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\begin{document}
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\title{Nearly Quadratic Modules}
\author{U. Meierfrankenfeld\\ B. Stellmacher\\{}\\
Department of Mathematics, Michigan State University, East
Lansing MI 48840\\ meier@math.msu.edu\\ Mathematisches Seminar, Christian-Albrechts- Universit\"at, D24098 Kiel\\stellmacher@math.uni-kiel.de}
\maketitle
\section{Introduction}
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All our actions are from the right. To save parenthesis we write $abc$
for $(ab)c$, $ab.cd$ for $(ab)(cd)$, $ab.cde.fg$ for
$((ab)(c(de)))(ef)$ and so on.
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\section{Cubic and Nearly Quadratic Action}
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In this section $A$ is a group, $\F$ is a field and $V$ an $\F A$-module.
\be{definition}{bdefinition cubic}
$V$ is
\bl
\li a{\em quadratic} $\F A$-module if $[V,A,A] = 0$,
\li b {\em cubic} $\F A$-module if $[V,A,A,A] = 0$,
\li c {\em nearly quadratic} $\F A$-module if $V$ is a cubic $\F A$-module such that
$$[V,A]+C_V(A) = [v\F,A] + C_V(A)\hbox{ for every }v \in V\setminus
[V,A]+C_V(A).$$
In the corresponding cases we also say that $A$ acts quadratically, cubically
and nearly quadratically on $V$.
\el
\en{definition}
\be{definition}{def:qva}
$Q_V(A)$ is the sum of all quadratic $\F A$-submodules of $V$ (and so
the largest quadratic $\F A$-submodule of $V$.
\en{definition}
\be{definition}{def:imp}
A system of imprimitivity for $A$ in $V$ is a set $\Delta$ of
$\F$-subspaces of $V$ such that
\bl[i]
\li 1 $|\Delta| > 1$ and $\Delta^A = \Delta$, and
\li 2 $V = \oplus_{W\in \Delta} W$.
\el
\en{definition}
\be{definition}{def:semiliner}
Let $\K$ be a field extension of $\F$ and $\sigma:\,A\to \Aut(\K)$ a
homomorphism. Then $V$ is a {\em semi-linear} $\K A$-module (with
respect to $\phi$) provided that
$vka =v.k\sigma.a$ for every $k \in K$, $a\in A$ and $v \in V$. Set $A_\K := \ker
\sigma$ and $\K_A := C_\K(A\sigma)$.
\en{definition}
\be{lemma}{quadratic implies nearly} Let $V$ be a quadratic $\F
A$-module. Then $V$ is a nearly quadratic $\F A$-module.
\en{lemma}
\proof Since $A$ is quadratic, $[V,A]\leq C_V(A)\leq [v\F,A]+C_V(A)$
for all $v\in V$.\qed
\be{lemma}{trivial property} Let $V$ be a nearly quadratic $\F A$-module and
$W$ be an $\F A$-submodule of $V$. Then
\bl
\li a Either $W\leq [V,A] + C_V(A)$ or $[V,A] \leq [W,A] + C_V(A)$.
\li b Either $Q_V(A) = V$ or $Q_V(A) = [V,A] + C_V(A)$.
\li c $W$ and $V/W$ are nearly quadratic $\F A$-modules.
\li d $A$ is quadratic on $W$ or on $V/W$.
\el
\en{lemma}
\proof
\rf a If $W\nleq [V,A]+C_V(A)$ the definition of nearly quadratic
implies $[V,A]\leq [W,A]+C_V(A)$.
\rf b Since $A$ is cubic, $[V,A]+C_V(A)\leq Q_V(A)$. Since $A$ is
quadratic on $Q_V(A)$ $[Q_V(A),A] \leq
C_V(A)$. By \rf a either $ [V,A]+C_V(A)\leq Q_V(A)$ or $[V,A]\leq
[Q_V(A),A]+C_V(A)$. In the first case $ [V,A]+C_V(A)= Q_V(A)$. In the
second case $[V,A]\leq C_V(A)$ and so $A$ acts quadratically on $V$ and
$V=Q_V(A)$.
\rf c and \rf d We first show that $\o V:=V/W$ is a nearly quadratic $\F
A$-module. Let $v\in V$ with $\o v\notin [\o V,A]+C_{\o V}(A)$. Since
$\o{[V,A]+C_V(A)}\leq [\o V,A]+C_{\o V}(A)$ we get that $v\notin
[V,A]+C_V(A)$. Thus $[V,A]\leq [v\F,A]+C_V(A)$ and $[\o V,A]\leq [\o v
\F,A]+C_{\o V}(A)$. Hence $\o V$ is nearly quadratic for $A$.
To show that $A$ is nearly quadratic on $W$ and is quadratic on $W$ or
$V/W$ we may assume that $A$ is not quadratic on $W$. Then by \rf a
$[V,A]\leq [W,A]+C_V(A)$. It follows that $ \o[V,A]\leq \o{C_V(A)}$
and $A$ is quadratic on $\o V$. Hence \rf d holds.
Moreover, $Q_V(A)=[V,A]+C_V(A)=[W,A]+C_V(A)$
and so $Q_W(A)=W\cap Q_V(A)=[W,A]+C_W(A)$. Hence if $w\in W$ with
$w\notin [W,A]+C_W(A)$, then $w\not Q_V(A)=[V,A]+C_V(A)$ and so
$[V,A]\leq [\F w,A]+C_V(A)$. Thus $[W,A]\leq [V,A]\cap W\leq [
w\F,A]+C_W(A)$ and $W$ is nearly quadratic. So also \rf c is proved.
\qed
\be{lemma}{belementary cubic prop} Let $V$ be a cubic $\F
A$-module and put $A_0=C_A(Q_V(A))$ Then the following hold:
\bl
\li b $A_0$ acts quadratically on $V$.
\li c For $z\in \mathbb Z$, $a \in A$ and $v\in V$ with $[v,a,a] = 0$,
$$z[v,a] = [v,a^z] = [zv,a].$$
\li a If $A$ acts quadratically on $V$, then $A/C_A(V)$ is an
elementary abelian $\chr F$-group. (Here an elementary abelian
$m$-group is an abelian group all of its non-trivial elements
have order $m$, if $m$ is a prime, and infinite order if $m=0$.)
\li d $A_0\nl A$, and $A/A_0$ and $A_0/C_A(V)$ are elementary abelian
$\chr F$-groups.
\li e If $\chr \F=0$, then all non-trivial elements in $A$ have
infinite order. If $\chr \F$ is a prime, then $A$ is a $\chr \F$-group.
\el
\en{lemma}
\proof
\rf b Since $A$ is cubic, $[V,A_0]\leq [V,A]\leq Q_V(A)\leq C_V(A_0)$.
\rf c: For $z = 0$ \rf c is obvious, and clearly $z[v,a] = [zv,a]$ for
every $z \in \mathbb Z$. Moreover, the property $[v,a,a] = 0$ and the
Three Subgroups Lemma give
$$[v,a^{-1}] = -[v,a] \hbox{ and }[v,a^z]^a = [v,a^z].$$
Now an elementary induction shows \rf c for $z \geq 0$. Then also the
case $z < 0$ follows using $[v,a^{-1}] = -[v,a]$.
\rf a Since $[V,A,A]=0=[A,V,A]$ the Three Subgroup Lemma gives
$[A,A,V]=0$. So $A/C_A(V)$ is abelian. Now let $a\in A$ and $v\in
V$ with $[v,a]\neq 0$. Let $i$ be a positive integer and put $p=\chr
F$. By \rf c
$[v,a^i]=i[v,a]$. If $p>0$ we conclude that $[v,a^p]=0$ and if
$p=0$, then $[v,a^i]\neq 0$. So $aC_A(V)$ has order $p$ if $p>0$
and infinite order if $p=0$.
\rf d: This follows from \rf a, since $A$ acts quadratically on
$Q_V(A)$ and $A_0$ acts quadratically on $V$.
\rf e follows immediately from \rf d.\qed
\be{lemma}{a2 and com} Let $a\in A$ and $v\in V$. Suppose that $\chr
\F=2$. Then $[v,a^2]=[v,a,a]$ and $[V,a^2]=[V,a,a]$.
\en{lemma}
\proof This follows for example since $(a-1)^2=a^2-1$ in $\End_\F(V)$.
\qed
\be{lemma}{binherit2} Let $V$ be a nearly quadratic, but not quadratic $\F A$-module and
$Y$ and $X$ be $\F A$-submodules of $V$ such
that $V = X \oplus Y$. Then at least one of the submodules $X$ and $Y$ is
centralized by $A$.
\en{lemma}
\proof Since $V = X
\oplus Y$ and $V$ is not quadratic at least one of the summands, say $X$, is not a quadratic $\F
A$-module. Then by \ref{trivial property}
$$[V,A] + C_V(A) = [X,A] + C_V(A) = Q_V(A).$$
In particular $[Y,A,A] \leq Y \cap [X,A] = 0$, so $Y \leq Q_V(A) =
[X,A] + C_V(A)$. Now the same argument gives $[Y,A] \leq Y \cap [X,A] =
0$. \qed
\be{lemma}{bimprimitivity1} Suppose that $V$ be an $\F A$-module and
$\Delta$ is a system of imprimitivity for $A$ in $V$. Let $\Delta_1$
be an orbit for $A$ on $\Delta$ and $\Delta_0\subseteq \Delta_1$. Then each
following conditions immplies that $\Delta_0 = \Delta_1$
\bl[1]
\li a $ \bigoplus\! \Delta_0\,\cap\, C_V(A) \ne 0$.
\li b $A$ is cubic and $\bigoplus\! \Delta_0\,\cap\, [V,A,A] \ne 0$.
\li c $A$ is quadratic and $\bigoplus\! \Delta_0\,\cap\, [V,A] \ne 0$.
\el
\en{lemma}
\proof Put $U=\bigoplus \Delta_0$. The cubic action of $A$ gives
$[V,A,A]\leq C_V(A)$, so \rf b implies \rf a. Similarly \rf c implies
\rf a. Hence, we may assume that $C_U(A) \ne 0$. For $X \in
\Delta$ let $\pi_X$ be the projection of $V$ onto $X$. Set
$$\Delta_2 := \{X \in \Delta_1 \mid C_U(A)\pi_X \ne 0\}.$$
Then $\Delta_2 \subseteq \Delta_0$, $\Delta_2\neq \emptyset$ and
$\Delta_2$ is $A$-invariant. Thus $\Delta_2=\Delta_1$ and so also
$\Delta_0=\Delta_1$ \qed
\be{lemma}{quadratic imprimitivity} Let $V$ be an quadratic $\F A$-module,
$\Delta$ a system of imprimitivity for $A$ in $V$, $\Delta_1$ a
non-trivial orbit for $A$ on $\Delta$, and $W\in \Delta_1$. Then
$|\Delta_1|=\chr \F=|A/C_A(\bigoplus \Delta_1)|=2$.
\en{lemma}
\proof Let $W\in \Delta_1$ and $B=N_A(W)$. Then $\{W\}\neq \Delta_1$
and so by \rf[bimprimitivity1] c, $W\cap [V,A]=0$. In particular,
$[W,B]=0$. Let $a\in A\setminus B$. Then $0\neq [W,a]\leq W+W^a$ and
so by \rf[bimprimitivity1] c, $\Delta_1=\{W,W^a\}$, $|A/B|=2$,
$B=C_A(\Delta_1)\n A$, $B$ centralizes $W+W^a$ and so
$|A/C_A(\bigoplus \Delta_1)|=2$. \ref{belementary cubic prop} gives
$\chr \F=2$.\qed
\be{lemma}{bimprimitivity2} Let $V$ be a cubic $\F A$-module,
$\Delta$ a system of imprimitivity for $A$ in $V$, $\Delta_1$ a
non-trivial orbit for $A$ on $\Delta$, and $W\in \Delta_1$
\bl
\li a $A/C_A(\bigoplus \Delta_1)$ is an elementary abelian $p$-group
for some prime $p$.
\li b $p=\chr \F\in\{2,3\}$.
\li c One of the following holds:
\bl[1]
\li a $|A/C_A(\bigoplus \Delta_1)|=|\Delta_1|\leq 4$ and
$N_A(W)=C_A(\bigoplus \Delta_1)=C_A(\Delta_1)$
\li b $p=|\Delta_1|=2$ and $N_A(W)=C_A(\Delta_1)$ acts quadratically on
$\bigoplus \Delta_1$.
\el
\el
\en{lemma}
\proof We may assume with loss that $V=\bigoplus \Delta_1$ and $V$ is
a
faithful $A$-module. If $A$ is quadratic on $V$, then
\ref{quadratic imprimitivity} implies that the lemma holds. Hence we may
assume
\bd 1 $A$ is not quadratic on $V$.\ed
Next we prove
\bd 2 Suppose $\chr \F=2$. Then $A$ is an elementary abelian
$2$-group.
\ed
Let $a\in A$ and suppose that $a^2\neq 1$. Pick $W\in \Delta_1$ with
$[W,a^2]\neq 0$. By \ref{a2 and com} $[W,a,a]=[W,a^2]\leq
W+W^{a^2}$. Hence \rf[bimprimitivity1] b implies that
$\Delta_1=\{W,W^{a^2}\}$. Thus $|\Delta_1|\leq 2$ and so $a^2$ acts
trivially on $\Delta_1$. But then $W=W^{a^2}$ and $\Delta_1=\{W\}$, a
contradiction.
So $a^2=1$ for all $a\in A$ and \rf 1 holds.
\bd 3 Let $W\in \Delta_1$ and $a,b\in V$ with $[W,b,a]\neq 0$. Then
$\Delta_1=\{W,W^b,W^a,W^{ab}\}$.
\ed
Note that $[W,b]\leq W+W^b$ and so also $[W,b,a]\leq
W+W^b+W^a+W^{ab}$. Hence \rf[bimprimitivity1] b implies that
$\Delta_1=\{W,W^b,W^a,W^{ab}\}$.
\bd 4 $|\Delta_1|\leq 4$\ed
By \rf 1 there exists $a,b\in A$ and
$[V,b,a]\neq 0$. Since $V=\bigoplus \Delta_1$ there exists $W\in
\Delta_1$ with $[W,b,a]\neq 0$ and so \rf 4 follows from \rf 3.
\bcase{NA(W) acts} Suppose that $[W,N_A(W)]\neq 0$ for some $W\in
\Delta_1$.
\ecase
Pick $b\in B:=N_A(W)$ with $[W,b]\neq 0$ and $a\in A\setminus
B$. Since $[W,b]\leq W$ we get $[W,b,a]\neq 0$. Since $W=W^b$, \rf 3
gives $\Delta_1=\{W,W^a\}$. Since $\Delta_1\neq \{W\}$,
\rf[bimprimitivity1] b gives that $[W,B,B]=0$ and so $B$ acts
quadratically on $V$. Since $A$ is cubic on $V$, $A$ is
quadratic on $[V,A]$ and so on $[W,AB\oplus [W^a,B]$. Hence
\ref{quadratic imprimitivity} implies
$p=2$. Thus \rf{c:b} holds. By
\rf 2, $A$ us elementary abelian and so the Lemma holds in this case.
\bcase{NA(W) trivial} Suppose that $[W,N_A(W)]= 0$ for all $W\in
\Delta_1$. \ecase
Put $p=\chr \F$. From the assumptions we have that
$C_A(\Delta_1)=1$. Thus $A$ is finite and so by \rf[belementary cubic
prop] e, $p>0$ and $A$ is a $p$-group. If $|\Delta_1|\leq 3$ we
conclude that $|\Delta_1|=p=|A|$ and the lemma holds. If
$|\Delta_1|\geq 4$, then by \rf 4, $|\Delta_1|=4$. So $p=2$ and by \rf
2, $A$ is elementary abelian. Since $A$ acts transitively and
faithfully on $\Delta_1$, this implies $|A|=4$ and
$N_A(W)=C_A(\Delta_1)=1$ for $W\in \Delta_1$. Again the lemma holds.\qed
\be{lemma}{binherit3} Let $V$ be a nearly quadratic $\F A$-module, and let $\Delta$ be a system of imprimitivity for
$A$ in $V$. Then one of the following holds:
\bl[1]
\li a There exists at most one $W \in \Delta$ with $[W,A] \ne 0$.
\li b $A$ acts trivially on $\Delta$ and quadratically on $V$.
\li c $A$ acts quadratically on $V$, $\chr \F=2$, and $|A/C_A(W)|\leq 2$ for
every $W\in \Delta\setminus C_\Delta(A)$.
\li d $A$ does not act quadratically on $V$, $A/C_A(V)$ is elementary abelian and there exists a unique
$A$-orbit $W^A\subseteq \Delta$ with $[W,A]\neq 0$. Moreover, $B: =N_A(W)$
acts quadratically on $V$, $B=C_A(\Delta)$ and one of the following holds:
\bl[1]
\li a $\chr \F=2$, $|W^A|=4$, $\dim_\F
W = 1$, $B = C_A(V)$, and $A/C_A(V)\cong C_2\times C_2$.
\li b $\chr \F=3$, $|W^A|=3$, $\dim_\F
W = 1$, $B = C_A(V)$, and $A/C_A(V)\cong
C_3$.
\li c $\chr \F=2$, $|W^A|=2$, and $C_A(W)=C_A(V)$. Moreover, $\dim_\F W/C_W(B)
= 1$ and $C_W(B)=[W,B]$.
\el
\el
\en{lemma}
\proof Suppose first that $A$ acts quadratically on $V$. Then
\ref{quadratic imprimitivity} gives that \rf b or \rf c holds.
Suppose next that $A$ does not act quadratically on $V$. Pick $W\in
\Delta$ with $[W,A]\neq 0$ and set
\[B:= N_A(W),\; U:= \bigoplus W^A \;\hbox{ and }U_0 :=
\bigoplus C_{\Delta}(A)E.\]
By \ref{binherit2} $[E,A]=0$ for all $E\in
\Delta\setminus W^A$. It follows that $V = U\oplus U_0$ and $[U_0,A] =
0$. Thus after replacing $V$ by $U$, we may assume that $\Delta =
W^A$.
If $W^A = W$ then \rf 1 holds. Thus we may
assume that $|W^A| \geq 2$.
We prove next that
\bd 1 $W$ is not the direct sum of two proper $\F B$ submodules.\ed
Suppose that $W=W_1\oplus W_2$ for some proper $\F B$-submodules $W_1$
and $W_2$. Then $V=\bigoplus W_1^A \oplus \bigoplus W_2^A$ and $A$
act non-trivial on both direct summands. But this contradicts
\ref{binherit2}.\medskip
Note that we can apply \ref{bimprimitivity2}. In particular
$A/C_A(V)$ is elementary abelian. If
\rf[bimprimitivity2]{c:a} holds, then $[W,B]=0$. Thus \rf 1 gives
$\dim_\F W=1$. Thus \rf{d:a} or \rf{d:b} hold in this case. So suppose that
\rf[bimprimitivity2]{c:b} holds. Hence $[W,B]\neq 0$ and $B$ is
quadratic on $W$. So $[W,B]\leq C_W(B)$. Pick an
$\F$-subspace $W_1\leq W$
with $W_1\cap C_W(B)=[W,B]$ and $W_1+C_W(B)=W$. Also pick an
$\F$-subspace $W_2\leq C_W(B)$ with $C_W(B)=W_2\oplus [W,B]$. Then $W=W_1\oplus W_2$ and $W_1$ and $W_2$ are $\F
B$-submodules of $W$. Thus by \rf 1, $W_2=0$ and so $C_W(B)= [W,B]$.
Let $a\in A\setminus B$ and $w\in W\setminus C_W(B)$.
Put $W_0=w\F+C_W(B)$ and $V_0=\$. Then $V=W+W^a$ and
$V_0=W_0+W_0^a$.
By \rf[trivial property] b, $Q_V(A)=[V,A]+C_V(A)$. Since
$[W_0,B]\neq 0$ we have $[V_0,B,A]\neq 0$. Thus $V_0\nleq
Q_V(A)=[V,A]+C_V(A)$ and so \rf[trivial property] a, $[V,A]\leq
V_0=W_0+W_0^a$. Thus
\[V=W\oplus W^a=W+[V,A]=W+W_0+W_0^a=W+W_0^a\]
Hence $W^a=W_0^a$, $W=W_0$ and $C_W(B)=[W,B]$ is an $\F$-hyperplane
in $W$. Thus \rf{d:c} holds.
\qed
\be{lemma}{bfield ext} Suppose $\K$ is a field with $1\neq A\leq \Aut(\K)$
and put $\E=C_\K(A)$. Suppose that $\K$ is a cubic $\E
A$-module. Then
\bl
\li a $p:=\chr \K\in \{2,3\}$.
\li b $A$ is an elemntary abelian $p$-group and $A=\Aut_\E(\K)$.
