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\begin{document}
\title {Isolated Subgroups in Finite Groups}
\author{Ulrich Meierfrankenfeld\\Christopher Parker \\Peter Rowley}
\address{U. Meierfrankenfeld\\
Department of Mathematics\\ Michigan State University\\ East
Lansing, MI 48824-1027\\USA}
\email{meier@math.msu.edu}
\address{C.W. Parker\\
School of Mathematics\\
University of Birmingham\\ Edgbaston\\ Birmingham B15 2TT, UK}
\email{C.W.Parker@bham.ac.uk}
\address{P.J. Rowley\\ School of Mathematics\\The University of Manchester\\PO Box
88\\Manchester M60 1QD, UK} \email{peter.j.rowley@manchester.ac.uk}
\maketitle
\centerline{\today}
\section{Introduction}
Suppose that $p$ is a prime, $P$ is a finite group and $S \in
\Syl_p(P)$. Then $P$ is \emph{$p$-minimal} if $S$ is not normal in
$P$ and $S$ is contained in a unique maximal subgroup of $P$. Now
suppose that $G$ is a finite group and $S \in \Syl_p(G)$. The set of
subgroups of $G$ which contain $S$ and are $p$-minimal will be
denoted by $\P_G(S)$. For $P\in \P_G(S)$ we put
$$L_G(P,S) := \<\P_G(S)\setminus \{P\},N_G(S)\>,$$ and next introduce
isolated $p$-minimal subgroups.
\be{definition}{def1} A $p$-minimal subgroup $P$ in $\P_G(S)$ is
$A$-isolated where $A$ is a normal $p$-subgroup of $L_G(P,S)$ if
$A\not\le O_p(P)$. We say that $P\in \P_G(S)$ is an isolated
subgroup of $G$ if it is $A$-isolated for some $A $. \en{definition}
Clearly { $P$ is an isolated subgroup of $G$ if and only if }
$O_p(L_G(P,S)) \not\le O_p(P)$. As a consequence $G\neq L_G(P,S)$.
Thus, as $G =\<\P_G(S),N_G(S)\>$ (see Lemma~\ref{L2.8}), any minimal
generating set of $G$ using subgroups from $\P_G(S)$, together with
$N_G(S)$, must include $P$.
%\red{I suggest to add: From now on $P$ will always denote a $p$-minimal subgroup of $G$
% containing $S$ and $L=L_G(P,S)$. And replace all $L_G(P,S)$ by $L$)}
The archetypical $p$-minimal subgroup is to be found in groups of
Lie type defined in characteristic $p$. If $G$ is such a group and
$R$ is a minimal parabolic subgroup of $G$, then $P=
O^{p^\prime}(R)$ is a $p$-minimal subgroup of $G$. { For such a $P$,
$L:=L_G(P,S)$ is
the unique maximal parabolic subgroup of $G$ not containing $R$ and $O_p(L)\nleq
O_p(P)$,} and so
in fact $P$ is an isolated subgroup of $G$. This in itself makes
isolated subgroups worthy of study yet there are other compelling
reasons for instigating a study of this type of subgroup. Such
subgroups arise in the on-going work whose aim is to classify finite
${\mathcal K}_p$-proper groups of local characteristic $p$. We
return to this important aspect of isolated subgroups shortly.
Our first theorem includes a statement about a special class of
$p$-minimal subgroups which we call narrow subgroups. Before
defining narrow subgroups, we first recall that a group $G$ acts
imprimitively on a $\GF(p)G$-module $V$ provided there exists a
non-trivial decomposition of $V$ into a direct sum of subspaces
$W_1,\dots, W_k$ which are permuted by the action of $G$. A group
which does not act imprimitively on a module $V$ is said to act
primitively on $V$.
\be{definition}{def2} Suppose that $P$ is a $p$-minimal group, $S\in
\Syl_p(P)$ and let $M$ be the unique maximal subgroup of $P$
containing $S$. Set $E:= O^p(P)F/F$ where $F$ is the core of $M$ in
$P$. Then we say that $P$ is \emph{narrow} if either $E$ is a simple
group or $E$ is elementary abelian and $M$ acts primitively on $E$.
\en{definition}
Again examples of narrow, isolated subgroups appear naturally in Lie
type groups of characteristic $p$.
Our first theorem shows that finite groups with isolated subgroups
have a restricted structure. Its proof consists {of} an analysis of
the over groups of Sylow $p$-subgroups and relies only on results
from classical group theory. This contrasts with the proof of
Theorem \ref{T1.5} where we are forced to draw upon specific
structural properties of certain finite simple groups.
\be{theorem}{T1.3} Suppose that $p$ is a prime, $G$ is a finite
group, $S \in \Syl_p(G)$ and $P\in \P_G(S)$ is an isolated subgroup
of $G$. Set $L:= L_G(P,S)$ and $Y := \$. Then $G=YL$, $P$
is an isolated subgroup of $YS$ and either
\bl \li a $Y= O^p(P)$; or
\li b $Y/O_p(Y)$ is a central product of quasisimple groups which
are transitively permuted by $S$ under conjugation and, if $P$ is
narrow, then $Y/O_p(Y)$ is quasisimple. {Furthermore, $Y/O_p(Y)$ has
order divisible by $p$.}\el
\en{theorem}
Of course, if $G$ is a quasisimple group and $G$ has an isolated
subgroup, then $Y=G$ and alternative (b) of Theorem~\ref{T1.3}
holds. If $G$ is a $p$-minimal group and $S\in \Syl_p(G)$, then
$L_G(G,S)$ is the unique maximal subgroup of $G$ containing $S$.
Hence, if $O_p(L_G(G,S))> O_p(G)$, $G$ is an isolated subgroup of
$G$ and the possibility in Theorem~\rf[T1.3]{a} trivially holds. As
an indication of the limitations of Theorem~\ref{T1.3}, we could
take $G = H_1 \times H_2$ with $P $ an isolated subgroup of $H_1S$
with $S\le P$ where $S \in \Syl_p(G)$. Then $P$ is an isolated subgroup of
$G$ (see Lemma~\ref{L2.85}) and of course $Y\le H_1$. So the direct
factor $H_2$ is invisible as far as $Y$ is concerned.
For a finite group $G$ and $p$ a prime, let $F_p^*(G)$ denote the
inverse image in $G$ of $F^*(G/O_p(G)))$, the generalized Fitting
subgroup of $G/O_p(G)$. Then we have the following corollary of
Theorem~\ref{T1.3}.
\be{corollary}{cor1} Suppose that $G$ is a finite group and $P\in
\P_G(S)$ is an isolated subgroup of $G$ where $S \in \Syl_p(G)$, $p$
a prime.
\bl \li a $O^p(P) \le F_p^*(G)$.
\li b If $Y := \$ is soluble, then $Y=O^p(P)$ and there is a
prime $t\neq p$ such that $Y/O_p(Y)$ is a $t$-group of class at most
$2$. \el \en{corollary}
Again casting an eye over the Lie type groups in characteristic $p$
we see that every $p$-minimal subgroup is an isolated subgroup. This
leads us to give
\be{definition}{completely} Let $G$ be a finite group, $p$ a prime
and $S \in \Syl_p(G)$. Then $G$ is called completely isolated if $P$
is isolated for all $P \in \P_G(S)$. \en{definition}
For $P_1, P_2 \in \P_G(S)$ write $P_1 \sim P_2$ whenever $\ = \<
O^p(P_2)^G \>$. Clearly $\sim$ is an equivalence relation on $\P_G(S)$.
Our second theorem describes the structure of completely
isolated finite groups.
\be{theorem}{completelypsing} If $G$ is a completely isolated finite
group, then $G=F_p^*(G)N_G(S)$, where $S \in \Syl_p(G)$. Moreover,
{$F^*(G/O_p(G))$} is a central product of the subgroups generated by
the $\sim$ equivalence classes of $\P_{G/O_p(G)}(S/O_p(G))$ and
$F(G/O_p(G))$ has class at most two.
%\red{You might want to rephrase this sentence. The first time a read
%it, I
% wasn't sure what it was saying. But by now I don't really know anymore what my
% problem was}
\en{theorem}
For $p$ a prime, we define
\begin{eqnarray*}
{\mathcal L}_1(p) &:=&\{O^p(X)\mid X \cong H/Z\mbox{ where } H \mbox{ is a universal
rank 1 Lie type group } \\&&\;\;\;\;\;\;\;\;\;\;\;\;\;\; \mbox{
defined in characteristic }p \mbox{ and }Z \le
Z(H)\}.\end{eqnarray*}
Observe that some of the groups in $\mathcal L_1(p)$ are smaller
than we might have expected. For example, $O^2(\SL_2(2))$ has order
$3$ and $O^2({}^2\B_2(2))$ has order $5$. So, to be explicit we list
the members of $\mathcal L_1(p)$. For $p \ge 5$, we have
$$\mathcal L_1(p) = \{\SL_2(p^a), \PSL_2(p^a),\SU_3(p^a),
\PSU_3(p^a)\mid a\ge 1\}.$$ Because of the non-simplicity of
$\SL_2(3)$ and ${}^2\GG_2(3)$, we have
$$\mathcal L_1(3)=\{\Q_8, 2^2,{}^2\GG_2(3)^\prime\cong \SL_2(8)\}
\cup \{\SL_2(3^a), \PSL_2(3^a),\SU_3(3^{a-1}),
{}^2\GG_2(3^{2a-1})\mid a\ge 2\}$$ and similarly, because of the
non-simplicity of $\SL_2(2)$, $\SU_3(2)$ and ${}^2\B_2(2)$, we have
$$\mathcal L_1(2)=\{3, 5,3^{1+2}_+,3^2\} \cup
\{\SL_2(2^a),\SU_3(2^a), \PSU_3(2^a), {}^2\B_2(2^{2a-1})\mid a\ge
2\}.$$
The groups in $\mathcal L_1(p)$ will be discussed further in
Section~3. Notice that if $O^p(P/O_p(P)) \in \mathcal L_1(p)$, $S
\in \Syl_p(P)$, and $S
> O_p(P)$, then $P$ is a narrow unless $p=2$ and
$O^2(P/O_p(P))\cong 3^2$ or $3^{1+2}_+$ in which case we have to be
rather more careful. Thus Theorem~\ref{T1.3} applies to narrow
isolated subgroups with the property that $O^p(P/O_p(P))\in \mathcal
L_1(p)$ and tells us that when $O_p(G)=1$, $Y:=\$ is
either quasisimple or equal to $O^p(P)$. In the latter case we have
that either $Y=O^p(P)$ is quasisimple or $Y$ is soluble and $O^p(P)
$ is isomorphic to one of $3$, $5$, $\Q_8$, $2^2$, $3^{1+2}_+$ and
$3^2$.
Our next theorem concerns certain isolated subgroups of $\mathcal{
CK}$-groups. We say that a finite simple group is a $\mathcal
K$-group if it is isomorphic to a cyclic group of prime order, an
alternating group of degree at least $5$, a simple group of Lie type
(including the Tits simple group) or one of the 26 sporadic simple
groups. A group is a $\ck$-group, if each of its composition factors
is a $\mathcal K$-group.
