# Classification of the non-trivial $2$-$(k^{2},k,\lambda)$ designs, with
$\lambda\mid k$, admitting a flag-transitive almost simple automorphism group

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

Non-trivial $2$-$(k^{2},k,\lambda)$ designs, with $\lambda\mid k$, admitting a
flag-transitive almost simple automorphism group are classified.

###### Key words and phrases:

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

###### 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 $2<k<v$. 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. A
_flag_ of D is a pair $(x,B)$ where $x$ is a point and $B$ is a block
containing $x$. If $G\leqslant\mathrm{Aut(\mathcal{D})}$ acts transitively on
the set of flags of $\mathcal{D}$, then we say that $G$ is _flag-transitive_
and that $\mathcal{D}$ is a _flag-transitive design_.

The $2$-$(v,k,\lambda)$ designs $\mathcal{D}$ admitting a flag-transitive
automorphism group $G$ have been widely studied by several authors. In 1990, a
classification of those with $\lambda=1$ and $G\nleq A\Gamma L_{1}(q)$ was
announced by Buekenhout, Delandtsheer, Doyen, Kleidman, Liebeck and Saxl in
[19] and proven in [18], [22], [23], [24], [38], [41] and [50]. Since then a
special attention was given to the case $\lambda>1$. A classification of the
flag-transitive $2$-designs with $\gcd(r,\lambda)=1$, $\lambda>1$ and $G\nleq
A\Gamma L_{1}(q)$, where $r$ is the replication number of $\mathcal{D}$, has
been announced by Alavi, Biliotti, Daneshkakh, Montinaro, Zhou and their
collaborators in [2] and proven in [3], [4], [5], [9], [10],[8], [16], [17],
[42], [52], [55], [56], [57], [58], [59], [60] and [61]. Moreover, recently
the flag-transitive $2$-designs with $\lambda=2$ have been investigated by
Devillers, Liang, Praeger and Xia in [26], where it is shown that apart from
the two known symmetric $2$-$(16,6,2)$ designs, $G$ is primitive of affine or
almost simple type. Moreover, a classification is provided when the socle of
$G$ is isomorphic to $PSL_{n}(q)\trianglelefteq G$ and $n\geqslant 3$.

The investigation of the flag-transitive $2$-$(k^{2},k,\lambda)$ designs, with
$\lambda\mid k$, has been recently started in [44]. The reason of studying
such $2$-designs is that they represent a natural generalization of the affine
planes in terms of parameters, and also because, it is shown in [45] that, the
blocks of imprimitivity of a family of flag-transitive, point-imprimitive
symmetric $2$-designs investigated in [48] have the structure of the
$2$-designs analyzed here. It is shown in [44] that, apart from the smallest
Ree group, a flag-transitive automorphism group $G$ of a
$2$-$(k^{2},k,\lambda)$ design $\mathcal{D}$, with $\lambda\mid k$, is either
an affine group or an almost simple classical group. Moreover, when
$G\cong~{}^{2}G_{3}(3)$, the smallest Ree group, $\mathcal{D}$ is isomorphic
either to the $2$-$(6^{2},6,2)$ design or to one of the three
$2$-$(6^{2},6,6)$ designs constructed in [44]. All the four $2$-designs have
the $36$ secants of a non-degenerate conic $\mathcal{C}$ of $PG_{2}(8)$ as a
point set and $6$-sets of secants in a remarkable configuration as a block
set. Clearly, $G\cong~{}^{2}G_{3}(3)$ is a special case of an almost simple
classical group, since $~{}^{2}G_{3}(3)\cong P\Gamma L_{2}(8)$.

The result contained in the present paper, together with that obtained in
[44], is a complete classification of $(\mathcal{D},G)$ when $G$ is almost
simple. More precisely, the following result is obtained:

######  Theorem 1.1.

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

  1. (1)

$\mathcal{D}$ is isomorphic to the $2$-$(6^{2},6,2)$ design constructed in
[44] and $PSL_{2}(8)\trianglelefteq G\leqslant P\Gamma L_{2}(8)$.

  2. (2)

$\mathcal{D}$ is isomorphic to one of the three $2$-$(6^{2},6,6)$ designs
constructed in [44] and $G\cong P\Gamma L_{2}(8)$.

  3. (3)

$\mathcal{D}$ is isomorphic to the $2$-$(12^{2},12,3)$ design constructed in
Example 2.1 and $G\cong PSL_{3}(3)$.

  4. (4)

$\mathcal{D}$ is isomorphic to the $2$-$(12^{2},12,6)$ design constructed in
Example 2.1 and $G\cong PSL_{3}(3):Z_{2}$.

A complete classification of $(\mathcal{D},G)$, with $G$ of affine type and
$G\nleq A\Gamma L_{1}(q)$, is contained in [43].

The proof of Theorem 1.1 is outlined as follows. In Lemma 3.2 it is shown that
$G$ acts point-primitively on $\mathcal{D}$ and the point-stabilizer $G_{x}$
is a large (maximal) subgroup of $G$. In Theorem 3.3 and in Lemma 3.4 it
proven that, if $G$ is non-abelian simple, $G$ is classical and
$\lambda\geqslant 2$ respectively. Then we use the results contained in [15],
[13] and [32] to show $Soc(G)_{x}$ is a large subgroup of $Soc(G)$. Finally,
we complete the proof by using the classification of the large maximal
subgroups of simple classical groups contained in [1] together with some
results arising from the geometry of the classical groups.

More details about the proof strategy are provided at the beginning of each
section.

##  2\. Examples

All the details about the examples of $2$-designs corresponding to cases (1)
and (2) of Theorem 1.1 are contained in [44]. Hence, in this small section we
focus on the constructions of the two examples of $2$-designs corresponding to
cases (3) and (4). The tools used for the constructions are the results about
tactical configurations contained in [25], [35], combined with some group-
theoretical information about $\mathrm{Aut(}PSL_{3}(3)\mathrm{)}$ contained in
[20].

Let $PSL_{3}(3)\trianglelefteq G\leqslant
PSL_{3}(3):\left\langle\sigma\right\rangle$, where $\sigma$ is the symmetric
polarity of $PG_{2}(3)$ defining the conic $\mathcal{C}:$ $XZ-Y^{2}=0$. Let
$P$ be the subgroup of $G^{\prime}$ generated by the following elements:

$\eta=\left(\begin{array}[]{ccc}1&1&-1\\\ 0&1&1\\\
1&1&0\end{array}\right)\;\textit{ and
}\;\psi=\left(\begin{array}[]{ccc}1&1&-1\\\ 0&1&-1\\\
0&0&1\end{array}\right)\text{.}$

It is easy to see that, $P$ is a Frobenius group of order $39$ with kernel
$\left\langle\eta\right\rangle$ and complement
$\left\langle\psi\right\rangle$. Then $N_{G}(P)$ is either $P$, or
$P:\left\langle\sigma\right\rangle$, according to whether $G$ is either
$PSL_{3}(3)$, or $PSL_{3}(3):\left\langle\sigma\right\rangle$, respectively.

Let $L=\left\langle\alpha,\gamma\right\rangle$, where

$\alpha=\left(\begin{array}[]{ccc}0&0&1\\\ 0&-1&0\\\
1&0&0\end{array}\right)\;\textit{ and
}\;\gamma=\left(\allowbreak\begin{array}[]{ccc}0&0&1\\\ 0&-1&-1\\\
1&-1&1\end{array}\right)\text{.}$

Then $L\cong\Omega_{3}(3)\cong A_{4}$ lies in the stabilizer in $PSL_{3}(3)$
of $\mathcal{C}$.

Let $\mathcal{D}=(\mathcal{P},\mathcal{B},\mathcal{I})$ be the incidence
structure, where $\mathcal{P}$ and $\mathcal{B}$ are the sets of the right
cosets of the subgroups $N_{G}(P)$ and $L$ in $G$, respectively, and where

$\mathcal{I}=\left\\{(N_{G}(P)x,Ly)\in\mathcal{P}\times\mathcal{B}:Px\cap
Ly\neq\varnothing\right\\}.$

######  Example 2.1.

The following hold:

  1. (1)

$\mathcal{D}$ is a $2$-$(12^{2},12,3)$ design admitting $G\cong PSL_{3}(3)$ as
the unique flag-transitive automorphism group.

  2. (2)

$\mathcal{D}$ is a $2$-$(12^{2},12,6)$ design admitting $G\cong
PSL_{3}(3):\left\langle\sigma\right\rangle$ as the unique flag-transitive
automorphism group.

###### Proof.

Assume that $G\cong PSL_{3}(3)$. Then $P$ is maximal in $G$ by [20], and hence
$N_{G}(\left\langle\eta\right\rangle)=P$. Then the incidence structure
$\mathcal{D}=(\mathcal{P},\mathcal{B},\mathcal{I})$ above defined is flag-
transitive with parameters $(v,b,k,r)=(144,468,12,39)$ by [35], Lemmas 1 and
2.

The group $P$ fixes exactly one element of $\mathcal{P}$, namely itself, and
$\left\langle\eta\right\rangle$ acts semiregularly on
$\mathcal{P-}\left\\{P\right\\}$. It follows from [20] that
$N_{G}(\left\langle\psi\right\rangle)\cong D_{18}$. Hence, the number of
elements in $\mathcal{P}$ fixed by $\left\langle\psi\right\rangle$ is $6$.
Since $P$ is a Frobenius group and since $\left\langle\eta\right\rangle$ acts
semiregularly on $\mathcal{P-}\left\\{P\right\\}$, the five points fixed by
$\left\langle\psi\right\rangle$ and distinct from $P$ lie in five distinct
$\left\langle\eta\right\rangle$-orbits by [47], Proposition 4.2. Thus
$\mathcal{P}=\left\\{Px:x\in G\right\\}$ is partitioned in one $P$-orbit of
length $1$, five ones of length $13$ and two ones of length $39$.

Each $P$-orbit is of the form $PyP$ with $y\in G$. If we set
$\beta=\alpha^{\gamma}$, then $\left\langle\alpha,\beta\right\rangle\cong
E_{4}$, and it is not difficult to check that

$P\alpha P\cap
L=\left\\{\alpha,\gamma^{\beta},\left(\gamma^{\beta}\right)^{-1}\right\\},\;P\beta
P\cap
L=\left\\{\beta,\gamma^{\alpha},\left(\gamma^{\alpha}\right)^{-1}\right\\}\text{
and }\;PxP\cap L=\left\\{x\right\\}$ (2.1)

for each
$x\in\left\\{1,\alpha\beta,\gamma,\gamma^{-1},\gamma^{\alpha\beta},\left(\gamma^{\alpha\beta}\right)^{-1}\right\\}$.
Thus, the double cosets $PyP$ with
$y\in\left\\{1,\alpha,\beta,\alpha\beta,\gamma,\gamma^{-1},\gamma^{\alpha\beta},\left(\gamma^{\alpha\beta}\right)^{-1}\right\\}$
are exactly all the eight $P$-orbits partitioning $\mathcal{P}$.

For each
$y\in\left\\{\alpha,\beta,\alpha\beta,\gamma,\gamma^{-1},\gamma^{\alpha\beta},\left(\gamma^{\alpha\beta}\right)^{-1}\right\\}$,
the incidence structure $\mathcal{K}_{y}=\left(PyP,LP\right)$ is a tactical
configuration by [25], 1.2.6. It follows from (2.1) that, the parameters
$(v_{y},b_{y},k_{y},r_{y})$ of $\mathcal{K}_{y}$ are either equal to
$(39,39,3,3)$, or to $(13,39,1,3)$, according to whether
$y\in\left\\{\alpha,\beta\right\\}$ or
$y\in\left\\{\alpha\beta,\gamma,\gamma^{-1},\gamma^{\alpha\beta},\left(\gamma^{\alpha\beta}\right)^{-1}\right\\}$
respectively. Thus $r_{y}=3$ in each case, and hence the number of blocks of
$\mathcal{D}$ incident with $P$ and $Px$, where $x\neq 1$, is $3$. Then $3$ is
the number of blocks incident with any two distinct points of $\mathcal{P}$,
as $\mathcal{D}$ is a flag-transitive tactical configuration. Therefore,
$\mathcal{D}=(\mathcal{P},\mathcal{B})$ is a $2$-$(12^{2},12,3)$ design
admitting $G$ as a flag-transitive automorphism group.

Assume that $\mathrm{Aut(\mathcal{D})}\neq G$. Then
$G\trianglelefteq\mathrm{Aut(\mathcal{D})}\leqslant\mathrm{Aut(}G\mathrm{)}$,
where $\mathrm{Aut(}G\mathrm{)}\cong G:\left\langle\sigma\right\rangle$ and
$\sigma$ is defined above, as a consequence of the O’Nan-Scott Theorem (e.g.
see [28], Theorem 4.1A), since $v=2^{4}\cdot 3^{2}$. Then
$\mathrm{Aut(\mathcal{D})}=\mathrm{Aut(}G\mathrm{)}$, since
$G\neq\mathrm{Aut(\mathcal{D})}$ by our assumption. By [54], one
$\mathrm{Aut(}G\mathrm{)}_{P}$-orbit on $\mathcal{P}$, say $\mathcal{O}$, is
of length $26$ and is union of two $G_{P}$-orbits of length $13$, the
remaining $\mathrm{Aut(}G\mathrm{)}_{P}$-orbits on $\mathcal{P}-\mathcal{O}$
are also $G_{P}$-orbit. Thus, if $\mathcal{B}_{P}$ denotes the set of blocks
of $\mathcal{D}$ incident with $P$, any block in $\mathcal{B}_{P}$ intersects
$\mathcal{O}$ in exactly $4$ points. Hence, $(\mathcal{O},\mathcal{B}_{P})$ is
a tactical configuration by [25], 1.2.6, and its parameters are
$(v_{0},b_{0},k_{0},r_{0})=(26,39,4,6)$. So, there are exactly $6$ blocks of
$\mathcal{D}$ incident with $P$ and with any fixed point of $\mathcal{O}$.
This is impossible, since $\mathcal{D}$ is a $2$-design with $\lambda=3$. Thus
$\mathrm{Aut(\mathcal{D})}=G$. Clearly, $G$ is also the minimal flag-
transitive automorphism group of $\mathcal{D}$, and (1) is proven.

Finally, set $A=\mathrm{Aut(}G\mathrm{)}$. Then $A$ has a unique primitive
permutation representation of degree $144$ and is the one on the set
$\mathcal{P}^{\prime}=\left\\{N_{A}(P)x:x\in A\right\\}$ by [20]. Also, $G$
permutes primitively the elements of $\mathcal{P}^{\prime}$ again by [20].
Hence, we may consider the isomorphic copy
$\mathcal{D}_{0}=(\mathcal{P}^{\prime},\mathcal{B}_{0})$ of the $2$-design
$\mathcal{D}$ constructed in (1), where $\mathcal{B}_{0}$ consists of suitable
$12$-subsets of $\mathcal{P}^{\prime}$. Note that,
$\mathcal{B}_{0}^{\sigma}\neq\mathcal{B}_{0}$ as
$\mathrm{Aut(\mathcal{D}}_{0}\mathrm{)}=G$ by (1). Thus,
$\mathcal{D}_{0}^{\sigma}=(\mathcal{P}^{\prime},\mathcal{B}_{0}^{\sigma})$ is
a isomorphic to $\mathcal{D}$, but $\mathcal{D}^{\sigma}\neq\mathcal{D}$.
Therefore,
$\mathcal{D}^{\prime}=(\mathcal{P}^{\prime},\mathcal{B}_{0}\cup\mathcal{B}_{0}^{\sigma})$
is a $2$-$(144,12,6)$ design admitting $A$ as a flag-transitive automorphism
group. Arguing as in (1), we see that $A$ is the unique flag-transitive
automorphism group of $\mathcal{D}^{\prime}$.

If $\mathcal{B}^{\prime}=\left\\{Ly:y\in A\right\\}$, the incidence structure
$(\mathcal{P}^{\prime},\mathcal{B}^{\prime},\mathcal{I}^{\prime})$, where

$\mathcal{I}=\left\\{(N_{A}(P)x,Ly)\in\mathcal{P}^{\prime}\times\mathcal{B}^{\prime}:N_{A}(P)x\cap
Ly\neq\varnothing\right\\}\text{,}$

is an isomorphic copy of $\mathcal{D}^{\prime}$ by [35], Lemmas 1, and we
obtain (2). ∎

Recall that, if $\Gamma_{1}$ and $\Gamma_{2}$ are subgroups of a group
$\Gamma$ such $\Gamma=\Gamma_{1}\Gamma_{2}\Gamma_{1}$, then
$(\Gamma,\Gamma_{1},\Gamma_{2})$ is said _triple factorization_. Moreover,
$(\Gamma,\Gamma_{1},\Gamma_{2})$ is _non-degenerate_ if
$\Gamma\neq\Gamma_{1}\Gamma_{2}$. See [12] for more details and results on
triple factorizations. As shown in the following corollary, Example 2.1 also
provides a non-degenerate triple factorization.

######  Corollary 2.2.

$\left(G,N_{G}(P),L\right)$ is a non-degenerate triple factorization.

###### Proof.

It follows from the proof of Example 2.1 that the block-$N_{G}(P)$-orbit
$LN_{G}(P)$ non-trivially intersect each point-$N_{G}(P)$-orbit. Thus
$G=N_{G}(P)LN_{G}(P)$ is a triple factorization of $G$ by [12], Proposition
3.1. Moreover, it is non-degenerate, since the cardinality of $N_{G}(P)L$ is
smaller that the order of $G$. ∎

##  3\. Preliminary Reductions

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

######  Lemma 3.1.

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

  1. (1)

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

  2. (2)

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

  3. (3)

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

######  Lemma 3.2.

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

  1. (1)

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

  2. (2)

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

  3. (3)

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

###### Proof.

The assertion (1) follows from [25], 2.3.7.c, since
$r=(k+1)\lambda>(k-3)\lambda$.

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

Let $y$ be any point of $\mathcal{D}$, $y\neq x$, and $B$ be any block of
$\mathcal{D}$ incident with $x$. Since $(y^{G_{x}},B^{G_{x}})$ is a tactical
configuration by [25], 1.2.6, it follows that
$\left|y^{G_{x}}\right|\lambda=r\left|B\cap y^{G_{x}}\right|$. Hence
$\left|y^{G_{x}}\right|=(k+1)\left|B\cap y^{G_{x}}\right|$ as
$r=(k+1)\lambda$. This proves (3). ∎

######  Theorem 3.3.

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

  1. (1)

$G$ is an affine group.

  2. (2)

$G$ is an almost simple classical group.

  3. (3)

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

  4. (4)

$\mathcal{D}$ is isomorphic to one of the three $2$-$(6^{6},6,6)$ designs
constructed in [44], and $G\cong$ $~{}^{2}G_{3}(3)$.

See [44] for a proof.

In this paper we classify the $2$-$(k^{2},k,\lambda)$ designs, with
$\lambda\mid k$, admitting a flag-transitive automorphism group when $G$ is as
in (2) of Theorem 3.3. Since $~{}^{2}G_{2}(3)^{\prime}\cong PSL_{2}(8)$, case
(3) and (4) of Theorem 3.3 are also special cases of (2).

From now on we assume that $\mathrm{Soc(}G\mathrm{)}$, the socle of $G$, is a
non-abelian simple classical group. $\mathrm{Soc(}G\mathrm{)}$ will simply be
denoted by $X$. Thus $X\trianglelefteq G\leqslant\mathrm{Aut(}X\mathrm{)}$,
where $X$ is isomorphic to one of the following groups:

  1. (1)

$PSL_{n}(q)$, with $n\geqslant 2$ and $(n,q)\neq(2,2),(2,3)$.

  2. (2)

$PSp_{n}(q)^{\prime}$, with $n$ even, $n\geqslant 2$ and
$(n,q)\neq(2,2),(2,3)$.

  3. (3)

$PSU_{n}(q^{1/2})$, with $n\geqslant 2$, $q$ a square and
$(n,q^{1/2})\neq(2,2),(2,3)$ or $(3,2)$.

  4. (4)

$P\Omega_{m}(q)$, with $n\geqslant 3$, $nq$ odd and $(n,q)\neq(3,3)$.

  5. (5)

$P\Omega_{n}^{\varepsilon}(q)$, with $\varepsilon=\pm$, $n$ even, $n\geqslant
4$ and $(n,\varepsilon)\neq(4,+)$.

Clearly, $q=p^{f}$, where $p$ is a prime and $f\geqslant 1$. It should be
noted that the case $PSp_{4}(2)^{\prime}\cong A_{6}$ is already ruled out in
Lemma 3.5 of [44]. Therefore, in case (2), we may actually assume that $X\cong
PSp_{n}(q)$.

######  Lemma 3.4.

$\lambda\geqslant 2$.

###### Proof.

Assume that $\lambda=1$. Then $\mathcal{D}$ is a translation plane of order
$k=p^{t}$, $t\geqslant 1$, and $X$ is an elementary abelian $p$-group of order
$k$ by [53], Theorems 2 and 4. This is a contradiction, since $X$ is a non-
abelian simple group by our assumption. Thus, $\lambda\geqslant 2$. ∎

In the sequel, we always assume that $\lambda\geqslant 2$ without recalling
Lemma 3.4.

######  Lemma 3.5.

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

###### Proof.

Let $x$ be any point of $\mathcal{D}$. If $y$ is a point of $\mathcal{D}$,
with $y\neq x$, then $\left|y^{X_{x}}\right|=\frac{\left|B\cap
y^{G_{x}}\right|\left(k+1\right)}{\mu}$, where
$\mu\left|y^{X_{x}}\right|=\left|y^{G_{x}}\right|$, by Lemma 3.2(3), as
$X_{x}\trianglelefteq G_{x}$. On the other hand, $\mu$ divides
$\left|\mathrm{Out(}X\mathrm{)}\right|$, as
$\mu=\frac{\left[G_{x}:X_{x}\right]}{\left[G_{x,y}:X_{x,y}\right]}$. Therefore
$\frac{k+1}{\gcd(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}$ divides
$\left|y^{X_{x}}\right|$ and hence $\left|X_{x}\right|$. ∎

Recall that, if $M$ is a maximal subgroup of an almost simple group $\Gamma$
such that $M\cap\mathrm{Soc(}\Gamma\mathrm{)}$ is a non-maximal in
$\mathrm{Soc(}\Gamma\mathrm{)}$, then $M$ is called a _novel_ maximal
subgroup, or, simply, _novelty_. If $K$ is a maximal subgroup of
$\mathrm{Soc(}\Gamma\mathrm{)}$ containing
$M\cap\mathrm{Soc(}\Gamma\mathrm{)}$ and such that
$N_{\Gamma}(K)\mathrm{Soc(}\Gamma\mathrm{)}\neq\Gamma$, then $M$ is called a
_type 1 novelty with respect to_ $K$. More details on types and properties of
novelties of almost simple groups can be found in [14].

