Flag-transitive non-symmetric $2$-designs with $(r,\lambda)=1$ and exceptional groups of Lie type

TL;DR

This paper classifies all flag-transitive non-symmetric 2-designs with specific parameters and exceptional Lie type automorphism groups, identifying five classes of such designs with Ree or Suzuki groups as socles.

Contribution

It provides a complete classification of non-symmetric 2-designs with (r,λ)=1 and exceptional Lie type automorphism groups, highlighting five distinct classes.

Findings

T must be a Ree or Suzuki group

Five classes of non-isomorphic designs identified

Complete classification of designs with given parameters

Abstract

This paper determined all pairs $(\mathcal{D},G)$ where $\mathcal{D}$ is a non-symmetric 2-$(v,k,\lambda)$ design with $(r,\lambda)=1$ and $G$ is the almost simple flag-transitive automorphism group of $\mathcal{D}$ with an exceptional socle of Lie type. We prove that if $T\trianglelefteq G\leq Aut(T)$ where $T$ is an exceptional group of Lie type, then $T$ must be the Ree group or Suzuki group, and there are five classes of non-isomorphic designs $\mathcal{D}$.

Tables

Table 1. Table 1 :

Table 2. Table 2 : The maximal subgroups of G 2 2 ​ ( q ) superscript subscript 𝐺 2 2 𝑞 {}^{2}G_{2}(q)

Group	Structure	Remarks
$M_1$	$Q:K$	the normalizer of $Q$ in ${}_{}{}^{2}G_2^{}\left(q\right)$
$M_2$	${\mathbb{Z}}_2\times L_2\left(q\right)$	the centralizer of an involution in ${}_{}{}^{2}G_2^{}\left(q\right)$
$M_3$	$\left({\mathbb{Z}}_2^2\times D_{\left(q+1\right)/2}\right):{\mathbb{Z}}_3$	the normalizer of a four-subgroup
$M_4$	${\mathbb{Z}}_{q+m+1}:{\mathbb{Z}}_6$	the normalizer of ${\mathbb{Z}}_{q+m+1}$
$M_5$	${\mathbb{Z}}_{q-m+1}:{\mathbb{Z}}_6$	the normalizer of ${\mathbb{Z}}_{q-m+1}$

Equations

$$|A:B|\ge |(A\cap C):(B\cap C)|.$$

$$({\alpha }_1,{\beta }_1,{\gamma }_1)({\alpha }_2,{\beta }_2,{\gamma }_2)=({\alpha }_1+{\alpha }_2,{\beta }_1+{\beta }_2-{\alpha }_1{\alpha }_2^m,{\gamma }_1+{\gamma }_2-{\alpha }_1^m{\alpha }_2^m-{\alpha }_2{\beta }_1+{\alpha }_1{\alpha }_2^{m+1}).$$

$$(\alpha ,\beta ,\gamma {)}^k=(k\alpha ,k^{1+m}\beta ,k^{2+m}\gamma ).$$

$$H\le N_{M_2}(Q_2^c)=N_{M_2}(Q_2{)}^c=H_2^c.$$

$$b=\frac{\lambda v(v-1)}{k(k-1)}=\frac{\lambda q^3(q^3+1)}{k(k-1)},$$

$$|T_B|=\frac{|T|}{b}=\frac{(q-1)k(k-1)}{\lambda }.$$

$$|B^{T_{\gamma }}|=|T_{\gamma }:T_{\gamma B}|\le |B^{G_{\gamma }}|=|G_{\gamma }:G_{\gamma B}|=r=q^3,$$

$$b=q^2(q^3+1),r=q^3,\lambda =q-1.$$

$$b=q(q^3+1),r=q^3,\lambda =q^2-1.$$

$$(v,b,r,k,\lambda )=(q^3+1,q^2(q^2-q+1),q^2,q+1,1)$$

$$(v,b,r,k,\lambda )=(q^2+1,q(q^2+1),q^2,q,q-1).$$

$$\mathcal{T}=\{{}^2{F}_4(q),{}^3{D}_4(q),G_2(q),F_4(q),E_6^{ϵ}(q),E_7(q),E_8(q)\},$$

$$v=\frac{(q^3+1)(q^4+1)(q^9-1)(q^6+1)(q^4+q^2+1)}{(q-1)}.$$

$$v=\frac{(q^8+q^4+1)(q^9-1)}{q-1}.$$

$$d=\frac{q(q^3+1)(q^8-1)}{(q-1)}\mathrm{and}{d}^{\prime }=\frac{q^8(q^4+1)(q^5-1)}{(q-1)}.$$

$$|T_{\alpha }|=q^{10}(q^2-1{)}^2(q^4-1)(q^6-1)\mathrm{and}v=q^{14}(q^4+1)(q^4+q^2+1)(q^6+1).$$

$$|T_{\alpha }|=q^{16}(q^2-1)(q^4-1)(q^6-1)(q^8-1)\mathrm{and}v=q^8(q^8+q^4+1).$$

Full text

Mathematics Subject Classification (2010): 05B05, 05B25, 20B25

Keywords: $2$-design; flag-transitive; exceptional group of Lie type

1 Introduction

A 2-$(v,k,\lambda)$ design $\mathcal{D}$ is a pair $(\mathcal{P},\mathcal{B})$, where $\mathcal{P}$ is a set of $v$ points and $\mathcal{B}$ is a set of $k$-subsets of $\mathcal{P}$ called blocks, such that any 2 points are contained in exactly $\lambda$ blocks. A flag is an incident point-block pair $(\alpha,B)$. An automorphism of $\mathcal{D}$ is a permutation of $\mathcal{P}$ which leaves $\mathcal{B}$ invariant. The design is non-trivial if $2<k<v-1$ and non-symmetric if $v<b$. All automorphisms of the design $\mathcal{D}$ form a group called the full automorphism group of $\mathcal{D}$, denoted by $Aut(\mathcal{D})$. Let $G\leq Aut(\mathcal{D})$, the design $\mathcal{D}$ is called point (block, flag)-transitive if $G$ acts transitively on the set of points (blocks, flags), and point-primitive if $G$ acts primitively on $\mathcal{P}$. Note that a finite primitive group is almost simple if it is isomorphic to a group $G$ for which $T\cong Inn(T)\leq G\leq Aut(T)$ for some non-abelian simple group $T$.

Let $G\leq Aut(\mathcal{D})$, and $r$ be the number of blocks incident with a given point. In [6], P. Dembowski proved that if $G$ is a flag-transitive automorphism group of a 2-design $\mathcal{D}$ with $(r,\lambda)=1$, then $G$ is point-primitive. In 1988, P. H. Zieschang [32] proved that if $\cal D$ is a 2-design with $(r,\lambda)=1$ and $G\leq Aut(\mathcal{D})$ is flag transitive, then $G$ must be of almost simple or affine type. Such 2-designs have been studied in [1, 2, 29, 31], where the socle of $G$ is a sporadic, an alternating group or elementary abelian $p$-group, respectively. In this paper, we continue to study the case that the socle of $G$ is an exceptional simple group of Lie type. We get the following:

Theorem 1

Let $\mathcal{D}=(\mathcal{P},\mathcal{B})$ be a non-symmetric $2$-$(v,k,\lambda)$ design with $(r,\lambda)=1$ and $G$ an almost simple flag-transitive automorphism group of $\mathcal{D}$ with the exceptional socle $T$ of Lie type in characteristic $p$ and $q=p^{e}$. Let $B$ be a block of $\mathcal{D}$. Then one of the following holds:

1. (1)

$T={{}^{2}G_{2}(q)}$* with $q=3^{2n+1}\geq 27$, and $\mathcal{D}$ is one of the following:*

1. (i)

a Ree unital with $G_{B}=\mathbb{Z}_{2}\times L_{2}(q)$; 2. (ii)

a $2$-$(q^{3}+1,q,q-1)$* design with $G_{B}=Q_{1}:K$;* 3. (iii)

a $2$-$(q^{3}+1,q,q-1)$* design with $G_{B}=Q_{2}:K$;* 4. (iv)

a $2$-$(q^{3}+1,q^{2},q^{2}-1)$* design with $G_{B}=Q^{\prime}:K$,*

where $Q\in Syl_{3}(T)$, and the definitions of $Q_{1},Q_{2}$ and $K$ refer to Section 3. 2. (2)

$T={{}^{2}B_{2}(q)}$* with $q=2^{2n+1}\geq 8$, and $\mathcal{D}$ is a $2$-$(q^{2}+1,q,q-1)$ design with $G_{B}=Z(Q):K$, where $Q\in Syl_{2}(T)$ and $K=\mathbb{Z}_{q-1}\cong\mathbb{F}_{q}^{*}$.*

2 Preliminary results

We first give some preliminary results about designs and almost simple groups.

