% EJC papers *must* begin with the following two lines. \documentclass[12pt]{article} \usepackage{amsfonts} \usepackage{e-jc} \specs{P2.6}{22(2)}{2015} % Please remove all other commands that change parameters such as % margins or pagesizes. % only use standard LaTeX packages % only include packages that you actually need % we recommend these ams packages \usepackage{amsthm,amsmath,amssymb} % we recommend the graphicx package for importing figures \usepackage{graphicx} % use this command to create hyperlinks (optional and recommended) \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} % use these commands for typesetting doi and arXiv references in the bibliography \newcommand{\doi}[1]{\href{http://dx.doi.org/#1}{\texttt{doi:#1}}} \newcommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{\texttt{arXiv:#1}}} % all overfull boxes must be fixed; % i.e. there must be no text protruding into the margins % declare theorem-like environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{fact}[theorem]{Fact} \newtheorem{observation}[theorem]{Observation} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{open}[theorem]{Open Problem} \newtheorem{problem}[theorem]{Problem} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{note}[theorem]{Note} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % if needed include a line break (\\) at an appropriate place in the title \title{\bf Flag-transitive non-symmetric 2-designs\\ with $(r,\lambda)=1$ and alternating socle} % input author, affilliation, address and support information as follows; % the address should include the country, and does not have to include % the street address \author{Shenglin Zhou\thanks{Supported by the NSFC (No.11471123) and the NSF of Guangdong Province (No.S2013010011928).}, Yajie Wang\\ \small School of Mathematics \\[-0.8ex] \small South China University of Technology\\[-0.8ex] \small Guangzhou, Guangdong 510640, P. R. China\\ \small\tt slzhou@scut.edu.cn\\ %\and %Forgotten Second Author \qquad Forgotten Third Author\\ %\small School of Hard Knocks\\[-0.8ex] %\small University of Western Nowhere\\[-0.8ex] %\small Nowhere, Australasiaopia\\ %\small\tt \{fsa,fta\}@uwn.edu.ao } % \date{\dateline{submission date}{acceptance date}\\ % \small Mathematics Subject Classifications: comma separated list of % MSC codes available from http://www.ams.org/mathscinet/freeTools.html} \date{\dateline{Sep 4, 2014}{Mar 25, 2015}{Apr 14, 2015}\\ \small Mathematics Subject Classifications: 05B25, 20B25} \begin{document} \maketitle % E-JC papers must include an abstract. The abstract should consist of a % succinct statement of background followed by a listing of the % principal new results that are to be found in the paper. The abstract % should be informative, clear, and as complete as possible. Phrases % like "we investigate..." or "we study..." should be kept to a minimum % in favor of "we prove that..." or "we show that...". Do not % include equation numbers, unexpanded citations (such as "[23]"), or % any other references to things in the paper that are not defined in % the abstract. The abstract will be distributed without the rest of the % paper so it must be entirely self-contained. \begin{abstract} This paper deals with flag-transitive non-symmetric 2-designs with $(r,\lambda)=1$. We prove that if $\mathcal D$ is a non-trivial non-symmetric $2$-$(v,k,\lambda)$ design with $(r,\lambda)=1$ and $G\leq Aut(\mathcal D)$ is flag-transitive with $Soc(G)=A_n$ for $n\geq 5$, then $\mathcal D$ is a $2$-$(6,3,2)$ design, the projective space $PG(3,2)$, or a $2$-$(10,6,5)$ design. % keywords are optional \bigskip\noindent \textbf{Keywords:} non-symmetric design; automorphism group; flag-transitive; alternating group \end{abstract} \section{Introduction} This paper is inspired by a paper of P. H. Zieschang \cite{ZIE} on flag-transitive 2-designs with $(r,\lambda)=1$. He proved in \cite[Theorem]{ZIE} that if $G$ is a flag-transitive automorphism group of a $2$-design with $(r,\lambda)=1$ and $T$ is a minimal normal subgroup of $G$, then $T$ is abelian, or simple and $C_G(T)=1$. From \cite[2.3.7(a)]{Dem}(see also Lemma \ref{Dem} below) we know that if $G\leq Aut(\mathcal D)$ is flag-transitive with $(r,\lambda)=1$ then $G$ acts primitively on $P$. It follows that $G$ is an affine or almost simple group. So it is possible to classify this type of designs by using the