######  Lemma 3.6.

Let $\Omega$ be a set of imprimitivity for $X$ on the point set of
$\mathcal{D}$ such that $X_{\Lambda}$ is maximal in $X$ for
$\Lambda\in\Omega$. If $k+1$ does not divide
$\left|\mathrm{Out(}X\mathrm{)}\right|$, then the followings hold:

  1. (1)

$X_{\Lambda}^{G_{x}}$ is split in at least two distinct conjugacy classes
under $X$.

  2. (2)

$G_{x}$ is a type 1 novelty with respect to $X_{\Lambda}$.

###### Proof.

Let $\Omega$ be a set of imprimitivity for $X$ on the point set of
$\mathcal{D}$ such that $X_{\Lambda}$ is maximal in $X$ for
$\Lambda\in\Omega$. Clearly, $X_{\Lambda}=N_{X}(X_{\Lambda})$. Let
$x\in\Lambda$, since $X_{x}\trianglelefteq G_{x}$ and $G_{x}$ is maximal in
$G$, it results $G_{x}=N_{G}(X_{x})$ and hence $X_{x}\leqslant
N_{X}(X_{x})\leqslant G_{x}$. Thus, $X_{x}=N_{X}(X_{x})$.

If $N_{G_{x}}(X_{\Lambda})=G_{x}$, then $G_{x}\leqslant X_{\Lambda}G_{x}$ and
hence either $X_{\Lambda}\leqslant G_{x}$ or $G=X_{\Lambda}G_{x}$, as $G_{x}$
is maximal in $G$. The former implies $X_{\Lambda}\leqslant X_{x}$, whereas
$X_{x}<X_{\Lambda}$, and this case is excluded. Thus, $G=X_{\Lambda}G_{x}$ and
hence $X_{\Lambda}\vartriangleleft G$, as $N_{G_{x}}(X_{\Lambda})=G_{x}$ by
our assumption. So, we get $X_{\Lambda}\vartriangleleft X$, which is a
contradiction, as $X$ is simple and $X_{x}<X_{\Lambda}<X$. Thus,
$\left[G_{x}:N_{G_{x}}(X_{\Lambda})\right]>1$.

Assume that $X_{y}<X_{\Lambda}$ for some $y\notin\Lambda$. Clearly, $\Lambda$
is a $X_{\Lambda}$-orbit and is union of $X_{y}$-orbits distinct from
$\left\\{y\right\\}$. Each $X_{y}$-orbit distinct from $\left\\{y\right\\}$ is
of length divisible by
$\frac{k+1}{(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}$ by Lemma 3.5.
Moreover, $\frac{k+1}{(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}>1$, since
$k+1$ does not divide $\left|\mathrm{Out(}X\mathrm{)}\right|$ by our
assumption. So $\frac{k+1}{(k+1,\left|\mathrm{Out(}X\mathrm{)}\right|)}$
divides $\left|\Lambda\right|$, and hence $k^{2}$, since
$k^{2}=\left[X:X_{\Lambda}\right]\left[X_{\Lambda}:X_{x}\right]=\left[X:X_{\Lambda}\right]\left|\Lambda\right|$,
a contradiction. Thus, for each $X_{x^{\prime}}<X_{\Lambda}$ if, and only if,
$x^{\prime}\in\Lambda$.

By (1), there is $\alpha\in G_{x}-N_{G_{x}}(X_{\Lambda})$ such that
$X_{\Lambda}^{\alpha}\neq X_{\Lambda}$. Also, $X_{x}\leqslant
X_{\Lambda}^{\alpha}\cap X_{\Lambda}$, with both $X_{\Lambda}^{\alpha}$ and
$X_{\Lambda}$ maximal in $X$. Suppose that there is $\beta\in X$ such that
$X_{\Lambda}^{\alpha}=X_{\Lambda}^{\beta}$. Then
$X_{\Lambda}^{\alpha}=X_{\Lambda^{\beta}}$ and hence $X_{\Lambda}^{\alpha}$
preserves the block of imprimitivity $\Lambda^{\beta}$. Then the above
argument implies $x\in\Lambda^{\beta}$ and hence $\Lambda^{\beta}=\Lambda$, as
$x\in\Lambda$ too. Then
$X_{\Lambda}^{\alpha}=X_{\Lambda^{\beta}}=X_{\Lambda}$, but $\alpha\in
G_{x}-N_{G_{x}}(X_{\Lambda})$, and so we obtain a contradiction. Thus
$X_{\Lambda}^{\alpha}$ and $X_{\Lambda}$ lies in distinct conjugacy classes
under $X$. This proves (1).

Assume that $G=N_{G}(X_{\Lambda})X$. Since
$\left[G_{x}:N_{G_{x}}(X_{\Lambda})\right]>1$, there are
$X_{x_{1}},X_{x_{2}}<X_{\Lambda}$ such that $X_{x_{1}}$ and $X_{x_{2}}$ are
conjugate in $N_{G}(X_{\Lambda})$ but not in $X_{\Lambda}$ by [14],
Proposition 1.3.10. Nevertheless, $x_{1},x_{2}\in\Lambda$ by the above
argument. Hence $X_{x_{1}}$ and $X_{x_{2}}$ are conjugate in $X_{\Lambda}$, as
this one acts transitively on $\Lambda$, and we reach a contradiction. Thus
$G\neq N_{G}(X_{\Lambda})X$ and hence $G_{x}$ is a type 1 novelty with respect
to $X_{\Lambda}$. Thus, we obtain (2). ∎

##  4\. Reductions for $X$ based on primitive prime divisors of its order

The first part of this section is devoted to the resolution of the case
$X\cong PSL_{2}(q)$, which is achieved by combining some group-theoretical
results on the structure of $PSL_{2}(q)$ given in [37], together with some
numerical constraints on the diophantine equation $k^{2}=[X:X_{x}]$ provided
in [49].

The second part focuses on the case $n\geqslant 3$, where some tools are
developed in order to control the structure of $X_{x}$. More precisely, it is
shown that $\left|X\right|$ is divisible by $\Phi_{ef}^{\ast}(p)$, the
primitive part of $p^{e}-1$ (more details on $\Phi_{ef}^{\ast}(p)$ are
provided below). Then, in Theorem 4.7, is proven the central result of this
section which states that either $X_{x}$ is large subgroup of $X$ and
$\left(\Phi_{ef}^{\ast}(p),\left|X_{B}\right|\right)>1$, or
$\left(\Phi_{ef}^{\ast}(p),k+1\right)>1$, or $\Phi_{ef}^{\ast}(p)\mid k^{2}$
and
$\left(\Phi_{ef}^{\ast}(p),\left|X_{B}\right|\right)=\Phi_{ef}^{\ast}(p)^{1/2}$.
In the final part of this section, a lower bound for $\left|X_{B}\right|_{p}$
is exhibited, which will play an important role to exclude the case
$\left(\Phi_{ef}^{\ast}(p),k+1\right)>1$.

Throughout the paper, if $m$ is any integer and $w$ is a prime, the symbols
$m_{w}$ and $m_{w^{\prime}}$ will denote the integers $\max_{i\geqslant
0}\\{w^{i}:w^{i}\mid m\\}$ and $m/m_{w}$ respectively.

######  Proposition 4.1.

If $X\cong PSL_{2}(q)$, then $q=8$ and one of the following holds:

  1. (1)

$\mathcal{D}$ is isomorphic to the $2$-$(6^{2},6,2)$ design constructed in
[44] and $PSL_{2}(8)\trianglelefteq G\leqslant P\Gamma L_{2}(8)$.

  2. (2)

$\mathcal{D}$ is isomorphic to one of the three $2$-$(6^{2},6,6)$ designs
constructed in [44] and $G\cong P\Gamma L_{2}(8)$.

###### Proof.

Let $X\cong PSL_{2}(q)$. Then
$\frac{k+1}{(k+1,(2,q-1)f)}\mid\left|X_{x}\right|$, as
$\left|\mathrm{Out(}X\mathrm{)}\right|=(2,q-1)f$, by Lemma 3.5. Moreover,
$q\neq 4,5,9$, since $PSL_{2}(4)\cong PSL_{2}(5)\cong A_{5}$ and
$PSL_{2}(9)\cong A_{6}$, and both these cases are ruled out in Lemma 3.5 of
[44].

Assume that $k+1$ divides $(2,q-1)f$. Thus, $k^{2}<(2,q-1)^{2}f^{2}$.
Moreover, $q>11$ and $f\geqslant 2$, as $\mathcal{D}$ is non-trivial. Hence
$q+1<(2,q-1)^{2}f^{2}$, since $q+1$ is the minimal primitive permutation
representation of $X$ for $q>11$ by [39], Theorem 5.2.2. So, $q=3^{3}$ and
$3\leqslant k\leqslant 5$. Actually, $k=5$ as $k+1$ must divide $(2,q-1)f$.
However, this case cannot occur, since $PSL_{2}(3^{3})$ has no transitive
permutation representations of degree less then $28$.

Assume that $\frac{k+1}{(k+1,(2,q-1)f)}>1$. Let $U$ be a Sylow $u$-subgroup of
$X_{x}$, where $u$ is a prime divisor of $\frac{k+1}{(k+1,(2,q-1)f)}$, and
assume that $U$ fixes a further point $y$ of $\mathcal{D}$. So,
$\left|y^{X_{x}}\right|$ is coprime to $u$. This is a contradiction, since
$\frac{k+1}{(k+1,(2,q-1)f)}$ divides $\left|y^{X_{x}}\right|$ by Lemma 3.5.
Therefore $x$ is the unique point fixed by $U$ and hence $N_{X}(U)\leqslant
X_{x}$. Then $X_{x}=N_{X}(U)$ and either $X_{x}\cong
U:Z_{\frac{q-1}{(2,q-1)}}$, or $X_{x}\cong D_{\frac{2(q+\epsilon)}{(2,q-1)}}$
and $\epsilon=\pm 1$, according to whether $u=p$ or $u\neq p$, respectively,
by [37], Satz 8.5.(a).

Assume that $X_{x}\cong U:Z_{\frac{q-1}{(2,q-1)}}$. Then
$k^{2}=\left[X:X_{x}\right]=q+1$ and hence $k=3$ and $q=8$ by [49], A5.1,
since $k\geqslant 3$. However, this case is ruled out in Lemma 3.6 of [44],
since $PSL_{2}(8)\cong$ ${}^{2}G_{2}(3)^{\prime}$.

Assume that $X_{x}\cong D_{\frac{2(q+\epsilon)}{(2,q-1)}}$, where
$\epsilon=\pm 1$. Then $k^{2}=\frac{q(q-\epsilon)}{2}$. If $q$ is even, then
$q/2=2^{f-1}$, with $f$ odd, and $z^{2}=2^{f}-\epsilon$ for some positive
integer $z$, since $k^{2}=\frac{q}{2}(q-\epsilon)$. Also $f>1$, since $X$ is
simple, and hence $f\geqslant 3$. Then $\epsilon=-1$ and $f=3$ by [49], A3.1
and A5.1. Thus $q=8$ and $k=6$, and hence assertions (1) and (2) follow from
Theorem 3.10 of [44], since $PSL_{2}(8)\cong$ ${}^{2}G_{2}(3)^{\prime}$.

Assume that $q$ is odd. Then $q$ is a square and $q^{1/2}$ is the highest
power of $p$ dividing $k$. Moreover, we saw that $q>9$. Clearly, $k+1$ divides
$\frac{q(q-\epsilon)}{2}-1$ and hence $(q-2\epsilon)\frac{(q+\epsilon)}{2}$.
By Lemma 3.2(3), $k+1$ divides $\left|G_{x}\right|$, which is
$2f\left(q+\epsilon\right)$, and hence it divides
$\frac{(q+\epsilon)}{2}f_{p^{\prime}}$, as $p\mid k$ and $q$ is odd. Then
there is an integer $c\geqslant 1$ such that

$2c(k+1)=\left(q+\epsilon\right)f_{p^{\prime}}\text{.}$ (4.1)

From (4.1) we derive that $q^{1/2}$ divides $2c-\epsilon f_{p^{\prime}}$, as
$q^{1/2}\mid k$.

Note that $2c>f_{p^{\prime}}$. Indeed, if $2c\leqslant f_{p^{\prime}}$ then
$k\geqslant q+\epsilon-1$ and hence
$\left(q+\epsilon-1\right)^{2}\leqslant\frac{q(q-\epsilon)}{2}$, which has no
solutions for $q>9$. Then $2c-\epsilon f_{p^{\prime}}=\theta q^{1/2}$ for some
$\theta\geqslant 1$, and (4.1) becomes $\left(\theta q^{1/2}+\epsilon
f_{p^{\prime}}\right)k+\theta q^{1/2}=qf_{p^{\prime}}$. Therefore, we obtain
$k=q^{1/2}\frac{q^{1/2}f_{p^{\prime}}-\theta}{\theta q^{1/2}+\epsilon
f_{p^{\prime}}}$.

If $\epsilon=1$, then $k<q^{1/2}f_{p^{\prime}}$ and hence
$k^{2}=\frac{q(q-1)}{2}$ implies $q-1<2f^{2}$, which has no solutions for
$q>9$. Therefore, we have $\epsilon=-1$ and
$k=q^{1/2}\frac{q^{1/2}f_{p^{\prime}}-\theta}{\theta q^{1/2}-f_{p^{\prime}}}$.
We have seen that $q-1\geqslant 2f^{2}$ as $q>9$. Then
$q^{1/2}>f_{p^{\prime}}$ and hence $\theta
q^{1/2}-f_{p^{\prime}}>(\theta-1)q^{1/2}$. So,
$k<q^{1/2}\frac{q^{1/2}f_{p^{\prime}}-\theta}{\left(\theta-1\right)q^{1/2}}<q^{1/2}f_{p^{\prime}}$
for $\theta>1$, and we obtain a contradiction as above. Hence, $\theta=1$.
Then $k=q^{1/2}\frac{q^{1/2}f_{p^{\prime}}-1}{q^{1/2}-f_{p^{\prime}}}$ which,
substituted in $k^{2}=\frac{q(q+1)}{2}$, yields
$f_{p^{\prime}}=\frac{q^{2}+q+2}{\sqrt{q}\left(q+3\right)}$. However, such
value of $f_{p^{\prime}}$ is not an integer as $q$ is odd and $q>1$. This
completes the proof. ∎

######  Remark 4.2.

From now on we assume that $X\ncong PSL_{2}(q)$. Bearing in mind the
isomorphisms $PSp_{2}(q)\cong PSL_{2}(q)$, $P\Omega_{3}(q)\cong PSL_{2}(q)$
for $q$ odd, $P\Omega_{4}^{-}(q)\cong PSL_{2}(q^{2})$, and
$PSU_{2}(q^{1/2})\cong PSL_{2}(q)$ when $q$ is a square, and the fact that
$P\Omega_{4}^{+}(q)$ is non-simple, in the sequel we assume that

  1. (1)

$n\geqslant 3$ for $X\cong PSL_{n}(q)$;

  2. (2)

$n\geqslant 3$ for $X\cong PSU_{n}(q^{1/2})$ for $q$ square and
$(n,q^{1/2})\neq(3,2)$;

  3. (3)

$n\geqslant 4$ for $X\cong PSp_{n}(q)$;

  4. (4)

$n\geqslant 5$ for $X\cong P\Omega_{5}(q)$;

  5. (5)

$n\geqslant 6$ for $X\cong P\Omega_{n}^{\pm}(q)$.

Now we want to transfer some of the Saxl’s arguments contained in [50] to our
context in order to rule out the parabolic case. In [50] the author shows that
$r\leqslant(v-1)_{p}$, where $r$ is the replication number of a linear space
with $v$ points admitting $G$ as a flag-transitive automorphism group,
whenever the conclusions of his Lemma 2.6 hold. Then, he proves that
$v\geqslant(v-1)^{2}_{p}\geqslant r^{2}$ and hence he rules out the parabolic
case when the conclusions of Lemma 2.6 hold. From a group-theoretical point of
view, Saxl shows that

$[G:G_{x}]<\left([G:G_{x}]-1\right)_{p}^{2},$ (4.2)

where $v=[G:G_{x}]$ and $G_{x}$ is a maximal parabolic subgroup of $G$, is
never fulfilled whenever the conclusions of his Lemma 2.6 hold, and hence no
linear spaces occur in these cases.
We may use the same argument of [50], with his $r$ replaced by our
$r/\lambda=k+1$, thus obtaining (4.2) and hence excluding the parabolic case.
This is possible since Lemma 2.1.(ii)–(iii) of [50] are replaced by our Lemmas
3.1(3) and 3.2(3) respectively.

######  Proposition 4.3.

Let $x$ be any point of $\mathcal{D}$, then $G_{x}$ is not a parabolic
subgroup of $G$ and $p\mid k$.

###### Proof.

Assume that $G_{x}$ is a parabolic subgroup of $G$. Suppose that $X$ is not
isomorphic to $PSL_{n}(q)$, or to $P\Omega_{n}^{+}(q)$ for $n/2$ odd. Then
$k+1=p^{s}$ for some $s\geqslant 1$ by [50], Lemma 2.6, and by Lemma 3.2(3).
Then $k+1$ divides $\left([G:G_{x}]-1\right)_{p}$, as $k^{2}=[G:G_{x}]$, and
hence we obtain (4.2) by Lemma 3.1(3).
Let $i$ be the dimension of the subspace of $PG_{n}(q)$ preserved by $G_{x}$.
In [50] it is shown that $\left([G:G_{x}]-1\right)_{p}$ is a divisor of one of
the following numbers:

  1. (1)

$q$ for $X\cong PSp_{n}(q)$, $P\Omega_{n}(q)\,$;

  2. (2)

$q$, $q^{2}$ or $q^{3}$ for $X\cong$ $PSU_{n}(q^{1/2})$, $n>3$ and
$(n,q)\neq(4,2)$, according to whether $i\neq[n/2]-1$, or $n$ even and
$i=(n-2)/2$, or $n$ odd and $i=(n-3)/2$ respectively;

  3. (3)

$q^{2}$ or $8$ for $P\Omega_{n}^{\varepsilon}(q)$, with $\varepsilon=\pm$, and
$n/2$ even and $n/2>4$ for $\varepsilon=+$;

  4. (4)

$(3,q)q$ or $8$ for $P\Omega_{8}^{+}(q)$.

Hence, the same argument used by Saxl in [50] can be deployed to show that
(4.2) is not fulfilled in cases (1)–(4). So these cases are ruled out.
Clearly, $X\cong PSU_{4}(2)$ cannot occur since $PSU_{4}(2)\cong PSp_{4}(3)$.
Finally, $X\cong$ $PSU_{3}(q^{1/2})$ cannot occur. Indeed, $k^{2}=q^{3/2}+1$
has no integer solutions by [49], A5.1, since $q^{1/2}\neq 2$.

Assume that $X\cong P\Omega_{n}^{+}(q)$ with $n/2$ odd. As pointed out in
[50], it is still true that $X$ has a subdegree which is a power of $p$,
provided that $G_{x}$ is not parabolic of type $n/2$. Hence we still obtain
(4.2). However, it is not fulfilled since $\left([G:G_{x}]-1\right)_{p}$ is
$q^{2}$ or $8$. Hence, the cases where $G_{x}$ is parabolic, but not of type
$n/2$, are ruled out.
Assume that $G_{x}$ is parabolic of type $n/2$. We may repeat the argument of
Saxl in [50], Section 5.(1), with $r/\lambda$ in the role of his $r$ and we
obtain that $k^{2}>q^{n/4(n/2-1)}$ and $k+1=r/\lambda\leqslant q(q^{n/2}-1)$,
and hence $n=10$, as $(r/\lambda)^{2}>k^{2}$. Thus

$k^{2}=\left(q+1\right)^{2}\left(q^{2}-q+1\right)\left(q^{2}+1\right)\left(q^{4}+1\right).$

Hence $q^{2}-q+1$, which is equal to $(q-1)^{2}+(q-1)+1$, is a square, as it
is coprime to the other factors of $\frac{k^{2}}{(q+1)^{2}}$. However this is
impossible by [49], A7.1, since $q$ is a power of a prime.

Finally, assume that $X\cong PSL_{n}(q)$. Again, proceeding as in [50],
Section 7.(1), with $r/\lambda$ in the role of $r$, we obtain $i\leqslant 2$.
Assume that $i=2$. Then $k^{2}=\frac{(q^{n}-1)(q^{n-1}-1)}{(q^{2}-1)(q-1)}$.
Since $\left(\frac{q^{n}-1}{q-1},\frac{q^{n-1}-1}{q-1}\right)=1$, either $n$
is odd and $\frac{q^{n}-1}{q-1}$ is a square, or $n$ is even and
$\frac{q^{n-1}-1}{q-1}$ is a square. Then either $(n,q)=(5,3)$ or
$(n-1,q)=(5,3)$, respectively, by [49], A8.1. However, both case are ruled out
as they yield $k^{2}=1210$ and $\allowbreak 11\,011$ respectively. Thus $i=1$,
$k^{2}=\frac{q^{n}-1}{q-1}$, and hence $(n,q,k)=(4,7,20)$ or $(5,3,11)$ again
by [49], A8.1. If $(n,q,k)=(4,7,20)$, then $b=21\cdot 20\cdot\lambda$ with
$\lambda\mid 20$ and $\lambda\geqslant 2$. However, $PSL_{4}(7)\unlhd
G\leqslant PGL_{4}(7)$ has no transitive permutation representations of degree
$b$ by [14], Tables 8.8–8.9. So, $(n,q,k)=(5,3,11)$, and hence $b=12\cdot
11^{2}$ since $\lambda\mid 11$ and $\lambda\geqslant 2$. As $G\cong
PSL_{5}(3)$ has no transitive permutation representations of degree $12\cdot
11^{2}$ by [14], Tables 8.18–8.19, this case is ruled out. Therefore $G_{x}$
is not a parabolic subgroup of $G$, and hence $[G:G_{x}]$ is divisible by $p$
by Tits’ Lemma (see [51], Theorem 1.6). Thus $p\mid k$. ∎

A divisor $r$ 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 [49], P1.7).

