# Criteria for supersolvability of saturated fusion systems

Fawaz Aseeri
Department of Mathematical Sciences, College of Applied Sciences, Umm Al-Qura
University,
Makkah 21955, Saudi Arabia
E-mail<EMAIL_ADDRESS>

Julian Kaspczyk111Corresponding author 222At the date of submission, this
author was not anymore affiliated with the Technische Universität Dresden.
However, the bulk of the work presented here was done when he still was a
postdoc at this university.
Institut für Algebra, Fakultät Mathematik, Technische Universität Dresden,
01069 Dresden, Germany
E-mail<EMAIL_ADDRESS>
###### Abstract

Let $p$ be a prime number. A saturated fusion system $\mathcal{F}$ on a finite
$p$-group $S$ is said to be supersolvable if there is a series $1=S_{0}\leq
S_{1}\leq\dots\leq S_{m}=S$ of subgroups of $S$ such that $S_{i}$ is strongly
$\mathcal{F}$-closed for all $0\leq i\leq m$ and such that $S_{i+1}/S_{i}$ is
cyclic for all $0\leq i<m$. We prove some criteria that ensure that a
saturated fusion system $\mathcal{F}$ on a finite $p$-group $S$ is
supersolvable provided that certain subgroups of $S$ are abelian and weakly
$\mathcal{F}$-closed. Our results can be regarded as generalizations of purely
group-theoretic results of Asaad [3].

22footnotetext: Keywords: fusion systems, supersolvable, weakly closed,
pronormal, weakly pronormal22footnotetext: Mathematics Subject Classification
(2020): 20D10, 20D20.

##  1 Introduction

In finite group theory, the term “fusion” refers to conjugacy relations
between $p$-elements and $p$-subgroups. The study of fusion in finite groups
has a long history, and many results concerning fusion in finite groups had a
significant impact on finite group theory. Some well-known examples are
Burnside’s fusion theorem [15, Lemma 5.12], Frobenius’ $p$-nilpotency
criterion [15, Theorem 5.26], Alperin’s fusion theorem [1] and Glauberman’s
$Z^{*}$-theorem [11].

A modern approach to study problems concerning fusion in finite groups is the
theory of fusion systems. The standard examples of fusion systems are the
fusion categories of finite groups over $p$-subgroups. Given a prime number
$p$, a finite group $G$ and a $p$-subgroup $S$ of $G$, the fusion category of
$G$ over $S$ is defined to be the category $\mathcal{F}_{S}(G)$ given as
follows: The objects of $\mathcal{F}_{S}(G)$ are the subgroups of $S$, the
morphisms in $\mathcal{F}_{S}(G)$ are the group homomorphisms between
subgroups of $S$ induced by conjugation in $G$, and the composition of
morphisms in $\mathcal{F}_{S}(G)$ is the usual composition of group
homomorphisms. Abstract fusion systems can be regarded as a generalization of
this concept. Given a prime number $p$ and a finite $p$-group $S$, a fusion
system on $S$ is a category $\mathcal{F}$ whose objects are the subgroups of
$S$ and whose morphisms behave as if they are induced by conjugation inside a
finite group containing $S$ as a $p$-subgroup (see [4, Part I, Definition
2.1]). A fusion system is said to be saturated if it satisfies certain
additional axioms (see [4, Part I, Definition 2.2]). Any fusion category of a
finite group over a Sylow subgroup is saturated (see [4, Part I, Theorem
2.3]), but not every saturated fusion system appears as the fusion category of
a finite group over a Sylow subgroup (see [4, Part III, Section 6]). If $S$ is
a Sylow $p$-subgroup of a finite group $G$ for some prime $p$, then we refer
to $\mathcal{F}_{S}(G)$ as the $p$-fusion system of $G$.

The book [4] provides a detailed introduction to the theory of fusion systems,
and the reader is asked to consult that book for any definitions on fusion
systems we do not explain here. For unfamiliar definitions on groups, the
reader is referred to [14, 15].

This paper is concerned with the concept of a supersolvable saturated fusion
system introduced in [24]. Given a prime $p$ and a finite $p$-group $S$, a
saturated fusion system $\mathcal{F}$ on $S$ is said to be supersolvable if
there is a series $1=S_{0}\leq S_{1}\leq\dots\leq S_{m}=S$ of subgroups of $S$
such that $S_{i}$ is strongly $\mathcal{F}$-closed for all $0\leq i\leq m$ and
such that $S_{i+1}/S_{i}$ is cyclic for all $0\leq i<m$ (see [24, Definition
1.2]). By [24, Proposition 1.3], for any prime $p$, the supersolvable
saturated fusion systems on finite $p$-groups are precisely the $p$-fusion
systems of supersolvable finite groups, and they are also precisely the
$p$-fusion systems of $p$-supersolvable finite groups. After their
introduction in [24], supersolvable saturated fusion systems (and related
concepts) were further studied in the papers [6, 22].

The goal of this paper is to obtain criteria for supersolvability of saturated
fusion systems. Our results are inspired from a current line of research in
finite group theory. Namely, there is currently very active research on
subgroup embedding properties, and a problem of particular interest is to
study the structure of a finite group $G$ under the assumption that some given
subgroups of $G$ satisfy a given embedding property. In this context, many
results of the following form were obtained: Given a fixed prime $p$ and a
finite group $G$, it is assumed that all (or at least sufficiently many)
$p$-subgroups of $G$ with some fixed order satisfy a certain embedding
property. Sometimes, a number of additional conditions are assumed, and the
conclusion usually is that $G$ is $p$-supersolvable or $p$-nilpotent. As an
example, we mention a result of Berkovich and Isaacs. Given a prime $p$, a
natural number $e\geq 3$ and a finite group $G$ with a noncyclic Sylow
$p$-subgroup of order exceeding $p^{e}$, they proved that $G$ is
$p$-supersolvable if any noncyclic subgroup of $G$ with order $p^{e}$ is
normal in $G$ (see [8, Theorem C]). Other results of this kind were obtained,
for example, in the papers [2, 3, 12, 17, 19, 26].

Our principal motivation is to prove supersolvability criteria of a similar
spirit for saturated fusion systems. Indeed, given a prime $p$ and a saturated
fusion system $\mathcal{F}$ on a finite $p$-group $S$, we will prove criteria
ensuring that $\mathcal{F}$ is supersolvable provided that all subgroups of
$S$ with some fixed order are “suitably embedded” in $S$ with respect to
$\mathcal{F}$. More precisely, we will assume that all subgroups of $S$ with
some fixed order are weakly $\mathcal{F}$-closed and that sufficiently many of
them are abelian.

Our results generalize some purely group-theoretic results of Asaad [3], and
before stating our results, we consider the corresponding results from [3].

First, let us recall some definitions. A subgroup $H$ of a group $G$ is said
to be pronormal in $G$ if $H$ and $H^{g}$ are conjugate in $\langle
H,H^{g}\rangle$ for each $g\in G$. By [19, Theorem 4.3], if $S$ is a Sylow
$p$-subgroup of a finite group $G$ for some prime $p$, then a subgroup $Q$ of
$S$ is pronormal in $G$ if and only if $Q$ is weakly
$\mathcal{F}_{S}(G)$-closed. Following Asaad [3], we say that a subgroup $H$
of a group $G$ is weakly pronormal in $G$ if there is a subgroup $K$ of $G$
such that $G=HK$ and such that $H\cap K$ is pronormal in $G$. Note that, if
$S$ is a Sylow $p$-subgroup of a finite group $G$ for some prime $p$, a
subgroup $Q$ of $S$ is weakly pronormal in $G$ if and only if there is a
subgroup $K$ of $G$ such that $G=QK$ and such that $Q\cap K$ is weakly
$\mathcal{F}_{S}(G)$-closed.

The following theorem is one of the main results of [3].

######  Theorem 1.1.

([3, Theorem 1.3]) Let $G$ be a nontrivial finite group, $p$ be the smallest
prime dividing $|G|$ and $S$ be a noncyclic Sylow $p$-subgroup of $G$. Suppose
that there is a subgroup $D$ of $S$ with $1<D<S$ such that any subgroup of $S$
with order $|D|$ or $p|D|$ is abelian and weakly pronormal in $G$. Then $G$ is
$p$-nilpotent.

Note that, because of Burnside’s $p$-nilpotency criterion [15, Corollary
5.14], the condition in Theorem 1.2 that $S$ is noncyclic is not necessary.

We will prove the following generalization of Theorem 1.1.

######  Theorem A.

Let $G$ be a finite group, $p$ be a prime and $S$ be a Sylow $p$-subgroup of
$G$. Suppose that there is a subgroup $D$ of $S$ with $1<D<S$ such that any
subgroup of $S$ with order $|D|$ or $p|D|$ is abelian and weakly pronormal in
$G$. Then $\mathcal{F}_{S}(G)$ is supersolvable.

By [24, Theorem 1.9 (a)], if $G$ is a finite group and if $S$ is a Sylow
$p$-subgroup of $G$ for some prime $p$ with $(|G|,p-1)=1$, then $G$ is
$p$-nilpotent provided that $\mathcal{F}_{S}(G)$ is supersolvable. Of course,
the condition $(|G|,p-1)=1$ is satisfied when $p$ is the smallest prime
divisor of $|G|$, and hence, Theorem 1.1 is covered by Theorem A.

When the hypotheses of Theorem A are satisfied, it is not necessarily true
that $G$ is $p$-supersolvable. For example, let $G:=A_{5}\times A_{5}$ and $S$
be a Sylow $5$-subgroup of $G$. Then $S\cong C_{5}\times C_{5}$ is abelian.
Also, any subgroup of $S$ with order $5$ and $S$ itself are complemented in
$G$ and hence weakly pronormal in $G$. However, $G$ is not $5$-supersolvable.

Our further results are supersolvability criteria for abstract saturated
fusion systems, and they are motivated by the following result of Asaad [3].

######  Theorem 1.2.

([3, Corollary 3.1]) Let $G$ be a nontrivial finite group, $p$ be the smallest
prime dividing $|G|$ and $S$ be a noncyclic Sylow $p$-subgroup of $G$. Suppose
that there is a subgroup $D$ of $S$ with $1<D<S$ such that any subgroup of $S$
with order $|D|$ is abelian and pronormal in $G$. If $S$ is a nonabelian
$2$-group, suppose moreover that any subgroup of $S$ with order $2|D|$ is
abelian and pronormal in $G$. Then $G$ is $p$-nilpotent.

Note that, because of Burnside’s $p$-nilpotency criterion [15, Corollary
5.14], the condition in Theorem 1.2 that $S$ is noncyclic is not necessary.

