# On flag-transitive $2$-$(k^{2},k,\lambda)$ designs with $\lambda\mid k$.

Alessandro Montinaro and Eliana Francot  Alessandro Montinaro and Eliana
Francot, Dipartimento di Matematica e Fisica “E. De Giorgi”, University of
Salento, Lecce, Italy<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

It is shown that, apart from the smallest Ree group, a flag-transitive
automorphism group $G$ of a $2$-$(k^{2},k,\lambda)$ design $\mathcal{D}$, with
$\lambda\mid k$, is either an affine group or an almost simple classical
group. Moreover, when $G$ is the smallest Ree group, $\mathcal{D}$ is
isomorphic either to the $2$-$(6^{2},6,2)$ design or to one of the three
$2$-$(6^{2},6,6)$ designs constructed in this paper. All the four $2$-designs
have the $36$ secants of a nondegenerate conic $\mathcal{C}$ of $PG_{2}(8)$ as
a point set and $6$-sets of secants in a remarkable configuration as a block
set.

###### Key words and phrases:

$2$-design; automorphism group; flag-transitive

###### MSC 2020: Mathematics Subject Classification:

05B05; 05B25; 20B25

Corresponding author: Alessandro Montinaro

##  1\. Introduction and Main Result

A $2$-$(v,k,\lambda)$ _design_ $\mathcal{D}$ is a pair
$(\mathcal{P},\mathcal{B})$ with a set $\mathcal{P}$ of $v$ points and a set
$\mathcal{B}$ of blocks such that each block is a $k$-subset of $\mathcal{P}$
and each two distinct points are contained in $\lambda$ blocks. We say
$\mathcal{D}$ is _nontrivial_ if $2<k<v$. All $2$-$(v,k,\lambda)$ designs in
this paper are assumed to be nontrivial. An automorphism of $\mathcal{D}$ is a
permutation of the point set which preserves the block set. The set of all
automorphisms of $\mathcal{D}$ with the composition of permutations forms a
group, denoted by $\mathrm{Aut(\mathcal{D})}$. For a subgroup $G$ of
$\mathrm{Aut(\mathcal{D})}$, $G$ is said to be _point-primitive_ if $G$ acts
primitively on $\mathcal{P}$, and said to be _point-imprimitive_ otherwise. A
_flag_ of D is a pair $(x,B)$ where $x$ is a point and $B$ is a block
containing $x$. If $G\leqslant\mathrm{Aut(\mathcal{D})}$ acts transitively on
the set of flags of $\mathcal{D}$, then we say that $G$ is _flag-transitive_
and that $\mathcal{D}$ is a _flag-transitive design_.

The $2$-$(v,k,\lambda)$ designs $\mathcal{D}$ admitting a flag-transitive
automorphism group $G$ have been widely studied by several authors. In 1990, a
classification of those with $\lambda=1$ and $G\nleq A\Gamma L_{1}(q)$ was
announced by Buekenhout, Delandtsheer, Doyen, Kleidman, Liebeck and Saxl in
[13] and proven in [12], [15], [16], [17], [24], [29] and [35]. Since then a
special attention was given to the case $\lambda>1$. A classification of the
flag-transitive $2$-designs with $\gcd(r,\lambda)=1$, $\lambda>1$ and $G\nleq
A\Gamma L_{1}(q)$, where $r$ is the replication number of $\mathcal{D}$, has
been announced by Alavi, Biliotti, Daneshkakh, Montinaro, Zhou and their
collaborators in [2] and proven in [3], [4], [5], [8], [10], [11], [30], [37],
[39], [40], [41], [42], [44], [45] and [46]. Moreover, recently the flag-
transitive $2$-designs with $\lambda=2$ have been investigated by Devillers,
Liang, Praeger and Xia in [19], where it is shown that apart from the two
known symmetric $2$-$(16,6,2)$ designs, $G$ is primitive of affine or almost
simple type. Moreover, a classification is provided when the socle of $G$ is
isomorphic to $PSL_{n}(q)\trianglelefteq G$ and $n\geqslant 3$.

The present paper represents a further contribution to the study of the flag-
transitive $2$-designs. More precisely, the flag-transitive
$2$-$(k^{2},k,\lambda)$ designs with $\lambda\mid k$ are investigated. The
reason of studying such $2$-designs is that they represent a natural
generalization of the affine planes in terms of parameters, and also because,
it is shown in [33] that, the blocks of imprimitivity of a family of flag-
transitive, point-imprimitive symmetric $2$-designs investigated in [34] have
the structure of the $2$-designs analyzed here. The following result is
obtained:

######  Theorem 1.1.

Let $\mathcal{D}$ be a $2$-$(k^{2},k,\lambda)$, with $\lambda\mid k$,
admitting a flag-transitive automorphism group $G$. Then $G$ is point
primitive and one of the following holds:

  1. (1)

$G$ is an affine group.

  2. (2)

$G$ is an almost simple classical group.

  3. (3)

$\mathcal{D}$ is isomorphic to the $2$-$(36,6,2)$ design constructed in
Example 3.9 and ${}^{2}G_{2}(3)^{\prime}\trianglelefteq G\leqslant$
${}^{2}G_{2}(3)$.

  4. (4)

$\mathcal{D}$ is isomorphic to one of the three $2$-$(36,6,6)$ designs
constructed in Example 3.9 and $G\cong$ ${}^{2}G_{3}(3)$.

Actually, (3) and (4) are special cases of (2), since ${}^{2}G_{3}(3)\cong
P\Gamma L_{2}(8)$. It worth noting that the example in (3) is not contained in
[19] and hence it is presumably new. A complete classification of (1) for
$G\nleq A\Gamma L_{1}(q)$, and of (2) are contained in [31] and [32]
respectively.

##  2\. Preliminary Reductions

We first collect some useful results on flag-transitive designs.

######  Lemma 2.1.

Let $\mathcal{D}$ be a $2$-$(k^{2},k,\lambda)$ design and let $b$ be the
number of blocks of $\mathcal{D}$. Then the number of blocks containing each
point of $\mathcal{D}$ is a constant $r$ satisfying the following:

  1. (1)

$r=\lambda(k+1)$;

  2. (2)

$b=\lambda k(k+1)$;

  3. (3)

$\left(r/\lambda\right)^{2}>k^{2}$.

######  Lemma 2.2.

If $\mathcal{D}$ is a $2$-$(k^{2},k,\lambda)$ design, with $\lambda\mid k$,
admitting a flag-transitive automorphism group $G$, then the following hold:

  1. (1)

$G$ acts point-primitively on $\mathcal{D}$.

  2. (2)

If $x$ is any point of $\mathcal{D}$, then $G_{x}$ is a large subgroup of $G$.

  3. (3)

$\left|y^{G_{x}}\right|=(k+1)\left|B\cap y^{G_{x}}\right|$ for any point $y$
of $\mathcal{D}$, with $y\neq x$, and for any block $B$ of $\mathcal{D}$
incident with $x$. In particular, $k+1$ divides the length of each
point-$G_{x}$-orbit on $\mathcal{D}$ distinct from $\left\\{x\right\\}$.

###### Proof.

The assertion (1) follows from [18], 2.3.7.c, since
$r=(k+1)\lambda>(k-3)\lambda$.

The flag-transitivity of $G$ on $\mathcal{D}$ implies
$\left|G\right|=k^{2}\left|G_{x}\right|$,
$\left|G_{x}\right|=\lambda\left(k+1\right)\left|G_{x,B}\right|$ and hence
$\left|G\right|<\left|G_{x}\right|^{3}$, which is the assertion (2).

Let $y$ be any point of $\mathcal{D}$, $y\neq x$, and $B$ be any block of
$\mathcal{D}$ incident with $x$. Since $(y^{G_{x}},B^{G_{x}})$ is a tactical
configuration by [18], 1.2.6, it follows that
$\left|y^{G_{x}}\right|\lambda=r\left|B\cap y^{G_{x}}\right|$. Hence
$\left|y^{G_{x}}\right|=(k+1)\left|B\cap y^{G_{x}}\right|$ as
$r=(k+1)\lambda$. This proves (3). ∎

The group $G$ is point-primitive on $\mathcal{D}$ by Lemma 2.2(1). The O’Nan-
Scott Theorem classifies primitive groups into five types: (i) Affine type;
(ii) Almost simple type; (iii) Simple diagonal type; (iv) Product type; (v)
Twisted wreath product type (see [27] for details). Hence, the first part of
the paper is devoted to prove that only families (i) and (ii) occur. The
result is achieved by adapting the techniques developed in [43] to the
$2$-designs investigated here.

######  Lemma 2.3.

$G$ is not of simple diagonal type.

###### Proof.

The proof is essentially that of [43], Propositions 3.1, but we use
$r=(k+1)\lambda$ instead of $\lambda\geqslant(r,\lambda)^{2}$.

Assume that $G$ is of diagonal type. Then

$G\leqslant W=\left\\{(a_{1},...,a_{m})\pi\mid a_{i}\in Aut(T),\pi\in
S_{m},a_{i}\equiv a_{j}\mod Inn(T)\text{ for all }i,j\right\\}$

and there is $x\in\mathcal{P}$ such that

$G_{x}\leqslant\left\\{(a,...,a)\pi\mid a\in Aut(T),\pi\in S_{m}\right\\}\cong
Aut(T)\times S_{m}$

and $M_{x}=D=\left\\{(a,...,a)\mid a\in Inn(T)\right\\}$ is a diagonal
subgroup of $M\cong T^{m}$. Put $\Sigma=\left\\{T_{1},...,T_{m}\right\\}$,
where $T_{i}$ is identified with $\left\\{(1,...,t,...,1)\pi\mid t\in
T\right\\}$ in the $i$-th position. Then $G$ acts on $\Sigma$ by [27].
Moreover, the set $\mathcal{P}$ can be identified with the set $M/D$ of the
cosets of $D$ in $M$ so that $x=D(1,...,1)$, $k^{2}=\left|T\right|^{m-1}$,
since $v=k^{2}$, and for $y=D(t_{1},...,t_{m})$, $\psi=(s_{1},...,s_{m})\in
M$, $\sigma\in Aut(T)$, $\pi\in S_{m}$, we have the actions

$y^{\psi}=D(t_{1}s_{1},...,t_{m}s_{m})\text{,
}y^{\sigma}=D(t_{1}^{\sigma},...,t_{m}^{\sigma})\text{ and
}y^{\pi}=D(t_{1\pi^{-1}},...,t_{m\pi^{-1}})\text{.}$

Since $M\trianglelefteq G$ and $G$ is primitive on $\mathcal{P}$, $M$ is
transitive on $\mathcal{P}$. Since $T_{1}\trianglelefteq M$, all
$T_{1}$-orbits on $\mathcal{P}$ have the same length $c>1$. Let $\Gamma_{1}$
be the $T_{1}$-orbit containing $x$. For any $t_{1}=(t,1...,1)\in T_{1}$, we
have $x^{t_{1}}=D(t,1...,1)$. So that

$\Gamma_{1}=x^{T_{1}}=\left\\{D(t,1...,1):t\in T\right\\}$

and $\left|\Gamma_{1}\right|=\left|x^{T_{1}}\right|=c$. Similarly, we define
$\left|\Gamma_{i}\right|=\left|x^{T_{i}}\right|$ for $1\leqslant i\leqslant
m$. Clearly, $\Gamma_{i}\cap\Gamma_{j}=\left\\{x\right\\}$ for $i\neq j$
provided that $m\geqslant 2$.

Chose a point-$G_{x}$-orbit $\Delta$ in $\mathcal{P}-\left\\{x\right\\}$ such
that $\left|\Delta\cap\Gamma_{1}\right|=d\neq 0$. Let
$m_{1}=\left[G_{x}:N_{G_{x}}(T_{1})\right]$. Since $G_{x}$ is isomorphic to a
subgroup of $Aut(T)\times S_{m}$, and $G^{\Sigma}$ acts transitively on
$\Sigma$, it follows that $m_{1}\leqslant m$ and hence

$\left|\Delta\right|\leqslant m_{1}d\leqslant m\left|T\right|\text{.}$

Then $k+1\leqslant\left|\Delta\right|\leqslant m\left|T\right|$ by Lemma
2.2(3). Since $v=k^{2}=\left|T\right|^{m-1}$, we have
$\left|T\right|^{(m-1)/2}<m\left|T\right|$ and hence
$60^{m-3}\leqslant\left|T\right|^{m-3}<m^{2}$. Therefore, $m\leqslant 3$.

