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\begin{center}

U.~Meierfrankenfeld\\ \medskip G.~Stroth

\end{center}}


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\today \end{center}}

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\begin{document}

\tit{$F$-modules for finite simple groups}
\prel
 
In this paper we are going to determine the irreducible faithful $F$-modules for the finite quasisimple groups. Here we just concern about $F$-modules over $GF(2)$. A module $V$ is called $F$ -module for $G$, if there is a nontrivial elementary abelian subgroup $A$ in $G$ such that $|V/C_V(A)| \leq |A|$.  The group $A$ then will be called an offending subgroup or an offender. More precisely we prove

\begin{maintheorem} Let $E(G)= F^*(G)$ be a quasisimple group and $V$ be an irreducible faithful $F$-module in characteristic two for $G= \langle A^G \rangle$, $A$ a quadratic offender. Then one of the following holds
\begin{enumerate}
\item[(i)] $E(G)/Z(E(G)) \cong A_n$ and one of the following is true
\begin{enumerate}
\item[($\alpha$)] $V$ is the natural module
\item[($\beta$)] $n \leq 8$ and $|V| = 16$
\item[($\gamma$)] $n = 6$, $|Z(E(G))| = 3$ and $|V| = 64$
\end{enumerate}
\item[(ii)] $E(G)/Z(E(G)) = G(q)$ is a group of Lie type , $q = 2^t$, and one of the following is true
\begin{enumerate}
\item[($\alpha$)] $G(q) \cong L_n(q)$, $Sp(2n,q)$, $\Omega^\pm(2n,q)$ or $U_n(q)$ and $V$ is the natural or dual module.
\item[($\beta$)] $G(q) \cong L_n(q)$ and $V$ is the exterior square of the natural or dual module
\item[($\gamma$)] $G(q) \cong Sp(6,q)$ and $V$ is the spin module
\item[($\delta$)] $G(q) \cong \Omega^+(8,q)$ or $\Omega^+(10,q)$ and $V$ is the half spin module
\item[($\epsilon$)] $G(q) \cong G_2(q)$ and $V$ is the natural module.
\end{enumerate}
\end{enumerate}
\end{maintheorem}

The theorem will follow from \ref{ffmodules1}, \ref{ffmodules2}, \ref{alternating}, \ref{sporadic}, and \ref{oddchar}.
\\

The $F$-modules for the sporadic groups, alternating groups and groups of Lie type in odd characteristic have been classified by M. Aschbacher \cite{Asch}. The $F$-modules for the groups of Lie type in characteristic two have been classified by B. Cooperstein \cite{Coop} and in an unpublished paper by B. Cooperstein and G. Mason \cite{CMFF}. Unfortunately in the last two papers never the case that $E(G) \not= G$ and $A \not\leq E(G)$ was handled. The basic idea in all these papers was to get lower bounds for $|V : C_V(t)|$ for involutions $t$ and irreducible modules $V$. This then gives lower bounds for the size of $|A|$ and so one either gets contradictions or is close to the $F$-module. In contrast to this we use basically the fact that there is always a quadratic offender. Then we apply the classification of quadratic modules in \cite{MeiStr}, \cite{MeiStr1} and \cite{Str} to end up with a very short list of modules and so it is easy to detect the $F$-modules.

If we drop the assumption that $V$ is irreducible for $G = \langle A^G \rangle$,
we get direct sums of the modules in (ii)($\alpha$), ($\beta$), provided $n$ is
large enough.

\section{Preliminaries}

\begin{lemma}\label{quad}
Let $G$ be a finite group and $V$ be a faithful $F$--module for $G$. Then there is some offending subgroup $A$ with $[V,A,A] = 1$.
\end{lemma}

\proof This is just Thompson replacement. \qed

\begin{lemma}\label{L2q}
Let $G = L_2(q)$, $q = 2^n$, and $V$ be a faithful module in characteristic two. Then for any involution $a \in G$ we have $|V : C_V(a)| \geq q$. If $G = Sz(q)$ or $U_3(q)$, then for any involution $a$ we get $|V : C_V(a)| \geq q^2$.
\end{lemma}

\proof Let $p$ be a Zsigmondy prime which divides $q+1$ if $G = L_2(q)$ and $q^2+1$ else. Then we may assume that $a$ inverts some element $\omega$ of order $p$ in $G$. As $|[V,\omega]| \geq q^2$, we get the assertion. Suppose there is no such prime, then $q=8$ and $q+1= 9$. Then $a$ inverts some element $\omega$ of order nine, whose commutator has to be of order 64 at least.
\qed

\begin{lemma}\label{L2qff} Let $V$ be an $F$--module for $G = L_2(q)$, $Sz(q)$ or $U_3(q)$. Then $G= L_2(q)$ and $V$ is the natural module.
\end{lemma}

\proof By \ref{L2q} we get $G = L_2(q)$ and an offending subgroup $A$ is a Sylow 2-subgroup of $G$. As there is some $g \in G$ with $G = \langle A, A^g \rangle$, we get $|V| = q^2$, and $V$ is the natural module by $\cite{Glaub}$.\qed
 

\begin{proposition}\label{stein} Let $G = G(q)$ be a group of Lie type and $V$ be an irreducible module over $GF(q)$. Then 
$$V = V_1^{\sigma_1} \otimes \cdots \otimes V_l^{\sigma_l}$$
where the $V_i$ are basic irreducible $GF(q)G$--modules. Further distinct $l$--tuples $(M_1, \ldots M_l)$ and $(M_1^\prime, \ldots , M_l^\prime)$ give nonisomorphic $GF(q)G$--modules.
\end{proposition}

\proof \cite[Th 41 and 43]{Stei} \qed

\begin{lemma}\label{split} Let $G = G(q)$ be a Chevalley group then $GF(q)$ is a splitting field for any irreducible module. 
\end{lemma}

\proof \cite[(7.5)]{Stei} \qed

\begin{lemma}\label{split1} Let $G = G(q)$ and $V$ an absolutely irreducible $KG$--module for some $K \subseteq \widetilde{GF(q)}$. Let $\{\sigma_1, \ldots \sigma_r\} = Gal_{GF(p)}(K)$. Then $K = GF(p)(\chi)$ iff $V^{\sigma_1}$, \ldots , $V^{\sigma_r}$ are pairwise nonisomorphic $KG$--modules. 
\end{lemma}

\proof Choose notation such that $\sigma_1 = id$. Let $\{f_1, \ldots f_r\}$ be a $GF(p)$--basis for $K$ and $\{v_1,\ldots ,v_n\}$ be a $K$--basis for $V$. Then 
$\{f_iv_j \mid i = 1, \ldots , r, j = 1, \ldots n \}$ is a $GF(p)$--basis for $V$. Define
$$\phi_k : K \otimes V  \longrightarrow V^{\sigma_k} \mbox{ by } f \otimes f_iv_j \longrightarrow f_i^{\sigma_k}fv_j.$$
Then $\phi_k$ are epimorphisms. Let $n \in {\rm Ker} \phi_k$ for all $k$. Set
$$n = \sum_{i,j} f_{ij} \otimes f_iv_j$$
Then $$\sum_{i,j}f_{ij}f_i^{\sigma_k}v_j = 0 \mbox{ for all } k$$
This gives
$$\sum_{i}f_i^{\sigma_k}f_{ij} = 0 \mbox{ for all } k,j$$
By linear independence of automorphisms we have $f_{ij} = 0$ for all $i,j$ and so $\bigcap_{k} {\rm Ker }\phi_k = 0$. Set
$$D_i = \bigcap_{k \not= i} {\rm Ker }\phi_k$$
Then
$$K \otimes V = \bigoplus_{i=1}^{r} D_i$$
Now as $K \otimes V/{\rm Ker }\phi_i \cong D_i$, we have $D_i \cong V^{\sigma_i}$. 

Let now $K = GF(p)(\chi)$. Then $({\rm  Ker }\phi_1)^{\sigma_k} = {\rm Ker }\phi_k$, so $Gal_{GF(p)}(K)$ permutes the $D_i$ regularly. Hence by \cite[70.15]{CurRei} the $D_i$ are pairwise nonisomorphic.

Assume $K > GF(p)(\chi)$. Then each $V^{\sigma_k}$ appears $|K : GF(p)(\chi)|$ times among the $D_i$. Hence the $V^{\sigma_i}$ are not pairwise nonisomorphic.
\qed
\\

We like to consider representations of twisted Chevalley groups as well. Here we fix notation as follows. We have $^\sigma G(q) \leq G(q^\sigma)$, where $G(q^\sigma)$ is the corresponding untwisted group, $G \not= F_4$ or $B_2$. Now following \cite[chapter 9]{Stei} we see that any basic module for $G(q^\sigma)$ reduced to $^\sigma G(q)$ remains irreducible. Moreover by \cite[9.3]{Stei} all irreducible modules are given by the tensor product theorem. Further by \cite[7.5]{Stei} $GF(q^\sigma)$ is a splitting field.

It remains the cases $G = F_4(q)$ or $B_2(q)$. Then there is a duality between the long and short roots. Just take all weights which vanish on all long roots. We call these modules restricted to $^\sigma G(q)$ the basic modules. Then again the tensor product theorem holds. If the rank is $\ell$ we now get $q^{\ell/2}$ modules. By \cite[12.2]{Stei} $GF(q)$ is the field of definition for all these modules. \\

\begin{proposition}\label{deff} Let $G = A_{\ell}$, $D_{\ell}$, $E_6$ or $D_4$ and $V$ be a basic module with high weight $\lambda$. Then the following holds for $\gamma$ the diagram automorphism
\begin{itemize}
\item[(1)] If $\lambda \not= \gamma (\lambda)$. Then $GF(q^\sigma)$ is the field of definition for $V$.
\item[(2)] If $\lambda = \gamma (\lambda)$ then $GF(q)$ is the field of definition for $V$.
\end{itemize}
\end{proposition}

\proof As $[\sigma \gamma, ^\sigma G] = 1$, we have that $V$ and $V^{\sigma \gamma}$ are isomorphic as $GF(q^\sigma)$-modules. So $V \cong V^\sigma$ iff $V \cong V^\gamma$. Now from \ref{stein}, the fact that $GF(q^\sigma)$ is a splitting field and \ref{split1} we get that $GF(q^\sigma)$ is the field of definition if $V \not\cong V^\sigma$ and $GF(q)$ is the field of definition otherwise. So we just have to determine, when $V \cong V^\gamma$. 

