns)(n-s-1r-1)n-2s-rr+s.

Moreover, χr,s restricts irreducibly to An unless either s=0 and n=2r+1, or s=1 and n=2r+2. Hence we have the degrees listed in Table 2 for An with n11. Observe that 4 divides χ1,1(1), χ2,1(1) and χ1,2(1) when n0(mod8), 4 divides χ2,0(1), χ3,0(1) and χ1,2(1) when n1(mod8), 4 divides χ2,0(1), χ3,0(1) and χ2,1(1) when n2(mod8), 4 divides χ3,0(1), χ1,1(1) and χ3,2(1) when n3(mod8), 4 divides χ2,1(1), χ3,2(1) and χ4,2(1) when n4(mod8), 4 divides χ1,2(1), χ3,2(1) and χ4,2(1) when n5(mod8), 4 divides χ2,1(1), χ8,1(1) and χ2,3(1) when n6(mod8), 4 divides χ3,2(1), χ3,0(1) and χ8,1(1) when n7(mod8). It is worth noting that if n6 or 7(mod8), then n18, and so χ8,1 restricts irreducibly to An because r=8, s=1 and n2r+2=18. If n6(mod8), then 2n, 22n-2, 2n-4, 23n-6, and so 4 divides χ8,1(1). Also, if n7(mod8), then 22n-3, 2n-5, 23n-7, and hence 4 divides χ8,1(1). It is easily verified that, since n11, these degrees of An are all distinct. Therefore, for n11, An does not have the property 𝒫32. We should note that A5PSL2(4)PSL2(5) and A6PSL2(9). These groups will be considered along with the groups PSL2(q). ∎

### Table 2

Degrees of An, n11.

 Char Degree χ2,0 (n-1)⁢(n-2)2 χ3,0 (n-1)⁢(n-2)⁢(n-3)2⋅3 χ1,1 n⁢(n-3)2 χ2,1 n⁢(n-2)⁢(n-4)3 χ1,2 n⁢(n-1)⁢(n-5)2⋅3 χ3,2 n⁢(n-1)⁢(n-3)⁢(n-4)⁢(n-7)22⋅5 χ4,2 n⁢(n-1)⁢(n-3)⁢(n-4)⁢(n-5)⁢(n-8)23⋅32 χ8,1 n⁢(n-2)⁢(n-3)⁢(n-4)⁢(n-5)⁢(n-6)⁢(n-7)⁢(n-8)⁢(n-10)24⋅34⋅5⋅7 χ2,3 n⁢(n-1)⁢(n-2)⁢(n-4)⁢(n-8)2⋅3⋅5

### Table 3

Three character degrees of sporadic simple groups divisible by a product of two primes.

 Grp Char Deg. M12 χ4 16 χ13 120 χ14 144 J1 χ3 56 χ4 76 χ9 120 J2 χ6 36 χ12 160 χ16 224 M23 χ3 45 χ14 990 χ16 1035 HS χ14 896 χ16 1056 χ18 1408 J3 χ9 816 χ10 1140 χ19 2432 M24 χ7 252 χ19 2024 χ21 3312 McL χ4 252 χ7 896 χ10 3520 He χ6 680 χ12 1920 χ13 4080 Ru χ6 3276 χ11 27000 χ15 34944 Suz χ3 364 χ4 780 χ6 3432 O⁢’⁢N χ2 10944 χ3 13376 χ4 25916 M22 χ7 154 χ8 210 χ10 280 Co3 χ6 896 χ9 2024 χ10 3520 Co2 χ6 2024 χ9 7084 χ15 23000 Fi22 χ3 429 χ4 1001 χ5 1430 HN χ4 760 χ5 3344 χ11 35112 Ly χ2 2480 χ7 120064 χ11 1534500 Th χ2 248 χ4 27000 χ6 30628 Fi23 χ3 3588 χ6 30888 χ7 60996 Co1 χ2 276 χ7 27300 χ14 376740 J4 χ11 1776888 χ21 95288172 χ23 259775040 Fi24′ χ13 48893768 χ14 74837400 χ17 112168056 B χ13 4275362520 χ17 13508418144 χ20 80426400000 M χ3 21296876 χ5 18538750076 χ15 2374124840062976

### Lemma 2.3.

Let S be one of the sporadic simple groups or an alternating simple group An with n7. If SGAut(S), then G satisfies the property P32 if and only if G=SM11, G=SA7, or SA7 and GS7.

### Proof.

If SM11 is one of the sporadic simple groups, then it follows from Lemma 2.1 that S and so G do not have the property 𝒫32 since |Aut(S):S|2 and none of character degrees of S given in Table 3 is twice another degree. Also, if SM11, then G=S since Aut(M11)M11, and hence G=S has the property 𝒫32 by using Lemma 2.1. If S=An with n8, then Lemma 2.2 implies that S does not have the property 𝒫32. Observing that Aut(S)Sn and since the degrees of An shown in Table 2 are in fact restrictions of irreducible characters of Sn, we deduce that Sn does not have the property 𝒫32. If S=A7, then G=A7 or G=S7, and so G has the property 𝒫32 as cd(A7)={1,6,10,14,15,21,35}, cd(S7)={1,6,14,15,20,21,35}. ∎

Since A5PSL2(4)PSL2(5) and A6PSL2(9), we will consider A5 and A6 as groups of Lie type rather than alternating groups.

### 3 Groups of Lie type

We prove Theorem A in this section. We consider the simple groups of Lie type, and we will prove that if S is a simple group of Lie type satisfying the property 𝒫32, then SPSL2(q) (q4), PSL3(3), PSU3(3), PSL3(4) or Sz(q2) (q2-1 is a prime). The classical groups are

A1(q)PSL2(q)forq2,3,
A(q)PSL+1(q),A2(q2)PSU+1(q2)for2,q2if=2,
B(q)Ω2+1(q),C(q)PSp2(q)for2,q2if=2,
D(q)PΩ2+(q),D2(q2)PΩ2-(q)for4,

and the groups of exceptional Lie type are

G2(q)forq2,2G2(q2)forq23,2B2(q2)forq22,F42(q2)forq22,2F4(2),F4(q),E6(q),E7(q),E8(q),E62(q2),3D4(q3).

The restrictions on and q are so that the groups will be simple and (generally) not isomorphic to others in the list. These restrictions will always be assumed in what follows.