Lemma 2.1

([29, Lemma 2.2])* For a $2$-$(v,k,\lambda)$ design $\mathcal{D}$, it is well known that*

1. (1)

$bk=vr$; 2. (2)

$\lambda(v-1)=r(k-1)$; 3. (3)

$v\leq\lambda v<r^{2}$; 4. (4)

if $G\leq Aut(\mathcal{D})$ is flag-transitive and $(r,\lambda)=1$, then $r\mid(|G_{\alpha}|,v-1)$ and $r\mid d$, for any non-trivial subdegree $d$ of $G$.

Lemma 2.2

Assume that $G$ and $\mathcal{D}$ satisfy the hypothesis of Theorem 1. Let $\alpha\in\mathcal{P}$ and $B\in\mathcal{B}$. Then

1. (1)

$G=TG_{\alpha}$* and $|G|=f|T|$ where $f$ is a divisor of $|Out(T)|$;* 2. (2)

$|G:T|=|G_{\alpha}:T_{\alpha}|=f$; 3. (3)

$|G_{B}|$* divides $f|T_{B}|$, and $|G_{\alpha B}|$ divides $f|T_{\alpha B}|$ for any flag $(\alpha,B)$.*

Proof.  Note that $G$ is an almost simple primitive group by [5]. So (1) holds and (2) follows from (1). Since $T\unlhd G$, then $|B^{T}|$ divides $|B^{G}|$ and $|(\alpha,B)^{T}|$ divides $|(\alpha,B)^{G}|$, hence $|G_{B}:T_{B}|$ divides $f$, and $|G_{\alpha B}:T_{\alpha B}|$ divides $f$, (3) holds. $\hfill\square$

Lemma 2.3

([6, 2.2.5])* Let $\mathcal{D}$ be a $2$-$(v,k,\lambda)$ design. If $\mathcal{D}$ satisfies $r=k+\lambda$ and $\lambda\leq 2$, then $\mathcal{D}$ is embedded in a symmetric $2$-$(v+k+\lambda,k+\lambda,\lambda)$ design. *

Lemma 2.4

([6, 2.3.8])* Let $\mathcal{D}$ be a $2$-$(v,k,\lambda)$ design and $G\leq Aut(\mathcal{D})$. If $G$ is $2$-transitive on points and $(r,\lambda)=1$, then $G$ is flag transitive. *

Lemma 2.5

Let $A$, $B$, $C$ be subgroups of group $G$. If $B\leq A$, then

[TABLE]

Lemma 2.6

([17])* Suppose that $T$ is a simple group of Lie type in characteristic $p$ and acts on the set of cosets of a maximal parabolic subgroup. Then $T$ has a unique subdegree which is a power of $p$ except $T$ is $L_{d}(q)$, $\Omega^{+}_{2m}(q)$ ($m$ is odd) or $E_{6}(q)$. *

Lemma 2.7

[26, 1.6]$\mathrm{(Tits~{}Lemma)}$* If $T$ is a simple group of Lie type in characteristic $p$, then any proper subgroup of index prime to $p$ is contained in a parabolic subgroup of $T$. *

In the following, for a positive integer $n$, $n_{p}$ denotes the $p$-part of $n$ and $n_{p^{\prime}}$ denotes the $p^{\prime}$-part of $n$, i.e., $n_{p}=p^{t}$ where $p^{t}\mid n$ but $p^{t+1}\nmid n$, and $n_{p^{\prime}}=n/n_{p}$.

Lemma 2.8

*Assume that $G$ and $\mathcal{D}$ satisfy the hypothesis of Theorem 1. Then $|G|<|G_{\alpha}|^{3}$ and if $G_{\alpha}$ is a non-parabolic maximal subgroup of $G$, then $|G|<|G_{\alpha}|{|G_{\alpha}|_{p^{\prime}}^{2}}$ and $|T|<|Out(T)|^{2}|T_{\alpha}|{|T_{\alpha}|_{p^{\prime}}^{2}}$. *

Proof.  From Lemma 2.1, since $r$ divides every non-trivial subdegree of $G$, then $r$ divides $|G_{\alpha}|$, and so $|G|<|G_{\alpha}|^{3}$. If $G_{\alpha}$ is not parabolic, then $p$ divides $v=|G:G_{\alpha}|$ by Lemma 2.7. Since $r$ divides $v-1$, $(r,p)=1$ and so $r$ divides $|G_{\alpha}|_{p^{\prime}}$. It follows that $r<|G_{\alpha}|_{p^{\prime}}$, and hence $|G|<|G_{\alpha}|{|G_{\alpha}|_{p^{\prime}}^{2}}$ by Lemma 2.1. Now by Lemma 2.2(2), we have that $|T|<|Out(T)|^{2}|T_{\alpha}|{|T_{\alpha}|_{p^{\prime}}^{2}}$. $\hfill\square$

Lemma 2.9

([20, Theorem 2, Table III])* If $T$ is a finite simple exceptional group of Lie type such that $T\leq G\leq Aut(T)$, and $G_{\alpha}$ is a maximal subgroup of $G$ such that $T_{0}=Soc(G_{\alpha})$ is not simple, then one of the following holds:*

1. (1)

$G_{\alpha}$* is parabolic;* 2. (2)

$G_{\alpha}$* is of maximal rank;* 3. (3)

$G_{\alpha}=N_{G}(E)$, where $E$ is an elementary abelian group given in **[4, Theorem 1 (II)]**; 4. (4)

$T=E_{8}(q)$* with $p>5$, and $T_{0}$ is either $A_{5}\times A_{6}$ or $A_{5}\times L_{2}(q)$;* 5. (5)

$T_{0}$* is as in Table 1.*

Lemma 2.10

([19, Theorem 3])* Let $T$ be a finite simple exceptional group of Lie type, with $T\leq G\leq Aut(T)$. Assume $G_{\alpha}$ is a maximal subgroup of $G$ and $Soc(G_{\alpha})=T_{0}(q)$ is a simple group of Lie type over $\mathbb{F}_{q}(q>2)$ such that $\frac{1}{2}rank(T)<rank(T_{0})$; assume also that $(T,T_{0})$ is not $(E_{8},{{}^{2}A_{5}(5)})$ or $(E_{8},{{}^{2}D_{5}(3)})$. Then one of the following holds:*

1. (1)

$G_{\alpha}$* is a subgroup of maximal rank;* 2. (2)

$T_{0}$* is a subfield or twisted subgroup;* 3. (3)

$T=E_{6}(q)$* and $T_{0}=C_{4}(q)(q~{}odd)$ or $F_{4}(q)$.*

Lemma 2.11

([22, Theorem 1.2])* Let $T$ be a finite simple exceptional group of Lie type such that $T\leq G\leq Aut(T)$, and $G_{\alpha}$ a maximal subgroup of $G$ with socle $T_{0}=T_{0}(q)$ a simple group of Lie type in characteristic $p$. Then if $rank(T_{0})\leq\frac{1}{2}rank(T)$, we have the following bounds:*

1. (1)

if $T=F_{4}(q)$, then $|G_{\alpha}|<4q^{20}\log_{p}q;$ 2. (2)

if $T=E_{6}^{\epsilon}(q)$, then $|G_{\alpha}|<4q^{28}\log_{p}q$; 3. (3)

if $T=E_{7}(q)$, then $|G_{\alpha}|<4q^{30}\log_{p}q$; 4. (4)

if $T=E_{8}(q)$, then $|G_{\alpha}|<12q^{56}\log_{p}q$.

*In all cases, $|G_{\alpha}|<12|G|^{\frac{5}{13}}\log_{p}q.$ *

The following lemma gives a method to check the existence of the design with the possible parameters.

Lemma 2.12

For the given parameters $(v,b,r,k,\lambda)$ and the group $G$, the conditions that there exists a design $\mathcal{D}$ with such parameters satisfying $G$ which is flag-transitive and point primitive is equivalent to the following four steps holding for some subgroup $H$ of $G$ with index $b$ and its orbit of size $k$:

1. (1)

$G$* has at least one subgroup $H$ of order $|G|/b$;* 2. (2)

$H$* has at least one orbit $O$ of length $k$;* 3. (3)

the size of $O^{G}$ is $b$; 4. (4)

the number of blocks incident with any two points is a constant.

