# The symmetric $2$-$(v,k,\lambda)$ designs, with
$k>\lambda\left(\lambda-3\right)/2$, admitting a flag-transitive, point
imprimitive automorphism group are known

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

The symmetric $2$-$(v,k,\lambda)$ designs, with
$k>\lambda\left(\lambda-3\right)/2$, admitting a flag-transitive, point-
imprimitive automorphism group are completely classified: they are the known
$2$-designs with parameters $(16,6,2),(45,12,3),(15,8,4)$ or $(96,20,4)$.

###### Key words and phrases:

Symmetric design; automorphism group; flag-transitive design

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

05B05; 05B25; 20B25

##  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 _non-trivial_ if $2<k<v$, and symmetric if $v=b$. All
$2$-$(v,k,\lambda)$ designs in this paper are assumed to be non-trivial. 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. In
this setting we also say that $\mathcal{D}$ is either _point-primitive_ or
_point-imprimitive_ respectively. A _flag_ of $\mathcal{D}$ 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_.

In 1987, Davies [7] proved that in a flag-transitive and point-imprimitive
$2$-$(v,k,\lambda)$ design, the block size is bounded for a given value of the
parameter $\lambda$, where $\lambda\geqslant 2$ by a result of Higman-
McLaughlin [13] dating back to 1961. In 2005, O’Reilly Regueiro [28] obtained
an explicit upper bound. Later that year, Praeger and Zhou [31] improved that
upper bound and gave a complete list of feasible parameters. In 2020, Mandić
and Šubasić [23] classified the flag-transitive point-imprimitive symmetric
$2$-designs with $\lambda\leqslant 10$ except for two possible numerical
cases. Recently, Montinaro [26] has classified those with
$k>\lambda\left(\lambda-3\right)/2$ and such that a block of the $2$-design
intersects a block of imprimitivity in at least $3$ points. In this paper we
complete the work started in [26] by classifying $\mathcal{D}$ with
$k>\lambda\left(\lambda-3\right)/2$ regardless the intersection size of a
block of $\mathcal{D}$ with a block of imprimitivity. More precisely, our
result is the following.

######  Theorem 1.1.

Let $\mathcal{D}=\left(\mathcal{P},\mathcal{B}\right)$ be a symmetric
$2$-$(v,k,\lambda)$ design admitting a flag-transitive, point-imprimitive
automorphism group. If $k>\lambda(\lambda-3)/2$ then one of the following
holds:

  1. (1)

$\mathcal{D}$ is isomorphic to one of the two $2$-$(16,6,2)$ designs.

  2. (2)

$\mathcal{D}$ is isomorphic to the $2$-$(45,12,3)$ design.

  3. (3)

$\mathcal{D}$ is isomorphic to the $2$-$(15,8,4)$ design.

  4. (4)

$\mathcal{D}$ is isomorphic to one of the four $2$-$(96,20,4)$ designs.

In 1945, Hussain [14] and in 1946 Nandi [27] independently proved the
existence of three symmetric $2$-$(16,6,2)$ designs. In 2005, O’Reilly
Regueiro [28] proved that exactly two of them are flag-transitive and point-
imprimitive. In the same paper O’Reilly Regueiro constructed a $2$-$(15,8,4)$
design. Such $2$-design was proved to be unique by Praeger and Zhou [31] in
2006. One year later, Praeger [30] constructed and proved there is exactly one
flag-transitive and point-imprimitive $2$-$(45,12,3)$ design. Finally, in
2009, Law, Praeger and Reichard [19] proved there are four flag-transitive and
point-imprimitive $2$-$(96,20,4)$ designs.

The outline of the proof is as follows. We start by strengthening the
classification result obtained in [23]. Indeed, in [23] it is proven that, if
$\lambda\leqslant 10$ then $\mathcal{D}$ is known except for two numerical
values for the parameters for $\mathcal{D}$. The two exceptions are ruled out
here in Theorem 2.3. Subsequently, we focus on the case $\lambda>10$. In
Proposition 2.5 it is shown that $G^{\Sigma}$ acts primitively on $\Sigma$ by
using the results contained in [26]. Then, it is proven in Theorem 3.1 that,
either $G$ acts point-quasiprimitively on $\mathcal{D}$ or $G(\Sigma)\neq 1$,
$G^{\Sigma}$ is almost simple and $\mathcal{D}$ has parameters
$(2^{d+2}(2^{d-1}-1)^{2},2\left(2^{d}-1\right)\left(2^{d-1}-1\right),2(2^{d-1}-1))$
where $d\geqslant 4$. Afterwards, by combining the O’Nan-Scott theorem for
quasiprimitive groups achieved in [29] with an adaptation of the techniques
developed by [41], in Theorem 4.1 we show that $G^{\Sigma}$ is almost simple
also in the quasiprimitive case. Moreover, if $L$ is the preimage in $G$ of
$\mathrm{Soc}(G^{\Sigma})$ and $\Delta\in\Sigma$, in Proposition 5.3,
Corollary 5.4 and Theorem 5.5 it is proven that, either $G(\Sigma)=1$,
$L_{\Delta}$ is contained in a semilinear $1$-dimensional group and
$\left|L\right|\leqslant
2\left|L_{\Delta}^{\Delta}\right|^{2}\left|\mathrm{Out}(L)\right|^{2}$, or
$G(\Sigma)=1$, $L_{\Delta}$ is a non-solvable $2$-transitive permutation group
of degree $\left|\Delta\right|$ and
$\left|L\right|\leqslant\left|L_{\Delta}\right|^{2}$, or $G(\Sigma)\neq 1$ and
a quotient of $L_{\Delta}^{\Sigma}$ is isomorphic either to $SL_{d}(2)$, or to
$A_{7}$ for $d=4$. In particular, in each case $L_{\Delta}^{\Sigma}$ is a
large subgroup of $L^{\Sigma}$. Finally, we use all the above mentioned
constraints on $L^{\Sigma}$ and on $L_{\Delta}^{\Sigma}$ together with the
results contained in [3] and [20] to precisely determine the admissible pairs
$(L^{\Sigma},L_{\Delta}^{\Sigma})$ and from these to prove that there are no
examples of $\mathcal{D}$ for $\lambda>10$. At this point, our classification
result follows from Theorem 2.3.

##  2\. Preliminary reductions

It is well know that, if $\mathcal{D}$ is a symmetric $2$-$(v,k,\lambda)$
design, then $r=k$, $b=v$ and $k(k-1)=(v-1)\lambda$ (for instance, see [8]).
Moreover, the following fact holds:

######  Lemma 2.1.

If $\mathcal{D}$ admits a flag-transitive automorphism group $G$ and $x$ is
any point of $\mathcal{D}$, then $\left|y^{G_{x}}\right|\lambda=k\left|B\cap
y^{G_{x}}\right|$ for any point $y$ of $\mathcal{D}$, with $y\neq x$, and for
any block $B$ of $\mathcal{D}$ incident with $x$.

###### Proof.

Let $x$, $y$ be points 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 [8, 1.2.6], it follows that
$\left|y^{G_{x}}\right|\lambda=k\left|B\cap y^{G_{x}}\right|$. ∎

The following theorem, which is a summary of [26] and some of the results
contained in [31], is our starting point.

######  Theorem 2.2.

Let $\mathcal{D}=(\mathcal{P},\mathcal{B})$ be a symmetric $2$-design
admitting a flag-transitive, point-imprimitive automorphism group $G$ that
leaves invariant a non-trivial partition
$\Sigma=\left\\{\Delta_{1},...,\Delta_{d}\right\\}$ of $\mathcal{P}$ such that
$\left|\Delta_{i}\right|=c$ for each $i=1,...,d$. Then the following hold:

  1. I.

There is a constant $\ell$ such that, for each $B\in\mathcal{B}$ and
$\Delta_{i}\in\Sigma$, the size $\left|B\cap\Delta_{i}\right|$ is either $0$
or $\ell$.

  2. II.

There is a constant $\theta$ such that, for each $B\in\mathcal{B}$ and
$\Delta_{i}\in\Sigma$ with $\left|B\cap\Delta_{i}\right|>0$, the number of
blocks of $\mathcal{D}$ whose intersection set with $\Delta_{i}$ coincides
with $B\cap\Delta_{i}$ is $\theta$.

  3. III.

If $\ell=2$ then $G_{\Delta_{i}}^{\Delta_{i}}$ acts $2$-transitively on
$\Delta_{i}$ for each $i=1,...,d$.

  4. IV.

If $\ell\geqslant 3$ then
$\mathcal{D}_{i}=\left(\Delta_{i},\left(B\cap\Delta_{i}\right)^{G_{\Delta_{i}}^{\Delta_{i}}}\right)$
is a flag-transitive non-trivial $2$-$\left(c,\ell,\lambda/\theta\right)$
design for each $i=1,...,d$.

Moreover, if $k>\lambda(\lambda-3)/2$ then one of the following holds:

  1. V.

$\ell=2$ and one of the following holds:

    1. 1.

$\mathcal{D}$ is a symmetric
$2$-$(\lambda^{2}(\lambda+2),\lambda(\lambda+1),\lambda)$ design and
$\left(c,d\right)=\left(\lambda+2,\lambda^{2}\right)$.

    2. 2.

$\mathcal{D}$ is a symmetric
$2$-$\left(\left(\frac{\lambda+2}{2}\right)\left(\frac{\lambda^{2}-2\lambda+2}{2}\right),\frac{\lambda^{2}}{2},\lambda\right)$
design,
$\left(c,d\right)=\left(\frac{\lambda+2}{2},\frac{\lambda^{2}-2\lambda+2}{2}\right)$,
and either $\lambda\equiv 0\pmod{4}$, or $\lambda=2w^{2}$, where $w$ is odd,
$w\geqslant 3$, and $2(w^{2}-1)$ is a square.

  2. VI.

$\ell\geqslant 3$ and one of the following holds:

    1. 1.

$\mathcal{D}$ is isomorphic to the $2$-$(45,12,3)$ design of [30, Construction
4.2].

    2. 2.

$\mathcal{D}$ is isomorphic to one of the four $2$-$(96,20,4)$ designs
constructed in [19].

We are going to focus on case (V) of the previous theorem. _Throughout the
paper, a $2$-design as in case (V.1) or (V.2) of Theorem 2.2 will be simply
called a $2$-design of type 1 or 2 respectively._

Let $\Delta\in\Sigma$ and $x\in\Delta$. Since $G(\Sigma)\trianglelefteq
G_{\Delta}$ and $G(\Delta)\trianglelefteq G_{x}$, it is immediate to verify
that $(G^{\Sigma})_{\Delta}=(G_{\Delta})^{\Sigma}$ and that
$\left(G_{\Delta}^{\Delta}\right)_{x}=(G_{x})^{\Delta}$. Hence, in the sequel
$(G^{\Sigma})_{\Delta}$ and $\left(G_{\Delta}^{\Delta}\right)_{x}$ will simply
be denoted by $G^{\Sigma}_{\Delta}$ and $G_{x}^{\Delta}$ respectively.
Moreover, the following holds:

$\frac{G^{\Sigma}_{\Delta}}{G(\Delta)^{\Sigma}}\cong\frac{G_{\Delta}}{G(\Delta)G(\Sigma)}\cong\frac{G^{\Delta}_{\Delta}}{G(\Sigma)^{\Delta}}.$
(2.1)

A further reduction is the following theorem.

######  Theorem 2.3.

Let $\mathcal{D}=\left(\mathcal{P},\mathcal{B}\right)$ be a symmetric
$2$-$(v,k,\lambda)$ design admitting a flag-transitive, point-imprimitive
automorphism group $G$. If $\lambda\leqslant 10$ then

  1. (1)

$\mathcal{D}$ is isomorphic to one of the two $2$-$(16,6,2)$ designs

  2. (2)

$\mathcal{D}$ is isomorphic to the $2$-$(45,12,3)$ design.

  3. (3)

$\mathcal{D}$ is isomorphic to the $2$-$(15,8,4)$ design.

  4. (4)

$\mathcal{D}$ is isomorphic to one of the four $2$-$(96,20,4)$ designs.

###### Proof.

This results is proven in [23, Theorem 1] with the following possible
exceptions of $(v,k,\lambda,c,d)=(288,42,6,8,36)$ or $(891,90,9,81,11)$. Note
that $k>\lambda(\lambda-3)/2$ in both exceptional cases. Actually, the latter
does not correspond to any case of Theorem 2.2(V–VI), and hence it cannot
occur. The former corresponds to Theorem 2.2(V.1) for $\lambda=6$. Also, if
$\Delta\in\Sigma$ then $G_{\Delta}^{\Delta}\cong AGL_{1}(8)$, $A\Gamma
L_{1}(8)$, $A_{8}$, $S_{8}$, $PSL_{2}(7)$ or $PGL_{2}(7)$ by [16, Lists (A)
and (B)] since $G_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$.

Assume that $u$ divides the order of $G(\Sigma)$, where $u$ is an odd prime,
and let $\psi$ be a $u$-element of $G(\Sigma)$. Then $\psi$ fixes at least a
point on each $\Delta\in\Sigma$ since $\left|\Delta\right|=8$. Thus $\psi$
preserves at least two distinct blocks of $\mathcal{D}$ by [18, Theorem 3.1],
say $B_{1}$ and $B_{2}$. Actually, $\psi$ fixes $B_{1}$ and $B_{2}$ pointwise
since any of these intersect each element of $\Sigma$ in $0$ or $2$ points and
$\psi\in G(\Sigma)$. Hence, $\psi$ fixes at least $2\cdot 42-6$ points of
$\mathcal{D}$, but this contradicts [18, Corollary 3.7]. Thus, the order of
$\left|G(\Sigma)\right|=2^{i}$ with $i\geqslant 0$.

Assume that $w$ divides the order of $G(\Delta)$, where $w$ is an odd prime,
$w\geqslant 7$, and let $\phi$ be a $w$-element of $G(\Delta)$. Then $\phi$
fixes the $6$ blocks incident with any pair of distinct points of $\Delta$.
Therefore, $\phi$ fixes at least $6\cdot\binom{8}{2}$ points of $\mathcal{D}$
by [18, Theorem 3.1], and we again reach a contradiction by [18, Corollary
3.7]. Thus, $G(\Delta)$ is a $\\{2,3,5\\}$-group.

Any Sylow $7$-subgroup of $G$ is of order $7$, since $\left|\Sigma\right|=36$,
$G_{\Delta}^{\Delta}$ is one of the groups listed above and the order of
$G(\Delta)$ is coprime to $7$. Then, by [10, Table B.4] one of the following
holds:

  1. (1)

$A_{9}\trianglelefteq G^{\Sigma}\leqslant S_{9}$ and $S_{7}\trianglelefteq
G^{\Sigma}_{\Delta}\leqslant S_{7}\times Z_{2}$;

  2. (2)

$PSL_{2}(8)\trianglelefteq G^{\Sigma}\leqslant P\Gamma L_{2}(8)$ and
$D_{14}\trianglelefteq G^{\Sigma}_{\Delta}\leqslant F_{42}$;

  3. (3)

$PSU_{3}(3)\trianglelefteq G^{\Sigma}\leqslant P\Gamma U_{3}(3)$ and
$PSL_{2}(7)\trianglelefteq G^{\Sigma}_{\Delta}\leqslant PGL_{2}(7)$;

  4. (4)

$G^{\Sigma}\cong Sp_{6}(2)$ and $G^{\Sigma}_{\Delta}\cong S_{8}$.

Since $G_{\Delta}^{\Delta}$ is one of the $2$-transitive-groups listed above,
(1) is immediately ruled out by (2.1). In (2) the group $G_{\Delta}$ is
solvable since $G^{\Sigma}_{\Delta}$ is solvable and $G(\Sigma)$ is a
$2$-group. Thus $G_{\Delta}^{\Delta}$ is solvable and hence it is isomorphic
to $AGL_{1}(8)$ or $A\Gamma L_{1}(8)$. Moreover, a quotient group of
$G_{\Delta}^{\Delta}$ must contain a subgroup isomorphic to $D_{14}$ by (2.1)
since $G(\Delta)$ is a $\\{2,3,5\\}$-group, but this is clearly impossible. It
follows that only (3) and (4) are admissible. Also, $G(\Sigma)=G(\Delta)=1$ in
both cases by comparing (2.1) with the possibilities for $G_{\Delta}^{\Delta}$
provided above.

Assume that (4) occurs. Then $G$ acts $2$-transitively on $\Sigma$. Let
$x\in\Delta$ and $\Delta^{\prime}\in\Sigma\setminus\\{\Delta\\}$. Then
$G_{x}\cong S_{7}$ and $G_{x,\Delta^{\prime}}\leqslant
G_{\Delta,\Delta^{\prime}}\cong(S_{4}\times S_{4}):Z_{2}$ by [6], hence
$G_{x}$ acts transitively on $\Sigma\setminus\\{\Delta\\}$. On the other hand,
$G_{x}=G_{B}$ for some block $B$ of $\mathcal{D}$ since $G$ has a unique
conjugate class of subgroups isomorphic to $S_{7}$ by [6]. Therefore, $G_{x}$
permutes transitively the $21$ elements of $\Sigma$ intersecting $B$, whereas
$G_{x}$ acts transitively on the $35$ elements $\Sigma\setminus\\{\Delta\\}$.
So this case is excluded. Finally, (3) is ruled out with the aid of GAP [42],
and the proof is thus completed. ∎

On the basis of Theorem 2.3, in the sequel we may assume that $\lambda>10$.

######  Lemma 2.4.

If $G$ preserves a further partition $\Sigma^{\prime}$ of the point set of
$\mathcal{D}$ in $d^{\prime}$ blocks of imprimitivity of size $c^{\prime}$.
Then $d^{\prime}=d$ and $c^{\prime}=c$, and one of the following holds:

  1. (1)

$\Sigma=\Sigma^{\prime}$;

  2. (2)

$\left|\Delta\cap\Delta^{\prime}\right|\leqslant 1$ for for each
$\Delta\in\Sigma$, $\Delta^{\prime}\in\Sigma^{\prime}$.

###### Proof.

Suppose there is a $G$-invariant partition $\Sigma^{\prime}$ of the point set
of $\mathcal{D}$. Let $B$ any block of $\mathcal{D}$ and let
$\Delta^{\prime}\in\Sigma^{\prime}$. If
$\left|B\cap\Delta^{\prime}\right|\geqslant 3$ then either
$(v,k,\lambda)=(45,12,3)$ or $(96,20,4)$ by [26, Theorem 1.1], whereas
$\lambda>10$ by our assumptions. Thus, $\left|B\cap\Delta^{\prime}\right|=2$.

If $\mathcal{D}$ is of different type with respect to $\Sigma$ and to
$\Sigma^{\prime}$, then $k=\lambda+2=\lambda^{2}/2$, which has no integer
solutions. Therefore $\mathcal{D}$ is of the same type with respect to
$\Sigma$ and to $\Sigma^{\prime}$, and hence $d^{\prime}=d$ and
$c^{\prime}=c$.

Let $\Delta\in\Sigma$ such that $\Delta\cap\Delta^{\prime}\neq\varnothing$. If
$\left|\Delta\cap\Delta^{\prime}\right|>1$ then $\Delta=\Delta^{\prime}$ since
$G$ induces a $2$-transitive group on $\Delta$ and
$\left|\Delta\right|=\left|\Delta^{\prime}\right|$. Therefore
$\Sigma=\Sigma^{\prime}$ which is (2). ∎

######  Proposition 2.5.

$G^{\Sigma}$ acts primitively on $\Sigma$. Moreover, and one of the following
holds:

  1. (1)

$G(\Sigma)\neq 1$ and $\mathrm{Soc}(G_{\Delta}^{\Delta})\trianglelefteq
G(\Sigma)^{\Delta}$;

  2. (2)

$G(\Sigma)=1$ and $G$ acts point-quasiprimitively on $\mathcal{D}$.

###### Proof.

It follows from Lemma 2.4 and [10, Theorem 1.5A] that $G^{\Sigma}$ acts
primitively on $\Sigma$.

Assume that $G(\Sigma)\neq 1$. If there is $\Delta^{\prime}\in\Sigma$ such
that $G(\Sigma)\leqslant G(\Delta^{\prime})$. Then $G(\Sigma)\leqslant
G(\Delta^{\prime\prime})$ for each $\Delta^{\prime\prime}\in\Sigma$, and hence
$G(\Sigma)=1$, since $G(\Sigma)\vartriangleleft G$ and $G$ acts transitively
on $\Sigma$. However, it contradicts the assumption $G(\Sigma)\neq 1$. Thus
$G(\Sigma)\nleq G(\Delta)$ for each $\Delta\in\Sigma$. Then $1\neq
G(\Sigma)^{\Delta}\trianglelefteq G_{\Delta}^{\Delta}$, and hence
$\mathrm{Soc}(G_{\Delta}^{\Delta})\trianglelefteq G(\Sigma)^{\Delta}$ by [10,
Theorem 4.3B] since $G_{\Delta}^{\Delta}$ is $2$-transitive on $\Delta$.

Assume that $G(\Sigma)=1$. Let $N$ be any normal subgroup of $G$. Then $N=G$
since $G(\Sigma)=1$ and since $G^{\Sigma}$ acts primitively on $\Sigma$.
Therefore, $G$ acts point-quasiprimitively on $\mathcal{D}$. ∎

######  Lemma 2.6.

Let $\Delta\in\Sigma$ and $\gamma\in G(\Delta)$, $\gamma\neq 1$. Then one of
the following holds:

  1. (1)

$\mathcal{D}$ is of type 1 and
$\left|\mathrm{Fix}(\gamma)\right|\leqslant(\lambda+2)\lambda$;

  2. (2)

$\mathcal{D}$ is of type 2 and
$\left|\mathrm{Fix}(\gamma)\right|\leqslant\lambda^{2}/2+\sqrt{\lambda^{2}/2-\lambda}$

###### Proof.

Let $\Delta\in\Sigma$ and $\gamma\in G(\Delta)$, $\gamma\neq 1$. It follows
from by [18, Theorem 3.1] that

$\left|\mathrm{Fix}(\gamma)\right|\leqslant\frac{\lambda}{k-\sqrt{k-\lambda}}\cdot\left|\Delta\right|\cdot\left|\Sigma\right|$
(2.2)

where
$\frac{\lambda}{k-\sqrt{k-\lambda}}\cdot\left|\Delta\right|\cdot\left|\Sigma\right|$
is either $(\lambda+1)\lambda$ or $\lambda^{2}/2+\sqrt{\lambda^{2}/2-\lambda}$
according to whether $\mathcal{D}$ is of type 1 or 2 respectively. ∎

######  Corollary 2.7.

The following hold:

  1. (1)

If $\mathcal{D}$ is of type 1, each prime divisor of $\left|G(\Delta)\right|$
divides $\lambda(\lambda-1)$.

