On Flag-Transitive, Point-Quasiprimitive Symmetric 2-(v,k,λ) Designs with λ Prime

Abstract

This paper contributes to the classification of flag-transitive symmetric 2-$(v,k,\lambda )$ designs with $\lambda$ prime. We investigate the structure of flag-transitive, point-quasiprimitive automorphism groups (G) of such 2-designs by applying the classification of quasiprimitive permutation groups. It is shown that the automorphism groups (G) have either an abelian socle or a non-abelian simple socle. Moreover, according to the classification of finite simple groups, we demonstrate that point-quasiprimitivity implies point-primitivity of G, except when the socle of G is $PS{L}_n\left(q\right)$.

1. Introduction

A symmetric 2-$(v,k,\lambda )$ design $\mathcal{D}=(\mathcal{P},\mathcal{B})$ consists of a set $\mathcal{P}$ of v points and a set $\mathcal{B}$ of k-element subsets of $\mathcal{P}$ (called blocks), such that every two points are contained in exactly $\lambda$ blocks and any two blocks intersect in precisely $\lambda$ points. The number of blocks passing through a point is a constant independent of the point’s choice, equaling k for symmetric designs. An automorphism of $\mathcal{D}$ is a permutation g of $\mathcal{P}$, which also satisfies $B^g\in \mathcal{B}$ for each $B\in \mathcal{B}$. All automorphisms of $\mathcal{D}$ form a group $\mathrm{Aut}\left(\mathcal{D}\right)$, called the full automorphism group of $\mathcal{D}$. Any subgroup G of $\mathrm{Aut}\left(\mathcal{D}\right)$ is called an automorphism group of $\mathcal{D}$. Naturally, G or $\mathcal{D}$ is said to be point (block, flag)-transitive if G acts transitively on points (blocks, flags), where flags are the incident point-block pairs. Similarly, G is said to be point-primitive (quasiprimitive, imprimitive) if G acts primitively (quasiprimitively, imprimitively) on points.

This paper investigates the structure of flag-transitive, point-quasiprimitive automorphism groups of symmetric 2-$(v,k,\lambda )$ designs with $\lambda$ prime. In [1], applying the O’Nan–Scott theorem, the first author and Zhou showed that flag-transitive, point-primitive automorphism groups of such symmetric designs have either an abelian socle or a non-abelian simple socle. Alavi et al., in [2], further completed the classification of these designs when the automorphism group has a non-abelian simple socle. In 1993, Praeger established an analog of the O’Nan–Scott theorem for quasiprimitive groups in [3], which implies the possibility of determining the socle of the point-quasiprimitive automorphism group of 2-designs. In [4], Camina and Praeger proved that the line-transitive point-quasiprimitive automorphism group G of a finite linear space (2-$(v,k,1)$ design) is of either affine type or almost simple type. Recently, the third author along with Chen and Zhou demonstrated in [5] that this conclusion also holds true for flag-transitive symmetric 2-$(v,k,\lambda )$ designs when $\lambda$ is smaller than 100. Notably, these works also show that for $\lambda \le 100$, there is no symmetric design with prime $\lambda$ admitting a flag-transitive, point-quasiprimitive but imprimitive automorphism group G. Inspired by this, the present paper studies the flag-transitive, point-quasiprimitive automorphism group of symmetric 2-designs, under the assumption that $\lambda$ is prime, leading to the following theorem:

Theorem 1. 

Let $\mathcal{D}=(\mathcal{P},\mathcal{B})$ be a nontrivial symmetric 2-$(v,k,\lambda )$ design with λ prime and G be a flag-transitive, point-quasiprimitive automorphism group of $\mathcal{D}$. Then the socle of G must be an abelian group or a non-abelian simple group. Moreover, the point-quasiprimitivity implies point-primitivity in this case, except when the socle of G is $PS{L}_n\left(q\right)$.

Remark 1.

 (i) 

As an exception, G is point-quasiprimitive but imprimitive when the socle is $PS{L}_n\left(q\right)$, and the point stabilizer, $G_{\alpha }$, is contained within the maximal parabolic subgroup $P_1$, which is the stabilizer of 1-subspaces.

(ii) 

When G is flag-transitive and point-primitive with a non-abelian simple socle, the classification of symmetric 2-designs with λ prime has been completed, as detailed in [2].

2. Preliminary Results

We first present some arithmetic results on the symmetric 2-designs and their automorphism groups.

Lemma 1.

For a symmetric 2-$(v,k,\lambda )$ design, the values of $v,k,\lambda$ must satisfy the following:

(i) 

$\lambda (v-1)=k(k-1)$;

(ii) 

$\lambda v<k^2$.

The flag-transitivity of G implies that k divides $|G_{\alpha }|$ since the point stabilizer $G_{\alpha }$ acts transitively on the k blocks containing $\alpha$. Thus, the lemma below is clear.

Lemma 2.

If G is a flag-transitive automorphism group of a symmetric 2-$(v,k,\lambda )$ design, then $k|\lambda ·gcd(v-1,\left|G_{\alpha }\right|)$.

The following lemmas concern the reduction of the flag-transitive, point-primitive symmetric 2-designs with some special values of $\lambda$.

Lemma 3.

([6]). If G is a flag-transitive automorphism group of a symmetric 2-design with $gcd(k,\lambda )=1$, and T is a minimal normal subgroup of G, then T is either abelian or simple with $C_G\left(T\right)=1$.

Lemma 4.

[1]) If $\mathcal{D}$ is a symmetric 2-$(v,k,\lambda )$ design with λ prime, and $G\le Aut\left(\mathcal{D}\right)$ is flag-transitive and point-primitive, then the socle of G is either an abelian group or a non-abelian simple group.

Lemma 5.

([5]) If $\mathcal{D}$ is a symmetric 2-$(v,k,\lambda )$ design with $\lambda \le 100$, and $G\le Aut\left(\mathcal{D}\right)$ is flag-transitive and point-quasiprimitive, then the socle of G is either an abelian group or a non-abelian simple group. Moreover, if G is point-quasiprimitive but imprimitive, then the value of λ ($\le 100$) cannot be a prime.

For a finite simple group T, the notation $Out\left(T\right)$ denotes the outer automorphism group of T, and $P\left(T\right)$ denotes the minimal degree of permutation representation of T. The following lemma pertains to the finite simple groups of Lie type, which is simple but necessary for our following discussion.

Lemma 6.

For a finite group T, let $p_0$ be a prime divisor of $\left|T\right|$. Then we have

(i) 

If T is a simple group of Lie type over the field ${\mathbb{F}}_q$, then ${\left|Out\left(T\right)\right|}^2<P\left(T\right)$ and $p_0\le P\left(T\right)$;

(ii) 

If T is a classical simple group of dimension n, then $n^2+3<P\left(T\right)$ for $n\ge$ 10;

(iii) 

If T is a finite exceptional simple group of Lie type, then $p_0^2<P\left(T\right)$, except when T is ${}^2{B}_2\left(q\right)$ with $P\left(T\right)=q^2+1$ or T is ${}^2{G}_2\left(q\right)$ with $P\left(T\right)=q^3+1$.