\li c $\dim_\E \K=|A|$ and $\K\cong \F A$ as an $\F A$-module.
\li d One of the following holds:
\bl[1]
\li 1 $|A|=2$, $A$ acts quadraticillay on $\K$ and $[\K,A]=\E$.
\li 2 $|A|>2$, $A$ does not act quadratically on $V$ and $[\K,A,A]=\E$.
\el
\el
\en{lemma}
\proof We consider first the case where $A$ is cyclic. Then $A=\<\sigma\>$ for some
$\sigma\in V$, $C_\K(\sigma)=\E$ and $C_\K(\sigma)$ is
$1$-dimensional over $\E$. Since $\sigma$ acts cubically on $V$ the
Jordan canonical form of $\sigma$ on $\K$ over $\E$ shows that $\sigma$
has a unique Jordan block and $\dim_\E\K\leq 3$.
Note that $\K$ over $\E$ is a finite Galois
extention since the fixed field of the Galois group $\Aut_\E(\K)$ is
$\E$, so $|A|=|\Aut_\E(\K)=\dim_\E\K\in \{2,3\}$. Since $A$ is cubic we conclude
from \rf[belementary cubic prop] e $\chr \E=|A|$. Let $k\in \K\setminus
[\K,A]$. Then it is easy to see that $k^A$ is an $\E$-basis for $\K$
and so $\K\cong \E A$ as an $\E A$-module.\medskip
We now consider the general case. Let $1\neq \sigma \in A$ and put
$\mb L=C_\K(\sigma)$. Then by the cyclic case $p=\chr \K=|\sigma|=\dim_{\mb L} \K$.
Suppose that $p=2$ and $A$ acts quadratically on $\K$ or that $p=3$. If $p=2$,
then $\mb L=[\K,\sigma]$ and if $p=3$ then $\mb L=[\K,\sigma,\sigma]$. So in any case $\mb L\leq C_\K(A)=\E$. Thus $\dim _\E \K$ =p and so also
$|\Aut_{\mb L}(\K)|=p$. Thus $A=\<\sigma\>$ and the lemma holds.
So suppose that $p=2$ and $A$ is not quadratic on $\K$. Then $A\neq
\<\sigma\>$ and there exists $1\neq \mu\in \Aut(\K)$ with $\mu\neq
\sigma$. Then $[\mb L,\mu]\leq [\K,A,A]\leq C_\K(A)=\E\leq \mb L$ and so
$\mb L^\mu=\mb L$. Since $\Aut_{\mb L}(\K)=\<\sigma\>$, $[\mb L,\mu]\neq 0$. The cyclic
case applied to $\mb L$ in place of $\K$ shows that $\dim_{\L,\mu]}\mb L=2$. We
conclude that $[\mb L,\mu]=\E$ and so $\dim_\E\K=4$. It follows that
$A=\Aut_\E(\K)$ and $|A|=4$. Since $\sigma^2=1$ for all $\sigma\in A$, $A$ is
elementary abelian. Finally if $k\in \K\setminus [\K,A]$, then $k^A$
is a $\E$-basis for $\K$ and so $K\cong \F A$ as an $\F A$-module. \qed
\be{lemma}{u:dimension commutator} Let $V$ a semi-linear, cubic $\K
A$-module. Put $\E=\K_A$ and suppose that $\E\neq \K$. Then one of the following holds.
\bl[1]
\li a $|A/C_A(V)|=\chr \E=\dim_\E \K=2$, $A_\K=C_A(V)$,$A$ is quadratic on $V$ and a
as an $\E A$-module, $V$ is the
direct sum of copies of $\K$.
\li b $ |A/C_A(V)|=\chr \E=\dim_\E\K=3$, $A_\K=C_A(V)$, $A$ is not quadratic on $V$ and as an $\E A$-module, $V$ is the
direct sum of copies of $\K$.
\li c $|A/C_A(V)\cong C_2\times C_2$, $A_\K=C_A(V)$, $\chr \E=2$, $\dim_\E \K=4$,
$A$ is not quadratic on $V$ and as an $\E A$-module, $V$ is the
direct sum of copies of $\K$.
\li d $|A/A_\K|=\chr \E=\dim_\E \K=2$, $A$ is not quadratic on $V$, $A/C_A(V)$
is elementary abelian and there
exists an $\E A$ submodule $W$ of $V$ such that $V\cong W\otimes_\E \K$ as an
$\E K$ module, $A=C_A(W)A_\K$ and $A_\K$ acts quadratically on $V$ and $W$.
\el
\en{lemma}
\proof We may assume that $A$ acts faithfully on $V$ and $V\neq 0$.
\bcase{ak=1} Suppose that $A_\K=1$.\ecase
Let $\ca B$ be a subset of $C_V(A)$ maximal with respect to being linearly
independent over $\K$. Note that $\ca B$ exists by Zorn's Lemma. Let $U$ be the
$\K$-span of $\ca B$. Since $A$ is cubic, $C_V(A)\neq 0$ and so $\ca B\neq
\emptyset$. Let $b\in \ca B$, then $b\K$ is isomorphic to $\K$ as an $\E
A$-module. So $A$ acts cubicly on $\K$ and we can apply \ref{bfield
ext}. It follows that $|A|\leq 4$ and $A$ is elementary abelian of order at most $4$. Moreover,
either $|A|=2$ and $[\K,A]=\E$ or $|A|>2$ and $[\K,A,A]=\E$.
Suppose that
$U\neq V$. Then $V/U$ has an $\E A$-submodule isomorphic to $\E$ and so
$[V/U,A]\neq 0$ and if $|A|>2$ then $[V/U,A,A]\neq 0$. So if $|A|=2$ we can $v\in
[V,A]$ with $v\notin U$ and if $|A|>2$ we can choose $v\in [V,A,A]$ with $v\not
in U$. If $|A|=2$, then $\chr \E=2$ and $A$ acts quadratically on $V$. So in any
case $v\in C_V(A)$. Since $U$ is a $\K$-subspace, $\ca B\cup\{v\}$ is linearly
independent over $\K$, a contradiction to the maximality of $\ca B$.
Thus $U=V$ and so $V=\bigoplus_{b\in \ca B} b\K$ is a direct sum of copies of
$\K$ as an $\E A$-module. Thus one of \rf a, \rf b and \rf c holds.
\bcase{ak neq 1} Suppose $A_\K\neq 1$.\ecase
Note that $[V,A_\K,A_\K]$ is a $\K$-subspace of $V$ centralized by $A$. Thus
$[V,A_\K,A_\K]=0$. Moreover, $A/A_\K$ acts quadratically and
faithfully on the non-trivial $\K$-space $[V,A_\K]$ and so \rf{ak=1} shows that $|A/A_\K|=2$ and $\dim_\E\K=2$. Let
$a\in A$. By \ref{a2 and com}
$[V,a^2]=[V,a,a]$. Since $a^2\in A_\K$ we conclude that $[V,a^2]$ is a $\K$-subspace
centralized by $A$. Thus $[V,a^2]=0$ and $A$ is elementary abelian. Let $a\in
A\setminus A_\K$. Put $W=C_V(a)$. Then $W$ is an $\E$-subspace of $V$. By
\rf{ak=1} applied to $\$, $W=[V,a]$. Hence $[W,A]=[V,a,A]\leq C_V(A)\leq W$
and so $W$ is an $\F A$-submodule of $V$. Since $a\in C_A(W)$,
$A=C_A(W)A_\K$. By the universal property of the tensor product, there exists an
$\F A$-homomorphism $\rho: W\otimes_\E\K\to V$ with $(w\otimes k)\rho=wk$ for
all $w\in W$ and $k\in \K$. By \rf{ak=1} applied to $\$, $\rho$ is a
bijection. Thus \rf d holds in this case.\qed
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%
\section{ Tensor Decomposition}
%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%
\be{lemma}{char scalar} Let $\K$ be field and $V$ a $\K$-space of
dimension at least $2$ and $\F$ a subfield of $\K$ and $\alpha\in
GL_\F(V)$ with $v\K\alpha=v \K$ for all $v\in V$. Then there exists
$k\in \K$ with $v\alpha=vk$ for all $v\in V$.
\en{lemma}
\proof Let $0\neq v \in V$. Then by assumption $v\alpha=vk_v$ for a
unique $k_v\in \K$. Let $0\neq w\in \K$. It suffices to show that $k_v=k_w$.
Suppose first that $v\K\neq w\K$. We have
\[vk_v+wk_w=v\alpha+w\alpha=(v+w)\alpha=(v+w)k_{v+w}=vk_{v+w}+wk_{v+w}\]
Since $v$ and $w$ are linearly independent over $\K$ we conclude that $k_v=k_{v+w}=k_w$.
Suppose next that $v\K=w\K$. Since $V$ is at least two dimensional
over $\K$ ther exists $u\in V\setminus v\K$. Thus by the preceeding
case $k_v=k_u=k_w$.\qed
\be{definition}{def:commutator dependent} Let $\K$ be a field, $G$
group and $V$ a quadratic $\K G$-module.
\bl
\li a We say that $G$ acts $\K$-commutator dependent on $V$ if
$[v,a]\K=[v,b]\K$ for all $a,b\in G\setminus C_V(G)$. Let
$\lambda:G\to (\K,+)$ be a homomorphism.
\li b We say that $G$ acts $\lambda$-dependent on $V$ if there exists
$\a\in \End_\K(V)$ with $\alpha^2=0$ and $[v,a]=v\a.a\lambda$ for all
$a\in G$ and $v\in V$.
\el
\en{definition}
\be{corollary}{dependent commutator} Let $\K$ be a field, $G$ a group
and $V$ a $\K G$-module. Then $G$ acts $\K$-commutator dependent on
$V$ iff $G$ acts $\lambda$-dependent on $V$ for some homomorphism
$\lambda:G\to (\K,+)$.
\en{corollary}
\proof If $G$ acts $\lambda$-dependent on $V$, then cleary $G$ is
$\K$-commutator dependent on $V$.
Suppose now that $G$ is $\K$-commutator dependent on $V$and fix $a\in G\setminus
V$. Define $\a:V\to V, v\mapsto [v,a]$. Since $G$ is quadratic on $V$,
$\alpha^2=0$.
Let $b\in G\setminus C_G(V)$. Then by assumption $[v,a]\K=[v,b]\K$
for all $v\in V$. Thus $C_V(a)=C_V(b)$ and $[V,a]=[V,b]$. Hence we
obtain $\K$-isomorphisms
\[\beta: V/C_V(a)\to [V,a], v+C_V(a)\to [v,a] \text{ and } \gamma:
V/C_V(a)\to [V,a], v+C_V(a)\to [v,b]\]
Put $\delta=\beta\gamma^{-1}$. From $[v,a]\K=[v,b]\K$ for all
$v\in V$ we conclude that $u\K\delta=u\K$ for all $u\in
V/C_V(a)$. Thus by \ref{char scalar} there exists $k_b\in \K$ with
$u\delta=uk_b$ for all $u\in V/C_V(A)$. Hence
$u\beta=uk_b\gamma=u\gamma k_b$ for all $u\in V/C_V(A)$ and so
$[v,b]=[v,a]k_b=v\alpha k_b$ for all $v\in V$. For $b\in C_G(V)$ put
$k_b=0$.
Define $\lambda:G\to \K, b\mapsto k_b$. Then for all $v\in V$ and
$b\in G$, $[v,b]=v\alpha.a\lambda$. Let $b,c\in G$. Using $v.bc=vbc$
we compute that $(bc )\lambda=b\lambda+c\lambda$. Thus $\lambda$ is a
homomorphism and $G$ acts $\lambda$-dependent on $V$.
\qed
\be{lemma}{char dependent} Let $\K$ be a field , $G$ a group,
$\lambda:G\to (\K,+)$ a homorphism and $V$ a $\lambda$-dependent $\K
G$-module. Let $W_\lambda$ be the $\K G$-module with $V=\K^2$ as
$\K$-space and $(k,l)a=(k,l+k.a\lambda)$. Then $V=W\oplus C$, where $W$
and $C$ are $\K G$ submodules of $V$ such that $G$ centralizes $C$ and
$W$ is isomorphic to a direct sum of copies of $W_\lambda$.
\en{lemma}
\proof By definition of $\lambda$-dependent there exists $\alpha\in
\End_\K(V)$ with $\a^2=0$ and $[v,a]=v\alpha.a\lambda$ for all $v\in
V, a\in G$. Choose $\ca V\subseteq V$ such that $(v+C_V(G))_{v\in \ca
V}$ is a basis for $V/C_V(A))$. For $v\in V$ put
$W_v=\_\K$. Let $C$ be a $\K$-subspace of $C_V(G)$ with
$C_V(G)=[V,G]\oplus C$. Then it is readily verfied that $W_v$ is a $\K$
subspace of $V$ isomorphic to $W^\lambda$ and $V=C\oplus \bigoplus_{v\in \ca
V} W_v$.\qed
\be{definition}{udef: tensor decomposition} Let $\F$ be a field and $V$ an
$\F$-space.
\bl
\li a A {\em tensor decomposition} $\ca V$ of $V$ is a tuple $(\Phi,\K,(V_i,i\in I),)$
where is a field extension of $\F$ with $\F\leq \K$, $(V_i,i\in I)$
is a finite
family of pairwise disjoint $\K$-spaces and
$\Phi:\bigotimes_{i\in I}^\K V_i \to V$ is an $\F$-isomorphism.
\li b A tensor decomposition $\ca V$ (as in \rf a) is called {\em proper} if $|I|>1$ and $\dim_\K V_i\geq 2$
for all $i\in I$.
\el
\en{definition}
\be{notation}{unot:tensor decomp} Let $\K$ be a field and $(V_i,i\in I)$ be a finite
family of pairwise disjoint $\K$-space. For $J\subseteq I$ let
$V_J:=\bigotimes^\K_{j\in J} V_j$ and $V^J:=\{\bigotimes^\K_{i\in
I\setminus J} V_i=V_{I\setminus J}$. $(u_i, i\in I)$ be a tuple of
elements such that there exists $\pi\in \Sym(I)$ with $u_{i\pi}\in
V_i$. The $\otimes_{i\in I} u_i$ denotes the element $\otimes_{i\in
I} u_{i\pi} $ in $V_I=\otimes^\K_{i\in I} V_i$. ( Note here that
since the $V_i$ are pairwise disjoint, $\pi$ is uniquely determined by
$(u_i,i\in I)$.) Often we also just wrote $\otimes u_i$ for
$\otimes_{\in I} u_i$. In the same spirit we identify $V_J\otimes V^J$
with $V_I$.
\en{notation}
\be{definition}{udef: g tensor decomposition}
Let $G$ be a group, $\F$ a field and $V$ an $\F G$-module.
\bl
\li a A $G$-invariant tensor decomposition $\ca T=(\Phi,\K,
(V_i,i\in I), \sigma, (g_i,g\in G,i\in I)($ of $V$ is a tuple consisting of
\bl[i]
\li i a tensor decomposition $(\Phi,\K,(V_i,i\in I))$ of $V$;
\li {ii} an action $I\times G\to I,(i,g)\to ig$ of $G$ on $I$;
\li{iii} an homomorphism $\sigma: G\to \Aut_\F(\K)$; and
\li{iv} for each $g \in G$ and $i\in I$ a $g\sigma$-linear map $g_i:\, V_i\to V_{ig}$.
\el
such that
\bl[i]
\li a $(\otimes_{i\in I} v_i)\Phi g=(\otimes_{i\in I}
v_{i}g_i)\Phi$ for all $g\in G$ and
\li b for each $g,h\in G$ and $i\in I$ there exists an element $\lambda_{i,g,h}\in \K$
with $g_ih_{ig}=(gh)_{i} \lambda_{i,g,h}$.
\el
\li b A $G$-invariant tensor decomposition as in \rf a is called strict if
$\lambda_{i,g,h}=1$ for all $i,g,h$, that is if
\[g_ih_{ig}=(gh)_{i}\] for all $g,h\in G$ and
$i\in I$.
\li c A $G$-invariant tensor decomposition is called regular if the action of $G$ on $I$ is trivial.
\li d A $G$-invariant tensor decomposition is called $\K$-linear if
$G\sigma=1$, that is if $G$ acts $\K$-linearly on $V$.
\li e A $G$-invariant tensor decomposition is ordinary if its
$\K$-linear, regular and strict.
\el
\en{definition}
\be{definition}{projective module} Let $G$ be a group, $\K$ a field,
$\sigma:G\to \Aut(\K)$ a homomorphism. A
projective $\sigma$-linear $\K G$--module is a $\K$-space $V$ together
with a map $V\times G\to V, (v,g)\to
vg$ such that
\bl[i]
\li{i}
For each $g\in G$, the map $V\to V, v\to vg$ is $g\sigma$-linear.
\li{ii} For each $g,h\in G$ there exists $\lambda_{g,h}\in \K$ with
\[v.gh=vgh\lambda_{g,h}\]
for all $v\in V$.
\el
In the case $\sigma=1$ (that is $g\sigma=\id_\K$ for all $g\in G$) a
$\sigma$-linear projective $\K G$-module is called projective $\K
G$-p-module.
\en{definition}
\be{lemma}{projective action} Let $G$ be a group, $\F$ a field
and $V$ an $\F G$-module an
\[\ca T=(\Phi,\K,(V_i,i\in I), \sigma, (g_i,g\in G,i\in I))\] a
$G$-invariant tensor decomposion. For $i\in I$ let
$\ca P_\K(V_i)=\{v_i\K\mid v_i\in V_i^\sharp\}$, be the set of
$1$-dimensional $\K$-subspaces of $V$. Then $G$ acts on
$\bigcup_{i\in I}\ca P_\K(V_i)$ via $v_i\K g=v_ig_i\K$ for all $v_i\in
V_i^\sharp$ and $g\in G$. In particular, for each $i\in I$, $V_i$ is a projective
$\sigma$- linear $\K C_G(i)$-module.
\en{lemma}
\proof This follows immediately from the definition of a $G$-invariant
tensor decomposition.
\qed
\section{Strict Tensor Decompositions}
Throughout this section we assume the following hypothesis:
\be{hypothesis}{hyp tensor} Let $G$ be a group, $\F$ a field, $V$ an $\F G$-module and \[\ca T=(\Phi,\K,(V_i,i\in I), \sigma, (g_i,g\in G,i\in I))\] a strict $G$-invariant tensor decompostion with $\Phi=\id_V$
\en{hypothesis}
\be{lemma}{induced action} Assume Hypothesis \ref{hyp tensor}.
\bl
\li a $G$ acts on $\bigcup_{i\in I} V_i$ via $vg:=vg_i$ for all $g\in G$ and $v\in \bigcup_{i\in I} V_i$, where $i$ is unique element in $I$ with $v\in V_i$.
\li b Let $i\in I$. Then $C_G(i)$ acts $\sigma$-semilinear on $V_i$.
\el
\en {lemma}
\proof This follows immediately from the definition of a strict tensor decomposition.\qed
\be{notation}{not tensor}
By \rf[induced action] a $G$ acts on $\bigcup_{i\in I} V_i$ and we can use the usual notation $C_W(B)$ and $C_B(W)$ where $W\subseteq \bigcup_{i\in I} V_i$ and $B\subseteq G$. Note that if $B$ fixes $v\in V_i$, then $B$ also fixes $i$.
By \rf[induced action] b, $V_j$ is an $\F C_G(i)$-module,
and we can use the usual notation $[V_j,B]$ where $B\subseteq C_G(i)$.
\en{notation}
\be{lemma}{elementary property} Assume Hypothesis \ref{hyp tensor}. For $i \in
I$ let $U_i$ be non-trivial $\F$-subspace of $V_i$. Suppose that there
exists $r \in J$ and $B\leq G$ such that
$$B\not\leq C_G(r) \hbox{ and }\{\otimes u_i \mid u_i \in U_i,i\in I\} \subseteq C_V(B).$$
Then $\dim_\K\< U_r\>_\K = 1$.
\en{lemma}
\proof Pick $0\ne u_i \in U_i$ and $a \in B$ with $ra \ne r$. Then
$$\otimes u_i = (\otimes u_i)a = \otimes u_ia_i$$
and so $u_ra_r = u_{ra} k$ for some $k \in \K^\sharp$.
Fixing $u_{ra}$ and
allowing $u_r$ to run through the elements of $U_r^\sharp$ shows that
$U_ra_r\leq u_{ra}\K$. Thus $\_\K=u_{ra}\K$ is
$1$-dimensional $\K$-space. Since $a_r$ is a $\K$-semilinear
isomorphism, also $\< U_r\>_\K$ is a $1$-dimensional $\K$-space. \qed
\be{lemma}{trivial action} Assume Hypothesis \ref{hyp tensor}. Suppose that
$\chr \K = p > 0$ and that $G$ is a finite $p$-group. Let $j\in I$ with
$\dim_\K V_j \geq 2$. Then $C_G(V) \leq C_G(V_j)$.