\be{theorem}{T1.5} Suppose that $p$ is a prime, $G$ is a finite
${\mathcal {CK}}$-group and $P\in \P_G(S)$. If $O^p(P/O_p(P)) \in
\mathcal L_1(p)$,
then either $O^p(P) \unlhd G$ or
$O_p(Z(L_G(P,S)))\leq O_p(P)$. \en{theorem}
In other words Theorem~\ref{T1.5} is asserting that if $P$ is
$O_p(Z(L_G(P,S)))$-isolated and $O^p(P/O_p(P)) \in \mathcal L_1(p)$,
then $O^p(P) $ is normal in $G$. We note that if we take $G= \J_2$
(Janko's second simple group) we have a narrow subgroup $P$ of $G$
with $P\cong \PSU_3(3)\in \mathcal L_1(3)$ and $L_G(P,S) \cong
3\udot\PGL_2(9)$. So $O_3(P)=1$, $O_3(L_G(P,S))$ has order $3$ and
just misses being in $Z(L_G(P,S))$.
We comment that not every finite group possesses an isolated subgroup.
However, as we have seen, if it does then it imposes global constraints
on the structure of the group. Another example of this is the
following corollary to \cite[Corollary 1.2]{MPR}. This asserts that
a simple ${\mathcal {CK}}$-group with an isolated subgroup $P$
satisfying $O^p(P/O_p(P)) \in \mathcal L_1(p)$ with $p\ge 11$ must
be a Lie type group in characteristic $p$.
Isolated subgroups, as mentioned earlier, also surface in the
project to classify finite ${\mathcal K}_p$-proper groups of local
characteristic $p$. That is finite groups all of whose $p$-local
subgroups $H$ satisfy $F^*(H)= O_p(H)$ and have {all simple
sections} from the list of finite simple groups itemized above. Let
$G$ to be such a group and let $ C$ be a maximal $p$-local subgroup
containing $N_G(\O_1(Z(S)))$ where $S \in \Syl_p(G)$. Then in this
context the $\widetilde P!$-Theorem \cite {ptilde} asserts that in
the presence of certain additional conditions there is an isolated
subgroup $\wt P$ of $G$ contained in $ \mathcal P_{C}(S)$. Moreover,
$O^p(\wt P/O_p(\wt P) )\in \mathcal L_1(p)$. Thus all of our
theorems apply to tell us that $Y:=\O_p(C)/O_p(C)$ is
a quasisimple group. In particular, as $C$ is a $\mathcal
{CK}$-group the results of \cite{MPR} which determine the finite
simple $\mathcal K$-groups possessing an isolated subgroup $P$ with
$O^p(P/O_p(P))\in \mathcal L_1(p)$, can be applied to limit the
possibilities for $Y$. For an account of the genealogy of this
(still in progress) programme as well as an overview of its aims see
\cite{MSS}.
Briefly this paper is organized as follows. In Section 2 we gather
some elementary background results that we require in this paper.
The following section contains a number of general properties of
$p$-minimal groups ranging from their generational properties, their
behaviour under quotients to the important compendium of structural
properties given in Lemma~\ref{L2.9}. Towards the end of Section 3,
in Lemmas~\ref{rank11} and ~\ref{rank1grps}, detailed information
about $p$-minimal isolated subgroups with $O^p(P/O_p(P)) \in
\mathcal L_1(p)$ is listed. These two results plus
Corollary~\ref{L2.7} are needed in our proof of Theorem~\ref{T1.5}.
We begin Section 4 by noting a number of basic facts about isolated
subgroups -- Lemma~\ref{L3.2} being a result we use frequently.
After establishing further preparatory results such as
Lemmas~\ref{L3.4.5}, ~\ref{L3.6} and ~\ref{simple}, we then prove
Theorems~\ref{T1.3} and ~\ref{completelypsing}. The remainder of
this section gives various consequences of Theorem~\ref{T1.3} which
will be of use in \cite{MPR}. Our final section is devoted to a
proof of Theorem~\ref{T1.5}.
If $\mathcal X$ is a collection of subsets of $G$, then $\bigcap
{\mathcal X}$ means the intersection of all the subsets in
${\mathcal X}$. Assume that $H$ is a subgroup of $G$. Then $H^G$
will denote the set of all $G$-conjugates of $H$. Hence $\bigcap
H^G$ is just the core of $H$ in $G$. For $p$ a { fixed} prime, we
sometimes write $Q_G$ for $O_p(G)$, the largest normal $p$-subgroup
of $G$. Also we denote the preimage of $F(G/Q_G)$ by $F_p(G)$ and
the preimage of $\Phi(G/Q_G)$ by $\Phi_p(G)$. The remainder of our
group theoretic notation is standard and can be found in
\cite{Aschbacher} or \cite{KS}. {\sc Atlas} \cite{Atlas} names and
conventions will be followed with a few exceptions which we now
note. We shall use $\Sym(n)$ and $\Alt(n)$ to denote, respectively,
the symmetric and alternating group of degree $n$. And $\Dih(n)$,
$\SD(n)$ stand, respectively, for the dihedral and semidihedral
group of order $n$ (so { $n$ is even} ).
The genesis of the work in this paper and \cite{MPR} occurred while
the authors were participants in the RiP (Research in Pairs)
programme at the Mathematische Forschungsinstitut Oberwolfach. The
completion of these papers was further supported by an LMS scheme 4
grant and funding from the Manchester Institute of Mathematical
Sciences -- the authors express their gratitude to all these
organizations.
\section{Preliminary Results}\label{two}
This section contains an assorted collection of background results.
The first two concern subnormal subgroups while Lemma~\ref{L2.3} is an
elementary criterion for a $p$-subgroup of a group $G$ to be
contained in $O_p(G)$. Lemmas~\ref{Frat1} and ~\ref{FratProd} look at the
Frattini subgroup and Lemma~\ref{L2.11.1} examines $\Phi_p(O^p(G))$.
And we end with Proposition~\ref{P2.5} which gives a property of
finite simple ${\mathcal K}$-groups with abelian Sylow
$p$-subgroups.
Our first lemma records a well-known property of components.
\be{lemma}{L2.1} Let $G$ be a finite group, $K$ a component of $G$
and $N$ a subnormal subgroup of $G$. Then either $K\leq N$ or
$[K,N]=1$. \en{lemma} \begin{proof}
\cite[31.4]{Aschbacher}.\end{proof}
In the proof of Lemma~\ref{L2.9} we shall need the following
result.
\be{lemma}{L2.2} Suppose that $G$ is a finite group, $M$ is a
maximal subgroup of $G$ and $N$ is a subnormal subgroup of $G$
with $N\leq M$. Then $N\leq \bigcap M^G$. \en{lemma}
\begin{proof} We may assume that $N$ is maximal, under inclusion, with the property that $N$ is
subnormal in $G$ and $N \le M$. Let $N=N_0\normal N_1\normal\dots
\normal N_{k}=G$ be a subnormal chain from $N$ to $G$. By
Wielandt's Subnormality Theorem \cite[6.7.1]{KS}, $\
\normal\normal G$. Because of the maximality of $N$, $\=N$ and
so $N \normal M$. Thus $N \normal\$. Also, by the maximality
of $N$, $N_1\not \le M$, whence $N \normal G$, and this proves the
lemma.
\end{proof}
\be{lemma}{L2.3} Suppose that $G$ is a finite group, $p$ is a prime,
and $A$ is a $p$-subgroup of $G$ for which $[A,G] \le N_G(A)$. Then
$A\le O_p(G)$ and $[A,G]$ is a $p$-group. \en{lemma}
\begin{proof} Since $$A \le O_p([A,G]A)= O_p(\langle A^G\rangle) \normal
G,$$ $A \le O_p(G)$ and $[A,G]$ is a $p$-group.\end{proof}
\be{lemma}{Frat1} Suppose that $G$ is a finite group and $N$ is a
normal subgroup of $G$. Then $\Phi(N) \le \Phi(G)$. \en{lemma}
\begin{proof} Assume that $M$ is a maximal subgroup of $G$
and that $\Phi(N)$ is not contained in $M$. Then, as $M$ is maximal
in $G$ and $\Phi(N)$ is normal in $G$, $G=\Phi(N)M$. Thus $$N =
N\cap G= N\cap \Phi(N)M = \Phi(N)(N\cap M).$$ Thus $N= N\cap M$ and
so $M \ge N \ge \Phi(N)$ which is not the case. Hence $\Phi(N) \le
\Phi(G)$ as claimed.
\end{proof}
\be{lemma}{Fratrec} Suppose that $G$ is a finite group. Then
\bl\li{a}$\Phi_p(G)$ is the intersection of all the maximal
subgroups of $G$ which contain a Sylow $p$-subgroup of $G$; \li{b}
$O_p(G)\in \Syl_p(\Phi_p(G))$; and\li{c} if $O_p(O^p(G))=1$, then
$\Phi_p(G)$ is nilpotent.\el \en{lemma}
\begin{proof} Let $F :=\Phi_p(G)$ and $H$ be the intersection of
all the maximal subgroups of $G$ which contain a Sylow $p$-subgroup
of $G$. Obviously $F \le H$ and $H$ is normal in $G$. Let $T \in
\Syl_p(H)$. Then the Frattini Argument gives $G=N_G(T)H$. Since $T$
has $p'$ index in $H$, we infer that $N_G(T)$ contains a Sylow
$p$-subgroup of $G$. Thus $N_G(T)= G$. Hence $T\le O_p(G)$ and so as
$O_p(G)\le F$, we have $O_p(G)= T$. Assume that $F F_X \cap M$ which together with the
maximality of $M$ in $X$ implies that $F_X\cap M $ is a normal
subgroup of $X$. Furthermore, because of the maximality of $M$ in
$X$, $F_X/(F_X\cap M)$ has no non-trivial proper $X$-invariant
subgroups. Hence $F_X/(F_X\cap M)$ is an elementary abelian
$t$-group for some prime $t$ with $t\neq p$. Let $R= \bigcap (F_X
\cap M)^G$. Assume that $R>1$. Then by induction we have $O^p(G/R)
\cap \Phi_p(G/R)= \Phi_p(O^p(G/R))$. Furthermore, as $R\le
\Phi_p(G)$,
$\Phi_p(G/R) \ge \Phi_p(G)/R$. Hence, using $X/R = O^p(G/R)$, we have
$$F_X/R =\Phi_p(G)/R\cap X/R \le \Phi_p(G/R)\cap O^p(G/R) =
\Phi_p(X/R).$$ But we know that $M$ has index a power of $t$ in $X$,
so $M/R \ge \Phi_p(X/R)$ by Lemma~\ref{Fratrec}(a). Thus $F_X/R \le
\Phi_p(X/R) \le M/R$, which is a contradiction to our choice of $M$.
Therefore, $R=1$ and, in particular, $F_X$ is an elementary abelian
$t$-group. Notice that $M/(F_X\cap M)$ is a complement to
$F_X/(F_X\cap M)$ in $X/(F_X\cap M)$. Choose $K$ of minimal order in
$M$ such that $X=KF_X$. Then $K\cap F_X $ is normalized by both
$F_X$ and $K$ and so is normal in $X$. Since $K\le M$, $K\cap F_X
\neq F_X$. We claim that $K$ is a complement to $F_X$ in $X$.