Now we consider almost simple groups SGAut(S), where SPSL2(q) and q=pf4. The character degrees of G have been worked out in detail by White [22]. The outer automorphism group of S is of order (2,q-1)f and is generated by a diagonal automorphism δ of order (2,q-1) and a field automorphism φ of order f.

### Theorem 3.1.

Let SPSL2(q), where q=pf4 for a prime p, A=Aut(S), and let SGA. Set H=PGL2(q) if δG, and H=S if δG, and let |G:H|=2am=d, m odd. If p is odd, let ε=(-1)q-12. Then

cd(G)={1,q,q+ε2}{(q-1)2ai:im}{(q+1)j:jd}

with the following exceptions:

1. If p is odd with G⩽̸Sφ or if p=2, then q+ε2 is not a degree of G.

2. If f is odd, p=3 and G=Sφ, then i1.

3. If f is odd, p=3 and G=A, then j1.

4. If f is odd, p=2, 3 or 5 , and G=Sφ, then j1.

5. If f2(mod4), p=2 or 3 , and G=Sφ or G=Sδφ, then j2.

### Proof.

This is [22, Theorem A]. ∎

### Corollary 3.2 ([22, Corollary 6.2]).

If PSL2(q)<GAut(PSL2(q)), q=pf, p an odd prime, then one of the following occurs:

1. δG so that PGL2(q)G and G=PGL2(q)φk for some kf with 1kf.

2. G=PSL2(q)φk for some kf with 1k<f.

3. G=PSL2(q)δφk for some kf with 1k<f and f/k even.

### Proof.

This is [22, Corollary 6.2]. ∎

### Lemma 3.3.

If 2f-1 is prime for some integer f>2, then 2f+1 is not prime.

### Proof.

It is easy to see that if 2f-1 is prime, then f is prime. If 2f+1 is also a prime number, then we obtain that f is a power of 2. This implies f=2 as f is a prime number, which is a contradiction. ∎

### Lemma 3.4.

Let SPSL2(q) with q=pf4, and let SGAut(S) be an almost simple group with socle S. Then G satisfies P32 if and only if G is one of the groups listed in Table 1.

### Proof.

If G is one of the groups listed in Table 1, then, by the character degree sets of these groups (in column 3 of Table 1), it is easy to see that the groups G in Table 1 have the property 𝒫32.

Conversely, let G be an almost simple group with socle SPSL2(q), where q=pf4. Suppose that G has the property 𝒫32. We prove that G is one of the groups listed in Table 1. We use the notation of Theorem 3.1. Assume first |G:H|=1. It follows that G=PSL2(q) or PGL2(q), and so

cd(G)=cd(PSL2(q))={1,2f-1,2f,2f+1},whereq=2f,
cd(G)=cd(PSL2(q))={1,q+ε2,q-1,q,q+1},whereqis odd,

or

cd(G)=cd(PGL2(q))={1,q-1,q,q+1},whereq5is odd,

which is case 1 of Table 1.

Thus we can assume |G:H|>1. Write |G:H|=2am with m odd and a0. We will consider the cases p odd and p even separately.

Case 1: p=2. In this case, since p=2, H=SPSL2(q) with q=2f. If f=2, then SPSL2(4)PSL2(5)A5, and hence GPGL2(5)S5 and cd(G)={1,4,5,6}, which is included in line 3 of case 1 of Table 1 (with q=5). So assume f>2. Hence Lemma 3.3 implies that at least one of q-1 or q+1 is not a prime number, and so at least one of them is divisible by a product of two primes.

Subcase 1: Assume that |G:H| is divisible by two distinct primes r and s, with r>s. If s>2, then it follows from Theorem 3.1 that

{2ar(q-1),2as(q-1),2ars(q-1),2ar(q+1),2as(q+1),2ars(q+1)}cd(G).

Therefore, either the degrees 2ar(q-1), 2as(q-1), 2ars(q-1)cd(G) are distinct with gcd(2ar(q-1),2as(q-1),2ars(q-1)) divisible by a product of two primes or the degrees 2ar(q+1), 2as(q+1), 2ars(q+1) are distinct with gcd(2ar(q+1),2as(q+1),2ars(q+1)) divisible by a product of two primes. We deduce that the group G does not have the property 𝒫32, which is a contradiction. Thus we should have s=2 and |G:H|=2arb with a,b1. We first assume a=b=1. If exceptions (iv) and (v) of Theorem 3.1 do not occur, then it follows from Theorem 3.1 that

cd(G)={1,q,2(q-1),2r(q-1),(q+1),2(q+1),r(q+1),2r(q+1)}.

Hence, if 2f+1 is not prime, then G has three distinct degrees q+1, 2(q+1), r(q+1)cd(G) with gcd(q+1,2(q+1),r(q+1)) divisible by a product of two primes, and so G does not have the property 𝒫32, which is a contradiction. Thus we must have 2f+1 a prime. But, then either G has three distinct degrees 2r(q-1), 2(q+1), 2r(q+1)cd(G) with gcd(2r(q-1),2(q+1),2r(q+1)) divisible by 2r if and only if r=q+1, or G has three distinct degrees 2(q-1), 2r(q-1), 2r(q+1)cd(G) with gcd(2(q-1),2r(q-1),2r(q+1)) divisible by 2r if and only if rq-1. As the group G has the property 𝒫32, we conclude that rq2-1, which is case 2 of Table 1.

If one of the exceptions (iv) and (v) of Theorem 3.1 occurs, then it follows from Theorem 3.1 that

cd(G)={1,q,2(q-1),2r(q-1),(q+1),r(q+1),2r(q+1)}

or

cd(G)={1,q,2(q-1),2r(q-1),2(q+1),r(q+1),2r(q+1)}.

In both cases, if 2f+1 is not prime, then G either has three distinct degrees q+1, r(q+1), 2r(q+1)cd(G) with gcd(q+1,r(q+1),2r(q+1)) divisible by a product of two primes, or G has three distinct degrees 2(q+1), r(q+1), 2r(q+1)cd(G) with gcd(2(q+1),r(q+1),2r(q+1)) divisible by a product of two primes, and so G does not have the property 𝒫32, which is a contradiction.