When we run through all possibilities of $H$ and its orbits with size $k$, then we found all designs with such parameters and admitting $G\leq Aut(\mathcal{D})$ is flag-transitive and point primitive. This is the essentially strategy adopted in [29].

We now give some information about the Ree group ${}^{2}G_{2}(q)$ with $q=3^{2n+1}$ and its subgroups, which from [8, 11, 15] and would be used later.

Set $m=3^{n+1}$, and so $m^{2}=3q$. The Ree group ${}^{2}G_{2}(q)$ is generated by $Q,K$ and $\tau$, where $Q$ is Sylow 3-subgroup of ${}^{2}G_{2}(q)$, $K=\{diag(t^{m},t^{1-m},t^{2m-1},1,t^{1-2m},t^{m-1},t^{-m})\,|\,t\in\mathbb{F}_{q}^{*}\}\cong\mathbb{Z}_{q-1}$ and $\tau^{2}=1$ such that $\tau$ inverts $K$, and $|^{2}G_{2}(q)|=(q^{3}+1)q^{3}(q-1)$.

Lemma 2.13

1. (1)

([15])* ${}^{2}G_{2}(q)$ is $2$-transitive of degree $q^{3}+1$.* 2. (2)

([7, p.252])* The stabilizer of one point is $Q:K$, and $N_{{}^{2}G_{2}(q)}(Q)=Q:K$.* 3. (3)

([11, p.292])* The stabilizer $K$ of two points is cyclic of order $q-1$ and the stabilizer of three points is of order*$2$. 4. (4)

([11, p.292])* The Sylow $2$-subgroup of ${}^{2}G_{2}(q)$ is elementary abelian with order $8$.*

Lemma 2.14

([8, Lemma 3.3] )* Let $M\leq{{}^{2}G_{2}(q)}$ and $M$ be maximal in ${}^{2}G_{2}(q)$. Then either $M$ is conjugate to $M_{6}:={{}^{2}G_{2}(3^{\ell})}$ for some divisor $\ell$ of $2n+1$, or $M$ is conjugate to one of the subgroups $M_{i}$ in the following table:*

Moreover, we see that from [8], the Sylow 3-subgroup $Q$ can be identified with the group consisting of all triples $(\alpha,\beta,\gamma)$ from $\mathbb{F}_{q}$ with multiplication:

[TABLE]

It is easy to check that $(0,0,\gamma)(0,\beta,0)=(0,\beta,\gamma)$. Set $Q_{1}=\{(0,0,\gamma)|\gamma\in\mathbb{F}_{q}\}$ and $Q_{2}=\{(0,\beta,0)|\beta\in\mathbb{F}_{q}\}$, then $Q_{1}\cong Q_{2}\cong\mathbb{Z}_{3}^{2n+1}$.

For a group $Q$, $Z(Q)$, $\Phi(Q)$, $Q^{\prime}$ denote the center, Frattini subgroup, and the derived subgroup of $Q$, respectively. Then $Q^{\prime}=\Phi(Q)=Q_{1}\times Q_{2}$, $Z(Q)=Q_{1}$, and $Q^{\prime}$ is an elementary abelian $3$-group. For any $(\alpha,\beta,\gamma)\in Q$ and $k\in K$,

[TABLE]

Lemma 2.15

([8, 15])* Let $Q$, $M$, $Q_{2}$, $M_{2}$ and $K$ as above, then*

1. (1)

the normalizer of any subgroup of $Q$ is contained in $M_{1}$; 2. (2)

for any $g\in{{}^{2}G_{2}(q)}$, either $Q^{g}=Q$ or $Q^{g}\cap Q=1$; 3. (3)

$Q_{2}$* is a Sylow $3$-subgroup of $M_{2}$ and $N_{M_{2}}(Q_{2})=2\times(Q_{2}:\langle k^{2}\rangle)$ with $\langle k\rangle=K$.*

Lemma 2.16

([8, Lemma 3.2])* The following hold for the cyclic subgroup $K$:*

1. (1)

$K$* is transitive on $Q_{1}\setminus\{1\}$ acting by conjugation;* 2. (2)

$K$* has two orbits $(0,1,0)^{K}$, $(0,-1,0)^{K}$ on $Q_{2}\setminus\{1\}$ acting by conjugation.*

From above lemmas, we have the following properties of the subgroups of ${}^{2}G_{2}(q)$.

Lemma 2.17

*If $H\leq M_{1}$ and $(q-1)\mid|H|$, then $K\leq H$. *

Proof.  Let $p$ be a prime divisor of $q-1$. If $P\in Syl_{p}(M_{1})$, then since $(p,3)=1$ and $Q\cap K=1$, we have $P\in Syl_{p}(K)$. Note that $K$ is cyclic, the Sylow $p$-subgroup of $K$ is unique, and so the Sylow $p$-subgroup of $M_{1}$ is unique. On the other hand, if $P_{0}\in Syl_{p}(H)$, since $H\leq M_{1}$, then $P_{0}=P\cap H$. Moreover, $|P_{0}|=|P|$ implies that $P=P_{0}\leq H$. Since $p$ is arbitrary, all Sylow subgroups of $K$ are contained in $H$, and so $K\leq H$. $\hfill\square$

Corollary 1

*Let $H\leq M_{1}$ and $|H|=q(q-1)$. Then $H=A:K$ where $A$ is the Sylow $3$-subgroup of $H$. *

Proof. Since $Q\unlhd M_{1}$, we have $A=H\cap Q$ and $A\unlhd H$. By Lemma 2.17, $K\leq H$. Now $A\cap K=1$, and so $H=A:K$. $\hfill\square$

Lemma 2.18

Let $Q_{2}$ be a Sylow $3$-subgroup of $M_{2}$ and $H_{2}:=N_{M_{2}}(Q_{2})$. If $Q_{2}\leq Q$ and $M_{1}=Q:K$, then the following hold:

1. (1)

$H_{2}=Q_{2}:K$* and $H_{2}\leq M_{1}$;* 2. (2)

for any $H\leq M_{2}$ satisfying $|H|=q(q-1)$, there exists $c\in M_{2}$ such that $H=H_{2}^{c}$ and $H\leq M_{1}^{c}$ .

Proof.  Clearly, (1) holds by Lemma 2.15(1) and Corollary 1. Let $H\leq M_{2}$ and $|H|=q(q-1)$. Note that $M_{2}\cong\mathbb{Z}_{2}\times L_{2}(q)$. Since $H\lesssim\mathbb{Z}_{2}\times L_{2}(q)$ and $H_{2}\lesssim\mathbb{Z}_{2}\times L_{2}(q)$, then by the list of maximal subgroups of $L_{2}(q)$, we know that $H\cong H_{2}\cong\mathbb{Z}_{2}\times([q]:Z_{\frac{q-1}{2}})$. Let $\sigma$ be an automorphism from $H_{2}$ to $H$. Then $Q_{2}^{\sigma}\unlhd H$ since $Q_{2}\unlhd H_{2}$. Moreover, since $q\mid|H|$, the Sylow 3-subgroup of $H$ is conjugate to $Q_{2}$ in $M_{2}$ and so $Q_{2}^{\sigma}=Q_{2}^{c}\unlhd H$ for $c\in M_{2}$. It follows that

[TABLE]

Therefore $H=H_{2}^{c}$.

Note that if $Q^{c}\neq Q$, then from $Q_{2}^{c}\leq Q^{c}$ and Lemma 2.15(1), we get $H=N_{M_{2}}(Q_{2}^{c})\leq M_{1}^{c}$, and so (2) holds.