Let $G$ be a nontrivial finite group, $p$ be the smallest prime divisor of
$|G|$ and $S$ be a Sylow $p$-subgroup of $G$. As remarked before Theorem 1.1,
a subgroup $Q$ of $S$ is pronormal in $G$ if and only if $Q$ is weakly
$\mathcal{F}_{S}(G)$-closed. Also, $G$ is $p$-nilpotent if and only if
$\mathcal{F}_{S}(G)$ is supersolvable (indeed, if $\mathcal{F}_{S}(G)$ is
supersolvable, then $G$ is $p$-nilpotent by the remark following Theorem A,
and conversely, if $G$ is $p$-nilpotent, then
$\mathcal{F}_{S}(G)=\mathcal{F}_{S}(S)$ is supersolvable). Consequently, we
can reformulate Theorem 1.2 as follows:

[25pt]0pt Let $G$ be a nontrivial finite group, $p$ be the smallest prime
dividing $|G|$ and $S$ be a noncyclic Sylow $p$-subgroup of $G$. Suppose that
there is a subgroup $D$ of $S$ with $1<D<S$ such that any subgroup of $S$ with
order $|D|$ is abelian and weakly $\mathcal{F}_{S}(G)$-closed. If $S$ is a
nonabelian $2$-group, suppose moreover that any subgroup of $S$ with order
$2|D|$ is abelian and weakly $\mathcal{F}_{S}(G)$-closed. Then
$\mathcal{F}_{S}(G)$ is supersolvable.

In view of this reformulation of Theorem 1.2, the following question naturally
arises.

######  Question 1.3.

Let $p$ be a prime number and $\mathcal{F}$ be a saturated fusion system on a
finite $p$-group $S$. Suppose that there is a subgroup $D$ of $S$ with $1<D<S$
such that any subgroup of $S$ with order $|D|$ is abelian and weakly
$\mathcal{F}$-closed. If $S$ is a nonabelian $2$-group, suppose moreover that
any subgroup of $S$ with order $2|D|$ is abelian and weakly
$\mathcal{F}$-closed. Is it true in general that $\mathcal{F}$ is
supersolvable?

Our next result positively answers Question 1.3 for the case $p=2$.

######  Theorem B.

Let $\mathcal{F}$ be a saturated fusion system on a finite $2$-group $S$.
Suppose that there is a subgroup $D$ of $S$ with $1<D<S$ such that any
subgroup of $S$ with order $|D|$ is abelian and weakly $\mathcal{F}$-closed.
If $S$ is nonabelian, suppose moreover that any subgroup of $S$ with order
$2|D|$ is abelian and weakly $\mathcal{F}$-closed. Then $\mathcal{F}$ is
supersolvable.

From [24, Proposition 1.3], we see that a saturated fusion system
$\mathcal{F}$ on a finite $2$-group $S$ is supersolvable if and only if
$\mathcal{F}$ is nilpotent, i.e. if and only if
$\mathcal{F}=\mathcal{F}_{S}(S)$. Therefore, we could replace the word
“supersolvable” by the word “nilpotent” in Theorem B.

The following result gives a positive answer to Question 1.3 for the case that
$p$ is odd and that $D$ is maximal in $S$. In fact, it shows that even more is
true.

######  Theorem C.

Let $p$ be an odd prime number and $\mathcal{F}$ be a saturated fusion system
on a finite $p$-group $S$. Suppose that any maximal subgroup of $S$ is weakly
$\mathcal{F}$-closed. If $S$ is not cyclic, suppose moreover that $S$ has more
than one abelian maximal subgroup. Then $\mathcal{F}$ is supersolvable.

In the statement of Theorem C, it would not be enough to assume that $S$ has
at least one abelian maximal subgroup when $S$ is not cyclic. This can be seen
from the following example.

######  Example 1.4.

Let $G$ be the group indexed in GAP [10] as `SmallGroup(324,160)`, $S$ be a
Sylow $3$-subgroup of $G$ and $\mathcal{F}:=\mathcal{F}_{S}(G)$. Then $S$ has
precisely four maximal subgroups, precisely one of them is abelian, and all of
them are weakly $\mathcal{F}$-closed. However, $\mathcal{F}$ is not
supersolvable. Indeed, since $G$ is solvable, the supersolvability of
$\mathcal{F}$ would imply that $G$ is $3$-supersolvable (see [24, Theorem 1.9
(b)]), but one can check by using GAP [10] that $G$ is not $3$-supersolvable.

The next result gives a complete positive answer to Question 1.3 for the case
that $p$ is odd.

######  Theorem D.

Let $p$ be an odd prime number and $\mathcal{F}$ be a saturated fusion system
on a finite $p$-group $S$. Suppose that there is a subgroup $D$ of $S$ with
$1<D<S$ such that any subgroup of $S$ with order $|D|$ is abelian and weakly
$\mathcal{F}$-closed. Then $\mathcal{F}$ is supersolvable.

We remark that our proof of Theorem D indirectly relies on the classification
of finite simple groups. More precisely, in our proof of Theorem D, we will
apply [16, Theorem 3.5], and the classification of finite simple groups was
used in the proof of that result.

We finish the introduction with the following open question that naturally
arises in view of Theorems C and D.

######  Question 1.5.

Let $p$ be an odd prime number and $\mathcal{F}$ be a saturated fusion system
on a finite $p$-group $S$. Suppose that there is a subgroup $D$ of $S$ with
$1<D<S$ such that any subgroup of $S$ with order $|D|$ is weakly
$\mathcal{F}$-closed. If $S$ is not cyclic, suppose moreover that $S$ has more
than one abelian subgroup with order $|D|$. Is it true in general that
$\mathcal{F}$ is supersolvable?

##  2 Preliminaries

In this section, we collect some lemmas needed for the proofs of our main
results.

######  Lemma 2.1.

Let $p$ be a prime number, $\mathcal{F}$ be a fusion system on a finite
$p$-group $S$ and $Q\leq R\leq S$. Suppose that $Q$ is strongly
$\mathcal{F}$-closed and that $R$ is weakly $\mathcal{F}$-closed. Then $R/Q$
is weakly $\mathcal{F}/Q$-closed.

###### Proof.

Let $\psi:R/Q\rightarrow S/Q$ be a morphism in $\mathcal{F}/Q$. We have to
show that $\psi(R/Q)=R/Q$. As an $\mathcal{F}/Q$-morphism, $\psi$ is induced
by an $\mathcal{F}$-morphism, i.e. there is a morphism
$\varphi\in\mathrm{Hom}_{\mathcal{F}}(R,S)$ such that $\psi(rQ)=\varphi(r)Q$
for all $r\in R$. Since $R$ is weakly $\mathcal{F}$-closed, we have
$\psi(R/Q)=\varphi(R)Q/Q=RQ/Q=R/Q$, as required. ∎

######  Lemma 2.2.

Let $p$ be a prime number, $\mathcal{F}$ be a fusion system on a finite
$p$-group $S$ and $Q\leq R\leq S$. Suppose that $Q\trianglelefteq\mathcal{F}$
and that $R/Q$ is strongly $\mathcal{F}/Q$-closed. Then $R$ is strongly
$\mathcal{F}$-closed.

###### Proof.

Let $P$ be a subgroup of $R$ and $\varphi:P\rightarrow S$ be a morphism in
$\mathcal{F}$. We have to show that $\varphi(P)\leq R$. Since $Q$ is normal in
$\mathcal{F}$ by hypothesis, $\varphi$ extends to a morphism
$\psi\in\mathrm{Hom}_{\mathcal{F}}(PQ,S)$ with $\psi(Q)=Q$. Let

$\overline{\psi}:PQ/Q\rightarrow S/Q,xQ\mapsto\psi(x)Q.$

Then $\overline{\psi}$ is a morphism in $\mathcal{F}/Q$. Since $R/Q$ is
strongly $\mathcal{F}/Q$-closed and $PQ/Q\leq R/Q$, it follows that
$\overline{\psi}(PQ/Q)\leq R/Q$. So, we have $\varphi(P)=\psi(P)\leq R$, as
required. ∎

######  Lemma 2.3.

Let $p$ be a prime number, $\mathcal{F}$ be a supersolvable saturated fusion
system on a finite $p$-group $S$ and $Q$ be a weakly $\mathcal{F}$-closed
subgroup of $S$. Then $Q$ is strongly $\mathcal{F}$-closed.

###### Proof.

Let $P$ be a subgroup of $Q$ and $\varphi:P\rightarrow S$ be a morphism in
$\mathcal{F}$. We have to show that $\varphi(P)\leq Q$. By [24, Proposition
2.3], $S$ is normal in $\mathcal{F}$. Hence, $\varphi$ extends to an
automorphism $\psi\in\mathrm{Aut}_{\mathcal{F}}(S)$. Since $Q$ is weakly
$\mathcal{F}$-closed, we have $\varphi(P)=\psi(P)\leq\psi(Q)=Q$, as required.
∎

######  Lemma 2.4.

Let $p$ be a prime number, $\mathcal{F}$ be a supersolvable saturated fusion
system on a finite $p$-group $S$ and $Q$ be a strongly $\mathcal{F}$-closed
subgroup of $S$. Then the following hold:

  1. (1)

There is a series

$1=S_{0}\leq S_{1}\leq\dots\leq S_{m}=S$

of subgroups of $S$ such that $S_{i}$ is strongly $\mathcal{F}$-closed for all
$0\leq i\leq m$, such that $S_{i+1}/S_{i}$ is cyclic for all $0\leq i<m$ and
such that $S_{j}=Q$ for some $0\leq j\leq m$.

  2. (2)

$\mathcal{F}/Q$ is supersolvable.

###### Proof.

By [24, Proposition 1.3], there is a $p$-supersolvable finite group $G$ with
$S\in\mathrm{Syl}_{p}(G)$ and $\mathcal{F}=\mathcal{F}_{S}(G)$. Since $S$ is
normal in $\mathcal{F}$ by [24, Proposition 2.3], we have
$\mathcal{F}=N_{\mathcal{F}}(S)=\mathcal{F}_{S}(N_{G}(S))$. Therefore, upon
replacing $G$ by $N_{G}(S)$, we may (and will) assume that $S\trianglelefteq
G$.

Since $Q$ is strongly $\mathcal{F}$-closed and $S\trianglelefteq G$, we have
$Q\trianglelefteq G$. Let $1=H_{0}\leq H_{1}\leq\dots\leq H_{n}=G$ be a chief
series of $G$ through $Q$ and $S$. Then $S=H_{m}$ for some $0\leq m\leq n$ and
$Q=H_{j}$ for some $0\leq j\leq m$.

Set $S_{i}:=H_{i}$ for all $0\leq i\leq m$. Then we have $1=S_{0}\leq
S_{1}\leq\dots\leq S_{m}=S$. For each $0\leq i\leq m$, the subgroup $S_{i}$ is
strongly $\mathcal{F}$-closed since $S_{i}\trianglelefteq G$. For each $0\leq
i<m$, the group $S_{i+1}/S_{i}$ is cyclic of order $p$ since it is a chief
factor of the $p$-supersolvable group $G$. Moreover, $S_{j}=H_{j}=Q$. The
proof of (1) is now complete.