Therefore, we must have 2f+1 a prime. Then, in both cases of cd(G), the group G has three distinct degrees 2(q-1), 2r(q-1) and 2r(q+1)cd(G) with gcd(2(q-1),2r(q-1),2r(q+1)) divisible by 2r if and only if rq-1. As the group G has the property 𝒫32, we conclude that rq-1.

Further, if cd(G)={1,q,2(q-1),2r(q-1),2(q+1),r(q+1),2r(q+1)}, then the group G has three distinct degrees 2r(q-1), 2(q+1), 2r(q+1)cd(G) with gcd(2r(q-1),2(q+1),2r(q+1)) divisible by 2r if and only if r=q+1. As the group G has the property 𝒫32, we conclude that rq+1 if

cd(G)={1,q,2(q-1),2r(q-1),2(q+1),r(q+1),2r(q+1)},

which is case 3 of Table 1.

The remaining case is

cd(G)={1,q,2(q-1),2r(q-1),(q+1),r(q+1),2r(q+1)},

where q+1 is a prime and rq-1, which is the case 4 of Table 1.

Now we assume a>1 (or b>1). We observe that G has three distinct degrees 2f, 2a(q-1) and 2ar(q-1) (or 2rb(q-1), rb(q+1) and 2rb(q+1)), which are all divisible by 4 (or r2). This is a contradiction as G has the property 𝒫32.

Subcase 2: Assume |G:H|=rc for some prime r and c1. In this case, since 2am=|G:H|=rc with m odd, we first assume r>2, and so a=0 and m=rc. If c>2, then the group G has three distinct degrees r2(q-1), r3(q-1), r2(q+1)cd(G) with gcd(r2(q-1),r3(q-1),r2(q+1)) divisible by r2, which is a contradiction as the group G has the property 𝒫32. Hence we should have c2.

If c=2 and exception (iv) of Theorem 3.1 does not occur, then it follows from Theorem 3.1 that

cd(G)={1,q,q-1,r(q-1),r2(q-1),q+1,r(q+1),r2(q+1)}.

Since, by Lemma 3.3, we know that at least one of the numbers q-1 or q+1 is not a prime, the group G does not have the property 𝒫32, which is a contradiction.

Thus exception (iv) should occur if c=2. This and Theorem 3.1 imply

cd(G)={1,q,q-1,r(q-1),r2(q-1),r(q+1),r2(q+1)}

as |G:H|=r2. Then either the group G has three distinct degrees q-1, r(q-1), r2(q-1)cd(G) with gcd(q-1,r(q-1),r2(q-1)) divisible by a product of two primes if and only if q-1 is not a prime, or G has three distinct degrees r(q±1), r2(q-1), r2(q+1)cd(G) with gcd(r(q±1),r2(q-1),r2(q+1)) divisible by r2 if and only if rq-1 or rq+1. Since the group G has the property 𝒫32, we deduce that if c=2, then exception (iv) occurs, q-1 is a prime number, rq-1 and rq+1, which is case 5 of Table 1.

If c=1, then |G:H|=r>2 is a prime number. Recall that q=2f with f>2, and we know that fr. Suppose that f=r>2. Then it follows from Theorem 3.1 that G=Sφ=Aut(S), and so

cd(G)={1,2f,2f-1,f(2f-1),f(2f+1)},

which is case 6 of Table 1. If f>r, then Theorem 3.1 implies G=Sφf/r, and so cd(G)={1,2f,2f-1,r(2f-1),2f+1,r(2f+1)}, which is case 6 of Table 1.

Now assume r=2 and so |G:H|=2c with c1. If c2, then G has three degrees q, 2c(q-1), 2c(q+1)cd(G), which are all divisible by 4, a contradiction. Hence we should have c=1. If f=r=2, then it follows Theorem 3.1 that G=PSL2(4)φPGL2(5) and cd(G)={1,4,5,6}, which is included in line 3 of case 1 of Table 1 (with q=5). If f>r=2, then G=PSL2(q)φf/2 and cd(G)={1,2f,2(2f-1),2f+1,2(2f+1)}, which is case 7 of Table 1.

Case 2: p>2. In this case, by Corollary 3.2, one of the following subcases holds.

Subcase 1: δG and G=PGL2(q)φk for some kf with 1kf. In this subcase, assuming the notation of Theorem 3.1, we have H=PGL2(q) and |G:H|=2am>1 (with m odd). We first assume that |G:H| is divisible by two distinct primes r>s2. Then it follows from Theorem 3.1 that the group G has three distinct degrees r(q+1), rs(q+1) and s(q+1), which are all divisible by q+1. Since 2q+1 is even, the group G does not have the property 𝒫32, which is a contradiction.

Thus we should have |G:H|=rc for some prime r and c1. If r=2 and c2, then, by using Theorem 3.1, we observe that the group G has at least three degrees 2c(q-1), 2(q+1) and 2c(q+1), which are all divisible by 4, which is a contradiction as the group G has the property 𝒫32. Thus if r=2, then c=1, and so G=PGL2(q)φf/2 and cd(G)={1,q,2(q-1),q+1,2(q+1)}, which is case 8 of Table 1. We note that, since p is odd and f is even, q=pf1(mod4), so q+1 is not divisible by 4.

If r>2 and c3, then Theorem 3.1 implies that the group G has c+1 degrees of the form rk(q-1) with 0kc, a contradiction since G has the property 𝒫32. Thus we should have c2. If c=2, then G has three degrees r(q-1), r2(q-1) and r2(q+1), which are all divisible by 2r, a contradiction. Hence c=1. If p=3 and G=Aut(S), that is, f=r, then

cd(G)=cd(PSL2(3f)φ)={1,3f,3f-1,f(3f-1),f(3f+1)},

which is case 9 of Table 1. If p=3 and GAut(S), or p>3, then either the group G has three degrees q-1, r(q-1) and r(q+1), which are divisible by 2r if and only if rq-1, or the group G has three degrees q+1, r(q-1) and r(q+1), which are divisible by 2r if and only if rq+1. Since G has the property 𝒫32, we deduce rq-1 and rq+1, which is case 10 of Table 1.