Since $r\mid\left|G_{x}\right|$ and $G_{x}$ is isomorphic to a subgroup of
$Aut(T)\times S_{m}$, it follows that
$(k+1)\lambda\mid\left|T\right|\left|Out(T)\right|m!$. On the other hand,
$k+1\mid\left|T\right|^{m-1}-1$, as $r/\lambda$ divides $k^{2}-1$. Thus
$k+1\mid\left|Out(T)\right|m!$ and hence
$\left|T\right|^{m-1}=k^{2}<\left|Out(T)\right|^{2}\left(m!\right)^{2}$ with
$m\leqslant 3$. At this point the final part of the proof of [43],
Propositions 3.1, can be applied to show that no cases occur. ∎

######  Lemma 2.4.

$G$ is not of twisted wreath product type

###### Proof.

We may apply the same argument of [43], Propositions 3.2, to show that there
is a point-$G_{x}$-orbit $\Delta$ in $\mathcal{P}-\left\\{x\right\\}$ such
$\left|\Delta\right|\leqslant m_{1}d\leqslant m\left|T\right|$ (this is shown
in [43], Propositions 3.2, without using the assumption
$\lambda\geqslant(r,\lambda)^{2}$). Then $k+1\leqslant m\left|T\right|$ by
Lemma 2.2(3). On the other hand, $k+1>\left|T\right|^{m/2}$, since
$k^{2}=v=\left|T\right|^{m}$. Then $\left|T\right|^{m/2}<m\left|T\right|$ and
hence $60^{m-2}\leqslant m$ and $m\leqslant 2$, whereas $m\geqslant 6$ by
[27]. ∎

######  Theorem 2.5.

$G$ is either of affine type or of almost simple type.

###### Proof.

The group $G$ is neither of simple diagonal type nor of twisted wreath product
type by Lemmas 2.3 and 2.4 respectively. Thus, in order to complete the proof,
we need to rule out the case where $G$ has a product action on $\mathcal{P}$.
Suppose the contrary. Then there is a group $K$ with a primitive action (of
almost simple or diagonal type) on a set $\Gamma$ of size $v_{0}\geqslant 5$,
such that $\mathcal{P}=\Gamma^{m}$ and $G\leqslant K^{m}\rtimes S_{m}$, where
$m\geqslant 2$. Let $x=(\gamma,...,\gamma)$ and $y=(\delta,...,\gamma)$ with
$\delta\neq\gamma$ and set $W=K^{m}$ and $H=W\rtimes S_{m}$. Then $W_{x}\cong
K_{\gamma}^{m}$, $W_{x,y}\cong K_{\gamma,\delta}\times K_{\gamma}^{m-1}$,
$H_{x}=W_{x}\rtimes S_{m}$ and
$K_{\gamma,\delta}\times\left(K_{\gamma}^{m-1}\rtimes S_{m-1}\right)\leqslant
H_{x,y}$. Suppose that $K$ has rank $s$ on $\Gamma$, $s\geqslant 2$. Then we
may choose $\delta$ such that
$\left[K_{\gamma}:K_{\gamma,\delta}\right]\leqslant\frac{v_{0}-1}{s-1}$.
Hence,

$\left|x^{H}\right|=\frac{\left|K_{\gamma}\right|^{m}\cdot
m!}{\left|K_{\gamma,\delta}\right|\left|K_{\gamma}\right|^{m-1}\cdot(m-1)!}=\left[K_{\gamma}:K_{\gamma,\delta}\right]m\leqslant\frac{v_{0}-1}{s-1}m\text{.}$

and, as $x^{G}\subseteq x^{H}$, we get

$v_{0}^{m/2}=v^{1/2}<k+1\leqslant\left|x^{G}\right|\leqslant\left|x^{H}\right|\leqslant
m\frac{v_{0}-1}{s-1}<mv_{0}\text{.}$

Then $m=2,3$ and $v_{0}<9$, as $m\geqslant 2$. If $m=3$, then
$k^{2}=v_{0}^{3}$ and hence $v_{0}=4$ and $s=3$, whereas $v_{0}\geqslant 5$.
Thus $m=s=2$. It follows that, $K$ acts $2$-transitively on $\Gamma$, and
$H=K^{2}\rtimes S_{2}$ has rank $3$ with $H_{x}$-orbits $1$, $2(k-1)$ and
$(k-1)^{2}$. Since each $H_{x}$-orbit is union $G_{x}$-orbit, and since each
$G_{x}$-orbit on $\mathcal{P}-\left\\{x\right\\}$ has length divisible by
$k+1$ by Lemma 2.2(3), we obtain $k+1\mid 2(k-1)$ and hence $k=v_{0}=3$. So,
we again reach a contradiction as $v_{0}\geqslant 5$. ∎

##  3\. Proof of Theorem 1.1

In this section $G$ is an almost simple group. Hence, $X\trianglelefteq
G\leqslant\mathrm{Aut(}X\mathrm{)}$, where $X$ is a non abelian simple group.
Moreover $X$, the socle of $G$, is either sporadic, or alternating, or an
exceptional group of Lie type, or classical. We analyze the first three cases
separately. The sporadic one is ruled out simply by filtering the groups
listed in [14] with respect to the constraints for $X$ to have a transitive
permutation representation of degree $k^{2}$, and when this occurs the
corresponding stabilizer of a point in $X$ to have the order divisible by
$\frac{k+1}{\gcd(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}$ (see Lemma 3.1).
The alternating case is settled as follows. We show that $X_{x}$, the
stabilizer in $X$ of a point $x$ of $\mathcal{D}$, is a large maximal subgroup
of $X$. Hence $X_{x}$ is listed in Theorem 2 of [9]. Then we combine some
group theoretical arguments, in particular those developed in [17], together
with some numerical properties of the binomial coefficients to exclude the
case. Finally, when $G$ is an exceptional group of Lie type, the reduction to
${}^{2}G_{2}(3)$ in its permutation representation of degree $36$ is settled
by transferring the arguments developed in [3] and in [8] to our context. The
key point of the analysis of the $2$-designs admitting ${}^{2}G_{2}(3)$ as a
flag transitive automorphism group is to see that ${}^{2}G_{2}(3)$ acts on
$PG_{2}(8)$ preserving a nondegenerate conic $\mathcal{C}$, since
${}^{2}G_{2}(3)\cong$ $P\Gamma L_{2}(8)$. Hence, its permutation
representation of degree $36$ is equivalent to that on the set of secants to
$\mathcal{C}$. Some geometry of $PG_{2}(8)$ is then used to complete the proof
of the case.

######  Lemma 3.1.

Let $\mathcal{D}$ be a $2$-$(k^{2},k,\lambda)$ design, with $\lambda\mid k$,
admitting a flag-transitive automorphism group $G$. If $x$ is any point of
$\mathcal{D}$, then
$\frac{k+1}{\gcd(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}$ divides
$\left|X_{x}\right|$.

###### Proof.

Let $x$ be any point of $\mathcal{D}$. If $y$ is a point of $\mathcal{D}$,
with $y\neq x$, then $\left|y^{X_{x}}\right|=\frac{\left|B\cap
y^{G_{x}}\right|\left(k+1\right)}{\mu}$, where
$\mu\left|y^{X_{x}}\right|=\left|y^{G_{x}}\right|$, by Lemma 2.2(3), as
$X_{x}\trianglelefteq G_{x}$. On the other hand, $\mu$ divides
$\left|\mathrm{Out(}X\mathrm{)}\right|$, as
$\mu=\frac{\left[G_{x}:X_{x}\right]}{\left[G_{x,y}:X_{x,y}\right]}$. Therefore
$\frac{k+1}{\gcd(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}$ divides
$\left|y^{X_{x}}\right|$ and hence $\left|X_{x}\right|$. ∎

######  Lemma 3.2.

$X$ is not sporadic.

###### Proof.

Assume that $X$ is sporadic. Then $X$ is listed in [14].

Assume that $X\cong M_{i}$, where $i=11,12,22,23$ or $24$. Since
$\left[X:X_{x}\right]=k^{2}$, it follows from [26], Table 5.1.C, that
$\lambda^{2}=2^{a_{1}}3^{a_{2}}$ for some $a_{1},a_{2}\geqslant 2$. Then
$k=12$ and either $X\cong M_{11}$ and $X_{x}\cong F_{55}$, or $X\cong M_{12}$
and $X_{x}\cong PSL_{2}(11)$ by [14]. However, these cases are ruled out by
Lemma 3.1, since
$\frac{k+1}{\gcd(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}=13$ does not
divide $\left|X_{x}\right|$.

Assume that $X\cong J_{i}$, where $i=1,2,3$ or $4$. Then $k^{2}$ divides
$2^{2}$, $2^{6}3^{2}5^{2}$, $2^{6}3^{4}$, or $2^{20}3^{2}11^{2}$,
respectively, by [26], Table 5.1.C. Then $i=2$ and either $k=10$ and
$X_{x}\cong PSU_{3}(3)$, or $k=60$ and $X_{x}\cong PSL_{2}(7)$ by [14].
However, these cases are ruled out as they contradict Lemma 3.1.

Assume that $X$ is isomorphic to one of the groups $HS$ or $McL$. By [26],
Table 5.1.C, $k^{2}$ divides $2^{8}3^{2}5^{2}$ or $2^{6}3^{6}5^{2}$
respectively. Then either $X\cong HS$, $X_{x}\cong M_{22}$ and $k=10$, or
$X\cong McL$, $X_{x}\cong M_{22}$ and $k=45$. The latter is ruled out by Lemma
3.1, since $\frac{k+1}{\gcd(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}=23$
does not divide $\left|X_{x}\right|$. The former yields $r=11\lambda$, where
$\lambda=1,2,5$ or $10$ as $\lambda\mid k$. If $B$ is any block incident with
$x$, then $\left[X_{x}:X_{x,B}\right]$ divides $r$. Then
$PSL_{3}(4)\trianglelefteq G_{x,B}\leqslant P\Sigma L_{3}(4)$, and hence
$\lambda=2$, by [14]. Thus, $\left|G_{B}\right|=10\left|G_{x,B}\right|$, since
$G_{B}$ is transitive on $B$, and hence $b=44$ or $88$. However,
$HS\trianglelefteq G\leqslant HS.Z_{2}$ has no such transitive representation
degrees by [14].

It is straightforward to check that the remaining cases are ruled out
similarly, as they do not have transitive permutation representations of
degree $k^{2}$ by [14] and [36]. ∎

######  Lemma 3.3.

If $X\cong A_{n}$, then $n\neq 6$ and $G=X$. Moreover, one of the following
holds:

  1. (1)

$X_{x}=\left(S_{t}\times S_{n-t}\right)\cap A_{n}$ where $1\leqslant t<n/2$.

  2. (2)

$X_{x}=\left(S_{t}\wr S_{h}\right)\cap A_{n}$ where $n=th$ and $2\leqslant
t\leqslant n/2$.

###### Proof.

Assume that $X\cong A_{n}$. If $n=6$, then $k^{2}=3^{2}$ or $6^{2}$, and the
former is ruled out by [14], whereas the latter yields $X_{x}\cong D_{10}$.
However this case cannot occur by Lemma 3.1, since
$\frac{k+1}{\gcd(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}=7$ does not
divide $\left|X_{x}\right|$. Thus $n\neq 6$, and hence
$\left|\mathrm{Out(}X\mathrm{)}\right|=2$ by [26], Theorem 5.1.3.

