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\setlength{\footskip}{1cm}
\parskip=10pt

%\section{xxx}
%\lemma{xxx}

%\includeonly{strong2}%Introduction
%\includeonly{strong3}%Section 0 
%\includeonly{strong4}%Section 1  
%\includeonly{strong5}%Section 2  
%\includeonly{strong6}%Section 3  
%\includeonly{strong7}%Section 4


\begin{document}
\title{\large\normalsize\bf Strong quadratic modules}
\author{
\normalsize G. Stroth}
\date{~~~~}
\maketitle



One of the main tools in the quite recent theory of parabolic systems
is the knowledge of $GF(p)$-modules $V$ for the groups involved
admitting quadratically acting $p$-groups.

Let $G$ be a finite simple group, $V$ a $GF(p)G$-module, $p$ a
prime. We call $V$ quadratic if there is a nontrivial $p$-subgroup $E$
of $G$ such that $[V,E,E]=0$. For $p\geq 5$ the groups $G$
possessing such a module have been studied by J. Thompson [Th]. For
$p=3$ results have been obtained by Ch.Ho [Ho]. The corresponding
modules for the groups of Lie type have been classified by 
A.A. Premet and I.D. Suprunenko [PS].
Similar results have been obtained by U. Meierfrankenfeld [Mei].
If $p=2$ the definition of a quadratic module 
is not any restriction at
all as for any involution $x$ we have $[V,x,x]=0$. So in this case we
have to add a further assumption. In [MeiStr1] and [MeiStr2] the case
of $G$ a sporadic simple group, an alternating group or a group of
Lie type over a field of odd characteristic was investigated under the
additional assumption that $|E|\geq 4$. Again this seems not to be
the right condition for groups of Lie type over a field of
characteristic $2$, as root groups tend to act quadratically on 
many modules. So a classification, which might be possible, would
contain a long list of modules defined over $GF(q)$  (if $G$ is
defined over $GF(q)$) and so would be just too long for interesting
applications. This paper is a first step in obtaining a satisfactory
result for quadratic modules for Chevalley groups over $GF(q)$,
$q=2^{n}$. We are going to prove the following theorem.

\bigskip

{\bf{Theorem.}}~~{\it{Let $G=G(q)$ be a Chevalley group over $GF(q)$,
$q=2^{n}$. Let $R$ be a long root group and
$Q=Z_{2}(O_{2}(N_{G}(R)))$. Let $V$ be a nontrivial irreducible
$GF(2)G$-module with $[V,R,Q]=0$. Then one of the following holds


\begin{itemize}
\item[(i)] $G\simeq (S)L(n,q)$ or $(S)U(n,q)$ and $V=V(\lambda)$ for
some fundamental weight $\lambda$.

\item[(ii)] $G\simeq \Omega^{\pm}(2n,q)$ or $Sp(2n,q)$ and $V$ is the
natural module or a spin module.

\item[(iii)] $G\simeq E_{6}(q)$ and $V=V(\lambda_{1})$ or
$V(\lambda_{6})$.

\item[(iv)] $G\simeq E_{7}(q)$ and $V=V(\lambda_{7})$.

\item[(v)] $G\simeq~\!^{2}\!E_{6}(q)$ or $F_{4}(q)$ and
$V=V(\lambda_{4})$.

\item[(vi)] $G\simeq G_{2}(q)$ or $^{3}D_{4}(q)$ and $V$ is the
natural module.
\end{itemize}}}

In section 1 we prove the Theorem, and then in sections 2 and 3 we
deduce the following corollaries. In fact corollary 1 had originally
inspired the work of this paper. 

\bigskip


{\bf{Corollary 1.}}~~{\it{Let $G=G(q)$ be a Chevalley group over
$GF(q)$, $q=2^{n}$, and $V$ be a nontrivial irreducible
$GF(2)G$-module. Let $X$ be some nontrivial elementary abelian
$2$-subgroup of $G$ such that $[V,X,X^{g}]=0$ whenever $[X,X^{g}]=1$,
$g\in G$. Then $V$ is one of the modules in the theorem or $G\simeq
(S)U(3,q)$ and $V$ is a basic module ($8$-dimensional module) or
$G\cong Sz(q)$ and $V$ is the natural module.}}

A question like the one of corollary 1 recently occured in a paper of
Timmesfeld [Ti]. If $E$ is a quadratic fours group we get the
following corollary.


\bigskip

{\bf{Corollary 2.}}~~{\it{Let $G=G(q)$ be a Chevalley group over
$GF(q)$, $q=2^{n}$, and $V$ be a nontrivial irreducible
$GF(2)G$-module. Let $E$ be a fours group such that $[V,E,E]=0$. If $E$
intersects some long root subgroup $R$ of $G$ nontrivially, but
$E\not\subseteq R$, then either $V$ is one of the modules in the
theorem or $G\cong Sp(2n,q)$ or $F_{4}(q)$ and $V=V(\lambda)$
for some fundamental weight $\lambda$.}}


I do not know whether $G=F_{4}(q)$ and $V=V(\lambda_{2})$ really
occurs.

For a Chevalley group $G$ let $q(G)$ be the minimum value of $n$ such
that the following is true: If $V$ is an irreducible module and $A$
is some elementary abelian subgroup of $G$ with $[V,A,A]=0$ and
$|A|=2^{n}$, then $V$ is one of the modules in corollary $2$. It
would be interesting to determine $q(G)$.


\vspace*{0.5cm}

\section*{0. Preliminaries}

\indent
\lemma{Let $G$ be a finite group and
$ \langle x,y\rangle \leq G $ be a fours group acting 
nontrivially on some
irreducible $ GF(2)G $--module $V$.  Let $ x \sim y \sim xy $ and $ G =
\langle \C_{G}(s), \C_{G}(t) \rangle  $ for any pair $ s,t $ 
from the triple $x,y,xy$
such that $ \langle s,t\rangle  =
\langle x,y\rangle  $.  Then $ [V,x] \cap [V,y] \neq 0 $.}

{\it{Proof.}}~~ As $ V $ is an irreducible module we have $ \C_{V}(x)
\not= \C_{V}(xy) $.  So $ [\C_{V}(xy),x] = [\C_{V}(xy),y]
\not= 0 $.

\bigskip

\lemma{Let $ W $ be a 2--group which carries the structure of a
2--dimensional vectorspace over $ GF(q) $, $ q = 2^{n} $.  Let $ V $
be a $ GF(2)W $--module on which $ q $ of the $ q+1 $ one-dimensional
subspaces of $ W $ act quadratically.  Then any one-dimensional
subspace of $ W $ acts quadratically.}

{\it{Proof.}}~~Let $ W = W_{1}W_{2} $ where $ W_{1} = \langle x_{1},\ldots,x_{n}\rangle  $, $ W_{2}
= \langle y_{1},\ldots,y_{n}\rangle  $ and $ W_{1},\, W_{2} $ are one--spaces which
act quadratically.  We may assume $ n \geq 2 $.  

Let $w_{1},w'_{1}\in W_{1}$, $w_{2},w'_{2}\in W_{2}$, $v\in V$. We
first show

\parbox{11cm}{\begin{eqnarray*}
[v,w_{1}w_{2},w'_{1}w'_{2}]=[v,w_{1},w_{2}']+[v,w'_{1},w_{2}]\\
                           =[v,w'_{2},w_{1}]+[v,w_{2},w'_{1}]
\end{eqnarray*}}\hfill
\parbox{1cm}{\begin{eqnarray}\end{eqnarray}}

For this we use freely the following general facts

\begin{quote}
$[v,ab]=[v,a]+[v,b]+[v,a,b]$ and\\
$[v,a,b]=[v,b,a]$ for commuting $a,b$.
\end{quote}


\noindent
We have

\hspace*{\fill}
$z=[v,w_{1}w_{2}]=[v,w_{1}]+[v,w_{2}]+[v,w_{1},w_{2}]=[v,w_{2}]+x
=[v,w_{1}]+y$
\hspace*{\fill}

\noindent
where $x\in [V,W_{1}]$ and $y\in [V,W_{2}]$.

\noindent
Now

\hspace*{\fill}
$[v,w_{1}w_{2},w'_{1}w'_{2}]=[z,w'_{1}w'_{2}]=[z,w'_{1}]+
[z_{1},w'_{2}]+[z,w'_{1},w'_{2}]$
\hspace*{\fill}

\noindent
We have

\begin{quote}
$[z,w'_{1}]=[v,w_{2},w'_{1}]$ as $[x,w'_{1}]=0$ by quadratic action\\
$[z,w'_{2}]=[v,w_{1},w'_{2}]$ as $[y,w'_{2}]=0$ by quadratic action\\
$[z,w'_{1},w_{2}']=
[[v,w_{2}],w'_{1},w'_{2}] = 
[[v,w_{2}],w'_{2},w_{1}']=0$ by quadratic action
\end{quote}

\noindent
This establishes (1).

For proving the lemma it is enough to show that for $ x_{1}y,\,
x_{2}z \in W_{0},\, W_{0} $ a one-space in $ W $, $ y,z \in W_{2} $,
we get $ [v,x_{1}y,\, x_{2}z] = 0 $ for any $v \in V$.