Subcase 2: G=PSL2(q)φk for some kf with 1k<f. We first assume that |G:H|=|φk|=f/k is divisible by two distinct primes r and s with r>s2. If s>2, then it follows from Theorem 3.1 that G has three degrees 2ar(q-1), 2as(q-1) and 2ars(q-1), which are all divisible by q-1, a contradiction since G has the property 𝒫32 and q4 with q-1 even. If s=2 and |G:H|=2arb, then Theorem 3.1 implies that G has degrees r(q+1), 2r(q+1) and either q+1 or 2(q+1), which are all divisible by q+1, a contradiction as G has the property 𝒫32 and q+14 is even.

Now we assume |G:H|=rc with c1. If r=2 and c2, then G has three degrees q+1, 2(q+1) and 4(q+1), which are all divisible by q+1, a contradiction. Hence we should have c=1 and G=PSL2(q)φf/2. If q=9, then G=PSL2(9)φS6 and cd(G)={1,5,9,10,16}, which is included in case 11 of Table 1. If q9, then cd(G)={1,q+ε2,q,q+1,2(q+1),2(q-1)}, where ε=(-1)(q-1)/2. If q-12 is odd, then q-12+1=q+12 is even, ε=-1 and cd(G)={1,q-12,q,q+1,2(q+1),2(q-1)}, and hence q+1, 2(q+1), 2(q-1)cd(G) are divisible by 4, which is a contradiction as G satisfies 𝒫32. Therefore, q-12 is even, and so ε=1, q+12 is odd, and

cd(G)={1,q+12,q,q+1,2(q+1),2(q-1)}.

Since the group G has the property 𝒫32, we conclude that q+12 should be a prime number, which is included in case 12 of Table 1.

Assume r>2 is an odd prime. If c2, then G has character degrees r2(q-1), r(q-1) and r(q+1), which are all divisible by 2r, a contradiction. Therefore, c=1. If f=r, then G=PSL2(q)φ. If p=3, then q=3f-1(mod4) and Theorem 3.1 (ii) and (iii) implies

cd(G)={1,3f,3f-12,(3f-1)f,(3f+1)f},

which is case 13 of Table 1.

If p=5, then q=5f1(mod4) and

cd(G)={1,5f,5f+12,5f-1,(5f-1)f,(5f+1)f},

which appears in case 14 of Table 1.

If p>5, then it follows from Theorem 3.1 that either G has three degrees q+1, f(q-1) and f(q+1), which are all divisible by 2f if and only if fq+1, or G has three degrees q-1, f(q-1) and f(q+1), which are all divisible by 2f if and only if fq-1. Since G has the property 𝒫32, we conclude that 2<r=fq2-1, which is included in case 15 of Table 1.

Finally, if f>r, then k>1 since r=|G:H|=f/k. As G=PSL2(q)φk with k>1, we observe that none of the exceptions (ii), (iii) and (iv) of Theorem 3.1 occurs. Therefore, it follows from Theorem 3.1 that

cd(G)={1,q,q+ε2,q-1,(q-1)r,q+1,(q+1)r}.

If rq-1, then q-1, r(q-1), r(q+1)cd(G) are divisible by 2r, and so G does not have the property 𝒫32. If rq+1, then q+1, r(q-1), r(q+1)cd(G) are divisible by 2r, and so G does not satisfy 𝒫32. Since the group G has the property 𝒫32, we have 2<rq2-1, which is included in case 16 of Table 1.

Subcase 3: G=PSL2(q)δφk for some kf with 1k<f and fk is even. In this subcase, since G⩽̸PSL2(q)φ, Theorem 3.1 (i) implies |G:H|=2am (so a1) is even and q+ε2 is not a degree of G. If |G:H|=2am with odd number m>1, then it follows from Theorem 3.1 that G has three degrees 2a(q-1), 2am(q-1) and 2am(q+1), which are divisible by 4, a contradiction. Thus we should have |G:H|=2a with a1. If a2, then G has three distinct degrees q+1, 2a(q+1) and 2a-1(q+1), which are divisible by 4, a contradiction. We deduce that we must have a=1 and |G:H|=2.

If p=3 and G=PSL2(q)δφ, then f=2 as |G:H|=2 and hence GM10 and cd(G)={1,9,10,16}, which is case 17 of Table 1.

Finally, if q9, then G=PSL2(q)δφk with fk=2 and

cd(G)={1,q,q+1,2(q+1),2(q-1)},

which appears in case 18 of Table 1. We note that, since p is odd and f is even, q=pf1(mod4), so q+1 is not divisible by 4. The proof is now complete. ∎

### Lemma 3.5.

Let SPSL3(q) with q=pf3, and let SGAut(S) be an almost simple group with socle S. Then G satisfies P32 if and only if

G=SPSL3(3).

### Proof.

Suppose that SPSL3(q) with q=pf3 and G is an almost simple group with the socle S. We know that |Aut(S):S|=2df, where d=(3,q-1), and so |Aut(S):S| divides 6f. Also, we assume that the group G has the property 𝒫32. We first prove that if q>4, then f=1 and q is a prime number. To prove this, if f2, then we choose irreducible characters θiIrr(S), 1i3, with degrees θ1(1)=q3, θ2(1)=q(q+1) and θ3(1)=q(q2+q+1). Let χi be an irreducible constituent of θiG for 1i3. Then χi(1)=aiθi(1) with ai2df. Since (q+1,q2+q+1)=1, if χ2(1)=χ3(1), then a2θ2(1)=a3θ3(1) and a2q(q+1)=a3q(q2+q+1), and hence q+1a3 and q2+q+1a2. Thus (q+1)(q2+q+1) divides |Aut(S):S| and so divides 6f. However, q>4 and q=pf2f (by [14, Lemma 3.2]). We obtain

(q+1)(q2+q+1)>q3q22f>6f,

a contradiction. Thus we must have χ2(1)χ3(1). Also, it is easy to see that χ1(1)χ2(1) and χ1(1)χ3(1). We deduce that χ1(1),χ2(1),χ3(1)cd(G) are three distinct degrees of G, which are divisible by q, which is a contradiction as q=pf with f2 and the group G has the property 𝒫32. Therefore, if q>4, then we must have f=1 and q is a prime number.

Assume q=3. Then SPSL3(3) and |Out(S)|=2, and hence G=S or G=Aut(S). By the Atlas [3],

eithercd(G)=cd(S)={1,12,13,16,26,27,39}
orcd(G)=cd(Aut(S))={1,12,13,26,27,32,39,52}.

