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\title{\bf Non-existence of point-transitive $2$-$(106, 6, 1)$ designs}

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\author{ Haiyan Guan, Shenglin Zhou\footnote{Corresponding author. Supported by the National Natural Science Foundation of China (No. 11071081) and the Natural Science Foundation
of Guangdong Province (No. S2013010011928).
 }\\
{\small  Department of Mathematics}\\
 {\small South China University of Technology}\\
 {\small Guangzhou, Guangdong 510640, P. R. China }\\
{\small \tt slzhou@scut.edu.cn}
}
% \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{Jun 28, 2013}{X X, 2013}\\
\small Mathematics Subject Classifications: 05B05, 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}
  Let $\S$ be a  linear space with 106 points,  with lines of  size 6,  and let $G$ be an automorphism group of $\S$.  We prove that $G$  cannot be point-transitive. In other words, there exists no point-transitive $2$-$(106, 6, 1)$ designs.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} linear space; design; point-transitive
\end{abstract}

\section{Introduction}
For positive integers $v$ and $b$ satisfying $b\geq v\geq 2,$ a
 {\it finite linear space} $\S$  is an incidence structure $(\P,\L)$
consisting  of a set $\P$ of $v$ points and a collection $\L$ of $b$ distinguished subsets of $\P$ called lines,
 with sizes $\geq 2$ such that any two points are
incident with exactly one line.
 Let $\al$ be a point of $\P,$ and $k$ be an positive integer. Then $r_{\al}^k$ denotes the number of lines  having size $k$
 through $\al,$ $b^k$  the number of lines of size $k,$ and $r_{\al}$ the number  of all lines through $\al,$ called the {\it degree} of $\al.$
 If all lines have a constant size $k,$  then we say that $\S$ is {\it regular}, so it is a $2$-$(v,k,1)$ design.

Let $\Delta$ be a subset of $\P$ with $|\Delta|\geq 2,$  $
\L_{\Delta}=\{\lam \cap \Delta: |\lam \cap \Delta|\geq 2 \ for \
\lam \in \L\}.$ Then  $(\Delta, \L_{\Delta})$ forms an incidence
structure, and the induced structure is a linear space. We are
interested in the case that $\Delta$ is $\F(g)$ (or
$\F(H)$), the set of fixed points of $g\in G$ (or
$H\leq G$ ) on $\P$.
  An {\it automorphism} of $\S$ is a  permutation  of $\P$ which leaves $\L$ invariant. The full automorphism group of $\S$ is
  denoted by $\Aut(\S)$ and any subgroup of $\Aut(\S)$ is called an {\it automorphism group} of $\S$.
 We say that the automorphism group $G$ of $\S$ is {\it point-transitive}  if
  $G$ acts transitively on the set of points. Similarly,  $G$ is said to be {\it line-transitive}  if
  $G$ acts transitively on the set  of lines.


Four $2$-$(91, 6, 1)$ designs have been found by Mills, McCalla
and Colbourn (\cite{CC1980,CC1981,Mills}). All of them have a
cyclic automorphism group of order $91,$ and so they have
point-transitive automorphism groups. In 1989, Camina and Di
Martino (\cite{91points}) proved that any automorphism group of a point-transitive
$2$-$(91, 6, 1)$ design is the natural split extension of a cyclic
group of order $91$ by a cyclic group of order $d$, where $d\mid
12.$   Later,  Janko and  Tonchev (\cite{JT1991}) showed that  any cyclic
$2$-$(91, 6, 1)$ design (i.e. have an automorphism group of order $91$) admitting an automorphism group whose
order is larger than 91 is one of the four known designs. Here
  we are going to discuss the $2$-$(106, 6, 1)$ designs, which point-number is also a product of two distinct primes and $6$ is the largest line-size for a non-trivial regular linear space with $106$ points (\cite{crc}).
The only known $2$-$(106, 6, 1)$ design found by Mills  (\cite{Mills106})  is not point-transitive, which is one of Miscellaneous Constructions (\cite{crc}) and  has a cyclic automorphism groups of order $53.$ It is a wonder whether there exist point-transitive $2$-$(106, 6, 1)$ designs, just like the $2$-$(91, 6, 1)$ designs.
In this paper,
we prove that there is no $2$-$(106, 6, 1)$ designs admitting a
point-transitive automorphism group.

\begin{theorem}\label{groups} Let $\S$ be a $2$-$(106, 6,1)$ design, and $G$  be an automorphism group of $\S$.  Then $G$
cannot be transitive on points of $\S.$
\end{theorem}

Our paper is organized  as follows. Section 2 presents some preliminary results and notation. In Section 3, by considering the number of fixed points of an involutive automorphism,
 we bound the size of the $2$-part of $|\Aut(\S)|$. In Section 4, we get a bound of $|\Aut(\S)|$ and prove Theorem \ref{groups}.