  2. (2)

If $\mathcal{D}$ is of type 2, each prime divisor of $\left|G(\Delta)\right|$
divides $\lambda(\lambda-1)(\lambda-2)(\lambda-3)$.

###### Proof.

Let $\gamma$ be $w$-element of $G(\Delta)$, where $w$ is a prime such that
$w\nmid\lambda$. Since $\gamma$ fixes $\Delta$ pointwise, $\gamma$ fixes at
least $\mu$ of the $\lambda$ blocks of $\mathcal{D}$ incident with any two
distinct points of $\Delta$, where $\mu=\lambda\pmod{w}$. Clearly, these fixed
blocks are pairwise distinct.

If $\mathcal{D}$ is of type 1 then $\gamma$ fixes
$\mu\frac{(\lambda+2)(\lambda+1)}{2}$ blocks of $\mathcal{D}$, and hence
$\left|\mathrm{Fix}(\gamma)\right|\geqslant\mu\frac{(\lambda+2)(\lambda+1)}{2}$.
Therefore $\mu=1$ by Lemma 2.6(1) and the assertion (1) follows.

If $\mathcal{D}$ is of type 2 then $\gamma$ fixes
$\mu\frac{\lambda}{4}\left(\frac{\lambda}{2}+1\right)$ blocks of
$\mathcal{D}$, and hence
$\left|\mathrm{Fix}(\gamma)\right|\geqslant\mu\frac{\lambda}{4}\left(\frac{\lambda}{2}+1\right)$.
Therefore $\mu\leqslant 3$ by Lemma 2.6(2) and the assertion (2) follows. ∎

######  Lemma 2.8.

Let $x$ be any point of $\mathcal{D}$. Then $G(\Sigma)_{x}$ lies in a Sylow
$2$-subgroup of $G(\Sigma)$.

###### Proof.

Let $x$ be any point of $\mathcal{D}$ and let $\varphi$ be any $w$-element of
$G(\Sigma)_{x}$, where $w$ is an odd prime. Then $\varphi$ fixes at least a
blocks $B$ of $\mathcal{D}$ by [18, Theorem 3.1]. Since $B$ intersects each
element of $\Sigma$ in $0$ or $2$ points, and since $\varphi$ is a $w$-element
of $G(\Sigma)$, it follows that $\varphi$ fixes $B$ pointwise. Therefore,
$\varphi$ fixes at least $k$ points of $\mathcal{D}$. Then $\varphi$ fixes at
least $k$ blocks of $\mathcal{D}$ again by [18, Theorem 3.1]. Let $B^{\prime}$
be further block fixed by $\varphi$. We may repeat the previous argument with
$B^{\prime}$ in the role of $B$ thus obtaining $\varphi$ fixing $B^{\prime}$
pointwise. Then $\varphi$ fixes at least $2k-\lambda$ points of $\mathcal{D}$,
as $\left|B\cap B^{\prime}\right|=\lambda$. Thus
$\left|\mathrm{Fix}(\alpha)\right|\geqslant 2k-\lambda$ and hence
$\left|\mathrm{Fix}(\alpha)\right|$ is greater than or equal to
$\lambda(2\lambda+1)$ or $\lambda^{2}-\lambda$ according to whether
$\mathcal{D}$ is of type 1 or 2 respectively. However, this contradicts Lemma
2.6, as $\lambda>2$. ∎

######  Lemma 2.9.

If $G(\Sigma)\neq 1$ and $v$ is odd, then the following hold:

  1. (1)

$G(\Sigma)=Soc(G_{\Delta})$ is an elementary abelian $p$-group, $p$ an odd
prime, acting regularly on $\Delta$;

  2. (2)

$G^{\Sigma}\leqslant GL_{d}(p)$, with $p^{d}=\left|\Delta\right|$ and
$\Delta\in\Sigma$.

###### Proof.

Assume that $G(\Sigma)_{x}\neq 1$, where $x$ is any point of $\mathcal{D}$.
Then $G(\Sigma)_{x}$ is a Sylow $2$-subgroup of $G(\Sigma)$ by Lemma 2.8.
Denote $G(\Sigma)_{x}$ simply by $S$. Then $S$ fixes the same number $t$ of
points in each $\Delta\in\Sigma$, where $t\geqslant 1$ since $v$ is odd.
Therefore, $\left|\mathrm{Fix}(S)\right|=t\left|\Sigma\right|$. Now, if
$\alpha$ is any non-trivial element of $S$ then
$\left|\mathrm{Fix}(\alpha)\right|\geqslant t\left|\Sigma\right|$, and hence
$t=1$ by Lemma 2.6. Thus $S$ fixes a unique point on each $\Delta$.

Suppose that $\left|S\right|\geqslant 4$ and let $B$ be any block preserved by
$S$ and $\Delta^{\prime}$ be such that $\left|B\cap\Delta^{\prime}\right|=2$.
Then there is a subgroup $S_{0}$ of $S$ of index at most $2$ fixing $y$ and
$y^{\prime}$ where $\left\\{y,y^{\prime}\right\\}=B\cap\Delta^{\prime}$. Then
$S_{0}\leqslant G(\Sigma)_{y}\cap G(\Sigma)_{y^{\prime}}$, where
$G(\Sigma)_{y}$ and $G(\Sigma)_{y^{\prime}}$ are two distinct Sylow
$2$-subgroups of $G(\Sigma)$, each of these fixing a unique point in
$\Delta^{\prime}$. Suppose there is a point $z$ of $\mathcal{D}$ such that
$z\in\mathrm{Fix}(G(\Sigma)_{y})\cap\mathrm{Fix}(G(\Sigma)_{y^{\prime}})$. The
$\left\langle G(\Sigma)_{y},G(\Sigma)_{y^{\prime}}\right\rangle\leqslant
G(\Sigma)_{z}$ and hence $G(\Sigma)_{y}=G(\Sigma)_{y^{\prime}}=G(\Sigma)_{z}$
since $G(\Sigma)_{z}$ is a Sylow $2$-subgroup of $G(\Sigma)$ by Lemma 2.8.
Then $G(\Sigma)_{y}$ fixes also $y^{\prime}$ in $\Delta^{\prime}$ with
$y^{\prime}\neq y$, and we reach a contradiction. Thus
$\mathrm{Fix}(G(\Sigma)_{y})\cap\mathrm{Fix}(G(\Sigma)_{y^{\prime}})=\varnothing$,
and hence $\left|\mathrm{Fix}(S_{0})\right|\geqslant 2\left|\Sigma\right|$
since $S_{0}\leqslant G(\Sigma)_{y}\cap G(\Sigma)_{y^{\prime}}$. Now, if we
use the above argument this time with $\alpha\in S_{0}$, $\alpha\neq 1$, we
reach a contradiction. Therefore $\left|S\right|=2$. Also,
$G(\Sigma)=O(G(\Sigma)).S$ by Proposition 2.5(1), and
$\left|\mathrm{Fix}(S)\right|=\left|\Sigma\right|$.

Let $\Lambda=\left\\{Fix(S)^{\gamma}:\gamma\in G(\Sigma)\right\\}$. Since
$S=G(\Sigma)_{x}$ is a Sylow $2$-subgroup of $G(\Sigma)$,
$G(\Sigma)\vartriangleleft G$ and $G$ acts point-transitively on
$\mathcal{D}$, it follows that $\Lambda$ is a $G$-invariant partition of the
point set of $\mathcal{D}$ in $\left|\Delta\right|$ blocks each of size
$\left|\Sigma\right|$. Then $\left|\Sigma\right|=\left|\Delta\right|$ by Lemma
2.4(2) and hence $\lambda=2$, but this contradicts our assumptions.

Assume that $G(\Sigma)_{x}=1$. Then $G(\Sigma)=O(G(\Sigma))$. Moreover,
$Soc(G_{\Delta}^{\Delta})\trianglelefteq G(\Sigma)^{\Delta}\cong G(\Sigma)$ by
Proposition 2.5(1). Then $G(\Sigma)=Soc(G_{\Delta}^{\Delta})$ is an elementary
abelian $p$-group for some odd prime $p$, since $G(\Sigma)$ acts regularly on
$\Delta$ and $\left|\Delta\right|$ is odd. Hence $G(\Sigma)$ is abelian, and
so $G(\Sigma)\trianglelefteq C_{G}(G(\Sigma))\trianglelefteq G$. If
$C_{G}(G(\Sigma))\neq G(\Sigma)$, then $G=C_{G}(G(\Sigma))$ since $G^{\Sigma}$
is primitive on $\Sigma$ by Proposition 2.5. This implies $G(\Sigma)\leqslant
Z(G)$ and hence $G_{x}\leqslant G(\Delta)$ for any $x\in\Delta$. This is a
contradiction since $G_{\Delta}^{\Delta}$ is $2$-transitive on $\Delta$.
Therefore $C_{G}(G(\Sigma))=G(\Sigma)=Soc(G_{\Delta}^{\Delta})$. Therefore
$G^{\Sigma}\leqslant\mathrm{Aut}(G(\Sigma))\cong GL_{d}(p)$. ∎

##  3\. Further Reductions

The aim of this section is to prove the following reduction result:

######  Theorem 3.1.

One of the following holds:

  1. (1)

$G(\Sigma)=1$ and $G$ acts point-quasiprimitively on $\mathcal{D}$.

  2. (2)

$G(\Sigma)$ is a non-trivial self-centralizing elementary abelian $2$-subgroup
of $G$ and the following hold:

    1. (a)

$\mathcal{D}$ is a symmetric $2$-$(2^{d+2}(2^{d-1}-1)^{2},\allowbreak
2\left(2^{d}-1\right)\left(2^{d-1}-1\right),2(2^{d-1}-1))$ design with
$d\geqslant 4$;

    2. (b)

Either $G_{x}^{\Delta}\cong SL_{d}(2)$, or $G_{x}^{\Delta}\cong A_{7}$ and
$d=4$.

    3. (c)

$G^{\Sigma}$ is almost simple.

Its proof is structured as follows. Case (1) is an immediate consequence of
Proposition 2.5(2), whereas the proof of (2) relies mainly on Corollary 2.7
and on O’Nan Scott theorem applied to $G^{\Sigma}$ for $v$ even, and on
Corollary 2.7 and on [4, Theorem 3.1] for $v$ odd.

###  3.1. Designs of type 1 and quasiprimitivity

######  Lemma 3.2.

Let $\mathcal{D}$ be of type 1. If $G(\Sigma)\neq 1$ and $v$ is even, then the
following hold:

  1. (1)

$\mathcal{D}$ is a symmetric $2$-$(2^{d+2}(2^{d-1}-1)^{2},\allowbreak
2\left(2^{d}-1\right)\left(2^{d-1}-1\right),2(2^{d-1}-1))$ design, $d\geqslant
4$;

  2. (2)

$G(\Sigma)$ is a non-trivial self-centralizing elementary abelian $2$-subgroup
of $G$. Also $G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}\cong G_{x}^{\Delta}$ and
one of the following holds:

    1. (a)

$G_{x}^{\Delta}\cong SL_{d}(2)$;

    2. (b)

$G_{x}^{\Delta}\cong A_{7}$ and $d=4$.

###### Proof.

Assume that $\mathcal{D}$ is of type 1, $G(\Sigma)\neq 1$ and $v$ is even.
Since $Soc(G_{\Delta}^{\Delta})\trianglelefteq G(\Sigma)^{\Delta}$ by
Proposition 2.5(1), it follows that
$Soc(G_{\Delta}^{\Delta})_{x}\trianglelefteq G(\Sigma)_{x}^{\Delta}$, where
$x\in\Delta$. Hence, $Soc(G_{\Delta}^{\Delta})_{x}$ is either trivial or a
$2$-group by Lemma 2.8. If $G_{\Delta}^{\Delta}$ is almost simple, then
$Soc(G_{\Delta}^{\Delta})$ acts $2$-transitively on $\Delta$, with
$Soc(G_{\Delta}^{\Delta})_{x}$ a $2$-group. However, this is impossible by
[16, List (A)]. Therefore, $G_{\Delta}^{\Delta}$ is of affine type, and hence
$\left|\Delta\right|=2^{d}$ since $v=\lambda^{2}(\lambda+2)$ is even. Then
$\lambda=2(2^{d-1}-1)$ and so $\left|\Sigma\right|=2^{2}(2^{d-1}-1)^{2}$,
where $d\geqslant 4$ since $\lambda>10$. In particular, $\mathcal{D}$ is a
symmetric $2$-design with parameters as in (1). Moreover, by [16, List (B)]
one of the following holds:

  1. (i)

$G_{x}^{\Delta}\leqslant\Gamma L_{1}(2^{d})$;

  2. (ii)

$SL_{d/h}(2^{h})\trianglelefteq G_{x}^{\Delta}\leqslant\Gamma L_{d/h}(2^{h})$
with $d/h>1$;

  3. (iii)

$Sp_{d/h}(2^{h})\trianglelefteq G_{x}^{\Delta}\leqslant\Gamma Sp_{d/h}(2^{h})$
with $d/h>1$ and $d/h$ even,

  4. (iv)

$G_{2}(2^{d/6})\trianglelefteq G_{x}^{\Delta}\leqslant G_{2}(2^{d/6}):Z_{d/6}$
with $d\equiv 0\pmod{6}$.

  5. (v)

$G_{x}^{\Delta}\cong A_{6}$ or $A_{7}$ and $d=4$.

Since $G_{x}$ is transitive on the $\lambda(\lambda+1)$ blocks incident with
$x$, it follows that
$2\left(2^{d-1}-1\right)\left(2^{d}-1\right)\mid\left|G_{x}\right|$. Suppose
that there is a prime $u$ dividing $\lambda/2=2^{d-1}-1$ but not dividing of
the order of $G_{x}^{\Delta}$. Hence, $u$ divides the order of $G(\Delta)$.
Let $U$ be a Sylow $u$-subgroup of $G(\Delta)$, then
$G_{x}=N_{G_{x}}(U)G(\Delta)$ by the Frattini argument.

Let $y\in\mathrm{Fix(U)}\setminus\\{x\\}$ and let $\Delta^{\prime}$ be the
element of $\Sigma$ containing $y$. Then $U\leqslant G_{y}$, and actually
$U\leqslant G(\Delta^{\prime})$ since $G_{x}$ and $G_{y}$ are $G$-conjugate
(clearly, $\Delta=\Delta^{\prime}$ is possible). Therefore
$\left|\mathrm{Fix}(U)\right|=t(\lambda+2)$, where $t\geqslant 2$ since
$\left|\Sigma\right|=\lambda^{2}$. Also, $U$ is a Sylow $u$-subgroup of
$G(\Delta,\Delta^{\prime})$. Thus
$G_{x,y}=N_{G_{x,y}}(U)G(\Delta,\Delta^{\prime})$ again by Frattini argument
since $G(\Delta,\Delta^{\prime})\vartriangleleft G_{x,y}$. Then

$\left|G_{x}:G_{x,y}\right|=\frac{\left|N_{G_{x}}(U):N_{G_{x,y}}(U)\right|\cdot\left|G(\Delta):G(\Delta,\Delta^{\prime})\right|}{\left|N_{G(\Delta)}(U):N_{G(\Delta,\Delta^{\prime})}(U)\right|}$

and hence $\lambda+1\mid\left|N_{G_{x}}(U):N_{G_{x,y}}(U)\right|$, since
$\lambda+1\mid\left|G_{x}:G_{x,y}\right|$ by Lemma 2.1, and since
$\left(\lambda+1,\left|G(\Delta)\right|\right)=1$ by Corollary 2.7 being
$\lambda$ even, as $v=\lambda^{2}(\lambda+2)$ is even. Therefore
$\lambda+1\mid\left|y^{N_{G_{x}}(U)}\right|$ and
$y^{N_{G_{x}}(U)}\subseteq\mathrm{Fix}(U)\setminus\\{x\\}$. Since
$\mathrm{Fix}(U)\setminus\\{x\\}$ is a disjoint union of
$N_{G_{x}}(U)$-orbits, it follows that
$\lambda+1\mid\left|\mathrm{Fix}(U)\right|-1$. Therefore, $\lambda+1\mid t-1$
since $\left|\mathrm{Fix}(U)\right|=t(\lambda+2)$. Therefore,
$\left|\mathrm{Fix}(\zeta)\right|\geqslant\left|\mathrm{Fix}(U)\right|\geqslant(\lambda+2)^{2}$
for any non-trivial element $\zeta\in U$ since $t\geqslant 2$, but this
contradicts Lemma 2.6(1). Thus, each prime divisor of $2^{d-1}-1$ divides
$\left|G_{x}^{\Delta}\right|$. Then $G_{x}^{\Delta}\cong SL_{7}(2)$ for $d=7$,
whereas $2^{d-1}-1$ admits a primitive prime divisor for $d\neq 7$ by [22,
Theorem II.6.2]. A this point, it is easy to check that only
$G_{x}^{\Delta}\cong SL_{d}(2)$, or $G_{x}^{\Delta}\cong A_{7}$ for
$d\geqslant 4$ by using [17, Proposition 5.2.15].

The two possibilities for $G_{x}^{\Delta}$ together with Proposition 2.5(1)
imply either $G(\Sigma)^{\Delta}=G_{\Delta}^{\Delta}$ or
$G(\Sigma)^{\Delta}=Soc(G_{\Delta}^{\Delta})$. The former yields
$G_{\Delta}=G(\Sigma)G(\Delta)$, and hence
$\lambda+1\mid\left|G(\Sigma)_{x}\right|$ since
$\left(\lambda+1,\left|G(\Delta)\right|\right)=1$ by Corollary 2.7. However,
this contradicts Lemma 2.8 since $\lambda+1$ odd. Thus
$G(\Sigma)^{\Delta}=Soc(G_{\Delta}^{\Delta})$, and hence
$\frac{G_{\Delta}^{\Sigma}}{G(\Delta)^{\Sigma}}\cong\frac{G_{\Delta}^{\Delta}}{G(\Sigma)^{\Delta}}\cong
G_{x}^{\Delta}$ by (2.1). Since $\frac{G(\Sigma)}{G(\Sigma)\cap
G(\Delta)}\cong Soc(G_{\Delta}^{\Delta})$, which is an elementary abelian
$2$-group, we have $\Phi(G(\Sigma))\leqslant G(\Sigma)\cap G(\Delta)$ for each
$\Delta\in\Sigma$. Thus $\Phi(G(\Sigma))$ fixes each point of $\mathcal{D}$,
hence $\Phi(G(\Sigma))=1$, and so $G(\Sigma)$ is an elementary abelian
$2$-group.

If $C_{G}(G(\Sigma))\neq G(\Sigma)$, then $G=C_{G}(G(\Sigma))$ since
$G^{\Sigma}$ is primitive on $\Sigma$ by Proposition 2.5. Thus
$G(\Sigma)\leqslant Z(G)$ acts transitively on each $\Delta$, and hence
$G_{x}\leqslant G(\Delta)$ for any $x\in\Delta$. However, this is a
contradiction since $G_{\Delta}^{\Delta}$ is $2$-transitive on $\Delta$.
Therefore $C_{G}(G(\Sigma))=G(\Sigma)$, and we obtain (2). ∎

######  Theorem 3.3.

If $\mathcal{D}$ is of type 1 and $v$ is even, then Theorem 3.1 holds.

###### Proof.

Assume that that $\mathcal{D}$ is of type 1. The assertion follows from
Proposition 2.5(2) for $G(\Sigma)=1$. Hence, assume that $G(\Sigma)\neq 1$.
Since $G^{\Sigma}$ acts primitively on $\Sigma$ again by Proposition 2.5 and
$\left|\Sigma\right|=2^{2}(2^{d-1}-1)^{2}$ with $d\geqslant 4$ by Lemma 3.2,
by the O’Nan-Scott theorem (e.g. see [10, Theorem 4.1A]) one of the following
holds:

  1. (i)

$Soc(G^{\Sigma})$ is almost simple.

  2. (ii)

$Soc(G^{\Sigma})\cong T^{2}$, where $T$ is a non-abelian simple group such
that $\left|T\right|=2^{2}(2^{d-1}-1)^{2}$.

  3. (iii)

$Soc(G^{\Sigma})\cong T^{2}$ and there is a non-abelian almost simple group
$Q$ with socle $T$ acting primitively on a set $\Theta$ of size $2(2^{d-1}-1)$
such that $\Sigma=\Theta^{2}$ and $G^{\Sigma}\leqslant Q\wr Z_{2}$.

Assume that (ii) holds. Then $T\leqslant G_{\Delta}^{\Sigma}\leqslant
Aut(T)\times Z_{2}$, and hence $G_{\Delta}^{\Sigma}/T$ is solvable. Therefore
$\left|G_{x}^{\Delta}\right|\mid\left|T\right|$ since
$G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}\cong G_{x}^{\Delta}$ with
$G_{x}^{\Delta}$ isomorphic either to $SL_{d}(2)$, or to $A_{7}$ for $d=4$ by
Lemma 3.2(2). However, this is impossible since
$\left|G_{x}^{\Delta}\right|\nmid 2^{2}(2^{d-1}-1)^{2}$, and (ii) is excluded.