It is easy to check that no product of two primes divides the greatest common divisor of any three members of the set {1,12,13,16,26,27,39}. Thus PSL3(3) has the property 𝒫32, which is case (iv) of Theorem A. But 412,32,52cd(Aut(S)), and so Aut(S) does not have the property 𝒫32.

If q=4, then SPSL3(4). By checking the Atlas [3], we have

cd(S)={1,20,35,45,63,64},

and the outer automorphism group of PSL3(4) is generated as a semidirect product by a diagonal automorphism δ of order (3,q-1)=3, a field automorphism φ of order f=2 and a graph automorphism τ of order 2. The diagonal automorphism is induced on SL3(4) by conjugation with a diagonal matrix in SL3(4) of order 3. The field automorphism is induced by raising the matrix entries to the power 2, and the graph automorphism by taking the inverse transpose of an element of SL3(4). Thus it follows from the Atlas [3] that

Out(S)=δ,φ,τδ3=φ2=τ2=[φ,τ]=1,δφ=δ-1,δτ=δ-12×S3.

Assume {θ1,θ2,θ3,θ4,θ5}Irr(S) with θ1(1)=20, θ2(1)=35, θ3(1)=45, θ4(1)=63, θ5(1)=64. Since SGAut(S), let {χ1,χ2,χ3,χ4,χ5}Irr(G) be such that χi is an irreducible constituent of θiG. Applying [8, Corollary 11.29], we obtain that χi(1)=aiθi(1) and ai divides 12. We know that θ5 is the Steinberg character, which can extend to G. First we assume that G/S is a nonabelian subgroup of 2×S3. Since cd(2×S3)={1,2}, it follows from Gallagher’s theorem [8, Corollary 6.17] that 128cd(G). We conclude that 20a1,64,128cd(G) are three distinct character degrees, which are divisible by 4, and so the group G does not have the property 𝒫32 if G/S is nonabelian.

Now we suppose that G/S is abelian. Since two almost simple groups with the socle S are isomorphic if and only if their images in Out(S) are conjugate, G is one of the following groups:

S,Sδ,Sφ,Sτ,Sφτ,Sδφτ,Sφτ.

By using the Atlas [3], we obtain

cd(Sφτ)={1,20,35,64,90,126},
cd(Sδ)={1,20,45,63,64,105},
cd(Sδφτ)={1,20,64,90,105,126},
cd(Sφ)={1,20,35,45,64,70,126},
cd(Sτ)={1,20,35,63,64,70,90}.

Thus, if G{S,Sφτ,Sδ,Sδφτ,Sφ}, then G has the property 𝒫32, which is case (vi) of Theorem A. Also, if G=Sτ, then 10 divides three degrees of G since cd(G)={1,20,35,63,64,70,90}, and so G does not have the property 𝒫32.

The remaining case is G=Sφτ. In this case, since Sφ and Sτ are normal subgroups of G=Sφτ, cd(Sφ)={1,20,35,45,64,70,126} and cd(Sτ)={1,20,35,63,64,70,90}, we deduce

{1,20a1,35a2,45a3,63a4,64,70a5,90a6,126a7}cd(G),

where ai divides 2 for 1i7. Hence 10 divides three distinct character degrees 20a1,70a5,90a6cd(G), and so the group G does not have the property 𝒫32.

Finally, we assume q=p5. If p=4k+3, then it follows by [21, Table 1] that SPSL3(q) has irreducible characters θi, 1i3, with θ1(1)=q(q+1), θ2(1)=(q-1)2(q+1) and θ3(1)=(q+1)(q2+q+1), which are all divisible by q+1=4k+4. Let χi be an irreducible constituent of θiG for 1i3. Then χi(1)=aiθi(1) for some ai2d. If χ1(1)=χ2(1), then

a1q(q+1)=a2(q-1)2(q+1),

and so a1q=a2(q-1)2, which implies qa2 and hence q2d. As (q,d)=1, we see that q=2, a contradiction. Also, if χ1(1)=χ3(1), then

a1q(q+1)=a3(q+1)(q2+q+1).

As (q,q2+q+1)=1, hence qa3; as above, this yields a contradiction. Now suppose that χ2(1)=χ3(1). Then a2(q-1)2(q+1)=a3(q+1)(a2+q+1), and thus

a2(q-1)2=a3(q2+q+1)
a2(4k+2)2=a3((4k+3)2+(4k+3)+1)
4a2(4k2+4k+1)=a3(16k2+28k+13)
4a342d,

which is a contradiction as d=(3,q-1).

We deduce that χ1(1),χ2(1),χ3(1)cd(G) are distinct and all divisible by q+1=4(k+1). We obtain that G does not satisfy the property 𝒫32, which is a contradiction.

If p=4k+1, then it follows from [21, Table 1] that SPSL3(q) has irreducible characters θi, 1i3, with θ1(1)=q(q+1), θ2(1)=(q-1)2(q+1) and θ3(1)=(q+1)(q2+q+1), which are all divisible by q+1=4k+2. Let χi be an irreducible constituent of θiG for 1i3. Then χi(1)=aiθi(1) for some ai2d. By similar arguments, we conclude that χ1(1),χ2(1),χ3(1)cd(G) are distinct and all divisible by q+1=2(2k+1). Therefore, G does not satisfy the property 𝒫32, which is a contradiction. ∎

### Lemma 3.6.

Let SPSU3(q) with q=pf3, and let SGAut(S) be an almost simple group with socle S. Then G satisfies P32 if and only if

G=SPSU3(3).

### Proof.

By the Atlas [3], we know that cd(PSU3(3))={1,6,7,14,21,27,28,32}, and hence SPSU3(3) satisfies 𝒫32.

Conversely, we suppose that SGAut(S) is an almost simple group with socle SPSU3(q), q3. If G satisfies 𝒫32, then we prove G=SPSU3(3). To do this, we know that |Out(S)|=2df, where q=pf and d=(3,q+1). If f2, then, by [21, Table 1], let θ1,θ2,θ3Irr(S) with degrees θ1(1)=q3, θ2(1)=q(q-1) and θ3(1)=q(q2-q+1). Let χi be an irreducible constituent of θiG for 1i3. Then χi(1)=aiθi(1) and ai2df. Since

(q-1,q2-q+1)=1,

then a2q(q-1)=a3q(q2-q+1) implies q-1a3 and q2-q+1a2. Thus (q-1)(q2-q+1)2df6f. Since q3, it follows from [14, Lemma 3.2] that q2f, and hence

(q-1)(q2-q+1)>(q-1)2q(3-1)2q=4q8f>6f,

which is a contradiction. Hence we have χ2(1)χ3(1). It is easy to observe that χ1(1)χ2(1),χ3(1). This implies that p2 divides three distinct character degrees of G, which is a contradiction as the group G satisfies 𝒫32. We conclude that f=1 and q=p is a prime number.