\section{Preliminary results and notation}

Let $\S$ be a linear space with $v$ points,   $K$ be a set of positive integers such that $v\geq k$ for every $k\in K$ and the set of line-sizes of $\S$ is contained in $K.$ Note that  it is not required that there is a line of size $k$ for any $k\in K.$ Let $\al$ be a point of $\P,$ then

\begin{equation}\label{v-1}
    \sum_{k\in K}(k-1)r_{\al}^k=v-1
\end{equation} and for each $k\in K,$

\begin{equation}\label{k}
    \sum_{\al\in \P}r_{\al}^k=k\cdot b^k.
\end{equation}

In particular, if $\S$ is a non-trivial finite regular linear space,  then the following result is well-known.
\begin{lemma} {\rm\cite[Lemma 2.1]{alternating}}\label{vk}  Let $\S$ be a non-trivial  finite  regular linear space.   Then
 $$
  r=\frac{v-1}{k-1},\,\,\, b=\frac{v(v-1)}{k(k-1)},$$
and $$k(k-1)+1\leq v,$$
where $k$ is the line size of $\S$, and $r$ is the number of lines through a point.
\end{lemma}

The following results are very useful for the proof of Theorem \ref{groups}.

\begin{lemma} {\rm\cite[Lemma 1]{AJ1989}}\label{nof} Let $\S$ be a finite regular linear space, $G$  an automorphism group of $\S,$ and $H\neq1$  a subgroup of  $G.$ Then $|\F(H)|\leq r$ unless every point lies on a fixed line and then
$|\F(H)|\leq r+k-3.$

\end{lemma}



\begin{lemma}{\rm \cite[Lemma 1]{91points}}\label{maxr}
  Let $\S$ be a linear space, $\al$  a point of $\S$, and $r_{\al}$ be the degree of $\alpha$. Then all lines of size $>r_{\al}$ contain $\al,$ and for any
  point $\be$ of $\P$, $\be\neq \al$,  the number of lines of size $>r_{\al}$ containing $\be$ is at most one.
\end{lemma}

Throughout this paper, we assume that the following hypothesis  holds:

\textbf{{\sc Hypothesis:}} Let $\S=(\P, \L)$ be a
$2$-$(106, 6,1)$ design,  and $G$ be a point-transitive subgroup of
$\Aut(\S)$. Let $N:Q$ be the semidirect product of groups $N$ by
$Q,$ $N\times Q$  the direct product of groups $N$ and $Q$,
$|G|_p$ the $p$-part of $|G|$, and $|G|_{p'}$  the $p'$-part of
$|G|.$
















\section{The $2$-part of $|\Aut(\S)|$}

In this section, our aim is to obtain the maximal size of the
$2$-part of $|\Aut(\S)|.$ We begin this section with some  information given in \cite{91points}
about the linear spaces. Assume that
$2\mid|\Aut(\S)|$ and $t$ is an involution of $\Aut(\S)$.  Let
$\D=(\F(t),\L_{\F(t)})$ be the linear space induced by $\F(t)$ and $K=\{2, 4, 6\}$ containing the set of its line sizes. In view of
 (\ref{v-1}), we get

\begin{equation}\label{F(t)-1}
    r_{\al}^2+3r_{\al}^4+5r_{\al}^6=|\F(t)|-1,
\end{equation}
 for  each $\al \in \F(t)$.
Since a non-fixed point of $t$ cannot be on two fixed lines of it,  all the non-fixed points $t$ on its fixed lines of $\S$  are distinct.
Thus
\begin{equation}\label{nonfixed}
  4b^2+2b^4\leq 106-|\F(t)|.
\end{equation}

Combing   (\ref{k}) and (\ref{nonfixed}), we obtain
\begin{equation}\label{nonfixed2}
    2\sum_{\al\in \F(t)}r_{\al}^2+\frac{1}{2}\sum_{\al\in \F(t)}r_{\al}^4\leq 106-|\F(t)|.
\end{equation}

Now for each point $\al\in \F(t),$  define the  {\it weight} (\cite{91points})
$\omega(\al)$ of $\al$
$$\omega(\al)=2r_{\al}^2+\frac{1}{2}r_{\al}^4 .$$
 So that (\ref{nonfixed2}) can be written
as
\begin{equation}\label{weight}
   \sum_{\al \in  \F(t)}\omega(\al)\leq106-|\F(t)|.
\end{equation}

If $r_{\al}^2=x, r_{\al}^4=y$ and $ r_{\al}^6=z,$ then we say that $\al$
 is of type $(x,y,z).$




\begin{lemma}\label{involution-1}Assume that the  {\sc Hypothesis} holds and
let $t$ be an involution of $\Aut{\S}.$  Then $|\F(t)|\neq 18, 20$ or $22.$

\end{lemma}


 \begin{proof}   We prove this Lemma by dealing with   the three cases separately.