The order of $X$ is clearly divisible by $\Phi_{ef}^{\ast}(p)$, where $e=n$
for $X$ isomorphic to $PSL_{n}(q)$, $PSp_{n}(q)$, $P\Omega_{n}^{-}(q)$, or
$PSU_{n}(q^{1/2})$, with $q$ square and $n$ odd, $e=n-1$ for $P\Omega_{n}(q)$,
with $nq$ odd, or $PSU_{n}(q^{1/2})$, with $q$ square and $n$ even, and
$e=n-2$ for $P\Omega_{n}^{+}(q)$.

######  Lemma 4.4.

$\Phi_{ef}^{\ast}(p)>1$.

###### Proof.

Assume that $\Phi_{ef}^{\ast}(p)=1$. Since $n\geqslant 3$ by Remark 4.2, it
follows that $(ef,p)=(6,2)$ (see above). Then $X$ is isomorphic to one of the
groups $PSL_{3}(4)$, $PSU_{4}(2)$, $PSL_{6}(2)$, $PSp_{6}(2)$,
$P\Omega_{6}^{-}(2)$ or $P\Omega_{8}^{+}(2)$.

Assume that $X\cong PSL_{3}(4)$. Then $35\mid\left|X_{x}\right|$, since
$\left[X:X_{x}\right]=k^{2}$, but $X$ does not contain such a group by [20].

Assume that $X\cong PSU_{4}(2)$. Then $PSU_{4}(2)\trianglelefteq G\leqslant
PSU_{4}(2):Z_{2}$. Also $5\mid\left|X_{x}\right|$ and
$\left[X:X_{x}\right]=k^{2}$, with $k\mid 72$ and $k+1\mid\left|G\right|$.
Easy computations show that $k=3,4,8$ or $9$. However, $X$ does not have
$k^{2}$ as a transitive permutation of degree for any such values of $k$ by
[20].

Assume that $X\cong PSL_{6}(2)$. Thus $2\cdot 5\cdot
31\mid\left|X_{x}\right|$, since $\left[X:X_{x}\right]=k^{2}$, and hence
either $X_{x}\cong SL_{5}(2)$ or $X_{x}\cong E_{2^{5}}:SL_{5}(2)$ by [14],
Tables 8.24 and 8.25, and by [20]. Since $\left|X_{x}\right|_{7}=7$ and
$\left|X\right|_{7}=7^{2}$, it follows that $\left[X:X_{x}\right]_{7}=7$, but
this contradicts $\left[X:X_{x}\right]=k^{2}$.

Assume that $X\cong PSp_{6}(2)$. Then $G=X$, since
$\mathrm{Out(}X\mathrm{)}=1$. Also $\left[G:G_{x}\right]=k^{2}$ implies
$2\cdot 5\cdot 7\mid\left|G_{x}\right|$. Then either $G_{x}=A_{7}$ and $k=24$,
or $G_{x}\cong S_{8}$ and $k=6$ by [20]. The former case is ruled out, since
it contradicts Lemma 3.2(3). Indeed, $k+1=5^{2}$ does not divide
$\left|G_{x}\right|$. Thus, $G_{x}\cong S_{8}$, $k=6$ and hence $\lambda=2,3$
or $6$, as $\lambda\mid k$ and $\lambda\geqslant 2$. However, $S_{8}$ does not
have transitive permutation representations of degree $r=14,21$ or $42$ by
[20]. Thus $X\cong PSp_{6}(2)$ is excluded.

Assume that $X\cong P\Omega_{6}^{-}(2)$. Then $5\mid\left|X_{x}\right|$, since
$\left[X:X_{x}\right]=k^{2}$, and hence $k\mid 2^{3}\cdot 3^{2}$. Also
$\frac{k+1}{(k+1,2)}$ divides $\left|X\right|$ by Lemma 3.5, since
$\mathrm{Out(}X\mathrm{)}\cong Z_{2}$. Easy computations show that $k=3,4,8$
or $9$. However, for these $k$ the group $X$ does not have a transitive
permutation representation of degree $k^{2}$ by [20], and so $X\cong
P\Omega_{6}^{-}(2)$ is ruled out too.

Assume that $X\cong P\Omega_{8}^{+}(2)$. Therefore $21\mid\left|X_{x}\right|$,
since $\left[X:X_{x}\right]=k^{2}$, and hence $k^{2}\mid 2^{12}\cdot
3^{4}\cdot 5^{2}$. Also $\left[X:X_{x}\right]$ must be divisible by $120$,
$135$ or $960$ by [20]. Then $15$ divides $\left[X:X_{x}\right]$ in each of
these cases. Actually, $15\mid k$ as $\left[X:X_{x}\right]=k^{2}$. Then
$k=15\theta$, with $\theta\mid 2^{6}\cdot 3$ and $\theta>1$, and
$\frac{k+1}{(k+1,2)}\mid\left|X\right|$ as $\mathrm{Out(}X\mathrm{)}\cong
S_{3}$. It is straightforward to check that
$\frac{k+1}{(k+1,2)}\nmid\left|X\right|$ for any $\theta\mid 2^{6}\cdot 3$ and
$\theta>1$, and hence $X\cong P\Omega_{8}^{+}(2)$ cannot occur. ∎

The following two lemmas play a central role in proving that $X_{x}$ is large
subgroup of $X$ when
$\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$.

######  Lemma 4.5.

Let $(x,B)$ be any flag of $\mathcal{D}$. Then $\left[G_{x}:X_{x}\right]<1+ef$
provided that one of the following holds:

  1. (1)

$\lambda\left|G_{x,B}\right|$ is odd;

  2. (2)

$\lambda\left|G_{x,B}\right|\equiv 2\pmod{4}$,
$\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$ and either
$X\cong PSU_{n}(q^{1/2})$, with $n$ even, or $X\cong P\Omega_{8}^{+}(q)$.

###### Proof.

Suppose that $\lambda\left|G_{x,B}\right|$ is odd. Let $\sigma$ be an
involution of $G$ and let $y$ be any point such that $y^{\sigma}\neq y$. Then
$\sigma$ permutes the $\lambda$ blocks incident with $y,y^{\sigma}$, and hence
$\sigma$ fixes one of them, say $C$, as $\lambda$ is odd. Thus $k$ is even,
since $\left|G_{C}\right|=k\left|G_{y,C}\right|$ and $\left|G_{y,C}\right|$ is
odd, and hence $\lambda(k+1)\left|G_{x,B}\right|$ is odd. Therefore,
$\left[G_{x}:X_{x}\right]$ is a divisor of
$\left|\mathrm{Out(}X\mathrm{)}\right|_{2^{\prime}}$, since
$\left|X_{x}\right|=\frac{\left|G_{x}\right|}{\left[G_{x}:X_{x}\right]}=\frac{\lambda(k+1)\left|G_{x,B}\right|}{\left[G_{x}:X_{x}\right]}$.
Since $\left|\mathrm{Out(}X\mathrm{)}\right|_{2^{\prime}}$ is less than, or
equal to, $nf$, $nf/2$, $3f$ or $f$ for $X\cong PSL_{n}(q)$,
$PSU_{n}(q^{1/2})$, $P\Omega_{8}^{+}(q)$ or $X$ as in the remaining cases,
respectively, by [39], Table 5.1.A, and as $n\geqslant 3$ by Remark 4.2, it
follows that $\left[G_{x}:H_{x}\right]<1+ef$ for any $X$.

Suppose that $\lambda\left|G_{x,B}\right|\equiv 2\pmod{4}$,
$\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$, and also
that either $X\cong PSU_{n}(q^{1/2})$, with $n$ even, or $X\cong
P\Omega_{8}^{+}(q)$. Assume that $k$ is odd. By [40], one of the following
holds:

  * (i).

$G_{x}$ is a parabolic subgroup of $G$, and $q$ is even.

  * (ii).

$G_{x}$ is the stabilizer of a non-singular subspace of $PG_{n-1}(q)$, and $q$
is odd.

  * (iii).

$G_{x}$ is a $\mathcal{C}_{4}$-subgroup of $G$.

  * (iv).

$G_{x}$ stabilizes an orthogonal decomposition into $t$ subspaces each of
dimension $n/t$ in its natural projective module $V$.

  * (v).

$X\cong P\Omega_{8}^{+}(q)$, $q$ is a prime such that $q\equiv\pm 3\pmod{8}$,
and $X_{x}$ is isomorphic either to $\Omega_{8}^{+}(2)$ or to $2^{3}\cdot
2^{6}\cdot PSL_{3}(2)$.

Case (i) is ruled out by Proposition 4.3. Since $\lambda\left|G_{x,B}\right|$
divides $\left|G_{x}\right|$, and since
$\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$, it follows
that $\left|G_{x}\right|$ is divisible by a primitive prime divisor $u$ of
$p^{ef}-1$. Thus, case (iii) is excluded by [39], Section 4.4 and Proposition
5.2.15.(i).
Assume that case (ii) holds. If $X\cong PSU_{n}(q^{1/2})$, with $n$ even, then
$G_{x}$ is the stabilizer of a non-singular point of $PG_{n-1}(q)$ by [39],
Proposition 4.1.4.(II), since $u\mid\left|G_{x}\right|$. Thus,

$k^{2}=q^{\left(n-1\right)/2}\frac{q^{n/2}-1}{q^{1/2}+1}$

and hence $f\equiv 0\pmod{4}$, as $q^{\left(n-1\right)/2}$ must be a square
and $n$ is even. So $\frac{q^{n/2}-1}{q^{1/2}+1}$ is even, as $q$ is odd,
whereas $k$ is odd.

If $X\cong P\Omega_{8}^{+}(q)$, then $G_{x}$ is either the stabilizer in $G$
of a non-singular point, or the stabilizer in $G$ of a non-singular line of
type $-$ of $PG_{7}(q)$ by [39], Propositions 4.1.6.(II), since $q$ is odd and
$\left(\Phi_{6f}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$. In the
former case, $k^{2}=q^{3}(q^{2}-1)\frac{q^{2}+1}{2}$ and hence $q^{2}-1$ is a
square. So, $q^{2}=z^{2}+1$ for some positive integer $z$, but this
contradicts [49], A3.1. Therefore, $G_{x}$ is the stabilizer in $G$ of a non-
singular line of type $-$ of $PG_{7}(q)$, then

$k^{2}=q^{6}\frac{(q-1)^{2}}{(4,q-1)}\allowbreak\left(q^{2}+1\right)\left(q^{2}+q+1\right).$

So, $q^{2}+q+1$ is a square, but this contradicts [49], A7.1.

Assume that case (iv) holds. Then $u\mid\frac{n}{t}$ by [39], Section 4.2, and
hence $1+ef\leqslant u\leqslant n/2$ by [39], Proposition 5.2.15.(i). This is
impossible, since $n\geqslant 4$ in both cases by Remark 4.2.
Finally, assume that case (v) occurs. Then $\left|G_{x}\right|\leqslant
24\left|X_{x}\right|$ and hence $\left|P\Omega_{8}^{+}(q)\right|\leqslant
24^{3}\left|X_{x}\right|^{3}$, as $G_{x}$ is a large subgroup of $G$ by Lemma
3.2(2). Then either $X_{x}\cong\Omega_{8}^{+}(2)$ and $p=3,5$ or $11$, or
$X_{x}\cong 2^{3}\cdot 2^{6}\cdot PSL_{3}(2)$ and $p=3$. However, all the
cases are ruled out, as they violate $\left[X:X_{x}\right]=k^{2}$. Thus, $k$
is even and hence $\lambda(k+1)\left|G_{x,B}\right|\equiv 2\pmod{4}$. Arguing
as above, we see that
$\left[G_{x}:X_{x}\right]\mid\lambda(k+1)\left|G_{x,B}\right|$, where
$\left[G_{x}:X_{x}\right]=\frac{\left|\mathrm{Out(}X\mathrm{)}\right|}{c}$ for
some $c\geqslant 1$ such that
$4\nmid\frac{\left|\mathrm{Out(}X\mathrm{)}\right|}{c}$.

If $X\cong PSU_{n}(q^{1/2})$, with $n$ even, then
$\frac{\left|\mathrm{Out(}X\mathrm{)}\right|}{c}\leqslant nf/4<1+(n-1)f$, as
$\left|\mathrm{Out(}X\mathrm{)}\right|=(q^{1/2}+1,n)f$ and as $f$ is even.
Then
$\left[G_{x}:H_{x}\right]=\left|\mathrm{Out(}X\mathrm{)}\right|/c<1+(n-1)f$
and we obtain the assertion in this case.

If $X\cong P\Omega_{8}^{+}(q)$, then
$\frac{\left|\mathrm{Out(}X\mathrm{)}\right|}{c}\leqslant 6f$, as
$4\nmid\frac{\left|\mathrm{Out(}X\mathrm{)}\right|}{c}$, and hence
$\left[G_{x}:H_{x}\right]\leqslant 6f<1+6f$. This completes the proof. ∎

######  Lemma 4.6.

Let $(x,B)$ be any flag of $\mathcal{D}$. If
$\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$, then
$\lambda\left|G_{x,B}\right|\geqslant\left[G_{x}:H_{x}\right]$.

###### Proof.

Suppose the contrary. Then
$\lambda\left|G_{x,B}\right|<\left[G_{x}:X_{x}\right]\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|$,
as $\left[G_{x}:H_{x}\right]$ divides $\left|\mathrm{Out(}X\mathrm{)}\right|$.
Also, $\lambda\left|G_{x,B}\right|=\rho\theta$, where
$\rho=\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)$ and
$\theta\geqslant 1$. Since $\rho>1$, $\rho=u_{1}^{m_{1}}\cdots
u_{\ell}^{m_{\ell}}$ with $u_{i}$ prime for $i=1,...,\ell$, and $u_{i}\neq
u_{j}$ for $i\neq j$. Furthermore, $u_{i}\equiv 1\pmod{ef}$ by [39],
Proposition 5.2.15.(i), and hence $u_{i}\geqslant 1+ef$. Therefore, either
$\rho\geqslant 3\left(1+ef\right)$ (actually, $\rho>6\left(1+6f\right)$ for
$P\Omega_{8}^{+}(q)$ by Remark 4.2), or $\rho=1+ef$ is a prime. Consequently,
$\left|G_{x,B}\right|\geqslant\mu(1+ef)\theta$, where $\mu\geqslant 3$, or
$\mu=1$ and $1+ef$ is a prime.

Assume that $X$ isomorphic to one of the groups $PSL_{n}(q)$ or
$PSU_{n}(q^{1/2})$. Then $\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant 2nf$
by [39], Table 5.1.A, and hence
$\mu\left(1+ef\right)\theta\leqslant\lambda\left|G_{x,B}\right|\leqslant 2nf$.
Then $\mu=1$, $\rho=1+ef$ is a prime and either $e=n$ and $\theta=1$, or
$e=n-1$, $X\cong PSU_{n}(q^{1/2})$, with $n$ even, and $\theta\leqslant 2$.

If $\theta=1$, then $\lambda\left|G_{x,B}\right|$ is odd and hence
$\rho=\lambda\left|G_{x,B}\right|\leqslant\left[G_{x}:H_{x}\right]<1+ef$ by
Lemma 4.5(1). However, this contradicts $\rho=1+ef$.

If $\theta=2$, then $X\cong PSU_{n}(q^{1/2})$, with $n$ even. It results
$2\rho=\lambda\left|G_{x,B}\right|\leqslant\left[G_{x}:H_{x}\right]<1+ef$ by
Lemma 4.5(2), and we again reach a contradiction.

Assume that $X$ isomorphic to one of the groups $PSp_{n}(q)$ or
$P\Omega_{n}(q)$. Then $\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant 2f$ by
[39], Table 5.1.A, and hence
$\theta\left(1+ef\right)\leqslant\lambda\left|G_{x,B}\right|\leqslant 2f$,
whereas $e=n-2\geqslant 2$ by Remark 4.2.

Assume that $X$ isomorphic to $P\Omega_{n}^{\varepsilon}(q)$, with
$\varepsilon=\pm$ and $(n,\varepsilon)\neq(8,+)$. Then
$\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant 8f$ by [39], Table 5.1.A, and
hence $\mu\left(1+ef\right)\theta=\lambda\left|G_{x,B}\right|\leqslant 8f$.
Then $\mu=\theta=1$, $n=6$ and $\rho=1+ef$ is a prime, and hence
$\lambda\left|G_{x,B}\right|=1+ef$, whereas $\lambda\left|G_{x,B}\right|<1+ef$
by Lemma 4.5.

Finally, assume that $X\cong P\Omega_{8}^{+}(q)$. Then
$\lambda\left|G_{x,B}\right|\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant
24f$ and hence $\lambda\left|G_{x,B}\right|=\theta\rho$, with $\theta\leqslant
3$ and $\rho=1+6f$ a prime. Moreover, either $\theta=1,3$ and
$\lambda\left|G_{x,B}\right|$ is odd, or $\theta=2$ and
$\lambda\left|G_{x,B}\right|\equiv 2\pmod{4}$. Then
$\lambda\left|G_{x,B}\right|\leqslant\left[G_{x}:H_{x}\right]<\rho$ by Lemma
4.5(2)–(3), and we again reach a contradiction. This completes the proof. ∎

The following theorem, which relies both on numerical properties of
$\Phi_{ef}^{\ast}(p)$ and on Lemmas 4.5 and 4.7, states that either $X_{x}$ is
a large subgroup of $X$ or $\Phi_{ef}^{\ast}(p)$ is a square, or
$\Phi_{ef}^{\ast}(p)$ has a non-trivial factor in common with $k+1$. As we
will see, the theorem is a tool that, together with [32] and [13], allows us
to control the structure of $X_{x}$.

######  Theorem 4.7.

Let $(x,B)$ be any flag in $\mathcal{D}$. Then one of the following holds:

  1. (1)

$\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)>1$ and one of the
following holds:

    1. (a)

$X_{x}$ is a large subgroup of $X$ and
$\left(\Phi_{ef}^{\ast}(p),\left|X_{B}\right|\right)>1$;

    2. (b)

$\left(\Phi_{ef}^{\ast}(p),k+1\right)>1$.

  2. (2)

$\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)=1$,
$\Phi_{ef}^{\ast}(p)\mid k^{2}$ and
$\left(\Phi_{ef}^{\ast}(p),\left|X_{B}\right|\right)=\Phi_{ef}^{\ast}(p)^{1/2}$.

###### Proof.

Since $r=\lambda(k+1)$ and $\left|G_{x}\right|=r\left|G_{x,B}\right|$, it
follows that

$\left|X_{x}\right|=\frac{\lambda(k+1)\left|G_{x,B}\right|}{\left[G_{x}:X_{x}\right]}\textit{.}$
(4.3)

Assume that $\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)>1$. If
$\left(\Phi_{ef}^{\ast}(p),k+1\right)=1$, then
$\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$ and hence
$\lambda\left|G_{x,B}\right|\geqslant\left[G_{x}:H_{x}\right]$ by Lemma 4.6.
Thus $\left|X_{x}\right|\geqslant k+1$ by (4.3), and hence $X_{x}$ is a large
subgroup of $X$, as $k^{2}=[X:X_{x}]$.

Since $\lambda\left|G_{x,B}\right|\mid\left|G_{B}\right|$ and
$\left(\Phi_{ef}^{\ast}(p),\lambda\left|G_{x,B}\right|\right)>1$, it follows
that $\left(\Phi_{ef}^{\ast}(p),\left|G_{B}\right|\right)>1$. Moreover, as
$\Phi_{ef}^{\ast}(p)\equiv 1\pmod{ef}$ by [39], Proposition 5.2.15.(i), it
follows that $\left(\Phi_{ef}^{\ast}(p),ef\right)=1$ and hence
$\left(\Phi_{ef}^{\ast}(p),\left|\mathrm{Out(}X\mathrm{)}\right|\right)=1$
(see [39], Table 5.1.A). Thus
$\left(\Phi_{ef}^{\ast}(p),\left|X_{B}\right|\right)>1$, as
$\left[G_{B}:X_{B}\right]\mid\left|\mathrm{Out(}X\mathrm{)}\right|$, which is
the assertion (1a).

Assume that $\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)=1$. Then
$\Phi_{ef}^{\ast}(p)\mid k^{2}$, since $\left[X:X_{x}\right]=k^{2}$, and hence
$\Phi_{ef}^{\ast}(p)\mid\left|X\right|$. Arguing as above, we obtain
$\left(\Phi_{ef}^{\ast}(p),\left|X_{B}\right|\right)=\Phi_{ef}^{\ast}(p)^{1/2}$,
which is the assertion (2). ∎

The following Lemma is the analogue of Corollary 3.10 of [11], but we do not
make any additional assumption on $X_{x}$ because of Proposition 4.3.

######  Lemma 4.8.

The following hold:

  1. (1)

$\left|X\right|<\left|\mathrm{Out(}X\mathrm{)}\right|^{2}\cdot\left|X_{x}\right|\cdot\left|X_{x}\right|_{p^{\prime}}^{2}$

  2. (2)

If $\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant\left|X_{x}\right|_{p}$,
then $X_{x}$ is a large subgroup of $X$.

###### Proof.

Since $p\mid k$ by Proposition 4.3, it follows that $(p,r/\lambda)=1$, as
$r/\lambda=k+1$, and hence
$r/\lambda\leqslant\left|G_{x}\right|_{p^{\prime}}\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|_{p^{\prime}}\left|X_{x}\right|_{p^{\prime}}$.
Since $k^{2}<\left(r/\lambda\right)^{2}$, it results
$\left|X\right|<\left|X_{x}\right|\left|\mathrm{Out(}X\mathrm{)}\right|_{p^{\prime}}^{2}\left|X_{x}\right|_{p^{\prime}}^{2}$,
and hence (1) holds. Moreover, if
$\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant\left|X_{x}\right|_{p}$, then
$\left|X\right|<\left|X_{x}\right|^{3}$ and hence $X_{x}$ is a large subgroup
of $X$, which is (2). ∎

The following lemma and the subsequent corollary provide a lower bound for
$\left|X_{B}\right|_{p}$.

