Flag-transitive $4$-designs and $PSL(2,q)$ groups

TL;DR

This paper classifies flag-transitive 4-designs with automorphism group PSL(2,q), identifying specific parameter sets and subgroup structures, and proving non-existence for certain parameters.

Contribution

It provides a classification of flag-transitive 4-designs with PSL(2,q) automorphism groups, detailing subgroup structures and parameter conditions, extending previous design theory results.

Findings

Classifies designs with λ between 5 and 10, listing specific parameters and subgroup types.

Shows non-existence of such designs when λ exceeds 10.

Identifies particular subgroup structures associated with the designs.

Abstract

This paper considers flag-transitive $4$-$(q+1,k,\lambda)$ designs with $\lambda\geq5$ and $q+1>k>4$. Let the automorphism group of a design $\cal D$ be a simple group $G=PSL(2,q)$. Depend on the fact that the setwise stabilizer $G_B$ must be one of twelve kinds of subgroups, up to isomorphism we get the following two results. (i) If $10\geq \lambda \geq 5$, then except $(G,G_x,G_B,k,\lambda)=(PSL(2,761),{E_{761}}\rtimes {C_{380}},S_4,24,7)$ or $(PSL(2,512),{E_{512}}\rtimes {C_{511}},{D_{18}},18,8)$ undecided, $\cal D$ is a $4$-$(24,8,5)$, $4$-$(9,8,5)$, $4$-$(8,6,6)$, $4$-$(10,9,6)$, $4$-$(9,6,10)$, $4$-$(9,7,10)$, $4$-$(12,11,8)$ or $4$-$(14,13,10)$ design with $G_B=D_8$, ${E_8}\rtimes {C_7}$, $D_6$, ${E_9}\rtimes {C_4}$, $PSL(2,2)$, $D_{14}$, ${E_{11}}\rtimes {C_{5}}$ or ${E_{13}}\rtimes {C_6}$ respectively. (ii) If $\lambda>10$, ${G_B}=A_4$, $S_4$, $A_5$, $PGL(2,q_0)$($g>1$ even) or $PSL(2,q_0)$, where ${q_0}^g=q$, then there is no such design.

Equations

$${q_0}^5-2>n\lambda ({q_0}^g-2)$$

$$(q_0-2)({q_0}^2-q_0-1)({q_0}^2-q_0-3)$$

$$=({q_0}^2-2)[(q_0-2)({q_0}^2-2q_0-1)]+(q_0-2)$$

$$=(q_0-4)({q_0}^4-2)+({q_0}^3+10{q_0}^2-3q_0-14),$$

$${q_0}^9-2>n\lambda ({q_0}^g-2)$$

$$({q_0}^3-q_0-1)({q_0}^3-q_0-2)({q_0}^3-q_0-3)$$

$$=({q_0}^2-2)({q_0}^7-{q_0}^5-6{q_0}^4+{q_0}^3+12q_0-6)+(13q_0-18)$$

$$=({q_0}^4-2)({q_0}^5-3{q_0}^3-6{q_0}^2+5q_0+12)+(4{q_0}^3-18{q_0}^2-q_0+18)$$

$$=({q_0}^6-2)({q_0}^3-3q_0-6)+(3{q_0}^5+12{q_0}^4+12{q_0}^3-6{q_0}^2-17q_0-18)$$

$$=q_0({q_0}^8-2)+(-3{q_0}^7-6{q_0}^6+3{q_0}^5+12{q_0}^4+10{q_0}^3-6{q_0}^2-9q_0-6),$$

$$({q_0}^3-q_0-n)({q_0}^3-q_0-2n)({q_0}^3-q_0-3n)$$

$$=({q_0}^2-2)[{q_0}^7-{q_0}^5-6n{q_0}^4+{q_0}^3+(11n^2+1)q_0-6n]+[(11n^2+2)q_0$$

$$-6n^3-12n]$$

$$=({q_0}^3-2)[{q_0}^6-3{q_0}^4-(6n-2){q_0}^3+3{q_0}^2+(12n-6)q_0+(11n^2-12n$$

$$+3)]+[-(6n-6){q_0}^2-(11n^2-24n+12)q_0-6n^3+22n^2-24n+6]$$

$$=({q_0}^4-2)({q_0}^5-3{q_0}^3-6n{q_0}^2+5q_0+12n)+[(11n^2-7){q_0}^3-18n{q_0}^2-$$

