# Sharp upper bounds on the minimal number of elements required to generate a
transitive permutation group

Gareth Tracey  Mathematical Institute, University of Oxford
<EMAIL_ADDRESS>
###### Abstract.

The purpose of this paper is to prove that if $G$ is a transitive permutation
group of degree $n\geq 2$, then $G$ can be generated by $\lfloor
cn/\sqrt{\log{n}}\rfloor$ elements, where $c:=\sqrt{3}/2$. Owing to the
transitive group $D_{8}\circ D_{8}$ of degree $8$, this upper bound is best
possible. Our new result improves a 2015 paper by the author, and makes use of
the recent classification of transitive groups of degree $48$.

###### 2010 Mathematics Subject Classification:

20B40, 05C30

The author was supported by the Engineering and Physical Sciences Research
Council (grant no. EP/T017619/1).

##  1\. Introduction

For an arbitrary group $G$, let $d(G)$ be the minimal size of a generating set
of $G$. In [13], the problem of finding numerical upper bounds for $d(G)$ for
an arbitrary transitive permutation group $G$ of degree $n$ is considered. It
had already been proved in [9] that $d(G)$ is at most $\frac{cn}{\sqrt{\log
n}}$ in this case, where $c$ is an absolute constant. This bound is shown to
be asymptotically best possible in [7] (that is, there exists constants
$c_{1}$, $c_{2}$, and an infinite family $(G_{n_{i}})_{i=1}^{\infty}$ of
transitive groups of degree $n_{i}$, with
$c_{1}\leq\frac{n_{i}}{d(G_{n_{i}})\sqrt{\log n_{i}}}\leq c_{2}$ for all $i$).

In [13] it is proved that, apart from a finite list of possible exceptions,
the bound $d(G)\leq\left\lfloor\frac{cn}{\sqrt{\log n}}\right\rfloor$ holds,
where $c:=\frac{\sqrt{3}}{2}$. This bound is best possible in the sense that
$d(G)=\frac{\frac{\sqrt{3}}{2}n}{\sqrt{\log n}}$ when $G=D_{8}\circ
D_{8}<\operatorname{\mathrm{S}}_{8}$ and $n=8$.

In this paper, we remove the possible exceptions listed in [13, Theorem 1.1].
More precisely, we prove:

######  Theorem 1.1.

Let $G$ be a transitive permutation group of degree $n$. Then

$d(G)\leq\left\lfloor\frac{cn}{\sqrt{\log{n}}}\right\rfloor$

where $c:=\frac{\sqrt{3}}{2}$.

There are a number of steps in the proof of Theorem 1.1. For this reason, we
will spend the next few paragraphs outlining our general strategy for the
proof.

First, by [13, Theorem 5.3], we only need to prove Theorem 1.1 when $G$ is
imprimitive with minimal block size $2$, and $n$ has the form $n=2^{x}3^{y}5$
with either $y=0$ and $17\leq x\leq 26$; or $y=1$ and $15\leq x\leq 35$. Thus,
in particular, $G$ may be viewed as a subgroup in a wreath product $2\wr
G^{\Sigma}$, where $\Sigma$ is a set of blocks for $G$ of size $2$. It follows
that $d(G)\leq d_{G^{\Sigma}}(M)+d(G^{\Sigma})$, where $M$ is the intersection
of $G$ with the base group of the wreath product, and $d_{G^{\Sigma}}(M)$ is
the minimal number of elements required to generate $M$ as a
$G^{\Sigma}$-module.

Of course, $d(G^{\Sigma})$ can be bounded using induction. The bulk of this
section will therefore be taken up with finding upper bounds on
$d_{G^{\Sigma}}(M)$. Since $d_{G^{\Sigma}}(M)\leq d_{H}(M)$ for any subgroup
$H$ of $G^{\Sigma}$, the strategy of the third author in [13] in this case
involved replacing $G^{\Sigma}$ by a convenient subgroup $H$ of $G^{\Sigma}$,
and then deriving upper bounds on $d_{H}(M)$, usually in terms of the lengths
of the $H$-orbits in $\Sigma$. This approach turns out to be particularly
fruitful when $H$ is chosen to be a soluble transitive subgroup of
$G^{\Sigma}$, whenever such a subgroup exists. When $G^{\Sigma}$ does not
contain a soluble transitive subgroup, however, the analysis becomes much more
complicated. This leads to less sharp bounds, and ultimately, the omitted
cases in [13, Theorem 1.1].

Our approach in this section involves a careful analysis of the orbit lengths
of soluble subgroups in a minimal transitive insoluble subgroup of $G$,
building on the work in [13] in the case $n=2^{x}3$.

The layout of the paper is as follows: First, Section 2 contains a necessary
discussion of subdirect products of finite groups. Then, in Section 3 we prove
a general theorem concerning minimal transitive groups of degree
$2^{x}3^{y}5^{z}$, with $0\leq y,z\leq 1$. Finally, this information will
allow us to prove Theorem 1.1, and we do so in Section 4.

Acknowledgements: It is a pleasure to thank Derek Holt and Gordon Royle for
helpful conversations and suggestions.

##  2\. Subdirect products of finite groups

As mentioned above, we begin preparations toward the proof of Theorem 1.1 with
some straightforward but necessary lemmas concerning subdirect products of
finite groups. First, we make the following remark concerning our notation and
terminology in direct products of finite groups:

######  Remark 2.1.

For a group $T$ we will write $T^{e}$ for the direct product of $e$ copies of
$T$. More generally, let $G_{1},\ldots,G_{r}$ be finite groups, and consider
the direct product $G:=G_{1}^{e_{1}}\times\ldots\times G_{r}^{e_{r}}$. Fix
subgroups $H_{i}\leq G_{i}$, for $1\leq i\leq r$, and set
$e:=\sum_{i=1}^{r}k_{i}$. Then we write $H_{1}^{e_{1}}\times\ldots\times
H_{r}^{e_{r}}$ for the subgroup $\\{(x_{1},\ldots,x_{e})\text{ : }x_{j}\in
H_{i}\text{ for }f_{i-1}+1\leq j\leq f_{i}\\}$ of $G$, where $f_{0}:=0$, and
$f_{i}:=\sum_{k=1}^{i}e_{k}$ for $i>0$.

Also, the subgroup
$\operatorname{\mathrm{S}}_{e_{1}}\times\ldots\times\operatorname{\mathrm{S}}_{e_{r}}$
of $\operatorname{\mathrm{S}}_{e}$ (in its natural intransitive action) acts
via automorphisms on $G$ (by permutation of coordinates). Thus we can, and do,
speak of
($\operatorname{\mathrm{S}}_{e_{1}}\times\ldots\times\operatorname{\mathrm{S}}_{e_{r}}$)-conjugates
of subgroups of $G$. This will be very useful for avoiding cumbersome
notation.

Finally, for a group $T$ we call a subgroup $G$ of $T^{e}$ a _diagonal
subgroup_ of $T^{e}$ if there exists automorphisms $\alpha_{i}$ of $T$ such
that $G=\\{(t^{\alpha_{1}},t^{\alpha_{2}},\ldots,t^{\alpha_{r}})\text{ : }t\in
T\\}$.

Suppose now that $G$ is a subgroup in a direct product
$G_{1}\times\ldots\times G_{e}$ of groups $G_{i}$, and write
$\pi_{i}:G\rightarrow G_{i}$ for the $i$th coordinate projections. Then $G$ is
a _subdirect product_ in $G_{1}\times\ldots\times G_{e}$ if $G\pi_{i}=G_{i}$
for all $i$. We also introduce the following non-standard definition:

######  Definition 2.2.

Let $G_{i}$ be as above, and let $n$ be a positive integer. We say that $G$ is
_a subdirect product of the form_ $G=\frac{1}{n}(G_{1}\times\ldots\times
G_{e})$ if each of the following holds:

  1. (a)

$G$ is a subdirect product in $G_{1}\times\ldots\times G_{e}$;

  2. (b)

$G$ contains $[G_{1},G_{1}]\times\ldots\times[G_{e},G_{e}]$; and

  3. (c)

$G$ has index $n$ in $G_{1}\times\ldots\times G_{e}$.

The following is a collection of easy facts concerning subdirect products in
direct products of finite groups:

######  Lemma 2.3.

Let $G_{1},\ldots,G_{e}$ be finite groups, and
$\pi_{i}:G_{1}\times\ldots\times G_{e}\rightarrow G_{i}$ be as above, and let
$K_{i}$ be the intersection of $G_{i}$ with the $i$th coordinate subgroup of
$G_{1}\times\ldots\times G_{e}$. Suppose that $G$ is a subdirect product in
$G_{1}\times\ldots\times G_{e}$. Then each of the following holds:

  1. (i)

$K_{i}\pi_{i}\unlhd G_{i}$ for all $i$.

  2. (ii)

If $e=2$ and $G_{1}$ and $G_{2}$ have no common nontrivial homomorphic images,
then $G=G_{1}\times G_{2}$.

  3. (iii)

Let $e=\sum_{i=1}^{r}e_{i}$ be a positive partition of $e$, and define
$f_{0}:=0$, and $f_{i}:=\sum_{k=1}^{i}e_{k}$ for $1\leq i\leq r$. Suppose that
for each $1\leq i\leq r$, there exists a finite simple group $T_{i}$ such that
$G_{j}$ is isomorphic to $T_{i}$ for all $f_{i-1}+1\leq j\leq f_{i}$. Then for
each $1\leq i\leq r$ there exists a partition $\sum_{k=1}^{t_{i}}d_{i,k}$ of
$e_{i}$ and diagonal subgroups $D_{i,k}$ of $T_{i}^{d_{i,k}}$ such that $G$ is
$(\operatorname{\mathrm{S}}_{e_{1}}\times\ldots\times\operatorname{\mathrm{S}}_{e_{r}})$-conjugate
to $(D_{1,1}\times\ldots\times
D_{1,t_{1}})\times\ldots\times(D_{r,1}\times\ldots\times D_{r,t_{r}})$.

  4. (iv)

Let $T$ be a nonabelian simple group, and suppose that $|G_{i}|<|T|$ for all
$i$. Then $G$ has no homomorphic image isomorphic to $T$.

  5. (v)

Fix $1<f<e$. Suppose that $G_{i}$ is nonabelian simple for $1\leq i\leq f$,
and that $|G_{j}|<|G_{i}|$ for all $1\leq i\leq f$ and $f+1\leq j\leq e$.
Assume also that $|G_{i}|$ does not divide the index of $G$ in
$G_{1}\times\ldots\times G_{e}$ for any $1\leq i\leq f$. Then
$G=G_{1}\times\ldots\times G_{f}\times L$, where $L$ is a subdirect product in
$G_{f+1}\times\ldots\times G_{e}$.

###### Proof.

Part (i) is clear, while part (ii) follows from the fact that if $e=2$, then

$\frac{G_{1}}{K_{1}\pi_{1}}\cong\frac{G_{2}}{K_{2}\pi_{2}}.$

Part (iii) is well-known (for example, see [11, Theorem 4.16(iii)]), so we
just need to prove (iv) and (v).

We begin with (iv). So assume that $T$ is a nonabelian finite simple group,
and that $|G_{i}|<|T|$ for all $i$. We prove the claim in (iv) by induction on
$e$, with the case $e=1$ being trivial. So assume that $e>2$, and write
$\hat{\pi}$ for the projection $\hat{\pi}:G\rightarrow G_{2}\times\ldots\times
G_{e}$. Suppose that $N$ is a normal subgroup of $G$ with $G/N\cong T$. Then
since $G\hat{\pi}/N\hat{\pi}$ is both a homomorphic image of $G/N\cong T$, and
the subdirect product $G\hat{\pi}\leq G_{2}\times\ldots\times G_{e}$, the
inductive hypothesis implies that $N\hat{\pi}=G\hat{\pi}$. Then since $K_{1}$
is the kernel of $\hat{\pi}$, we have $|G|=|K_{1}||G\hat{\pi}|$ and
$|N|=|K_{1}\cap N||G\hat{\pi}|$. Hence,
$|G_{1}|\geq|K_{1}|\geq|K_{1}/K_{1}\cap N|=|T|$ – a contradiction. This proves
(iv).

Finally, assume that the hypotheses in (v) hold. Then $G$ may be viewed as a
subdirect product in $R\times L$, where $R$ is a subdirect product in
$G_{1}\times\ldots\times G_{f}$, and $L$ is a subdirect product in
$G_{f+1}\times\ldots\times G_{e}$. It then follows from (ii), (iii), and (iv)
that $G=R\times L$. Moreover, since $|G_{1}\times\ldots\times G_{f}:R|$
divides $|G_{1}\times\ldots\times G_{e}:G|$, we must have
$R=G_{1}\times\ldots\times G_{f}$, by (iii). This completes the proof of (v),
whence the lemma. ∎

##  3\. Minimal transitive groups of degree $2^{x}3^{y}5^{z}$

In this subsection, we investigate the structure of minimal transitive groups
of degree $2^{x}3^{y}5^{z}$, with $0\leq y,z\leq 1$. Our main result reads as
follows:

######  Theorem 3.1.

Let $G$ be an insoluble minimal transitive permutation group of degree
$n=2^{x}3^{y}5^{z}$, with $0\leq y,z\leq 1$. Then:

  1. (i)

$G$ has at most $y+z$ nonabelian chief factors;

  2. (ii)

each nonabelian chief factor of $G$ is isomorphic to a direct product of
copies of one of $\operatorname{\mathrm{A}}_{5}$;
$\operatorname{\mathrm{A}}_{6}$; $\operatorname{\mathrm{S}}_{4}(4)$;
$\operatorname{\mathrm{O}}^{+}_{8}(2)$; $\operatorname{\mathrm{A}}_{8}$;
$\operatorname{\mathrm{A}}_{16}$; or $\operatorname{\mathrm{L}}_{2}(q)$, where
$q$ has the form $q=2^{x_{1}}3^{y_{1}}5^{z_{1}}-1$ with $0\leq y_{1},z_{1}\leq
1$;

  3. (iii)

if $G$ has two nonabelian chief factors, then they are isomorphic to
$T_{1}^{e_{1}}$ and $\operatorname{\mathrm{L}}_{2}(p)^{e_{2}}$, where
$T_{1}\in\\{\operatorname{\mathrm{A}}_{5},\operatorname{\mathrm{A}}_{6}\\}$,
and $p$ is a Mersenne prime; and

  4. (iv)

$G$ has a soluble subgroup $F$ whose orbit lengths are given in Table 1.

Now, Table 1 requires some explanation: In Theorem 3.1, we prove that a
minimal transitive permutation group of degree $2^{x}3^{y}5^{z}$ with $0\leq
y,z\leq 1$, has at most two nonabelian chief factors, which have to be
isomorphic to direct powers of one of $\operatorname{\mathrm{A}}_{5}$;
$\operatorname{\mathrm{A}}_{6}$; $\operatorname{\mathrm{A}}_{8}$;
$\operatorname{\mathrm{A}}_{16}$; or $\operatorname{\mathrm{L}}_{2}(q)$, where
$q$ has the form $q=2^{x_{1}}3^{y_{1}}5^{z_{1}}-1$. The first column of Table
1 will be a group $T^{e}$, where $T$ is one of these simple groups. This means
that if $G$ is the minimal transitive group under consideration, then $G/R(G)$
has a minimal normal subgroup $L/R(G)$ which is isomorphic to a direct product
of $e$ copies of $T$ (recall that $R(G)$ is the soluble radical of $G$). The
second column gives the degree of $G$ as a permutation group, while the third
column gives the orbit lengths of a soluble subgroup $F$ of $G$. Moreover,
these orbit lengths will be given in one of the following ways:

Rows 23–30:

In these cases, the orbit lengths are given in the form of a list with entries
$\sum_{i=1}^{f}k_{i}\times l_{i}$, with the entry $\sum_{i=1}^{f}k_{i}\times
l_{i}$ meaning that $F$ has $k_{i}$ orbits of size $l_{i}$ for each $1\leq
i\leq f$.

Rows 1–11 and 13–21:

In these cases, the orbit lengths are given in the form $\alpha X_{i}$, $\beta
Y_{j}$. Here, $\\{X_{i}\\}_{i}$ and $\\{Y_{j}\\}_{j}$ are positive unordered
partitions whose sum is given in the fourth column of the table. For example,
if $F$ is as in row 1, then $\\{X_{i}\\}_{i}$ is a positive unordered
partition of $4$ (there are five such partitions) and $\\{Y_{j}\\}_{j}$ is a
positive unordered partition of $2$ (there are two such partitions). Thus,
there are ten possibilities for the orbit lengths of $F$. For instance, the
orbit sizes ($40a$, $100a$, $100a$) correspond to the partitions $4=4$ and
$2=1+1$ of $4$ and $2$, respectively.

Rows 12 and 22:

The orbits lengths in these cases are given in the form
$\sum_{i=0}^{e_{2}}\sum_{j,k}C^{(i)}_{j}\times 2^{b}3p^{i}5X^{(i,j)}_{k}$,
$0\leq i\leq e_{2}$. By this notation we mean that $(C^{(i)}_{j})_{j}$ is a
positive ordered partition of a particular number $C$ for each $i$;
$(X^{(i,j)}_{k})_{k}$ is a positive ordered partition of a particular number
$X$ for each $i,j$; and $F$ has $C^{(i)}_{j}$ orbits of size
$2^{b}3p^{i}5X^{(i,j)}_{k}$ for each $i$, $j$, $k$. The numbers $C$, $X$, $b$,
and $p$ are given in the fourth column of the table (in fact,
$C:=\binom{e_{2}}{i}$ in each case).

As an example, assume that $G$ is a minimal transitive group of degree
$n=2^{15}15$. If a minimal normal subgroup of $G$ is isomorphic to a direct
product of copies of $\operatorname{\mathrm{A}}_{5}$, then the first section
of Table 1 gives us the orbit lengths of a soluble subgroup $F$ of $G$. For
example, if $F$ is as in the first line of the table, then $F$ has orbit
lengths $2^{12}5X_{i}$, $2^{13}25Y_{j}$, where
$(X_{i})_{i}\in\\{(4),(3,1),(2,2),(1,1,1,1)\\}$, and
$(Y_{j})_{j}\in\\{(2),(1,1)\\}$. Thus, the possible $F$-orbit lengths in this
case are: ($2^{14}5$, $2^{14}25$); or ($2^{14}5$, $2^{13}25$, $2^{13}25$); or
($2^{12}15$, $2^{12}5$, $2^{14}25$); or ($2^{12}15$, $2^{12}5$, $2^{13}25$,
$2^{13}25$); or ($2^{13}5$, $2^{13}5$, $2^{14}25$); or ($2^{13}5$, $2^{13}5$,
$2^{13}25$, $2^{13}25$); or ($2^{12}5$, $2^{12}5$, $2^{12}5$, $2^{12}5$,
$2^{14}25$); or ($2^{12}5$, $2^{12}5$, $2^{12}5$, $2^{12}5$, $2^{13}25$,
$2^{13}25$). The rest of the possibilities for the orbit lengths of $F$ can be
computed in an entirely analogous way. (Although these computations seem
tedious, they can be done very quickly using a computer).