Proof. 

The lemma follows immediately from ([7] Tables 5.1.A–5.1.C, 5.2.A, Corollary 5.2.3, and [8] Table 4), with some routine arithmetic. Here, we use $G_2\left(q\right)$ as an example. When T is $G_2\left(q\right)$, we know that $\left|T\right|=q^6(q^2-1)(q^6-1)$, $\left|Out\right(T\left)\right|=e$ or $2e$ with $q=p^e$ from ([7] Table 5.1 B), and $P\left(T\right)=\frac{q^6-1}{q-1}$ from ([8] Table 4). Therefore, $p_0\le \frac{q^3-1}{q-1}$, so (i) and (iii) are clear. □

3. Proof of Theorem 1

From Lemmas 3 and 4, Theorem 1 holds when G is point-primitive or when $gcd(k,\lambda )=1$. Therefore, we need to only consider the case when G is point-quasiprimitive but imprimitive and $\lambda$ is a prime divisor of k. In addition, we assume that $k\le \frac{\lambda (\lambda -3)}{2}$ since the flag-transitive, point-imprimitive symmetric 2-designs with $k>\frac{\lambda (\lambda -3)}{2}$ have already been discussed in [9]. To summarize, we propose the following hypothesis for the rest of our discussion.

Hypothesis 1 (H1).

Let $\mathcal{D}$ be a symmetric 2-$(v,k,\lambda )$ design, where λ is a prime divisor of k, $k\le \frac{\lambda (\lambda -3)}{2}$ and let G be an automorphism group of $\mathcal{D}$. Assume that G is flag-transitive and point-quasiprimitive, but point-imprimitive.

Since the automorphism group G acts imprimitively on $\mathcal{P}$, there exists a nontrivial partition

$$\Theta =\{{\Delta }_1,{\Delta }_2,\dots ,{\Delta }_d\}$$

of $\mathcal{P}$ with $|{\Delta }_i|=c>1$. Obviously, $v=cd.$ By the flag transitivity of G, the size of the intersection of a design block B and the imprimitive block ${\Delta }_i$ is a constant, say ℓ. Now, for a given point $\alpha$, counting the number of flags $(\beta ,B)$, such that $\alpha ,\beta$ are in the same imprimitive block and incident with the same block, we arrive at the following equation:

$$\lambda (c-1)=k(\ell -1).$$

(1)

As a result, the following lemmas are immediate:

Lemma 7.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then $d(\ell -1)<k\le \frac{\lambda (\lambda -3)}{2}$.

Proof. 

Equation (1) gives $c=\frac{k(\ell -1)+\lambda }{\lambda }$, which, together with $\lambda (cd-1)=k(k-1)$, yields $\lambda (d-1)=k[k-1-d(\ell -1\left)\right]$. Clearly, the left is a positive integer, so $d(\ell -1)<k$. Also, from Hypothesis 1, we have $k\le \frac{\lambda (\lambda -3)}{2}$. □

Lemma 8.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then there exists a maximal subgroup M of G, such that

(i) 

$|G:M|\le d<\frac{{\lambda }^2}{2}$;

(ii) 

$P\left(T\right)\le |G:M|$ when G is almost simple with socle T.

Proof. 

Since G is point-imprimitive, there exists a maximal subgroup M of G, such that $G_{\alpha }<G_{\Delta }\le M$, where $\alpha \in \Delta$. Moreover, for any partition $\Theta =\{{\Delta }_1,{\Delta }_2,\dots ,{\Delta }_d\}$ of $\mathcal{P}$, the point-transitivity of G implies $|G:G_{\Delta }|=d$. Hence, $|G:M|$ divides $|G:G_{\Delta }|=d$, and so (i) holds by Lemma 7. If G has a non-abelian simple socle T, then $P\left(T\right)\le P\left(G\right)\le |G:M|$ and (ii) holds. □

Lemma 9.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then λ divides $|G_{\alpha }|$. Furthermore, λ divides $|T_{\alpha }|$, where T is the simple socle of G.

Proof. 

Since $\lambda$ is a prime divisor of k, then the flag-transitivity of G indicates that $\lambda$ divides $|G_{\alpha }|$. If $\lambda$ divides $\left|Out\right(T\left)\right|$, then we have $P\left(T\right)<\frac{{\left|Out\left(T\right)\right|}^2}{2}$ by Lemma 8, which contradicts Lemma 6. Hence, $\lambda$ divides $|T_{\alpha }|$ by $G\le T.Out\left(T\right)$. □

3.1. The Abelian Socle or Simple Socle of G

Here, we assume that the flag-transitive automorphism group G is point-quasiprimitive. An analog of the O’Nan–Scott theorem for quasiprimitive groups has been established by Praeger, who showed that each quasiprimitive group is permutationally equivalent to one of the following eight types: the (i) Holomorph affine type, (ii) holomorph simple type, (iii) holomorph compound type, (iv) almost simple type, (v) twisted wreath type, (vi) simple diagonal type, (vii) compound diagonal type, and (viii) product action type. See [3,10] for more details. In this section, our objective is to prove the following result based on the classification of quasiprimitive permutation groups.

Proposition 1.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then G is of holomorph affine type or almost simple type.

Proof. 

If G is of holomorph simple type or holomorph compound type, then G is primitive and the proposition holds by Lemma 4.

Assume that G is of twisted wreath product type, then $G=T^m⋊G_{\alpha }$ with $v={\left|T\right|}^m$ and the point stabilizer $G_{\alpha }$ is embedded into $Sym\left(m\right)$ with $m>2$. Since $\lambda$ divides $|G_{\alpha }|$, we have $\lambda \le m$. However, Lemma 1(ii) and Lemma 7 give ${60}^m\le \left|T^m\right|<\frac{m^3}{4}$, which is impossible.

Assume that G is of simple diagonal type, then we have

$$T^m\le G\le T^m.(Out\left(T\right)\times S_m)$$

and $v={\left|T\right|}^{m-1}$ with $m>2$. Moreover, $\lambda | |G_{\alpha }|$ implies $\lambda \le m$ or $\lambda | |Aut\left(T\right)|$. The former is impossible by Lemma 1(ii) and Lemma 7. For the latter, it implies ${\left|T\right|}^{m-1}<\frac{{\left|T\right|}^{\frac{3}{2}}}{4}$, which is also impossible when $m>2$.

Clearly, G cannot be of the compound diagonal type.