Suppose that $K \cap F_X
>1$. Then, as $R=1$, there exists $g\in G$ such that $K^g\cap
F_X\not\le F_X\cap M$. It follows that $K^g(F_X\cap M)=K^g(F_X\cap
K^g)(F_X\cap M)= K^gF_X=X$ as $(F_X\cap M)(F_X\cap K^g)$ is normal
in $X$. Therefore, as $K^g/(K^g\cap (F_X \cap M)) \cong X/(F_X\cap
M)$ and $X/(F_X\cap M)$ splits over $F_X/(F_X\cap M)$, $K^g/(K^g\cap
F_X \cap M)$ splits over $(K^g\cap F_X)/(K^g\cap F_X \cap M)$ and
thus $K$ contains a proper subgroup $K_0$ such that $K= K_0(K\cap
F_X)$. But then $K_0F_X = KF_X = X$ and we have a contradiction to
the minimal choice of $K$. Therefore, $X$ splits over $F_X$. By
\cite[17.7]{Aschbacher} the total number of complements to $F_X$ in
$X$ is $|F_X/C_{F_X}(X)||H^1(X,F_X)|=t^a$ for some integer $a$.
Since $S$ permutes (by conjugation) the complements to $F_X$ in $X$
and $S$ is
a $p$-group, we get that there exists a complement $K$ to $F_X$ in
$X$ which is normalized by $S$. So $KS$ is a subgroup of $G$ and
$F_XKS= XS = G$. Since $KS \ge Q_G$, we infer that $KS = G$. As
$F_X$ is a $p'$-group, we get $F_X=1$ and we have a contradiction to
$F_X \not\le \Phi(X)$. This concludes the proof of
Lemma~\ref{L2.11.1}
\end{proof}
\be{proposition}{P2.5} Suppose that $p$ is a prime, $G$ is a finite
non-abelian simple ${\mathcal K}$-group and $S$ is a Sylow
$p$-subgroup of $G$. If $S$ is abelian, then $N_G(S)$ acts
irreducibly on $\Omega_1(S)$.
\en{proposition} \begin{proof} See \cite[12-1,pg 158]{GL} or
\cite[7.8.1]{GLS3}.\end{proof}
\section{$p$-minimal groups}
The first result of this section is elementary to prove but is
never-the-less a fundamental result for our work on $p$-minimal
subgroups.
\be{lemma}{L2.8} For $G$ a finite group, $p$ a prime and $S$ a Sylow
$p$-subgroup of $G$, $G= \<\P_G(S)\>N_G(S)$.\en{lemma}
\begin{proof} This is also proved in \cite[1.3]{PPS}.
By induction on $|G|$.
If $S$ is contained in a unique maximal subgroup of $G$, then either
$S$ is
normal in $G$ or $G \in
\P_G(S)$ and the lemma holds. Otherwise we can choose different
maximal subgroups $M_1$ and $M_2$ of $G$ which contain $S$. Since
$G= \$, applying induction to $M_1$ and $M_2$ gives $G=
\<\P_G(S)\>N_G(S)$.
\end{proof}
We now move on to study the structure of $p$-minimal groups -- here
is the main structural result for such groups.
\be{lemma}{L2.9} Suppose that $p$ is a prime, $P$ is $p$-minimal and
$S \in \Syl_p(P)$. Let $M$ be the (unique) maximal subgroup of $P$
containing $S$ and set $F :=\bigcap M^P$. Then the following hold.
\bl
\li a $Q_P\in \Syl_p(F)$.
\li d $F=\Phi_p(P)$ and, in particular, if $O_p(O^p(P)) =1$, then
$F$ is nilpotent.
\li c If $N$ is a subnormal subgroup of $P$ contained in $M$, then
$N\cap S\leq Q_P$.
\li b If $O^p(P)$ is $p$-closed, then $P$ is a $\{t,p\}$-group for
some prime $t\neq p$.
\li e For $N\unlhd P$, either $O^p(P)\leq N$ or $N\leq F$. In
particular, $P=O^{p^\prime}(P)$.
\li z $O^p(P)/(F\cap O^p(P))$ is a minimal normal subgroup of
$P/(F\cap O^p(P))$.
\li y If $P$ is soluble, then $O^p(P)$ is $p$-closed and $P$ is a
$\{t,p\}$-group for some prime $t\neq p$.
\li f $O^p(P)=[O^p(P),P]$.
\el \en{lemma}
\begin{proof} Parts \rf a and \rf d follow immediately from
Lemma~\ref{Fratrec}(a) and (b).
\rf c From Lemma~\ref{L2.2}, $N \le \bigcap M^P =F$. Hence by
\rf{a}, $N \cap S \le Q_P$.
\rf b For $r$ a prime divisor of $|O^p(P)|$ with $r\neq p$, $O^p(P)$
possesses a Sylow $r$-subgroup $S_r$ such that $O_p(O^p(P))S_r$ is
$S$-invariant. If \rf b were false, then we would get $SS_r\le M$
for all prime divisors $r$ of $|O^p(P)|$, which then forces $M=P$.
Thus \rf b holds.
\rf e Clearly $S \le SN \le P$ and so either $SN=P$ or $SN \le M$.
The former alternative yields $O^p(P) \le N$ and the latter $N \le
\bigcap M^P =F$, as required. If
$O^{p^\prime}(P)\leq F$, then by \rf a $Q_P=S$ a contradiction. Thus
$O^{p^\prime}(P)\nleq F$ and so $P=SO^p(P)\leq O^{p^\prime}(P)$.
\rf z Note that $O^p(P) > F\cap O^p(P)$, as $O^p(P) \le F$ would
give $FS = P$ whereas $FS \le M$. Then \rf z follows immediately
from \rf e.
\rf y Suppose that $P$ is soluble. Hence, by \rf z, $O^p(P)/(F\cap
O^p(P))$ is a $t$-group, $t$ a prime with $t\neq p$. Since $F$ is
$p$-closed by \rf a, $F\cap O^p(P)$ is also $p$-closed and so
$O^p(P)$ is $p$-closed. Now \rf{b} gives the result.
\rf f As $[O^p(P),P]\normal P$, applying \rf e gives either $O^p(P)
\le [O^p(P),P]$ or $[O^p(P),P] \le F$. So to prove \rf e we must
show that $[O^p(P),P] \le F$ cannot occur. Suppose that $[O^p(P),P]
\le F$. Then $[O^p(P),M] \le F$ and, as $O^p(P)M=P$, we infer that
$M \normal P$. Hence $M=F$ and so, by \rf a, $S = Q_P$. But then $S
\normal P$ contrary to $P$ being $p$-minimal. Thus $[O^p(P),P] \le
F$ cannot hold.
\end{proof}
It is of course important to know about the quotients of a
$p$-minimal group $P$. For $N$ a normal subgroup of $P$, having
$O^p(P) \le N$ is the kiss of death as far as $P/N$ being a
$p$-minimal group (because a $p$-group is not $p$-minimal). However,
as we see next, this is the only bad case.
\be{lemma}{L2.10} Suppose that $p$ is a prime, $P$ is a
$p$-minimal group and let $N$ be a normal subgroup of $P$ with
$O^p(P)\nleq N$. Put $\ov P := P/N$. Then \bl \li a $\ov P$ is a
$p$-minimal group;
\li b $P/\Phi_p(P)\cong \ov P/\Phi_p(\ov P)$; and
\li c if $P$ is narrow, then so is $\ov P$. \el \en{lemma}
\begin{proof} Let $S \in \Syl_p(P)$ and let $M$ be the unique maximal subgroup of $P$
containing $S$. By Lemma~\rf[L2.9]{e}, as $O^p(P) \not \le N$, $N
\le M$ and hence $\ov{M}$ is the unique maximal subgroup of $\ov P$
containing $\ov S$. Suppose that $\ov S \normal \ov P$. Then, by the
Frattini argument, $P = N_P(S)N \le M$, which is impossible.
Therefore $\ov{S}$ is not normal in $\ov P$ and so $\ov P$ is a
$p$-minimal group. Clearly \rf{b} follows from Lemma~\rf[L2.9]{d}
and \rf c follows from \rf a and \rf b.
\end{proof}
\be{lemma}{Opprime} $\P_G(S)=\P_{O^{p^\prime}(G)}(S)$ and
$O^{p^\prime}(G)= \<\P_G(S)\>$.\en{lemma}
\begin{proof} Set $X:= O^{p^\prime}(G)$.
The inclusion $\P_{X}(S)\subseteq \P_G(S)$ is obvious. Assume that
$P\in \P_G(S)$. Then, by Lemma~\rf[L2.9]{e}, $P= O^{p'}(P) \le P
\cap X$. Thus $\P_{G}(S)\subseteq \P_X(S)$ and so $\P_G(S)=\P_X(S)$.
Since $\<\P_{X}(S)\>=\<\P_{G}(S)\>$ is normalized by $N_G(S)$,
Lemma~\ref{L2.8} implies that
$\<\P_{X}(S)\>$ is a normal subgroup of
$G$ which contains $S$. Thus $\<\P_{X}(S)\>=O^{p^\prime}(G)$.
\end{proof}
We next see that $p$-minimal subgroups are nicely located within
central products.
\be{lemma}{L2.85} Assume that $G$ has subgroups $H_1$ and $H_2$ such
that $[H_1,H_2]=1$ and $G = H_1H_2$. Then
$$\P_G(S) = \P_{H_1S}(S)\cup\P_{H_2S}(S).$$ \en{lemma}
\begin{proof} For $i=1,2$, set $S_i := S\cap H_i$ and let $P \in \P_G(S)$.
Then, by Lemma~\rf[L2.9]{e}, $P = \\$. {Let $M$ be
the
unique maximal subgroup of $P$ containing $S$. We may assume $\\nleq
M$. So, using Lemma~\ref{L2.9}(e), $O^p(P)\leq \\leq H_1$ and $P\leq H_1S$. Thus $P\in
\P_{H_1S}(S)$.}
\end{proof}
We continue this section by illuminating, in the next two lemmas,
the structure of narrow groups with $O^p(P/Q_P) \in \mathcal
L_1(p)$.
\be{lemma}{rank11} Suppose that $p$ is a prime, $P$ is a narrow
group and
$O^p(\ov P)\in \mathcal L_1(p)$ where $\ov{P} =P/Q_P$. Then exactly
one of the following holds.
\bl \li a $p^a \ge 4$ and $\ov P$ is isomorphic to $\SL_2(p^a)$ or
$\PSL_2(p^a)$ perhaps extended by field automorphisms of order a
power of $p$.
\li b $p^a \ge 3$ and $\ov P$ is isomorphic to $\SU_3(p^a)$ or
$\PSU_3(p^a)$ perhaps extended by field automorphisms of order a
power of $p$.
\li c $p=2$, $a >1$ and $\ov P$ is isomorphic to ${}^2\B_2(2^a)$.
\li d $p=3$, $a>1$ and $\ov P$ is isomorphic to ${}^2\G_2(3^a)$
perhaps extended by field automorphisms of order a power of $3$.