Assume that (iii) holds. Since $G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}\cong
G_{x}^{\Delta}$ by Lemma 3.2(2), then $\lambda+1=2^{d}-1$ divides the order of
$G_{\Delta}^{\Sigma}$. If $d=6$ then $\lambda=2\cdot 31$,
$\left|\Sigma\right|=2^{2}31^{2}$ and $G_{x}^{\Delta}\cong SL_{6}(2)$. Since
$\lambda(\lambda+1)\mid\left|G_{\Delta}\right|$, it follows that
$31^{3}\mid\left|G\right|$. On the other hand, $G^{\Sigma}\leqslant Q\wr
Z_{2}$, where $\mathrm{Soc}(Q)$ is isomorphic to one of the groups $A_{62}$ or
$PSL_{2}(61)$ by [10, Table B.2], and hence $31$ divides the order of
$G(\Sigma)$, which is not the case by Lemma 3.2(2). Thus $d\neq 6$, and hence
there is a primitive prime divisor of $\lambda+1=2^{d}-1$, say $z$, by [22,
Theorem II.6.2]. If $z\mid\left|G_{\Delta,\Delta_{1}}\right|$ for some
$\Delta_{1}\in\Sigma\setminus\\{\Delta\\}$, it follows that there is a
$u$-element $\varphi$ of $G_{\Delta,\Delta_{1}}$ fixing a point $x$ in
$\Delta$ and $x^{\prime}$ in $\Delta_{1}$. Therefore, it fixes at least one of
the $\lambda=2(2^{d-1}-1)$ blocks incident with $x$ and $x^{\prime}$, say $B$.
Hence, $\varphi$ fixes a further element in $B\cap\Delta$. Thus, $\varphi\in
G(\Delta)$ by [12, Theorem 3.5] since $G_{\Delta}^{\Delta}$ is of affine type.
However, this is impossible by Corollary 2.7(1) since $u\mid\lambda+1$. Then
$z\nmid\left|G_{\Delta,\Delta_{1}}\right|$ and hence
$z\mid\left|G_{\Delta}^{\Sigma}:G_{\Delta,\Delta^{\prime}}^{\Sigma}\right|$
for each $\Delta^{\prime}\in\Sigma\setminus\\{\Delta\\}$ since $G(\Sigma)$ is
a $2$-group by Lemma 3.2(2). Actually,
$z\mid\left|G_{\Delta}^{\Sigma}:G(\Delta)^{\Sigma}G_{\Delta,\Delta^{\prime}}^{\Sigma}\right|$
again by Corollary 2.7(1), hence
$z\mid\left|G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}:G(\Delta)^{\Sigma}G_{\Delta,\Delta^{\prime}}^{\Sigma}/G(\Delta)^{\Sigma}\right|$.
Thus

$P(G_{x}^{\Delta})\leqslant\left|G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}:G(\Delta)^{\Sigma}G_{\Delta,\Delta^{\prime}}^{\Sigma}/G(\Delta)^{\Sigma}\right|\leqslant\left|G_{\Delta}^{\Sigma}:G_{\Delta,\Delta^{\prime}}^{\Sigma}\right|\text{,}$

where $P(G_{x}^{\Delta})$ is the minimal primitive permutation representation
of $G_{x}^{\Delta}$, since $G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}\cong
G_{x}^{\Delta}$ by Lemma 3.2(2). We may choose
$\Delta^{\prime}\in\Sigma\setminus\\{\Delta\\}$ such that

$P(G_{x}^{\Delta})\leqslant\left|G_{\Delta}^{\Sigma}:G_{\Delta,\Delta^{\prime}}^{\Sigma}\right|\leqslant
2\frac{2(2^{d-1}-1)}{s-1}$

where $s$ denotes the rank of $Q$ on $\Theta$ since $G^{\Sigma}\leqslant Q\wr
S_{2}$ and $\Sigma=\Theta^{2}$. If $d\neq 4$ then $P(G_{x}^{\Delta})=2^{d}-1$
by [17, Theorem 5.2.2], hence $s=2$. Then $Q$ acts $2$-transitively on the set
$\Theta$, with $\left|\Theta\right|=2(2^{d-1}-1)$, and hence $Q\wr Z_{2}$ acts
as a primitive rank $3$ group on $\Sigma=\Theta^{2}$. Moreover, the
$\left(Q\wr Z_{2}\right)_{\Delta}$-orbits on
$\Sigma\setminus\left\\{\Delta\right\\}$ are two of length
$\left|\Theta\right|-1$, one of $\left(\left|\Theta\right|-1\right)^{2}$. Each
of these orbits is a union of $G_{\Delta}^{\Sigma}$-orbits, and since each
$G_{\Delta}^{\Sigma}$-orbits distinct from $\left\\{\Delta\right\\}$ is
divisible by $z$, then $z\mid\left|\Theta\right|-1$. So $z\mid 2^{d}-3$,
whereas $z$ is a divisor of $2^{d}-1$. Therefore $d=4$, $s=2,3,4$ and
$G_{x}^{\Delta}\cong A_{m}$, where $m=7,8$, since $SL_{4}(2)\cong A_{8}$.

Suppose that $s\geqslant 3$. Then
$\left|G_{\Delta}^{\Sigma}:G_{\Delta,\Delta^{\prime}}^{\Sigma}\right|\leqslant\frac{28}{s-1}\leqslant
14$, and hence
$\left|G_{\Delta}^{\Sigma}:G_{\Delta,\Delta^{\prime}}^{\Sigma}\right|=m$ by
[6]. Then $m\mid\left|\Sigma\right|-1$ with $m=7,8$, and we reach a
contradiction since $\left|\Sigma\right|-1=\allowbreak 3\cdot 5\cdot 13$. Thus
$s=2$, and hence $Q$ acts $2$-transitively on $\Theta$. As above, $Q\wr Z_{2}$
acts as a primitive rank $3$ group on $\Sigma=\Theta^{2}$, and the $\left(Q\wr
Z_{2}\right)_{\Delta}$-orbits on $\Sigma\setminus\\{\Delta\\}$ are two of
length $13$ and one of length $13^{2}$. None of these lengths is divisible by
$z=5$, and so this case is excluded. Thus only (i) occurs, which is the
assertion. ∎

######  Remark 3.4.

If $\mathcal{D}$ is of type 1 with $v$ even and $G(\Sigma)\neq 1$, it follows
from Lemma 3.2(2) and Theorem 3.3 that, $G^{\Sigma}$ is an almost simple
subgroup of $GL_{d+t}(2)$ , where $\left|\Delta\right|=2^{d}$ with
$\Delta\in\Sigma$, $\left|G(\Sigma)\right|=2^{d+t}$ and $\left|G(\Sigma)\cap
G(\Delta)\right|=2^{t}$. However, it is not easy to fully exploit the previous
property because it is not easy to control the order of $G(\Sigma)\cap
G(\Delta)$ in this case. This motivates our choice of an alternative proof in
which the embedding of $G^{\Sigma}$ in $GL_{d+t}(2)$ is partially considered.

######  Theorem 3.5.

If $\mathcal{D}$ is of type 1 and $v$ is odd, then $G$ acts point-
quasiprimitively on $\mathcal{D}$.

###### Proof.

Assume that $\mathcal{D}$ is of type 1. Recall that $G(\Sigma)$ is an
elementary abelian $p$-group acting regularly on $\Delta$, $p$ is and odd
prime and $G^{\Sigma}\leqslant GL_{d}(p)$ by Lemma 2.9. Thus
$\lambda+2=\left|\Delta\right|=p^{d}$, $d\geqslant 1$, and hence
$\lambda=p^{d}-2$. Therefore, $(p^{d}-1)(p^{d}-2)$ divides
$\left|G_{x}\right|$, and hence $\left|G\right|$, as
$k\mid\left|G_{x}\right|$. Also,
$(p^{d}-1)(p^{d}-2)^{3}\mid\left|G^{\Sigma}\right|$ since
$\left|\Sigma\right|=(p^{d}-2)^{2}$.

If $d=2$ then $(p^{2}-1)(p^{2}-2)^{3}\mid\left|G^{\Sigma}\right|$ with
$G^{\Sigma}\leqslant GL_{2}(p)$, which is a contradiction. Thus $d>2$, and
hence $\Phi_{d}^{\ast}(p)>1$ by [22, Theorem II.6.2] since $p$ is odd. Then
$G^{\Sigma}$ is an irreducible subgroup of $GL_{d}(p)$ by [12, Theorem
3.5(iv)]. For each divisor $m$ of $d$ the group $\Gamma L_{d/m}(p^{m})$ has a
natural irreducible action. We may choose $m$ to be minimal such that
$G^{\Sigma}\leqslant\Gamma L_{d/m}(p^{m})$. Set $K^{\Sigma}=G^{\Sigma}\cap
GL_{d/m}(p^{m})$. Then
$\Phi_{d}^{\ast}(p)\frac{(p^{d}-2)^{3}}{((p^{d}-2)^{3},d)}\mid\left|K^{\Sigma}\right|$
by [17, Proposition 5.2.15.(ii)]. Easy computations show that the order of
$GL_{d/m}(p^{m})$, and hence that of $K^{\Sigma}$, is not divisible by
$p^{d}\Phi_{d}^{\ast}(p)\frac{(p^{d}-2)}{((p^{d}-2),d)}$ for
$(d/m,p^{m})=(3,5^{2}),(4,3),(6,3),(6,5),(9,3)$. So these cases are excluded.
Thus, bearing in mind the minimality of $m$ and the fact that $p$ is odd, [4,
Theorem 3.1] implies that $K^{\Sigma}$ contains a normal subgroup isomorphic
to one of the groups
$SL_{d/m}(p^{m}),Sp_{d/m}(p^{m}),\Omega_{d/m}^{-}(p^{m})$, or
$SU_{d/m}(p^{m/2})$ with $d/m$ odd. Since
$\left|G^{\Sigma}:G_{\Delta}^{\Sigma}\right|=\lambda^{2}$, and
$\lambda=p^{d}-2$ with $p$ odd, it follows that $G_{\Delta}^{\Sigma}$ is a
maximal parabolic subgroup of $G^{\Sigma}$ by Proposition 2.5 and [33, Theorem
1.6]. Also, $\Phi_{d}^{\ast}(p)\mid\left|G_{\Delta}^{\Sigma}\right|$ since
$\left(\Phi_{d}^{\ast}(p),p^{d}-2\right)=1$, but this contradicts [12, Theorem
3.5(iv)] applied to $G_{\Delta}^{\Sigma}$. Thus $G(\Sigma)=1$, and hence $G$
acts point-quasiprimitively on $\mathcal{D}$ by Proposition 2.5(2). ∎

###  3.2. Designs of type 2 and quasiprimitivity

######  Lemma 3.6.

If $\mathcal{D}$ is of type 2 with $\lambda=2w^{2}$, where $w$ is odd,
$w\geqslant 3$, and $2(w^{2}-1)$ is a square, then $G$ acts point-
quasiprimitively on $\mathcal{D}$.

###### Proof.

Suppose that $\mathcal{D}$ is of type 2 with $\lambda=2w^{2}$, where $u$ is
odd, $u\geqslant 3$, and $2(w^{2}-1)$ is a square. Then
$\left|\Delta\right|=\lambda/2+1=w^{2}+1$ is even. If
$\mathrm{Soc}(G_{\Delta}^{\Delta})$ is an elementary abelian $2$-group, then
$w^{2}+1=2^{s}$ for some $s\geqslant 1$. However, it has no integer solutions
for [32, B1.1]. Thus $\mathrm{Soc}(G_{\Delta}^{\Delta})$ is non-abelian
simple, and hence $\mathrm{Soc}(G_{\Delta}^{\Delta})$ is isomorphic to one of
the groups $A_{w^{2}+1}$, $PSL_{d}(s)$ with $w^{2}+1=\frac{s^{d}-1}{s-1}$,
$d\geqslant 2$ and $(d,s)\neq(2,2),(2,3)$, or $PSU_{3}(w^{2/3})$ by [16, List
(A)] since $\left|\Delta\right|=w^{2}+1$. In the second case one has
$w^{2}=s\frac{s^{d-1}-1}{s-1}$, and so $s$ is an even power of an odd prime.
Therefore $\left(\frac{w}{s^{1/2}}\right)^{2}=\frac{s^{d-1}-1}{s-1}$, and
hence $d=2$ by [32, A7.1, A8.1 and B1.1], and so
$\mathrm{Soc}(G_{\Delta}^{\Delta})\cong PSL_{2}(w^{1/2})$. Also, $w^{2}-1\neq
2^{t}$ for some $t\geqslant 1$. Indeed, if not so, then $t=w=3$ by [32, B1.1],
and hence $(\lambda,\left|\Delta\right|,\left|\Sigma\right|)=(18,10,145)$ and
$\mathrm{Soc}(G_{\Delta}^{\Delta})\cong A_{5}$ and
$\mathrm{Soc}(G^{\Sigma})\cong A_{145}$ by [10, Table B.4]. Then
$A_{144}\trianglelefteq G_{\Delta}^{\Sigma}\leqslant S_{144}$ but this
contradicts (2.1) since $G_{\Delta}^{\Delta}\leqslant S_{5}$. Thus in each
case $\mathrm{Soc}(G_{\Delta}^{\Delta})$ contains elements of order an odd
prime divisor of $w^{2}-1$, and each of these elements fixes at least two
points on $\Delta$.

If $G(\Sigma)\neq 1$ then $\mathrm{Soc}(G_{\Delta}^{\Delta})\trianglelefteq
G(\Sigma)^{\Delta}$ by Proposition 2.5(1). Now, let $\eta\in G(\Sigma)$ be an
element of order an odd prime divisor of $w^{2}-1$, which exists by the
previous argument. Then, for each $\Delta\in\Sigma$ either $\eta\in G(\Delta)$
or $\eta$ induces an element of $G(\Sigma)^{\Delta}$ fixing at least two
points of $\Delta$. Therefore $\left|\mathrm{Fix}(\eta)\right|\geqslant
2\left|\Sigma\right|=\lambda^{2}-2\lambda+2$, but his contradicts Lemma
2.6(2). Thus $G(\Sigma)=1$, and hence $G$ acts point-quasiprimitively on
$\mathcal{D}$ by Proposition 2.5(2). ∎

######  Theorem 3.7.

If $\mathcal{D}$ is of type 2 then $G$ acts point-quasiprimitively on
$\mathcal{D}$.

###### Proof.

Suppose that $\mathcal{D}$ is of type 2. If $\lambda=2w^{2}$, where $w$ is
odd, $w\geqslant 3$, and $2(w^{2}-1)$ is a square, the assertion follows from
Lemma 3.6. Hence, we may assume that $\lambda\equiv 0\pmod{4}$ by Theorem 2.2.
Hence, $v$ is odd.

Suppose that $G(\Sigma)\neq 1$. Then $G(\Sigma)\cong Soc(G_{\Delta})$, where
$\Delta\in\Sigma$, is an elementary abelian $p$-group of order acting
regularly on $\Delta$, where $p$ is odd, and $G^{\Sigma}\leqslant GL_{d}(p)$,
with $p^{d}=\left|\Delta\right|$ by Lemma 2.9 since $v$ is odd. Then
$\lambda=2(p^{d}-1)$ and $k=\lambda^{2}/2=2(p^{d}-1)^{2}$. Then
$\left(p^{d}-1\right)^{2}\mid\left|G_{B}\right|$, where $B$ is any block of
$\mathcal{D}$, since $G_{B}$ acts transitively on $B$. Then
$\left(p^{d}-1\right)^{2}$ divides $\left|G^{\Sigma}\right|$ since
$\left|G(\Sigma)\right|=p^{d}$. Now, we may proceed as in Theorem 3.5 and we
see that no admissible groups arise in this case as well. Thus $G(\Sigma)=1$,
and hence $G$ acts point-quasiprimitively on $\mathcal{D}$ by Proposition
2.5(2). ∎

###### Proof of Theorem 3.1.

The assertion follows from Theorems 3.3 and 3.5 for $\mathcal{D}$ of type 1,
Theorem 3.7 for $\mathcal{D}$ of type 2. ∎

##  4\. Reduction to the almost simple case

In this section we analyze the case where $G$ acts point-quasiprimitively on
$\mathcal{D}$. In this case $G=G^{\Sigma}$. The main tool used to tackle this
case is the O’Nan-Scott Theorem for quasiprimitive groups proven in [29] and
reported below for reader’s convenience. We investigate each of the seven
possibilities for $G^{\Sigma}$ provided in the above mentioned theorem by
adapting the techniques developed in [41], and we show that
$\mathrm{Soc}(G^{\Sigma})$ is a non-abelian simple group. This fact, together
with the conclusions of Theorem 3.1, yields the following result.

######  Theorem 4.1.

$G^{\Sigma}$ is an almost simple group acting primitively on $\Sigma$.
Moreover, one of the following holds:

  1. (1)

$G(\Sigma)=1$ and $G$ acts point-quasiprimitively on $\mathcal{D}$.

  2. (2)

$G(\Sigma)$ is a self-centralizing elementary abelian $2$-subgroup of $G$.
Also, the following hold:

    1. (a)

$\mathcal{D}$ is a symmetric $2$-$(2^{d+2}(2^{d-1}-1)^{2},\allowbreak
2\left(2^{d}-1\right)\left(2^{d-1}-1\right),2(2^{d-1}-1))$ design with
$d\geqslant 4$;

    2. (b)

$G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}\cong G_{x}^{\Delta}$ and either
$G_{x}^{\Delta}\cong SL_{d}(2)$, or $G_{x}^{\Delta}\cong A_{7}$ and $d=4$.

We only need to prove that $G^{\Sigma}$ is almost simple, the remaining parts
of Theorem 4.1 have been already proven in Theorem 3.1

In the sequel we denote $\mathrm{Soc}(G)$ simply by $L$ and let
$x\in\mathcal{P}$. Then $L\cong T^{h}$ with $h\geqslant 1$, where $T$ is a
simple group. By [29, Theorem 1], one of the following holds:

  1. I.

_Affine groups._ Here $T\cong Z_{p}$ for some prime $p$, and $L$ is the unique
minimal normal subgroup of $G$ and is regular on $\mathcal{P}$ of size
$p^{h}$. The set $\mathcal{P}$ can be identified with $L\cong Z_{p}^{h}$ so
that $G\leqslant AGL_{h}(p)$ with $L$ the translation group and $G_{x}=G\cap
GL_{h}(p)$ acting irreducibly on $L$. Moreover, $G$ acts primitively on
$\mathcal{P}$.

  2. II.

_Almost simple groups._ Here $h=1$, $T$ is a non-abelian simple group,
$T\trianglelefteq G\leqslant\mathrm{Aut}(T)$ and $G=TG_{x}$.__

  3. III.

In this case $L\cong T^{h}$ with $h\geqslant 2$ and $T$ is a non-abelian
simple group. We distinguish three types:

  4. III(a).

_Simple diagonal action._ Define

$W=\left\\{(a_{1},...,a_{h})\cdot\pi:a_{i}\in\mathrm{Aut}(T),\pi\in
S_{h},a_{i}\equiv a_{j}\pmod{\mathrm{Inn}(T)}\text{ for all
}i,j\right\\}\text{,}$

where $\pi\in S_{h}$ just permutes the components $a_{i}$ naturally. With the
usual multiplication, $W$ is a group with socle $L\cong T^{h}$, and
$W=L.(\mathrm{Out}(T)\times S_{h})$. The action of $W$ on $\mathcal{P}$ is
defined by setting

$W_{x}=\left\\{(a,...,a)\cdot\pi:a\in\mathrm{Aut}(T),\pi\in
S_{h}\right\\}\text{.}$

Thus $W_{x}\cong\mathrm{Aut}(T)\times S_{h}$, $L_{x}\cong T$ and
$\left|\mathcal{P}\right|=\left|T\right|^{h-1}$.

For $1\leqslant i\leqslant h$ let $T_{i}$ be the subgroup of $L$ consisting of
the $h$-tuples with $1$ in all but the $i$-th component, so that $T_{i}\cong
T$ and $L\cong T_{1}\times\cdots\times T_{h}$. Put
$\mathcal{T}=\left\\{T_{1},...,T_{h}\right\\}$, so that $W$ acts on
$\mathcal{T}$. We say that subgroup $G$ of $W$ is of type III(a) if
$L\leqslant G$ and, letting $P$ the permutation group of $G^{\mathcal{T}}$,
one of the following holds:

    1. (i)

$P$ is transitive on $\mathcal{T}$;

    2. (ii)

$h=2$ and $P=1$.

We have $G_{x}\leqslant\mathrm{Aut}(T)\times P$ and $G\leqslant
L.(\mathrm{Out}(T)\times P)$. Moreover, in case (i) $L$ is the unique minimal
normal subgroup of $G$ and $G$ is primitive on $\mathcal{P}$ if and only if
$P$ is primitive on $\mathcal{T}$. In case (ii) $G$ has two minimal normal
subgroups $T_{1}$ and $T_{2}$, both regular on $\mathcal{P}$, and $G$ is
primitive on $\mathcal{P}$.

  5. III(b).

_Product action._ Let $H$ be a quasiprimitive permutation group on a set
$\Gamma$ of type II or III(a). For $l>1$, let $W=H\wr S_{l}$, and take $W$ to
act $\Lambda=\Gamma^{l}$ in its natural product action. Then for $y\in\Gamma$
and $z=(y,...,y)\in\Lambda$ we have $W_{z}=H_{y}\wr S_{l}$ and
$\left|\Lambda\right|=\left|\Gamma\right|^{l}$. If $K$ is the socle $H$, then
the socle $L$ of $W$ is $K^{l}$.

Now $W$ acts naturally on the $l$ factors in $K^{l}$, and we say that a
subgroup $G$ of $W$ is of type III(b) if $L\leqslant G$, $G$ acts transitively
on these $l$ factors, and one of the following holds:

    1. (i)

$H$ is of type II, $K=T$, $h=l$, and $L$ is the unique minimal normal subgroup
$G$; further $\Lambda$ is a $G$ invariant partition of $\mathcal{P}$ and, for
$x$ in the part $z\in\Lambda$, $L_{z}=T_{y}^{h}<L$ an for some non-trivial
normal subgroup $R$ of $T_{y}$, $L_{x}$ is a subdirect product of $R^{k}$,
that is $L_{x}$ projects surjectively on each of the direct factors $R$.

    2. (ii)

$H$ is of type III(a), $\mathcal{P}=\Lambda$, $K\cong T^{h/l}$ and both $G$
and $H$ have $m$ minimal normal subgroups where $m\leqslant 2$ . If $m=2$ then
each of the two minimal normal subgroups of $G$ is regular on $\mathcal{P}$.