(1)\,\, If $|\F(t)|= 22,$ then according to inequality (\ref{weight}), $\F(t)$ has a point $\al_0$ such that $\omega(\al_0)\leq \frac{106-22}{22}=\frac{42}{11}.$
Thus $$2r_{\al_0}^2+\frac{1}{2}r_{\al_0}^4\leq\frac{42}{11}.$$
Recall that for any $\al \in \F(t)$, we have $r_{\al}^2+3r_{\al}^4+5r_{\al}^6=21$ from (\ref{F(t)-1}).
This implies that $\al_0$ is of type $(0, 7, 0), (0, 2, 3)$ or $(1, 0, 4).$

If $\al_0$ is of type $(1, 0, 4),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t)$ by Lemma \ref{maxr}. Precisely, the $20$ points   on the lines of size $6$ through $\al_0$ are of type $(x, y, 1),$ where $x$ and $y$ are non-negative integers.  Thus all these $20$ points have weight  $\geq \frac{9}{2},$  and they contribute at least
$20\cdot \frac{9}{2}=90$ to the total weight, contradicting  (\ref{weight}). Similar discussion leads to have a contradiction for  type $(0, 2, 3)$. Therefore, there must be a point $\al_0\in\F(t)$ of type $(0, 7, 0),$ but there is no point of type $(0, 2, 3)$ or $(1, 0, 4).$
Moreover, $\omega(\al)\geq \frac{9}{2}$ for any $\al \in \F(t)$ with $r_{\al}^6\geq 1.$

If all points of $\F(t)$ are of type $(0, 7, 0)$, then $\D$ is a $2$-$(22, 4, 1)$ design. But then $b=\frac{v(v-1)}{k(k-1)}$ is not an integer. Thus, there exists a point $\be\in \F(t)$ that is not of type $(0, 7, 0).$

(1.1)\,\, Suppose that $r_{\be}^6=0.$ Then $r_{\be}^2\geq3$ and
$\omega(\be)\geq9.$ Let $\be_1, \be_2$ and $\be_3$ be three
distinct points such that $\{\be,\be_i\}$ $(i=1,2,3)$ are lines of
size $2.$  For $i=1,2 $ and $3,$  $\omega(\be_i)\geq9$ if
$r_{\be_i}^6=0,$ and $\omega(\be_i)\geq\frac{9}{2}$ if
$r_{\be_i}^6\neq 0.$  If $r_{\be_i}^6=0$ for  all $i=1,2$ and $3,$
then
$$\sum_{\al\in \F(t)}\omega(\al)\geq \omega(\al_0)+\omega(\be)+\sum_{i=1}^3\omega(\be_i)+17\cdot\frac{7}{2}=99,$$ contrary to inequality (\ref{weight}).
 If there exists one point, say $\be_1$, such that $r_{\be_1}^6\neq0,$ let $\lam$ be a line of size $6$ through $\be_1.$ Then $\omega(\al)\geq \frac{9}{2}$ for all $\al \in \lam.$ Hence
$$\sum_{\al\in \F(t)}\omega(\al)\geq \omega(\al_0)+\omega(\be)+\sum_{\al\in \lam}\omega(\al)+14\cdot\frac{7}{2}=88+\frac{1}{2},$$ which is impossible.

(1.2)\,\, Suppose that $r_{\be}^6\neq0.$  If  $r_{\be}^2=0,$  then
$\be$ is of type $(0, 2, 3),$ a contradiction. Thus there is at
least one line $\{\be, \be_1\}$ of size $2$ and one line $\lam $
of size $6$ through $\be.$ If $r_{\be_1}^6\neq0,$  let $\lam_1$ be
a line of size $6$ through $\be_1.$ Then $$\sum_{\al\in
\F(t)}\omega(\al)\geq \omega(\al_0)+\sum_{\al \in
\lam\cup\lam_1}\omega(\al)+10\cdot\frac{7}{2}>84,$$ which is a
contradiction. If $r_{\be_1}^6=0,$ then $\omega(\be_1)\geq 9$  and
$$\sum_{\al\in \F(t)}\omega(\al)\geq \omega(\al_0)+\omega(\be_1)+\sum_{\al\in \lam}\omega(\al)+14\cdot\frac{7}{2}=88+\frac{1}{2},$$ which is impossible.
Therefore, $|\F(t)|\neq 22.$

(2)\,\, If $|\F(t)|=20,$ then $\F(t)$ has a point $\al_0$ such that $\omega(\al_0)\leq \frac{106-20}{20}=\frac{43}{10}.$
From (\ref{F(t)-1}), we have
 $\al_0$ is of type $(0, 3, 2)$ or $(1, 1, 3).$

If $\al_0$ is of type $(0, 3, 2),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t)$ by Lemma \ref{maxr}. Exactly, the $10$ points on the lines of size $6$ through $\al_0$ are of type $(m, n, 1)$ having weight at least $ 6,$  and the  $9$ points   on the lines of size $4$ through $\al_0$ are of type $(x, y, 0)$ having weight at least $ 5,$ where $m,n, x$ and $y$ are non-negative integers.
Then $$\sum_{\al\in \F(t)}\omega(\al)\geq \omega(\al_0)+10\cdot 6+9\cdot 5=106+\frac{1}{2},$$  and we have a contradiction.
Similarly, we can prove that $\al_0$ is not of type  $(1, 1, 3).$ Therefore, $|\F(t)|\neq 20.$