\en{lemma}
\proof Pick $h \in C_G(V)$ and for $i\in I$ pick $0\ne v_i \in V_i$. Then
$$\otimes v_i = (\otimes v_i)h = \otimes v_ih_i$$
so $v_jh_j \in v_{ih}\K$.
If $j \ne jh$, then $V_jh_j
\leq v_{jh}\K$, and since $h_j$ is a $\K$-semilinear isomomorphism, $\dim_\K V_j = 1$, a
contradiction. Hence $j = jh$ and by \ref{char scalar} $h$ acts via
scalar multiplication by a fixed
scalar $\lambda \in \K$ on $V_j$.
On the other hand, since $\chr \K = p$ and $G$ is a finite $p$-group,
$C_{V_i}(G) \ne 0$ and so $\lambda = 1$. \qed
\be{lemma}{centralizer index} Assume Hypothesis \ref{hyp
tensor}. Suppose $\ca T$ is ordinary and that
$|I|\geq 2$. Suppose that there exists $r \in I$ such that $G$ acts
non-trivially on $\ca P_\K(V_r)$. Then
$$\dim_\K \otimes_{r\neq i\in I}^\K V_ i\leq \dim_\K V/C_V(G).$$
\en{lemma}
\proof Put $W:= V^r=\otimes_{r\neq i\in I}^\K V_i$. Then $V$ and
$V_r \otimes_\K W$ are isomorphic $\K G$-modules.
Since $G$ acts non-trivially on $\ca P_\K(V_r)$ there exist $a\in G$ and $v\in V_r$ with $v\K\neq va\K$. Hence
\[(v\otimes W)a\cap v\otimes W=va_r\otimes W\cap v\otimes W=0.\]
In particular, $v\otimes W\cap C_{V_r\otimes W}(a)=0$ and
\[\dim V/C_V(G)\geq \dim V/C_V(a)\geq \dim v\otimes W=\dim W.\]
\be{lemma}{tensor and imprimitivity} Assume Hypothesis \ref{hyp
tensor}. Suppose $G$ is transitive on $I$ and $|I| \geq 2$.
Fix $r \in I$ and let $X_r$ be a proper $C_G(r)$-invariant $\K$-subspace of
$V_r$. For $h \in G$ put $X_{rh} := X_rh$ and
$$X := \otimes_\K^IX_i,\, U_i := V_i\otimes (\otimes_{j\ne i} X_j),\,U :=
\sum_{i\in I} U_i,\,\Delta := \{U_i/X \mid i \in I\}.$$
Then
\bl
\li a $U$ and $U/X$ are semi-linear $\K G$-modules,
\li b $\Delta $ is a system of imprimitivity for $G$ in $U/X$.
\li c $G$ acts transitively on $\Delta$.
\el
\en{lemma}
\proof Observe that $X_{rh} = X_{rg}$ for $g \in G_rh$ since $X_r$ is
$C_G(r)$-invariant. So $X_{rh}$ is well-defined. For $h \in G$
$$Xh = (\otimes X_i)h = \otimes X_{ih} = \otimes X_i = X$$
and similarly
$$U_ih = V_{i}h_i\otimes \bigotimes_{j\neq i}
X_{j}h_j=V_{ih} \otimes \bigotimes_{j\neq i}
X_{jh}= U_{ih}.$$
This shows that $X$ and $U$ are $G$-invariant, so \rf a holds, and that $G$
acts on $\Delta$, so \rf c holds.
Observe that $\sum_{j\neq i} U_j\leq X_i\otimes V^i$ and that $(X_i\otimes V^i)\cap U_i=X$. Thus also $U_i\cap \sum_{j\neq i} U_j=X$ and \rf b holds.\qed
\be{lemma}{quadratic and tensor1} Assume Hypothesis \ref{hyp tensor} and in
addition:
\bl[i]
\li i $|I|=2$.
\li{ii} $\ca T$ is ordinary.
\li{iii} $G$ acts quadratically on $V$ and $V_i\ne C_{V_i}(G) \ne 0$ for every $i \in I$
\el
Then
\bl[a]
\li a For all $i\in I$, $G$ acts quadratically on $V_i$ and $C_G(V)=C_G(V_i)$
\li b There exists a homomorphism $\lambda:G\to \K$ such that $G$
acts $\lambda$-dependent on each $V_i, i\in I$.
\el
\en{lemma}
\proof Let $I=\{i,j\}$. Note that
$$[V_i \otimes C_{V_j}(G),G] = [V_i,G]\otimes C_{V_j}(G) \leq C_V(G),$$
so $[V_i,G,G]\otimes C_{V_i}(G) = 0$ and thus $[V_i,G,G] = 0$. This \rf a.
Let $a,b \in G\setminus C_G(V_i)$ and $x \in
V_j$. Then $V_i\neq C_{V_i}(a)\cup C_{V_i}(b)$ and so there
exists $y \in V_i$ with $[y,a] \ne 0$ and $[y,b] \ne 0$. Then
\[[x\otimes y,a] = [x,a]\otimes [y,a] + x\otimes [y,a] + [x,a]\otimes
y\leqno(1)\]
Taking commutators with $b$ and using that $G$ acts quadratcially on $V_i,V_j$ and $V$ we get
$$0 = [x\otimes y,a,b] = [x,b]\otimes [y,a] + [x,a]\otimes
[y,b].\leqno (2)$$
By choice of $y$, $[y,a]\neq 0\neq [y,b]$ and so (2) implies
$C_{V_j}(a) = C_{V_j}(b) = C_{V_j}(G) $ and
that $ [x,a]\K= [x,b]\K$. \ref{dependent commutator} now implies
that $V$ acts $\lambda_j$-dependent with respect to $\alpha_j$ on
$V_j$ for some homomorphism $\lambda_j:G\to (\K,+)$ and some
$\alpha_j\in \End_\K(V_j)$ with $\alpha_j^2=0$. By symmetry the same
holds for $i$ in place of $j$. Without loss
$a\lambda_i=1=\a\lambda_j$. Substitution into (2) yields,
$b\lambda_i=-b\lambda_j$. In the case $a=b$ we have $1=-1$ and so
$\chr \K=2$. It follows that $\lambda_i=\lambda_j$ and the lemma is proved.
\qed
\be{lemma}{quadratic and tensor2} Assume Hypothesis \ref{hyp tensor} and in
addition:
\bl[i]
\li i $\ca T$ is proper and $\K$-linear.
\li{ii} $G$ acts transitively on $I$.
\li{iv} $|G| > 2$ and $V$ is a a faithful quadratic $\K G$-module.
\el
Then $\chr \F = 2$, $|I| = 2$ and for $i\in I$, $\dim_\K V_i = 2$,
and $[V_i,C_G(I)] = C_{V_i}(C_G(I))$ is a $1$-dimensional subspace of $V_i$.
\en{lemma}
\proof Recall from \rf[belementary cubic prop]e that $G$ is an elementary
abelian $\chr \F$-group. Put $B:= C_G(I)$ and fix $r\in I$. Then $B = C_G(r)$ since
$G$ is abelian.
Let $X_r$ be an $1$-dimensional $\K B$-subspace of $V_r$. We apply
\ref{tensor and imprimitivity} with the notation given there. Then
$\Delta$ is a system of imprimitivity for $G$ in $U/X$, so we can
apply \ref{binherit3}. Since $G$ acts transitively on $I$ and
quadratically on $U/X$, we are in case (\ref{binherit3:c}) of
\ref{binherit3}, so $\chr \F=2$, $|I| = 2$, say $I =\{1,2\}$, and
$[U_i,B] \leq X$. As $|G| > 2$ we also get that $B \ne 1$. Since the
$\K B$-modules $U_j/X$ and $V_j/X_j$ are isomorphic, $[V_j,B] = X_j$.
Pick $1\ne b \in B$ and $a \in G\setminus B$, and put $C_1:= C_{V_1}(b)$. Then
by the quadratic action of $G$,
$$[C_1\otimes V_2,b] = C_1\otimes X_2 \leq C_V(G).$$
Hence \ref{elementary property} shows that $\dim_\K C_1 = \dim_\K X_2 = 1$, so
also $\dim_\K X_1 = 1$. The quadratic action of $b$ on $V_1$ gives $C_1 = X_1$
and $\dim_\K V_1 = 2$. \qed
%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
\section{Tensor Decompositions of Homogeneous Modules}
%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%
\be{lemma}{simple tensor} Let $\F$ be a finite field of characteristic
$p$, $G$ a group and $V$ a finite dimensional $\F G$-module. Let $D$ and $E$ be subgroups of $G$ such that $[D,E] = 1$.
Suppose that $V$ is a homogeneous $\F D$-module and $X$ is a simple
$\F D$-submodule of $V$. Then the following hold, where $Y:= \Hom_{\F
D}(X,V)$ and $\K:=\End_{\F D}(X)$:
\bl\li a $\K$ is a finite field and
$Y$ is a $\K E$-module via
$$\a k:\, x \mapsto (xk)\a \;\hbox{ and }\a e:\, x\mapsto (x\a)e\quad
(x\in X,\,\a\in Y,\, k\in \K,\, e\in E).$$
\li b $X$ is an absolutely
simple $\K D$-module.\footnote{An $\F G$-module $V$ is absolutely simple if
$\End_{\F G}(V) = \F$.}
\li d There exists an $\F(D\times E)$-module isomorphism
$\Phi: X\otimes_\K Y \to V$ with $(x\otimes\alpha)\Phi=x\alpha$ for all $x\in
X$ and $\alpha\in Y$.
\li c For $\K$-subspaces $Z \leq Y$ the map $Z\mapsto(X\otimes Z)\Phi$ is an
$E$-invariant
bijection between the $\K$-subspaces $Z$ of $Y$ and the $\F D$-submodules of
$V$
with inverse $U\mapsto\Hom_{\F D}(X,U)$, $U$ an $\F D$-subspace of $V$.
\li e $V$ is a simple $\F DE$-module iff
$Y$ is a simple $\K E$-module.
\li f $\End_{\F D}(V)\cong \End_\K(Y)$ and
$\K\cong \op Z(\End_{\F D}(V))$.
\el
\en{lemma}
\proof \rf a: By Schur's Lemma $\K $
is a divisionring. As $\dim_\F X$ is finite, also $\dim_\F \K$ is
finite. Now the finiteness of $\F$ shows that $\K$ is finite, so by
Wedderburn's Theorem $\K$ is a finite field. The remaining assertions in \rf
a are readily verified.
\rf b: Since $\F$ and $\K$ are commutative we have $\F\leq \K$ and $\K\leq
\End_{\K D}(X)$, so $\K \leq \End_{\K D}(X)\leq \End_{\F D}(X)=\K$.
\rf d: Let $x\in X, \a\in Y$ and $k\in \K$. By definition of $\alpha k$ we have $x.\alpha k=x\a k$ and so the map $X\times Y\to V, (x,\a)\mapsto x\alpha$ is $\K$-balanced. Hence by the universal property of tensor products there exists a unique $\mb Z$-linear map
$$\Phi:\, X\otimes_\K Y \to V \hbox{ with }x\otimes \alpha \mapsto x\alpha.$$
It is evident that this map is $\F$-linear.
Let $(d,e)\in D\times E$. Note that $V$ is an $\F(D\times E)$-module via $v(d,e)=v.de$. Then
\[(x\otimes
\alpha)(d,e)\Phi=(xd\otimes \alpha e)\Phi=xd.\a e = xd\a e = x\a de =
x\a.de= (x\otimes \alpha)\Phi.de=(x\otimes \alpha)\Phi(d,e),\]
so $\Phi$ is $\F(D\times E)$-linear.
\rf c: Let $W$ be an $\F D$-submodule of $V$.
Since $V$ is
homogenous as an $\F D$-module there exist simple $\F D$-submodules
$X_1,\ldots X_n$ with $W=\bigoplus_{i=1}^n X_i$, $X_i\cong X$ as an $\F D$
module. Choose $\alpha_i\in Y$ with $X\alpha_i=X_i$ and set
$Z :=\sum_{i=1}^n \alpha_i\K\leq Y$. Then $(X\otimes\alpha_i) \Phi=X_i$ and so
$(X\otimes_{\K}
Z)\Phi=W$. Moreover,\[Z\leq \Hom_{\F D}(X,W)\cong \bigoplus_{i=1}^n \Hom_{\F
D}(X,X_i)\cong\bigoplus_{i=1}^n \Hom_{\F D}(X,X)\cong \K^n.\]
It is evident that the elements $\a_1,\ldots,\a_n$ are $\K$-linearly independent in
$Y$. Hence $\dim_\K Z = n$, so $Z\cong \K^n$ and $Z = \Hom_{\F D}(X,W)$. This
shows that $Z\mapsto (X\otimes Z)\Phi$ gives a a bijection between the
$\K$-subspaces of $Y$ and the $\F D$-submodules of $V$. Because of \rf d this
bijection is $E$-invariant.
\rf e: This is a direct consequence of \rf c since a $\K$-subspace of $Y$ is
$E$-invariant if and only if the corresponding $\F D$-submodule of $V$ is
$E$-invariant.
\rf f: Let $x \in X$, $\alpha\in Y$ and $\beta\in \End_{\F D}(V)$. Then the
map
$$\a \b:\, x \mapsto x\a\b$$
is in $Y$, and $Y$ is an $\End_{\F D}(V)$-module. Moreover, for $k \in \K$
\[ x.\alpha\beta k =xk.\a \b = xk\a\b = x.\a k.\b = x.\a k\b.\]
Hence $\alpha\beta k=\alpha k \beta$ and $\End_{\F D}(V)$ acts $\K$-linearly on $Y$. Hence we obtain a ring
homomorphism $\End_{\F D}(V)\to \End_\K Y$.
Observe that $X\otimes_\K Y$ is an $\End_\K(Y)$-module via
$$(x\otimes \a)\delta=x\otimes \a\delta \quad (x\in X,\,\a\in Y,\, \delta\in
\End_\K(Y)).$$
So by \rf d $V$ is an $\End_{\K}(Y)$-module with $(x\otimes
\alpha)\Phi\delta := (x\otimes \a)\delta\Phi$. Since $\Phi$ is an
$\F(D\times E)$-module homomorphism this action of $\End_{\K}(Y)$ on
$V$ is $\F D$-linear. Hence, we obtain a ring homomorphism
$\End_\K(Y)\to\End_{\F D}(V)$ which is inverse to the one from the
previous paragraph. Thus the first statement in \rf f holds. Since
$\op Z(\End_\K(Y))\cong \K$, the second statement follows from the first.
\qed
\be{lemma}{pure tensors} Let $\F$ be a finite field of characteristic
$p$, $G$ a group and $V$ a finite dimensional, simple $\F G$-module. Let
$(D_i,\,i\in I)$ be a finite family of subgroups of $G$ with
$G=\$ and $[D_i,D_j]=1$ for all $i,j\in I$. Put $\K:=\End_{\F G}(V)$. Then the following hold:
\bl
\li a For every $i \in I$ there exists an absolutely simple $\K D_i$-module
$V_i$ isomorphic to a $\K D_i$-submodule of $V$, and there exists a
$\K(\kreuz_{i\in I}D_i)$-isomorphism $\Phi: \otimes^{I}_\K V_i\to V$ (where
$D_i$ acts trivially on $V_j$ for $i\ne j$).
\li b For $0\neq v\in V$ the following two statements are equivalent:
\bl[1]
\li 1 For all $i\in I$, $v\K D_i$ is a simple $\K D_i$-submodule of $V$.
\li 2 There exist $0\neq v_i\in V_i$ with $v=(\otimes v_i)\Phi$.
\el
\el
\en{lemma}
\proof Replacing $\F$ be $\K$ we may assume that $V$ is an absolutely simple
$\F G$-module and so $\F=\K$. By induction on $|I|$ we may also assume that
$|I|=2$, say $I=\{1,2\}$. Let $V_1$ be a simple $\K D_1$-submodule of $V$ and
put $V_2:= \Hom_{\K D_1}(V_1,V)$.
\rf a: Let $\E :=\End_{\K D_1}(V_1)$. By \ref{simple tensor} there
exists a $\K(D_1\times D_2)$-isomorphism
$$\Phi:\, V_1\otimes_\E V_2\to V\hbox{ with }(w\otimes \a)\Phi=w\a\quad (w \in
V_1,\,\a \in V_2).$$
It follows that $\End_{\K(D_1\times D_2)}(V_1 \otimes_\E V_2) \cong \End_{\K
G}(V)$.
Since $V$ is absolutely simple and $\K \leq \E \leq \End_{\K(D_1\times
D_2)}(V_1 \otimes_\E V_2)$, we get that $\K = \E$, and $V_1$ is an
absolutely simple $\K \D_1$-module. By symmetry any simple $\K D_2$ submodule
of $V$ is an absolutely simple module.
Let
$0\neq v_1\in V_1$. Then, again by \ref{simple tensor}, $V_2$ is isomorphic
to the $\K D_2$-submodule $(v_1\otimes V_2)\Phi$ of $V$, and $V_2$ is a
simple $\K D_2$-module since $V$ is a simple $\K G$-module. It follows that $(v_1\otimes V_2)\Phi$
and so also $V_2$ is an absolutely simple $\K D_2$-module.
\rf b: Let $0\neq v\in V$. Suppose first that \rf{b:1} holds. Since $V$ is a
homogeneous $\K D_1$-module, there exists an $\K D_1$-isomorphism $\a \in V_2$
such that $V_1\a = v\K D_1$. Put $v_1=v\alpha^{-1}$. Then $(v_1\otimes
\a)\Phi=v$.
Suppose next that \rf{b:2} holds. Then $v\K
D_1=(v_1\otimes v_2)\Phi\K D_1=(V_1\otimes v_2)\Phi\cong V_1$ as $\K
D_1$-module. By symmetry $v\K D_2\cong V_2$ as $\K D_2$-module. Thus \rf{b:1}
holds.\qed
\be{lemma}{tensor decomposition for homogenous} Let $\F$ be a finite field of
characteristic $p$, $G$ be a finite group,$V$ is a finite dimensional $\F G$-module and $I$ a finite $G$-set. Further
let $T$ be a $p$-subgroup of $G$ and $(D_i,\,i\in I)$ be a family of
subgroups of $G$. Put $D:=
\$. Suppose that
\bl[i]
\li{i} $D_i^h=D_{ih}$ and $[D_i,D_j]=1$ for all $i\neq j\in I,h\in H$, and
\li{ii} $V$ is homogeneous as an $\F D$-module.
\el
Put $\K := \op Z(\End_{\F D}(V))$, If $V$ is a simple $\F D$-module let
$J=I$, otherwise let $J:=I\uplus\{0\}$. View $J$ has a $G$-set
with $G$ fixing $0$. Then there exist
$\K D_i$-modules $V_i,\, i\in I$, a finite dimensional
$\K$-space $V_0$ and a
tensor decomposition $\ca T=(\Phi,\K, (V_j,j\in J), \sigma,(g_j,j\in J,g\in
G)) $ such that the following hold:
\bl
\li a $V_j$ is an absolutely simple $ \K D_j$-module for $j \ne 0$, and
$V_0$ is a trivial $\K D$-module. Moreover, every simple
$D_j$-submodule of $V$ is isomorphic to $V_j$ as a $\K D_j$-module.
\li b $\Phi: \otimes_\K^J V_j\to V$ is a $\K(\kreuz_{i\in
I}D_i)$-module isomorphism (where $D_i$ acts trivially on $V_j$ for $j\neq i$).
\li c $\ca T$ restricted to $T$ is
strict.
\el
\en{lemma}
\proof To simplify notation we assume without loss that $V$ is a faithful
$\F G$-module and that $G$ is subgroup of $\GL_\F(V)$.
Let $D_0:=\GL_{\F D}(V)$, and for $j\in J$ let $R_j$ be the subring
of $\End_\F(V)$ spanned by $\K$ and $D_j$, $j\in J$.
By \rf[simple tensor] f,(\ref{simple tensor:e}) (with $D_0$ in place of $E$)
$V$ is a simple $\F DD_0$-module and so $V$ is an absolutely simple $\K
DD_0$-module. Thus \ref{pure tensors} implies:
\bd 1
There exist absolutely simple $\K D_j$-modules $V_j$ and a $\K(\kreuz_{i\in
J} D_j)$-isomorphism
$$\Phi: \otimes_{\K}^{J} V_j\to V.$$\ed
Let $\a_j$ be canonical ring homomorphism from $\K D_j$ onto $ R_j$.