It remains to prove (2). We have

$1=S_{j}/Q\leq S_{j+1}/Q\leq\dots\leq S_{m}/Q=S/Q.$

For each $j\leq k\leq m$, we have $S_{k}/Q\trianglelefteq G/Q$, and since
$\mathcal{F}/Q=\mathcal{F}_{S/Q}(G/Q)$, it follows that $S_{k}/Q$ is strongly
$\mathcal{F}/Q$-closed. Moreover, for each $j\leq k<m$, the quotient
$(S_{k+1}/Q)/(S_{k}/Q)\cong S_{k+1}/S_{k}$ is cyclic of order $p$. It follows
that $\mathcal{F}/Q$ is supersolvable, and so the proof of (2) is complete. ∎

To state the next lemma, which will play a key role in the proofs of all our
main results, we introduce the following notation.

######  Notation 2.5.

Let $p$ be a prime number and $\mathcal{F}$ be a saturated fusion system on a
finite $p$-group $S$. We set

$\mathcal{E}_{\mathcal{F}}^{*}:=\\{Q\leq S\mid\textnormal{$Q$ is
$\mathcal{F}$-essential, or $Q=S$}\\}.$

######  Lemma 2.6.

Let $p$ be a prime number and $\mathcal{F}$ be a saturated fusion system on a
finite $p$-group $S$. Suppose that, for each
$Q\in\mathcal{E}_{\mathcal{F}}^{*}$, the fusion system $N_{\mathcal{F}}(Q)$ is
supersolvable. Then $\mathcal{F}$ is supersolvable.

###### Proof.

Let $Q\in\mathcal{E}_{\mathcal{F}}^{*}$. Since $N_{\mathcal{F}}(Q)$ is
supersolvable,
$\mathrm{Aut}_{N_{\mathcal{F}}(Q)}(Q)=\mathrm{Aut}_{\mathcal{F}}(Q)$ is
$p$-closed by [24, Proposition 1.3]. Hence, $\mathrm{Out}_{\mathcal{F}}(Q)$ is
$p$-closed. From [4, Proposition A.7 (c)], we see that a $p$-closed finite
group cannot possess a strongly $p$-embedded subgroup. Consequently,
$\mathrm{Out}_{\mathcal{F}}(Q)$ does not possess a strongly $p$-embedded
subgroup, and so $Q$ is not $\mathcal{F}$-essential. It follows that
$\mathcal{E}_{\mathcal{F}}^{*}=\\{S\\}$. Applying [4, Part I, Proposition
4.5], we conclude that $S\trianglelefteq\mathcal{F}$. The proof is now
complete since $N_{\mathcal{F}}(S)=\mathcal{F}$ is supersolvable by
hypothesis. ∎

######  Lemma 2.7.

Let $p$ be a prime number, $\mathcal{F}$ be a saturated fusion system on a
finite $p$-group $S$ and $Q\trianglelefteq\mathcal{F}$. Suppose that there is
a series $1=Q_{0}\leq Q_{1}\leq\dots\leq Q_{n}=Q$ of subgroups of $Q$ such
that $Q_{i}$ is strongly $\mathcal{F}$-closed for all $0\leq i\leq n$ and such
that $Q_{i+1}/Q_{i}$ is cyclic for all $0\leq i<n$. Suppose moreover that
$\mathcal{F}/Q$ is supersolvable. Then $\mathcal{F}$ is supersolvable.

###### Proof.

Since $\mathcal{F}/Q$ is supersolvable, there is a series $Q=V_{0}\leq
V_{1}\leq\dots\leq V_{m}=S$ of subgroups of $S$ such that $V_{i}/Q$ is
strongly $\mathcal{F}/Q$-closed for all $0\leq i\leq m$ and such that
$(V_{i+1}/Q)/(V_{i}/Q)\cong V_{i+1}/V_{i}$ is cyclic for all $0\leq i<m$. By
Lemma 2.2, $V_{i}$ is strongly $\mathcal{F}$-closed for all $0\leq i\leq m$.
Set $S_{i}:=Q_{i}$ for all $0\leq i\leq n$ and $S_{n+\ell}:=V_{\ell}$ for all
$1\leq\ell\leq m$. Then $1=S_{0}\leq S_{1}\leq\dots\leq S_{n+m}=S$. Also,
$S_{i}$ is strongly $\mathcal{F}$-closed for all $0\leq i\leq n+m$, and
$S_{i+1}/S_{i}$ is cyclic for all $0\leq i<n+m$. Consequently, $\mathcal{F}$
is supersolvable. ∎

For the next lemma, we recall that the supersolvable hypercentre of a finite
group $G$, denoted by $Z_{\mathcal{U}}(G)$, is the largest normal subgroup of
$G$ such that every $G$-chief factor below it is cyclic.

######  Lemma 2.8.

Let $p$ be a prime number and $P$ be a nontrivial normal $p$-subgroup of a
finite group $G$. Suppose that $P$ has a subgroup $D$ with $1<D<P$ such that
every subgroup of $P$ of order $|D|$ is normal in $G$. If $P$ is a nonabelian
$2$-group and $|D|=2$, suppose moreover that every cyclic subgroup of $P$ of
order $4$ is normal in $G$. Then $P\leq Z_{\mathcal{U}}(G)$.

###### Proof.

This follows from [23]. ∎

######  Lemma 2.9.

Let $G$ be a finite group, $p$ be a prime and $S$ be a Sylow $p$-subgroup of
$G$. Suppose that, for any proper subgroup $H$ of $G$ with $O_{p}(G)<S\cap H$
and $S\cap H\in\mathrm{Syl}_{p}(H)$, the fusion system $\mathcal{F}_{S\cap
H}(H)$ is supersolvable. Suppose moreover that $O_{p}(G)\leq
Z_{\mathcal{U}}(G)$. Then $\mathcal{F}_{S}(G)$ is supersolvable.

###### Proof.

Set $\mathcal{F}:=\mathcal{F}_{S}(G)$, $\overline{G}:=G/O_{p}(G)$ and
$\overline{\mathcal{F}}:=\mathcal{F}_{\overline{S}}(\overline{G})$.

Since $O_{p}(G)\leq Z_{\mathcal{U}}(G)$, there is a series $1=U_{0}\leq
U_{1}\leq\dots\leq U_{n}=O_{p}(G)$ of subgroups of $O_{p}(G)$ such that
$U_{i}\trianglelefteq G$ for all $0\leq i\leq n$ and such that $U_{i+1}/U_{i}$
is cyclic for all $0\leq i<n$. In particular, $U_{i}$ is strongly
$\mathcal{F}$-closed for all $0\leq i\leq n$. So, by Lemma 2.7, it suffices to
show that $\mathcal{F}/O_{p}(G)=\overline{\mathcal{F}}$ is supersolvable. This
is clear when $S=O_{p}(G)$, and so we assume that $O_{p}(G)\neq S$.

Let $O_{p}(G)<Q\leq S$ such that $\overline{Q}$ is fully
$\overline{\mathcal{F}}$-normalized. Then
$N_{\overline{S}}(\overline{Q})=\overline{N_{S}(Q)}$ is a Sylow $p$-subgroup
of $N_{\overline{G}}(\overline{Q})=\overline{N_{G}(Q)}$. Hence, $N_{S}(Q)$ is
a Sylow $p$-subgroup of $N_{G}(Q)$. Since $O_{p}(G)<Q$, we have $N_{G}(Q)<G$.
Moreover, $O_{p}(G)<Q\leq S\cap N_{G}(Q)$, and $S\cap N_{G}(Q)=N_{S}(Q)$ is a
Sylow $p$-subgroup of $N_{G}(Q)$. So, by hypothesis, the fusion system
$N_{\mathcal{F}}(Q)=\mathcal{F}_{N_{S}(Q)}(N_{G}(Q))$ is supersolvable. Lemma
2.4 (2) implies that
$N_{\mathcal{F}}(Q)/O_{p}(G)=\mathcal{F}_{\overline{N_{S}(Q)}}(\overline{N_{G}(Q)})=N_{\overline{\mathcal{F}}}(\overline{Q})$
is supersolvable.

Now, let $O_{p}(G)\leq R\leq S$ such that
$\overline{R}\in\mathcal{E}_{\overline{\mathcal{F}}}^{*}$. Then $\overline{R}$
is $\overline{\mathcal{F}}$-centric and fully
$\overline{\mathcal{F}}$-normalized. Since $\overline{S}\neq 1$, we have
$\overline{R}\neq 1$ and thus $O_{p}(G)<R$, and the preceding paragraph
implies that $N_{\overline{\mathcal{F}}}(\overline{R})$ is supersolvable.
Lemma 2.6 yields that $\overline{\mathcal{F}}$ is supersolvable, as required.
∎

######  Lemma 2.10.

Let $p$ be a prime number and $\mathcal{F}$ be a saturated fusion system on a
finite $p$-group $S$. Suppose that, whenever $R$ is a subgroup of $S$ with
$O_{p}(\mathcal{F})<R$, any proper saturated subsystem of $\mathcal{F}$ on $R$
is supersolvable. Suppose moreover that there is a series $1=S_{0}\leq
S_{1}\leq\dots\leq S_{n}=O_{p}(\mathcal{F})$ of subgroups of
$O_{p}(\mathcal{F})$ such that $S_{i}$ is strongly $\mathcal{F}$-closed for
all $0\leq i\leq n$ and such that $S_{i+1}/S_{i}$ is cyclic for all $0\leq
i<n$. Then $\mathcal{F}$ is supersolvable.

###### Proof.

Set $\overline{S}:=S/O_{p}(\mathcal{F})$ and
$\overline{\mathcal{F}}:=\mathcal{F}/O_{p}(\mathcal{F})$. By Lemma 2.7, it
suffices to show that $\overline{\mathcal{F}}$ is supersolvable. This is clear
when $O_{p}(\mathcal{F})=S$, and so we assume that $O_{p}(\mathcal{F})\neq S$.

Let $O_{p}(\mathcal{F})<Q\leq S$ such that $\overline{Q}$ is fully
$\overline{\mathcal{F}}$-normalized. Then, by [9, Proposition 5.58 (iii)], $Q$
is fully $\mathcal{F}$-normalized. Hence, $N_{\mathcal{F}}(Q)$ is a saturated
subsystem of $\mathcal{F}$ on $N_{S}(Q)$. Since $O_{p}(\mathcal{F})<Q$, we
have $O_{p}(\mathcal{F})<N_{S}(Q)$ and $N_{\mathcal{F}}(Q)\neq\mathcal{F}$.
So, by hypothesis, the fusion system $N_{\mathcal{F}}(Q)$ is supersolvable.
Lemma 2.4 (2) implies that $N_{\mathcal{F}}(Q)/O_{p}(\mathcal{F})$ is
supersolvable. By [9, Exercise 5.11], we have
$N_{\mathcal{F}}(Q)/O_{p}(\mathcal{F})=N_{\overline{\mathcal{F}}}(\overline{Q})$,
and so $N_{\overline{\mathcal{F}}}(\overline{Q})$ is supersolvable.