Let $\mu=\left[G_{x}:X_{x}\right]$. Since $G=G_{x}X$, it follows that
$G_{x}/X_{x}\cong G/X\leqslant\mathrm{Out(}X\mathrm{)}$ and hence
$\mu\leqslant 2$. Assume that $\mu=2$. Let $M$ be a maximal subgroup of $X$
containing $X_{x}$. Then $x^{M}$ is a block of imprimitivity for $X$ and hence
$\left|x^{M}\right|\mid k^{2}$. Since $x^{M}-\left\\{x\right\\}$ is union of
$X_{x}$-orbit, and each $X_{x}$-orbit distinct from $\left\\{x\right\\}$ is of
length divisible by $\frac{k+1}{\gcd(k+1,2)}$ Lemma 3.1, it follows that
$\frac{k+1}{\gcd(k+1,2)}\mid\left|x^{M}\right|-1$. Then
$\left|x^{M}\right|=c\frac{k+1}{\gcd(k+1,2)}+1$, for some $c\geqslant 1$, and
hence $k^{2}=d\left(c\frac{k+1}{\gcd(k+1,2)}+1\right)$ for some $d\geqslant
1$. Thus

$dc\frac{k+1}{\gcd(k+1,2)}+d-1=k^{2}-1$ (3.1)

and hence

$d=\theta\frac{k+1}{\gcd(k+1,2)}+1$ (3.2)

for some $\theta\geqslant 1$. Now, substituting (3.2) in (3.1) we obtain

$\theta
c\left(k+1\right)+\theta\gcd(k+1,2)<\left(k-1\right)\left(\gcd(k+1,2)\right)^{2}$

and hence $\theta c<\left(\gcd(k+1,2)\right)^{2}$. Therefore, $k$ is odd and
$\left(\theta,c\right)=(1,1),(1,2),(1,3)$ or $(3,1)$, which, substituted in
(3.1) and (3.2), yield $(k,d,c)=(15,9,3)$ or $(3,3,1)$. Then
$n\leqslant\left[A_{n}:M\right]=9,3$, respectively. Actually, $n=9$ by [14],
as $n\geqslant 5$ and $n\neq 6$. However, $A_{9}$ has no transitive
permutation representations of degree $15^{2}$. Thus $\mu=1$, $G=X$, and hence
$X_{x}$ is a large, maximal subgroup of $X$ by Lemma 2.2(2). The last part of
Lemma’s statement follows from [9], Theorem 2, since
$\left[X:X_{x}\right]=k^{2}$. ∎

######  Lemma 3.4.

Case (2) of Lemma 3.3 cannot occur.

###### Proof.

Assume that $X_{x}=\left(S_{t}\wr S_{h}\right)\cap A_{n}$, where $n=th$ and
$2\leqslant t\leqslant n/2$. Then

$k^{2}=\left[X:X_{x}\right]=\frac{th!}{\left(t!\right)^{h}h!}\text{.}$ (3.3)

By [17], there is a $G_{x}$-orbit (namely a $2$-cycle) of length either
$h(h-1)$ or $t^{2}\frac{h(h-1)}{2}$ according to whether $t=2$ or $t>2$
respectively. Then $k+1$ divides the length of such a orbit by Lemma 2.2(3),
as $G=X$ by Lemma 3.3. Thus, in both cases we have
$\frac{th!}{\left(t!\right)^{h}h!}<\left(th\right)^{2}$. The inequality

$\frac{h^{ht}}{\left(ht\right)^{h}\cdot
h}=\frac{e\left(\frac{th}{e}\right)^{th}}{e^{h}t^{h}\left(\frac{t}{e}\right)^{th}\cdot
eh\left(\frac{h}{e}\right)^{h}}\leqslant\frac{th!}{\left(t!\right)^{h}h!}<\left(th\right)^{2}$
(3.4)

is determined by using the known bound $e\left(\frac{f}{e}\right)^{f}\leqslant
f!\leqslant ef\left(\frac{f}{e}\right)^{n}$ for $f\in\mathbb{N}$, where $e$ is
the Napier’s constant. Thus $h^{ht}<\left(ht\right)^{h+2}\cdot
h<\left(ht\right)^{h+3}$.

Assume that $h^{t}\geqslant(ht)^{2}$, then $\left(ht\right)^{2h}\leqslant
h^{ht}<\left(ht\right)^{h+3}$ and hence $h=2$, as $2\leqslant t\leqslant n/2$.
Then (3.4) becomes
$2^{t}\leqslant\binom{2t}{t}=\frac{2t!}{\left(t!\right)^{2}}<8t^{2}$ and so
$t\leqslant 9$. However, (3.3) is not fulfilled for $h=2$ and any of these
values of $t$.

Assume that $h^{t}<(ht)^{2}$. Then $2^{t-2}\leqslant h^{t-2}<t^{2}$ and hence
either $t=2$, or $3\leqslant t\leqslant 8$ and $h\leqslant 9$. Actually, the
pairs $(h,t)$ in latter case do not fulfill (3.3). Hence $t=2$. Since
$h!\geqslant 2^{h}$, being $h\geqslant 2$, and (3.4) yields
$2^{h}\leqslant\binom{2h}{h}\leqslant\frac{2h!}{2^{h}h!}<\left(2h\right)^{2}$
and hence $h\leqslant 8$. However, (3.3) is not fulfilled for $t=2$ and any of
these values of $h$. ∎

######  Lemma 3.5.

$X$ is not isomorphic to $A_{n}$.

###### Proof.

In order to prove the assertion we need to rule out case (1) of Lemma 3.3,
since case (2) has been ruled out in Lemma 3.4. Hence, assume that
$X_{x}=\left(S_{t}\times S_{n-t}\right)\cap A_{n}$ where $1\leqslant t<n/2$.
Then the action on the point set of $\mathcal{D}$ and on the $t$-subsets of
$\left\\{1,...,n\right\\}$ are equivalent. Thus $k^{2}=\binom{n}{t}$. Then
either $t\leqslant 2$, or $n=50$ and $t=3$ by [1], Chapter 3, since
$1\leqslant t<n/2$.

Assume that $t=1$. Then $k^{2}=n\geqslant 9$, $X_{x}\cong A_{n-1}$ and hence
$X$ acts point-$2$-transitively on $\mathcal{D}$. Then $X_{x,B}$ is a subgroup
of $X_{x}$ of index $r=\lambda(\sqrt{n}+1)$, where $\lambda\mid\sqrt{n}$.
Since $X_{x}$ has no subgroups of index less than $n-1$, as $n\geqslant 9$, it
follows that $\lambda=\sqrt{n}$. Then $r<\binom{n-1}{2}$ and hence $X_{x,B}$
is one of the subgroups of $A_{n-1}$ listed in [20], Theorem 5.2.A, as
$n\geqslant 9$. Assume that $X_{x,B}$ preserves an $s$-subset of
$\left\\{1,...,n-1\right\\}$, where $s=1,2$. Then
$\binom{n-1}{s}=\sqrt{n}(\sqrt{n}+1)$, a contradiction. Then $n$ is odd and
$r=\left[X_{x}:X_{x,B}\right]=\frac{1}{2}\binom{n-1}{n/2}$, since $G=X$ by
Lemma 3.3. Therefore

$2^{n/2-1}\leqslant\frac{1}{2}\binom{n-1}{n/2}=\sqrt{n}(\sqrt{n}+1)<2n$

and hence $n=9$, $X_{x}\cong A_{8}$ and $X_{x,B}\cong AGL_{3}(2)$, since $n$
is a square and $n$ is odd. However, this case is ruled out since $r\neq 15$.

Assume that $t=2$. Then $G$ has rank $3$ and, if $x$ is any point of
$\mathcal{D}$, the $G_{x}$-orbits, say $\mathcal{O}_{i}$, $i=1,2,3$, have
length $1,\allowbreak 2n-4$ and $\binom{n-2}{2}$, respectively (see [17]).
Then $k+1$ divides the length of each of such orbits by Lemma 2.2(3), as $G=X$
by Lemma 3.3. Then $2\left(n-2\right)=c(k+1)$ for some $c\geqslant 1$ and
hence

$\left(\frac{2\left(n-2\right)}{c}-1\right)^{2}=\frac{n(n-1)}{2}\text{.}$
(3.5)

Then (3.5) yields $c=2$, $n=9$, $k=6$ and $X_{x}\cong S_{7}$. Note that,
$\lambda>1$ since there are no affine planes of order $6$. Also $\lambda\neq
6$, since $S_{7}$ has no transitive permutation representations of degree $63$
by [14]. Thus, either $\lambda=2$, $X_{x,B}\cong A_{6}$ and
$X_{B}\cong\left(A_{6}\times Z_{3}\right):Z_{2}$, or $\lambda=3$,
$X_{x,B}\cong Z_{2}\times S_{5}$ and $X_{B}\cong S_{4}\times S_{5}$ again by
[14], since $\lambda\mid k$. Then the actions on the point-set and on the
block-set of $\mathcal{D}$ are equivalent to the actions on the sets of
$2$-subsets and $(\lambda+1)$-subsets of
$\mathbb{N}_{9}=\left\\{1,...,9\right\\}$ respectively. Then we may identify
the point-set and the block-set of $\mathcal{D}$ with these sets,
respectively, in a way that the incidence relation is the set-theoretic
inclusion, as $X_{x,B}$ is isomorphic either to $A_{6}$ or to $Z_{2}\times
S_{5}$ according to whether $\lambda=2$ or $3$ correspondingly. So,
$k\leqslant\lambda+1\leqslant 4$ and we reach a contradiction as $k=6$.

Assume that $t=3$ and $n=50$. Then $X_{x}\cong\left(S_{3}\times
S_{47}\right)\cap A_{50}$, $k=140$ and $r=141\lambda$, where $\lambda\mid 140$
and $b=141\cdot 140\cdot\lambda$. Then $b\leqslant 141\cdot
140^{2}<\binom{50}{6}$ and $b\neq\frac{1}{2}\binom{50}{25}$ and hence
$A_{50-\ell}\leqslant X_{B}\leqslant\left(S_{\ell}\times
S_{50-\ell}\right)\cap A_{50}$, where $\ell<6$, by [20], Theorem 5.2.A.
Moreover $\left|X_{B}\right|$ is coprime to $47$, as $b=141\cdot
140\cdot\lambda$, and hence $\ell=4,5$. Thus $f\binom{50}{\ell}=141\cdot
140\cdot\lambda$, where $f$ is the index of $X_{B}$ in $\left(S_{\ell}\times
S_{50-\ell}\right)\cap A_{50}$. If $\ell=5$, then $23$ divides $\binom{50}{5}$
and hence $\lambda$, whereas $\lambda\mid 140$. Therefore, $\ell=4$,
$A_{46}\leqslant X_{B}\leqslant\left(D_{8}\times S_{46}\right)\cap A_{50}$ and
hence $f=3\mu$ and $\lambda=35\mu$ with $\mu\leqslant 4$. Then $X_{B}$
preserves a $4$-set $Y$ of $\mathbb{N}_{50}=\left\\{1,...,50\right\\}$,
whereas $X_{x}$ preserves a $3$-set $Z$ of $\mathbb{N}_{50}$. Then $X_{x,B}$
preserves $Z\cup Y$. Set $w=\left|Z\cup Y\right|$, then $4\leqslant w\leqslant
7$, and $\binom{50}{w}$ must divide $\left[X:X_{x,B}\right]$, which is equal
to $\binom{50}{3}\cdot 141\cdot 35\cdot\mu$. Thus $w=4$ and hence $Z\subseteq
Y$. Then $A_{46}\leqslant X_{x,B}$, and so $k\mid 8$, as $A_{46}\leqslant
X_{B}\leqslant\left(D_{8}\times S_{46}\right)\cap A_{50}$. This is a
contradiction, as $k=6$. ∎

######  Lemma 3.6.

If $X$ is isomorphic to socle a finite exceptional group of Lie type, then
$\mathcal{D}$ is a $2$-$(36,6,\lambda)$ design, where $\lambda\mid 6$ and
$\lambda>1$, and $X$ is isomorphic either to $G_{2}(2)^{\prime}$ or to
${}^{2}G_{2}(3)^{\prime}$.

###### Proof.

Recall that an exceptional group of Lie type is simple apart from
${}^{2}B_{2}(2)$, $G_{2}(2)$, ${}^{2}G_{2}(3)$, or ${}^{2}F_{4}(2)$ by [26],
Theorem 5.1.1. Thus, either $X$ is isomorphic to an exceptional simple group
of Lie type, or $X$ is isomorphic to one of the groups ${}^{2}B_{2}(2)$,
$G_{2}(2)$, ${}^{2}G_{2}(3)$, or ${}^{2}F_{4}(2)$. If the latter occurs, since
$G$ has a primitive permutation representation of degree $k^{2}$, then the
unique admissible cases to be analyzed are either $G_{2}(2)$ and $k=6$, or
${}^{2}G_{2}(3)$ and $k=3,6$ by [14]. Suppose that $k=3$. Then $\lambda=1,3$
as $\lambda\mid k$. If $\lambda=1$, then $\mathcal{D}\cong AG_{2}(3)$ and
hence $G\leqslant AGL_{2}(3)$, which is impossible. Then $\lambda=3$, $r=12$
and hence $G\cong$ $P\Gamma L_{2}(8)$ and $G_{x}\cong F_{56}:Z_{3}$. So
$\left|G_{x,B}\right|=14$ and $G_{x,B}\leqslant F_{56}$, a contradiction. Thus
$k=6$ and hence $\lambda=1,2,3$ or $6$ as $\lambda\mid k$. Also, $\lambda>1$
since there are no affine planes of order $6$. Therefore, $k=6$, $\lambda\mid
6$ and $\lambda>1$, and $X$ is isomorphic either to $G_{2}(2)^{\prime}$ or to
${}^{2}G_{2}(3)^{\prime}$.