Let $C$ be the Cartan subgroup of $G$ normalized by $\gamma$. Let $C$ induce the character $\lambda$ and $\tilde{\lambda}$ on the weight space of $V$ and $V^\gamma$, respectively. Then we see that for $c_{\alpha_i} \in C$ we have $\tilde{\lambda}(c_{\alpha_i}) = \lambda (c_{\gamma (\alpha_i)}) = \lambda^\gamma (c_{\alpha_i})$. Hence $V^\gamma$ has high weight $\lambda^\gamma$, and so $V \cong V^\gamma$ precisely if $\lambda \cong \lambda^\gamma$. 
\qed

\begin{lemma}\label{o2} Let $G = G(q), q = p^f,$ be a Chevalley group and $V$ be an irreducible module in characteristic $p$. Let $P$ be a parabolic and $V_P = C_V(O_p(P))$. Then $V_P$ is an irreducible $P-module$.
\end{lemma}

\proof This is \cite{Sm}. \qed


\begin{lemma}\label{factor} Let $V$ be an $F$-module for $G$ and $V_1$ be an invariant subspace. Then $V_1$ is a trivial subspace or $V_1$ is an $F$-module too.
\end{lemma}

\proof Choose $A \in {\cal P}(G,V)$. Set $B = C_A(V_1)$. Then $|V_1C_V(A)||B| \leq |C_V(A)||A|$. Hence $|V_1 : C_{V_1}(A)| \leq |A|/|B|$ and so $A/B$ offends on $V_1$. \qed

\begin{Def}\label{pgv} {\rm Let $V$ be a faithful $GF(p)$-module for $G$. Define ${\cal P}(G,V)$ to be the set of all non-identity $p$-subgroups $X$ of $G$ such that $|X||C_V(X)| \geq |Y||C_V(Y)|$ for all $Y \leq X$ (including $Y = 1$).}
\end{Def}

If $V$ is an $F$ - module for $G$ then there is always some elementary abelian $A$ in ${\cal P}(G,V)$.
 
\begin{lemma}\label{closure} Let $V$ be an $F$-module for $G$.

(a) Let $A,B \in {\cal P}(G,V)$ be with $|C_V(A)||A| = |C_V(B)||B|$ be maximal, then $AB \in {\cal P}(G,V)$ and $|C_V(AB)||AB| = |C_V(A)||A|$.

(b) Let $A \in {\cal P}(G,V)$, $A \leq O_p(G)$. If $|C_V(A)||A|$ is maximal among all such $A$, then $\langle A^G \rangle$ is a (maybe nonabelian) offender on $V$.

(c) Assume (b). If $\langle A^G \rangle$ is abelian then there is an abelian quadratic offender, which is normal in $G$.

(d) There is always an $A \in {\cal P}(G,V)$, which is quadratic.
\end{lemma}

\proof (a) We have $|C_V(AB)||AB| = |C_V(AB)||A||B|/|A \cap B|$. Furthermore $$C_V(AB) = C_V(A) \cap C_V(B).$$ So we have 
\begin{eqnarray*} |C_V(AB)||AB|& = &|C_V(A) \cap C_V(B)||A||B|/|A \cap B|\\ &= &|C_V(A)||C_V(B)||A||B|/(|C_V(A)C_V(B)||A \cap B|).
\end{eqnarray*}
 As $C_V(A)C_V(B) \leq C_V(A \cap B)$, we get $$|C_V(\langle A,B \rangle)||\langle A, B \rangle| \geq |C_V(AB)||AB| \geq |C_V(A)||A||C_V(B)||B|/(|C_V(A \cap B)||A \cap B|).$$ So $\langle A, B \rangle \in {\cal P}(G,V)$. By maximality we get equality everywhere, the assertion.

(b) This follows by repeated application of (a).

(c) This is Timmesfeld replacement.

(d) This is \ref{quad}. \qed
\\

Let $G$ be a Chevalley group with root system $\Phi$ and
fundamental roots  $\Pi$. Let
$l=|\Pi|$, $I=\{1,2,\ldots l\}$ and $\Pi=\{\alpha_1,\alpha_2,\ldots
\alpha_l\}$ where we choose the labeling of the Dynkin diagram as follows:


$$ \anl$$

$$\cnl$$

$$\dnl$$

$$\esixl$$


$$ \esevenl$$

$$\eeightl$$

$$\ffourl$$

$$\gtwol$$


For the remainder of this paper we fix the following notation. Let $V$ be some irreducible $GF(2)$ - module for some group of Lie type over $GF(q)$. Then $V \otimes GF(q)$ is a direct sum of algebraic conjugates of some irreducible $GF(q)$ - module $\hat{V}$. If this is some algebraic conjugate of some fundamental module $V(\lambda)$ we write $V = V(\lambda)$. If this is a tensor product we call $V$ the corresponding tensor product.


\begin{lemma}\label{lambda2} Let $G = L_n(q)$, $n \geq 5$ and $V = V(\lambda_2)$. Then $V$ has an unique offender, which is a transvection group of order $q^{n-1}$. If $n \leq 4$, then an offender intersects a transvection group nontrivially.
\end{lemma}

\proof We prove the assertion by induction on $n$. Let first $n = 5$. Then $V$ is a $10$--dimensional module for $L_5(q)$. Let $G_4$ be the parabolic which has the following chief series in $V$
$$1  < V_1 < V,$$
where $V_1$ is the $6$--dimensional $L_4(q)$--module and $V/V_1$ the $4$--dimensional one. It means that $O_2(G_4)$ offends. Let $x \in O_2(G_4)$. We may assume that $C_G(x) = G_{14}$, so $|[V,x]| = q^3$, as all involutions of $O_2(G_4)$ are conjugate. Let now $A_1 \leq O_2(G_4)$, $|A_1| \geq q^2$. If $C_V(a) = C_V(A_1)$ for all $a \in A_1^\sharp$, then $C_V(A_1)$ is invariant under $\langle C_G(a) \mid a \in A_1^\sharp \rangle = O^\prime(G_4)$. But as $C_V(a) \not= V_1$, this is impossible. So we have shown that in fact $V_1 = C_V(A_1)$. Let now $A$ be any offender. As one can check easily $|V_1 : C_{V_1}(A)| \geq |A/A \cap O_2(G_4)|$. Hence we see $A \cap O_2(G_4) \not= 1$. So we get $|C_V(A \cap O_2(G_4)) : V_1| \leq q$. But then we see that $|A \cap O_2(G_4)| \geq q^3$. Now by the above we have $V_1 = C_V(A \cap O_2(G_4))$. This shows $A \geq O_2(G_4)$ and so $A = O_2(G_4)$. 

It remains $n \leq 4$. For $n \leq 3$ there is nothing to prove. So assume $n = 4$. Then $V$ is the natural $\Omega^+(6,q)$--module. Hence $|V : C_V(A)| \geq q^2$. If $|A| > q^2$, we are done. So assume $|A| = q^2$. So we may assume $A \cap O_2(G_1) = 1$. Now $G_1$ induces on $C_V(O_2(G_1))$ the natural module and on $V/C_V(O_2(G_1))$ the dual. As $A$ corresponds to a transvection group in $G_1/O_2(G_1)$, we get $|V : C_V(A)| \geq q^3$, a contradiction. \qed

\begin{lemma}\label{lambdan} Let $G = U_4(q) \cong \Omega^-(6,q)$, $Sp_6(q)$ or $\Omega^+(10,q)$ and $V = V(\lambda_n)$. Let $A$ be a quadratic offender, then $A$ is $O_2(G_1)$ if $G \not\cong Sp_6(q)$, while if $G \cong Sp_6(q)$, then $|A| \geq q^4$ and $|C_W(A)| \leq q^2$, where $W$ is the narural $Sp_6(q)$--module. In all cases $A$ contains some root element.
\end{lemma}

\proof Obviously $O_2(G_1)$ is a quadratic offender. So let first $A$ be a quadratic offender which lies in $G_1$ but not in $O_2(G_1)$. Then by \ref{factor} we may assume that $A/A \cap O_2(G_1)$ offends on  $C_V(O_2(G_1))$. Let $G \cong U_4(q)$. Then $|A/A \cap O_2(G_1)| = q^2$ by \ref{L2q} and so
$|A \cap O_2(G_1)| \leq q$. Hence $|A| \leq q^3$. But as $V$ involves two natural $L_2(q^2) \cong (G_1/O_2(G_1)^\prime$--modules we see $|V : C_V(A)| \geq q^4$, a contradiction.

Let now $G \cong Sp_6(q)$. Let $A/A\cap O_2(G_1)$ be a transvection group. If $C_V(A \cap O_2(G_1)) = C_V(O_2(G_1))$, then we get $|A| = q^5$. Hence $A$ induces a transvection group on $O_2(G_1)$ and $C_V(O_2(G_1))$ as well. But this contradicts the action of $Sp_6(q)$ on the spin module. So let now $|A/A \cap O_2(G_1))| > q$. As there is exactly one elementary abelian subgroup of order $q^6$ in a Sylow 2-subgroup, which does not offend on the spin module, we get $|C_{C_V(O_2(G_1))}(A)| = q^2$. Let $R$ be the root group normal in $G_1$. Then $O_2(G_1)/R \not\cong C_V(R)$ as an $Sp_4(q)$--module. So we get $|A/A \cap O_2(G_1)| = q^3$ and $|A \cap O_2(G_1)| = q$, where $A \cap O_2(G_1)$ is a root subgroup. Now we see that in the natural module $W$ we have $|C_W(A)| = q^2$.

Let finally $G \cong \Omega^+(10,q)$. Let $1 \not= x \in A \cap O_2(G_1)$. Then $|[V,x]| = q^4$ or $q^8$. If $|[V,x]| = q^8$, then $[V,x] = C_V(O_2(G_1))$ and as $C_G(C_V(O_2(G_1))) = O_2(G_1)$, we have a contradiction. So we get $|[V,x]| = q^4$ and $A$ is in the centralizer of this group. This shows $|A/A \cap O_2(G_1)| \leq q^6$ and $|V : C_V(A)| \geq q^8$. But $|O_2(G_1) : C_{O_2(G_1)}(A)| \geq q|A/A \cap O_2(G_1)|$, a contradiction. Hence $A \cap O_2(G_1) = 1$ and so $|A| \leq q^6$ and $|C_V(O_2(G_1)) : C_{C_V(O_2(G_1))}(A)| \leq q^3$. This shows that we must have equality everywhere. Now $C_V(O_2(G_1))A$ contains an elementary abelian subgroup of order $q^{11}$. But $C_V(O_2(G_1))\Omega^+(8,q)$ is isomorphic to a subgroup of $\Omega^+(10,q)$ and the latter group contains no elementary abelian 2-subgroups of order greater than $q^{10}$, a contradiction.

So we are left with $A \leq O_2(G_1)$. If $A$ contains some element of type $c_2$, we get the assertion. So assume not. Then $|A| \leq q,q^4, q^2$ according to $G \cong U_4(q)$, $\Omega^+(10,q)$, or $Sp_6(q)$. The first case is not possible. In the other two cases we see that $C_V(A) = C_V(a)$ for all $a \in A^\sharp$ and $\langle C_G(a) \mid a \in A^\sharp \rangle = G$, a contradiction. \qed

\begin{lemma}\label{tensor1q} Let $G = L_n(q)$ or $U_n(q)$ and $V = V(\lambda_1 + \lambda_n)$. Assume $A \leq G$ with $[V,A,A] = 1$. Then $|A| \leq q$.
\end{lemma}

\proof We have that $V$ is the adjoint module. Let first $n = 3$ and $G = L_3(q)$. then any elementary abelian subgroup intersects a root group. As $V$ is not strong quadratic \cite{Str}, we get the assertion. Let now $G = L_n(q)$. By induction as $V$ involves the adjoint module for $G_1/O_2(G_1)$ we have $|A/A \cap O_2(G_1)| \leq q$. So either we have the assertion or $A \cap O_2(G_1) \not= 1$. But then $V$ would be strong quadratic, a contradiction.