Assume that G is of product action type, then there exists a G-invariant partition $\Theta$ of $\mathcal{P}$, such that the blocks of $\Theta$ are of size c. Then $\Theta$ can be identified with the Cartesian power ${\Lambda }^m$ of a set $\Lambda$ with size $\omega$$(m\ge 2)$, such that G acts on $\Theta$ as a subgroup of the wreath product $H\wr S_m$ in product action, where H is a quasiprimitive group on $\Lambda$ with a non-abelian simple socle T. Obviously, we have $v=c{\omega }^m$. Moreover, from Lemma 8, there is

$$d={\omega }^m<\frac{{\lambda }^2}{2}.$$

(2)

Since $\lambda$ is a prime divisor of $\left|G\right|$, we have $\lambda |m$ or $\lambda | |Aut\left(T\right)|$. The former implies $5^m\le {\omega }^m<\frac{m^2}{2}$, which is impossible. For the latter, we have $P{\left(T\right)}^m\le {\omega }^m<{\lambda }^2$ by Lemma 8, and it follows that $P\left(T\right)<\lambda$, contradicting Lemma 6.

To summarize, G is of holomorph affine type or almost simple type. □

3.2. The Primitivity of G

By Proposition 1, we know that the socle T of G is a non-abelian simple group or an abelian group. When T is abelian, the quasiprimitive group G is primitive. Hence, we need only to consider the case where T is a non-abelian simple group. In the following, for all cases, except for one possibility of T, we show that no maximal subgroup of G satisfies $G_{\alpha }<G_{\Delta }\le M$ and $|G:M|<\frac{{\lambda }^2}{2}$, so that the flag-transitive, point-quasiprimitive automorphism group G must be primitive by Lemma 8.

Lemma 10.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle of G cannot be an alternating group $A_t$.

Proof. 

Assume that the socle T of the point-quasiprimitive G is an alternating group $A_t$. Then the maximal subgroups M of G are known in [11]. We prove that no maximal subgroup M satisfies $G_{\alpha }<G_{\Delta }\le M$ and $|G:M|<\frac{{\lambda }^2}{2}$.

First, assume that the maximal subgroup M is intransitive on $\{1,2,\dots ,t\}$. In this case, M is isomorphic to $(S_i\times S_{t-i})\cap G$ with $2\le i<t/2$ and $|G:M|=\left(\genfrac{}{}{0pt}{}{t}{i}\right)$. Since $G_{\alpha }<M$ and $\lambda | |G_{\alpha }|$, we have $\lambda | |M|$ and, hence, $\lambda \le t-i$. However, by $|G:M|<\frac{{\lambda }^2}{2}$, we have $\left(\genfrac{}{}{0pt}{}{t}{i}\right)<\frac{{(t-i)}^2}{2}$, which is impossible for $i\ge 2$. Then, assume that M is transitive but imprimitive on $\{1,2,\dots ,t\}$. In this case, M is isomorphic to $(S_s\wr S_i)$, with $i\ge 2$, $s\ge 2$ and $t=is$. Obviously, $\lambda <\max \{s,i\}$. Moreover, since $\frac{\left(\genfrac{}{}{0pt}{}{t}{s}\right)}{i}=\left(\genfrac{}{}{0pt}{}{is-1}{s-1}\right)=\frac{is-1}{s-1}·\frac{is-2}{s-2}\dots \frac{is-(s-1)}{1}>i^{s-1},$ we know that $\frac{{(i!)}^{(s-1)}}{2}<\frac{\left(\genfrac{}{}{0pt}{}{is}{s}\right)\left(\genfrac{}{}{0pt}{}{(i-1)s}{s}\right)...\left(\genfrac{}{}{0pt}{}{3s}{s}\right)\left(\genfrac{}{}{0pt}{}{2s}{s}\right)}{i!}=|G:M|<\frac{{\lambda }^2}{2}$. If $s\le i$, then $\frac{{(i!)}^{(s-1)}}{2}<\frac{1}{2}{i}^2,$ which induces $\lambda \le i\le 3$. If $i<s$, $\lambda \le s\le 6$. However, both cases are ruled out by Lemma 5. Lastly, assume that M is primitive on $\{1,2,\dots ,t\}$. Since $\lambda \le t$, then by ([12] Theorem 14.2), we have

$$\frac{\left[\frac{t+1}{2}\right]!}{2}\le |G:M|<\frac{t^2}{2},$$

(3)

which implies $\lambda \le t\le 8$, and this can also be ruled out by Lemma 5. □

Lemma 11.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle of G cannot be a Sporadic group.

Proof. 

If the socle of G is a Sporadic group, then by $\lambda | |G|$, we have $\lambda <100$. The conclusion follows immediately from Lemma 5. □

Lemma 12.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle of G cannot be a simple exceptional group of Lie type.

Proof. 

Assume that G is point-imprimitive and the socle T is a simple exceptional group of Lie type. Then the facts $P\left(T\right)<d<\frac{{\lambda }^2}{2}$ and $\lambda | |G|$, together with Lemma 6, imply that T is ${}^2{B}_2\left(q\right)$ with $P\left(T\right)=q^2+1$, or ${}^2{G}_2\left(q\right)$ with $P\left(T\right)=q^3+1$.

If T is ${}^2{B}_2\left(q\right)$, from [13], we know that the order $\left|M\right|$ is either $f{q}^2(q-1)$, $2f(q-1)$, $4f(q\pm \sqrt{2q}+1)$, or $f(q_0^2+1)q_0^2(q_0-1)$ with $q_0^i=q$, and i is odd, where $f | |Out\left(T\right)|$. Since $q^2+1=P\left(T\right)<\frac{{\lambda }^2}{2}$, we have $\lambda >\sqrt{2}q$. However, for all cases, except $\left|M\right|=4f(q+\sqrt{2q}+1)$, the prime divisors of $\left|M\right|$ are smaller than $\sqrt{2}q$. When $\left|M\right|=4f(q+\sqrt{2q}+1)$, we have $|G:M|=\frac{q^2(q-\sqrt{2q}+1)(q-1)}{4}$ and $\lambda \le (q+\sqrt{2q}+1)$, which contradicts $|G:M|<\frac{{\lambda }^2}{2}$.

If T is ${}^2{G}_2\left(q\right)$ with $q=3^{2n+1}$, then $P\left(T\right)=q^3+1$. From ([14] Lemma 3.3), we know that the order $\left|M\right|$ is either $f{q}^3(q-1)$, $2f(q+1)$, $fq(q^2-1)$, $6f(q\pm \sqrt{3q}+1)$, or $f{q}_0^3(q_0^3+1)(q_0-1)$ with $q_0^i=q$ and an odd i, where $f | |Out\left(T\right)|$. However, the fact that $\lambda | |M|$ forces $\lambda <2q$ for all cases, which is contradictory to ${\lambda }^2<2P\left(T\right)$. □

In what follows, we will handle the case when T is a classical simple group. Herein, we adopt the notation of [7]. For example, $PS{L}_n\left(q\right)$, $PS{p}_n\left(q\right)$, $PS{U}_n\left(q\right)$, and $P{\Omega }_n^{\pm }\left(q\right)$ denote the linear, symplectic, unitary, and orthogonal groups on finite vector spaces of dimension n, respectively. The key tool used here is the well-known Aschbacher theorem, which divides the maximal subgroups of classical groups into nine classes. Roughly speaking, the first eight classes consist of subgroups that preserve some kind of geometric structure, while the ninth class consists of subgroups that are not of geometric type—that are almost simple, taking into account the subgroup of scalar matrices on a modulo basis. For more detail, see [15].