Assume that $X$ is isomorphic to an exceptional simple group of Lie type.
Suppose that $G_{x}$ is not parabolic. Then $G_{x}$ is one of the groups
listed in [8], Theorem 1.6, or equivalently in Tables 2 and 3 of [3], since
$G_{x}$ is a large maximal subgroup of $G$ by Lemma 2.2(2). In [3], Alavi
points out that the first and the second column of Table 2 contains $X$ and
$X_{x}$, respectively, the third one contains a lower bound $\ell_{v}$ for
$v=\left[X:X_{x}\right]$, and the fourth one contains an upper bound $u_{r}$
for $r$, determined by using the fact that $r$ is a common divisor of $v-1$,
of $\left|G_{x}\right|$ and of the subdegrees of $G_{x}$ by his Lemma 4. Then,
the author shows that $u_{r}^{2}<\ell_{v}$ for each case contained in Table 2,
hence $r^{2}<v$, and so all the cases in Table 2 are ruled out in his paper.

Our aim is to transfer Alavi’s argument in order to rule out the possibility
for $G_{x}$ to not be a parabolic subgroup of $G$. Clearly, the first three
columns of Table 2 have the same meaning as in our paper, where $v=k^{2}$ for
us. The role of $r$ and of Lemma 4 of [3] are played by $r/\lambda=k+1$ and by
our Lemma 2.2(3) respectively. Thus the upper bound $u_{r}$ for $r$ in [3]
becomes an upper bound for $r/\lambda$ in our context. Therefore the
inequality $u_{r}^{2}<\ell_{v}$ implies $\left(r/\lambda\right)^{2}<k^{2}$ but
this is impossible in our context, since $r/\lambda=k+1$. Thus all the groups
listed in Tables 2 of [3] cannot occur.

It is even easier to rule out the groups listed in Tables 3 of [3] as they are
filtered with respect to the property that $v=k^{2}$. Indeed, we obtain the
following admissible cases:

  1. (1)

$X\cong G_{2}(3)$, $X_{x}\cong 2^{3}:PSL_{3}(2)$ and $k^{2}=3528=\allowbreak
2^{3}3^{2}7^{2}$;

  2. (2)

$X\cong G_{2}(4)$, $X_{x}\cong PSL_{2}(13)$ and $k^{2}=230400=\allowbreak
2^{10}3^{2}5^{2}$.

Then $k+1$ is $43$ or $481$ respectively, but none of these divides the order
of the corresponding $G_{x}$. So, these cases violate Lemma 2.2(3) and hence
they are ruled out.

Assume that $G_{x}$ is a maximal parabolic subgroup of $G$. Assume that
$E_{6}(q)$ is not contained in $G$. Then $G$ has a subdegree of order $p^{t}$
(e.g. see [3], Lemma 3, or [35], Lemma 2.6). Then $\frac{r}{\lambda}\mid
p^{t}$ and so $k+1=p^{s}$ for some $s\leqslant t$. Then $k=p^{s}-1$ and hence
$k^{2}=(p^{s}-1)^{2}$. Then $s\leqslant\zeta_{p}(G)$, where $\zeta_{p}(G)$ is
defined in [26] (5.2.4), and is determined in Proposition 5.2.17.(i) and Table
5.2.C. If $s=\zeta_{p}(G)$, then
$\left(p^{\zeta_{p}(G)}-1\right)^{2}\mid\left|X\right|$. On the other hand,
$\left|X\right|$ is listed in [26], Table 5.1.B, and hence none of these
groups admits $\left(p^{\zeta_{p}(G)}-1\right)^{2}$ as a divisor. Then
$s<\zeta_{p}(G)$. Then $G_{x}$ contains a Sylow $u$-subgroup $G$, where $u$ is
a primitive prime divisor of $p^{\zeta_{p}(G)}-1$, since
$(p,\zeta_{p}(G))\neq(2,6)$ being $X\ncong G_{2}(2)^{\prime}$. On the other
hand, $G_{x}$ can be obtained by deleting the $i$-th node in the Dynkin
diagram of $X$, and we see that none of these groups is of order divisible by
$u$. Indeed, for instance, if $F_{4}(q)\trianglelefteq G$, $q=p^{f}$, then
$\zeta_{p}(G)=12f$ and hence $G_{x}$ contains a Sylow $u$-subgroup of $G$,
where $u$ is a primitive prime divisor of $p^{12f}-1$, whereas the maximal
parabolic subgroups are of type $B_{3}(p^{f})$, $C_{3}(p^{f})$ or
$A_{1}(p^{f})\times A_{2}(p^{f})$ and none of these is divisible by $u$. As
stressed out in [3], Remark 1, even in the case $E_{6}(q)$, when $G$ contains
a graph automorphism or $G_{x}$ is parabolic of type $1,2$ or $4$, then $G$
has a subdegree of order $p^{t}$, and hence these cases are excluded by the
above argument. For the remaining maximal parabolic subgroups we may use the
same argument as [3] at pp.1012–1013, with $r/\lambda$ and Lemma 2.2(3) in the
role of $r$ and Lemma 4 of [3], respectively, to see that
$\left(r/\lambda\right)^{2}>v=k^{2}$ is violated. Hence
$E_{6}(q)\trianglelefteq G$ cannot occur and the proof is thus completed. ∎

######  Lemma 3.7.

$X$ is not isomorphic to $G_{2}(2)^{\prime}$.

###### Proof.

Suppose that $\mathcal{D}$ is a $2$-$(36,6,\lambda)$ design, where
$\lambda\mid 6$ and $\lambda>1$, admitting a flag-transitive automorphism
group $X\trianglelefteq G\leqslant\mathrm{Aut(}X\mathrm{)}$, where $X\cong
G_{2}(2)^{\prime}$. Therefore $PSL_{2}(7)\trianglelefteq G_{x}\leqslant
PGL_{2}(7)$ by [14], and $\lambda=2,3$ or $6$.

Assume that $\lambda=2$. Then $\left|G_{x,B}\right|=12$ and hence either $G=X$
and $G_{x,B}\cong A_{4}$, or $G\neq X$ and $G_{x,B}\cong A_{4}$, $D_{12}$ by
[14]. Then $72$ divides $\left|G_{B}\right|$ and hence
$G_{B}\leqslant\left(U:Z_{8}\right).Z_{2}$, where $U$ is a Sylow $3$-subgroup
of $G$. Then $U:Z_{8}\leqslant G_{B}$ since $36$ divides
$\left|G_{B}\cap\left(U:Z_{8}\right)\right|$ and since $U/Z(U):Z_{8}$ is a
Frobenius group (for instance, see [23], Satz II.10.12, since
$G_{2}(2)^{\prime}\cong PSU_{3}(3)$). So $9$ divides $\left|G_{x,B}\right|$, a
contradiction.

Assume that $\lambda=3$ or $6$. Then $4\mid\left|X_{x,B}\right|$ as $r=21$ or
$63$. If $G\neq X$, then the set of points of $\mathcal{D}$ distinct from $x$
is partitioned into two $G_{x}$-orbits, say $\mathcal{O}_{1}$,
$\mathcal{O}_{2}$, of length $14$ and $21$ respectively by [38]. If $G=X$,
then $\mathcal{O}_{1}$ is split into two $X_{x}$-orbits $\mathcal{O}_{1j}$,
$j=1,2$, each of length $7$, whereas $\mathcal{O}_{2}$ is also a
$X_{x}$-orbit. Hence, $X_{x,y}\cong D_{8}$ for $y\in\mathcal{O}_{2}$.

Let $B$ be any block incident with $x$. Then
$\left|\mathcal{O}_{i}\right|=7\left|\mathcal{O}_{i}\cap B\right|$, where
$i=1,2$, by Lemma 2.2(3) and hence $\left|\mathcal{O}_{1}\cap B\right|=2$ and
$\left|\mathcal{O}_{2}\cap B\right|=3$. Then $X_{x,B}$ fixes
$\mathcal{O}_{1}\cap B$ pointwise, since $\left|\mathcal{O}_{1j}\cap
B\right|=1$ for each $j=1,2$. Then there is a non trivial subgroup $W$ of
$X_{x,B}$ of index at most $2$, such that
$B\subseteq\mathrm{Fix(}W\mathrm{)}$, since $4\mid\left|X_{x,B}\right|$.

If $\lambda=3$, then $r=21$, $b=126$ and $X_{x,B}\cong D_{8}$. Hence, $W$ is
isomorphic to $Z_{4},E_{4}$ or to $D_{8}$, and $N_{X_{x}}(W)$ is isomorphic to
$D_{8}$, $S_{4}$ or to $D_{8}$ respectively. The number of points in
$\mathcal{O}_{2}$ fixed by $W$ is given by the well known formula
$\left|N_{X_{x}}(W)\right|\left|X_{x,y}\cap
W^{X}\right|/\left|X_{x,y}\right|$. Thus, it is easy to see that, $W$ fixes
$6$ or $1$ points in $\mathcal{O}_{2}$ according to whether $W$ is or is not
isomorphic to $E_{4}$ respectively, since $X_{x,y}\cong D_{8}$. Therefore,
$W\cong E_{4}$, since $B\subseteq\mathrm{Fix(}W\mathrm{)}$ and
$\left|\mathcal{O}_{2}\cap B\right|=3$. Then $W$ lies in the kernel $N$ of the
action of $G_{B}$ on $B$. Also
$\mathrm{Fix(}W\mathrm{)}\neq\mathrm{Fix(}X_{x,B}\mathrm{)}$ as $X_{x,B}\cong
D_{8}$ fixes exactly one point on $\mathcal{O}_{2}$, and hence $N\cap
X_{B}=W$. Therefore $W\trianglelefteq G_{B}$ and hence $G_{B}\leqslant
N_{G}(W)$, where $N_{G}(W)$ is isomorphic either to
$\left(Z_{4}\right)^{2}.S_{3}$ or to $\left(Z_{4}\right)^{2}.D_{12}$ according
as $G=X$ or $G\neq X$, respectively, by [14]. Note that,
$\left[N_{G}(W):G_{B}\right]=2$, since $\left[G:N_{G}(W)\right]=63$ and
$b=126$. Thus, $G_{B}\vartriangleleft N_{G}(W)$ and $3$ divides
$\left|N_{X_{x}}(W)\cap G_{B}\right|$ as $N_{X_{x}}(W)\cong S_{4}$. Hence,
$G_{x,B}$ contains a Sylow $3$-subgroup of $N_{G}(W)$, since these are cyclic
of order $3$, but this contradicts $[G_{B}:G_{x,B}]=6$.

If $\lambda=6$, then $r=42$, $b=252$ and hence either $X_{x,B}\cong Z_{4}$ or
$X_{x,B}\cong E_{4}$. Therefore, $W$ is isomorphic to $Z_{2}$, $E_{4}$ or to
$Z_{4}$. Actually, $W\ncong Z_{4}$ since $Z_{4}$ fixes one point on
$\mathcal{O}_{2}\cap B$, whereas $B\subseteq\mathrm{Fix(}W\mathrm{)}$ and
$\left|\mathcal{O}_{2}\cap B\right|=3$.

Assume that $W\cong E_{4}$. Then $W=X_{x,B}$ and hence $W=X_{B}\cap N$, where
$N$ is defined above. Therefore $W\trianglelefteq G_{B}$ and hence
$G_{B}\leqslant N_{G}(W)$. Moreover, $\left[N_{G}(W):G_{B}\right]=4$, since
$\left[G:N_{G}(W)\right]=63$ and $b=252$. So, $G\cong G_{2}(2)$,
$N_{G}(W)\cong\left(Z_{4}\right)^{2}.D_{12}$ and
$G_{B}\cong\left(Z_{4}\right)^{2}.Z_{3}$ by [14]. Then $G_{B}\vartriangleleft
N_{G}(W)$, $3$ divides $\left|N_{X_{x}}(W)\cap G_{B}\right|$ and we reach a
contradiction as above.