\li e $p=2$, $O^2(\ov P) \cong 3$ and $\ov P \cong \SL_2(2)\cong
\Sym(3).$
\li f $p=3$, $O^3(\ov P) \cong \Q_8$ or $2^2$ and, respectively,
$\ov P \cong \SL_2(3)\cong 2\udot \Alt(4)$ or $\PSL_2(3) \cong
\Alt(4)$.
\li g $p=2$, $O^2(\ov P) \cong 3^{1+2}_+$ or $3^2$ and $P/O^2(P)Q_P
\cong \SD(16)$, $8$ or $\Q_8$.
\li h $p=2$, $O^2(\ov P) \cong 5$ and $\ov P \cong \Dih(10)$ or
${}^2\B_2(2)$.
\li i $p=3$ and $\ov P$ is isomorphic to ${}^2\G_2(3)^\prime \cong
\SL_2(8)$ or ${}^2\G_2(3) \cong \SL_2(8):3$.
\el
\en{lemma}
\begin{proof} The structure of $O^p(\ov P)$ follows directly from
the structure of the rank one Lie type groups and then the rest
follows from the structure of the automorphism groups of $O^p(\ov
P)$ and the fact that $P=O^{p'}(P)$ is $p$-minimal. {However, we
note in detail the special case of $\ov{P} \cong \SU_3(2)$ or
$\PSU_3(2)$. In these cases we have that $O^2(P) \cong 3_+^{1+2}$ or
$3^2$ respectively. Thus $\Out(O^2(\ov P)) \cong \GL_2(3)$ and so
$X:=P/O^2(P)Q_P$ is isomorphic to a subgroup of $\SD(16)$ (which is
isomorphic to a Sylow $2$-subgroup of $\GL_2(3)$). Since $\ov{P}$
is $2$-minimal, we have that $X$ acts irreducibly on $O^2(\ov
P)/\Phi(O^2(\ov P))\cong 3^2$. Thus $X \cong \SD(16)$, $8$, $\Q_8$,
$\Dih(8)$ or $4$. However, the latter two groups act imprimitively
on $O^2(\ov P)/\Phi(O^2(\ov P))$ and so these two possibilities do
not arise in \rf g.}
%
%
%
%
%\red{I suggest to add a few more details. At least one should point that
% 'narrow' is used to rule out $P/O^2(P)Q_P\cong \Dih(8), 4$ in \rf g}
\end{proof}
\be{lemma}{rank1grps} Suppose that $p$ is a prime, $P$ is narrow and
that $O^p(\ov P)\in {\mathcal L}_1(p)$ where $\ov{P}:=P/Q_P$. Let
$\ov{S} \in \Syl_p(\ov P)$, $\ov M$ be the unique maximal subgroup
of $\ov P$ containing $\ov S$ and $\ov R := N_{O^p(\ov P)}(\ov S\cap
O^p(\ov P))$.
\bl
\li {a} If $\ov P $ has abelian Sylow $p$-subgroups, then either
\bl \li [(i)] i
$\ov P\cong \PSL_2(p^a)$ or $\SL_2(p^a)$ for some $a\ge 1$; \li [(ii)]{ii} $p=3$ and $\ov P \cong {}^2\G_2(3)^\prime \cong
\SL_2(8)$;
\li [(iii)]{iii} $p=2$ and $\ov {P}\cong {}^2\B_2(2)$ or $\Dih(10)$; or
\li [(iv)]{iv}$p=2$ and {$\ov P \cong 3^{1+2}_+:8$ or $3^{2}:8$.}
\el
\li{b} If $\ov P$ has cyclic Sylow $p$-subgroups, then $\ov P\cong
\PSL_2(p)$, $\SL_2(p)$, ${}^2\G_2(3)^\prime$, ${}^2\B_2(2)$,
$\Dih(10)$, $3^{1+2}_+:8$, and $3^{2}:8$.
\li{c} If $\ov D\le \ov S$ is normal in $\ov M$ with {$\ov D$
abelian but not
elementary abelian,} then $\ov D$ is cyclic and either $\ov P$
is soluble, or $p=3$ and $\ov P^\prime\cong {}^2\G_2(3)^\prime$.
\li {d-1} $\ov R$ is soluble.
\li{d} Either $C_{\ov S}(\ov R)=1$ or $\ov P$ is soluble.
\li{e} If $\ov R$ normalizes a non-trivial cyclic subgroup of $\ov
S$, then either $|\ov S|\le p^3$, or $p=2$ and $\ov S \cong
\SD(16)$.
\li {f} If $P$ is soluble, then $p\le 3$.
\el
\en{lemma}
\begin{proof}
{Suppose first that $\ov P$ is soluble. Then Lemma~\ref{rank11}
implies that $p=2$ or $3$ and gives an explicit description of $\ov
P$. With this information it is easy to verify all the claims in the
lemma in this case. So assume that $\ov P$ is not soluble. Let $\ov
X = O^p(\ov P)$ and $\ov T = \ov S \cap \ov X$. If $\ov X=
{}^2\GG_2(3)^\prime\cong \SL_2(8)$. Then $\ov P$ has cyclic Sylow
$3$-subgroups if and only if $\ov P= \ov X$. In this case we have
$\ov S$ is cyclic of order $9$ and is inverted by $\ov R$. Thus (a),
(b), (c) and (d) hold in this case. Since $\ov S$ is either cyclic
or extraspecial, and $\ov R$ inverts $\ov T$, $C_{\ov S}(\ov R)
=1$. Hence \rf d is true and, as $|\ov S| \le 3^3$, \rf e holds as
well. So we may assume that $\ov X \not \cong {}^2\GG_2(3)^\prime$.
Hence we now consider the cases when $\ov X$ is not soluble and
$$\ov X \in \{\SL_2(p^a), \PSL_2(p^a),
\SU_3(p^a),\PSU_3(p^a),{}^2\B_2(2^{2a+1}), {}^2\GG_2(3^{2a+1})\mid a
\ge 1\}.$$ In particular, we note that $p$ divides $|\ov X|$. If
$\ov P$ has abelian Sylow $p$-subgroups, then so does $\ov X$. Thus
the candidates for the groups $\ov P$ with abelian Sylow
$p$-subgroups are as listed in (a) and (b). Now for (a) we note that
if $\ov{P} \not = \ov X$, then, by Lemma~\ref{rank11} (a) and (i),
$\ov{P}$ includes field automorphisms of $\SL_2(p^a)$. Since the
field automorphism of $\SL_2(p^a) $ of order $p$ centralizes a
subgroup of $\ov X$ which has Sylow $p$-subgroups of order
$p^{a/p}$, we have $\ov{S}$ is abelian if and only if $\ov P =
\ov X$. Hence (a) holds and (b) follows from (a).
Suppose that $\ov D \le \ov S$ where $\ov D$ is
abelian and is normalized by $\ov M= \ov R\ov S$. We will prove \rf
c by showing that $\ov D$ is elementary abelian. Since $[\ov R,\ov
D] \le \ov R \cap \ov D \le \ov T$, we infer that $\ov D \le \ov T$
as in all the cases we are considering the field automorphisms of
$\ov X$ do not centralize $\ov R/\ov T$.
If $\ov X \cong
\SL_2(p^a)$ or $\PSL_2(p^a)$, we get $\ov D$ is elementary
abelian. Now suppose that $\ov X \cong \SU_3(p^a)$ or $\PSU_3(p^a)$. We use
\cite[II 10.12]{Huppert} to deduce the required facts about the
structure of $\ov X$. If $\ov D \not \le Z(\ov T)$, then, as $\ov R$
acts irreducibly on $\ov T/Z(\ov T)$, we have $\ov T = \ov D Z(\ov
T)$. Since $Z(\ov T)= \Phi (\ov T)$, this means $\ov D =\ov T$ is
not abelian which is impossible. Thus $\ov D \le Z(\ov T)$ is
elementary abelian. In the case when $\ov X = {}^2\B_2(2^{2a+1})$, a
similar argument to the one above shows that $\ov D$ is elementary
abelian . Suppose that $\ov X \cong {}^2\GG_2(3^{2a+1})$ with $a \ge
1$. We use \cite[13.2]{HuppertB3} or \cite{Ward} to extract facts
about the structure of $\ov T$. This time $\ov R$ acts irreducibly
on $\ov T/\Phi(\ov T)$, $\Phi(\ov T)/Z(\ov T)$ and on $Z(\ov T)$ and
each of these groups has order $3^{2a+1}$. Furthermore $\Phi(\ov T)$
is elementary abelian. If $\ov D \not \le \Phi(\ov T)$, then $\ov T
= \ov D \Phi(\ov T)$ and so $\ov D = \ov T$, again a contradiction.
Thus $\ov D \le \Phi(\ov T)$ and so $\ov D$ is elementary abelian.
This completes the proof of \rf c. In particular, we note from the
above proof that the irreducible action of $\ov R$ implies that $\ov
D \ge Z(\ov T)$ for all abelian subgroups of $\ov S$ normalized by
$\ov R$. Thus if $\ov R$ normalizes a non-trivial cyclic subgroup
$\ov D$ of $\ov S$, we must have $\O_1(\ov D) = Z(\ov T)$ has order
$p$. It follows that $|\ov S| =|\ov T|\le p^3$ and thus \rf e
holds.
We know from the above sources that $\ov{R}/\ov T$ is a cyclic
group. Thus $\ov R$ is soluble and (d) holds.
Finally for part \rf{d} we cite \cite[(5.1)(e)]{DS}.}
\end{proof}
For a $p$-group $S$, $p$ a prime, we use $J(S)$ to denote the
elementary abelian version of the Thompson subgroup of $S$. That is,
$J(S) := \< \mathfrak A_e(S)\>$ where $\mathfrak A_e(S)$ is defined
to be the set of elementary abelian subgroups of $S$ which have
maximal rank.
\be{theorem}{T2.6} Suppose that $p$ is a prime, $P$ is a $p$-minimal
group, $C_P(Q_P)\le Q_P$, $S\in \Syl_p(P)$ and $P> \langle
C_P(\O_1(Z(S))),N_P(J(S))\rangle$. Let $V:= \O_1(Z(Q_P))$ and $\ov P
:= P/C_P(V)$. Then there exist subgroups $E_1,\dots, E_n $ of $P$
such that
\bl \li a $S$ acts transitively on $\{E_1,\dots,E_n\}$ by
conjugation; \li b $[\ov{E_i},\ov{E_j}]=1 $ for $i\neq j$;
\li c either $\ov{E_i} \cong \SL_2(p^n)$ for some $n \ge 1$ or $p=2$
and $\ov{E_i} \cong \Sym(2^m+1)$ for some $m \ge 1$; and
\li d $V/C_V(E_i)$ is a {corresponding} natural $\ov E_i$-module
and, for $i\neq j$, $V/C_V(E_i)$ is centralized by $E_j$.
\el \en{theorem}
\begin{proof} This is the $C^{**}(G,T)$-Theorem from \cite{Bundyetal}.