We first deal with the case where the maximal subgroup M is almost simple.

Lemma 13.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1. If the socle T of G consists of classical simple groups of dimension n, then the maximal subgroup M of G cannot be almost simple.

Proof. 

Assuming that M is an almost simple irreducible subgroup with socle $M_0$, then ([16] Theorem 4.1) shows that either $M_0$ is an alternating group $A_t$ with $n=t-1,t-2$, or $\left|M\right|\le q^{3n}$. If the former holds, then $\lambda \le n+2$. For $n\ge 10$, Lemmas 6 and 8 imply

$$n^2+3\le P\left(T\right)\le |G:M|<\frac{{(n+2)}^2}{2},$$

which is a contradiction. For $n<10$, it can also be ruled out by Lemma 5. Therefore, the latter case must hold, where $\left|M\right|\le q^{3n}$. For this case, when T is $PS{L}_n\left(q\right)$ or $PS{U}_n\left(q\right)$, we have $q^{n^2-3}<\left|T\right|$ by ([17] Corollary 4.3) and $\lambda <\frac{q^n-1}{q-1}<2q^{n-1}$ by $\lambda | |G|$. These, together with $|G:M|<\frac{{\lambda }^2}{2}$, imply $q^{n^2-3}<\left|G\right|<\frac{{\lambda }^2}{2}\left|M\right|<q^{5n-1}$. It follows that $n\le 5$ and all possibilities of M are listed in ([18] Section 8.2). Since $\lambda | |M|$, one can easily find that $\lambda <100$ for all possibilities of M, which is ruled out by Lemma 5. When T is the remaining classical simple group, the fact $\lambda | |G|$ implies $\lambda \le q^{[n/2]}+1$, and so $\frac{{\lambda }^2}{2}<q^n$. Moreover, from ([17] Corollary 4.3), we have $\frac{1}{8}{q}^{\frac{1}{2}n(n-1)}<\left|T\right|$. This, together with $\left|G\right|<\frac{{\lambda }^2}{2}\left|M\right|$, implies $n\le 9$. However, from ([18] Section 8.2), we find that $\frac{{\lambda }^2}{2}<|G:M|$ or $\lambda <100$ for all possibilities of M, which contradicts Lemma 5. □

Next, we address the cases where M is of geometric type. All possibilities of such an M can be found in ([7] Tables 3.5.A–3.5.F). The discussions below rely heavily on the minimal degree $P\left(T\right)$, which is presented in ([7] Table 5.2.A), and ([8] Table 4).

Lemma 14.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1. If the socle T of G is $PS{L}_n\left(q\right)$ with $n\ge 2$, then the point stabilizer $G_{\alpha }$ is contained in the maximal parabolic subgroup.

Proof. 

Suppose that T is $PS{L}_n\left(q\right)$ with $n\ge 2$. Then we know that

$$P\left(T\right)=\frac{q^n-1}{q-1}$$

from ([8] Table 4). All possibilities of subgroup M are listed in ([7] Table 3.5.A), and will be discussed case-by-case in the following.

(1) If M is of type $P_i$, the maximal parabolic subgroup, then

$$|G:M|=\frac{(q^n-1)\dots \left(q^{n-i+1}\right)}{(q^i-1)\dots (q-1)}$$

and $\lambda \le \frac{q^{n-i}-1}{q-1}$ since $\lambda | |M|$. So, by Lemma 8, we have

$$q^{i(n-i)}<|G:M|<\frac{{\lambda }^2}{2}\le \frac{{(q^{n-i}-1)}^2}{2{(q-1)}^2},$$

which implies $i=1$. Hence, M is of type $P_1$, the stabilizer of 1-subspaces of G. If M is of type $G{L}_i\times G{L}_{n-i}$ with $1\le i<n/2$, then

$$q^{2i(n-i)}<|G:M|=\frac{q^{i(n-i)}(q^n-1)\dots (q^{i+1}-1)}{(q^{n-i}-1)\dots (q-1)}$$

and $\lambda \le \frac{q^{n-i}-1}{q-1}$. By $|G:M|<\frac{{\lambda }^2}{2}$, we have $q^{2i(n-i)}<\frac{{(q^{n-i}-1)}^2}{2{(q-1)}^2}$, which is impossible for any $i\ge 1$. If M is of type $P_{i,n-i}$ with $1\le i<n/2$, then

$$q^{i(2n-3i)}<|G:M|=\frac{(q^n-1)\dots \left(q^{n-2i+1}\right)}{{(q^i-1)}^2\dots {(q-1)}^2}.$$

By $\lambda | |M|$, we have $\lambda \le \frac{q^i-1}{q-1}$ for $n<3i$ and $\lambda \le \frac{q^{n-2i}-1}{q-1}$ for $3i<n$. However, both cases contradict $P\left(T\right)<\frac{{\lambda }^2}{2}$.

(2) Assume that M is of type $G{L}_{n/i}\wr S_i$ with $i\ge 2$. We first consider the case $n=i$. In this case, $\left|M\right|=\frac{n!{(q-1)}^{n-1}}{(n,q-1)}$, which implies $\lambda \le n$ or $\lambda \le q-1$. However, the condition $P\left(T\right)<\frac{{\lambda }^2}{2}$ implies $\frac{q^n-1}{q-1}<\frac{n^2}{2}$ for $\lambda \le n$ and $\frac{q^n-1}{q-1}<\frac{{(q-1)}^2}{2}$ for $\lambda \le q-1$. The former is impossible and the latter forces $n=2$. For $n=2$, we have $|G:M|=\frac{q(q+1)}{2}$ and $\lambda \le q-1$, which contradicts $|G:M|<\frac{{\lambda }^2}{2}$. Now, we assume that $2\le n/i\le n/2$. In this case, $\lambda \le \frac{(q^{n/i}-1)}{(q-1)}<2q^{n/i-1}$, which implies $\frac{{\lambda }^2}{2}<2q^{2(n/i-1)}<q^{n-1}<P\left(T\right)$, which is, again, a contradiction.

(3) Assume that M is of type $G{L}_{n/i}\left(q^i\right)$ with i prime. If $n=i$, then $\left|M\right|=\frac{n(q^n-1)}{(n,q-1)(q-1)}$ and $\lambda \le \frac{q^n-1}{q-1}$. However, inequality $q^{\frac{n(n-1)}{2}}\le |G:M|<\frac{{\lambda }^2}{2}<\frac{{(q^n-1)}^2}{2{(q-1)}^2}$ holds only when $n\le 3$. Note that i is a prime, so $n=i=3$ and $\lambda \le \frac{q^3-1}{q-1}$. It follows that $|G:M|=\frac{1}{3}{q}^3(q^2-1)$, which contradicts Lemma 8. When $2\le n/i\le n/2$, we have $q^{n^2\left(\frac{i-1}{2i}\right)}\le |G:M|$ and $\lambda \le \frac{q^n-1}{q^i-1}<2q^{n-i-1}$. So, $q^{n^2\left(\frac{i-1}{2i}\right)}<2q^{2(n-i)}$ by $|G:M|<\frac{{\lambda }^2}{2}$, which is impossible.