######  Lemma 4.9.

$\left|X\right|_{p}\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|_{p}^{3}\left|X_{B}\right|_{p}^{3}$.

###### Proof.

Since $\left[G:G_{B}\right]=\lambda k(k+1)$, it follows that
$\left|G\right|_{p}=\lambda_{p}k_{p}(k+1)_{p}\left|G_{B}\right|_{p}$ by
Proposition 4.3. Thus,
$\left|G\right|_{p}=\lambda_{p}k_{p}^{2}\left|G_{x,B}\right|_{p}$.
If $\lambda_{p}\leqslant\left|G_{x,B}\right|_{p}$, then
$\left|G\right|_{p}\leqslant\left|G_{B}\right|_{p}^{2}$ and hence
$\left|X\right|_{p}\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|_{p}^{2}\left|X_{B}\right|_{p}^{2}$,
as $\left[G_{B}:X_{B}\right]$ divides $\left|\mathrm{Out(}X\mathrm{)}\right|$,
whereas if $\lambda_{p}>\left|G_{x,B}\right|_{p}$, then
$\left|G\right|_{p}\leqslant\left|G_{B}\right|_{p}^{3}$ and hence
$\left|X\right|_{p}\leqslant\left|G\right|_{p}\leqslant\left|G_{B}\right|_{p}^{3}\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|_{p}^{3}\left|X_{B}\right|_{p}^{3}$.
∎

It should be stressed out that, the following corollary works under the
assumptions of Remark 4.2. Furthermore, the lower bounds for
$\left|X_{B}\right|_{p}$ are not necessarily integers.

######  Corollary 4.10.

The following hold:

  1. (i).

$\left|X_{B}\right|_{p}\geqslant
q^{\frac{\left(n+2\right)\left(n-3\right)}{6}}$ for $X\cong PSL_{n}(q)$.

  2. (ii).

$\left|X_{B}\right|_{p}\geqslant q^{\frac{n^{2}-12}{12}}$ for $X\cong
PSp_{n}(q)$

  3. (iii).

$\left|X_{B}\right|_{p}\geqslant
q^{\frac{\left(n+2\right)\left(n-3\right)}{12}}$ for $X\cong
PSU_{n}(q^{1/2})$.

  4. (iv).

$\left|X_{B}\right|_{p}\geqslant q^{\frac{(n-1)^{2}-12}{12}}$ for $X\cong
P\Omega_{n}(q)$.

  5. (v).

$\left|X_{B}\right|_{p}\geqslant q^{\frac{n(n-2)-12}{12}}$ for $X\cong
P\Omega_{n}^{\varepsilon}(q)$.

In each case $\left|X_{B}\right|_{p}\geqslant p$

###### Proof.

Assume that $X$ is isomorphic to $PSL_{n}(q)$. Then
$\left|X\right|_{p}=q^{\frac{n\left(n-1\right)}{2}}$ and
$\left|\mathrm{Out(}X\mathrm{)}\right|_{p}=(2f)_{p}$. Since $q=p^{f}\geqslant
2f\geqslant(2f)_{p}$, it follows that $\left|X_{B}\right|_{p}\geqslant
q^{\frac{\left(n+2\right)\left(n-3\right)}{6}}$ by Lemma 4.9.

Assume that $X\cong\mathrm{PSp}_{n}(q)$. Hence,
$\left|X\right|_{p}=q^{\frac{n^{2}}{4}}$ and
$\left|\mathrm{Out(}X\mathrm{)}\right|_{p}=f_{p}$. Then
$\left|X_{B}\right|_{p}^{3}\geqslant q^{\frac{n^{2}}{4}-3}$ by Lemma 4.9 and
hence $\left|X_{B}\right|_{p}\geqslant q^{\frac{n^{2}-12}{12}}$.

Assume that $X\cong PSU_{n}(q^{1/2})$. Then
$\left|X\right|_{p}=q^{\frac{n(n-1)}{4}}$ and
$\left|\mathrm{Out(}X\mathrm{)}\right|_{p}=f_{p}$. Since $q^{1/2}\geqslant f$,
arguing as above, we obtain $\left|X_{B}\right|_{p}\geqslant
q^{\frac{\left(n+2\right)\left(n-3\right)}{12}}$.

Assume that $X\cong\Omega_{n}(q)$, with $nq$ odd. Hence,
$\left|X\right|_{p}=q^{\frac{(n-1)^{2}}{4}}$ and
$\left|\mathrm{Out(}X\mathrm{)}\right|_{p}=f_{p}$. Then
$\left|X_{B}\right|_{p}^{3}\geqslant q^{\frac{(n-1)^{2}}{4}-3}$ by Lemma 4.9
and hence $\left|X_{B}\right|_{p}\geqslant q^{\frac{(n-1)^{2}-12}{12}}$.

Assume that $X\cong P\Omega_{n}^{\varepsilon}(q)$. Then
$\left|X\right|_{p}=q^{\frac{n(n-2)}{4}}$ and
$\left|\mathrm{Out(}X\mathrm{)}\right|_{p}=\left(3f\right)_{p}$ or
$\left(2f\right)_{p}$ according as $(n,\varepsilon,p)$ is or is not $(8,+,3)$
respectively. In both cases we have
$q\geqslant\left|\mathrm{Out(}X\mathrm{)}\right|_{p}$, and hence
$\left|X_{B}\right|_{p}\geqslant q^{\frac{n(n-2)-12}{12}}$, again by Lemma
4.9.
In each case, $\left|X_{B}\right|_{p}>1$ and hence
$\left|X_{B}\right|_{p}\geqslant p$. ∎

In the sequel, we denote by $A^{\char 94\relax}$ the pre-image of any group
$A$ in the corresponding linear group.

Recall that $e=n,n-1$ or $n-2$. If $e\leqslant n/2$, then either $n=2$ and
$e=1$, or $n=4$ and $e=2$. However, such cases cannot occur by Remark 4.2.
Thus, $n/2<e\leqslant n$ for each $X$, and hence, if $H$ is any subgroup of
$X$ such that $(\Phi_{ef}^{\ast}(p),\left|H\right|)>1$, then $H^{\char
94\relax}$ is classified in the Main Theorem of [32], and therefore $H$ is
known.

Now, we are in position to tackle cases (1) and (2) of Theorem 4.7. They are
investigated in separate sections.

##  5\. Reduction to the case
$\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)>1$

The aim of this section is to rule out case (2) of Theorem 4.7. The proof
strategy for doing so is as follows. By using [32], [39] and [31] we determine
the structure of $X_{B}^{\char 94\relax}$. From this we derive an upper bound
for $\left|X_{B}\right|_{p}$. Then, we show that such a bound is in contrast
with the lower bound for $\left|X_{B}\right|_{p}$ determined in Corollary
4.10, and hence no cases arise.

Assume $\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)=1$,
$\Phi_{ef}^{\ast}(p)\mid k^{2}$ and
$\left(\Phi_{ef}^{\ast}(p),\left|X_{B}\right|\right)=\Phi_{ef}^{\ast}(p)^{1/2}$.
Then
$\Phi_{ef}^{\ast}(p)^{1/2}\mid\left(\left|X_{B}\right|,\left[X:X_{B}\right]\right)$.
Also, $p\mid\left|X_{B}\right|$ by Corollary 4.10 for $n\geqslant 4$. Then
$\Phi_{ef}^{\ast}(p)^{1/2}\mid\left(\left|X_{B}^{\char
94\relax}\right|,\left[SL_{n}(q):X_{B}^{\char 94\relax}\right]\right)$, with
$\Phi_{ef}^{\ast}(p)>1$ a square. Moreover, $p\mid\left|X_{B}^{\char
94\relax}\right|$, as $\left|X_{B}^{\char
94\relax}\right|_{p}=\left|X_{B}\right|_{p}$. Then $X_{B}^{\char 94\relax}$ is
isomorphic to one of the subgroups classified in [32] and satisfying the
previous additional constraints.

Let $Z$ denote the center of $GL_{n}(q)$. A subgroup $Y$ of $GL_{n}(q)$ is
_nearly simple_ if $S\leqslant Y/(Y\cap Z)\leqslant Aut(S)$, for some non-
abelian simple group $S$, and if $N$ is the preimage of $S$ in $Y$, then $N$
is absolutely irreducible on $V_{n}(q)$ and $N$ is not a classical group
defined over a subfield of $GF(q)$ (in its natural representation).

######  Lemma 5.1.

Let $\Phi_{ef}^{\ast}(p)>1$ be a square. If $Y$ is a nearly simple subgroup of
$SL_{n}(q)$ such that $p\mid\left|Y\right|$ and
$\Phi_{ef}^{\ast}(p)^{1/2}\mid\left(\left|Y\right|,\left[SL_{n}(q):Y\right]\right)$,
then one of the following holds:

  1. (1)

$e=n=4$, $q=7$, $\Phi_{4}^{\ast}(7)=5^{2}$ and $Z_{2}.A_{7}\unlhd Y\leqslant
Z_{2}.S_{7}$.

  2. (2)

$e=n=5$, $q=3$, $\Phi_{5}^{\ast}(3)=11^{2}$ and either $Y\cong PSL_{2}(11)$ or
$Y\cong M_{11}$.

###### Proof.

Assume that $S\cong A_{\ell}$, $\ell\geqslant 5$. Then $Y$ is one of the
groups listed in Example 2.6.(a)–(c) of [32]. If $Y$ is one of the groups in
Example 2.6.(b)–(c), then $\ell\leqslant 10$ and $\Phi_{ef}^{\ast}(p)^{1/2}$
is one of the primes $5$ or $7$, as $n\geqslant 3$ (see Tables 2–3 in [32]).
Then $\Phi_{ef}^{\ast}(p)$ is either $5^{2}$ or $7^{2}$. Actually, only the
former is admissible and occurs for $(ef,p)=(4,7)$ by [31], Theorem 3. Then
$Z_{2}.A_{7}\trianglelefteq Y$ by Tables 2–3 in [32], since
$p\mid\left|Y\right|$, and hence $Z_{2}.A_{7}\unlhd Y\leqslant Z_{2}.S_{7}$ by
[54].

Suppose that $Y$ is one of the groups in Example 2.6.(a). Then $V_{n}(q)$,
$q=p$, is the fully deleted permutation module for $S$ and $A_{\ell}\leqslant
Y\leqslant S_{\ell}\times Z_{p-1}$, where $\ell=n+\varepsilon$,
$\varepsilon=1$ or $2$ according as $p$ does not divide or does divide $n$
respectively. Let $u$ be any prime divisor of $\Phi_{ef}^{\ast}(p)$. Then
$u^{i}\mid\left|Y\right|$, and hence $u^{i}\mid\left|S\right|$, for some
$i\geqslant 1$. Then $i<\frac{\ell-1}{u}$ by [28], Exercise 2.6.8. On the
other hand, $u=1+te$, for some $t\geqslant 1$, by [39], Proposition
5.2.15.(i). Thus $1+i(1+te)<n+\varepsilon$, where $e=n,n-1$ or $n-2$. Then
$e\neq n$, $i=1$, and either $t=1$, or $\varepsilon=2$, $e=1$, $n=p=3$ and
$t=2$. The latter implies $u=p=3$, and hence a contradiction, whereas the
former yields $(e,q,u)=(4,7,5)$ by [31], Theorem 3, since
$\Phi_{e}^{\ast}(q)=(e+1)^{2}$. Then $(\ell,n,\varepsilon)=(7,6,1)$, since
$\ell=n+\varepsilon$, $e=n-1,n-2$ and $7\mid\left|Y\right|$. However, $A_{7}$
does not afford an irreducible representation on $V_{4}(7)$ by [54], and this
case is excluded.

Assume that $S$ is sporadic. Then $S$ is one of the groups listed in Table 5
of [32]. Note that, the order of each of these groups is divisible by exactly
one primitive prime divisor of $p^{e}-1$, with $e=n,n-1$ or $n-2$. Thus
$\Phi_{ef}^{\ast}(p)$ is the square of a prime, and hence $(ef,p,u)=(4,7,5)$
or $(5,3,11)$ by [31], Theorem 3. Only the latter yields a group that is
listed in Table 5 and this is $Y^{\prime}\cong M_{11}$. Then $Y\cong M_{11}$,
since $\mathrm{Out(}M_{11}\mathrm{)}=1$ by [54].

Assume that $S$ is a Lie type simple group in characteristic different from
$p$. Then $Y$ can be determined from Tables 7–8 of [32]. As in the sporadic
case, the order of the groups in these two tables is divisible by exactly one
primitive prime divisor of $p^{e}-1$, with $e=n,n-1$ or $n-2$. Then
$\Phi_{ef}^{\ast}(p)$ is the square of a prime and hence $(ef,p,u)=(4,7,5)$ or
$(5,3,11)$ as above. If $S$ is listed in Table 7, then $S\cong Sp_{4}(3)$, but
no groups $Y$ occur with order divisible by $7$. Hence, assume $S$ is one of
the groups listed in Table 8. If $u=5$, then $e=4$ and it is easy to see that
$S$ is isomorphic to one of the groups $PSp_{4}(3)$, or $PSL_{2}(9)$. However,
there are no corresponding $Y$ with order divisible by $7$. Thus, $u=11$,
$u=2e+1$ and hence $s=11$ and $S\cong PSL_{2}(11)$. Then $Y\cong PSL_{2}(11)$
by [54].

Assume that $S$ is a Lie type simple group in characteristic $p$. Then $Y$ is
one of the groups in Example 2.8, and hence in Table 6, of [32]. Let
$GF(q_{0})$ be a subfield of $GF(q)$ for which $Y\leqslant GL_{n}(q_{0})\circ
Z_{q-1}$, and $Y\cap GL_{n}(q_{0})$ cannot be realized modulo scalar over a
proper subfield of $GF(q_{0})$. Let $u$ be a prime divisor of
$\Phi_{ef}^{\ast}(p)$, then $u$ is a primitive prime divisor of $p^{ef}-1$ and
this forces $q_{0}=q$ by [39], Proposition 5.2.15.(ii).
Assume that $n=8$, $e=6$ and $Y^{(\infty)}\cong SL_{2}(q^{3})$. Then
$\left[SL_{8}(q):Y\right]$ divides

$\left[SL_{8}(q):Y^{(\infty)}\right]=q^{25}\prod_{i=2,i\neq 6}^{8}(q^{i}-1)$

and hence $\Phi_{6f}^{\ast}(p)^{1/2}$ does not divide
$\left[SL_{8}(q):Y\right]$, a contradiction. A similar reasoning rules out the
remaining cases in Table 6. ∎

######  Lemma 5.2.

Let $\Phi_{ef}^{\ast}(p)>1$ be a square. If
$\Phi_{ef}^{\ast}(p)^{1/2}\mid\left(\left|Y\right|,\left[SL_{n}(q):Y\right]\right)$
and $p\mid\left|Y\right|$, then $Y$ is not isomorphic to any of the groups
listed in Examples 2.1, 2.3 or 2.5 of [32].

###### Proof.

Assume that $Y$ is isomorphic to one of the groups defined as in Example 2.1
of [32]. Then $\Phi_{ef}^{\ast}(p)$ divides the order of $Y$, but it is
coprime to $\left[SL_{n}(q):Y\right]$, and this contradicts our assumptions.

Assume that $Y$ is isomorphic to one of the groups defined as in Example 2.3
of [32]. Then $Y\leqslant GL_{1}(q)\wr S_{n}$ and $u=e+1\leqslant n$ for each
prime divisor $u$ of $\Phi_{ef}^{\ast}(p)^{1/2}$. Then $e=n-1$ and $u=n$ or
$e=n-2$ and $u=n-1$. Thus $\Phi_{ef}^{\ast}(p)=u^{2}=(e+1)^{2}$ and hence
$(ef,p,u)=(4,7,5)$ by [31], Theorem 3. Therefore, $f=1$, $u=n=5$ and hence
$Y\leqslant GL_{1}(7)\wr S_{5}$. However, this group cannot occur since
$7\nmid\left|Y\right|$.

Assume that $Y$ is isomorphic to one of the groups defined as in Example 2.5
of [32]. Let $u$ be a prime divisor of $\Phi_{ef}^{\ast}(p)$. Then $e=n=2^{m}$
and either $u=n+1$ is a Fermat prime or $u=n-1$ is a Mersenne prime. Moreover,
$\Phi_{ef}^{\ast}(p)^{1/2}=u$ by Table 1 of [32]. Hence
$\Phi_{ef}^{\ast}(p)=u^{2}$ and, as above, $(e,q,u)=(4,7,5)$ since $n$ is
even. Thus $Y\leqslant\left((D_{8}\circ Q_{8})\cdot S_{5}\right)\circ Z_{6}$,
which cannot occur since $7$ does not divide $\left|Y\right|$. This completes
the proof ∎

######  Proposition 5.3.

Let $\Phi_{ef}^{\ast}(p)>1$ be a square. If
$\Phi_{ef}^{\ast}(p)^{1/2}\mid\left(\left|Y\right|,\left[SL_{n}(q):Y\right]\right)$
and $p\mid\left|Y\right|$, then one of the following holds:

  1. (1)

$e=n$ and one of the following holds:

    1. (a)

$Y\leqslant GL_{1}(q^{n})\cdot Z_{n}$ and $p\mid n$.

    2. (b)

$e=n=4$, $q=7$, $\Phi_{4}^{\ast}(7)=5^{2}$ and $Z_{2}.A_{7}\unlhd Y\leqslant
Z_{2}.S_{7}$.

    3. (c)

$e=n=5$, $q=3$, $\Phi_{5}^{\ast}(3)=11^{2}$ and either $Y\cong PSL_{2}(11)$ or
$Y\cong M_{11}$.

  2. (2)

$e=n-1$ and $Y$ lies in a parabolic subgroup of type $P_{1}$ of $SL_{n}(q)$.

  3. (3)

$e=n-2$ and $Y$ lies in a parabolic subgroup of type $P_{2}$ or $P_{1,n-1}$ of
$SL_{n}(q)$.

###### Proof.

It follows from Lemmas 5.1 and 5.2 that either assertions (1.b) or (1.c) hold,
or $Y$ is isomorphic to one of the groups listed in Examples 2.2 and 2.4 of
[32].

Assume that $e=n$. If $Y$ is isomorphic to one of the groups listed as in
Example 2.2 of [32], then $Y$ is the stabilizer of a subspace or a quotient of
a subspace $V_{n}(q)$ of dimension $m\geqslant e$, and hence this case cannot
occur for $e=n$. Then $Y$ is listed in Example 2.4 of [32] and hence
$Y\leqslant GL_{n/c}(q^{c})\cdot Z_{c}$ for some divisor $c$ of $n$ with
$c>1$. Clearly, $\Phi_{nf}^{\ast}(p)$ divides the order of
$GL_{n/c}(q^{c})\cdot Z_{c}$. Then $\Phi_{nf}^{\ast}(p)$ divides the order of
$SL_{n/c}(q^{c})$, as $\Phi_{nf}^{\ast}(p)\equiv 1\pmod{nf}$ by [39],
Proposition 5.2.15.(i). Also $\Phi_{nf}^{\ast}(p)$ is coprime to
$\left[SL_{n}(q):SL_{n/c}(q^{c})\right]$, and hence
$\Phi_{nf}^{\ast}(p)^{1/2}$ divides the order of $Y\cap SL_{n/c}(q^{c})$,
since $Y\leqslant GL_{n/c}(q^{c})\cdot Z_{c}$.

Set $f^{\prime}=cf$. Clearly,
$\Phi_{nf}^{\ast}(p)=\Phi_{\frac{n}{c}f^{\prime}}^{\ast}(p)$ and
$\Phi_{\frac{n}{c}f^{\prime}}^{\ast}(p)^{1/2}$ divides $\left|Y\cap
SL_{n/c}(q^{c})\right|$ and $\left[SL_{n/c}(q^{c}):Y\cap
SL_{n/c}(q^{c})\right]$. Hence, we may apply Lemmas 5.1 and 5.2 with $Y\cap
SL_{n/c}(q^{c})$ and $SL_{n/c}(q^{c})$ in the role $Y$ and $SL_{n}(q)$
respectively, thus obtaining $Y\cap SL_{n/c}(q^{c})\leqslant
GL_{n/c^{\prime}}(q^{c^{\prime}})\cdot Z_{c^{\prime}}$ with $c\mid
c^{\prime}$. Indeed, no cases arise from Lemma 5.1, since $c>1$. We may
iterate the previous argument and eventually obtain that $Y\leqslant
GL_{1}(q^{n})\cdot Z_{n}$ and $p\mid n$, as $p\mid\left|Y\right|$, which is
(1.a).

Assume that $e=n-1$. If $Y$ is a group listed in Example 2.2 of [32], then $Y$
is the stabilizer of a subspace or a quotient of a subspace $V_{n}(q)$ of
dimension $n-1$ and we obtain (2).

Assume that $Y$ is listed in Example 2.4 of [32]. Let $u$ is prime divisor of
$\Phi_{(n-1)f}^{\ast}(p)$, then $u=n$ and $Y\leqslant GL_{1}(q^{n-1})\cdot
Z_{n-1}$ by Example 2.4(a) of [32]. Since $Y$ contains a normal Sylow
$u$-subgroup $U$ which stabilizes a unique hyperplane of $V_{n}(q)$, we obtain
(2).

Assume that $e=n-2$. If $Y$ is one of the groups listed in Example 2.2 of
[32], then $Y$ is the stabilizer of a subspace or a quotient of a subspace $Q$
of $V_{n}(q)$ of dimension $m=n-1$ or $n-2$. In the latter case $Y$ lies in a
subgroup of type $P_{2}$ of $SL_{n}(q)$, and hence (3) holds.

Assume that $m=n-1$. Suppose that $\dim Q=1$. Then
$Y\leqslant[q^{n-1}]:GL_{n-1}(q)$, with $[q^{n-1}]$ fixing $Q$ pointwise, by
[39], Proposition 4.1.17.(II). Set $W=Y/(Y\cap[q^{n-1}])$ and
$n^{\prime}=n-1$. Hence $e=n-2=n^{\prime}-1$ and $W$ acts on $V_{n}(q)/Q$
inducing a subgroup of $GL_{n^{\prime}}(q)$. Moreover, arguing as in the
$e=n-1$ case, we see that $W$ is the stabilizer of a subspace or a quotient of
a subspace $V_{n}(q)/Q$ of dimension $n^{\prime}-1$. Hence $Y$ lies in a
parabolic subgroup of type $P_{2}$ or $P_{1,n-1}$ of $SL_{n}(q)$, which is
(3). The same conclusions hold for $\dim Q=n-1$.

