On the odd order composition factors of finite linear groups

Theorem 1.1.

Let G be a finite solvable group, and let V≠0 be a finite, faithful, completely reducible G-module with char⁡(V)=p>0. Then

1. |G|≤|V|α/λ.

2. If 2∤|G| or if 3∤|G|, then |G|≤|V|2/λ.

3. If 2∤|G| and p≠2, then |G|≤|V|3/2/λ.

Theorem 1.2.

Let G be a finite group acting on V faithfully and completely reducibly, where V is of characteristic p. Then the following hold.

1. a⁢(G)≤|V|2/λ.

2. If p≠2, then a⁢(G)≤|V|3/2/λ.

Proposition 2.1.

Let G be a finite solvable group, and let V≠0 be a finite, faithful, completely reducible G-module with char⁡(V)=p>0. Let H be a subgroup of G.

1. If 2∤|H| or if 3∤|H|, then |H|≤|V|2/λ.

2. If 2∤|H| and p≠2, then |H|≤|V|3/2/λ.

Proof.

Since G is solvable, we only need to consider the Hall 2′-subgroup or the Hall 3′-subgroup of G. The proof follows the arguments in [9, Theorem 3.5] with some slight adjustments in each of the steps. For consistency, we will adopt the notation used in [9, Theorem 3.5]. Step 1 shows that V is irreducible and the argument here is unchanged. Step 2 shows that V is quasi-primitive, and the argument there is unchanged as well. Step 3 shows that if we set |V|=pn, then we may assume that n≥2 and pn≥16. The calculation remains the same.

In step 4, we show that G≰Γ⁢(pn), n>3, and if p=2, then n≥8. All the arguments are the same with the exception of proving n>3 for statement (2). Assume n=2; we note that e=2, 2∣p-1 and p≥5. We have

When n=3, we have e=3, p≥7, and thus |V|≥p3. Thus

We observe that |G|2′≤p-12⋅27≤p4.5/λ≤|V|3/2/λ.

In step 5, by examining the proof of [9, Theorem 3.5] carefully, we only need to check a few cases when e is small for case (2).

1. If e=2, then |G| is divisible by 8 and A/F≤GL⁡(2,2). Thus |A/F|2′≤3. Since |V|≥81, we have

   |G|2′=(|G/A|⁢|A/F|⁢|F/T|⁢|T|)2′≤3⁢|U|2≤3⁢|V|≤|V|3/23.

2. If e=3, then |A/F|≤GL⁡(2,3), p≥4. Thus |A/F|2′≤3. Since |V|≥256, we have

   |G|2′=(|G/A|⁢|A/F|⁢|F/T|⁢|T|)2′≤27⋅|U|2≤27⋅|V|2/3≤|V|3/23.

3. If e=4, we note that |A/F|2′≤15 and |V|≥81. Thus we have

   |G|2′≤(|G/A|⁢|A/F|⁢|F/T|⁢|T|)2′≤15⋅|U|2≤15⋅|V|1/2≤|V|3/23.

This completes the proof. ∎

Lemma 3.1.

Let G be a finite non-abelian simple group, and let r be a fixed prime. Then there exists a solvable subgroup H of G such that

Proof.

We now go through the Classification of Finite Simple Groups.

(1) Let G be one of the alternating groups 𝖠n, n≥5. It is well known that |Out⁡(𝖠n)|=2 except when n=6 and |Out⁡(𝖠6)|=4. Thus |Out⁡(𝖠n)|2′=1 . Since 5∣|𝖠n| and 3∣|𝖠n|, the result follows.

(2) Let G be one of the sporadic or Tits groups. Then |Out⁡(G)|≤2, and the result can be confirmed by [2].

For simple groups of Lie type, we go through various families of Lie type. To illustrate the method, [8, Proposition 4.1] gives a detailed analysis for An⁢(q) and shows how to handle most cases. For those finitely many exceptional cases, we will check that the required inequalities hold by direct calculation. Since these arguments are similar, for the remaining families of simple groups of Lie type, there is a table in [8, Proposition 4.1] that handles all the exceptional cases.

(3) Let G=A1⁢(q), where q=pf. We have

where d=(2,q-1) and |Out⁡(G)|=d⁢f.

Case (a). Suppose that q is even. Then d=1 and |Out⁡(G)|=f.