$\overline{L}$ |  degree of $G$ |  orbit lengths of a soluble subgroup $F$ |  notes
---|---|---|---
$\operatorname{\mathrm{A}}_{5}^{e}$ |  $240a$ |  $10aX_{i}$, $100aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(4,2)$
|  $120a$ |  $5aX_{i}$, $50aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(4,2)$
|  $120a$ |  $10aX_{i}$, $50aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,2)$
|  $60a$ |  $5aX_{i}$, $25aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,2)$
|  $60a$ |  $5aX_{i}$, $50aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,1)$
|  $30a$ |  $5aX_{i}$, $25aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(1,1)$
|  $2^{b}10a$ |  $2^{b}5aX_{i}$ |  $\sum_{i}X_{i}=2$; $b\in\\{0,1\\}$; $G/L$ soluble
|  $60a$ |  $10a$, $50a$ |
|  $30a$ |  $5aX_{i}$, $10aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,2)$
|  $120a$ |  $5aX_{i}$, $25aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(4,4)$
|  $60a$ |  $5aX_{i}$, $25aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,2)$
|  $2^{b}(p+1)^{e_{2}}30a$ |  $\sum_{i=0}^{e_{2}}\sum_{j,k}C^{(i)}_{j}\times 2^{b}3p^{i}5aX^{(i,j)}_{k}$ |  $\sum_{k}C^{(i)}_{j}=\binom{e_{2}}{i}$ and $\sum_{j}X^{(i,j)}_{k}=2$ for each $i,j$; $b\in\\{0,1\\}$; $G/L$ insoluble; $p$ a Mersenne prime; $1\leq e_{2}\leq e$
$\operatorname{\mathrm{A}}_{6}^{e}$ |  $240a$ |  $10aX_{i}$, $50aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(4,4)$
|  $120a$ |  $5aX_{i}$, $25aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(4,4)$
|  $60a$ |  $5aX_{i}$, $25aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,2)$
|  $120a$ |  $10aX_{i}$ |  $\sum_{i}X_{i}=12$
|  $15a$ |  $5aX_{i}$ |  $\sum_{i}X_{i}=3$
|  $40a$ |  $10aX_{i}$ |  $\sum_{i}X_{i}=4$; $G/L$ soluble
|  $2^{b}5a$ |  $5aX_{i}$ |  $\sum_{i}X_{i}=2^{b}$; $b\in\\{1,2\\}$; $G/L$ soluble
|  $60a$ |  $5aX_{i}$, $10aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(4,4)$
|  $30a$ |  $5aX_{i}$, $10aY_{j}$ |  $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,2)$
|  $2^{b}X(p+1)^{e_{2}}15a$ |  $\sum_{i=0}^{e_{2}}\sum_{j,k}C^{(i)}_{j}\times 2^{b}3p^{i}5aX^{(i,j)}_{k}$ |  $\sum_{j}C^{(i)}_{j}=\binom{e_{2}}{i}$, $\sum_{k}X^{(i,j)}_{k}=X$ for each $i,j$; $(X,b)\in\\{(4,1),(4,0),(2,0)\\}$; $G/L$ insoluble; $p$ a Mersenne prime; $1\leq e_{2}\leq e$
$T^{e}$ |  $2^{b}120a$ |  $2^{b}24a+2^{b}96a$ |  $b\in\\{0,3\\}$, $T\in\\{\operatorname{\mathrm{S}}_{4}(4),\operatorname{\mathrm{O}}^{+}_{8}(2)\\}$
$\operatorname{\mathrm{A}}_{8}^{e}$ |  $8^{e_{2}}960a$ |  $\sum_{j=1}^{3}\sum_{i=0}^{e_{2}}as_{j}\binom{e_{2}}{i}\times 7^{i}l_{j}$ |  $0\leq e_{2}\leq e$; $1\leq j\leq 3$, where $(s_{1},l_{1})=(1,1)$, $(s_{2},l_{2})=(11,7)$, and $(s_{3},l_{3})=(42,21)$.
|  $8^{e_{2}}120a$ |  $\sum_{j=1}^{3}\sum_{i=0}^{e_{2}}as_{j}\binom{e_{2}}{i}\times 7^{i}l_{j}$ |  $0\leq e_{2}\leq e$; $1\leq j\leq 3$, where $(s_{1},l_{1})=(1,1)$, $(s_{2},l_{2})=(5,7)$, and $(s_{3},l_{3})=(4,21)$.
|  $8^{e_{2}}15a$ |  $\sum_{j=1}^{2}\sum_{i=0}^{e_{2}}as_{j}\binom{e_{2}}{i}\times 7^{i}l_{j}$ |  $0\leq e_{2}\leq e$; $1\leq j\leq 2$, where $(s_{1},l_{1})=(1,1)$ and $(s_{2},l_{2})=(2,7)$.
$\operatorname{\mathrm{A}}_{16}^{e}$ |  $15\frac{16^{e_{2}+1}}{2^{b}}a$ |  $\sum_{i=0}^{e_{2}}a\frac{16}{2^{b}}\binom{e_{2}}{i}\times 15^{i+1}$ |  $b\in\\{0,1\\}$, $0\leq e_{2}\leq e$
$\operatorname{\mathrm{L}}_{2}(q)^{e}$ |  $15a(q+1)^{e_{2}+2}$ |  $\sum_{i=0}^{e_{2}}a\binom{e_{2}}{i}\times 15q^{i}+2a\binom{e_{2}}{i}\times 15q^{i+1}$ |  $q$ a Mersenne prime, $0\leq e_{2}\leq e$
|  $3^{y_{0}}5^{z_{0}}a(q+1)^{e_{2}}$ |  $\sum_{i=0}^{e_{2}}a\binom{e_{2}}{i}\times 3^{y_{0}}5^{z_{0}}q^{i}$ |  $q$ a Mersenne prime, $1\leq e_{2}\leq e$
|  $2^{x^{\prime}}15a$ |  $aq3^{1-y^{\prime}}5^{1-z^{\prime}}+a3^{1-y^{\prime}}5^{1-z^{\prime}}$ |  $q=2^{x^{\prime}}3^{y^{\prime}}5^{z^{\prime}}-1$, $\\{y^{\prime},z^{\prime}\\}=\\{1,0\\}$
Table 1. Orbit lengths of a certain soluble subgroup of a minimal transitive
group $G$ with $\overline{L}:=L/R(G)$ a minimal normal subgroup of $G/R(G)$.

######  Remark 3.2.

Let $G$ be as in the statement of Theorem 3.1, and suppose that we have proved
the theorem for all degrees less than $n$. Let $K$ be the kernel of the action
of $G$ on a block system $\Sigma=\\{\Delta^{g}\text{ : }g\in G\\}$ for $G$ in
$\operatorname{\mathrm{O}}$. Then $G^{\Sigma}\cong G/K$ is minimal transitive.
Thus, we may choose a soluble subgroup $F/K$ of $G/K\cong G^{\Sigma}$
satisfying Theorem 3.1(iv) with $G$ replaced by the minimal transitive group
$G^{\Sigma}$

If $K$ is soluble and acts transitively on $\Delta$, then $F$ is soluble and
the $F$-orbits have lengths $|\Delta||\Sigma_{i}|$, as $\Sigma_{i}$ runs over
the $F^{\Sigma}$-orbits. Suppose that the $\Sigma_{i}$ are as in row k in
Table 1. Then since $|\Delta|$ divides $n$, it is easy to see that by
replacing $a$ by $|\Delta|a$, $F\leq G$ has the required orbit lengths in
order to fulfil the conclusion of (iv). (For example, if $F/K$ is as in row 2
of Table 1, then $G$ has degree $120|\Delta|a$, and $F$ is soluble with orbit
sizes $5|\Delta|aX_{i}$, $50|\Delta|aY_{j}$.) This is a useful inductive tool
which we will use throughout the section.

We begin preparations toward the proof of Theorem 3.1 with a reduction lemma:

######  Lemma 3.3.

Let $G$ be a counterexample to Theorem 3.1 of minimal degree. Then all minimal
normal subgroups of $G$ are nonabelian. Moreover, if $L$ is a minimal normal
subgroup of $G$ and $K$ denotes the kernel of the action of $G$ on the set of
$L$-orbits, then $K/L$ is soluble.

###### Proof.

Let $L$ be a minimal normal subgroup of $G$. Also, let $\Delta$ be an
$L$-orbit, and let $\Sigma:=\\{\Delta^{g}\text{ : }g\in G\\}$ be the set of
$L$-orbits in $[n]$, so that $G^{\Sigma}\cong G/K$. Then $n=|\Delta||\Sigma|$,
and $G^{\Sigma}$ is minimal transitive. The minimality of $G$ as a
counterexample then implies that $G^{\Sigma}$ satisfies the conclusion of
Theorem 3.1.

Let $E/L$ be a subgroup of $G/L$ which is minimal with the property that
$(E/L)(K/L)=G/L$. Then $(E/L)\cap(K/L)\leq\Phi(E/L)$, so $(E/L)\cap(K/L)$ is
soluble. But also, $EK=G$. Since $E$ contains $L$ and $E^{\Sigma}=G^{\Sigma}$
is transitive, we deduce that $E$ is transitive, whence $E=G$, since $G$ is
minimal transitive. Thus, $K/L$ is soluble. This proves the second part of the
lemma.

Finally, assume that $L$ is abelian. Then $K$ is soluble. It follows that the
set of nonabelian chief factors of $G$ is same as the set of nonabelian chief
factors of $G/K$. Thus, $G=G^{\operatorname{\mathrm{O}}}$ satisfies (i), (ii),
and (iii) in Theorem 3.1. Moreover, since the theorem holds in
$G^{\Sigma}\cong G/K$, we may choose a soluble subgroup $F/K$ of $G/K$
satisfying Theorem 3.1(iv). Then $F$ is soluble, since $K$ is soluble, and the
$F$-orbits in $[n]$ have lengths $|\Delta||\Sigma_{i}|$, as $\Sigma_{i}$ runs
over the $F^{\Sigma}$-orbits in $\Sigma$. Then $F$ is a subgroup of $G$
satisfying the conclusion of Theorem 3.1(iv) (see Remark 3.2). Thus, (i)–(iv)
in Theorem 3.1 hold in $G$, contradicting the assumption that $G$ is a
counterexample. ∎

By Lemma 3.3, a minimal normal subgroup $L$ of a minimal counterexample to
Theorem 3.1 must be a direct product of nonabelian simple groups. Since the
subnormal subgroups of a transitive permutation group of degree $n$ have orbit
lengths dividing $n$, this motivates us to study the pairs ($T$, $H$) where
$T$ is a nonabelian simple group, $H$ is a proper subgroup of $T$, and $|T:H|$
has the form $2^{x}3^{y}5^{z}$, for some $0\leq y,z\leq 1$.

Our main tool in this regard is a paper of Liebeck, Praeger, and Saxl [8],
where the finite simple groups $T$ with a maximal subgroup $M$ satisfying
$\pi(|M|)=\pi(|T|)$ are classified. In order to use this result, we will first
need to determine the finite simple groups $T$ with a subgroup $H$ of index
$2^{x}3^{y}5^{z}$ as above, such that $\pi(|M|)\neq\pi(|T|)$ for some maximal
subgroup $M<T$ containing $H$.

######  Proposition 3.4.

Let $T$ be a nonabelian finite simple group, and suppose that $T$ has a
subgroup $H$ of index $2^{x}3^{y}5^{z}$, with $0\leq y,z\leq 1$. Let $M$ be a
maximal subgroup of $T$ containing $H$, and assume that either
$\pi(|T|)\neq\pi(|M|)$, or that $T$ is an alternating group of degree less
than $10$. Then one of the following holds:

  1. (i)

$T=\operatorname{\mathrm{A}}_{n}$ is an alternating group and either

    1. (a)

$n=5$ and $H$ is any subgroup of $T$;

    2. (b)

$n=6$ and $|H|=2^{i}3$, for some $i\leq 3$;

    3. (c)

$n=7$, and either $H=\operatorname{\mathrm{GL}}_{3}(2)$ (two classes) or
$H=7:3<\operatorname{\mathrm{GL}}_{3}(2)$;

    4. (d)

$n=8$ and either $H=2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$ (two classes); or
$H=2^{3}:(7:3)<2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$ (two classes); or
$H\cong\operatorname{\mathrm{GL}}_{3}(2)<2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$
(three classes); or $H\cong 7:3<2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$; or

    5. (e)

$n=9$ and $H=\operatorname{\mathrm{L}}_{2}(8).3$ (two classes).

  2. (ii)

$T=\operatorname{\mathrm{L}}_{n}(q)$ and either:

    1. (a)

$n=2$, $q$ has the form $q=2^{x}3^{y_{1}}5^{z_{1}}-1$, and $H$ is a subgroup
of a maximal parabolic subgroup $M$ of $T$, with
$|M:H|=3^{y-y_{1}}5^{z-z_{1}}$; or

    2. (b)

$(n,q)=(4,3)$, and $H=3^{3}:\operatorname{\mathrm{L}}_{3}(3)$ (two classes).

  3. (iii)

$T=\operatorname{\mathrm{S}}_{4}(3)$ with $|T:H|=40$ (two classes); or
$T=\operatorname{\mathrm{S}}_{6}(2)$, and $|T:H|\in\\{120,960\\}$.

###### Proof.

Assume first that $T=\operatorname{\mathrm{A}}_{n}$. We claim that in order
for $|T:M|$ to have the form $|T:M|=2^{x_{1}}3^{y_{1}}5^{z_{1}}$ with $0\leq
y_{1},z_{1}\leq 1$ and $\pi(|T|)\neq\pi(|M|)$, we must have $n\leq 9$. To
prove this, we first list the the possibilities for $M$:

  * •

$M=(\operatorname{\mathrm{A}}_{k}\times\operatorname{\mathrm{A}}_{n-k}).2$,
for some $k\leq\frac{n}{2}$.

  * •

$M=\operatorname{\mathrm{A}}_{n}\cap(\operatorname{\mathrm{S}}_{r}\wr\operatorname{\mathrm{S}}_{s})$,
where $r,s>1$ and $rs=n$.

  * •

$M=\operatorname{\mathrm{A}}_{n}\cap X$, with
$X\leq\operatorname{\mathrm{S}}_{n}$ primitive in its natural action.

In particular, if $|M|$ is odd, then $M$ must be as in the third listed case
with $n$ an odd prime, $\frac{n-1}{2}$ odd, and $M=C_{n}\rtimes
C_{\frac{n-1}{2}}$. However, in this case $|T:M|=(n-2)!$, which has the form
$|T:M|=2^{x_{1}}3^{y_{1}}5^{z_{1}}$ with $0\leq y_{1},z_{1}\leq 1$ if and only
if $n\leq 7$. So we may assume that $|M|$ is even. But then either $|T|_{3}=3$
or $|T|_{5}=5$, since $\pi(|T|)\neq\pi(|M|)$. Thus $n\leq 9$, as claimed. We
can now use direct computation to quickly determine the possibilities for $H$.

So we may assume that $T$ is not an alternating group. If
$T\in\\{\operatorname{\mathrm{L}}_{2}(8),\operatorname{\mathrm{L}}_{3}(3),\operatorname{\mathrm{U}}_{3}(3),\operatorname{\mathrm{S}}_{4}(8)\\}$
then direct computation quickly shows that $T$ has no maximal subgroups $M$
with index of the form $|T:M|=2^{x_{1}}3^{y_{1}}5^{z_{1}}$ with $0\leq
y_{1},z_{1}\leq 1$. So we may assume that
$T\not\in\\{\operatorname{\mathrm{L}}_{2}(8),\operatorname{\mathrm{L}}_{3}(3),\operatorname{\mathrm{U}}_{3}(3),\\\
\operatorname{\mathrm{S}}_{4}(8)\\}$.

Suppose next that $T=\operatorname{\mathrm{L}}_{2}(p)$ with $p=2^{x}-1$ a
Mersenne prime. Then clearly $M$ must contain a Sylow $p$-subgroup of $T$.
Hence, $M$ is a maximal parabolic subgroup of $T$; $|T:M|=p+1=2^{x}$ and
$|M:H|=3^{y}5^{z}$ (since $|M|$ is odd). So assume also that $T$ is not of the
form $T=\operatorname{\mathrm{L}}_{2}(p)$ with $p$ a Mersenne prime. Then by
[8, Corollary 6], there exists a set $\Pi$ of odd prime divisors of $|T|$ such
that $\Pi$ intersects $\pi(|T|)\setminus\pi(|M|)\subseteq\\{2,3,5\\}$ non-
trivially. Moreover, the proof of [8, Corollary 6] shows that either $T$ is in
[8, Table 10.1] and $\Pi$ is as in the third column of [8, Table 10.1]
(ignoring the third row for
$T\in\\{\operatorname{\mathrm{S}}_{n}(q),\operatorname{\mathrm{O}}_{n}(q)\\}$
and the second row for $T=\operatorname{\mathrm{O}}^{-}_{n}(q)$); or $T$ is in
[8, Table 10.3] and $\Pi$ is as in the third column of [8, Table 10.3]; or $T$
is in [8, Table 10.4] and $\Pi$ is as in the second column of [8, Table 10.4]
(except when $T=\operatorname{\mathrm{U}}_{5}(2)$, where $\Pi:=\\{3,5,11\\}$);
or $T$ is in [8, Table 10.5] and $\Pi$ is as in the second column of [8, Table
10.5] (except when $T=G_{2}(q)$, where $\Pi:=\\{p,7,13\\},\\{5,7,13\\}$, or
$\\{7,13\\}$ according to whether $q\geq 5$, $q=4$, or $q=3$, respectively);
or $T$ is in [8, Table 10.6] and $\Pi$ is as in the second column of [8, Table
10.6] (except when $T=M_{11},M_{12}$, or $M_{24}$, where
$\Pi:=\\{3,11\\},\\{3,11\\}$, or $\\{7,23\\}$, respectively). Fix $r$ to be a
prime in $\Pi\cap(\pi(|T|)\setminus\pi(|M|))$, so that $r\in\\{3,5\\}$. This
implies in particular that $|T|_{r}=r$.