From $(\otimes v_jd_j)\Phi=(\otimes v_j) \Phi.\prod d_j$ for all $v_j\in
V_j,d_j\in D_j$ we conclude that $(\otimes v_ja_j)\Phi=(\otimes v_j)
\Phi.\prod a_j\alpha_j$ for all $v_j\in V_j, a_j\in \K D_j$. This
implies $\ker\a_j=\Ann_{\K D_j}(V_j)$ and we conclude that
\bd{1.1} $V_j$ can be view as simple $R_j$-module such that $(\otimes
v_jr_j)\Phi=(\otimes v_j) \Phi.\prod r_j$ for all $v_j\in V_j, r_j\in
R_j$.\ed
The number of elements of the form $(\otimes v_j) \Phi$, $0\neq v_j\in V_j$,
is not divisible by $p$ and so we
can choose $0\ne w_j\in V_j, j\in J$, such that $w:=(\otimes w_j)\Phi$
is centralized by $T$.
Let $g\in G$ and $j\in J$. By \rf[pure tensors] b, $wR_j$ is a simple $R_j$-module. Since $g$ normalizes $\K$ and
$D_j^g=D_{jg}$ we have $R_j^g=R_{jg}$ and so $wR_jg$ is a simple
$R_{jg}$-module. Also $wR_jg=wgR_{jg}$ and so for $j\in J$, $wgR_{jg}$ is a
simple $R_{jg}$-module. Thus by \rf[pure tensors]b, there exist $0\neq u_j\in
V_j$, $j\in J$, with $wg=(\otimes u_j)\Phi$.
If $g\in T$ then $wg=w$ and so we can choose $u_j=w_j$.
Let $v_j\in V_j$. Since $V_j$ is a simple
$R_j$-module there exists $r_j\in R_j$ with $v_j=w_jr_j$. Next we show:
\bd 2 Let $r_j,s_j\in R_j$ with $w_j r_j = w_js_j$. Then $u_{jg}r_j^g = u_{jg}s_j^g$.\ed
Put $t_j=r_j-s_j$. Then $w_jt_j=0$ and by \rf{1.1} $wt_j=0$. Thus
$(\otimes u_j)\Phi t_j^g=wgt_j^g=wt_jg=0$. Since $t_j^g\in R_{jg}$
we conclude from \rf{1.1} that $u_{jg}t_j^g=0$ and so $u_{jg}r_j^g=u_{jg}s_j^g$.
\medskip
We now define
\[g_j:\, V_j\to V_{jg}\hbox{ with }v_j\to u_{jg}r_j^g,\hbox{ where }r_j
\in R_j \hbox{ and }v_j = w_jr_j.\]
Using \rf 2 we get
\bd 3 $g_j$ is independent from the choice of $r_j$.\ed
Clearly $g_j$ is a homomorphism between the additive group $V_j$ and $V_{jg}$. Next we
define a homomorphism $\sigma:\, G \to \Aut(\K)$. Observe that for
fixed $g \in G$,
$$g\sigma:\,\K \to \K \hbox{ with } k\mapsto k^g$$
is an element of $\Aut_\F(\K)$ and so $g \mapsto g\sigma$ defines the desired
homomorphism $\sigma$.
Let $k\in\K$. Since $v_jk=w_jr_jk=w_j.r_jk$ and
$r_jk\in R_j$, the definition of $g_j$ shows that
$$v_jkg_j=u_{jg}
(r_jk)^g=u_{jg}r_j^gk^g=v_jg_j(k.g\sigma).$$ Hence $g_j$
is $g\sigma$-linear.
To verify \rf[udef: g tensor decomposition]{a} we
compute
$$\begin{array}{lllll}
(\otimes v_jg_j)\Phi &= (\otimes u_{jg}r_j^g) \Phi
= (\otimes u_j) \Phi \prod r_j^g
= wg (\prod r_j)^g = w (\prod
r_j) g\\ &= (\otimes w_j)\Phi (\prod r_j)g = (\otimes w_jr_j)\Phi g =
(\otimes v_j)\Phi g.
\end{array}\leqno{(*)}
$$
This is \rf[udef: g tensor decomposition]{a}.
Let $g,h\in G$ and put $v:=\otimes v_j$. Note that $v\Phi.gh=v\Phi gh$ and so
using (*) three times
\[ (\otimes v_j(gh)_j)\Phi=v\Phi.gh=v\Phi gh = (\otimes v_jg_j)\Phi h=(\otimes v_jg_jh_{jg})\Phi\]
Since $\Phi$ is bijective, this implies:
\[ \otimes v_j(gh)_j= \otimes v_jg_jh_{jg}\]
Thus $v_j(gh)_j\K=v_jg_jh_{jg}\K$. Fix
$i,g$ and $h$ and put $\delta=g_jh_{jg}(gh)_j^{-1}$. Then
$\delta:V_j\to V_j$ is $\K$-linear and $v_j\delta\K=v_j\K$ for all
$v_j\in V_j$. If $\dim_\K V_j\geq 2$ we conclude from \ref{char
scalar} $\delta$ acts as a scalar $\mu$ on $V_j$. Obviously
the same is true if $\dim_\K V_j=1$. Thus
$g_jh_{jg}=\mu(gh)_j=(gh)_j\lambda$ where
$\lambda=\mu.gh\sigma$. Hence \rf[udef: g tensor decomposition]{b}
holds.
Therefore $\ca T= (\Phi,\K,(V_j,j\in J),\sigma,(g_j,j\in J,g\in G)$ is a $G$-invariant tensor decomposition of $V$.
Let $g\in T$. Recall that we chose $w_j = u_j$ in this case. Hence for $v_j=w_j$ we can choose $r_j=1$ and so $w_jg_j=w_{jg}$. For $a,b\in T$ we conclude
$$w_j a_jb_{ja}=w_{ja}b_{ja} = w_{jab} =
w_j(ab)_j,$$
and $\lambda_{k,a,b} = 1$. Thus \rf c holds.
\qed
We remark that \rf[tensor decomposition for homogenous] c maybe false if $\K$ is infinite. Indeed, let $\F$ be a finite field of
characteristic $2$ and $\E=\F(t)$ with $t$ transcendental over
$\E$. Put $\K=\E(t^2)$, $V=\E\otimes_\K\E$ and let $\alpha\in \GL_\K(V)$
with $(k\otimes l)\alpha=kt\otimes lt^{-1}$ for all $k,l\in \E$. Then
$\alpha^2=1$ and so $\<\alpha\>$ has order two. Moreover, it is easy
to verify that the tensor decomposition $\E\otimes_\K\E$ is not
strict for $\<\alpha\>$.
%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%
\section{Tensor Products and Nearly Quadratic Modules}
%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%
Throughout this section we assume
\be{hypothesis}{hyp:strict tensor} $p$ is a prime, $\K$ is a field of characteristic
$p$, $A$ is a finite $p$-group, $J$ is a finite
$A$-set, $V$ is an $\F A$-module and $\ca T=(\Phi,\K, (V_j,j\in J), \sigma, (g_j,\mid, g\in G,j\in J))$ is a strict tensor decomposition with $\Phi=\id_V$.
The fixed field of $A$ in $\K$ is
denoted by $\K_A$.
\en{hypothesis}
\be{lemma}{nearly quadratic and tensor} Suppose $\ca T$ is proper
and ordinary, that $J=\{1,2\}$ and that $V$ is a nearly quadratic $\F
A$-module but not a quadratic $\F A$-module. Then the following hold for $j
\in J$:
\bl
\li z $A$ acts quadratically and non-trivially on $V_j$. In
particular, $A$ is elementary abelian.
\li a $[V_j,A]=C_{V_j}(A)$ is a $\K$-hyperplane of $V_j$.
\li b $[z\F ,A]=[V_j,A]$ for all $z\in V_j\setminus [V_j,A]$.
\li c $Q_V(A) =[V_1,A]\otimes V_2+V_1\otimes[V_2,A]$ is a $\K$-hyperplane of
$V$.
\li d One of the following holds:
\bl[1]
\li a $C_V(A)=[V_1,A]\otimes [V_2,A]$ and if $\F=\F_p$ then $A=C_A(V_1)C_A(V_2)$.
\li b $\dim_\K C_V(A)=2$, $\chr \F\neq 2$, $\dim_\K V_1=2=\dim_\K V_2$, and $V_1$ and
$V_2$ are isomorphic as $\K A$-modules.
\el
\li e If $\F=\F_p$ then $A$ induces on $C_{\SL_\K(V_j)}([V_j,A])$ on $V_j$.
\el
\en{lemma}
\proof As $A$ is not quadratic on $V$, by \ref{trivial property} $Q_V(A) =
[V,A]+C_V(A)$ is the largest quadratic $\F A$-submodule of $V$ and $V\ne
Q_V(A)$. Observe that $Q_V(A)$ is $\K$-subspace of $V$.
Let $J=\{i,j\}$ and put $C_i=C_{V_i}(A)$.
\bd 0 $V_i\otimes C_j$ is as $\K A$-module is a direct sum of $\dim_\K C_j$ copies of $V_i$.\ed
Here with a copy of $V_j$ we mean a $\K A$-module isomorphic to
$V_i$. Let $\ca B$ be a $\K$-basis for $C_j$. Then $V_i\otimes C_j=\bigoplus_{b\in \ca B} V_i\otimes b$ and each $V_i\otimes b$ is isomorphic to $V_i$ as an $\K A$-module.
\bd 1 $A$ acts non-trivially on $V_j$.\ed
Suppose $A$ centalizes $V_j$. Then $C_j=V_j$ and since $\dim_\K V_j\geq 2$ we conclude that from \rf 0 and \ref{binherit2}, that $A$ centralizes $V_i$ and $V$, a contradiction.
\bd{2} $V_i$ is not the direct sum of two proper $\K A$-submodules.
\ed
Suppose $V_i=X\oplus Y$ with $X$ and $Y$ proper $\K A$ submodules of $V_i$. Then $V=X\otimes V_j\oplus X\otimes V_j$ and so by \ref{binherit2}, $A$ centralizes on the summands, say $X\otimes V_j$. So by \ref{trivial action}, $A$ centralizes $V_j$, a contradiction to \rf 1
\bd 3 $A$ acts quadratically on $V_i$. So $[V_i,A]\leq C_i$ and \rf a
holds.\ed
Since $\dim_\K V_j\geq 2$, there exists a proper $\K A$-submodule $X$
of $V_j$. By \rf[trivial property] d, $A$ acts quadratically on
$V/(V_i\otimes X)$ or on $V_i\otimes X$. Since $V/(V_i\otimes X)\cong
V_i\otimes( V_j/X)$ we conclude that $G$ acts qudratically on $V_i\otimes
Y$ where $Y=X$ and $Y=V_j/X$, respectively. If $\dim _\K Y=1$, then
$V_i\cong V_i\otimes Y$ and so \rf 3 holds. If $\dim Y\geq 2$, then
\rf[quadratic and tensor1] a shows that $G$ acts quadratically on
$V_i$.
\bd 4 $C_i=[V_i,A]$\ed
By \rf 3, $[V_i,A]\leq C_i$. Let $C_i=X\oplus [V_i,A]$ and
$V_i/[V_i,A]=C_i/[V_i,A]\oplus Y/[V_i,A]$ for some $\K$-subspaces $X$
and $Y$ of
$V_i$ with $[V_i,A]\leq Y$. Then $V_i=X\oplus Y$, $Y\neq 0$ and both $X$ and $Y$
are $\K A$-submodules of $V_i$. So by \rf 2, $X=0$ and so $C_i=[V_i,A]$.
\bd 5 $C_1\otimes V_2+V_1\otimes C_2\leq Q_V(A)$.\ed
By \rf 3, $A$ is quadratic on $V_i$ and so by \rf 0 also on
$V_i\otimes C_j$. Thus $V_i\otimes C_j\leq Q_V(A)$.
\bd 6 Let $v_1\in V_1,v_2\in V_2$ and $a\in A$. Then
\[[v_1\otimes v_2,a] = [v_1,a]\otimes [v_2,a] + v_1\otimes[v_2,a] + [v_1,a]\otimes v_2.\]\ed
This is readily verified.\medskip
Since $V\neq Q_V(A)$ there exists $x_i\in V_i$ with $x_1\otimes x_2\notin Q_V(A)$. By \rf 5 we have $x_i\notin C_i$.
\bd 7 $Q_V(A)=x_1\otimes C_2+C_1\otimes x_2+C_V(A)$\ed
By \rf 5 the right-hand-side is contained in the left-hand-side of \rf 7. By the definition of nearly quadratic we have $Q_V(A)=[(x_1\otimes x_2)\F,A]+C_V(A)$. Since $[V_i,A]\leq C_i$ we conclude from \rf 6 that the left-hand-side is contained in the right-hand-side.
\bd 8 Let $t_1\in V_1$ with $t_1\otimes x_2\in Q_V(A)$. Then $t_1\in C_1$.\ed
By \rf 7 there exist $c \in C_V(A)$, $c_1\in C_1$ and $c_2\in C_2$ such that
\[t_1\otimes x_2 = x_1 \otimes c_2 + c_1 \otimes x_2 + c.\]
Taking commutators with $a$ on both sides and using \rf 6 we conclude
\[ [t_1,a]\otimes [x_2,a] + t_1\otimes[x_2,a] + [t_1,a]\otimes x_2=[x_1,a]\otimes c_2+c_1\otimes [x_2,a]\]
Hence \rf{3} gives $[t_1,a] \otimes x_2 \in V_1\otimes C_2$. Since $x_2 \notin C_2$ we
get $[t_1,a] = 0$ and thus $t_1 \in C_{V_1}(A) = C_1$.
\bd 9 $C_i$ is a $\K$-hyperplane of $V_i$\ed
Since $[V,A,A]\neq 0$ there exists a $\K$-hyperplane $H_i$ of $C_i$
with $[V,A,A]\nleq H_i\otimes V_j$.
Put $\o V_i=V_i/H_i$. Hence by \rf[trivial property] c, $V/(H_1\otimes
V_2)\cong (\o V_1)\otimes V_2$ is a nearly quadtric, but not quadratic
$\F A$-module. Thus by \rf 4,
\[\o C_i=\o{[V_i,A]}=[\o V_i,A]=C_{\o V_i}(A)\] and so we may replace
$V_i$ by $\o V_i$ and assume that $\dim_\K C_i=1$. Similarly we may
assume that $\dim_\K V_i=1$.
Assume for a contradiction that $\dim_\K V_i \geq 3$.
By \rf 2, $V_i\otimes C_j\leq Q_V(A)$ and by \rf 7, $\dim_\K
Q_V(A)/(C_i\otimes x_j
+ C_V(A)) \leq 1$. Hence $R:= V_i\otimes C_j \cap (C_i\otimes x_j + C_V(A))$
contains a
hyperplane of $V_i\otimes C_j$. Moreover, $C_i\otimes C_j \leq R$ since
$C_i\otimes C_j
\leq C_V(A)$. As $\dim_\K V_i\geq 3$, there exists $t_i \in
V_i\setminus C_j$ and
$c_j \in C_j$ such that $t_i\otimes c_j \leq R$. As $C_j$ is
$1$-dimensional, also $t_i \otimes C_j \leq R$.
From \rf{8} we get that $t_i\otimes x_j \notin Q_V(A)$, so \rf 7 applies with
$t_i$ in place of $x_i$; i.e., $Q_V(A) = t_i\otimes C_j + C_i\otimes
x_j + C_V(A)$. Since $t_i\otimes C_j\leq R$ we get $Q_V(A) = C_i\otimes x_j + C_V(A)$,
and $(V_i\otimes C_j)
\cap C_V(A)$ is a hyperplane of $V_i\otimes C_j$ containing $C_i\otimes C_j$. It
follows that $C_{V_i}(A) = C_i$ is a hyperplane of $V_i$ contradicting $\dim_\K
C_i = 1$ and $\dim_K V_i \geq 3$.
\bd{10} \rf a and \rf c hold.\ed
Claim \rf a follows from \rf 9 and \rf{4}. In particular, $C_1\otimes
V_2+V_1\otimes C_2$ is a $\K$-hyperplane of $V$. So \rf 5 implies \rf c.
\bd{11} Suppose that $C_V(A)= C_1\otimes C_2$. Then \rf b, \rf{d:a} and \rf
e hold.\ed
To prove \rf b let $z_1 \in V_1\setminus C_1$. According to \rf{a} we may
assume that $z_1 = x_1$. From \rf 5, \rf 6 and since $A$ is nearly
quadratic we get
$$C_1\otimes x_2 \leq Q_V(A) = [(x_1\otimes x_2)\F,A] + C_1\otimes C_2
\leq [x_1\F,A]\otimes x_2 + V_1\otimes C_2$$
Since $x_2 \notin C_2$ we get that $C_1\otimes x_2 = [ x_1\F,A]\otimes x_2$ and $C_1 = [
x_1\F,A]$. This is \rf b.
The first part of \rf{d:a} is true by
assumption, so we can assume that $\F=\F_p$. Since $\F=\F_p$ and $A$ is quadratic on
$V/C_V(A)$ we have $[(x_1\otimes x_2)\F,A]+C_V(A)=\{[x_1\otimes
x_1,a]\mid a\in A\}+C_V(A)$ and so
$$C_1\otimes x_2+x_1\otimes C_2 \leq [x_1\otimes
x_2\F,A]+C_1\otimes C_2=\{[x_1,a]\otimes x_2+x_1\otimes [x_2,a]\mid a\in A\}+C_1\otimes C_2.$$
Hence, for every $c_1\in C_1$ and $c_2\in C_2$, there exists $a\in A$ with
$[x_1,a]=c_1$ and $[x_2,a]=c_2$. The particular case when $c_1 = 0$ (or $c_2 = 0$) gives
\rf{d:a}. Moreover, \rf e follows.
\bd{12} Suppose that $C_V(A)\neq C_1\otimes C_2$. Then \rf b, \rf{d:b} and
\rf e hold.\ed
By \rf a and \rf c,
\[Q_V(A)=C_1\otimes C_2+ C_1\otimes x_2+ x_1\otimes C_2.\]
Since $C_1\otimes C_2 \lneq C_V(A)$ we
conclude that there exist $c_1\in C_1$ and $c_2\in C_2$ with $0\neq
c_1\otimes x_2-x_1\otimes c_2 \in C_V(A)$. Hence \rf 6 implies that
for all $a\in A$
%
$$c_1\otimes [x_2,a]=[x_1,a]\otimes c_2.\leqno{(*)}$$
%
Suppose that $c_1=0$. By the choice of $x_1$, there exists $a \in A$ with $[x_1,a]
\ne 0$, so $c_2=
0$, which contradicts $c_1\otimes x_2-x_1\otimes c_2\neq 0$. Hence $c_1\ne 0$ and
similarly also $c_2 \ne 0$. Then ($*$) implies that $[x_1,a]\in c_1\K$ for all
$a\in A$. So by \rf a $\dim_\K C_1=1$ and and $\dim_\K V_1=2$. By
symmetry $\dim_\K C_2 = 1$ and $\dim_\K V_2 = 2$.
Define
$\lambda_i:A \to \K$ by $[x_i,a]=c_i.a\lambda_i $ for all $a\in A$. Then by
($*$), $a\lambda_1=a\lambda_2$. So $\lambda:=\lambda_1=\lambda_2$ and
$V_1$ and $V_2$ are
isomorphic as $\F A$-modules. Moreover,
%
$$C_V(A)=C_1\otimes C_2 + (c_1\otimes x_2-x_1\otimes c_2)\K.$$
%
Let $L$ be the $\F$-subspace of $\K$ spanned by $A\lambda=\{a\lambda \mid a\in
A\}$. Then
%
$$Q_V(A)=[(x_1\otimes x_2)\F,A]+C_V(A)=(c_1\otimes x_2+ x_1\otimes
c_2)L+(c_1\otimes
x_2-x_1\otimes c_2)\K + C_1\otimes C_2.$$
%
Let $k\in \K$. Then $x_1k\otimes c_2\in Q_V(A)$, so there exists
$\ell\in L$ and $s\in \K$ with
%
$$x_1k\otimes c_2\in (c_1\otimes x_2+ x_1\otimes c_2)\ell+(c_1\otimes
x_2-x_1\otimes c_2)s+C_1\otimes C_2=c_1(\ell+s)\otimes x_2+x_1\otimes
c_2(\ell-s)+C_1\otimes C_2.$$
%
This implies $s=-\ell$ and $k=2\ell$. Since $k\in \K$ was arbitray we
conclude that $\chr \F\neq 2$ and $L=\K$. Thus \rf {d:b} and \rf{b}
holds.
If $\F=\F_p$, then $\K=L=A\lambda$ and so also \rf e
is proved.\qed
\be{lemma}{bnearly quadratic fieldext} Let $V$ ba a nearly quadratic $\F
A$-module. Suppose that there exists a field $\K$
with $\F\leq \K$ such that $V$ is a semi-linear, but not linear $\K
A$-module. Then $A/C_A(V) $ is elementary abelian and one of the following holds:
\bl[1]
\li a $[V,A,A] =0$, $[V,A_\K] =0$, and $\chr \K
=2=|A/A_\K|$.
