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\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 18, 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
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\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 < k<v-1 $. So that we have
$$b>v \,\, \mbox{and}\,\, r>k.$$



 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 $|\Delta|<s$  we have $G_{(\Delta)}\leq K \leq G_{\{\Delta\}}$;
\item[\rm(ii)] $n=2m$ is even, $K$ is imprimitive with two blocks of size $m$, and
$|G:K|=\frac{1}{2}\binom{n}{m}$; or
\item[\rm(iii)] one of six exceptional cases hold where:
\begin{enumerate}
\item[\rm(a)] $K$ is imprimitive on $\Omega$ and $(n,s,|G:K|) = (6, 3, 15)$;
\item[\rm(b)] $K$ is primitive on $\Omega$ and $(n,s,|G:K|, K)$ $= (5, 2, 6, 5:2)$,
$(6, 2, 6, PSL_2(5))$, $(7, 2, 15, PSL_3(2))$, $(8, 2, 15,
AGL_3(2))$, or $(9, 4, 120, P\Gamma L_2(8))$.
\end{enumerate}
\end{enumerate}
\end{lemma}


\begin{remark}\label{remark}
 {\rm (1)}\, From part (i) of Lemma \ref{Alt} we know that  $K$ contains the alternating group $G_{(\Delta)}=Alt(\Omega\setminus\Delta)$
of degree $n-s+1$.

{\rm (2)}\, A result similars to Lemma \ref{Alt} holds for the
finite symmetric groups $Sym(\Omega)$ which can be found in
{\rm\cite[Theorem 5.2B]{JD}}.
\end{remark}


We will also need some elementary inequalities.

\begin{lemma}\label{inequ1}  Let $s$ and $t$ be two positive integers.
\begin{enumerate}
 \item[\rm(i)]  If $t>s\geq 7$, 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}2<k<r;
\end{equation}
\begin{equation} b=\frac{vr}{k};
\end{equation}
\begin{equation}
\lambda=\frac{bk(k-1)}{v(v-1)};
\end{equation}
\begin{equation}
(r,\lambda)=1;
\end{equation}
\begin{equation}
r\mid |H|.
\end{equation}
 We obtain three possible parameters
$(v,b,r,k,\lambda)$ as follows:
$$
\begin{array}{lllllllllll}
  & (10, 30, 9, 3, 2); &  (10, 18, 9, 5, 4); & (10, 15, 9, 6, 5).
\end{array}
$$
Now we consider the existence of flag-transitive non-symmetric
designs with above possible parameters. For the parameters $(10,
15, 9, 6, 5)$, we will consider the existence of its complement
design which has parameters $(10,15,6,4,2)$.

Suppose that there exists a $2$-$(10,k,\lambda)$ design $\cal D$ with
 flag-transitive automorphism group $G$, where
the block size $k$ is $3, 5$ or $4$, and $\lambda=2,4$ or $2$
respectively. Here $v=10$. Let $P=\{1,2,3,4,5,6,7,8,9,10\}$, the
group $G\cong M_{10}$, $PGL_2(9)$ or $P\Gamma L_2(9)$ as the
primitive permutation group of degree 10 acting on $P$, has the
following generators respectively (\cite[p.828]{Handbook}):
\begin{align*}
M_{10}&\cong
\langle(1,6,10,9,3,8,4,5)(2,7),(1,7,2,6,5,9,4,10)(3,8)\rangle,\\
PGL_2(9)&\cong
\langle(1,2,3,4,5,6,7,8,9,10),(1,2,5,8,9,7,4,10,6,3)\rangle,\\
P\Gamma L_2(9)&\cong
\langle(1,2,3,4,5,6,7,8,9,10),(1,8,6,2,9,5,3,10)(4,7)\rangle.
\end{align*}

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 $G=M_{10}$, $PGL_2(9)$ or $P\Gamma L_2(9)$
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.

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
$2<k<r$ and $(r,\lambda)=1$ are $(15, 35, 14, 6, 5)$, $(15, 21,
14, 10, 9)$.

If there is a design  $\cal D$ with parameters $(15, 21, 14, 10,
9)$, then the complement design $\overline{\cal D}$ with
parameters $(15, 21, 7, 5, 2)$ also exists. However, by
\cite[Theorem 5.2]{Con}, we know that $2$-$(15,5,2)$ design does
not exist. So the parameters $(15, 21, 14, 10, 9)$ cannot occur.