(4) If M is of type $G{L}_i\left(q\right)\otimes G{L}_{n/i}\left(q\right)$ with $2\le i<\left[\sqrt{n}\right]$, then $\lambda \le \frac{q^{n/i}-1}{q-1}<2q^{n/i-1}$, too small to satisfy $P\left(T\right)<\frac{{\lambda }^2}{2}$.

(5) If M is of type $G{L}_n\left(q_0\right)$, where $q=q_0^i$, and i is a prime, then $\lambda \le \frac{q_0^n-1}{q_0-1}$, which is too small to satisfy $P\left(T\right)<\frac{{\lambda }^2}{2}$.

(6) If M is of type $n_0^{2i}.S{p}_{2i}\left(n_0\right)$, where $n=n_0^i$, and $n_0\ne p$ is a prime divisor of $q-1$, then $\lambda \le n_0^i+1=n+1$, and it is too small to satisfy Lemma 8 when $n_0\ge 3$. If M is $2^{2i}.S{p}_{2i}\left(2\right)$ and $n=2^i\ge 4$, then $\lambda \le 2^i+1$, and $\lambda$ is too small, since $\frac{{\lambda }^2}{2}<\frac{{(2^i+1)}^2}{2}<2^{2^i}-1=P\left(T\right)$. If $n=2$, then $\lambda \le 3$, which has already been ruled out by Lemma 5.

(7) If M is of type $G{L}_i\left(q\right)\wr S_j$, a tensor product subgroup, with $n=i^j$ and $i\ge 3$, then $\lambda \le \frac{q^i-1}{q-1}$, which is also too small to satisfy $P\left(T\right)<\frac{{\lambda }^2}{2}$.

(8) If M is of type $P{s}_n\left(q\right)$ with n$(\ge 4)$ being even, then $q^{\frac{n^2-2n}{4}}<|G:M|$ and $\lambda \le q^{\frac{n}{2}}+1$. However, $q^{\frac{n^2-2n}{4}}<|G:M|<\frac{{\lambda }^2}{2}<\frac{{(q^{\frac{n}{2}}+1)}^2}{2}$ holds only when $n=4$. For $n=4$, $|G:M|=\frac{q^2(q^3-1)}{(2,q-1)}$ and $\lambda \le q^2+1$, which implies $\frac{{\lambda }^2}{2}<|G:M|$, which is a contradiction. If M is of type $O_n^{ϵ}$, with q being odd and $n(\ge 4)$ being even, then $q^{i^2}(q^i-1)<|G:M|$, where $i=n/2$ and $\lambda \le q^i+1$. However, $q^{i^2}(q^i-1)<|G:M|<\frac{{\lambda }^2}{2}<\frac{q^{2i}+2q^i+1}{2}$, which is impossible. If M is of type $O_{2i+1}$, where $n=2i+1$, and q is odd, then $q^{i^2+i}(q^n-1)<|G:M|$ and $\lambda \le q^i+1$. It follows that $q^{i^2+i}(q^n-1)<|G:M|<\frac{{\lambda }^2}{2}<\frac{q^{2i}+2q^i+1}{2}$, which is also impossible. If M is of type $U_n\left(q_0\right)$, with $q=q_0^2$, then $|G:M|=\frac{(4,q+1)}{(4,q-1)}{q}_0^{\frac{n(n-1)}{2}}(q_0^n+1)\dots (q_0^2+1)$ and $\lambda \le \frac{q_0^n-{(-1)}^n}{q_0-{(-1)}^n}$. We can easily find that $\lambda$ is too small to satisfy $P\left(T\right)<\frac{{\lambda }^2}{2}$.

In conclusion, when T is $PS{L}_n\left(q\right)$, the only possible maximal subgroup containing $G_{\alpha }$ is $P_1$, which is the maximal parabolic subgroup. □

Lemma 15.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle T of G cannot be $PS{U}_n\left(q\right)$ with $n\ge 3$.

Proof. 

Assume that T is $PS{U}_n\left(q\right)$ with $n\ge 3$. If $n\ne 4$, $(n,q)\ne (2m,2)$ or $(3,5)$ (the exceptional cases will be discussed later), from ([8] Table 4), we have

$$P\left(T\right)=\frac{(q^n-{(-1)}^n)(q^{n-1}-{(-1)}^{n-1})}{q^2-1}.$$

Moreover, the facts $\lambda | |PS{U}_n\left(q\right)|$ and $P\left(T\right)<\frac{{\lambda }^2}{2}$ imply

$$\frac{1}{2}{q}^{n-2}<\frac{q^{n-1}-{(-1)}^{n-1}}{q-{(-1)}^{n-1}}<\lambda .$$

(4)

Similar to the above discussion, we will check all possibilities of M on a case-by-case basis, as listed by ([7] Table 3.5.B), and show that none satisfy $P\left(T\right)\le |G:M|<\frac{{\lambda }^2}{2}$.

(1) If M is of type $P_i$ with $i\le [n/2]$, then since $\lambda$ divides $\left|M\right|$, $\lambda$ is a prime divisor of $|U_{n-2i}\left(q\right)|$ or $|L_i\left(q^2\right)|$. By (4), the former is impossible, and the latter implies $\lambda \le \frac{q^{2i}-1}{q^2-1}$. This inequality leads to $\frac{{\lambda }^2}{2}\le \frac{{(q^{2i}-1)}^2}{2{(q^2-1)}^2}\le \frac{{(q^n-1)}^2}{2{(q^2-1)}^2}<\frac{(q^n-1)(q^{n-1}+1)}{q^2-1}=P\left(T\right)$, which is a contradiction. In the case when M is of type $U_i\left(q\right)\times U_{n-i}\left(q\right)$ with $1\le i<\frac{n}{2}$, we have $\lambda \le \frac{q^{n-1}-{(-1)}^{n-1}}{q-{(-1)}^{n-1}}$, which contradicts (4).

(2) Assume that M is of type $G{U}_i\left(q\right)\wr S_j$, where $n=ij\ge 2i$. If $i=1$, then $\lambda \le (q+1)$ or $\lambda \le n$, and (4) cannot hold. If $2\le i\le [n/2]$, then $\lambda \le \frac{q^i-{(-1)}^i}{q-{(-1)}^i}$, and $\lambda$ is too small to satisfy $P\left(T\right)<\frac{{\lambda }^2}{2}$. If M is of type $G{L}_{n/2}\left(q^2\right).2$ with n being even, then $\lambda \le \frac{{\left(q^2\right)}^{n/2}-1}{q^2-1}$, which implies that $\lambda$ is too small to satisfy $p\left(T\right)<\frac{{\lambda }^2}{2}$.