Let $O_{p}(\mathcal{F})\leq R\leq S$ such that
$\overline{R}\in\mathcal{E}_{\overline{\mathcal{F}}}^{*}$. Then $\overline{R}$
is fully $\overline{\mathcal{F}}$-normalized and
$\overline{\mathcal{F}}$-centric. Since $\overline{S}\neq 1$, we have
$\overline{R}\neq 1$ and thus $O_{p}(\mathcal{F})<R$, and the preceding
paragraph implies that $N_{\overline{\mathcal{F}}}(\overline{R})$ is
supersolvable. Lemma 2.6 yields that $\overline{\mathcal{F}}$ is
supersolvable, as required. ∎

######  Lemma 2.11.

Let $p$ be an odd prime number and $SL_{2}(p)\leq H\leq GL_{2}(p)$. Let $V$
denote the group consisting of all row vectors
$\begin{pmatrix}x&y\end{pmatrix}$, where $x,y\in\mathbb{F}_{p}$, together with
the componentwise addition. Moreover, let $G=V\rtimes H$ be the outer
semidirect product of $V$ and $H$ with respect to the natural action of $H$ on
$V$, i.e. $G$ is the group consisting of all pairs $(h,v)$ with $h\in H$ and
$v\in V$, together with the multiplication given by

$(h_{1},v_{1})\cdot(h_{2},v_{2}):=(h_{1}h_{2},v_{1}h_{2}+v_{2})$

for all $h_{1},h_{2}\in H$, $v_{1},v_{2}\in V$. Let

$U:=\left\\{\begin{pmatrix}1&a\\\ 0&1\end{pmatrix}\ :\
a\in\mathbb{F}_{p}\right\\}\leq SL_{2}(p)\leq H$

and

$S:=\\{(u,v)\ :\ u\in U,v\in V\\}.$

Then $S$ is a Sylow $p$-subgroup of $G$, and there is a maximal subgroup of
$S$ which is not weakly $\mathcal{F}_{S}(G)$-closed.

###### Proof.

Since $|SL_{2}(p)|=(p^{2}-1)p$ by [14, Kapitel II, Hilfssatz 6.2 (2)] and
since $|H:SL_{2}(p)|$ is not divisible by $p$, the largest power of $p$
dividing $|H|$ is $p$. Therefore, $p^{3}$ is the largest power of $p$ dividing
$|G|$. Clearly, $S$ is a subgroup of $G$ of order $p^{3}$. Thus
$S\in\mathrm{Syl}_{p}(G)$.

Let

$A:=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}\in U,\ \ \ \
B:=I_{2}=\begin{pmatrix}1&0\\\ 0&1\end{pmatrix}\in U,$

and

$a:=\begin{pmatrix}1&0\end{pmatrix}\in V,\ \ \
\,b:=\begin{pmatrix}0&1\end{pmatrix}\in V.$

Set $r:=(A,a)\in S$ and $s:=(B,b)\in S$. A direct calculation shows that
$|\langle r\rangle|=|\langle s\rangle|=p$, $rs=sr$ and $\langle
r\rangle\cap\langle s\rangle=1$. Therefore, $S_{1}:=\langle r,s\rangle=\langle
r\rangle\langle s\rangle$ has order $p^{2}$, whence $S_{1}$ is maximal in $S$.

We show now that $S_{1}$ is not weakly $\mathcal{F}_{S}(G)$-closed. Let

$M:=\begin{pmatrix}-1&1\\\ 0&-1\end{pmatrix}\in SL_{2}(p)\leq H$

and $g:=(M,0_{V})\in G$. Then

$g^{-1}rg=(A,b-a)\in S$

and

$g^{-1}sg=(B,-b)\in S.$

Thus $g^{-1}S_{1}g=\langle g^{-1}rg,g^{-1}sg\rangle\leq S$. Assume that
$S_{1}=g^{-1}S_{1}g$. Then $g^{-1}rg\in S_{1}$, and so there exist $0\leq
i,j<p$ with $g^{-1}rg=r^{i}s^{j}$. Then $A=A^{i}$ and hence $i=1$. It follows
that $(A,b-a)=g^{-1}rg=rs^{j}=(A,a+jb)$, whence
$\begin{pmatrix}-1&1\end{pmatrix}=b-a=a+jb=\begin{pmatrix}1&j\end{pmatrix}$
and thus $-1=1$. This is a contradiction since $p$ is odd. So we have
$S_{1}\neq g^{-1}S_{1}g\leq S$, and this shows that $S_{1}$ is not weakly
$\mathcal{F}_{S}(G)$-closed. ∎

The next lemma is certainly well-known, but we include a proof for the
convenience of the reader.

######  Lemma 2.12.

Let $p$ be an odd prime number and $n\in\\{2,3\\}$. Then the following hold:

  1. (1)

The Sylow $p$-subgroups of $GL_{n}(p)$ have exponent $p$.

  2. (2)

If $H$ is a nontrivial cyclic $p$-subgroup of $GL_{n}(p)$, then $H\cong
C_{p}$.

###### Proof.

By [14, Kapitel II, Hilfssatz 6.2 (1)], we have
$|GL_{2}(p)|=(p^{2}-1)(p^{2}-p)=p(p^{2}-1)(p-1)$. Hence, the Sylow
$p$-subgroups of $GL_{2}(p)$ have order $p$. In particular, they have exponent
$p$, and so (1) holds for $n=2$.

Again by [14, Kapitel II, Hilfssatz 6.2 (1)], we have
$|GL_{3}(p)|=(p^{3}-1)(p^{3}-p)(p^{3}-p^{2})=p^{3}(p^{3}-1)(p^{2}-1)(p-1)$.
Hence, the Sylow $p$-subgroups of $GL_{3}(p)$ have order $p^{3}$. Set

$U:=\left\\{\begin{pmatrix}1&a&b\\\ 0&1&c\\\ 0&0&1\end{pmatrix}\ :\
a,b,c\in\mathbb{F}_{p}\right\\}.$

Then $U$ is easily seen to be a subgroup of $GL_{3}(p)$ with order $p^{3}$. In
other words, $U$ is a Sylow $p$-subgroup of $GL_{3}(p)$. Let
$a,b,c\in\mathbb{F}_{p}$ and

$x:=\begin{pmatrix}1&a&b\\\ 0&1&c\\\ 0&0&1\end{pmatrix}\in U.$

One can show by induction that

$x^{k}=\begin{pmatrix}1&ka&kb+\frac{(k-1)k}{2}ac\\\ 0&1&kc\\\
0&0&1\end{pmatrix}$

for every positive integer $k$. In particular, $x^{p}$ is the identity matrix
$I_{3}$. Since $x$ was an arbitrarily chosen element of $U$, it follows that
$U$ has exponent $p$. Hence, any Sylow $p$-subgroup of $GL_{3}(p)$ has
exponent $p$, and so (1) holds for $n=3$.

Now, let $n\in\\{2,3\\}$ and $H$ be a nontrivial cyclic $p$-subgroup of
$GL_{n}(p)$. Then $H=\langle y\rangle$ for some nontrivial $p$-element $y$ of
$GL_{n}(p)$. By (1), $y$ has order $p$. Thus $H=\langle y\rangle\cong C_{p}$,
and so (2) holds. ∎

Recall that a group $G$ is called minimal nonnilpotent if any proper subgroup
of $G$ is nilpotent, while $G$ itself is not nilpotent.

######  Lemma 2.13.

Let $G$ be a minimal nonnilpotent finite group and $S$ be a nonnormal Sylow
subgroup of $G$. Then $S$ is cyclic.

###### Proof.

By [14, Kapitel III, Satz 5.2], we have $|G|=p^{a}q^{b}$ with distinct prime
numbers $p$, $q$ and positive integers $a$, $b$, where $G$ has a normal Sylow
$p$-subgroup and cyclic Sylow $q$-subgroups. Since $S$ is not normal in $G$,
we have that $S$ is a Sylow $q$-subgroup of $G$. Consequently, $S$ is cyclic.
∎

The next lemma is due to Oliver [20]. It will play an important role in the
proof of Theorem C.

######  Lemma 2.14.

([20, Lemma 1.11]) Let $p$ be a prime number, $A$ be finite abelian $p$-group
and $G$ be a subgroup of $\mathrm{Aut}(A)$. Suppose that the following hold:

  1. (1)

The Sylow $p$-subgroups of $G$ have order $p$ and are not normal in $G$.

  2. (2)

For each $x\in G$ of order $p$, the group $[x,A]$ has order $p$.

Set $H:=O^{p^{\prime}}(G)$, $A_{1}:=C_{A}(H)$ and $A_{2}:=[H,A]$. Then $G$
normalizes $A_{1}$ and $A_{2}$, $A=A_{1}\times A_{2}$, and $A_{2}\cong
C_{p}\times C_{p}$.

Let $p$ be a prime number and $S$ be a finite $p$-group. Following [7, §65],
we say that $S$ is an $\mathcal{A}_{2}$-group if $S$ contains a nonabelian
maximal subgroup, while any subgroup of $S$ with index $p^{2}$ is abelian.

######  Lemma 2.15.

Let $p$ be a prime number and $S$ be a finite $p$-group. Suppose that $S$ is a
nonmetacyclic $\mathcal{A}_{2}$-group and that $|S|>p^{4}$. Then the following
hold:

  1. (1)

([7, Proposition 71.4]) If $S$ possesses precisely one abelian maximal
subgroup and if $S^{\prime}\leq Z(S)$, then $Z(S)=\Phi(S)$.

  2. (2)

([7, Proposition 71.5]) If any maximal subgroup of $S$ is nonabelian and if
$p$ is odd, then $|S|=p^{5}$.

##  3 Proofs of the main results

###### Proof of Theorem A.

Suppose that the theorem is false, and let $G$ be a minimal counterexample. We
will derive a contradiction in several steps. Set
$\mathcal{F}:=\mathcal{F}_{S}(G)$.

(1) Let $H<G$ with $S\cap H\in\mathrm{Syl}_{p}(H)$ and $|S\cap H|\geq p|D|$.
Then $\mathcal{F}_{S\cap H}(H)$ is supersolvable.

By hypothesis, any subgroup of $S\cap H$ with order $|D|$ or $p|D|$ is abelian
and weakly pronormal in $G$. Applying [3, Lemma 2.2 (2)], we conclude that any
subgroup of $S\cap H$ with order $|D|$ or $p|D|$ is abelian and weakly
pronormal in $H$. The minimality of $G$ implies that $\mathcal{F}_{S\cap
H}(H)$ is supersolvable.

(2) Let $Q\in\mathcal{E}_{\mathcal{F}}^{*}$. Then $|Q|\geq p|D|$, and if $Q$
is not normal in $G$, then $N_{\mathcal{F}}(Q)$ is supersolvable.

