$ and $P^\epsilon=\O_p(P^\epsilon)(P^+\cap P^-)$. It follows that $L\unlhd L(P^+\cap P^-)=\

=G$,
and since $G$ is simple, $G=L$.
\ep
\be{lemma}{generate g2} Let $G\cong G_2(q)$, $p=q^k$, $P$ a Lie-parabolic subgroup of $G$ with
$\Z(\O^{p^\prime}(P))=1$ and $A\unlhd P$ with $|A|=q^3$. Then $G=\$ for some $t\in G$.
\en{lemma}
\bp Choose a root system $\Phi$ for $G$ such that $P$ is a Lie-parabolic with respect to $\Phi$ and
let $N/H$ be the corresponding Weyl-group. Let $\ca R_l$ ( $\ca R_s$) be set root subgroups in $G$
corresponding to the long (short) roots in
$\Phi$. Put $L=\<\ca R_l\>$. Then $L$ is a genuine group of Lie-type of type $A_2$ and $P\cap L$
is a Lie-parabolic subgroup of $L$ with $L\cap A=\O_p(P\cap L)$. Since $N/H\cong D_{12}$ we can
choose $t\in
N\setminus H$ with $[t,N]\leq H$. Put $K=\$. Since $(P\cap L)^t$ is opposite to $P\cap L$ in $L$,
\ref{generate lie} implies that $L=\__$. Let $i,j\in I$ with
$i 1$ and by induction that $Z_{e_i-e_{j-1}}\leq U$. Thus using $(*)$,
\[Z_{e_i-e_j}= [Z_{e_i-e_{j-1}}, Z_{e_{j-1}-e_j}]\leq U.\]
Next we will show by downwards induction on $j-i$, then $Z_{e_j-e_i}\leq L$. If $j-i=n-1$, then
$j=n$ and $i=1$ and so this holds by definition on $L$. So suppose $j-i__

$ contains all the root subgroups from $\Phi$. So
$\$ and $Z:=Q\cap Q^g$. Then
$H/\C_H(Y)\cong \SL_2(q)$, and $H$ acts transitively on the $1$-dimensional
subspaces of $Y$. Thus $H =\

$; in particular, $T$ normalizes
$H$. Moreover, $Q\O_p(HT) = T \in \Syl_p(HT)$, and using \rf 0:
\bd{0.1} If $n\geq 5$, then $\o{\C_{Q^g}(Y)}=\O_p(\C_{\o L}(Y/U_0))$, and $Z$ is a $1$-dimensional singular subspace of $Q$.\ed
\rf a: Put $\K:=\End_K(W)$. By Smith's Lemma \ref{smith} applied to $W$ and its
dual,
$\C_W(Q)$ and $W/[W,Q]$ are simple $\K L$-modules. Since $[W,Q]\leq \C_W(Q)$ we
conclude that $[W,Q]=\C_W(Q)$. Suppose that $n=3$ or $4$. Then $Q=T$ and so
$\C_W(Q)$ and $W/[W,Q]$ are 1-dimensional over $\K$. Thus $\dim_\K(W)=2$.
If $n= 3$ or $(n,\epsilon)=(4,+)$ then $W$ is
a natural $\SL_2(q)$-module. If $(n,\epsilon)=(4,-)$, then $W$ is a natural
$\SL_2(q^2)$-module. These are the (half-)spin
modules for these groups, so \rf a follows in this case.
Suppose now that $n\geq 5$, so we are allowed to use the subgroups $Y$,
$H$ and $Z$
constructed above. Since $[W,Z,H]=0$ and $Z\neq 0$ we conclude that $\C_W(HT)\neq
0$. By Smith's Lemma \ref{smith} $\C_W(T)$ is $1$-dimensional over $\K$ and so
$\C_W(T)=\C_W(TH)$. Since $K=\

**$, so
\[[W,a,d]\leq [W,a,\ ]=\<[W,a,b]^{\C_Q(a)}\>=0.\]
Since $A$ is elementary abelian, $d \in \ \leq \C_L(a)$ and
so $[a,d,W] = 0$. Hence by the Three Subgroups Lemma also $[W,d,a] = 0$, and
$D:=\$ satisfies \rf 1.
\bd 2 There exists $B\leq Q$ and $1\neq e\in B$ such that $[W,B,B]=0$,
$h(e)=0$ and $B\nleq \F_q\hspace{0.7pt}e$.
\ed
Let $D$ be as in \rf 1. Pick $1\neq b\in D\cap Q$, and put $E
:=\**