If q=p5, by applying [21, Table 1], we choose characters θ1,θ2,θ3Irr(S) with degrees

θ1(1)=q(q-1),θ2(1)=(q-1)(q2-q+1),θ3(1)=(q+1)2(q-1),

which are divisible by 4k or 4k+2 as either q=p=4k+1 or q=p=4k+3. Let χi be an irreducible constituent of θiG for 1i3. Then χi(1)=aiθi(1) and ai2df. If χ1(1),χ2(1),χ3(1) are distinct, then they are divisible by 4k or 4k+2=2(2k+1), and hence the group G does not have the property 𝒫32, which is a contradiction. Therefore, it suffices to show that χ1(1),χ2(1),χ3(1) are distinct. To do this, if χ1(1)=χ2(1), then a1q(q-1)=a2(q-1)(q2-q+1) and a1q=a2(q2-q+1). Since (q,q2-q+1)=1, hence qa22d, and so q2 as d=(3,q+1) and (q,d)=1. We obtain q=2, a contradiction. Thus χ1(1)χ2(1). If χ1(1)=χ3(1), then a1q=a3(q+1)2 and qa32d. Since (q,d)=1, we observe that q2, which is a contradiction. If χ2(1)=χ3(1), then a2(q2-q+1)=a3(q+1)2. Since (q2-q+1,(q+1)2)=1 or 3, this implies either (q+1)2a2 or (q+1)2/3a2. But a22df=2d6 as d=(3,q+1). We obtain either (q+1)26 or (q+1)2/36, which are contradictions as q5 and (q+1)236. Therefore, χ2(1)χ3(1) and so χ1(1),χ2(1),χ3(1)cd(G) are distinct which is the desired conclusion.

Thus we must have q=3 and SPSU3(3). Then |Out(S)|=2, and so G=S or G=Aut(S). If G=Aut(S), then, since S has a unique irreducible character of degree 14, three irreducible characters of degree 21 and two irreducible characters of degree 28, either 14,42,28cd(Aut(S)) or 14,42,56cd(Aut(S)), which are divisible by 14. This implies that G=Aut(S) does not satisfy 𝒫32, which is a contradiction. We conclude that G=SPSU3(3), which is the desired conclusion and is case (v) of Theorem A. ∎

### Theorem 3.7.

Let SB22(q2), q2=22f+1, f1, and let SGAut(S) with |G:S|=d. Then the set of irreducible character degrees of G is

cd(G)={1,q4,2q(q2-1)2}{(q4+1)a,(q2-1)(q2-2q+1)b,(q2-1)(q2+2q+1)c;a,b,cd}

with the following exceptions:

1. If G=Aut(S), then a1.

2. If f1 or 2(mod4) and G=Aut(S), then b1 and c3.

3. If f0 or 3(mod4) and G=Aut(S), then b3 and c1.

### Proof.

This is the main theorem of [4]. ∎

### Lemma 3.8.

Let SB22(q2)=Sz(q2), where q2=22f+1 with f1, and let SGAut(S) be an almost simple group with socle S. Then the group G satisfies P32 if and only if one of the following holds:

1. G=S, and q2-1=22f+1-1 and 2f+1 are prime numbers. In this case,

cd(G)=cd(S)={1,q4,2f(q2-1),q4+1,(q2-1)(q2-2q+1),(q2-1)(q2+2q+1)}.
2. G=Aut(S), q2-1=22f+1-1, 2f+1 and 22f+1+2f+1+1 are prime numbers, f1 or 2(mod4), and if f>1, then

cd(G)={1,q4,2f(q2-1),(q4+1)(2f+1),(q2-1)(q2-2q+1)(2f+1),(q2-1)(q2+2q+1),(q2-1)(q2+2q+1)(2f+1)},

and if f=1, then

cd(G)={1,2×7,3×5×7,26,3×5×13,7×13}.
3. G=Aut(S), q2-1=22f+1-1, 2f+1 and 22f+1-2f+1+1 are prime numbers, f0 or 3(mod4), and

cd(G)={1,q4,2f(q2-1),(q4+1)(2f+1),(q2-1)(q2-2q+1),(q2-1)(q2-2q+1)(2f+1),(q2-1)(q2+2q+1)}.

### Proof.

If the group G is any of the cases above, then, by its character degree set, it is easy to see that the group G satisfies 𝒫32. Conversely, we assume that the group G has the property 𝒫32, and we prove that one of the cases above holds. Since SGAut(S), SB22(q2), q2=22f+1, f1, it follows from Theorem 3.7 that

2f(q2-1),(q2-1)(q2-2q+1)b,(q2-1)(q2+2q+1)ccd(G),

where b and c are some divisors of |G:S|. Hence q2-1 divides three distinct character degrees of G. As the group G satisfies the property 𝒫32, we obtain that q2-1=22f+1-1 must be a prime number. Thus 2f+1=|Aut(S):S| is a prime number, and so either G=S or G=Aut(S).

If G=S, since q2-1=22f+1-1 and 2f+1 are prime numbers, then

cd(G)=cd(S)={1,q4,2f(q2-1),q4+1,(q2-1)(q2-2q+1),(q2-1)(q2+2q+1)},

and so case 1 holds.

Thus we can assume G=Aut(S). We prove that one of the cases 2 or 3 holds. To do this, first suppose f>1 and so 2f+15. It follows from [5, Lemma 2.9] that at least one of the numbers q2-2q+1 and q2+2q+1 is divisible by 5. Since q2+2q+1, q2-2q+1>5 (otherwise, q2-2q+1=5, and so f=1, a contradiction), we deduce that either q2+2q+1 is not prime or q2-2q+1 is not prime. If none of the exceptions (ii) and (iii) of Theorem 3.7 occurs, then, since 2f+1 is prime and G=Aut(S),

cd(G)={1,q4,2f(q2-1),(q4+1)(2f+1),(q2-1)(q2-2q+1),(q2-1)(q2-2q+1)(2f+1),(q2-1)(q2+2q+1),(q2-1)(q2+2q+1)(2f+1)}.