  6. III(c).

_Twisted wreath action._ Here $G$ is a twisted wreath action $T\wr_{\phi}P$
defined as follows. Let $P$ have a transitive action on
$\left\\{1,...,h\right\\}$ and let $Q$ be the stabilizer $P_{1}$ of the point
$1$ in this action. We suppose that there is an homomorphism
$\phi:Q\rightarrow\mathrm{Aut}(T)$ such that
$\mathrm{core}_{P}\left(\phi^{-1}(\mathrm{Inn}(T))\right)=\cap_{x\in
P}\phi^{-1}(\mathrm{Inn}(T))^{x}=\left\\{1\right\\}$. Define

$L=\left\\{f:P\rightarrow T:f(\alpha\beta)=f(\alpha)^{\phi(\beta)}\text{ for
all }\alpha\in P\text{, }\beta\in Q\right\\}\text{.}$

Then $B$ is a group under the pointwise multiplication, and $L\cong T^{l}$.
Let $P$ act on $L$ by

$f(\gamma)=f(\alpha\gamma)\text{ for }\alpha,\gamma\in P\text{.}$

Define $T\wr_{\phi}P$ to be the semidirect product of $L$ by $P$ with this
action, and define the action on $\mathcal{P}$ by setting $G_{x}=P$. Then
$\left|\mathcal{P}\right|=\left|T\right|^{h}$, and $L$ is the unique minimal
normal subgroup of $G$ and acts regularly on $\mathcal{P}$. We say that $G$ is
of type III(c).

######  Theorem 4.2.

If $\mathcal{D}$ is of type 1 then $G$ is almost simple.

###### Proof.

Suppose that $\mathcal{D}$ is of type 1. Case (I) is ruled out since $G$ acts
imprimitively on $\mathcal{P}$ by our assumptions. Suppose that $G$ is of type
III(a) or III(c). Then
$(\lambda+2)\lambda^{2}=\left|\mathcal{P}\right|=\left|T\right|^{j}$ where $T$
is non-abelian simple and $j=h-1$ or $h$, with $h\geqslant 2$, respectively.
Hence, $\lambda$ is even.

If $j=1$, then $h=j+1=2$ and hence $G$ acts primitively on $\mathcal{P}$ (see
III(a)), which is a contradiction. Thus $j>1$. Note that
$\left|x^{T_{i}}\right|=\left|T\right|$ since $T_{i}$ acts semiregularly on
$\mathcal{P}$, where $T_{i}$ is the subgroup of $L$ consisting of the
$h$-tuples with $1$ in all but the $i$-th component, $T_{i}\cong T$. Moreover,
$x^{T_{i_{1}}}\cap x^{T_{i_{2}}}=\left\\{x\right\\}$ for each $i_{1},i_{2}$
such that $i_{1}\neq i_{2}$. Let $w\in x^{T_{1}}\setminus\left\\{x\right\\}$,
then $w^{G_{x}}$ is the disjoint union of $w^{G_{x}}\cap x^{T_{i}}$ for
$1\leqslant i\leqslant h$. Therefore
$\left|w^{G_{x}}\right|=\left|w^{G_{x}}\cap x^{T_{i}}\right|h$ since $G_{x}$
permutes transitively $T_{1},...,T_{h}$. On the other hand
$\lambda+1\mid\left|w^{G_{x}}\right|$ by Lemma 2.1, hence
$\lambda+1\leqslant\left|w^{G_{x}}\cap
x^{T_{i}}\right|h\leqslant\left|T\right|h$. Thus
$(\lambda+2)\lambda^{2}=\left|T\right|^{j}$ implies

$\left|T\right|^{j}\leqslant(\left|T\right|h-1)^{2}(\left|T\right|h+1)\leqslant\left|T\right|^{3}h^{3},$

and hence $j=2$ or $3$, or $j=4$ and $T\cong A_{5}$, as $j>1$ and
$\left|T\right|\geqslant 60$. The latter is ruled out since it does not
provide integer solutions for $(\lambda+2)\lambda^{2}=\left|T\right|^{4}$. If
$j=3$ then $\left|T\right|>\lambda>\left|T\right|-2$ since
$(\lambda+2)\lambda^{2}=\left|T\right|^{3}$, and hence
$\lambda=\left|T\right|-1$, but this contradicts $\lambda$ even. Thus, $j=2$.
If $h=3$ then $G$ acts primitively on $\mathcal{P}$. Indeed, $G$ is as in
III(a) and $Z_{3}\leqslant P\leqslant S_{3}$. Thus $j=h=2$ and $G$ is as in
III(c). Moreover, $\lambda+1\mid\left|w^{G_{x}}\cap x^{T_{1}}\right|$ for each
$w\in x^{T_{1}}\setminus\left\\{x\right\\}$ since $\lambda$ is even. Then
there is $\theta\geqslant 1$ such that $\left|T\right|=\theta(\lambda+1)+1$
since $\left|x^{T_{1}}\right|=\left|T\right|$. Then
$\lambda^{2}(\lambda+2)=\left|T\right|^{2}$ implies

$\left(\lambda+1\right)\left(\left(\lambda+1\right)\lambda-1\right)=\theta^{2}(\lambda+1)^{2}+2\theta(\lambda+1)$
(4.1)

and hence $\lambda(\lambda+1)-1=\theta^{2}(\lambda+1)+2\theta$. Then

$\theta=\frac{(\lambda+1)t-1}{2}$

for some $t\geqslant 1$. If $t\geqslant 2$ then $\theta>\lambda$ and we reach
a contradiction, thus $t=1$ and hence $\theta=\lambda/2$ which substituted in
(4.1) yields a contradiction too. Therefore, $G$ is not of type III(a) or
III(c).

Suppose that $G$ is of type III(b.ii). Then $G_{z}\leqslant W_{z}=H_{y}\wr
S_{l}$ where $z=(y,y,...,y)$, and denoted by $\mu$ the number of
$W_{z}$-orbits on $\Delta$, we see that $\mu\geqslant 4$ by [9, Corollary
1.9].

Let $y_{1}\in\Delta$, $y_{1}\neq y$ such that
$\left|y_{1}^{H}\right|\leqslant\frac{\left|\Delta\right|-1}{\mu-1}$, and let
$z_{1}=(y_{1},y,\dots,y)$. Since $\left(H_{y,y_{1}}\times
H_{y}\times\cdots\times H_{y}\right):S_{l-1}\leqslant W_{z,z_{1}}$, it results

$\left|z_{1}^{G_{z}}\right|\leqslant\left|z_{1}^{W_{z}}\right|\leqslant\frac{\left|H_{y}\right|^{l}\left(l!\right)}{\left|H_{y,y_{1}}\right|\left|H_{y}\right|^{l-1}\left((l-1)!\right)}=l\left|y_{1}^{H}\right|\leqslant\frac{l\left(\left|\Delta\right|-1\right)}{\mu-1}\text{,}$

and since $\lambda+1\mid\left|z_{1}^{G_{z}}\right|$ by Lemma 2.1, it follows
that

$\lambda+1\leqslant\frac{l\left(\left|\Delta\right|-1\right)}{\mu-1}\text{.}$

Then $\lambda$ is even since
$(\lambda+2)\lambda^{2}=\left|\mathcal{P}\right|=\left(\left|T\right|^{h/l-1}\right)^{l}=\left|T\right|^{h-l}$,
$l>1$. Thus

$\left|T\right|^{l(h/l-1)/3}-1\leqslant\frac{l\left(\left|T\right|^{h/l-1}-1\right)}{\mu-1}<\frac{h\left(\left|T\right|^{h/l-1}-1\right)}{3}\text{.}$

and hence $l=2,3$, as $l>1$. If $l=3$ then
$\left|T\right|^{h/3-1}>\lambda>\left|T\right|^{h/3-1}-2$ and hence
$\lambda=\left|T\right|^{h/3-1}-1$ with $h>l$, whereas $\lambda$ is even. Thus
$l=2$, and since $\lambda$ is even and
$\left(\Delta\setminus\\{y\\}\times\\{y\\}\right)\cup\left(\\{y\\}\times\Delta\setminus\\{y\\}\right)$
is union of some non-trivial $G_{z}$-orbits, being $G_{z}\leqslant H_{z}\wr
S_{l}$, it follows from Lemma 2.1 that
$\lambda+1\mid\left|T\right|^{h/2-1}-1$. Now, we may apply the final argument
used to rule out III(a) with $\left|T\right|^{h/2-1}$ in the role of
$\left|T\right|$ to exclude this case as well.

Finally, assume that $G$ is of type III(b.i). Then $h=l$ and $T^{h}$ is the
unique minimal normal subgroup of $G$. In this case $\Sigma$ can be identified
with the Cartesian product $\Gamma^{h}$, hence each $\Delta\in\Sigma$
corresponds to a unique $h$-tuple of elements of $\Gamma$. Therefore, it
results that $\left|\Sigma\right|=\lambda^{2}=\left|\Gamma\right|^{h}$. Let
$y\in\Gamma$ and $\Delta=(y,...,y)\in\Sigma$. Then

$\bigcup_{i=1}^{h}\left(\\{y\\}\times\cdots\times\Gamma_{i}\setminus\\{y\\}\times\cdots\times\\{y\\}\right)$

is a union of some non-trivial $G_{\Delta}$-orbits and ultimately of some non-
trivial $G_{x}$-orbits, where $x\in\Delta$. Then
$\lambda+1\mid(\lambda+2)h\left(\left|\Gamma\right|-1\right)$ by Lemma 2.1.
Thus $\lambda+1\mid(\lambda+2)h\left(\lambda^{2/h}-1\right)$, and hence
$\lambda+1\mid h\left(\lambda^{2/h}-1\right)$.

If $h=2$ then $\lambda+1\mid 4$, whereas $\lambda>10$. Thus $h\geqslant 3$ and
hence $\lambda^{1/3}\leqslant\lambda^{1-2/h}\leqslant h$. Therefore
$5^{h/3}\leqslant\left|T:T_{\Delta}\right|^{h/3}\leqslant h^{2}$ since $T$ is
a non-abelian simple group and $\left|T:T_{\Delta}\right|=\lambda^{2}$, and so
$h\leqslant 7$. Actually, $\left|T:T_{\Delta}\right|^{h/2(1-2/h)}\leqslant h$
implies $h=3$, and hence $\lambda+1\mid 3\left(\lambda^{2/3}-1\right)$ which
is impossible since $\lambda>10$. This completes the proof. ∎

######  Theorem 4.3.

If $\mathcal{D}$ is of type (2) then $G$ is almost simple.

###### Proof.

If $\mathcal{D}$ is of type (2) then $v\not\equiv 0\pmod{4}$ (see Theorem
2.2). Case (I) is ruled out since $G$ acts imprimitively on $\mathcal{P}$ by
our assumptions. Also $v\neq$ $\left|T\right|^{j}$ with $j\geqslant 1$, where
$T$ is non-abelian simple, and hence $G$ is not of type III(a) or III(c).
Assume that $G$ is of type III(b.i). Arguing as in Theorem 4.2 we see that

$\frac{\lambda^{2}-2\lambda+2}{2}=\left|\Sigma\right|=\left|\Gamma\right|^{h}\text{
and }\lambda+1\mid(\lambda/2+1)h\left(\left|\Gamma\right|-1\right)$

where $h\geqslant 2$. It follows that $\lambda+1\mid
h\left(\left|\Gamma\right|-1\right)$. Since
$\left|\Gamma\right|^{l}-1=\allowbreak\frac{1}{2}\lambda\left(\lambda-2\right)$,
it follows $\lambda+1\mid 3h$ and hence $2\cdot 60^{h}\leqslant
2\left|\Gamma\right|^{h}<9h^{2}$, which is impossible for $h\geqslant 2$.

Assume that $G$ is of type III(b.ii). Then

$\left(\frac{\lambda+2}{2}\right)\left(\frac{\lambda^{2}-2\lambda+2}{2}\right)=\left|\mathcal{P}\right|=\left|\Gamma\right|^{l}\text{,}$
(4.2)

where $\left|\Gamma\right|=\left|T\right|^{k/l-1}$ with $k/l\geqslant 2$, and
from (4.2) we derive that $v\equiv 0\pmod{4}$, a contradiction. This completes
the proof. ∎

###### Proof of Theorem 4.1.

The assertion immediately follows from Theorems 3.1, 4.2 and 4.3. ∎

##  5\. Reduction to the case $\lambda\leqslant 10$

Let $L$ be the preimage in $G$ of $\mathrm{Soc}(G^{\Sigma})$. Hence
$L^{\Sigma}\trianglelefteq G^{\Sigma}\leqslant\mathrm{Aut}(L^{\Sigma})$ with
$L^{\Sigma}$ non-abelian simple by Theorem 4.1. Moreover, $G=G^{\Sigma}$ and
$L=L^{\Sigma}$ when $G(\Sigma)=1$. The first part of this section is devoted
to prove that either $L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$,
where $\Delta\in\Sigma$, or $G(\Sigma)=1$,
$Soc(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$, where
$u^{h}=\left|\Delta\right|$. Then we prove that either
$\left|L^{\Sigma}\right|\leqslant\left|L_{\Delta}^{\Sigma}\right|^{2}$ or
$\left|L\right|\leqslant
4\left|L_{\Delta}^{\Delta}\right|^{2}\left|\mathrm{Out}(L)\right|^{2}$
respectively. As we will see, these constraints on $L_{\Delta}^{\Sigma}$ are
combined with [3] and [20] allows us to to completely classify $\mathcal{D}$.
Finally, the analysis of $2$-designs $\mathcal{D}$ of type 1 and 2 is carried
out in separate subsections.

In the sequel the minimal primitive permutation representation of a non-
abelian simple group $\Gamma$ will be denoted by $P(\Gamma)$. It is known that
$P(A_{\ell})=\ell\geqslant 5$, whereas $P(\Gamma)$ is determined in [6], [17,
Theorem 5.2.2] and in [35, 36, 37] according to whether $\Gamma$ is sporadic,
classical or exceptional of Lie type respectively. The following technical
lemma is useful to show that $L_{\Delta}^{\Delta}$ acts $2$-transitively on
$\Delta$ provided that $G_{\Delta}^{\Delta}$ is not a semilinear
$1$-dimensional group.

######  Lemma 5.1.

If $\Gamma$ is a non-abelian simple group non-isomorphic to $PSL_{2}(q)$ and
such that
$P(\Gamma)<2\left(\left|\mathrm{Out}(\Gamma)\right|+1\right)\left|\mathrm{Out}(\Gamma)\right|$,
then one of the following holds:

  1. (1)

$\Gamma\cong A_{\ell}$ with $\ell=7,8,9,10,11$;

  2. (2)

$\Gamma\cong M_{11}$;

  3. (3)

$\Gamma\cong PSL_{3}(q)$ with $q=4,7,8,16$;

  4. (4)

$\Gamma\cong PSp_{4}(3)$;

  5. (5)

$\Gamma\cong PSU_{n}(q)$ with $(n,q)=(3,5),(3,8),(4,3)$;

  6. (6)

$\Gamma\cong P\Omega_{8}^{+}(3)$.

###### Proof.

Assume that $\Gamma\cong A_{\ell}$ with $\ell\geqslant 5$. Then $\ell\neq 5,6$
since $A_{5}\cong PSL_{2}(5)$ and $A_{6}\cong PSL_{2}(9)$. Then
$7\leqslant\ell=P(\Gamma)<12$ and we obtain (1).

If $\Gamma$ is sporadic then $P(\Gamma)<12$, and hence $\Gamma\cong M_{11}$ by
[6], which is (2).

If $\Gamma$ is a simple exceptional group of Lie type then $P(\Gamma)$ is
provided in [35, 36, 37], and it is easy to check that no cases arise.

Finally, suppose that $\Gamma$ is a simple classical group. Assume that
$\Gamma\cong PSL_{n}(q)$, where $q=p^{f}$, $f\geqslant 1$, and $n\geqslant 3$.
We may also assume that $(n,q)\neq(4,8)$ since $PSL_{4}(2)\cong A_{8}$ was
analyzed above. Then $P(\Gamma)=\frac{q^{n}-1}{q-1}$ by [17, Theorem 5.2.2]
since $n\geqslant 3$, and hence

$\frac{q^{n}-1}{q-1}<2(2(n,q-1)f+1)\cdot 2(n,q-1)f\text{.}$ (5.1)

Assume that $p^{f}\geqslant 2f(f+1)$. Then $\frac{q^{n}-1}{q-1}<4q(q-1)^{2}$
and hence $n\leqslant 4$. Actually, $(n,q)=(3,7)$ by (5.1) since $n\geqslant
3$.

Assume that $p^{f}<2f(f+1)$. Then either $p=2$ and $2\leqslant f\leqslant 6$,
or $p=3$ and $f=1,2$. Hence, $n=3$ and $q=4,8,16$ by (5.1) since $n\geqslant
3$ and since $PSL_{3}(2)\cong PSL_{2}(7)$ cannot occur. Thus, we get (3).

We analyze the remaining classical groups by proceeding as in the
$PSL_{n}(q)$-case. It is a routine exercise to show that one obtains
$PSp_{4}(2)^{\prime}$, $PSp_{4}(3)$, $PSU_{n}(q)$ with
$(n,q)=(3,5),(3,8),(4,2),(4,3)$, and $P\Omega_{8}^{+}(3)$. Since
$PSp_{4}(2)^{\prime}\cong PSL_{2}(9)$ this case is excluded, whereas the
remaining ones yield (4), (5) and (6) respectively, bearing in mind that
$PSU_{4}(2)\cong PSp_{4}(3)$. ∎

######  Lemma 5.2.

Let $x\in\Delta$, if $G(\Sigma)=1$ and
$\mathrm{Soc}(G^{\Delta}_{\Delta})\cong(Z_{u})^{h}$, where $u$ is an odd
prime, then $u^{h}\neq 3^{2},5^{2},7^{2},11^{2}$,
$19^{2},23^{2},29^{2},59^{2}$ and $3^{4}$

###### Proof.

Suppose the $u^{h}=3^{2},5^{2},7^{2},11^{2},19^{2},23^{2},29^{2},59^{2}$ or
$3^{4}$. Then $\lambda=u^{h}-2$ or $2(u^{h}-1)$ according to whether
$\mathcal{D}$ is of type $i=1$ or $2$ respectively. Now, Table 1 contains the
admissible pairs $(\lambda,\left|\Sigma\right|)_{i}$, where $i=1,2$.

Table 1. Admissible $(\lambda,\left|\Sigma\right|)_{i}$ for $\mathcal{D}$ of type $i=1,2$ | $3^{2}$ | $5^{2}$ | $7^{2}$ | $11^{2}$ | $19^{2}$
---|---|---|---|---|---
$(\lambda,\left|\Sigma\right|)_{1}$ | $(7,7^{2})$ | $(23,23^{2})$ | $(47,47^{2})$ | $(7\cdot 17,7^{2}\cdot 17^{2})$ | $(17,17^{2})$
$(\lambda,\left|\Sigma\right|)_{2}$ | $(16,113)$ | $(48,1105)$ | $(96,4513)$ | $(7\cdot 103,53\cdot 4877)$ | $(240,13^{4})$
| $23^{2}$ | $29^{2}$ | $59^{2}$ | $3^{4}$ |
$(\lambda,\left|\Sigma\right|)_{1}$ | $(17\cdot 31,17^{2}\cdot 31^{2})$ | $(3^{3},3^{6})$ | $(3\cdot 19,3^{2}\cdot 19^{2})$ | $(79,79^{2})$ |
$(\lambda,\left|\Sigma\right|)_{2}$ | $(1056,556513)$ | $(41^{2},17\cdot 82913)$ | $(2^{4}\cdot 3\cdot 5\cdot 29,101\cdot 149\cdot 1609)$ | $(160,12641)$ |

From Table 1 we deduce that $\left|\Sigma\right|$ is either a power of a prime
or a squarefree number. It is straightforward to check that no admissible
cases occur by [11, Theorem 1] and [15, Theorem 1] respectively since $L$ is
almost simple by Theorem 4.1, being $G(\Sigma)=1$, and since
$G^{\Delta}_{\Delta}$ must act $2$-transitively on $\Delta$, where
$\Delta\in\Sigma$. ∎

######  Proposition 5.3.

Let $\Delta\in\Sigma$, then $Soc(G_{\Delta}^{\Delta})\trianglelefteq
L_{\Delta}^{\Delta}$ and $G=G_{x}L$, where $x$ is any point of $\mathcal{D}$.
Moreover, one of the following holds:

  1. (1)

$G(\Sigma)=1$, $\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$, where
$u^{h}=\left|\Delta\right|$, and
$\left|G_{\Delta}^{\Delta}:L_{\Delta}^{\Delta}\right|\mid\left|\mathrm{Out}(L)\right|$.

  2. (2)

$L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$.

###### Proof.

Since $L_{\Delta}\trianglelefteq G_{\Delta}$ and $G_{\Delta}^{\Delta}$ acts
primitively on $\Delta$, either $L_{\Delta}=L(\Delta)$ or
$Soc(G_{\Delta}^{\Delta})\trianglelefteq L_{\Delta}^{\Delta}$ by [10, Theorem
4.3B]. Moreover, $G=G_{\Delta}L$ since $L^{\Sigma}=\mathrm{Soc}(G^{\Sigma})$
and $G^{\Sigma}$ acts primitively on $\Sigma$ by Theorem 4.1, and hence
$G/L=G_{\Delta}L/L\cong G_{\Delta}/L_{\Delta}$. Also, it results that
$G/G(\Sigma)L\cong G^{\Sigma}/L^{\Sigma}\leqslant\mathrm{Out}(L^{\Sigma})$.

Assume that $G(\Sigma)\neq 1$. Since $(G/L)/(G(\Sigma)L/L)\cong G/G(\Sigma)L$
is solvable and since $G(\Sigma)L/L\cong G(\Sigma)/L(\Sigma)$ is a $2$-group
by Theorem 4.1(2), it follows that $G/L$ is solvable. Therefore,
$G_{\Delta}/L_{\Delta}$, and hence
$G^{\Delta}_{\Delta}/L^{\Delta}_{\Delta}\cong G_{\Delta}/G(\Delta)L_{\Delta}$
is solvable. Then $L_{\Delta}^{\Delta}=G_{\Delta}^{\Delta}$ since
$G_{\Delta}^{\Delta}\cong E_{2^{d}}:G_{x}^{\Delta}$, where $G_{x}^{\Delta}$ is
either $SL_{d}(2)$, or $G_{x}^{\Delta}\cong A_{7}$ for $d=4$ by Theorem
4.1(2.b), and the assertion (2) follows in this case.

Assume that $G(\Sigma)=1$. Then $G=G^{\Sigma}$ and $L=L^{\Sigma}$, and hence
$G_{\Delta}/L_{\Delta}$ is isomorphic to a subgroup of $\mathrm{Out}(L)$.