Now, we prove that $Q^{c}\neq Q$. If $Q^{c}=Q$, then $Q_{2}^{c}\leq Q$, and so $H\leq M_{1}$. By Corollary 1, we have $H=Q_{2}^{c}:K$ and $H_{2}=Q_{2}:K$. Since $Q_{2}\unlhd Q^{\prime}$, $Q_{2}^{c}\unlhd Q^{\prime}$. Recall that $Q^{\prime}=Q_{1}\times Q_{2}$ is an elementary abelian 3-group, so $Q_{2}^{c}\cap Q_{1}\neq 1$ or $Q_{2}^{c}\cap Q_{2}\neq 1$. Now suppose that $(0,\beta,0)\in Q_{2}^{c}\cap Q_{2}$, since $Q_{2}^{c}\cap Q_{2}\leq Q_{2}$, we have $(0,\beta,0)^{-1}=(0,-\beta,0)\in Q_{2}^{c}\cap Q_{2}$. This, together with $K\leq H$ and $K\leq H_{2}$, implies $(0,\beta,0)^{K}\cup(0,-\beta,0)^{K}=Q_{2}\setminus\{1\}=Q_{2}^{c}\setminus\{1\}$. Hence $Q_{2}^{c}=Q_{2}$, a contradiction. Similarly, if $Q_{2}^{c}\cap Q_{1}\neq 1$, we have $Q_{2}^{c}=Q_{1}$, a contradiction. $\hfill\square$

Lemma 2.19

*Suppose that $H\leq{{}^{2}G_{2}(q)}$ and $|H|=q(q-1)$. Then $H$ is conjugate to $H_{1}=Q_{1}:K$ or $H_{2}=Q_{2}:K$, and there are only two conjugacy classes of subgroups of order $q(q-1)$ in ${}^{2}G_{2}(q)$. *

Proof.  Let $H\leq{{}^{2}G_{2}(q)}$ and $|H|=q(q-1)$. By Lemma 2.14, $H$ must be contained in a conjugacy of $M_{1}$ or $M_{2}$. Firstly, if $H^{g^{-1}}\leq M_{1}$, then by Corollary 1, $H^{g^{-1}}=A:K$ where $A$ is a Sylow 3-subgroup of $H^{g^{-1}}$. We now show that $A\leq Q^{\prime}$. Assume that $F$ is a maximal subgroup of $Q$ such that $A\leq F$. If $A\cap Q^{\prime}=1$, then by Lemma 2.5 and the fact $Q^{\prime}\leq F$, we have $|F:A|\geq|F\cap Q^{\prime}:A\cap Q^{\prime}|=q^{2},$ and so $|F|\geq q^{3}$, a contradiction. Therefore, there exists an element $(0,\beta,\gamma)\in A\cap Q^{\prime}$, which implies that $A\setminus\{1\}=(0,\beta,\gamma)^{K}\subseteq Q^{\prime}\setminus\{1\}$ and hence $A\leq Q^{\prime}$. It follows that $A\cap Q_{1}\neq 1$ or $A\cap Q_{2}\neq 1$. Similar to the proof of Lemma 2.18, if $A\cap Q_{1}\neq 1$, then $A=Q_{1}$ and so $H^{g^{-1}}=H_{1}$, and if $A\cap Q_{2}\neq 1$, then $A=Q_{2}$ and so $H^{g^{-1}}=H_{2}$. Secondly, if $H$ contained in a conjugacy of $M_{2}$, then $H$ is conjugate to $H_{2}$ by Lemma 2.18(2).$\hfill\square$

Lemma 2.20

*Let $H\leq{{}^{2}G_{2}(q)}$ and $|H|=q^{2}(q-1)$. Then $H$ is conjugate to $Q^{\prime}:K$, and there are only one conjugacy class of subgroups of order $q^{2}(q-1)$ in ${}^{2}G_{2}(q)$. *

Proof.  Since $Q^{\prime}~{}char~{}Q\unlhd M_{1}$, so $Q^{\prime}:K$ is a subgroup of $M_{1}$ with order $q^{2}(q-1)$. Suppose that $H\leq{{}^{2}G_{2}(q)}$ and $|H|=q^{2}(q-1)$. By Lemma 2.14, we have $H^{g^{-1}}\leq M_{1}$. Similarly as the proof of Corollary 1, we get that $H^{g^{-1}}$ has the structure $A:K$ where $A$ is the Sylow 3-subgroup of $H^{g^{-1}}$. Let $F$ be a maximal subgroup of $Q$ satisfying $A\leq F$. Since $|F:A|\geq|F\cap Q_{i}:A\cap Q_{i}|$, we have $|A\cap Q_{i}|>1$, which implies $Q_{i}=Q_{i}^{K}\leq A^{K}=A$ for $i=1,2$. So $Q^{\prime}\leq A$, and it follows that $Q^{\prime}=A$ and $H^{g^{-1}}=Q^{\prime}:K$ in $M_{1}$. $\hfill\square$

Similarly, we have the following result on the Suzuki group ${}^{2}B_{2}(q)$ by [9] and [7, p.250].

Lemma 2.21

*Suppose that $Q$ is the Sylow $2$-subgroup of ${}^{2}B_{2}(q)$ and $M_{1}=Q:K$ is the normalizer of $Q$. Let $H\leq{{}^{2}B_{2}(q)}$ and $|H|=q(q-1)$. Then $H$ is conjugate to $Z(Q):K$. There exists a unique conjugacy class of subgroups of order $q(q-1)$ in ${}^{2}B_{2}(q)$. *

3 Proof of Theorem 1

3.1 T is the Ree group

Proposition 3.1

Suppose that $G$ and $\mathcal{D}$ satisfy hypothesis of Theorem 1. Let $B$ be a block. If $T={}^{2}G_{2}(q)$ with $q=3^{2n+1}$, then $\cal D$ is the Ree unital or one of the following:

1. (1)

$\mathcal{D}$* is a $2$-$(q^{3}+1,q,q-1)$ design with $G_{B}=Q_{1}:K$ or $Q_{2}:K$;* 2. (2)

$\mathcal{D}$* is a $2$-$(q^{3}+1,q^{2},q^{2}-1)$ with $G_{B}=Q^{\prime}:K$.*

This proposition will be proved into two steps. We first assume that there exists a design satisfying the assumptions and obtain the possible parameters $(v,b,r,k,\lambda)$ in Lemma 3.1, then prove the existence of the designs using Lemma 2.12.

Lemma 3.1

*Suppose that $G$ and $\mathcal{D}$ satisfy the hypothesis of Theorem 1. If $T={}^{2}G_{2}(q)$ with $q=3^{2n+1}$, then $(v,b,r,k,\lambda)=(q^{3}+1,q^{2}(q^{3}+1),q^{3},q,q-1)$ or $(q^{3}+1,q(q^{3}+1),q^{3},q^{2},q^{2}-1)$ or $\cal D$ is the Ree unital. *

Proof.  Let $T_{\alpha}:=G_{\alpha}\cap T$. Since $G$ is primitive on $\mathcal{P}$, then $T_{\alpha}$ is one of the cases in Lemma 2.14 by [13]. First, the cases that $T_{\alpha}=\mathbb{Z}_{2}^{2}\times D_{(q+1)/2}$ and $\mathbb{Z}_{q\pm m+1}:\mathbb{Z}_{6}$ are impossible by Lemma 2.8. If $T_{\alpha}=\mathbb{Z}_{2}\times L_{2}(q)$, then $v=q^{2}(q^{2}-q+1)$ and $(|G_{\alpha}\cap T|,v-1)=(q(q^{2}-1),q^{4}-q^{3}+q^{2}-1)=q-1$. But since $r$ divides $f(|G_{\alpha}\cap T|,v-1)$, which is too small to satisfy $v<r^{2}$. Similarly, $T_{\alpha}$ cannot be ${{}^{2}G_{2}(3^{\ell})}$.

We next assume that $T_{\alpha}=Q:K$, and so $v=q^{3}+1$. Moreover, from [7, p.252], $T$ is 2-transitive on $\mathcal{P}$, so $T$ is flag-transitive by Lemma 2.4. Hence we may assume that $G=T={}^{2}G_{2}(q)$. The equations in Lemma 2.1 show

[TABLE]

then by the flag-transitivity of $T$, we have

[TABLE]

Let $M$ be a maximal subgroup of $T$ such that $T_{B}\leq M$. Then since $|T_{B}|\mid|M|$ and $q\geq 27$, $M$ must be $M_{1}$ or $M_{2}$ shown in Lemma 2.14.