$$(11n^2-10)q_0-6n^3+24n]$$

$$=({q_0}^5-2)({q_0}^4-3{q_0}^2-6n{q}_0+3)+[(12n+2){q_0}^4+(11n^2-1){q_0}^3-(6n$$

$$+6){q_0}^2-(11n^2+12n)q_0-6n^3+6]$$

$$=({q_0}^6-2)({q_0}^3-3q_0-6n)+[3{q_0}^5+12n{q_0}^4+(11n^2+1){q_0}^3-6n{q_0}^2-$$

$$(11n^2+6)q_0-6n^3-12n]$$

$$=({q_0}^7-2)({q_0}^2-3)+[-6n{q_0}^6+3{q_0}^5+12n{q_0}^4+(11n^2-1){q_0}^3-(6n-2)$$

$${q_0}^2-11n^2{q}_0-6n^3-6]$$

$$=q_0({q_0}^8-2)+[-3{q_0}^7-6n{q_0}^6+3{q_0}^5+12n{q_0}^4+(11n^2-1){q_0}^3-6n{q_0}^2-$$

$$(11n^2-2)q_0-6n^3],$$

Full text

Flag-transitive $4$-designs and $PSL(2,q)$ groups

Huili Dong111 Corresponding author:011141@htu.edu.cn

*College of Mathematics and Information Science, Henan Normal University,

Xinxiang, Henan 453007, P. R. China

Abstract This paper considers flag-transitive $4$-$(q+1,k,\lambda)$ designs with $\lambda\geq 5$ and $q+1>k>4$. Let the automorphism group of a design $\cal D$ be a simple group $G=PSL(2,q)$. Depend on the fact that the setwise stabilizer $G_{B}$ must be one of twelve kinds of subgroups, up to isomorphism we get the following two results. (i) If $10\geq\lambda\geq 5$, then except $(G,G_{x},G_{B},k,\lambda)=(PSL(2,761),{E_{761}}\rtimes{C_{380}},S_{4},24,7)$ or $(PSL(2,512),{E_{512}}\rtimes{C_{511}},{D_{18}},18,8)$ undecided, $\cal D$ is a $4$-$(24,8,5)$, $4$-$(9,8,5)$, $4$-$(8,6,6)$, $4$-$(10,9,6)$, $4$-$(9,6,10)$, $4$-$(9,7,10)$, $4$-$(12,11,8)$ or $4$-$(14,13,10)$ design with $G_{B}=D_{8}$, ${E_{8}}\rtimes{C_{7}}$, $D_{6}$, ${E_{9}}\rtimes{C_{4}}$, $PSL(2,2)$, $D_{14}$, ${E_{11}}\rtimes{C_{5}}$ or ${E_{13}}\rtimes{C_{6}}$ respectively. (ii) If $\lambda>10$, ${G_{B}}=A_{4}$, $S_{4}$, $A_{5}$, $PGL(2,q_{0})$($g>1$ even) or $PSL(2,q_{0})$, where ${q_{0}}^{g}=q$, then there is no such design.

Keywords $4$-Design, flag-transitive, $PSL(2,q)$

MSC(2000) Subject Classification 05B05, 05B25, 20B25

1 Introduction

A $t$-$(v,k,\lambda)$ design with a simple group $G=PSL(2,q)$ as an automorphism group is interesting. Many classification results about flag-transitive designs have been got. In 1986, $2$-$(v,k,1)$ designs are classified by Delandtsheer[1]. In 2016, $2$-$(v,k,\lambda)$ symmetric designs with the sole of $G$ equal to $PSL(2,q)$ are studyed by Alavi, Bayat and Daneshkhah[2]. In 2018, all nonsymmetric $2$-designs with $(r,\lambda)=1$ are determined by Zhan and Zhou[3]. In 2019, $2$-designs with $k=4$ are considered by Zhan, Ding and Bai[4]. For $3$-designs, there are also many results given in[5, 6, 7, 8]. After Dai and Li[9, 10] classified flag-transitive $4$-designs with $\lambda=3$ or 4, we continue this work, and get the following two theorems:

Theorem 1.1**.**

Let $\cal D$ be a flag-transitive $4$-$(q+1,k,\lambda)$ design with $10\geq\lambda\geq 5$ and $q+1>k>4$, $G=PSL(2,q)$ be an automorphism simple group. Then up to isomorphism except $(G,G_{x},G_{B},k,\lambda)=(PSL(2,761),{E_{761}}\rtimes{C_{380}},S_{4},24,7)$ or $(PSL(2,512),{E_{512}}\rtimes{C_{511}},{D_{18}},18,8)$ undecided, $\cal D$ is a $4$-$(24,8,5)$, $4$-$(9,8,5)$, $4$-$(8,6,6)$, $4$-$(10,9,6)$, $4$-$(9,6,10)$, $4$-$(9,7,10)$, $4$-$(12,11,8)$ or $4$-$(14,13,10)$ design with $G_{B}=D_{8}$, ${E_{8}}\rtimes{C_{7}}$, $D_{6}$, ${E_{9}}\rtimes{C_{4}}$, $PSL(2,2)$, $D_{14}$, ${E_{11}}\rtimes{C_{5}}$ or ${E_{13}}\rtimes{C_{6}}$ respectively.

Theorem 1.2**.**

Let $\cal D$ be a flag-transitive $4$-$(q+1,k,\lambda)$ design with $\lambda>10$ and $q+1>k>4$, $G=PSL(2,q)$ be a simple automorphism group of $\cal D$, ${G_{B}}=A_{4}$, $S_{4}$, $A_{5}$, $PGL(2,q_{0})$($g>1$ even) or $PSL(2,q_{0})$, where ${q_{0}}^{g}=q$. Then there is no such design.

2 Preliminaries

We give some useful results. The notations $n$, $G_{B}$ and $G_{xB}$ denote $\gcd(2,q-1)$, the setwise stabilizer and $G_{x}\cap G_{B}$ respectively, where $x\in B$.

Lemma 2.1**.**

Let $\cal D$ be a $4$-$(q+1,k,\lambda)$ design with an automorphism group $G=PSL(2,q)$ flag-transitively acting on $\cal D$. Then

(i) $\lambda(q-2)=\frac{k(k-1)(k-2)(k-3)}{n|G_{B}|}$;

(ii) $q=\frac{(k-1)(k-2)(k-3)}{\lambda n|G_{xB}|}+2$;

(iii) $k\mid(\lambda n(q-2)|G_{xB}|+6)$;

(iv) $k\mid\gcd(\frac{q(q^{2}-1)}{n},\lambda n(q-2)|G_{xB}|+6)$.

Proof.  Since $G$ is flag-transitive, then by [11] $G$ is $2$-transitive, therefore, $G$ is also block-transitive and point-primitive. From $|G|=|G_{B}||B^{G}|=|G_{xB}||(x,B)^{G}|$, we get $b=\frac{|G|}{|G_{B}|}=\frac{|G|}{|G_{xB}|k}$. By [12], $\frac{\lambda q(q+1)(q-1)(q-2)}{k(k-1)(k-2)(k-3)}=\frac{q(q^{2}-1)}{n|G_{B}|}=\frac{q(q^{2}-1)}{nk|G_{xB}|}$. Obviously (i) and (ii) hold. Since $(k-1)(k-2)(k-3)=(k^{2}-6k+11)k-6$, $k\mid|G_{B}|$ and $|G_{B}|\mid|G|$, (iii) and (iv) hold.

3 Proof of two theorems

First we assume that $\cal D$ is a $4$-$(q+1,k,\lambda)$ design with $\lambda\geq 5$ and $q+1>k>4$, $G=PSL(2,q)$ is a flag-transitive automorphism group of $\cal D$, where $q=p^{f}\geq 4$ and $p$ is a prime. Since $\cal D$ is flag-transitive, we get that $k$ is one orbit length of $G_{B}$ acting on $\cal D$. By [10, 13], up to conjugation $G_{B}$ must be one of twelve kinds of subgroups, then several results can be got.

Lemma 3.1**.**

$G_{B}\neq A_{4}$, $S_{4}$ or $A_{5}$ except $(G,G_{x},G_{B},k,\lambda)=(PSL(2,761),$ ${E_{761}}\rtimes{C_{380}},S_{4},24,7)$ undecided.