Suppose that $|Q|<p|D|$. Then there is a subgroup $R$ of $S$ such that
$|R|=p|D|$ and $Q<R$. By hypothesis, $R$ is abelian, and so we have $R\leq
C_{S}(Q)$. As a member of $\mathcal{E}_{\mathcal{F}}^{*}$, the subgroup $Q$ is
$\mathcal{F}$-centric. It follows that $R\leq C_{S}(Q)\leq Q$, a
contradiction. Thus $|Q|\geq p|D|$.

Suppose now that $Q$ is not normal in $G$. Hence, $N_{G}(Q)$ is a proper
subgroup of $G$. As $Q$ is fully $\mathcal{F}$-normalized, we have $S\cap
N_{G}(Q)=N_{S}(Q)\in\mathrm{Syl}_{p}(N_{G}(Q))$. Also, $|N_{S}(Q)|\geq|Q|\geq
p|D|$. Now, (1) implies that
$N_{\mathcal{F}}(Q)=\mathcal{F}_{N_{S}(Q)}(N_{G}(Q))$ is supersolvable.

(3) $|O_{p}(G)|\geq p|D|$.

Assume that there is no $Q\in\mathcal{E}_{\mathcal{F}}^{*}$ with
$Q\trianglelefteq G$. Then, for each $Q\in\mathcal{E}_{\mathcal{F}}^{*}$, the
fusion system $N_{\mathcal{F}}(Q)$ is supersolvable by (2), and Lemma 2.6
implies that $\mathcal{F}$ is supersolvable. This contradiction shows that
there is some $Q\in\mathcal{E}_{\mathcal{F}}^{*}$ with $Q\trianglelefteq G$.
Applying (2), we conclude that $|O_{p}(G)|\geq|Q|\geq p|D|$.

(4) $O_{p}(G)\leq Z_{\mathcal{U}}(G)$.

Let $U$ be a subgroup of $O_{p}(G)$ with order $|D|$ or $p|D|$. By hypothesis,
$U$ is weakly pronormal in $G$. Hence, there is a subgroup $K$ of $G$ such
that $G=UK$ and such that $U\cap K$ is pronormal in $G$. By [19, Theorem 4.3],
$U\cap K$ is weakly $\mathcal{F}$-closed. For each $g\in G$, we have $(U\cap
K)^{g}\leq O_{p}(G)\leq S$, and so it follows that $(U\cap K)^{g}=U\cap K$.
Consequently, $U\cap K$ is normal in $G$, and therefore, $U$ is
$c$-supplemented in $G$ in the sense of [2, 5].

Since $U$ was arbitrarily chosen, any subgroup of $O_{p}(G)$ with order $|D|$
or $p|D|$ is $c$-supplemented in $G$. Applying [2, Theorem 3.2 and Corollary
3.4], we conclude that $O_{p}(G)\leq Z_{\mathcal{U}}(G)$.

(5) The final contradiction.

If $H$ is a proper subgroup of $G$ with $O_{p}(G)<S\cap H$ and $S\cap
H\in\mathrm{Syl}_{p}(H)$, then $\mathcal{F}_{S\cap H}(H)$ is supersolvable by
(1) and (3). Also, $O_{p}(G)\leq Z_{\mathcal{U}}(G)$ by (4). So $\mathcal{F}$
is supersolvable by Lemma 2.9. This contradiction completes the proof. ∎

###### Proof of Theorem B.

Suppose that the theorem is false, and let $\mathcal{F}$ be a counterexample
such that $|\mathcal{F}|$, the number of morphisms in $\mathcal{F}$, is
minimal. We will derive a contradiction in several steps. In some parts of the
proof, we argue similarly as in the proof of Theorem A.

(1) Let $R$ be a subgroup of $S$ with $|R|\geq 2|D|$ and $\mathcal{F}_{0}$ be
a proper saturated subsystem of $\mathcal{F}$ on $R$. Then $\mathcal{F}_{0}$
is supersolvable.

By hypothesis, any subgroup of $R$ with order $|D|$ is abelian and weakly
$\mathcal{F}$-closed, and if $R$ is nonabelian, then we moreover have that any
subgroup of $R$ with order $2|D|$ is abelian and weakly $\mathcal{F}$-closed.
Clearly, any weakly $\mathcal{F}$-closed subgroup of $R$ is weakly
$\mathcal{F}_{0}$-closed. Consequently, the fusion system $\mathcal{F}_{0}$
satisfies the hypotheses of the theorem, and the minimality of $\mathcal{F}$
implies that $\mathcal{F}_{0}$ is supersolvable.

(2) Let $Q\in\mathcal{E}_{\mathcal{F}}^{*}$. Then $|Q|\geq 2|D|$, and if $Q$
is not normal in $\mathcal{F}$, then $N_{\mathcal{F}}(Q)$ is supersolvable.

Suppose that $|Q|<2|D|$. Then there is a subgroup $R$ of $S$ such that
$|R|=2|D|$ and $Q<R$. By hypothesis, $R$ is abelian, and so we have $R\leq
C_{S}(Q)$. As a member of $\mathcal{E}_{\mathcal{F}}^{*}$, the subgroup $Q$ is
$\mathcal{F}$-centric. It follows that $R\leq C_{S}(Q)\leq Q$, a
contradiction. Thus $|Q|\geq 2|D|$.

Suppose now that $Q$ is not normal in $\mathcal{F}$. Then $N_{\mathcal{F}}(Q)$
is a proper saturated subsystem of $\mathcal{F}$ on $N_{S}(Q)$. Since
$|N_{S}(Q)|\geq|Q|\geq 2|D|$, it follows from (1) that $N_{\mathcal{F}}(Q)$ is
supersolvable.

(3) $|O_{2}(\mathcal{F})|\geq 2|D|$.

Assume that there is no $Q\in\mathcal{E}_{\mathcal{F}}^{*}$ with
$Q\trianglelefteq\mathcal{F}$. Then, for each
$Q\in\mathcal{E}_{\mathcal{F}}^{*}$, the fusion system $N_{\mathcal{F}}(Q)$ is
supersolvable by (2), and Lemma 2.6 implies that $\mathcal{F}$ is
supersolvable. This contradiction shows that there is some
$Q\in\mathcal{E}_{\mathcal{F}}^{*}$ with $Q\trianglelefteq\mathcal{F}$.
Applying (2), we conclude that $|O_{2}(\mathcal{F})|\geq|Q|\geq 2|D|$.

(4) There is a series $1=S_{0}\leq S_{1}\leq\dots\leq
S_{n}=O_{2}(\mathcal{F})$ of subgroups of $O_{2}(\mathcal{F})$ such that
$S_{i}$ is strongly $\mathcal{F}$-closed for all $0\leq i\leq n$ and such that
$S_{i+1}/S_{i}$ is cyclic for all $0\leq i<n$.

Let $N:=O_{2}(\mathcal{F})$, $A:=\mathrm{Aut}_{\mathcal{F}}(N)$ and
$G:=N\rtimes A$ be the outer semidirect product of $N$ and $A$ with respect to
the natural action of $A$ on $N$. Identifying each element of $N$ with its
canonical image in $G$, we may regard $N$ as a normal subgroup of $G$.
Likewise, identifying each element of $A$ with its canonical image in $G$, we
may regard $A$ as a subgroup of $G$. Then $G=NA$, $N\cap A=1$, and for all
$x\in N$ and $\alpha\in A$, we have $\alpha^{-1}x\alpha=\alpha(x)$.

Let $P\leq N$ such that $P$ is weakly $\mathcal{F}$-closed. We show that
$P\trianglelefteq G$. Since $P$ is weakly $\mathcal{F}$-closed, we have
$P\trianglelefteq S$ and thus $N\leq N_{G}(P)$. For all $\alpha\in A$, we have
$\alpha^{-1}P\alpha=\alpha(P)=P$, where the latter equality holds since $P$ is
weakly $\mathcal{F}$-closed. Thus $A\leq N_{G}(P)$, and it follows that
$G=NA\leq N_{G}(P)$. So we have $P\trianglelefteq G$, as wanted.

By the preceding paragraph and by hypothesis, any subgroup of $N$ with order
$|D|$ is normal in $G$, and if $N$ is nonabelian, then we moreover have that
any subgroup of $N$ with order $2|D|$ is normal in $G$. Lemma 2.8 implies that
$N\leq Z_{\mathcal{U}}(G)$. Hence, there is a series $1=S_{0}\leq
S_{1}\leq\dots\leq S_{n}=N$ of subgroups of $N$ such that
$S_{i}\trianglelefteq G$ for all $0\leq i\leq n$ and such that $S_{i+1}/S_{i}$
is cyclic for all $0\leq i<n$. To complete the proof of (4), we show that
$S_{i}$ is strongly $\mathcal{F}$-closed for all $0\leq i\leq n$. Let $P$ be a
subgroup of $S_{i}$ for some $0\leq i\leq n$, and let $P_{1}$ be a subgroup of
$S$ such that there is an $\mathcal{F}$-isomorphism $\varphi:P\rightarrow
P_{1}$. Clearly, $N$ is strongly $\mathcal{F}$-closed, and therefore, we have
$P_{1}\leq N$. Since $N$ is normal in $\mathcal{F}$, the isomorphism $\varphi$
extends to an automorphism $\alpha\in\mathrm{Aut}_{\mathcal{F}}(N)=A$. We have
$P_{1}=\varphi(P)=\alpha(P)\leq\alpha(S_{i})=\alpha^{-1}S_{i}\alpha=S_{i}$,
where the last equality holds since $S_{i}$ is normal in $G$. We have shown
that each $\mathcal{F}$-conjugate of $P$ lies in $S_{i}$. Therefore, $S_{i}$
is strongly $\mathcal{F}$-closed, as required.

(5) The final contradiction.

Applying Lemma 2.10, we deduce from (1), (3) and (4) that $\mathcal{F}$ is
supersolvable. This contradiction completes the proof. ∎

######  Remark 3.1.

Another proof of Theorem B works as follows: One assumes that Theorem B is
false and considers a counterexample $\mathcal{F}$ such that $|\mathcal{F}|$
is minimal. Arguing as in the above proof of Theorem B, one shows that there
is some $Q\in\mathcal{E}_{\mathcal{F}}^{*}$ with
$Q\trianglelefteq\mathcal{F}$. Then $\mathcal{F}$ is constrained, and the
model theorem [4, Part III, Theorem 5.10] implies that $\mathcal{F}$ is the
$2$-fusion system of a finite group $G$. Theorem 1.2 shows that $G$ is
$2$-nilpotent, and so it follows that $\mathcal{F}$ is nilpotent, a
contradiction. This proof of Theorem B has the advantage of being shorter than
the above proof of Theorem B, but it has the disadvantage of being less
elementary since it depends on the model theorem, which is a quite deep
result.

