/O_p(\

) \cong
\SL_3(p^b)$ or $\PSL_3(p^b)$ for some integer $b$, then is either
$X$ is a Lie type group defined in characteristic $p$ or $p=2$ and
$X$ is one of the following sporadic simple groups .....
\en{corollary}
Suppose that $p$ is a prime. Then we use ${\mathcal R}_p$ to
denote the set of group $G$ for which there is $P\in \P_p(G)$
which is narrow and $p$-restricted. One of our goals is to
determine those simple groups which are in ${\mathcal R}_p$.
For a group $H$, we denote the preimage of
$F(H/O_p(H))$ by $F_p(H)$ and the preimage of $\Phi(H/O_p(H))$ by
$\Phi_p(H)$. The remainder of our group theoretic notation is
standard as can be found in \cite{Aschbacher}.
\section{Groups with isolated $p$-minimal parabolic subgroups}\label{two}
Recall from the introduction that for a prime $p$, a group $P$ is
called \emph{$p$-minimal} if for a Sylow $p$-subgroup $S$ of $P$,
$S$ is not normal in $P$ and $S$ is contained in a unique
maximal subgroup of $P$. For a group $G$ and $S \in \Syl_p(G)$, we
denote the set of $p$-minimal parabolic subgroups of $G$ (which
contain $S$) by $\P_G(S)$. For an arbitrary subgroup $R$ of $G$,
we set $Q_R=O_p(R)$.
The following result is elementary to prove.\co{ See for example \cite{PPS}}
\be{lemma}{gen} $G= \<\P_G(S)\>N_G(S)$.\en{lemma}
We shall also need the following general result.
\be{lemma}{no subnormal} Let $M$ be a maximal subgroup of the
finite group $H$ and let $N$ be a subnormal subgroup of $H$ with
$N\leq M$. Then $N\leq \bigcap M^H$. \en{lemma}
\proof Suppose that $H$ is a counterexample to the statement
\co{$H$ being a counterexample is not used, remove}and
select $N$ subnormal in $H$ with $N \le M$ of maximal order. Let
$N=N_0\normal N_1\normal\dots \normal N_k\normal N_{k+1}=H$ be a
subnormal chain from $N$ to $H$. By Wielandt's subnormal lemma,
$\**$} we have that $Q_BY
\normal G$ and $Q_BY$ centralizes $E(G)$. But then $Q_BY \le
C_G(E(G))=F(G)$ by assumption. Thus $Q_BY$ is nilpotent and this
contradicts $Q_B \not\le Q_P$. \qed
\be{lemma}{quot} Suppose that $P\in \P_G(S)$ is isolated in $G$.
Then $C_B(Y)$ is normal in $G$ and if $X \le C_B(Y)$ is normal in
$G$, then \bl
\li a $PX/X \in \P_G(SX/X)$ is isolated in $G/X$; and
\li b if $P$ is narrow, then $PX/X$ is narrow.
\el \en{lemma}
\proof We first of all note that $G = BY$.\co{$G=\ $} So $C_B(Y) = B \cap
C_G(Y)$ is normalized by $G=BY$.\co{$G=\$} Now suppose that $X\le C_B(Y)$ is
normal in $G$. We claim that $PX/X$ is a $p$-minimal parabolic
subgroup which is narrow and isolated in $G/X$. Suppose that $\bar
R \in \P_{G/X}(SX/X)$ is not contained in $B/X$. Let $R$ be the
full preimage of $R$ in $G$. Then $R \ge S$ and $R\not \le B$ so
$P\le R$ by \rf[quasi]{a-1}. Furthermore, as $\bar R \in
\P_{G/X}(SX/X)$ and $B \ge X$, $B \cap R$ is the maximal subgroup
$R$ containing $SX$. Since $P\not \le B$, we infer that $R=PX$ and
consequently $\bar R=PX/X$
is the unique $p$-minimal parabolic
subgroup of $G/X$ not contained in $B/X$. Let $U$ be the full
preimage of $O_p(\ov {R})$
\co{ $\bar R$} and let $U_p = U\cap S$. We have
$R=N_R(U_p)U $ by the Frattini lemma. In particular, as $R\not \le
B$ and $U \le XS \le B$, $N_R(U_p) \not \le B$. Since $S \le
N_R(U_p)$, it follows that $P \le N_R(U_p)$ and so $U_p \le Q_P$.
Since $Q_B \not \le Q_P$, it follows that $Q_B \not \le U_p =
U\cap S$. Therefore, $Q_B X/X not\le UX/X$\co{$\nleq$} and $PX/X$ is isolated
in $G/X$. This proves \rf a.
Suppose that $P$ is narrow. Then $PX/X \in \P_G(S)$ by \rf{a}. We
have so $PX/X \cong P/P\cap X$ and, putting $F = \bigcap (B\cap
P)^P$, we have $P\cap X \le F$. It follows that $O^p(P)F/F \cong
\O^p(P/X)(FX/X)/(FX/X)$
\co{$O^p$} and so $PX/X$ is narrow.\qed
\be{corollary} {quot2}Assume that $P\in \P_G(S)$ is narrow and
$p$-restricted. Let $Y =\**