Let $ y = \prod y_{i}^{m_{i}} $ then set $ \overline{y} = \prod
x_{i}^{m_{i}} $.  So to $ x_{1}y,\, x_{2}z $ there correspond $ y_{1}
\overline{y} $ and $ y_{2}\overline{z} $. Looking at $W$ as a two
dimensional vectorspace over $GF(q)$ implies that there is an
automorphism $\rho$ of order $q-1$ (the multiplication by field
elements) which acts on $W_{1}$ and $W_{2}$ in the same way. Now the
$q+1$ one dimensional subspaces of $W$ are just the $q+1$ $\langle
\rho\rangle$-invariant subgroups of order $q$. As $x_{1}y$ and
$x_{2}z$ are in $W_{0}$ there is some power of $\rho$ mapping $x_{1}$
onto $x_{2}$ and $y$ onto $z$. But then the same power maps $y_{1}$
onto $y_{2}$ and $\bar{y}$ onto $\bar{z}$, so
$y_{1}\bar{y},y_{2}\bar{z}$ are contained in some subspace $W_{3}$.
We may
assume that $ W_{3} $ acts quadratically.  Denote by $ W_{4} =
\langle x_{1}
y_{1},\ldots , x_{n} y_{n} \rangle  $, a diagonal of $ W_{1} W_{2} $.  We
may also assume that $ W_{4} $ acts quadratically.  As $ y\overline{y} $, 
$ z\overline{z} $ are in $ W_{4} $ we get with (1)
\begin{equation}
0 = [v,y\overline{y},\, z\overline{z}] = [v,y,\overline{z}]+
[v,\overline{y},z] \; \mbox{ for any pair }\; y,z \in W\, ,\,v\in W.
\end{equation}
Hence in our situation we have
\begin{equation}
[v,y_{1},\overline{z}] = [v,x_{1},z]\; \mbox{ and }\;
[v,\overline{y},y_{2}] = [v,y,x_{2}]\, ,\, v\in V
\end{equation}

\noindent
and


\begin{equation}
0 = [v,y_{1}\overline{y},\, y_{2}\overline{z}] 
\begin{array}[t]{c}
=\\[-0.2cm]
{\scriptstyle{(1)}}\end{array}
[v,y_{1},\overline{z}] + [v,\overline{y},y_{2}]
\end{equation}

\noindent
So by (2) and (3) we have

\begin{equation}
[v,x_{1},z]=[v,y,x_{2}],v\in V
\end{equation}

\noindent
This gives $ [v,x_{1}y,x_{2}z] 
= [v,x_{1},z] + [v,y,x_{2}] = 0 $ by (5). 

\bigskip


\lemma{([S]). Let $ G $ be a Chevalley group over $ GF(q) $
and $ V $ be an irreducible $ GF(2)G $--module.  Then $ \C_{V}(K) $
is an irreducible $ GF(2)P $--module for any parabolic $ P $ of $ G
$, where $ K = O_{2}(P) $.}

\bigskip


\lemma{(a). Let $ G $ be a Chevalley group over $ GF(q)
$, $ P_{1},\ldots ,P_{n} $ be the set of minimal parabolics of $ G $
containing a given Sylow 2--subgroup.  Set $ K_{i} = O_{2}(P_{i}) $, 
$ i = 1, \ldots, n $.  Let $ V $ be an irreducible module over $
GF(q) $.  Then $ V $ is uniqueley determined by the action of $ P_{i} $
on $ \C_{V}(K_{i}) $, $ i = 1,\ldots, n $. 

(b) If
$V=V(\lambda),~\lambda=\sum\limits_{i=1}^{n}a_{i}\lambda_{i},\lambda_{i}$
fundamental weights, then (a) says that whenever $P_{i}$ acts
nontrivially on $\C_{V}(K_{i})$, we get $a_{i}\neq 0$, otherwise
$a_{i}=0$.

(c) Let $V$ be as in (b). If all $\C_{V}(K_{i})$ are trivial for all
but one $i=i_{0}$ and $\C_{V}(K_{i_{0}})$ is a natural module, then
$V=V(\lambda)$ for some fundamental weight $\lambda$.}


{\it{Proof.}}~~(a) is [RS], while (b) and (c) are easy consequences
from (a).


\noindent
A consequence of (0.4) we will use quite often is


\lemma{\it Let $G$ be a Chevalley group over $GF(q)$, $G\not\cong
L_{n}(q)$, and $R$ be some long root subgroup. Let
$Q=O_{2}(N_{G}(R)),~S\in Syl_{2}(N_{G}(R))$ and $P_{i}$ the minimal
parabolic with $S\subseteq P_{i},~P_{i}\not\subseteq
N_{G}(R),~O^{2'}(P_{i}/O_{2}(P_{i}))\cong L_{2}(q)$. Let $V$ be some
irreducible nontrivial module for $G$ over $GF(q)$ with $[V,R,Q]=0$.
Suppose that there is some $g\in G$ such that $R^{g}\subseteq
Q,~R^{g}\neq R$ and $(RR^{g})\subseteq x^{G}$ for $x\in
R^{\sharp}$. Then $O^{2'}(P_{i})$ has a fixed point on $V$.
Furthermore if
$V=V(\lambda),\lambda=\sum\limits_{i=1}^{n}a_{i}\lambda_{i}$,
$\lambda_{i}$ fundamental weights, then the coefficient $a_{i}$ of
$\lambda_{i}$ corresponding to $P_{i}$ is zero.}

{\it{Proof.}}~~As $G\not\cong L_{n}(q)$, we have that $N_{G}(R)$ is a
maximal parabolic of $G$. Now we are going to apply (0.1) to $\langle
x,x^{g}\rangle,x\in R^{\sharp}$. As $\C_{G}(x)=\C_{G}(R)$ we get 
$G=\langle \C_{G}(x),\C_{G}(x^{g})\rangle$. So by (0.1) $0\neq
[V,x]\cap [V,x^{g}]=U$. Now $\langle Q,Q^{g}\rangle$ centralizes $U$
and so $S\langle Q,Q^{g}\rangle=O^{2'}(P_{i})$ has a fixed point on
$V$, i.e. $a_{i}=0$ by (0.4).


\lemma{Let $ G \simeq Sz(q) $, $ V $ an irreducible $ GF(2)G
$--module.  Then $ [V,Z(S),S] \not= 0 $, for $ S \in Syl_{2}(G) $.}

{\it{Proof.}}~~Suppose false.  Then $ [V,Z(S),Z(S)] = 0 $.  
Any two conjugates of $
Z(S) $ generate $ G $. By [BHu;XI,\S\,3] there is some 
involution $i\in Z(S)$
inverting some element $\omega$ of order $5$ in $G$. As 
$G=\langle Z(S),Z(S)^{\omega}\rangle=\langle
i,Z(S)^{\omega}\rangle$ we get

\hspace*{\fill}
$V=[V,Z(S)]\oplus [V,Z(S)^{\omega}]=
\C_{V}(Z(S))\oplus \C_{V}(Z(S)^{\omega})$
\hspace*{\fill}

\noindent
and

\hspace*{\fill}
$\C_{V}(i)\cap \C_{V}(Z(S)^{\omega})=0$.
\hspace*{\fill}


\noindent
So $\C_{V}(i)=\C_{V}(Z(S))$ and $\C_{V}(i)\cap \C_{V}(\omega)=0$. As
$i$ normalizes $\langle \omega \rangle$ we get $\C_{V}(\omega)=0$.
By [Mar] $ V $ is the natural module. 
But then $ [V,Z(S),S] \not= 0 $, [BHu;XI,\S\,3].

\bigskip


\lemma{Let $ G = G_{2}(q) $, $ q = 2^{n} $, $V$ be a nontrivial $ GF(2)G
$--module, $ R $ a long root subgroup of $ G $.  Let $ G_{1} = N_{G}
(R) $, $ G_{2} $ be the other minimal parabolic containing a Sylow
2--subgroup $ S $ of $ G_{1} $.  Set $ V_{i} = \C_{V} (O_{2}
(G_{i})) $.  If $ [V_{2}, O^{2'} (G_{2})] = 0 $ and $ [V,R,R] = 0 $,
then $ [V_{1},S,S] = 0 $.}

{\it{Proof.}}~~Set $ Q = \bigcap\limits_{g\in G_{2}} O_{2} (G_{1})^{g} $.  Then $ |Q| =
q^{3} $.  Set $ V_{3} = \C_{V}(Q) $.

We have $ V_{1} \subseteq V_{3} $.  As $ \langle G_{1},G_{2}\rangle  = G $, $ V_{1}
\not= V_{3} $.  We study the action of $ G_{2} $ on $ V_{3} $.  Let $
g \in G $ such that $ S = O_{2} (G_{2}) R^{g} $.  By assumption $
[V_{3},R^{g},R^{g}] = 0 $.  As $ S/Q $ is isomorphic to a Sylow
2--subgroup of $ L_{3}(q) $, there is an elementary abelian group $ W $
of order $ q^{2} $ in $ S/Q $ such that $ S/Q = W(O_{2}(G_{2})/Q) $
and $ R^{g} Q/Q \subseteq W $.  As $ W $ carries the structure of a
2--dimensional vectorspace over $ GF(q) $ and $ R^{g} Q/Q $
corresponds to a one--space, we get with (0.2) that $ W \cap O_{2}
(G_{2})/Q $ acts quadratically on $ V_{3} $.  Now also $
O_{2}(G_{2})/Q $ carries the structure of a 2--dimensional
vectorspace over $ GF(q) $ and $ W \cap O_{2} (G_{2})/Q $ is a
one--space.  Now any one--space in $ O_{2} (G_{2})/Q $ acts
quadratically on $ V_{3} $ as all one--spaces in $ O_{2}(G_{2})/Q $
are conjugate under $ G_{2} $.  There is a subgroup $ T $ of $
O_{2} (G_{2}) $ such that $ T Q/Q $ is a one--space in $ O_{2}
(G_{2})/Q $ and $ T O_{2} (G_{1}) = S $.  Hence $ [V_{1},T,T]
\subseteq [V_{3},T,T] = 0 $ and so $ [V_{1},S,S] = 0 $.

For what follows we fix the following:  Let $ M $ be some irreducible $
GF(2) $--module for $ G = G(q) $.  Then $ M \otimes GF(q) $ is a
direct sum of algebraic conjugates of some irreducible $ GF(q)
$--module $ V $.  If $ M \otimes GF(q) $ is an algebraic conjugate of
a fundamental module for
some fundamental weight $ \lambda $ we write $ M = V(\lambda) $.

As the assumptions of the theorem and the corollaries are independent
of the field, it is enough to prove the theorems for $ V $ to be an
irreducible $ GF(q) $--module, where $ G $ is defined over $ GF(q) $.

\vspace*{0.5cm}

\section{Some special modules}

Hypothesis 1:  Let $ G $ be a Chevalley group over $ GF(q) $, $ q =
2^{m} $.  Let $ R $ be some root group and $ Q = Z_{2} (O_{2}(N_{G}
(R))) $.  In case $ G \cong Sp (2n,q) $ or $ F_{4}(q) $ we assume
in addition $ R $ to be long.  Let $ V $ be a nontrivial irreducible $
GF(q)G $--module.  Assume $ [V,R,Q] = 0 $.