So let now $G \cong U_n(q)$. If $n = 3$ there is nothing to prove. Let $n = 4$. If $A \leq O_2(G_1)$, we get that $V$ is strong quadratic or $|A| \leq q$. So let $A \not\leq O_2(G_1)$. Then we have $|A /A \cap O_2(G_1)| \leq q$. Let $A \cap O_2(G_1) \not= 1$. As $V$ is not strong quadratic we get $|A| \leq q^2$ and $A \leq O_2(G_2)$. So let $x \in A^\sharp$. Then $[\langle A^{C_G(x)} \rangle, x] = 1$, and we get a quadratic fours group which intersects a root group and contains $x$, a contradiction. Let now $n \geq 5$. Then by induction $|A : A \cap O_2(G_1)| \leq q$. As $V$ is not strong quadratic, we see that $|A \cap O_2(G_1)| \leq q$ and $A$ induces $GF(q^2)$--transvections on $O_2(G_1)/Z(O_2(G_1))$ with $A \cap O_2(G_1) \leq [A, O_2(G_1)]$. But now $\langle A, [A,O_2(G_1)] \rangle \leq U_4(q)$, and so as before we get strong quadratic action, a contradiction. \qed

\begin{lemma}\label{tensor2q} Let $X = V \otimes W$ be a faithful $GF(q)$--modules and $A$ be quadratic on $X$ . Then $|A| \leq q$. 
\end{lemma}

\proof Let $v \otimes w \in V \otimes W$, $a \in A$. Then $[v \otimes w, a] = [v,a] \otimes w + v \otimes [w,a] + [v,a] \otimes [w,a]$. Let $b \in A$ then $0 = [v \otimes w,a,b] = [v,a] \otimes [w,b] + [v,b] \otimes [w,a]$, hence $[v,a] = \lambda [v,b]$ and $[w,b] = \lambda^{-1} [w,a]$, for some $\lambda \in GF(q)^\sharp$. In particular $|A| \leq q$. \qed

 \begin{lemma}\label{tensorff} Let $X = V \otimes W$, be a faithful $GF(q)$--modules. Then $X$ is not an $F$--module.
\end{lemma}

\proof Suppose by way of contradiction that $X$ is an $F$--module. By \ref{quad} we have a quadratic offender $A$. By \ref{tensor2q} we have $|A| \leq q$. Now we get that for $a \in A^\sharp$ the codimension of $C_X(a)$ is at least twice the codimension of the centralizers in the modules $V$ and $W$. This shows that the codimension is at least two. Hence we get $|X : C_X(a)| \geq q^2$, a contradiction. \qed


\begin{lemma}\label{transv} (a) Let $G = G(q)$ be of Lie rank at least two and $A \leq G$, $|A| \leq q$. Let $V$ be an irreducible module with $|V : C_V(A)| \leq |A|$, then $V$ is strong quadratic.

(b) Let $R$ be a root group in $G$ with $|[V,R]| \leq q$ for some irreducible module $V$, then $G \cong L_n(q)$ or $Sp(2n,q)$ and $V$ is the natural module.
\end{lemma}

\proof (a) Let $R$ be a root group with $[R,A] = 1$. Suppose first $R \geq A$, then obviously $[C_G(a)^\prime,[a,V]] = 1$ for $a \in A^\sharp$ and so $V$ is strong quadratic. So let $R \not\geq A$. If $A \leq O_2(C_G(R))$ but $A \not\leq Z(O_2(C_G(R)))$, then there is some $a \in A$ and $r \in R$ such that $a \sim ar$. So $|[V,r]| \leq q^2$. As $C_G(r)$ acts on $[V,r]$ we see again that $V$ is strong quadratic. So let now $A \leq Z(O_2(C_G(R)))$. Then $G \cong F_4(q)$ or $Sp(2n,q)$. But then there is some $g \in G$ and $a \in A$ such that $a^g \in C_G(R) \setminus O_2(C_G(R))$. So we may assume that $A \not\leq O_2(C_G(R))$. Then there is even some $h \in C_G(R)$ and some $a \in A$ such that $a^h \in P \setminus O_2(P)$, where $P$ is some minimal parabolic of $C_G(R)$. (Recall that for $G = L_3(q)$ of course $A$ intersects some root subgroup nontrivially and so $V$ is strong quadratic. ) Now $P$ acts on $[V,R]$ and $V/C_V(R)$. But by \ref{L2q} we have that $P/O_2(P)$ can just induce one nontrivial module in $V$, so we see that $[[V,R], O^2(P/O_2(P))] = 1$. Hence $V$ is strong quadratic. 

(b) We may argue by induction. By (a) we have that $V$ is strong quadratic. So by \cite{Str} $V = V(\lambda)$ for some weight $\lambda$. Further as $|[V,R]| = q$, we see that $[[V,R],C_G(R)^\prime] = 1$, so $\lambda$ is the weight corresponding to $C_G(R)$, or $G \cong L_n(q)$ and $\lambda = \lambda_1$ or $\lambda_{n-1}$. But the latter are the natural modules and we are done. So assume the former. By induction we have that any maximal parabolic with connected diagram, which does not belong to $\lambda$ is of type $L_m$ or $Sp_{2m}$. Hence we get that $G \cong Sp(2n,q)$ and $\lambda = \lambda_1$ or $\lambda_2$, $G \cong F_4(q)$, $\lambda = \lambda_1$ or $\lambda_4$, or $G \cong G_2(q)$ or $^3D_4(q)$. If $G \cong G_2(q)$, then $V$ is the natural $Sp(6,q)$--module and $R$ corresponds to an element of type $a_2$, i.e. $|[V,R]| = q^2$. If $G \cong ~^3D_4(q)$, then $V$ is the natural $\Omega^+(8,q^3)$-module and so $|[V,R]| \geq q^3$. For 
$G \cong F_4(q)$ we may assume by duality that $V = V(\lambda_4)$. If $G \cong Sp(2n,q)$ we may assume $V = V(\lambda_2)$, as $V(\lambda_1)$ is the natural module. Now we have $G_1/O_2(G_1)$ contains $Sp(2m,q)$ for some $m$. By induction $C_V(O_2(G_1))$ is the natural module. Let $R_1$ be the root group normal in $G_1$. As there is some conjugate of $R$ in $G_1 \setminus O_2(G_1)$, we see that $[V,R_1] \leq C_V(O_2(G_1))$ and $[O^2(C_G(R_1)), V] \leq C_V(O_2(G_1))$, a contradiction. \qed


\section{$F$--modules in even characteristic for Lie type groups}

Throughout this chapter we will assume that $G$ is a group of Lie type over 
a field with $q = 2^n$ elements. Further $V$ is an $F$-module and $A \in {\cal P}(G,V)$ is an offending subgroup. By \ref{closure} we may assume that $A$ acts quadratically. If $G_i$ is a maximal parabolic in $G$ we set $Q_i = O_p(G_i)$.

\begin{lemma}\label{Vlambda2} Let $G = G(q)$ be classical or $F_4(q)$ and $V$ be an $F$-module. If $V = V(\lambda_2)$, then $G \cong L_n(q)$, $\Omega^+(6,q)$ or $Sp_4(q)$.
\end{lemma}

\proof If $G \cong Sp_{2n}(q)$, $n > 2$ or $\Omega^{\pm}(2n,q)$, $n > 3$, then $V$ is the adjoint module. So for $G_n/Q_n$, we have some adjoint module involved too. Now $C_V(Q_n) \cong V(\lambda_2)$ for $G_n/Q_n$. Suppose $A \not\leq Q_n$. Then by \ref{factor} and \ref{lambda2} we have that $AQ_n/Q_n$ is a transvection group in $G_n/Q_n$ or $G_n/Q_n$ induces $L_4(q)$ or $L_3(q)$. Further $AQ_n/Q_n$ intersects a root group. Assume that $|AQ_n/Q_n| > q$. Now $A$ acts quadratically on any $G_n/Q_n$--chieffactor in $V$. This contradicts \ref{tensor1q}.  So we are left with $G = Sp_6(q)$ or $\Omega^-(8,q)$ and $|AQ_n/Q_n| = q$. If $G = Sp_6(q)$ we so get $|A| \leq q^5$. But the codimension of the centralizer of $A$ in the adjoint module for $L_3(q)$ is four and so
$|V : C_V(A)| \geq q^6$, a contradiction. Let now $G = \Omega^-(8,q)$. Again $|V :C_V(A)| \geq q^6$. As $V$ is not strong quadratic, we get $A \not\leq Q_1$. So we see with \ref{tensor1q} that $|AQ_1/Q_1| \leq q$ and so again $|A| \leq q^5$, a contradiction.

So $A \leq Q_n$. Then by \ref{closure} we may assume that $A \unlhd G_n$. Then there is a root group $R$ in $A$, where $R \unlhd G_2$. From this by quadratic action we see that $[[V,R], \langle A^{G_2}\rangle] = 1$, i.e. $[V,R] \leq C_V(Q_2)$. But $|C_V(Q_2)| = q$, contradicting \ref{transv}.

Let now $G \cong U_n(q)$. Then $G$ is embedded in $L_n(q^2)$ and the modules $V(\lambda_2)$ are the same for both groups. Let $n \geq 5$. By \ref{lambda2} $L_n(q^2)$ has an unique offender, the transvection group on $V(\lambda_2)$. But this group is not in $U_n(q)$. 

So let finally $G \cong F_4(q)$. Then $C_V(Q_4) \cong V(\lambda_2)$ for $Sp_6(q)$. So we have $|C_V(Q_4)| = q^{14}$ and $A \leq Q_4$, as $V(\lambda_2)$ is not an $F$--module for $Sp_6(q)$. By \ref{transv} we have $|A| > q$. Let $|A| = q^7$, $A \unlhd G_4$. Then $|AQ_1/Q_1| = q$. But $C_V(Q_1)$ is the spin module for $G_1/Q_1$. Now we get a contradiction with \ref{factor} and \ref{transv}.
So we have $\langle A^G_4 \rangle = Q_4$ and then $|V : C_V(Q_4)| \leq |Q_4|$ by \ref{closure}. This shows $|V| \leq q^{29}$. As $|V : C_V(Q_4^\prime)| = q^{14}$, we get $|V| = q^{28}$ or $q^{29}$. As $Q_4/Q_4^\prime$ is not irreducible and $V/C_V(Q_4^\prime) \cong C_V(Q_4)$, we see $[V,Q_4,Q_4] = 1$, a contradiction as $Q_4$ contains elements of order four. \qed

\begin{lemma}\label{lambdai} Let $G = G(q)$ and $V$ be an $F$--module. If $C_V(Q_i) = C_V(Z(S))$ for some $i$, then $V = V(\lambda_i)$.
\end{lemma}

\proof By \ref{stein} we have $V \cong V(\lambda_i)^{\sigma_1} \otimes V(\lambda_i)^{\sigma_2} \otimes \cdots \otimes V(\lambda_i)^{\sigma_k}$, for some field automorphisms $\sigma_1$, $\ldots , \sigma_k$. By \ref{tensorff} we get $k = 1$. \qed

\begin{lemma}\label{classicalff} Let $G = G(q) \not\cong L_n(q)$ be classical and $V$ be an $F$--module with $A \leq Q_1$. Then $V \cong V(\lambda_1)$ or $G \cong Sp_4(q)$, $Sp_6(q)$, $\Omega^+(8,q)$, $\Omega^+(10,q)$ or $\Omega^-(6,q)$ and $V \cong V(\lambda_n)$.
\end{lemma}

\proof Application of \ref{transv} shows that we may assume that $|A| > q$. Now application of \ref{closure} shows that $Q_1$ offends on $V$. So $|V : C_V(Q_1)| \leq |Q_1|$. Furthermore by 
\ref{lambdai} we may assume that $C_V(Q_1)$ is a nontrivial $G_1/Q_1$--module. If $G \cong \Omega^{\pm}(2n,q)$, then again by \ref{closure} $Q_1$ acts quadratically on $V$. So we have that $V$ is strongly quadratic and so by \cite{Str} $V \cong V(\lambda_n)$. But now also $|V/C_V(Q_1)| = |C_V(Q_1)|$ and
$|C_V(Q_1)$ is the spin module for $G_1/Q_1$. So $|C_V(Q_1)| \leq |Q_1|$, which shows $G \cong \Omega^+(2n,q)$, $n =3,4,5$. 