Proof.  Assume that $G_{B}=A_{4}$, $S_{4}$ or $A_{5}$, then $|G_{B}|=12$, $24$ or 60. Since $k\mid|G_{B}|$, we have $(|G_{B}|,k)=(12,12)$, $(12,6)$, $(24,24)$, $(24,12)$, $(24,8)$, $(24,6)$, $(60,60)$, $(60,30)$, $(60,20)$, $(60,15)$, $(60,12)$, $(60,10)$, $(60,6)$ or $(60,5)$. Note that $q>k-1$, $q$ is a power of a prime and $\lambda\geq 5$, by Lemma 2.1(i) and (ii), we get that $(|G_{B}|,k,q,\lambda)=(12,12,32,33)$, $(12,12,13,45)$, $(12,12,17,33)$, $(12,12,47,11)$, $(12,12,101,5)$, $(12,6,8,5)$, $(24,24,25,231)$, $(24,24,71,77)$, $(24,24,79,69)$, $(24,24,163,33)$, $(24,24,233,23)$, $(24,24,761,$ $7)$, $(24,8,16,5)$, $(24,8,9,5)$,$(60,60,61,1653)$, $(60,60,89,1121)$, $(60,60,179,$ $551)$, $(60,60,1123,87)$, $(60,30,128,87)$, $(60,30,31,189)$, $(60,30,89,63)$, $(60,$ $30,263,21)$, $(60,30,191,29)$, $(60,20,53,19)$, $(60,20,59,$ $17)$, $(60,15,16,39)$, $(60,15,23,13)$, $(60,15,41,7)$, $(60,12,13,9)$ or $(60,10,16,$ $6)$. By the lengths of orbits listed in [10, 13], we need consider $(|G_{B}|,k,q,\lambda)=(12,12,17,33)$, $(12,12,47,11)$, $(12,12,101,5)$, $(24,24,71,77)$, $(24,24,79,69)$, $(24,24,233,2$ $3)$, $(24,24,761,7)$, $(60,60,89,1121)$, $(60,60,179,551)$ or $(60,30,89,63)$. Using Magma[14] we can rule out all cases except case 7. The possible parameters are listed in Table 1, where the notation $m^{n}$ means that the degree $m$ appears $n$ times. For example, in case 1, the group PrimitiveGroup(18,1) denotes the primitive permutation group $G=PSL(2,17)$ acting on the set $\Omega=\{1,2,\ldots,18\}$. And $G$ have two conjugacy classes of subgroups of order $12$ and the possible lengths of all orbits of $G_{B}$ acting on $\Omega$ are both $6$, 12. There are 2 possible basic blocks $B_{0}$ of designs. However, the command Design$<$4,18$|$D$>$ where ${\cal D}={B}^{G}$ returns both structures are not 4-designs, then case 1 can be ruled out. For case 7, from $|G_{x}|=\frac{|G|}{v}=289180$, we know $G_{x}={E_{761}}\rtimes{C_{380}}$. There are 64 possible basic blocks $B_{0}$ of designs. However, $b=\frac{|G|}{|G_{B}|}=9181465$ is too large to use Magma. This case is undecided.

Table 1: Possible parameters with $G_{B}=A_{4}$, $S_{4}$ or $A_{5}$

[TABLE]

Lemma 3.2**.**

$G_{B}=PGL(2,q_{0})$, where ${q_{0}}^{g}=q$ and $g>1$ even.

Proof. Assume that $G_{B}=PGL(2,q_{0})$, where ${q_{0}}^{g}=q$ and $g>1$ even. Then $k=q_{0}+1$, $q_{0}(q_{0}-1)$ or $q_{0}({q_{0}}^{2}-1)$.

If $k=q_{0}+1$, then by Lemma 2.1(i) $n\lambda({q_{0}}^{g}-2)={q_{0}}-2$. However, there is no solution of this equation.

If $k=q_{0}(q_{0}-1)$, then $q_{0}\geq 3$ and

[TABLE]

Thus $g=2$ or 4. Since

[TABLE]

we get $({q_{0}}^{2}-2)\mid({q_{0}}-2)$ or $({q_{0}}^{4}-2)\mid({q_{0}}^{3}+10{q_{0}}^{2}-3{q_{0}}-14)$. However, this is impossible.

If $k=q_{0}({q_{0}}^{2}-1)$, then

[TABLE]

Thus $g=2$, 4, 6 or 8. Since

[TABLE]

we get $({q_{0}}^{2}-2)\mid(13{q_{0}}-18)$, $({q_{0}}^{4}-2)\mid(4{q_{0}}^{3}-18{q_{0}}^{2}-{q_{0}}+18)$, $({q_{0}}^{6}-2)\mid(3{q_{0}}^{5}+12{q_{0}}^{4}+12{q_{0}}^{3}-6{q_{0}}^{2}-17{q_{0}}-18)$ or $({q_{0}}^{8}-2)\mid(3{q_{0}}^{7}+6{q_{0}}^{6}-3{q_{0}}^{5}-12{q_{0}}^{4}-10{q_{0}}^{3}+6{q_{0}}^{2}+9{q_{0}}+6)$. Therefore, $(g,{q_{0}})=(2,2)$, $(2,3)$. However, this is contrary with $q>k-1$.