In this chapter we always work under hypothesis 1.  Remember that if $
G $ is not $ L(2,q), Sp(4,q) $ or $ ^{2}F_{4}(q) $ then $ R = Q'$. 
Furthermore in all cases but $ ^{2}F_{4}(q) $ we have $ Q = O_{2}
(N_{G}(R)) $.

\bigskip

\lemma{Let $ H $ be a maximal parabolic of $ G $ such that $
[V,R,O_{2}(H)] = 0 $.  If there is some $ g \in G $ such that $ R^{g}
\subseteq H - O_{2}(H) $ and $ [R^{g},S] \subseteq O_{2}
(H) $ for some $ S \in Syl_{2}(H) $ and $ [\C_{V}(O_{2}(H)), R^{g}]
\neq 0$, then for the minimal parabolic $ P $ of $ G $ containing $
S $ but $ P \not\subseteq H $ we get $ [\C_{V}(S),O^{2'}(P)] = 0
$. So if
$V=V(\lambda),\lambda=\sum\limits^{n}_{i=1}a_{i}\lambda_{i}$,
$\lambda_{i}$ fundamental weights, then $a_{i}=0$, where $P=P_{i}$.}

{\it{Proof.}}~~Set $ X = \C_{V} (O_{2}(H)) $.  Then by assumption $ [R^{g},X]
\not= 0 $.  As $ [S,R^{g}] \subseteq O_{2}(H) $, we have that $
[R^{g},X] $ is invariant under $ S $.  This gives $ \C_{V}(S) \cap
[R^{g},X] \not= 0 $.  By (0.3) $ \C_{V}(S) \subseteq [R^{g},X] $. 
So $ \C_{V}(S) $ is centralized by $ \langle S,O_{2}(H)^{g}\rangle 
\not\subseteq H $.

\lemma{Let $ G = SL(n,q) $, $ n \geq 2 $, then $ V = V(\lambda) $ for
some fundamental weight $ \lambda $.}

{\it{Proof.}}~~Let first $ n = 2 $.  Then by hypothesis 1 we have $ [V,S,S] = 0 $,
for some Sylow 2--subgroup $ S $ of $ G $.  So the assertion follows
with [Hig].

The general case we treat by induction on $ n $.  We have $ n \geq 3
$.  Let $ G_{1} $ be the stabilizer of a point and $ G_{n-1} $ be the
stabilizer of a hyperplane incident to the point for $G_{1}$
in the natural representation of $ G $. 
Then $ N_{G}(R) = G_{1} \cap G_{n-1} $.  Let $ K_{i} = O_{2}(G_{i}) $
and $ H_{i} $ a complement of $ K_{i} $ in $ O^{2'}(G_{i}) $, $ i =
1, n-1 $.  Set $ V_{i} = \C_{V}(K_{i}) $.  As there are conjugates
of $ R^{g} $ contained in $ H_{1} $, $ H_{n-1} $ respectively, with $
Q^{g} \cap H_{i} = Z_{2}(O_{2}(N_{H_{i}}(R^{g}))) $ we get by
induction that $ V_{i} $ are trivial modules or fundamental modules.

Let $V=V(\lambda),\lambda=\sum\limits^{n-1}_{i=1}a_{i}\lambda_{i}$.
Now as $V_{1}$ is trivial or fundamental, we get with (0.4) that for
$j>1$ there is at most one $a_{j}\neq 0$. Similarly for $V_{n-1}$.
Together we have either just one $a_{i}$ is nonzero hence $1$ and so
$V=V(\lambda),\lambda$ a fundamental weight, or
$\lambda=a_{1}\lambda_{1}+a_{n-1}\lambda_{n-1}$. In the latter
$ [V,R,H_{1} \cap H_{n-1}] = 0 $.  Application of (1.1) now
yields the contradiction $a_{1}=0=a_{n-1}$.

\bigskip

\lemma{Let $ G $ be $ \Omega^{+} (2n,q) $, $ n \geq 4 $, then $ V $
is the natural or spin module.}

{\it{Proof.}}~~Let $ G_{1} $, $ G_{n-1} $, $ G_{n} $ be the three maximal parabolics
with connected diagram, i.e. $ O^{2'} (G_{1}/K_{1}) \simeq \Omega^{+}
(2n-2,q) $, $ O^{2'}(G_{n-1}/K_{n-1}) \simeq O^{2'} (G_{n}/K_{n})
\simeq SL(n,q) $, where $ K_{i} = O_{2}(G_{i}) $.  Set $ V_{i} =
\C_{V} (K_{i}) $.

As there are conjugates of $ R $ in $ G_{i} $ intersecting $ K_{i} $
trivially, $ i = 1,n,n-1 $, we may apply (1.2) or use induction.  
Let $V=V(\lambda)$, then this puts restrictions on the coeffizients
other than $a_{i}$, for
$\lambda=\sum\limits_{i=1}^{n}a_{i}\lambda_{i}$.
So we have that $ V_{1} $ is either trivial, a natural module or a
spin module.

Suppose first that $ V_{1} $ is trivial.  Then by (1.2) both $ V_{n-1}
$, $ V_{n} $ are natural modules and so by (0.4) $ V $ is the natural
module.

So we may assume that $ V_{1} $ is nontrivial.  Let $ g \in G $ such
that $ R^{g} \leq Q - R $.  Then  by (0.5)
$O^{2'} (P_{2}) $ has a fixed point on $ V $ and $a_{2}=0$.
This shows that $ V_{1} $ has to be a
spin module.  So we may choose notation such that 
$a_{n}=0$. Then we get with
(1.2) that $ V_{n} $ is the natural module and so $ V_{n-1} $ is a
trivial module.  Hence we have that $ V $ is a spin module.

\bigskip

\lemma{Let $ G $ be $ \Omega^{-} (2n,q) $, $ n \geq 3 $.  Then $ V $
is the natural module or the spin module.}

{\it{Proof.}}~~We first treat the case $ n = 3 $.  Let $ G_{1},\, G_{2} $ be the two
maximal parabolics containing a Sylow 2--subgroup $ S $,  where we
choose notation such that $ G_{2} = N_{G}(R) $.  As $ R $ acts
quadratically and $ S = R^{g} O_{2}(G_{2}) $ for some $ g \in G $, we
get that $ V_{2} = \C_{V}(O_{2}(G_{2}))$ is either a trivial $ O^{2}
(G_{2}/O_{2}(G_{2})) $ module or the natural $ SL(2,q) $--module.

Suppose first that $ V_{2} $ is not a trivial module.  We may assume $
R^{g} \leq O_{2}(G_{1}) $.  Now we have $ W = [V,R] \cap [V,R^{g}]
\not= 0 $.  This gives that $ W $ is centralized by $ \langle Q,Q^{g}\rangle  $. 
Thus $ O^{2'}(G_{1}) = \langle Q,Q^{g}\rangle  S $ has a fixed point on $ V $,
where $ S \in Syl_{2}(G_{1}) $.  So we have that $ V $ is the natural
module.

Suppose now that $ V_{2} = [V,R] $ is a trivial module for $ O^{2'}
(G_{2}) $.  Then $ [V,R,O_{2}(G_{1})] = 0 $ and so $ [V,O_{2}(G_{1}),
O_{2}(G_{1})] = 0 $, as $ O_{2}(G_{1}) $ is generated by $ G_{1} $ --
conjugates of $ R $.  Now there is some $ h \in G $ such that $ O_{2}
(G_{1}) (O_{2}(G_{1})^{h} \cap G_{1}) = S $.  Hence a Sylow
2--subgroup of $ SL(2,q^{2}) $ acts quadratically on $ V_{1} $, which
gives that $ V_{1} $ is the natural $ SL(2,q^{2}) $--module.  So by
(0.4) $ V $ is the spin module.

So assume now $ n \geq 4 $.  Let $ G_{1} $ be the maximal parabolic
such that $ O^{2'} (G_{1}/K_{1}) \simeq \Omega^{-} (2n-2,q),K_{1} =
O_{2}(G_{1}) $.  Set $ V_{1} = \C_{V}(K_{1}) $.  Then by induction $
V_{1} $ is trivial, a natural module or a spin module for $ G_{1} $. 
Let $ G_{n-1} $ be the maximal parabolic containing a Sylow
2--subgroup of $ G_{1} $ such that $ O^{2'} (G_{n-1}/K_{n-1}) \simeq
SL(n-1,q) $, $ K_{n-1} = O_{2}(G_{n-1}) $.  Set $ V_{n-1} = \C_{V}
(K_{n-1}) $.  Then by (0.3) and (1.2) $ V_{n-1} $ is either trivial
or a fundamental module. In particular (0.4) puts restrictions on
the coefficients $a_{i}$, $V=V(\lambda)$,
$\lambda=\sum\limits_{i=1}^{n}a_{i}\lambda_{i}$.

Suppose first that $ V_{1} $ is trivial.  Then we have that $ V_{n-1} $
has to be the natural module. So just $a_{1}$ is nonzero and hence $V$
is the natural module by (0.4).

So let $ V_{1} $ be nontrivial.  Let $ g \in G $ such that $
R^{g} \leq Q - R $.  Then by (0.5)
we have $a_{2}=0$. Hence $ V_{1} $ is the spin module.
So $a_{i}=0$ for $i>1$, $i\neq n-1$. If 
now $ V_{n-1} $ is a
trivial module we get $a_{1}=0$ and so 
$ V $ is the spin module.  So assume that $
V_{n-1} $ is nontrivial.  Then $a_{1}\neq 0$ and so $V_{n-1}$
has to be the natural module.  Now
by (1.1) $ \C_{V}(S) $ is centralized by $ O^{2'}(P_{n-1}) $,
so $a_{n-1}=0$, contradicting the fact that $ V_{1} $ is a spin module.

\bigskip

\lemma{Let $ G $ be $ Sp(2n,q) $, $ n \geq 2 $, and $ R $ be a long
root group. Then $ V $ is the natural module or the spin module.}

{\it{Proof.}}~~Let first $ n = 2 $.  Let $ G_{1} = N_{G}(R) $.  Then as $ S = QR^{g}
\in Syl_{2}(G_{1}) $ for suitable $ g \in G $, we get that $ \C_{V}
(Q) $ is either a trivial or a natural $ SL(2,q) $--module for $
O^{2'}(G_{1}/Q) \simeq SL(2,q) $.