By [8, Corollary 6] and inspection of simple group orders, the case $r=3$
occurs only if $T=\operatorname{\mathrm{L}}_{2}(p)$ with $p$ of the form
$p=2^{x_{1}}3^{y_{1}}-1$, with $0\leq y_{1}\leq 1$. In this case, one can see
from [2, Table 8.1] that $M$ is a maximal parabolic subgroup of $T$. Since
$2^{x_{1}}$ is the largest power of $2$ dividing $|T|$ in this case, we deduce
that $x=x_{1}$ and that $H$ must have index dividing $3^{y-y_{1}}5^{z}$ in
$M$.

So we may assume, for the remainder of the proof, that $r=5$. In what follows,
for a prime power $q$ and a positive integer $n$ we will write $q_{n}$ for an
arbitrary primitive prime divisor of $q^{n}-1$. By Zsigmondy’s theorem these
exists for $n>1$ and $(q,n)\neq(2,6)$. By Fermat’s Little Theorem we have

(3.1) $n\mid q_{n}-1$.

Suppose first that $G$ is a finite simple classical group, and write
$T=X_{n}(q)$, where
$X\in\\{\operatorname{\mathrm{L}},\operatorname{\mathrm{U}},\operatorname{\mathrm{S}},\operatorname{\mathrm{O}}^{\pm},\operatorname{\mathrm{O}}\\}$,
and $q=p^{f}$ with $p$ prime. Then using (3.1) we can deduce from [8, Tables
10.1–10.6] that $T\in\\{\operatorname{\mathrm{L}}_{n}(q)\text{ }(n\leq
5),\operatorname{\mathrm{U}}_{n}(q)\text{ }(n\leq
6),\operatorname{\mathrm{S}}_{n}(q)\text{ }(n\leq
8),\operatorname{\mathrm{O}}^{\pm}_{n}(q)\text{ }(n\leq
10),\operatorname{\mathrm{O}}_{n}(q)\text{ }(n\leq 11)\\}$. Below, we examine
these remaining cases.

Suppose first that $T=\operatorname{\mathrm{L}}_{n}(q)$, with $n\leq 5$. If
$n\in\\{4,5\\}$, then (3.1) and [8, Tables 10.1 and 10.3] imply that either
$5$ is a primitive prime divisor of $q^{4}-1$ or $p=5$. If $n\in\\{2,3\\}$,
then $5$ is a primitive prime divisor of $q^{2}-1$. Suppose first that $M$ is
parabolic. Then we see from [2, Tables 8.2, 8.3, 8.8, and 8.18] that $|T:M|$
has the form $\frac{q^{n}-1}{(q-1)}$ or
$\frac{(q^{n}-1)(q^{n-1}-1)}{(q-1)(q^{2}-1)}$. A routine exercise then shows
that only the former can take the form $2^{x_{1}}3^{y_{1}}5^{z_{1}}$ with
$0\leq y_{1},z_{1}\leq 1$, and this can only happen if $n\in\\{2,4\\}$ and $q$
has the form $q=2^{a}3^{b}5^{c}-1$. It is then easy to see that the only
possibilities are either $(n,q)=(4,3)$ or
$(n,q)=(2,2^{x_{1}}3^{y_{1}}5^{z_{1}}-1)$ with $0\leq y_{1},z_{1}\leq 1$. If
$(n,q)=(4,3)$, then the only possibility is
$M=H=3^{3}:\operatorname{\mathrm{L}}_{3}(3)$. If
$(n,q)=(2,2^{x_{1}}3^{y_{1}}5^{z_{1}}-1)$ then $2^{x_{1}}$ is the largest
power of $2$ dividing $|T|$, so $x=x_{1}$ and $H$ has index dividing
$3^{y-y_{1}}5^{z-z_{1}}$ in $M$. So we may assume that $M$ is not parabolic.
Then $p$ divides $|T:M|$, so $p\in\\{2,3,5\\}$. Also, as mentioned above,
$r=5$ does not divide $|M|$. We can then quickly see from [2, Tables 8.2, 8.3,
8.8, and 8.18] that the only possibility is $(n,q)=(3,2)$ with $H$ having
order $7$ or $21$. Since
$\operatorname{\mathrm{L}}_{3}(2)\cong\operatorname{\mathrm{L}}_{2}(7)$, this
case is already listed in (ii)(a). This completes the proof in the case
$T=\operatorname{\mathrm{L}}_{n}(q)$.

The other classical cases are entirely similar: Suppose that
$T=\operatorname{\mathrm{U}}_{n}(q)$, with $3\leq n\leq 6$. If $M$ is
parabolic, then $|T:M|$ is divisible by either
$\frac{(q^{n}-(-1)^{n})(q^{n-1}-(-1)^{n-1})}{q^{2}-1}$ or
$\prod_{i=1}^{\left\lfloor\frac{n}{2}\right\rfloor}(q^{2i+1}+1)$. It is easy
to see, however, that neither $q^{3}+1$ not $q^{5}+1$ can ever have the form
$2^{x_{1}}3^{y_{1}}5^{z_{1}}$ with $0\leq y_{1},z_{1}\leq 1$. So $M$ must be
non-parabolic. Then $p$ divides $|T:M|$, so $p\in\\{2,3,5\\}$. Also, $5$ does
not divide $|M|$. We can then use the tables in [2, Chapter 8] to quickly find
that the only case of a maximal non-parabolic subgroups $M$ of $T$ with
$|T:M|=2^{x_{1}}3^{y_{1}}5^{z_{1}}$ and $0\leq y_{1},z_{1}\leq 1$ occurs when
$(n,q)=(4,2)$. Since
$\operatorname{\mathrm{U}}_{4}(2)\cong\operatorname{\mathrm{S}}_{4}(3)$, this
case is accounted for in (iii). This completes the proof in the case
$T=\operatorname{\mathrm{U}}_{n}(q)$.

The arguments in the remaining classical cases are almost identical, so we now
move on to the case where $G$ is an exceptional group of Lie type. Then $r=5$
is one of the primes occurring in the second column of the row for $T$ in [8,
Table 10.5]. By using (3.1), we see that the only possibilities are
$T\in\\{{}^{2}F_{4}(q)^{\prime},{}^{2}B_{2}(q)\\}$, and $5$ is a primitive
prime divisor of $q^{4}-1$. In each of these cases, the maximal subgroups of
$T$ are known: see [10] for $T={}^{2}F_{4}(q)^{\prime}$ and [12] for
$T={}^{2}B_{2}(q)$. In each case, we can quickly check that no maximal
subgroup $M$ of $T$ with $\pi(|T|)\neq\pi(|M|)$ can have $|T:M|$ of the form
$2^{x_{1}}3^{y_{1}}5^{z_{1}}$, for $0\leq y_{1},z_{1}\leq 1$.

Finally, assume that $G$ is a sporadic simple group. Then $T=J_{2}$, by [8,
Table 10.6]. However, a quick check of the Web Atlas [14] shows that no
maximal subgroup $M$ of $J_{2}$ with $\pi(|T|)\neq\pi(|M|)$ can have $|T:M|$
of the form $2^{x}3^{y}5^{z}$, for $0\leq y,z\leq 1$. This completes the
proof. ∎

We are now ready to determine the possibilities for the pairs ($T$,$H$), with
$T$ simple, $|T:H|$ dividing $2^{x}15$, and $H$ contained in a maximal
subgroup $M$ of $T$ with $\pi(|T|)=\pi(|M|)$.

######  Proposition 3.5.

Let $T$ be a nonabelian finite simple group, and suppose that $T$ has a
subgroup $H$ of index $n=2^{x}3^{y}5^{z}$, with $0\leq y,z\leq 1$. Suppose
also that $T$ is not an alternating group of degree less than $10$. Let $M$ be
a maximal subgroup of $T$ containing $H$, and assume that $\pi(|T|)=\pi(|M|)$.
Then one of the following holds:

  1. (i)

$T=\operatorname{\mathrm{A}}_{n}$, and either
$H=\operatorname{\mathrm{A}}_{n-1}$, or $n=16$ and
$\operatorname{\mathrm{A}}_{14}\leq H\leq\operatorname{\mathrm{A}}_{14}.2$.

  2. (ii)

$T=\operatorname{\mathrm{S}}_{2m}(2^{f})$ with $(m,f)\in\\{(4,1),(2,2)\\}$,
and $\operatorname{\mathrm{O}}^{-}_{2m}(2^{f})\unlhd H$.

  3. (iii)

$T=\operatorname{\mathrm{S}}_{4}(q)$, with $q\in\\{2,4\\}$, and
$\operatorname{\mathrm{S}}_{2}(q^{2})\unlhd H$.

  4. (iv)

$T=M_{11}$ and $H=\operatorname{\mathrm{L}}_{2}(11)$;

  5. (v)

$T=\operatorname{\mathrm{O}}^{+}_{8}(2)$ and $H=\operatorname{\mathrm{A}}_{9}$
(three classes, fused by a triality automorphsim); or

  6. (vi)

$T=M_{12}$ or $M_{24}$, and $H=M_{11}$ or $H=M_{23}$, respectively.

###### Proof.

We can quickly check using direct computation that $T$ can have no maximal
subgroup $M$ of the required index when $T$ is
$\operatorname{\mathrm{L}}_{2}(8)$, $\operatorname{\mathrm{L}}_{3}(3)$,
$\operatorname{\mathrm{U}}_{3}(3)$ or $\operatorname{\mathrm{S}}_{4}(8)$. So
we may assume that $T$ is not one of these groups. Suppose first that
$T\neq\operatorname{\mathrm{L}}_{2}(p)$ for a Mersenne prime $p$. Then using
[8, Corollary 6 and Table 10.7], the possibilities for $T$ and $M$ are as
follows.

  1. (1)

$T=\operatorname{\mathrm{A}}_{n}$ and
$M=\operatorname{\mathrm{A}}_{k}\cap(\operatorname{\mathrm{S}}_{k}\times\operatorname{\mathrm{S}}_{n-k})$
for some $k\leq n-1$ with $k\geq p$ for all primes $p\leq n$. Then
$|\operatorname{\mathrm{A}}_{n}:\operatorname{\mathrm{A}}_{n}\cap(\operatorname{\mathrm{S}}_{k}\times\operatorname{\mathrm{S}}_{n-k})|=\binom{n}{k}$
divides $|T:H|=2^{x}3^{y}5^{z}$. By a well-known theorem of Sylvester and
Schur (see [4]), $\binom{n}{k}$ has a prime divisor exceeding
$\min\left\\{k,n-k\right\\}$. Thus $k\in\\{n-1,n-2,n-3,n-4\\}$, since $k\geq
10$. We then quickly see that the only possibilities are
$H=\operatorname{\mathrm{A}}_{n-1}$; or $n=16$ and $k=14$, which gives us what
we need.

  2. (2)

$T=\operatorname{\mathrm{S}}_{2m}(q)$ ($m$, $q$ even) or
$\operatorname{\mathrm{O}}_{2m+1}(q)$ ($m$ even, $q$ odd), and
$M=N_{T}(\operatorname{\mathrm{O}}^{-}_{2m}(q))$. Now,
$|N_{T}(\operatorname{\mathrm{O}}^{-}_{2m}(q)):\operatorname{\mathrm{O}}^{-}_{2m}(q)|\leq
2$ using [6, Corollary 2.10.4(i) and Table 2.1.D]. Hence,
$|T:\operatorname{\mathrm{O}}^{-}_{2m}(q)|$ divides $2^{x+1}3^{y}5^{z}$. Also,
for each of the two choices of $T$ we get
$|T:\operatorname{\mathrm{O}}^{-}_{2m}(q)|=q^{m}(q^{m}-1)$. But
$q^{m}(q^{m}-1)$ has the form $2^{x_{1}}3^{y_{1}}5^{z_{1}}$ for $0\leq
y_{1},z_{1}\leq 1$ if and only if $q=2^{f}$ with $fm\in\\{2,4\\}$, since $m$
is even. Since
$\operatorname{\mathrm{S}}_{4}(2)\cong\operatorname{\mathrm{S}}_{6}$ is not
simple, we deduce that ($m,f$)$\in\\{(4,1),(2,2)\\}$.

  3. (3)

$T=\operatorname{\mathrm{O}}^{+}_{2m}(q)$ ($m$ even, $q$ odd) and
$M=N_{T}(\operatorname{\mathrm{O}}_{2m-1}(q))$. By [6, Corollary 2.10.4 Part
(i) and Table 2.1.D], we have
$|N_{T}(\operatorname{\mathrm{O}}_{2m-1}(q)):\operatorname{\mathrm{O}}_{2m-1}(q)|\leq
2$. It follows that
$\frac{1}{2}q^{m-1}(q^{m}-1)=|T:\operatorname{\mathrm{O}}_{2m-1}(q)|$ divides
$2^{x+1}3^{y}5^{z}$. However, since $q$ is odd, $m\geq 4$, and $0\leq y,z\leq
1$, this cannot be the case.

  4. (4)

$T=\operatorname{\mathrm{S}}_{4}(q)$ and
$M=N_{T}(\operatorname{\mathrm{S}}_{2}(q^{2}))$. Then [6, Corollary 2.10.4
Part (i) and Table 2.1.D] gives
$|N_{T}(\operatorname{\mathrm{S}}_{2}(q^{2})):\operatorname{\mathrm{S}}_{2}(q^{2})|\leq
2$. It follows that $q^{2}(q^{2}-1)=|T:\operatorname{\mathrm{S}}_{2}(q^{2})|$
divides $2^{x+1}3^{y}5^{z}$. Thus, $q\in\\{2,4\\}$. We then use direct
computation to see that the only possibility is
$\operatorname{\mathrm{S}}_{2}(q^{2})\unlhd H$.

  5. (5)

In each of the remaining cases (see [8, Table 10.7]), we are given a pair
($T$, $Y$), where $T$ is $\operatorname{\mathrm{L}}_{2}(8)$,
$\operatorname{\mathrm{L}}_{3}(3)$, $\operatorname{\mathrm{L}}_{6}(2)$,
$\operatorname{\mathrm{U}}_{3}(3)$, $\operatorname{\mathrm{U}}_{3}(3)$,
$\operatorname{\mathrm{U}}_{3}(5)$, $\operatorname{\mathrm{U}}_{4}(2)$,
$\operatorname{\mathrm{U}}_{4}(3)$, $\operatorname{\mathrm{U}}_{5}(2)$,
$\operatorname{\mathrm{U}}_{6}(2)$, $\operatorname{\mathrm{S}}_{4}(7)$,
$\operatorname{\mathrm{S}}_{4}(8)$, $\operatorname{\mathrm{S}}_{6}(2)$,
$\operatorname{\mathrm{O}}^{+}_{8}(2)$, $G_{2}(3)$, ${}^{2}F_{4}(2)^{\prime}$,
$M_{11}$, $M_{12}$, $M_{24}$, $HS$, $McL$, $Co_{2}$ or $Co_{3}$, and $Y$ is a
subgroup of $T$ containing $H$. Apart from when
$T=\operatorname{\mathrm{O}}^{+}_{8}(2)$, $M_{11}$, $M_{12}$, and $M_{24}$, we
find that $|T:Y|$ does not divide $2^{x}3^{y}5^{z}$, so we get a contradiction
in all other cases. When $T=\operatorname{\mathrm{O}}^{+}_{8}(2)$, the only
possibilities are $H=\operatorname{\mathrm{A}}_{9}$ of index $2^{6}15$ (three
$T$-conjugacy classes fused under the triality automorphism). When $T=M_{11}$,
$M_{12}$, or $M_{24}$, the only possibilities for $H$ are
$H=\operatorname{\mathrm{L}}_{2}(11)\leq M_{11}$ (of index $12$),
$H=M_{11}\leq M_{12}$ (of index $12$), or $H=M_{23}\leq M_{24}$ (of index
$24$).

Finally, assume that $T=\operatorname{\mathrm{L}}_{2}(p)$, for some Mersenne
prime $p$, and let $M$ be a maximal subgroup of $T$ containing $H$. Then,
since $|T:M|$ divides $|T:H|=2^{x}3^{y}5^{z}$ with $0\leq y,z\leq 1$, $M$ must
be parabolic (see [2, Table 8.1]). But then $\pi(|T|)\neq\pi(|M|)$, since
$p+1$ is the highest power of $2$ dividing $|T|$ – a contradiction. This
completes the proof.∎

This completes our analysis of the pairs $(T,H)$ with $T$ a nonabelian simple
group and $H$ a subgroup of $T$ of index $|T:H|=2^{x}3^{y}5^{z}$, with $0\leq
y,z\leq 1$. The possibilities are listed in Proposition 3.4(i)–(iii) and
Proposition 3.5(i)–(vi).

In particular, if $G$ is a counterexample to Theorem 3.1 of minimal degree,
$L=T^{e}$ is a minimal normal subgroup of $G$, and $H$ is a point stabilizer
in $L$, then we know the possibilities for the group $T$, and the
intersections of $H$ with the $i$th coordinate subgroups in $L$. Our next task
is to use this information to find out more about the structure of $H$.

One of our main tools for doing this is the _Frattini argument_. The proof is
well-known, but we couldn’t find a reference so we include a proof here.

######  Lemma 3.6.

Let $G$ be a finite group, and let $L$ be a normal subgroup of $G$. Let
$A\leq\mathop{\mathrm{Aut}}(L)$ be the image of the induced action of $G$ on
$L$ (by conjugation). Suppose that $H$ is a subgroup of $L$ with the property
that $H$ and $H^{\alpha}$ are $L$-conjugate for each $\alpha\in A$. Then
$G=N_{G}(H)L$.

###### Proof.

Let $g\in G$. Then $H^{g}=H^{l}$ for some $l\in L$, by hypothesis. Hence,
$gl^{-1}\in N_{G}(H)$, so $g\in N_{G}(H)L$, and this completes the proof.∎

With the Frattini argument in mind, the next lemma will be crucial.