Lemma 3.3**.**

Let $G_{B}=PSL(2,q_{0})$, where ${q_{0}}^{g}=q$. Then $\cal D$ is a $4$-$(9,6,10)$ design with $G_{B}=PSL(2,2)$, denoted by ${\cal D}_{1}$.

Proof. Assume tht $G_{B}=PSL(2,q_{0})$, where ${q_{0}}^{g}=q$. Then $k=q_{0}+1$, $q_{0}(q_{0}-1)$ if $g$ is even or $\frac{{q_{0}}({q_{0}}^{2}-1)}{\gcd(2,{q_{0}}-1)}$.

If $k=q_{0}+1$, then $\lambda({q_{0}}^{g}-2)=q_{0}-2$. However, there is no such $q_{0}>3$ satisfying the equation.

If $k=q_{0}(q_{0}-1)$, then $q_{0}\geq 3$ and ${q_{0}}^{5}-2>\lambda({q_{0}}^{g}-2)=({q_{0}}-2)({q_{0}}^{2}-{q_{0}}-1)({q_{0}}^{2}-{q_{0}}-3)$. Thus $g=2$ or 4. The same as in Lemma 3.2, this is impossible.

If $k=\frac{{q_{0}}({q_{0}}^{2}-1)}{\gcd(2,{q_{0}}-1)}$, then ${q_{0}}^{9}-2>{n^{4}}\lambda({q_{0}}^{g}-2)=({q_{0}}^{3}-{q_{0}}-n)({q_{0}}^{3}-{q_{0}}-2n)({q_{0}}^{3}-{q_{0}}-3n)$. Note that $q+1>k>4$, we get $8\geq g\geq 2$.

Since

[TABLE]

we get $(g,{q_{0}})=(3,2)$. Thus $(q,k,\lambda)=(8,6,10)$. The generators of $G=PSL(2,8)$ acting on the set $\Omega=\{1,2,\ldots,9\}$ are as follows: $g_{1}=(1,8)(2,4)(3,7)(5,6)$, $g_{2}=(2,7)(3,6)(4,5)(8,9)$, $g_{3}=(1,2,3,4,5,6,7)$. Then all orbits of $G_{B}$ acting on $\Omega$ are $\Omega_{1}=\{4,6,8\}$ and $\Omega_{2}=\{1,2,3,5,7,9\}$. Take $B=\Omega_{2}$ as a possible basic block of a design, then it returns a $4$-$(9,8,5)$ design ${\cal D}_{1}$. Clearly, ${\cal D}_{1}$ is also flag-transitive.

Next we consider $G_{B}$ is one of the remaining four kinds of subgroups and assume that $10\geq\lambda\geq 5$.

Lemma 3.4**.**

Let $G_{B}={C_{c}}$ or $D_{2c}$, where $c\mid\frac{q\pm 1}{n}$, then $(\lambda,|G_{B}|,c,k,q)=(5,8,4,8,23)$, $(8,18,9,18,512)$, $(10,6,3,6,8)$, $(7,8,8,8,17)$, $(7,8,4,8,17)$, $(10,14,7,7,8)$ or $(6,6,3,6,7)$.

Proof. Assume that $G_{B}={C_{c}}$ or $D_{2c}$, where $c\mid\frac{q\pm 1}{n}$. Then $k=c$ or 2c$$(|G_{B}|=2c).

Let $c\mid\frac{q+1}{n}$, then by Lemma 2.1(iii) $c\mid\gcd(\lambda n(q-2)|G_{xB}|+6,\frac{q+1}{n})$, that is, $c\mid\gcd(3\lambda n|G_{xB}|-6,\frac{q+1}{n})$. By Lemma 2.1(ii), $(\lambda,|G_{B}|,c,k,q)=(5,8,4,8,23)$, $(8,18,9,18,512)$ or $(10,6,3,6,8)$.