\li b $[V,A,A] \ne 0$, $[V,A_\K]=C_V(A_\K)$, $\dim_\K V/C_V(A_\K)=1$,
$\F = \K_A$, and $\chr \K =2=|A/A_\K|=\dim_\F\K$.
\li c $[V,A,A] \ne 0$, $[V,A_\K] = 0$, $\F = \K_A$, $\dim_\K V=1$, and
$\chr F=3=|A/A_\K|=\dim_\F \K$.
\li d $[V,A,A] \ne 0$, $[V,A_\K] = 0$, $\F = \K_A$, $\dim_\K V=1$,
$\chr \F=2$, $A/A_\K\cong C_2\times C_2$, $\dim_\F \K=4$ and $\F$ is infinite.
\el
\en{lemma}
\proof Without loss $V$ is a faithful $\F A$-module. Put
$\E=C_\K(A)$. Since $A$ is cubic on $V$ we can apply
\ref{u:dimension commutator} If $A$ is quadratic on $V$ we conclude that \rf a
holds. So we may assume that $A$ is not quadratic.
Suppose first that $[V,A_\K]=0$. Then $V$ is the direct
sum of copies of $\K$. Hence by \ref{binherit2}, $V\cong \K$ as an $\F
A$-module. Thus by \ref{bfield ext}, $V\cong \E A$ as an $\E A$-module. As an
$\F A$-module, $\E A$ is a direct sum of $\dim_\F\E$-copies of $\F A$
and so \ref{binherit2} gives $\F=\E$. Thus \rf c or \rf d holds.
Suppose next that $[V,A_\K]\neq 0$. Then by \ref{u:dimension commutator} $p=|A/A_\K|=\dim_\E \K=2$ and there exists an $\E
A$-module $W$ such that $V\cong W\otimes_\E\K$ as an $\E A$-module and
$A=A_\K C_A(W)$. Hence we can
apply \ref{nearly quadratic and tensor}. Note that $[\K,A]=\E$. By
\rf[nearly quadratic and tensor] c, $[\K,A]=[z\F,A]$ for some $z\in
\K$. Since $|A/A_\K|=2$ this implies that $[\K,A]$ is 1-dimensional
over $\F$. Hence $\E=\F$. Also by \rf[nearly quadratic and tensor] c
$[W,A]=C_W(A)$ is an $\E$-hyperplane of $V$. Since $A=A_\K C_A(W)$,
$[W,A_\K]=C_W(A_\K)$ is a $\E$-hyperplane of $W$. Since $V\cong
W\otimes_\E \K$ we conclude that $[V,A_\K]=C_V(A_\K)$ is an $\K$-hyperplane of
$V$. Thus \rf b holds.\qed
\be{lemma}{tensor and cubic} Suppose that $\ca T$ is proper and $\K$-linear ,
$[V_j,A]\neq 1$ for all $j\in J$, $\chr \F=2$ and $A$ acts cubically on
$V$. Then $A/C_A(V)$ is elementary abelian.
\en{lemma}
\proof We may assume that $C_A(V)=1$. Suppose $A$ is not elementary abelian. Then there exists
$a\in A$ with $a^2\neq 1$. Since $\chr \F=2$, \ref{a2 and com} gives $[V,a^2]=[V,a,a]$ and
since $A$ is cubic we conclude that
\[ [V,a^2]\leq C_V(A)\leqno (*)\]
If $A$ does not act transitively on $I$ let $J$ be be a non-trivial orbit
for $A$ on $I$ and $K=I\setminus J$. If $A$ is transitively on $I$, let $(J,K)$ be a maximal
system of imprimitivity for $A$ on $I$. In both cases we obtain a
strict tensor decompsoition $V_J\otimes \K V_K$ for $A$
on $V$. So we may assume that $|I|=2$ and say $I=\{1,2\}$. In
particular $a^2$ acts trivially on $I$. Without loss $[V_1,a^2]\neq
0$. In particular
\[ [V_1\otimes C_{V_2}(a^2),a^2]=[V_1,a^2]\otimes C_{V_2}(a^2),\]
%
and so by (*)
%
\[ [V_1,a^2]\otimes C_{V_2}(a^2)\leq C_V(A).\leqno (**)\]
%
Suppose that $a$ acts trivially on $I$. Then (**) and \ref{trivial
action} imply $C_{V_2}(a^2)=C_{V_2}(a)$. By \ref{a2 and com},
$C_{V_2}(a^2)=Q_V(a)$ and so $Q_{V_2}(a)=C_{V_2}(a)$. Thus $a$
centralizes $V_2$.
So (**) implies that also $C_A(J)$ centralizes $V_2$. It follows that
$A\neq C_A(J)$ and since $C_A(J)\unlhd A$, $C_A(J)$ centralizes
$V_1$ and $V$. Thus $|A|=2$, a contradiction.
Thus $a$ acts non-trivial on $J$. Let $x_1\in V_1$ with $[x_1,a^2]\neq 0$ and
$[x_1,a^2,a^2]=0$. Put $x_2=x_1a_1$. Then
\[[x_1\otimes x_2,a]=(x_1\otimes x_2)^a-x_1\otimes
x_2=x_2a\otimes x_1a=x_1\otimes x_2=x_1{a^2}\otimes x_2-x_1\otimes x_2=[x_1,a^2]\otimes x_2\leqno(***)\]
Since $A$ acts quadratically on $[V_1\otimes V_2,A]$, $a^2$ centralizes
$[V_1\otimes V_2,A]$ . Since $a^2$ also centralizes $[x_1,a^2]$ we
conclude from (***) that $a^2$ centralizes $x_2$. But then $a^2$ also
centralizes $x_1=x_2a^{-1}$, a contradiction.\qed
\be{proposition}{nearly quadratic and tensor decompositions} Suppose that
\bl[i]
\li i $|A| >2$, $\ca T$ is proper, and
\li{ii} $V$ is a faithful, nearly quadratic $\F A$-module.
\el
Then $A$ acts $\K$-linearly on $V$, $A$ is elementary abelian
$chr \F$-group and
one of the following holds, where $B :=C_A(I)$:
\bl[1]
\li 2 $A$ is quadratic on $V$, and there exists $j\in J$ such that $A$
centralizes $V_i$ for all $i\in J\setminus\{j\}$.
\li 3 $\chr F=2$, $A$ is quadratic on $V$, and there exists an
$A$-invariant subset
$J_0$ in $J$ with $|J_0|=2$ such that $A$ centralizes $V_i$
for all $i\in J\setminus J_0$. Moreover
one of the following holds:
\bl[1]
\li 1 $A$ acts trivially on $J_0$ and there exists a homomorphism $\lambda: G\to
(\K,+)$ such that $V_j$ is a $\lambda$-dependent $\K A$-module for all $j\in J_0$.
\li 2 $A$ acts non-trivially on $J_0$, $\dim_\K V_j=2$ and $C_B(V_j)=C_B(V)$
for all $j\in J$.
\el
\li 4 $|J|=2$, $A$ is not quadratic on
$V$ and for $j\in J$, $[V_j,B]=C_{V_j}(B)$ is a $\K$-hyperplane
of $V_j$. Moreover, one of the following holds:
\bl[1]
\li 1 $A$ acts trivially on $J$, and $[V_j,A] = [ v_j\F,A]$ for all
$v_j\in V_j \setminus
[V_j,A]$ and $j \in J$.
\li 2 $A$ acts non-trivially on $J$, $\chr \F=2$, $\F =\K$, and $C_B(V_j)=C_B(V)$
for all $j\in J$.
\el
\el
\en{proposition}
\proof The proof is by induction on $|A|$ and $|J|$. Note that
$\dim_\K V\geq 4$ since $|J|\geq 2$ and $\dim_\K V_j \geq 2$. First we show:
\bd{11} $A$ acts $\K$-linearly on $V$.\ed
Assume that $A\ne A_\K$. Then.
we can apply \ref{bnearly quadratic fieldext}. Since $|A|>2$ and
$\dim_\K V\neq 1$ we conclude that $A_\K\neq 1$. Hence $p = 2$, $[V,A_\K] = C_V(A_\K)$,
and $ \dim_\K V/C_V(A_\K) = 1$. If $|A_\K| = 2$, then $\dim_\K
[V,A_\K] = 1$ and
so $\dim_\K V
= 2$, which contradicts $\dim_\K V \geq 4$. If $|A_\K| > 2$, then we can apply
induction with $A_\K$ in place of $A$ since $A_\K$ acts quadratically on
$V$. Hence one of the cases \rf 2 or \rf 3 holds for $A_\K$. In both cases
$|A_\K/A_\K\cap B| \leq 2$, so $[V_r,A_\K\cap B] \ne 0$ for some $r \in
J$. Hence
\ref{centralizer index} applied to $A_\K\cap B$ shows that $C_V(A_\K\cap B)$
is not a $\K$-hyperplane of
$V$. This contradiction shows that $A$ acts $\K$-linearly on
$V$.
\bcase{non-transitive} $A$ is not transitive on $J$.\ecase
Let $L$ be an orbit of $A$ on $J$. We choose $L$ in such a way that $|L|$ is
minimal and that $A$ centralizes $V_j$ for $j \in L$ if this is possible.
Put $J:=I\setminus L$, . Then $V=V_L\otimes_\K V_I$ is a strict
tensor decompsotion of $V$ for $A$, By \rf{11} $A$ induces $\K$-linear
transformations on $V_L$ and $V_J$.
\bd{22} $A$ acts quadratically on $V_{L}$ and $V_{J}$. \ed
If $A$ acts quadratically on $V$, then by \rf[quadratic and tensor1]a $A$ also
acts quadratically on $V_{L}$ and
$V_{J}$. If $A$ is not quadratic on $V$, then \ref{nearly quadratic and tensor} shows
that $A$ acts quadratically on $V_{L}$ and $V_{J}$.
\bd{33} Suppose that $A$ centralizes $V_L$. Then \rf 2 or \rf 3 of
the proposition holds.\ed
Note that by our choice of $L$ and \ref{trivial action}, $|L| = 1$. Moreover, the faithful action of
$A$ on $V$ shows that $A$ acts faithfully on $V_J$.
If $|J| = 1$, then \rf 2 holds. If
$|J|>1$, then by induction on $|I|$ we see that \rf 2 or \rf 3 holds for $V_J$
since $A$ is quadratic on $V_J$. But then the same case also holds for $V$.
\bd{44} Suppose that $A$ acts non-trivially on $V_L$ and $V_J$. Then \rf{3:1}
or \rf{4:1} holds. \ed
By our choice of $L$, $A$ does not centralize any $V_i$ for $i \in I$.
Assume first that $A$ acts quadratically on $V$. Then $A$ is
elementary abelian and by by \ref{quadratic and tensor1} there exists a homomorphism $\lambda:G\to (\K,+)$ such that $V_L$ and
$V_J$ are $\lambda$-dependent as $\K A$-modules. Suppose for a contradiction
that $|J|\geq 2$. Then by induction $|J|=2$ and \rf{3:1} or \rf{3:2} holds for
$V_J$. Let $J=\{i,k\}$.
Suppose that \rf{3:1} holds. Then there exists a homomorphism $\mu:G\to (\K,+)$ such that $V_L$ and
$V_I$ are $\mu$-dependent as $\K A$-module. Let $1\neq a,b \in A$. Without loss $a\mu=1$. Put $\xi=b\mu$. For $j\in J$ let
$x_j\in V_j\setminus C_{V_j}(A)$ and put $y_j=[x_j,a]$. Then
\[[x_i\otimes x_j,a]=x_i\otimes y_k+y_i\otimes x_k+y_i\otimes y_k\]
and
\[[x_i\otimes x_j,b]=x_i\otimes y_k\xi +y_i\otimes x_k\xi +y_i\otimes y_k\xi^2\]
Since $A$ acts $\lambda$-dependent on $V_J$ this implies $\xi=\xi^2$. So
$\xi=1$ and $a=b$. Thus $|A|=2$, a contradiction to the assumptions.
Suppose next that \rf{3:2} holds. Let $a\in A\setminus C_A(J)$. Since $|A|>2$
there also exists $1\neq b\in C_A(i)$. Since $|J|=2$, $a$ interchanges $i$ and
$k$ while $b$ fixes $i$ and $k$. Let $v_i\in V_i\setminus C_{V_i}(b)$ and
$c_i:=[v_i,b]$. Put $c_k:=c_ia_i=c_ia$ and $v_k:=v_ia_i=v_ia$. By
\rf[induced action] a and since $A$ is abelian, $[v_k,a]=c_k$. Since
$\chr \K=2$ we have $a^2=1$, $v_ka=v_i$ and $c_ka=c_i$. Thus
\[ [v_i\otimes c_k,b]=c_i\otimes c_k\] and
\[[v_i\otimes
c_k,a]=(c_ka_k\otimes v_ia_i)-(v_i\otimes c_k)=c_i\otimes v_k-v_i\otimes c_k\]
Since $c_j$ and $v_j$ are $\K$-linearly independent we conclude that $[v_i\otimes c_k,b]\K\neq [v_i\otimes
c_k,a]\K$, a contradiction since $A$ acts $\lambda$-dependent on $V_J$.
Thus $|J|=1$ and the minimal choice of $|L|$ gives $|L|=1$. Thus \rf{3:1} holds.\medskip
Assume now that $A$ is not quadratic on $V$. Then by \rf[nearly quadratic
and tensor]a (with $V_1 = V_{L}$ and $V_2 = V_J$), $C_{V_{J}}(A)=[V_J,A]$ is a
$\K$-hyperplanes of $V_{I}$. Suppose for a contradiction that $|J|\geq 2$. If
$[C_A(J),V]\neq 1$, then by
\ref{centralizer index} (applied to $V_J$ and $C_A(V_J)$, $\dim_\K
V_J/C_{V_J}(C_A(V_J)>1$, a contradiction. Thus $C_A(V_J)=C_A(J)$
and so $|A/C_A(V_J)|=2$. But then $C_{V_{J}}(A)=[V_J,A]$ is
$1$-dimensional and $\dim_\K V_J=2$, a contradiction. Hence $|J| =
1$, and thus by our choice of $L$ also $|L| =
1$. Now \rf[nearly quadratic and tensor]b gives \rf{4:1}.
\bcase{transitive} $A$ is transitive on $J$.\ecase
Let $X_1$ be a $1$-dimensional $\K B$-submodule of $V_1$.
We apply \ref{tensor and imprimitivity} (with $A$ in place of $G$) and use the
notation introduced there.
\bd{55} Either $|I| \geq 3$ and \ref{binherit3} (\ref{binherit3:d:a}) or
(\ref{binherit3:d:b}) holds for $\Delta$ and $A$, or $|I| = 2$ and
\ref{binherit3} (\ref{binherit3:c}) or
(\ref{binherit3:d:c}) holds for $\Delta$ and $A$. In particular $[V_1,B,B]
\leq X_1$.\ed
This follows from \ref{binherit3} using the transitivity of $A$ on $\Delta$.
\bd{66} $|I| = 2=\chr \K$, in particular $B\ne 1$.\ed
Suppose that $|I| \geq 3$. Then by \rf{55} $A$ is not quadratic on $U/X$. Let $V_i^*$ be
the $\K$-dual of $V_i$. Then $V^*$ is canonically isomorphic to $\bigotimes V_i^*$
via $(\otimes v_i).(\otimes \phi_i)=\prod v_i\phi_i$. Define $\tilde X=\bigotimes
X_i^\perp$, $\tilde U_i=V_i\otimes\bigotimes_{j\neq i} X_j^\perp$, $\tilde
U=\sum_{i\in I} \tilde U_i$. Then by symmetry, $A$ is not quadratic on $\tilde
U/\tilde X$. From $|I|\geq 3$ we have $U_i\perp \tilde U_j$ for all $i,j\in
I$. Thus $U\leq \tilde U^\perp$. Since $\tilde X^\perp/\tilde U^\perp
$ is isomorphic to the dual of
$\tilde X/\tilde U$ we conclude that $A$ is not quadratic on $\tilde
X^\perp/\tilde U^\perp $. Hence $A$ is neither quadratic on $U$ nor on
$V/U$, which contradicts \ref{trivial property}.
Thuse $|I|=2$. It follows that $|A/B| =
|J| = 2$ and thus $B\ne 1$ since $|A| \geq 3$.
\bd{66.3} $A$ is elementary abelian.\ed
By \rf{66} $\chr \F=2$. So by \ref{tensor and cubic}, $A$ is
elementary abelian.
\bd{66.5} Suppose that $A$ is quadratic on $V$. Then \rf{3:2} holds. \ed
This follows from \ref{quadratic and tensor2}.
\bd{77} Suppose that $C_{V_1}(B) \ne X_1$. Then \rf{3:2} or \rf{4:2} holds. \ed
There exists a $1$-dimensional $\K$-subspace $X_1^\prime$ of $C_{V_1}(B)$ different
from $X_1$. Hence by \rf{55} $[V_j,B,B] \leq X_1 \cap X_1^\prime = 0$, so $B$
acts quadratically on $V_1$. By \rf{66} $|I| = 2$, say $I = \{1,2\}$, $|A/B|
= 2$ and $B\ne 1$.
Assume first that $C_{V_1}(B) \ne [V_1,B]$. Then there exists a non-zero
a $\K B$-submodule $Z_1 \leq C_{V_1}(B)$ with $C_{V_1}(B)=Z_1\oplus [V_1,B$. Hence,
there also
exists a $\K B$-submodule $Y_1 \leq V_1$ with $C_{V_1}(B)\cap
Y_1=[V_1,B]$ and $V_1 = Y_1\oplus Z_1$. Pick $a \in A\setminus B$ and
put
%
$$Z_2 := Z_1a_1,\;Y_2:= Y_1a_1,\; Y:= Z_1\otimes Y_2 +
Y_1\otimes Z_2,\; D:= Y_1\otimes Y_2.$$
%
By \rf[induced action] a, $Y$, $D$ and $Z_1\otimes Z_2$ are $\K
A$-submodules of $V$. Note that
$$V = (Z_1\oplus Y_1)\otimes (Z_2\oplus Y_2) = (Z_1\otimes Z_2) \oplus
(Z_1\otimes Y_2) \oplus (Y_1\otimes Z_2) \oplus (Y_1\otimes Y_2) = (Z_1\otimes
Z_2)\oplus Y \oplus D.$$ \ref{trivial action}
implies that $A$ neither centralizes $Y$ nor $D$. Hence, \ref{binherit2} shows
that $A$ is quadratic on $V$. \rf{66.5} now implies
\rf{77}.
Assume now that $C_{V_1}(B) = [V_1,B]$. Then $X_1\leq [V_1,B]$. We apply \rf{55}. If \ref{binherit3}
(\ref{binherit3:d:c}) holds for $\Delta$ then $[U_1/X,B]$ is a
$\K$-hyperplane of $U_1/X$. Hence also $[V_1,B]$ is a $\K$-hyperplane
of $V_1$ and \rf{4:2} follows. If \ref{binherit3}
(\ref{binherit3:c}) holds for $\Delta$ then $[V_1,B] = X_1$ and so
$C_{V_1}(B) = X_1$, a contradiction.
\bd{88} Suppose that $C_{V_1}(B) = X_1$. Then \rf{3:2} or \rf{4:2} holds.\ed
If $\dim_\K V_1 = 2$, then\rf{3:2} or \rf{4:2} follows. Hence we may assume that $\dim_\K
V_1 \geq 3$. Let $Y$ be a $3$-dimensional $\K B$-submodule of
$V_1$.
Since $\chr \K=2$, the elementary abelian $2$-subgroups of $GL_3(\K)$
are quadratic. Hence $[Y,B,B]=0$ and since $C_{V_1}(B)=X_1$ we conclude that
$[Y,B]=X_1$. Fix $1\ne b \in B$ and $a \in A\setminus B$. Then $\dim_\K
V_1/C_{V_1}(b) = \dim_\K X_1 = 1$,
so there exists $x,y,z \in V_1$ such that
$$ Y= \langle x,y,z\rangle_\K,\; X_1= x\K,\; [z,b] = 0,\; [y,b] = x.$$
By \rf{66.3} $A$ is abelian and so by \ref{induced action}
the map
$$V_1 \to V_2 \hbox{ with }v_1\mapsto v_1a =:v_1^\prime$$
is a $\K B$-module isomorphism.