If $T_{B}\leq M_{1}$, then $k(k-1)\mid\lambda q^{3}$. Furthermore, since $(r,\lambda)=1$ and so $\lambda\mid(k-1)$ by Lemma 2.1(2). Therefore $\lambda=k-1$, and it follows that $r=v-1=q^{3}$ and $k\mid q^{3}$. Note that $M_{1}$ is point stabilizer of $T$ in this action. So there exists $\alpha$ such that $M_{1}=T_{\alpha}$ and $T_{B}\leq T_{\alpha}$. However, the flag-transitivity of $T$ implies $\alpha\notin B$. For any point $\gamma\in B$, $T_{\gamma B}\leq T_{\alpha\gamma}$. By Lemma 2.13, $|T_{\alpha\gamma}|=q-1$, and so $|T_{\gamma B}|\mid(q-1)$. On the other hand, from

[TABLE]

we have $T_{\gamma B}=T_{\alpha\gamma}$ and so $B^{T_{\alpha\gamma}}=B$. Since the stabilizer of three points is of order 2 by Lemma 2.13, so the size of $T_{\alpha\gamma}$-orbits acting on $\mathcal{P}\setminus\{\alpha,\gamma\}$ is $q-1$ or $\frac{1}{2}(q-1)$. This , together with $B^{T_{\alpha\gamma}}=B$ and $\alpha\notin B$, implies that $k-1=a\frac{(q-1)}{2}$ for an integer $a$. Recall that $k\mid q^{3}$ and $k<r$, we get $k=q$ or $k=q^{2}$. If $k=q$, then

[TABLE]

If $k=q^{2}$, we have

[TABLE]

Now we deal with the case that $T_{B}\leq M_{2}$ by the similar method in [12, Theorem 3.2].

If $T_{B}$ is a solvable subgroup of $M_{2}\cong\mathbb{Z}_{2}\times L_{2}(q)$, then $T_{B}$ must map into either $\mathbb{Z}_{2}\times A_{4}$, $\mathbb{Z}_{2}\times D_{q\pm 1}$ or $\mathbb{Z}_{2}\times([q]:\mathbb{Z}_{\frac{q-1}{2}})$. Obviously, the former two cases are impossible. For the last case, $T_{B}\lesssim\mathbb{Z}_{2}\times([q]:\mathbb{Z}_{\frac{q-1}{2}})$. Since $T_{B}\leq M_{2}$, by Lemma 2.18, this can be reduced to the case $T_{B}\leq M_{1}$.

If $T_{B}$ is non-solvable, then it embeds in $\mathbb{Z}_{2}\times L_{2}(q_{0})$ with $q_{0}^{\ell}=q=3^{2n+1}$. The condition that $|T_{B}|$ divides $|\mathbb{Z}_{2}\times L_{2}(q_{0})|$ forces $q_{0}=q$ and so $T_{B}$ is isomorphic to $\mathbb{Z}_{2}\times L_{2}(q)$ or $L_{2}(q)$.

If $T_{B}\cong\mathbb{Z}_{2}\times L_{2}(q)$, then $T_{B}=M_{2}$ and so $b=q^{2}(q^{2}-q+1)$. Hence, from Lemma 2.1, we have $k\mid q(q+1),~{}q^{2}\mid r$ and $r\mid q^{3}.$ Since $k\geq 3$, then the fact that the stabilizer of three points is of order 2 implies that $T_{B}$ cannot acting trivially on the block $B$. Moreover, since $q+1$ is the smallest degree of any non-trivial action of $L_{2}(q)$, we have $k=\frac{\lambda(v-1)}{r}+1\geq q+1$.

If the design $\mathcal{D}$ is a linear space, then $\mathcal{D}$ is the Ree unital (see [12]) with parameters

[TABLE]

and $T$ is flag-transitive with the block stabilizer $M_{2}$.

If $\lambda>1$, we claim that $\lambda=k-1$. Clearly, $\lambda\mid(k-1)$ as $(r,\lambda)=1$ by Lemma 2.1(2). If $3\mid(k-1)$ and $(k,3)=1$, then since $k\mid q(q+1)$ and $k\geq q+1$, we have $k=q+1$ and so $\lambda\mid q$, which contradicts $(r,\lambda)=1$ as $q^{2}\mid r$. Hence $(k-1,3)=1$. Moreover, $(k-1)\mid\lambda q^{3}$ implies that $(k-1)\mid\lambda$. So we have $\lambda=k-1$.

Let $\Delta_{1}$, $\Delta_{2}$,$\ldots$, $\Delta_{t}$ be the orbits of $M_{2}$. Since $M_{2}$ is the block stabilizer of the Ree unital, it has an orbit of size $q+1$. Without loss of generality, suppose that $|\Delta_{1}|=q+1$. On the one hand, recall that $k\mid q(q+1)$ and $T$ is flag transitive, $T_{B}=M_{2}$ has at least one orbit with size less than $q(q+1)$. On the other hand, we show that $|\Delta_{i}|>q(q+1)$ for $i\neq 1$ in the following and we obtain the desired contradiction. Assume that $\delta\in\mathcal{P}\setminus\Delta_{1}$, we claim that $({M_{2}})_{\delta}$ is a 2-group. Let $p$ be a prime divisor of $|({M_{2}})_{\delta}|$ and $P$ be a Sylow $p$-subgroup of $({M_{2}})_{\delta}$. If $p\neq 2$ and $p\neq 3$, then since $({M_{2}})_{\delta}\leq T_{\delta}$, we have $p\mid(q-1)$. Obviously, since $\Delta_{1}$ is an orbit of $M_{2}$ and $P\leq({M_{2}})_{\delta}$, and so $P$ acts invariantly on $\Delta_{1}$ and $\mathcal{P}\setminus\Delta_{1}$. Note that the length of a $P$-orbit is either 1 or divided by $p$, so $P$ fixes at least two points in $\Delta_{1}$. Moreover, $P$ also fixes $\delta$. Therefore $P$ fixes at least three points of $\mathcal{P}$, which is impossible as the order of the stabilizer of three points is 2 by Lemma 2.13(3). If $p=3$, since $P$ fixes the point $\delta\in\mathcal{P}\setminus\Delta_{1}$ and $|\mathcal{P}\setminus\Delta_{1}|=q^{3}-q$, then $P$ fixes at least three points in $\mathcal{P}\setminus\Delta_{1}$, which is also impossible. As a result, $({M_{2}})_{\delta}$ is a 2-group. The fact that the Sylow 2-subgroup of $T$ is of order 8 implies that the sizes of the $M_{2}$-orbits $\Delta_{i}$ ($i\neq 1$) are at least $\frac{q(q^{2}-1)}{8}$ and hence larger than $q(q+1)$, which contradicts the fact $k\mid q(q+1)$. Therefore, $T_{B}\not\cong\mathbb{Z}_{2}\times L_{2}(q)$. Similarly, $T_{B}\not\cong L_{2}(q)$. Thus $T_{B}$ is not a non-solvable subgroup in $M_{2}$. $\hfill\square$

Proof of Proposition 3.1. We use Lemma 2.12 to prove the existence of the design with parameters listed in Lemma 3.1.

Assume that $(v,b,r,k,\lambda)=(q^{3}+1,q^{2}(q^{3}+1),q^{3},q,q-1)$. Then from Lemma 2.19 we known that there are only two conjugacy classes of subgroups of order $q(q-1)$ in $T$ and $H_{1}=Q_{1}:K\leq T_{\alpha}$ and $H_{2}=Q_{2}:K\leq T_{\alpha}$ as representatives, respectively.

First, we consider the orbits of $H_{1}$. Let $\gamma\neq\alpha$ be the point fixed by $K$. Since $K\leq H_{1}$, then $K_{\gamma}=K\leq(H_{1})_{\gamma}\leq T_{\alpha\gamma}=K,$ which implies ${(H_{1})}_{\gamma}=T_{\alpha\gamma}$ and so $|H_{1}:(H_{1})_{\gamma}|=|\gamma^{H_{1}}|=q$. It is easy to see that $|\delta^{H_{1}}|\neq q$ for any point $\delta\neq\alpha,\gamma$. Therefore, $H_{1}$ has only one orbit of size $q$. Let $B_{1}=\gamma^{H_{1}}$.

Now we show that $H_{1}=T_{B_{1}}$, which implies $|B_{1}^{T}|=b$. Since $H_{1}\leq T_{B_{1}}$ and $B_{1}=\gamma^{H_{1}}=\gamma^{T_{B_{1}}}$, then $|H_{1}:(H_{1})_{\gamma}|=|T_{B}:T_{\gamma B_{1}}|=q.$ If $K=(H_{1})_{\gamma}<T_{\gamma B_{1}}$, then 3 divides $|T_{\gamma B_{1}}:T_{\delta\gamma B_{1}}|$ for any $\delta\in B_{{1}}\setminus\{\gamma\}$ by Lemma 2.13(3). It follows that $3\mid(q-1)$, a contradiction. As a result, $K=(H_{1})_{\gamma}=T_{\gamma B_{1}}$ and so $H_{1}=T_{B_{1}}$. Let ${\mathcal{B}}_{1}:=B_{1}^{T}$. Therefore $|{\mathcal{B}}_{1}|=|T:H_{1}|=b$. Let ${\mathcal{B}}_{1}$ be the set of blocks.