If $L_{\Delta}=L(\Delta)$ then
$G_{\Delta}^{\Delta}=G_{\Delta}/G(\Delta)\cong\left(G_{\Delta}/L_{\Delta}\right)/\left(G(\Delta)/L_{\Delta}\right)$.
Therefore $G_{\Delta}^{\Delta}$ is isomorphic to a quotient group of
$G_{\Delta}/L_{\Delta}$, and hence $\mathrm{Out}(L)$ contains a subgroup with
a quotient isomorphic to $G_{\Delta}^{\Delta}$. Then
$\left|\Delta\right|\left(\left|\Delta\right|-1\right)\mid\left|\mathrm{Out}(L)\right|$
since $G_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$. Therefore

$P(L)\leqslant\left|L:L_{\Delta}\right|=\left|\Sigma\right|\leqslant
2\left|\Delta\right|\left(\left|\Delta\right|-1\right)\leqslant
2\left|\mathrm{Out}(L)\right|\text{,}$ (5.2)

where $P(L)$ is the minimal primitive permutation representation of $L$. Then
$L\cong PSL_{2}(q)$ by Lemma 5.1. If $q>11$ then $q+1\leqslant 4(2,q-1)\log q$
which has no solutions. So $q=4,5,7,8,9,11$, which are ruled out since their
corresponding value of $\left|\Delta\right|$ does not fulfill
$\left|\Delta\right|\left(\left|\Delta\right|-1\right)\mid\left|\mathrm{Out}(L)\right|$.
Thus $L_{\Delta}=L(\Delta)$ is ruled out. Therefore
$Soc(G_{\Delta}^{\Delta})\trianglelefteq L_{\Delta}^{\Delta}$, and hence $L$
acts transitively on $\Delta$. Then $L$ acts point-transitively on
$\mathcal{D}$ since $L$ acts transitively on $\Sigma$. Thus, $G=G_{x}L$ where
$x$ is any point of $\mathcal{D}$.

Since the assertion (2) immediately follows if $Soc(G_{\Delta}^{\Delta})$ is
non-abelian simple, we may assume that $Soc(G_{\Delta}^{\Delta})$ is an
elementary abelian $u$-group for some prime $u$. Suppose that
$L_{\Delta}^{\Delta}=Soc(G_{\Delta}^{\Delta})$. Since $G/L=G_{\Delta}L/L\cong
G_{\Delta}/L_{\Delta}$, $G/L\leqslant\mathrm{Out}(L)$ and
$G_{\Delta}/G(\Delta)L_{\Delta}\cong G_{\Delta}^{\Delta}/L_{\Delta}^{\Delta}$,
it follows that $G_{\Delta}^{\Delta}/L_{\Delta}^{\Delta}$ is isomorphic to a
quotient subgroup of $\mathrm{Out}(L)$. Thus
$\left|\Delta\right|-1\mid\left|\mathrm{Out}(L)\right|$. Moreover,

$P(L)\leqslant\left|L:L_{\Delta}\right|=\left|\Sigma\right|<2\left|\Delta\right|\left(\left|\Delta\right|-1\right)\leqslant
2\left(\left|\mathrm{Out}(L)\right|+1\right)\left|\mathrm{Out}(L)\right|$
(5.3)

Then either $L\cong PSL_{2}(q)$, or $L$ is as in Lemma 5.1. In the latter case
$L$ is neither isomorphic to $A_{\ell}$ with $7\leqslant\ell\leqslant 11$, nor
to $M_{11}$, $PSp_{4}(3)$ or $PSU_{3}(5)$ since
$\left|\Delta\right|-1\leqslant\left|\mathrm{Out}(L)\right|$ (we use this
weaker constraint rather than
$\left|\Delta\right|-1\mid\left|\mathrm{Out}(L)\right|$ in order to apply the
argument here to Theorem 5.5 as well), where $\left|\Delta\right|$ is
$\lambda+2$ or $\lambda/2+1$ according to whether $\mathcal{D}$ is of type 1
or 2 respectively, and since $\lambda>10$ by our assumption. The unique
admissible group among the remaining ones listed in Lemma 5.1 is $L\cong
PSU_{4}(3)$ and $(\left|\Delta\right|,\left|\Sigma\right|)=(20,18^{2})$ or
$(38,36^{2})$ by [6] and [5, Tables 8.3–8.4] since $\left|\Sigma\right|$ is
not of the form as in Theorem 2.2(V). However, both exceptional cases are
ruled out since $\left|\Delta\right|>\left|\mathrm{Out}(PSU_{4}(3))\right|$.
Thus $L\cong PSL_{2}(q)$ and hence $\left|\Delta\right|\leqslant(2,q-1)f$.
Then $f\geqslant 3$ for $q$ odd and $f\geqslant 6$ for $q$ even since
$\lambda>10$ and since $\left|\Delta\right|$ is $\lambda+2$ or $\lambda/2+1$
according to whether $\mathcal{D}$ is of type 1 or 2 respectively. Thus
$P(L)=q+1$, hence (5.3) becomes $p^{f}+1\leqslant
2(2,q-1)f((2,q-1)f+1)\leqslant 4f(2f+1)$ which yields $q=2^{6},3^{3},3^{4}$
and $\mathcal{D}$ is of type 2 with $\lambda=12,12,16$ respectively. Thus
$\left|\Sigma\right|=61,61,113$ respectively, but $L\cong PSL_{2}(q)$ with
$q=2^{6},3^{3},3^{4}$ has no such transitive permutation degrees. Thus
$Soc(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}$ and
$\left|G_{\Delta}^{\Delta}:L_{\Delta}^{\Delta}\right|\mid\left|\mathrm{Out}(L)\right|$.

If $G_{\Delta}^{\Delta}\not\leqslant A\Gamma L_{1}(u^{h})$, then
$G_{\Delta}^{\Delta}$ contains as a normal subgroup one of the groups
$SL_{n}(u^{h/n})$, $Sp_{n}(u^{h/n})$, $G_{2}^{\prime}(2^{h/6})$ with $h\equiv
0\pmod{6}$, $A_{6}$ or $A_{7}$ for $(u,h)=(2,4)$, or $SL_{2}(13)$ for
$(u,h)=(3,6)$ by [16, List(B)] since $u^{h}\neq 3^{2},5^{2},7^{2},11^{2}$,
$19^{2},23^{2},29^{2},59^{2}$ and $3^{4}$ by Lemma 5.2, and
$L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$ in these cases. This
completes the proof. ∎

######  Corollary 5.4.

If $G(\Sigma)\neq 1$, then $\lambda=2(2^{d-1}-1)$,
$\left|\Sigma\right|=\lambda^{2}$ and a quotient group of
$L_{\Delta}^{\Sigma}$ is isomorphic either to $SL_{d}(2)$ for $d\geqslant 4$,
or to$A_{7}$ for $d=4$.

###### Proof.

If $G(\Sigma)\neq 1$, then $G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}\cong
G_{x}^{\Delta}$ where either $G_{x}^{\Delta}\cong SL_{d}(2)$ for $d\geqslant
4$, or $G_{x}^{\Delta}\cong A_{7}$ for $d=4$ by Theorem 4.1. On the other
hand, $G_{x}/L_{x}\cong G_{x}L/L=G/L$ is isomorphic to a subgroup of
$\mathrm{Out}(L)$ as a consequence of Proposition 5.3. Therefore
$G_{x}/L_{x}$, and hence $G_{x}/G(\Delta)L_{x}$ is solvable. Since
$G_{x}/G(\Delta)L_{x}\cong G_{x}^{\Delta}/L_{x}^{\Delta}$ and $G_{x}^{\Delta}$
is non-abelian simple, it follows that $L_{x}^{\Delta}=G_{x}^{\Delta}$. Thus
$L_{\Delta}^{\Sigma}\not\leqslant G(\Delta)^{\Sigma}$, and hence
$L_{\Delta}^{\Sigma}/\left(L_{\Delta}^{\Sigma}\cap
G(\Delta)^{\Sigma}\right)\cong G_{\Delta}^{\Sigma}/G(\Delta)^{\Sigma}\cong
G_{x}^{\Delta}$, which is the assertion. ∎

######  Theorem 5.5.

Let $\Delta\in\Sigma$ then $L_{\Delta}^{\Sigma}$ is a large subgroup of
$L^{\Sigma}$. Moreover one of the following holds:

  1. (1)

$\left|L^{\Sigma}\right|\leqslant\left|L_{\Delta}^{\Sigma}\right|^{2}$.

  2. (2)

$G(\Sigma)=1$, $Soc(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$, where
$u^{h}=\left|\Delta\right|$. Furthermore, the following holds:

    1. (a)

$L_{\Delta}$ does not act $2$-transitively on $\Delta$;

    2. (b)

$\left|L\right|\leqslant
4\left|L_{\Delta}^{\Delta}\right|^{2}\left|\mathrm{Out}(L)\right|^{2}$;

    3. (c)

$\left|L(\Delta)\right|<2\left|\mathrm{Out}(L)\right|<\left|L_{\Delta}\right|$.

###### Proof.

Suppose that $G(\Sigma)=1$. Then $G=G^{\Sigma}$ and $L=L^{\Sigma}$. Assume
that $L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$. If $\mathcal{D}$
is of type 1 then
$\left|L:L_{\Delta}\right|=\left|\Sigma\right|\leqslant\left|\Delta\right|\left(\left|\Delta\right|-1\right)\leqslant\left|L_{\Delta}^{\Delta}\right|$,
and hence $\left|L\right|\leqslant\left|L_{\Delta}\right|^{2}$, which is the
assertion (1) in this case.

If $\mathcal{D}$ is of type 2, since $G$ acts flag-transitively on
$\mathcal{D}$, it follows that $\lambda^{2}/2\mid\left|G_{x}\right|$ and hence
$\lambda^{2}/2\mid\left|L_{x}\right|\left|\mathrm{Out}(L)\right|$. On the
other hand,
$\left|L_{\Delta}\right|=\left(\frac{\lambda}{2}+1\right)\frac{\lambda}{2}\left|L_{x,y}\right|$
since $L_{\Delta}$ induces a $2$-transitive group on $\left|\Delta\right|$. In
particular, $\left|L_{x}:L_{x,y}\right|=\frac{\lambda}{2}$ and so
$\lambda\mid\left|L_{x,y}\right|\left|\mathrm{Out}(L)\right|$. If
$\lambda\mid\left|\mathrm{Out}(L)\right|$ then
$P(L)\leqslant\left|L:L_{\Delta}\right|=\frac{\lambda^{2}-2\lambda+2}{2}<\left|\mathrm{Out}(L)\right|^{2}$.
Then $L\cong PSL_{2}(q)$ or $L$ is one of the groups listed in Lemma 5.1.
Actually, in the latter case only $L\cong PSL_{3}(q)$ with
$(q,\lambda)=(4,12),(16,12),(16,24)$ are admissible since
$\lambda\mid\left|\mathrm{Out}(L)\right|$ and $\lambda>10$. Then
$\left|\Sigma\right|=61$ or $265$ respectively, but none of these divides the
order of the corresponding $L$. So these cases are excluded, and hence $L\cong
PSL_{2}(q)$ with $q=p^{f}$ and $f\geqslant 6$, as
$\lambda\mid\left|\mathrm{Out}(L)\right|$ and $\lambda>10$ and
$\left|\mathrm{Out}(L)\right|=(2,p^{f}-1)f$. However,
$p^{f}+1\leqslant\left|L:L_{\Delta}\right|\leqslant 2f^{2}-2f+1$ has no
admissible solutions for $f\geqslant 6$. Then
$\left(\left|\mathrm{Out}(L)\right|,\lambda\right)<\lambda$ and hence
$\left|L_{x,y}\right|\geqslant 2$ since
$\lambda\mid\left|L_{x,y}\right|\left|\mathrm{Out}(L)\right|$. Then
$\left|L_{\Delta}\right|\geqslant\left(\frac{\lambda}{2}+1\right)\lambda>\left|\Sigma\right|=\left|L:L_{\Delta}\right|$
and we obtain the assertion (1) also in this case.

Assume that $L_{\Delta}^{\Delta}$ does not act $2$-transitively on $\Delta$.
Then $Soc(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}$ and $G_{\Delta}^{\Delta}$
is a $2$-transitive subgroup of the semilinear $1$-dimensional group by
Proposition 5.3. Note that
$\left|L:L_{\Delta}\right|=\left|G:G_{\Delta}\right|\leqslant
2\left|G_{\Delta}^{\Delta}\right|$. Also, it results that
$G_{\Delta}/L_{\Delta}\cong G_{\Delta}L/L=G/L\leqslant\mathrm{Out}(L)$ and
$G_{\Delta}/G(\Delta)L_{\Delta}\cong G_{\Delta}^{\Delta}/L_{\Delta}^{\Delta}$.
Hence, $G_{\Delta}^{\Delta}/L_{\Delta}^{\Delta}$ is isomorphic to a quotient
group of a subgroup of $\mathrm{Out}(L)$. Therefore,
$\left|L:L_{\Delta}\right|\leqslant
2\left|L_{\Delta}^{\Delta}\right|\left|\mathrm{Out}(L)\right|$.

Assume that $\left|L(\Delta)\right|\geqslant 2\left|\mathrm{Out}(L)\right|$.
Then
$2\left|L_{\Delta}^{\Delta}\right|\left|\mathrm{Out}(L)\right|\leqslant\left|L_{\Delta}\right|$,
and we still obtain (1).

Assume that $\left|L(\Delta)\right|<2\left|\mathrm{Out}(L)\right|$. Then
$\left|L\right|\leqslant
4\left|L_{\Delta}^{\Delta}\right|^{2}\left|\mathrm{Out}(L)\right|^{2}$.
Suppose that $\left|L_{\Delta}\right|\leqslant\left|\mathrm{Out}(L)\right|$.
Then $2\left|\Delta\right|\leqslant\left|L_{\Delta}\right|\leqslant
2\left|\mathrm{Out}(L)\right|$ since
$\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}$. Then
$\left|\Delta\right|\leqslant\left|\mathrm{Out}(L)\right|$, and hence
$P(L)\leqslant\left|L:L_{\Delta}\right|\leqslant
2\left|\mathrm{Out}(L)\right|(\left|\mathrm{Out}(L)\right|-1)$. Then
$L_{\Delta}$ is isomorphic either to $PSL_{2}(q)$ or to one of the groups
listed in Lemma 5.1. However, the same argument used in Proposition 5.3 can be
also used here to rule out all these groups since
$\left|\Delta\right|\leqslant\left|\mathrm{Out}(L)\right|$. Thus
$\left|L_{\Delta}\right|>2\left|\mathrm{Out}(L)\right|$ and hence
$\left|L:L_{\Delta}\right|<\left|L_{\Delta}\right|^{2}$, which means that
$L_{\Delta}$ is a large subgroup of $L$, and we obtain assertions (2a)–(2c).

Suppose that $G(\Sigma)\neq 1$. Then $\lambda=2(2^{d-1}-1)$,
$\left|\Sigma\right|=\lambda^{2}$ and a quotient group of
$L_{\Delta}^{\Sigma}$ is isomorphic either to $SL_{d}(2)$ for $d\geqslant 4$,
or to $A_{7}$ for $d=4$ by Corollary 5.4. In either case one has
$\left|\Delta\right|\left(\left|\Delta\right|-1\right)\leqslant\left|L_{\Delta}^{\Sigma}\right|$
since $\left|\Delta\right|=2^{d}$. Since $G(\Sigma)\trianglelefteq
L_{\Delta}<L$ and since $G^{\Sigma}$ acts primitively on $\Sigma$, it follows
that
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\left|L:L_{\Delta}\right|=\left|\Sigma\right|\leqslant\left|\Delta\right|\left(\left|\Delta\right|-1\right)\leqslant\left|L_{\Delta}^{\Sigma}\right|$
and the assertion (1) follows in this case.

∎

##  6\. Classification of the $2$-designs of type 1

In this section we mainly use the constraints for $L^{\Sigma}$ provided in
Proposition 5.3, Corollary 5.4 and Theorem 5.5 to prove Theorem 6.1 stated
below. It is worth noting that, when $L^{\Sigma}$ is a Lie type simple group
we show that $L^{\Sigma}_{\Delta}$ is a large subgroup of $L^{\Sigma}$ of
order divisible by a suitable primitive prime divisor of $p^{\zeta}-1$, where
$\zeta$ is determined in [17, Proposition 5.2.16]. We combine this constraints
on $L^{\Sigma}_{\Delta}$ to show that a small number of groups listed in [3]
are admissible. These groups are then ruled out by exploiting the
$2$-transitivity of $G_{\Delta}^{\Delta}$ on $\Delta$.

######  Theorem 6.1.

If $\mathcal{D}$ is a symmetric
$2$-$\left(\left(\lambda+2\right)\lambda^{2},(\lambda+1)\lambda,\lambda\right)$
design admitting a flag-transitive and point-imprimitive automorphism group,
then $\lambda\leqslant 10$.

We analyze the cases where $L^{\Sigma}$ is sporadic, alternating, a Lie type
simple classical or exceptional group separately.

######  Lemma 6.2.

$L^{\Sigma}$ is not sporadic.

###### Proof.

Either $L_{\Delta}^{\Sigma}$ is maximal in $L^{\Sigma}$, or
$\mathrm{Out}(L)\cong Z_{2}$ and $G_{\Delta}^{\Sigma}$ is a novelty. In the
latter case, the unique admissible case is $G^{\Sigma}\cong M_{12}:Z_{2}$ and
$G_{\Delta}^{\Sigma}\cong PGL_{2}(11)$ since $\left|\Sigma\right|$ is a
square. Hence, $\left|\Sigma\right|=144$ by [38, Table 1] (see also [39] for
the cases $L^{\Sigma}\cong Fi_{22}$ or $Fi_{24}^{\prime}$). Then
$\lambda+2=14$ and hence $G(\Sigma)=1$ by Theorem 4.1. Since $G_{\Delta}\cong
PGL_{2}(11)$ has no transitive permutation representations of degree $14$, the
case is ruled out by Proposition 5.3. Therefore, $L_{\Delta}$ is maximal in
$L$.

Assume that $L^{\Sigma}\cong M_{i}$, where $i=11,12,22,23$ or $24$. Since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\lambda^{2}$, it follows from
[17, Table 5.1.C] that, $k^{2}=2^{a_{1}}3^{a_{2}}$ for some
$a_{1},a_{2}\geqslant 2$. Then $\lambda=12$ and either $L\cong M_{11}$ and
$L_{\Delta}\cong F_{55}$, or $L^{\Sigma}\cong M_{12}$ and $L^{\Sigma}\cong
PSL_{2}(11)$ by [6]. However, these cases are ruled out by Proposition 5.3
since $\lambda+2$ does not divide the order of $L_{\Delta}$.

Assume that $L^{\Sigma}\cong J_{i}$, where $i=1,2,3$ or $4$. Then
$\lambda^{2}$ divides $2^{2}$, $2^{6}3^{2}5^{2}$, $2^{6}3^{4}$, or
$2^{20}3^{2}11^{2}$, respectively, by [17, Table 5.1.C]. Then $i=2$ and either
$\lambda=10$ and $L_{\Delta}^{\Sigma}\cong PSU_{3}(3)$, or $\lambda=60$ and
$L_{\Delta}^{\Sigma}\cong PSL_{2}(7)$ by [6]. However these cases are ruled
out, as $L_{\Delta}^{\Sigma}$ does not have a $2$-transitive permutation
representation of degree $12$ or $62$ respectively, and this contradicts
Proposition 5.3.

Assume that $L^{\Sigma}$ is isomorphic to one of the groups $HS$ or $McL$. By
[17, Table 5.1.C] $\lambda^{2}$ divides $2^{8}3^{2}5^{2}$ or $2^{6}3^{6}5^{2}$
respectively. Then either $L^{\Sigma}\cong HS$, $L_{\Delta}^{\Sigma}\cong
M_{22}$ and $\lambda=10$, or $L^{\Sigma}\cong McL$, $L_{\Delta}^{\Sigma}\cong
M_{22}$ and $\lambda=45$. Both these cases are excluded since they contradict
Proposition 5.3.

It is straightforward to check that the remaining cases are ruled out
similarly, as they do not have transitive permutation representations of
degree $\lambda^{2}$ by [6] and [39]. ∎

######  Lemma 6.3.

If $L^{\Sigma}\cong A_{\ell}$, $\ell\geqslant 5$, then $L^{\Sigma}$ acts
primitively on $\Sigma$.

###### Proof.

Assume that $L\cong A_{6}$. Then $A_{6}\cong PSL_{2}(9)\trianglelefteq
G\leqslant P\Gamma L_{2}(9)$, hence $G$ does not have a primitive permutation
representation of degree $\lambda^{2}$ with $\lambda>10$ by [6]. Thus
$\ell\neq 6$ by Theorem 4.1, and hence $\mathrm{Out}(L)\cong Z_{2}$.

Suppose the contrary of the statement. Then there is a subgroup $M$ of $L$
containing $L_{\Delta}$ such that $L_{\Delta}^{\Sigma}<M^{\Sigma}<L^{\Sigma}$
with $M^{\Sigma}$ maximal in $L^{\Sigma}$. Let $x\in\Delta$, then $x^{M}$ is a
union of $\theta$ elements of $\Sigma$, where
$\theta=\left|M^{\Sigma}:L^{\Sigma}_{\Delta}\right|$. Therefore
$\left|x^{M}\right|=\theta(\lambda+2)$ with $\lambda^{2}=a\theta$ for some
$a\geqslant 1$. Then $x^{M}\setminus\left\\{x\right\\}$ is a union of
$L_{x}$-orbits since $L_{x}<L_{\Delta}<L$. Therefore
$\frac{\lambda+1}{\eta}\mid\theta(\lambda+2)-1$ by Lemma 2.1, where
$\eta=(\lambda+1,2)$ since $\mathrm{Out}(L)\cong Z_{2}$. Then
$\theta=f\frac{\lambda+1}{\eta}+1$ for some $f\geqslant 1$, hence

$\left(f\frac{\lambda+1}{\eta}+1\right)a-1=\lambda^{2}-1$ (6.1)

and so $a=e\frac{\lambda+1}{\eta}+1$ for some $e\geqslant 1$. Then (6.1)
becomes

$ef\left(\frac{\lambda+1}{\eta}\right)^{2}+(e+f)\left(\frac{\lambda+1}{\eta}\right)+1=\lambda^{2}$

Then $ef<\eta^{2}=(\lambda+1,2)^{2}$, and hence $\eta=2$ and $\lambda$ is odd.
Therefore $ef=3$ and $\lambda=15$, $\left|\Delta\right|=17$ and
$\left|\Sigma\right|=225$ since $\lambda>10$. Moreover, $G(\Sigma)=1$ by
Theorem 4.1. Then $L_{\Delta}\cong A_{224}$ by [10, Table B.4]. However, this
case cannot occur by Proposition 5.3 since $A_{224}$ has no quotient groups
with a transitive permutation representations of degree $17$. ∎

######  Lemma 6.4.

$L^{\Sigma}$ is not isomorphic to $A_{\ell}$, $\ell\geqslant 5$.