######  Lemma 3.7.

Let $T$ be a nonabelian finite simple group, and suppose that $T$ has a
subgroup $H$ of index $n:=2^{x}3^{y}5^{z}$, with $0\leq y,z\leq 1$. Assume
also that:

  * •

If $T=\operatorname{\mathrm{A}}_{5}$, then $H$ is either transitive in its
natural action, or $H=\operatorname{\mathrm{A}}_{4}$;

  * •

if $T=\operatorname{\mathrm{A}}_{6}$, then $H$ is either transitive in its
natural action, or $H=\operatorname{\mathrm{A}}_{5}$ is a natural point
stabilizer in $T$;

  * •

if $T=\operatorname{\mathrm{L}}_{2}(q)$, with $q$ of the form
$q=2^{x_{1}}3^{y_{1}}5^{z_{1}}-1$, then $y=y_{1}$ and $z=z_{1}$;

  * •

if $T=\operatorname{\mathrm{S}}_{4}(4)$, then
$H\neq\operatorname{\mathrm{S}}_{2}(16)$ and
$H\neq\operatorname{\mathrm{S}}_{2}(16).2$;

  * •

if $T=\operatorname{\mathrm{O}}^{+}_{8}(2)$, then
$H\neq\operatorname{\mathrm{A}}_{9}$ (three classes);

  * •

if $T=\operatorname{\mathrm{A}}_{16}$, then
$H\neq\operatorname{\mathrm{A}}_{14}$ and
$H\neq\operatorname{\mathrm{A}}_{14}.2$; and

  * •

$(T,H)$ is not $(\operatorname{\mathrm{A}}_{8},H)$, with $H$ of shape $7:3$,
$2^{3}:(7:3)$, $\operatorname{\mathrm{GL}}_{3}(2)$, or
$2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$.

Denote by $\Sigma$ the set of right cosets of $H$ in $T$. Then there exists a
proper subgroup $S_{0}$ of $T$ with the following properties:

  1. (i)

$S_{0}$ and $S_{0}^{\alpha}$ are conjugate in $T$ for each
$\alpha\in\mathop{\mathrm{Aut}}(T)$;

  2. (ii)

$N_{T}(S_{0})^{\Sigma}$ is transitive;

  3. (iii)

$N_{T}(S_{0})$ acts transitively on the set of cosets of any subgroup of $T$
of $2$-power index.

###### Proof.

Note first that if $T=\operatorname{\mathrm{A}}_{n}$ and $K$ is a subgroup of
$T$ of $2$-power index, then $K=\operatorname{\mathrm{A}}_{n-1}$ and $n$ is a
power of $2$, by Proposition 3.4 and 3.5. Thus, if
$T=\operatorname{\mathrm{A}}_{n}$, with $n$ not a power of $2$, and $H$ is
transitive, then setting $S_{0}$ to be a natural point stabilizer
$S_{0}:=\operatorname{\mathrm{A}}_{n-1}$ suffices. Also, if $|T:H|$ is a power
of $2$, then setting $S_{0}$ to be a Sylow $2$-subgroup of $T$ gives us what
we need. So we may assume that we are in neither of these cases.

By Proposition 3.4 and 3.5, the remaining possibilities for the pair $(T,H)$
are as follows:

  1. (1)

$(T,H)=(\operatorname{\mathrm{A}}_{n},\operatorname{\mathrm{A}}_{n-1})$, with
$n=2^{x}3^{y}5^{z}$ and $y+z\neq 0$. Set
$S_{0}=\langle(1,\ldots,3^{y}5^{z}),(3^{y}5^{z}+1,\ldots,2\times
3^{y}5^{z}),\ldots,(n-3^{y}5^{z}+1,\ldots,n)\rangle$. Then
$N_{T}(S_{0})^{\Sigma}$ is transitive, and $S_{0}$ and $S_{0}^{\alpha}$ are
$T$-conjugate for all $\alpha\in\mathop{\mathrm{Aut}}(T)$ (this includes the
case $n=6$, when $\mathop{\mathrm{Out}}{(\operatorname{\mathrm{A}}_{6})}$ has
order $4$). Thus, (i) and (ii) are satisfied. Property (iii) is vacuously
satisfied since $T$ has no proper subgroup of $2$-power index in this case, as
mentioned at the beginning of the proof.

  2. (2)

$T=\operatorname{\mathrm{S}}_{8}(2)$, and
$\operatorname{\mathrm{O}}^{-}_{8}(2)\unlhd H$. Note that property (iii) is
vacuously satisfied in each case, since $T$ has no proper subgroup of
$2$-power index. Moreover, $T$ has a maximal parabolic subgroup $S_{0}$ of
shape $2^{10}.GL_{4}(2)$ satisfying (i) and (ii).

  3. (3)

$T=\operatorname{\mathrm{L}}_{4}(3)$,
$H=3^{3}:\operatorname{\mathrm{L}}_{3}(3)$ (two classes). In this case, take
$S_{0}$ to be the unique maximal subgroup of $T$ of shape
$\operatorname{\mathrm{L}}_{2}(9).4.2$. Then it is easily seen that (i) and
(ii) hold, for either of the two non-conjugate choices for $H$. Property (iii)
holds since $T$ has no proper subgroups of $2$-power index.

  4. (4)

$T=\operatorname{\mathrm{S}}_{4}(3)$, and $|T:H|=40$ (two classes). Here, the
unique maximal subgroup $S_{0}$ of $T$ of index $27$ has properties (i) and
(ii) for each of two non-conjugate choices for $H$. Property (iii) again holds
since $T$ has no proper subgroups of $2$-power index.

  5. (5)

$T=\operatorname{\mathrm{S}}_{6}(2)$, and $|T:H|\in\\{120,960\\}$. Taking
$S_{0}$ to be the unique maximal subgroup of $T$ of index $36$ gives us (i)
and (ii) in either of the cases $|T:H|=120$ and $|T:H|=960$. Property (iii)
holds since $T$ has no proper subgroups of $2$-power index.

  6. (6)

$(T,H)=(M_{12},M_{11})$ or $(M_{24},M_{23})$: In each case, let $S_{0}$ be a
subgroup of $T$ generated by a fixed point free element of order $3$. When
$T=M_{12}$,
$N_{T}(S_{0})\cong\operatorname{\mathrm{A}}_{4}\times\operatorname{\mathrm{S}}_{3}$
(see [14]) is a maximal subgroup of $T$, and acts transitively on the cosets
of $H$. Thus, (ii) holds. Also, outer automorphisms of $M_{12}$ fix the set of
$T$-conjugates of $S_{0}$, so $S_{0}$ and $S_{0}^{\alpha}$ are $T$-conjugate
for all $\alpha\in\mathop{\mathrm{Aut}}(T)$. Hence, (i) holds. When
$T=M_{24}$, $N_{T}(S_{0})$ has order $1008$, and acts transitively on the
cosets of $H$ (using Magma [1], for example). Also,
$\mathop{\mathrm{Out}}{(T)}$ is trivial. Thus, (i) and (ii) are again
satisfied. Property (iii) is vacuously satisfied in each case, since $T$ has
no proper subgroup of $2$-power index.

  7. (7)

$T=\operatorname{\mathrm{L}}_{2}(q)$, with $q$ of the form
$q=2^{x_{1}}3^{y_{1}}5^{z_{1}}-1$, $0\leq y_{1},z_{1}\leq 1$. Since we are
assuming that $y=y_{1}$ and $z=z_{1}$, $H$ must be a maximal parabolic
subgroup of $T$ (see [2, Table 8.1]). Let $S_{0}$ be a $D_{q+1}$ maximal
subgroup of $T$. Then $S_{0}$ and $S_{0}^{\alpha}$ are $T$-conjugate for all
$\alpha\in\mathop{\mathrm{Aut}}(T)$, so (i) holds. Clearly, (ii) also holds.
The group $T$ only has a proper subgroup of $2$-power index if
$y_{1}=z_{1}=0$, and in this case the only such subgroup is $H$. Thus,
property (iii) is also satisfied.

∎

######  Corollary 3.8.

Let $G$ be a minimal transitive group of degree $n=2^{x}3^{y}5^{z}$, where
$0\leq y,z\leq 1$. Suppose that $G$ has a nonabelian minimal normal subgroup
$L=T^{e}$, and let $H$ be the intersection of $L$ with a point stabilizer in
$G$. Write $|L:H|=2^{x_{0}}3^{y_{0}}5^{z_{0}}$, where $x_{0}\leq x$,
$y_{0}\leq y$, $z_{0}\leq z$. Then:

  1. (i)

Either $y_{0}$ or $z_{0}$ is non-zero;

  2. (ii)

Suppose that $H=H\pi_{1}\times\ldots\times H\pi_{e}$, and all but one of the
groups $H\pi_{i}$, say $H\pi_{e}$, has $2$-power index in $T$. Then
($T$,$H\pi_{e}$) cannot satisfy the hypothesis of Lemma 3.7.

###### Proof.

Suppose that either $|L:H|$ is a power of $2$; or
$H=H\pi_{1}\times\ldots\times H\pi_{e}$ where all but one of the groups
$H\pi_{i}$, say $H\pi_{e}$, has $2$-power index in $T$, and ($T$,$H\pi_{e}$)
satisfies the hypothesis of Lemma 3.7. Then we claim that there exists a
proper subgroup $S$ of $L$ with $S$ acting transitively on the cosets of $H$
in $L$, and $LN_{G}(S)=G$ ($\ast$). In particular, $N_{G}(S)$ is a proper
subgroup of $G$, since $L$ is a minimal normal subgroup of $G$. But then
$N_{G}(S)<G$ acts transitively both on an $L$-orbit, and on the set of
$L$-orbits, so $N_{G}(S)$ is transitive. This contradicts the minimal
transitivity of $G$. Thus, it will suffice to prove the existence of such a
subgroup $S$.

Now, if $|L:H|$ is a power of $2$, then $S\in\mathop{\mathrm{Syl}}_{2}(L)$
satisfies ($\ast$), by Lemma 3.6 and the fact that $(|L:H|,|L:S|)=1$. So
assume that $H=H\pi_{1}\times\ldots\times H\pi_{e}$, and all but one of the
groups $H\pi_{i}$, say $H\pi_{e}$, has $2$-power index in $T$. Assume that the
pair ($T$,$H\pi_{e}$) satisfies the hypothesis of Lemma 3.7, and let $S_{0}$
be the subgroup of $T$ exhibited therein. Set $S:=S_{0}^{e}<L$. Then $S$ acts
transitively on the cosets of $H$ in $L$. Moreover, since
$\mathop{\mathrm{Aut}}(L)\cong\mathop{\mathrm{Aut}}(T)\wr\mathrm{Sym}_{e}$,
Lemma 3.7(i) implies that $S$ and $S^{\alpha}$ are $L$-conjugate for all
$\alpha\in\mathop{\mathrm{Aut}}(L)$. Thus, $LN_{G}(S)=G$ by Lemma 3.6, so $S$
satisfies ($\ast$), and this completes the proof. ∎

In what follows, recall from Definition 2.2 that _a subdirect product of the
form_ $G=\frac{1}{n}(G_{1}\times\ldots\times G_{e})$ is a subdirect product of
index $n$ in $G_{1}\times\ldots\times G_{e}$ containing
$[G_{1},G_{1}]\times\ldots\times[G_{e},G_{e}]$.

######  Corollary 3.9.

Let $G$ be a minimal transitive group of degree $n=2^{x}3^{y}5^{z}$, where
$0\leq y,z\leq 1$. Suppose that $G$ has a nonabelian minimal normal subgroup
$L=T^{e}$, and write $\pi_{i}:L\rightarrow T$ for the $i$th coordinate
projections. Let $H$ be the intersection of $L$ with a point stabilizer in
$G$. Write $|L:H|=2^{x_{0}}3^{y_{0}}5^{z_{0}}$, where $x_{0}\leq x$,
$y_{0}\leq y$, $z_{0}\leq z$. Then one of the following holds:

  1. (1)

$T=\operatorname{\mathrm{A}}_{n}$ with $n\in\\{5,6\\}$ and one of the
following holds:

    1. (a)

$H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e-2}\times H_{e-1}\times H_{e}$,
where $H_{e-1}$ and $H_{e}$ are subgroups of $T$ with
$|T:H_{e-1}|=2^{x_{e-1}}3$, $|T:H_{e}|=2^{x_{e}}5$, and $x_{0}=x_{e-1}+x_{e}$;

    2. (b)

$H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e-1}\times H_{e}$, where $H_{e}\leq
T$ is not transitive in its natural action; $H_{e}$ is not $T$-conjugate to a
natural point stabilizer $\operatorname{\mathrm{A}}_{n-1}$; and
$|T:H_{e}|=|L:H|$;

    3. (c)

$T=\operatorname{\mathrm{A}}_{5}$ and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e-2}\times J$, where $J<T^{2}$ is a subdirect product of the form
$J=\frac{1}{2}(\operatorname{\mathrm{S}}_{3}\times D_{10})<T^{2}$; or

    4. (d)

$T=\operatorname{\mathrm{A}}_{5}$ and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e-2}\times J$, where $J$ is a diagonal subgroup of $T^{2}$.

  2. (2)

$T=\operatorname{\mathrm{A}}_{8}$ and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e_{1}}\times H_{e_{1}+1}\times\ldots\times H_{e-1}\times H_{e}$, where
$H_{e}$ is one of the subgroups of shape $7:3$, $2^{3}:(7:3)$,
$\operatorname{\mathrm{GL}}_{3}(2)$, or
$2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$ in $T$; $H_{i}$ is $T$-conjugate to
$\operatorname{\mathrm{A}}_{7}$ for all $e_{1}+1\leq i\leq e-1$; and $0\leq
e_{1}\leq e-1$.

  3. (3)

$T=\operatorname{\mathrm{A}}_{16}$ and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e_{1}}\times H_{e_{1}+1}\times\ldots\times H_{e-1}\times H_{e}$, where
$H_{e}$ is one of the subgroups of shape $\operatorname{\mathrm{A}}_{14}$ or
$\operatorname{\mathrm{A}}_{14}.2$ in $T$; $H_{i}$ is $T$-conjugate to
$\operatorname{\mathrm{A}}_{15}$ for all $e_{1}+1\leq i\leq e-1$; and $0\leq
e_{1}\leq e-1$.

  4. (4)

$T=\operatorname{\mathrm{L}}_{2}(q)$, with $q$ of the form
$q=2^{x^{\prime}}3^{y^{\prime}}5^{z^{\prime}}-1$, $0\leq
y^{\prime},z^{\prime}\leq 1$, and one of the following holds:

    1. (a)

$y^{\prime}=z^{\prime}=0$; $5\mid q-1$; $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e_{1}}\times P_{1}\times\ldots\times P_{e_{2}}\times H_{e-1}\times H_{e}$,
where each $P_{i}$ is a maximal parabolic subgroup of $T$; $H_{e-1}$
[respectively $H_{e}$] is a subgroup of index $3$ [resp. $5$] in a maximal
parabolic subgroup of $T$; and $e_{1}+e_{2}=e-2$;

    2. (b)

$y^{\prime}=z^{\prime}=0$ and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e_{1}}\times L_{1}$, where $L_{1}$ is a subdirect product of the form
$L_{1}=\frac{1}{t}(L_{2}\times H\pi_{e})$; $L_{2}$ is a subdirect product of
the form $L_{2}=\frac{1}{s}(P_{1}\times\ldots\times P_{e_{2}})$; each $P_{i}$
is a maximal parabolic subgroup of $T$; $H\pi_{e}$ is a subgroup of index $r$
in a maximal parabolic subgroup of $T$; $rst=3^{y_{0}}5^{z_{0}}$; and
$e_{1}+e_{2}=e-1$; or

    3. (c)

$y^{\prime}+z^{\prime}=1$, and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e-1}\times\left(q:\frac{q-1}{2r}\right)$, where $r>1$ divides $q-1$, and
$3^{y^{\prime}}5^{z^{\prime}}r=3^{y_{0}}5^{z_{0}}$.

  5. (5)

$T=\operatorname{\mathrm{S}}_{4}(4)$ and $H$ is $\mathrm{Sym}_{e}$-conjugate
to $T^{e-1}\times H_{e}$, where
$H_{e}\in\\{\operatorname{\mathrm{S}}_{2}(16)\text{ }\mathrm{(}\text{two
}T\text{-classes}\mathrm{)},\\\ \operatorname{\mathrm{S}}_{2}(16).2\text{
}\mathrm{(}\text{two }T\text{-classes}\mathrm{)}\\}$.

  6. (6)

$T=\operatorname{\mathrm{O}}^{+}_{8}(2)$ and $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e-1}\times H_{e}$, where
$H_{e}=\operatorname{\mathrm{A}}_{9}$ (three $T$-classes, fused under a
triality automorphism).

###### Proof.

By Corollary 3.8, either $y_{0}$ or $z_{0}$ is non-zero. Now, set
$H_{i}:=(T_{i}\cap H)\pi_{i}$, for $1\leq i\leq e$. Then $H_{i}\unlhd
H\pi_{i}$, and $|T:H_{i}|$ has the form $2^{a_{i}}3^{b_{i}}5^{c_{i}}$ for each
$i$. Note also that since $H$ is a subgroup of $H\pi_{1}\times\ldots\times
H\pi_{e}\leq L$, we have that $\prod_{i=1}^{e}|T:H\pi_{i}|$ divides $|L:H|$.
Thus, $|T:H\pi_{i}|$ has the form $2^{x_{i}}3^{y_{i}}5^{z_{i}}$ for each $i$,
and $y_{i}$ [respectively $z_{i}$] is $1$ for at most $y_{0}$ [resp. $z_{0}$]
values of $i$. It follows that $|T:H\pi_{i}|$ is a power of $2$ for all but at
most two values of $i$. Let $\Lambda:=\\{i\text{ : }|T:H\pi_{i}|\text{ is a
power of }2\\}$, so that $|\Lambda|\geq t-2$. For a subset $E$ of
$\\{1,\ldots,e\\}$, let $\pi_{E}:L\rightarrow\prod_{i\in E}T_{i}$ be the
natural projection homomorphism onto the ‘$E$-part’ of $L$. Then since $H$ is
a subgroup of $H\pi_{\Lambda}\times\prod_{i\not\in\Lambda}H\pi_{i}$ (up to
permutation of coordinates), we have that

(3.2) $|T^{|\Lambda|}:H\pi_{\Lambda}|\prod_{i\not\in\Lambda}|T:H\pi_{i}|$
divides $|L:H|$.