Let now $G \cong Sp_{2n}(q)$ or $U_n(q)$. Let $R \unlhd G_1$ be a root group. Then we see that $V/C_V(Q_1)$ involves some module isomorphic to $C_V(Q_1)$. As $G$ acts strongly quadratic and $V \not\cong V(\lambda_2)$ by \ref{Vlambda2}, we get with \cite{Str} that $C_V(Q_1)$ is not the natural $G_1/Q_1$--module. Hence $V/C_V(Q_1)$ is not the natural module and so $[V,Q_1,Q_1] = 1$, in particular $G \not\cong U_n(q)$. Let now $R_1$ be a root group in $Q_1$, $R_1 \not= R$. Then we may assume that $C_G(R_1) \leq G_2$. As we have by \cite{Str} $n \geq 4$,
we see that $V/C_V(Q_1)$ involves at least one nontrivial $Sp_{2n-4}(q)$--module. If $C_V(R_1)$ covers this module we get $|V : C_V(R_1)| \leq q^3$. Now there is a conjugate $R_2$ of $R_1$ in $G_1 \setminus Q_1$. This shows that $|C_V(Q_1) : C_{C_V(Q_1)}(R_2)| \leq q$. But now application of \ref{transv} gives a contradiction. \qed

\begin{lemma}\label{rank2} Let $G = G(q)$ of rank at most two and $V$ be an $F$--module. Then $G \cong L_2(q)$, $L_3(q)$, $U_4(q)$, $\Omega^-(6)$ or $G_2(q)$ and $V$ is the natural module.
\end{lemma}

\proof If $G$ is of rank one the assertion follows with \ref{L2qff}. So let $G$ be of rank two. If $G \cong L_3(q)$, then by \ref{transv} $V$ is strong quadratic and the assertion follows with \cite{Str}. 

Let $G \cong Sp_4(q)$, $U_4(q)$, $U_5(q)$, or $\Omega^-(6,q)$. Then by 
\ref{classicalff} $A \not\leq Q_1$. Furthermore by \ref{lambdai} $V_1 = C_V(Q_1)$ is a nontrivial $G_1/Q_1$--module. So it is the natural module in all cases. In particular $|A : A \cap Q_1| = q$ and $G \not\cong U_5(q)$. If $G \cong Sp_4(q)$, we have $A \leq Q_2$ and we get the assertion with \ref{classicalff}.
Now in all cases we have by \cite{Str} that $V$ is not strongly quadratic and so $G \cong U_4(q)$, $q < |A| \leq q^2$. Further $A \not\leq Q_2$. Now we get by \ref{factor} that $|A| = q^2$ and $A \cap Q_2 = 1$. But then $C_V(Q_2)$ is the natural $G_2/Q_2$--module and $[Q_2,V] \leq C_V(Q_2)$, i.e. $Q_2$ acts quadratically, contradicting the fact that $V$ is not strongly quadratic.

Let now $G \cong G_2(q)$. If $V$ is strongly quadratic then by \cite{Str} $V \cong V(\lambda_1)$, the assertion. So we may assume that $V$ is not strongly quadratic. Now by \ref{transv} $|A| > q$. Hence $A \not\leq Q_1$. So we get $|A| \leq q^2$. If $[A,Z(Q_2)] = 1$, we have $A \cap Z(Q_2) \not= 1$, and so $V$ would be strong quadratic. So we have that $|A : C_A(Z(Q_2))| = q$ and $A \cap Z(Q_2) = 1$. Then there is some $x \in Z(Q_2)$ and $a \in A^\sharp$ with $[a,x] \not= 1$. Now for $b \in C_A(Z(Q_2))^\sharp$ we get that $\langle [a,x], b \rangle$ is quadratic on $V$ and so $V$ is strong quadratic, a contradiction.

Let finally $G \cong \,^3D_4(q)$ or $^2F_4(q)$. Then in both cases $|A| \leq q^5$. If $V$ is strongly quadratic, we get with \cite{Str} that $G \cong \,^3D_4(q)$ and $V$ is the natural $\Omega^+(8,q^3)$--module. But in this case $|[V,t]| \geq q^6$ for any involution $t \in G$, contradicting $|A| \leq q^5$. So we have that $V$ is not strongly quadratic. In particular by \ref{transv} $|A| > q$ and then $A \not\leq Q_1$. Let $C_V(Q_1)$ be a trivial $G_1$--module. Let $R$ be a root group normal in $G_1$. Then we see that $[V,R]$ involves at least the eight-dimensional $L_2(q^3)$--module for $G \cong \, ^3D_4(q)$ or twice the natural $Sz(q)$--module for $G \cong \,^2F_4(q)$. Now $|[V,t]| \geq q^8$ for involutions $t \in G_1 \setminus Q_1$, contradicting $|A| \leq q^5$. 
So we have with \ref{factor} that $C_V(Q_1)$ is an $F$--module for $G_1/Q_1$, and so by \ref{L2qff} we get $G \cong \, ^3D_4(q)$ and $C_V(Q_1)$ is the natural $L_2(q^3)$--module. But then there is exactly one such module involved, i.e. $[V,G_1] \leq C_V(Q_1)$ and so $V$ would be strongly quadratic, a contradiction. \qed


\begin{lemma}\label{Vlambdan} Let $G = G(q) \not\cong L_n(q)$ of rank at least three be classical and $V = V(\lambda_n)$ be an $F$--module. Then $G \cong Sp_6(q)$, $\Omega^+(8,q)$ or $\Omega^+(10,q)$.
\end{lemma}

\proof By \ref{classicalff} we may assume $A \not\leq Q_1$. So by \ref{factor} it is enough to prove that $V(\lambda_n)$ is not an $F$--module for $G \cong Sp_8(q)$, $\Omega^+(12,q)$, $\Omega^-(8,q)$, $U_5(q)$ and $U_6(q)$. The case $G \cong U_5(q)$ follows from \ref{rank2}. As the spin modules for $Sp_8(q)$ and $\Omega^-(8,q)$ are the same, it is enough to handle $G \cong Sp_8(q)$, $\Omega^+(12,q)$ and $U_6(q)$. Now by \ref{lambdan} we have $AQ_1/Q_1 = O_2(G_{12}/Q_1)$ or $G \cong Sp_8(q)$ and $|AQ_1/Q_1| \geq q^4$. Let first $G \cong \Omega^+(12,q)$. Then $|A| \leq q|A/A \cap Q_1|$ and $|C_V(Q_1) : C_{C_V(Q_1)}(A)| = |A/A \cap Q_1|$. Hence we see that $[G_1,V] = C_V(Q_1)$. But then $[Q_1,V] = 1$ as $C_V(Q_1) \not\cong Q_1$ as $G_1$--modules.

So let now $G \cong Sp_8(q)$ or $U_6(q)$. Then we see that $|C_{Q_1}(A)| \leq q^3$ and so $|A| < |A/A \cap Q_1|^2$. Let $R$ be the root group normal in $G_1$. We have $C_V(Q_1) \leq [V,R]$, so there are at least two such modules involved. But $|C_V(Q_1) : C_{C_V(Q_1)}(A)| = |A/A \cap Q_1|$ or $G \cong Sp_8(q)$ and $AQ_1/Q_1 = O_2(G_{12})/Q_1$, in which case we have $|C_V(Q_1) : C_{C_V(Q_1)}(A)| = |A/A \cap Q_1|q^{-1}$ and $|A| \leq q^2|A/A \cap Q_1|$. In both cases we see that $|V : C_V(A)| > |A|$, a contradiction. \qed

\begin{lemma}\label{Omega} Let $G \cong \Omega^-(2n,q)$, $n > 3$, and $V$ be an $F$--module. Then $V \cong V(\lambda_1)$. 
\end{lemma}

\proof By \cite{Str} it is enough to prove that $V$ is strongly quadratic. So assume not. Then by \ref{closure} $A \not\leq Q_1$. If $A \cap Q_1 \not= 1$, there is some $t \in A \setminus Q_1$ such that $[t,Q_1]$ contains some root element and so $V$ is strongly quadratic. Hence we may assume $A \cap Q_1 = 1$.
If there are $s,t \in A^\sharp$ with $C_{Q_1}(s) \not= C_{Q_1}(t)$, then there is some $g \in C_{Q_1}(s)$, say, with $t^g = t$ and $s^g = sa \not= s$. Now $\langle t,a \rangle$ acts quadratically and if $a$ is not in a root subgroup there is $h \in Q_1$ with $t^h = tb \not= t$, $b$ in a root subgroup. Now $\langle a, b \rangle $ acts quadratically and so $V$ is strongly quadratic. So we may assume that $C_{Q_1}(A) = C_{Q_1}(t)$ for all $t \in A^\sharp$. Hence $|A| \leq q^{2n-4}$. So $|V : C_V(t)| \leq q^{2n-4}$. For a root element $a$ we then get $|V : C_V(a)| \leq q^{4n-8}$. But as $V$ is not strongly quadratic, we get $|[V,a]| > |Q_2/Q_2^\prime| = q^{4n-8}$, a contradiction. \qed


\begin{lemma}\label{sp2n} Let $G \cong Sp_{2n}(q)$, $n = 3,4$, and $V \cong V(\lambda_1 + \lambda_n)$. Then $V$ is not an $F$--module.
\end{lemma}