It is easy to calculate that
$$x\otimes x^\prime \in C_V(A), \;y\otimes y^\prime \in C_V(a) \hbox{ and }
[(y\otimes z^\prime),a] = y\otimes z^\prime + z \otimes y^\prime.$$
This shows that
$$[y\otimes z^\prime,a,b] = [y\otimes z^\prime + z \otimes y^\prime,b] =
x\otimes z^\prime + z \otimes x^\prime \ne 0,$$
Thus $y\otimes z^\prime\notin Q_V(A)$. Since $V$ is a nearly
quadratic $\F A$-module we get for $Y^\prime:= Ya_1$
\[\begin{array}{*3l} Q_V(A) &=& [y\otimes z^\prime,A]\F +
C_V(A)=[y\otimes z^\prime,a]\F+[y\otimes z^\prime,B]\F+C_V(A)\\
& \leq& (y\otimes z^\prime + z
\otimes y^\prime)\F + x\otimes Y^\prime +Y\otimes x^\prime + C_V(A).\end{array}\leqno{(*)}\]
If $y\otimes y^\prime \not\in Q_V(A)$ then $Q_V(A)=[y\otimes y^\prime,A]\F +
C_V(A)\leq C_V(a)$, since
$y\otimes y^\prime \in C_V(a)$. But then $[V,A,a] = 0$, which contradicts
$[y\otimes z^\prime,a,b] \ne 0$. Thus we have $y\otimes y^\prime \in Q_V(A)$,
so $(*)$ shows that there exist $u,w \in Y$, $t \in \F$ and $c \in C_V(A)$
such that
$$y\otimes y^\prime =(y\otimes
z^\prime + z
\otimes y^\prime)t+ x\otimes u^\prime + w\otimes x^\prime + c.$$
Taking the commutator with $b$ on both sides gives
$$
x \otimes x + x \otimes y^\prime + y \otimes x^\prime = (x \otimes
z^\prime + z \otimes x)t+x \otimes [u^\prime,b] + [w,b]\otimes x^\prime $$
and
$$x \otimes y^\prime + y \otimes x^\prime \equiv (x \otimes
z^\prime + z \otimes x)t \mod (x\otimes x)\K.$$
But then $x,y,z$ are not linearly independent in $V_1$ which contradicts
$\dim_\K Y = 3$. This contradiction shows \rf{88} and
completes the proof of \ref{nearly quadratic and
tensor decompositions}. \qed
%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%
\section{The Nearly Quadratic Subgroup Theorem}
%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%
\be{definition}{def:reduced} Let $H$ be a group, $\F$ a field and $V$
an $\F H$-module. We say that $V$ is $H$-reduced if $[V,N]=0$ whenever
$N$ is a normal subgroup of $H$ acting nilpotently on $V$.
\en{definition}
\be{lemma}{simple slm} Let $\F$ be a field, $V$ a finite dimensional $\F$-space,
and $G\leq SL_\F(V)$. Suppose that $V$ is $G$-reduced
and
$M: = \langle C_{\SL_\F(V)}(V/U)^G\rangle\neq 1$
for some $1$-dimensional subspace $U$ of $V$. Then $M =
SL_\F(V)$.
\en{lemma}
\proof We may assume that that $\dim_\F V > 1$ since otherwise $M = \SL_\F(V)
= 1$. Let $\ca P=\ca P_\K(V)$ be the set of $1$-dimensional subspaces of $V$, and let
$L(X):= C_{\SL_\F(V)}(V/X)$ for $X\in \ca P$ and
$$\ca P(M):= \{ X \in \ca P \mid L(X) \leq M\}.$$
As $\SL_\F(V) = \langle L(X) \mid X \in \ca P\rangle$, it suffices to show that
$\ca P(M) = \ca P$.
Since $[V,L(U)]=U$ we get $[V,M]=\sum_{U\in \ca P(M)} U$. If
$[V,M]\neq V$, then $1\neq C_{L(U)}([V,M])\leq C_M([V,M])) \cap
C_M(V/[V,M])$, a contradiction since the latter group is normal in
$M$ and acts nilopotently on $V$.
Thus $V=[V,K]=\sum_{U\in \ca P(M)} U$.
Let $U_1,U_2\in \ca
P(M)$. Then $L(U_1)$ acts transitively on the $1$-dimensional subspaces of
$U_1+U_2$ unequal to $U_1$. Hence $\ca P(M)$ contains all the
$1$-dimensional subspaces of $U_1+U_2$. Since $V = \sum_{U\in \ca P(M)} U$ we
conclude that $V=\sum_{U\in \ca P(M)} U = \bigcup_{U\in \ca P(M)} U$, and $\ca
P(M)$ contains all the $1$-dimensional subspaces of $V$. \qed
\bigskip
{\it Remark. Let $L(U)$ be defined as in the previous lemma. Then $U =
C_V(L(U))$ and by the dimension formula
$$\dim U + \dim [V^*,L(U)] = \dim V,$$
where $V^*$ is the dual vector space. Hence passing to the dual space $V^*$
transforms \ref{simple slm} into a statement about reduced
subgroups $G \leq
SL_\F(V)$ and $M = \langle C_{SL_\F(V)}(U)^G\rangle$
for
some hyperplane $U \leq V$. We will refer to this version as the ``dual
version of \ref{simple slm}''.}
\be{theorem}{nearly quadratic decomposition} Let $\F$ be field, $H$ a group and
$V$ be a faithful semisimple $\F H$-module. Let $\ca Q$ be the set of nearly quadratic, but
not quadratic subgroups of $H$. Suppose that $H=\<\ca Q\>$. Then there exists a partition $(\ca
Q_i)_{i\in I}$ of $\ca Q$ such that
\bl
\li a $H=\bigoplus_{i\in I} H_i$, where $H_i=\<\ca Q_i\>$.
\li b $V=C_V(H)\oplus \bigoplus_{i\in I} [V,H_i]$.
\li c For each $i\in I$, $[V,H_i]$ is a faithful simple $\F H_i$-module.
\el
\en{theorem}
\proof Since $V$ is semisimple, $V=\bigoplus_{j\in J} V_j$ for some simple $\F
H$-submodules $V_j,j\in J$. Let $I=\{i\in J\mid [V_i,H]\neq 0\}$. Then clearly
\[V=C_V(H)\oplus \bigoplus_{i\in I} V_i\]
For $i\in I$ let $\ca Q_i=\{A\in \ca Q\mid [V_i,A]\neq 0\}$. Then by
\ref{binherit2} each $A\in \ca Q$ is contained in a unique $\ca Q_i$ and so
$(Q_i)_{i\in I}$ is a partition of $\ca Q$. Observe that $H_i$ centralizes $V_j$
for all $i\in I$ and $j\in J$ with $i\neq j$. In particular $V_i$ is a faithful
$H_i$-module and $ H_i\cap \=1$. Thus \rf a holds. Since $V_i$ is a simple $\F H$ module, we conclude
that $V_i$ is a simple $\F H$. Since $0\neq [V,H_i]\leq V_i$ we get
$V_i=[V,H_i]$ and so \rf b and \rf c hold.\qed
\be{lemma}{nh(a)} Let $\K$ be a finite field of charteristic $2$, $\F\leq \K$
a subfield with $|\F|\geq 4$, $V$ a 3-dimensional $\K$-space, $L\leq GL_\K(V)$ such that $L\cong SL_2(\F)$ and $V\cong W\otimes_\F\K$,
where $W$ is a natural $\Omega_3(\F)$-module for $\F L$. Let $A\in
\Syl_2(L)$. Put $H=N_{GL_\K(V)}(A)$, $B=C_{SL_\K(V)}(C_V(A))$,
$Z=\op Z(\GL_\K(V))$, $V_1=[C_V(A),H\cap L]$ and $V_2=C_V(L)$. Then
$C_V(A)=V_1\oplus V_2$. Moreover:
\bl
\li a If $|\F|> 4$, then $V_1$ and $V_2$ are $H$-invariant and
$H=(H\cap L)ZB=C_H(V_2)Z$.
\li b If $\F|=4$, then $V_1^H=\{V_1,V_2\}$ and $N_H(V_1)=(H\cap
L)ZB=C_H(V_2)Z$.
\el
\en{lemma}
\proof Let $q$ be the $L$-invariant quadratic form on $W$ and $s$ the
corresponding bilinear form. Let $(w_1,w_2,w_3)$ be an $\F$-basis for $W$ such that $w_2\in
C_W(L)$, $(w_1,w_3)s=1$, $w_1q=0$, $w_2q=1$ and $w_3q=0$ and $w_1\in
C_V(A)$. Let $a\in A$ then $w_3aq=0$ and so
$w_3a=w_1\lambda^2+w_2\lambda+w_3$ for some $\lambda\in \F$. We
denote this element of $A$ by $a_\lambda$. Let $v_i$ be the image of
$w_i\otimes 1$ in $V$ under the isomorphism from $W\otimes_\F\K$ to
$V$. Then $(v_1,v_2,v_3)$ is an $\K$-basis for $V$ and the matrix of
$a_\lambda$ with respect to this basis is
\[a_\lambda\leftrightarrow
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
\lambda^2 & \lambda & 1
\end{pmatrix}
\]
Note that $B$ is abelian and $A\leq B$. Thus $ZB\leq H$. Since $ZB$
acts transitively on $V\setminus C_V(A)$ we have $H=C_H(v_3)ZB$. Let
$h\in C_H(v_3)$. Then
\[h\leftrightarrow
\begin{pmatrix}
a & b & 0 \\
c & d & 0 \\
0 & 0 & 1
\end{pmatrix}
\]
for some $a,b,c,d\in \K$ with $ad-cb\neq 0$. Let $\lambda\in \F$. Since $h\in N_H(A)$,
$a_\lambda h=ha_\mu$ for some $\mu\in \F$. We have
\[a_\lambda h\leftrightarrow
\begin{pmatrix}
a & b & 0 \\
c & d & 0 \\
\lambda^2 a+\lambda c & \lambda^2b+\lambda d & 1
\end{pmatrix}\text{ and }
ha_\mu\leftrightarrow
\begin{pmatrix}
a & b & 0 \\
c & d & 0 \\
\mu^2 & \mu & 1
\end{pmatrix}
\]
Hence
\[ \lambda^2b+\lambda d=\mu \text{ and } \lambda^2 a+\lambda c
=\mu^2=\lambda^4b^2+\lambda^2 d^2\]
Thus
\[ \lambda c+\lambda^2(a+d^2)+\lambda^4 b^2=0\text{ for all }
\lambda\in \F\]
Suppose that $|\F|>4$ and consider the polynomial
$f=cx+(a+d^2)x^2+b^2x^4$. Then each $\lambda\in \F$ is a root of
$f$. Since $\deg f\leq 4<|\F|$ we conclude that $f$ is the zero
polynomial. Hence $c=0, b=0$ and $a=d^2$. From $\mu=\lambda^2b+\lambda
d=\lambda d$ we conclude that $d\in \F$. Moreover,
\[h \leftrightarrow
\begin{pmatrix}
d^2 & 0 & 0 \\
0 & d & 0 \\
0 & 0 & 1
\end{pmatrix}=
\begin{pmatrix}
d & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & d^{-1}
\end{pmatrix}
\begin{pmatrix}
d & 0 & 0 \\
0 & d & 0 \\
0 & 0 & d
\end{pmatrix}\]
And so $h\in (H\cap L)Z$. Thus $H=(H\cap L)BZ$. Since $(H\cap L)B\leq
C_H(V_2)$ we see that \rf{b} holds.
Suppose next that $|F|=4$. Then $\lambda^4=\lambda$ for all $\lambda$
in $\F$. Thus $\lambda(c+b^2)+\lambda^2(a+d^2)=0$. Hence
$f=(c+b^2)x+(a+d^2)x^2$ is the zero polynomial and so $c=b^2$ and
$a=d^2$. From $\lambda^2 b+\lambda d=\mu\in \F$ for all $\lambda\in
\F$ we conclude that $b,d\in \F$. Moreover, $0\neq
ad-bc=a^3-b^3$. Since $u^3=1$ for all $0\neq u\in \F$ we have $a=0$ or
$b=0$. If $a=0$, then $v_1h=v_2b\in V_2$ and if $b=0$ then
$v_1h=v_1a\in V_1$. Thus $V_1^H=\{V_1,V_2\}$. Also if $b=0$ then as
above $h\in (L\cap H)BZ\leq C_H(V_2)Z$ and so \rf b holds.
\qed
\be{proposition}{char slm} Let $H$ be a finite group, $\K$ a finite field and $V$
faithful finite dimensional $\K H$-module. Put
\[\ca H=\{A\leq H\mid C_V(A)=[V,A]\text{ and } \dim V/C_V(A)=1\}\]
Suppose that $H=\<\ca H\>$. Then $C_V(H)$ is the unique maximal $\K
H$-submodule of $V$ and $C_H(V/C_V(H))=O_p(H)$. Moreover, put $\o V=V/C_V(H)$, $\w
H=H/C_H(\o V)$, $\p=\chr \K$ and $n=\dim_\K(\o V)$. Then one of the following holds:
\bl[1]
\li a $p=2$, $n=2$ and $\w H\cong D_{2m}$ for some $odd$ integer
$m$ with $m>3$
\li b $p=3$, $n=2$, $\F_9\leq \K$ and $\w H\cong \SL_2(5)$.
\li c $p=2$, $n=3$ and $\w H\cong 3.\Alt(6)$.
\li d $\w H \cong \SL_n(\F)$ for some subfield $\F$ of $\K$. Moreover,
$\o V\cong W\otimes_\F \K$ for a natural $\F \tilde H$-module $W$.
\el
\en{proposition}
\proof The proof is by induction on $n$. We first show that
\bd 1 $C_V(H)$ is the unique maximal $\K H$-submodule.\ed
Let $U$ be a $\K H$-submodule of $V$ with $U\nleq C_V(H)$ and pick
$A\in \ca H$ with $[U,A]\neq 0$. Since $\dim_\K V/C_V(A)=1$ we have
$V=U+C_V(A)$. Thus $C_V(A)=[V,A]=[U,A]\leq U$ and so $V=U$. Hence \rf 1
holds.
\bd 3 $\o V$ is a simple $\K H$-module and $O_p(H)= C_H(\o
V)$.\ed
The first statement follows immediately from \rf 1. In particular, $O_p(H)\leq C_H(\o
V)$. Since $C_H(\o V)$ centralizes $C_V(H)$ and $V/C_V(H)$ we get that
$C_H(\o V)$ is a $p$-group and so \rf 3 holds.
\bd 2 Let $\ca A$ be the set of maximal elements of $V$ with respect to
inclusion. Then $H=\<\ca A\>$, each $A$ in $H$ is weakly closed and
$H$ acts transitively on $\ca A$.\ed
Since each $B\in \ca H$ is contained in some $A\in \ca A$, $H=\<\ca
A\>$. Let $A,B\in \ca A$ such that $B$ normalizes $A$. By \rf 1
applied to $AB$ in place of $H$, $C_V(A)\leq C_V(AB)$. Thus $C_V(A)=C_V(B)$ and so
$AB\in \ca H$. By maximality of $A$ and $B$ we get $A=AB=B$. Thus $A$
is weakly closed in $H$ and any Sylow $p$-subgroup of $H$ contains a
unique member of $\ca A$. So \rf 2 holds.\medskip
If $n=1$, then \rf
d holds with $\K=\F$. If $n=2$ the Proposition follows from Dickson's
List of subgroups of $\SL_2(\K)$ \cite[II.8.27]{Huppert}. Replacing $V$ be $\o V$ and $H$ by $\tilde H$ we
may assume from now on that
\bd {3.5} $V$ is a faithful simple $\K H$ module and $n\geq 3$.\ed
For $L\leq H$ let $\ca A_L=\{A \in \ca A\mid A\leq L\}$. Since
$C_V(H)=0$ and $\dim V/C_V(A)=1$ for all $A\in \ca A$, there exists
$L\leq H$ with $L=\<\ca A_L\>$ and $\dim C_V(L)=1$. Let $\ca L$ be the
set of such subgroups of $H$.
\bd 4 Suppose that $O_p(L)=1$ for some $L\in \ca L$. Then $n=3$,
$p=2$, $L\cong SL_2(\F)$ for some subfield $\F$ of $\K$, $\ca
A_L=\Syl_p(L)$ and $V\cong W\otimes_\F \K$ for a natural $\F
\Omega_3(\F)$-module $W$ for $L$. \ed
By induction the theorem holds for $L$ in place of $H$. By \rf 1 applied to $L$, $V$ is indecomposable as $\K L$-module. In
particular, $O_{p^\prime}(L)=1$. We conclude that \rf d holds for
$L$ and so $L\cong \SL_{n-1}(\F)$. Moreover, if $n=3$, then $p=2$. Let $A\in \ca A_L$ and put
$P^*=N_L(C_V(A))$ and
$P=C_{P^*}(V/C_V(A))$. Note that $P^*$ acts simply on
$O_p(P^*)$. Since $A$ is
weakly closed in $H$ and $A\leq O_p(P^*) $ we conclude that $A\unlhd P^*$ and
$A=O_p(P^*)$. If $n=3$ we conclude \rf 4 holds. So suppose that
$n>3$. Then $A=O_p(P^*)$ is natural module for $P/O_p(P^*)\cong
\SL_{n-2}(\F)$. Let $x\in V\setminus C_V(A)$. Then $[x,A]\cong
A/C_A(x)\cong A$
as an $\F_pP$-module. Thus $[x,A]$ is a nontrivial simple module for
$P$. Since this holds for any such $x$, $[V,A]$ is a sum of non
-trivial simple $\F_pP$-module and so $C_{[V,A]}(P)=0$. But this is a
contradiction to $C_V(L)\leq C_V(A)=[V,A]$.
\bd 5 Suppose $L\in \ca L$ with $O_p(L)=1$ and let $A\in \ca A_L$.
\bl
\li{5:a} Let $L\leq R\in \ca L$. Then $L=R$.
\li{5:b} Let $\ca L_-(A)=\{R\in \ca L\mid A\leq R, O_p(R)=1\}$. If
$|A|=4$, then $|\ca L_-(A)\leq 2$ and if $|A|>4$, then $\ca L_-(A)=\{L\}$.
\li{5:c} There exists $R\in \ca L$ with $O_p(R)\neq 1$.
\el
\ed Let $L\leq R\in \ca L$. Since $A\leq O_p(R)\leq O_p(L)\leq 1$ and
$A$ is weakly closed, $[A,O_p(R)]=1$. On the other hand since
$V/C_V(R)$ is a simple $\K R$-module, $O_p(R)$ is as an $\F_p
R$-module the direct sum of copies of $V/C_V(R)$. So $[O_p(R),A]=1$,
implies $O_p(R)=1$. So by \rf 4, both $L$ and $R$ are isomorphic to
$SL_2(|A|)$ and so $L=R$. Thus \rf{5:a} holds.\medskip
Put $V_1:=[C_V(A),N_L(A)]$ and $V_2:=C_V(L)$. Let $R\in \ca L_-(A)$.
Since $N_R(A)=O^2(N_R(A))$, \ref{nh(a)} implies that $N_R(A)$
normalizes $V_1$ and $V_2$. Since $\K N_R(A)$ has exactly proper
submodule in $C_V(A)$ we conclude that $C_V(R)=V_1$ or
$C_V(R)=V_2$. In the second case $\\in \ca L$ and so by \rf{5:a},
$R=L$. So suppose $C_V(R)=V_1$, then $V_2=[C_V(A),N_R(A)]$. Put
$P=N_H(A)\cap N_H(V_2)$. From the action of $R\cong \SL_2(|A|)$ on $V$ we conclude
that $|P/C_P(V_2)|\geq |A|-1$. On
the other hand since $P\leq SL_\K(V)$ and $V_2=C_V(L)$, \ref{nh(a)} implies that
$P\leq \op Z(\SL_\K(V))C_P(V_2)$ and so $P/C_P(V_2)|\leq 3$. Thus
$|A|=4$. Moreover by \rf{5:a}, $R$ is unique in $\ca L$ with
$C_V(R)=V_1$ and so \rf{5:b} holds.
Suppose that $O_p(R)=1$ for all $R\in \ca L$. Let $B\in \ca A$ with
$B\nleq L$. Then $B\neq A$ and so $C_V(A)\neq C_V(B))$. Hence $R:=\\in \ca L$. Since $R\neq L$, \rf{5:b}
gives that $|A|=4$ and $R$ is unique detemined. Hence $\ca A=\ca
A_L\cup \ca A_R$. Since $R\cong L\cong SL_2(4)$ we conclude that $|\ca
A|=1+4+4=9$. Put $m=|\cal L|$. Then there exists $5m$ pairs $(B,P)$
with $B\leq \ca A, P\in \ca L$ and $B\leq P$. On the otherhand each of
the 9 members of $\ca A$ is contained in exactly 2 members of $\ca L$
and so $5m=2\cdot 9=18$, a contradiction. Thus \rf{5:c} holds.
\bd{6} Put $\ca H^*=\{D\leq H\mid [V,D]=C_V(D)\}$, put $H^*=\<\ca A^*\>$ and
let $\ca A^*$ the set of maximal elements of $\ca H^*$ under
inclusion. Let $L\in \ca L$ with $O_p(L)\neq 1$ and $A\in \ca L$. Then
\bl
\li{6:a} $H=H^*$.