Assume $W\cong Z_{2}$,
$\left|\mathrm{Fix(}W\mathrm{)}\cap\mathcal{O}_{2}\right|=5$ and
$B\subset\mathrm{Fix(}W\mathrm{)}$. As above, $W$ lies in the kernel $N$ of
the action of $G_{B}$ on $B$. Then
$\mathrm{Fix(}W\mathrm{)}\neq\mathrm{Fix(}X_{x,B}\mathrm{)}$. Indeed, it is
true for $X_{x,B}\cong Z_{4}$, as any $Z_{4}$ fixes exactly one point on
$\mathcal{O}_{2}$, and it is still true for $X_{x,B}\cong E_{4}$ as it follows
from the above arguments where it is shown that a subgroup isomorphic to
$E_{4}$ cannot fix $B$ pointwise. Consequently, we have that $N\cap X_{B}=W$.
Thus $G_{B}\leqslant C_{G}(W)$, where $C_{G}(W)$ is isomorphic either to
$GU_{2}(3)$ or to $\Gamma U_{2}(3)$ according as $G=X$ or $G\neq X$,
respectively, by [14]. Moreover, $\left[C_{G}(W):G_{B}\right]=4$, as $b=252$,
and hence $\left|X_{B}\right|=24$. Therefore, $X_{B}\cong Z.S_{3}$ or
$X_{B}\cong SL_{2}(3)$ since $C_{X}(W)\cong\left(Z\circ
SL_{2}(3)\right).Z_{2}$, where $Z$ is the center of $GU_{2}(3)$. Assume the
former occurs. Since $\left[X_{B}:X_{x,B}\right]$ divides $6$, it follows that
$Z\leqslant X_{x,B}$. Then $Z$ fixes $B$ pointwise, as $Z$ is normal in
$G_{B}$, and hence $Z\leqslant N\cap X_{B}=W$. This is impossible, since
$W\cong Z_{2}$ whereas $Z\cong Z_{4}$. Thus $X_{B}\cong SL_{2}(3)$. Therefore,
either $G\cong G_{2}(2)^{\prime}$ and hence $G_{B}\cong SL_{2}(3)$, or $G\cong
G_{2}(2)$ and $G_{B}\cong GL_{2}(3)$, $Z_{4}\circ SL_{2}(3)$, as
$\left[C_{G}(W):G_{B}\right]=4$. However, it is easily seen with the aid of
[47] that, no $2$-designs occur. ∎

Before analyzing the remaining case involving $G_{3}(3)$. Recall some useful
facts about the action of this group in the desarguesian plane of order $8$.

Let $G\cong G_{3}(3)$. It is well known that $G\cong$ $P\Gamma L_{2}(8)$ acts
on $PG_{2}(8)$ preserving a regular hyperoval, namely a $10$-arc, consisting
of a nondegenerate conic $\mathcal{C}$ and it nucleus $N$ (see [21], Section
8). Moreover, $G$ acts primitively on the set $\mathcal{S}$ of the $36$
secants to $\mathcal{C}$ by [14]. If $\ell$ is any secant to $\mathcal{C}$ and
$\left\\{P_{1},P_{2}\right\\}=\ell\cap\mathcal{C}$, then $G_{\ell}\cong
F_{42}$ and the set $\mathcal{S}(\ell)$ consisting of the $14$ secant lines to
$\mathcal{C}$ incident with $P_{1}$ or $P_{2}$ is a $G_{\ell}$-orbit, as $G$
acts $3$-transitively on $\mathcal{C}$. The complementary set
$\mathcal{S}-\mathcal{S}(\ell)$ consisting of $21$ secants intersecting $\ell$
in a point different from $P_{1},P_{2}$ is also a $G_{\ell}$-orbit.

By [14], $G$ contains two conjugacy classes of subgroups of order $3$, one
contained in $G^{\prime}$ the other in $G-G^{\prime}$. If
$\left\langle\eta\right\rangle$ and $\left\langle\gamma\right\rangle$ are the
representatives of such classes, then $C_{G}(\eta)\cong S_{3}$ and
$C_{G}(\gamma)\cong Z_{3}\times S_{3}$. Moreover,
$C_{G}(\eta)=C_{G^{\prime}}(\eta)$ and $C_{G^{\prime}}(\gamma)\cong S_{3}$. We
may choose $\eta$ and $\gamma$ to belong to the same Sylow $3$-subgroup of $G$
in a way that $C_{G}(\gamma)=\left\langle\gamma\right\rangle\times
C_{G}(\eta)$. Note that, $C_{G}(\eta)=\left\langle\eta,\sigma\right\rangle$,
where $\sigma$ is an involutory elation of $PG_{2}(8)$ with center $C$ not in
$\mathcal{C}$ and axis $a$ tangent to $\mathcal{C}$ (a detailed description of
the collineations of the desarguesian plane can be found in [21] and in [22]).

Set $K=\left\langle\gamma,\sigma\right\rangle$. Then $K\cong Z_{6}$ is a self-
normalizing subgroup of $G$ by [14]. Moreover, $K$ fixes exactly one point $F$
on $\mathcal{C}$, and $\left\langle\gamma\right\rangle$ fixes two further
points switched by $\sigma$, say $W$ and $W^{\sigma}$. Then $C=W^{\sigma}W\cap
F_{0}N$, $a=F_{0}N$. The set $\left\\{F,W,W^{\sigma}\right\\}$ is a
$C_{G}(\gamma)$-orbit on $\mathcal{C}$. The set
$\mathcal{C}\mathbf{-}\left\\{F,W,W^{\sigma}\right\\}$ is split into two
$\left\langle\gamma\right\rangle$-orbits,
$\left\\{P,P^{\gamma},P^{\gamma^{2}}\right\\}$ and
$\left\\{P^{\sigma},P^{\sigma\gamma},P^{\sigma\gamma^{2}}\right\\}$, and these
are also $\left\langle\gamma,\eta\right\rangle$-orbits. Moreover,
$\left\\{P,P^{\gamma},P^{\gamma^{2}},P^{\sigma},P^{\sigma\gamma},P^{\sigma\gamma^{2}}\right\\}$
is both a $K$-orbit and a $C_{G}(\gamma)$-orbit on $\mathcal{C}$.

######  Lemma 3.8.

The following hold:

  1. (1)

$\left(PP^{\gamma}\right)^{C_{G}(\gamma)}$ is the unique $C_{G}(\gamma)$-orbit
on $\mathcal{S}$ of length $6$.

  2. (2)

$\left(PP^{\gamma}\right)^{C_{G}(\eta)}$ and
$\left(FP^{\gamma^{i}}\right)^{C_{G}(\eta)}$, where $i=0,1,2$, are the unique
$C_{G}(\eta)$-orbits on $\mathcal{S}$ of length $6$.

  3. (3)

$\left(P^{\gamma}P\right)^{K}$, $\left(P^{\gamma}P^{\sigma}\right)^{K}$,
$\left(PW\right)^{K}$, $\left(P^{\sigma}W\right)^{K}$, and
$\left(PF\right)^{K}$ are the unique $K$-orbits on $\mathcal{S}$ of length
$6$.

In particular,
$\left(PP^{\gamma}\right)^{C_{G}(\gamma)}=\left(PP^{\gamma}\right)^{C_{G}(\eta)}=\left(P^{\gamma}P\right)^{K}$.

###### Proof.

Since $\left\\{P,P^{\gamma},P^{\gamma^{2}}\right\\}$ and
$\left\\{P^{\sigma},P^{\sigma\gamma},P^{\sigma\gamma^{2}}\right\\}$ are
$\left\langle\gamma,\eta\right\rangle$-orbits, it follows that
$\left|\left(PP^{\gamma}\right)^{C_{G}(\gamma)}\right|\mid 6$. On the other
hand $\left|\left(PP^{\gamma}\right)^{K}\right|=6$ and
$\left(PP^{\gamma}\right)^{K}\subseteq\left(PP^{\gamma}\right)^{C_{G}(\gamma)}$.
Thus $\left(PP^{\gamma}\right)^{C_{G}(\gamma)}$ is a $C_{G}(\gamma)$-orbit on
$\mathcal{S}$ of length $6$.

Since $\left\\{F,W,W^{\sigma}\right\\}$ is a $C_{G}(\gamma)$-orbit on
$\mathcal{C}$. It follows that
$\left|\left(FW\right)^{C_{G}(\gamma)}\right|=3$. Moreover,
$\left|\left(FP\right)^{K}\right|=6$ and hence
$\left|\left(FP\right)^{C_{G}(\gamma)}\right|=18$, as
$\left\\{P,P^{\gamma},P^{\gamma^{2}},P^{\sigma},P^{\sigma\gamma},P^{\sigma\gamma^{2}}\right\\}$
is both a $K$-orbit and a $C_{G}(\gamma)$-orbit on $\mathcal{C}$.

Since $\sigma$ fixes $PP^{\sigma}$, it follows that
$\left(PP^{\sigma}\right)^{C_{G}(\gamma)}$ is of odd length. Assume that
$\left|\left(PP^{\sigma}\right)^{C_{G}(\gamma)}\right|=3$. Then there is an
element $\vartheta$ in $\left\langle\gamma,\eta\right\rangle$ preserving
$PP^{\sigma}$ and hence fixing both $P$ and $P^{\sigma}$. Then $\vartheta$
fixes $\mathcal{C}\mathbf{-}\left\\{F,W,W^{\sigma}\right\\}$ pointwise, since
$\left\langle\gamma,\eta\right\rangle$ is an elementary abelian group of order
$9$ acting transitively both on $\left\\{P,P^{\gamma},P^{\gamma^{2}}\right\\}$
and on $\left\\{P^{\sigma},P^{\sigma\gamma},P^{\sigma\gamma^{2}}\right\\}$.
However, this is impossible. Hence
$\left|\left(PP^{\sigma}\right)^{C_{G}(\gamma)}\right|=9$. As
$\left(FW\right)^{C_{G}(\gamma)}\cup\left(FP\right)^{C_{G}(\gamma)}\cup\left(PP^{\sigma}\right)^{C_{G}(\gamma)}$
covers $\mathcal{C}-\left(PP^{\gamma}\right)^{C_{G}(\gamma)}$, the assertion
(1) follows.

Since $C_{G}(\eta)\trianglelefteq C_{G}(\gamma)$, each $C_{G}(\eta)$-orbit on
$\mathcal{S}$ of length $6$ lies either in
$\left(PP^{\gamma}\right)^{C_{G}(\gamma)}$ or in
$\left(FP\right)^{C_{G}(\gamma)}$. Since $C_{G}(\eta)=C_{G^{\prime}}(\eta)$,
and $\left\langle\eta\right\rangle$ acts semiregularly on $\mathcal{C}$, it
results that $\left\langle\eta\right\rangle$ does not fix secants to
$\mathcal{C}$. On the other hand, since $C_{G}(\gamma)_{PP^{\gamma}}\cong
Z_{3}$, $C_{G}(\gamma)_{PP^{\gamma}}\cap G^{\prime}=1$ and
$C_{G}(\gamma)_{FP}=1$, it follows that
$\left(PP^{\gamma}\right)^{C_{G}(\eta)}=\left(PP^{\gamma}\right)^{C_{G}(\gamma)}$
and that $\left(FP\right)^{C_{G}(\gamma)}$ is split into three
$C_{G}(\eta)$-orbits each of length $6$, namely,
$\left(FP^{\gamma^{i}}\right)^{C_{G}(\eta)}$ where $i=0,1,2$.

Finally, it is easy to check that $W^{\sigma}W$, $WF$, $P^{\sigma}P$, $PF$,
$PW$, $P^{\sigma}W$, $P^{\gamma}P$ and $P^{\gamma}P^{\sigma}$ are
representatives of all $K$-orbits on $\mathcal{S}$ and these have lengths $1$,
$2$, $3$, $6$, $6$, $6$, $6$, $6$ respectively. Thus (3) holds. ∎

######  Example 3.9.

Let $\ell_{1}=PP^{\gamma}$, $\ell_{2}=P^{\gamma}P^{\sigma}$, $\ell_{3}=PW$ and
$\ell_{4}=P^{\sigma}W$, and let $B_{1}=\ell_{1}^{C_{G}(\gamma)}$ and
$B_{i}=\ell_{i}^{K}$ for $i=2,3,4$. Then the following hold:

  1. (1)

$\mathcal{D}_{1}=(\mathcal{S},B_{1}^{G})$ is a $2$-$(36,6,2)$ design admitting
$G$ as the full flag-transitive automorphism group. Moreover, $G^{\prime}$
acts flag-transitively on $\mathcal{D}_{1}$.