\end{proof}
\be{corollary}{L2.7} Suppose that $P$ is as in Theorem~\ref{T2.6}
(and use the notation there). Let $M$ be the unique maximal subgroup
of $P$ containing $S$, and set $E :=\langle E_i\mid 1\le i\le
n\rangle$. Assume there is a non-trivial subgroup $A$ of $V$ which
is centralized by $M$ and such that $A\cap C_V(E)=1$. Then $p=2$,
$\ov{E_i}\cong \Sym(2^m+1)$ and $AC_V(E)/C_V(E) = C_{V/C_V(E)}(S)$
has order $2$. \en{corollary}
\begin{proof} As $A$ is centralized by $S$ and $S$ is transitive on
$\{E_1,\dots,E_n\}$, $A\cap C_V(E_1) \le A\cap C_V(E)=1$. Therefore,
$|A|= |AC_V(E_1)/C_V(E_1)|$. Now $M\cap E_1$ is the unique maximal
subgroup of $E_1$ containing $S\cap E_1\in \Syl_p(E_1)$ and so
$C_{V/C_V(E_1)}(M\cap E_1)=1$ whenever $V/C_V(E_1)$ is a natural
module for $\ov{E_1} \cong \SL_2(p^n)$ with $p^n \neq 2$. Since
$AC_V(E_1)/C_V(E_1)\le C_{V/C_V(E_1)}(M\cap E_1)$, we infer that
$p=2$, $\ov{E_1} \cong \Sym(2^m+1)$ and then that
$|AC_V(E_1)/C_V(E_1)|=2$ by direct calculation in the natural
$\Sym(2^m+1)$-module.
\end{proof}
\section{Finite groups with isolated subgroups}
Throughout this section $G$ is assumed to be a finite group, $p$ a
prime and $S\in Syl_p(G)$. Our first three results give some
elementary properties of isolated subgroups, Lemma~\rf[L3.1]{a}
having already been mentioned in Section 1.
\be{lemma}{Opprime2} Suppose that $P\in \P_G(S)$ is an $A$-isolated
subgroup of $G$. Then $P$ is an $A$-isolated subgroup of
$O^{p^\prime}(G)$. \en{lemma}
\begin{proof} Set $X := O^{p^\prime}(G)$. Suppose that $P\in \P_G(S)$ is $A$-isolated. Then, by
Lemma~\ref{Opprime}, $P\in \P_X(S)$, $A$ is normal in $L_X(P,S)$
and, of course, $A\not\le Q_P$. Thus $P$ is $A$-isolated in $X$.
\end{proof}
\be{lemma}{L3.1} Let $P\in \P_G(S)$ be an isolated subgroup of $G$.
Then \bl \li a $P\not \le L_G(P,S)$ and $L_G(P,S) \neq G$; and \li
b$N_G(S)\leq N_G(P)$.\el
\en{lemma} \begin{proof} Set $L:= L_G(P,S)$. If $P\le L$, then $Q_L \le Q_P$, contrary to $P$ being an isolated subgroup. So $P\not \le
L$, and \rf{a} holds. For $g\in N_G(S)$, $P^g \in \P_G(S)$ and, as
$N_G(S) \le L$ and $P\not\le L$, $P^g \not\le L$. Since $P^g$ is also $p$-minimal,
$P=P^g$
and the lemma holds. \end{proof}
The next lemma is the initial structural result about groups which
possess an isolated subgroup.
\be{lemma}{L3.2} Assume that $P\in \P_G(S)$ is an $A$-isolated
subgroup, and set $L:=L_G(P,S)$.
\bl
\li{a} Suppose that $S \le H \le G$ and $H \not \le L$. Then
$P \le H$, $L_H(P,S)=L\cap H$ is a maximal subgroup of $H$ and $P$ is an $A$-isolated subgroup of $H$.
\li{b} $P\cap L$ is the unique maximal subgroup of $P$
containing $S$.
\li{c} $L$ is a maximal subgroup of $G$.
\el
\en{lemma}
\begin{proof} From Lemma~\ref{L2.8}, $H = \<\P_H(S)\>N_H(S)$ and so,
as $H\not\le L$, $P\in \P_H(S)$. Thus $P\le H$. By
Lemma~\rf[L3.1]{a} and the definition of $L$, $\P_H(S)\setminus\{P\}
=\P_{L\cap H}(S)$. Since $N_H(S) \le L\cap H$, using
Lemma~\ref{L2.8} gives
$$L_H(P,S) = \<\P_{H}(S)\setminus\{P\}\> N_H(S)=L\cap H.$$
If $L\cap H N_K(S)=\<\P_H(S)\> N_H(S)=H.$$
Thus $L\cap H$ is a maximal subgroup of $H$. Since $A$ is a normal
$p$-subgroup of $L\cap H=L_H(P,S)$, $P$ is an $A$-isolated subgroup
of $H$ and this proves \rf{a}. Parts \rf{b} and \rf{c} follow from
\rf{a} by taking, respectively, $H=P$ and $H=G$.
\end{proof}
We now obtain further restrictions on the structure of $P$.
\be{lemma}{L3.4.5} Suppose that $P\in \P_G(S)$ is an $A$-isolated
subgroup of $G$ and $M$ is the unique maximal subgroup of $P$
containing $S$. Put $X:=O^p(P)$ and $F := \bigcap M^P$. Then
\bl \li a $XA = \$;
\li b$[XA,X\cap F]\le O_p(X)$;
\li c $X/O_p(X)$ is either a central product of
quasisimple groups transitively permuted by $S$ or there is a prime $t$
such that $X/O_p(X)$ is a $t$-group of
class at most $2$; and
\li d $\Phi_p(X)=X \cap F$.
\el \en{lemma}
\begin{proof} Since $F$ is $p$-closed by Lemma~\rf[L2.9]{a} and
$A\not \le Q_P$, Lemma~\rf[L2.9]{e} implies that $O^p(P)\le \<
A^P\>$. So \rf{a} holds.
Set $L:=L_G(P,S)$. Then $(F\cap X)S \le M= P\cap L$
by Lemma~\rf[L3.2]{b}. Thus $F\cap X$ normalizes $A$.
Hence, by Lemma~\rf[L2.9]{a},
$$[A,F\cap X] \le Q_L \cap F \cap X \le Q_P \cap X=
O_p(X).$$ Hence \rf b follows from \rf a.
From Lemma~\rf[L2.9]{z} $X/(X \cap F)$ is a minimal normal subgroup
of $P/(X \cap F)$. Hence $X/(X \cap F)$ is isomorphic to a direct
product of simple groups. If $X/(X \cap F)$ is an elementary abelian
$t$-group ($t$ a prime), then $X$ is $p$-closed, whence $X/O_p(X)$
is a $t$-group by Lemma~\rf[L2.9]{b}. Since $[X,X \cap F] \le
O_p(X)$, $X/O_p(X)$ will also have class at most two. In the case
when $X/(X \cap F)$ is a direct product of non abelian simple
groups, $[X,X \cap F] \le O_p(X)$ and $P$ being $p$-minimal force
$X/(X \cap F)$ to be a central product of quasisimple groups
transitively permuted by $S$. So \rf c holds.
Finally, combining Lemmas~\ref{L2.11.1} and ~\rf[L2.9]{d} gives
$$\Phi_p(X)=X \cap \Phi_p(P) = X \cap F.$$
\end{proof}
\be{lemma}{L3.3} Suppose that $P\in \P_G(S)$ is an $A$-isolated
subgroup of $G$. \bl \li a If $A \le T \normal S$, then $N_G(T) \le
L_G(P,S)$.\li b If $S \le H \le G$ and $Q_H \le Q_P$, then $P\le
H$.\el \en{lemma}
\begin{proof} Put $L:= L_G(P,S)$. Assume that $A \le T \normal S$ and $N_G(T) \not \le
L$. Since $S \le N_G(T)$, applying Lemma~\rf[L3.2]{a} yields $P\le
N_G(T)$. But then $$A \le T \le O_p(N_G(T)) \le Q_P,$$ which is not
the case. Thus $N_G(T)\le L$, and we have \rf{a}. Now assume that $S
\le H \le G$ with $Q_H \le Q_P$. If we have $H\le L$ then $$A \le
Q_L \le Q_H \le Q_P, $$ contrary to $P$ being an $A$-isolated
subgroup. Therefore $H \not \le L$ and, again by Lemma~\rf[L3.2]{a},
$P\le H$, so proving the lemma.
\end{proof}
As we shall see in Lemma~\ref{L3.4} the property of having an
isolated subgroup is preserved when taking certain quotients.
\be{lemma}{L3.4}
Suppose that $P\in \P_G(S)$ and let $N$ be a
normal subgroup of $G$ with $O^p(P)\nleq N$. Put $L:= L_G(P,S)$,
$Y:= \$ and $\ov G := G/N$. Then the following hold.
\bl \li a $P\cap N\leq \Phi_p(P)$.
\li{a+} $G=LY$.
\li c If $P$ is an isolated subgroup of $G$, then $N\leq \bigcap
L^G$ and
$$\bigcap L^G=C_G(\/Q_G)=C_L(YQ_G/Q_G).$$
\li b If $P$ is an $A$-isolated subgroup of $G$, then $\ov
P$ is an $\ov A$-isolated subgroup of $\ov G$.
\el
\en{lemma}
\begin{proof} Part \rf{a} follows immediately from
Lemmas~\rf[L2.9]{d} and \rf[L2.9]{e}, and part \rf{a+} follows from
Lemma~\ref{L2.8} as $LY$ contains {$N_G(S)$ and} all the $p$-minimal
subgroups containing $S$.
Set $ Z := \bigcap L^G$. Since $O^p(P)\nleq N$, $P\nleq NS$ and so
$NS\leq L$ by Lemma~\rf[L3.2]{a}. Hence $N\leq Z$. Put $J:=
C_G(\/Q_G)$ and $K :=C_L(YQ_G/Q_G)$. Because $P$ is an
isolated subgroup of $G$, Lemma~\ref{L3.4.5}(a) gives $O^p(P) \le
\$ and so {$Y\leq \$. Since $Q_L\normal JS$, $P\nleq
JS$ and so $JS\leq L$. Therefore $J\le
K$. Because $Z\leq L$, $Z$ normalizes $Q_L$ and so $[Q_L,Z]\le Q_L
\cap Z=
Q_Z\le Q_G$. So we have $Z\le J\le K$. Since $Q_G\leq K$, $Y$
normalizes $K$ and so $K\normal
\ =G$.} So, as $K\le L$, $K\le Z$. Thus $Z=J=K$ and \rf c
holds.
{Since $\ov P=PN/N \cong P/(P\cap N)$, Lemma~\rf[L2.10]{a} implies
that $\ov P \in \P_{\ov{G}}(\ov S)$.} By (c) we also have $\ov P
\not\le \ov L$. Now suppose that $\ov R \in \P_{\ov G}(\ov S)$ with
$\ov R \not \le \ov L$. Then, as $N \le L$ by \rf c, $R \not \le L$.
Since $R \ge NS \ge S$, we infer from Lemma~\rf[L3.2]{a} that $P \le
R$ and that, as $\ov R$ is $p$-minimal, $L \cap R$ is the unique
maximal subgroup of $R$ containing $NS$. But $P \ge S$ and so $PN
\ge NS$. Since $P \not \le L$, we deduce that $PN= R$ and that
$\ov{R} = \ov{P}$. Hence, as $N_{\ov G}(\ov S) =\ov{N_G(S)}\le\ov
L$, $\ov L = L_{\ov{G}}(\ov{P},\ov{S})$. Finally assume that $\ov{A}
\le Q_{\ov{P}}$. Let $K$ denote the inverse image in $G$ of $Q_{\ov
P}$. Evidently $\ \le K$. {By Lemma~\rf[L3.4.5] a, $O^p(P) \le
\$.} So $O^p(P) \le K$ and consequently, as $K/N$ is a
{$p$-group}, we obtain $O^p(P) \le N$ whereas $O^p(P) \not \le N$.