classification of finite primitive permutation groups, especially for the case of $G$ is almost simple, with alternating socle. A 2-design with $\lambda=1$ is also called a finite linear space. In 2001, A. Delandtsheer \cite{Dela01} classified flag-transitive finite linear spaces, with alternating socle, see Lemma \ref{de} below. Recently, in \cite{ZGZ}, Zhu, Guan and Zhou have classified flag-transitive symmetric designs with $(r,\lambda)=1$ and alternating socle. This paper is a continuation of \cite{Dela01,ZGZ} and a contribution to the case where $\mathcal D$ is a non-symmetric 2-design. A {\it $2$-$(v,k,\lambda)$ design} ${\mathcal D}$ is a pair $(P, \mathcal B)$ where $P$ is a $v$-set and $\mathcal B$ is a collection of $b$ $k$-subsets (called blocks) of $P$ such that each point of $P$ is contained in exactly $r$ blocks and any $2$-subset of $P$ is contained in exactly $\lambda$ blocks. The numbers $v,b, r,k,\lambda$ are parameters of the design. It is well known that \begin{align*} bk&=vr, \\ b&\geq v, \end{align*} and so $$r\geq k.$$ The complement $\overline{\cal D}$ of a $2$-$(v,k,\lambda)$ design ${\cal D}=(P, \cal B)$ is a $2$-$(v,v-k,b-2r+\lambda)$ design $(P, \overline{\cal B})$, where $\overline{\cal B}=\{P\setminus B\,|\,B\in\cal B\}$. A $2$-$(v,k,\lambda)$ design is symmetric if $b=v$ (or equivalently, $r=k$), otherwise is non-symmetric. This paper deals only with non-trivial non-symmetric designs, those with $2 < kv \,\, \mbox{and}\,\, r>k.$$ An {\it automorphism} of ${\mathcal D}$ is a permutation of $P$ which leaves $\mathcal B$ invariant. The full automorphism group of $\mathcal D$, denoted by $Aut({\mathcal D})$, is the group consisting of all automorphisms of $\mathcal D$. A $flag$ of ${\mathcal D}$ is a point-block pair $(x, B)$ such that $x\in B$. For $G\leq Aut(\mathcal D)$, $G$ is called {\it flag-transitive} if $G$ acts transitively on the set of flags, and {\it point-primitive} if $G$ acts primitively on $P$. A design $\cal D$ is antiflag transitive if $G\leq Aut(\mathcal D)$ acts transitively on the set $\{(x,B)\, |\, x\notin B\}\subseteq { P}\times \mathcal{B}$ of antiflags of $\cal D$. It is easily known that $G\leq Aut(\cal D)$ is flag-transitive on $\cal D$ if and only if $G$ is antiflag transitive on $\overline{\cal D}$. Flag-transitivity is one of many conditions that can be imposed on the automorphism group $G$ of a design $\mathcal D$. Lots of work have been done on flag-transitive symmetric 2-designs, see \cite{PraegerZhou,Re05,TianZhou,ZhouDong10a}, for example. Although there exists large families of non-symmetric 2-designs, less is known when $\mathcal D$ is non-symmetric admitting a flag-transitive automorphism group. The aim of this paper is to classify the flag-transitive non-symmetric 2-designs with $(r,\lambda)=1$, whose automorphism group is almost simple with an alternating group as socle. This can be viewed as a first step towards a classification of non-symmetric 2-designs with $(r,\lambda)=1$. The main result of this paper is the following. \begin{theorem}\label{theorem 1} Let $\mathcal D$ be a non-symmetric $2$-$(v,k,\lambda)$ design with $(r,\lambda)=1$, where $r$ is the number of blocks through a point. If $G\leq Aut(\mathcal D)$ is flag-transitive with alternating socle, then up to isomorphism $({\mathcal D}, G)$ is one of the following: \begin{enumerate} \item[\rm(i)] \, $\cal D$ is a unique $2$-$(15,3,1)$ design and $G=A_7$ or $A_8$. \item[\rm(ii)] \, $\cal D$ is a unique $2$-$(6,3,2)$ design and $G=A_5$. \item[\rm(iii)] \, $\cal D$ is a unique $2$-$(10,6,5)$ design and $G=A_6$ or $S_6$. \end{enumerate} \end{theorem} This, together with \cite[Theorem 1.1]{ZGZ}, yields the following. \begin{corollary} If $\mathcal D$ is a $2$-$(v,k,\lambda)$ design with $(r,\lambda)=1$, which admits a flag-transitive automorphism group $G$ with alternating socle, then $\mathcal D$ is a $2$-$(6,3,2)$ design, a $2$-$(10,6,5)$ design, the projective space $PG(3,2)$ or $PG_2(3,2)$. \end{corollary} The paper is organized as follows. In Section 2, we introduce some preliminary results that are important for the remainder of the paper. In Section 3, we complete the proof of Theorem \ref{theorem 1} in three parts. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Preliminaries} \begin{lemma}\label{Dem}{\rm(\cite[2.3.7(a)]{Dem})} Let ${\mathcal D}$ be a $2$-$(v,k,\lambda)$ design with flag-transitive automorphism group $G$. If $(r,\lambda)=1$ then $G$ is point-primitive. \end{lemma} The following result due to A. Delandtsheer \cite{Dela01} gives the classification of flag-transitive finite linear spaces with alternating socle. \begin{lemma}\label{de} Let $\mathcal S$ be a finite non-trivial linear space having an automorphism group $G$ which acts flag-transitively on $\mathcal S$. If $A_n\trianglelefteq G\leq Aut(A_n)$ with $n\geq 5$, then ${\mathcal S}=PG(3,2)$ and $G\cong A_7$ or $A_8\cong PSL_4(2)$. \end{lemma} \begin{lemma}\label{k1} Let ${\mathcal D}$ be a $2$-$(v, k, \lambda)$ design. Then \begin{enumerate} \item[\rm(i)] \, $bk(k-1)=\lambda v (v-1)$. \item[\rm(ii)] \, $r=\frac{\lambda(v-1)}{k-1}$. In particular, if $(r,\lambda)=1$ then $r\mid v-1$ and $(r,v)=1$. \end{enumerate} \end{lemma} {\bf Proof}. Counting in two ways triples $(\alpha,\beta,B)$, where $\alpha$ and $\beta$ are distinct points and $B$ is a block incident with both of them, gives (i). Part (ii) follows from the basic equation $bk=vr$ and Part (i). $\hfill\square$ \begin{lemma}\label{k2} If ${\mathcal D}$ is a non-symmetric $2$-$(v, k, \lambda)$ design and $G$ is a flag-transitive point-primitive automorphism group of $\mathcal D$, then \begin{enumerate} \item[\rm(i)] \, $r^2>\lambda v$, and $|G_x|^3>\lambda |G|$, where $x\in P$; \item[\rm(ii)] \, $r\mid \lambda d_i$, where $d_i$ is any subdegree of $G$. Furthermore, if $(r,\lambda)=1$ then $r\mid d_i$. \end{enumerate} \end{lemma} {\bf Proof}. (i) The equality $r=\frac{\lambda(v-1)}{k-1}$ implies $\lambda v=r(k-1)+\lambda < r(r-1)+\lambda=r^2-r+\lambda$, and by $r> k>\lambda$ we have $r^2>\lambda v$. Combining this with $v=|G:G_x|$ and $r\leq |G_x|$ gives $|G_x|^3>\lambda |G|$. Part (ii) was proved in \cite{Davies}. $\hfill\square$ \begin{lemma}\label{three}{\rm(\cite[p.366]{LIE87})} \,If $G$ is $A_n$ or $S_n$, acting on a set $\Omega$ of size $n$, and $H$ is any maximal subgroup of $G$ with $H\neq A_n$, then $H$ satisfies one of the following: \begin{enumerate} \item[\rm(i)] $H=(S_k\times S_{\ell})\cap G$, with $n=k+\ell$ and $k\neq \ell$ {\rm (}intransitive case{\rm)}; \item[\rm(ii)] $H=(S_k\wr S_{\ell})\cap G$, with $n=k\ell$, $k>1$ and $\ell >1$ {\rm(}imprimitive case{\rm)}; \item[\rm(iii)] $H=AGL_k(p)\cap G$, with $n=p^k$ and $p$ prime {\rm(}affine case{\rm)}; \item[\rm(iv)] $H=(T^k.(OutT\times S_k))\cap G$, with $T$ a nonabelian simple group, $k\geq2$ and $n=|T|^{k-1}$ {\rm (}diagonal case{\rm )}; \item[\rm(v)] $H=(S_k\wr S_{\ell})\cap G$, with $n=k^{\ell}$, $k\geq5 $ and $\ell>1$ {\rm(}wreath case{\rm)}; \item[\rm(vi)] $T\unlhd H\leq Aut(T)$, with $T$ a nonabelian simple group, $T\neq A_n$ and $H$ acting primitively on $\Omega$ {\rm(}almost simple case{\rm)}. \end{enumerate} \end{lemma} \begin{remark} This lemma does not deal with the groups $M_{10}$, $PGL_2(9)$ and $P\Gamma L_2(9)$ that have $A_6$ as socle. These exceptional cases will be handled in the first part of Section 3. \end{remark} \begin{lemma}\label{odd}{\rm \cite[Theorem (b)(I)]{Lie85}}\, Let $G$ be a primitive permutation group of odd degree $n$ on a set $\Omega$ with simple socle $X:=Soc(G)$, and let $H=G_x$, $x\in \Omega$. If $X\cong A_c$, an alternating group, then one of the following holds: \begin{enumerate} \item[\rm(i)] $H$ is intransitive, and $H=(S_a\times S_{c-a} )\cap G$ where $1\leq a<\frac{1}{2}c$; \item[\rm(ii)] $H$ is transitive and imprimitive, and $H=(S_a\wr S_{c/a})\cap G$ where $a>1$ and $a\mid c$; \item[\rm(iii)] $H$ is primitive, $n=15$ and $G\cong A_7$. \end{enumerate} \end{lemma} \begin{lemma}\label{Alt}{\rm \cite[Theorem 5.2A]{JD}}\,\, Let $G:=Alt(\Omega)$ where $n:= |\Omega|\geq 5$, and let $s$ be an integer with $1\leq s\leq \frac{n}{2}$. Suppose that, $K\leq G$ has index $|G:K|<\binom{n}{s}$. Then one of the following holds: \begin{enumerate} \item[\rm(i)] For some $\Delta\subset \Omega$ with $|\Delta|s\geq 7$, then $\binom{s+t}{s}>t^4>s^2t^2$. \item[\rm(ii)] If $s\geq 6$ and $t\geq 2$, then $2^{(s-1)(t-1)}>s^4\binom{t}{2}^2$ implies $2^{s(t-1)}>(s+1)^4\binom{t}{2}^2$. \item[\rm(iii)] If $t\geq 6$ and $s\geq 2$, then $2^{(s-1)(t-1)}>s^4\binom{t}{2}^2$ implies $2^{(s-1)t}>s^4\binom{t+1}{2}^2$. \item[\rm(iv)] If $t\geq 4$ and $s\geq 3$, then $\binom{s+t}{s}>s^2t^2$ implies $\binom{s+t+1}{s}>s^2(t+1)^2$. \end{enumerate} \end{lemma} {\bf Proof}. (i) It is necessary to prove that $\binom{s+t}{s}>t^4$ holds. Since $t>s\geq 7$ then $7\leq s\leq [\frac{s+t}{2}]$, it follows that $\binom{s+t}{s}\geq \binom{t+7}{7}>t^4$. (ii) Suppose that $s\geq 6,t\geq 2$ and $2^{(s-1)(t-1)}>s^4\binom{t}{2}^2$. Then $$2^{s(t-1)}=2^{(s-1)(t-1)} 2^{t-1}>s^4\binom{t}{2}^2 2^{t-1}=(s+1)^4\binom{t}{2}^2\Big(1-\frac{1}{s+1}\Big)^4 2^{t-1}.