This implies that q2+2q+1 and q2-2q+1 divide three distinct character degrees of G, and so the group G does not satisfy 𝒫32, which is a contradiction.

We conclude that one of the exceptions (ii) or (iii) of Theorem 3.7 occurs. If exception (ii) holds, then f1 or 2(mod4), and, because f>1,

cd(G)={1,q4,2f(q2-1),(q4+1)(2f+1),(q2-1)(q2-2q+1)(2f+1),(q2-1)(q2+2q+1),(q2-1)(q2+2q+1)(2f+1)}.

This implies that the group G satisfies 𝒫32 if and only if

q2+2q+1=22f+1+2f+1+1

is a prime number because

q4+1=(q2-2q+1)(q2+2q+1).

Therefore, case 2 holds.

If exception (iii) holds, then f0 or 3(mod4), and

cd(G)={1,q4,2f(q2-1),(q4+1)(2f+1),(q2-1)(q2-2q+1),(q2-1)(q2-2q+1)(2f+1),(q2-1)(q2+2q+1)(2f+1)}.

As the group G satisfies 𝒫32, this forces that q2-2q+1=22f+1-2f+1+1 is a prime number since

q4+1=(q2-2q+1)(q2+2q+1).

Hence case 3 holds.

If f=1, then 2f+1=3, and so, by Theorem 3.7, we have

cd(G)={1,2×7,3×5×7,26,3×5×13,7×13}

because G=Aut(S). We obtain that case 2 holds, and we are done. ∎

### Lemma 3.9.

If

SB2(q)C2(q)PSp4(q),q=pf3,

and SGAut(S), then G does not satisfy P32.

### Proof.

By [2, Theorem 13.8], S has unipotent irreducible characters labeled by the symbols

(1 20),(0 21),(0 1 2-)

of degrees q(q2+1)/2, q(q+1)2/2 and q(q-1)2/2, respectively. They are extendible to G by [15, Theorems 2.4 and 2.5] except for the character of degree q(q2+1)/2, which implies that G has an irreducible character of degree eq(q2+1)/2 with e>1. Thus eq(q2+1)/2, q(q+1)2/2, q(q-1)2/2cd(G), and it is easy to check that these degrees are pairwise distinct. If q=2f and f3, then 4 divides these three degrees and so G does not satisfy 𝒫32. If q=pf is odd, then 2q divides these degrees, and hence G does not have the property 𝒫32.

Thus the only remaining case is q=4, and so SB2(4)C2(4)PSp4(4). Then it follows from the Atlas [3] that

cd(S)={1,18,34,50,51,85,153,204,225,255,256,340}.

Let θiIrr(S), 1i3, be irreducible characters of S of degrees θ1(1)=44, θ2(1)=340, θ3(1)=12×17. Assume χiIrr(Gθi) with 1i3. Since |Aut(S):S|=4, χi(1)=aiθi(1), where ai4. It is obvious that χ1(1), χ2(1), χ3(1)cd(G) are distinct and divisible by 4. Thus G does not satisfy 𝒫32, which is the desired conclusion. ∎

### Lemma 3.10.

If SG2(q), q=pf3, and SGAut(S), then G does not satisfy P32.

### Proof.

Since SG2(q), q=pf3, we know that |Aut(S):S|=fg, where g=2 if p=3, and g=1 if p3. First assume f2 if p2, and f3 if p=2. Then, by Table 4, suppose that θ1,θ2,θ3Irr(S) with degrees θ1(1)=q6, θ2(1)=q(q-1)(q3-1)/2 and θ3(1)=q(q+1)(q3+1)/2. Also, suppose that χiIrr(Gθi), 1i3. Then it follows from [14, Lemma 3.4 (type G2)] that χ1(1), χ2(1), χ3(1) are distinct and divisible by p2. This implies that the group G does not satisfy 𝒫32.

If p=2 and f=2, then SG2(4) has unique irreducible characters of degrees 1300, 4096, 4160. Since |Aut(G2(4)):G2(4)|=2, G/S is cyclic, and hence these unique irreducible characters are extendible to G. Therefore, 1300, 4096, 4160cd(G) are divisible by 4. This shows that the group G does not have the property 𝒫32.

The only remaining case is f=1, and so q=p3 is a prime number. If q=p=4k+1, by Table 4, we choose irreducible characters μ1,μ2,μ3Irr(S) with degrees μ1(1)=qΦ12Φ6/6, μ2(1)=qΦ12Φ3/2, μ3(1)=qΦ12Φ22/3. Furthermore, suppose that ψiIrr(Gμi), 1i3. Then ψi(1)=aiμi(1), where ai|Aut(S):S|=f, 1i3. It is easy to see that ψ1(1),ψ2(1),ψ3(1)cd(G) are distinct and divisible by Φ12=(q-1)2=16k2. This implies that the group G does not satisfy 𝒫32.

### Table 4

Degrees of exceptional groups of Lie type.