If $ \C_{V}(Q) $ is the natural module, we get that $ 0 \not= [V,R]
\cap [V,R^{g}] $ is centralized by $ \langle Q,Q^{g}\rangle  $, and so $ \C_{V}
(S) $ is centralized by $ S\langle Q,Q^{g}\rangle  = O^{2'}(G_{2}) $, where $ G_{2} $
is the other parabolic containing $ S $.  Hence $ V $ is the natural
module by (0.4).

Let $ \C_{V}(Q) $ be a trivial module.  Then $ [V,R,O_{2}(G_{2})]
= 0 $ and so $ [V,O_{2}(G_{2}),O_{2}(G_{2})] = 0 $, as $ O_{2}(G_{2}) $
is generated by $ G_{2} $ -- conjugates of $ R $.  As $ S =
O_{2}(G_{2})(O_{2}(G_{2})^{h} \cap G_{2}) $ for suitable $ h \in G $,
we get that $ V_{2} = \C_{V}(O_{2}(G_{2})) $ is the natural module
for $ O^{2'} (G_{2}/O_{2}(G_{2})) \simeq SL(2,q) $.  Then $ V $ is
the spin module.

Let now $ n \geq 3 $.  Let $ G_{1},\; G_{n} $ be the maximal
parabolics containing a given Sylow 2--subgroup $ S $ such that $
O^{2'} (G_{1}/K_{1}) \simeq Sp(2n-2,q) $ and $ O^{2'} (G_{n}/K_{n})
\simeq SL(n,q) $, where $ K_{i} = O_{2}(G_{i}) $, $ i = 1,n $.  Set $
V_{i} = \C_{V}(K_{i}) $, $ i = 1,n $.  By induction we have that $
V_{1} $ is a trivial module, a natural module or a spin module. 
By (1.2) $ V_{n} $ is a fundamental module. Let $V=V(\lambda)$,
$\lambda=\sum\limits_{i=1}^{n}a_{i}\lambda_{i}$. If $V_{1}$ is
trivial we get with (0.4) $a_{i}=0$ for $i\neq 1$. So $a_{1}\neq 0$
and $ V $ is the natural module.

So suppose that $ V_{1} $ is nontrivial.  Choose $ R^{g} \leq Q-R $. 
Then (0.5) yields $a_{2}=0$. In particular we
get that $ V_{1} $ is a spin module, implying $a_{n}\neq 0$.
If $ V_{n} $ is trivial, we get
that $ V $ is the spin module.  So assume that $ V_{n} $ is the
natural module.  But as in (1.4) we get that $ \C_{V}(S) $ is
centralized by $ O^{2'} (P_{n}) $ and so $a_{n}=0$, a contradiction.

\bigskip

\lemma{Let $ G $ be $ SU(n,q) $, or $ Sp(2n,q) $, $n \geq 3 $.  Then $
V = V(\lambda) $ for some fundamental weight $ \lambda $.}

{\it{Proof.}}~~Suppose first $ G \cong SU(n,q) $.  Let $ n = 3 $.  Then we have $ Q
\in Syl_{2}(G) $.  Hence by (0.3) $\C_{V}(Q)$ is an irreducible
module for the torus. This gives that $\C_{V}(Q)$ is one
dimensional over $GF(q^{2})$ and as $[V,R]\leq \C_{V}(Q)$ we get 
$ |[V,R]| \leq q^{2} $.  As there
are $ g,h \in G $ such that $ G = \langle R,R^{g},R^{h}\rangle  $, we get $ |V|
\leq q^{6} $ and so $ V $ has to be the natural module.  For $ n = 4
$, we have $ U(4,q) \simeq \Omega^{-} (6,q) $ and the assertion
follows with (1.4).

Assume $ n \geq 5 $ or $ G$ a symplectic group.  In the latter the
assertion follows for $ R $ long with (1.5).  So we may assume that $
R $ is short.  Let $ G_{1} = N_{G}(R) $.  Then by induction $ V_{1} =
\C_{V}(Q) $ is a trivial module for $ G_{1} $ or some fundamental
module.  Let $ V_{1} $ be nontrivial.  There is some $ R^{g} \leq
G_{1} -Q $ such that $ [S,R^{g}] \leq Q $, where $ S \in
Syl_{2}(G_{1}) $.  Now we get with (1.1) that $ \C_{V}(S) $ is
centralized by $ \langle S,Q^{g}\rangle  = O^{2'}(P_{1}) $.  
Further as $V_{1}$ is fundamental we get with (0.4) that
$V=V(\lambda)$, $\lambda=\Sigma a_{i}\lambda_{i}$, where $a_{i}=0$
for all but one $i,i>1$.  As $\C_{V}(S)$ is centralized by
$O^{2'}(P_{1})$ we get $a_{1}=0$ and so $V$ is fundamental. 
If $ V_{1} $ is trivial, then let $ G_{m} $
be the other maximal parabolic containing $ S $ with connected
diagram.  As $ [V,R,\C_{G}(R)] = 0 $, we get $ [V,R,O_{2}(G_{m})] = 0
$.  Now $ R \subseteq O_{2}(G_{m}) $ and so $
[V,\langle R^{G_{m}}\rangle ,\langle R^{G_{m}}\rangle ] = 0 $.  Hence a Sylow 2--subgroup of $
P_{1}/O_{2}(P_{1}) $ acts quadratically on $ \C_{V}(O_{2}(P_{1}))
= W $.  So $ W $ is the natural module by [Hig]. Let $V=V(\lambda)$,
$\lambda=\Sigma a_{i}\lambda_{i}$. We have by (0.4) $a_{1}\neq 0$ and
$a_{i}=0$ for $i>1$. So $V$ is the natural module. 


\bigskip

\lemma{Let $ G $ be $ E_{n}(q) $, $ n = 6,7,8 $, with Dynkin diagram
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Then one of the following holds
\begin{itemize}
\item[(i)]  $ n = 6 $ and $ V = V(\lambda_{1}) $ or $ V(\lambda_{6}) $
\item[(ii)]  $ n = 7 $ and $ V = V(\lambda_{7}) $.
\end{itemize}}

{\it{Proof.}}~~(i) Let $ G = E_{6} (q) $.  Then $ N_{G}(R) = G_{4} $.  Furthermore $
O^{2'} (G_{4}/Q) \simeq SL(6,q) $.  So by (1.2)  $ V_{4} =
\C_{V}(Q) $ is trivial or a fundamental module.  Let $ R^{g} \leq
Q-R $.  Then by (0.5) $\C_{V}(S) $ is centralized by 
$ O^{2'}(P_{4}) $ for $ S \in Syl_{2}
(P_{4}) $, so if $V=V(\lambda)$,
$\lambda=\sum\limits_{i=1}^{6}a_{i}\lambda_{i}$, then $a_{4}=0$. Now
some $a_{i}$, $i\neq 4$, has to be nonzero.

We have that $ O^{2'} (G_{1}/K_{1}) \simeq O^{2'} (G_{6}/K_{6})
\simeq \Omega^{+}(10,q) $, where $ K_{i} = O_{2}(G_{i}) $, $ i = 1,6 $. 
Now by (1.3) $ V_{i} = \C_{V}(G_{i}) $ is trivial, a natural
module or a spin module for $ O^{2'} (G_{i}/K_{i}) $, $ i = 1,6 $. 
Now $V_{1}$ implies that $a_{3}=a_{5}=0$ and at most one of $a_{2}$
or $a_{6}$ is nonzero. Furthermore $V_{6}$ implies that
$a_{2}=a_{3}=0$ and at most one of $a_{1}$ or $a_{5}$ is nonzero.
This shows that $a_{2}=a_{3}=a_{4}=a_{5}=0$. As $V_{4}$ is
fundamental we get exactly one of $a_{1}$ or $a_{6}$ is nontrivial,
the assertion. 


(ii)  Let $ G \simeq E_{7}(q) $.  Then $ G_{1} = N_{G}(R) $.  We have $
O^{2'} (G_{1}/Q) \simeq \Omega^{+} (12,q) $.  So by (1.3) $ V_{1} =
\C_{V}(Q) $ is a trivial, natural or spin module for $ \Omega^{+}
(12,q) $.  Application of (0.5) shows that $ [\C_{V}(S), O^{2'}
(P_{1})] = 0 $. Let $V=V(\lambda)$,
$\lambda=\sum\limits_{i=1}^{7}a_{i}\lambda_{i}$, $\lambda_{i}$
fundamental weights. We have $a_{1}=0$ by (0.5) and in any case
$a_{6}=0$. Further exactly one of $a_{i}$, $i\neq 1,6$, is nonzero.
Hence $ V_{1} $ is nontrivial.  
As there is some conjugate $
R^{g} \leq G_{7}- O_{2}(G_{7}) $, we may apply (i).  As
$a_{1}=a_{6}=0$ we get $[\C_{V}(S), O^{2'}(G_{7})] = 0 $ and so 
$a_{i}=0$ for $i\neq 7$ and then $V=V(\lambda_{7})$.



(iii) Let now $ G \simeq E_{8}(q) $.  Then $ G_{8} =N_{G}(R) $.  By
(0.5) we again have that $ \C_{V}(S) $ is centralized by $ O^{2'}
(P_{8}) $. By (ii) we have that $
V_{8} = \C_{V}(Q) $ is a trivial $ O^{2'} (G_{8}) $--module.  
Let $V=V(\lambda)$, $\lambda=\sum\limits_{i=1}^{8}a_{i}\lambda_{i}$,
$\lambda_{i}$ fundamental weights. By (0.5) $a_{8}=0$. By (ii)
$a_{i}=0$ for $i\neq 7$. With (1.3) applied to
$V_{1}=\C_{V}(O_{2}(G_{1}))$ we get $a_{7}=0$, a contradiction. 