Finally, assume that $Y$ is one of the groups in Example 2.4 of [32]. Then $n$
is even and $Y\leqslant GL_{n/2}(q^{2})\cdot Z_{2}$. If $f^{\prime}=2f$, then
$\Phi_{\left(n/2-1\right)f^{\prime}}^{\ast}(p)^{1/2}\mid\left(\left|Y\right|,\left[SL_{n/2}(q^{2}):Y\right]\right)$
and $p\mid\left|Y\right|$. Arguing as in the $e=n-1$ case, we get that $Y$
fixes subspace or a quotient of a subspace $V_{n/2}(q^{2})$ of dimension
$n/2-1$. Then $Y$ is lies in a parabolic subgroup of type $P_{2}$ of
$SL_{n}(q)$, and we obtain (3). ∎

######  Corollary 5.4.

$X_{B}^{\char 94\relax}$ is isomorphic to one of the groups listed in
Proposition 5.3.

###### Proof.

Since $\Phi_{ef}^{\ast}(p)^{1/2}\mid\left(\left|X_{B}^{\char
94\relax}\right|,\left[SL_{n}(q):X_{B}^{\char 94\relax}\right]\right)$ and
$q\mid\left|X_{B}^{\char 94\relax}\right|$, the assertion follows. ∎

Now, we analyze the cases $e=n,n-1$ or $n-2$ separately.

######  Proposition 5.5.

If $\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)=1$, then $e\neq n$.

###### Proof.

Assume that $\left(\Phi_{nf}^{\ast}(p),\left|X_{x}\right|\right)=1$. If $n=3$,
then $X$ is isomorphic to one of the groups $PSL_{3}(q)$ or
$PSU_{3}(q^{1/2})$. Then $\left|X\right|_{p}=p^{3f}$,
$\left|\mathrm{Out(}X\mathrm{)}\right|_{p}\leqslant(2f)_{p}$ and
$\left|X_{B}\right|_{p}\leqslant(3,p)$. Then
$p^{3f}\leqslant(2f)_{p}^{3}(3,p)^{3}$ by Lemma 4.9, and hence
$p^{f}\leqslant(2f)_{p}(3,p)$. If $p\neq 3$, then $p^{f}\leqslant 2f$ and
hence $p=2$ and $f=1,2$. Then $X$ is isomorphic either $PSL_{3}(2)$ or to
$PSL_{3}(4)$. However, these cases cannot occur by Lemmas 4.1 and 4.4,
respectively. Thus $p=3$, $3^{f}\leqslant 3f$ and hence $f=1$. Then $X\cong
PSL_{3}(3)$, and we get a contradiction since $\Phi_{3}^{\ast}(3)=13$ is not a
square. Therefore, $n\geqslant 4$.

Assume that $X_{B}^{\char 94\relax}$ is nearly simple. If $n=5$ and $q=3$,
then $X\cong PSL_{5}(3)$ and either $X_{B}\cong PSL_{2}(11)$ or $X_{B}\cong
M_{11}$. However, these groups cannot occur since
$\left|X_{B}\right|_{3}=3^{2}$, whereas $\left|X_{B}\right|_{3}\geqslant
3^{3}$ by Corollary 4.10(1). Then $n=4$, $q=7$, $X$ is isomorphic to one of
the groups $PSL_{4}(7)$ or $PSp_{4}(7)$ (see Remark 4.2), and $X_{B}\cong
A_{7}$.
If $X\cong PSp_{4}(7)$, then $b=[X:X_{B}]=54880$, since $A_{7}$ is maximal in
$PSp_{4}(7)$, and this one has a unique conjugacy class of subgroups
isomorphic to $A_{7}$, by [14], Tables 8.13–8.13. Then $\lambda(k+1)k=54880$,
with $2\leqslant\lambda\leqslant k\leqslant 165$ and $\lambda\mid k$, since
$b=\lambda(k+1)k$ and $\lambda\mid k$. However, the previous equation does not
have solutions, and hence this case is ruled out.
If $X\cong PSL_{4}(7)$, then $[X:X_{B}]=919679040$. Hence $b=919679040\cdot
t$, for some $t$ divisor of $4$, since
$b=\frac{[G:X]}{[G_{B}:X_{B}]}[X:X_{B}]$ and $\frac{[G:X]}{[G_{B}:X_{B}]}$
divides $\left|\mathrm{Out(}X\mathrm{)}\right|$, and since
$\left|\mathrm{Out(}X\mathrm{)}\right|=4$. Then $\lambda(k+1)k=919679040\cdot
t$, with $2\leqslant\lambda\leqslant k\leqslant 21443\cdot t$ and $\lambda\mid
k$, since $b=\lambda(k+1)k$ and $\lambda\mid k$. However, the previous
equation does not have solutions, and hence this case is excluded.
Assume that $X_{B}^{\char 94\relax}\leqslant GL_{1}(q^{n})\cdot Z_{n}$ and
$p\mid n$. Then $\left|X_{B}\right|_{p}=\left|X_{B}^{\char
94\relax}\right|_{p}\leqslant n_{p}\leqslant n$ by Proposition 5.3(1). On the
other hand, by Corollary 4.10(i)–(iii), either
$\left|X_{B}\right|_{p}\geqslant q^{\frac{n^{2}}{12}}$ for $X\cong
PSp_{n}(q)$, or $\left|X_{B}\right|_{p}\geqslant
q^{\frac{\left(n+2\right)\left(n-3\right)}{6\theta}}$ with $\theta=1$ for
$X\cong PSL_{n}(q)$ and $\theta=2$ for $X\cong PSU_{n}(q^{1/2})$ and $n$ odd.
In the former case, $q^{\frac{n^{2}}{12}}\leqslant n$ and hence $(q,n)=(2,4)$
but this case was already ruled out (see the remark before Lemma 3.4). Thus
$q^{\frac{\left(n+2\right)\left(n-3\right)}{6\theta}}\leqslant n$ and hence
either $\theta=1$ and $(n,q)=(4,2),(4,3),(4,4),(4,5),(5,2)$, or $\theta=2$ and
$(q,n)=(4,5),(9,5),(16,5)$. However, $\Phi_{nf}^{\ast}(p)$ is a square for
none of these pairs, and hence they are excluded. ∎

######  Proposition 5.6.

If $\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)=1$, then $e\neq n-1$.

###### Proof.

Assume that $\left(\Phi_{(n-1)f}^{\ast}(p),\left|X_{x}\right|\right)=1$. Then
$X$ is isomorphic either to $\Omega_{n}(q)$ with $nq$ odd, or to
$PSU_{n}(q^{1/2})$ with $n$ even. Suppose that the latter occurs. Then
$X_{B}^{\char 94\relax}$ fixes a non-degenerate $1$-dimensional subspace of
$V_{n}(q)$ by Proposition 5.3(2) and by [39], Propositions 4.1.4 and 4.1.20,
since its order is divisible by $\Phi_{(n-1)f}^{\ast}(p)^{1/2}$. Hence
$X_{B}^{\char 94\relax}\leqslant GU_{n-1}(q^{1/2})$. Then $X_{B}^{\char
94\relax}\cap SU_{n-1}(q^{1/2})$ is a subgroup of $SL_{n-1}(q)$ such that
$\Phi_{n-1f}^{\ast}(p)$ divides the order of $X_{B}^{\char 94\relax}\cap
SU_{n-1}(q^{1/2})$ and the index of this one in $SL_{n-1}(q)$. Also $p$
divides the order of $X_{B}^{\char 94\relax}\cap SU_{n-1}(q^{1/2})$. Then we
may apply Proposition 5.3, with $X_{B}^{\char 94\relax}\cap SU_{n-1}(q^{1/2})$
in the role of $Y$ and $SL_{n-1}(q)$ in the role of $SL_{n}(q)$, and we obtain
that $X_{B}^{\char 94\relax}\cap SU_{n-1}(q^{1/2})$ is contained in
$GL_{1}(q^{n-1})\cdot Z_{n-1}$, with $p\mid n-1$, since $q$ is a square. Then
$\left|X_{B}^{\char 94\relax}\right|_{p}\leqslant(n-1)_{p}\leqslant n-1$, as
$\left|X_{B}^{\char 94\relax}\cap
SU_{n-1}(q^{1/2})\right|_{p}=\left|X_{B}^{\char 94\relax}\right|_{p}$. On the
other hand, $\left|X_{B}\right|_{p}\geqslant
q^{\frac{\left(n+2\right)\left(n-3\right)}{12}}$ by Corollary 4.10(iii).
Therefore $q^{\frac{\left(n+2\right)\left(n-3\right)}{12}}\leqslant n-1$ and
hence $n=4$ and $q=4$ or $9$, but in none of these cases
$\Phi_{2(n-1)}^{\ast}(p)$ is a square.

Assume that $X\cong$ $\Omega_{n}(q)$ with $nq$ odd. Then $X_{B}^{\char
94\relax}$ preserves a non-degenerate $1$-dimensional subspace of $V_{n}(q)$
by Proposition 5.3 and by [39], Propositions 4.1.6 and 4.1.14, since
$\Phi_{(n-1)f}^{\ast}(p)^{1/2}$ divides $\left(\left|X_{B}^{\char
94\relax}\right|,\left[SL_{n}(q):X_{B}^{\char 94\relax}\right]\right)$ and
since $p$ divides $\left|X_{B}^{\char 94\relax}\right|$ by Corollary 4.10.
Thus $X_{B}^{\char 94\relax}\leqslant\Omega_{n-1}^{-}(q).Z_{2}<SL_{n-1}(q)$
and hence $X_{B}^{\char 94\relax}$ is a subgroup of $SL_{n-1}(q)$ such that
$\Phi_{(n-1)f}^{\ast}(p)^{1/2}$ divides $\left(\left|X_{B}^{\char
94\relax}\right|,\left[SL_{n-1}(q):X_{B}^{\char 94\relax}\right]\right)$. Also
$p\mid\left|X_{B}^{\char 94\relax}\right|$. Then we may apply Proposition 5.3,
with $X_{B}^{\char 94\relax}$ in the role of $Y$ and $SL_{n-1}(q)$ in that of
$SL_{n}(q)$, thus obtaining either $q=7$, $n-1=4$ and $Z_{2}.A_{7}\unlhd
X_{B}^{\char 94\relax}\leqslant Z_{2}.A_{7}$, or $X_{B}^{\char
94\relax}\leqslant GL_{1}(q^{n-1})\cdot Z_{n-1}$ and $p\mid n-1$. The former
case cannot occur, since $X\cong\Omega_{5}(7)\cong\mathrm{PSp}_{4}(7)$
contains $A_{7}$ but not its covering group $Z_{2}.A_{7}$. Thus $X_{B}^{\char
94\relax}\leqslant GL_{1}(q^{n-1})\cdot Z_{n-1}$ and $p\mid n-1$, and hence
$\left|X_{B}\right|_{p}\leqslant(n-1)_{p}\leqslant n-1$. On the other hand,
$q^{\frac{(n-1)^{2}-12}{12}}\leqslant\left|X_{B}\right|_{p}$ by Corollary
4.10(3). Then $q^{\frac{(n-1)^{2}-12}{12}}\leqslant n-1$ and hence $n=5$ and
$q<4^{3}$, since $n\geqslant 5$ by Remark, 4.2. Then $q=7$, since
$\Phi_{4f}^{\ast}(p)$ must be a square, but $7$ does not divide $n-1=4$. So,
this case cannot occur and the proof is completed. ∎

######  Theorem 5.7.

$\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)>1$.

###### Proof.

Suppose the contrary. Then $e=n-2$ by Lemmas 5.5 and 5.6, and hence $X\cong
P\Omega_{n}^{+}(q)$. By Proposition 5.3, $X_{B}^{\char 94\relax}$ lies in a
parabolic subgroup of type $P_{2}$ or $P_{1,n-1}$ of $SL_{n}(q)$. If the
former occurs, then $X_{B}^{\char 94\relax}$ preserves a non-degenerate
$2$-dimensional subspace of $V_{n}(q)$ of type $-$ by [39], Propositions 4.1.6
and 4.1.20, since $n\geqslant 6$ by Remark 4.2.
If $X_{B}^{\char 94\relax}$ lies in a parabolic subgroup of type $P_{1,n-1}$
of $SL_{n}(q)$, then $X_{B}^{\char 94\relax}$ preserves two subspaces $U$, $W$
of $V_{n}(q)$ such that $U<W$ and $\dim U=1$ and $\dim W=n-1$. Also, $U$ is
non-degenerate by [39], Proposition 4.1.20, since
$\Phi_{(n-2)f}^{\ast}(q)^{1/2}$ divides the order of $X_{B}^{\char 94\relax}$.
Thus $W^{\perp}\neq U$, since $W^{\perp}<U^{\perp}$, and hence $X_{B}^{\char
94\relax}$ preserves $W^{\perp}$ with $\dim W^{\perp}=1$. Then
$W^{\perp}\oplus U$ is a non-degenerate $2$-dimensional subspace of $V_{n}(q)$
of type $-$ preserved by $X_{B}^{\char 94\relax}$. Therefore, $X_{B}^{\char
94\relax}$ preserves a non-degenerate $2$-dimensional subspace of $V_{n}(q)$
of type $-$ in each case.

The group $X_{B}^{\char 94\relax}$ is contained in a maximal subgroup $M$ of
$X^{\char 94\relax}$, where $M$ is either
$\left(Z_{\frac{q+1}{(q-1,2)}}\times\Omega_{n-2}^{-}(q)\right).Z_{2}$ or
$\left(Z_{\frac{q+1}{(q-1,2)}}\times\Omega_{n-2}^{-}(q)\right).[4]$ according
to whether $q$ is even or odd respectively, by [39], Propositions 4.1.6. Thus

$\left|X_{B}\right|_{p}=\left|X_{B}^{\char
94\relax}\right|_{p}=(2,q)\left|X_{B}^{\char
94\relax}\cap\Omega_{n-2}^{-}(q)\right|_{p}.$ (5.1)

Note that, $\Phi_{(n-2)f}^{\ast}(q)^{1/2}$ divides the order of $X_{B}^{\char
94\relax}\cap\Omega_{n-2}^{-}(q)$ and the index of it in $SL_{n-2}(q)$. Also
$p\mid\left|X_{B}^{\char 94\relax}\right|$ by Corollary 4.10, and hence $p$
divides the order of $X_{B}\cap\Omega_{n-2}^{-}(q)$ by (5.1), since $X\ncong
P\Omega_{6}(2)^{+}$ by Lemma 4.4. Then we may apply Proposition 5.3, with
$X_{B}\cap\Omega_{n-2}^{-}(q)$ in the role of $Y$ and $SL_{n-2}(q)$ in the
role of $SL_{n}(q)$, thus obtaining either $q=7$, $n-2=4$ and $X_{B}^{\char
94\relax}\cap\Omega_{n-2}^{-}(q)\cong Z_{2}.A_{7}$, or $X_{B}^{\char
94\relax}\cap\Omega_{n-2}^{-}(q)\leqslant GL_{1}(q^{n-2})\cdot Z_{n-2}$ with
$p\mid n-2$. Since $X\cong P\Omega_{6}^{+}(7)\cong PSL_{4}(7)$, we may apply
the same argument used in the proof of Proposition 5.6 to rule out this case.
Thus $X_{B}^{\char 94\relax}\cap\Omega_{n-2}^{-}(q)\leqslant
GL_{1}(q^{n-2})\cdot Z_{n-2}$ and hence $\left|X_{B}^{\char
94\relax}\cap\Omega_{n-2}^{-}(q)\right|_{p}=(n-2)_{p}\leqslant(n-2)$.
Therefore

$q^{\frac{n(n-2)-12}{12}}\leqslant\left|X_{B}\right|_{p}\leqslant(2,q)(n-2)$
(5.2)

by Corollary 4.10(v), where $n\geqslant 6$ by Remark 4.2. It is
straightforward to see that, only $n=6$ and $q=2,4$ or $8$ fulfill (5.2).
However, all these cases are ruled out. Indeed, $q=2$ cannot occur by Lemma
4.4, whereas $q=4$ or $8$ cannot occur since $\Phi_{8}^{\ast}(2)=17$ and
$\Phi_{12}^{\ast}(2)=13$ respectively. ∎

##  6\. $X_{x}$ is a large subgroup of $X$

The aim of this section is to show that
$\left(\Phi_{ef}^{\ast}(p),k+1\right)>1$ implies that $X_{x}$ is a large
subgroup of $X$. This allows us to treat Cases (1.a) and (1.b) of Theorem 4.7
simultaneously.

The proof strategy is as follows. Since
$\left(\Phi_{ef}^{\ast}(p),k+1\right)>1$, and since $k+1$ divides the length
of each $G_{x}$-orbit distinct from $\left\\{x\right\\}$ by Lemma 3.2(3), we
derive that $\Phi_{ef}^{\ast}(p)\mid\left|X_{x}\right|$ and, moreover, that
$X_{x}$ contains a Singer cyclic subgroup of $X$ for $e=n$ by using simple
group-theoretical arguments. Then $X_{x}^{\char 94\relax}$ is a subgroup of
$SL_{n}(q)$ classified in Theorem 3.1 of [13] for $e<n$. For $e=n$, we use
instead the classification for $X_{x}^{\char 94\relax}$ contained in [15]. By
combining the fact that the candidates $X_{x}$ must fulfill
$\left[X:X_{x}\right]=k^{2}$ with the information contained in Lemma 3.13 of
[11], we show that
$\left|\mathrm{Out(}X\mathrm{)}\right|\geqslant\left|X_{x}\right|_{p}$. Hence,
$X_{x}$ is a large subgroup of $X$ by Lemma 4.8(2).

######  Lemma 6.1.

If $\left(\Phi_{ef}^{\ast}(p),k+1\right)>1$, then
$\Phi_{ef}^{\ast}(p)\mid\left|X_{x}\right|$. Moreover, if $e=n$, then $X_{x}$
contains a Singer cyclic subgroup of $X$.

###### Proof.

Let $w$ be any primitive prime divisor $w$ of $q^{e}-1$ dividing $k+1$ and let
$W$ be a Sylow $w$-subgroup of $G_{x}$. Then $w\equiv 1\pmod{ef}$ and hence
$(u,ef)=1$. Thus $w$ is coprime to $\left|\mathrm{Out(}X\mathrm{)}\right|$
(see [39], Table 5.1.A). Therefore, $W\leqslant X_{x}$.

Assume that $W$ fixes a further point $y$ of $\mathcal{D}$. Then
$\left|y^{G_{x}}\right|$ has order prime to $w$. However, this is impossible,
since $k+1$ divides $\left|y^{G_{x}}\right|$ by Lemma 3.2(3), and since $u\mid
k+1$. Thus $x$ is the unique point fixed by $W$. Then $N_{X}(W)\leqslant
N_{G}(W)\leqslant G_{x}$, and hence $N_{X}(W)\leqslant X_{x}$, as $W\leqslant
X_{x}$.
The group $W$ preserves a unique $(e-1)$-subspace $\mathcal{H}$ of
$PG_{n-1}(q)$ and acts faithfully and irreducibly on it (see [33], Theorem
3.5). So does $N_{X}(W)$. It is not difficult to see that $W$ lies in a cyclic
subgroup $Q$ of $X_{\mathcal{H}}$ inducing a Singer cyclic subgroup $\bar{Q}$
on $\mathcal{H}$. The order of $\bar{Q}$ is well known and, for instance, can
be found in [15],[34],[37], or determined in [39], Section 4.3. Thus
$\Phi_{ef}^{\ast}(p)$ divides the order of $\bar{Q}$, and hence that of $Q$.
Then $\Phi_{ef}^{\ast}(p)$ divides $\left|N_{X}(W)\right|$, and hence
$\left|X_{x}\right|$, and as $Q\leqslant N_{X}(W)$. Finally, if $e=n$, clearly
$\mathcal{H}=PG_{n-1}(q)$ and hence $Q$ is a Singer cyclic subgroup of $X$. ∎

Now, case $e=n,n-1$ and $n-2$ are analyzed separately.

######  Lemma 6.2.

If $\left(\Phi_{nf}^{\ast}(p),k+1\right)>1$, then $X_{x}$ is a large maximal
$\mathcal{C}_{3}$-subgroup of $X$.

###### Proof.

Assume that $\left(\Phi_{nf}^{\ast}(p),k+1\right)>1$. Then $X_{x}$ contains a
Singer cyclic subgroup of $X$ by Lemma 6.1. Let $Y$ be a maximal subgroup of
$X$ containing $X_{x}$. It follows from the Main Theorem of [15] that, either
$Y\in\mathcal{C}_{3}(X)$ or, since $\Phi_{nf}^{\ast}(p)>1$ and
$\Phi_{nf}^{\ast}(p)\mid\left|Y\right|$, one of the following holds:

  1. (1)

$X\cong PSU_{n}(q)$, where $(n,q)=(3,3),(3,5)$ or $(5,2)$.

  2. (2)

$X\cong PSp_{n}(q)$, where either $q$ is even and $\Omega_{n}^{-}(q)\leqslant
Y$, or $(n,q)=(8,2)$.

  3. (3)

$X\cong P\Omega_{6}^{-}(3)$.

The case $X\cong PSU_{3}(3)$ is ruled out in Lemma 3.6 of [44], as
$PSU_{3}(3)\cong G_{2}(2)^{\prime}$. The group $PSU_{3}(5)$ is ruled out by
[20], as it does not have a transitive permutation representation of square
degree. The same conclusion holds for $PSp_{8}(2)$ by [14], Tables 8.48–8.49.
If $X$ is isomorphic either to $PSU_{5}(2)$ or to $P\Omega_{6}^{-}(3)$, then
$X_{x}\cong PSL_{2}(11)$ and $k=2^{4}\cdot 3^{2}$, or $X_{x}\cong A_{7}$ and
$k=2^{4}\cdot 3^{2}$ , respectively, by [20]. However, these cases are ruled
out since $k+1$ does not divide the order of $\mathrm{Aut(}X\mathrm{)}$ and
hence that of $G$.