(3)\,\, If $|\F(t)|=18,$ then $\F(t)$ has a point $\al_0$ such that $\omega(\al_0)\leq \frac{106-18}{18}=\frac{44}{9}.$
According to equation (\ref{F(t)-1}),
$\al_0$ is of type $(0, 4, 1), (1,2,2)$ or $(2, 0, 3).$

If $\al_0$ is of type $(0, 4, 1),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t)$ by Lemma \ref{maxr}. Or rather, the $5$ points  on the line of size $6$ through $\al_0$ are of type $(m, n, 1)$ with weight $\geq 2,$ and the  $12$ points  on the lines of size $4$ through $\al_0$ are of type $(x, y, 0)$ with weight $\geq \frac{13}{2},$ where $m,n, x$ and $y$ are non-negative integers.
Then $$\sum_{\al\in \F(t)}\omega(\al)\geq 6\cdot 2+12\cdot  \frac{13}{2}=90,$$ a contradiction. Thus there is no point of type $(0, 4, 1)$ and $\omega(\al)\geq \frac{15}{2}$ for any $\al\in \F(t)$ with $r_{\al}^6\geq1.$

If $\al_0$ is of type $(1, 2, 2),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t)$ by Lemma \ref{maxr}. In particular, the $10$ points on the lines of size $6$ through $\al_0$ are of type $(m, n, 1)$ with weight $\geq \frac{15}{2},$  and the  $7$ points  which do not lie on the lines of size $6$ through $\al_0$ are of type $(x, y, 0)$ with weight $\geq \frac{13}{2},$ where $m,n, x$ and $y$ are non-negative integers.
This implies  $$\sum_{\al\in \F(t)}\omega(\al)\geq 3+10\cdot  \frac{15}{2}+7\cdot  \frac{13}{2}=123+\frac{1}{2},$$  which is impossible.
 Similarly,  $\alpha_0$ cannot be of type $(2, 0, 3)$. Thus $|\F(t)|\neq 18.$

\end{proof}

\begin{lemma}\label{involution-2}Assume that the  {\sc Hypothesis} holds and
  let $t$ be an involution of $\Aut{(\S)}.$ If there is a line $\lam$ of $\S$ contained in $\F(t),$
then $|\F(t)|\neq 12,14$ or $ 16.$
\end{lemma}

\begin{proof} Since there is a line $\lam\in \S$ such that $\lam \subseteq\F(t),$
 the linear space $\D$ induced by $\F(t)$ has at least one line of size $6.$

(1) If $|\F(t)|=16,$  then $\F(t)$ has a point $\al_0$ such that $\omega(\al_0)\leq \frac{106-16}{16}=\frac{45}{8}.$
Thus $$2r_{\al_0}^2+\frac{1}{2}r_{\al_0}^4\leq\frac{45}{8}.$$ Recall that $r_{\al}^2+3r_{\al}^4+5r_{\al}^6=15$ for all $\al \in \F(t)$ from equation (\ref{F(t)-1}). So $\al_0$ is of type $(0, 0, 3), (0, 5, 0),$ $ (1, 3, 1)$ or $(2, 1, 2).$


Since $\D$ has at least one line of size $6,$   there is no point of
type $(0, 5, 0)$ by Lemma \ref{maxr}. If $\al_0$ is of type $(0, 0,
3),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6=1$
for any other point $\al\in \F(t).$  More precisely, all the $15$
points  are of type $(10, 0, 1)$ having weight $20.$ Then
$$\sum_{\al\in \F(t)}\omega(\al)=0+ 15\cdot 20=300,$$  which is
impossible.

If $\al_0$ is of type $(1, 3, 1),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t).$  In detail, the $5$ points on the line of size $6$ through $\al_0$ are of type $(x, y, 1)$ with  weight $\geq \frac{7}{2},$ where $x$ and $y$ are non-negative integers. Let $\{\al_0,\be_0\}$ be the line of size $2$ through $\al_0,$  and so $\omega(\be_0)\geq8.$
 If $\al \neq\al_0$ lies on one of lines of size $4$ through $\al_0,$   then $r_{\al}^4\leq4$ since there is no point of type $(0, 5, 0),$ thus $\omega(\al)\geq8.$
 Hence $$\sum_{\al\in \F(t)}\omega(\al)\geq 6\cdot \frac{7}{2}+10\cdot 8=101.$$  It follows that there is no point of type $(1, 3, 1)$ and $\omega(\al)\geq 9$ for any $\al\in \F(t)$  with $r_{\al}^6\geq 1$. Similar analysis as above, we known that $\al_0$ cannot be of type $(2,1,2).$ Therefore, $|\F(t)|\neq 16.$


(2) If $|\F(t)|=14,$  then $\F(t)$ has a point $\al_0$ such that $\omega(\al_0)\leq \frac{106-14}{14}=\frac{46}{7}.$
According to  (\ref{F(t)-1}),
 $\al_0$ is of type $(0, 1, 2), (1, 4, 0), (2, 2, 1)$ or $(3, 0, 2).$