###### Proof of Theorem C.

Suppose that the theorem is false, and let $\mathcal{F}$ be a counterexample
such that $|\mathcal{F}|$ is minimal. We will derive a contradiction in
several steps.

(1) $S$ is not normal in $\mathcal{F}$.

Assume that $S\trianglelefteq\mathcal{F}$. We claim that there is a $p$-closed
finite group $G$ with $S\in\mathrm{Syl}_{p}(G)$ and
$\mathcal{F}=\mathcal{F}_{S}(G)$. Indeed, the normality of $S$ in
$\mathcal{F}$ implies that $\mathcal{F}$ is constrained, and so the existence
of such a group $G$ follows from the model theorem [4, Part III, Theorem
5.10]. Alternatively, without using the model theorem, one can argue as
follows: Let $A:=\mathrm{Aut}_{\mathcal{F}}(S)$. Since $\mathcal{F}$ is
saturated, $\mathrm{Inn}(S)$ is a Sylow $p$-subgroup of $A$, whence
$|\mathrm{Inn}(S)|$ and $|A:\mathrm{Inn}(S)|$ are coprime. Since
$\mathrm{Inn}(S)\trianglelefteq A$, the Schur-Zassenhaus theorem [15, Theorem
3.8] implies that $\mathrm{Inn}(S)$ has a complement $H$ in $A$. Let
$G=S\rtimes H$ be the outer semidirect product of $S$ and $H$ with respect to
the natural action of $H$ on $S$. Identifying each element of $S$ with its
canonical image in $G$, we may regard $S$ as a normal subgroup of $G$. Then we
have $S\in\mathrm{Syl}_{p}(G)$ since $|H|$ is not divisible by $p$, and it is
not hard to show that $\mathcal{F}=\mathcal{F}_{S}(G)$.

Since $S$ is normal in $G$ and since any maximal subgroup of $S$ is weakly
$\mathcal{F}$-closed, we have that any maximal subgroup of $S$ is normal in
$G$. Lemma 2.8 implies that $S\leq Z_{\mathcal{U}}(G)$. Hence, there is a
series $1=S_{0}\leq S_{1}\leq\dots\leq S_{m}=S$ of subgroups of $S$ such that
$S_{i}\trianglelefteq G$ for all $0\leq i\leq m$ and such that $S_{i+1}/S_{i}$
is cyclic for all $0\leq i<m$. In particular, $S_{i}$ is strongly
$\mathcal{F}$-closed for all $0\leq i\leq m$, and so it follows that
$\mathcal{F}$ is supersolvable. This contradiction shows that $S$ is not
normal in $\mathcal{F}$.

(2) $S$ possesses more than one abelian maximal subgroup, and any abelian
maximal subgroup of $S$ contains $Z(S)$. Moreover, we have $|S^{\prime}|=p$
and $|S:Z(S)|=p^{2}$.

Assume that $S$ is abelian. Then $S\trianglelefteq\mathcal{F}$ by [4, Part I,
Corollary 4.7 (a)], which contradicts (1). So $S$ must be nonabelian. In
particular, $S$ is not cyclic, and so $S$ has more than one abelian maximal
subgroup by hypothesis. If $R$ is an abelian maximal subgroup of $S$, then
$Z(S)\leq R$ because otherwise $S=RZ(S)$ would be abelian. We see from [20,
Lemma 1.9] that $|S^{\prime}|=p$ and $|S:Z(S)|=p^{2}$.

(3) $O_{p}(\mathcal{F})$ is an abelian maximal subgroup of $S$, we have
$\mathcal{E}_{\mathcal{F}}^{*}=\\{O_{p}(\mathcal{F}),S\\}$, and the subgroups
$O_{p}(\mathcal{F})$ and $S$ are precisely the $\mathcal{F}$-centric,
$\mathcal{F}$-radical subgroups of $S$.

First, we argue that any member of
$\mathcal{E}_{\mathcal{F}}^{*}\setminus\\{S\\}$ is an abelian maximal subgroup
of $S$. Let $Q\in\mathcal{E}_{\mathcal{F}}^{*}\setminus\\{S\\}$. Then, since
$Q$ is $\mathcal{F}$-centric, we have $Z(S)<Q<S$. As $|S:Z(S)|=p^{2}$ by (2),
it follows that $Q$ is a maximal subgroup of $S$. Also, $Q$ is abelian since
$|Q:Z(S)|=p$.

Let $Q\in\mathcal{E}_{\mathcal{F}}^{*}$ such that $Q$ is not normal in
$\mathcal{F}$. Then $N_{\mathcal{F}}(Q)$ is a proper saturated subsystem of
$\mathcal{F}$ on $S$, and it is easy to see that $N_{\mathcal{F}}(Q)$
satisfies the hypotheses of the theorem. So $N_{\mathcal{F}}(Q)$ is
supersolvable by the minimality of $\mathcal{F}$. If no member of
$\mathcal{E}_{\mathcal{F}}^{*}$ is normal in $\mathcal{F}$, then
$N_{\mathcal{F}}(Q)$ is supersolvable for each
$Q\in\mathcal{E}_{\mathcal{F}}^{*}$, whence $\mathcal{F}$ is supersolvable by
Lemma 2.6. Since this is not the case, there must exist some
$Q\in\mathcal{E}_{\mathcal{F}}^{*}$ with $Q\trianglelefteq\mathcal{F}$. We
have $Q\neq S$ by (1), and so the preceding paragraph implies that $Q$ is an
abelian maximal subgroup of $S$. Since $Q\leq O_{p}(\mathcal{F})$ and
$O_{p}(\mathcal{F})\neq S$, we have $O_{p}(\mathcal{F})=Q$. Hence, we have
shown that $O_{p}(\mathcal{F})$ is an abelian maximal subgroup of $S$.

Clearly, $S$ is $\mathcal{F}$-centric and $\mathcal{F}$-radical. Since
$O_{p}(\mathcal{F})$ is $\mathcal{F}$-essential, we also have that
$O_{p}(\mathcal{F})$ is $\mathcal{F}$-centric and $\mathcal{F}$-radical (see
[4, Part I, Proposition 3.3 (a)]). Conversely, if $R$ is an
$\mathcal{F}$-centric, $\mathcal{F}$-radical subgroup of $S$, then
$O_{p}(\mathcal{F})\leq R$ by [4, Part I, Proposition 4.5], and the maximality
of $O_{p}(\mathcal{F})$ in $S$ implies that $R=O_{p}(\mathcal{F})$ or $R=S$.
Consequently, the subgroups $O_{p}(\mathcal{F})$ and $S$ are precisely the
$\mathcal{F}$-centric, $\mathcal{F}$-radical subgroups of $S$.

Since any member of $\mathcal{E}_{\mathcal{F}}^{*}$ is $\mathcal{F}$-centric
and $\mathcal{F}$-radical, it follows that
$\mathcal{E}_{\mathcal{F}}^{*}\subseteq\\{O_{p}(\mathcal{F}),S\\}$. The other
inclusion also holds, and so we have
$\mathcal{E}_{\mathcal{F}}^{*}=\\{O_{p}(\mathcal{F}),S\\}$.

(4) There is no nontrivial subgroup of $Z(S)$ which is normal in
$\mathcal{F}$.

Assume that there is a subgroup $1\neq Z\leq Z(S)$ with
$Z\trianglelefteq\mathcal{F}$. From Lemma 2.1, we see that any maximal
subgroup of $S/Z$ is weakly $\mathcal{F}/Z$-closed. By (2), $S$ possesses more
than one abelian maximal subgroup, and each of them contains $Z$. Hence, $S/Z$
has more than one abelian maximal subgroup. Consequently, the fusion system
$\mathcal{F}/Z$ satisfies the hypotheses of the theorem, and so
$\mathcal{F}/Z$ is supersolvable by the minimality of $\mathcal{F}$.

Let $R_{1}$ and $R_{2}$ be two distinct abelian maximal subgroups of $S$.
Since $R_{1}/Z$ and $R_{2}/Z$ are weakly $\mathcal{F}/Z$-closed and since
$\mathcal{F}/Z$ is supersolvable, we see from Lemma 2.3 that $R_{1}/Z$ and
$R_{2}/Z$ are in fact strongly $\mathcal{F}/Z$-closed. Lemma 2.2 implies that
$R_{1}$ and $R_{2}$ are strongly $\mathcal{F}$-closed. Since $R_{1}$ and
$R_{2}$ are abelian, it follows from [4, Part I, Corollary 4.7 (a)] that
$R_{1}$ and $R_{2}$ are normal in $\mathcal{F}$. Hence, $S=R_{1}R_{2}$ is
normal in $\mathcal{F}$, which contradicts (1). Therefore, there is no
subgroup $1\neq Z\leq Z(S)$ with $Z\trianglelefteq\mathcal{F}$.

(5) The Sylow $p$-subgroups of
$\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F}))$ have order $p$ and are not
normal in $\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F}))$.

As $\mathcal{F}$ is saturated and $O_{p}(\mathcal{F})$ is fully
$\mathcal{F}$-normalized, $\mathrm{Aut}_{S}(O_{p}(\mathcal{F}))$ is a Sylow
$p$-subgroup of $\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F}))$. Since
$O_{p}(\mathcal{F})$ is $\mathcal{F}$-centric, abelian and maximal in $S$ by
(3), we have $\mathrm{Aut}_{S}(O_{p}(\mathcal{F}))\cong
S/O_{p}(\mathcal{F})\cong C_{p}$. Since $O_{p}(\mathcal{F})$ is
$\mathcal{F}$-radical by (3), we also have
$1=\mathrm{Inn}(O_{p}(\mathcal{F}))=O_{p}(\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F})))$.
Hence, $\mathrm{Aut}_{S}(O_{p}(\mathcal{F}))$ is not normal in
$\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F}))$.

(6) Assume that $\alpha\in\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F}))$ has
order $p$. Then $[\alpha,O_{p}(\mathcal{F})]$ has order $p$.

Since $\mathrm{Aut}_{S}(O_{p}(\mathcal{F}))$ is a Sylow $p$-subgroup of
$\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F}))$, there is some
$\beta\in\mathrm{Aut}_{\mathcal{F}}(O_{p}(\mathcal{F}))$ such that
$\gamma:=\beta^{-1}\alpha\beta\in\mathrm{Aut}_{S}(O_{p}(\mathcal{F}))$. A
direct calculation shows that
$[\gamma,O_{p}(\mathcal{F})]=\beta([\alpha,O_{p}(\mathcal{F})])$, whence
$|[\gamma,O_{p}(\mathcal{F})]|=|[\alpha,O_{p}(\mathcal{F})]|$. Therefore, we
may assume without loss of generality that
$\alpha\in\mathrm{Aut}_{S}(O_{p}(\mathcal{F}))$. Then there is some $s\in S$
with $\alpha(x)=x^{s}$ for all $x\in O_{p}(\mathcal{F})$, and we have
$[\alpha,O_{p}(\mathcal{F})]=[s,O_{p}(\mathcal{F})]\leq S^{\prime}$. Since
$S^{\prime}$ has order $p$ by (2) and since $\alpha$ acts nontrivially on
$O_{p}(\mathcal{F})$, it follows that $[\alpha,O_{p}(\mathcal{F})]$ has order
$p$, as wanted.