\li{6:b} $O_p(L)\in \ca A^*$ and $|O_p(L)|=|A|$.
\li{6:c} $L/O_p(L)\cong SL_{n-1}(\F)$ for some subfield $\F$ of $\K$.
\li{6:d}d $O_p(L)$ is a natural module for $\F_p \SL_{n-1}(\F)$,
$|A|=|\F|^{n-1}$ and $|A\cap O_p(L)|=|\F|$.
\el
\ed
Observe first that by \rf 5\rf{5:c} there exists $L\in \ca L$ with
$O_p(L)\neq 1$. Let $B$ be a simple $\F_pL$-submodule of
$O_p(L)$. Then $B$ is dual to an simple $\F_p L$-quotient of
$V/C_V(L)$. By \rf 1 $V/C_V(L)$ is a simple $\K L$-module and
$C_V(B)=C_V(O_p(L))=C_V(L)=[V,B]=[V,O_p(B)]$. So both $B$ and $O_p(L)$
are in $\ca H^*$. Let $D^*\in \ca A^*$ with $O_p(L)\leq D^*$ . Put
$\F=\End_L(B)$. Then $V/C_V(L)\cong B^* \otimes_\F \K$ as an $\F_p
L$-module. Since $[V/C_V(L),B]=C_{V/C_V(L)}(B)]$ is a $\K$-hyperplane
of $V/C)V(L)$ it follows that $[B,A]=C_B(A)$
is $1-$dimensional over $\F$. In particular, $|B/C_B(A)|\geq
|A/C_A(B)|$. Observe that $C_A(B)=C_A(V/C_V(L)=A\cap O_p(L)$ and since
$A$ is weakly closed $C_B(A)=[B,A]\leq A$. Hence $C_B(A)=A\cap B$. Thus
%
\[ |A|=|A/C_A(B)||C_A(B)|\leq |B/C_B(A)||A\cap O_p(L)|=|B/B\cap
A||A\cap O_p(L)|\leq |O_p(L)/O_p(L)\cap A||A\cap O_p(L)|=|O_p(L)|.\]
Since $V$ is a simple $\K H$-module and $1\neq H^*\unlhd
H$, $V=[V,H^*]$ and so $C_{V^*}(H^*)=0$, where $V^*$ is the dual of
$V$ has a $\K H$-module. So we can apply the results we proved about $H$ and $V$ to $H^*$ and
$V^*$. Put
\[\ca L^*:=\{L^*\leq H^*\mid L^*=\<\ca A^*_{L^*}\>, \dim_\K C_{V^*}(L^*)=1\}.\]
By \rf 5\rf{5:c} there exists $L^*$ with $O_p(L^*)\neq 1$. By the result
proved in \rf 6 so far ( applied to $L^*$ and $V^*$), $O_p(L^*)\in
\ca A$ and $|D^*|\leq |O_p(L^*)|$. Let $D\in \in \ca A$ with $O_p(L^*)\leq
D$. Then
\[|A|\leq |O_p(L)|\leq |\D^*|\leq |O_p(L^*)|\leq |D|=|A|\]
It follows that $|A|=|D^*|$, $O_p(L)=D^*$ and $O_p(L^*)=D$. From
$|A|=|O_p(L)|$ we get $B=O_p(L)$ and $|B/C_B(A)|=|A/C_B(A)$. Since the
theorem holds for $L$ and $V/C_V(L)$ in place of $H$ and $V$ we
conclude that $L/O_p(L)\cong SL_{n-1}(\F)$ and $B=O_p(L)$ is natural
module for $L$. Thus $|O_p(L)|=|\F|^{n-1}$. Since $D\in \ca A$ and $H$
acts transitively on $\ca A$, $H=\<\ca A\>=\\leq H^*$ and so
$H=H^*$. Thus all parts of \rf 6 are proved.
\bd {6.5} Let $A\in \ca A$ and put $L^*_A:=\<\ca A^*_{N_H(A)}\>$. Then
$L^*_A\in \ca L^*$ and $A=O_p(L^*_A)$.
\ed
By \rf 5\rf{5:c} there exists $R\in \ca L^*$ with $O_p(R^*)\neq 1$. By \rf 6, $O_p(R)\in \ca A$
and so we may assume that $O_p(R^*)=A$. Thus $A\leq R^*\leq L^*_A$. Note that $A\leq O_p(L^*_A)$. Since
$L^*_A$ normalizes $[V,A]$ we have $[V,L^*_A]=[V,A]$ and so $L^*_A\in
\ca L^*$. Thus by \rf 6, $|R^*|=|L^*_A|$ and so $R^*=L^*_A$ and $A=O_p(L^*_A)$.
\bd 7 Let $A,B\in \ca A$ with $A\neq B$. Then exactly one of the
follwing holds.
\bl[1]
\li {7:a}There exists $D\in \ca A^*$ with $D\leq N_H(A)$ and $B\leq
N_H(D)$.
\li {7:b}$\\in \ca L$ and $O_p(\)=1$.
\el
\ed
Pick $L\in \ca L$ with $\\leq L$.
Suppose that $D:=O_p(L)\neq
1$. Then by \rf 6, $D\in \ca A^*$. Clearly $B\leq N_H(D)$ and since
$AD$ is a $p$-group and $A$ is weakly closed, $D\leq N_H(A)$. Thus
\rf{7:b} holds.
Suppose that $O_p(L)=1$. Then by \rf 4, $L=\$ and so \rf{7:b}
holds.
Suppose for a contradiction that \rf{7:a} and \rf{7:b} holds. Since
$A$ is weakly closed, $A\leq N_H(D)$ and so $L\leq N_H(D)$. Since
$O_p(L)=1$, $[A,D]\leq A\cap D\leq L\cap D\leq O_p(L)=1$. But
$C_H(D)=\op Z(H)D$ and so $A\leq D$ and $A=1$, a contradiction. So
\rf{7} is proved.\medskip
We now divide the proof into two cases.
\bcase{op(l)= 1} Suppose that $O_p(L)= 1$ for some $L\in \ca
L$. Then \rf c holds.
\ecase
By \rf 4 $n=3$, $p=2$ and $L\cong \SL_2(\F)$ subfield $\F$ of $\K$ with $|\F|=|A|$.
Let $A \in \ca A_L$. By \rf 5 there exists $R\in \ca L$ with
$O_p(L)\neq 1$ and by \rf 2 we can choose $R$ such that $A\leq R$.
Suppose that $|A\cap O_p(R)|>2$. Then the action of $L$ on $V$ shows
that $[V,A\cap O_p(R]=[V,A]$. But this is a contradiction since
$[V,O_p(R)]$ is $1$-dimensional and $[V,A]$ is a hyperplane. Thus
$|A\cap O_p(R)|=2$. By \rf
6, $R/O_p(R)\cong \SL_{2}(\E)$ for some subfield $\E$ of
$\K$. Morever, $|A|=|\E|^{n-1}=|\E|^2$ and $|A\cap O_p(R)|=|\E|$. Thus $|\E|=2$
and $|\F|=|A|=4$.
By \rf{6.5}, $N_H(A)\neq N_H(C_V(L))$. So $N_H(A)$ does not normalizes
$L$. Thus \rf{5}\rf{5:b} implies that $|\ca L_-(A)|=2$. Each $T\in \ca
L_-(A)$ is isomorphic to $\SL_2(4)$ and so contains four elements of
$\ca A$ other than $A$. So there exists eight elements of $\ca A$
fulfilling \rf 7\rf{7:b}.
By \rf {6.5} and \rf 7, $L_A^*\cong \Sym(4)$ and so there exist
exactly three $D\in \ca A^*$ with $D\leq N_H(A)$. Similarly there
exists three elements $B\in \ca A$ with $B\leq N_H(D)$, two of which
are different from $A$. If $B\neq A$, then $\=L_D$ and so
$D=O_p(\)$ is uniquely determined by $A$ and $B$. Thus there are
$6=2\cdot 3$ elements of $\ca A$ fulfilling \rf 7\rf{7:a}. This shows that
$|\ca A|=1+6+8=15$. Since $C_H(C_V(A))=A$, we conclude from
\ref{nh(a)} that $N_H(A)=\op Z(H)L_A^*$. Thus $|N_H(A)/\op Z(H)|=24$ and
$|H/\op Z(H)|=24\cdot 15=360$. Since $|L\op Z(H)/\op Z(H)|=|L|=60$ we have
$|H/L\op Z(H)|=6$ and so $H/\op Z(H)\cong \Alt(6)$. The elements of order
three in $L_A^*$ acts fixed-point freely on $C_V(A)$, but the elements
of order three in $N_L(H)$ do not. Thus
$N_L(A)\leq L_A^*$. Hence $\op Z(H)\neq 1$ and $|\op Z(H)|=3$. Therefore
$H\sim 3.\Alt(6)$ and \rf c holds in this case.
\bcase{opl neq 1} Suppose that $O_p(L)\neq 1 $ for all $L\in \ca L$. Then \rf
d holds.\ecase
Let $\F$ be as in \rf 6. By \rf 6\rf{6:d}, $|\F|$ and so also $\F$ is
independent of the chooice of $L\in \ca L$. Let
\[ \ca W =\{x\in V\mid x\in C_V(A^*) \text{ for some } A^*\in \ca A^*
\}\]
Let $\ca W^\sharp=\ca W\setminus \{0\}$. For $x\in \ca W^\sharp$ let $A^*_x\in \ca A^*$ with $x\in C_V(A^*_x)$ and observe
that $A^*_x$ is uniquely determined by $x$. Define the relation
'$\sim$' on $\ca W^\sharp$ by $x\sim y$ if either $x\F=y\F$ or $y\in
[x,A^*_y]$. Since $x\F=x\F$, $\sim$ is reflexive.
\bd 8 Let $x,y\in \ca W^\sharp$ with $x\sim y$. Then $y\sim x$,
$x\F+y\F\subseteq \ca W$. If $x\F\neq y\F$, then $x\F+y\F$ is a natural
$\F SL_2(|F)$-module for $\$ and $y\F=[x,A^*_y]$.
\ed
If $x\F=y\F$ then clearly $y\sim x$ and $x\F+y\F\subseteq \ca W$. So
suppose $y\in [x,A^*_y]$. Put $R^*=\$ and pick $L^*\in \ca L^*$ with
$R^*\leq L^*$. Put $V_1:=x\K+y\K$ and observe that $R^*$ normalizes $V_1$. From
\rf 7 we conclude that $R^*/C_R^*(V_1)\cong \SL_2(\F)$ and $V_1\cong
W_1\otimes_\F \K$ for some natural $\F SL_2(\F)$-module $W_1$ for
$R^*$. Let $z\in \{x,y\}$. Since $z\in C_{V_1}(A^*_z)$, $x\leftrightarrow
w_z\otimes k_z$ for some $w_z\in W_1$ and $k_z\in \K$. Since $y\in
[x,A_y]$ and since $W\otimes k_x$ is invariant under $A_y$ we
conclude that $w_y\otimes k_y\in W\otimes k_x$. Thus
$x\F+y\F\leftrightarrow W\otimes k_x$. Thus $x\F+y\F$ is invariant
under $R^*$,$x\F+y\F$ is natural $SL_2(\F)$-module for $R^*$, $R^*$ acts
tranistively on $(x\F+y\F)^\sharp$, $x\F+y\F\subseteq \ca W$, $[x,A_y^*]=y\F$ and
$x\in [y,A_x]$. So \rf 8 holds.
\bd 9 $\sim$ is an equivalence relation on $\ca W$.\ed
We already proved that $\sim $ is reflexive and symmetic. To show that
$\sim$ is tranistive, let $x,y,z\in \ca W^\sharp$ wirth $x\sim y$ and
$y\sim z$. If $x\F=y\F$ and $y\F=z\F$, then $x\F=y \F$. If
$x\F=y \F$ and $y\F\neq z\F$, then $x\in y\F\in y\F+z\F$ and by
\rf 8, $[x,A^*_z]=z\F$ and so $x\sim z$. So we may assume that
$x\F\neq y\F$ and similarly that $y\F\neq z\F$.
Put $V_1:=x\K+y\K$, $R^*:=\$. and $N=N_H(V_1)$. Then $\<\ca A^*_N\>\leq L^*$ for some
$L^*\in \ca L^*$ and from \rf 6 gives that $\<\ca A^*_N\>=R^*$. By \rf 8 $\=x\F+y\F$.
Suppose first that $z\in V_1$. Then $V_1=z\K+y\K$ and by summetry $R=\<\ca
A^*_N\>=\{A^*_z,A^*_x\>$ and $\=z\F+y\F$. Hence by \rf 8, $x\sim z$.
Suppose next that $z\notin V_1$. Note that
$C_V(C_{A^*_y}(x))=V_1$. Thus $C_V(C_{A^*_y}(x))\neq
C_V(C_{A^*_y}(z))$. Since both $C_V(C_{A^*_y}(x))$ and $
C_V(C_{A^*_y}(z))$ are $\F$-hyperplanes of $A_y^*$ we conclude that
$A^*_y=C_{A^*_y}(x)C_{A^*_y}(z)$. Thus
\[[z+V_1,C_{A^*_y}(V_1)]=[z,C_{A^*_y}(x)]=[z,A^*_y]=y\F.\]
By \rf 8 there exists $g\in R^*$ with $y^g=x$. Since $[V,R^*]\leq V_1$
we have
$(z+V_1)^g=z+V_1$. So conjugating the preceeding line by $g$ gives
\[[z+V_1,C_{A^*_x}(y)]=x\F\]
Hence $x\in [z,A^*_x]$ and $x\sim z$. Thus \rf 9 holds.
\bd {10} Let $W$ be an equivallence class of $\sim$. Then $W$ is an
$\F$-subspace of $V$, $H$ normalizes $W$, $H$ induces $SL_\F(W)$ on $V$ and
$V\cong W\otimes_\F \K$ as an $\K H$-module.
\ed
It follows from \rf 8 that $W$ is an $\F$-subspace of $V$. Let $x\in
W^\sharp$, $A\in \ca A$ and $y\in [x,A]^\sharp$. Then $A_y=A$ and
$x\sim y$. So $y\in W$ and $H$ normalizes $W$. If $w\in W\setminus
x\F$, then $[w,A_y]=x\F$. Hence by the dual version of \ref{simple slm}, $H$ induces
$SL_\F(W)$ on $W$. By the universal property of the tensor product
there exists a unique $\F$-linear map $\alpha: W\otimes_\F \K\to V$ with
$(w\otimes k)\alpha=wk$ for all $w\in W, k\in \K$. Clearly this map is a $\K H$-homorphism. Since
both $W\otimes_\F \K$ and $V$ are simple $\K H$-modules, $\alpha$ is
an isomorphism. Hence \rf{10} holds. \medskip
\rf{10} completes the proof for \rf{opl neq 1}.\qed
\be{definition}{def:aug wreath}
Let $L$ be a group and $m$ an integer with $m>1$. Then $\op{Wr}(K,m)$ denotes the
augmented wreath-product of $K$ with $\Sym(m)$, that is $\op{Wr}(L,m)$ is
the normal closure of $\Sym(m)$ in $L\wr \Sym(m)$. So $L\wr
\Sym(m)/\op{Wr}(L,n)\cong L/L^\prime$ and if $L$ is
perfect, $\op{Wr}(L,n)=L\wr \Sym(n)$.
\en{definition}
\be{theorem}{nearly quadratic module theorem}
Let $H$ be a finite group,
and $V$ a faithful simple $\F_p H$-module. Suppose that
$H$ is generated by nearly quadratic, but
not quadratic subgroups of $H$.
Let $W$ a Wedderburn -component for
$\F_p\op F^*(H)$-submodule on $V$ and $\K:=\op Z(\End_{\op F^*(H)}(W)).$
Then $W$ is a simple $\F_p \F^*(H)$-module and one of the follwing
holds for $H,V,W,\K$ and ( if $V=W$) $H/C_H(\K)$
\newcounter{count}
\newcommand{\nc}{\stepcounter{count}\thecount.}
\[\begin{array}{|r|*6{c|}}\hline
&H &V & W & \K & H/C_H(\K)& \\\hline
\nc&(C_2\wr \Sym(m))^\prime & \F_3^m& \F_3& \F_3 & -& m\geq 3, m\neq
4\\
\nc &\SL_n(\F_2)\wr \Sym(m) & (\F_2^n)^m & \F_2^n& \F_2&- &m\geq 2, n\geq 2\\
\nc &\op{Wr}(\SL_2(\F_2),m) & (\F_2^n)^m & \F_2^n& \F_4 & -& m\geq 2\\\hline
\nc &\Frob(39) & \F_{27} & V & \F_{27} & C_3&\\
\nc &\Gamma \GL_n(\F_4)& \F_4^n& V & \F_4& C_2 & n\geq 2\\
\nc &\Gamma \SL_n(\F_4)& \F_4^n& V & \F_4& C_2& n\geq 2\\
\nc &\SL_2(\F_2)\times \SL_n(\F_2)& \F_2^2\otimes \F_2^n& V & \F_4& C_2& n\geq 3\\
\nc &3\udot \Sym(6) & \F_4^3 & V & \F_4 & C_2 & \\\hline
\nc &\SL_n(\K)\circ \SL_m(\K)& \K^n\otimes \K^m& V &\text{any}& 1&n,m\geq 3 \\
\nc &\SL_2(\K)\circ \SL_m(\K)& \K^2\otimes \K^m& V & \K\neq \F_2&
1&m\geq 2 \\
\nc &\SL_n(\F_2)\wr C_2& \F_2^n\otimes \F_2^n& V &\F_2& 1&n\geq 3\\\hline
\nc &(C_2\wr \Sym(4))^\prime & \F_3^4& V& \F_3 & 1&\\
\nc & \op{SU}_3(2)^\prime & \F_4^3 & V & \F_4 & 1 & \\\hline
\nc & F^*(H)=\op Z(H)K, K\text{ quasisimple}& ? & V & ? & 1&\\\hline
\end{array}\]
Moreover, in case 13., $H$ is not generated by abelian, nearly
quadratic subgroups.
\en{theorem}
\proof Let $\ca Q$ be the set of nearly quadratic but not quadratic
subgroups of $H$. To simplify notation we view $H$ as a subgroup of $GL_\F(V)$. By assumption $H=\<\ca Q\>$.
Let $\Delta$ be the set of Wedderburn components of $F$ on $V$ and for $A\in \ca Q$ let
$$\Delta_A:= \{W \in\Delta \mid [W,A] \ne 0\}.$$
Since $V$ is a simple $H$-module, Clifford Theory ensures that
$V=\bigoplus \Delta$.
Observe that $\Delta$ is a system of imprimitivity if $|\Delta| > 1$.
Let $W\in \Delta$ and $A\in \ca Q$. Put
\[N:= N_G(W),\, \w N:=N/C_G(W),\, A_W:=N_A(W),\, E:=\.\]
By Clifford Theory $W$ is a simple $\F N$-module. By \ref{simple tensor} $\K$ is a finite field. We now devide the proof into
several cases.
\bcase{w neq i} The case $V\neq W$. \ecase
Since $H$ acts transitively on $\Delta$ and $H=\<\ca Q\>$ there exists
$A\in \ca Q$ with $A\neq A_W$. Hence $|\Delta_A|>1$ and \ref{binherit3} applies. Since $A$ is not quadratic on
$V$, we are in case \rf[binherit3]d.
Clearly $|W| > 2$ since $C_V(F) = 0$. Hence \rf[binherit3]d shows
that $A$ is elementary abelian and either
\bl[i]
\li i $p = 3$, $A_W = 1$ and $|W| = |A| = |\Delta_A| = 3$; or
\li{ii} $p = 2$, $|A/A_W| = |\Delta_A| =2$, $C_W(A_W)=[W,A_W]$, and
$|W/C_W(A_W)| = 2$.
\el
Put $m=|\Delta|$ and let $a\in A\setminus A_W$. Suppose that \rf{i} holds. Then $A$ acts as $3$-cycle on
$\Delta$. It follows that $H/C_H(\Delta)$ is a transitive subgroups of
$H$ generated by 3-cycles and so $H/C_H(\Delta)\cong
\Alt(\Delta)$ and $m\geq 3$. Let $D=N_{\GL_\F}(\Delta)$. Then $D\cong C_2\wr \Sym(m)$
and $O^{3^\prime}(D)=D^\prime\sim 2^{m-1}\Alt(m)$. Put
$X:=C_{D^\prime}(\Delta)$. Then $X$ is an elementary abelian
$2$-group, $X=[X,D^\prime]$, $D^\prime$ acts simple on
$X/C_{X}(D)$, $F\leq X$ and $D^\prime=XH$. Since $F\nleq Z(H)$ we
conclude that $F=X$ and $H=D^\prime$. If $m=4$, the
$O_2(D^\prime)\nleq X$. So $m\neq 4$ and the first case of the Theorem holds.