Suppose that $X\cong PSp_{n}(q)$, with $q$ even, and
$\Omega_{n}^{-}(q)\leqslant Y$. Then $Y\cong SO_{n}^{-}(q)$ by [39],
Proposition 4.8.6. If $X_{x}\neq Y$, then $X_{x}\cong\Omega_{n}^{-}(q)$ again
by Main Theorem of [15] but applied to $Y$. Thus, $X_{x}$ is isomorphic either
to $\Omega_{n}^{-}(q)$ or to $SO_{n}^{-}(q)$. If $\rho$ is the index of
$X_{x}$ in $Y$, then $\rho=1,2$ and hence
$k^{2}=\frac{q^{n(n+2)/4}}{\rho}(q^{n/2}-1)$, which has no solutions by [49],
A3.1, since $n\geqslant 4$ by Remark 4.2. Thus, $Y\in\mathcal{C}_{3}(X)$.

The group $X$ contains a unique conjugacy class of subgroups isomorphic to
$Y$, except for $X\cong P\Omega_{n}^{-}(q)$ and $O_{n/2}(q^{2})$ with $qn/2$
odd by [39], Section 4.3, since $\Phi_{nf}^{\ast}(p)\mid\left|X_{x}\right|$
and since $n\geqslant 6$ (see Proposition 4.3.20.(I) for the exception). Then
$X_{x}=Y$ by Lemma 3.6(1) when $Y$ is none of the exceptions. In the
exceptional cases, $X_{x}$ is a type 1 novelty with respect to $Y$ by Lemma
3.6(2). However, this is impossible by [39], Tables 3.5.H–I, for $n\geqslant
13$ and [14] for $3\leqslant n\leqslant 12$. Thus $X_{x}=Y$ and hence $X_{x}$
is a maximal $\mathcal{C}_{3}$-subgroup of $X$.

Suppose that $X_{x}$ is not a large subgroup of $X$. Then
$\left|X_{x}\right|_{p}<\left|\mathrm{Out(}X\mathrm{)}\right|$ by Lemma 4.8.
Hence, either $X\cong PSL_{n}(q)$ and $X_{x}$ is of type $GL_{1}(q^{n})$, or
$X\cong PSU_{n}(q^{1/2})$, with $n$ odd, and $X_{x}$ is of type
$GU_{1}(q^{n/2})$, or $X\cong P\Omega_{6}^{-}(q)$ by [11], Lemma 3.13, since
$\Phi_{nf}^{\ast}(p)\mid\left|X_{x}\right|$.

Assume that $X\cong PSL_{n}(q)$ and $X_{x}$ is of type $GL_{1}(q^{n})$. Then

$q^{n^{2}-2}<\left|X\right|\leqslant\left|G\right|\leqslant\left((q^{n}-1)nf\right)^{3}\leqslant
q^{3n+3}n^{3}$

since $G_{x}$ is a large subgroup of $G$ normalizing a Singer cyclic subgroup
of $X$ and by [1], Corollary 4.3.(i). Hence $n=3,4$. If $n=4$, then

$k^{2}=\frac{q^{6}(q-1)^{2}(q^{2}-1)(q^{2}+q+1)}{4}$ (6.1)

by [39], Proposition 4.3.6.(II). On the other hand, $\frac{k+1}{(k+1,f)}\mid
4\frac{q^{4}-1}{(q-1)(4,q-1)}$ by Lemma 3.5, and hence $k+1\leqslant
4f\frac{q^{4}-1}{(q-1)(4,q-1)}$, which, compared to (6.1), yields $q\leqslant
5$. However, none of these values fulfills (6.1). Thus $n=3$ and hence
$\left|X_{x}\right|=3\frac{q^{2}+q+1}{(3,q-1)}$ again by [39], Proposition
4.3.6.(II). Therefore,

$k^{2}=\frac{q^{3}(q-1)^{2}(q+1)}{3}.$

If $q\equiv 1\pmod{3}$, then both $q$ and $q+1$ are squares, but this
contradicts [49], A5.1. If $q\equiv 0\pmod{3}$, then both $q^{3}/3$ and $q+1$
are squares and hence $q=3$ again by [49], A5.1. So
$\left|X_{x}\right|_{p}=3$, whereas $\left|\mathrm{Out(}X\mathrm{)}\right|=2$.
Therefore, $q\equiv 2\pmod{3}$ and hence both $q$ and $\frac{q+1}{3}$ are
squares, but this is impossible by [49], P5.5.

Assume $X\cong PSU_{n}(q^{1/2})$, with $n$ odd, and $X_{x}$ is of type
$GU_{1}(q^{n/2})$. Then

$(q^{1/2}-1)q^{\frac{n^{2}-3}{2}}<\left|X\right|\leqslant\left|G\right|\leqslant\left((q^{n/2}+1)nf\right)^{3}\leqslant(q^{n/2}+1)^{3}q^{3/2}n^{3}$

since $G_{x}$ is a large subgroup of $G$ normalizing a Singer cyclic subgroup
of $X$, since $f$ is even and hence $q^{1/2}\geqslant f$, and by [1],
Corollary 4.3.(ii). Thus either $n=3$, or $n=5$ and $q^{1/2}=2$, since $n$ is
odd. We have seen above that $PSU_{5}(2)$ cannot occur, hence the latter is
ruled out. Thus $n=3$ and hence
$\left|X_{x}\right|=3\frac{q-q^{1/2}+1}{(3,q^{1/2}+1)}$ and

$k^{2}=\frac{q^{3/2}(q^{1/2}-1)(q^{1/2}+1)^{2}}{3}\text{.}$ (6.2)

by [39], Proposition 4.3.6.(II). Then $q^{1/2}\equiv 0,1\pmod{3}$ by [49],
A3.1. Since $\frac{k+1}{(k+1,f)}\mid 3\left(q-q^{1/2}+1\right)$ by Lemma 3.5,
it follows that $k+1\leqslant 3f\left(q-q^{1/2}+1\right)$, which, compared to
(6.2), yields $q^{1/2}=3,4$. However, none of these fulfills (6.2).

Finally, assume that $X\cong P\Omega_{6}^{-}(q)$. Then
$\left|\mathrm{Out(}X\mathrm{)}\right|=(4,q^{3}+1)2f$ and $X_{x}$ is a
subgroup of $\frac{1}{(2,q-1)}GU_{3}(q)$ by [39], Propositions 4.3.18.(II) and
4.3.20.(II), as $\Phi_{6f}^{\ast}(p)\mid\left|X_{x}\right|$. Since
$q\mid\left|X_{x}\right|_{p}$ by Corollary 4.10(v), and since
$\left|X_{x}\right|_{p}<(4,q^{3}+1)2f$, it follows that $q<8f$ and hence
$q=2,3,5$ or $7$. Actually, $q\neq 2,3$ as $P\Omega_{6}^{-}(2)$ is ruled out
in Lemma 4.4, since $\Phi_{6}^{\ast}(2)=1$, and $P\Omega_{6}^{-}(3)$ is ruled
out above in this proof. Since $\left[X:X_{x}\right]=k^{2}$, it is not
difficult to obtain that $2\cdot 7\cdot 13$ or $43$ divides
$\left|X_{x}\right|$ according to whether $q=5$ or $7$ respectively. However,
no cases arise by [14], Tables 8.10–8.11, since $X_{x}$ is a maximal
$\mathcal{C}_{3}$-subgroup of $X$. ∎

######  Lemma 6.3.

If $\left(\Phi_{(n-1)f}^{\ast}(p),k+1\right)>1$, then $X_{x}$ is a large
subgroup of $X$.

###### Proof.

Assume that $\left(\Phi_{(n-1)f}^{\ast}(p),k+1\right)>1$. If $X\cong
PSU_{n}(q^{1/2})$, with $n$ even, then either $n=4$ and
$PSL_{3}(2^{t})\leqslant X_{x}$ with $(t,q)=(1,3),(1,5),(2,3)$, or $X_{x}$
preserves either a point or a hyperplane of $PG_{n-1}(q)$ by [13], Theorem
4.2. If $(t,q)=(2,3)$, then $X_{x}\cong PSL_{3}(4)$ by [20] and hence
$k^{2}=162$, a contradiction. If $(t,q)=(1,3)$, then $PSL_{3}(2)\leqslant
X_{x}\leqslant M$, where $M$ is one of the following maximal subgroups of $X$
by [20]: $PSL_{3}(4)$, $PSU_{3}(3)$ or $A_{7}$. Either $X_{x}\cong PSL_{3}(2)$
or $X_{x}=M$, since $PSL_{3}(2)$ is maximal in $M$. Actually, the unique
admissible case is $X_{x}\cong A_{7}$ and $k=36$, since
$\left[X:X_{x}\right]=k^{2}$. However, this case is ruled out, since $k+1$
does not divide the order of $P\Gamma U_{4}(3)$, and hence that of $G$.

Assume that $X_{x}$ preserves either a point or a hyperplane of $PG_{n-1}(q)$.
Actually, $G_{x}$ is the stabilizer of a non-isotropic point $P$ of
$PG_{n-1}(q)$ by Proposition 4.3 and by [39], Proposition 4.1.6.(II), since
$\Phi_{ef}^{\ast}(p)\mid\left|X_{x}\right|$. Hence, the actions of $G$ on the
point set of $\mathcal{D}$ and on the set of non-singular points of
$PG_{n-1}(q)$ are equivalent. Thus,
$k^{2}=q^{\left(n-1\right)/2}\frac{q^{n/2}-1}{q^{1/2}+1}$ and hence $f\equiv
0\pmod{4}$ as $q^{\left(n-1\right)/2}$ must be a square, being $n$ is even.
Then

$k^{2}=s^{2\left(n-1\right)}\frac{\left(s^{n}+1\right)\left(s^{n}-1\right)}{s^{2}+1}\text{,}$
(6.3)

where $s=q^{1/4}$. Then $s$ is odd by [49], A3.1. Moreover, if we consider a
further non-singular point $y$ of $PG_{n-1}(q)$ such that $y\in x^{\perp}$, we
see that $\left|y^{G_{x}}\right|$ divides $(s^{2}+1)(s^{2(n-1)}+1)$. Then
$k+1\mid$ $(s^{2}+1)(s^{2(n-1)}+1)$ by Lemma 3.2(3). Actually, $k+1\mid$
$\frac{(s^{2}+1)(s^{2(n-1)}+1)}{4}$ as $s$ is odd, which, compared to (6.3),
leads to a contradiction as $n\geqslant 4$. Thus, $X\ncong PSU_{n}(q^{1/2})$
for $\left(\Phi_{(n-1)f}^{\ast}(p),k+1\right)>1$.

Assume that $X\cong P\Omega_{n}(q)$, with $nq$ odd. Assume also that
$\left|X_{x}\right|_{p}<\left|\mathrm{Out(}X\mathrm{)}\right|$. Then
$\left|X_{x}\right|_{p}<2f$, as $\left|\mathrm{Out(}X\mathrm{)}\right|=2f$.
Since $X_{x}<X<PSL_{n}(q)$ and $\Phi_{(n-1)f}^{\ast}(p)$ divides
$\left|X_{x}\right|$, it follows that $X_{x}^{\char 94\relax}$ is one of the
groups listed in Theorem 3.1 of [13].

Assume that $X_{x}^{\char 94\relax}$ is a nearly simple subgroup of
$SL_{n}(q)$. Suppose that $\mathrm{Soc(}X_{x}\mathrm{)}$ is alternating. Then
$X_{x}$ arises from the action of $A_{\ell}$ on its fully deleted permutation
module, since $nq$ is odd. Then $q$ is prime, $n=\ell-1$ or $\ell-2$ according
as $q$ does not divide or does divide $n$ respectively. In both cases, we have
$\Phi_{n-1}^{\ast}(q)=n$, with $n=4,6$, whereas $n$ is odd.

Assume that $\mathrm{Soc(}X_{x}\mathrm{)}$ is a sporadic group. Then
$\mathrm{Soc(}X_{x}\mathrm{)}\cong M_{11}\trianglelefteq X_{x}$, $n=5$ and
$q=3$. So $\left|X_{x}\right|_{3}=3^{2}$, whereas
$\left|\mathrm{Out(}X\mathrm{)}\right|=2$.

Assume that $\mathrm{Soc(}X_{x}\mathrm{)}$ is a simple Lie type group in cross
characteristic. $\left|X_{x}\right|_{p}<\left|\mathrm{Out(}X\mathrm{)}\right|$
implies $n=7$, $q=5$ and $\mathrm{Soc(}X_{x}\mathrm{)}$ isomorphic to one of
the groups $PSL_{2}(7)$, $PSL_{2}(8)$ or $PSU_{3}(3)$. However, these cases
are ruled out. Indeed, $31$ must divide $\left|X_{x}\right|$, since
$\left|P\Omega_{7}(5)\right|$ is divisible by $31$, and not by $31^{2}$, and
since $\left[X:X_{x}\right]=k^{2}$.

Assume that $\mathrm{Soc(}X_{x}\mathrm{)}$ is a simple group of Lie type in
characteristic $p$. Then $n=7$, $q$ is odd, with $q$ dividing
$\left|X_{x}\right|$ by Theorem 3.1 of [13]. So
$q<\left|\mathrm{Out(}X\mathrm{)}\right|=2f$, which has no solutions for $q$
odd.

Clearly $X_{x}^{\char 94\relax}$ is not a Symplectic Type Example. Also,
$X_{x}^{\char 94\relax}$ is not a Classical Example since $X_{x}<X\cong
P\Omega_{n}(q)$. If $X_{x}^{\char 94\relax}$ is an Imprimitive Example, then
$q=p$ and $X_{x}^{\char 94\relax}\leqslant GL_{1}(p)\wr S_{n}$, where either
$q=3$ and $n=5,7$, or $q=5$ and $n=7$ by Theorem 3.1 of [13], being $n$ odd.
The case $q=3$ and $n=5$ cannot occur, since $P\Omega_{5}(3)\cong PSU_{4}(2)$
has been ruled out above. The groups $P\Omega_{7}(3)$ and $P\Omega_{7}(5)$ are
ruled out, as they do not have a transitive permutation representation of
square degree by [14], Tables 8.39 and 8.40. Therefore, $X_{x}$ fixes either a
point or a hyperplane of $PG_{n-1}(q)$. So does $G_{x}$, as
$X_{x}\trianglelefteq G_{x}$. Then $G_{x}$ should be one of the groups listed
in [11], Lemma 3.13, since
$\left|X_{x}\right|_{p}<\left|\mathrm{Out(}X\mathrm{)}\right|$, whereas it is
not, and we obtain a contradiction. Thus,
$\left|X_{x}\right|_{p}\geqslant\left|\mathrm{Out(}X\mathrm{)}\right|$ and
hence $X_{x}$ is a large subgroup of $X$ by Lemma 4.8(2). ∎

######  Theorem 6.4.

$X_{x}$ is a large subgroup of $X$ such that
$\left(\Phi_{ef}^{\ast}(p),\left|X_{x}\right|\right)>1$.

###### Proof.

Since the assertion follows from Lemma 6.2 and 6.3 for $e=n$ or $e=n-1$,
respectively, in order to prove the theorem, it remains to tackle the case
$e=n-2$ and $X\cong P\Omega_{n}^{+}(q)$. Clearly,
$\left(\Phi_{(n-2)f}^{\ast}(p),\left|X_{x}\right|\right)>1$.

Suppose that $X_{x}$ is not a large subgroup of $X$. Then
$\left|X_{x}\right|_{p}<\left|\mathrm{Out(}X\mathrm{)}\right|$ by Lemma
4.8(2). Recall that $n\geqslant 6$ by Remark 4.2, and that
$\left|\mathrm{Out(}X\mathrm{)}\right|$ is either $2(4,q^{n/2}-1)f$ or
$6(4,q^{n/2}-1)f$ according as $n\neq 8$ or $n=8$ respectively.

Since $X_{x}<X<PSL_{n}(q)$ and $\Phi_{(n-2)f}^{\ast}(p)$ divides
$\left|X_{x}\right|$, the group $X_{x}^{\char 94\relax}$ is one the groups
listed in Theorem 3.1 of [13].

Assume that $X_{x}^{\char 94\relax}$ is a nearly simple subgroup of
$SL_{n}(q)$. Suppose that $\mathrm{Soc(}X_{x}\mathrm{)}$ is alternating. If
$X_{x}^{\char 94\relax}$ arises from the action of $A_{\ell}$ on its fully
deleted permutation module, then $q$ is a prime (hence $f=1$), $n=e+2$,
$A_{e+4}\leqslant X_{x}\leqslant S_{e+4}\times Z_{2}$ where
$(p,e)=(2,4),(2,10),(2,12),(2,18),(3,4),(3,6),(5,6)$. If $(p,e)\neq(3,4)$,
then $\left|X_{x}\right|_{2}\geqslant 2^{6}$, $\left|X_{x}\right|_{3}\geqslant
9$ and $\left|X_{x}\right|_{5}\geqslant 25$, whereas
$\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant 24$ in the first case and
$\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant 8$ in the remaining ones. So,
these cases are excluded. Then $(p,e)=(3,4)$, $X\cong P\Omega_{8}^{+}(3)$ and
$A_{8}\leqslant X_{x}\leqslant S_{8}\times Z_{2}$. However, this is impossible
since $13$ must divide $\left|X_{x}\right|$, since $P\Omega_{8}^{+}(3)$ is
divisible by $13$, and not by $13^{2}$, and since
$\left[X:X_{x}\right]=k^{2}$.

Suppose that $\mathrm{Soc(}X_{x}\mathrm{)}$ does not arise from the action of
$A_{\ell}$ on its fully deleted permutation module. Then $n=8$ and either
$q=5$ and $\ell=7,8,9,10$, or $q=3$ and $\ell=9,8$. The above argument
excludes $q=3$. Case $q=5$ is ruled out too. Indeed, $P\Omega_{8}^{+}(5)$ is
divisible by $31$ and not by $31^{2}$.

Assume that $\mathrm{Soc(}X_{x}\mathrm{)}$ is either a sporadic group or a
simple Lie type group in cross characteristic. Set
$S\cong\mathrm{Soc(}X_{x}\mathrm{)}$. By Theorem 3.1 of [13], one of the
following holds:

  1. (1)

$(n,q)=(6,3)$ and $S\cong M_{12},PSL_{3}(4)$;

  2. (2)

$(n,q)=(8,3)$ and $S\cong P\Omega_{8}^{+}(2)$, $PSp_{6}(2)$;

  3. (3)

$(n,q)=(8,5)$ and $S\cong PSL_{3}(4)$, $P\Omega_{8}^{+}(2)$, $PSp_{6}(2)$, or
$Sz(8)$;

  4. (4)

$(n,q)=(14,2)$ and $S\cong PSL_{2}(13),$ $G_{2}(3)$;

  5. (5)

$(n,q)=(20,2)$ and $S\cong J_{2},PSL_{2}(19)$.

$M_{12},PSL_{3}(4)$ are not subgroups of $P\Omega_{6}^{+}(3)$, hence Case (1)
cannot occur. Cases (2) and (3) cannot occur by Lemma 4.4. Finally, $127$
divides the order of $P\Omega_{14}^{+}(2)$ and of $P\Omega_{20}^{+}(2)$. Then
$127^{2}$ must divide the order of $S$, since $\left[X:X_{x}\right]=k^{2}$,
and since $S\trianglelefteq X_{x}\leqslant\mathrm{Aut(}S\mathrm{)}$. However,
this is impossible for $S$ as in Cases (4) or (5), and hence these are ruled
out.

The group $X_{x}$ cannot be a Classical Example by [39], Section 4.8, as
$X_{x}<X$. The same argument used above to rule out Case (1) excludes the
possibility for $X_{x}$ to be a Symplectic Type Example as well. Now, assume
that is an Imprimitive Example. Then $q=p$, $e=n-2$, $\Phi_{e}^{\ast}(p)=e$
and $X_{x}^{\char 94\relax}\leqslant GL_{1}(q)\wr S_{n}$ with
$(n,q)=(6,2),(12,2),(14,2),(20,2),(6,3),(8,3),(8,5)$. Actually,
$(n,q)\neq(6,2)$ since $P\Omega_{6}^{+}(2)\cong A_{8}$ has been ruled out in
Lemma 3.5 of [44]. In the remaining cases, there is a prime $w$ such that
$w\mid\left|X\right|$, $w^{2}\nmid\left|X\right|$. Such a prime $w$ is $31$,
$127$, $257$, $13$, $13$, $31$, respectively. Then $w\mid\left|X_{x}\right|$,
since $\left[X:X_{x}\right]=k^{2}$, whereas $w>n$ and $w\nmid q-1$. So,
$X_{x}$ cannot be an Imprimitive Example. Thus, again by Theorem 3.1 of [13],
either $X_{x}$ preserves a point or a hyperplane of $PG_{n-1}(q)$, or
$X_{x}^{\char 94\relax}$ is an Extension Field Example.

Assume that $X_{x}^{\char 94\relax}$ is an Extension Field Example. Since
$X_{x}^{\char 94\relax}$ preserves an orthogonal form on $V_{n}(q)$, Theorem
3.1 of [13] implies that either $n\equiv 2\pmod{4}$, $q$ is odd and
$X_{x}^{\char 94\relax}\leqslant\Gamma O_{n/2}(q^{2})$, or $n\equiv 0\pmod{4}$
and $X_{x}^{\char 94\relax}\leqslant\Gamma U_{n/2-1}(q)$ respectively. Thus
$X_{x}$ lies in a maximal member $Y$ of $\mathcal{C}_{3}(X)$ of type
$O_{n/2}(q^{2})$ or $GU_{n/2-1}(q)$ respectively. If $X_{x}<Y$, then $X_{x}$
is a type 1 novelty with respect to $Y$ by Lemma 3.6(2). However, this is
impossible by [39], Tables 3.5.H–I, for $n\geqslant 13$ and [14] for
$6\leqslant n\leqslant 12$. Therefore, $X_{x}$ is a maximal
$\mathcal{C}_{3}$-subgroup of $X$. Thus, $\left|X_{x}\right|_{p}\geqslant
q^{2}$ for $n\neq 8$, and $\left|X_{x}\right|_{p}\geqslant q^{3}$ for $n=8$.
Hence, $\left|X_{x}\right|_{p}<\left|\mathrm{Out(}X\mathrm{)}\right|$ implies
$n=8$ and $q=2$, as $\left|\mathrm{Out(}X\mathrm{)}\right|$ is either
$2(4,q^{n/2}-1)f$ or $6(4,q^{n/2}-1)f$ according as $n\neq 8$, or $n=8$,
respectively. However, $X\ncong P\Omega_{8}^{+}(2)$ by Lemma 4.4.