###### Proof.

Since $L_{\Delta}^{\Sigma}$ is a large maximal subgroup of $L^{\Sigma}$ by
Theorem 5.5 and Lemma 6.3, and since $\left|\Sigma\right|=\lambda^{2}$ with
$\lambda>10$, only the following cases are admissible by [3, Theorem 2]:

  1. (i)

$L_{\Delta}^{\Sigma}\cong\left(S_{t}\times S_{\ell-t}\right)\cap A_{\ell}$
with $1\leqslant t\leqslant\ell/2$;

  2. (ii)

$L_{\Delta}^{\Sigma}\cong\left(S_{t}\wr S_{\ell/t}\right)\cap A_{\ell}$ with
$2\leqslant t\leqslant\ell/2$.

Suppose that (i) holds. Then $\binom{\ell}{t}=\lambda^{2}$, and hence either
$t=1,2$, or $t=3$ and $\ell=50$ by [1, Chapter 3].

Assume that $t=1$. Since $L_{\Delta}^{\Sigma}\cong A_{\lambda^{2}-1}$ with
$\lambda>10$, no quotient groups of $L_{\Delta}^{\Sigma}$ are isomorphic to
$SL_{d}(2)$ for any $d\geqslant 4$. Also, the minimal transitive permutation
representation of $A_{\lambda^{2}-1}$ is strictly greater than $\lambda+2$.
Thus, this case cannot occur by Proposition 5.3 and Corollary 5.4.

Assume that $t=2$. Then $L_{\Delta}^{\Sigma}\cong\left(S_{2}\times
S_{\ell-2}\right)\cap A_{\ell}$. Suppose that $G(\Sigma)\neq 1$. Then a
quotient group of $G_{\Delta}^{\Sigma}$ is isomorphic to $SL_{d}(2)$ with
$d>4$ by Theorem 4.1 since $\binom{\ell}{2}\neq 14^{2}$, but this is clearly
impossible. Thus $G(\Sigma)=1$ and hence a quotient group of
$L_{\Delta}\cong\left(S_{2}\times S_{\ell-2}\right)\cap A_{\ell}$ must have a
$2$-transitive permutation representation of degree $\ell-2=\lambda+2$ by
Proposition 5.3 since $\lambda>10$. However,
$\binom{\lambda+4}{2}=\lambda^{2}$ has no integer solutions. Thus, this case
is excluded.

Assume $t=3$ and $\ell=50$. Then $\lambda+2=142$, as $\lambda>10$, and hence
$G(\Sigma)=1$ by Theorem 4.1. Then $\lambda+2\mid\left|G_{\Delta}\right|$,
whereas $G_{\Delta}=\left(S_{3}\times S_{47}\right)\cap G$, which is a
contradiction.

Suppose that (ii) holds. Hence, $L_{\Delta}^{\Sigma}\cong\left(S_{\ell/t}\wr
S_{t}\right)\cap A_{\ell}$ where $s/t,t>1$. Then
$\left|\Sigma\right|=\frac{\ell!}{((\ell/t)!)^{t}t!}$ and $A_{\ell/t}\wr
A_{t}\leqslant L_{\Delta}^{\Sigma}\leqslant\left(S_{\ell/t}\wr
S_{t}\right)\cap L$. Easy computations show that $\ell>25$ since
$\left|\Sigma\right|=\lambda^{2}$. Moreover,
$\left(A_{\ell/t}\right)^{t}\trianglelefteq L_{\Delta}^{\Sigma}$ and
$A_{t}\leqslant
L_{\Delta}^{\Sigma}/\left(A_{\ell/t}\right)^{t}\leqslant(Z_{2})^{t}:S_{t}$,
where $(Z_{2})^{t}$ is a permutation module for $A_{t}$. Thus, by [17, Lemma
5.3.4], $L_{\Delta}^{\Sigma}/\left(A_{\ell/t}\right)^{t}$ is isomorphic to one
of the following groups:

  1. (1)

$A_{t},S_{t}$;

  2. (2)

$A_{t}\times Z_{2},S_{t}\times Z_{2}$;

  3. (3)

$(Z_{2})^{t-1}:A_{t},(Z_{2})^{t-1}:S_{t}$;

  4. (4)

$(Z_{2})^{t}:A_{t},(Z_{2})^{t}:S_{t}$.

Suppose that $G(\Sigma)\neq 1$. Then a quotient group of $L_{\Delta}^{\Sigma}$
is isomorphic either to $SL_{d}(2)$ for $d\geqslant 4$, or to $A_{7}$ for
$d=4$ by Corollary 4.1. Matching such information with (1)–(4) one obtains
$d=4$, $t=7,8$ and $\lambda=14$. Then
$((\ell/t)!)^{t-1}<\frac{\ell!}{((\ell/t)!)^{t}t!}=196$ as shown in [24,
(3.4)] and hence $\ell=14,16$. However, $\left|\Sigma\right|$ is not a square
for such values of $t$ and $\ell$.

Suppose that $G(\Sigma)=1$. Then $L_{\Delta}/\left(A_{\ell/t}\right)^{t}$ is
one of the groups in (1)–(4). Assume that $L_{\Delta}^{\Delta}$ does not act
$2$-transitively on $\Delta$. Then
$\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(2^{h})$, where $\lambda+2=2^{h}$
and $\left|G_{\Delta}^{\Delta}:L_{\Delta}^{\Delta}\right|\leqslant 2$ by
Proposition 5.3, as $\ell>6$, which force $L_{\Delta}^{\Delta}$ to act
$2$-transitively on $\Delta$, and we reach a contradiction. Therefore,
$L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$.

If $L_{\Delta}^{\Delta}$ is of affine type then $\lambda+2=2^{i}$ with
$i\leqslant t$. Therefore
$((\ell/t)!)^{t-1}<\frac{\ell!}{((\ell/t)!)^{t}t!}<2^{2t}$ and hence
$\ell/t=2$, and $t>12$ since $\ell>25$. However, this is impossible by [16,
List (B)] since $A_{t}\leqslant L_{\Delta}/\left(A_{\ell/t}\right)^{t}$. Thus
$L_{\Delta}^{\Delta}$ is almost simple and hence $A_{t}\trianglelefteq
L_{\Delta}^{\Delta}\leqslant S_{t}$ with $t\geqslant 5$ by (1)–(4). Therefore,
$t=\lambda+2>12$ and
$2^{t-1}\leqslant((\ell/t)!)^{t-1}<\frac{\ell!}{((\ell/t)!)^{t}t!}<t^{2}$,
which is a contradiction. This completes the proof. ∎

######  Lemma 6.5.

$L^{\Sigma}\ncong PSL_{2}(q)$.

###### Proof.

Assume that $L^{\Sigma}\cong PSL_{2}(q)$, $q=p^{f}$, $p$ prime and $f\geqslant
1$. Then $q\neq 7$ since $PSL_{2}(7)$ does not have transitive permutation
representations of square degree, and $q\neq 9$ by Lemma 6.4 since
$PSL_{2}(9)\cong A_{6}$. Moreover, if $q=p$, then
$p\mid\left|L_{\Delta}^{\Sigma}\right|$ since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\lambda^{2}$. Thus no novelties
occur by [5, Table 8.1], hence $L_{\Delta}^{\Sigma}$ is maximal in
$L^{\Sigma}$. Moreover, no quotients groups of $L_{\Delta}^{\Sigma}$ are
isomorphic to $SL_{d}(2)$ with $d\geqslant 4$ or to $A_{7}$ and $d=4$ for
$G(\Sigma)\neq 1$, hence $G(\Sigma)=1$ by Corollary 5.4. Thus, $L_{\Delta}$ is
maximal in $L$.

Assume that $L_{\Delta}$ is isomorphic to any of the groups $A_{4},S_{4}$ or
$A_{5}$. Then $\lambda+2\mid\left|L_{\Delta}\right|$, hence $\lambda+2$ is not
a power of a prime since $\lambda>10$. Thus $L_{\Delta}$ is non-solvable and
acts $2$-transitively on $\Delta$ by Proposition 5.3, hence $L_{\Delta}\cong
A_{5}$ and $\lambda+2=6$, whereas $\lambda>10$. Thus, these groups are ruled
out.

Assume that $L_{\Delta}$ is isomorphic to $D_{\frac{q\pm 1}{(2,q-1)}}$. Then
$\lambda\leqslant(q-1)/(2,q-1)$ since $\lambda+2\mid\left|L_{\Delta}\right|$,
and hence $q(q\mp
1)=\left|\Sigma\right|\leqslant\frac{(q-1)^{2}}{(2,q-1)^{2}}$, which has no
admissible solutions.

Assume that $PSL_{2}(q^{1/m})\trianglelefteq L_{\Delta}\leqslant
PGL_{2}(q^{1/m})$. Then $L_{\Delta}$ acts $2$-transitively on $\Delta$ by
Proposition 5.3. Moreover, $\lambda+2=q^{1/m}+1$, since $q^{1/m}>11$ being
$\lambda>10$. Thus $\lambda=q^{1/m}-1$, and hence $L_{\Delta}$ must contain a
Sylow $p$-subgroup of $L$, which is a contradiction.

Finally, assume that $L_{\Delta}\cong E_{q}:Z_{\frac{q-1}{(q-1,2)}}$. Then
$\lambda^{2}=q+1$, which has no solutions by [32, A5.1] since $\lambda>10$. ∎

######  Lemma 6.6.

$L^{\Sigma}$ is not isomorphic to one of the groups $PSL_{3}(4)$,
$PSU_{4}(2)$, $PSL_{6}(2)$, $PSp_{6}(2)$, $P\Omega_{8}^{+}(2)$,
$G_{2}(2)^{\prime}$, ${}^{2}G_{2}(3)^{\prime}$ or ${}^{2}F_{4}(2)^{\prime}$.

###### Proof.

Assume that $L^{\Sigma}\cong PSL_{3}(4)$. Then
$35\mid\left|L_{\Delta}^{\Sigma}\right|$ since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\lambda^{2}$, but $L^{\Sigma}$
does not contain such a group by [6].

Assume that $L^{\Sigma}\cong PSU_{4}(2)$. Then
$5\mid\left|L_{\Delta}^{\Sigma}\right|$ since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\lambda^{2}$, and hence
$L_{\Delta}^{\Sigma}\cong S_{6}$ and $\lambda=6$ by [6], whereas $\lambda>10$
by our assumptions.

Assume that $L^{\Sigma}\cong PSL_{6}(2)$. Thus $2\cdot 5\cdot
31\mid\left|L_{\Delta}^{\Sigma}\right|$ since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\lambda^{2}$, and hence either
$L_{\Delta}^{\Sigma}\cong SL_{5}(2)$ or $L_{\Delta}^{\Sigma}\cong
E_{2^{5}}:SL_{5}(2)$ by [5, Tables 8.24 and 8.25]. However, both cases are
ruled out since $7$ divides $\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|$ but
$7^{2}$ does not.

Assume that $L^{\Sigma}\cong PSp_{6}(2)$. Then $\mathrm{Out}(L^{\Sigma})=1$,
$L_{\Delta}^{\Sigma}\cong S_{8}$ and $\lambda=6$ by [6], whereas $\lambda>10$.

Assume that $L^{\Sigma}\cong P\Omega_{8}^{+}(2)$. Therefore
$21\mid\left|L_{\Delta}^{\Sigma}\right|$ since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\lambda^{2}$, and hence
$\lambda^{2}\mid 2^{12}\cdot 3^{4}\cdot 5^{2}$. Then
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|$ must be divisible by one among
its primitive degrees $120$, $135$ or $960$ by [6]. Thus $15^{2}$ divides
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|$ in each case, and hence
$\lambda=15j$ for some $j\geqslant 1$. If $G(\Sigma)\neq 1$ then
$\lambda=15j=2^{d}-2$ with $d\geqslant 4$ by Theorem 4.1. Easy computations
show that no admissible cases occur. Therefore $G(\Sigma)=1$, and hence $G$ is
a subgroup of $P\Omega_{8}^{+}(2).S_{3}$. Moreover, the order of $G$ is
divisible by $(15j)^{3}(15j+2)(15j+1)$ since $G_{\Delta}^{\Delta}$ acts
$2$-transitively on $\Delta$, which has size $15j+2$, and since $G$ acts flag-
transitively on $\mathcal{D}$ and $k=\lambda(\lambda+1)$. Since no groups
occur, this case is excluded.

The case $L^{\Sigma}\cong G_{2}(2)^{\prime}$ is ruled out in Lemma 6.5 since
$G_{2}(2)^{\prime}\cong PSL_{2}(8)$. Also, if $L^{\Sigma}\cong$
${}^{2}G_{2}(3)^{\prime}\cong PSU_{3}(3)$ then $L_{\Delta}^{\Sigma}\cong
PSL_{2}(7)$ and $\left|\Sigma\right|=36$ by [6]. So $\lambda=6$, whereas
$\lambda>10$.

Finally, if $L^{\Sigma}\cong$ ${}^{2}F_{4}(2)^{\prime}$ then
$L_{\Delta}^{\Sigma}\cong PSL_{3}(3):Z_{2}$ and $\left|\Sigma\right|=1600$ by
[6]. Then $\lambda=40$ and hence $\lambda+1=41$ must divide the order of $G$
by Lemma 2.1. However, this is impossible since
$G^{\Sigma}\leqslant$${}^{2}F_{4}(2)$ and the order of $G(\Sigma)$ is either
$1$ or a power of $2$ by Theorem 4.1. ∎

A divisor $s$ of $q^{e}-1$ that is coprime to each $q^{i}-1$ for $i<e$ is said
to be a _primitive divisor_ , and we call the largest primitive divisor
$\Phi_{e}^{\ast}(q)$ of $q^{e}-1$ the _primitive part_ of $q^{e}-1$. One
should note that $\Phi_{e}^{\ast}(q)$ is strongly related to cyclotomy in that
it is equal to the quotient of the cyclotomic number $\Phi_{e}(q)$ and
$(n,\Phi_{e}(q))$ when $e>2$. Also $\Phi_{e}^{\ast}(q)>1$ for $e>2$ and
$(q,e)\neq(2,6)$ by Zsigmondy’s Theorem (for instance, see [32, P1.7]).

Let $p$ be a prime, $w$ a prime distinct from $p$, and $m$ an integer which is
not a power of $p$. Also let $\Gamma$ be a group which is not a $p$-group.
Then we define

$\displaystyle\zeta_{p}(w)$ $\displaystyle=$
$\displaystyle\min\left\\{z:z\geqslant 1\text{ and }p^{z}\equiv
1\pmod{w}\right\\}$ $\displaystyle\zeta_{p}(m)$ $\displaystyle=$
$\displaystyle\max\left\\{\zeta_{p}(w):w\text{ prime, }w\neq p\text{ and
}w\mid m\right\\}$ $\displaystyle\zeta_{p}(X)$ $\displaystyle=$
$\displaystyle\zeta_{p}(\left|X\right|)\text{.}$

If $L^{\Sigma}$ is isomorphic to a simple group of Lie type over $GF(q)$,
$q=p^{f}$, then $\zeta_{p}(L^{\Sigma})$ is listed in [17, Proposition 5.2.16
and Table 5.2.C]. In the sequel we will denote $\zeta_{p}(L^{\Sigma})$ simply
by $\zeta$. It is worth noting that $\Phi_{\zeta}^{\ast}(p)>1$ by Lemmas 6.5
and 6.6.

######  Lemma 6.7.

$L_{\Delta}^{\Sigma}$ is a large subgroup of $L^{\Sigma}$ such that
$\left(\Phi_{\zeta}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$.

###### Proof.

Suppose
$\left(\Phi_{\zeta}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)=1$. Note
that
$\left|G_{\Delta}^{\Sigma}:L_{\Delta}^{\Sigma}\right|\mid\left|\mathrm{Out}(L^{\Sigma})\right|$
since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\left|G^{\Sigma}:G_{\Delta}^{\Sigma}\right|$,
as $G^{\Sigma}$ acts primitively on $\Sigma$. Thus
$\left(\Phi_{\zeta}^{\ast}(p),\left|G_{\Delta}^{\Sigma}\right|\right)=1$ since
$\left(\Phi_{\zeta}^{\ast}(p),\left|\mathrm{Out}(L^{\Sigma})\right|\right)=1$
by [17, Proposition 5.2.15(ii)]. Therefore
$\Phi_{\zeta}^{\ast}(p)\mid\left|G^{\Sigma}:G_{\Delta}^{\Sigma}\right|$, and
hence $\Phi_{\zeta}^{\ast}(p)\mid\lambda^{2}$ since
$\left|G^{\Sigma}:G_{\Delta}^{\Sigma}\right|=\lambda^{2}$. Then
$(\Phi_{\zeta}^{\ast}(p),\left|G_{x}\right|)>1$, where $x$ is any point of
$\mathcal{D}$, since $\lambda(\lambda+1)\mid\left|G_{x}\right|$ being $G$
flag-transitive on $\mathcal{D}$. Therefore
$(\Phi_{\zeta}^{\ast}(p),\left|G_{\Delta}\right|)>1$, where $\Delta$ is the
element of $\Sigma$ containing $x$, and hence
$(\Phi_{\zeta}^{\ast}(p),\left|G_{\Delta}^{\Sigma}\right|)>1$ which is a
contradiction. Thus
$\left(\Phi_{\zeta}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$, and
$L_{\Delta}^{\Sigma}$ is large by Theorem 5.5. ∎

######  Lemma 6.8.

$L^{\Sigma}$ is not isomorphic to an exceptional simple group of Lie type.

###### Proof.

Let $M$ be a subgroup of $L$ such that $M^{\Sigma}$ is a maximal subgroup of
$L^{\Sigma}$ containing $L_{\Delta}^{\Sigma}$. Also $M^{\Sigma}$ is large
since $L_{\Delta}^{\Sigma}$ is so by Lemma 5.5. Therefore $M^{\Sigma}$ is one
of the groups classified in [3, Theorem 5]. Moreover,
$\left(\Phi_{\zeta}^{\ast}(p),\left|M^{\Sigma}\right|\right)>1$ by Lemma 6.7.

Assume that $M^{\Sigma}$ is parabolic. If $L^{\Sigma}$ is untwisted then
$M^{\Sigma}$ can be obtained by deleting the $i$-th node in the Dynkin diagram
of $L^{\Sigma}$, and we see that none of these groups is of order divisible by
a prime factor of $\Phi_{\zeta}^{\ast}(p)$. Indeed, for instance consider the
Levi factors of the maximal parabolic subgroups of $L^{\Sigma}\cong F_{4}(q)$,
$q=p^{f}$ 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 has order divisible by a prime factor of
$\Phi_{12f}^{\ast}(p)$. If $L^{\Sigma}$ is twisted, that is $L^{\Sigma}$ is
centralized by an automorphism $\gamma$ of the corresponding untwisted group
and $\gamma$ induces a non-trivial symmetry $\rho$ on the Dynkin diagram. In
this case the $M^{\Sigma}$ exist only when deleting the resulting subset
obtained by deleting the $i$-th node in the Dynkin diagram of corresponding
untwisted group is $\rho$-invariant. The Levi factor of $M^{\Sigma}$ is
obtained by taking the fixed points of the automorphism $\gamma$ on the Levi
factor of the corresponding untwisted subgroup. Also in the twisted case the
order of any maximal subgroups of $L^{\Sigma}$ is not divisible by a prime
factor of $\Phi_{\zeta}^{\ast}(p)$. Indeed, for instance, the Levi factors of
the maximal parabolic subgroups of ${}^{2}E_{6}(q)$, $q=p^{f}$, are of types
${}^{2}A_{5}(q)$, ${}^{2}D_{4}(q)$, $A_{1}(q)\times A_{2}(q^{2})$ and
$A_{1}(q^{2})\times A_{2}(q)$, and none of these has order divisible by a
prime factor of $\Phi_{18f}^{\ast}(p)$.

Assume that $M^{\Sigma}$ is not parabolic. Then $(L^{\Sigma},M^{\Sigma})$ is
listed in [3, Table 2]. Since
$\left(\Phi_{\zeta}^{\ast}(p),\left|M^{\Sigma}\right|\right)>1$, only the
groups contained in Table 2 are admissible by [21].

Table 2. Admissible $(L^{\Sigma},M^{\Sigma})$ $L^{\Sigma}$ | $M^{\Sigma}$ | Conditions
---|---|---
$E_{7}(q)$ | $(3,q+1).(^{2}E_{6}(q)\times\frac{q+1}{(3,q+1)}).(3,q+1).{2}$ |
$E_{6}(q)$ | $F_{4}(q)$ |
|  $(q^{2}+q+1\times$ ${}^{3}D_{4}(q)).3$ |
$F_{4}(q)$ | ${}^{3}D_{4}(q).3$ |
| ${}^{2}F_{4}(q)$ |
$G_{2}(q)$ | $SU_{3}(q):2$ |
| ${}^{2}G_{2}(q)$ | $q=3^{2e+1}>1$
| $G_{2}(2)$ | $q=5$
| $PSL_{2}(13)$ | $q=3,4$
| $2^{3}.SL_{3}(2)$ | $q=3$
${}^{2}B_{2}(q)$ | $13:4$ | $q=8$
${}^{3}D_{4}(q)$ | $G_{2}(q)$ |

$L^{\Sigma}$ is not isomorphic to any groups $G_{2}(3),G_{2}(5)$ and
${}^{2}B_{2}(8)$ since these do not have a transitive permutation
representation of square degree by [6]. Also, if $L^{\Sigma}\cong G_{2}(4)$
then $L_{\Delta}^{\Sigma}\cong PSL_{2}(13)$ again by [6] and hence
$\lambda=480$. Then $G(\Sigma)=1$ by Corollary 5.4, and hence
$PSL_{2}(13)\trianglelefteq G_{\Delta}^{\Delta}\leqslant PGL_{2}(13)$.
However, $\left|\Delta\right|=\lambda+2=482$ does not divide the order of
$G_{\Delta}^{\Delta}$, and we reach a contradiction.