Recall that $\D$ has at least one line of size $6.$  Thus there is no point of type $(1, 4, 0)$ by Lemma \ref{maxr}.
If $\al_0$ is of type $(0, 1, 2),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t).$   Particularly, the $10$ points on the lines of size $6$ through $\al_0$ are of type $(8, 0, 1)$ with weight $16,$ and   the $3$ points on the line of size $4$ through $\al_0$ are of type $(13, 0, 0)$ with weight $ 26.$
Then $$\sum_{\al\in \F(t)}\omega(\al)= \frac{1}{2}+10\cdot 16+3\cdot 26=238+\frac{1}{2},$$ which is impossible.


If $\al_0$ is of type $(2, 2, 1),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t).$
Let $ \al_1$ and $ \al_2$ be the two points that joined to $\al_0$ form the two lines of size $2.$  Suppose that there is a line of $\D$ having size $4$ not containing $\al_0.$  Then this line must contain one point of $\{\al_1, \al_2\}.$ Thus $r_{\al}^4\leq 2$ for each point $\al$ which is on the line of size $6$ through $\al_0,$ and $r_{\al}^4\leq 3$ for each point $\al$  which do not lie on the line of size $6$ through $\al_0.$
This implies  $$\sum_{\al\in \F(t)}\omega(\al)\geq 5+5\cdot 5+8\cdot  \frac{19}{2}=106,$$  a contradiction.

If $\al_0$ is of type $(3, 0, 2),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t).$
Let $ \al_1, \al_2$ and $ \al_3$ be the  points each of which  forms a line of size $2$ with $\al_0.$ Suppose that there is a line  of $\D$ having size $4.$   Then this line must contain two points of $\{\al_1, \al_2, \al_3\}.$ Thus $r_{\al}^4\leq 1$ for each point $\al\in \F(t).$
Hence $$\sum_{\al\in \F(t)}\omega(\al)\geq 6+10\cdot  \frac{21}{2}+3\cdot  \frac{41}{2}>92.$$  Thus $\F(t)$ does not have a point of type of $(3, 0, 2).$
It follows that $|\F(t)|\neq 14.$

(3) If $|\F(t)|=12,$  then $\F(t)$ has a point $\al_0$ such that $\omega(\al_0)\leq \frac{106-12}{12}=\frac{47}{6}.$
So $\al_0$ is of type $(0, 2, 1), (1, 0, 2), (2, 3, 0)$ or $(3, 1, 1)$ from equation (\ref{F(t)-1}).


Since $\D$ has at least one line of size $6,$   there is no point of type $(2, 3, 0)$ by Lemma \ref{maxr}.
If $\al_0$ is of type $(0, 2, 1),$ then $\al_0$ lies on all lines of size $6,$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t).$   Precisely, the $5$ points  on the line of size $6$ through $\al_0$ are of type $(6, 0, 1)$  having weight $12,$ and   the $6$ points  on the lines of size $4$ through $\al_0$ are of type $(8, 1, 0)$ having weight $ \frac{33}{2}.$
Then $$\sum_{\al\in \F(t)}\omega(\al)= 1+5\cdot 12+6\cdot \frac{33}{2}=160,$$ which is impossible.


If $\al_0$ is of type $(1, 0, 2),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t).$   The $10$ points which lie on the lines of size $6$ through $\al_0$ are of type $(6, 0, 1)$ having weight $ 12,$ and the point $\al_1$ such that $\{\al_0,\al_1\}$  is   the line of size $2$ is of type $(11, 0, 0)$ having weight $22.$
Then $$\sum_{\al\in \F(t)}\omega(\al)= 2+10\cdot 12+1\cdot 22=144,$$ a contradiction.



If $\al_0$ is of type $(3, 1, 1),$ then $\al_0$ lies on all lines of size $6$ and $r_{\al}^6\leq1$ for any other point $\al\in \F(t).$
Let $ \al_1, \al_2$ and $ \al_3$ be the  points that joined to $\al_0$ form the there lines of size $2.$  Suppose that there is a line of size $4$ not containing $\al_0.$  Then this line must contain two points of $\{\al_1, \al_2, \al_3\}.$ Thus $r_{\al}^4\leq 1$ for the point $\al$  which do not lie on one line of size $4$ through $\al_0,$ and $r_{\al}^4\leq 2$ for the  point $\al$  which  lie on the line of size $4$ through $\al_0.$
Hence $$\sum_{\al\in \F(t)}\omega(\al)\geq  \frac{13}{2}+5\cdot  \frac{13}{2}+3\cdot  \frac{33}{2}+3\cdot  11>94.$$  Thus $\F(t)$ does not have a point with type $(3, 1, 1).$
Therefore $|\F(t)|\neq 12.$