Let $c\mid\frac{q-1}{n}$. Assume that $(\lambda,n,|G_{B}|,k)=(6,1,2c,2c)$, then $6(2^{f}-2)=(2c-1)(2c-2)(2c-3)=2c(4c^{2}-12c+11)-6$, therefore, $3\frac{2^{f}-1}{c}=4c^{2}-12c+11$. From $3\mid(4c^{2}-12c+11)$, we have that $c=3l+1$ or $3l+2$, where $l$ is a positive integer. If $c=3l+1$, then $2^{f}=l(6l+1)(6l-1)+2$, therefore, $8\mid(l(6l+1)(6l-1)+2)$, we get $l\equiv 2(\mod 8)$. Let $l=8m+2$, then $2^{f-1}=(4m+1)(48m+13)(48m+11)+1$. Obviously, $m\neq 0$, therefore, $2^{f-1}\geq 5\cdot 61\cdot 59+1>2^{14}$. Thus $2^{14}\mid((4m+1)(48m+13)(48m+11)+1)$, however, this is impossible. Assume that $(\lambda,n,|G_{B}|,k)=(6,1,c,c)$, then $6(2^{f}-2)=(c-1)(c-2)(c-3)=c(c^{2}-6c+11)-6$, therefore, $6\frac{2^{f}-1}{c}=c^{2}-6c+11$. From $6\mid(c^{2}-6c+11)$, we have that $c=6l+1$ or $6l+5$, where $l$ is an integer. If $c=6l+1$, then $2^{f-1}=l(6l-1)(3l-1)+1$, therefore, $4\mid(l(6l-1)(3l-1)+1)$, that is, $4\mid(18l^{3}-9l^{2}+l+1)$. However, this is impossible. If $c=6l+5$, then $2^{f-1}=(3l+2)(2l+1)(3l+1)+1$, therefore, $4\mid((3l+2)(2l+1)(3l+1)+1)$, that is, $4\mid(18l^{3}+27l^{2}+13l+3)$. However, this is also impossible. Now we consider that $(\lambda,n,|G_{B}|,k)\neq(6,1,2c,2c)$ and $(6,1,c,c)$, then $c\mid\gcd(\lambda n(q-2)|G_{xB}|+6,\frac{q-1}{n})$, that is, $c\mid\gcd(\lambda n|G_{xB}|-6,\frac{q-1}{n})$. Thus $(\lambda,|G_{B}|,c,k,q)=(7,8,8,8,17)$, $(7,8,4,8,17)$, $(7,16,8,16,197)$, $(10,14,7,7,8)$, $(6,6,3,6,7)$. By employing the command Subgroups(G:OrderEqual:=b), we rule out $(\lambda,|G_{B}|,c,k,q)=(7,16,8,16,197)$.

Lemma 3.5**.**

Let $G_{B}=E_{q_{0}}$ or ${E_{q_{0}}}\rtimes{C_{c}}$, where $q_{0}\mid q$, $c\mid(q_{0}-1)$ and $c\mid(q-1)$, then $(\lambda,i,k,q)=(5,1,8,8)$, $(6,2,9,9)$, $(8,2,11,11)$ or $(10,2,13,13)$ where $k=q_{0}$ and $c=\frac{k-1}{i}$.

Proof. Let $G_{B}=E_{q_{0}}$ or ${E_{q_{0}}}\rtimes{C_{c}}$, where $q_{0}\mid q$, $c\mid(q_{0}-1)$ and $c\mid(q-1)$. Then $k=q_{0}$ or $c{q_{0}}$.

Let $k=q_{0}$, then $k\mid\gcd(\lambda n(q-2)|G_{xB}|+6,q)$, that is, $k\mid\gcd(2\lambda n|G_{xB}|-6,q)$. If $|G_{B}|=q_{0}$, then $(\lambda,|G_{B}|,k,q)=(7,8,8,32)$. If $|G_{B}|=c{q_{0}}$, then $|G_{xB}|=c$, therefore, $k\leq 2\lambda nc-6$, so $\frac{k-1}{2\lambda n}<\frac{k+6}{2\lambda n}\leq c$. From $c\mid(k-1)$, we let $c=\frac{k-1}{i}$, where $i=1$, 2, $\cdots$, $2\lambda n-1$. Then from $k\mid\gcd(\frac{2\lambda n(k-1)}{i}-6,q)$, we get $k\mid\gcd(2\lambda n+6i,q)$. All possible parameters of $(\lambda,i,k,q)$ are $(5,1,8,8)$, $(5,1,13,13)$, $(6,2,9,9)$, $(6,3,7,7)$, $(7,1,17,17)$, $(8,1,19,19)$, $(8,2,11,11)$, $(10,1,23,23)$ or $(10,2,13,13)$. By employing the command Subgroups(G:OrderEqual:=b), we rule out $(\lambda,i,k,q)=(5,1,13,13)$, $(6,3,7,$ $7)$, $(7,1,17,17)$, $(8,1,19,19)$ and $(10,1,23,23)$.