\proof By \cite{Str} we know that $V$ is not strong qudartic. Let first $A \cap Q_n \not= 1$. There is $U \unlhd G_n$, $U$ is the natural $L_3(q)$--module or the natural $\Omega^+(6,q)$--module. So by \ref{closure} we have $A \not\leq Q_n$, as otherwise $U \leq A$, and $U$ contains root elements. But now there is some $t \in A \setminus Q_n$, with $[t,U]$ contains a root element. As $A$ does not contain root elements, we see that $V$ is strongly quadratic. So we have $A \cap Q_n = 1$. Now we have $C_{Q_n}(A) = C_{Q_n}(a)$ for all $a \in A^\sharp$. 
This shows $|A| \leq q^2$ for $n = 3$ and $|A| \leq q$ for $n = 4$. But as $V$ is not strongly quadratic we get with \ref{transv} that $n = 3$. Now for a root element $r$ we have $|[V,r]| \leq q^4$. As $C_V(O_2(C_G(r)))$ is a non trivial module for $G_2$, we see that $C_{O_2(C_G(r))}([V,r]) > R$, $R$ the root group containing $r$, which shows that $V$ is strongly quadratic, a contradiction. \qed

\begin{proposition}\label{fmodulesc} Let $G = G(q)$ be classical and $V$ be an $F$--module. Then one of the following holds
\begin{enumerate}
\item[(i)] $G \cong L_n(q)$, $V \cong V(\lambda_1)$, $V(\lambda_2)$, $V(\lambda_{n-2})$, $V(\lambda_{n-1})$
\item[(ii)] $G \cong Sp_{2n}(q)$, $\Omega^{\pm}(2n,q)$, or $U_n(q)$ and $V \cong V(\lambda_1)$
\item[(iii)] $G \cong Sp_6(q)$, $V \cong V(\lambda_3)$
\item[(iv)] $G \cong \Omega^+(2n,q)$, $n = 4,5$ and $V \cong V(\lambda_n)$.
\end{enumerate}
\end{proposition}

\proof By \ref{rank2} we may assume that the rank of $G$ is at least three. By 
\ref{Omega} we have $G \not\cong \Omega^-(2n,q)$. By \ref{classicalff} and symmetry for $G \cong L_n(q)$ we may assume that $A \not\leq Q_1$. By \ref{lambdai} we have that $C_V(Q_1)$ is an $F$--module. So by induction we first assume that $C_V(Q_1)$ is the natural module. Let $A \leq Q_n$, then by \ref{transv} and \ref{closure} we have that $V$ is strongly quadratic and so $V \cong V(\lambda_2)$. If $A \not\leq Q_n$, we have that $C_V(Q_n)$ is an $F$--module. As $G_n/Q_n$ is linear we get $V \cong V(\lambda_2)$ again. Now the assertion follows with \ref{Vlambda2}.

So we may assume that $C_V(Q_1)$ is not natural. Let first this module correspond to $\lambda_n$. If $A \leq Q_n$, we get $V \cong V(\lambda_n)$. If $A \not\leq Q_n$ and $C_V(Q_n)$ is the trivial module, we get $V \cong V(\lambda_n)$ with \ref{lambdai} again. So by \ref{Vlambdan} we have the assertion.

So let $C_V(Q_n)$ be an $F$--module too. Then we get $V \cong V(\lambda_1 + \lambda_n)$. Now \ref{tensor1q} and \ref{transv} yield $G \cong Sp_{2n}(q)$ or $\Omega^+(2n,q)$. If $G \cong Sp_{2n}(q)$ then by induction we have $n = 3$ or $4$ and we get a contradiction with \ref{sp2n}. So let $G \cong \Omega^+(2n,q)$. Now for $G_{n-1}$ we get that $C_V(Q_{n-1})$ is $V(\lambda_1 + \lambda_n)$ for $G_{n-1}/Q_{n-1}$. As $V$ is not strongly quadratic by \cite{Str} we get with \ref{closure} that $A \not\leq Q_{n-1}$. But now we have a contradiction with \ref{factor} and induction. 

So by induction the only case left is $G \cong L_n(q)$ and $C_V(Q_1)$ corresponds to $\lambda_{n-2}$ or $\lambda_3$. assume the former. If $V$ is strongly quadratic we get with \cite{Str} that $V \cong V(\lambda_{n-2})$. So let $V$ not be strongly quadratic. Then $C_V(Q_{n-1})$ is not the natural module. In particular it is not an $F$--module. Hence by \ref{factor} $A \leq Q_{n-1}$ which contradicts \ref{closure} and the fact that $V$ is not strongly quadratic.

So let $C_V(Q_1)$ correspond to $\lambda_3$. Then either $C_V(Q_{n-1})$ is an $F$--module or $V$ is strong quadratic. In both cases we get $V \cong V(\lambda_3)$. Further we may assume that $n > 5$, otherwise $\lambda_3 = \lambda_{n-2}$ or $\lambda_{n-1}$. Now by \ref{lambda2} we have $|A : A \cap Q_1| = q^{n-2}$. Further $|A \cap Q_1| \leq q$. As $|C_V(Q_1) : C_{C_V(Q_1)}(A)| = |A/A \cap Q_1|$, we get that either $|A \cap Q_1| = q$ or $C_V(Q_1) = [V,G_1]$. As $C_V(Q_1) \not\cong Q_1$ as $G_1/Q_1$--modules, the latter is not possible. Now for $r \in A \cap Q_1$ we see that $C_V(Q_1) \leq [V,r]$. Hence ${n-2 \choose 2} \leq n-1$, a contradiction. \qed

\begin{proposition}\label{En} There is no $F$--module for $G = E_n(q)$, $n = 6,7,8$.  \end{proposition}

\proof Suppose false. Let first $G = E_6(q)$. If $A \subseteq Q_1$, then by \ref{closure} we get $A = Q_1$ and $V$ is strongly quadratic. Now by \cite{Str} we have $V \cong V(\lambda_1)$ or $V(\lambda_6)$. Let $A \not\leq Q_1$ and $V \not\cong V(\lambda_1)$. Then $C_V(Q_1)$ is an $F$--module by \ref{factor} for $\Omega^+(10,q)$. By \ref{fmodulesc} this is the natural or half spin module. Suppose it is not the natural one. Then we get that $V$ is not strong quadratic and so $A \not\leq Q_6$. Hence $C_V(Q_6)$ is an $F$--module, which now has to be a half spin module and we have $V \cong V(\lambda_4)$. 

So let now $C_V(Q_1)$ be the natural module. If $V \not\cong V(\lambda_6)$, we may assume by symmetry that also $C_V(Q_6)$ is the natural module and so $V \cong V(\lambda_1 + \lambda_6)$. But now $C_V(Q_4)$ is not an $F$--module by \ref{fmodulesc}, and so $A \leq Q_4$, which shows that $V$ is strongly quadratic, contradicting \cite{Str}. 

So in any case we have that either $V \cong V(\lambda_1)$ or $V(\lambda_4)$. Let first $V \cong V(\lambda_4)$. Then $V$ is the adjoint module, $V$ is not strongly quadratic and so $A \not\leq Q_1$. For $G_1$ we have that $C_V(Q_1)$ is the half spin module. By \ref{Vlambdan} we know $A/A\cap Q_1$. So $A/A \cap Q_1$ acts strongly quadratic on any $G_1/Q_1$--module involved in $V$. But as an adjoint module it involves $V(\lambda_2)$ for $G_1/Q_1$, which is not strongly quadratic by \cite{Str}. 

So let $V \cong V(\lambda_1)$. We may assume that we have the following $G_1$--series on $V$
$$1 < V_1 < V_2 < V,$$
where $V_1$ is a trivial module, $V_2/V_1 \cong Q_1$ and $V/V_2$ is the natural module. As $Q_1$ does not offend, we have $A \not\leq Q_1$ by \ref{closure}. Now in $V/V_2$ there are two $G_1$--classes of vectors $v_1V_2$, $v_2V_2$. We have that $v_1V_2$ is centralized by $H_1 \cong Sp_8(q)$ in $G_1/Q_1$ and $H_1$ induces the spin module on $Q_1$, so $C_{\langle v_1, V_2 \rangle/V_1}(Q_1) = V_2/V_1$. Now for $x \in Q_1$ we have that $C_V(x)V_2/V_2$ just contains vectors of type $v_2V_2$. In particular $|C_V(x)V_2/V_2| \leq q^5$. By \ref{Vlambdan} we have $|V_2/V_1 : C_{V_2/V_1}(A/A \cap Q_1)| \geq |A/A \cap Q_1|$, we see that $|A \cap Q_1| \geq q^5$. Now $v_2V_2$ is centralized by some $H_2 \cong \Omega^+(8,q)$ and $Q_1$ is an extension of an 8--dimensional $H_2$--module by an 8--dimensional one, i.e. $|C_{Q_1}(\langle v_2, V_2 \rangle/V_1)| \leq q^8$. Let $W$ be a fours group of vectors of type $v_2V_2$ in $V/V_2$. Then $W$ is centralized by $\Omega^+(6,q)$, which shows that the elements in $W$ cannot have the same centralizer. Let $W_1$ be the preimage of $W$ in $V/V_1$. then we see that $|C_{Q_1}(W_1)| = q^4$. Hence $|C_V(A \cap Q_1)V_2/V_2| \leq q$. This shows $|A \cap Q_1| \geq q^9$. But now also $|V_2 : C_{V_2}(A \cap Q_1)| \geq q^9$, and so $|A| \geq q^{18}$, a contradiction to the 2-rank of $E_6(q)$.

Let $G \cong E_7(q)$. Let first  $A \leq Q_7$. If $|A| > q$, then $V$ is strongly quadratic and the same holds for $|A| \leq q$ by \ref{transv}. By \cite{Str} $V \cong V(\lambda_7)$. Assume $A \not\leq Q_7$. As $E_6(q)$ does not posses $F$--modules, we get with \ref{factor} that $C_V(Q_7)$ is a trivial $G_7$--module. Now by \ref{lambdai} $V \cong V(\lambda_7)$ again. As a $G_7/Q_7$--module we have the following series
$$1 < V_1 < V_2 < V_3 < V,$$
where $V_1$ and $V/V_3$ are trivial modules and $V_2/V_1$, $V_3/V_2$ are 27--dimensional $E_6(q)$--modules. As $[Q_7,V_2] = V_1$, we see that $A \not\leq Q_7$. But on $V_3/V_2$ we have $|V_3/V_2 : C_{V_3/V_2}(A/A \cap Q_7)| > |A/A \cap Q_7|$. Hence as $|V_2 : C_{V_2}(A \cap Q_7)| = |A \cap Q_7|$, we see that $|V : C_V(A)| > |A|$, a contradiction.

Let finally $G \cong E_8(q)$. If $A \leq Q_8$, then $V$ is strongly quadratic, contradicting \cite{Str}. So $A \not\leq Q_8$. As $E_7(q)$ does not posses an $F$--module we get with \ref{factor} and \ref{lambdai} that $V \cong V(\lambda_8)$. As $V$ is not strong quadratic, we have $|A \cap Q_8| \leq q$. Now for the root group $R$ normal in $G_8$, we have that $[V,R]$ and $V/C_V(R)$ involve nontrivial $E_7(q)$--modules. Hence $|V : C_V(A/A \cap Q_8)| \geq q^2|A/A \cap Q_8|^2$. But then $q \geq q|A/A \cap Q_8|$, a contradiction. \qed

\begin{proposition}\label{twistff} The groups $^2E_6(q)$ and $F_4(q)$ do not posses $F$--modules. 
\end{proposition}

\proof Suppose false. If $V$ is strongly quadratic we may assume $V \cong V(\lambda_1)$ or $V(\lambda_2)$ in case of $F_4(q)$ by \cite{Str}. So assume first that $V$ is not strong quadratic. If $A \leq Q_1$ then by \ref{transv}, we have $|A| > q$. Now we see that $V$ is strongly quadratic, a contradiction, where we use \ref{closure} in case of $G \cong F_4(q)$ and $A \leq Z(Q_1)$. So we have $A \not\leq Q_1$. Further $|A \cap Q_1| \leq q$ for $G \cong \,^2E_6(q)$ and $A \cap Q_1 \leq Z(Q_1)$ for $G \cong F_4(q)$. So for $G \cong F_4(q)$ we have some $t \in A \setminus Q_1$ and $x \in Q_1$ with $t^x = ta$, $a \in Q_1 \setminus Z(Q_1)$. If $A \cap Q_1 \not=1$, we get that $V$ is strongly quadratic. So we may assume that $A \cap Q_1 = 1$ in that case. 