Therefore $\ov P$ is $\ov A$-isolated in $\ov G$ and so \rf{b}
holds.
\end{proof}
\be{lemma}{L3.5} Suppose that $P\in \P_G(S)$ is an $A$-isolated
subgroup of $G$. If $O^p(G)$ is $p$-closed, then $G = L_G(P,S)P$ and
$\ = \$. \en{lemma}
\begin{proof} We set $L:=L_G(P,S)$, and prove that $G=LP$.
Because of Lemma~\rf[L3.4]{b}, without loss of generality we may
assume that $Q_G=1$. So $O^p(G) $ is a $p^\prime$-group, and hence
$O^p(P)$ is also a $p^\prime$-group. Therefore, by
Lemma~\ref{L2.9}(d), $P$ is a $\{t,p\}$-group for some prime $t\neq
p$. For each prime $r$ dividing $|O^p(G)|$, $O^p(G)$ has an
$S$-invariant Sylow $r$-subgroup, denoted $S_r$, and we may select
$S_t$ so that $P \le S S_t$. Therefore $P= (P\cap S_t)S$. If $r$ is
a prime divisor of $|O^p(G)|$ with $r\neq t$, then evidently $P \not
\le SS_r$ and consequently $SS_r \le L$ by Lemma~\rf[L3.2]{a}. Thus
$O^p(G) = (O^p(G) \cap L)S_t$. Since $SS_t \not\le L$,
Lemma~\rf[L3.2]{a} implies that $SS_t \cap L= S(S_t \cap L)$ is a
maximal subgroup of $SS_t$. Now $N_{S_t}(S_t \cap L)$ is
$S$-invariant and so $SN_{S_t}(S_t \cap L)$ is a subgroup of $SS_t$.
Thus $S_t \cap L \normal S_t$ and $S$ acts irreducibly on $S_t/(S_t
\cap L)$. As $S_t \cap P \not \le S_t \cap L$, this gives $S_t =
(S_t \cap L)(S_t \cap P)$. Hence we have
\begin{eqnarray*}
G&=& O^p(G)S = (O^p(G) \cap L)S_tS\\&=& (O^p(G) \cap L)(S_t \cap
L)(S_t \cap P)S = LP.
\end{eqnarray*}
Finally it follows that $\ = \ = \$ and this
completes the proof of Lemma~\ref{L3.5}.
\end{proof}
{ \be{corollary}{divp} Suppose that $P\in \P_G(S)$ is an
$A$-isolated subgroup of $G$. If $Y:=\$ is $p$-closed,
then $Y= O^p(P)$. \en{corollary}
\begin{proof} We have that $P$ is an $A$-isolated subgroup of
$H:=SY$. Since $O^p(H) \le Y$ and $Y$ is $p$-closed, $O^p(H)$ is
$p$-closed. Therefore as $A$ is normalized by $L:= L_G(P,S)$,
Lemmas~\rf[L3.4]{a+} and \ref{L3.5} imply
$$\ = \=\= \.$$ By
Lemma~\rf[L3.4.5]{a}, $O^p(P)$ is a characteristic subgroup of
$\$ and so $O^p(P)$ is normal in $G$. Thus $Y=\ =
O^p(P)$, as claimed.
\end{proof}}
\be{lemma}{L3.6} Suppose that $P\in \P_G(S)$ is an $A$-isolated
subgroup of $G$, and set $L:=L_G(P,S)$, $Z:= \bigcap L^G$ and $Y :=
\< O^{p}(P)^G \>$.\bl
\li {a} If $N\normal G$ and $N\not \le L$, then $Y\le
N$.
\li{b} If $N$ is a proper characteristic subgroup of $Y$, then
$N\le Z$ and $[Y,N]\leq Q_G$.
%\li{c} $[Z,Y]\le Q_G$.
\li{d} Either $Y= O^p(P)$ and there exists a prime $t\neq p$ such that
$YQ_G/Q_G$ is a $t$-group of class at most $2$, or $YQ_G/Q_G$ is a central
product of quasisimple groups.
\li{e} Assume that $Y\neq O^p(P)$, and let $K \le Y$ be such
that $KQ_G/Q_G$ is a component in $YQ_G/Q_G$. Then $Y Q_G/Q_G=\langle K^S\rangle Q_G/Q_G$ and
$K\cap P\not\le L$.
\li{f} $Y \le F_p^*(G)$.
\el \en{lemma}
\begin{proof} \rf{a} Since $N \normal G$ and $N \not\le L$, $S \le NS
\le G$ and $NS \not\le L$. Thus $P\le NS$ by Lemma~\rf[L3.2]{a} and
so $O^p(P) \le O^p(NS)\le N$. Therefore $Y=\ \le N$.
\rf{b} Because $Y \not\le N$, part \rf{a} implies that $N \le L$.
Hence $N \le \bigcap L^G = Z$. {Thus, by Lemma~\rf[L3.4]{c},
$[N,Y]\leq Q_G$.}
For the proof of parts \rf{d} and \rf{e}, by Lemma~\rf[L3.4]{b}, we
may suppose without loss of generality that $Q_G=1$.
%
%\rf{c} Now, as $Z\le L$, $Z$ normalizes $A$ and therefore $A\cap
%Z\le Q_Z \le Q_G =1$. Hence $[A,Z]\le A\cap Z=1$, which yields
%$[Z,\]=\<[Z,A]^G\>=1$. Zrom Lemma~\rf[L2.9]{a} and
%\rf[L2.9]{e} $\\ge O^p(P)$ whence $\ \ge \ =
%Y$. Thus $[Z,Y]=1$ and \rf{c} holds.
\rf{d} Suppose that $E(Y) \neq Y$. Then $[E(Y),Y]=1$ by {\rf{b}} and
so $E(Y)=1$. Hence $F^*(Y) = F(Y)$. Since $C_Y(F^*(Y))\le F^*(Y)$
using {\rf{b}} again we see that $Y= F^*(Y) = F(Y)$. So $Y$ is
nilpotent and in particular is $p$-closed. Applying
Corollary~\ref{divp} gives $Y= O^p(P)$. Therefore either $Y= E(Y)$
or $Y= O^p(P)$. Finally, in the latter case, Lemma~\rf[L3.4.5]{c}
implies that $Y$ is a $t$-group of class at most $2$ for some prime
$t \neq p$.
\rf{e} Assume that $Y \neq O^p(P)$ and let $K$ be a component of
$Y$. Then $K \normal Y$ by \rf{d}. Hence, as $G=YL$,
$$\ = \ =\. $$ Since $[Z,Y]=1$
by Lemma~\rf[L3.4]{c}, we infer that $K \not\le L$. Therefore
$\ \ge O^p(P)$ by Lemma~\rf[L3.2]{a}. If $K_1 $ is a component
of $Y$ which is not contained in $\$, then using {\rf{b}} we
get
$$O^p(P) \le \ \cap \\le Z(Y) \le Z\le L,$$ which is
impossible. Therefore $Y=\$ and the first part of \rf{e}
holds.
Assume that $K \cap P\leq L$, and argue for a contradiction. So
$K\cap P \le L\cap P$ which is the unique maximal subgroup of $P$ containing $S$.
Since $(K\cap P)\normal\normal P$,
Lemma~\rf[L2.9] {c} gives $K \cap S = (K\cap P) \cap S \le Q_{P}$.
Hence, as $Y= \$, $Y \cap S = Y\cap Q_{ P}\normal P$ and, in
particular, $O^p( P)$ is $p$-closed. Set $R := N_Y(Y\cap S)S$. Then
$O^p(R)$ is $p$-closed and $R \ge P$. Therefore, $R= (R\cap L)P$ and
$\=\\le P$, using Lemma~\ref{L3.5}. Suppose that $A \le
N_G(K)$. Then
$$[A,K\cap R] \le K \cap \ \le K \cap P\le L$$ which implies
that $[A,K\cap R]$ normalizes $A$. It follows from Lemma~\ref{L2.3}
that $[A,K\cap R]$ is a $p$-group. Consequently, as $K\cap R
=N_K(K\cap S)$ is $p$-closed,
$$[A,K\cap R]\leq Q_P\cap K.$$ Thus $[A,R] =
\<[A,K\cap R]^S\> \le Q_P$. But then $\=\ \ge O^p(P)$ is
a $p$-group, a contradiction. Hence we conclude that $A \not \le
N_G(K)$. So $Q_L \not\le N_G(K)$. Since $[K\cap L,Q_L] $ is a
$p$-group we infer that $(K\cap L)Z(Y)/Z(Y)$ is a $p$-group which
yields that $K\cap L$ is nilpotent. Hence $K\cap S = O_p(K\cap L)$
and therefore $Y\cap S=O_p(Y\cap L)$. Thus $$Y\cap Q_P=Y \cap S
\normal \< P,Y\cap L\> = Y,$$ and we conclude that $Y$ is
$p$-closed. Now a final application of Corollary~\ref{divp} gives
$Y=O^p(Y)=O^p( P)$. By assumption $Y \neq O^p(P)$ and so this is the
desired contradiction. Therefore $K\cap P\not\le L$.
Part \rf f follows from {\rf d}.
\end{proof}
\be{lemma}{simple} Suppose that $P\in \P_G(S)$ is $A$-isolated and
narrow and set $Y: = \< O^{p}( P)^G \>$. Then either $Y=O^p(P)$ or
$Y Q_G/Q_G$ is quasisimple. \en{lemma}
\begin{proof} As usual we set $L:=L_G(P,S)$.
We assume that $Y\neq O^p(P)$ and aim to show that $YQ_G/Q_G$ is
quasisimple. Because of Lemma~\rf[L3.4]{b} there is no loss in
assuming that $\bigcap L^G=1$. {By Lemma~\rf[L3.6]{d}, $Z(Y)Q_Y< Y$
and so, by
Lemma~\rf[L3.6]{b}, $Z(Y)Q_Y\le \bigcap L^G=1$.}
Thus we seek to show that $Y$ is simple. Put $F:= \Phi_p(P)$ and $X
:= O^p(P)$. By Lemma~\rf[L3.6]{d} and \rf[L3.6]{e}, $Y$ is
semisimple and for any component $K$ of $Y$, $Y=\$ and $K\cap
P\not \le L$. In particular, if $K$ is normalized by $S$, then
$Y=\=K$ is simple and we are done. Hence we assume that $K$ is
not normalized by $S$ and look for a contradiction.
Suppose first that $XF/F$ is a non-abelian simple group. Observe
that $(K\cap P)F/F\cap XF/F \neq 1$. For if $(K\cap P)F/F\cap XF/F =
1$, then $(K\cap P)F/F$ is a $p$-group and so, as a result, $(K\cap
P)F \le FS\le L$ whereas $K\cap P \not\le L$. Then, as $X\leq Y$,
$(K\cap P)F/F$ is normalized by $XF/F$ and so $(K\cap P)F \ge XF$.