\bigskip

\lemma{Let $ G $ be $ ^{2}E_{6}(q) $ or $ F_{4}(q) $ with
Dynkin diagram
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%
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%
\end{picture}
$$
Choose notation such that $ N_{G}(R) = G_{1} $, then $ V =
V(\lambda_{4}) $.}

{\it{Proof.}}~~By (0.5) we have that $ [\C_{V}(S), O^{2'} (P_{1})] = 0 $, $ S \in
Syl_{2}(G_{1}) $.  So $ V_{1} $ is a nontrivial module.  Furthermore
there is some conjugate $ R^{g} \in G_{1}-Q $, such that $ R^{g} \leq
Z(O_{2}(G_{4})) $.  This gives $ O^{2'} (G_{4}) = \langle Q,Q^{g}\rangle  S $.  
As $V_{1}\subseteq [V,R]$ we have $[V,R]\cap [V,R^{g}]\neq 0$ and so 
we get that $ \C_{V}(O_{2}(G_{4})) $
is a trivial module.  Now by (1.5), (1.6) we have that $ V_{1} $ is
the natural module, so $ V = V(\lambda_{4}) $.


\lemma{Let $ G $ be $ G_{2}(q) $.  Then $ V $ is the natural module.}

{\it{Proof.}}~~The assertion is clear for $ q = 2 $.  So assume $ q \geq 4 $.  Let $
G_{1} = N_{G}(R) $ and $ G_{2} $ be the other maximal parabolic
containing a Sylow 2--subgroup $ S $ of $ G $.  Let $ V_{i} =
\C_{V}(O_{2}(G_{i})) $, $ i = 1,2 $.  Then by (0.5) we have that $
V_{2} $ is a trivial module.  By (0.7) we have that $ \C_{V}(Q) $
is the natural $ G_{1}/Q $--module. Now (0.4) yields the assertion. 

\bigskip

\lemma{Let $ G $ be $ ^{3}D_{4}(q) $.  Then $ V $ is the natural
module.}

{\it{Proof.}}~~Let $ G_{1},\, G_{2} $ be the two minimal parabolics containing a
Sylow 2--subgroup $ S $ of $ G $.  Choose notation such that $ G_{2}
= N_{G}(R) $.  Let $ K_{i} = O_{2}(G_{i}) $ and $ V_{i} = \C_{V}
(K_{i}) $, $ i = 1,2 $.  Then by (0.5) we have that $ V_{1} $ is a
trivial module.

Choose conjugates $ R^{g} $, $ R^{h} $ such that $ \langle
R,R^{g}\rangle \unlhd G_{1}
$, $ R^{h} \cap K_{1} = 1 $, $ R^{h} \leq K_{2} $.  Choose $x\in R$,
$y\in R^{g}$. We have 

$$
[\C_{V}(xy),x]=[\C_{V}(xy),y]$$

\noindent
and so 

\begin{itemize}
\item[] $|\C_{V}(xy):
\C_{V}(\langle x,y\rangle)|=|\C_{V}(xy):\C_{\C_{V}(xy)}(x)|=$

\item[] $|[\C_{V}(xy),x]\cap [\C_{V}(xy),y]|\leq
|[V,x]\cap[V,y]|\leq |[V,R]\cap [V,R^{g}]|$

\end{itemize}

\noindent
As $xy\sim x$ we get 

$|\C_{V}(x):\C_{V}(\langle x,y\rangle|\leq
|[V,R]\cap [V,R^{g}]|$ \hfill{(1)}


Now choose $u\in R^{h}$. As $O^{2'}(G_{1})$ acts $2$-transitively on
$R^{G_{1}}$ and $[[V,R]\cap [V,R^{g}],O^{2'}(G_{1})]=0$, we get that
$[V,R]\cap [V,R^{g}]=[V,(R^{g})^{u}]\cap [V,R^{g}]=[V,R^{g}]^{u}\cap
[V,R^{g}]$. This shows $|[[V,R^{g}],u]|=|[V,R]:[V,R^{g}]\cap [V,R]|$.
As $R^{h}\sim R^{g}$ in $G_{2}$, we get for $x\in R$ by (1) using
$[V,R^{g},x]=0$: 

\noindent
$|[V,R]\cap [V,R^{g}]|\geq |\C_{V}(x):\C_{V}(\langle x,u\rangle
|\geq |[V,R^{g}]:\C_{[V,R^{g}]}(u)|
=$\\
$|[[V,R^{g}],u]|=|[V,R]:[V,R^{g}]\cap [V,R]|$.  
{\raisebox{0.3cm}{\hspace*{9cm}(2)}}


As $ [V,R] \cap [V,R^{g}] $ is normalized by $ G_{1} $ and
centralized by $ O^{2'}(G_{1}) $, we get that $ |[V,R] \cap [V,R^{g}]|
\leq V_{1} $. As $V_{1}$ is trivial for $O^{2'}(G_{1})$ and
irreducible for a torus we get $|V_{1}|\leq q^{3}$ and so 
$|[V,R]\cap [V,R^{g}]|\leq q^{3}$. Furthermore $[V,R]$ is a nontrivial
module for $L_{2}(q^{3})$ which shows 
$ |[V,R]| \geq q^{6} $.  Hence by (2) $ |[V,R]| =
q^{6} $ and $ |[V,R] \cap [V,R^{g}]| = q^{3} $.  If $ q \not= 2 $,
then we have $ \C_{[V,R]}(S) = \C_{[V,R]}(s) $ for any $ s \in
S^{\#} $.  If $ q = 2 $ the same is true as any involution in $ S $
inverts an element of order 7 in $ SL(2,8) \simeq G_{2}/K_{2} $.  So $
S $ acts quadratically on $ [V,R] $.  Hence $ V_{2} $ is the natural $
SL(2,q^{3}) $--module. So by (0.4) we get the assertion.

\lemma{$ G \not\simeq~\!^{2}F_{4}(q) $.}

{\it{Proof.}}~~Suppose false.  The proof follows [MeiStr;(2.17)] where the case of
the Tits group has been treated.  Let $ G_{1},\, G_{2} $ be the two
minimal parabolics containing a Sylow 2--subgroup $ S $.  Choose
notation such that $ G_{1} = N_{G}(R) $.  Let $ \Gamma $ be the coset
graph with respect to $ G_{1} $ and $ G_{2} $.  Let $ (1,2,3,4,5) $
be a path of length 4 in $ \Gamma $ with ``$G_{1} = G_{1}$'' and
``$G_{2} = G_{2}$''.  Set $ Z_{2} = \langle R^{G_{2}}\rangle  $ and $ U_{1} =
\langle Z_{2}^{G_{1}}\rangle  $, then $ U_{1} = Q $.  We know that 
$ [V,R,U_{1}] =0 $. As $U_{1}\not\leq O_{2}(G_{3})\not\geq U_{5}$ we
get $\langle U_{1},U_{5}\rangle O_{2}(G_{3})\geq O^{2}(G_{3})$. As
$U_{1}\cap U_{5}$ is centralized by $\langle U_{1},U_{5}\rangle$, we
see $|U_{1}\cap U_{5}|\leq q$. Now $|U_{1}\cap O_{2}(G_{3})|=q^{4}$,
which gives $U_{1}\cap O_{2}(G_{3})\not\leq U_{3}$. So we get
$\langle U_{1},U_{5}\rangle \geq O^{2}(G_{3})$
and $ (O^{2}(G_{3})
\cap G_{1}) R = G_{1} \cap G_{2} $.  Moreover $ \langle U_{1},U_{5}\rangle  $
centralizes $ [\C_{V}(U_{1}), Z(O_{2}(G_{5}))] $ and hence $ G_{1}
\cap G_{2} $ centralizes $ [U,Z(O_{2}(G_{5}))] $ for any chief factor $
U $ of $ G_{1} $ in $ \C_{V}(U_{1}) $.  As $ O^{2'}(G_{1}/O_{2}
(N_{G}(R))) \simeq Sz(q) $ we get with (0.6) that $ [U,O^{2'}(G_{1})]
= 1 $.  Hence we have that $ [\C_{V}(Q),O^{2'}(G_{1})] = 1 $.

Choose now $ R^{g} \leq Q-R $.  Then (0.5) 
shows that $ \C_{V}(S) $ is centralized by $
O^{2'}(G_{2}) $.  But then $ \C_{V}(S) $ is
centralized by $ G $, a contradiction.

\noindent
Now (1.1) -- (1.11) and (0.6) prove  the Theorem.

\vspace*{0.5cm}


\section{Strong quadratic modules}

In this chapter we are going to prove Corollary 1.  So we assume

\noindent
Hypothesis 2:  Let $ G $ be a Chevalley group over $ GF(q) $, $ q =
2^{n} $, $ V $ a nontrivial irreducible $ GF(q)G $--module and $ A $
be a nontrivial 2--subgroup of $ G $ such that $ [V,A,A^{g}] = 0 $
for any $ g \in G $ with $ [A,A^{g}] = 1 $.

Choose $ A $ maximal satisfying hypothesis 2.  Let $ h \in G $ such
that $ [A,A^{h}] = 1 $.  Set $ B = \langle A,A^{h}\rangle $.  Then for $ g \in G $
with $ [B,B^{g}] = 1 $ we get $ [V,\langle A,A^{h}\rangle ,\langle A^{g},A^{hg}\rangle ] = 0 $. 
By maximality of $ A $ this implies $ A = B $, i.e. $ A $ is weakly
closed in $ \C_{G}(A) $ with respect to $ G $.

For the remainder of this chapter let $ G $, $V$ be as in hypothesis
2 and $A\leq G$ be maximal satisfying hypothesis 2.

\bigskip

\lemma{Let $G\not\simeq Sp(4,q),~G_{2}(2)$ and $R$ be a long (if
$G\simeq F_{4}(q)$ any) root group. Suppose there is some quadratic
fours group $W$, (with $\langle W^{N_{G}(R)}\rangle$ nonabelian in case
$G\simeq SL(n,q)$), contained in $N_{G}(R)$ such that one of the
following holds

\begin{itemize}
\item[(i)] $|W\cap Z(O_{2}(N_{G}(R)))|=|W\cap R|=2$.

\item[(ii)] $W\cap Z(O_{2}(N_{G}(R)))=1$ and there is some $g\in G$
such that $w^{g}_{1}=w_{1}$, $w^{g}_{2}=w_{2}r$, for $1\neq r\in
R$ and $W=\langle w_{1},w_{2}\rangle$.