If $G \cong F_4(q)$, then $V$ involves at least four nontrivial irreducible $G_1/Q_1$--modules, at most two of them are natural. So $|V : C_V(A)| > q^6$, a contradiction.

If $G \cong \,^2E_6(q)$, we get that there are at least two nontrivial $G_1/Q_1$--modules involved, which are isomorphic to $Q_1/Z(Q_1)$, or at least four nontrivial irreducible modules are involved. As $Q_1/Z(Q_1)$ is not an $F$--module for $U_6(q)$, we would get $q|A/A \cap Q_1| \geq |V :C_V(A)| \geq q^2|A/A \cap Q_1|^2$ in the former, a contradiction. In the latter we get for the natural module $V_1$ that $|V_1 : C_{V_1}(A/A \cap Q_1)| < q^3$. This shows $|A/A \cap Q_1| \leq q$. But then $|A| \leq q^2$, a contradiction again.

Let $V = V(\lambda_2)$ and $G = F_4(q)$. Then $C_V(Q_1)$ is the $V(\lambda_2)$--module for $Sp(6,q)$. Now by \ref{fmodulesc} we have $A \leq Q_1$. Let $R$ be the root group normal in $G_1$. Then $[V,R] \geq C_V(R)$ and so $|[V,R]| \geq q^{14}$. Hence $R \not\leq A$. By \ref{closure} we have $\langle A^{G_1} \rangle = Q_1$ and so $Q_1$ offends. Hence $|V : C_V(Q_1)| \leq |Q_1|$, and so $|V| \leq q^{29}$, a contradiction.

So we have that $V \cong V(\lambda_1)$. Now $G \leq E_6(q^2)$ or $E_6(q)$ and the module $V(\lambda_1)$ of those groups restricted to $G$ is either irreducible, i.e. $V$, or a direct sum of a trivial module with $V$. Now by \ref{En} $V$ is not an $F$--module.  \qed
\\

So we have shown

\begin{proposition}\label{ffmodules1} Let $G = G(q)$ be of Lie type, $V$ be an $F$--module with offending subgroup $A \leq G$. Then one of the following holds
\begin{enumerate}
 \item[(i)] $G \cong L_n(q)$, $V \cong V(\lambda_1)$, $V(\lambda_2)$, $V(\lambda_{n-2})$, $V(\lambda_{n-1})$
\item[(ii)] $G \cong Sp_{2n}(q)$, $\Omega^{\pm}(2n,q)$, or $U_n(q)$ and $V \cong V(\lambda_1)$
\item[(iii)] $G \cong Sp_6(q)$, $V \cong V(\lambda_3)$
\item[(iv)] $G \cong \Omega^+(2n,q)$, $n = 4,5$ and $V \cong V(\lambda_n)$.
\item[(v)] $G \cong G_2(q)$ and $V$ is the natural module.
\end{enumerate}
\end{proposition}


  
\section{$F$-modules for Lie type groups in characteristic two with offending subgroup not in $E(G)$}

In this section we will prove the following 

\begin{proposition}\label{ffmodules2} Let $G = G(q), q = 2^t$, be of Lie type, $V$ be an irreducible $F$-module with quadratic offending subgroup $A \in {\cal P}(G,V)$, A
$ \not\leq E(G)$. Then $V$ is the natural module for $E(G) = \Omega^\pm(2n,q)$.
\end{proposition}

We will establish this proposition in a series of lemmas.  

\begin{lemma}\label{irred} We have that $V$ is irreducible for $E(G)$.
\end{lemma}


\proof Suppose there is $t \in A$ such that $V = V_1 \oplus V_1^t$. Then by quadratic action we have $|A| = 2$. Now $|V| = 4$, a contradiction. \qed

\begin{lemma}\label{orth1} Let $G = \Omega^\pm(2n,q)$, $V$ be the natural module and $A \leq G$ be a 2-group such that $C_V(a) = C_V(A)$ for all $a \in A^\sharp$, then $|A| \leq |V : C_V(A)|$.
\end{lemma}

\proof Let $W = C_V(a) = C_V(A)$. Then $[V,a] = W^\perp$. This shows $[V,a] = [V,b]$ for all $a,b \in A^\sharp$, hence $|A| \leq |W^\perp| = |V : C_V(A)|$. \qed

\begin{lemma}\label{orth2} Let $G = \Omega^\pm(2n,q)$, $G_1$ be the point stabilizer in the natural representation of $G$ and $E = O_2(G_1)$. Let $\langle a,b \rangle$ be a fours group in $G_1$  with $E \cap \langle a,b \rangle = 1$. Suppose that $\langle a,b \rangle$ acts quadratically on some $GF(2)$--module $V$. If $C_E(a) \not= C_E(b)$ then $V$ is strong quadratic.
\end{lemma}

\proof We may assume that there is some $e \in E^\sharp$ with $[e,a] = 1 \not= [e,b]$. Then $\langle a, [e,b]b \rangle$ acts quadratically and so by the 3-subgroup lemma also $\langle a, [e,b] \rangle$ acts quadratically. Now as $[E, b]$ always contains root elements, we get the assertion. \qed
\\

From now on let $G$ be a minimal counterexample to \ref{ffmodules2}. By \ref{irred} we may assume that $V$ is an irreducible $E(G)$--module.

\begin{lemma}\label{rank21} Let $E(G)$ be of rank at most two, then $E(G) \cong L_2(4)$ and $V$ is the natural $\Omega^-(4,2)$-module, or $E(G) \cong \Omega^-(6,q)$ and $V$ is the natural module, or $G \cong \Sigma_6$ and $V$ is the irreducible permutation module or its dual.
\end{lemma}

\proof Let first $E(G)$ be of rank one. Then $E(G) \cong L_2(q)$, $Sz(q)$, or $U_3(q)$. If $E(G) \not\cong L_2(q)$, then for each $a \in A \cap E(G)^\sharp$ we have $|V : C_V(a)| \geq q^2$ by \ref{L2q}. So suppose $A \cap E(G) \not= 1$, then we get $E(G) \cong L_2(q)$. By \ref{L2q} we have $|V : C_V(a)|~ \geq q$ for $a \leq A \cap E(G)^\sharp$. As $|A| \leq 2 \sqrt{q}$, we see $q = 4$ and we have the first assertion. So let $A \cap E(G) = 1$, then $|A| = 2$. Then for any involution $x \in E(G)$, we get $|V : C_V(x)| \leq 4$. This with \ref{L2q} shows $q =4$ and $E(G) \cong L_2(4)$, the conclusion.

Let next $E(G) \cong $ $^3D_4(q)$, $^2F_4(q)$, or $G_2(q)$. As $^2F_4(q)$ does not have an outer automorphism of order two, the second case does not occur. Now $t$ induces a field automorphisms. So by induction and \ref{factor} we get $q = 4$. Furthermore as the $L_2(4)$--module is a tensor product we see that $V$ is a tensor product. Hence by \ref{tensor2q}, we see $|A \cap E(G)| \leq 4$, so $|A| \leq 8$. So for a root element $r$ we have $|V : C_V(r)| \leq 64$. If $E(G) \cong $ $^3D_4(q)$, then this would imply that $[[V,r],O_2(C_G(r))] = 1$, and so $V$ is strong quadratic, contradicting \cite{Str}. So we have $E(G) \cong G_2(4)$. Further we get $|V : C_V(r)| = 64$ and so $|A| = 8$. Finally $A \cap E(G)$ just contains short root elements. Let $s$ be such an element, then $C_{E(G)}(s) \cong S \times E_{16}L_2(4)$, $|S| = 4$. As $|[V,s]| \leq 8$, we get $[[V,s],E_{16}L_2(4)] = 1$, and so $V$ is strong quadratic, a contradiction.

 

Let $E(G) \cong L_3(q)$. Then $V$ is not strong quadratic and so $A \cap E(G) \leq R$, where $R$ is a root group. Suppose that $t$ induces a graph automorphism, then $C_{E(G)}(t) \cong L_2(q)$. By the $A \times B$-lemma and \ref{L2q}, we see that $|V : C_V(t)| \leq 2$. But then for $r \in R^\sharp$ we have $|V : C_V(r)| \leq 4$ and so by \ref{L2q} $q \leq 4$. Suppose $t$ does not induce a graph automorphism, then $|A \cap E(G)| \leq \sqrt{q}$ and so we see with \ref{L2q} that $q \leq 4$ again. Let $q = 2$, then $t$ inverts an element of order 7, a contradiction. Let $q = 4$. Then $|[V,r]| \leq 4$ and so an elementary abelian group of order 8 centralizes $[V,r]$, whence $V$ is strong quadratic, a contradiction.

Let $E(G) \cong Sp(4,q)$. Let first $q = 2$. Then $E(G) \cong A_6$. If $t$ induces an outer automorphism on $\Sigma_6$, then $|A| \leq 4$. As $t$ inverts some element of order 5, we get $|A| = 4$ and $C_V(A) = C_V(t)$. But some $a \in A$ inverts the element of order 5 centralized by $t$, contradicting the A $\times$ B - Lemma. So we have $G \cong \Sigma_6$. If $A \cap E(G) = 1$, we have the assertion. So let $A \cap E(G) \not = 1$. As the dimensions for $\Sigma_6$-modules are 1,4,4,16, $|V : C_V(t)| \leq 8$ and $G$ is generated by 5 conjugates of $t$, we get $|V| = 16$. This is the assertion. 

So let now $q > 2$. Suppose first that $t$ normalizes both parabolics. Then by induction and \ref{factor} we get $q  = 4$. Further $V$ is a tensorproduct and so by \ref{tensor2q} we see $|A \cap E(G)| \leq 4$. Let $c \in A \cap E(G)^\sharp$. Then $|[V,c]| \leq 8$. So we see that there is some Sylow 2 -subgroup $S$ such that $[[V,c], Z(S)] = 1$. But then $V$ is strong quadratic, a contradiction. So we have $|A| = 2$ and so for some root element $r$ we get $|[V,r]| \leq 4$, a contradiction as before. 

So we may assume that $t$ does not normalize the parabolics. Then $|A \cap E(G)| \leq q$. Let $A \cap E(G) \not= 1$. Then elements in $A \cap E(G)^\sharp$ invert elements of order $q^2+1$ in $G$. So $|V : C_V(a)| \geq q^2$. Hence again we see $q = 4$ and then $|A : A \cap E(G)| = 4$. But this is not possible. 