Now selecting $s\in S$ such that $K^s \not = K$, we have
$$X=[X,X] \le [XF,XF] \le [(K\cap P)F,(K^s\cap P)F]\le
[K, K^s]F =F,$$ which is a contradiction as $X $ is not contained in
$F$. Therefore, $XF/F$ is an elementary abelian $t$-group for some
prime $t\neq p$ and $P$ is a $\{t,p\}$-group. Moreover, by
Lemma~\rf[L2.9]{z}, $SF$ is a maximal subgroup of $P$. We have that
$O^p(K\cap P) \le O^p(P)= X$ and so, as $K\cap P\not\le L$,
$X=\$. Set $D:= O^p(K\cap P)$ and note that, as $Y$
is semisimple,
$$X = {\kreuz_{T \in D^S}}T.$$
By Lemma~\rf[L3.4.5]{d}, $\Phi_p(X)=X\cap F$. On the other hand, by
Lemma~\ref{FratProd}, $\Phi_p(X) = {\kreuz}_{T \in D^S}\Phi_p(T)$
and so we conclude that
$$XF/F \cong X/(X\cap F) = {\kreuz_{T \in D^S}}T/\Phi_p(T).$$ Since $P$
is narrow, it must be that $D^S=\{D\}$ and so $X=D$. Hence $K$ is
normalized by $S$, a contradiction.\end{proof}
It is now a simple matter to deduce Theorems~\ref{T1.3} and
~\ref{completelypsing} and Corollary~\ref{cor1}.
\begin{proof}[Proof of Theorem~\ref{T1.3}] Combining
{Corollary~\ref{divp},}
Lemma~\rf[L3.6]{d}, {(\ref{L3.6:e}) and (\ref{L3.6:f})} together with
Lemma~\ref{simple} yields Theorem~\ref{T1.3}. For part (b) we note
that $p$ divides $|Y/Q_Y|$ else $Y$ would be $p$-closed and so, by
part (a), $Y=O^p(P)$.
\end{proof}
\begin{proof}[Proof of {Corollary}~\ref{cor1}] This follows from
Theorem~\ref{T1.3} and Lemma~\rf[L3.6]{d}.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{completelypsing}]By Lemma~\rf[L3.6]{f},
$\<\P_G(S)\> \le F_p^*(G)$ and hence, by Lemma~\ref{L2.8}, $G =
F_p^*(G)N_G(S)$. The next lemma completes the proof of
Theorem~\ref{completelypsing}.\end{proof}
\be{lemma}{normal commute}
Suppose that $Q_G=1$ and $P_1,P_2\in P_G(S)$ are both isolated
subgroups of $G$. Then either $\=\$ or
$[\,\]=1$.
\en{lemma}
\begin{proof} For $i=1,2$, set $Y_i=\$ and $L_i= L_G(P_i,S)$.
If $Y_1$ and $Y_2$ are products of components of $G$ then
Theorem~\ref{T1.3} guarantees that either $Y_1=Y_2$ or
$[Y_1,Y_2]=1$. So suppose that $Y_1 = O^p(P_1)$ {and without loss
that $P_1\neq P_2$. Then
$P_1\leq L_2$ and so
\[[Q_{L_2},Y_1]=[Q_{L_2},O^p(P_1)]\leq O_p(O^p(P_1))\leq Q_G=1.\] Since
$O^p(P_2)\leq \$ by Lemma~\ref{L3.4.5}(a), we get $Y_2\leq \$ and
conclude that
$[Y_1,Y_2]=1$.}
\end{proof}
In the companion paper \cite{MPR} we will investigate specific
simple groups with an eye to showing that they have or do not have
an isolated narrow subgroup. The remaining results in this section
will be applied to proper subgroups of such groups. And in all these
results for $P\in \P_G(S)$ an isolated subgroup of $G$ we set
$L:=L_G(P,S)$ and $Y:= \langle O^p(P)^G\rangle$.
\be{lemma} {qs} Suppose that $P\in
\P_G(S)$ is an isolated subgroup in $G$. Then either $P$ is soluble
or $O^p(P)Q_P/Q_P$ is a central product of quasisimple groups. If,
additionally, $P$ is narrow, then either $P$ is soluble or
$O^p(P)Q_P/Q_P$ is quasisimple. \en{lemma}
\begin{proof} Let $M:= L \cap P$. Then, by Lemma~\rf[L3.2]{b}, $M$ is the unique
maximal subgroup of $P$ containing $S$. Put $F :=\bigcap M^P$. Then
$F$ is normal in $P$ and contained in $L$. Therefore $[Q_L,F] \le
Q_L\cap F \le S\cap F= Q_P$. Thus $F/Q_P$ is centralized by
$\ \ge O^p(P)$ and so $FQ_P/Q_P\le Z(O^p(P)Q_P/Q_P)$. Since
$P\in \P_G(S)$, we have that either $P$ is soluble or
$O^p(P)Q_P/Q_P$ is a central product of quasisimple groups. If $P$
is narrow, the latter case implies that $O^p(P)Q_P/Q_P$ is
quasisimple. \end{proof}
\be{lemma}{qs2}
Suppose that $O^p(G)$ is quasisimple and set $\ov G := G/Z(O^p(G))$.
If $P\in \P_G(S)$ is an $A$-isolated subgroup of $G$, then $\ov P$ is an $\ov{A}$-isolated subgroup of $\ov G$.
\en{lemma}
\begin{proof} Set $Z := Z(O^p(G))$ and $L:=L_G(P,S)$. If $O^p(P) \not\le Z$, then the
result follows from Lemma~\rf[L3.4]{b}. So assume $O^p(P)\le Z$.
Then $[O^p(P),O^p(L)]=1$ which, as $G=\$, means that
$O^p(L) \normal G$. But $O^p(L) \le O^p(G)$ and $O^p(G)$ being
quasisimple implies that $O^p(L)\le Z$ or $O^p(L) = O^p(G)$. Both
possibilities are impossible as, in the first case, we get that
$G=SZ$ which is soluble and, in the second case, we get
$P=O^p(P)S\le O^p(G)S =O^p(L)S =L$.
\end{proof}
\be{lemma} {comps} Suppose that $Q_G=1$, $F(G)=C_G(E(G))$ and $G$
operates transitively by conjugation on the set of components of
$G$. If $P\in \P_G(S)$ is narrow and isolated in $G$, then $E(G)=Y$
is quasisimple. \en{lemma}
\begin{proof} Assume that $P$ is a narrow isolated subgroup of $G$. Suppose first
that $Y$ is soluble. Then Corollary~\ref{cor1}(b) implies that
$Y=O^p(P)$. As $G = LY$, we have that $Q_LY \normal G$ and this
subgroup is also soluble. Hence $Q_LY$ centralizes $E(G)$. But then
$Q_LY \le C_G(E(G))$ which is nilpotent by assumption. Thus $Q_LY$
is nilpotent and this contradicts $Q_L\not\le Q_P$. Thus $Y$ is not
soluble and so Lemma~\ref{simple} implies that $Y$ is a component of
$G$. Since $G$ acts transitively on the set of components of $G$, we
obtain $Y= E(G)$ and this proves the lemma.
\end{proof}
\be{lemma} {Opnormal} Suppose that $Q_G=1$, $N$ is a normal subgroup
of $ G$ with $N$ soluble and $C_G(N)\le N$. If $P\in \P_G(S)$ is
isolated in $G$, then $O^p(P) \normal G$ and $P$ is soluble.
\en{lemma}
\begin{proof} Suppose that $N \le L$. Then $O^p(P)\not\le N$ and hence $Y \not\le
N$. However, Lemma~\rf[L3.4]{c} implies that $Y \le C_G(N) \le N$,
which is a contradiction. Therefore $N \not \le L$. Hence $P\le NS$
by Lemma~\rf[L3.2]{a}. But then $O^p(P)\le N$ and, as $N$ is
soluble, Lemma~\rf[L3.6]{d} gives the result.
\end{proof}
Lemma~\ref{ngs maximal} exploits the fact that, as observed in
Lemma~\rf[L3.1]{b}, an isolated $p$-minimal subgroup is normalized
by $N_G(S)$.
\be{lemma}{ngs maximal} Suppose that $P\in \P_G(S)$ is an isolated
subgroup of $G$. Assume that one of the following holds. \bl \li 1
$N_G(S)$ acts irreducibly on $S$. \li 2 $N_G(S)=L$.
\li 3 $N_G(S)$ is contained in a unique maximal subgroup of $G$.
\el
Then $ P\normal G$.
\en{lemma} \begin{proof} Suppose \rf 1 holds. Since $N_G(S)\leq L$,
we get that
$Q_L=S$
and so \rf 2 holds. Now suppose \rf 2 holds. Then, by Lemma~\rf[L3.2]{c}, $N_G(S)$ is a
maximal subgroup of $G$ and hence \rf 3 holds. So we may assume that
\rf 3 holds. Since $N_G(S)\leq L$, we get that $L$ is the unique
maximal subgroup of $G$ containing $N_G(S)$. Because $N_G(S) P\leq
N_G(P)$ and $P\nleq L$, we must have $G=N_G( P)$ and so $P\unlhd G$.
\end{proof}
\be{lemma}{reduce} Suppose that $G$ is an almost simple group, and
set $X:=F^*(G)$. If $P\in \P_G(S)$ is $A$-isolated, then $P\in
\P_{XS}(S)$ is $A$-isolated. \en{lemma}
\begin{proof} Note that if $H \ge XS$, then $O_p(H)\le C_G(X)= C_G(F^*(G))=1$. Thus $L_G(P,S)\not \ge XS$, whence $P\le XS$.
\end{proof}
\section{The Centre of $L_G(P,S)$}
The main purpose of this section is to prove Theorem~\ref{T1.5}.
\be{theorem}{central} Suppose that $p$ is a prime {and} $G$ a finite
${\mathcal {CK}}$-group. If $P\in \P_G(S)$ with $O^p(P/Q_P)\in
\mathcal L_1(p)$, then either $O^p(P) \unlhd G$ or
$O_p(Z(L_G(P,S)))\leq Q_P$. \en{theorem}
\begin{proof} Suppose the theorem is false and let $G$ be a minimal counterexample. Again set $L:= L_G(P,S)$ and
$Y:= \$. So we have $O^p(P)$ is not normal in $G$ and
$D:=O_p(Z(L)) \not\le Q_P$. {Thus $P$ is $D$-isolated.} The
minimality of $G$ gives
\bd 1 $Q_G=1$.\ed
Thus, by Theorem~\ref{T1.3},
\bd 2 {$Y$ is a central product of quasisimple simple groups each
of which has order divisible by $p$.} \ed
Since $Q_P\leq L$, we have $[Q_P,D]=1$ and so,
by Lemma~\rf[L3.4.5]{a},
\bd{2.5} $[Q_P,O^p(P)]=1$.\ed
{We next show that}
\bd{14} $P$ is soluble and $O^p(P)$ is a $p^\prime$-group.\ed
Since $P\cap L$ centralizes $DQ_P/Q_P$, \cite[(5.1)e]{DS} implies
that $P$ is soluble. Hence \rf{14} now follows from \rf{2.5}.