$$ Combing this with the fact $(1-\frac{1}{s+1})^4 2^{t-1}\geq2\times(\frac{6}{7})^4 >1$ gives (ii). (iii) Suppose that $t\geq 6,s\geq 2$ and $2^{(s-1)(t-1)}>s^4\binom{t}{2}^2$. Then $$2^{(s-1)t}=2^{(s-1)(t-1)} 2^{s-1}>s^4\binom{t}{2}^2 2^{s-1}=s^4\binom{t+1}{2}^2\Big(1-\frac{2}{t+1}\Big)^2 2^{s-1}.$$ Combing this with the fact $(1-\frac{2}{t+1})^2 2^{s-1}\geq2\times(\frac{5}{7})^2 >1$ gives (iii). (iv) Suppose that $\binom{s+t}{s}>s^2t^2$. Then $$\binom{s+t+1}{s}=\binom{s+t}{s}\frac{(s+t+1)}{(t+1)} >s^2t^2\frac{(s+t+1)}{(t+1)}=s^2(t+1)^2\frac{(s+t+1)t^2}{(t+1)^3}.$$ This, together with the inequality $(s+t+1)t^2>(t+1)^3$, yields (iv). $\hfill\square$ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Proof of Theorem \ref{theorem 1}} Throughout this paper, we assume that the following hypothesis holds. {\sc Hypothesis}: Let $\mathcal D$ be a non-symmetric $2$-$(v, k, \lambda)$ design with $(r,\lambda)=1$, $G\leq Aut(\mathcal D) $ be a flag-transitive automorphism group $G$ with $Soc(G)=A_n$. Let $x$ be a point of $P$ and $H=G_x$. By Lemma \ref{Dem}, $G$ acts primitively on $P$. So that $H$ is a maximal subgroup of $G$ by \cite[Theorem 8.2]{Hw} and $v=|G:H|$. Furthermore, by the flag-transitivity of $G$, we have that $b$ divides $G$, $r$ divides $|H|$, and $r^2 >v$ by Lemma \ref{k2}(i). Suppose first that $n=6$ and $G\cong M_{10}$, $PGL_2(9)$ or $P\Gamma L_2(9)$. Each of these groups has exactly three maximal subgroups with index greater than 2, and their indices are precisely 45, 36 and 10. By using the computer algebra system {\sf GAP} \cite{GAP}, for $v=45$, 36 or 10, we will compute the parameters $(v,b,r,k,\lambda)$ that satisfy the following conditions: \begin{equation} r\mid v-1; \end{equation} \begin{equation}2b$ for any $k$-element subset $B$. So $G$ cannot act block-transitively on $\cal D$, a contradiction. Now we consider $G=A_n$ or $S_n$ with $n\geq 5$. The point stabilizer $H=G_{x}$ acts both on $P$ and the set $\Omega_{n}: =\{1,2,\cdots, n\}$. Then by Lemma \ref{three} one of the following holds: \begin{itemize} \item $H$ is primitive in its action on $\Omega_{n}$; \item $H$ is transitive and imprimitive in its action on $\Omega_{n}$; \item $H$ is intransitive in its action on $\Omega_{n}$. \end{itemize} The proof of Theorem \ref{theorem 1} consists of three subsections. \subsection{$H$ acts primitively on $\Omega_{n}$.} \begin{proposition}\label{primitive} Let $\mathcal D$ and $G$ satisfy {\sc Hypothesis}. Let the point stabilizer act primitively on $\Omega_{n}$. Then $\mathcal D$ is a $2$-$(6,3,2)$ design or the projective space $PG(3,2)$. \end{proposition} {\bf Proof}. Suppose first that $r$ is even, since $r\mid v-1$ then $v$ is odd. Thus by Lemma \ref{odd}, $v=15$ and $G=A_7$, and then $r=14$. The possible parameters $(v,b,r,k,\lambda)$ such that $235$ for any $6$-element subset $B$. So $A_7$ cannot act block-transitively on $\cal D$, a contradiction. Then $r$ is odd. Let $p$ be an odd prime divisor of $r,$ then $(p,v)=1$ according to Lemma \ref{k1}(ii). Thus $H$ contains a Sylow $p$-subgroup $R$ of $G$. Let $g\in G$ be a $p$-cycle, then there is a conjugate of $g$ belongs to $H$. This implies that $H$ acting on $\Omega_{n}$ contains an even permutation with exactly one cycle of length $p$ and $n-p$ fixed points. By a result of Jordan \cite[Theorem 13.9]{Hw}, we have $n-p\leq 2$. Therefore $n-2\leq p\leq n$, $p^2\nmid |G|$, and so $p^2\nmid r$. It follows that $r$ is either a prime, namely $n-2, n-1, n$, or the product of two twin primes, namely $r=(n-2)n$. Moreover, since the primitivity of $H$ acting on $\Omega_{n}$ and $H \ngeqslant A_n$ implies that $v\geq \frac{[\frac{n+1}{2}]!}{2}$ by \cite[Theorem 14.2]{Hw}, combining this with $r^2> v$ gives $$r^2>\frac{[\frac{n+1}{2}]!}{2}.$$ Therefore, $(n,r)=(5,5),(5,15), (6,5), (7,5), (7,7), (7,35), (8,7)$ or (13,143). From Lemmas \ref{k1}, \ref{k2}, the facts $v\geq \frac{[\frac{n+1}{2}]!