  2. (2)

$\mathcal{D}_{i}=(\mathcal{S},G_{i}^{G})$, where $i=2,3,4$, is a
$2$-$(36,6,6)$ design admitting $G$ as the unique flag-transitive automorphism
group.

  3. (3)

$\mathcal{D}_{2}$, $\mathcal{D}_{3}$ and $\mathcal{D}_{4}$, are pairwise non
isomorphic.

###### Proof.

Let $Z$ denote $C_{G}(\gamma)$ for $i=1$ and $K$ for $i>1$. Clearly,
$Z\leqslant G_{B_{i}}$. Assume that $Z\neq G_{B_{i}}$. Then $1\neq
G_{\ell_{i},B_{i}}\leqslant F_{42}$ and $G_{B_{i}}$ does not contain elements
of order $7$, as $\left|B_{i}\right|=6$ and as the unique element of $G$
fixing more than three points on $\mathcal{C}$ is the identity. Thus
$G_{\ell_{i},B_{i}}\leqslant Z_{6}$. Suppose that
$\left|G_{\ell_{i},B_{i}}\right|$ is even. Then either
$36\mid\left|G_{B_{1}}\right|$, or $12\mid\left|G_{B_{i}}\right|$ for $i>1$.
In the former case we obtain $G=G_{B_{1}}$ by [14], but this is impossible.
Hence $i>1$ and either $G_{B_{i}}\cong A_{4}$ or $G_{B_{i}}$ contains a Sylow
$2$-subgroup of $G$. Both are ruled out. Indeed, the former cannot occur since
$Z_{6}\cong K\leqslant G_{B_{i}}$ but $A_{4}$ does not contain such groups. In
the latter case $G_{B_{i}}$ contains an involution $\alpha$ fixing
$\ell_{i}\cap B_{i}$ pointwise. However, this is impossible since the
involutions are elations of $PG_{2}(8)$ and their unique fixed point is the
tangency point of their axis, whereas $\ell_{i}$ is a secant to $\mathcal{C}$.
Thus $G_{\ell_{i},B_{i}}\cong Z_{3}$ and hence $\left|G_{B_{i}}\right|=18$.
Therefore, $G_{B_{i}}=C_{G}(\gamma)$. This is clear for $i=1$, whereas, for
$i>1$ it follows from $Z_{6}\cong K\leqslant G_{B_{i}}$ and $K\cap
G_{\ell_{i},B_{i}}=1$. However, $G_{B_{1}}=C_{G}(\gamma)$ contradicts the
assumption. Thus $i>1$ and hence $B_{i}=B_{1}$ by Lemma 3.8(1), but we still
obtain a contradiction since $B_{1}$ is also a $K$-orbit by Lemma 3.8(3). Thus
$G_{B_{i}}=Z$ for each $i=1,2,3,4$. Therefore, by [18], 1.2.6,
$\mathcal{D}_{i}=(\mathcal{S},B_{i}^{G})$ is a flag-transitive tactical
configuration with parameters $(v,b,k,r)=(36,84,6,14)$ or $(36,252,6,42)$
according as $i=1$ or $i=2,3,4$ respectively.

In order to prove that $\mathcal{D}_{i}$ is a $2$-design with $\lambda=2$ for
$i=1$ and $\lambda=6$ for $i>1$, bearing in mind that $G$ acts transitively on
$\mathcal{S}$, and for any $\ell\in\mathcal{S}$ the group $G_{\ell}$ acts
transitively both on $\mathcal{S}(\ell)$ and on
$\mathcal{S}-\mathcal{S}(\ell)$, it is enough to prove that there are
precisely $\lambda$ elements of $B^{G}$ containing $\ell_{i}$ and any
$m_{i}\in B_{i}-\left\\{\ell_{i}\right\\}$, $i$ fixed.

  1. (i).

$\mathcal{D}_{1}=(\mathcal{S},B_{1}^{G})$ is a $2$-$(36,6,2)$ design admitting
$G$ as the full flag-transitive automorphism group of
$\mathrm{\mathcal{D}}_{1}$. Moreover, $G^{\prime}$ acts flag-transitively on
$\mathrm{\mathcal{D}}_{1}$.

Let $m_{1}\in B_{1}$. Assume that $\ell_{1}\cap m_{1}\in\mathcal{C}$. Then
$m_{1}\in\mathcal{S}(\ell_{1})$ and hence
$\left|m_{1}^{G_{\ell_{1}}}\right|=14$. Clearly
$\left|B_{1}^{G_{\ell_{1}}}\right|=14$ as $G_{\ell_{1},B_{1}}\cong Z_{3}$.
Moreover, $\left|B_{1}\cap m_{1}^{G_{\ell_{1}}}\right|=2$ and hence
$\left(m_{1}^{G_{\ell_{1}}},B_{1}^{G_{\ell_{1}}}\right)$ is a tactical
configuration with parameters $(v_{1},b_{1},k_{1},r_{1})=(14,14,2,2)$ by [18],
1.2.6. Hence, the number of secants in $B_{1}^{G_{\ell_{1}}}$ containing both
$\ell_{1}$ and $m_{1}$ is $2$. Then the number of secants in $B_{1}^{G}$
containing both $\ell_{1}$ and $m_{1}$ is $2$, as $\mathcal{D}_{1}$ is a flag-
transitive tactical configuration.

If $\ell_{1}\cap m_{1}\notin\mathcal{C}$. Then
$m_{1}\in\mathcal{S}-\mathcal{S}(\ell_{1})$ and hence
$\left|m_{1}^{G_{\ell_{1}}}\right|=21$. Moreover,
$\left|B_{1}^{G_{\ell_{1}}}\right|=14$ and $\left|B_{1}\cap
m_{1}^{G_{\ell_{1}}}\right|=3$. Indeed, $B_{1}\cap
m_{1}^{G_{\ell_{1}}}=m_{1}^{\left\langle\gamma\right\rangle}$ and hence
$\left(m_{1}^{G_{\ell_{1}}},B_{1}^{G_{\ell_{1}}}\right)$ is a tactical
configuration with parameters $(v_{1},b_{1},k_{1},r_{1})=(21,14,3,2)$. Hence,
the number of elements in $B^{G}$ containing both $\ell_{1}$ and $m_{1}$ is
$2$, as $\mathcal{D}_{1}$ is a flag-transitive tactical configuration. Thus,
there are precisely $2$ elements of $B^{G}$ containing both $\ell_{1}$ and
$m_{1}$ regardless $\ell_{1}\cap m_{1}$ lies or does not lie in $\mathcal{C}$.
Therefore, $\mathcal{D}_{1}=(\mathcal{S},B_{1}^{G})$ is a $2$-$(36,6,2)$
design admitting $G$ as a flag-transitive automorphism group.

Note that, $\mathrm{Aut(\mathcal{D}}_{1}\mathrm{)}=G$ as a consequence of the
O’Nan-Scott Theorem (e.g. see [20], Theorem 4.1A), since $v=2^{2}\cdot 3^{2}$,
$G=\mathrm{Aut(}G\mathrm{)}$ and $G\leqslant$
$\mathrm{Aut(\mathcal{D}}_{1}\mathrm{)}$. Thus $G$ is the full flag-transitive
automorphism group of $\mathrm{\mathcal{D}}_{1}$.

Since $r=14$, $\left(G^{\prime}\right)_{\ell}\cong D_{14}$, and
$\left(G^{\prime}\right)_{\ell,B_{1}}\leqslant G_{\ell,B_{1}}\cong Z_{3}$, it
follows that $\left(G^{\prime}\right)_{\ell,B_{1}}=1$. Therefore,
$\left[\left(G^{\prime}\right)_{\ell}:\left(G^{\prime}\right)_{\ell,B_{1}}\right]=14$
and hence $G^{\prime}\cong PSL_{2}(8)$ acts flag-transitively on
$\mathrm{\mathcal{D}}_{1}$.

  1. (ii).

$\mathcal{D}_{2}=(\mathcal{S},B_{2}^{G})$ is a $2$-$(36,6,6)$ design admitting
$G$ as the unique flag-transitive automorphism group.

Let $m_{2}\in B_{2}$. Then
$\left(m_{2}^{G_{\ell_{2}}},B_{2}^{G_{\ell_{2}}}\right)$ is a tactical
configuration with parameters $(v_{2},b_{2},k_{2},r_{2})$ equal either to
$(14,42,2,6)$ or to $(21,42,3,6)$ according as $\ell_{2}\cap m_{2}$ lies or
does not lie in $\mathcal{C}$ respectively. Therefore, $\mathcal{D}_{2}$ is a
$2$-$(36,6,6)$ design admitting $G$ as a flag-transitive automorphism group.

Arguing as in (i), we see that $G$ is the full flag-transitive automorphism
group of $\mathrm{\mathcal{D}}_{2}$. Let $H$ be the minimal flag-transitive
automorphism group of $\mathrm{\mathcal{D}}_{2}$. Then $H\leqslant G$ and
hence $H=G$ by [14], since $2^{3}\cdot 3^{3}\cdot
7\mid\left[G:G_{\ell_{2},B_{2}}\right]$. Thus $G$ is the unique flag-
transitive automorphism group of $\mathcal{D}_{2}$.

  1. (iii).

$\mathcal{D}_{i}=(\mathcal{S},B_{i}^{G})$, $i=3,4$, is a $2$-$(36,6,6)$ design
admitting $G$ as the unique flag-transitive automorphism group.

Let $m_{3}\in B_{3}$ and suppose that $\ell_{3}\cap m_{3}\in\mathcal{C}$.
Since $G$ is $3$-transitive on $\mathcal{C}$, we may assume that $\ell_{3}\cap
m_{3}=\left\\{P\right\\}$. Then $G_{\ell_{3},m_{3}}$ fixes the vertices of the
triangle inscribed in $\mathcal{C}$ having $\ell_{3},m_{3}$ as two of its
three sides. Hence $G_{\ell_{3},m_{3}}\cong Z_{3}$ since $G$ is $3$-transitive
on $\mathcal{C}$. Thus, $\left|m_{3}^{G_{\ell_{3}}}\right|=14$. Since
$G_{B_{1}}$ acts regularly on $B_{1}$ it follows that
$\left|B^{G_{\ell_{3}}}\right|=42$. Finally, $B$ contains exactly $3$ lines of
$m_{3}^{G_{\ell_{3}}}$ including $\ell_{3}$. Indeed, $G_{\ell_{3}}$ contains a
cyclic group of order $7$ acting regularly on $\mathcal{C}-\ell_{3}$. Thus
$\left|B\cap m_{3}^{G_{\ell_{3}}}\right|=2$ and hence
$\left(m_{3}^{G_{\ell_{3}}},B^{G_{\ell_{3}}}\right)$ is a tactical
configuration with parameters $(v_{3},b_{3},k_{3},r_{3})=(14,42,2,6)$. Thus
the number of blocks containing both $\ell_{3}$ and $m_{3}$ is $6$ as
$\mathcal{D}_{3}$ is a flag-transitive tactical configuration.

Suppose that $\ell_{3}\cap m_{3}\notin\mathcal{C}$. We may assume that
$W\in\ell_{3}$ and $W^{\sigma}\in m_{3}$. Then $G_{\ell_{3},m_{3}}$ is
generated by the unique elation of $PG_{2}(8)$ lying in $G$ and with center
$\ell_{3}\cap m_{3}$. Thus $G_{\ell_{3},m_{3}}\cong Z_{2}$ and hence
$\left|m_{3}^{G_{\ell_{3}}}\right|=21$. As above
$\left|B^{G_{\ell_{3}}}\right|=42$. Also $\left|B\cap
m_{3}^{G_{\ell_{3}}}\right|=3$. Indeed $G_{\ell_{3},W^{\sigma}}\cong Z_{6}$
acts regularly on $\mathcal{C}-(\ell_{3}\cup\left\\{W^{\sigma}\right\\})$.
Therefore $\left(m_{3}^{G_{\ell_{3}}},B^{G_{\ell_{3}}}\right)$ is a tactical
configuration with parameters
$(v_{3}^{\prime},b_{3}^{\prime},k_{3}^{\prime},r_{3}^{\prime})=(21,42,3,6)$.
Thus the number of blocks containing both $\ell_{3}$ and $m_{3}$ is $6$ as
$\mathcal{D}_{3}$ is a flag-transitive tactical configuration. Therefore,
$\mathcal{D}_{3}=(\mathcal{S},B_{2}^{G})$ is a $2$-$(36,6,6)$ design admitting
$G$ as a flag-transitive automorphism group. Arguing as in (i) and (ii), we
see that $G$ is the unique flag-transitive automorphism group of
$\mathrm{\mathcal{D}}_{3}$. The statement (iii) for $\mathrm{\mathcal{D}}_{4}$
is proven similarly.