Now assume that  $(v,b,r,k,\lambda)=(15, 35, 14, 6, 5)$. Let
$P=\{1,2,3,4,5,6,7,8,9,10,11$, $12,13,14,15\}$. Suppose that there
exists a $2$-$(15,6,5)$ design $\cal D$ with flag-transitive
automorphism group $A_7$. By \cite[p.829]{Handbook}, we know
$$A_7\cong \langle(1,4,7,10,13)(2,5,8,11,14)(3,6,9,12,15),(2,6,3,8,12,4,9)(5,7,13,11,10,14,15)\rangle.$$
There are totally $5005=\binom{15}{6}$ $6$-element subsets of $P$.
For any $6$-element subset $B\subseteq P$, using {\sf GAP}
\cite{GAP}, we calculate the length of the $A_7$-orbit $B^{A_7}$.
It follows that $|B^{A_7}|>35$ 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
$G_1=\langle(2,6,3,8,12,4,9)(5,7,13,11,10,14,15)$,
$(3,8)(4,11)(5,6)(7,9)(10,14)(12,15)\rangle \cong PSL_3(2)$ or
$\langle(2,5,8,3,15,11,7)(4,14,10,12,6,9,13)$,
$(3,14)(7,12)(8,10)(9,15)\rangle$ $\cong AGL_3(2)$ 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}...\binom{3s}{s}\binom{2s}{s}}{t!}\notag\\
&=\binom{ts-1}{s-1}\binom{(t-1)s-1}{s-1}...\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)}< v<r^2\leq s^4\binom{t}{2}^2,
\end{equation*}
and so
\begin{equation}\label{inequality4}
      2^{(s-1)(t-1)}< s^4\binom{t}{2}^2.
\end{equation}

We will calculate all pairs $(s,t)$ satisfying the inequality
(\ref{inequality4}). Since
$2^{(s-1)(t-1)}=2^{25}>s^4\binom{t}{2}^2=2^4\cdot3^6\cdot5^2$, 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 $\Delta \subseteq \Omega_{10}=\{1,2,...,10\}$
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 $|S|=s<n/2$ by
Lemma \ref{three}(i). By the flag-transitivity of $G$, $H$ is
transitive on the blocks through $x$, and so $H$ fixes exactly one
point in $P$. Since $H$ stabilizes only one $s$-subset of
$\Omega_{n}$, we can identify the point $x$ with $S$. As the orbit
of $S$ under $G$ consists of all the $s$-subsets of $\Omega_{n}$, we
can identify $P$ with the set of $s$-subsets of $\Omega_{n}$. So
that $v=\binom{n}{s}$, $G$ has rank $s+1$ and the subdegrees  are:
 \begin{equation}\label{sub}
 n_0=1, n_{i+1}=\binom{s}{i}\binom{n-s}{s-i}, i=0,1, ..., s-1.
\end{equation}

First, we claim that $s\leq 6$. Since $r \mid n_i$ for any
subdegree $n_i$ of $G$ by Lemma \ref{k2}(ii) and $n_{s}=s(n-s)$ is
a subdegree of $G$ by (\ref{sub}), then $r\mid s(n-s)$. Combining
this with $r^2> v$, we have $s^2(n-s)^2> \binom{n}{s}$. Since the
condition $s<\frac{n}{2}$ equals to $s<t: = n-s$, we have
\begin{equation}\label{st}
s^2 t^2> \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 $2<k\leq v-2$, $G$
acts $k$-transitively on $P$. Then $b=$$|\mathcal
B|$$=|B^G|=\binom{n}{k}$ for every block $B\in \mathcal B$. From
the equality $bk=vr$ we can obtain $\binom{n}{k}k=nr$. On the one
hand, by Lemma \ref{k1}(ii) we have $r\leq n-1$, it follows that
$\binom{n}{k}k\leq n(n-1)$. On the other hand, by $2<k\leq n-2$,
we have $n-i\geq k-i+2>k-i+1$ for $i=2, ..., k-1$. 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 $\frac{n(n-1)}{2}=v<r^2\leq
\big(\frac{n-2}{2}\big)^2,$ which is impossible.

Assume that $r\mid (n-2)$. From Lemma \ref{k2}(i),
$\lambda v=\frac{\lambda n(n-1)}{2}<r^2\leq (n-2)^2$
which forces $\lambda=1$. It follows from Lemma \ref{de} that $\mathcal{D}=PG(3,2)$ and $G\cong A_7$ or $A_8$. This contracts the assumption that $v=\binom{n}{2}=21$ or 28.

Hence $r\mid 2(n-2)$.
From above analysis we know that $r$ is even. By Lemma \ref{k2}(i),
$\lambda v=\frac{\lambda n(n-1)}{2}<r^2\leq 4(n-2)^2$
which implies $1\leq\lambda\leq7$.
Recall that $(r, \lambda)=1$,
so that $\lambda$ is odd.  Since Lemma \ref{de} implies that $\lambda \neq 1$, we assume that $\lambda=3$, $5$ or $7$ in the following.

Let $r=\frac{2(n-2)}{u}$ for some integer $u$. From Lemma \ref{k2}(i), we have  $\frac{4(n-2)^2}{u^2}>\frac{\lambda n(n-1)}{2}$. 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 $P=\{1,2,3,...,21\},$ 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
$r^2<v$ which is impossible. If $s=5$, then $n=11,12,13$ or 14,
$v=462,792,1287$ or 2002 and $d=6,7,8$ or 9, respectively. It is
easy to check that $r^2<v$ for every case. This is the final
contradiction.    $\hfill\square$


Propositions \ref{primitive}-\ref{intransitive} finish the proof
of Theorem \ref{theorem 1}.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
The authors wish to thank the referees for their valuable comments and suggestions which lead to the improvement of the paper.

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