Let next $E(G) \cong U_5(q)$. Then by induction we get $q = 2$. Further $|A| \leq 16$. As $V$ is not strongly quadratic we even get $|A| \leq 8$. Let $P$ be the normalizer of a root element $r$. We have $|V : C_V(r)| \leq 64$. But then $[[V,r] , O_2(P)] = 1$, and so $V$ is strong quadratic, a contradiction.

Let finally $E(G) \cong \Omega^-(6,q)$. If $V$ is a tensor product, then by \ref{tensor2q} we get $|A \cap E(G)| \leq q$. If there is a root element in $A$, then $V$ is strongly quadratic, contradicting \cite{Str}. Let some $c \in A$ of type $c_2$. Then $c$ inverts some element of order $q^2+1$, so $|[V,c]| \geq q^2$ and then $q = 2$. Now $|A| \leq 8$ and so $|[V,c]| \leq 8$. If $A \cap E(G)| = 1$, then for some $c \in E(G)$, $c$ of type $c_2$, we have $|[V,c]| \leq 4$. Let $G_1$ be the stabilizer of an isotropic vector in the natural representation. Then $[G_1, C_V(O_2(G_1))] \not=1$. Hence this is an irreducible $A_5$-module. As $|[V,c]| \leq 8$ for any involution in $A_5$, we see that $[O^2(G_1), V] = C_V(O_2(G_1))$. But then $O_2(G_1)$ acts quadratically , and so $V$ is strong quadratic, contradicting \cite{Str}. 

So we have that $V$ is not a tensor product. Then by induction and \ref{factor} we see that $C_V(O_2(G_1))$ (notation as before) has to be an $F$-module. As $V$ is not a tensor product it has to be the trivial module, so $V$ is the natural module, the conclusion. \qed  
\\

First we study the action of $A$ on the Dynkin diagram.

\begin{lemma}\label{reduction} Let the rank of $E(G)$ be at least three and $G_1$ be one of the maximal parabolics with connected Dynkin diagram. Assume that $[A, G_1] \leq G_1$. Let $Q_1 = O_2(G_1)$. Then either $C_V(Q_1)$ is a trivial $G_1/Q_1$--module or $A$ induces inner automorphisms on $G_1/Q_1$.
\end{lemma}

\proof Suppose false. Then by induction we may assume that $G_1/Q_1$ is $\Omega^\pm(2n-2,q)$. This shows that $E(G) \cong \Omega^\pm(2n,q)$, $E_n(q)$, $U_6(q)$ or $^2E_6(q)$. Now in any case the action of $A$ on $E(G)$ is known. Suppose $E(G) \not\cong \Omega^\pm(2n,q)$. Then there is a second parabolic on which $A$ induces an outer automorphism. But then by induction we have a contradiction. So we have $E(G) \cong \Omega^\pm(2n,q)$. Further $C_V(Q_1)$ is the natural $G_1/Q_1$--module. Hence by \cite{Str} we have that $V$ is not strongly quadratic. This shows that $A \cap Q_1 = 1$. By \ref{orth1} and \ref{orth2} we have that $|C_V(Q_1) : C_{C_V(Q_1)}(A \cap E(G))| \geq |A \cap E(G)|$. Recall that $C_V(Q_1)$ is isomorphic to $Q_1$ as $G_1/Q_1$--module. So we see that $[V,A] \leq C_V(Q_1)$. Then $[V,Q_1,Q_1] = 1$, a contradiction. \qed

\begin{lemma}\label{inner} Let the notation be as in \ref{reduction}. Then $A$ induces inner automorphisms on $G_1/Q_1$.
\end{lemma}

\proof Suppose false. Then by \ref{reduction} we have that $C_V(Q_1)$ is the trivial module. Let $G_2$ be a second maximal parabolic with connected diagram with $[G_2,A] \leq G_2$. Then by \ref{reduction} we have that $A$ induces inner automorphisms on $G_2/Q_2$. In particular we see that $A$ does not induce field automorphisms on $E(G)$. This shows that $E(G) \cong \Omega^\pm(2n,q)$ or $E_6(q)$. 

Suppose first that $V \not\cong V(\lambda)$ for some fundamental weight $\lambda$. Then it is a tensor product and by \ref{tensor2q} we have $|A \cap E(G)| \leq q$, so $|A| \leq 4q$. This shows that for some root element $r$ we have $|[V, r]| \leq q^8$, As  the $q$ - rank of $L_8(q)$ is 16 but in $E_6(q)$ we have $|O_2(C_G(r))/Z(O_2(C_G(r))| = q^{20}$, we see that $V$ would be strong quadratic, a contradiction to \cite{Str}. Hence $E(G) \not\cong E_6(q)$. In the same way we see that for $E(G) \cong \Omega^\pm(2n,q)$ we have $n \leq 6$. But in that case we get $|A| \leq q^2$ and so $|[V,r]| \leq q^4$ and then $n \leq 3$. This shows $G \cong \Omega^\pm(6,q)$, $q \leq 4$. But then for some $w \in E(G)$, $w$ of type $c_2$ we see $|V : C_V(w)| \leq q^2$, which now easily shows that $V$ is strongly quadratic a contradiction. 

So $V = V(\lambda)$, and so we have $G = E_6(q)$ and $V = V(\lambda_4)$. But now for a root group $R$ we have $|[V,R]| = q^{21}$ and so $|C_V(R) : C_{C_V(R)}(A \cap Q_1)| \geq |A \cap Q_1|$, hence $V/C_V(R)$ has to be an $F$-module for $G_1/Q_1 \cong L_6(q):2$, a contradiction. \qed

\begin{lemma}\label{fix} Let the rank of $E(G)$ be at least three. Then $A$ does not fix any maximal parabolic with connected diagram.
\end{lemma}

\proof Suppose false. Then by \ref{inner} $A$ has to induce inner automorphisms on any such parabolic. This shows that $A$ does not induce field automorphisms. Hence we see that $A$ induces graph or graph-field automorphisms, which have to be trivial on any such parabolic which is left invariant, a contradiction. \qed
\\

So for the remainder of this section we have to consider $E(G) \cong L_n(q)$ or $F_4(q)$ and $A$ induces a diagram automorphism.

\begin{lemma}\label{F4} Let $E(G) \cong L_n(q)$, $n \geq 4$, or $F_4(q)$ and assume that $A$ contains some $t$ which acts nontrivially on the Dynkin diagram. Then $G \cong L_4(q) \cong \Omega^+(6,q)$ and $V$ is the natural orthogonal module.
\end{lemma}

\proof Let first $E(G) \cong L_4(q)$. Suppose that $V$ is not strong quadratic. Then $|A \cap E(G)| \leq q^2$. Choose $s \in A \cap E(G)$, $s \not= 1$. Then $|[V,s]| \leq 2q^2$. If $q \not= 2$ we see that $V$ has to be strong quadratic. 
If $A \cap E(G) = 1$, then $|A| \leq q$ and so again there is some $s \in E(G)$ with $|[V,s]| \leq q^2$. Hence in any case we may assume that $q=2$ and so $G \cong \Sigma_8$ and $|V : C_V(x)| = 8$ for all $x \in A \cap E(G)^\sharp$. But then $A \cap E(G)$ just contains short involutions and so $A$ contains a transposition $t$. Now $[C_V(t), C_{E(G)}(t)'] = 1$, contradicting the $A \times B$-lemma. So we have that $V$ is strongly quadratic and the assertion follows with \cite{Str}.

Let next $E(G) \cong L_5(q)$. Then by \cite{Str} $V$ is not strongly quadratic. So for any involution $s \in E(G)$ we have $|[V,s]| \geq q^5$. From this we see that $A \cap E(G)$ does not contain root elements and so $|A \cap E(G)| \leq q^4$. Let $s \in A \cap E(G)^\sharp$. Then $C_{E(G)}(s)$ contains an extension of a class two group of order $q^8$ by $L_2(q)$. This shows that $|[V,s]| \geq q^5$. Now we get $q = 2$ and $|A \cap E(G)| = 16$.  Let $r$ be a root element in $E(G)$, which is centralized by $A$. Then $|A \cap O_2(C_G(r))|\geq 8$, but then there is a quadratic fours group in $E(G)$ which contains $r$, a contradiction.

Let next $E(G) \cong L_6(q)$ and $V = V(\lambda_3)$. Then dim $V = 20$ and for the parabolic $G_{15}$ we get the following series
$$1 < V_1 < V_2 < V_3 < V,$$
dim $V_1 =$ dim $V/V_3 = 6$, dim $V_2/V_1 = $ dim $V_3/V_2 = 4$. Further $[t,V_1] \leq V_1$, $[t,V/V_3] \leq V/V_3$ and $[t,V_2/V_1] = V_3/V_2$. This shows $|[V,t]| \geq q^6$. Further we see that $A \cap E(G) \leq Q_{15}$. This shows $|A| \leq 2q^5$. Hence $q = 2$ and $|A| = 64$. Further $Z(G_{15}) \leq A$, but $[t,[V,Z(G_{15})]] = [t, V_1] \not= 1$, a contradiction.

Let now $E(G) \cong L_n(q)$, $n \geq 6$. Assume first that $V$ is not strongly quadratic. Then $[G_{1,n-1}, C_V(Q_{1,n-1})] \not= 1$, and so by induction we see $E(G_{1,n-1}/Q_{1,n-1}) \cong L_4(q)$ and $C_V(Q_{1,n-1})$ is the natural module. So we have $n=6$ and $V \cong V(\lambda_3)$, a contradiction.
So we have that $V$ is not strongly quadratic.    

Let $\langle c,d \rangle \leq A \cap E(G)$ with $c$ of type $a_2$ and $d \not\in Z(C_{E(G)}(c))$. We have that $$O^{2^\prime}(C_{E(G)}(c)/O_2(C_{E(G)}(c))) \cong L_{n-4}(q) \times L_2(q)$$ and 
$O_2(C_{E(G)}(c))$ is an eytension of $Z(O_2(C_{E(G)}(c)))$ by a group of order $q^{4(n-4)}$, which is a sum of tensor products of the two natural modules for $L_2(q)$ and $L_{n-4}(q)$. As $C_{E(G)}(c)$ acts indecomposably on $Z(O_2(C_{E(G)}(c)))$ for $q > 2$ and $V$ is not strongly quadratic, we get that $q = 2$ and $d \in Z(O_2(C_{E(G)}(c))) = E$.  

Set $X = \langle P_1, P_{n-1}\rangle$. Let $E \unlhd X$, where $E$ is as before. Let first $A \cap O_2(X) \not=1$. As $E = Z(O_2(X))$, we see that $[t,E]$ contains some element $1 \not= c \in E$. As $V$ is not strongly quadratic, this is not a transvection. So it is of type $a_2$. Now by the above we get $q= 2$ and $A \cap O_2(X) \leq E$. In particular $|A \cap O_2(X)| \leq 4$. Let $s \in A \cap X$, $s \not\in O_2(X)$. Then $[s,E]$ contains transvections, hence $A \cap O_2(X) = 1$. So we have $|A| \leq 8$ or $A \cap O_2(X) = 1$. Let $|A| \leq 8$. Then $|[V,c]| \leq 8$ and so we get $n \leq 4$, a contradiction. 