Furthermore, we can determine the subgroups of $2$-power index in the simple
groups $T$ from Propositions 3.4 and 3.5: we see that either

  1. (I)

$H\pi_{i}=T$ for all $i\in\Lambda$

  2. (II)

$T=\operatorname{\mathrm{A}}_{2^{x_{i}}}$ and $H\pi_{i}$ is $T$-conjugate to
either $T$ or $\operatorname{\mathrm{A}}_{2^{x_{i}}-1}$ for all $i\in\Lambda$;
or

  3. (III)

$T=\operatorname{\mathrm{L}}_{2}(p)$, with $p=2^{x_{i}}-1$, and $H\pi_{i}$ is
either $T$ or a maximal parabolic subgroup of $T$, for all $i\in\Lambda$.

Assume first that $|\Lambda|=t-2$, and without loss of generality, write
$[i]\setminus\Lambda=\\{e-1,e\\}$. Then we may assume, again without of loss
of generality, that $|T:H\pi_{e-1}|=2^{x_{e-1}}3$ and
$|T:H\pi_{e}|=2^{x_{e}}5$. It then follows from (3.2) that

(3.3) $\displaystyle|T^{e-2}:H\pi_{\Lambda}|\text{ is a power of }2.$

Next, from Propositions 3.4 and 3.5, we can determine all finite simple groups
$T$ which contain both a subgroup of index $2^{x_{e-1}}3$ and a subgroup of
index $2^{x_{e}}5$: we see that
$T\in\\{\operatorname{\mathrm{L}}_{2}(p),\operatorname{\mathrm{A}}_{5}$,
$\operatorname{\mathrm{A}}_{6}\text{ : }p\text{ a Mersenne prime, }15\mid
p-1\\}$. Suppose first that $T=\operatorname{\mathrm{A}}_{n}$, for
$n\in\\{5,6\\}$. Then since $T$ is the only subgroup of $T$ of $2$-power
index, we have $H\pi_{i}=T$ for all $i\in\Lambda$. Thus, $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e-2}\times J$, where $J$ is a subdirect
subgroup of $H\pi_{e-1}\times H\pi_{e}$, by Lemma 2.3 and (3.3). We can now
quickly determine, using Magma [1] for example, all subgroups $J$ of $T^{2}$
where $J\pi_{e-1}$, $J\pi_{e}<T$, and $|T^{2}:J|$ has the form
$2^{A}3^{B}5^{C}$. This yields cases (1)(a) and (1)(c) in the statement of the
corollary.

Suppose next that $T=\operatorname{\mathrm{L}}_{2}(p)$, where $p$ is a
Mersenne prime. Then the only subgroups of $T$ of $2$-power index are $T$
itself, together with the maximal parabolic subgroup $P<T$, of index $p+1$. It
follows from (3.3) and Lemma 2.3(v) that $H\pi_{\Lambda}=T^{e_{1}}\times
L_{2}$ (up to permutation of coordinates), where $L_{2}$ is a subdirect
product in $P_{1}\times\ldots\times P_{e_{2}}$, each $P_{i}$ is $T$-conjugate
to $P$, and $e_{1}+e_{2}=e$. Also, since $P$ has odd order, (3.3) implies that
that $L_{2}=P_{1}\times\ldots\times P_{e_{2}}$. Hence,
$H\pi_{\lambda}=T^{e_{1}}\times P_{1}\times\ldots\times P_{e_{2}}$ (up to
permutation of coordinates). Thus, $H$ is a subdirect product in
$T^{e_{1}}\times P_{1}\times\ldots\times P_{e_{2}}\times J$, where
$J:=H\pi_{\\{e-1,e\\}}\leq H{\pi_{e-1}}\times H\pi_{e}$. Now, the only
subgroup of $T$ (up to $T$-conjugacy) with index $2^{x_{e-1}}3$ is the unique
subgroup $Y_{3}$ of $P$ with $|P:Y|=3$ , $|T:Y|=3(p+1)$; while the only
subgroup of $T$ (up to $T$-conjugacy) with index $2^{x_{e-1}}5$ is the unique
subgroup $Y_{5}$ of $P$ with $|P:Y|=5$ , $|T:Y|=5(p+1)$. Thus, $J$ is a
subdirect product in $H\pi_{e-1}\times H{\pi_{e}}\cong p:\frac{p-1}{6}\times
p:\frac{p-1}{10}$. Since $|T^{2}:J|$ has the form $2^{A}3^{b}5^{C}$ and $|J|$
is odd, we must have $J=H\pi_{e-1}\times H\pi_{e-1}\cong p:\frac{p-1}{6}\times
p:\frac{p-1}{10}$. Moreover, since $|T_{e-1}\times T_{e}:H\cap(T_{e-1}\times
T_{e})|$ has the form $2^{A}3^{B}5^{C}$, we must have $J=H\pi_{e-1}\times
H\pi_{e}=H\cap(T_{e-1}\times T_{e})$. It follows that $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times P_{1}\times\ldots\times
P_{e_{2}}\times H_{e-1}\times H_{e}$, and this yields case (4)(a) in the
statement of the corollary.

Suppose next that $|\Lambda|=t-1$. Without loss of generality, we may assume
that $[i]\setminus\Lambda=\\{e\\}$. Hence, $|T:H\pi_{e}|=2^{x_{e}}r$, with
$r\in\\{3,5,15\\}$. It then follows from (3.2) that

(3.4) $|T^{e-1}:H\pi_{\Lambda}|2^{x_{e}}r$ divides $|L:H|$ for some
$r\in\\{3,5,15\\}$.

In particular, $|T|$ does not divide $|T^{e-1}:H\pi_{\Lambda}|$. Suppose that
case (I) occurs (recall that (I) is defined at the end of the first paragraph
of this proof). Then $H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e-1}\times
H_{e}$ by (3.4) and Lemma 2.3(v). Suppose now that we are in case (II). Then
$H\pi_{\Lambda}$ is a subdirect product in $T^{e_{1}}\times
H_{e_{1}+1}\times\ldots\times H_{e-1}$, where $0\leq e_{1}\leq e-1$, and each
$H_{i}$ is $T$-conjugate to $\operatorname{\mathrm{A}}_{2^{x_{i}}-1}$. The
analogous argument to the above then shows that $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times H_{e_{1}+1}\times\ldots\times
H_{e}$, where $0\leq e_{1}\leq e-1$. By Corollary, 3.8, the pair ($T$,$H_{e}$)
cannot satisfy the hypothesis of Lemma 3.7, in either of the cases (I), (II).
Since $|T:H_{e}|$ has the form $2^{A}3$, $2^{A}5$, or $2^{A}15$, Propositions
3.4 and 3.5 then imply that one of the following holds:

  * •

$T=\operatorname{\mathrm{A}}_{n}$ with $n\in\\{5,6\\}$, and $H_{e}$ is both
intransitive and not conjugate to $\operatorname{\mathrm{A}}_{4}$;

  * •

$T=\operatorname{\mathrm{A}}_{8}$, and $H_{e}$ has shape $7:3$, $2^{3}:(7:3)$,
$\operatorname{\mathrm{GL}}_{3}(2)$, or
$2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$;

  * •

$T=\operatorname{\mathrm{A}}_{16}$, and
$H_{e}\in\\{\operatorname{\mathrm{A}}_{14},\operatorname{\mathrm{A}}_{14}.2\\}$;

  * •

$T=\operatorname{\mathrm{S}}_{4}(4)$, and either
$H_{e}\cong\operatorname{\mathrm{S}}_{2}(16)$ or
$H_{e}\cong\operatorname{\mathrm{S}}_{2}(16).2$;

  * •

$T=\operatorname{\mathrm{O}}^{+}_{8}(2)$, and
$H\cong\operatorname{\mathrm{A}}_{9}$; or

  * •

$T=\operatorname{\mathrm{L}}_{2}(q)$ with
$q=2^{x^{\prime}}3^{y^{\prime}}5^{z^{\prime}}-1$; $|L:H|=r(q+1)$ for some
$r\in\\{3,5\\}$ (so either $y^{\prime}$ or $z^{\prime}$ is $0$), and
$H_{e}\cong q:\frac{q-1}{2r}$.

The first five of these are cases (1)(b), (2), (3), (5), and (6),
respectively, in the statement of the corollary. The last is case (4)(b) (with
$s=t=1$) if $y^{\prime}=z^{\prime}=0$, and case (4)(c) otherwise. Finally,
assume that case (III) holds. Then using (3.4) and Lemma 2.3 again, we see
that $H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times L_{1}$, where
$L_{1}$ is a subdirect product in $L_{2}\times H\pi_{e}$; $L_{2}$ is a
subdirect product in $P_{1}\times\ldots\times P_{e_{2}}$; each $P_{i}$ is a
maximal parabolic subgroup of $T$; and $e_{1}+e_{2}=e-1$. Moreover,
$H\pi_{\Lambda}=T^{e_{1}}\times L_{2}$ (up to permutation of coordinates). In
particular, $|H|=|T|^{e_{1}}|L_{2}||H_{e}|$. Thus,
$|L:H|=|T^{e_{2}}:L_{2}||T:H\pi_{e}||H\pi_{e}:H_{e}|$. Let $s$ and $t$ be the
odd parts of $|T^{e_{2}}:L_{2}|$ and $|H\pi_{e}:H_{e}|$, respectively. Then
$s$ is the index of $L_{2}$ in $P_{1}\times\ldots\times P_{e_{2}}$,
$rst=3^{y_{0}}5^{z_{0}}$, and one of the following must hold:

  * •

$s=t=1$ and $H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times
L_{2}\times H\pi_{e}$, $L_{2}=P_{1}\times\ldots\times P_{e_{2}}$.

  * •

$s\in\\{3,5\\}$, $t=1$, and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e_{1}}\times L_{2}\times H\pi_{e}$, where $L_{2}$ has the form
$L_{2}=\frac{1}{s}(P_{1}\times\ldots\times P_{e_{2}})$.

  * •

$t\in\\{3,5\\}$, $s=1$, and $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e_{1}}\times L_{1}$, where $L_{1}$ has the form
$L_{1}=\frac{1}{t}(P_{1}\times\ldots\times P_{e_{2}}\times p:\frac{p-1}{2r})$.

These are the cases in (4)(b) in the statement of the corollary.

Finally, assume that $|\Lambda|=t$. Suppose first that case (I) holds (see the
first paragraph above), so that $H$ is a subdirect product in $L=T^{e}$. Then
$|T|$ divides $|L:H|$ by Lemma 2.3(iii), since $H\neq L$. Thus, we must have
$T=\operatorname{\mathrm{A}}_{5}$, since $|L:H|$ divides $2^{x_{0}}15$.
Moreover, $H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e-2}\times J$, where $J$
is a diagonal subgroup of $T^{2}$, since $|T|^{2}$ cannot divide $|L:H|$. This
is case (1)(d) in the statement of the corollary. Suppose next that case (II)
holds. Then since neither $|\operatorname{\mathrm{A}}_{2^{x_{i}}}|$ nor
$|\operatorname{\mathrm{A}}_{2^{x_{i}}-1}|$ can divide $|L:H|$ in this, case
another application of Lemma 2.3 yields that $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times Y_{1}\times\ldots\times
Y_{e_{2}}$, where $Y_{i}$ is $T$-conjugate to
$\operatorname{\mathrm{A}}_{2^{x_{i}}-1}$ for each $i$, and $e=e_{1}+e_{2}$.
But then $|L:H|$ is a power of $2$ – a contradiction. Suppose next that case
(III) holds. Then arguing as in the $|\Lambda|=t-1$ case above we get $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times L_{2}$, where $L_{2}$ is a
subdirect product of the form
$L_{2}=\frac{1}{3^{y_{0}}5^{z_{0}}}(P_{1}\times\ldots\times P_{e_{2}})$, with
$P_{i}$ a maximal parabolic subgroup of $T$ for all $1\leq i\leq e_{2}$, and
$e_{1}+e_{2}=e$. If (III) holds with $3^{y_{0}}5^{z_{0}}=1$, then $|L:H|$ is a
power of $2$ – a contradiction. Thus, if case (III) holds then
$3^{y_{0}}5^{z_{0}}\neq 1$. This is case (4)(b) (with $r=1$) in the statement
of the corollary. The proof is complete. ∎

Corollary 3.9 gives us information about the possible minimal normal subgroups
$L$ (and their actions) in a minimal counterexample to Theorem 3.1. We would
now like to analyse the orbit lengths of certain soluble subgroups of $L$.

Since a point stabilizer $H$ in $L$ in this case is a direct product of
subgroups $X$ which are themselves subdirect products of the form
$X=\frac{1}{n}(X_{1}\times\ldots\times X_{f})$ (see Definition 2.2), our
analysis will have two parts: First, we will show how to determine the orbit
lengths of certain subgroups $F<L$ acting on the cosets of a subdirect product
$H$ of the form $H=\frac{1}{n}(P_{1}\times\ldots\times P_{e})$, where each
$P_{i}$ is a subgroup of $T$. Then, we will show how to determine the orbit
lengths of certain subgroups $F<L$ acting on the cosets of a subgroup $H<L$ of
the form $H=H_{1}\times\ldots\times H_{r}$, where $H_{i}\leq T^{e_{i}}$, and
$\sum_{i=1}^{r}e_{i}=e$ (i.e. when $L$ has ‘product action’ type). These two
bits of information, together with Corollary 3.9, will then allow us to prove
Theorem 3.1(iv). We begin with the first part:

######  Lemma 3.10.

Let $L=G_{1}\times\ldots\times G_{e}$, where each $G_{i}$ is a finite group,
and fix proper subgroups $P_{i}$, $S_{i}$ of $G_{i}$. Suppose that $H\leq
P_{1}\times\ldots\times P_{e}$ is a subdirect product of the form
$H=\frac{1}{n}(P_{1}\times\ldots\times P_{e})$, and that $H$ is core-free in
$L$. Assume that $S:=S_{1}\times\ldots\times S_{e}$ has orbit lengths $m_{1}$,
$\ldots$, $m_{r}$ in its action on the cosets of $P:=P_{1}\times\ldots\times
P_{e}$ in $L$, where $\sum_{i}m_{i}=|L:P|$. Assume also that $(S_{i}\cap
P_{i}^{x_{i}})[P_{i}^{x_{i}},P_{i}^{x_{i}}]=P_{i}^{x_{i}}$ for all $x_{i}\in
G_{i}$. Then $S$ has $r$ orbits in its action on the cosets of $H$ in $L$,
with lengths $nm_{1},\ldots nm_{r}$.

###### Proof.

Note first that since $H$ is core-free in $L$, $L$ may be viewed as a
transitive permutation group acting on the set $\Omega$ of cosets of $H$ in
$L$. Moreover, $H<P$, so the set of cosets of $P$ in $L$ is permutation
isomorphic to a set $\Sigma$ of blocks for $P$ in $\Omega$. Note that the
blocks in $\Sigma$ have size $n$, $H\unlhd P$, and $P/H$ is abelian of order
$n$. It follows that $L$ is isomorphic to a subgroup of the wreath product
$n\wr L^{\Sigma}$.

Now, write $\Sigma_{i}$, $1\leq i\leq r$, for the $S$-orbits in $\Sigma$,
where $|\Sigma_{i}|=m_{i}$. Also, fix a block $\Delta_{i}\in\Sigma_{i}$, for
each $i$. If $\mathop{\mathrm{Stab}}_{S}(\Delta_{i})^{\Delta_{i}}$ is
transitive for each $i$, then it follows that the
$S^{\operatorname{\mathrm{O}}}$-orbits have sizes
$|\Delta||\Sigma_{i}|=nm_{i}$, for each $i$. Thus, we just need to prove that

(3.5) ${\mathop{\mathrm{Stab}}}_{S}(\Delta_{i})^{\Delta_{i}}$ is transitive
for each $i$.

To this end, fix $1\leq i\leq e$. Then
$\mathop{\mathrm{Stab}}_{S}(\Delta_{i})^{\Delta_{i}}=S\cap P^{\alpha_{i}}$ for
some $\alpha_{i}\in L$. To show that
$\mathop{\mathrm{Stab}}_{S}(\Delta_{i})^{\Delta_{i}}$ is transitive, we just
need to show that $H^{\alpha_{i}}(S\cap P^{\alpha_{i}})=P^{\alpha_{i}}$. But
$\alpha_{i}=(x_{1},\ldots,x_{e})$, for some $x_{j}\in G_{j}$. Thus, $S\cap
P^{\alpha_{i}}=(S_{1}\cap P_{1}^{x_{1}})\times\ldots\times(S_{e}\cap
P_{e}^{x_{e}})$. Since $H^{\alpha_{i}}$ contains
$[P_{1}^{x_{1}},P_{1}^{x_{1}}]\times\ldots\times[P_{e}^{x_{e}},P_{e}^{x_{e}}]$,
we have $H^{\alpha_{i}}(S\cap P^{\alpha_{i}})=P^{\alpha_{i}}$ by hypothesis.
This proves (3.5), whence the lemma. ∎

Next, we prove a general lemma concerning the orbits of subgroups in a
transitive permutation group with ‘product action’.

######  Lemma 3.11.

Let $L=G_{1}\times\ldots\times G_{e}$, where each $G_{i}$ is a finite group,
and fix proper subgroups $H_{i}$, $S_{i}$ of $G_{i}$. Set
$S=S_{1}\times\ldots\times S_{e}$ and $H=H_{1}\times\ldots\times H_{e}$. Then:

  1. (i)

Suppose that $S_{i}$ acts transitively on the cosets of $H_{i}$ in $G_{i}$ for
$1\leq i\leq e-1$, and that $S_{e}$ has $r$ orbits, of lengths $l_{1}$,
$\ldots$, $l_{r}$ say, in its action on the cosets of $H_{e}$ in $G_{e}$. Then
$S$ has $r$ orbits in its action on the cosets of $H$ in $L$, of lengths
$l_{1}m,\ldots,l_{r}m$, where $m:=\prod_{i=1}^{e-1}|G_{i};H_{i}|$.

  2. (ii)

Assume that there exists $0\leq e_{1}\leq e-1$ such that $S_{i}$ acts
transitively on the cosets of $H_{i}$ in $G_{i}$ for $i\leq e_{1}$; that
$S_{i}$ has $2$ orbits of lengths $k_{1}$ and $k_{2}$ in each of the cosets
spaces $H_{i}\backslash G_{i}$, for $e_{1}+1\leq i\leq e-1$ (in particular,
$|G_{i}:H_{i}|=|G_{j}:H_{j}|=k_{1}+k_{2}$ for all $e_{1}+1\leq i,j\leq e-1$ in
this case); and that $S_{e}$ has $r$ orbits, of lengths $l_{1}$, $\ldots$,
$l_{r}$ say, in its action on the cosets of $H_{e}$ in $G_{e}$. Then $S$ has
$2^{e-e_{1}-1}\times r$ orbits in its action on the cosets of $H$ in $G$, with
$\binom{e-e_{1}-1}{i}$ orbits of length $l_{j}k_{1}^{i}k_{2}^{e-e_{1}-i}$ for
each $0\leq i\leq e-e_{1}-1$, and each $1\leq j\leq r$.