(7) $S$ is extraspecial of order $p^{3}$ and exponent $p$.

Set $A:=O_{p}(\mathcal{F})$, $G:=\mathrm{Aut}_{\mathcal{F}}(A)$,
$H:=O^{p^{\prime}}(G)$, $A_{1}:=C_{A}(H)$ and $A_{2}:=[H,A]$. By (3), (5), (6)
and Lemma 2.14, $G$ normalizes $A_{1}$ and $A_{2}$, $A=A_{1}\times A_{2}$ and
$A_{2}\cong C_{p}\times C_{p}$.

Since $\mathrm{Aut}_{S}(A)$ is a Sylow $p$-subgroup of $G$, we have
$\mathrm{Aut}_{S}(A)\leq H$. Consequently, $\mathrm{Aut}_{S}(A)$ centralizes
$A_{1}$. In other words, we have $A_{1}\leq Z(S)$.

We show now that $A_{1}\trianglelefteq\mathcal{F}$. Since $G$ normalizes
$A_{1}$ and since $\mathcal{E}_{\mathcal{F}}^{*}=\\{A,S\\}$ by (3), it
suffices to show that $A_{1}$ is $\mathrm{Aut}_{\mathcal{F}}(S)$-invariant
(see [4, Part I, Proposition 4.5]). But this follows from the fact that $G$
normalizes $A_{1}$ because any $\mathcal{F}$-automorphism of $S$ restricts to
an $\mathcal{F}$-automorphism of $A$.

We have shown that $A_{1}$ is a subgroup of $Z(S)$ which is normal in
$\mathcal{F}$. Therefore, by (4), $A_{1}$ must be trivial. It follows that
$A=A_{2}\cong C_{p}\times C_{p}$. Since $A$ is maximal in $S$ by (3), it
further follows that $|S|=p^{3}$. By (2), $S$ is nonabelian, and since any
nonabelian $p$-group of order $p^{3}$ is extraspecial, it follows that $S$ is
extraspecial.

It remains to show that $S$ has exponent $p$. Assume that this is not the
case. Then $S$ has exponent $p^{2}$, and hence, $S$ has a cyclic maximal
subgroup. So $S$ is metacyclic, and [6, Theorem C] implies that $\mathcal{F}$
is supersolvable. This contradiction shows that $S$ has exponent $p$.

(8) The final contradiction.

By (7), $S$ is extraspecial of order $p^{3}$ and exponent $p$. As a
consequence of (3), $S$ has precisely one elementary abelian
$\mathcal{F}$-centric and $\mathcal{F}$-radical subgroup. Applying [21,
Theorem 1.1], or only [21, Lemma 4.7], we conclude that $\mathcal{F}$ is
isomorphic to one of the fusion systems considered in Lemma 2.11.
Alternatively, this can be seen from [9, Lemma 9.2]. Therefore, by Lemma 2.11,
there is a maximal subgroup of $S$ which is not weakly $\mathcal{F}$-closed.
On the other hand, we have by hypothesis that every maximal subgroup of $S$ is
weakly $\mathcal{F}$-closed. With this contradiction, the proof is complete. ∎

For the proof of Theorem D, we need the following lemma. It verifies Theorem D
for the case that the fusion system $\mathcal{F}$ in the statement of the
theorem is realized by a finite group.

######  Lemma 3.2.

Let $G$ be a finite group, $p$ be an odd prime number, $S$ be a Sylow
$p$-subgroup of $G$ and $\mathcal{F}:=\mathcal{F}_{S}(G)$. Suppose that there
is a subgroup $D$ of $S$ with $1<D<S$ such that any subgroup of $S$ with order
$|D|$ is abelian and weakly $\mathcal{F}$-closed. Then $\mathcal{F}$ is
supersolvable.

###### Proof.

Suppose that the lemma is false, and let $G$ be a minimal counterexample. We
will derive a contradiction in several steps.

(1) Let $H<G$ with $S\cap H\in\mathrm{Syl}_{p}(H)$ and $|S\cap H|>|D|$. Then
$\mathcal{F}_{S\cap H}(H)$ is supersolvable.

By hypothesis, any subgroup of $S\cap H$ with order $|D|$ is abelian and
weakly $\mathcal{F}$-closed. Clearly, any weakly $\mathcal{F}$-closed subgroup
of $S\cap H$ is weakly $\mathcal{F}_{S\cap H}(H)$-closed. Hence, any subgroup
of $S\cap H$ with order $|D|$ is abelian and weakly $\mathcal{F}_{S\cap
H}(H)$-closed. The minimality of $G$ implies that $\mathcal{F}_{S\cap H}(H)$
is supersolvable.

(2) $C_{G}(O_{p}(G))\leq O_{p}(G)$.

Let $Q\in\mathcal{E}_{\mathcal{F}}^{*}$. Since $C_{S}(Q)\leq Q$ and since any
subgroup of $S$ with order $|D|$ is abelian, we have $|Q|\geq|D|$. This
implies that $|N_{S}(Q)|>|D|$. Indeed, if $Q\neq S$, then
$|N_{S}(Q)|>|Q|\geq|D|$, and if $Q=S$, then $|N_{S}(Q)|=|S|>|D|$.

Assume that $Q$ is not normal in $G$. Hence $N_{G}(Q)<G$. We have $S\cap
N_{G}(Q)=N_{S}(Q)\in\mathrm{Syl}_{p}(N_{G}(Q))$ since $Q$ is fully
$\mathcal{F}$-normalized and $|S\cap N_{G}(Q)|=|N_{S}(Q)|>|D|$ by the
preceding paragraph. So, by (1), the fusion system
$N_{\mathcal{F}}(Q)=\mathcal{F}_{N_{S}(Q)}(N_{G}(Q))$ is supersolvable.

If no member of $\mathcal{E}_{\mathcal{F}}^{*}$ is normal in $G$, then
$N_{\mathcal{F}}(Q)$ is supersolvable for each
$Q\in\mathcal{E}_{\mathcal{F}}^{*}$, whence $\mathcal{F}$ is supersolvable by
Lemma 2.6. Since this is not the case, there must exist some
$Q\in\mathcal{E}_{\mathcal{F}}^{*}$ with $Q\trianglelefteq G$. Hence,
$\mathcal{F}$ is constrained, and [13, Proposition 8.8] shows that the model
of $\mathcal{F}$ is isomorphic to $G/O_{p^{\prime}}(G)$.

If $O_{p^{\prime}}(G)\neq 1$, then
$\mathcal{F}_{SO_{p^{\prime}}(G)/O_{p^{\prime}}(G)}(G/O_{p^{\prime}}(G))$ is
supersolvable by the minimality of $G$, whence $\mathcal{F}$ is supersolvable.
Thus $O_{p^{\prime}}(G)=1$. Consequently, $G$ is the model of $\mathcal{F}$,
and so we have $C_{G}(O_{p}(G))\leq O_{p}(G)$, as wanted.

(3) $|O_{p}(G)|=|D|$.

Assume that $|O_{p}(G)|>|D|$. By hypothesis, any subgroup of $O_{p}(G)$ with
order $|D|$ is weakly $\mathcal{F}$-closed. Clearly, any weakly
$\mathcal{F}$-closed subgroup of $O_{p}(G)$ is normal in $G$. Therefore, any
subgroup of $O_{p}(G)$ with order $|D|$ is normal in $G$. So we have
$O_{p}(G)\leq Z_{\mathcal{U}}(G)$ by Lemma 2.8. If $H$ is a proper subgroup of
$G$ with $O_{p}(G)<S\cap H$ and $S\cap H\in\mathrm{Syl}_{p}(H)$, then
$\mathcal{F}_{S\cap H}(H)$ is supersolvable by (1). So, by Lemma 2.9,
$\mathcal{F}$ is supersolvable. This contradiction shows that
$|O_{p}(G)|\leq|D|$.

Because of (2), we also have $|O_{p}(G)|\geq|D|$. Thus $|O_{p}(G)|=|D|$, as
desired.

(4) $O_{p}(G)$ is elementary abelian.

Assume that $O_{p}(G)$ is not elementary abelian. Then $\Phi(O_{p}(G))\neq 1$
by [15, Lemma 4.5]. Set $\overline{G}:=G/\Phi(O_{p}(G))$ and
$\overline{\mathcal{F}}:=\mathcal{F}/\Phi(O_{p}(G))=\mathcal{F}_{\overline{S}}(\overline{G})$.
By hypothesis and by Lemma 2.1, any subgroup of $\overline{S}$ with order
$\frac{|D|}{|\Phi(O_{p}(G))|}$ is abelian and weakly
$\overline{\mathcal{F}}$-closed. The minimality of $G$ implies that
$\overline{\mathcal{F}}$ is supersolvable.

By Lemma 2.4 (1), there is a series $\Phi(O_{p}(G))=V_{0}\leq
V_{1}\leq\dots\leq V_{m}=S$ of subgroups of $S$ such that $\overline{V_{i}}$
is strongly $\overline{\mathcal{F}}$-closed for all $0\leq i\leq m$, such that
the quotient $\overline{V_{i+1}}/\overline{V_{i}}$ is cyclic for all $0\leq
i<m$ and such that $V_{j}=O_{p}(G)$ for some $0<j<m$. Since
$\overline{V_{i}}\trianglelefteq\overline{G}$ for all $0\leq i\leq j$, it
follows that $\overline{O_{p}(G)}\leq Z_{\mathcal{U}}(\overline{G})$. Applying
[25, Chapter 1, Theorem 7.19], we conclude that $O_{p}(G)\leq
Z_{\mathcal{U}}(G)$.

By (1) and (3), for any proper subgroup $H$ of $G$ with $O_{p}(G)<S\cap H$ and
$S\cap H\in\mathrm{Syl}_{p}(H)$, the fusion system $\mathcal{F}_{S\cap H}(H)$
is supersolvable. Lemma 2.9 implies that $\mathcal{F}$ is supersolvable. This
contradiction shows that $O_{p}(G)$ is elementary abelian.

(5) If $O_{p}(G)\leq H<G$, then $H$ is $p$-closed.

Let $O_{p}(G)\leq H<G$. Without loss of generality, we assume that $S\cap
H\in\mathrm{Syl}_{p}(H)$. Clearly, $O_{p}(G)\leq S\cap H$, and if
$O_{p}(G)=S\cap H$, then $H$ is $p$-closed. Assume now that $O_{p}(G)<S\cap
H$. Then $\mathcal{E}:=\mathcal{F}_{S\cap H}(H)$ is supersolvable by (1) and
(3). So, by [24, Proposition 2.3], $S\cap H$ is normal in $\mathcal{E}$. Thus
$\mathcal{E}=N_{\mathcal{E}}(S\cap H)=\mathcal{F}_{S\cap H}(N_{H}(S\cap H))$.
For each $h\in H$, let $c_{h}$ denote the automorphism of $O_{p}(G)$ induced
by conjugation with $h$, i.e.