Finally, since $T$ is 2-transitive on $\mathcal{P}$, the number of blocks which incident with two points is a constant. Hence ${\mathcal{D}}_{1}=(\mathcal{P},{\mathcal{B}}_{1})$ is a 2-$(q^{3}+1,q,q-1)$ design admitting $T$ as a flag transitive automorphism group by Lemma 2.12.

In a similar way, we get the design ${\mathcal{D}}_{2}$ satisfying all hypothesis when the subgroup is $H_{2}=Q_{2}:K$. Furthermore, since $H_{1}$ is not isomorphic to $H_{2}$, so $\mathcal{D}_{1}$ is not isomorphic to $\mathcal{D}_{2}$ by [6, 1.2.17].

Similarly , if $(v,b,r,k,\lambda)=(q^{3}+1,q(q^{3}+1),q^{3},q^{2},q^{2}-1)$, we can construct the design with these parameters. $\hfill\square$

3.2 T is the Suzuki group

Proposition 3.2

*Suppose that $G$ and $\mathcal{D}$ satisfy hypothesis of Theorem 1. If $T={}^{2}B_{2}(q)$ with $q=2^{2n+1}$, then $\cal D$ is a $2$-$(q^{2}+1,q,q-1)$ design with $G_{B}=Z({Q}):K$ where $Q\in Syl_{2}(T)$ and $K=Z_{q-1}$. *

Proof.  Suppose that $T={}^{2}B_{2}(q)$ with order $(q^{2}+1)q^{2}(q-1)$. Then $|G|=f(q^{2}+1)q^{2}(q-1)$ where $f$ divides $|Out(T)|$. By [9] or [27], the order of $G_{\alpha}$ is one of the following:

1. (1)

$fq^{2}(q-1)$; 2. (2)

$2f(q-1)$; 3. (3)

$4f(q\pm\sqrt{2q}+1)$; 4. (4)

$f(q_{0}^{2}+1)q_{0}^{2}(q_{0}-1)$ with $q_{0}^{\ell}=q$.

Since $|G|<|G_{\alpha}|^{3}$, we first have that $|G_{\alpha}|\neq 2f(q-1)$. If $|G_{\alpha}|=4f(q\pm\sqrt{2q}+1)$, from the inequality $|G|<|G_{\alpha}|^{3}$, we get $f(q^{2}+1)q^{2}(q-1)<(4f)^{3}(2q)^{3}$, and so $q^{2}+q+1\leq 4^{3}f^{2}2^{3}$. Since $f\leq|Out(T)|=e$ and $q=p^{e}$, hence $q+1<4^{3}2^{3}$ and $q=2^{7}$, $2^{5}$ or $2^{3}$. If $q=2^{7}$, then $|G|=f2^{14}(2^{14}-1)(2^{7}-1)>f^{3}4^{3}(2^{7}+2^{4}+1)^{3}=|G_{\alpha}|^{3}$ where $f=7$ or 1, a contradiction. If $q=2^{5}$, then $v=198400$ or 325376 for $|G_{\alpha}|=4f(q+\sqrt{2q}+1)$ or $4f(q-\sqrt{2q}+1)$ respectively. By calculating $(|G_{\alpha}|,v-1)$, since $r$ divides $(|G_{\alpha}|,v-1)$, we know that $r$ is too small. Similarly, we get $q\neq 2^{3}$.

If $|G_{\alpha}|=f(q_{0}^{2}+1)q_{0}^{2}(q_{0}-1)$ with $q_{0}^{\ell}=q$, then the inequality $|G|<|G_{\alpha}||G_{\alpha}|_{p^{\prime}}^{2}$ forces $m=3$. So $v=(q_{0}^{4}-q_{0}^{2}+1)q_{0}^{4}(q_{0}^{2}+q_{0}+1)$. Since $r$ divides $(|G_{\alpha}|_{p^{\prime}},v-1)$, then $r\leq|G_{\alpha}|_{p^{\prime}}\leq fq_{0}^{3}<q_{0}^{9/2}$. From $v<r^{2}$, we get $(q_{0}^{4}-q_{0}^{2}+1)q_{0}^{4}(q_{0}^{2}+q_{0}+1)<r^{2}<q_{0}^{9}$, which is impossible.

Now assume that $|G_{\alpha}|=fq^{2}(q-1)$. Then $v=q^{2}+1$ and $T$ is 2-transitive by [7, p.250]. Hence, $T$ is flag-transitive by Lemma 2.4. Similarly, we have $|T_{B}|=\frac{|T|}{b}=\frac{k(k-1)(q-1)}{\lambda}.$ Let $M$ be the maximal subgroup of $T$ such that $T_{B}\leq M$ as in Lemma 3.1. The fact that $|T_{B}|$ divides $|M|$ implies that $|M|=q^{2}(q-1)$ and $k(k-1)$ divides $\lambda q^{2}$. Similar to the proof of Lemma 3.1, we have $T_{\gamma B}=T_{\alpha\gamma}$ with the order $q-1$. Furthermore, we get

[TABLE]

Next we prove the existence of the design with above parameters by Lemma 2.12. Firstly, from Lemma 2.21 we know that the Suzuki group has a unique conjugacy class of subgroups of order $q(q-1)$, let $H:=Z(Q):K\leq T_{\alpha}$ as the representative.

Note that $K$ is the stabilizers of two points in ${}^{2}B_{2}(q)$ by [11, p.187]. Let $\gamma\neq\alpha$ be the point fixed by $K$ and $B=\gamma^{H}$. Then similar as the proof of Proposition 3.1 we get that $B$ is the only $H$-orbit of length $q$ and $H=T_{B}$. Let $\mathcal{B}=B^{T}$ be the set of blocks. Finally, since $T$ is 2-transitive on $\mathcal{P}$, the number of blocks which incident with two points is a constant. Hence $\mathcal{D}=(\mathcal{P},\mathcal{B})$ is a 2-$(q^{2}+1,q,q-1)$ design admitting $T$ be a flag transitive automorphism group by Lemma 2.12. $\hfill\square$

3.3 T is one of the remaining families

In this subsection, let

[TABLE]

we will prove that there are no new design arise when $T\in\mathcal{T}$.

First, we show that $G_{\alpha}$ cannot be a parabolic subgroup of $G$ for any $T\in\mathcal{T}$.

Lemma 3.2

*Suppose that $G$ and $\mathcal{D}$ satisfy hypothesis of Theorem 1. If $T\in\mathcal{T}$, then $G_{\alpha}$ cannot be a parabolic subgroup of $G$. *

Proof.  By Lemma 2.6, for all cases that $T\in\mathcal{T}\setminus E_{6}(q)$, there is a unique subdegree which is a power of $p$, so $r$ is a power of $p$ by Lemma 2.1(4). We can easily check that $r$ is too small and the condition $r^{2}>v$ cannot be satisfied. Now, assume that $T=E_{6}(q)$. If $G$ contains a graph automorphism or $G_{\alpha}\cap T$ is $P_{2}$ or $P_{4}$, then there is also a unique subdegree which is a power of $p$ and so $r$ is too small again. If $G_{\alpha}\cap T$ is $P_{3}$ with type $A_{1}A_{4}$, then

[TABLE]

Since $r$ divides $(|G_{\alpha}|,v-1)$, we have $r\mid eq(q-1)^{5}(q^{5}-1)$ and so $r$ is too small to satisfy $r^{2}>v$. If $G_{\alpha}\cap T$ is $P_{1}$ with type $D_{5}$, then

[TABLE]

From [16], we know that there exists two non-trivial subdegrees:

[TABLE]

Since $(d,d^{\prime})=q(q^{4}+1)$, we have $r\mid q(q^{4}+1)$ by Lemma 2.1(4), which contradicts with $r^{2}>v$.$\hfill\square$

Let $\mathcal{T}_{1}=\{F_{4}(q),E_{6}^{\epsilon}(q),E_{7}(q),E_{8}(q)\}.$

Lemma 3.3

*Suppose that $G$ and $\mathcal{D}$ satisfy the hypothesis of Theorem 1. If $T\in\mathcal{T}_{1}$ and $G_{\alpha}$ is non-parabolic, then $G_{\alpha}$ cannot be a maximal subgroup of maximal rank. *

Proof.  If $G_{\alpha}$ is non-parabolic and of maximal rank, then for any $T\in\mathcal{T}_{1}$, we have a complete list of $T_{\alpha}:=G_{\alpha}\cap T$ in [18, Tables 5.1-5.2]. All subgroups in [18, Table 5.2] and some cases in [18, Table 5.1] can be ruled out by the inequality $|T|<|Out(T)|^{2}|T_{\alpha}|{|T_{\alpha}|_{p^{\prime}}^{2}}$. Since $r$ divides $(|G_{\alpha}|,v-1)$, for the remaining cases we have that $r^{2}<v$, a contradiction.