}{2}$ and $[b,v]\mid |G|$, where the condition $[b,v]\mid |G|$ is a consequence of $v\mid |G|$ and $b\mid |G|$, we obtain 3 possible parameters $(v,b,r,k,\lambda)$ which listed in the following: $$(6,10,5,3,2), (15,21,7,5,2), (15,35,7,3,1). $$ {\bf Case (1):}\, $(v,b,r,k,\lambda)=(6,10,5,3,2)$. By \cite[p.27, p.36]{Handbook}, we know that there is up to isomorphism a unique $2$-$(6,3,2)$ design ${\mathcal D}=(P, \mathcal B)$ where $$ \begin{array}{ll} P=&\{1,2,3,4,5,6\};\\ \mathcal B=&\{\{ 1, 2, 3\}, \{ 1, 2, 5 \}, \{ 1, 3, 4 \}, \{ 1, 4, 6 \}, \{ 1, 5, 6 \}, \\ &\,\,\,\{ 2, 3, 6 \}, \{ 2, 4, 5 \}, \{ 2, 4, 6 \}, \{ 3, 4, 5 \}, \{ 3, 5, 6 \}\}.\\ \end{array} $$ Here $G=A_5, S_5, A_6$ or $S_6$, the stabilizer $G_x=D_{10}, AGL_1(5), A_5$ or $S_5$ respectively. Assume first that $G=S_5$, $A_6$ or $S_6$. As the primitive permutation group of degree 6, $S_5$, $A_6$ or $S_6$ is 3-, 4- or 6-transitive respectively, so $G$ is 3-transitive. If we choose $B=\{1,2,3\}$, then $|B^G|=20>b=10$, a contradiction. Hence $G=A_5$. Without loss of generality, assume that $A_5=\langle (24)(56),(123)(456)\rangle$. Let $B=\{1,2,3\}$ be a block. Then $B^{A_5}=\mathcal B$. Now $G_1=\langle (24)(56), (35)(46)\rangle\cong D_{10}$, and $$B^{G_1}=\{\{ 1, 2, 3 \}, \{ 1, 2, 5 \}, \{ 1, 3, 4 \}, \{ 1, 4, 6 \}, \{ 1, 5, 6 \}\},$$ which is the $G_1$-orbit on $\mathcal{B}$ containing $B$. So that $G_1$ is transitive on 5 blocks through 1, note that $G$ is also transitive on $P$, and hence $\cal{D}$ is flag-transitive. {\bf Case (2):}\, $(v,b,r,k,\lambda)=(15,21,7,5,2)$. This can be ruled out by \cite[Theorem 5.2]{Con}. {\bf Case (3):}\, $(v,b,r,k,\lambda)=(15,35,7,3,1)$. Here $\lambda=1$, by \cite{Dela01} or Lemma \ref{de}, we know that ${\mathcal D}$ is the projective space $PG(3,2)$ and $G=A_7$ or $A_8\cong PSL_4(2)$. For completeness, the structure of the design and the proof of flag-transitivity are given below. Let $P=\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\},$ the group $G\cong A_7$ or $A_8$, as the primitive group of degree 15 acting on $P$, has the following generators respectively (\cite[p.829]{Handbook}): $$ \begin{small} \begin{array}{lll} A_7&\cong& \langle(1,4,7,10,13)(2,5,8,11,14)(3,6,9,12,15), (2,6,3,8,12,4,9)(5,7,13,11,10,14,15)\rangle,\\ A_8&\cong& \langle(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15),(1,5)(6,13)(7,8)(10,12)\rangle.\\ \end{array} \end{small} $$ There are totally $455=\binom{15}{3}$ $3$-element subsets of $P$. For any $3$-element subset $B\subseteq P$, using {\sf GAP} \cite{GAP}, we calculate the length of the $G$-orbit $B^{G}$. It follows that up to isomorphism there exists a unique $2$-$(15,3,1)$ design ${\mathcal{D}}=(P, \mathcal{B})$ where $$ \begin{array}{ll} %P=&\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\};\\ \mathcal{B}&=\{\{ 1, 2, 13 \}, \{ 1, 4, 5 \}, \{ 1, 6, 11 \}, \{ 4, 7, 8 \}, \{ 1, 7, 9 \}, \{ 4, 9, 14 \}, \{ 1, 3, 10 \}, \\ &\{ 7, 10, 11 \}, \{ 9, 12, 13 \}, \{ 4, 10, 12 \},\{ 2, 7, 12 \}, \{ 2, 9, 15 \}, \{ 4, 6, 13 \}, \{ 1, 8, 14 \},\\ &\{ 10, 13, 14 \},\{ 1, 12, 15 \},\{ 2, 4, 11 \}, \{ 7, 13, 15 \}, \{ 5, 10, 15 \}, \{ 3, 5, 12 \},\{ 2, 5, 6 \},\\ &\{ 3, 9, 11 \}, \{ 11, 14, 15 \}, \{ 3, 4, 15 \}, \{ 5, 7, 14 \}, \{ 6, 9, 10 \}, \{ 5, 11, 13 \}, \{ 3, 8, 13 \}, \\ &\{ 6, 8, 15 \}, \{ 5, 8, 9 \},\{ 3, 6, 7 \}, \{ 6, 12, 14 \}, \{ 2, 8, 10 \}, \{ 2, 3, 14 \}, \{ 8, 11, 12 \} \}.\\ \end{array} $$ Let $B=\{1, 2, 13\}$ be a block. Then it is easily known that $B^{G}=\mathcal B$. Now \begin{align*} G_1 &=\langle(2,6,3,8,12,4,9)(5,7,13,11,10,14,15), (3,8)(4,11)(5,6)(7,9)(10,14)(12,15)\rangle\\ & \cong PSL_3(2) \end{align*} or \begin{align*} G_1& =\langle(2,5,8,3,15,11,7)(4,14,10,12,6,9,13), (3,14)(7,12)(8,10)(9,15)\rangle\cong AGL_3(2) \end{align*} with $G=A_7$ or $A_8$ respectively. Then $B^{G_1}$ containing $7$ blocks: $ \{ 1, 2, 13 \}$, $\{ 1, 6, 11 \}$, $\{ 1, 3, 10 \}$, $\{ 1, 4, 5 \}$, $\{ 1, 8, 14 \}$, $\{ 1, 7, 9 \}$, $\{1, 12, 15 \}.$ So that $G_1$ is transitive on $r$ blocks through 1, note that $G$ is also transitive on $P$, and hence $G$ is flag-transitive and ${\cal D}= PG(3,2)$. $\hfill\square$ \subsection{ $H$ acts transitively and imprimitively on $\Omega_{n}$.} \begin{proposition}\label{transitive-imprimiitve} Let $\mathcal D$ and $G$ satisfy {\sc Hypothesis}. Let the point stabilizer acts transitively and imprimitively on $\Omega_{n}$, then $\mathcal D$ is a $2$-$(10,6,5)$ design with $Soc(G)=A_6$. \end{proposition} {\bf Proof}. Suppose on the contrary that $\Sigma:=\{\triangle_0,\triangle_1,\ldots,\triangle_{t-1}\}$ is a nontrivial partition of $\Omega_n$ preserved by $H$, where $|\triangle_i|=s, 0\leq i\leq t-1, s,t\geq2$ and $st=n$. Then \begin{align}\label{v} v&=\frac{\binom{ts}{s}\binom{(t-1)s}{s}\ldots\binom{3s}{s}\binom{2s}{s}}{t!