\end{proof}


\begin{lemma}\label{2part}Assume that the  {\sc Hypothesis} holds and
  let $T\neq 1$ be a Sylow $2$-subgroup of $\Aut{(\S)}.$  Then $|T|$ divides $2^{15}.$
\end{lemma}

\begin{proof} Since $b=371,$ then there exists a line $\lam\in \L$ such that $T$ fixes $\lam$ setwise. Thus $T/T_{(\lam)}\leq S_6$ and $|T:T_{(\lam)}|$ divides $ 2^4.$
Assuming that $S=T_{(\lam)}\neq 1,$ we obtain
$$100=\sum_{i=1}^r|S:S_{\al_i}|,$$ where $\al_1, \al_2,\ldots, \al_m$ are all the representatives of the orbits of $S$ on $\P\setminus \F(S).$
If $S$ is semiregular on $\P\setminus\F(S),$ then $|S|$ divides $ 100,$ and hence $|T|$ divides $ 2^4\cdot 2^2.$ If  $S$ is not semiregular on $\P\setminus\F(S),$ then $\P\setminus\F(S)$ has at least one point $\al_j$$(1\leq j \leq m)$ such that $S_{\al_j}\neq1$ and $2^6$ is  divisible  by $|S:S_{\al_j}|.$ In view of Lemmas \ref{involution-1} and \ref{involution-2},  $|\F(S_{\al_j})|=8, 10 $ or $24.$

Let $\lam_1\in\L$ be the line containing $\al_j$ that has non-empty intersection with $\lam.$  Suppose that $1\neq S_0$ is the kernel of the action of $S_{\al_j}$ on the points of $\lam_1,$
  then $|\F(S_0)|=24$, $2^4$ is divisible by $|S_{\al_j}:S_0|$ and  $S_0$
  acts semiregularly on $\P\setminus\F(S_0)$ since $|\F(t)|\leq24$ for any involution $t$ of $\Aut(\S).$ Thus $|S_0|$ divides $82,$
  and it follows that $|T|$ divides $2^4\cdot 2^6\cdot 2^4\cdot2=2^{15}.$ \end{proof}


\section{The $2'$-part of $|\Aut(\S)|$ and proof of Theorem \ref{groups}}
In this section, we prove a bound  on the $2'$-part of $|\Aut(\S)|.$ First, we list a result which is important for our discussion.



\begin{lemma}{\rm \cite{91points}}\label{3and6}
  If $\S$ is a linear space having lines of size $3$ and $6$ $($with at least one line of size $3$ and one line of size $6$$)$.  Then $v=16$ or $18,$ provided that $v<21.$
\end{lemma}

\begin{lemma}\label{p>4}Assume that the  {\sc Hypothesis} holds. Then
   $|\Aut(\S)|_{2'}$ divides $3^a\cdot5^3 \cdot7 \cdot 53,$ where $a$ is a non-negative integer. Furthermore,
any element of order $53$ in $\Aut(\S)$ is fixed-point-free.

\end{lemma}

 \begin{proof} Let $p\geq 5$ be a prime divisor of $|\Aut(\S)|,$ and $g $  an element  of order $p$ in $\Aut(\S).$ Then $|\F(g)\cap \lam|=0,1$ or $6.$

Suppose that $\F(g)$ is not contained in a line of $\S,$ then $\F(g)$ induces a regular linear space $2$-$(|\F(g)|, 6, 1).$ Thus $|\F(g)|\geq k(k-1)+1=31$ by Lemma \ref{vk}. But $|\F(g)|\leq k+r-3=24$ according to Lemma  \ref{nof}, which is a contradiction. Hence there exists a line $\lam\in \L$ such that $\F(g)\subseteq \lam$ and $|\F(g)|=0, 1$ or $6.$ Therefore,
the possible values of $p$ are $5,7$ and $53$ since $106-|\F(g)|\equiv0\pmod p.$



Let $P$ be a Sylow $5$-subgroup of $\Aut(\S)$ and $P\neq 1.$ Then $P$ fixes one line $\lam\in \L$ and $|P|\mid 5\cdot|P_{(\lam)}|$ since $P/P_{(\lam)}\leq S_6.$  If $P_{(\lam)}\neq 1, $ then $P_{(\lam)}$ acts semiregularly on $\P\setminus \lam.$  Thus $|P_{(\lam)}|\mid 100.$ So $|P_{(\lam)}|\mid 5^2,$ and $|P|\mid 5^3.$

Let $P$ be a Sylow $p$-subgroup of $\Aut(\S),$  where $p=7$ or $53.$  Suppose  $P\neq1,$ then $|\F(P)|=1$ if $p=7,$ and $|\F(P)|=0$ if $p=53.$
 Moreover, $P$ is semiregular on $\P\setminus \F(P).$  So $|P|=7$ or $53.$
\end{proof}
\begin{lemma}\label{p=3}Assume that the  {\sc Hypothesis} holds. Then
 $|\Aut(\S)|_{3}$ divides $3^7.$

\end{lemma}


 \begin{proof} If $1\neq T\in Syl_3(G),$ then $\S$ has a line $\lam\in \L$ such that $T\leq G_{\lam} $ since $b=371.$
Thus $T/T_{(\lam)}\leq S_6$ and $|T|$ divides $ 3^2\cdot |T_{(\lam)}|. $ Let $S=T_{(\lam)},$ then $|\F(S)|\equiv 1 \pmod 3.$

If $S\neq 1,$ then $\F(S)$ induces a linear space $\D$ with sizes of lines from $K=\{3,6\}.$
Since $|\F(S)|\leq 24$ by Lemma \ref{nof} and $v\geq 31$ for a $2$-$(v, 6, 1)$ design by Lemma \ref{vk} ,  $\D$ has at least one line of size $3$ and one line of size $6.$
It follows that $|\F(S)|=16$ or $22$ by Lemma \ref{3and6}.