(3) If M is of type $G{U}_i\left(q^j\right)$, where $n=ij$, and j is an odd prime, then $\lambda \le \frac{{\left(q^j\right)}^i-{(-1)}^i}{q^j-{(-1)}^i}\le \frac{q^n-1}{q^3-1}$, which contradicts (4).

(4) The case when M is of type $G{U}_i\left(q\right)\otimes G{U}_j\left(q\right)$ with $n=ij$ and $2\le i<j\le n/2$ is impossible by $\lambda >\frac{q^{n-1}-{(-1)}^{n-1}}{q-{(-1)}^{n-1}}$.

(5) The case when M is of type $G{U}_n\left(q_0\right)$ with $q=q_0^i$ (i an odd prime) is impossible; otherwise, $\lambda \le \frac{q_0^n-{(-1)}^n}{q_0-{(-1)}^n}<q^{n/3}+1$, which contradicts (4). If M is of type $O_n$ with $nq$ being odd, then $\frac{q^{n-1}-1}{q-1}<\lambda \le q^{\frac{n-1}{2}}+1$, which is a contradiction. If M is of type $O_n^{ϵ}$, where q is odd, and n is even or of type $S{p}_n\left(q\right)$, with n being even, then we have $\frac{q^{n-1}+1}{q+1}<\lambda \le q^{n/2}+1$, which implies $n=4$. However, when $n=4$, $P\left(T\right)<\frac{{\lambda }^2}{2}$ gives $\frac{(q^4-1)(q^3+1)}{q^2-1}<\frac{{(q^2+1)}^2}{2}$, which is a contradiction.

(6) Assume that M is of type $n_0^{2m}.S{p}_{2m}\left(n_0\right)$, where $n=n_0^m$, and $n_0\ne p$ is a prime. If $n_0$ is an odd prime, then n is odd; thus, $2^{n-1}-1\le \frac{q^{n-1}-1}{q-1}<\lambda \le n_0^m+1=n+1$, which is a contradiction. For the case where $n_0=2$, and M is of type $2^{2m}.S{p}_{2m}\left(2\right)$, where $n=2^m\ge 4$, we have $\lambda \le 2^m+1=n+1$. Note that $n_0\ne p$, so $q\ge 3$, and $\frac{3^{n-1}+1}{4}\le \frac{q^{n-1}+1}{q+1}<\lambda \le n+1$, which is impossible.

(7) If M is of type $G{U}_i\left(q\right)\wr S_j$, a tensor product subgroup, where $n=i^j$ with $i\ge 3$ and $j\ge 2$, then $\lambda \le q^i+1$, which is contrary to (4).

Now, we consider the remaining cases, where $n=4$, $(n,q)=(3,5)$, or $q=2$, with $n(\ge 6)$ being even. For $n=4$, T is $PS{U}_4\left(q\right)$, and we have $P\left(T\right)=(q^3+1)(q+1)$. Note that $\lambda | |T|$; thus, $\lambda \le q^2+1$, which is contrary to $P\left(T\right)<\frac{{\lambda }^2}{2}$. If T is $PS{U}_3\left(5\right)$ with $P\left(T\right)=50$, then $\lambda | |M|$ and $P\left(T\right)<\frac{{\lambda }^2}{2}$ imply that $\lambda =31$ and $\left|M\right|\le 93\times 6$. It contradicts $|G:M|<\frac{{\lambda }^2}{2}$. We assume that T is $PS{U}_n\left(2\right)$ with $n(\ge 6)$ being even. Then, $P\left(T\right)=\frac{2^{n-1}(2^n-1)}{3}$. Since $\lambda | |T|$, $\lambda \le \frac{2^{n-1}+1}{3}$, this implies that $P\left(T\right)<\frac{{\lambda }^2}{2}$ cannot be satisfied. □

Lemma 16.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle T of G cannot be $PS{p}_{2m}\left(q\right)$ with $m\ge 2$.

Proof. 

Assume that T is $PS{p}_{2m}\left(q\right)$ with $m\ge 2$. If $q\ge 3$, then from ([8] Table 4), we have

$$P\left(T\right)=\frac{q^{2m}-1}{q-1}.$$

Moreover, the facts that $\lambda | |PS{p}_{2m}\left(q\right)|$ and $P\left(T\right)<\frac{{\lambda }^2}{2}$ yield the following:

$$q^{m-1}+1\le \frac{q^m-1}{q-1}<\lambda .$$

(5)

We discuss each maximal subgroup case M of G in turn, as listed in ([7] Table 3.5.C), and show that none satisfy $P\left(T\right)\le |G:M|<\frac{{\lambda }^2}{2}$.

(1) If M is of type $P_i$, with $1\le i\le m$, then $\lambda$ is a prime divisor of $|PG{L}_i\left(q\right)|$ or $|PS{p}_{n-2i}\left(q\right)|$. This implies that $\lambda \le \frac{q^i-1}{q-1}$ or $\lambda \le q^{\frac{n-2i}{2}}+1$, respectively, and both cases contradict (5) since $1\le i\le m$. If M is of type $S{p}_{2i}\left(q\right)\perp S{p}_{2(m-i)}\left(q\right)$ with $1\le i<m/2$, then $\lambda$ is a prime divisor of $|PS{p}_{2i}\left(q\right)|$ or $|PS{p}_{2(m-i)}\left(q\right)|$. It follows that $q^{m-1}+1<\lambda \le q^{m-i}+1$, which is a contradiction.

(2) If M is of type $S{p}_{2i}\left(q\right)\wr S_j$, where $m=ij\ge 2i$, then $q^{m-1}+1<\lambda \le q^i+1$. If M is of type $G{L}_m\left(q\right).2$ with q being odd, then $\lambda \le \frac{q^m-1}{q-1}$. If M is of type $D_{2(q\pm 1)}\wr S_2$, where $m=2$, and q is even, then $\lambda \le q+1$. For all cases, $\lambda$ is too small to satisfy (5).

(3) Assume that M is of type $S{p}_{2i}\left(q^j\right)$ with j prime and $m=ij$. If $j=2$, then $q^{2i^2}(q^2-1)\le |G:M|$ and $\lambda \le {\left(q^2\right)}^i+1$. However, the inequality $|G:M|\le \frac{{\lambda }^2}{2}$ cannot be satisfied. If j is an odd prime, then $\frac{q^m-1}{q-1}<\lambda \le \frac{{\left(q^j\right)}^i-1}{q^i-1}\le \frac{q^m-1}{q^i-1}$, which is a contradiction. If M is of type $G{U}_m\left(q\right)$ with q being odd, then $\lambda \le \frac{q^m-{(-1)}^m}{q-{(-1)}^m}$, which is a contradiction.