Finally, assume that $X_{x}$ preserves either a point or a hyperplane of
$PG_{n-1}(q)$. Note that $X_{x}$ does not lie in a maximal parabolic subgroup
of $X$ by Proposition 4.3. Then $X_{x}$ preserves either a non-degenerate
point, or a non-degenerate line of $PG_{n-1}(q)$ of type $-$ by [39],
Proposition 4.1.6.(II), since $\Phi_{n-2}^{\ast}(p)\mid\left|X_{x}\right|$. So
does $G_{x}$. Then $X\cong P\Omega_{6}^{+}(q)$ by [11], Lemma 3.13. However,
this is impossible by Lemma 6.2, since $X\cong PSL_{4}(q)$. This completes the
proof. ∎

##  7\. Proof of Theorem 1.1

This final section is devoted to the completion of the proof of Theorem 1.1.
The proof strategy is as follows. $X_{x}$ lies in a large maximal subgroup $Y$
of $X$ by Theorem 4.7. Moreover, $Y$ is classified in [1]. We filter the
$Y$-candidates with respect to
$\left(\Phi_{ef}^{\ast}(p),\left|Y\right|\right)>1$. Then, by comparing the
information contained in [39] and [14] on the number of conjugacy classes of
the subgroups of $X$ isomorphic to $Y$, with those contained in Lemma 3.6, we
obtain $X_{x}=Y$. Hence, $X_{x}$ is a known large maximal subgroup of $X$. At
this point, we can precisely evaluate the diophantine equation
$k^{2}=[X:X_{x}]$, and, by combining some of the results contained in [49]
with properties arising from the geometry of the classical groups, we reduce
to the case $X\cong PSL_{3}(3)$, $k=12$ and hence to Cases (3) and (4) of
Theorem 1.1.

######  Theorem 7.1.

$X_{x}$ lies in a large maximal geometric subgroup of $X$.

###### Proof.

Let $Y$ be a maximal subgroup of $X$ containing $X_{x}$. Then $Y$ is a large
subgroup of $X$ such that $\left(\Phi_{ef}^{\ast}(p),\left|Y\right|\right)>1$
by Theorem 6.4. Assume that $Y$ is a member of $\mathcal{S}(X)$. Then the
pairs $(X,Y)$ are listed in [1], Proposition 4.28. Filtering such pairs
$(X,Y)$ with respect to the property
$\left(\Phi_{ef}^{\ast}(p),\left|Y\right|\right)>1$, one of the followings
occurs:

  1. (1)

$X\cong P\Omega_{8}^{+}(q)$ and $Y$ is isomorphic either to $\Omega_{7}(q)$ or
to $Sp_{6}(q)$ according to whether $q$ is odd or even respectively.

  2. (2)

$X\cong\Omega_{7}(q)$ and $Y\cong G_{2}(q)$.

  3. (3)

$X\cong Sp_{6}(q)$, $q$ even, and $Y\cong G_{2}(q)$.

  4. (4)

$X\cong PSp_{4}(q)$, $q$ an odd power of $2$, $q>2$, and $Y\cong Sz(q)$.

  5. (5)

$(X,Y)$ are listed in the Table 1.

Table 1. Admissible pairs $(X,Y)$ and significant primes dividing the order of $X$. $X$ | $(n,q,e)$ | $\left|\mathrm{Out(}X\mathrm{)}\right|$ | $\Phi_{ef}^{\ast}(p)$ | $Y$ | $s$
---|---|---|---|---|---
$PSL_{n}(q)$ | $(5,3,5)$ | $2$ | $11^{2}$ | $M_{11}$ | $13$
| $(4,2,4)$ | $2$ | $5$ | $A_{7}$ | $-$
| $(4,7,4)$ | $2$ | $5^{2}$ | $PSU_{4}(2)$ | $19$
$PSp_{n}(q)$ | $(12,2,12)$ | $1$ | $13$ | $S_{14}$ | $31$
| $(6,5,6)$ | $2$ | $7$ | $J_{2}$ | $31$
| $(4,7,4)$ | $2$ | $5^{2}$ | $A_{7}$ | $-$
$PSU_{n}(q)$ | $(4,3,3)$ | $8$ | $5$ | $A_{7},PSL_{3}(4)$ | $-$
$P\Omega_{n}^{-}(q)$ | $(18,2,12)$ | $2$ | $19$ | $A_{20}$ | $257$
| $(12,2,12)$ | $2$ | $13$ | $A_{13}$ | $31$
| $(10,2,10)$ | $2$ | $11$ | $A_{12}$ | $17$
$P\Omega_{n}(q)$ | $(7,3,6)$ | $2$ | $7$ | $PSp_{6}(2),S_{9}$ | $13$
| $(7,5,6)$ | $2$ | $7$ | $PSp_{6}(2)$ | $31$
$P\Omega_{n}^{+}(q)$ | $(14,2,12)$ | $2$ | $13$ | $A_{16}$ | $127$
| $(8,3,6)$ | $24$ | $7$ | $P\Omega_{8}^{+}(2)$ | $13$
| $(8,5,6)$ | $24$ | $7$ | $P\Omega_{8}^{+}(2)$ | $31$

Assume that Cases (1)–(4) hold. Since $X_{x}$ is a large subgroup of $X$, it
follows that $\left|X_{x}\right|>\left|X\right|^{1/3}$. By using the lower
bounds for $\left|X\right|$ provided in Lemma 4.3 of [1], we see that
$\left|X_{x}\right|$ is greater than $q^{28/3}/2$, $q^{7}/\sqrt[3]{4}$,
$q^{7}/\sqrt[3]{2}$ and $q^{10/3}/\sqrt[3]{2}$ in Cases (1)–(4) respectively.

Assume that $X\cong P\Omega_{8}^{+}(q)$. Then $Y$ is isomorphic either to
$\Omega_{7}(q)$, or to $Sp_{6}(q)$, according to whether $q$ is odd or even
respectively. Note that, $q=2$ is ruled out by Lemma 4.4. Also, $q=3$ cannot
occur. Indeed, in this case $13\mid\left|X\right|$,
$13^{2}\nmid\left|X\right|$, but $13$ does not divide $\left|X_{x}\right|$ and
$k^{2}$. Hence $q>3$.
Assume that $X_{x}\neq Y$. Then either
$X_{x}\cong\Omega_{6}^{-}(q).Z_{\epsilon}$, where $\epsilon=1,2$, or
$X_{x}\cong G_{2}(q)$, or $X_{x}\cong Sp_{2}(q^{3})\cdot Z_{3}$ with $q$ even
and $2<q<2^{8}$ by [14], Tables 8.28–8.30, 8.33–8.34 and 8.39–8.42, since
$\left|X_{x}\right|>$ $q^{28/3}/2$ and
$\left(\Phi_{6f}^{\ast}(p),\left|X_{x}\right|\right)>1$. It is not difficult
to check that, $k^{2}=[X:X_{x}]$ has no solutions for $X_{x}\cong
Sp_{2}(q^{3})\cdot Z_{3}$ with $q$ even and $2<q<2^{8}$. So, this case is
ruled out.

If $X_{x}\cong\Omega_{6}^{-}(q).Z_{\epsilon}$, then

$k^{2}=\frac{q^{6}(q+1)(q^{2}+1)(q-1)^{2}(q^{2}+q+1)}{(4,q^{4}-1)\epsilon}$

and hence $q^{2}+q+1$ is a square, but this is impossible by [49], A7.1.

If $X_{x}\cong G_{2}(q)$, then $k^{2}=\frac{q^{6}(q^{4}-1)^{2}}{(4,q^{4}-1)}$
and hence $k=\frac{q^{3}(q^{4}-1)}{(2,q-1)}$. Therefore
$k+1=q^{3}(q^{2}-1)\frac{q^{2}+1}{(2,q-1)}+1$. Set
$\theta=(k+1,6(4,q^{4}-1)f)$. Then $\frac{k+1}{\theta}\mid\left|X_{x}\right|$
by Lemma 3.5. Furthermore, $\theta=(k+1,3f)$, since $k+1$ is odd. Thus
$\frac{k+1}{\theta}\mid\frac{q^{6}-1}{q-1}$, since $\frac{k+1}{\theta}$ is
coprime to $q^{2}(q^{2}-1)^{2}$. Then
$q^{3}(q^{2}-1)\frac{q^{2}+1}{(2,q-1)}<\frac{q^{6}-1}{q-1}\cdot 3f$, which has
no solutions for $q>3$.
Finally, assume that $X_{x}=Y$. We treat $X_{x}\cong\Omega_{7}(q)$ and
$X_{x}\cong Sp_{6}(q)$ at the same time. Then
$k^{2}=q^{3}(q^{2}-1)\frac{(q^{2}+1)}{(2,q-1)}$, and since
$\frac{(q^{2}+1)}{(2,q-1)}$ is odd, it follows that $q^{2}-1$ is a square.
However, this is impossible by [49], A3.1.
Assume that $X$ is either $\Omega_{7}(q)$, or $Sp_{6}(q)$ for $q$ even, and
$Y\cong G_{2}(q)$. Assume that $X_{x}\neq Y$. Since
$\left|X_{x}\right|>q^{7}/\sqrt[3]{2}$ and
$\left(\Phi_{6f}^{\ast}(p),\left|X_{x}\right|\right)>1$, it follows from [14],
Tables 8.30 and 8.41–8.43, that $X_{x}\cong SU_{3}(q):Z_{\epsilon}$, where
$\epsilon=1,2$, or $q=3^{2m+1}$, $m\geqslant 1$, and $X_{x}\cong$
${}^{2}G_{2}(q)$. Thus,

$k^{2}=\frac{q^{6}(q^{2}-1)^{2}(q^{2}+q+1)(q^{2}+1)}{(2,q-1)\theta}\text{,}$

where either $\theta=(q+1)\epsilon$, or $\theta=1$ respectively. So,
$q^{2}+q+1$ is a square in both cases, but this is impossible by [49], A7.1.
Thus $X_{x}=Y$ and hence $k^{2}=\frac{q^{3}(q^{4}-1)}{(2,q-1)}$, which, as we
have already seen, does not have solutions.

Assume that $X\cong PSp_{4}(q)$ and $Y\cong Sz(q)$, where $q$ is an odd power
of $2$, $q>2$. Since $\left|X_{x}\right|>q^{10/3}/\sqrt[3]{2}$ and
$\left(\Phi_{4f}^{\ast}(p),\left|X_{x}\right|\right)>1$, it follows that
$X_{x}=Y$ by [14], Table 8.16. Hence
$k^{2}=q^{2}\left(q-1\right)\left(q+1\right)^{2}$, but this contradicts [49],
A3.1, since $q$ is an odd power of $2$ and $q>2$. So far, we have ruled out
Cases (1)–(4).

Assume that Case (5) holds. Then $(X,Y)$ are listed in Table 1, in which the
symbol $s$ in Column 6, when recorded, denotes the largest prime divisor of
$\left|X\right|$ such that $s\nmid\Phi_{ef}^{\ast}(p)$, $s\nmid\left|Y\right|$
and $s^{2}\nmid\left|X\right|$. The previous constraints imply
$s\mid\left[X:X_{x}\right]$ and hence $s^{2}\mid\left|X\right|$, which is a
contradiction. Hence, only the groups, whose corresponding cell in Column 6 of
Table 1 is empty are admissible, and these are in Lines 2 and 6–7 of the same
table. The group $X\cong PSL_{4}(2)\cong A_{8}$ is ruled out in Lemma 3.5 of
[44]. If $X\cong PSp_{4}(7)$ and $X_{x}\cong A_{7}$, then $7^{3}$ is the
highest power of $7$ dividing $\left[X:X_{x}\right]$, a contradiction.
Finally, if $X\cong PSU_{4}(3)$ and $X_{x}\cong A_{7},PSL_{3}(4)$, then
$k^{2}=1296$ or $162$. The latter is ruled out since it is not a square,
whereas the former yields $k=36$. Then $k+1=37$ must divide the order of $G$
by Lemma 3.2(3). However, this is impossible since $G\leqslant P\Gamma
U_{4}(3)$, and the order of $P\Gamma U_{4}(3)$ has not such a divisor. This
completes the proof. ∎

######  Lemma 7.2.

If $X\cong PSL_{n}(q)$, then $n=q=3$ and $k=12$.

###### Proof.

Assume that $X\cong PSL_{n}(q)$. Then $n\geqslant 3$ by Remark 4.2, and
$X_{x}$ lies in a large maximal geometric subgroup $Y$ of $X$ by Theorem 7.1.
Then $Y$ is one of the groups listed in Proposition 4.7 of [1]. It follows
from Theorem 3.5 of [33] and Proposition 5.2.15.(i) of [39], that
$Y\notin\mathcal{C}_{1}(X)$ and
$Y\notin\mathcal{C}_{2}(X)\cup\mathcal{C}_{5}(X)$, respectively, since
$\left(\Phi_{nf}^{\ast}(p),\left|Y\right|\right)>1$. For the same reason
$(X,Y)$ is not equal to $(PSL_{4}(5),2^{4}.A_{6})$ or to
$(PSL_{3}(4),PSU_{3}(2))$. Thus, one of the following holds:

  1. (i).

$Y$ is a $\mathcal{C}_{3}$-subgroup of type $GL_{n/t}(q^{t})$, where $t=2$, or
$t=3$ and either $q\in\left\\{2,3\right\\}$, or $q=5$ and $n$ is odd.

  2. (ii).

$Y\in\mathcal{C}_{8}(X)$.

Assume that Case (i) holds. Then $Y\cong
SL_{n/t}(q^{t})\cdot(q^{t}-1)(q-1)^{-1}\cdot t$ by [39], Proposition
4.3.6.(II). Also $X_{x}=Y$ by Lemma 3.6(1) and by [39], Proposition 4.3.6.(I),
for $n\geqslant 13$ and by [14] for $n<13$.

Assume that $t=3$. Note that
$\left|X\right|\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|^{2}\cdot\left|X_{x}\right|\cdot\left|X_{x}\right|_{p^{\prime}}^{2}$
by Lemma 4.8(1). Also, since $2f\leqslant q\,$ implies
$\left|\mathrm{Out(}X\mathrm{)}\right|\leqslant q(q-1)$, we get
$\left|X\right|\leqslant
q^{2}(q-1)^{2}\cdot\left|X_{x}\right|\cdot\left|X_{x}\right|_{p^{\prime}}^{2}$.
This in turn yields the inequality (4.11) of [11] (see part (3) in the proof
of Proposition 4.2), which leads to $n/t=q=2$, or to $n=t=3$ and $q=2,3$ or
$5$. The former is ruled out, since $X$ is not isomorphic to $PSL_{6}(2)$ by
Lemma 4.4, whereas $(n,q)=(3,2)$ is ruled out in Proposition 4.1, since
$PSL_{3}(2)\cong PSL_{2}(7)$. Finally, in the remaining cases, $[X:X_{x}]=144$
and $4000$ for $q=3$ or $5$ respectively. Thus, only the former is admissible,
since $[X:X_{x}]$ is a square, and hence $k=12$.

Assume that $t=2$. Then $n$ is even and $X_{x}\cong
SL_{n/2}(q^{2})\cdot(q+1)\cdot 2$. Then $\left|X_{x}\right|\leqslant
2(q+1)\frac{q^{n^{2}/2}}{q^{4}-1}\leqslant 2q^{n^{2}/2-2}$ and
$\left|X\right|\geqslant q^{n^{2}-2}$ by [1], Corollary 4.3.(i), hence
$k^{2}=[X:X_{x}]>q^{n^{2}/2}/2$. On the other hand, if $n\geqslant 8$ then
$k+1$ divides $(q^{n}-1)(q^{n-2}-1)$ by Lemma 3.2(3) and by [11], Lemma 3.6.
Therefore, $q^{n^{2}/2}/2<k^{2}<q^{2n-2}$ and hence $q^{(n-2)^{2}}<4$, which
has no solutions for $n\geqslant 8$.

If $n=6$, then

$k^{2}=\frac{\allowbreak
q^{9}\left(q-1\right)\left(q^{3}-1\right)\left(q^{5}-1\right)}{2(6,q-1)}\text{.}$
(7.1)

On the other hand,
$k+1\mid\left|X_{x}\right|\cdot\left|\mathrm{Out(}X\mathrm{)}\right|$ by Lemma
3.5, and hence $k+1\mid(q^{2}+1)(q^{3}+1)(q+1)^{2}\cdot 3f$, with $k+1$ odd.
So, we get

$k+1\leqslant(q^{2}+1)(q^{3}+1)(q+1)q\text{.}$ (7.2)

However, there are no solutions in (7.1) by comparing it with (7.2).

If $n=4$, then

$k^{2}=(q^{2}+q+1)\frac{q^{4}(q-1)^{2}}{2(4,q-1)}$

and hence $q^{2}+q+1$ is a square, but this contradicts [49], A7.1.

Assume that Case (ii) holds. Then, by [39], Propositions 4.8.3.(II), 48.4.(II)
and 4.8.5.(II), one of the following holds:

  1. (I).

$Y\cong PSp_{n}(q)\cdot[\frac{(q-1,n/2)(q-1,2)}{(q-1,n)}]$ with $n\geqslant
4$.

  2. (II).

$Y\cong PSU_{n}(q^{1/2})\cdot[\frac{(q^{1/2}+1,n)c}{(q-1,n)}]$, with
$c=\frac{q-1}{(q^{1/2}+1,(q-1)/(q-1,n))}$, $n$ odd and $n\geqslant 3$.

  3. (III).

$Y\cong PSO_{n}^{-}(q)\cdot 2$, with $q$ odd, and $n\geqslant 4$.

If $X_{x}<Y$, then $G_{x}$ is a type 1 novelty with respect to $Y$ by Lemma
3.6(2). However, this is impossible by [39], Tables 3.5.H–I for $n\geqslant
13$ and by [14] for $n<13$. Thus $X_{x}=Y$ in Cases (I)–(III).

Assume that case (I) holds. Then $\left|X_{x}\right|\leqslant 4q^{n(n+1)/2}$
and $\left|X\right|>q^{n^{2}-2}$, by [1], Corollary 4.3.(i) and (iii)
respectively. Hence $k^{2}>q^{(n^{2}-n-4)/2}$. On the other hand, if
$n\geqslant 8$ then $k+1$ divides $(q^{n}-1)(q^{n-2}-1)$ by Lemma 3.2(3) and
by [11], Lemma 3.6. So $q^{(n^{2}-n-4)/2}/4<k^{2}<q^{2n-2}$ and we again reach
a contradiction. Thus, $3\leqslant n<8$ and hence $n=4$ or $6$ as $n$ is even.
If $n=6$, then

$k^{2}=\frac{q^{6}\left(q-1\right)^{2}\left(q^{2}+q+1\right)\left(q^{4}+q^{3}+q^{2}+q+1\right)}{(q-1,3)}$
(7.3)

and hence $q^{4}+q^{3}+q^{2}+q+1$ is a square. So $q=3$ by [49], A8.1.
However, it does not fulfill (7.3).

If $n=4$, then $k^{2}=\frac{\allowbreak q^{2}\left(q^{3}-1\right)}{(q-1,2)}$.
On the other hand, $k+1$ divides $2f\cdot q(q^{2}-1)$ by Lemma 3.2(3), and
hence $k+1$ divides $2f\cdot(q+1)$ as $k$ is coprime to
$q\frac{q-1}{(2,q-1)}$. So, $k<2f\cdot(q+1)$, which, compared to
$k^{2}=\frac{\allowbreak q^{2}\left(q^{3}-1\right)}{(q-1,2)}$ yields a
contradiction.

Assume that Case (II) holds. Then
$\left|X_{x}\right|<\frac{(q^{1/2}+1,n)(q^{1/2}-1)}{(q^{1/2}+1,n)}q^{\frac{n^{2}-1}{2}}<2q^{n^{2}/2}$
and $\left|X\right|\geqslant q^{n^{2}-2}$, by [1], Lemma 4.2.(ii) and
Corollary 4.3.(i) respectively. Hence $k^{2}>q^{n^{2}/2-2}/2$. On the other
hand, if $n\geqslant 5$ then $k+1$ divides $q^{n/2}+1$ by Lemma 3.2(3) and by
[11], Lemma 3.6, as $n$ is odd. Thus $q^{n^{2}/2-2}/2<q^{n/2}+1$, which has no
solutions for $n\geqslant 5$. Therefore, $n=3$ and hence

$k^{2}=q^{3/2}(q+q^{1/2}+1)\frac{(q^{1/2}-1,3)(q+1)}{(q-1,3)}.$ (7.4)

So $q+q^{1/2}+1$ is a square, but this contradicts [49], A7.1.

Finally, assume that Case (III) holds. Then

$k^{2}=\frac{q^{n^{2}/4}(4,q^{n/2}+1)\left(q^{n/2}-1\right)}{2\rho(q-1,n)}\prod_{j=1}^{n/2-1}(q^{2j+1}-1)\text{,}$
(7.5)

with $\rho=1,2$, and hence $q^{n-1}-1$ divides $k^{2}$. Also
$\Phi_{(n-1)f}^{\ast}(p)\mid k^{2}$, where $\Phi_{(n-1)f}^{\ast}(p)>1$, by
[49], P1.7, since $n\geqslant 4$ and $n$ is even. Thus
$\Phi_{(n-1)f}^{\ast}(p)^{1/2}\mid\left|G_{B}\right|$, as
$k\mid\left|G_{B}\right|$.