\cong \

\cong \PSp_4(3)$ and $\

\cong \GU_3(r^a)$. The embedding of $\PSp_4(3)$ into $\SU_4(r^a)$ stems from the fact that $\SU_4(r^a) \cong \Omega_6^-(r^a)$ and $\PSp_4(3)$ has index $2$ in the Weyl group of type $\E_6$. Finally we note that the two subgroups $\

$ and $\

$ are conjugate in $\GU_4(r^a)$. Because of the situation for $\SU_5(2)$ we need to individually inspect $\SU_6(2)$ and $\SU_7(2)$. \be{lemma}{u62andu72} Suppose that $X \cong \SU_6(2)$ or $\SU_7(2)$ and $P$ is $3$-restricted in $G$. Then $X \cong \SU_6(2)$, $|G/X | \le 3$, $P \cap X \cong 3^5:\PSL_2(9) \cong 3^5:\Alt(6)$ and $B\cap X \sim 3^{1+4}_+. (\Q_8\times \Q_8):3$. \en{lemma} \proof For $G \cong \SU_6(2)$, we just inspect the Atlas\cite{Atlas}. So suppose that $X \cong \SU_7(2)$. Then $L\cong 3^6 :\Sym(7)$. Since $\Sym(7)$ is not $3$-restricted, we have $B \ge LS$. Setting $\ov{L_1} = \GU_6(2)\times \GU_1(2)\cap \ov{G}$. We have that $P\le L_1S$. From the $\PSU_6(2)$ example we now read that $P \sim 3^6:\Alt(6)\le B$, which is a contradiction.\qed \be{lemma} {dimp2} Suppose that $p\ge 5$ and $n \le p$. If $X \not \cong \PSL^\eps_n(r^a)$, then $G$ is not $p$-restricted. \en{lemma} \proof Suppose for a contradiction that $P$ is $p$-restricted in $G$. If $n