(4) If M is of type $S{p}_{2i}\left(q\right)\otimes O_{m/i}^{ϵ}\left(q\right)$ with $m/i\ge 3$, then we obtain a contradiction by

$$q^{m-1}+1<\lambda \le max\{q^i+1,q^{\left[\frac{m}{2i}\right]}+1\}.$$

(5) If M is of type $S{p}_n\left(q_0\right)$ with $q=q_0^i$ and i is prime, then $\lambda \le q_0^m+1\le q^{m-1}+1$, which contradicts (5).

(6) If M is of type $2^{1+2i}.O_{2i}^{-}\left(2\right)$ with $n=2^i$ and $q=p\ge 3$, then $\frac{3^m-1}{2}<\lambda \le 2^i+1=n+1$ holds only when $(n,q)=(4,3)$. It follows that $\lambda \le 5$, which can be ruled out by Lemma 5.

(7) If M is of type $SP{s}_i\left(q\right)\wr S_j$, where $n=i^j$, and $qj$ is odd, then $q^{m-1}+1<\lambda \le q^{\frac{i}{2}}+1$, which is a contradiction.

(8) If M is of type $O_{2m}^{+}\left(q\right)$ with q being even, then $\lambda \le \frac{q^m-1}{q-1}$, which is a contradiction. If M is of type $O_{2m}^{-}\left(q\right)$ with q being even, then $|G:M|=\frac{q^m(q^m-1)}{2}$. Moreover, the facts that $\lambda | |M|$ and $\lambda >\frac{q^m-1}{q-1}$ imply that $\lambda |q^m+1$. If $\lambda \ne q^m+1$, then $\lambda \le \frac{q^m+1}{3}$ and ${\lambda }^2\le \frac{{(q^m+1)}^2}{9}<\frac{q^m(q^m-1)}{2}$, contradicting $|G:M|<\frac{{\lambda }^2}{2}$. Thus, we have $\lambda =q^m+1.$ Note that $d=|G:G_{\Delta }|=|G:M | |M:G_{\Delta }|$. If $G_{\Delta }<M$, then by $d<\frac{{\lambda }^2}{2}$, we have $q^m(q^m-1)\le d<\frac{{(q^m+1)}^2}{2}$, which is impossible. Therefore, $G_{\Delta }=M$ and $d=\frac{q^m(q^m-1)}{2}$. This implies $d=\frac{(\lambda -1)(\lambda -2)}{2}$ and $k\le \frac{\lambda (\lambda -3)}{2}<d$, contradicting $(\ell -1)d<k$ in Lemma 7.

We now consider the case when T is $S{p}_{2m}\left(2\right)$. We have $P\left(T\right)=2^{m-1}(2^m-1)$ and $\lambda \le 2^m+1$. If $\lambda =2^m+1$, then $P\left(T\right)=\frac{(\lambda -1)(\lambda -2)}{2}$. The facts that $\lambda | |T|$ and $k<\frac{\lambda (\lambda -3)}{2}$ imply $k<P\left(T\right)\le d$, resulting in a contradiction. Hence, $\lambda \le 2^m-1$, and ${\lambda }^2<{(2^m-1)}^2<2P\left(T\right)$, which is contrary to $P\left(T\right)<\frac{{\lambda }^2}{2}$. □

Lemma 17.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle T of G cannot be $P{\Omega }_{2m+1}\left(q\right)$ with $m\ge 3$.

Proof. 

Suppose that T is $P{\Omega }_{2m+1}\left(q\right)$. Note that $P{\Omega }_5\left(q\right)\cong PS{p}_4\left(q\right)$, so we have $m\ge 3$. If $q\ge 5$, then from ([8] Table 4), we have

$$P\left(T\right)=\frac{q^{2m}-1}{q-1}.$$

Moreover, considering the facts that $\lambda | |P{\Omega }_{2m+1}\left(q\right)|$ and $P\left(T\right)<\frac{{\lambda }^2}{2}$, we obtain

$$q^{m-1}+1\le \frac{q^m-1}{q-1}<\lambda .$$

(6)

The maximal subgroups M of G are listed in ([7] Table 3.5.D), we will discuss them case-by-case.

(1) If M is of type $P_i$ with $1\le i\le m$, then $\lambda$ is a prime divisor of $|O_{n-2i}^{ϵ}\left(q\right)|$ or $|G{L}_i\left(q\right)|$. It follows that

$$\frac{q^m-1}{q-1}<\lambda \le max\{q^{m-i}+1,\frac{q^i-1}{q-1}\},$$

a contradiction. Assume that M is of type $O_i\left(q\right)\perp O_{n-i}^{\pm }\left(q\right)$, where i is odd and $1\le i<n$, it follows that $i<n$, and i is an odd force $i\le 2m-1$. If $\lambda$ is a prime divisor of $|O_i\left(q\right)|$, then $\lambda \le q^{\frac{i-1}{2}}+1\le q^{m-1}+1$, and $\lambda$ is too small to satisfy (6). If $\lambda$ is a prime divisor of $|O_{n-i}^{+}\left(q\right)|$, then $\lambda \le \frac{q^{\frac{n-i}{2}}-1}{q-1}$, and $\lambda$ is too small again. If $\lambda$ is a prime divisor of $|O_{n-i}^{-}\left(q\right)|$, then $\lambda \le \frac{q^{\frac{n-i}{2}}+1}{2}$, which implies that $i=1$. When $i=1$, and the type of M is $|O_{2m}^{-}\left(q\right)|$, we have $|G:M|=\frac{q^m(q^m-1)}{2}$. Since $\lambda | |M|$, and q is odd, $\lambda \le \frac{q^m+1}{2}$, and $|G:M|<\frac{{\lambda }^2}{2}$ cannot be satisfied.

(2) If M is of type $O_i\left(q\right)\wr S_j$ with $n=ij\ge 2i$ and $i\ge 3$ being odd, then $\lambda \le \frac{q^{\frac{i-1}{2}}+1}{2}<q^{m-1}$. If M is of type $O_1\left(q\right)\wr S_n$ with $q=p$, then $\frac{q^{\frac{n-1}{2}}-1}{q-1}<\lambda \le n$ when $q\ge 5$. Both cases imply that $\lambda$ is too small to satisfy (6).

(3) Assume that M is of type $O_j\left(q^i\right)$, where $n=ij$, and i is an odd prime. Obviously, j is odd. Then $\lambda \le \frac{{\left(q^i\right)}^{\frac{j-1}{2}}+1}{q^{\frac{j-1}{2}}+1}\le q^{\frac{n-3}{2}}$, which is contrary to $q^{m-1}<\lambda$.

(4) If M is of type $O_i\left(q\right)\otimes O_j^{ϵ}\left(q\right)$ with $n=ij$ and $3\le i<j$, then $\lambda \le q^{\frac{j-1}{2}}+1<q^{m-1}$, which is contrary to $q^{m-1}<\lambda$.