Assume there exists a Zsigmondy prime L1 for p2⁢f-1. Then L1∣p2⁢f-1, and thus L1∣pf+1 and L1≥2⁢f=2⁢|Out⁡(G)|≥2⁢|Out⁡(G)|2′.

Assume there exists a Zsigmondy prime L2∣pf-1, where L2≥f. It is clear that L1≠L2. If L2≥2⁢f, then we are done. Otherwise, if L22∣pf-1, we consider the Sylow L2-subgroup L. Then |L|≥2⁢f. However, we have the following exceptions by [8, Lemma 3.1].

1. f=4, thus |Out⁡(G)|=4, and |Out⁡(G)|2′=1. Since 24+1=17 and 24-1=15=3⋅5, we may choose L1=17 and L2=5.

2. f=6, thus |Out⁡(G)|=6 and |Out⁡(G)|2′=3. Since 26+1=65=5⋅17 and 26-1=63=7⋅32, we may choose L1=13 and L2=7.

3. f=12, thus |Out⁡(G)|=12 and |Out⁡(G)|2′=3. Since

   212+1=4097=17⋅241 and 212-1=4095=32⋅5⋅7⋅13,

   we may choose L1=17 and L2=7.

Case (b). Suppose that q is odd. Then d=2 and |Out⁡(G)|=2⁢f, implying that |Out⁡(G)|2′=f. We apply the same idea as before,

There exists an L2∣pf-1, where L2≥f and L1≠L2. If L2≥2⁢f, then we are done. Otherwise, if L22∣pf-1, we consider the Sylow L2-subgroup L. Then |L|≥2⁢f. The following case is the exception by [8, Lemma 3.1]:

1. When p=3, f=4, thus |Out⁡(G)|2′=1. Since 34+1=82=2⋅41 and 34-1=80=24⋅5, we may choose L1=41 and L2=5.

(4) Let G=An⁢(q), where q=pf and n≥2. Set m=∏i=1n(qi+1-1). Then |G|=d-1⁢qn⁢(n+1)/2⁢m, |Out⁡(G)|=2⁢f⁢d, where d=(n+1,q-1).

With the exception of a finite number of cases, there exists a Zsigmondy prime L1 for pf⁢(n+1)-1 such that L1≥2⁢f⁢(n+1) or L12∣pf⁢(n+1)-1. It follows that L12≥2⁢f⁢(n+1). Let H1 be a Sylow L1-subgroup G. By [8, Lemma 3.2], with the exception of a finite number of cases, there exists a Zsigmondy prime L2 for pf⁢n-1 such that L2≥3⁢f⁢n≥2⁢f⁢(n+1) or L22∣pf⁢n-1. This implies that L22≥3⁢f⁢n≥2⁢f⁢(n+1). Let H2 be a Sylow L2-subgroup of G. Notice that L1≠L2.

Since |Out⁡(G)|=2⁢f⁢d, |Out⁡(G)|2′≤f⁢d. Also, n+1≥d=(n+1,q-1), and |H1|,|H2|≥2⁢|Out⁡(G)|2′. Therefore, the result follows. The exceptions are listed in Table 1 (by [8, Lemma 3.2]).

(5) Let G=An2⁢(q2), where n≥2. Note that if n=2, then q>2. Set

Then |G|=d-1⁢m⁢qn⁢(n+1)/2, and |Out⁡(G)|=d⁢f. By [4, Theorem A] and [8, Lemma 3.2], there exists a Zsigmondy prime L1∣pf⁢(n+1)/2-(-1)n+1 such that L1≥2⁢(n+1)⁢f≥2⁢d⁢f or

Moreover, by [4, Theorem A] and [8, Lemma 3.2], with the exception of a finite number of cases, there exists a Zsigmondy prime L2∣pf⁢n/2-(-1)n+1 such that L2≥52⁢(n+1)⁢f≥2⁢(n+1)⁢f≥2⁢d⁢f or

For all the exceptional cases, we can check the results via direct calculation and by considering the order of the group G and Out⁡(G) (see Table 2).

We now provide a table (Table 3) for the simple groups of Lie types other than An⁢(q) and A2⁢(q2). The second and the third columns in the table are two large prime divisors that correspond to Sylow subgroups. ∎

Proposition 4.1.

Let G be a group of permutations on a set Ω of order n. Then a⁢(G) is at most 2n-1.

Proof.

We first check that the result is true for n≤4 (|𝖲2|2′≤2, |𝖲3|2′≤4, |𝖲4|2′≤8). We may assume that n≥5. We shall proceed by induction on |G|.