###### Proof.

The proof here is routine: if $S_{i}$ is a group acting on a set $\Omega_{i}$,
then the orbits of $S_{1}\times\ldots\times S_{e}$ in its product action on
$\Omega_{1}\times\ldots\times\Omega_{e}$ are precisely the sets
$\Delta_{i_{1},1}\times\ldots\times\Delta_{i_{f},f}$, where the
$\Delta_{1,j},\ldots,\Delta_{k_{j},j}$ are the orbits of $S_{j}^{\Omega_{j}}$,
for $1\leq j\leq e$. ∎

We will also need the following corollary, which is notationally heavy, but
routine.

######  Corollary 3.12.

Let $L$, $H$, and $S$ be as in Lemma 3.11(i), and assume that $G_{i}=T$ for
$1\leq i\leq e-1$, and $G_{e}=T^{k}$ where $T$ is a fixed nonabelian finite
simple group. Assume also that $S_{i}=S_{0}$ for all $1\leq i\leq e-1$, and
$S_{e}=S_{0}^{k}$ where $S_{0}$ is a fixed subgroup of $T$. Write
$n_{1},\ldots,n_{u}$ for the distinct orbit lengths of $S_{e}$ acting on the
cosets of $H_{e}$ in $T^{k}$, and suppose that $S_{e}$ has $f(n_{j})$ orbits
of length $n_{j}$. Let $G\leq\mathrm{Sym}(\Omega)$ be a finite transitive
permutation group in which $L$ is a minimal normal subgroup. Suppose that $H$
is the intersection of $L$ with a point stabilizer in $G$, and let $K$ be the
kernel of the action of $G$ on the set of $L$-orbits. Let $R/K$ be a soluble
subgroup of $G/K$ with orbits of lengths $a_{1}$, $\ldots$, $a_{t}$ on the set
of $L$-orbits. Assume also that:

  1. (i)

$S_{0}$ is $T$-conjugate to $S_{0}^{\alpha}$ for all automorphisms $\alpha$ of
$T$; and

  2. (ii)

$N_{L}(S)$ is soluble (equivalently, $N_{T}(S_{0})$ is soluble).

Then $F$ has orbit lengths $a_{i}X^{(i,j)}_{k}n_{j}m$ for some positive
partitions $f(n_{j})=\sum_{k=1}^{r_{i,j}}X^{(i,j)}_{k}$ of $f(n_{j})$, where
$1\leq j\leq u$, $1\leq i\leq t$.

###### Proof.

Let $\omega\in\Omega$ such that $H=L\cap\mathop{\mathrm{Stab}}_{G}(\omega)$,
and let $\Delta$ be the $L$-orbit containing the point $\omega$. Let
$\Sigma:=\\{\Delta^{g}\text{ : }g\in G\\}$ be the set of $L$-orbits. Fix a
subgroup $B$ of $G$, and assume that $B^{\Sigma}$ has orbits
$\Sigma_{1},\ldots,\Sigma_{r}$, of lengths $b_{1}$, $\ldots$, $b_{r}$,
respectively. Let $\Delta_{1},\ldots,\Delta_{r}\in\Sigma$ be representatives
of these orbits. Then $B$ acts on each of the sets $\bigcup_{x\in
B}\Delta_{i}^{x}$. Suppose that
$\mathop{\mathrm{Stab}}_{B}(\Delta_{i})^{\Delta_{i}}$ has orbits lengths
$x_{i,1},\ldots,x_{i,k_{i}}$. Then $B^{\Omega}$ has orbit lengths
$b_{i}x_{i,1},\ldots,b_{i}x_{i,k_{i}}$.

Now, as in the proof of Corollary 3.8, $S_{0}$ being $T$-conjugate to
$S_{0}^{\alpha}$ for all automorphisms $\alpha$ of $T$ implies that
$S=S_{0}^{e}$ is $L$-conjugate to $S^{\alpha}$ for all automorphisms
$\alpha\in\mathop{\mathrm{Aut}}(T)\wr\operatorname{\mathrm{S}}_{e}\cong\mathop{\mathrm{Aut}}(L)$.
Hence, $LN_{G}(S)=G$, by Lemma 3.6. Thus, we may choose a subgroup $F$ of
$N_{G}(S)$ containing $N_{L}(S)$ with $LF/L=R/L$. It follows that $F^{\Sigma}$
has orbits $\Sigma_{1}$, $\ldots$, $\Sigma_{t}$ of lengths
$a_{1},\ldots,a_{t}$, respectively. Fix $\Delta_{i}\in\Sigma_{i}$. Then since
$F$ normalizes $S$ and $S$ acts trivially on $\Sigma$, each
$\mathop{\mathrm{Stab}}_{F}(\Delta_{i})^{\Delta_{i}}$-orbit is a union of
$S$-orbits of the same length. Thus, by Lemma 3.11 the orbit lengths of
$\mathop{\mathrm{Stab}}_{F}(\Delta_{i})^{\Delta_{i}}$ are
$X^{(i,j)}_{1}n_{j}m$, $\ldots$, $X^{(i,j)}_{r_{i}}n_{j}m$, where
$\sum_{k=1}^{r_{i}}X^{(i,j)}=f(n_{j})$, for each $i$, $j$. Using the first
paragraph above, we deduce that $F^{\operatorname{\mathrm{O}}}$ has orbits of
lengths $a_{i}X^{(i,j)}_{k}n_{j}m$, for $1\leq k\leq r_{i}$, $1\leq j\leq u$,
$1\leq i\leq t$.

Finally, since both $FL/L=R/L$ and $F\cap L=N_{L}(S)$ are soluble, the group
$F$ is soluble. This completes the proof. ∎

We are now ready to exhibit orbit lengths of particular soluble subgroups of
minimal counterexamples to Theorem 3.1.

######  Lemma 3.13.

Let $G$ be a minimal counterexample to Theorem 3.1, and let $L=T^{e}$ be a
minimal normal subgroup of $G$, where $T=\operatorname{\mathrm{A}}_{5}$. Let
${H}$ be the intersection of $L$ with a point stabilizer for $G$, and let $a$
be the number of $L$-orbits. Then $G$ has a soluble subgroup $F$ such that one
of the following holds:

  1. (i)

If $H$ is as in case (1)(a) of Corollary 3.9, then $F$ has orbit lengths
$2^{b_{1}}5aX_{i}$, $2^{b_{2}}50aY_{j}$, where

$(\sum_{i}X_{i},\sum_{j}Y_{j},b_{1},b_{2})\in\\{(4,2,0,0),(4,2,1,1),(2,2,1,0),(2,2,0,-1),(2,1,0,0),(1,1,0,-1)\\}.$

  2. (ii)

If $H$ is as in case (1)(b) of Corollary 3.9, and $G/L$ is soluble, then $F$
either has orbit lengths $2^{b}5aX_{i}$, where

$(\sum_{i}X_{i},b)\in\\{(2,1),(2,0)\\},$

or $5aX_{i}$, $10aY_{j}$, where $(\sum_{i}X_{i},\sum_{j}Y_{j})=(2,2)$.

  3. (iii)

If $H$ is as in case (1)(c) of Corollary 3.9, then $F$ has orbit lengths
$5aX_{i}$, $25aY_{i}$, where

$(\sum_{i}X_{i},\sum_{j}Y_{j})=(4,4).$

  4. (iv)

If $H$ is as in case (1)(d) of Corollary 3.9, then $F$ has precisely two
orbits, of lengths $10a$ and $50a$.

###### Proof.

Set $S_{0}:=D_{10}$, $S:=S_{0}^{e}$, and $F:=N_{G}(S)$. We note first that
$N_{T}(S_{0})$ is soluble, and $D_{10}$ is
$\operatorname{\mathrm{A}}_{5}$-conjugate to $S_{0}^{\alpha}$ for all
$\alpha\in\operatorname{\mathrm{S}}_{5}$. Thus, (i) and (ii) in Corollary 3.12
hold. We will also show that $G$ modulo the kernel $K$ of the action of $G$ on
the set of $L$-orbits is soluble in each case, so we can take $R=G$ in
Corollary 3.12. Since either $H$ is $\mathrm{Sym}_{e}$-conjugate to
$T^{e-2}\times J$ with $J\leq T^{2}$, or $H$ is $\mathrm{Sym}_{e}$-conjugate
to $T^{e-1}\times J$ with $J\leq T$, we therefore just need to compute the
orbit lengths for $S_{0}^{2}$ [respectively $S_{0}$] in the case $e=2$ [resp.
$e=1$] and then apply Lemma 3.11 and Corollary 3.12.

Suppose first that we are in case (i). Then either
$|T_{i}:H\pi_{e-1}|=2^{x_{1}}3$, and $|T_{i}:H\pi_{e}|=2^{x_{2}}5$, or vice
versa, where $x_{1}+x_{2}=x$. Thus, $G/K$, being a minimal transitive group of
$2$-power degree, is soluble (so we can indeed take $G=R$ in Corollary 3.12).
Moreover, $H_{e-1}\in\\{C_{5},D_{10}\\}$ and
$H_{e}\in\\{\operatorname{\mathrm{A}}_{3},\operatorname{\mathrm{S}}_{3},\operatorname{\mathrm{A}}_{4}\\}$.
The orbit lengths of $S_{0}$ acting on the cosets of $H_{e-1}$ and $H_{e}$ are
then given in Table 2.

$J$ | Orbit lengths for $S_{0}=D_{10}<\operatorname{\mathrm{A}}_{5}$ acting on the cosets of $J<\operatorname{\mathrm{A}}_{5}$
---|---
$\operatorname{\mathrm{A}}_{3}$ | two of length $10$
$\operatorname{\mathrm{S}}_{3}$ | two of length $5$
$\operatorname{\mathrm{A}}_{4}$ | one of length $5$
$C_{5}$ | two of length $1$, one of length $10$
$D_{10}$ | one of length $1$, one of length $5$
Table 2. Orbit lengths for $S=D_{10}<\operatorname{\mathrm{A}}_{5}$ acting on
the cosets of a subgroup $J<\operatorname{\mathrm{A}}_{5}$

If $S_{0}$ has orbit lengths $c_{1},\ldots,c_{k}$ on the cosets of $H_{e-1}$,
and $d_{1},\ldots,d_{l}$ on the cosets of $H_{e}$, then $S_{0}^{2}$ has orbit
lengths $c_{i}d_{j}$, $1\leq i\leq k$, $1\leq j\leq l$, on the cosets of
$H_{e-1}\times H_{e}$. Thus, we can use the table above and Lemma 3.11 and
Corollary 3.12 to deduce the orbit lengths of the soluble group $F=N_{G}(S)$.
For example, if $H_{e-1}=\operatorname{\mathrm{A}}_{3}$ and $H_{e}=C_{5}$,
then $S_{0}^{2}$ has four orbits of length $10$, and two orbits of length
$100$ in its action on the cosets of $H_{e-1}\times H_{e}$ in $T^{2}$. By
Lemma 3.11(i), $S_{0}^{e}$ then has orbits of the same lengths in its action
on the cosets of $H$ in $L$ ($m=1$ in this case). Thus, by Corollary 3.12, $F$
has orbit lengths $10aX_{i}$, $100aY_{j}$, where $X_{i},Y_{j}\in\mathbb{N}$,
and $\sum_{i}X_{i}=4$, $\sum_{j}Y_{j}=2$.

We now move on to cases (ii), (iii) and (iv). In case (ii), $K\geq L$, so
$G/K$ is soluble by hypothesis. In cases (iii) and (iv), $|L:H|=2^{x}15$, so
$G/K$ is a minimal transitive group of $2$-power degree, whence soluble. Thus,
we can take again $G=R$ (and hence $t=1$, $a=a_{1}$) in Corollary 3.12.

For part (ii), the computations of the possible orbit lengths is completely
analogous to the computations in case (i) above (except that the cases
$H_{e}\in\\{\operatorname{\mathrm{A}}_{4},C_{5},D_{10}\\}$ cannot occur, and
we also need to compute the orbit lengths of the subgroups of
$\operatorname{\mathrm{A}}_{5}$ of orders $1$, $2$, and $4$). For part (iii),
one can use Magma [1] for example, to see that $S_{0}^{2}$ has four orbits of
length $5$ and four orbits of length $25$, in its action on the cosets of
$J=\frac{1}{2}(\operatorname{\mathrm{S}}_{3}\times D_{10})$ in $T^{2}$. For
part (iv), $S_{0}^{2}$ has one orbit of length $10$ and one orbit of length
$50$ in its action on the cosets of any diagonal subgroup in $T^{2}$. We now
apply Corollary 3.12 in each case to complete the proof. ∎

######  Lemma 3.14.

Let $G$ be a minimal counterexample to Theorem 3.1, and let $L=T^{e}$ be a
minimal normal subgroup of $G$, where $T=\operatorname{\mathrm{A}}_{6}$. Let
${H}$ be the intersection of $L$ with a point stabilizer for $G$, and let $a$
be the number of $L$-orbits. Then $G$ has a soluble subgroup $F$ such that one
of the following holds:

  1. (i)

If $H$ is as in case (1)(a) of Corollary 3.9, then $F$ has orbit lengths
$2^{b}5aX_{i}$, $2^{b}50aY_{j}$, where

$(\sum_{i}X_{i},\sum_{j}Y_{j},b)\in\\{(4,4,1),(4,4,0),(2,2,0)\\}.$

  2. (ii)

If $H$ is as in case (1)(b) of Corollary 3.9, and $G/L$ is soluble, then $F$
either has orbit lengths $2^{b}5aX_{i}$, where

$(\sum_{i}X_{i},b)\in\\{(4,1),(12,1),(4,0),(3,0),(2,0)\\},$

or $5aX_{i}$, $10aY_{j}$, where
$(\sum_{i}X_{i},\sum_{j}Y_{j})\in\\{(4,4),(2,2)\\}$.

###### Proof.

Set
$S_{0}:=D_{10}<\operatorname{\mathrm{A}}_{5}<\operatorname{\mathrm{A}}_{6}$,
$S:=S_{0}^{e}$, and $F:=N_{G}(S)$. Then $S_{0}$ has soluble normalizer in
$\operatorname{\mathrm{A}}_{6}$, and $D_{10}$ is
$\operatorname{\mathrm{A}}_{6}$-conjugate to $S_{0}^{\alpha}$ for all
$\alpha\in\mathop{\mathrm{Aut}}(\operatorname{\mathrm{A}}_{6})$. Thus, (i) and
(ii) in Corollary 3.12 hold. Note also that, as in the proof of Lemma 3.13,
$G$ modulo the kernel $K$ of the action of $G$ on the set of $L$-orbits is
soluble in each of (i) and (ii). Thus, we can take $R=G$ in Corollary 3.12
(and hence $t=1$, $a=a_{1}$). Since either $H$ is $\mathrm{Sym}_{e}$-conjugate
to $T^{e-2}\times J$ with $J=H_{e-1}\times H_{e}\leq T^{2}$, or $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e-1}\times J$ with $J=H_{e}\leq T$, we
therefore just need to compute the orbit lengths for $S_{0}^{2}$ [respectively
$S_{0}$] in the case $e=2$ [resp. $e=1$] and then apply Lemma 3.11 and
Corollary 3.12.

Suppose first that we are in case (i). Then, without loss of generality, we
may assume that $|T_{i}:H\pi_{e-1}|=2^{x_{1}}3$ and
$|T_{i}:H\pi_{e}|=2^{x_{2}}5$, where $x_{1}+x_{2}=x$. Hence, $H_{e-1}$ is in
one of the two conjugacy classes of $\operatorname{\mathrm{A}}_{5}$ in
$\operatorname{\mathrm{A}}_{6}$, while
$H_{e}\in\\{\operatorname{\mathrm{A}}_{3}\times\operatorname{\mathrm{A}}_{3},(\operatorname{\mathrm{A}}_{3}\times\operatorname{\mathrm{A}}_{3}).2,(\operatorname{\mathrm{A}}_{3}\times\operatorname{\mathrm{A}}_{3}).2.2\\}$.
The orbit lengths of $S_{0}$ acting on the cosets of $H_{e-1}$ and $H_{e}$ are
then given in Table 3.

$J$ | Orbit lengths for $S=D_{10}<\operatorname{\mathrm{A}}_{6}$ acting on the cosets of $J<\operatorname{\mathrm{A}}_{6}$
---|---
$\operatorname{\mathrm{A}}_{3}\times\operatorname{\mathrm{A}}_{3}$ | four of length $10$
$(\operatorname{\mathrm{A}}_{3}\times\operatorname{\mathrm{A}}_{3}).2$ | four of length $5$
$(\operatorname{\mathrm{A}}_{3}\times\operatorname{\mathrm{A}}_{3}).2.2$ | two of length $5$
$\operatorname{\mathrm{A}}_{5}$ (transitive) | one of length $1$, one of length $5$
$\operatorname{\mathrm{A}}_{5}$ (intransitive) | one of length $1$, one of length $5$
Table 3. Orbit lengths for $S=D_{10}<\operatorname{\mathrm{A}}_{6}$ acting on
the cosets of a subgroup $J<\operatorname{\mathrm{A}}_{6}$

The computations of the possible orbit lengths are then completely analogous
to the computations in the proof of Lemma 3.13.

If we are in case (ii), then $H_{e}$ is either one of the groups $J$ in Table
3, with $J$ intransitive, or $|H_{e}|\in\\{3,6,12,24\\}$. We can quickly
compute the orbit lengths of $S_{0}$ in these latter cases, and part (ii)
follows as in the proof of Lemma 3.13. ∎

######  Lemma 3.15.

Let $G$ be a minimal counterexample to Theorem 3.1, and let $L=T^{e}$ be a
minimal normal subgroup of $G$, where
$T\in\\{\operatorname{\mathrm{S}}_{4}(4),\operatorname{\mathrm{O}}^{+}_{8}(2)\\}$.
Let ${H}$ be the intersection of $L$ with a point stabilizer for $G$, and let
$a$ be the number of $L$-orbits. Then $G$ has a soluble subgroup $F$ such that
either $n=120a$ and $F$ has two orbits of lengths $24a$ and $96a$; or $n=960a$
and $F$ has two orbits of length $192a$ and $768a$.