\medskip
\bd 3 $D\cap Q_P=1$.\ed
By \rf{2.5} $D\cap Q_P$ is normalized by
$\=G$. Thus \rf 1 gives \rf{3}.
\medskip
Pick $1\neq a\in D$ of order $p$ and set $A:=\$. Then, by
\rf{3}, $A\not\le Q_P$. Since $L$ is a maximal subgroup of $G$, \rf
1 implies
\bd 4 $N_G(A)=L=C_G(A)$.\ed
\bd 5 $A$ is not weakly closed in $G$ (with respect to $S$).\ed
Suppose that $A$ is weakly closed in $G$. If there exists $g\in G$
such that $a^g \in S$, then $A^g \le S$ whence $A=A^g$. So $g \in
N_G(A)$ and hence, by \rf{4}, $a^g =a$. Therefore $a$ is weakly
closed in $G$ and, in particular, in $AY$ (with respect to $S\cap
AY$). Now, as $G$ is a $\mathcal {CK}$-group, \cite[Remark
7.8.3]{GLS3} implies that $[A,Y]$ is a $p'$-group. {Hence by \rf{2},
$[A,Y] \le Z(Y)$ and so the Three Subgroup Lemma and \rf{2} once
more give $[A,Y]=1$.} Hence, using \rf{4}, $O^p(P)\le Y\le L$, a
contradiction and so \rf{5} holds.
\medskip
By \rf 5 there exists a $p$-subgroup $R$ of $G$ with $A\leq R$ and
$N_G(R)\nleq N_G(A)=L$. So there also exists a subgroup $H$ of $G$
with $H\nleq L$ and $A\leq Q_H$. Choose such an $H$ with $|H\cap
L|_p$ maximal and then with $|H|$ minimal. Let $T\in \Syl_p(H\cap
L)$. Since $A\unlhd L$ we may assume that $T\leq S$. If $T=S$, then,
as $H\not \le L$, $P\le H$ and then $A \le Q_H\le Q_P$, a
contradiction. Therefore,
\bd 8 $S > T$.\ed
{Next we show}
\bd 9 Let $U$ be a $p$-group with $T\leq U$. Then $U\leq N_G(U)\leq
L$. In particular, $T \in \Syl_p(H)$ and $H\neq N_H(T)$. \ed
We have $A\leq T\leq O_p(N_G(U\cap L))$ and {either $U\cap L > T$
or, by \rf{8},
$$|N_G(U\cap L)\cap L|_p> |U\cap L|= |T|.$$ Hence $|N_G(U\cap L)\cap L|_p > |T|=|H\cap L|_p$ and so, by the maximal
choice of $H$, $N_G(U\cap L)\leq L$. In particular}, $U=U\cap L$
and \rf 9 holds.
\medskip
\bd 7 $L\cap H$ is the unique maximal subgroup of $H$ containing
$T$. In particular, $H \in \P_H(T)$.\ed
Indeed let $T\leq M< H$. Then $A\leq Q_H\leq Q_M$, $|M\cap
L|_p=|T|=|H\cap L|_p$ and {thus} the minimal choice of $H$ implies
$M\leq L$. So $M \le L\cap H$. By \rf 9, $T\nnormal H$ and so $H\in
\P_H(T)$.
\medskip
\bd{8.5} $C_G(O^p(H)) \cap Z(S)=1$.\ed
Suppose that $E := C_G(O^p(H)) \cap Z(S)>1$ and put $C:= C_G(E)$.
Then, by \rf 1, $G>C \ge \$. {Note that $ S \le C$ and
$C \nleq
L$. Therefore, by Lemma~\ref{L3.2}(a), $P \le C$. Also $L\cap C = L_C(P,S)$ and $D\leq Z(L\cap C)$.
Thus $O_p(Z(L\cap C)) \not\le Q_P$ and hence} $O^p(P) \normal C$ by
induction. {We now have} $C = (L\cap C)O^p(P)$. So $$\< A^{O^p(H)}\>
\le \ = \ =\\le O^p(P)A.$$
Since also $\langle A^{O^p(H)}\> \le Q_H$, we get that $\langle
A^{O^p(H)}\> \le O^p(P)A\cap Q_H = A$ by \rf{14}. But then, by
\rf{4}, $O^p(H) \le N_G(A)= L$, a contradiction.
\medskip
By the maximal choice of $H$, $T \in \Syl_p(N_G(Q_H))$. {Hence
$Z(S)\le T\le H$. Since $H \in \P_H(T)$ by \rf 7, Lemma~\rf[L2.9]{a}
and (e) imply that either $C_{H}(Q_H)\ge O^p(H)$ or $Q_H
\in\Syl_p(C_H(Q_H))$ and $C_H(Q_H)$ is $p$-closed. In the former
case, $O^p(H)\le C_G(A) =L$ which is not the case. Thus, as $Z(S)\le
C_{H}(Q_H)$, we have}
\bd {7.5} {$C_H(Q_H) \le Q_H$ and} $Z(S) \le Q_H$.\ed
\bd{aa} $p=2$ and $\O_1(Z(S))=A$ has order $2$. \ed
Since $A \le Z(T)$ and $H$ is $p$-minimal by \rf 7, we have
$C_H(\O_1(Z(T))) \le H\cap L$. Also, since $T~~ T$
and so the maximal choice of $H$ shows that $J(T)$ is not normal in
$H$. It follows that $H > H \cap L \ge \langle
N_H(J(T)),C_H(\O_1(Z(T))) \rangle$. Therefore, since $\O_1(Z(S)) \le
\O_1(Z(Q_H))$ and $H\cap L$ centralizes $A$, we infer from
Corollary~\ref{L2.7}, \rf{8.5} and \rf{7.5} that $p=2$ and $|A|\le
|\O_1(Z(S))|=2$.
\bd{16} $Q_P=1$.\ed
Since $A\cap Q_P=1$ by \rf 3, \rf{aa} implies $Q_P\cap
\Omega_1(Z(S))=1$. Hence also $Q_P=1$.
\medskip
\bd{17} $Q_H$ is a fours group, $T\cong \Dih(8)$, $H/C_H(Q_H)\cong
\Sym(3)$, $S\cong \SD(16)$ and $O^2(P)\cong 3^2$ or $3_+^{1+2}$. \ed
From \rf{14}{, \rf{16}} and Lemma~\ref{rank11}, $|S|\leq 2^4$. {Note
that $1 < A< Q_H \cong \Dih(8)$. Together with Lemma~\ref{rank11} we
infer that \rf{17} holds.}
\bd{18} $G$ has a unique conjugacy class of
involutions.\ed
By \rf{17} all involutions in $S$ are conjugate {in $S$} to an
involution in the fours group $Q_H$. Since $H$ {acts transitively by
conjugation on} $Q_H^\sharp$, \rf{18} follows.\medskip
From \rf{17} and \rf{18} either $G$ is simple or $G^\prime$ is
simple and $T\in \Syl_2(G^\prime)$.
Suppose first that $G$ is simple. Then, since $S\cong \SD(16)$, $G\cong
\Mat_{11}$ or $G\cong \PSL^\epsilon_3(q)$ with $q+\epsilon\equiv 4 \pmod 8$ by \cite{ABG}.
In the first case there exists $S\leq K\leq G$ with $K\cong
\Mat_{10}$. So $K\in \P_G(S)$, but neither $P=K$ nor $K\leq L$, a
contradiction. If $\eps = +$, then $L = C_G(A) \cong \GL_2(q)$ is
not a maximal subgroup of $G$. So $\eps = -$. In this case, as the Sylow $3$-subgroups of
$\PSL^-_3(q)$ are abelian, we have $O^2(P) = 3^2$ and $P\cong
3^2:\SD(16)$. Since the Sylow $3$-subgroups of $G$ are not cyclic, we infer that $3$ divides
$q+1$ and that $|Z(\SL^-_3(q))|=3$. It follows that the preimage $J$ of
$O^2(P)$ in $\SL_3^-(q)$ is extraspecial and that
$N_{\SL_3^-(q)}(J)/J \cong \Q_8$. {This contradicts $S\leq
N_G(O^2(P))$ and $|S|=2^4$.} Therefore $G\not\cong \PSL_3^-(q)$. So
$G$ is not simple.
{ Hence} $G^\prime$ is simple and by \cite{GW}, $G^\prime \cong
\Alt(7)$ or $G^\prime\cong \PSL_2(q)$ with $q\cong 7,9\pmod {16}$.
The former possibility is ruled out as a Sylow $2$-subgroup of
$\Sym(7)$ is not isomorphic to $\SD(16)$. Hence $G^\prime\cong
\PSL_2(q)$. If $q \neq 3^a$, then $G$ has cyclic Sylow $3$-subgroups
and this contradicts $O^2(P)\le G^\prime$. Thus $q= 3^a$. In this
case $P\cap G^\prime$ normalizes a $3$-subgroup and hence a Sylow
$3$-subgroup $U$ of $G^\prime$. But $N_{G^\prime}(U)/U$ is cyclic,
$T\leq G^\prime$ and $P \ge T \cong \Dih(8)$. This contradiction
completes the proof of Theorem~\ref{central}.
\end{proof}
Our final observation turns out to be very useful in \cite{MPR}.
\be{corollary}{qb cap x} Let $p$ be a prime and $G$ be a finite
${\mathcal {CK}}$-group with $Q_G=1$. Suppose that $P\in \P_G(S)$ is
an isolated subgroup of $G$ with $O^p(P/Q_P) \in \mathcal L_1(p)$
and that $O^p(P)$ is not normal $G$. Then the following hold.
\bl \li a If $[Q_P,O^p(P)]=1$, then $C_{Q_L}(O^p(L))=1$.
\li b $Q_L\cap O^p(G)\neq 1$. \li c $L\cap O^p(G)$ is a $p$-local
subgroup of $O^p(G)$.
\el \en{corollary}
\begin{proof} Put $D:=O_p(Z(L))$ where $L:=L_G(P,S)$. By Theorem~\ref{central},
$D\leq Q_P$ and so $[D,O^p(P)]=1$ and $D\unlhd \=G$.
Thus $D=1$ and, since {$C_{C_{Q_L}(O^p(L))}(S)\le D$}, \rf a holds.
For \rf b, suppose that $Q_L\cap O^p(G)=1$. Then
$$[Q_L,O_p(O^p(P))]\leq Q_L\cap O^p(P)\leq Q_L\cap O^p(G)=1.$$
Therefore $[\,O_p(O^p(P))]=1$. Since $O^p(P) \le \$
by Lemma~\ref{L3.4.5}(a), we get $[O^p(P),O_p(O^p(P))]=1$. Now
$$[Q_P,O^p(P)]\le Q_P\cap O^p(P) = O_p(O^p(P)),$$ and so
$[Q_P,O^p(P),O^p(P)]=1$. Hence $[Q_P,O^p(P)]=1$. Thus
$C_{Q_L}(O^p(L))=1$ by part (a), Further $$[Q_L,O^p(L)] \le Q_L \cap
O^p(L) \le Q_L \cap O^p(G)=1$$ and consequently $Q_L \le
C_{Q_L}(O^p(L))=1$, a contradiction. Hence $Q_L \cap O^p(G)\not=1$
and (b) holds.
\end{proof}
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