1$, then $X$ has abelian Sylow $p$-subgroups and \ref{abelian} delivers a contradiction. Therefore, $n=p$ and $d=1$. So $\ov{L^*}\cong (r^a-\eps)\wr \Sym(p)$. Using $p\ge 5$, \ref{monomial3} and \ref{QBnX=1} we have that $B = LS$ and $Q_P=1\cap X$. Since $Q_B$ is normalized by $N_P(O^p(P)\cap S)$ and $Q_B\cap X = O_p(LS)\cap X$ is abelian, we have that $O^p(P) \cong \SL_2(p^b)$, $\PSL_2(p^b)$ for some $a$ and $Q_B\cap O^p(P) \in \Syl_p(O^p(P))$ is elementary abelian. Since $Q_P \cap X=1$, we then have $b=p-2$ and $P\cap X = O^p(P)$ because $\SL_2(p^{p-2})$ has no field automorphisms. This contradicts the fact that $X$ has non-abelian Sylow $p$-subgroups.\qed We at last come to the induction argument. \be{lemma} {LU-generic} Suppose that $X\cong \PSL_n^\eps(r^a)$ and assume one of the following conditions hold: \bl \li {a} if $p=2$ and $r^a >3$ or $X \cong \PSL_n(r^a)$, then $n \ge 5$; \li {a1} if $p=2$ and $X\cong \PSU_n(3)$, then $n \ge 8$; \li b if $p=3$ and $r^a >2$ or $X\cong \PSL_n(r^a)$, then $n \ge 6$; \li {b2} if $p=3$ and $X\cong \PSU_n(2)$, then $n \ge 8$; \li c if $p\ge 5$, then $n >p$. \el Then $G$ is not $p$-restricted. \en{lemma} \proof Furthermore, assume that $n$ as defined in the statement of the lemma and then is chosen minimally so that $G$ is $p$-restricted. Recall that $\ov{L^*} = \GL^\eps_d(r^a)\wr \Sym (s)\times \GL_{n-ds}^\eps(r^a)$ where $s = \left[\frac n d\right]$ and $L$ is normalized by $S$. \bd{1} $ds=n$.\ed Assume that $ds < n$. Plainly $d >1$. Then $S\cap X$ centralizes an $n-ds$ dimensional subspace of $V$ and so, in particular, it centralizes a $1$-dimensional subspace $W$ of $V$. If $\eps =+$ and $G/X$ consist of diagonal or graph automorphisms or $\eps = -$ and $W$ is singular, then $S\cap X$ is contained in a parabolic subgroup of $X$ and we have a contradiction via \ref{para-arg}. In the other cases we have $W$ is non-degenerate and, setting $\bar H = \GL_{n-1}^\eps(r^a)$ (fixing $W$), $S\cap X \le H$ and $H$ is normalized by $S$. Because of the choice of $n$, \ref{4dim}, \ref{su63}, \ref{4dim2}, \ref{u62andu72} and \ref{dimp2} imply that $H $ is not $p$-restricted. Hence $B \ge H$. Since $Q_B>1$, we infer that $p$ divides $r^a-\eps$ and so $d=1$ when $p$ is odd, a contradiction. So $p=2$. In this case, since $d>1$, $|Q_B|=2$, and $n-1$ is even. \ref{norm} implies that $P/Q_P \cong \SL_2(2)$, (a subgroup of) $\PSU_3(2)$, (a subgroup of) $\SU_3(2)$ or (a subgroup of) ${}^2\B_2(2)$. Since $d=2$, $Q_B \le [S\cap X, S\cap X, S\cap X]\le Q_P$ (from the structure of a Sylow $2$-subgroup in $\GL_2(r^a)$), which is a contradiction. \bd{2} $s\ge p$.\ed If $s