Suppose first that $|\F(S)|=16.$ Then $$90=\sum_{i=1}^r|S:S_{\al_i}|,$$ where $\al_1, \al_2,\ldots, \al_m$ are all the representatives of the orbits of $S$ on $\P\setminus \F(S).$
If $S$ is semiregular on $\P\setminus \F(S),$ then $|S|\mid90,$ this implies that $|S|$ divides $3^2.$ If there exists $\al_j\in \P\setminus \F(S) $$(1\leq j \leq m)$  such that
$S_{\al_j}\neq 1,$ then $|\F(S_{\al_j})|=22$ and $S_{\al_j}$ acts semiregularly on $\P\setminus\F(S_{\al_j}).$ Hence $|S_{\al_j}|$ divides $ 3,$  so $|S|$ divides $ 3^5$ since $3^4$ is divisible by $|S:S_{\al_j}|.$

Now suppose that $|\F(S)|=22.$ Then $S$ is semiregular on
$\P\setminus\F(S).$ Hence $|S|$ divides $3.$ Therefore, $|T|$ divides $3^7.$
\end{proof}

 According to the  discussion above,  we have $|G|$ divides $2^{15}\cdot3^7\cdot5^3\cdot7\cdot53.$

\begin{lemma}\label{mn}
 Assume that the  {\sc Hypothesis} holds. If $N$ is a minimal normal subgroup of $G,$ then $N=\Soc(G)\cong Z_{53}.$
\end{lemma}
\begin{proof} Since $N\unlhd G$ and $G$ is point-transitive,  $N$ is $\frac{1}{2}$-transitive on $\P$, and the common  length of orbits is $2, 53$ or $106.$
Assume that $N$ is not elementary abelian. Then  $N \cong T^{\ell}$ is a direct product of $\ell\geq1$ copies of non-abelian simple groups $T.$ So $ |N|$ is divisible by  $3$ or $5$ (\cite[Remarks 3.7 a)]{HU1982}).
Suppose that the orbit-length of $N$ on the points is $2$ and let $g\in N$ be of order $3$ or $5.$
Then $g$ fixes every orbit of $N$ on $\P,$ which implies $|\F(g)|=106,$ a contradiction. Thus the common length must be divisible by $53,$ so $53$ divides $|N|$ and $N=T=\Soc(G)$ is simple  since $53^2\nmid |G|.$

Using the list of non-abelian simple groups, it is easy  to check that there is
no  non-abelian simple group $N$ such that $53$ divides $ |N|$ and $|N|$ divides $
2^{15}\cdot3^7\cdot5^3\cdot7\cdot53.$ Hence $N$ is
abelian and  $N\cong Z_{53}.$ \end{proof}

\begin{lemma}\label{first}
  Assume that the  {\sc Hypothesis} holds. Then $G$ is solvable and $G\cong Z_{53}:Z_4.$
\end{lemma}

\begin{proof} According to Lemma \ref{mn}, $N=\Soc(G)\cong Z_{53}.$

If $N\neq C_G(N),$ then by Schur-Zassenhaus'theorem,  $C_G(N)$ has a normal subgroup $M$ such that $C_G(N)=M\times N,$ since $N$ is a normal Hall-subgroup of  $C_G(N).$  Since  $M$ is characteristic in $C_G(N)$  and $C_G(N)\unlhd G,$  then there exists another minimal normal subgroup of $G$ contained in $M,$ but this is impossible. Therefore $N=C_G(N),$ and $G/C_G(N)\lesssim \Aut(N) \cong Z_{52}.$ This implies that $G$ is solvable  and $G\lesssim Z_{53} : Z_4 $  since $13\nmid |G|.$

If $|G|=106,$ then $G$ is regular on $\P.$ Thus any involution of
$G$ has no fixed point. This implies that $k\mid v,$ a
contradiction. Therefore, $G\cong Z_{53}:Z_4.$ \end{proof}


\begin{proof}[{\rm Proof of Theorem~\ref{groups}}]
 According to Lemma
\ref{first}, $G\cong Z_{53}:Z_4$ which is a metacyclic group. Let
$G=\langle x, y | x^{53}=1, y^4=x^h, x^{y}=x^{\ell}\rangle,$ where $h$ and $\ell$ are non-negative integers
satisfying  \begin{center}
  $\ell^4\equiv1\pmod {53}, h(\ell-1)\equiv0\pmod{53}$
\end{center} by
\cite{metacyclic}. It follows that $ \ell\equiv 1, 23, 30 \,
{\mbox {or}} \, 52 \pmod {53}.$  If $\ell\equiv 1 \pmod {53},$ then $h\equiv 1\pmod {53} $ and  if $ \ell\equiv 23, 30 \,
{\mbox {or}} \, 52 \pmod {53},$  then $h\equiv 0\pmod {53} .$ Suppose first $\ell\equiv 1 \, {\mbox {or}} \, 52 \pmod{53},$ then $G$
has a normal subgroup of order $2$, a contradiction.