We first suppose that G is intransitive on Ω. Write Ω=Γ1∪Γ2 for nonempty subsets Γi of Ω such that G permutes Γ1 and also Γ2. Write ni=|Γi| for i=1,2, so clearly n1+n2=n. Let L be the kernel of G on Γ1 so that L acts faithfully on Γ2. By induction, we then have a⁢(G/L)≤2n1-1 and a⁢(L)≤2n2-1, and so a⁢(G)≤2n1-1⋅2n2-1=2n-2<2n-1, as desired. So now, we may assume that G is transitive on Ω.

Suppose that G is imprimitive on Ω. Then there is a nontrivial decomposition Ω=⋃iΩi with G permuting X={Ω1,…,Ωr} and NG⁢(Ω1) acting primitively on Ω1, where |Ω1|=m and n=m⁢r. Let π be the permutation representation of G on X and K=ker⁡π. Set

We note that a⁢(G)=a0⁢a1⁢…⁢ar, where a0=a⁢(G/K) and ai=a⁢(Ki-1/Ki) for i≥1. By induction, a0≤2r-1 and ai≤2m-1 for i≥1. It is easy to see that a⁢(G)≤2n-1.

Thus we may assume G is primitive; then we know that G either contains 𝖠n or is one of the groups in the exceptional list by [10, Corollary 1.4]. If G contains 𝖠n, then the result is clear since n≥5. Otherwise, G is one of the groups in the exceptional list; we verify the result in Table 4. ∎

Proof of Theorem 1.2.

Let S be the maximal normal solvable subgroup of G. Consider G¯=G/S. It is easy to see that F⁢(G¯)=1. Therefore,

Moreover, we know that Z⁢(E⁢(G¯)) is trivial; otherwise, S is not maximal. Since E⁢(G¯)/Z⁢(E⁢(G¯)) is the unique largest semi-simple subgroup of G¯, E⁢(G¯) is the product of simple non-abelian subgroups.

Let C¯=CG¯⁢(E⁢(G¯)). Since F*⁢(G) is self-centralizing, we have C¯<F*⁢(G). Let K=G¯/C¯. Then K acts faithfully on E⁢(G¯). We may assume that K acts transitively on L1=E11×…×E1⁢k1, where E11,…,E1⁢k1 are non-abelian simple components of E⁢(G¯). Let K1=CK⁢(L1). We may assume that K1 acts transitively on L2=E21×…×E2⁢k2, where E21,…,E2⁢k2 are non-abelian simple components of E⁢(G¯). Let K2=CK1⁢(L2), and inductively, we may define L3,K3,…,Lt,Kt. Then E⁢(G¯)=L1×…×Lt, and Ki-1/Ki acts transitively on Li.

Since G acts on V completely reducibly and S is a normal subgroup of G, we know that S acts on V completely reducibly. Since E1,…,Em are non-abelian simple subgroups, there exists a solvable subgroup H=H1×…×Hm by Lemma 3.1, where Hi<Ei such that (Hi,p)=1 and |Hi|2′≥2⁢|Out⁡(Ei)|2′. Therefore, |H|2′=∏i|Hi|2′≥2m⁢|Out⁡(E⁢(G¯))|2′.

Moreover, K is a permutation group permuting E1,…,Em. By Proposition 4.1, the product of the orders of all the odd order composition factors of K is less than 2m-1. Thus |H|2′≥2m⁢|Out⁡(E⁢(G¯))|2′, which is greater than the product of the orders of all the odd order composition factors of K and the 2′ part of the outer automorphism of E⁢(G¯).

Let ϕ:G→G/S be the canonical homomorphism. Since (|H|,p)=1, we know that ϕ-1⁢(H) acts on V completely reducibly by the generalized Maschke theorem (cf. [7, Problem 1.8]). Therefore, by Proposition 2.1, we have

and if p≠2, then |ϕ-1⁢(H)|2′≤|V|3/2/λ.

We observe that the odd order composition factors of G are distributed in the maximal normal solvable subgroup S, the outer automorphism of the direct product of simple groups E⁢(G¯), and the odd order composition factors of the permutation group G¯/C¯. Therefore, a⁢(G)≤|ϕ-1⁢(H)|2′, and the result follows. ∎

Corollary 4.2.

Let G be a finite group acting on V faithfully and completely reducibly (V is possibly of mixed characteristic). Then a⁢(G)≤|V|2/λ.