If $k=cq_{0}$, then $|G_{xB}|=1$. From $k\mid(\lambda n(q-2)+6)$, we have that ${q_{0}}\mid\gcd(2\lambda n-6,q)$. Since $c\mid(q_{0}-1)$. We get no possible parameters.

Lemma 3.6**.**

Let $G_{B}={C_{c}}$, $D_{2c}$ or $E_{q_{0}}$, ${E_{q_{0}}}\rtimes{C_{c}}$, where $c\mid\frac{q\pm 1}{n}$ or $q_{0}\mid q$, $c\mid(q_{0}-1)$ and $c\mid(q-1)$. Then except $(G,G_{x},G_{B},k,\lambda)=(PSL(2,512),{E_{512}}\rtimes{C_{511}},{D_{18}},18,8)$ undecided, up to isomorphism $\cal D$ is a $4$-$(24,8,5)$, $4$-$(9,8,5)$, $4$-$(8,6,6)$, $4$-$(10,9,6)$, $4$-$(9,6,10)$, $4$-$(9,7,10)$, $4$-$(12,11,8)$ or $4$-$(14,13,10)$ design with $G_{B}=D_{8}$, ${E_{8}}\rtimes{C_{7}}$, $D_{6}$, ${E_{9}}\rtimes{C_{4}}$, $PSL(2,2)$, $D_{14}$, ${E_{11}}\rtimes{C_{5}}$ or ${E_{13}}\rtimes{C_{6}}$ respectively.

Proof.  We consider the cases appearing in Lemma 3.4 and 3.5 and list the results in Table 2. The lengths of all orbits and the number of the designs are listed in column 5 and column 6 respectively, where twice or five times denote such lengths of orbits appearing twice or five times.

Table 2: Possible values with $G_{B}={C_{c}}$, $D_{2c}$, $E_{q_{0}}$ or ${E_{q_{0}}}\rtimes{C_{c}}$

[TABLE]

For case 2, from $|G_{x}|=\frac{|G|}{v}=261632$, we know $G_{x}={E_{761}}\rtimes{C_{380}}$. There are 28 possible basic blocks $B_{0}$ of designs. However, $b=\frac{|G|}{|G_{B}|}=7456512$ is too large to use Magma.

For the remaining cases, the same as before, we can deal with them by the same method. For example, the generators of the group $G=PSL(2,23)$ acting on the set $\Omega=\{1,2,\ldots,24\}$ are listed as follows:

$g_{1}=(1,8)(2,10)(3,16)(4,24)(5,15)(6,21)(7,23)(9,20)(11,22)(12,14)\\ (13,18)(17,19)$,

$g_{2}=(1,19,24,6)(2,13,22,20)(3,12,23,5)(4,17,8,21)(7,14,16,15)\\ (9,11,18,10)$,

$g_{3}=(1,24)(2,22)(3,23)(4,8)(5,12)(6,19)(7,16)(9,18)(10,11)(13,20)\\ (14,15)(17,21)$.

$G$ has only one conjugacy class of subgroups of order $8$ and all the orbits of $G_{B}$ acting on the set $\Omega$ are listed as follows:

$\Omega_{1}=\{1,4,6,8,17,19,21,24\},\\ \Omega_{2}=\{2,9,10,11,13,18,20,22\}$,

$\Omega_{3}=\{3,5,7,12,14,15,16,23\}$.

First take $B_{0}=\Omega_{1}$ as a possible basic block of a design. However, the structure is not 4-design. Now take $B_{0}=\Omega_{2}$ or $\Omega_{3}$ as a possible basic block of a design, then we construct two flag-transitive 4-$(24,8,5)$ designs ${\cal D}_{2}$ and ${\cal D}_{3}$. By[15], it shows that both designs are isomorphic. It need to note that the design corresponding to case 3 is isomorphic to the design ${\cal D}_{1}$.

This completes the proof of Theorem 1.1 and 1.2.

AcknowledgementsThis work was supported by the National Natural Science Foundation of China(No:11301158,11601132).