Note that $\left(\Phi_{(n-1)f}^{\ast}(p),\left|X_{x}\right|\right)=1$, as
$X_{x}\cong PSO_{n}^{-}(q)\cdot 2$, with $q$ odd and $n\geqslant 4$. Also,
$\left(\Phi_{(n-1)f}^{\ast}(p),\left|\mathrm{Out(}X\mathrm{)}\right|\right)=1$,
as $\Phi_{(n-1)f}^{\ast}(p)\equiv 1\pmod{(n-1)f}$ by [39], Proposition
5.2.15.(i). Thus, $\left(\Phi_{(n-1)f}^{\ast}(p),\left|G_{x}\right|\right)=1$
and $\left(\Phi_{(n-1)f}^{\ast}(p),\left[G_{B}:X_{B}\right]\right)=1$, as both
$\left[G_{x}:X_{x}\right]$ and $\left[G_{B}:X_{B}\right]$ divide
$\left|\mathrm{Out(}X\mathrm{)}\right|$. Then
$\Phi_{(n-1)f}^{\ast}(p)^{1/2}\mid\left|X_{B}\right|$ and
$\Phi_{(n-1)f}^{\ast}(p)^{1/2}\mid\left[X:X_{B}\right]$, as
$\left|G_{x}\right|=\lambda(k+1)\left|G_{x,B}\right|$ and
$\left|G_{B}\right|=k\left|G_{x,B}\right|$. Therefore,
$\Phi_{(n-1)f}^{\ast}(p)^{1/2}$ divides both $\left|X_{B}^{\char
94\relax}\right|$ and $\left[SL_{n}(q):X_{B}^{\char 94\relax}\right]$.
Moreover, $q$ divides $\left|X_{B}^{\char 94\relax}\right|$ by Corollary
4.10(i) for $n\geqslant 4$, since $\left|X_{B}\right|_{p}=\left|X_{B}^{\char
94\relax}\right|_{p}$. Then $X_{B}^{\char 94\relax}$ lies in a parabolic
subgroup of type $P_{1}$ of $SL_{n}(q)$ by Proposition 5.3(2). Hence $X_{B}$
fixes a point of $PG_{n-1}(q)$. Therefore, $\frac{q^{n}-1}{q-1}$ divides
$\left[X:X_{B}\right]$ and hence $b$. Thus $\Phi_{nf}^{\ast}(p)\mid b$. It
follows from (7.5) that, $\Phi_{nf}^{\ast}(p)$ is coprime to $k$. Thus
$\Phi_{nf}^{\ast}(p)\mid k+1$, since $b=k(k+1)\lambda$ and $\lambda\mid k$.
Then $X_{x}\in\mathcal{C}_{3}(X)$ by Lemma 6.2, as $X_{x}$ is maximal in $X$.
However, this is impossible, since $X_{x}\in\mathcal{C}_{8}(X)$. ∎

Before analyzing the case $X\cong PSL_{3}(3)$ and $k=12$, we recall some
useful facts about $PSL_{3}(3):\left\langle\sigma\right\rangle$ and about the
action of some of its subgroups on $PG_{2}(3)$.

The group $PSL_{3}(3)$ has two conjugacy classes of subgroups of order $3$,
say $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$ by [20]. If
$H_{1}\in\mathcal{K}_{1}$, then $H_{1}$ is a group of elations of $PG_{2}(3)$
having the same center $C$ and the same axis $\ell$ (e.g. see [36]). Hence,
$H_{1}$ fixes seven flags of $PG_{2}(3)$. If $H_{2}\in\mathcal{K}_{2}$, then
$H_{2}$ normalizes six suitable Sylow $13$-subgroups of $PSL_{3}(3)$, and
fixes a unique flag of $PG_{2}(3)$. No further points or lines of $PG_{2}(3)$
are fixed by $H_{2}$. Finally, $\sigma$ is a symmetric polarity of
$PG_{2}(3)$. Thus $\sigma$ switches points and lines of $PG_{2}(3)$ and hence
preserves both $\mathcal{K}_{1}$ and $\mathcal{K}_{2}$. Therefore,
$\mathcal{K}_{1}$ and $\mathcal{K}_{2}$ are also conjugacy
$PSL_{3}(3):\left\langle\sigma\right\rangle$-classes.

######  Lemma 7.3.

The following hold:

  1. (1)

If $H_{1}\in\mathcal{K}_{1}$, then $N_{PSL_{3}(3)}(H_{1})\cong
3_{+}^{1+2}:E_{4}$ and
$N_{PSL_{3}(3):\left\langle\sigma\right\rangle}(H_{1})=3_{+}^{1+2}:D_{8}$.

  2. (2)

If $H_{2}\in\mathcal{K}_{2}$, then $N_{PSL_{3}(3)}(H_{2})\cong E_{3}\times
S_{3}$ and $N_{PSL_{3}(3):\left\langle\sigma\right\rangle}(H_{2})\cong
S_{3}\times S_{3}$.

###### Proof.

Let $H_{1}\in\mathcal{K}_{1}$. Then $H_{1}$ is $(C,\ell)$-elation group of
$PG_{2}(3)$ and hence $N_{PSL_{3}(3)}(H_{1})$ is the stabilizer in
$PSL_{3}(3)$ of the flag $(C,\ell)$. Thus $N_{PSL_{3}(3)}(H_{1})\cong
U:E_{4}$, and hence
$N_{PSL_{3}(3):\left\langle\sigma\right\rangle}(H_{1})=U:D_{8}$, where $U$ is
a Sylow $3$-subgroup of $PSL_{3}(3)$, since $\mathcal{K}_{1}$ and
$\mathcal{K}_{2}$, are conjugacy
$PSL_{3}(3):\left\langle\sigma\right\rangle$-classes. This proves (1).

Let $H_{2}\in\mathcal{K}_{2}$ and let $\mathcal{C}$ be any non-degenerate
conic of $PG_{2}(3)$. It is well known that $PSL_{3}(3)_{\mathcal{C}}\cong
PGL_{2}(3)\cong S_{4}$ and that $PSL_{3}(3)_{\mathcal{C}}$ does not contain
elations of $PG_{2}(3)$ (e.g. see [36], Theorem 2.37). Then any Sylow
$3$-subgroup of $PSL_{3}(3)_{\mathcal{C}}$ lie in $\mathcal{K}_{2}$. Hence, we
may assume that $H_{2}$ is such a group. Then $S_{3}\leqslant
N_{PGL_{2}(3)}(H_{2})$ and hence $N_{PGL_{2}(3)}(H_{2})\cong E_{3}\times
S_{3}$, since $C_{PSL_{3}(3)}(H_{2})\cong E_{9}$ by [20]. Now, it is easy to
see that $N_{PGL_{2}(3):\left\langle\sigma\right\rangle}(H_{2})\cong
S_{3}\times S_{3}$, which is (2). ∎

######  Lemma 7.4.

Let $T\leqslant PSL_{3}(3):\left\langle\sigma\right\rangle$ . Then the
followings hold:

  1. (1)

If $T$ contains at least any two elements of $\mathcal{K}_{1}$, and
$\left|T\right|\mid 24$, then $T\leqslant PSL_{3}(3)$ and $T\cong SL_{2}(3)$.

  2. (2)

If $T$ contains at least any two elements of $\mathcal{K}_{2}$, and
$\left|T\right|=12$ or $24$, then $A_{4}\trianglelefteq T\cap PSL_{3}(3)$ and
$T\cong A_{4}$, $S_{4}$ or $A_{4}\times Z_{2}$.

###### Proof.

Let $H,K$ be any two elements of $\mathcal{K}_{1}$ contained $T$. Then they
are elation groups of $PG_{2}(3)$ with center $C_{i}$ and axis $\ell_{i}$,
where $i=1,2$, respectively. If either $C_{1}=C_{2}$ or $\ell_{1}=\ell_{2}$,
then $\left\langle H,K\right\rangle$ is elementary abelian of order $9$ by
[36], Theorem 4.14, but this contradicts $\left|T\right|\mid 24$. Therefore,
$C_{1}\neq C_{2}$ and $\ell_{1}\neq\ell_{2}$. Then $\left\langle
H,K\right\rangle$ fixes the point $\ell_{1}\cap\ell_{2}$ and the line
$C_{1}C_{2}$. If $\ell_{1}\cap\ell_{2}\in C_{1}C_{2}$, then
$\ell_{1}\cap\ell_{2}\in\left\\{C_{1},C_{2}\right\\}$, as
$\ell_{1}\neq\ell_{2}$. Suppose that
$\ell_{1}\cap\ell_{2}=\left\\{C_{1}\right\\}$, then $K$ fixes $C_{1}$ and
there is $\gamma\in K$ such that $\ell_{1}^{\gamma}\neq\ell_{1}$ as $C_{1}\neq
C_{2}$. So $\left\langle H,H^{\gamma}\right\rangle\leqslant T$, with
$\left\langle H,H^{\gamma}\right\rangle$ elementary abelian of order $9$ again
by [36], Theorem 4.14, and we reach a contradiction. The case
$\ell_{1}\cap\ell_{2}=\left\\{C_{2}\right\\}$ is ruled out similarly. Thus,
$\ell_{1}\cap\ell_{2}\notin C_{1}C_{2}$. Then
$SL_{2}(3)\trianglelefteq\left\langle H,K\right\rangle\leqslant GL_{2}(3)$, as
the stabilizer in $PSL_{3}(3)$ of $\ell_{1}\cap\ell_{2}$ and of $C_{1}C_{2}$
is $GL_{2}(3)$. Therefore, $T=\left\langle H,K\right\rangle\cong SL_{2}(3)$,
as $\left\langle H,K\right\rangle\leqslant T$ and $\left|T\right|\mid 24$.
This proves (1).

Assume that $T$ contains at least two elements $L,M$ of $\mathcal{K}_{2}$.
Then $T_{0}$ contains $L,M$, where $T_{0}=T\cap PSL_{3}(3)$. By [20], either
$T_{0}\leqslant W:GL_{2}(3)$, where $W$ is an elementary abelian group of
order $9$ consisting of elations of $PG_{2}(3)$, or $T_{0}\leqslant S_{4}$ and
$T_{0}$ preserves a conic of $PG_{2}(3)$. If $T_{0}\leqslant W:GL_{2}(3)$,
then $T_{0}\cap W=1$, since the Sylow $3$-subgroups of $T_{0}$ belong to
$\mathcal{K}_{2}$. Hence, $T_{0}$ is isomorphic to a subgroup of $GL_{2}(3)$.
Since $GL_{2}(3)$ does not contain subgroups of order $12$ with at least two
distinct Sylow $3$-subgroups, and since $\left|T\right|=12$ or $24$, it
follows that $T_{0}\cong SL_{2}(3)$. Then $T_{0}$ fixes an anti-flag of
$PG_{2}(3)$, as $Z(T)\cong Z_{2}$ is an homology of $PG_{2}(3)$. However this
case is ruled out, since any element of $\mathcal{K}_{2}$ fixes just a flag
and no more points or lines of $PG_{2}(3)$. Thus, $T_{0}\leqslant S_{4}$,
$T_{0}$ preserves a conic of $PG_{2}(3)$ and hence $A_{4}\trianglelefteq
T_{0}\leqslant S_{4}$, since $\left\langle H,K\right\rangle\leqslant T_{0}$.
Then either $T_{0}=T$ and $T\cong A_{4}$, $S_{4}$, or $T_{0}\neq T$,
$T_{0}\cong A_{4}$ and $T\cong S_{4}$, $A_{4}\times Z_{2}$, since
$\left|T\right|=12$ or $24$. This proves (2). ∎

######  Proposition 7.5.

If $X\cong PSL_{3}(3)$, then $\mathcal{D}$ is isomorphic to one of the
$2$-designs constructed in Example 2.1.

###### Proof.

Assume that $G$ is either isomorphic to $PSL_{3}(3)$ or to
$PSL_{3}(3):\left\langle\sigma\right\rangle$, where $\sigma$ is of order $2$,
and that $k=12$. If $(x,B)$ is any flag of $\mathcal{D}$, then either
$G_{x}\cong F_{39}$, or $G_{x}\cong F_{78}$, respectively, by [20]. Moreover
$b=12\cdot 13\cdot\lambda$ and the cyclic subgroups of order $3$ lying in
$G_{x}$ belong to the conjugacy class $\mathcal{K}_{2}$ defined above. Hence
$\left|G_{B}\right|=\frac{36c}{\lambda}$, where $c=\left[G:PSL_{3}(3)\right]$.
Then $\lambda\mid 3c$, as $12\mid\left|G_{B}\right|$, with $\lambda\geqslant
2$. Thus, $\left(\lambda,c\right)=(2,2),(3,1)$, $(3,2)$ or $(6,2)$.

  1. (i).

$\left(\lambda,c\right)\neq(2,2)$.

Assume that $\left(\lambda,c\right)=(2,2)$. Then $G\cong
PSL_{3}(3):\left\langle\sigma\right\rangle$, $\left|G_{B}\right|=36$, and
$G_{x,B}\cong Z_{3}$, with $G_{x,B}\in\mathcal{K}_{2}$. Then the
point-$G_{x}$-orbits on $\mathcal{D}$ are $\left\\{x\right\\}$, one of length
$13$, two ones of length $26$ and two ones of length $39$ by [54]. Then $B$
intersects each of these orbits in $1,1,2,2,3,3$ points respectively by Lemma
3.2(3). Hence, $G_{x,B}$ fixes at least six points on $B$. On the other hand,
Lemma 7.3(2) implies $\left[N_{G}(G_{x,B}):N_{G_{x}}(G_{x,B})\right]=6$.
Therefore, $\left|\mathrm{Fix(}G_{x,B}\mathrm{)}\right|=6$. Thus,
$\mathrm{Fix(}G_{x,B}\mathrm{)}\subset B$ and hence the number of blocks of
$\mathcal{D}$ containing $\mathrm{Fix(}G_{x,B}\mathrm{)}$ is at most $2$, as
$\lambda=2$. Thus either $r=\left|\mathcal{K}_{2}\right|$ or
$r=2\left|\mathcal{K}_{2}\right|$ and we reach a contradiction, as $r=24$ but
$\left|\mathcal{K}_{2}\right|=312$.

  1. (ii).

If $\left(\lambda,c\right)=(3,1)$, then $\mathcal{D}$ is isomorphic to the
$2$-design constructed in Example 2.1(1).

Assume that $\left(\lambda,c\right)=(3,1)$. Then $G\cong PSL_{3}(3)$ and
$G_{B}$ acts regularly on $B$. Let $H$ be a Sylow $3$-subgroup of $G_{B}$. By
Lemma 7.3(1)–(2), either $H\in\mathcal{K}_{1}$ and $G_{B}\cong D_{12}$, or
$H\in\mathcal{K}_{2}$ and $G_{B}\cong A_{4}$, according to whether $H$ is or
is not normal in $G_{B}$ respectively. With the aid of [62] we see that, no
flag-transitive $2$-designs occur with $G_{B}\cong D_{12}$, whereas, up to
isomorphism, only one flag-transitive $2$-$(12^{2},12,3)$ design occurs with
$G_{B}\cong A_{4}$. Thus, such a $2$-design is necessarily isomorphic to that
constructed in Example 2.1(1).

  1. (iii).

$\left(\lambda,c\right)\neq(3,2)$.

Assume that $\left(\lambda,c\right)=(3,2)$. Then $G\cong
PSL_{3}(3):\left\langle\sigma\right\rangle$, $G_{x,B}\cong Z_{2}$ and $r=39$.
Since $(G^{\prime})_{x}\cong F_{39}$ by [20], it follows that
$(G^{\prime})_{x,B}=1$. Therefore, $(G^{\prime})_{x}$ acts transitively on the
set of blocks of $\mathcal{D}$ incident with $x$. Also, $G^{\prime}$ acts
point-transitively on $\mathcal{D}$, as $G$ acts point-primitively on
$\mathcal{D}$. Thus, $G^{\prime}$ acts flag-transitively on $\mathcal{D}$.
Therefore, $\mathcal{D}$ is isomorphic to the $2$-design constructed in
Example 2.1(1) by (ii), and hence $G^{\prime}$ is its full flag-transitive
automorphism group. However, this is impossible, since $G\cong
PSL_{3}(3):\left\langle\sigma\right\rangle$ acts flag-transitively on
$\mathcal{D}$ by our assumption.

  1. (iv).

If $\left(\lambda,c\right)=(6,2)$, then $\mathcal{D}$ is isomorphic to the
$2$-design constructed in Example 2.1(2).

If $\left(\lambda,c\right)=(6,2)$, then $G\cong
PSL_{3}(3):\left\langle\sigma\right\rangle$, $r=78$, $G_{x,B}=1$ and
$\left|G_{B}\right|=12$. By using Lemma 7.3(1)–(2), it is easy to see that,
either $G_{B}$ is isomorphic to $Z_{12}$, $D_{12}$ (three conjugacy classes)
or to $A_{4}$. With the aid of [62] we see that, no flag-transitive
$2$-designs occur with $G_{B}\cong Z_{12}$ or $D_{12}$, whereas, up to
isomorphism, only one flag-transitive $2$-$(12^{2},12,6)$ design occurs with
$G_{B}\cong A_{4}$. Thus, such $2$-design is necessarily isomorphic to that
constructed in Example 2.1(2). ∎

######  Lemma 7.6.

$X$ is not isomorphic to $PSp_{n}(q)$ for $n\geqslant 4$.

###### Proof.

Assume that $X\cong PSp_{n}(q)$ with $n\geqslant 4$. Then $X_{x}$ lies in a
large maximal geometric subgroup $Y$ of $X$ by Theorem 7.1. Hence, $Y$ is one
of the groups listed in [1], Proposition 4.22. It follows from Theorem 3.5 of
[33] and Proposition 5.2.15.(i) of [39], that $Y\notin\mathcal{C}_{1}(X)$ and
$Y\notin\mathcal{C}_{2}(X)\cup\mathcal{C}_{5}(X)$, respectively, since
$\left(\Phi_{nf}^{\ast}(p),\left|Y\right|\right)>1$. For the same reason
$(X,Y)\neq(PSp_{8}(3),2^{6}.\Omega_{6}^{-}(2))$,
$(PSp_{4}(5),2^{4}.\Omega_{4}^{-}(2))$. Thus, one of the following holds:

  1. (1)

$Y$ is a $\mathcal{C}_{8}$-subgroup of $X$;

  2. (2)

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

  3. (3)

$(X,Y)$ is either $(PSp_{4}(7),2^{4}.O_{4}^{-}(2))$ or
$(PSp_{4}(3),2^{4}.\Omega_{4}^{-}(2))$.

Assume that Case (1) holds. Then $Y\cong O_{n}^{-}(q)$ with $q$ even by [39],
Proposition 4.8.6.(II), as
$\left(\Phi_{nf}^{\ast}(p),\left|Y\right|\right)>1$. Also $X_{x}=Y$ by Lemma
3.6(1) and by [39], Proposition 4.8.6.(I). Then $k^{2}=q^{n/2}(q^{n/2}-1)$,
since $k^{2}=\left[X:X_{x}\right]$, and hence $q^{n/2}-1$ is a square.
However, this is impossible by [49], A3.1, since $n\geqslant 4$.

Assume that Case (2) holds. If $Y$ is a $\mathcal{C}_{3}$-subgroup of $X$ of
type $Sp_{2i}(q^{t})$, $t=2,3$, where $n=2it$, then $Y\cong
PSp_{2i}(q^{t})\cdot Z_{t}$ by [39], Proposition 4.3.10.(II). Also $X_{x}=Y$
by Lemma 3.6(1) and by [39], Proposition 4.3.10.(I). It follows from Lemma
4.8(1) that
$\left|X\right|\leqslant\left|\mathrm{Out(}X\mathrm{)}\right|^{2}\cdot\left|X_{x}\right|\cdot\left|X_{x}\right|_{p^{\prime}}^{2}$,
which in turn yields the first (centered) inequality at page 21 of [11] (see
part (4) in the proof of Proposition 4.4). Then $i=1$ for $t=3$ and
$i\leqslant 2$ for $t=2$. The case $t=3$ is clear, whereas some remarks should
be done for $t=2$. In [11] the authors assert that the replication number of
the $2$-designs they investigate divides
$\left|\mathrm{Out(}X\mathrm{)}\right|(q^{n/t}-1)$ by making use of their
their Lemmas 3.6 and 3.7.(c)–(d). In our context, Lemmas 3.6 of [11] still
works. Indeed, it is a general group-theoretical result. The role of Lemma
3.7(c) is replaced by the inequality $\left(r/\lambda\right)^{2}>v$, since
$r=(k+1)\lambda$ and $v=k^{2}$. Finally, the role of Lemma 3.7.(d) is replaced
by Lemma 3.2(3). Thus, we get
$r/\lambda\mid\left|\mathrm{Out(}X\mathrm{)}\right|(q^{4i}-1)$. At this point
we may proceed as in [11] to gain
$q^{4i^{2}/8}<k^{2}<\left(r/\lambda\right)^{2}<2q^{8i+1}$, as
$\left|\mathrm{Out(}X\mathrm{)}\right|^{2}\leqslant 2q$. Therefore $i\leqslant
2$.

Assume that $t=3$ and $i=1$. Then $X\cong PSp_{6}(q)$ and $X_{x}\cong
PSp_{2}(q^{3})\cdot Z_{3}$. Hence, $k^{2}=q^{3}(q^{2}-1)^{2}(q^{2}+1)/3$.
Since $3$ does not divide $q^{2}+1$, this one is a square. However, this is
impossible by [49], A3.1.

Assume that $t=2$ and $i=2$. Then $X\cong PSp_{8}(q)$ and $X_{x}\cong
PSp_{4}(q^{2})\cdot Z_{2}$. Then
$k^{2}=q^{8}\left(q^{2}-1\right)^{2}\left(q^{4}+q^{2}+1\right)/2$, but this is
impossible by [49], A7.1.

Assume that $t=2$ and $i=1$. Then $X\cong PSp_{4}(q)$ and $X_{x}\cong
PSp_{2}(q^{2})\cdot Z_{2}$. Then $k^{2}=q^{2}(q^{2}-1)/2$. Then $q$ is odd by
[49], A3.1. Then $\frac{q\pm 1}{2}=z_{1}^{2}$ and $2(q\mp 1)=z_{2}^{2}$ for
some positive integers $z_{1},z_{2}$, according to whether $q$ is equivalent
to $1$ or $3$ modulo $4$ respectively. So $z_{1}^{2}\mp 1=(z_{2}/2)^{2}$,
which has no solutions again by [49], A3.1.

Finally, assume that $Y$ is a $\mathcal{C}_{3}$-group of type $GU_{n/2}(q)$.
Then $X_{x}=Y\cong Z_{\frac{q-1}{2}}.PGU_{n/2}(q).Z_{2}$ by Lemma 3.6(1) and
by [39], Proposition 4.3.7.(I)–(II). Moreover, $n/2$ is odd, since
$\left(\Phi_{nf}^{\ast}(p),\left|Y\right|\right)>1$, and hence $n\geqslant 6$.
Then