}\notag\\ &=\binom{ts-1}{s-1}\binom{(t-1)s-1}{s-1}\ldots\binom{3s-1}{s-1}\binom{2s-1}{s-1}. \end{align} Moreover, the set $O_j$ of $j$-cyclic partitions with respect to $X$ (a partition of $\Omega_n$ into $t$ classes each of size $s$) is an union of orbits of $H$ on $P$ for $j=2,\ldots,t$ (see \cite{Dela01,ZhouDong10a} for definitions and details). (1) Suppose first that $s=2$, then $t\geq 3$, $v=(2t-1)(2t-3)\cdots 5\cdot 3$, and $$d_j=|O_j|=\frac{1}{2}\binom{t}{j}\binom{s}{1}^j=2^{j-1}\binom{t}{j}.$$ We claim that $t<7$. If $t\geq 7$, then it is easy to know that $v=(2t-1)(2t-3)\cdots 5\cdot 3>t^2(t-1)^2$, and as $r$ divides $d_2=t(t-1)$ it follows that $t(t-1)\geq r$, hence $v>t^2(t-1)^2\geq r^2$ which is a contradiction. Thus $t< 7$. For $t=3,4,5$ or 6, we calculate $d=\gcd(d_2,d_3)$ which listed in Table \ref{tab1} below. \begin{table}[h] \begin{center} \caption{Possible $d$ when $s=2$}\label{tab1} \vspace{3mm} \begin{tabular}{cccccc} \hline $t$ & $n$ & $v$ & $d_2$ & $d_3$ & $d$ \\ \hline $3$ & $6$ & $15$ & $6$ & $4$ & $2$ \\ $4$ & $8$ & $105$ & $12$ & $16$ & $4$ \\ $5$ & $10$ & $945$ & $20$ & $40$ & $20$ \\ $6$ & $12$ & $10395$ & $30$ & $80$ & $10$ \\ \hline \end{tabular} \end{center} \end{table} In each case $r\leq d$ which contradicts to the fact $r^2 > v$. (2) Suppose second that $s\geq3,$ then $O_j$ is an orbit of $H$ on $P$, and $d_j=|O_j|=\binom{t}{j}\binom{s}{1}^j=s^j\binom{t}{j}.$ In particular, $d_2=\binom{t}{2}\binom{s}{1}^2=s^2\binom{t}{2}$ and $r\mid d_2$. Moreover, from $ \binom{is-1}{s-1}=\frac{is-1}{s-1}\cdot\frac{is-2}{s-2} \cdots\frac{is-(s-1)}{1}>i^{s-1}$, for $i=2,3,\ldots,t$, we have $v>2^{(s-1)(t-1)}$. Then \begin{equation*} 2^{(s-1)(t-1)}< vs^4\binom{t}{2}^2=2^4\cdot3^6\cdot5^2$, i.e. the pair $(s,t)=(6,6)$ does not satisfy the inequality (\ref{inequality4}) but satisfies the conditions of Lemma \ref{inequ1} (ii) and (iii). Thus, we must have $s<6$ or $t<6$. It is not hard to get 32 pairs $(s,t)$ satisfying the inequality (\ref{inequality4}) as follows: $$ \begin{array}{lllllllllll} &(3, 2), &(3, 3), & (3, 4), &(3, 5), & (3, 6), &(3, 7), &(3, 8), & (3,9),&(4, 2), & (4, 3), \\ & (4, 4),& (4, 5),&(4, 6), &(5, 2), & (5, 3), & (5, 4), & (6, 2), &(6, 3), &(6, 4), & (7, 2), \\ & (7, 3), & (8, 2), &(8, 3), &(9, 2), &(10, 2), &(11, 2), & (12, 2), & (13, 2), & (14, 2), &(15, 2),\\ & (16, 2), & (17, 2).& &&&&&&& \\ \end{array} $$ For each $(s,t)$, we calculate the parameters $(v,b,r,k,\lambda)$ satisfying Eq.(\ref{v}), Lemmas \ref{k1}, \ref{k2} and $r\mid d_2$. Then we obtain five possible parameters $(v,b,r,k,\lambda)$ corresponding to $(s,t)$ are the following: (2.1)\, $(s,t)=(3,2)$: $(10,30,9,3,2)$, $(10,18,9,5,4)$, $(10,15,9,6,5)$; (2.2)\, $(s,t)=(5,2)$: $(126,525,25,6,1)$, $(126,150,25,21,4)$. {\bf Case (2.1):} $(s,t)=(3,2)$. Then $n=6, v=10$ and $G\cong A_6$ or $S_6$. Let $P=\{1,2,3,4,5,6,7,8,9,10\},$ the group $G\cong A_6$ or $S_6$, as the primitive permutation group of degree 10 acting on $P$, has the following generators respectively (\cite[p.828]{Handbook}): $$ \begin{array}{lll} A_6&\cong& \langle(1,10,4,7,5)(2,8,6,9,3),(1,3,4,5,7)(2,10,9,8,6)\rangle,\\ S_6&\cong& \langle(1,8,6)(2,3,7,9,10,5),(1,8,9,3,5,6)(2,7,4)\rangle.\\ \end{array} $$ Assume first $(v,b,r,k,\lambda)=(10,30,9,3,2)$ or $(10,18,9,5,4)$, and there exists a $2$-$(10,k,\lambda)$ design $\cal D$ with flag-transitive automorphism group $G$, where $k=3$ or $5$, and $\lambda=2$ or $4$, respectively. There are totally $\binom{v}{k}$ $k$-element subsets of $P$. For any $k$-element subset $B\subseteq P$, we calculate the length of the $G$-orbit $B^G$ where $G=A_6$ or $S_6$ respectively. By using {\sf GAP} \cite{GAP}, we found that $|B^{G}|>b$ for any $k$-element subset $B$. So $G$ cannot act block-transitively on $\cal D$, a contradiction. Therefore, $(v,b,r,k,\lambda)=(10,15,9,6,5)$. Here $G\cong A_6$ or $S_6$ acts transitively on $P$. There are totally $210=\binom{10}{6}$ $6$-element subsets of $P$. For any $6$-element subset $B\subseteq P$, using {\sf GAP} \cite{GAP}, we calculate the length of the $G$-orbit $B^{G}$. It follows that up to isomorphism there exists a unique $2$-$(10,6,5)$ design ${\mathcal{D}}=(P, \mathcal{B})$ where $$ \begin{array}{ll} \mathcal{B}=&\{\{1, 2, 3, 4, 5, 8\}, \{1, 2, 6, 7, 8, 10\}, \{3, 4, 5, 6, 7, 10\},\{4, 5, 6, 8, 9, 10\}, \{1, 2, 3, 6, 9, 10\}, \\ &\,\,\,\{1, 2, 4, 5, 7, 9\},\{1, 3, 4, 6, 7, 9\}, \{2, 5, 6, 7, 8, 9\}, \{2, 3, 4, 8, 9, 10\}, \{1, 3, 5, 7, 8, 10\}, \\ &\,\,\, \{2, 3, 5, 7, 9, 10\}, \{1, 3, 5, 6, 8, 9\},\{2, 3, 4, 6, 7, 8\}, \{1, 2, 4, 5, 6, 10\}, \{1, 4, 7, 8, 9, 10\}\}.