Suppose that $\left|M^{\Sigma}\right|^{2}<\left|L^{\Sigma}\right|$. Then
$G(\Sigma)=1$, $L_{\Delta}^{\Delta}$ is solvable, $\left|L\right|\leqslant
4\left|\mathrm{Out}(L)\right|^{2}\left|L_{\Delta}^{\Delta}\right|^{2}$ and
$\left|L(\Delta)\right|<2\left|\mathrm{Out}(L)\right|$ by Theorem 5.5. Thus
$\left|M\right|^{2}<\left|L\right|\leqslant
4\left|\mathrm{Out}(L)\right|^{2}\left|L_{\Delta}\right|^{2}$, and hence
$\left|M:L_{\Delta}\right|<2\left|\mathrm{Out}(L)\right|$. In the remaining
admissible groups of Table 2 the last term of the derived series
$M^{(\infty)}$ of $M$ is non-abelian simple. Let $P(M^{(\infty)})$ be the
minimal primitive permutation representation of $M^{(\infty)}$. If
$M^{(\infty)}\not\leqslant L_{\Delta}$, then

$P(M^{(\infty)})\leqslant\left|M^{(\infty)}:L_{\Delta}\cap
M^{(\infty)}\right|\leqslant\left|M:L_{\Delta}\right|<2\left|\mathrm{Out}(L)\right|,$

and we reach a contradiction by Lemma 5.1 Therefore $M^{(\infty)}\leqslant
L_{\Delta}$, and hence $M^{(\infty)}\leqslant L(\Delta)$ since
$L_{\Delta}^{\Delta}$ is solvable. Then
$P(M^{(\infty)})\leqslant\left|L(\Delta)\right|<2\left|\mathrm{Out}(L)\right|$
and we again reach a contradiction by Lemma 5.1.

Suppose that $\left|L^{\Sigma}\right|\leqslant\left|M^{\Sigma}\right|^{2}$.
Then one of the following holds by [21]:

  1. (1)

$L^{\Sigma}\cong E_{7}(q)$ and
$L_{\Delta}^{\Sigma}=M^{\Sigma}\cong(3,q+1).(^{2}E_{6}(q)\times(q-1)/(3,q+1)).(3,q+1).2$;

  2. (2)

$L^{\Sigma}\cong E_{6}(q)$ and $L_{\Delta}^{\Sigma}=M^{\Sigma}\cong F_{4}(q)$;

  3. (3)

$L^{\Sigma}\cong F_{4}(q)$ and $L_{\Delta}^{\Sigma}=M^{\Sigma}\cong$
${}^{3}D_{4}(q).Z_{3}$;

  4. (4)

$L^{\Sigma}\cong G_{2}(q)$ and $L_{\Delta}^{\Sigma}=M^{\Sigma}\cong
SU_{3}(q):Z_{2}$.

Then $G(\Sigma)=1$ by Corollary 5.4. If $L_{\Delta}^{\Delta}$ does not act
$2$-transitively on $\Delta$, then $L_{\Delta}^{\Delta}$ is solvable by
Proposition 5.3, hence $\left|L(\Delta)\right|<2\left|\mathrm{Out}(L)\right|$
by Theorem 5.5. However, this is impossible in cases (1)–(4). Then
$L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$, and hence only (4)
occurs with $\lambda+2=q^{3}+1$. Then $\left|\Sigma\right|=(q^{3}-1)^{2}$,
whereas $L_{\Delta}^{\Sigma}$ is a maximal non-parabolic subgroup of
$L^{\Sigma}$. So this case is excluded, and the proof is completed. ∎

Now, it remains to analyze the case where $L^{\Sigma}$ is a simple classical
group.

######  Proposition 6.9.

$L_{\Delta}^{\Sigma}$ is a large maximal geometric subgroup of $L^{\Sigma}$.
Moreover, it results that
$\left(\Phi_{\zeta}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$.

###### Proof.

Recall that $L_{\Delta}^{\Sigma}$ is a large subgroup of $L^{\Sigma}$ such
that $\left(\Phi_{\zeta}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$
by Lemma 6.7. Let $M$ be a subgroup of $L$ such that $M^{\Sigma}$ is a maximal
subgroup of $L^{\Sigma}$ containing $L_{\Delta}^{\Sigma}$, then $M^{\Sigma}$
is large and $\left(\Phi_{nf}^{\ast}(p),\left|M^{\Sigma}\right|\right)>1$. If
$\left(L^{\Sigma},L_{\Delta}^{\Sigma}\right)$ is not
$\left(P\Omega^{+}_{8}(q),G_{2}(q)\right)$ and $\left(PSU_{4}(3),A_{7}\right)$
then we may use the same argument of [25, Theorem 7.1], with
$L^{\Sigma},L^{\Sigma}_{\Delta}$ and $M^{\Sigma}$ in the role of $X,X_{x}$ and
$Y$ respectively, to prove that $M^{\Sigma}$ is a geometric subgroup of
$L^{\Sigma}$.

Assume that $L^{\Sigma}\cong P\Omega^{+}_{8}(q)$ and $L_{\Delta}^{\Sigma}\cong
G_{2}(q)$. Then $G(\Sigma)=1$ by Corollary 5.4. If $L_{\Delta}^{\Delta}$ does
not act $2$-transitively on $\Delta$, then
$\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$, where $u^{h}=\lambda+2$.
Then $\lambda<f$, where $q=p^{f}$, and hence
$p^{2f}\leqslant\left|L:L_{\Delta}\right|<f^{2}$ a contradiction. Thus
$L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$, and hence $q=2$.
However, this is impossible by Lemma 6.6.

Assume that $L^{\Sigma}\cong PSU_{4}(3)$ and $L_{\Delta}^{\Sigma}\cong A_{7}$.
Then $\lambda=36$, and hence $G(\Sigma)=1$ by Corollary 5.4. Therefore, one
obtains $G\leqslant P\Gamma U_{4}(3)$. However, this is impossible by Lemma
2.1 since $\lambda+1=37$ does not divide the order of $G$. Thus, we have
proven that $M^{\Sigma}$ is a geometric subgroup of $L^{\Sigma}$ containing
$L_{\Delta}^{\Sigma}$.

If $L_{\Delta}^{\Sigma}\neq M^{\Sigma}$, then $G_{\Delta}^{\Sigma}$ is a
novelty, and hence $L_{\Delta}^{\Sigma}$ is listed in [17, Tables 3.5.H–I] for
$n\geqslant 13$ and in [5] for $3\leq n\leqslant 12$. Now, the candidates for
$L_{\Delta}^{\Sigma}$ must be large subgroups of $L^{\Sigma}$ and their order
must have a factor in common with $\Phi_{\zeta}^{\ast}(p)$. For instance, if
$L^{\Sigma}\cong PSL_{n}(q)$, $q=p^{f}$, then $\zeta=nf$ by [17, Proposition
5.2.16], and the unique admissible case is when $L_{\Delta}^{\Sigma}$ lies in
a maximal member of $\mathcal{C}_{1}(L^{\Sigma})$. However, this is impossible
by [12, Theorem 3.5(iv)]. Therefore, no novelties occur when $L^{\Sigma}\cong
PSL_{n}(q)$. The remaining simple classical groups are analyzed similarly and
it is straightforward to check that there are no novelties which are
compatible with the constraints on $L_{\Delta}^{\Sigma}$. Thus
$L_{\Delta}^{\Sigma}=M^{\Sigma}$, which is the assertion. ∎

######  Lemma 6.10.

$L^{\Sigma}$ is not isomorphic to $PSL_{n}(q)$.

###### Proof.

Assume that $L^{\Sigma}\cong PSL_{n}(q)$. Then $n\geqslant 3$ by Lemma 6.5,
and $L_{\Delta}^{\Sigma}$ is a large maximal geometric subgroup of
$L^{\Sigma}$ such that
$\left(\Phi_{nf}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$ by
Proposition 6.9 and by [17, Proposition 5.2.16]. Then
$L_{\Delta}^{\Sigma}\notin\mathcal{C}_{1}(L^{\Sigma})$ by [12, Theorem
3.5(iv)], and hence one of the following holds by [3, Propositions 4.7]:

  1. (i)

$\mathrm{Soc}(L_{\Delta}^{\Sigma})$ is one of the groups
$PSp_{n}^{\prime}(q)$, $PSU_{n}(q^{1/2})$ and $n$ odd, or
$P\Omega_{n}^{-}(q)$;

  2. (ii)

$L_{\Delta}^{\Sigma}$ is a $\mathcal{C}_{3}$-group of type $GL_{n/t}(q^{t})$,
where $t=2$, or $t=3$ and either $q=2,3$ or $q=5$ and $n$ is odd.

Assume that (i) holds. Then $G(\Sigma)=1$ by Corollary 5.4. If
$L_{\Delta}^{\Delta}$ does not act $2$-transitively on $\Delta$, then
$\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$, where $u^{h}=\lambda+2$ by
Proposition 5.3. If $\mathrm{Soc}(L_{\Delta}^{\Sigma})\cong PSU_{n}(q^{1/2})$
then $\lambda+2\leqslant q-1$ by [17, Proposition 4.8.5(II)], and hence
$q^{2}+q+1\leqslant P(L)\leqslant\left|\Sigma\right|\leqslant(q-3)^{2}$ since
$n\geqslant 3$, which is a contradiction. Thus, $L_{\Delta}^{\Delta}$ acts
$2$-transitively on $\Delta$ in this case. The previous argument together with
[17, Propositions 4.8.3–4.8.4] can be used to show that $L_{\Delta}^{\Delta}$
acts $2$-transitively on $\Delta$ also in the remaining cases. Furthermore,
$(n,q)\neq(4,2),(6,2)$ by Lemmas 6.4 and 6.9 respectively, and $n\geqslant 3$.
Then one of the following holds by [17, Propositions 4.8.3–4.8.5]:

  1. (1)

$n>6$, $q=2$, $L_{\Delta}\cong Sp_{n}(2)$ and $\lambda=2^{2n-1}\pm 2^{n}-2$;

  2. (2)

$n=3$, $\mathrm{Soc}(L_{\Delta})\cong PSU_{3}(q^{1/2})$ and
$\lambda=q^{3/2}-1$;

  3. (3)

$n=4$, $\mathrm{Soc}(L_{\Delta})\cong P\Omega_{4}^{-}(q)\cong PSL_{2}(q^{2})$
and $\lambda=q^{2}-1$.

Cases (2) and (3) are immediately ruled out since $(\lambda,p)=1$ but
$L_{\Delta}$ does not contain a Sylow $p$-subgroup of $L$. In case (1)
$2^{n(n-1)/2-2}$ must divide the order of a Sylow $2$-subgroup of $L_{\Delta}$
which is $2^{n^{2}/4}$, and we reach a contradiction since $n>6$.

Assume that (ii) holds. Then $L_{\Delta}^{\Sigma}\cong
Z_{a}.PSL_{n/t}(q^{t}).Z_{e}.Z_{t}$, where
$a=\frac{(q-1,n/t)(q^{t}-1)}{(q-1)(q-1,n)}$ and
$e=\frac{(q^{t}-1,n/t)}{(q-1,n/t)}$ by [17, Proposition 4.3.6.(II)], and
$t=2,3$. Then $G(\Sigma)=1$ by Corollary 5.4. Recall that
$\mathrm{Soc}(G_{\Delta}^{\Delta})\trianglelefteq L_{\Delta}^{\Delta}$ by
Proposition 5.3.

If $G_{\Delta}^{\Delta}$ is of affine type, then $L_{\Delta}^{\Delta}\leqslant
Z_{e}.Z_{t}$ for $n>t$. Thus $\lambda+2\leqslant e\leqslant n/t$, and hence

$\frac{q^{n}-1}{q-1}=P(L)\leqslant\left|L:L_{\Delta}\right|=\left|\Sigma\right|\leqslant
n^{2}/2$ (6.2)

by [17, Theorem 5.2.2] since $n\geqslant 3$ and $(n,q)\neq(4,2)$. However,
(6.2) has no admissible solutions. Thus $n=t=3$ since $t=2,3$ and $n\geqslant
3$. Moreover, $L_{\Delta}\cong Z_{\frac{q^{2}+q+1}{(3,q-1)}}.Z_{3}$ and hence
$\left|\Sigma\right|=\frac{1}{3}q^{3}(q+1)(q-1)^{2}$. On the other hand, since
$\lambda+2$ is a power of prime and divides the order of $L_{\Delta}$, it
follows that $\lambda\leqslant q^{2}+q-1$ and
$\left|\Sigma\right|\leqslant(q^{2}+q-1)^{2}$, and we reach a contradiction.

If $G_{\Delta}^{\Delta}$ is almost simple, then $t<n$ and
$L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$ by Proposition 5.3.
Thus $L_{\Delta}^{\Delta}\cong PSL_{n/t}(q^{t}).Z_{e}.Z_{t}$ and hence either
$(n/t,q^{t})=(2,9)$ and $\lambda+2=6$, or $\lambda+2=\frac{q^{n}-1}{q^{t}-1}$
since $t=2,3$. The former contradicts $\lambda>10$, the latter yields
$\lambda=\frac{q^{n}-2q^{t}+1}{q^{t}-1}$ and hence $\left(\lambda,p\right)=1$.
Then $L_{\Delta}^{\Sigma}$ must contain a Sylow $p$-subgroup of $L^{\Sigma}$,
which is a contradiction. ∎

######  Lemma 6.11.

$L^{\Sigma}$ is not isomorphic to $PSU_{n}(q)$.

###### Proof.

Assume that $L^{\Sigma}\cong PSU_{n}(q)$. Then $n\geqslant 3$ by Lemma 6.5.
Moreover, $L_{\Delta}^{\Sigma}$ is a large maximal geometric subgroup of
$L^{\Sigma}$ such that
$\left(\Phi_{\zeta}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$,
where $\zeta$ is either $nf$ or $(n-1)f$ according to whether $n$ is even or
odd respectively, by Proposition 6.9 and [17, Proposition 5.2.16]. Then
$L_{\Delta}^{\Sigma}\notin\mathcal{C}_{1}(L^{\Sigma})$ by [12, Theorem
3.5(iv)], and hence one of the following holds by [3, Propositions 4.17]:

  1. (i)

$L_{\Delta}^{\Sigma}$ is a $\mathcal{C}_{1}$-subgroup of $L^{\Sigma}$;

  2. (ii)

$L_{\Delta}^{\Sigma}$ is a $\mathcal{C}_{3}$-subgroup of $L^{\Sigma}$ of type
$GU_{n/3}(3^{3})$ with $n$ odd.

Assume that (i) holds. Then $n$ is even by [12, Theorem 3.5(iv)], and hence
$L_{\Delta}^{\Sigma}\cong Z_{\frac{q+1}{(q+1,n)}}.PSU_{n-1}(q).Z_{(q+1,n-1)}$
is the stabilizer of a non-isotropic point of $PG_{n-1}(q^{2})$ by [17,
Propositions 4.1.4.(II)-4.1.18.(II)]. Also, $G(\Sigma)=1$ by Corollary 5.4. If
$L_{\Delta}^{\Delta}$ does not act $2$-transitively on $\Delta$, then
$\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$ with $u^{h}=\lambda+2$ by
Proposition 5.3. Then $\lambda\leqslant(q+1,n-1)-2$ by [17, Propositions
4.1.4.(II)], and hence $q^{n-1}\leqslant\left|L:L_{\Delta}\right|<q$, which is
impossible for $n\geqslant 4$. Then $L_{\Delta}^{\Delta}$ acts
$2$-transitively on $\Delta$, and hence $n=4$ and $\lambda^{2}=q^{3}(q-1)$. So
$q-1$ is a square, which is impossible by [32, B1.1].

Assume that (ii) holds. Then $L_{\Delta}^{\Sigma}\cong
Z_{7}.PSU_{n/3}(3^{3}).Z_{(n/3,7)}.Z_{3}$ by [17, Proposition 4.3.6(II)].
Hence, $G(\Sigma)=1$ by Corollary 5.4. Also, $L_{\Delta}^{\Delta}$ is forced
to act $2$-transitively on $\Delta$ by Proposition 5.3 since $\lambda>10$.
Therefore, $n=9$ and $\lambda=3^{9}-2$. However,
$\left|L:L_{\Delta}\right|\neq\lambda^{2}$ in this case, which is then ruled
out. ∎

######  Lemma 6.12.

$L^{\Sigma}$ is not isomorphic to $PSp_{n}(q)^{\prime}$.

###### Proof.

Assume that $L^{\Sigma}\cong PSp_{n}(q)^{\prime}$. Then $n\geqslant 4$ by
Lemma 6.5 since $n$ is even. Also $(n,q)\neq(4,2)$ by Lemma 4.2 since
$PSp_{4}(2)^{\prime}\cong A_{6}$. Thus $L^{\Sigma}\cong PSp_{n}(q)$. Moreover,
$L_{\Delta}^{\Sigma}$ is a large maximal geometric subgroup of $L^{\Sigma}$
such that
$\left(\Phi_{nf}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$ by
Proposition 6.9 and [17, Proposition 5.2.16]. Then
$L_{\Delta}^{\Sigma}\notin\mathcal{C}_{1}(L^{\Sigma})$ by [12, Theorem
3.5(iv)], and hence one of the following holds by [3, Propositions 4.22]:

  1. (1)

$L_{\Delta}^{\Sigma}$ is a $\mathcal{C}_{8}$-subgroup of $L^{\Sigma}$;

  2. (2)

$L_{\Delta}^{\Sigma}$ is a $\mathcal{C}_{3}$-subgroup of $L^{\Sigma}$ of type
$Sp_{n/2}(q^{2})$, $Sp_{n/3}(q^{3})$ or $GU_{n/2}(q)$;

  3. (3)

$\left(L^{\Sigma},L_{\Delta}^{\Sigma}\right)$ is either
$(PSp_{4}(7),2^{4}.O_{4}^{-}(2))$ or $(PSp_{4}(3),2^{4}.\Omega_{4}^{-}(2))$.

Assume that Case (i) holds. Then $L_{\Delta}^{\Sigma}\cong
O_{n}^{\varepsilon}(q)$, with $\varepsilon=\pm$ and $q$ even, by [17,
Proposition 4.8.6.(II)]. Then
$\lambda^{2}=\frac{q^{n/2}}{2}(q^{n/2}+\varepsilon)$ since
$\lambda^{2}=\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|$, and hence
$q^{n/2}+\varepsilon$ is a square. Then $(n,q)=(6,2)$ by [32, B1.1] since
$n\geqslant 4$. However, $L^{\Sigma}\cong PSp_{6}(2)$ cannot occur by Lemma
6.6.

Assume the Case (ii) holds. Then either $L_{\Delta}^{\Sigma}$ is isomorphic
either to $PSp_{n/t}(q^{t}).Z_{t}$ with $t=2,3$ or
$Z_{(q+1)/2}.PGU_{n/2}(q).Z_{2}$ with $q$ odd by [17, Propositions 4.3.7(II)
and 4.3.10.(II)]. Also, in both cases it results $G(\Sigma)=1$ by Corollary
5.4.

If $L_{\Delta}^{\Delta}$ does not act $2$-transitively on $\Sigma$, then
$Z_{(q+1)/2}.PSU_{n/2}(q)\leqslant L(\Delta)$ and
$\left|L_{\Delta}^{\Delta}\right|\mid 4f(n/2,q+1)$ since $\lambda>10$. By [3,
Corollary 4.3(ii)–(iii)] we obtain

$p^{f\frac{n^{2}+2n}{4}}\leqslant\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|\leqslant
16f^{2}(n/2,q+1)^{2},$

which has no solutions for $q$ odd and $n\geqslant 4$.

If $L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Sigma$. Then
$L_{\Delta}^{\Sigma}$ is isomorphic either to $PSp_{2}(q^{t}).Z_{t}$ with
$t=2,3$ or $Z_{(q+1)/2}.PGU_{3}(q).Z_{2}$ with $q$ odd. Then either
$\lambda=q^{t}-1$ with $t=2,3$ or $q^{3}-1$ respectively. In each case
$(\lambda,q)=1$, and hence $L_{\Delta}^{\Sigma}$ must contain a Sylow
$p$-subgroup of $L^{\Sigma}$, which is a contradiction.

Finally, Case (iii) cannot occur since
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|$ is a non-square. ∎

######  Lemma 6.13.

$L^{\Sigma}$ is not isomorphic to simple classical group.

###### Proof.

In order to prove the assertion we only need to tackle the case
$L^{\Sigma}\cong P\Omega_{n}^{\varepsilon}(q)$, where
$\varepsilon\in\left\\{\circ,\pm\right\\}$ since the remaining classical
groups are ruled out in Lemmas 6.10, 6.11 and 6.12. Hence, assume that
$L^{\Sigma}\cong P\Omega_{n}^{\varepsilon}(q)$, where
$\varepsilon\in\left\\{\circ,\pm\right\\}$. Then $n\neq 4$ by Lemma 6.5.
Moreover $n\neq 5,6$ by Lemmas 6.10, 6.11 and 6.12, since
$P\Omega_{5}^{\circ}(q)\cong PSp_{4}(q)$ with $q$ odd,
$P\Omega_{6}^{+}(q)\cong PSL_{4}(q)$ and $P\Omega_{6}^{-}(q)\cong PSL_{6}(q)$.
Thus $n\geqslant 7$. Moreover $(n,q,\varepsilon)\neq(8,2,+)$ by Lemma 6.6.
Finally, $L_{\Delta}^{\Sigma}$ is a large maximal geometric subgroup of
$L^{\Sigma}$ such that
$\left(\Phi_{\zeta}^{\ast}(p),\left|L_{\Delta}^{\Sigma}\right|\right)>1$ by
Proposition 6.9, where $\zeta$ is either $nf$, $(n-1)f$ or $(n-2)f$ by [17,
Proposition 5.2.16] according to whether $\varepsilon$ is $-,\circ$ or $+$
respectively. Thus, one of the following holds by [3, Proposition 4.23]:

  1. (i)

$L_{\Delta}^{\Sigma}$ is $\mathcal{C}_{1}$-subgroup of $L^{\Sigma}$;

  2. (ii)

Either $(n,q)=(7,3),(7,5)$ or $(n,q,\varepsilon)=(8,3,+)$ and
$L_{\Delta}^{\Sigma}$ is of type $O_{1}(q)\wr S_{n}$;

  3. (iii)

$L_{\Delta}^{\Sigma}$ is a $\mathcal{C}_{3}$-subgroup of $L^{\Sigma}$.
Moreover, its type is either $O_{n/2}^{\varepsilon^{\prime}}(q^{2})$ with
$(\varepsilon,\varepsilon^{\prime})=(-,-)$ and $n/2$ even or
$(\varepsilon,\varepsilon^{\prime})=(+,\circ)$ and $n/2$ odd, or $GU_{n/2}(q)$
with $\varepsilon=-$ and $n/2$ odd or $\varepsilon=+$ and $n/2$ even;

  4. (iv)

$L^{\Sigma}\cong P\Omega^{+}_{8}(3)$ and $L_{\Delta}^{\Sigma}\cong
2^{6}.\Omega_{6}^{+}(2)$.