 Group Label Degree D43⁢(q3) ϕ1,6 q12 ϕ2,2 12⁢q3⁢Φ22⁢Φ12 ϕ1,3′′ q7⁢Φ12 F42⁢(q2) ε q24 ε′ q2⁢Φ12⁢Φ24 ε′′ q10⁢Φ12⁢Φ24 F4⁢(q) ϕ1,24 q24 ϕ9,2 q2⁢Φ32⁢Φ62⁢Φ12 ϕ9,10′′ q10⁢Φ32⁢Φ62⁢Φ12 E6⁢(q) ϕ1,36 q36 ϕ20,2 q2⁢Φ4⁢Φ5⁢Φ8⁢Φ12 ϕ6,25 q25⁢Φ8⁢Φ9 E62⁢(q2) ϕ1,24 q36 ϕ4,13 q20⁢(q6+1)⁢Φ10⁢Φ8 ϕ2,16′′ q25⁢Φ8⁢Φ18 E7⁢(q) ϕ1,63 q63 ϕ21,2 q2⁢Φ32⁢Φ62⁢Φ9⁢Φ12⁢Φ18 ϕ7,46 q46⁢Φ7⁢Φ12⁢Φ14 E8⁢(q) ϕ1,120 q120 ϕ35,2 q2⁢Φ5⁢Φ7⁢Φ10⁢Φ14⁢Φ14⁢Φ15⁢Φ20⁢Φ30 ϕ8,91 q91⁢Φ42⁢Φ8⁢Φ12⁢Φ20⁢Φ24 G2⁢(q), q≠2 ϕ1,6 q6 ϕ2,1 16⁢q⁢Φ22⁢Φ3 G2⁢[1] 16⁢q⁢Φ12⁢Φ6 ϕ2,2 12⁢q⁢Φ22⁢Φ6, 12⁢q⁢Φ12⁢Φ3, 13⁢q⁢Φ12⁢Φ22 G22⁢(q2), q2≠3 ε q6 cusp 13⁢q⁢Φ1⁢Φ2⁢Φ4 cusp 12⁢3⁢q⁢Φ1⁢Φ2⁢Φ12′ cusp 12⁢3⁢q⁢Φ1⁢Φ2⁢Φ12′′

If q=p=4k+3, by Table 4, we get irreducible characters γ1,γ2,γ3Irr(S) with degrees γ1(1)=qΦ22Φ3/6, γ2(1)=qΦ22Φ6/2, γ3(1)=qΦ12Φ22/3. Also, let φiIrr(Gγi), 1i3. Then φ1(1)=e1qΦ22Φ3/6, φ2(1)=e2qΦ22Φ6/2, φ3(1)=e3qΦ12Φ22/3cd(G), where ei are some divisors of 2f, are distinct and divisible by Φ22=(q+1)2=16(k+1)2. We conclude that G does not have the property 𝒫32, which is the desired conclusion. ∎

### Table 5

Degrees of groups of classical Lie type.

 Group Label Degree Aℓ⁢(q), ℓ⩾3 st qℓ⁢(ℓ+1)/2 (1,1,ℓ-1) q3⋅(qℓ-1-1)⁢(qℓ-1)(q2-1)⁢(q-1) (1,1,1,ℓ-2) q6⋅(qℓ-1)⁢(qℓ-1-1)⁢(qℓ-2-1)(q3-1)⁢(q2-1)⁢(q-1) Aℓ2⁢(q2), ℓ⩾3 st qℓ⁢(ℓ+1)/2 (1,1,ℓ-1) q3⋅(qℓ-1-(-1)ℓ-1)⁢(qℓ-(-1)ℓ)(q+1)⁢(q2-1) (1,1,1,ℓ-2) q6⋅(qℓ-2-(-1)ℓ-2)⁢(qℓ-1-(-1)ℓ-1)⁢(qℓ-(-1)ℓ)(q3+1)⁢(q2-1)⁢(q+1) Bℓ⁢(q), Cℓ⁢(q), ℓ⩾3 st qℓ2 (1 2⁢ℓ0 1) 12⁢q4⋅(qℓ-2-1)⁢(q2⁢(ℓ-1)-1)⁢(qℓ+1)(q2-1)2 (1⁢ℓ-11) q3⋅(q2⁢ℓ-1)⁢(q2⁢(ℓ-2)-1)(q2-1)2 Dℓ⁢(q), ℓ⩾4 st qℓ⁢(ℓ-1) (1⁢ℓ-10 2) q3⋅(q2⁢(ℓ-2)-1)⁢(qℓ-1)⁢(qℓ-2+1)(q2-1)2 (1 2⁢ℓ0 1 2) q6⋅(q2⁢(ℓ-2)-1)⁢(q2⁢(ℓ-1)-1)(q2-1)⁢(q4-1) Dℓ2⁢(q2), ℓ⩾5 st qℓ⁢(ℓ-1) (0 1 2⁢ℓ1 2) q6⋅(q2⁢(ℓ-2)-1)⁢(q2⁢(ℓ-1)-1)(q2-1)⁢(q4-1) (1 2 3⁢ℓ-10 1) 12⁢q7⋅(qℓ+1)⁢(q2⁢(ℓ-2)-1)⁢(qℓ-3-1)⁢(qℓ-4-1)⁢(qℓ-1+1)(q+1)⁢(q2-1)2⁢(q3-1) D42⁢(q2) st q12 (0 1 2 41 2) q6⋅(q6-1)(q2-1) (2 30 1) 12⁢q3⋅(q4+1)⁢(q2-q+1)⁢(q2+1)(q2-1)

### Lemma 3.11.

If SG2(q), q3, is one of the exceptional groups of Lie type and SGAut(S), then G does not satisfy P32.

### Proof.

In Table 4, we list three irreducible characters and their degrees of the simple groups D43(q3), E6(q), E62(q2), E7(q), E8(q), F4(q), F42(q2) (q22), G22(q2) (q23). These characters are all unipotent characters and hence are extendible to G by [15, Theorems 2.4 and 2.5]. In the last case SG22(q2), q23, the degrees of these characters are divisible by qΦ1Φ2, and so the group G does not satisfy 𝒫32. In the other cases of S, the degrees of these characters are divisible by q2, and hence the group G does not have the property 𝒫32, which is the desired result. ∎

### Lemma 3.12.

If S is one of the remaining simple classical groups of Lie type, and SGAut(S), then G does not satisfy P32.

### Proof.

The characters of S given in Table 5 are unipotent characters which can be found in [2, Theorem 13.8]. They are extendible to Aut(S) by [15, Theorems 2.4 and 2.5]. It is easy to see that the given character degrees are pairwise distinct. So, G has three distinct degrees, which are all divisible by q2. Thus G does not have property 𝒫32. ∎

### Proof of Theorem A.

The result follows from Lemmas 2.3, 3.4, 3.5, 3.6, 3.8, 3.9, 3.10, 3.11, 3.12. ∎

Communicated by Robert M. Guralnick

Funding source: Institute for Research in Fundamental Sciences

Award Identifier / Grant number: 93200042

Funding statement: This research was partially supported by a grant from IPM (No. 93200042).

## Acknowledgements

The authors would like to express their sincere thanks to the referee for the helpful comments and suggestions throughout the manuscript. Also, the first author would like to thank the University of Tabriz for supporting this work.

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