###### Proof.

In these cases, $H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e-1}\times H_{e}$,
where $(T,H_{e})$ is one of the pairs
$(\operatorname{\mathrm{S}}_{4}(4),\operatorname{\mathrm{S}}_{2}(16))\text{
(}2\text{classes),
}(\operatorname{\mathrm{O}}^{+}_{8}(2),\operatorname{\mathrm{A}}_{9})\text{
(}3\text{ classes), or
}(\operatorname{\mathrm{O}}^{+}_{8}(2),\operatorname{\mathrm{O}}_{7}(2))\text{
(}3\text{ classes)}$. Thus, $|T:H|$ is $120$, $960$, or $120$, respectively.
If $T=\operatorname{\mathrm{S}}_{4}(4)$, then let $S_{0}$ be the maximal
parabolic subgroup of $T$ of index $425$. If
$T=\operatorname{\mathrm{O}}^{+}_{8}(2)$, then let $S_{0}$ be the maximal
parabolic subgroup of $T$ of index $1575$. Then in each case, $N_{T}(S_{0})$
is soluble, and $S_{0}$ and $S_{0}^{\alpha}$ are $T$-conjugate for all
$\alpha$ in $\mathop{\mathrm{Aut}}(T)$. That is, (i) and (ii) in Corollary
3.12 hold. Moreover, $S_{0}$ has two orbits in its action on the cosets of
$H_{e}$ in $T$, of lengths $24$ and $96$ when $|T:H|=120$, and lengths $192$
and $768$ when $|T:H|=960$.

Let $\Sigma$ be the set of $L$-orbits. Then since $15$ divides $|L:H|$ in each
case, $G^{\Sigma}$ is a $2$-group. Thus, as in the proof of Lemmas 3.13 and
3.14, we see that either $n=120a$ and $F$ has two orbits of lengths $24a$ and
$96a$, or $n=960a$ and $F$ has two orbits of length $192a$ and $768a$. ∎

In the next three lemmas we deal with the cases
$T\in\\{\operatorname{\mathrm{A}}_{8},\operatorname{\mathrm{A}}_{16},\operatorname{\mathrm{L}}_{2}(q)\text{
: }q=2^{x^{\prime}}3^{y^{\prime}}5^{z^{\prime}}-1\\}$. Apart from the case
$q=2^{x^{\prime}}3^{y^{\prime}}5^{z^{\prime}}-1$ with
$\\{y^{\prime},z^{\prime}\\}=\\{1,0\\}$, we will find a soluble subgroup $S$
of $L$ with convenient orbit lengths, and then simply set $F:=S$ (we do not
need to trouble to find a soluble subgroup with fewer orbits, as we did in the
$T\in\\{\operatorname{\mathrm{A}}_{5},\operatorname{\mathrm{A}}_{6},\operatorname{\mathrm{S}}_{4}(4),\operatorname{\mathrm{O}}^{+}_{8}(2)\\}$
cases).

######  Lemma 3.16.

Let $G$ be a minimal counterexample to Theorem 3.1, and let $L=T^{e}$ be a
minimal normal subgroup of $G$, where $T=\operatorname{\mathrm{A}}_{8}$. Let
${H}$ be the intersection of $L$ with a point stabilizer for $G$, so that
${H}$ is of type (2) in Corollary 3.9. That is, $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times Y_{1}\times\ldots\times
Y_{e_{2}}\times H_{e}$, where $e_{1}+e_{2}=e-1$; $Y_{i}$ is $T$-conjugate to
$\operatorname{\mathrm{A}}_{7}$ for each $i$; and $H_{e}$ has shape $7:3$,
$2^{3}:(7:3)$, $\operatorname{\mathrm{GL}}_{3}(2)$, or
$2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$. Let $a$ be the number of
$L$-orbits, let $S_{0}:=7:3<T$, and set $F:=S_{0}^{e}<L$. Then

  1. (i)

Suppose that $H_{e}=7:3$, and define $(s_{1},l_{1}):=(1,1)$;
$(s_{2},l_{2}):=(11,7)$; and $(s_{3},l_{3}):=(42,21)$. Then $F$ has
$as_{j}\binom{e_{2}}{i}$ orbits of size $7^{i}l_{j}$, for each $1\leq i\leq
e_{2}$, $1\leq j\leq 3$.

  2. (ii)

Suppose that $H_{e}\cong 2^{3}:(7:3)$ or
$K\cong\operatorname{\mathrm{GL}}_{3}(2)$, and define $(s_{1},l_{1}):=(1,1)$;
$(s_{2},l_{2}):=(5,7)$; and $(s_{3},l_{3}):=(4,21)$. Then $F$ has
$as_{j}\binom{e_{2}}{i}$ orbits of size $7^{i}l_{j}$, for each $1\leq i\leq
e_{2}$, $1\leq j\leq 3$.

  3. (iii)

Suppose that $H_{e}=2^{3}:\operatorname{\mathrm{GL}}_{3}(2)$, and define
$(s_{1},l_{1}):=(1,1)$; and $(s_{2},l_{2}):=(2,7)$. Then $F$ has
$as_{j}\binom{e_{2}}{i}$ orbits of size $7^{i}l_{j}$, for each $1\leq i\leq
e_{2}$, $1\leq j\leq 2$.

###### Proof.

Note that $S_{0}$ has orbits of lengths $7$ and $1$ in its action on the
cosets of $\operatorname{\mathrm{A}}_{7}$ in $T$, while $S_{0}$ has $s_{i}$
orbits of length $l_{i}$ in its action on the cosets of $H_{1}$ in $T$, where
$s_{i}$ and $l_{i}$ are as defined in each of the listed cases. We then apply
Lemma 3.11(ii) to find the orbit lengths of $F$ in its action on the cosets of
$H$ in $L$: we see that $F$ has $s_{j}\binom{e_{2}}{i}$ orbits of size
$7^{i}l_{j}$ in each case.

Finally, since this holds for each $G$-conjugate of $H$ in $L$, and $F$ fixes
each $L$-orbit, we deduce that $F<G$ has $as_{j}\binom{e_{2}}{i}$ orbits of
size $7^{i}l_{j}$ in $[n]$, as required. ∎

######  Lemma 3.17.

Let $G$ be a minimal counterexample to Theorem 3.1, and let $L=T^{e}$ be a
minimal normal subgroup of $G$, where $T=\operatorname{\mathrm{A}}_{16}$. Let
${H}$ be the intersection of $L$ with a point stabilizer for $G$, so that
${H}$ is of type (3) in Corollary 3.9. That is, $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times Y_{1}\times\ldots\times
Y_{e_{2}}\times H_{e}$, where $e_{1}+e_{2}=e-1$; $Y_{i}$ is $T$-conjugate to
$\operatorname{\mathrm{A}}_{15}$ for each $i$; and $H_{e}$ has shape
$\operatorname{\mathrm{A}}_{14}.2^{b}$, for some $b\in\\{0,1\\}$. Let $a$ be
the number of $L$-orbits, let $S_{0}:=C_{15}<T$, and set $F:=S_{0}^{e}<L$.
Then $F$ has $\frac{16}{2^{b}}\binom{e_{2}}{i}a$ orbits of size $15^{i+1}$,
for each $0\leq i\leq e_{2}$.

###### Proof.

The proof is identical to the $\operatorname{\mathrm{A}}_{8}$ case above:
$S_{0}$ has orbits of lengths $15$ and $1$ in its action on the cosets of
$\operatorname{\mathrm{A}}_{15}$ in $T$, while $S_{0}$ has $\frac{16}{2^{b}}$
orbits of length $15$ in its action on the cosets of $H_{e}$ in $T$ We then
apply Lemma 3.11 and argue as in the proof of Lemma 3.16 to complete the
proof. ∎

######  Lemma 3.18.

Let $G$ be a minimal counterexample to Theorem 3.1, and let $L=T^{e}$ be a
minimal normal subgroup of $G$, where $T=\operatorname{\mathrm{L}}_{2}(q)$,
with $q$ an odd prime power of the form
$q=2^{x^{\prime}}3^{y^{\prime}}5^{z^{\prime}}-1$, $0\leq
y^{\prime},z^{\prime}\leq 1$. Let ${H}$ be the intersection of $L$ with a
point stabilizer for $G$, so that ${H}$ is of type (4) in Corollary 3.9. Let
$a$ be the number of $L$-orbits, and set $F:=S_{0}^{e}<L$, where $S_{0}$ is a
maximal parabolic subgroup of $T$. Then

  1. (i)

Suppose that $H$ has type (4)(a) from Corollary 3.9, so that $q=p$ is a
Mersenne prime with $15$ dividing $p-1$, and $H$ is
$\operatorname{\mathrm{S}}_{e}$-conjugate to $T^{e_{1}}\times
Y_{1}\times\ldots\times
Y_{e_{2}}\times(p:\frac{p-1}{6})\times(p:\frac{p-1}{10})$, where each $Y_{i}$
is a maximal parabolic subgroup of $T$. Then $F:=S$ has $a\binom{e_{2}}{i}$
orbits of size $15p^{i}$; $2a\binom{e_{2}}{i}$ orbits of size $15p^{i+1}$; and
$a\binom{e_{2}}{i}$ orbits of size $15p^{i+2}$, for each $0\leq i\leq e_{2}$.

  2. (ii)

Suppose that $H$ has type (4)(b) from Corollary 3.9, so that $q=p$ is a
Mersenne prime and $H$ is $\mathrm{Sym}_{e}$-conjugate to $T^{e_{1}}\times
L_{1}$, where $L_{1}$ is a subdirect product of the form
$L_{1}=\frac{1}{t}(L_{2}\times H\pi_{e})$; $L_{2}$ is a subdirect product of
the form $L_{2}=\frac{1}{s}(P_{1}\times\ldots\times P_{e_{2}})$; each $P_{i}$
is a maximal parabolic subgroup of $T$; $H\pi_{e}$ is a subgroup of index $r$
in a maximal parabolic subgroup of $T$; $rst=3^{y_{0}}5^{z_{0}}$ is the odd
part of $|L:H|$; and $e_{1}+e_{2}=e-1$. Then $F:=S$ has $a\binom{e_{2}+1}{i}$
orbits of size $3^{y_{0}}5^{z_{0}}p^{i}$ for each $0\leq i\leq e_{2}+1$.

  3. (iii)

Suppose that $H$ has type (4)(c) from Corollary 3.9, so that $H$ is
$\mathrm{Sym}_{e}$-conjugate to $T^{e-1}\times(q:\frac{q-1}{2r})$, where
$y^{\prime}+z^{\prime}=1$; $0\in\\{y^{\prime},z^{\prime}\\}$; and
$r=3^{1-y^{\prime}}5^{1-z^{\prime}}>1$. Then $F:=N_{G}(S)$ is soluble, and has
precisely two orbits, of sizes $ar$ and $aqr$.

###### Proof.

We need to determine the orbit lengths of $S_{0}$ in its action of the cosets
of the subgroups
$X\in\\{q:\frac{q-1}{2},q:\frac{q-1}{6},q:\frac{q-1}{10}\text{
}\mathrm{(}5\mid q-1\mathrm{)},q:\frac{q-1}{15}\text{ }\mathrm{(}5\mid
q-1\mathrm{)}\\}$ for part (ii); while we need to determine the orbit lengths
of $S_{0}^{2}$ in its action of the cosets of the subgroup
$(p:\frac{p-1}{6})\times(p:\frac{p-1}{10})<T^{2}$ for part (i).

We first consider the orbits of the action of $S_{0}\cong q:\frac{q-1}{2}$ on
the cosets of itself in $T$. This is equivalent to the action of $S_{0}$ on
the set of one dimensional subspaces in $\mathbb{F}_{q}^{2}$. It is well known
that $S_{0}$ has one orbit of size $1$ and one orbit of size $q$ in this
action. It follows easily that if $X$ has shape $q:\frac{q-1}{2r}$ with $r$ a
divisor of $\frac{q-1}{2}$, then $S_{0}$ has one orbit of size $r$, and one
orbit of size $rq$. Next, when $p=q$, $S_{0}^{2}$ has four orbits on the
cosets of $(p:\frac{p-1}{6})\times(p:\frac{p-1}{10})$, of lengths $15$, $15q$,
$15q$, and $15q^{2}$. Lemma 3.11(ii) then implies that $F$ has
$\binom{e_{2}}{i}$ orbits of size $15p^{i}$; $2a\binom{e_{2}}{i}$ orbits of
size $15p^{i+1}$; and $a\binom{e_{2}}{i}$ orbits of size $15p^{i+2}$ in its
action on the cosest of $H$ in $L$, for each $0\leq i\leq e_{2}$. Since our
arguments hold with $H$ replaced by any $G$-conjugate of $H$, we deduce that
$F$ has $a\binom{e_{2}}{i}$ orbits of size $15p^{i}$; $2a\binom{e_{2}}{i}$
orbits of size $15p^{i+1}$; and $a\binom{e_{2}}{i}$ orbits of size $15p^{i+2}$
in $[n]$, for each $0\leq i\leq e_{2}$.

Suppose now that we are in case (ii). Note that

(3.6) $(S_{0}\cap P_{i}^{x})[P_{i}^{x_{i}},P_{i}^{x_{i}}]=P_{i}^{x_{i}}$ for
all $x_{i}$ in $T$.

Thus, by the arguments above, together with Lemma 3.10, we deduce that
$S_{0}^{e_{2}}$ has $\binom{e_{2}}{i}$ orbits of length $sq^{i}$ on the cosets
of $L_{2}$ in $T^{e_{2}}$, for each $0\leq i\leq e_{2}$. Hence, since $S_{0}$
has two orbits, of sizes $r$ and $qr$, on the cosets of $H\pi_{e}$ in $T$, we
deduce from Lemma 3.11 that $S_{0}^{e_{2}+1}$ has $\binom{e_{2}}{i}$ orbits of
length $srq^{i}$ and $\binom{e_{2}}{i}$ orbits of length $srq^{i+1}$ in its
action on the cosets of $L_{2}\times H\pi_{e}$ in $T^{e_{2}+1}$. By (3.6), we
can then use Lemma 3.10 again: we get that $S_{0}^{e_{2}+1}$ has
$\binom{e_{2}}{i}$ orbits of length $rstq^{i}$ and $\binom{e_{2}}{i}$ orbits
of length $rstq^{i+1}$ in its action on the cosets of $L_{1}$ in
$T^{e_{2}+1}$. Note that $rst=3^{x_{0}}5^{x_{0}}$ and
$\binom{e_{2}}{i}+\binom{e_{2}}{i-1}=\binom{e_{2}+1}{i}$ for $i>0$. We can
then deduce from Lemma 3.11(i) (with $m=1$) that $F$ has $\binom{e_{2}+1}{i}$
orbits of size $3^{y_{0}}5^{z_{0}}p^{i}$ in its action on the cosets of $H$ in
$L$, for each $0\leq i\leq e_{2}+1$. Since our arguments are independent of
the choice of $G$-conjugate of $H$ in $L$, we see as above that $F$ has
$a\binom{e_{2}+1}{i}$ orbits of size $3^{y_{0}}5^{z_{0}}p^{i}$ in $[n]$, for
each $0\leq i\leq e_{2}+1$.

Finally, the proof in case (iii) is identical to the proof of Lemma 3.15. ∎

######  Lemma 3.19.

Let $G$ be a minimal counterexample to Theorem 3.1, and let $L=T^{e}$ be a
minimal normal subgroup of $G$, where $T=\operatorname{\mathrm{A}}_{m}$ with
$m\in\\{5,6\\}$, and $G/L$ is insoluble. Let ${H}$ be the intersection of $L$
with a point stabilizer for $G$, and let $a$ be the number of $L$-orbits. Then

  1. (i)

$H$ is as in case (1)(b) of Corollary 3.9;

  2. (ii)

$|T:H_{e}|=2^{x_{1}}5$ and $a=2^{x_{2}}3$;

  3. (iii)

$G/L$ has a unique nonabelian chief factor $L^{\prime}$ such that
$L^{\prime}\cong\operatorname{\mathrm{L}}_{2}(p)^{f}$, where $p$ is a Mersenne
prime; and

  4. (iv)

There exist positive integers $X$, $b$, and $e_{2}\leq e$; positive partitions
$\binom{e_{2}}{i}=\sum_{j}C^{(i)}_{j}$ and $X=\sum_{k}X^{(i,j)}_{k}$ of
$\binom{e_{2}}{i}$ and $X$ (for each $i,j$); and a soluble subgroup $F$ of
$G$, such that one of the following holds:

     * •

$T=\operatorname{\mathrm{A}}_{5}$, $(X,b)\in\\{(2,1),(2,0)\\}$, and $F$ has
$C^{(i)}_{j}$ orbits of length $2^{b}3p^{i}5X^{(i,j)}_{k}$, for $0\leq i\leq
e_{2}$, and each $j,k$.

     * •

$T=\operatorname{\mathrm{A}}_{6}$, $(X,b)\in\\{(4,1),(4,0),(2,0)\\}$, and $F$
has $C^{(i)}_{j}$ orbits of length $2^{b}3p^{i}5X^{(i,j)}_{k}$, for $0\leq
i\leq e_{2}$, and each $j,k$.

###### Proof.

Let $K$ be the kernel of the action of $G$ on the set $\Sigma$ of $L$-orbits,
and fix an $L$-orbit $\Delta$. Then since $G/K$ is an insoluble minimal
transitive group, its degree $a$ cannot be a power of $2$. Hence, either $3$
or $5$ must divide $a$. Thus, the odd part of $|\Delta|=|L:H|$ is either $3$
or $5$, and it follows from Corollary 3.9, and by inspection of the subgroups
of $\operatorname{\mathrm{A}}_{5}$ of $\operatorname{\mathrm{A}}_{6}$, that
the only possibility is that $H$ is as in case (1)(b) of Corollary 3.9. Part
(i) follows.

Now, by Corollary 3.9(1)(b), $H_{e}$ is not transitive, and not equal to a
natural point stabilizer
$\operatorname{\mathrm{A}}_{m-1}<T=\operatorname{\mathrm{A}}_{m}$. Thus, going
through the subgroups of $\operatorname{\mathrm{A}}_{5}$ and
$\operatorname{\mathrm{A}}_{6}$, we see that $|T:H_{e}|$ must be of the form
$|T:H_{e}|=2^{x_{1}}5$. Hence, $G/K$ is an insoluble minimal transitive group
of degree $2^{x_{2}}3$. This proves (ii). Part (iii) then follows immediately
from [13, Theorem 1.8].