For the cases $\ell\equiv 23  \, {\mbox {and}} \, 30 \pmod{53},$ the
groups are the same up to isomorphic.  Without loss of generality we assume  that $\ell\equiv 23 \pmod{53}$ in the following.
 Since $G$ has only one conjugacy class of involutions, $G$ has exactly
 one transitive permutation representation on $106$ points, which is equivalent to the permutation representation of $G$ on $H=\langle y^2 \rangle.$

Since $G=\{x^my^n | 0\leq m\leq 52, 0\leq n\leq 3\},$  the
right cosets of  $H$ in $G$ are $H$, $ Hx$, $ Hx^2$,$\ldots$, $Hx^{52}$,
$Hy$, $Hxy$, $\ldots$, $Hx^{52}y$. Let $Hx^m=i_{m+1},
Hx^my=j_{m+1},$ where $0\leq m\leq 52.$  Then the permutation
representation of $G$ on $\P=\{i_1,i_2,\cdots,
i_{53},j_1,j_2,\cdots,j_{53}\}$ is $P(G)=\langle
g_1,g_2\rangle$, where  $$g_1=(i_1i_2\cdots
i_{53})(j_{53}j_{52}\cdots j_1)$$ and
$$g_2=(i_1j_1)(i_2j_2i_{53}j_{53})(i_3j_3i_{52}j_{52})\cdots(i_{27}j_{27}i_{28}j_{28}).$$

Assume that $\lam$ is the line through $i_1$ and $j_1.$ Then
$\lam^{g_2}=\lam.$ It follows that $\lam$ is a union of orbits of
$g_2$ on $\P.$ Thus there exists an integer $e$  $(2\leq e\leq 27)$ such
that
$$\lam=\{i_1,j_1,i_e,j_e,i_{55-e},j_{55-e}\}.$$
However,
$\lam^{g_1^{54-e}}=\{i_{55-e},j_{e},i_1,j_{2e-1},i_{56-2e},j_1\},$
which is impossible.  This completes the proof of Theorem
\ref{groups}.  \end{proof}


\subsection*{Acknowledgements}
Thanks to Professor Sanming Zhou at University of Melbourne for corrections and some useful discussion which lead to the improvement of the paper.

\begin{thebibliography}{10}
\bibitem{metacyclic}B G Basmaji.
On the isomorphisms of two metacyclic groups. Pro. Amer. Math.
Soc., 22: 175-182, 1969.



\bibitem{designtheory}T Beth, D Jungnickel, and H Lenz. Design Theory.  Bibliographisches Institut Mannheim,
Wien, Zurich, 1985.



\bibitem{91points} A R Camina, and L Di Martino. The group of automorphisms of a transitive $2$-$(91, 6, 1)$ design. Geometriae Dedicata, 31: 151-163, 1989.
\bibitem{AJ1989}A R Camina, and J Siemons. Block transitive automorphism groups of $2$-$(v,k,1)$ block designs.  J. Combinatorial Theory Ser. A,  51(2): 268-276, 1989.
\bibitem{alternating} A R Camina, P M Neumann, and C E Praeger. Alternating groups acting on finite linear spaces.  London Math. Soc., 87(3): 29-53, 2003.


\bibitem{CC1980} C J Colbourn, and M J Colbourn. Cyclic steiner systems having multiplier automorphisms. Utilitas Math., 17: 127-149, 1980.
\bibitem{CC1981}C J Colbourn, and M J Colbourn. On cyclic Steiner systems S(2, 6, 91). Abstracts Amer. Math. Soc.,  2: 463, 1981.


\bibitem{crc} C J Colbourn, J H Dinitz. Handbook of Combinatorial Designs
 (2nd edition).
Boca Raton, FL: CRC press, 2006



\bibitem{HU1982} B Hupper, N Blackburn.  Endliche Gruppen III, Springer-Verlag, Berlin, 1982.


\bibitem{JT1991} Z Janko, and V D Tonchev. Cyclic $2$-$(91, 6, 1)$ designs with multiplier automorphisms. Discrete Mathematics, 97(1-3): 265-268, 1991.
  \bibitem{Mills106} W H Mills. A new block design. Congr. Numer, 14: 461-465, 1975.
    \bibitem{Mills} W H Mills. Two new block designs. Utilitas Math., 7: 73-75, 1975.


\end{thebibliography}


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