\\ \end{array} $$ Let $B=\{1, 2, 3, 4, 5, 8\}$ be a block. Then it is easily known that $B^{G}=\mathcal B$. Now $$G_1=\langle(2,9,10,3)(4,7,8,5), (2,5,8,10)(3,7,6,4)\rangle\cong 3^2: 4$$ or $$\langle(2,10,8,3,9,4)(5,6,7), (3,9,7,8)(4,10,5,6)\rangle\cong 3^2:D_8$$ with $G=A_6$ or $S_6$ respectively. Then $B^{G_1}$ containing $9$ blocks: \begin{align*} &\{1, 2, 3, 4, 5, 8\}, \{1, 2, 4, 5, 7, 9\}, \{1, 3, 5, 7, 8, 10\},\\ &\{1, 4, 7, 8, 9, 10\}, \{1, 3, 5, 6, 8, 9\}, \{1, 2, 6, 7, 8, 10\},\\ &\{1, 2, 3, 6, 9, 10\}, \{1, 2, 4, 5, 6, 10\}, \{1, 3, 4, 6, 7, 9\}. \end{align*} So that $G_1$ is transitive on the blocks through 1, note that $G$ is also transitive on $P$, and hence $\cal{D}$ is flag-transitive. {\bf Case (2.2):} $(s,t)=(5,2)$. Then $n=10$, $v=126$ and $G\cong A_{10}$ or $S_{10}$. For the parameters $(126,525,25,6,1)$ we have $\lambda =1$, it can be ruled out by Lemma \ref{de}. Suppose that $(v,b,r,k,\lambda)=(126,150,25,21,4)$. Since $G$ is flag-transitive, then it is block-transitive and point-transitive. So that $G$ must has subgroups with index 126 and 150. By using {\sc Magma} \cite{Magma} we know that $G$ has 126 subgroups with index 126. Let $B\in \cal B$, so that $|G:G_B|=b=150$ and $|G_B|=12096$ or $24192$ with $G\cong A_{10}$ or $S_{10}$ respectively. Clearly, $|G:G_B|<\binom{10}{4}$. Then by Lemma \ref{Alt} and \cite[Theorem 5.2B]{JD}, one of 3 cases holds. If Case (i) holds, there exists some $\Delta \subseteq \Omega_{10}=\{1,2,\ldots,10\}$ with $|\Delta|<4$. We have $A_7\leq G_B$ by Remark \ref{remark}. However, $|A_7|=7!/2$ does not divide $|G_B|$, a contradiction. Case (ii) can be ruled out by $|G:G_B|=150\neq \frac{1}{2}\binom{10}{5}$ and Case (iii) can be ruled out by $n=10$. Hence, $G$ has no subgroups with index 150, a contradiction. So that $G\not\cong A_{10}$ or $S_{10}$. $\hfill\square$ \subsection{$H$ acts intransitively on $\Omega_{n}$.} \begin{proposition}\label{intransitive} Let $\mathcal D$ and $G$ satisfy {\sc Hypothesis}. Then the point stabilizer cannot be intransitive on $\Omega_{n}.$ \end{proposition} {\bf Proof}. Suppose on the contrary that $H$ acts intransitively on $\Omega_{n}$. We have $H=(Sym(S)\times Sym(\Omega\setminus S))\cap G$, and without loss of generality assume that $|S|=s v$, we have $s^2(n-s)^2> \binom{n}{s}$. Since the condition $s<\frac{n}{2}$ equals to $s \binom{s+t}{s}. \end{equation} Combining it with Lemma \ref{inequ1} (i), we get $s\leq 6$. {\bf Case (1):} If $s=1$, then $v=n\geq 5$ and the subdegrees are $1, n-1$. The group $G$ is $(v-2)$-transitive on $P$. Since $2k-i+1$ for $i=2, \ldots, k-1$. So that $$\binom{n}{k}k=n(n-1)\cdot\frac{n-2}{k-1}\cdot\frac{n-3}{k-2}\cdot\cdot\cdot\frac{n-k+1}{2} >n(n-1),$$ a contradiction. {\bf Case (2):} If $s=2$, then the subdegrees are $1, ~\binom{n-2}{2},~2(n-2)$, and $G$ is a primitive rank 3 group acting on $P$. By Lemma \ref{k2}(ii), $r\mid \Big(~\binom{n-2}{2},2(n-2)\Big)=\frac{n-2}{2}$, $n-2$, or $2(n-2)$ with $n\equiv2\pmod 4$, $n\equiv1\pmod 4$, or $n\equiv 3\pmod 4$ respectively. Assume first that $r\mid \frac{n-2}{2}$. Then $\frac{n(n-1)}{2}=v\frac{\lambda n(n-1)}{2}$. It follows that $8>\frac{8(n-2)^2}{n(n-1)}>\lambda u^2$ which forces $u=1$. Therefore, $r=2(n-2)$. By Lemma \ref{k1}, we get $k=\frac{\lambda(n+1)}{4}+1$, and $b=\frac{vr}{k}=\frac{4n(n-1)(n-2)}{\lambda(n+1)+4}$, where $\lambda=3$, $5$ or $7$. If $\lambda=3$, then $(3n+7)\mid 4n(n-1)(n-2)$ since $b\in \mathbb{N}$. And since $(3,3n+7)=1$, we have $(3n+7)\mid 36n(n-1)(n-2)$. Recall that $n\equiv3\pmod 4$, and from $$3^2b=\frac{36n(n-1)(n-2)}{3n+7}=12n^2-64n+173-\frac{1211-n}{3n+7}\in \mathbb{N}$$ we have $(3n+7)\mid (1211-n)$, so $n=7$, $11$, $15$, $91$, $119$ or $171$. Similarly, if $\lambda=5$, then $5^2b=\frac{100n(n-1)(n-2)}{5n+9}=20n^2-96n+213-\frac{n+1917}{5n+9}\in \mathbb{N}$. Now the facts $(5n+9)\mid (n+1917)$ and $n\equiv3\pmod 4$ imply $n=15$, $99$, $135$ or $211$. If $\lambda=7$, then $7^2 b=\frac{196n(n-1)(n-2)}{7n+11}=28n^2-128n+257-\frac{2827-n}{7n+11}\in \mathbb{N}$. Now the facts $(7n+11)\mid (2827-n)$ and $n\equiv3\pmod 4$ imply $n=7$, $11$, $27$, $55$, $127$ or $187$. For each value of $n$, we compute the possible parameters $(n,v,b,r,k)$ satisfying Lemma \ref{k1} and the condition $b>v$ which listed in the following. $$ \begin{array}{llllll} \lambda=3: &(7, 21, 30, 10, 7), &(11, 55, 99, 18, 10),\\ &(15, 105, 210, 26, 13), &(91, 4095, 10413, 178, 70),\\ &(119, 7021, 18054, 234,91), &(171, 14535,37791, 338, 130).