(5) If M is of type $O_n\left(q_0\right)$ with $q=q_0^i$ and i is a prime, then $\lambda \le q_0^m+1\le \frac{q^m-1}{q-1}$, which is contrary to $\frac{q^m-1}{q-1}<\lambda$.

(6) If M is of type $O_i\left(q\right)\wr S_j$ with $n=i^j\ge i^2$, and i is odd, then $\lambda \le q^{\frac{i-1}{2}}+1$, and $\lambda$ is too small.

We now consider the case where T is $P{\Omega }_{2m+1}\left(3\right)$, when $q=3$. In this case, $P\left(T\right)=\frac{3^m(3^m-1)}{2}$ and $\lambda \le \frac{3^m+1}{2}$. It follows that ${\lambda }^2<{\frac{(3^m+1)}{4}}^2<2P\left(T\right)$, which contradicts $P\left(T\right)<\frac{{\lambda }^2}{2}$. □

Lemma 18.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle T of G cannot be $P{\Omega }_{2m}^{-}\left(q\right)$ with $m\ge 3$.

Proof. 

Suppose that T is $P{\Omega }_{2m}^{-}\left(q\right)$ with $m\ge 3$. Note that $P{\Omega }_6^{-}\left(q\right)\cong PS{U}_4\left(q\right)$, so we have $m\ge 4$. Then, from ([8] Table 4), we know that

$$P\left(T\right)=\frac{(q^m+1)(q^{m-1}-1)}{q-1}.$$

Furthermore, considering $\lambda | |P{\Omega }_{2m}^{-}\left(q\right)|$ and $P\left(T\right)<\frac{{\lambda }^2}{2}$, we have

$$q^{m-1}+1<\lambda .$$

(7)

We will discuss all types of M on a case-by-case basis, as listed in ([7] Table 3.5.F).

(1) If M is of type $P_i$ with $1\le i\le m-1$, then $\lambda$ is a prime divisor of $|O_{n-2i}^{ϵ}\left(q\right)|$ or $|G{L}_i\left(q\right)|$. It follows that $\lambda \le q^{m-i}+1$ or $\lambda \le \frac{q^i-1}{q-1}$. If M is of type $O_i^{ϵ}\left(q\right)\times O_{n-i}^{-ϵ}\left(q\right)$, with $2\le i\le m$ when i is even and $1\le i<m$ when i is odd, then since $i\le n-i$, we have $\lambda \le q^{\frac{n-i}{2}}+1\le q^{m-1}+1$ when i is even and $\lambda \le q^{\frac{n-i-1}{2}}+1\le q^{m-1}+1$ when i is odd. If M is of type $P{s}_{n-2}\left(q\right)$ with q being even, then $q^{m-1}+1<\lambda \le q^{m-1}+1$. Both cases contradict (7).

(2) If M is of type $O_i^{-}\left(q\right)\wr S_{n/i}$ with i being even and $2\le i\le n/2$, then $q^{m-1}+1<\lambda \le q^{\frac{i}{2}}+1<q^{m-1}+1$, which is a contradiction. If M is of type $O_i\left(q\right)\wr S_j$ with i being odd and $2\le i\le n/2$, then $q^{m-1}+1<\lambda \le q^{\frac{i-1}{2}}+1<q^{m-1}+1$, which is a contradiction. If M is of type $O_1\left(q\right)\wr S_n$ with $q=p\ge 3$, then $3^{m-1}+1<\lambda \le 2m+1$, which is impossible when $m\ge 4$. If M is of type $O_m{\left(q\right)}^2$ and $mq$ is odd, then $q^{m-1}+1<\lambda \le q^{\frac{m-1}{2}}+1$; this is a contradiction.

(3) If M is of type $G{U}_{n/2}\left(q\right)$ with $m=n/2$ being odd, then $\lambda \le \frac{q^m+1}{q+1}<q^{m-1}+1$. If M is of type $O_m\left(q^2\right)$ with $mq$ being odd, then $\lambda \le {\left(q^2\right)}^{\frac{m-1}{2}}+1=q^{m-1}+1$. Both cases contradict (7). Assume that M is of type $O_{n/i}^{-}\left(q^i\right)$ with i being prime and $n/i\ge 3$. When i is odd, we have $\lambda \le \frac{{\left(q^i\right)}^{\frac{n}{2i}}+1}{q^{\frac{n}{2i}}+1}<q^m+1$, which is a contradiction by (7). When $i=2$, we have $q^{\frac{m^2}{2}}<|G:M|$ and $\lambda \le {\left(q^2\right)}^{\frac{n}{4}}+1=q^m+1$. This implies that $q^{\frac{m^2}{2}}<\frac{{(q^m+1)}^2}{2}$, which is impossible when $m\ge 4$.

(4) If M is of type $O_i\left(q\right)\otimes O_{n/i}^{-}\left(q\right)$ with i being odd and $n/i\ge 4$, then $\lambda \le q^{\frac{i-1}{2}}+1<q^{m/2-1}$ by $i\le n/4=m/2$ or $\lambda \le q^{m/i}+1<q^{m-1}+1$, which is contrary to (7).

(5) If M is of type $O_n^{-}\left(q_0\right)$ with $q=q_0^i$ and i is an odd prime, then $\lambda \le q_0^m+1<q^{m-1}+1$, which is contrary to (7). □

Lemma 19.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1, then the socle of G cannot be $P{\Omega }_{2m}^{+}\left(q\right)$ with $m\ge 4$.

Proof. 

Assume that T is $P{\Omega }_{2m}^{+}\left(q\right)$ with $m\ge 4$. If $q\ge 4$, then from ([8] Table 4), we have

$$P\left(T\right)=\frac{(q^m+1)(q^{m-1}-1)}{q-1}.$$

Moreover, since $\lambda$ is a prime divisor of $|P{\Omega }_{2m}^{+}\left(q\right)|$, we have $\lambda \le \frac{q^m-1}{q-1}$, which contradicts $P\left(T\right)<\frac{{\lambda }^2}{2}$. If $q=3$, we have $\lambda \le \frac{3^m-1}{2}$ and $P\left(P{\Omega }_{2m}^{+}\left(3\right)\right)=\frac{3^{m-1}(3^m-1)}{2}$, which contradicts $P\left(T\right)<\frac{{\lambda }^2}{2}$. For $q=2$, we have $\lambda \le 2^m-1$ and $P\left(P{\Omega }_{2m}^{+}\left(2\right)\right)=2^{m-1}(2^m-1)$, which is a contradiction again. Therefore, the socle of G cannot be $P{\Omega }_{2m}^{+}\left(q\right)$ with $m\ge 4$. □

Proposition 2.

Assume that $\mathcal{D}$ and G satisfy Hypothesis 1. If the socle T of G is a non-abelian simple group, then T must be $PS{L}_n\left(q\right)$.

Proof. 

For Lemmas 10–19, we draw a conclusion. □

Proof of Theorem 1.

It follows immediately from Propositions 1 and 2. □