So we have $A \cap O_2(X) = 1$. If $q > 2$, we see that $C_V(O_2(X))$ is trivial, as it has to be an $F$-module. Now by induction applied to $G_{1,n-1}$ we get again $n = 6$ and $V = V(\lambda_3)$, a contradiction. So we are left with $q = 2$ and $|A| \leq 4$. As $n \geq 6$, we see that $t$ inverts an element of order 7 in $\langle P_1,P_2, P_{n-2},P_{n-1} \rangle = T$. Hence we get that $C_V(O_2(T))$ is a trivial module. Now again induction on $G_{1,n-1}$, yields the contradiction.

So let now $E(G) \cong F_4(q)$. By \cite{Str} we have that $V$ is not strongly quadratic. By \ref{rank21} we have that $G_{1,4}$ acts trivially on $C_V(O_2(G_{1,4}))$. Again set $X = \langle P_1, P_4 \rangle$. Let first $A \cap O_2(X) = 1$. Then $|A| \leq 2q$. Hence for some root element $r \in E(G)$ we get $|V : C_V(r)| \leq 16q^4 \leq q^8$. From this we see that $[O_2(C(r)), [V,r]] = 1$, contradicting $V$ to be not strongly quadratic. So we have $A \cap O_2(X) \not = 1$. Hence again there is some quadratic fours group $\langle c,d \rangle \leq O_2(X)$ with $c \leq Z(O_2(X))$. Let $r$ be a root element with $Z(O_2(X)) \leq O_2(C(r)) = Y$. If $d \not\in Y$, we easily get a strongly quadratic module. So $d \in Y$ and then in fact $d \in Z(Y)$. But in $Z(O_2(X))$ there are root elements $r$ and $s$ of both types. So the same is true for $Y_1 = O_2(C(s))$. Hence $d \in Z(Y) \cap Z(Y_1)$, which is of order $q^2$. As $d$ is not a root element, 
we get that $\langle c,d \rangle$ is in $Z(Y) \cap Z(Y_1)$ and so $d \in Z(C_{E(G)}(c))$, a contradiction. \qed
\\
 
Now the proposition follows from \ref{rank21}, \ref{fix} and \ref{F4}.


\section{$F$-modules for alternating, sporadic and Lie type groups in odd characteristic}

Throughout this chapter we will assume that $E(G)$ is a perfect central extension of an alternating group, a sporadic group or a group of Lie type over a field of odd characteristic, which is not a group of Lie type over a field of characteristic 2 too. 
 Further $V$ is an $F$-module and $A \in {\cal P}(G,V)$ a quadratic offending subgroup. 

\begin{lemma}\label{alternating} Let $E(G) = A_n$ or $E(G) = 3A_6$, then either $V$ is the permutation module or
\begin{enumerate}
\item[(i)] $G \cong A_8 \cong L_4(2)$ and $V$ is the natural $L_4(2)$-module.
\item[(ii)] $G \cong A_7$ and $V$ is as in (i).
\item[(iii)] $E(G) \cong 3A_6$ and $|V| = 2^6$.
\item[(iv)] $E(G) \cong A_5 \cong L_2(4)$ and $V$ is the natural $L_2(4)$-module.
\end{enumerate}
\end{lemma}

\proof Suppose first $|A| = 2$. Then $A$ induces transvections. Hence $A$ just inverts elements of order three. So either $A = \langle (1,2) \rangle$ and so $V$ is the permutation module, or $A = \langle (1,2)(3,4) \cdots (n-1,n) \rangle$, $n$ is even, and $n \leq 8$. As $A_8 \cong L_4(2)$ and $A_6 \cong Sp_4(2)^\prime$, we have handled these cases before. So let $E(G) \cong 3A_6$. But then $A$ inverts the center of $E(G)$, which has to act fixed point freely on $V$, a contradiction. 

So let $|A| \geq 4$. Let furthermore $E(G) \not\cong 3A_6$. By \cite{MeiStr} we have that $V$ is either the natural module or the spin module. So assume $V$ to be the spin module. Let $a\in A^{\sharp}$. Let $k$ be the number of
fixed points of $a$. Then there is 
$K\leq C_{G}(a), K\cong \Sigma_{k}$. Furthermore 
$C_{C_G(a)}(K^\prime)$ is an extension 
of a 2--group by $\Sigma_{m}$, $m = (n-k)/2$. Now choose 
$a \in A$ with $m > 2$ if possible. Suppose first 
$[A, C_{C_G(a)}(K^\prime)] \not= 1$. 
If $m \ge 5$, then $\Sigma_m$ is nonsolvable and so 
$C_{C_G(a)}([V,a])$ contains an elementary abelian subgroup of $O_2(C_G(a))$ of order $2^{m-1}$. 
But then this group contains a conjugate of (12)(34) which 
contradicts \cite[(4.3)]{MeiStr}.\\
\indent
Let $m = 4$. Then $a \sim (12)(34)(56)(78)$. Furthermore as we may 
assume that there is no $x \sim (12)(34)$ which is contained in 
$\langle A^{C_G(a)} \rangle$ we see that $A$ is conjugate to a 
subgroup of 
$\langle (12)(34)(56)(78), (13)(24)(57)(68), (15)(26)(37)(48) \rangle$. \\
\indent
Let $m = 3$. Then $C(K^\prime) \le \Sigma_6$ and $a \sim (12)(34)(56)$. 
Then $\langle A^{C_G(a)} \rangle$ contains some $x \sim (12)(34)$, 
contradicting \cite[(4.3)]{MeiStr}.\\
\indent
So let $[A, O^{2^\prime}(C_{C_G(a)}(K^\prime))] = 1$. 
If $[A, K^\prime] \not= 1$, then $[K^\prime, [V,a]] = 0$. 
If $k \ge 4$, then $K^\prime$ contains some $
x \sim (12)(34)$. This again contradicts \cite[(4.3)]{MeiStr}. 
Let $k \le 3$. As $[A, O^{2^\prime}(C_{C_G(a)}(K^\prime))] = 1$ 
and $m > 2$, there is $x \sim (12)$ in $A$, a contradiction. 
So we are left with 
$[A, K^\prime] = 1 = [A, C_{C_G(a)}(K^\prime)]$. But this is 
impossible with $m > 2$.\\
\indent
So we have $m \le 2$ for all $a \in A^\sharp$. As there is no fours 
group of transvections we may assume $a = (12)(34) \in A$. Now 
$A \ge \langle a, b \rangle$, $b = (13)(24)$, $(12)(56)$ or $(34)$. Let 
$[b,K^\prime] \not= 1$. Then $b = (12)(56)$ and so $K^\prime$ contains 
no involutions by \cite[(4.3)]{MeiStr}. This shows $k \le 3$ and so 
$A \le \Sigma_7$. But this group 
$A = \langle (12)(34), (12)(56) \rangle$ does not act quadratically on 
the four dimensional module.\\
\indent
Assume now $b = (34)$. Then $[a, E(C_G(b))] \not= 1$. Now 
$E(C_G(b)) \cong \Sigma_{n-2}$, which is nonsolvable. But then 
$\langle (34), (12)(56) \rangle$ acts quadratically, a contradiction.

So we have seen that $|A| \leq 8$ and so $|V : C_V(a)| \leq 8$ for all $a \in A^\sharp$. Furthermore for $n > 8$, we see that there is some $a \in A^\sharp$ which inverts $(1,2,3) = \rho$. But $\rho$ is fixed point freely on $V$. This shows $|V| \leq 2^6$, a contradiction. So we have $n \leq 8$.. The case of $n = 5,6,8$ was done before. So we have $n = 7$. Then $|A| = 4$, and $|V| = 16$, which is (ii).

So we are left with $E(G) \cong 3A_6$ and $|A| = 4$, $A \leq E(G)$. Now we see that $E(G)$ is generated by three conjugates of $A$ and so $|V| \leq 2^6$. As $Z(E(G))$ has to act fixed point freely, we see $|V| = 2^6$.\qed

\begin{lemma}\label{sporadic} There are no $F$-modules for $E(G)$ sporadic.
\end{lemma}

\proof If there is a transvection, then there is also a quadratic fours group. So by \cite{MeiStr} we just have to handle the cases $E(G) \cong M_{12}$, $M_{22}$, $M_{24}$, $3Suz$, $3M_{22}$, $J_2$, $Co_1$, $Co_2$. Furthermore $V$ is known. In all cases besides $E(G) \cong 3M_{22}$ we have $|A| \leq 4$.

Let $|A| = 2$, then any fours group containing $A$ would be quadratic, which contradicts \cite{MeiStr}. So we have $|A| = 4$ and if $E(G) \not\cong M_{12}$ or
$M_{22}$, we see that for each $a \in A^\sharp$ we have $[O^2(C_{E(G)}(a)), [V,a]] = 1$. But then we get to many quadratic fours groups which contradicts \cite{MeiStr} again. So we are left with $E(G) \cong M_{12}$ or $M_{22}$ and $C_{E(G)}(a)$ induces $\Sigma_3$ on $[V,a]$. In both cases there is some $a \in A^\sharp$ which inverts some element $\omega$ of order 5 in $E(G)$ which acts on some element $\nu$ of order eleven. Hence $|[V,a]| \geq 2^4$, a contradiction.

So we are left with $E(G) \cong 3M_{22}$ and $|V| = 2^{12}$. As can be seen from the action described in \cite{MeiStr} $|A| \not = 16$. As $V$ is defined over $GF(4)$, we see that the elements in $A$ induce $GF(4)$-transvections on $V$. But any $a$ inverts some element $\omega$ of order 5 which in turn normalizes some element $\nu$ of order 11. Hence $|[V,\omega]| = 2^8$ and so $|[V,a]| = 2^4$, a contradiction. \qed

\begin{lemma}\label{oddchar} If $E(G)$ is some group of Lie type in odd characteristic then it is a group of Lie type in even characteristic too.
\end{lemma}

\proof Suppose false. We have a quadratic fours group. So by \cite{MeiStr1} we get $E(G) \cong 3U_4(3)$ and $|V| = 2^{12}$. As $V$ is defined over $GF(4)$, we see that $A \cap E(G) \not= 1$. For $a \in A \cap E(G)^\sharp$ , we have $|V : C_V(a)| = 16$, so we have $|A| \geq 16$. As $C_{E(G)}(a)$ is a maximal subgroup in $E(G)$, we see that $|A| = 2^5$ and so $A$ is uniquely determined. But now $A \not\leq E(G)$ and so $|V : C_V(A)| = 2^6$, a contradiction.  \qed 
 



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\end{thebibliography}


Department of Mathematics

Michigan State University

East Lansing, Michigan 48824

USA

meier@math.msu.edu\bs


Fachbereich Mathematik und Informatik

Institut f\"{u}r Algebra und Geometrie

Martin-Luther-Universit\"{a}t Halle Wittenberg

06099 Halle

Germany

stroth@coxeter.mathematik.uni-halle.de

\end{document}