\end{itemize}}

\noindent
{\it Then $(G,V)$ satisfies the assertion of corollary 1.}

{\it{Proof.}}~~  Suppose (ii), then $\langle w_{1},r\rangle$ satisfies
(i). So we may assume that we have (i). Set
$Q=Z_{2}(O_{2}(N_{G}(R)))$. Now we get $\langle W^{{\C}_{G}(R\cap
W)}\rangle \geq Q$ and as $[V,R\cap W,W^{\C_{G}(R\cap W)}]=0$ we get
$[V,R\cap W,Q]=0$. Now $[V,R,Q]=[V,(R\cap
W)^{N_{G}(R)},Q^{N_{G}(R)}]=0$. Application of the theorem gives that
$(G,V)$ is as in the corollary 1. 


\bigskip

\lemma{Let $G\not\simeq SL(n,q),U(3,q),Sp(4,q),Sz(q)$ or $G_{2}(2)$.
Let $R$ be a long root group. If $A\cap R\neq 1$, then corollary 1
holds.}

{\it{Proof.}}~~As $G\not\cong SL(2,q)$, $U_{3}(q)$, or $Sz(q)$, we
have $R$ is not weakly closed in $\C_{G}(R)$. So $A\not\leq R$. If
$A\not\leq Z(O_{2}(N_{G}(R)))$, then $A$ contains a fours group
satisfying (2.1)(i). (Notice $G\not\cong SL(n,q))$. So (2.1) yields
the assertion. Let $A\subseteq Z(O_{2}(N_{G}(R))$. So we have
$Z(O_{2}(N_{G}(R)))\neq R$ and then $G\simeq Sp(2n,q)$ or
$F_{4}(q)$. As $N_{G}(R)$ acts indecomposably on $Z(O_{2}(N_{G}(R)))$
and $A$ is weakly closed in $\C_{G}(A)$ we get 
$A=Z(O_{2}(N_{G}(R)))$. In case of $F_{4}(q)$ there is
another root group $L$ such that $L\cap A\neq 1$. Now the same
argument shows $A=Z(O_{2}(N_{G}(L)))$. As $Z(O_{2}(N_{G}(R)))\neq
Z(O_{2}(N_{G}(L)))$ in $F_{4}(q)$ we get a contradiction.

We are left with $G\simeq Sp(2n,q),~n\geq 3,~|A|=q^{3}$. Let $L$ be a short
root group $L\leq A$. Then $\langle
A^{N_{G}(L)}\rangle=O^{2}(N_{G}(L))$. This gives
$[V,L,O^{2'}(N_{G}(L))]=0$. By (1.6) we get that $V$ is the natural
module and we are done. 

For the remainder of this chapter we fix the following notation: Let
$R$ be a long root group centralized by $A$ and $Q=O_{2}(N_{G}(R))$.
Let $S$ be a Sylow $2$-subgroup of $N_{G}(R)$ containing $A$. 

\bigskip

\lemma{If $G\not\simeq
(S)L(n,q),~^{2}F_{4}(q),~Sp(2n,q),~F_{4}(q),~(S)U(3,q),~Sz(q)$ or
$G_{2}(2)$, then corollary 1 holds. }

{\it{Proof.}}~~Let $A\leq Q$. In all Chevalley groups not excluded by
assumption we have $[A,Q]\leq R$ and if $a\in A - R$, then
$[a,Q]=R$. So as $A$ is weakly closed in $\C_{G}(A)$, we get either
$A\subseteq R$ or $R\leq A$. Now the assertion follows with (2.2). 


Let now $A\not\leq Q$. Suppose $A\cap Q=1$. Then $[N_{Q}(A),A]=1$. As
$\C_{G}(Q)\subseteq Q$, we have $N_{Q}(A)\neq Q$. Now for $x\in
Q-N_{Q}(A),[x,A]\subseteq N_{Q}(A)$, we have $[A^{x},A]=1$.
But $A$ is weakly closed in $\C_{G}(A)$, a contradiction. 
So $A\cap Q\neq 1$. 



Let $1\neq x\in
A\cap Q$. Set $Y={\C}_{Q}(x)$. 
Choose $a\in A^{\sharp}$, $g\in Y$ with $a^{g}=ay$, $y\neq 1$. Then
$\langle x,y\rangle$ acts quadratically on $V$. If $x\in R$, then
(2.2) yields the assertion. Suppose $x\not\in R$. If $y\not\in Z(Y)$,
then $\langle x,y\rangle \cap Z(O_{2}(N_{G}(R)))=1$. Furthermore
there is some $u\in Y$ with $y^{u}=yr$, $r\in R^{\sharp}$, as
$R=O_{2}(N_{G}(R))'$. So (2.1)(ii) applies and corollary 1 holds. We
are left with $y\in Z(Y)$. Let $Y_{1}=[a,Y]\subseteq Z(Y)$. We have
$|Q:Y|=q$  and $|Q:\C_{Q}(t)|=q$ for any $t\in Q-Z(Q)$. If
$|Y_{1}|>q$, then there is some $t\in Q-Y$ with $x^{t}=xr$,
$r\in R^{\sharp}$, but $y^{t}_{1}=y_{1}$ for some $y_{1}\in
Y^{\sharp}_{1}$. So either (2.1)(ii) applies to $\langle
y_{1},x\rangle$ or $y_{1}\in R$ and so (2.1)(i) applies as $x\not\in
R$. In any case we get the assertion.

So as $[Q,A]\leq Y$, we get $|[Q/R,a]|\leq q^{2}$ for any $a\in
A-Q$. In particular $G\not\cong E_{n}(q)$, $n=6,7,8$,
$^{2}E_{6}(q)$, or $^{3}D_{4}(q)$, by using Chevalley commutator
formulas. So we have $G\cong (S)U(n,q)$, $G_{2}(q)$ or
$\Omega^{\pm}(2n,q)$.



Let $G\simeq (S)U(n,q)$. Then $Q/R$ is the natural module for
$SU(n-2,q)$. So $A$ is a group of transvections on the natural module
implying $|A:A\cap Q|\leq q$ and there is some
$g\in G$ such that $AQ\leq QR^{g}$. Furthermore by (2.1) 
we may assume $A\cap Q\leq
Z(Y)$, otherwise $\langle x,y\rangle,x\in A\cap Q,y\not\in Z(Y)$
satisfies (2.1)(ii). 

We have $[Z(Y),R^{g}]=1$. Let $a\in A-Q$, then $a=uv$,
$u\in Y$, $v\in R^{g}$, $[u,v]=1$. Suppose $u\not\in
Z(\C_{Q}(R^{g}))$. Then there is some $s\in \C_{Q}(R^{g})$ with
$u^{s}=ur$, $r\in R^{\sharp}$. Now $a^{s}=ar$ but $x^{s}=x$. Hence
$\langle x,a\rangle$ satisfies (2.1)(ii) and so we have the
assertion. Assume now $u\in Z(\C_{Q}(R^{g}))$. Then $uv$ is in a
conjugate of $R^{g}$ and so we have the assertion with (2.2).

Let $G\cong G_{2}(q)$ and $L\neq N_{G}(R)$ be the other parabolic
containing $S$. Set $E=O_{2}(L)$. As $[E,O^{2'}(L)]\neq 1$, we get
as before that $A\cap E\neq 1$. As $Z(E)^{\sharp}\subseteq R^{G}$, we
get with (2.2) that $A\cap Z(E)=1$ or the conclusion holds. So assume
$A\cap Z(E)=1$. Then $[A,Z(E)]=1$ and so $A\subseteq E$. As $E'=Z(E)$
this contradicts the fact that $A$ is weakly closed in $\C_{G}(A)$.

So assume finally that $G\cong \Omega^{\pm}(2n,q)$. Let $L$ be the
parabolic with $O^{2'}(L/E)\cong \Omega^{\pm}(2n-2,q)$, $E=O_{2}(L)$.
Again $A\cap E\neq 1$. If $A\subseteq E$, then by weak closure $A=E$,
which with (2.2) yields the assertion. Let $A\not\leq E$. As $E$ is
the natural module for $\Omega^{\pm}(2n-2,q)$, there is some $t\in
A-E$ and $s\in E$ such that $[t,s]=f$ is contained in 
a some long root group
$F$. So $\langle f,e\rangle,e\in E\cap A$ is a quadratic fours group.
As $e\not\in T$ and $Z(O_{2}(N_{G}(F)))=F$, 
$\langle f,e\rangle$ satisfies (2.1)(i). So we get the assertion.

In what follows we analyze the cases left in (2.3) one by one.

\bigskip

\lemma{If $G\cong Sp(2n,q)$, corollary 1 holds.}

{\it{Proof.}}~~Suppose false. Let $L$ be a short root subgroup contained in $Z(S)$
and $M=O_{2}(N_{G}(L))$. Suppose first $n=2$. Obviously we may assume
$q\neq 2$. Then $A\leq Q$ or $M$ as any elementary abelian subgroup
of $S$ is in $Q$ or $M$. As $N_{G}(Q)$ and $N_{G}(L)$ act irreducibly
on $Q,M$, respectively, we get $A=Q$ or $A=M$ by weak closure. Now
the assertion follows with the theorem. 



So we have $n\geq 3$. Let $A\leq Q$. If $A\cap R\neq 1$ the assertion
follows with (2.2). So we may assume $A\cap R=1$. Now $[A,Q]\subseteq
R$. If $A\not\subseteq Z(Q)$, then $\langle A^{Q}\rangle=\langle
A,R\rangle$, contradicting $A^{Q}=A$ and $R\cap A=1$. So $A\subseteq
Z(Q)$. If $q>2$, then $N_{G}(Q)$ acts indecomposably on $Z(Q)$. As $A$
is weakly closed in $Z(Q)$, we get $A=Z(Q)$, contradicting $A\cap
R=1$. So we have $q=2$ and $A$ is a complement of $R$ in $Z(Q)$, i.e.
$|A|=4$. Let $U$ be the maximal parabolic containing $S$ such that
$O^{2'}(U/O_{2}(U))\cong L_{n}(2)$. Then $\langle A^{U}\rangle$ is the
natural module $(n\geq 3)$, a contradiction to the fact that $A$ is
weakly closed in $\C_{G}(A)$.