Assume that (i) holds. Then one of the following cases occurs by [17,
Propositions 4.1.6(II), 4.1.7(II) and 4.1.20(II)]:

  1. (1)

$L_{\Delta}^{\Sigma}$ is the stabilizer in $L^{\Sigma}$ of a non-singular
point of $PG_{n-1}(q)$:

    1. (a)

$\varepsilon=\circ$ and $L_{\Delta}^{\Sigma}\cong\Omega_{n-1}^{-}(q).Z_{2}$

    2. (b)

$\varepsilon=+$ and $L_{\Delta}^{\Sigma}\cong\Omega_{n-1}(q)$ with $q\equiv
1\pmod{4}$, or $q\equiv 3\pmod{4}$ and $n/2$ even;

    3. (c)

$\varepsilon=+$ and $L_{\Delta}^{\Sigma}\cong\Omega_{n-1}(q).Z_{2}$ with
$q\equiv 3\pmod{4}$ and $n/2$ odd;

    4. (d)

$\varepsilon=+$ and $L_{\Delta}^{\Sigma}\cong Sp_{n-2}(q)$ with $q$ even.

  2. (2)

$\varepsilon=+$ and $L_{\Delta}^{\Sigma}$ is the stabilizer in $L^{\Sigma}$ of
a non-singular line of type “$-$” of $PG_{n-1}(q)$:

    1. (a)

$L_{\Delta}^{\Sigma}\cong\left(Z_{\frac{q+1}{(q+1,2)}}\times\Omega_{n-2}^{-}(q)\right).Z_{2}$
with $q$ even or $q\equiv 1\pmod{4}$;

    2. (b)

$L_{\Delta}^{\Sigma}\cong\left(Z_{\frac{q+1}{2}}\times\Omega_{n-2}^{-}(q)\right).[4]$
with $q\equiv 3\pmod{4}$ and $n/2$ odd;

    3. (c)

$L_{\Delta}^{\Sigma}\cong Z_{2}.\left(Z_{\frac{q+1}{4}}\times
P\Omega_{n-2}^{-}(q)\right).[4]$ with $q\equiv 3\pmod{4}$ and $n/2$ even.

In each case $G(\Sigma)=1$ by Corollary 5.4 since $n\geqslant 7$. Hence,
$L=L^{\Sigma}$ and $G=G^{\Sigma}$. Moreover, $L_{\Delta}^{\Delta}$ acts
$2$-transitively on $\Delta$ by Proposition 5.3 since $\lambda>10$. Then
$\varepsilon=+$, $q=2$, $L_{\Delta}^{\Sigma}\cong Sp_{n-2}(2)$ and
$\lambda=2^{2(n/2-1)}\pm 2^{n/2-2}-2$. On the other hand,

$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|=\lambda^{2}=2^{n/2-1}\left(2^{n/2}-1\right)$

and hence $2^{2}(2^{2(n/2-1)-1}\pm
2^{n/2-1}-1)^{2}=2^{n/2-1}\left(2^{n/2}-1\right)$, which has no admissible
integer solutions for $n\geqslant 8$.

Assume that (iii) holds. The possibilities for $L_{\Delta}^{\Sigma}$ are
provided in [17, Propositions 4.3.16(II), 4.3.18(II) and 4.3.20(II)]. Thus
$G(\Sigma)=1$ by Corollary 5.4. If $L_{\Delta}^{\Delta}$ does not act
$2$-transitively on $\Delta$, then
$\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}\leqslant
G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$ with $u^{h}=\lambda+2$ by
Proposition 5.3. Thus $L_{\Delta}$ is forced to be of type $GU_{n/2}(q)$ with
$\left|L_{\Delta}^{\Delta}\right|\mid(n/2,2,q)(q+1,n/2)$ since $\lambda>10$.
Then $\lambda\leqslant n-2$ and so
$q^{\frac{1}{8}n\left(n+2\right)}\leqslant\left|L:L_{\Delta}\right|\leqslant(n-2)^{2}$,
which has no solutions for $n\geqslant 8$. Thus $L_{\Delta}^{\Delta}$ acts
$2$-transitively on $\Delta$, and hence $L\cong P\Omega^{-}_{8}(q)$ and
$L_{\Delta}\cong P\Omega^{-}_{4}(q).Z_{4}\cong PSL_{2}(q^{4}).Z_{4}$ since
$n\geqslant 8$. Therefore $\lambda=q^{4}-2$. If $q$ is odd then $L_{\Delta}$
must contain a Sylow $p$-subgroup of $L$ since
$\left|L:L_{\Delta}\right|=\lambda^{2}$, which is not the case. So $q$ is even
and $\left|\Sigma\right|=q^{12}(q^{6}-1)(q^{2}-1)$ which is different from
$(q^{4}-2)^{2}$.

Finally, it is easy to check that
$\left|L^{\Sigma}:L_{\Delta}^{\Sigma}\right|$ is a non-square in (ii) and
(iv), hence these are ruled out. This completes the proof. ∎

###### Proof of Theorem 6.1.

Since $L^{\Sigma}$ is almost simple by Theorem 4.1, the assertion follows from
Lemmas 6.2, 6.4, 6.8 or 6.13 respectively. ∎

##  7\. Classification of the $2$-designs of type 2

In this section we assume that $\mathcal{D}$ is of type $2$. Recall that
$G(\Sigma)=1$ and $G$ is an almost simple group acting point-quasiprimitively
on $\mathcal{D}$ by Theorem 4.1. Thus, $G=G^{\Sigma}$ and $L=L^{\Sigma}$ where
$L=\mathrm{Soc}(G)$. Further constraints for $L$ are provided in Proposition
5.3 and Theorem 5.5 which are then combined with the results contained in [3,
20]. An important restriction is provided in Lemma 7.5 where it is proven
that, if $L$ is Lie type simple group, either $L_{\Delta}$ lies in a maximal
parabolic subgroup of $L$ or $L_{\Delta}^{\Delta}$ is a non-solvable group
acting $2$-transitively on $\Delta$. We use all these information to prove the
following result.

######  Theorem 7.1.

If $\mathcal{D}$ is a symmetric
$2$-$\left(\left(\frac{\lambda+2}{2}\right)\left(\frac{\lambda^{2}-2\lambda+2}{2}\right),\frac{\lambda^{2}}{2},\lambda\right)$
design admitting a flag-transitive and point-imprimitive automorphism group,
then $\lambda\leqslant 10$.

We analyze the cases where $L^{\Sigma}$ is sporadic, alternating, classical or
exceptional of Lie type separately.

Recall that, when $\mathcal{D}$ is of type (2) either $\lambda\equiv
0\pmod{4}$ and hence $\left|\Delta\right|=\lambda/2+1$ is odd, or
$\lambda=2w^{2}$, where $w$ is odd, $w\geqslant 3$, $2(w^{2}-1)$ is a square
and $\left|\Delta\right|=w^{2}+1$. In both cases it results that
$\left|\Delta\right|\not\equiv 0\pmod{4}$.

A preliminary filter in the study of the the $2$-designs of type 2 is the
following lemma.

######  Lemma 7.2.

If $\mathcal{D}$ is of type 2, then the following hold:

  1. (1)

$\left|\Sigma\right|$ is odd and $2\left|\Sigma\right|-1$ is a square.

  2. (2)

If $u$ is any prime divisor of $\left|\Sigma\right|$, then $u\equiv
1\pmod{4}$.

  3. (3)

If $\lambda=2w^{2}$, $w$ odd, $w\geqslant 3$ and such that $2(w^{2}-1)$ is a
square, then $\mathrm{Soc}(G_{\Delta}^{\Delta})$ is isomorphic to one of the
groups $A_{w^{2}+1}$, $PSL_{2}(w^{2})$ or $PSU_{3}(w^{2/3})$.

###### Proof.

$\left|\Sigma\right|$ is clearly odd. If $\lambda=2w^{2}$, where $w$ is odd,
$w\geqslant 3$, and $2(w^{2}-1)=x^{2}$ then
$2w^{4}-2w^{2}t+\left(1-\left|\Sigma\right|\right)=0$, whereas if $\lambda=4t$
for some $t\geqslant 1$, then
$16t^{2}-8t+\left(2-2\left|\Sigma\right|\right)=0$. In both cases
$y^{2}=2\left|\Sigma\right|-1$ for some positive integer $y$. Therefore,
$\left|\Sigma\right|=\frac{y^{2}+1}{2}$ is odd. Moreover, if $u$ is any prime
divisor of $\left|\Sigma\right|$ then $y^{2}\equiv-1\pmod{u}$ and hence
$u\equiv 1\pmod{4}$. Thus, we obtain (1) and (2).

Finally, (3) follows from the first part of the proof of Lemma 3.6. ∎

######  Lemma 7.3.

$L$ is not isomorphic to a sporadic group.

###### Proof.

Assume that $L$ is sporadic. Then the possibilities for $G$ and $G_{x}$, where
$x$ is any point of $\mathcal{D}$, and hence for
$\left|\Sigma\right|=\left|G:G_{x}\right|$ are provided in [2]. It is easy to
see that $2\left|\Sigma\right|-1$ is never square if $\left|\Sigma\right|$ is
any of such degrees. Thus $L$ cannot be a sporadic simple group by Lemma
7.2(2). ∎

######  Lemma 7.4.

If $L$ is not isomorphic to $A_{s}$ with $s\geqslant 5$.

###### Proof.

Assume that $L\cong A_{s}$, where $s\geqslant 5$. Then one of the following
holds by [20]:

  1. (1)

$G_{\Delta}\cong\left(S_{t}\times S_{s-t}\right)\cap G$, $1\leqslant t<s/2$;

  2. (2)

$G_{\Delta}\cong(S_{s/t}\wr S_{t})\cap G$, $s/t,t>1$;

  3. (3)

$G_{\Delta}\cong A_{7}$ and $\left|\Sigma\right|=15$.

Assume that (i) holds. Then $\left|\Sigma\right|=\binom{s}{t}$ and
$A_{t}\times A_{s-t}\trianglelefteq L_{\Delta}\leqslant\left(S_{t}\times
S_{s-t}\right)\cap L$. Suppose that $t\geqslant 5$. Then $s-t>t\geqslant 5$.
Hence, both $A_{t}$ and $A_{s-t}$ are simple groups. Moreover,
$L_{\Delta}^{\Delta}$ is either $A_{t}$ or $A_{s-t}$ by Proposition 5.3 since
$\lambda>10$. If $L_{\Delta}^{\Delta}\cong A_{t}$ then either
$\left|\Delta\right|=t$, or $t=6$ and $\left|\Delta\right|=10$, or $t=7,8$ and
$\left|\Delta\right|=15$ since $t\geqslant 5$. Actually, $t=6$ and
$\left|\Delta\right|=10$ imply $\lambda=2w^{2}=18$ and
$\left|\Sigma\right|=145$, which is not of the form $\binom{s}{6}$ and hence
it cannot occur. Also, if $t=7,8$ and $\left|\Delta\right|=15$ then
$\binom{s}{t}=\left|\Sigma\right|=365$, and we reach a contradiction. Thus,
$\lambda=2(t-1)$, and hence $\left|\Sigma\right|=\allowbreak 2t^{2}-6t+5$.
Therefore, we have

$2^{t}<\left(\frac{s}{t}\right)^{t}\leqslant\binom{s}{t}=2t^{2}-6t+5\text{,}$

which is impossible for $t\geqslant 5$. We reach the same contradiction for
$L_{\Delta}^{\Delta}\cong A_{s-t}$.

Assume that $1\leqslant t\leqslant 4$. If $L_{\Delta}^{\Delta}$ is non-
solvable, then $s-t\geqslant 5$ and $L_{\Delta}^{\Delta}\cong A_{s-t}$ and the
previous argument rules out this case. Thus $L_{\Delta}^{\Delta}$ is solvable,
and hence $\left|\Delta\right|=3$ by Proposition 5.3 since
$L_{\Delta}\leqslant\left(S_{t}\times S_{s-t}\right)\cap L$ and
$\left|\Delta\right|$ is odd by Lemma 7.2(3). Then $\lambda=4$, whereas
$\lambda>10$ by our assumptions.

Assume that (ii) holds. Then $\left|\Sigma\right|=\frac{s!}{((s/t)!)^{t}(t!)}$
and $A_{s/t}\wr A_{t}\leqslant L_{\Delta}\leqslant\left(S_{s/t}\wr
S_{t}\right)\cap L$. Moreover, $\left(A_{s/t}\right)^{t}\trianglelefteq
L_{\Delta}$ and $A_{t}\leqslant
L_{\Delta}/\left(A_{s/t}\right)^{t}\leqslant(Z_{2})^{t}:S_{t}$, where the
action of $A_{t}$ on $(Z_{2})^{t}$ and on its permutation module are
equivalent. Thus $L_{\Delta}/\left(A_{s/t}\right)^{t}$ is isomorphic to one of
the groups $A_{t}$, $S_{t}$, $Z_{2}\times A_{t}$, $Z_{2}\times S_{t}$,
$(Z_{2})^{t-1}(2):A_{t}$, $(Z_{2})^{t-1}(2):S_{t}$, $(Z_{2})^{t}:A_{t}$ or
$(Z_{2})^{t}:S_{t}$ by [17, Lemma 5.3.4]. If $t\leqslant 4$ then
$L_{\Delta}^{\Delta}$ is solvable, and hence
$\lambda/2+1=\left|\Delta\right|=3$ by Proposition 5.3 since
$\left|\Delta\right|$ is odd by Lemma 7.2(3). However, this is impossible
since $\lambda>10$. Thus $t\geqslant 5$. Also, $L_{\Delta}^{\Delta}$ is non-
solvable otherwise we reach a contradiction as above. Then
$L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$ by Proposition 5.3.
Therefore $L_{\Delta}^{\Delta}\cong A_{t}$, and hence either
$\left|\Delta\right|=t$ for $t\geqslant 5$, or $t=6$ and
$\left|\Delta\right|=10$, or $t=7,8$ and $\left|\Delta\right|=15$. On the
other hand,
$2^{t-1}\leqslant((s/t)!)^{t-1}<\frac{s!}{((s/t)!)^{t}(t!)}<2t^{2}$ as shown
in [24, (34)]. Thus $(t,s)=(3,6),(5,10),(7,14)$ and hence
$\left|\Sigma\right|=15$, $945$, $135135$, which are ruled out since they
violate Lemma 7.2(1). Then either $t=6$, $\left|\Delta\right|=15$ and
$\left|\Sigma\right|=41$, or $t=7,8$, $\left|\Delta\right|=15$ and
$\left|\Sigma\right|=365$. However both cases cannot occur since
$\frac{(2t)!}{(2)^{t}(t!)}>\left|\Sigma\right|$.

Finally, (iii) is excluded by Lemma 7.2(1), as $2\left|\Sigma\right|-1$ is not
a square. ∎

######  Lemma 7.5.

The following hold:

  1. (1)

If either $p\mid\lambda-\mu$ for some $\mu\in\left\\{0,1,2\right\\}$, or
$p\mid\lambda-3$ and $p\neq 5$, then $L_{\Delta}$ lies in a maximal parabolic
subgroup of $L$.

  2. (2)

If $L_{\Delta}$ does not lie in a maximal parabolic subgroup of $L$ then
$L_{\Delta}^{\Delta}$ is a non-solvable group acting $2$-transitively on
$\Delta$.

###### Proof.

Since $\left|\Sigma\right|=\frac{\lambda^{2}-2\lambda+2}{2}$ it is immediate
to see that $\left(\left|\Sigma\right|,\lambda-\mu\right)=1$ for either
$\mu=0,1,2$, or $\mu=3$ and $p\neq 5$. In these cases $L_{\Delta}$ contains a
Sylow $p$-subgroup of $L$, and hence $L_{\Delta}$ lies in a maximal parabolic
subgroup of [33, Theorem 1.6] since $L$ is a non-abelian simple group acting
transitively on $\Sigma$, and (1) holds.

Suppose to the contrary that $L_{\Delta}$ does not lie in a maximal parabolic
subgroup of $L$ and that $L_{\Delta}^{\Delta}$ is solvable. Then
$p\mid\left|\Sigma\right|$ by [33, Theorem 1.6]. Also $p\neq 2,3$ by Lemma
7.2(1)–(2), and
$\mathrm{Soc}(G_{\Delta}^{\Delta})<L_{\Delta}^{\Delta}<G_{\Delta}^{\Delta}\leqslant
A\Gamma L_{1}(u^{h})$, where $\frac{\lambda}{2}+1=\left|\Delta\right|=u^{h}$
for some prime $u$ by Proposition 5.3. Also, $u$ is odd since either
$\lambda\equiv 0\pmod{4}$, or $\lambda=2w^{2}$ with $w$ odd and $w\geq 3$ by
Theorem 2.2. Then, $\lambda=2(u^{h}-1)$ with $h>1$ since $\lambda>10$.

If $p=5$ divides $\lambda-3$, then $u=p$ since
$\left(\left|\Sigma\right|,\left|\Delta\right|\right)=\left(\left|\Sigma\right|,\lambda-3\right)=\left(\left|\Delta\right|,\lambda-3\right)\mid
5$. Thus $\left|\Sigma\right|=\allowbreak 2\cdot 5^{2h}-6\cdot 5^{h}+5$, and
$5^{2}\nmid\left|\Sigma\right|$ since $h>1$. Hence, $L_{\Delta}$ contains a
subgroup of index $p$ of a Sylow $p$-subgroup of $L$ with $p=5$. It is a
straightforward check that none of the groups listed in [20] fulfills the
previous constraint. So this case is excluded.

If $p\nmid\lambda-\mu$ for some $\mu\in\left\\{0,1,2,3\right\\}$. Then
$\left(p,\left|L(\Delta)\right|\right)=1$ by Corollary 2.7(2) since $p\neq
2,3$. Thus $\left|L_{\Delta}\right|_{p}=\left|L_{\Delta}^{\Delta}\right|_{p}$,
and hence $p\mid\frac{\lambda}{2}\left(\frac{\lambda}{2}+1\right)h$ since
$L_{\Delta}^{\Delta}<G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$.
Actually, $p\nmid\frac{\lambda}{2}$ by our assumption. If
$p\mid\frac{\lambda}{2}+1$, then $p=5$ divides $\lambda-3$ since
$\left(\left|\Sigma\right|,\left|\Delta\right|\right)=\left(\left|\Sigma\right|,\lambda-3\right)=\left(\left|\Delta\right|,\lambda-3\right)\mid
5$ which we saw being impossible. Therefore, $\left|L_{\Delta}\right|_{p}\mid
h$ and any Sylow $p$-subgroup of $L_{\Delta}$ is cyclic since
$L_{\Delta}^{\Delta}<G_{\Delta}^{\Delta}\leqslant A\Gamma L_{1}(u^{h})$.
However, this is impossible by [20]. Thus $L_{\Delta}^{\Delta}$ is non-
solvable, and hence $L_{\Delta}^{\Delta}$ acts $2$-transitively on $\Delta$ by
Proposition 5.3. ∎

######  Lemma 7.6.

$L$ is not a simple exceptional group of Lie type.

###### Proof.

Assume that $L_{\Delta}$ is parabolic. Then $L\cong E_{6}(q)$ and
$L_{\Delta}\cong[q^{16}].d.(P\Omega_{10}^{+}(q)\times(q-1)/ed).d$, where
$d=(2,q-1)$ e $e=(3,q-1)$ by [20, Table 1]. If $L_{\Delta}^{\Delta}$ is
solvable then the order of $L_{\Delta}^{\Delta}$ must divide $(q-1)/e$. So
does $\left|\Delta\right|$, and hence

$\frac{q^{9}-1}{q-1}\cdot(q^{8}+q^{4}+1)=\left|\Sigma\right|\leqslant
2\left|\Delta\right|^{2}\leqslant 2\frac{(q-1)^{2}}{e^{2}},$

which is clearly impossible. Thus $L_{\Delta}^{\Delta}$ is non-solvable acting
$2$-transitively on $\Delta$ by Proposition 5.3, and this is impossible too.

Assume that $L_{\Delta}$ is not parabolic. Then $L_{\Delta}^{\Delta}$ is non-
solvable acting $2$-transitively on $\Delta$ by Lemma 7.5 with
$\left|\Delta\right|$ odd. Also,
$\left|L\right|\leqslant\left|L_{\Delta}\right|^{2}$ by Theorem 5.5. All these
constraints together with [20, Table 1] lead to the following admissible
cases:

  1. (1)

$L\cong E_{7}$ and $\mathrm{Soc}(L_{\Delta}^{\Delta})\cong PSL_{2}(q)$;

  2. (2)

$L\cong$ ${}^{3}D_{4}(q)$ and $\mathrm{Soc}(L_{\Delta}^{\Delta})$ is
isomorphic to one of the groups $PSL_{2}(q^{j})$ with $j=1$ or $3$,
$PSL_{3}(q)$ or $PSU_{3}(q)$;

  3. (3)

$L\cong$ ${}^{2}G_{2}(q)$, $q=3^{2m+1}$, $m\geqslant 1$, and
$\mathrm{Soc}(L_{\Delta}^{\Delta})\cong PSL_{2}(q)$.