$c_{h}:O_{p}(G)\rightarrow O_{p}(G),x\mapsto x^{h}.$

Let $h\in H$. Then $c_{h}$ is a morphism in $\mathcal{E}=\mathcal{F}_{S\cap
H}(N_{H}(S\cap H))$, and so we have $c_{h}=c_{u}$ for some $u\in N_{H}(S\cap
H)$. Then $hu^{-1}\leq C_{G}(O_{p}(G))\leq O_{p}(G)\leq N_{H}(S\cap H)$ by
(2). It follows that $h\in N_{H}(S\cap H)$, and since $h$ was an arbitrarily
chosen element of $H$, we can conclude that $S\cap H\trianglelefteq H$. Hence,
$H$ is $p$-closed, as desired.

(6) $S/O_{p}(G)$ is cyclic.

Set $\overline{G}:=G/O_{p}(G)$. Let $O_{p}(G)\leq H<G$. Then $H$ is $p$-closed
by (5). Hence, $\overline{H}$ is $p$-closed. The group $\overline{G}$ is not
$p$-closed because otherwise $S$ would be normal in $G$, which is not true
because of (3). We have shown that $\overline{G}$ is a non-$p$-closed group
all of whose proper subgroups are $p$-closed. In other words, $\overline{G}$
is minimal non-$p$-closed in the sense of [18]. Applying [18, Lemma 1], we
conclude that $\overline{G}$ is minimal nonnilpotent or that
$\overline{G}/\Phi(\overline{G})$ is nonabelian simple.

If $\overline{G}$ is minimal nonnilpotent, then $\overline{S}$ is cyclic by
Lemma 2.13.

We assume now that $\widehat{G}:=\overline{G}/\Phi(\overline{G})$ is
nonabelian simple, and we show that $\overline{S}$ is cyclic also in this
case. Since $\Phi(\overline{G})$ is nilpotent and $O_{p}(\overline{G})=1$, we
have that $\Phi(\overline{G})$ is a $p^{\prime}$-group. Thus, $\overline{S}$
is isomorphic to the Sylow $p$-subgroups of $\widehat{G}$, and therefore, it
suffices to show that $\widehat{G}$ has cyclic Sylow $p$-subgroups.

Let $O_{p}(G)\leq L<G$ such that $\overline{L}=\Phi(\overline{G})$. If $L\leq
H<G$, then $H$ is $p$-closed by (5), whence $H/L$ is $p$-closed. Consequently,
any proper subgroup of $G/L\cong\widehat{G}$ is $p$-closed. Since
$\widehat{G}$ is simple, the only proper quotient of $\widehat{G}$ is
$\widehat{G}/\widehat{G}$, which is of course $p$-closed. The simplicity of
$\widehat{G}$ also implies that $\widehat{G}$ is not $p$-closed. We have shown
that $\widehat{G}$ is a non-$p$-closed group all of whose proper subgroups and
proper quotients are $p$-closed. Hence, $\widehat{G}$ is minimal
non-$p$-closed in the sense of [16] (note that the definition of a minimal
non-$p$-closed group used in [16] does not coincide with the one used in
[18]). Applying [16, Theorem 3.5], we conclude that $\widehat{G}$ has cyclic
Sylow $p$-subgroups, as required.

(7) $O_{p}(G)$ is not maximal in $S$.

Assume that $O_{p}(G)$ is maximal in $S$. Then, by (3) and by hypothesis, any
maximal subgroup of $S$ is abelian and weakly $\mathcal{F}$-closed. Theorem C
implies that $\mathcal{F}$ is supersolvable, which is a contradiction. Hence,
$O_{p}(G)$ is not maximal in $S$.

(8) $|O_{p}(G)|\geq p^{4}$.

By (2) and (4), we have $C_{S}(O_{p}(G))=O_{p}(G)$. Hence, $S/O_{p}(G)$ is
isomorphic to a $p$-subgroup of $\mathrm{Aut}(O_{p}(G))$. By (6), $S/O_{p}(G)$
is cyclic. We have $|S/O_{p}(G)|\geq p^{2}$ since $O_{p}(G)\neq S$ by (3) and
since $O_{p}(G)$ is not maximal in $S$ by (7).

Now, let $e\in\mathbb{N}$ such that $|O_{p}(G)|=p^{e}$. We have to show that
$e\geq 4$. As $|O_{p}(G)|=|D|>1$ by (3), we have $e\neq 0$. Since $O_{p}(G)$
is elementary abelian by (4), we have $\mathrm{Aut}(O_{p}(G))\cong GL_{e}(p)$.
If $e=1$, then $\mathrm{Aut}(O_{p}(G))\cong GL_{1}(p)\cong C_{p-1}$ does not
have any nontrivial $p$-subgroups, and if $e\in\\{2,3\\}$, then we see from
Lemma 2.12 (2) that $\mathrm{Aut}(O_{p}(G))\cong GL_{e}(p)$ does not have any
cyclic $p$-subgroups of order greater than $p$. Since $S/O_{p}(G)$ is
isomorphic to a cyclic $p$-subgroup of $\mathrm{Aut}(O_{p}(G))$ of order at
least $p^{2}$, it follows that $e\geq 4$, as required.

(9) The final contradiction.

Since $|S/O_{p}(G)|\geq p^{2}$ by (3) and (7), there exists $O_{p}(G)<T\leq S$
with $|T/O_{p}(G)|=p^{2}$. Then $O_{p}(G)$ is properly contained in a maximal
subgroup $T_{1}$ of $T$, and since $C_{T_{1}}(O_{p}(G))\leq
C_{G}(O_{p}(G))\leq O_{p}(G)$ by (2), we have that $T_{1}$ is nonabelian. By
(3) and by hypothesis, any subgroup of $T$ with index $p^{2}$ is abelian.
Hence, $T$ is an $\mathcal{A}_{2}$-group in the sense of the definition given
before Lemma 2.15. We have $|T|>|O_{p}(G)|\geq p^{4}$ by (8).

Assume that $T$ is metacyclic. Then $O_{p}(G)$ is metacyclic as well. At the
same time, $O_{p}(G)$ is elementary abelian by (4). It is rather easy to see
that an elementary abelian finite $p$-group can only be metacyclic when its
order is at most $p^{2}$. So it follows that $|O_{p}(G)|\leq p^{2}$. This
contradicts (8), and therefore, $T$ is not metacyclic.

Assume that $T$ has no abelian maximal subgroups. Then $|T|=p^{5}$ by Lemma
2.15 (2) and hence $|O_{p}(G)|=p^{3}$, which is a contradiction to (8). Thus,
there exists an abelian maximal subgroup of $T$.

Assume that there is only one abelian maximal subgroup of $T$, say $U$. We
claim that $T^{\prime}\leq Z(T)$. Because of (2), we have $O_{p}(G)\not\leq U$
and hence $T=O_{p}(G)U$. Set $Z:=O_{p}(G)\cap U$. Since $O_{p}(G)$ and $U$ are
abelian, the subgroup $Z$ is centralized by both $O_{p}(G)$ and $U$. This
implies that $Z\leq Z(T)$. We have $O_{p}(G)/Z\cong O_{p}(G)U/U=T/U\cong
C_{p}$, and so $Z$ is maximal in $O_{p}(G)$. As a consequence of (2), $Z(T)$
is a proper subgroup of $O_{p}(G)$. So we have $Z\leq Z(T)<O_{p}(G)$, and the
maximality of $Z$ in $O_{p}(G)$ implies that $Z=Z(T)$. Therefore,
$O_{p}(G)/Z(T)$ is a normal subgroup of $T/Z(T)$ with order $p$ and hence a
central subgroup of $T/Z(T)$. The corresponding factor group
$(T/Z(T))/(O_{p}(G)/Z(T))\cong T/O_{p}(G)$ is cyclic by (6). It follows that
$T/Z(T)$ is abelian. Thus $T^{\prime}\leq Z(T)$, as claimed above. Applying
Lemma 2.15 (1), we conclude that $\Phi(T)=Z(T)<O_{p}(G)$. So $T/O_{p}(G)$ is
elementary abelian by [15, Lemma 4.5], and since $|T/O_{p}(G)|=p^{2}$, it
follows that $T/O_{p}(G)$ is not cyclic. This is a contradiction to (6).
Therefore, $T$ has more than one abelian maximal subgroup.

Now, applying [20, Lemma 1.9], we conclude that $Z(T)$ has index $p^{2}$ in
$T$. Since $O_{p}(G)$ also has index $p^{2}$ in $T$ and $Z(T)\leq
C_{G}(O_{p}(G))\leq O_{p}(G)$ by (2), it follows that $O_{p}(G)=Z(T)$. In
particular, we have $C_{G}(O_{p}(G))\not\leq O_{p}(G)$. This contradicts (2),
and the proof is complete with this contradiction. ∎

With Lemma 3.2 at hand, we are able to prove Theorem D in few lines. Our proof
strongly relies on the model theorem [4, Part III, Theorem 5.10].

###### Proof of Theorem D.

Suppose that the theorem is false, and let $\mathcal{F}$ be a counterexample
such that $|\mathcal{F}|$ is minimal.

Let $Q$ be a member of $\mathcal{E}_{\mathcal{F}}^{*}$ which is not normal in
$\mathcal{F}$. Since $C_{S}(Q)\leq Q$ and since any subgroup of $S$ with order
$|D|$ is abelian, we have $|Q|\geq|D|$. This implies that $|N_{S}(Q)|>|D|$.
Hence, $N_{\mathcal{F}}(Q)$ is a proper saturated subsystem of $\mathcal{F}$
on $N_{S}(Q)$ satisfying the hypotheses of the theorem. Thus,
$N_{\mathcal{F}}(Q)$ is supersolvable by the minimality of $\mathcal{F}$.

If no member of $\mathcal{E}_{\mathcal{F}}^{*}$ is normal in $\mathcal{F}$,
then $\mathcal{F}$ is supersolvable by the preceding paragraph and Lemma 2.6.
Therefore, there must be a member of $\mathcal{E}_{\mathcal{F}}^{*}$ which is
normal in $\mathcal{F}$. Hence, $\mathcal{F}$ is constrained. The model
theorem [4, Part III, Theorem 5.10] implies that there is a finite group $G$
with $S\in\mathrm{Syl}_{p}(G)$ and $\mathcal{F}=\mathcal{F}_{S}(G)$. Then
Lemma 3.2 implies that $\mathcal{F}$ is supersolvable. This contradiction
completes the proof. ∎

Declarations of interest: none.

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