So we may assume $A\not\leq Q$. We have $O^{2'}(N_{G}(R)/Q)\cong
Sp(2n-4,q)\times L_{2}(q)$. By (2.2) we may assume $A\cap R=1$.
Suppose $A\cap Z(Q)\neq 1$. Let $a\in A\cap Z(Q)^{\sharp},x\in
A-Q$. Then $\langle a,x\rangle$ acts quadratically on $V$.
So $[V,a,x^{\C_{N_{G}(R)}(a)}]=0$. Hence $[V,a,Q]=0$ because any
normal subgroup of $\C_{G}(Z(Q))$ not contained in $Q$ has to contain
$Q$. So there is a fours group $\langle ba,a\rangle,ba\in R$ acting
quadratically on $V$. Let $a\in L$. If $A\leq M$, then by weak
closure $A=M$. But then $A\cap R\neq 1$, a contradiction. So we have
$A\not\leq M$. But as $A\cap M\not\subseteq L$ by weak closure, we
get some $t\in M\cap A,t$ in a long root group, which with (2.2) yields the
assertion.

So we are left with $a\not\in L$. Then $a$ is an involution  of type
$c_{2}$. So we may assume $b\in L$. As $a\in N_{G}(L)$ we get
$a^{\C_{G}(b)}\geq M$ and so $[V,b,M]=0$. Now $[V,b,M\cap Q]=0$. As
$[V,a,Q]=0$, we get $[V,ab,M\cap Q]=0$. But then $[V,ab,(M\cap
Q)^{\C_{G}(ab)}]=0$ and so $[V,ab,Q]=0$, and the assertion holds by
the theorem.

So we just have to treat the case $A\cap Z(Q)=1$. Now weak closure
implies $A\subseteq \C(Z(Q))$.
As in (2.3) we get that
$|[Q/Z(Q),t]|\leq q^{2}$ for any $t\in A-Q$. This shows $|A:A\cap
Q|\leq q$ and $A$ induces transvections on both $Sp(2n-4,q)$-modules
involved in $Q/Z(Q)$. Now there is some $\rho\in N_{G}(R)$,
$o(\rho)=q+1$, with $[A,\rho]\subseteq Q$. As $A\subseteq N(M)$, we
get $A\subseteq N(M^{\rho})$ and so $A\cap M\neq 1\neq A\cap
M^{\rho}$. But $M\cap M^{\rho}=(Q\cap M)\cap (Q\cap
M^{\rho})\subseteq Z(Q)$. This shows that there is some $\langle
x,y\rangle \subseteq A$, $\langle x,y\rangle \cap Z(Q)=1$, $x\in M$,
$y\in M^{\rho}$. Hence there is some $g\in M\cap Q$ with $y^{g}=yr$,
$r\in R^{\sharp}$. Application of (2.1)(ii) gives the assertion.

\bigskip

\lemma{\it $G\cong F_{4}(q)$, corollary 1 holds.}

{\it{Proof.}}~~Let $A\leq Q$. If $A\not\leq Z(Q)$, then $\langle
A^{Q}\rangle=AR$. As $A$ is weakly closed in $\C_{G}(A)$ we get
$R\subseteq A$. Now (2.2) yields the assertion. So assume $A\subseteq
Z(Q)$. By weak closure we get $A=Z(Q)$ and so again $R\leq A$ and
(2.2) yields the assertion. So we may assume $A\not\leq Q$. Now we
show that we may assume that $A$ induces
transvections on $Z(Q)$. For this choose $t\in A-Q$ such that
$|[Z(Q),t]|\geq q^{2}$. As $A\cap Z(Q)\neq 1$, there is a foursgroup
$\langle a,b\rangle\leq Z(Q),~a\in A$, such that $\langle a,b\rangle$
acts quadratically on $V$, $b$ is a root element, and $\langle
a,b\rangle \leq O_{2}({\C}_{G}(b))-Z({\C}_{G}(b))$. But now (2.1)(i)
yields the assertion.

So assume that $A$ induces transvections on $Z(Q)$. This shows
$|A:A\cap Q|\leq q$ and
$|[A,Q/Z(Q)]|=q^{4}$. If $|A\cap Q:A\cap Z(Q)|> q$, then there is
some foursgroup $\langle x,y\rangle \subseteq A\cap Q$ such that
$\langle x,y\rangle \cap Z(Q)=1$ and for some $h\in Q$, $x^{h}=x$,
$y^{h}=yr$, $r\in R^{\sharp}$, as $|Q:\C_{Q}(t)|=q$ for any $t\in
Q-Z(Q)$. Now (2.1)(ii) yields the assertion. So assume
$|A\cap Q:A\cap Z(Q)|\leq q$. Let $a\in A\cap Q-Z(Q)$. Then the 
action of $A$ on $Q$ shows that
there is some $g\in {\C}_{Q}(a)-Z(Q)$, such that $b^{g}=bc$, for
$b\in A-Q$ and $c$ not in the root group determined by $a$. But then
$\langle a,c\rangle$ acts quadratically on $V$, and (2.1)(i) yields
the assertion. So assume  $A\cap Q=A\cap Z(Q)$. Let $L$ be the other
type of root groups such that $LR\subseteq [Z(Q),A]$. Set
$M=O_{2}(N_{G}(L))$. Then as $A$ induces transvections on $Z(Q)$, we
get $AQ\subseteq Z(M)Q$. Now $(Z(M)Q)'\subseteq Z(M)$, so
$[A,Q]\subseteq Z(M)$. As $|[a,Q/Z(Q)]|=q^{4}$ for $a\in A-Q$, we get 
that $|[a,Q]L/L|=q^{5}$ and $[a,Q,a]=1$. But then
$[a,Q]L/L$ is centralized by $a$ and $[[A,Q],A]=1$. Hence $\langle
A^{Q}\rangle$ is abelian. Now weak closure gives $|A\cap Z(M)|\geq
q^{5}$. But then as $|Z(M):Z(M)\cap Q|=q$, we get $|A\cap Q|=|A\cap
Z(Q)|\geq q^{4}$. So $|A\cap Z(Q)\cap Z(M)|\geq q^{4}$, but this
contradicts $|Z(Q)\cap Z(M)|=q^{2}$.

\bigskip

\lemma{If $G\cong SL(n,q)$, corollary 1 holds.}

{\it{Proof.}}~~Let $Q=E_{1}E_{2}$, $E_{i}\unlhd N_{G}(R)$, $E_{i}$
elementary abelian, $i=1,2$. Then $A\leq N_{G}(E_{1})$. 
As in all cases before we get $A\cap E_{1}\neq 1$ by weak closure.
But all elements in $E^{\sharp}_{1}$ are conjugate and so we may
assume $A\cap R\neq 1$.

If $A\subseteq E_{1}$ we get by weak closure $A=E_{1}$ and then the
assertion with [MeiStr1;(1.6)]. So assume $A\not\subseteq E_{1}$.
Choose $x\in R\cap A$, $y\in A-E_{1}$. Then $\langle
x,y\rangle$ is a quadratic fours group and $U=\langle\langle
x,y\rangle^{N_{G}(R)}\rangle\geq E_{1}$. So $U'\neq 1$. Now we may
apply (2.1) (i) to $\langle x,y\rangle$ which implies the assertion.


\bigskip

\lemma{$G\not\simeq~\!^{2}\!F_{4}(q)$.}


{\it{Proof.}}~~Suppose false. By (2.2) $A\cap R=1$. So by weak closure we have
$A\not\leq O_{2}(N_{G}(R))$. Let $t\in A-O_{2}(N_{G}(R))$. Then
$|[t,Q]|=q^{2}$. As $A$ is weakly closed in ${\C}(A)$ we get that
either $|A\cap Q|\geq q^{2}$ or there is some $a\in A\cap Q,~b\in Q$
such that $t^{b}=tr$, where $r$ is not in the root group belonging
to $a$. Hence $\langle a,r\rangle$ is a quadratic fours group. So in
both cases we get a quadratic fours group in $Q$ which is not
contained in a root subgroup, which contradicts (2.1).

\bigskip

\lemma{If $G\cong Sz(q)$, then $V$ is the natural module.}

{\it{Proof.}}~~We have $A=R$. As $G=\langle R,R^{g}\rangle$ for any
$g\in G-N_{G}(R)$, we get $V=[V,R]\oplus [V,R^{g}]$. By [Mar] $V$ is
the natural module.

\bigskip

\lemma{If $G\cong (S)U(3,q)$, then $V$ is a basic module as in
corollary 1.}

{\it{Proof.}}~~We have $R=A$. Let $X=\langle R,R^{g}\rangle\simeq
SL(2,q)$ for some $g\in G$. Then we know that $V$ involves just
natural $X$ - modules by (2.6). As any irreducible $G$-module 
is a restriction
of some irreducible $SL(3,q^{2})$-module, we may assume that $V$ is
irreducible for $H=(S)L(3,q)$ and $X\leq G\leq H$. By Steinberg's
tensor product lemma any irreducible $H$-module is a tensorproduct of
algebraic conjugates of the basic modules. We may assume that all
this basic modules are nontrivial. Then $V$ involves tensorproducts
of algebraic conjugates of the natural $X$-module. The quadratic
action of $X$ now gives that the tensor product just has one factor,
i.e. $V$ is an algebraic conjugate of a basic module. Over $GF(2)$
this means that $V$ is a basic module.


\bigskip

\lemma{If $G\cong G_{2}(2)$, then $V$ is the natural module.}

{\it{Proof.}}~~We have a quadratic fours group. It is easy to see
that $V$ has to be the natural $6$-dimensional module.

\bigskip

Now Corollary 1 follows from (2.3) - (2.10).



\vspace*{0.5cm}

\section{Quadratic fours groups}

\noindent
{\bf{Hypothesis 3:}}~~Let $G$ be a Chevalley group over
$GF(q),~q=2^{n}$, $V$ a nontrivial irreducible $GF(q)G$-module. Let
$R$ be a long root subgroup of $G$ and $E$ be a fours group with
$|E\cap R|=2$. Assume $[V,E,E]=0$.

For the remainder of this section we are working under hypothesis 3.
We are going to prove Corollary 2.