  1. (iv).

$\mathcal{D}_{2}$, $\mathcal{D}_{3}$ and $\mathcal{D}_{4}$, are pairwise non
isomorphic.

Since $G_{B_{i}}=K$ is self-normalizing in $G$, where $i=1,2,3$, it follows
that $B_{i}$ is the unique block of $\mathcal{D}_{i}$ preserved by
$G_{B_{i}}$. Clearly, $\mathcal{D}_{2}\ncong\mathcal{D}_{3}$ and
$\mathcal{D}_{2}\ncong\mathcal{D}_{4}$, since none of the $6$ secants lying in
$B_{2}$ contains a point fixed by $\left\langle\gamma\right\rangle$, whereas
$B_{3}$ and $B_{4}$ do.

Suppose that $\Phi$ is an isomorphism from $\mathcal{D}_{3}$ onto
$\mathcal{D}_{4}$. Since $G$ acts point-transitively on $\mathcal{D}_{i}$,
$i=3,4$, we may assume that $\Phi$ fixes $F$. Also $G^{\Phi}=G$, since $G$ is
the full flag-transitive automorphism group of $\mathcal{D}_{i}$. Then
$\left[\Phi,G\right]=1$, since $\mathrm{Aut(}G\mathrm{)}=G$, and hence
$\Phi=1$ as $\Phi$ fixes $F$. Then $\mathcal{D}_{3}=\mathcal{D}_{4}$. Then
there is $\delta\in G$ such that $B_{1}^{\delta}=B_{2}$. Hence $\delta\in
N_{G}(G_{B_{3}})$, as $G_{B_{3}}=G_{B_{4}}=K$. Then $\delta\in G_{B_{3}}$,
since $K$ is self-normalizing in $G$, and hence $B_{3}=B_{4}$, a
contradiction. Thus $\mathcal{D}_{2}\ncong\mathcal{D}_{3}$. ∎

######  Theorem 3.10.

The following hold:

  1. (1)

If $\mathcal{D}$ is a $2$-$(36^{2},6,\lambda)$ design, with $\lambda\mid 6$,
admitting $G$ as a flag-transitive automorphism group, then either $\lambda=2$
and $\mathcal{D}$ is isomorphic to $\mathcal{D}_{1}$, or $\lambda=6$ and
$\mathcal{D}$ is isomorphic to one of the $\mathcal{D}_{i}$, where $i=2$, $3$
or $4$.

  2. (2)

If $\mathcal{D}$ is a $2$-$(36^{2},6,2)$ design admitting $G^{\prime}$ as a
flag-transitive automorphism group, then $\mathcal{D}$ is isomorphic to
$\mathcal{D}_{1}$.

###### Proof.

Let $\mathcal{D}=(\mathcal{P},\mathcal{B})$ be a $2$-$(36^{2},6,\lambda)$,
with $\lambda\mid 6$, admitting $G$ a flag-transitive automorphism group. Then
$G$ acts point-primitively on $\mathcal{D}$ by Lemma 2.2(1). On the other
hand, $G$ has a unique permutation representation of degree $36$ by [14] and
this one is equivalent to that on $\mathcal{S}$, the set of secants to the
nondegenerate conic $\mathcal{C}$ of $PG_{2}(8)$ preserved by $G$. Hence, we
may identify the point set of $\mathcal{D}$ with $\mathcal{S}$. Thus, any
block $B$ of $\mathcal{D}$ consists of $6$ secants to $\mathcal{C}$.

Assume that $\lambda=2$ and that $G^{\prime}$ acts flag-transitively on
$\mathcal{D}$. Set $J=G^{\prime}$. Then $\left|J_{B}\right|=6$ and hence
$J_{B}$ is a $J$-conjugate of $C_{G}(\eta)$ (recall that
$C_{G}(\eta)=C_{J}(\eta)$). Without loss, we may assume that
$J_{B}=C_{G}(\eta)$. Then $B$ is one of the following $C_{G}(\eta)$-orbits
$\left(PP^{\gamma}\right)^{C_{G}(\eta)}$ and
$\left(FP^{\gamma^{i}}\right)^{C_{G}(\eta)}$, where $i=0,1,2$, by Lemma
3.8(3).

Assume that $B=\left(FP\right)^{C_{G}(\eta)}$. Set $\ell=FP$. Since
$C_{G}(\eta)$ acts transitively on $\left\\{F,W,W^{\sigma}\right\\}$, it
follows that
$B=\left\\{\ell,\ell^{\sigma},\ell^{\eta},\ell^{\sigma\eta},\ell^{\eta^{2}},\ell^{\sigma\eta^{2}}\right\\}$,
where $\ell\cap\ell^{\sigma}=\left\\{F\right\\}$,
$\ell^{\eta}\cap\ell^{\sigma\eta}=\left\\{W\right\\}$ and
$\ell^{\eta^{2}}\cap\ell^{\sigma\eta^{2}}=\left\\{W^{\sigma}\right\\}$.
Moreover, as $C_{G}(\eta)$ acts regularly on
$\mathcal{C}-\left\\{F,W,W^{\sigma}\right\\}$, any two secants lying in $B$ do
not intersect in $\mathcal{C}-\left\\{F,W,W^{\sigma}\right\\}$. Thus, through
any point of $\mathcal{C}-\left\\{F,W,W^{\sigma}\right\\}$ there is exactly
one secant to $\mathcal{C}$ lying in $B$ and incident with the point.

Let $m\in B$, $m\neq\ell$. Assume that $\ell\cap m\in\mathcal{C}$. Then
$m=\ell^{\sigma}\in\mathcal{S}(\ell)$ and hence $\left|m^{J_{\ell}}\right|=12$
as $\mathcal{S}(\ell)$ is also a $J_{\ell}$-orbit, being $J_{\ell}\cong
D_{14}$. Clearly $\left|B^{J_{\ell}}\right|=14$ as $J_{\ell,B}=1$. If there is
$e\in B\cap m^{J_{\ell}}$, with $e\neq m$, then $e\in\mathcal{S}(\ell)$ and
hence $e\cap\ell\in\mathcal{C}$. Then
$e\cap\ell\in\mathcal{C}-\left\\{F,W,W^{\sigma}\right\\}$, as $e\neq
m=\ell^{\sigma}$, and we obtain a contradiction, since through any point of
$\mathcal{C}-\left\\{F,W,W^{\sigma}\right\\}$ there is exactly one secant to
$\mathcal{C}$ lying in $B$ and incident with the point. Thus $\left|B\cap
m^{J_{\ell}}\right|=1$ and hence $\left(m^{J_{\ell}},B^{J_{\ell}}\right)$ is a
tactical configuration with parameters
$(v^{\prime},b^{\prime},k^{\prime},r^{\prime})=(14,14,1,1)$ by [18], 1.2.6.
Hence the number of elements in $B^{J_{\ell}}$ containing both $\ell_{1}$ and
$m_{1}$ is $1$. Then the number of $B^{J}$ containing both $\ell_{1}$ and
$m_{1}$ is $1$, as $\mathcal{D}$ is flag-transitive by our assumption.
However, that is impossible as it contradicts the assumption $\lambda=2$. The
cases $B=\left(FP^{\gamma^{i}}\right)^{C_{G}(\eta)}$, with $i=1,2$, are
similarly ruled out. Then $B=\left(PP^{\gamma}\right)^{C_{G}(\eta)}$ and hence
$B=\left(P^{\gamma}P\right)^{C_{G}(\gamma)}$ by Lemma 3.8. Thus,
$\mathcal{D}\cong\mathcal{D}_{1}$ by Example 3.9(1).

Assume that $\lambda=2$ and that $G$ acts flag-transitively on $\mathcal{D}$.
Let $\ell\in B$, then $G_{\ell,B}\cong Z_{3}$ and hence
$\left|G_{B}\right|=18$. Moreover, $G_{\ell,B}\cap G^{\prime}=1$ and
$G_{B}\cap G^{\prime}\cong S_{3}$, since $G^{\prime}\cong PSL_{2}(8)$, and
hence $G_{B}\cong Z_{3}\times S_{3}$. Since the centralizer in $G$ of any
subgroup of order $3$ of $G^{\prime}$ is of order $27$ by [14], it follows
that $G_{B}=C_{G}(\rho)$ for some element $\rho$ of order $3$ lying in
$G-G^{\prime}$. We may assume that $G_{B}=C_{G}(\gamma)$, since the subgroups
of order $3$ of $G$ intersecting $G^{\prime}$ in $1$ lies in one conjugacy
class under $G$ again by [14]. Thus, $B=B_{1}$ by Lemma 3.8(1) and so
$\mathcal{D}\cong\mathcal{D}_{1}$ by Example 3.9(1).

Assume that $\lambda=3$. Then $G_{\ell,B}\cong Z_{2}$ and
$\left|G_{B}\right|=12$, as $k=6$, and hence $G_{B}\cong A_{4}$ by [14]. Let
$\alpha,\beta,\delta\in G_{B}$ such that
$\left\langle\alpha,\beta\right\rangle\cong E_{4}$ and $o(\delta)=3$. Since
$G_{B}$ preserves $\mathcal{C}$, the group
$\left\langle\alpha,\beta\right\rangle$ consists of elations of $PG_{2}(8)$
having the same axis $a$ tangent to $\mathcal{C}$ and distinct centers
$C_{\alpha}$, $C_{\beta}$ and $C_{\alpha\beta}$. Furthermore,
$\left\langle\delta\right\rangle$ fixes $a$ and permutes transitively
$\left\\{C_{\alpha},C_{\beta},C_{\alpha\beta}\right\\}$. Then the block $B$
consists of two secants incident with $C_{\alpha}$, two ones incident with
$C_{\beta}$ and two ones incident with $C_{\alpha\beta}$. We may assume that
$C_{\alpha}\in\ell$. Let $m\in B-\left\\{\ell\right\\}$ such that
$C_{\alpha}\in m$. Then $\left(m^{G_{\ell}},B^{G_{\ell}}\right)$ is a tactical
configuration with parameters
$(v^{\prime\prime},b^{\prime\prime},k^{\prime\prime},r^{\prime\prime})=(21,21,1,1)$
by [18], 1.2.6. Hence the number of elements in $B^{G_{\ell}}$ containing both
$\ell$ and $m$ is $1$. Then the number of $B^{G}$ containing both $\ell$ and
$m$ is $1$, as $\mathcal{D}$ is flag-transitive. However, that is impossible
as it contradicts the assumption $\lambda=3$.

Assume that $\lambda=6$. Then $\left|G_{B}\right|=6$. If $G_{B}\leqslant
G^{\prime}$, then $G_{B}\cong S_{3}$ and hence is a $G$-conjugate of
$C_{G}(\eta)$ by [14]. Without loss, we may assume that $G_{B}=C_{G}(\eta)$.
Then $B$ is one of the $C_{G}(\eta)$-orbits
$\left(PP^{\gamma}\right)^{C_{G}(\eta)}$ and
$\left(FP^{\gamma^{t}}\right)^{C_{G}(\eta)}$, where $t=0,1,2$, by Lemma
3.8(2). If $B=\left(PP^{\gamma}\right)^{C_{G}(\eta)}$, then
$B=\left(P^{\gamma}P\right)^{C_{G}(\gamma)}$ again by Lemma 3.8, and hence
$\mathcal{D}\cong\mathcal{D}_{1}$ by Example 3.9(1), whereas $\lambda=6$. So,
$B\neq\left(PP^{\gamma}\right)^{C_{G}(\eta)}$ and hence
$B_{t}=\left(FP^{\gamma^{t}}\right)^{C_{G}(\eta)}$ for some $t=0,1,2$.