\\ \lambda=5: &(15, 105, 130, 26, 21), &(99, 4851, 7469, 194, 126),\\ &(135, 9045, 14070, 266, 171), &(211, 22155, 34815, 418, 266).\\ \lambda=7: &(55, 1485, 1590, 106, 99), &(127, 8001, 8890, 250, 225),\\ &(187, 17391, 19499, 370, 330). \end{array} $$ Assume first that $(n,v,b,r,k)=(7, 21, 30, 10, 7)$ and there exists a $2$-$(21,7,3)$ design $\cal D$ with flag-transitive automorphism group $G\cong A_7$ or $S_7$. Let $P=\{1,2,3,\ldots,21\},$ the group $G\cong A_7$ or $S_7$, as the primitive permutation group of degree 21 acting on $P$, has the following generators respectively (\cite[p.830]{Handbook}): $$ \begin{array}{lll} A_7&\cong \langle(1,2,3,4,5,6,7)(8,9,10,11,12,13,14)(15,16,17,18,19,20,21),\\ &(1,4)(2,11)(3,9)(5,15)(7,20)(8,13)(10,21)(12,14)\rangle,\\ S_7&\cong \langle(1,2,3,4,5,6,7)(8,9,10,11,12,13,14)(15,16,17,18,19,20,21),\\ &(1,21,4,10)(2,9,11,3)(5,12,15,14)(7,13,20,8)(17,18)\rangle.\\ \end{array} $$ There are totally $\binom{21}{7}$ $7$-element subsets of $P$. For any $7$-element subset $B\subseteq P$, we calculate the length of the $G$-orbit $B^G$ where $G=A_7$ or $S_7$ respectively. By using {\sf GAP} \cite{GAP}, we found that $|B^{G}|>30$ for any $7$-element subset $B$. So $G$ cannot act block-transitively on $\cal D$, a contradiction. For all other possible parameters $(n,v,b,r,k)$ listed above, $G\cong A_{n}$ or $S_{n}$. On the one hand, by the block-transitivity, we have $b=|G:G_B|$ where $B\in \mathcal{B}$. On the other hand, since $|G: G_B|=b<\binom{n}{3}$ for each case, by Lemma \ref{Alt} and \cite[Theorem 5.2B]{JD}, it is easily known that $G$ has no subgroups of index $b$, a contradiction. {\bf Case (3):} Suppose that $3\leq s \leq 6$. Now for each value of $s$, using the inequality (\ref{st}) and Lemma \ref{inequ1}(iv), we know that $t$ (and hence $n$) is bounded. For example, let $s=3$, since $\binom{3+48}{3}>3^2\cdot48^2$, we must have $4\leq t\leq 47$ by Lemma \ref{inequ1}(iv), and so $7\leq n\leq 50$. The bound of $n$ listed in Table \ref{tab3} below. Here the last column denotes the arithmetical conditions which we used to ruled out each line. \begin{table}[h] \caption{Bound of $n$ when $3\leq s\leq 6$} \label{tab3} \begin{center} \begin{tabular}{lllc} \hline $s$ & \quad\quad $t$& \quad\quad $n$ & Reference\\ \hline 3 & $4\leq t\leq 47$&$7\leq n\leq 50$ &(1)-(4), Lemma \ref{k2}\\ 4 & $5\leq t\leq 14$&$9\leq n\leq 18$& Lemma \ref{k2}\\ 5 & $6,7,8,9$ & $11, 12, 13, 14$ &Lemma \ref{k2}\\ 6 & $7$ &$13$& Lemma \ref{k2}\\ \hline \end{tabular} \end{center} \end{table} Note that $v=\binom{n}{s}$, and $n_1=\binom{n-s}{s}$, $n_{s}=s(n-s)$ are two subdegrees of $G$ acting on $P$. Therefore, the 6-tuple $(v,b,r,k,\lambda, n)$ satisfies the following arithmetical conditions: (1)-(4), $(r,\lambda)=1$, $r^2>v$ (Lemma \ref{k2}(i)), and \begin{equation} r\mid d, \,\,{\mbox{where}}\,\, d=\gcd(n_1,n_{s}). \end{equation} If $s=3$, by using {\sf GAP} \cite{GAP}, it outputs five 6-tuples satisfying above arithmetical conditions: $$ \begin{array}{llllll} &(364, 1001, 33, 12, 1, 14), &(1540, 3135, 57, 28, 1, 22), &(4960, 7440, 87, 58, 1, 32),\\ &(19600, 19740, 141, 140, 1, 50),&(1540, 1596, 57, 55, 2, 22). & \end{array} $$ The four parameters with $\lambda=1$ can be ruled out by Lemma \ref{de}. For the parameters $(1540$, $1596$, $57$, $55$, $2$, $22)$, we have $\lambda=2$, $G\cong A_{22}$ or $S_{22}$. By the block-transitivity, $b=|G:G_B|=1596$ where $B\in \mathcal{B}$. However, since $|G: G_B|=1596<\binom{22}{4}$, by Lemma \ref{Alt} and \cite[Theorem 5.2B]{JD}, it is easily known that $A_{22}$ and $S_{22}$ has no subgroups of index 1596. So the parameters $(1540, 1596, 57, 55, 2, 22)$ cannot occur. If $s=4, 5$ or 6, by using {\sf GAP} \cite{GAP}, there are no parameters $(v,k,n)$ satisfying the conditions. For example, if $s=6$, then $n=13$, $v=1716$, $d=7$. It follows that $r\leq 7$ by (10), then $r^2