We fix the following notation:

\begin{itemize}
\item[(i)] $\langle e\rangle=E\cap R$.
\item[(ii)] $X=\langle E^{{\C}_{G}(e)}\rangle$.
\end{itemize}

\lemma{If $X\supseteq Z_{2}(O_{2}(N_{G}(R)))$, then the conclusion of
corollary 2 holds.}

{\it{Proof.}}~~As $[V,e,X]=0$ and $\langle e^{N_{G}(R)}\rangle=R$,
the assertion follows from the main theorem.

\bigskip

\lemma{If $G\not\cong SL(n,q),~Sp(2n,q),~F_{4}(q)$ or $G_{2}(q)$ the
conclusion of corollary 2 holds.}

{\it{Proof.}}~~In all other cases ${\C}_{G}(R)$ acts irreducibly on
$Z_{2}(O_{2}(N_{G}(R)))/R$ and $Z_{2}(O_{2}(N_{G}(R)))/R$ is the only
minimal normal subgroup of ${\C}_{G}(R)/R$. So
$Z_{2}(O_{2}(N_{G}(R)))\subseteq X$. The assertion follows with
(3.1).

\bigskip

\lemma{Let $G\cong SL(n,q)$, then $V=V(\lambda)$ for some fundamental
weight $\lambda$.}

{\it{Proof.}}~~By (3.1) we may assume $O_{2}(N_{G}(R))\not\subseteq
X$. Let $O_{2}(N_{G}(R))=E_{1}E_{2}$, $E_{i}$ elementary abelian of
order $q^{n-1}$, $E_{i}\triangleleft N_{G}(R),~i=1,2$. Then we may
assume $X\subseteq E_{1}$. Assume first $X=E_{1}$. As $X=\langle
e^{N_{G}(x)}\rangle$, we get $0=[V,\langle
e^{N_{G}(R)}\rangle,X]=[V,X,X]$. As $X$ is weakly closed in
${\C}_{G}(X)$ with respect to $G$, the assertion follows with
corollary 1.

So $X\neq E_{1}$. This is only possible for $n=3$. We have that $R$
acts quadratically on $V$. Furthermore $S=E_{1}R^{g}\in
Syl_{2}(N_{G}(E_{1}))$ for suitable $g\in G$. Set
$V_{1}={\C}_{V}(E_{1})$.

If $[R^{g},V_{1}]\neq 0$, then ${\C}_{V}(S)$ is centralized by
$\langle S,X^{g}\rangle$. As $X^{g}\subseteq E^{g}_{1}$ and
$E^{g}_{1}\cap N_{G}(E_{1})=R$, we get $\langle
S,X^{g}\rangle=O^{2'}(N_{G}(E_{2}))$. As by [Hig] $V_{1}$ is a
natural $SL(2,q)$-module, we get with (0.4) that $V$ is the natural
module, as $V_{1},V_{2}$ are like in the natural module.





If $[V_{1},O^{2'}(N_{G}(E_{1}))]=0$, then ${\C}_{V}(E_{2})$ has to be
a natural module by [Hig] as $S=E_{2}R^{h}$ for suitable $h\in G$,
$S\in Syl_{2}(N_{G}(E_{2}))$. Now again by (0.3) $V$ is a natural
module.

\bigskip

\lemma{Let $G\cong Sp(2n,q)$, then $V=V(\lambda)$ for some
fundamental weight $\lambda$.}

{\it{Proof.}}~~By (3.1) we may assume $X=Z(O_{2}(N_{G}(R)))$. Let
$f\in X$, $f\in \tilde{R}$, $\tilde{R}$ a short root subgroup. Then
$[V,\langle e,f\rangle,\langle e,f\rangle]=0$. Set $Y=\langle \langle
e,f\rangle^{{\C}_{G}(f)}\rangle$. As $e\in
O_{2}({\C}_{G}(\tilde{R}))$, we get $Y=O_{2}({\C}_{G}(\tilde{R}))$.
Furthermore $\langle f^{N_{G}(\tilde{R})}\rangle=\tilde{R}$ and so
$[V,\tilde{R},Y]=0$. Now the assertion follows with (1.6).

\bigskip

\lemma{Let $G\cong F_{4}(q)$, then $V=V(\lambda)$ for some
fundamental weight $\lambda$.}

{\it{Proof.}}~~By (3.1) we may assume that $X=Z(O_{2}(N_{G}(R)))$. So
we may assume $E=\langle e,f\rangle$, $e\in R$, $f\in \tilde{R}$,
$\tilde{R}\subseteq X$, $\tilde{R}$ a short root subgroup. As $G$
possesses an outer automorphism $\alpha$ with $R^{\alpha}=\tilde{R}$,
we may also assume that $\langle
E^{{\C}_{G}(f)}\rangle=Z(O_{2}(N_{G}(\tilde{R})))$. So $E\subseteq
Z(O_{2}(N_{G}(\tilde{R})))\cap Z(O_{2}(N_{G}(R)))=R\tilde{R}$.
Furthermore we have $\langle e^{N_{G}(R)}\rangle=R$ and $\langle
f^{N_{G}(\tilde{R})}\rangle=\tilde{R}$.

\noindent
This gives

\begin{itemize}
\item[$(\ast)$]
$[V,R,Z(O_{2}(N_{G}(R)))]=0=
[V,\tilde{R},Z(O_{2}(N_{G}(\tilde{R})))]$
\end{itemize}


Now we fix a Sylow 2-subgroup $S$ of $N_{G}(R)$ with
$Z(S)=R\tilde{R}$. Let $G_{1},G_{2},G_{3},G_{4}$ be the four maximal
parabolic subgroups of $G$ containing $S$. Set $K_{i}=O_{2}(G_{i})$
and $V_{i}={\C}_{V}(K_{i})$.

Let $V_{i}$ be a nontrivial module for $O^{2'}(G_{i})$ for some $i$.
There is some $g\in G$ such that for $T=R^{g}$ or $\tilde{R}^{g}$ the
following holds:

\begin{itemize}
\item[(i)] $T\subseteq G_{i}\backslash O_{2}(G_{i})$.
\item[(ii)] $[T,S]\subseteq O_{2}(G_{i})$.
\item[(iii)] $[V_{i},T]\neq 0$.
\end{itemize}

By (0.3), (ii) and (iii) ${\C}_{V}(S)\subseteq [V_{i},T]$ and so by
$(\ast)$ ${\C}_{V}(S)$ is centralized by $\langle
S,Z(O_{2}(N_{G}(T)))\rangle$. As $Z(O_{2}(N_{G}(T)))\not\subseteq
G_{i}$, we get that ${\C}_{V}(S)$ is centralized by $O^{2'}(P_{i})$,
where $P_{i}$ is the minimal parabolic subgroup containing $S$ but
not contained in $G_{i}$. So we have 

\begin{itemize}
\item[$(\ast\ast)$] If $[V_{i},O^{2'}(G_{i})]\neq 0$ for some $i\in
\{1,2,3,4\}$, then $[{\C}_{V}(S),O^{2'}(P_{i})]=0$, where $S\subseteq
P_{i}$, $P_{i}\not\subseteq G_{i}$.
\end{itemize}

From $(\ast\ast)$, we get that there is exactly one $j\in\{1,2,3,4\}$
such that $[{\C}_{V}(S),O^{2'}(P_{j})]\neq 0$. As there is some $h\in
G$ such that for $U=R^{h}$ or $\tilde{R}^{h}$ we have
$S=O_{2}(P_{j})U$ and $[V,U,U]=0$ by $(\ast)$, we get with [Hig] and
(0.3) that ${\C}_{V}(O_{2}(P_{j}))$ is the natural module for
$O^{2'}(P_{j}/O_{2}(P_{j}))\cong SL(2,q)$. Now $V=V(\lambda_{j})$ by
(0.4).

\bigskip

\lemma{Let $G\cong G_{2}(q)$, then $V$ is the natural module.}

{\it{Proof.}}~~By (3.1) we may assume $X\subseteq
O_{2}(N_{G}(R)),~X\neq O_{2}(N_{G}(R))$. Let $G_{1},G_{2}$ be the two
maximal parabolics containing $S\in Syl_{2}(N_{G}(R))$, where
$G_{1}=N_{G}(R)$. There is some $g\in G$ such that
$R^{g}O_{2}(G_{2})=S$. As $R\subseteq X$, we have $[V,R,R]=0$.

Suppose first that $[{\C}_{V}(O_{2}(G_{2})),~O^{2'}(G_{2})]\neq 0$.
Then ${\C}_{V}(S)\subseteq [{\C}_{V}(O_{2}(G_{2})),R^{g}]$ and so
${\C}_{V}(S)$ is centralized by $S$ and $X^{g}$. But
$X^{g}\not\subseteq G_{2}$ and so $\langle
S,X^{g}\rangle=O^{2'}(G_{1})$ centralizes ${\C}_{V}(S)$. By (0.3) and
[Hig] ${\C}_{V}(O_{2}(G_{2}))$ is the natural
$O^{2'}(G_{2})/O_{2}(G_{2})\cong SL(2,q)$ module. So by (0.4)(c)
$V=V(\lambda_{2})$. 


Let $f\in Z(O_{2}(G_{2}))-R$, then $f\sim e\sim ef$. By (0.1) 
$W=[V,f]\cap [V,e]\neq 0$. Let $g\in G_{2}$ with $e^{g}=f$, then $W$
is centralized by $\langle X,X^{g}\rangle$.

We have that $W$ is invariant under $O_{2}(G_{2})$ and so $W\cap
{\C}_{V}(O_{2}(G_{2}))\neq 0$. This shows that there is some $v\in
{\C}_{V}(O_{2}(G_{2}))^{\sharp}$ such that $v$ is centralized by some
element $\nu\neq 1$ of odd order in $O^{2'}(G_{2})$. (Notice that
$\langle X^{N_{G}(S)}\rangle=O_{2}(G_{1})$ and so $X\not\subseteq
O^{2}(G_{2})$). But ${\C}_{V}(O_{2}(G_{2}))$ is the natural module, a
contradiction.

So we have that $[{\C}_{V}(S),O^{2'}(G_{2})]=0$. As $R\subseteq X$,
we get with (0.6) that ${\C}_{V}(O_{2}(G_{1}))$ is the natural module
for $O^{2'}(G_{1})/O_{2}(G_{1})$ and so by (0.3) $V$ is the natural
module for $G$.

Now Corollary 2 follows from (3.1) - (3.6).

\vspace*{0.2cm}

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