Let $t=0$ and let $m=\ell^{\sigma}$. Then $m\cap\ell\in\mathcal{C}$. A similar
argument to that used for the case $\lambda=2$ shows that
$\left(m^{G_{\ell}},B^{G_{\ell}}\right)$ is a tactical configuration with
parameters
$(v^{\prime\prime\prime},b^{\prime\prime\prime},k^{\prime\prime\prime},r^{\prime\prime\prime})=(14,14,1,1)$
by [18], 1.2.6, since through any point of
$\mathcal{C}-\left\\{F,W,W^{\sigma}\right\\}$ there is exactly one secant to
$\mathcal{C}$ lying in $B$ and incident with the point. Hence the number of
elements in $B^{G_{\ell}}$ containing both $\ell$ and $m$ is $1$. Then the
number of $B^{G}$ containing both $\ell$ and $m$ is $1$, as $\mathcal{D}$ is
flag-transitive. So $\lambda=1$, but this contradicts our assumption. The
cases $t=1,2$ are excluded similarly.

Assume that $G_{B}\nleq G^{\prime}$. Then $G_{B}\cong Z_{6}$ and hence is a
$G$-conjugate of $K$. Thus, without loss, we may assume that $G_{B}=K$. Then
$B$ is one of the following $K$-orbits on $\mathcal{S}$:
$\left(P^{\gamma}P\right)^{K}$, $\left(P^{\gamma}P^{\sigma}\right)^{K}$,
$\left(PW\right)^{K}$, $\left(P^{\sigma}W\right)^{K}$, and
$\left(PF\right)^{K}$ by Lemma 3.8(3). Then
$B=\left(P^{\gamma}P\right)^{K}=\left(P^{\gamma}P\right)^{C_{G}(\gamma)}$
implies $\mathcal{D}\cong\mathcal{D}_{1}$ by Example 3.9(1), and we again
reach a contradiction as $\lambda=6$. Thus,
$B\neq\left(P^{\gamma}P\right)^{K}$. Also $B\neq\left(PF\right)^{K}$,
otherwise all the secants lying in any block of $\mathcal{D}$ intersects in a
point, whereas for any two secants $s$ and $s^{\prime}$ to $\mathcal{C}$ such
that $s\cap s^{\prime}\notin\mathcal{C}$, then there are no blocks of
$\mathcal{D}$ incident with them. Thus, $B$ is either
$\left(P^{\gamma}P^{\sigma}\right)^{K}$ or $\left(PW\right)^{K}$, or
$\left(P^{\sigma}W\right)^{K}$, and we obtain
$\mathcal{D}\cong\mathcal{D}_{i}$, where $i=2$, $3$ or $4$, respectively, by
Example 3.9(2). ∎

###### Proof of Theorem 1.1.

By Theorem 2.5, $Soc(G)$, the socle of $G$, is either an elementary abelian
$p$-group for some prime $p$ or a non abelian simple group.

Assume that the latter occurs. Then $X$ is neither sporadic nor an alternating
group by Lemmas 3.2 and 3.5 respectively. If $X$ is classical, then assertion
(1) is immediate, but also (2b)–(2c) follow by Theorem 3.10, since
$PSL_{2}(8)\cong$ ${}^{2}G_{2}(3)^{\prime}$. Finally, if $X$ is isomorphic to
the socle a finite exceptional group of Lie type, then $2$-$(36,6,\lambda)$
design, where $\lambda=2,3$ or $6$, and $X\cong$ ${}^{2}G_{2}(3)^{\prime}$ by
Lemmas 3.6 and 3.7. Then the assertions (2b)–(2c) follow again from Theorem
3.10. This completes the proof. ∎

## References

  * [1] M. Aigner, G. M. Ziegler, Proofs from THE BOOK, (5th ed.). Berlin, New York: Springer-Verlag. 2014
  * [2] S. H. Alavi , M. Bayat, M. Biliotti, A. Daneshkhah, E. Francot, H. Guan, A. Montinaro, F. Mouseli, P. Rizzo, D. Tian, Y. Wang, X. Zhan, Y. Zhang, S. Zhou, Y. Zhu, Block designs with flag-transitive automorphism groups. submitted
  * [3] S. H. Alavi, Flag-Transitive Block Designs and Finite Simple Exceptional Groups of Lie Type, Graphs and Combinatorics 36 (2020) 1001–1014.
  * [4] S. H. Alavi, A note on two families of 2-designs arose from Suzuki-Tits ovoid, Algebra and Discrete Math. Accepted.
  * [5] S. H. Alavi, M. Bayat, A. Daneshkhah, Symmetric designs admitting flag-transitive and point-primitive automorphism groups associated to two dimensional projective special groups, Designs, Codes and Cryptography 79 (2016) 337–351.
  * [6] S. H. Alavi, M. Bayat, A. Daneshkhah, Flag-transitive block designs and unitary groups, Monatshefte für Mathematik 193 (2020) 535–553.
  * [7] S. H. Alavi, M. Bayat, A. Daneshkhah, Correction to: Flag-transitive block designs and unitary groups, Monatshefte für Mathematik 195 (2021) 371–376.
  * [8] S. H. Alavi, M. Bayat, and A. Daneshkhah, Finite exceptional groups of Lie type and symmetric designs. Submitted arxiv:1702.01257.
  * [9] S. H. Alavi, T. C. Burness, Large subgroups of simple groups, J. Algebra 421 (2015) 187–233.
  * [10] M. Biliotti, A. Montinaro, On flag-transitive symmetric designs of affine type. J. Combin. Des. 25 (2017) 85–97.
  * [11] M. Biliotti, A. Montinaro, P. Rizzo, $2$-$(v,k,\lambda)$ designs, with $(r,\lambda)=1$, admitting a solvable flag-transitive automorphism group of affine type. J. Combin. Des 27 (2019) 784–800.
  * [12] F. Buekenhout, A. Delandtsheer, J. Doyen, Finite linear spaces with flag-transitive sporadic groups, Unpublished note.
  * [13] F. Buekenhout, A. Delantsheer, J. Doyen, P. B. Kleidman, M. Liebeck, and J. Saxl, Linear Spaces with flag-transitive automorphism groups, Geom. Dedicata 36 (1990) 89-94.
  * [14] J. H. Conway, R. T. Curtis, R. A. Parker, R. A. Wilson, An Atlas of Finite Groups, Clarendon Press, Oxford, 1985.
  * [15] D. Davies, Automorphisms of designs, Phd thesis, University of East Anglia (1987).
  * [16] A. Delandtsheer, Flag-transitive finite simple groups, Arch. Math. (Basel) 47 (1986) 395–400.
  * [17] A. Delandtsheer, Finite Flag-transitive Linear Spaces with Alternating Socle, pp. 79–88 in Algebraic combinatorics and Applications, Eds. A. Betten, A Kohnert, R. Laue, A. Wassermann, Proc.of theEuroconference ALCOMA, held in Gö$\beta$wienstein, Germany, Springer, 1999.
  * [18] P. Dembowski, Finite Geometries, Springer, Berlin, Heidelberg, New York, 1968.
  * [19] A. Deviller, H. Liang, C. E. Praeger, B. Xia, OnGalg-transitive $2$-$(v,k,2)$ designs, J. Combin. Theory Ser. A 177 (2021) 105309.
  * [20] J. D. Dixon, B. Mortimer, Permutation Groups. Springer Verlag, New York (1966).
  * [21] J. W. P. Hirshfeld, Projective Geometry Over Finite Fields. Second Edition. Clarendon Press, Oxford, 1998.
  * [22] D. R. Hughes, F.C. Piper, Projective Planes. Springer, New York, Berlin (1973).
  * [23] B. Huppert, Endliche gruppen I, Die Grundlehren der Mathematischen Wissenschaften, Band 134, Springer-Verlag, Berlin–New York, 1967\.
  * [24] P. B. Kleidman, The finite flag-transitive linear spaces with an exceptional automorphism group, 461 in: Finite geometries and combinatorial designs (Lincoln, NE, 1987), vol. 111 of Contemp. Math., 462 Amer. Math. Soc., Providence, RI, 1990, pp. 117–136.
  * [25] W. M. Kantor, Classification of 2-transitive symmetric designs, Graphs Combin. 1 (1985) 165–450.
  * [26] P. Kleidman, M. Liebeck, The subgroup structure of the finite classical groups. London Mathematical Society Lecture Note Series, 129. Cambridge University Press, Cambridge, 1990.
  * [27] M. W. Liebeck, C. E. Praeger, J. Saxl, The O’Nan Scott theorem for finite primitve permutation groups, J. Austral. Math. Soc. 44 (1988) 389–396.
  * [28] M. W. Liebeck, J. Saxl, and G. M. Seitz, On the overgroups of irreduciblensubgroups of the finite classical groups, Proc. London Math. Soc. 55 (1987) 507–537.
  * [29] M. W. Liebeck, The classification of finite linear spaces with flag-transitive automorphism groups of affine type, J. Combin. Theory Ser. A 84 (1998) 196–235.
  * [30] A. Montinaro, M. Biliotti, E. Francot, Classification of $2-(v,k,\lambda)$ designs with $(r,\lambda)=1$ and $\lambda>1$, admitting a non-solvable flag-transitive automorphism group of affine type, J. Algebr. Comb. (2021). https://doi.org/10.1007/s10801-021-01075-1.
  * [31] A. Montinaro, Classification of the $2$-$(k^{2},k,\lambda)$ design, with $\lambda\mid k$, admitting a flag-transitive automorphism group $G$ of affine type. Submitted.
  * [32] A. Montinaro, Classification of the $2$-$(k^{2},k,\lambda)$ design, with $\lambda\mid k$, admitting a flag-transitive automorphism group $G$ of almost simple type. Submitted.
  * [33] A. Montinaro, Flag-transitive, point-imprimitive symmetric $2$-$(v,k,\lambda)$ designs with $k>\lambda(\lambda-3)/2$. Submitted.
  * [34] C. E. Praeger, S. Zhou, Imprimitive flag-transitive symmetric designs. J. Comb. Theory Ser. A, 113 (2006) 1381–1395.
  * [35] J. Saxl. On nite linear spaces with almost simple flag-transitive automorphism groups. J. Combin. Theory Ser. A, 100 (2002) 322–348.
  * [36] R. A. Wilson, Maximal subgroups of sporadic groups, arXiv:1701.02095v2
  * [37] D. Tian, S. Zhou, Flag-transitive $2-(v,k,\lambda)$ symmetric designs with sporadic socle, Journal of Combinatorial Designs 23 (2015) 140–150.
  * [38] R. Wilson, P. Walsh, J. Tripp, I. Suleiman, R. Parker, S. Norton, S. Nickerson, S. Linton, J. Bray, R. Abbott, An Atlas of Finite Groups Representation- Version 3, available at http://brauer.maths.qmul.ac.uk/Atlas/v3/
  * [39] P. H. Zieschang, Flag transitive automorphism groups of $2$-designs with $(r,\lambda)=1$, Journal of Algebra 118 (1988) 369–375.
  * [40] X. Zhan, S. Ding, Comments on “Flag-transitive block designs and unitary groups”, Monatshefte für Mathematik 195 (2021) 177–180.
  * [41] X. Zhan, S. Zhou, Flag-transitive non-symmetric $2$-designs with $(r,\lambda)=1$ and sporadic socle, Des. Codes Cryptography 81 (2016) 481–487.
  * [42] X. Zhan, S. Zhou, Non-symmetric 2-designs admitting a two-dimensional projective linear group, Designs, Codes and Cryptography 86 (12) (2018) 2765–2773.
  * [43] S. Zhou, X. Zhan, Flag-transitive automorphism groups of $2$-designs with $\lambda\geqslant(r,\lambda)^{2}$ and an application to symmetric designs, Ars Math. Contemp. 14 (2018) 187–195.
  * [44] Y. Zhang, S. Zhou, Flag-transitive non-symmetric $2$-designs with $(r,\lambda)=1$ and Exceptional Groups of Lie type, The Electronic Journal of Combinatorics 27 (2020) P2.9.
  * [45] S. Zhou, Y. Wang, Flag-transitive non-symmetric $2$-designs with $(r,\lambda)=1$ and alternating socle, Electron. J. Combin. 22 (2015) Paper 2.6, 15.
  * [46] Y. Zhu, H. Guan, S. Zhou, Flag-transitive $2$-$(v,k,\lambda)$ symmetric designs with $(r,\lambda)=1$ and alternating socle, Frontiers of Mathematics in China 10 (2015) 1483–1496.
  * [47] The GAP Group, GAP ”Groups, Algorithms, and Programming, Version 4.11.0, 2020, available at http://www.gap.system.org