1$, then $B \ge LS$ and either \bl
\li i $p=3$, $d=r^a=2$ and $\ov{L^*}\cong \GL^+_2(2)\wr\Sym(n/2)$;
\li {ii} $p=2$, $d=2$, $r^a=3$ and $\ov{L^*}\cong \GL^+_2(3)\wr
\Sym(n/2)$.
\el\ed
If $B \ge LS$, then by \ref{QBX} $Q_B \cap X >1$ and \ref{sln1} implies that one of
\rf{i} or \rf{ii} holds. So for a contradiction assume that $B$
does not contain $ LS$. Then $P\le LS$. If $\GL_d^\eps(r^a)$, is
not soluble, we apply \ref{comps} to obtain $s=1$ and this
contradicts \rf{2}. Therefore $\GL_d(r^a)$ is soluble. Using
\rf{Opnormal} we have that $O^p(P)\le F_p(LS)$. Suppose that $d\ge
2$ and $r^a=2$. If $O_p(LS) =1$, then $F_p(LS)$ is a $3$-group and
we have $O^p(P)$ is a $3$-group from which we infer that
$p=2=r^a$, a contradiction. Therefore, $p=3$, and
$\ov{L^*}/O_3(\ov{L^*}) \sim 2\wr \Sym(n/2)$ and $n \ge 6$ by
\rf{b} or $ \Q_8:3\wr \Sym(n/3)$ and $n \ge 8$ by \rf{b2}.
Furthermore, from \rf[quasi]{c} we have that $LS = (B\cap
LS)O^p(P)$, $|Q_BO_3(LS)/O_3(LS)|=3$ and $O^p(P) O_3(LS)/O_3(LS)$
is either elementary abelian of order $4$ or isomorphic or $\Q_8$
and is normal in $LS/O_3(LS)$. The only possibility is that
$n/2=3$ and that $LS/O_3(LS) \cong 2\wr \Sym(3) \cong 2 \times
\Sym(4)$. So $L \cong \SL_2(2)\wr \Sym(3) \le X=G =\SL_6(2)$. Now
in this case $L$ can be embedded in $H \cong \Sp_6(2)$ and so we
must have that $\Sp_6(2)$ is $3$-restricted. However, a look in
the Atlas \cite{Atlas} confirms that a Sylow $3$-subgroup of
$\Sp_6(2) $ is contained in a unique maximal subgroup and
consequently $\Sp_6(2)$ is a $2$-minimal parabolic group, a
contradiction. Suppose that next that $r^a=3$. Then just as above
we argue that $p=2$. Then $\ov{L^*}/O_2(\ov{L^*})\sim \Sym(3)\wr
\Sym(n/2)$ and $O^2(P)O_2(L)/O_2(L) $ has order $3$. Considering
again the centralizer of $Q_B$ in $L/O_2(L)$, we obtain a
contradiction. This proves \rf{3}.
\bd{4} If $d=1$, then $LS\le B$. \ed
Suppose that $B \not \ge LS$. Then \ref{monomial1} gives either
$O^p(P)$ is soluble and $n \le 3$, which contradicts $n \ge 4$, or
$r^a-\eps$ is a power of $p$ and $LS/O_p(LS)$ is $p$-restricted.
So $LS/O_p(LS)\in \Rp$. If $p=2$, then as $n \ge 6$, we have
$LS/O_2(LS) \cong \Sym(6)$, $\Sym(8)$ or $\Sym(12)$ by
\ref{smallalts} and \ref{altcase}. Suppose that $LS/O_2(LS) \cong
\Sym(6)$ and define $\ov{L_1} =\GL_2^\eps(r^a)\wr \Sym(3) \cap G$.
Then $S\cap G$ is contained in $L_1$. If $r^a>3$, then \ref{comps}
implies that $L_1S$ is not $2$-restricted. Thus, if $r^a >3$, then
$B \ge L_1S$ and we get $Q_B \le O_p(L)\le
Q_P$, a contradiction. Therefore, $r^a=3$. But then $L_1$ is
soluble and so, if $P\le L_1S$, $O^2(P)(L_1)O_2(L_1)/O_2(L_1)$ has
order $3$ and $L_1/O_2(L_1) =
C_{B/O_2(L_1)}(O^2(P)O_2(L_1)/O_2(L_1) \times
Q_BO^2(P)O_2(L_1)/O_2(L_1)$ has a direct factor isomorphic to
$\Sym(3)$, a contradiction. Hence $B \ge L_1S$. Finally consider,
$\ov{L_2} =(\GL^\eps_2(3)\times \GL^\eps_4(3)) \cap G$. Then as $B
\ge L_1$, $P\le L_2S$. And furthermore, letting $K$ be the
component in $L_2$, we have $P \le K S$ is $3$-restricted. It
follows that $\eps=-$ but then we contradict our supposition on
the size of $n $ as given in \rf{a1}.
Next assume that $n=8$ and set $\ov{L_1} = \GL_4^\eps(r^a)\wr
\Sym(2)\cap \ov G$. By \ref{comps}, $B \ge L_1S$ and so $P\le LS$
and $Q_B \le Q_P$, a contradiction. So suppose that $n=12$ and set
$\ov{L_1} = \GL_4^\eps(r^a)\times \GL^\eps_8(r^a) \cap \ov{G}$. If
$P\le L_1S$, then we must have $P\le \GL_8(r^a)$ and this
contradicts the minimal choice of $n$. Thus $B \ge L_1S$ and again
$Q_B \le Q_P$, a contradiction. Thus $p>2$ and consequently $n\le
6$. Hence $p\ge 5$. This then forces $n = 5$ and $L/O_5(L) \cong
\Sym(5)$ and this configuration has been considered in
\ref{dimp2}. This proves \rf{3}.
\bd {5} $p$ does not divide $s$.\ed
Suppose that
$p$ divides $s$. Assume additionally that $d =1$. Set $\ov{L_1} =
\GL_{dp}^\eps(r^a)\wr \Sym(s/p)\cap \ov{G}$. Then, as $s>p$,
\ref{comps} implies that $L_1S \le B$. Therefore \rf{4} gives $B
\ge \