For example, if $T=F_{4}(q)$ with order $q^{24}(q^{2}-1)(q^{6}-1)(q^{8}-1)(q^{12}-1)$. Then $T_{\alpha}$ is one of the following: (1) $2.(L_{2}(q)\times PSp_{6}(q)).2$ ($q$ odd); (2) $d.\Omega_{9}(q)$; (3) $d^{2}.P\Omega^{+}_{8}(q).S_{3}$; (4) ${}^{3}D_{4}(q).3$; (5) $Sp_{4}(q^{2}).2$ ($q$ even); (6) $(Sp_{4}(q)\times Sp_{4}(q)).2$($q$ even); (7) $h.(L^{\epsilon}_{3}(q)\times L^{\epsilon}_{3}(q)).h.2$, with $d=(2,q-1)$ and $h=(3,q-\epsilon)$.

If $T_{\alpha}=2.(L_{2}(q)\times PSp_{6}(q)).2$ with $q$ odd, then

[TABLE]

Since $(q^{2}+1)\mid v$ and $(q^{4}+q^{2}+1)\mid v$, then $(|G_{\alpha}|,v-1)\mid|Out(T)|(q^{2}-1)^{4}$ and so $r^{2}<q^{9}<v$, a contradiction.

If $T_{\alpha}=2.P\Omega_{9}(q)$ with $q$ odd, then

[TABLE]

Since $q\mid v$, $(q^{4}+q^{2}+1)\mid v$, $v-1\equiv 2{\pmod{q^{4}-1}}$, we get $r$ divides $2^{4}|Out(T)|(q^{4}+1)$ and so $r^{2}<v$, a contradiction.

Cases (3)-(6) can be ruled out similarly, and Case (7) cannot occur because of $|T|<|Out(T)|^{2}|T_{\alpha}|{|T_{\alpha}|_{p^{\prime}}^{2}}$. $\hfill\square$

Lemma 3.4

*Suppose that $G$ and $\mathcal{D}$ satisfy the hypothesis of Theorem 1. If $T\in\mathcal{T}_{1}$ and $G_{\alpha}$ is non-parabolic, then $T_{0}=Soc(G_{\alpha}\cap T)$ is simple and $T_{0}=T_{0}(q_{0})\in Lie(p)$. *

Proof.  Assume that $T_{0}=Soc(G_{\alpha}\cap T)$ is not simple. Then by Lemma 2.9 and Lemma 3.3, one of the following holds:

1. (1)

$G_{\alpha}=N_{G}(E)$, where $E$ is an elementary abelian group given in [4, Theorem 1(II)]; 2. (2)

$T=E_{8}(q)$ with $p>5$, and $T_{0}$ is either $A_{5}\times A_{6}$ or $A_{5}\times L_{2}(q)$; 3. (3)

$T_{0}$ is as in Table 1.

From [4, Theorem 1(II)], we check that all subgroups in Case (1) are local and too small to satisfy $|T|<|Out(T)|^{2}|T_{\alpha}|{|T_{\alpha}|_{p^{\prime}}^{2}}$.

The order of subgroup in Case (2) is too small.

For Case (3), since $G_{\alpha}$ is not simple and not local by [4, Theorem 1], $G_{\alpha}$ is of maximal rank by [25, p.346], which has already been ruled out in Case (1). Therefore, $T_{0}$ is simple.

Now assume that $T_{0}=T_{0}(q_{0})\not\in Lie(p)$. Then for all $T$, we find the possibilities of $T_{0}$ in [21, Table 1]. Some cases can be ruled out by the inequality $|T|<|Out(T)|^{2}|T_{\alpha}|{|T_{\alpha}|_{p^{\prime}}^{2}}$. In each of the remaining cases, since $r$ must divides $(|G_{\alpha}|,v-1)$, $r$ is too small to satisfy $v<r^{2}$. For example, assume that $T=F_{4}(q)$. If $T_{0}\not\in Lie(p)$, then according to [21, Table 1], it is one of the following: $A_{5-10}$, $L_{2}(7)$, $L_{2}(8)$, $L_{2}(13)$, $L_{2}(17)$, $L_{2}(25)$, $L_{2}(27)$, $L_{3}(3)$, $U_{3}(3)$, $U_{4}(2)$, $Sp_{6}(2)$, $\Omega_{8}^{+}(2)$, ${{}^{3}D_{4}(2)}$, $J_{2}$, $J_{2}$, $A_{11}(p=11)$, $L_{3}(4)(p=3)$, $L_{4}(3)(p=2)$, ${{}^{2}B_{2}(8)}(p=5)$, $M_{11}(p=11)$. The possibilities of $T_{0}$ such that $|G|<|G_{\alpha}|^{3}$ are $A_{9}(q=2)$, $A_{10}(q=2)$, $Sp_{6}(2)(q=2)$, $\Omega_{8}^{+}(2)(q=2,3)$, ${{}^{3}D_{4}(2)}(q=2,3)$, $J_{2}(q=2)$, $L_{4}(3)(q=2)$. However, since $r\mid(|G_{\alpha}|,v-1)$, we have $r^{2}<v$ for all these cases, which is a contradiction.$\hfill\square$

Lemma 3.5

*Suppose that $G$ and $\mathcal{D}$ satisfy the hypothesis of Theorem 1. If $T_{0}=T_{0}(q_{0})$ is a simple group of Lie type and $G_{\alpha}$ is non-parabolic, then $T\not\in\mathcal{T}_{1}.$ *

Proof.  First assume that $T=F_{4}(q)$. If ${\rm rank}(T_{0})>\frac{1}{2}{\rm rank}(T)$, then by Lemma 2.10 and Lemma 3.3, the only possible cases of $G_{\alpha}\cap T$ satisfying $|G|<|G_{\alpha}|^{3}$ are $F_{4}(q^{\frac{1}{2}})$ and $F_{4}(q^{\frac{1}{3}})$ when $q_{0}>2$. If $G_{\alpha}\cap T=F_{4}(q^{\frac{1}{2}})$, then $v=q^{12}(q^{6}+1)(q^{4}+1)(q^{3}+1)(q+1)>q^{26}.$ Since $q$, $q+1$, $q^{2}+1$ and $q^{3}+1$ are factors of $v$, then $r\mid 2e(q-1)^{2}(q^{3}-1)^{2}$ by $r\mid(|G_{\alpha}|,v-1)$, which implies that $r^{2}<v$, a contradiction. If $G_{\alpha}\cap T=F_{4}(q^{\frac{1}{3}})$, since $p\mid v$, then $r$ divides $|G_{\alpha}|_{p^{{}^{\prime}}}$, which also implies $r^{2}<v$. When $q_{0}=2$, the subgroups $T_{0}(2)$ with ${\rm rank}(T_{0})>\frac{1}{2}{\rm rank}(T)$ that satisfy $|G|<|G_{\alpha}|^{3}$ are $A_{4}^{\epsilon}(2)$, $B_{3}(2)$, $B_{4}(2)$, $C_{3}(2)$, $C_{4}(2)$ or $D_{4}^{\epsilon}(2)$. But in each case, $r\mid(|G_{\alpha}|,v-1)$ forces $r^{2}<v$, a contradiction. If ${\rm rank}(T_{0})\leq\frac{1}{2}{\rm rank}(T)$, then from Lemma 2.11, we have $|G_{\alpha}|<4q^{20}\log_{p}q$. Looking at the orders of groups of Lie type, we see that if $|G_{\alpha}|<4q^{20}\log_{p}q$, then $|G_{\alpha}|_{p^{\prime}}<q^{12}$, and so $|G_{\alpha}||G_{\alpha}|_{p^{\prime}}^{2}<|G|$, contrary to Lemma 2.8.