We now prove (iv). Set $S_{0}:=D_{10}<T$, and $S:=S_{0}^{e}$. Then as shown in
Lemmas 3.13 and 3.14, the hypotheses in Corollary 3.12 hold with this choice
of $S_{0}$. Thus we can, and do, apply Corollary 3.12, with $R/K$ chosen to be
the soluble subgroup of $G/K$ exhibited in Lemma 3.18(ii), with
$p=2^{x_{1}}-1$ and $rst=3$ . We can then deduce (iv) by taking the possible
orbit lengths of $S_{0}$ acting on the cosets of $H_{e}$ from Tables 2 and 3
(noting again that $H_{e}$ is intransitive, and that
$H_{e}\neq\operatorname{\mathrm{A}}_{4}$ in the
$\operatorname{\mathrm{A}}_{5}$ case). ∎

Finally, we are ready to prove Theorem 3.1.

###### Proof of Theorem 3.1.

Let $G$ be a counterexample to the theorem of minimal degree, and let $L\cong
T^{e}$ be a minimal normal subgroup of $G$, with $T$ simple. Then $T$ is
nonabelian, by Lemma 3.3. Let $H$ be the intersection of $L$ with a point
stabilizer in $G$, and write $|L:H|=2^{x_{0}}3^{y_{0}}5^{z_{0}}$, where
$x_{0}\leq x$, $y_{0}\leq y$, and $z_{0}\leq z$. Thus, if $K$ denotes the
kernel of the action of $G$ on the set $\Sigma$ of $L$-orbits, then
$G^{\Sigma}\cong G/K$ is minimal transitive of degree
$2^{x-x_{0}}3^{y-y_{0}}5^{z-z_{0}}$. Also, either $y_{0}$ or $z_{0}$ is non-
zero, by Corollary 3.8(i). Thus, since $K/L$ is soluble by Lemma 3.3, the
minimality of $G$ as a counterexample implies that $G$ has at most
$1+y-y_{0}+z-z_{0}\leq y+z$ nonabelian chief factors. That is, (i) holds.

Now, $T$ is one of the groups in part (ii) of Theorem 3.1, by Corollary 3.9,
so (ii) also holds. Next, we prove (iii). So assume that $G$ has two
nonabelian chief factors, and let $L^{\prime}\cong T^{\prime e^{\prime}}$ be a
nonabelian chief factor of $G/L$. Then the minimal transitive group
$G^{\Sigma}\cong G/K$ has $L^{\prime}$ as a chief factor, since $K/L$ is
soluble. Since minimal transitive groups of $2$-power degree are $2$-groups,
we deduce that $G^{\Sigma}\cong G/K$ is minimal transitive of degree either
(1) $2^{x-x_{0}}5$ or (2) $2^{x-x_{0}}3$. The size of an $L$-orbit in these
cases is $2^{x_{0}}3$ or $2^{x_{0}}5$, respectively. Now, by Corollary 3.9,
the only possibility for $T$ when the size of an $L$-orbit is $2^{x_{0}}3$ is
$T=\operatorname{\mathrm{L}}_{2}(p)$ with $p$ a Mersenne prime; while the only
possibility for $T$ when the size of an $L$-orbit is $2^{x_{0}}5$ is
$T\in\\{\operatorname{\mathrm{A}}_{5},\operatorname{\mathrm{A}}_{6},\operatorname{\mathrm{L}}_{2}(p)\text{
: }p\text{ a Mersenne prime}\\}$. This gives us what we need. Moreover,
inspection of Table 1 and the inductive hypothesis imply that in case (1), we
must have
$T^{\prime}\in\\{\operatorname{\mathrm{A}}_{5},\operatorname{\mathrm{A}}_{6},\operatorname{\mathrm{L}}_{2}(p^{\prime})\text{
: }p\text{ a Mersenne prime}\\}$; while in case (2), we have
$T^{\prime}=\operatorname{\mathrm{L}}_{2}(p^{\prime})$ with $p^{\prime}$ a
Mersenne prime. This proves (iii).

Finally, in Lemmas 3.13, 3.14, 3.15, 3.16, 3.17, 3.18, and Lemma 3.19, we
exhibit a soluble subgroup $F$ of $G$ with particular orbit lengths. These
orbit lengths are as listed in Table 1, and this completes the proof. ∎

##  4\. The proof of Theorem 1.1

In this section, we prove Theorem 1.1. First, we require the following
definitions and lemma from [13].

######  Definition 4.1.

For a positive integer $s$ with prime factorisation
$s=p_{1}^{r_{1}}p_{2}^{r_{2}}\ldots p_{t}^{r_{t}}$, set
$\omega{(s)}:=\sum r_{i}$, $\omega_{1}{(s)}:=\sum r_{i}p_{i}$,
$K(s):=\omega_{1}{(s)}-\omega(s)=\sum r_{i}(p_{i}-1)$ and

${\widetilde{\omega}}(s)=\frac{s}{2^{K(s)}}\binom{K(s)}{\left\lfloor\frac{K(s)}{2}\right\rfloor}.$

For a prime $p$, write $s_{p}$ for the $p$-part of $s$.

######  Definition 4.2.

Let $p$ be prime, and let $s$ be a positive integer. We define

$E_{sol}(s,p):=\min\left\\{{\widetilde{\omega}}(s),s_{p}\right\\}.$

######  Lemma 4.3.

[13, Corollary 4.25(ii)(b)] Let $G$ be a transitive permutation group of
degree $n$, and suppose that $G$ is imprimitive with a minimal block of size
$2$. Let $\Sigma$ be the associated system of blocks of size $\frac{n}{2}$, so
that $G$ may be viewed as a subgroup of the wreath product $2\wr G^{\Sigma}$.
Let $F$ be a soluble subgroup of $G^{\Sigma}$, and suppose that $F$ has orbit
lengths $s_{1},\ldots,s_{r}$. Then

$d(G)\leq\sum_{i=1}^{r}E_{sol}(s_{i},2)+d(G^{\Sigma}).$

######  Corollary 4.4.

Let $G$ be a transitive permutation group of degree $n$, and suppose that $G$
is imprimitive with a minimal block of size $2$. Let $\Sigma$ be the
associated system of blocks of size $\frac{n}{2}$, so that $G$ may be viewed
as a subgroup of the wreath product $2\wr G^{\Sigma}$. Suppose that
$\frac{n}{2}$ has the form $\frac{n}{2}=2^{x}3^{y}5^{z}$, with $0\leq y,z\leq
1$, and that $G^{\Sigma}$ contains no soluble transitive subgroups. Let $F$ be
a soluble subgroup of $G^{\Sigma}$ with orbit lengths as exhibited in Table 1.

  1. (1)

If $F$ is as in row 12 of Table 1, then $d(G)\leq
2^{e_{2}+b+1}a+d(G^{\Sigma})$.

  2. (2)

If $F$ is as in row 22 of Table 1, then $d(G)\leq
2^{e_{2}+b+2}a+d(G^{\Sigma})$.

  3. (3)

If $F$ is as in row 24, 25, or 26 of Table 1, then $d(G)\leq
2^{e_{2}}54a+d(G^{\Sigma})$; $d(G)\leq 2^{e_{2}}10a+d(G^{\Sigma})$; or
$d(G)\leq 2^{e_{2}}3a+d(G^{\Sigma})$, respectively.

  4. (4)

If $F$ is as in row 27 of Table 1, then $d(G)\leq
2^{e_{2}+4-b}a+d(G^{\Sigma})$.

  5. (5)

If $F$ is as in row 28 of Table 1, then $d(G)\leq 2^{e_{2}}3a+d(G^{\Sigma})$.

  6. (6)

If $F$ is as in row 29 of Table 1, then $d(G)\leq 2^{e_{2}}a+d(G^{\Sigma})$.

  7. (7)

If $F$ is as in row 30 of Table 1, then $d(G)\leq 2a+d(G^{\Sigma})$.

###### Proof.

Write $s_{1}$, $\ldots$ $s_{r}$ for the $F$-orbit lengths. Then by Lemma 4.3
and the definition of $E_{sol}$, we have

(4.1) $\displaystyle d(G)\leq\sum_{i=1}^{r}(s_{i})_{2}+d(G^{\Sigma}).$

Parts (3), (4), (5), and (6) now follow immediately from (4.1) and inspection
of the relevant row in Table 1.

Suppose now that $F$ is as in row 12 or 22 of Table 1. Then (4.1) implies that

$\displaystyle d(G)\leq
2^{b}\sum_{i,j,k}C^{(i)}_{j}(X^{(i,j)}_{k})_{2}+d(G^{\Sigma}).$

Since $(X^{(i,j)}_{k})$ is an ordered positive partition of either $2$ or $4$,
the result now follows easily by going through each of the possibilities for
$(X^{(i,j)}_{k})_{k}$, and noting that
$\sum_{i,j}C^{(i)}_{j}=\sum_{i}\binom{e_{2}}{i}=2^{e_{2}}$. ∎

We are now ready to prove Theorem 1.1.

###### Proof of Theorem 1.1.

By [13, Theorem 1.1], we may assume that $n$ has the form $n=2^{x}3^{y}5$,
where $y\in\\{0,1\\}$. Furthermore, we may assume that $17\leq x\leq 26$ if
$y=0$, and $15\leq x\leq 35$ if $y=1$; that $G$ has a block system $\Sigma$
consisting of blocks of size $2$; and that $G^{\Sigma}$ contains no soluble
transitive subgroup.

It then follows from Theorem 3.1 that $G^{\Sigma}$ has a soluble subgroup $F$
whose orbit lengths occur in one of the rows of Table 1. We can then use this
fact, together with Lemma 4.3 to prove the theorem. Indeed, suppose that we
have proved the theorem for all degrees properly dividing $n$, and that the
possible orbit lengths for the soluble subgroup $F$ are
$s_{1,j},s_{2,j},\ldots,s_{r_{j},j}$, for $1\leq j\leq t$. We then compute the
maximum value of $\sum_{i=1}^{r_{j}}E_{sol}(s_{i,j},2)$ over $1\leq j\leq t$,
and add it to our previously obtained bound for $d(G^{\Sigma})$. By Lemma 4.3,
this gives a bound for $d(G)$.

Let us illustrate this strategy with some examples. Suppose first that
$n=2^{17}5$. From Table 1, we see that
$G^{\Sigma}\leq\operatorname{\mathrm{S}}_{2^{16}5}$ has a soluble subgroup
whose orbit lengths lie in one of the rows 7, 18, 19, or 29 (with $y_{0}=0$,
$z_{0}=1$ in the last case). Computing as in the $n=2^{15}15$ example given at
the beginning of the section, we see that the lengths of the $F$-orbits are
one of the following:

  * •

$2^{i}$ orbits of equal lengths, $i\in\\{1,2\\}$;

  * •

$2$ orbits of length $2^{14}5$ and one orbit of length $2^{15}5$;

  * •

$1$ orbit of length $2^{14}5$ and $1$ orbit of length $2^{14}15$; or

  * •

$\binom{e_{2}}{i}a$ orbits of size $5p^{i}$ and $\binom{e_{2}}{i}a$ orbits of
size $5p^{i+1}$, for each $0\leq i\leq e_{2}$, where the possibilities for
$e_{2}$ and $p$ are $(e_{2},p)\in\\{(A,2^{5}-1),(1,2^{13}-1)\text{ : }1\leq
A\leq 2\\}$, and $\frac{n}{2}=5a(p+1)^{e_{2}}$.

We make some remarks about the last point above. Since $(p+1)^{e_{2}}$ must
divide $\frac{n}{2}$, we only need to consider the Mersenne primes $p=2^{u}-1$
such that $p+1=2^{u}<2^{k}:=(\frac{n}{2})_{2}$, and those $e_{2}$ with $1\leq
e_{2}\leq\left\lfloor\frac{k}{u}\right\rfloor-1$. And since we are assuming,
throughout this proof, that $n_{2}\leq 2^{35}$, the only Mersenne primes under
consideration are $2^{u}-1$, for $u$ a prime less than or equal to $31$. In
the case $n=2^{17}5$ above, we must also have that $5$ divides $p-1$. So only
the cases $u\in\\{5,13\\}$ can occur.

We now proceed to compute a bound for $d(G)$ (for $n=2^{17}5$) in each case,
by using Table 1 and Corollary 4.3, as explained above. By [13, Table A.1], we
have $d(G^{\Sigma})\leq 65538$. If $F$ has $2$ orbits of length $2^{15}5$ then
we get $d(G)\leq 2E_{sol}(2^{15}5,2)+65538=123274$. Similarly, if $F$ has $4$
orbits of length $2^{14}5$ then we get $d(G)\leq 126313$; if $F$ has $2$
orbits of length $2^{14}5$ and $1$ orbit of length $2^{15}5$ then $d(G)\leq
124793$; and if $F$ has $1$ orbit of length $2^{14}5$ and $1$ orbit of length
$2^{14}15$ then $d(G)\leq 97115$. Suppose now that $F$ has $\binom{e_{2}}{i}d$
orbits of size $5p^{i}$ and $\binom{e_{2}}{i}d$ orbits of size $5p^{i+1}$, for
each $0\leq i\leq e_{2}$, where
$(e_{2},p)\in\\{(1,2^{5}-1),(2,2^{5}-1),(1,2^{13}-1)\\}$ and
$a=\frac{n}{10(p+1)^{e_{2}}}$. If $(e_{2},p)=(1,2^{5}-1)$, then $F$ has
$2^{11}$ orbits of size $5$, and $2^{11}$ orbits of size $5\times 31=155$.
Therefore, we get $d(G)\leq
2^{11}E_{sol}(5,2)+2^{11}E_{sol}(155,2)+65538=69634$. The other cases are
entirely similar, and we see that $d(G)$ is always bounded above by $126313$.
Since $\left\lfloor\frac{c2^{17}5}{\sqrt{\log{2^{17}5}}}\right\rfloor=129117$,
this proves Theorem 1.1 in the case $n=2^{17}5$.

Next, we will give an example to show how the required upper bounds can be
derived in the cases $n=2^{x}15$. More precisely, we will go through the
necessary computations in the case $n:=2^{15}15$. In order to do this, we will
first need an upper bound on $d(G^{\Sigma})$. Since $G^{\Sigma}$ is a
transitive permutation group of degree $2^{14}15$, we can deduce from [13,
Table A.1] that $d(G^{\Sigma})\leq 49156$. Using the classification of the
transitive groups of degree $48$ from [5], however, we can improve this bound.
In order to avoid repetition of details, we will simply refer to [13, proof of
Lemma 5.12, case 2, page 40], where the proofs of the bounds in [13, Table
A.1] are given. The author uses the classification of the transitive
permutation groups of degree up to $32$ [3], together with the bounds in Lemma
4.3 above, to derive the upper bound on $d(G^{\Sigma})\leq 49156$. Within this
computation, the estimate $d(X)\leq 16$ for transitive groups
$X\leq\operatorname{\mathrm{S}}_{48}$ is used. We can now use the exact bound
$d(X)\leq 10$ for transitive $X\leq\operatorname{\mathrm{S}}_{48}$ from [5,
Section 5]. Performing the computations in [13, proof of Lemma 5.12, case 2,
page 40] again with this new bound, we quickly deduce that

(4.2) $\displaystyle d(G^{\Sigma})\leq 49150\text{ whenever
}G^{\Sigma}\leq{\operatorname{\mathrm{S}}}_{2^{14}15}\text{ is transitive.}$

Using (4.2) and Lemma 4.3, we can now determine an upper bound for $d(G)$ when
$n=2^{15}15$. At the beginning of the section, we listed the possible
$F$-orbit lengths in the case when $G$ has a unique nonabelian chief factor,
and this chief factor is a direct product of copies of
$\operatorname{\mathrm{A}}_{5}$ (these cases correspond to rows 1–11 in Table
1). Entirely analogous calculations give us the possible $F$-orbit lengths
when $G$ has a unique nonabelian chief factor, and this chief factor is a
direct product of copies of $\operatorname{\mathrm{A}}_{6}$ (these cases
correspond to rows 13–21 in Table 1) or $\operatorname{\mathrm{S}}_{4}(4)$ or
$\operatorname{\mathrm{O}}^{+}_{8}(2)$ (row 23). When $F$ has orbit lengths as
in rows 12, 22, and 24–30 of 1, we use Corollary 4.4 to bound $d(G)$. Taking
the maximum of these bounds for $d(G)$, we get $d(G)\leq 97401$ in each case.
This proves the theorem in this case, since
$\left\lfloor\frac{c2^{15}15}{\sqrt{\log{2^{15}15}}}\right\rfloor=97895$.

The remaining cases with $n$ of the form $n=2^{x}15$, and $16\leq x\leq 35$,
are entirely similar, except that we use the bound
$d(G^{\Sigma})\leq\left\lfloor\frac{c2^{x-1}15}{\sqrt{\log{2^{x-1}15}}}\right\rfloor$
for $d(G^{\Sigma})$ (which holds by the inductive hypothesis). We note in
particular that we get $d(G^{\Sigma})\leq 189053$ when $x=16$ (this will be
important). The result now follows in each case, except when $x=17$. So assume
that $x=17$. By [13, Theorem 1.1], we may assume that $G^{\Sigma}$ is
imprimitive with a minimal block of size $2$. Let $K_{1}$ be the kernel of the
action of $G^{\Sigma}$ on a set $\Sigma_{1}$ of blocks of size $2$ in
$\Sigma$. Assume first that $G^{\Sigma}\cap K_{1}\neq 1$. If
$(G^{\Sigma})^{\Sigma_{1}}\cong G^{\Sigma}/K_{1}$ contains a soluble
transitive subgroup $S/K_{1}$, then $S$ is a soluble transitive subgroup of
$G^{\Sigma}$ – a contradiction. Thus, $(G^{\Sigma})^{\Sigma_{1}}$ contains no
soluble transitive subgroups. Because of this, we can use our previously
obtained bound $d(G^{\Sigma})\leq 189053$ in this case. If $G^{\Sigma}\cap
K_{1}=1$, then
$d(G^{\Sigma})=d(G^{\Sigma_{1}})\leq\left\lfloor\frac{c2^{15}15}{\sqrt{\log{2^{15}15}}}\right\rfloor=98547$,
so in either case, we get $d(G^{\Sigma})\leq 189053$. By computing the upper
bound for $d(G)$ as we did in the case $n=2^{15}15$ above, we now get
$d(G)\leq 371369$. This gives us what we need, since
$\left\lfloor\frac{c2^{17}15}{\sqrt{\log{2^{17}15}}}\right\rfloor=372380$. ∎

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