For $T=E_{6}^{\epsilon}(q)$, if ${\rm rank}(T_{0})>\frac{1}{2}{\rm rank}(T)$, then when $q_{0}>2$, by Lemma 2.10 the only possibilities are $E^{\epsilon}_{6}(q^{\frac{1}{2}})$, $E^{\epsilon}_{6}(q^{\frac{1}{3}})$, $C_{4}(q)$ and $F_{4}(q)$. In all these cases $r$ are too small. When $q_{0}=2$, the possibilities $T_{0}(2)$ satisfying $|G|<|G_{\alpha}|^{3}$ with order dividing $|E_{6}^{\epsilon}(2)|$ are $A^{\epsilon}_{5}(2)$, $B_{4}(2)$, $C_{4}(2)$, $D_{4}^{\epsilon}(2)$ and $D_{5}^{\epsilon}(2)$. However, since $r\mid(|G_{\alpha}|,v-1)$, for all these cases we obtain $r^{2}<v$, a contradiction. If ${\rm rank}(T_{0})\leq\frac{1}{2}{\rm rank}(T)$, then from Lemma 2.11, we have $|G_{\alpha}|<4q^{28}\log_{p}q$. By further check the orders of groups of Lie type, we see that $|G_{\alpha}|_{p^{\prime}}<q^{17}$, and so $|G_{\alpha}||G_{\alpha}|_{p^{\prime}}^{2}<|G|$, a contradiction.

Assume that $T=E_{7}(q)$. If $rank(T_{0})\leq\frac{1}{2}rank(T)$, then by Lemma 2.11 $|G_{\alpha}|^{3}\leq|G|$, a contradiction. If ${\rm rank}(T_{0})>\frac{1}{2}{\rm rank}(T)$, then when $q_{0}>2$, B by Lemma 2.10, the only cases $T\cap G_{\alpha}$ satisfying $|G|<|G_{\alpha}|^{3}$ are $G_{\alpha}\cap T=E_{7}(q^{\frac{1}{s}})$, where $s=2$ or $3$. But in all cases we have $r^{2}<v$. If $q_{0}=2$, then the possible subgroups such that $|G|<|G_{\alpha}|^{3}$ with order dividing $|E_{7}(2)|$ are $A_{6}^{\epsilon}(2)$, $A_{7}^{\epsilon}(2)$, $B_{5}(2)$, $C_{5}(2)$, $D_{5}^{\epsilon}(2)$ and $D_{6}^{\epsilon}(2)$. However in all of these cases, since $r\mid(|G_{\alpha}|,v-1)$ we have $r^{2}<v$, a contradiction.

Assume that $T=E_{8}(q)$. If ${\rm rank}(T_{0})\leq\frac{1}{2}{\rm rank}(T)$, then by Lemma 2.11 we get $|G_{\alpha}|^{3}<|G|$, a contradiction. Therefore, ${\rm rank}(T_{0})>\frac{1}{2}{\rm rank}(T)$. If $q_{0}>2$, then Lemma 2.10 implies $G_{\alpha}\cap T=E_{8}(q^{\frac{1}{s}})$, with $s=2$ or $3$. However in both cases we get a small $r$ with $r^{2}<v$, a contradiction. If $q_{0}=2$, then ${\rm rank}(X_{0})\geq 5$. All subgroups satisfying $|G_{\alpha}|^{3}>|G|$ are $A_{8}^{\epsilon}(2)$, $B_{7}(2)$, $B_{8}(2)$, $C_{7}(2)$, $C_{8}(2)$, $D_{8}^{\epsilon}(2)$ and $D_{7}^{\epsilon}(2)$. But for all these cases we have $r^{2}<v$.$\hfill\square$

Lemma 3.6

*If $T=G_{2}(q)$ with $q=p^{e}>2$, then $G_{\alpha}$ cannot be a non-parabolic maximal subgroup of $G$. *

Proof.  Suppose that $T=G_{2}(q)$ with $q>2$ since $G_{2}(2)^{\prime}=PSU_{3}(3)$. All maximal subgroups of $G$ can be found in [13] for odd $q$ and in [3] for even $q$.

Assume that $G_{\alpha}$ be a non-parabolic maximal subgroup of $G$. First we deal with the case where $G_{\alpha}\cap T=SL_{3}^{\epsilon}(q).2$ with $\epsilon=\pm$. Then we have $v=\frac{1}{2}q^{3}(q^{3}+\epsilon 1).$ By Lemma 2.1 and [25, Section 8] we conclude that $r$ divides $\frac{(q^{3}-\epsilon 1)}{2}$ for odd $q$ (cf. [25, Section 4, Case 1, $i=1$]) and $r$ divides $(q^{3}-\epsilon 1)$ for even $q$ (cf. [25, Section 3, Case 8]). The case that $q$ odd is ruled out by $v<r^{2}$. If $q$ is even, then $r=q^{3}-\epsilon 1$. This, together with $k<r$, implies $k-1=\lambda\frac{q^{3}+\epsilon 2}{2}$, and so $\lambda=1$ or $\lambda=2$. From the result of [25] we known that $\lambda\neq 1$. If $\lambda=2$, then since $k<r$, we have $\epsilon=-$. It follows that $k=q^{3}-1$ and $r=q^{3}+1$. This is impossible by Lemma 2.3 and [24, Theorem 1].

Now, if $G_{\alpha}\cap T={{}^{2}G_{2}(q)}$ with $q=3^{2n+1}\geq 27$, then $v=q^{3}(q+1)(q^{3}-1).$ Note that $q\mid v$ and $(q^{2}-1,v-1)=1$, we have $(|G_{\alpha}|,v-1)\mid e(q^{2}-q+1)$, and it follows that $r^{2}<v$, a contradiction.

The cases that $G_{\alpha}\cap T$ is $G_{2}(q_{0})$ or $(SL_{2}(q)\circ SL_{2}(q))\cdot 2$ can be ruled out similarly.

Using the inequality $|G|<|G_{\alpha}|^{3}$ and the fact that $r$ divides $(|G_{\alpha}|,v-1)$, we find $r$ too small to satisfy $r^{2}>v$ for every other maximal subgroup. $\hfill\square$

Lemma 3.7

*If $T={{}^{2}F_{4}(q)}$, then $G_{\alpha}$ cannot be a non-parabolic maximal subgroup. *

Proof.  Let $T={{}^{2}F_{4}(q)}$ and $G_{\alpha}$ be a non-parabolic maximal subgroup of $G$. Then from the list of the maximal subgroups of $G$ in [23], there are no subgroups $G_{\alpha}$ satisfying $|G|<|G_{\alpha}|{|G_{\alpha}|_{p^{\prime}}^{2}}$, except for the case $q=2$. For the case $q=2$, $G_{\alpha}\cap T$ is $L_{3}(3).2$ or $L_{2}(25)$. However in each case, since $r$ divides $(|G_{\alpha}|,v-1)$, and so $r$ is too small.$\hfill\square$

Lemma 3.8

*If $T={{}^{3}D_{4}(q)}$, then $G_{\alpha}$ cannot be a non-parabolic maximal subgroup. *

Proof.  If $T={{}^{3}D_{4}(q)}$ and $G_{\alpha}$ is a non-parabolic maximal subgroup of $G$, then all possibilities of $G_{\alpha}\cap T$ are listed in [14]. However, for all cases, the fact that $r$ divide $(|G_{\alpha}|,v-1)$ give a small $r$ which cannot satisfy the condition $v<r^{2}$. For example, if $G_{\alpha}\cap T$ is $G_{2}(q)$ of order $q^{6}({q}^{2}-1)({q}^{6}-1)$, then $v={q}^{6}({q}^{8}+q^{4}+1)$. Since $q\mid v$ and $(q^{4}+q^{2}+1)\mid v$, then $r\mid 3e(q^{2}-1)^{2}$, which contradicts with $v<r^{2}$. $\hfill\square$

Lemma 3.9

*Suppose that $G$ and $\mathcal{D}$ satisfy the hypothesis of Theorem 1. If the socle $T\in\mathcal{T}$, then $G_{\alpha}$ cannot be a non-parabolic maximal subgroup. *

Proof.  It is follows from Lemmas 3.3$-$3.8. $\hfill\square$

Proof of Theorem 1.  Now Theorem 1 is an immediate consequence of Propositions 3.1-3.2 and of Lemmas 3.2 and 3.9.