Suppose that \rf{ii} holds. Then $H/C_H(\Delta)$ is a transitive
subgroup of $\Sym(\Delta)$ generated by 2-cycles and so $H/C_H(\Delta)\cong
\Sym (\Delta)$. Also $|A_W| = |[w,A_W]| = |[W,A_W]|$ for $w \in
W\setminus C_W(A_W)$ and thus $|A_W| = |C_W(A_W)|$. Since $W$ is a
simple $\F N$-module, the dual version of \ref{simple slm} implies that $\w E \cong
SL_{\F_2}(W)$. Since $C_V(F)\neq 0$, $\w F\neq 1$. Since
$\w E$ normalizes $\w F$ we conclude that either $ \dim_{\F_2}W > 2$ and $\w
F=\w F$ or $ \dim_{\F_2}W > 2$ and $\w F=w E^\prime\cong C_3$.
Assume first that $n:=\dim_{\F_2}W > 2$. Then there exists a
component $K$ of $H$ such that $F = KC_F(W)$ and $C_F(W) = C_F(K)$.
Suppose that $K$ is $A$-invariant, then $KA_W$ also induce
$SL_{\F_2}(W^a)$ on $W^a$. Moreover,
$C_F(W) = C_F(K)=C_F(W^a))$ and $W$ and $W^a$ are
isomorphic $F$-modules since $A_W$ centralizes a hyperplane in both. This
contradicts the fact that $W$ and $W^a$ are distinct Wedderburn
components. Thus, we have that $K\neq K^a$. It follows that $F\cong
K^m\cong SL_n(\F_2)^m$ and the second case of the theorem holds.
Assume now that $n = 2$. Put $D=N_{\GL_\F}(\Delta)$. Then $D\cong
SL_2(2)\wr \Sym(m)$. Let $D_1$ be the inverse image of $
\op{Wr}(\SL_2(2),m)$ in $D$ under this isomorphism. Put
$B=C_D(\Delta)\cong \SL_2(2)^m$ and $B_1=B\cap D_1$. Since $A$ centralizes
each $W_0\in \Delta\setminus \Delta_A$ and $A$ is elementary abelian
we have that $A\leq D_1$.
Each $2$-subgroup of $B$ acts quadratically on $V$ and so each member of $ \ca
Q$ acts non-trivially on $\Delta$. Thus $H\leq D_1$ and
$D_1=B_1H$. It follows that $B_1=B^\prime\$ and
$D_1=B^\prime H$. Note that $1\neq F\leq B^\prime$. We claim that
$F=B^\prime$. If $m>2$, then
$D$ and so also $H$ acts simply on $B^\prime$. Hence
$B^\prime=F$. Since $C_3$ has a unique non-trivial simple module
over $\F_2$ and $W$ is a Wedderburn component for $F$, $|F|>3$. If
$m=2$, then $|B^\prime|=3^2$ and so again $F=B^\prime$. From
$F=B^\prime$ we conclude that $H=D_1$ and so the third case of the
theorem holds.\medskip
\bcase{homogeneous} The case $V = W$ and $H$ is not $\K$-linear on $V$.\ecase
Since $H=\<\ca Q\>$, there exists $A\in \ca K$ such that $A$ is not
$\K$-linear on $V$. Hence we can apply \ref{bnearly quadratic
fieldext}. Since $A$ is not
quadratic on $V$, we are either in case \rf[bnearly quadratic fieldext]b or
case (\ref{bnearly quadratic
fieldext:c}). If \rf[bnearly quadratic fieldext]c holds, then it
is easy to see that the third case of the Theorem holds.
Assume now that \rf[bnearly quadratic fieldext]b holds. Then
$$|\K|=4,\; [V,A_\K]=C_V(A_\K),\;|A/A_\K| = 2,\hbox{ and
}|V/C_V(A_\K)|=4.\leqno{(*)}$$
In particular, $A_\K$ acts quadratically on $V$ and $A_\K \ne 1$. This last
property excludes the case $\dim_\K V= 1$. Let $E_1=\$. From
\ref{char slm} we conclude that either $E_1\cong \SL_{\F_4}(V)$, $E_1\cong
\SL_{\F_2}(U)$ where $U$ is an $\F_2$-space with $V\cong U\otimes_{\F_2}\F_4$, or
$\dim_{\F_4} V=3$ and $E\cong 3\udot\Alt(6)$. In the first case
$\Gamma \GL_\K(V)/E_1\cong \Sym(3)$ and so either $H=\Gamma \GL_\K(V)$
or $|H/E_1|=2$ $H=\Gamma \SL_\K(V)$. In the second case
$N_{\Gamma_{\F_2}(V)}(E_1)\cong \SL_2(\F_2)\times \SL_{\F_2}(U)$ and so
$H\cong \SL_2(\F_2)\times \SL_{F_2}(U)$. In the last case
$N_{\Gamma GL_{\K}(V)}(E_1)\cong 3\udot\Sym(6)$. It follows that Case 5,6,7
or 8 of the Theorem holds. \medskip
We may assume from now on that $V=W$ and $H$ acts $\K$-linear on $V$. Let $\ca I$ be the set of all $X\nl F$ such that either
$X$ is a component of $G$ or $X = O_q(G)$ with $X^\prime\neq 1$,
$q$ a prime.
We show next that
\bd {1.5} $\op Z(F)=\op Z(H)=F\cap \K$ and $F=\op Z(H)\<\ca I\>$ \ed
The definition of $\ca I$ implies that $F=\op Z(F)\<\ca K\>$. Moreover
$\op Z(F)\leq C_{GL_\F(V)}(F)\cap F\subseteq \K$. Since $H$ is $\K$-linear we
conclude that $\op Z(F)=\op Z(H)=F\cap \K$ and so \rf{1.5} holds.
\bd 2 Let $A\in \ca Q$. Then there exists $D\in \ca I$ such that $[D,A] \ne 1$.\ed
Otherwise \rf{1.5} implies that $[F,A]= 1$ and so $A\leq \op Z(F)$ and
$A\leq O_p(F)=1$, a contradiction.
\bigskip
\bcase{klinear} The case $V= W$, $H$ is $\K$-linear on $V$ and $V$ is
not a simple $\F F$-module.\ecase
We apply \ref{tensor decomposition for homogenous} with $I=\{1\}$ and
$D_1=F$. Hence there exists a tensor decompsoition
$$\ca T=(\Phi,\otimes_\K^{J}V_j,\sigma,(g_j,j\in J,g\in H))$$
where $J=\{0,1\}$, $H$ acts trivially on $J$, $V_1$ is a simple $\K F$-module, $V_0$ is a trivial
$\K F$=module and $\Phi: V_0\otimes V_1\to V$ is a $\K
F$-isomorphism. Moreover, $\ca T$ is strict when restricted to $A$. Since $V$ is a simple $\F_p H$-module we conclude that
$V_0$ and $V_1$ are simple projective $\K H$-module. By \rf 2, $F$
is not abelian and so $\dim_\K V_1\geq 2$. Since $V$ is not a simple
$\F H$-module, $\dim_\K V_0\geq 2$. Thus $\ca T$ is proper, regular
and $\K$-linear. So \ref{nearly quadratic and tensor} applies. It
follows that for $j=0,1$, $[V_j,A]$ is a $\K$-hyperplane of $V_j$ and $A$ induces
$C_{SL_{\K}(V_i)}([V_j,A]) $ on $V_j$. Thus the dual
version of \ref{simple slm} shows that $\$ induces
$\op{PSL}_\K(V_j)$ on $\cap P_\K(V_j)$. Since $F$ is normal in $H$ and acts faithfully
on $V_1$ we conclude that $F\cong SL_\K(V_1)^\prime$. Let $Z=C_G(\ca
P_\K(V_1))$. Then $[Z,F]\leq Z\cap F\leq Z(F)\leq \op Z(H)$. But
$C_H(F/\op Z(H))\leq F$ and so $Z=\op Z(F)=\op Z(H)$. Hence $H/\op Z(H)$ is isomorphic
to a subgroup of $\op P\Gamma \GL_\K(V_1)$ containing $\op{PSL}_\K(V_1)$. Since
$F$ acts trivially on $\ca P_\K(V_0)$ we see that $H/C_H(\ca
P_\K(V_0)$ is isomorphic to a section of $\op P\Gamma
\GL_\K(V_1)/\op{PSL}_\K(V_1)^\prime$. Therefore $H/C_H(\ca
P_\K(V_0)$ is solvable. On the otherhand $H/C_H(\ca
P_\K(V_0)$ contains a subgroup isomorphic to $\op{PSL}_\K(V_0)$. Hence
$p=2$ or $3$, $\K=\F_p$ and $\dim_\K V_0=2$. In particular, $\op P\Gamma
\GL_\K(V_1) =\op P\GL_\K(V_1)$. Since
$PSL_\K(V_0)$ is not abelian, we conclude that
$PGL_\K(V_1)/SL_\K(V_1)^\prime$ is not abelian. Hence $\K=\F_3$,
$\dim_\K V_1)= $ and $PGL_\K(V_1)/SL_\K(V_1)^\prime\cong
\Sym(3)$. Thus $H/C_H(\ca
P_\K(V_0))$ has order at most 6. But
$\op P\SL_\K(V_0)\cong \op P\SL_2(3)$ has order $12$, a contradiction.
This contradiction shows that case \rf{klinear} does not occur.
\bcase{simple} The case $V$ is a simple $\F_p F$-module, $H$ is $\K$-linear and
$|\ca I|\geq 2$.
\ecase
Let $D_1,\ldots,D_r$ be the elements of the set $\ca I$ and put $I :=
\{1,\ldots,r\}$. Since $H$ acts on $\ca I$ by conjugation, $I$ is an $N$-set
with respect to the induced action; i.e., $D_{ig} := D_i^g$ for $g \in
G$. Hence, the module $V$, $I$ and $\ca I$ satisfy the conditions
(\ref{tensor decomposition for homogenous:i})
and (\ref{tensor decomposition for homogenous:ii}) of \ref{tensor
decomposition for homogenous} (with $\F = \F_p$ and $H$ in place of $G$).
Thus there exists a tensor decomposition
$$\ca T=(\Phi,\otimes_\K^{I}V_i,\sigma,(g_i,i\in I,g\in H))$$
where $V_i$ is a
simple $\K D_i$-module for $i\in I$ and $V_j$ is a trivial $\K D_i$
for $D\neq i\in I$. Moreover
$$\Phi: \,\otimes_\K^JV_j \to V$$
is a $\K(D_1\times \ldots \times D_r)$-module isomomorphism
and the decomposition restricted to $A$ is strict. Since each
$D_i$ is non-abelian,
$\dim_\K V_i\geq 2$ for each $i\in I$. By assumption of the current case, $|I|\geq
2$. So $\ca T$ is proper. Thus we are allowed to apply \ref{nearly quadratic and
tensor decompositions}. Since $A$ is not quadratic we conclude that
one of the following holds:
\bl
\li[\rf{klinear}a]{ca} $| I| = 2$, $I$ is a trivial $A$-set, and
$[V_j,A] = C_{V_j}(A) = [v_j,A]$ is a
$\K$-hyperplane of $V_j$ for $v_j \in V_J\setminus [V_j,A]$ and $j \in
J$.
\li[\rf{klinear}b]{cb} $|I| = 2$, $J$ is a non-trivial $A$-set, $\K =
\F_2$,
and $[V_j,B] = C_{V_j}(B)$ is a $\K$-hyperplane of $V_j$ for $j \in J$,
where $B := C_A(J)$.
\el
Let $H_0=C_H(J)$, $B=C_A(J)$ and $I=\{i,j\}$. Since $V_i$ is a simple $\K
D_i$-module, $V_i$ is a simple projective $\K H_0$-module.
Thus the dual
version of \ref{simple slm} shows $\**$ induces $\op{PSL}_\K(V_i)$
on $\ca P_\K(V_i)$. Since $H_0$ normalizes $D_i$ and $D_i$ acts faithfully on
$V_i$ we conclude that $D_i\cong \SL_\K(V_i)^\prime$. If $\SL_\K(V_i)$
is perfect for $i=1$ and $2$ we see that one of the case 9,10 and 11
of the Theorem holds. So suppose $\SL_\K(V_i)$ is not perfect for some
$i\in I$. Since
$D_i$ is not abelian, $\K=\F_3$ and $\dim_\K V_i=2$. Thus
$D_i=O_3(H)$. Hence $D_j\neq O_3(H)$ and so $\dim_\K V_j\geq 3$ and
$D_j\cong \SL_\K(V_j)$ It
follows that Case 10 of the Theorem holds.
\bcase{quasimple} The case $V$ is a simple $\F_p H$-module, $H$ is
$\K$-linear on $V$ and $\ca I=\{D\}$ with $D$ not solvable.
\ecase
Since $D$ is not solvable, $D$ is a component of $G$. By \rf{1.5}
$F=\op Z(H)D$ and so Case 14 of the Theorem holds.
\bcase{solvable} The case $V$ is a simple $\F_p F$-module, $H$ is
$\K$-linear on $V$ and $\ca I=\{D\}$ with $D$ solvable. \ecase
By definition of $\ca I$ there exists a prime $r$ such that $D = O_r(G)$. Put $K=[D,A]$. By
coprime action $D=C_D(A)K$ and since $F=\op Z(H)D$, $F=C_F(A)K$. So
$K=[K,A]$ and $KA=\=\$.
We choose a normal subgroup $R$ of $FA$ that is minimal with $[R,A]\neq
1$, so $R \leq K$.
\bd{14.5} $R = [R,KA]$, and either $R$ is elementary abelian, or $C_R(A) =
\op Z(R)$ and $A$ acts simply on $R/\op Z(R)$.\ed
Note $[R,\] $ is a normal subgroups of $FA$ contained in
$R$. Since $[R,A,A]=[R,A]\neq 1$, the minimality of $R$ gives
$R=[R,\]=[R,KA]$.
As $[R,F]=[R,D]< R$ the minimality of $R$ gives $[R,F,A]=1$. So
$[R,F]\leq C_R(A)$ and $C_R(A)$ is a normal subgroup of $FA$. Since
$F$ centralizes $R/C_R(A)$, the
minimality of $R$ implies that $A$ acts simply on $R/C_R(A)$. Since $FA$
normalizes $C_R(A)$, $\=AK$ centralizes $C_R(A)$ and so
$C_R(A)=C_R(KA)\leq \op Z(R)$.
If $R$ is non-abelian, the
minimality of $R$ yields $C_R(A) = \op Z(R)$. If $R$ is abelian, then
$R=[R,A]\times C_R(A)$. Hence $[\Omega_1(R), A]\neq 1$ and
$\Omega_1(R) = R$. So $R$ is elementary abelian.
\bd{15} $1\neq C_{R}(F)\subseteq \K$; in particular $|\K|>2$.\ed
Since $R$ is a normal subgroup of $O_r(F)$, $C_R(F)=C_R(D) \ne 1$. By the
definition of $\K$, $C_R(F)\subseteq \K$ so $|\K|>2$.
\bd{14} $V$ is a simple $\F_p K A$-module.\ed
Let $V_1$ be a simple $\F_p \ A$-submodule of $V$. Since $V$ is
simple $\F_p F$-module and $F=C_F(A)K$ we get $V=\$ acts transitively on $\Delta$, so $KA/C_{KA}(\Delta)\cong
\Alt(\Delta)$. The solvability of $KA$ gives $|\Delta|=3$ or $4$,
and since $|\Delta| $ is a power of $r$, $r=2$ and $|\Delta|=4$. Thus
$FA$ is a subgroup of $N_{GL_{\K}(V)}(\Delta)\cong C_2\wr
\Sym(4)$. Since $R\leq C_H(\Delta)$ and
by \rf{14.5} $R=[R,KA]$ we conclude that $R=[R,KA]\cong C_2^3$ and $KA
\cong 2^3:\Alt(4)$. This shows that $K \cong Q_8\circ Q_8$. Hence
$F\leq C_H(\Delta)K$ and $F\cap \SL_\K(V)=K$. Thus $K\unlhd H$ and
$H\leq N_{\GL_\K(V)}(K)\sim (\GL_2(3)\circ \GL_2(3)).2$. Since
$H=\<\ca Q\>=O^{3^\prime}(H)$
we conclude that either $H=KA$ and Case 12 of the Theorem holds or
$H\cong SL_2(3)\circ SL_2(3)$ and Case 10 holds.
\bigskip
We may assume from now on that $V$ is a homogeneous $\F_p R$-module. By
\ref{simple tensor} $\E:=\op Z(\End_R(V))$ is a field, so $\K\leq \E$, and $F
A$ acts $\E$-semi-linearly on $I_A$.
\bd{17} $A$ acts $\E$-linearly on $V$, and $R$ is not abelian.\ed
Suppose $A$ does not act $\E$-linearly on $V$. Since $A$ is not quadratic
on
$V$, \rf[bnearly quadratic fieldext]b or (\ref{bnearly quadratic fieldext:c}) holds. In both
cases $\dim_{\F_p}\E=p$ and so $|\Aut(\E)|=p$. Since $D$ is an
$r$-group and $F=D(D\cap \K)$, $\E \leq \End_F(V) = \K$, and $A$ acts
$\E$-linearly on $V$, a contradiction.
Thus $A$ is $\E$-linear. If $R$ is
abelian, then $R \leq \End_R(V)\cap R\subseteq \E$ and $[R,A]=1$, a contradiction.
\bd{18} Suppose $V$ is a homogenous but not simple $\F_p R$-module.
Then Case 10 or 12 of the Theorem hold.
\ed
By \ref{tensor decomposition for homogenous} there exists a tensor
decomposition $(\Phi, V_0\otimes_\E V_1, \sigma,(h_j \mid j=0,1,\,h \in
FA))$ such that $V \cong V_0\otimes_\E V_1$ as an $\E R$-module,
$V_1$ is a simple $\E R$-module and $V_0$ is a trivial $\E R$-module. Since $R$ is not
abelian, $\dim_\E V_1>1$, and since $V$ is not a simple $R$-module,
$\dim_\E V_0>1$. Moreover, by \rf{17} $A$ acts $\E$-linearly on $V_0\otimes_\E
V_1$. Hence one of the two cases in \rf[nearly quadratic and tensor
decompositions]{4} hold.
In both cases the dual version of \ref{simple slm} shows $KA=\$
induces $PSL_\E(V_i)$ on $\ca P_\E(V_i)$. Since $R$ is solvable, $p=2$ or
$3$, $\E=\F_p$ and $\dim_\E V_i=2$. Since $R$ is non-abelian, $p=3$ and $R\cong
Q_8$. Hence $FA\leq N_{\GL_\K(V)}(R)\cong
\GL_2(3)\circ\GL_2(R)$. Since $A$ is not quadratic on $V$, $A$ is not
contained in any of the normal $GL_2(R)$. Since $A$ normalizes $F$
and $F$ is simply on $V$, $F\cong O_2(\GL_2(3)\circ\GL_2(3)\cong Q_8\circ
Q_8$. Thus $H\leq N_{\GL_\K(V)}(F)\cong
(\GL_2(3)\circ\GL_2(3)).2$. It follows that Case 10 or 12 of the
Theorem holds.
\bd{19} Suppose $V$ is a simple $\F_p R$-module. Then Case 13 of the
Theorem holds. \ed
By \rf{17} $R$ is non-abelian, and by \rf{14.5} $\o R:=R/\op Z(R)$ is a simple
$A$-module. Let
$X$ be maximal abelian normal subgroup of $A$. Put
\[\ca Y=\{C_X(U)\mid U \text{ a simple } X-\text{ submodule of } \o
R\}\]
For $Y\in \ca Y$ let
\[\o R_Y=\< Y\mid Y \text{ a simple } X\ \text{ submodule of } \o
R \text{ with } C_X(U)=Y\>\]
Then $\o R_Y$ is a sum of Wedderburn components for $X$ on $\o R$ and
so
\[ \o R=\bigoplus_{Y\in \ca Y } \o R_Y\]
Suppose for a contradiction that $|\ca Y|\geq 2$. For $Y\in \ca Y$ let
$R_Y$ be the inverse image of $\o R_Y$ in $R$. Let $Y,Z\in \ca Y$ with
$Y\neq Z$. We claim that $[R_Y,R_Z]=1$. For this we may assume $Y\nleq Z$. Hence
$\o R_Z=[\o R_Z,Y]$ and thus $R_Y\leq Z(R)[R,Y]$. Since $R^\prime\leq
Z(R)\leq C_R(Y)$, $[R,C_R(Y),Y]=1$. Also $[C_R(Y),Y,R]=1$ and so the Three Subgroup
Lemma implies $[[Y,R],C_Y(R)]=1$. So the claim holds. Note that $R=\$. If $R_Y$ is abelian we would conclude $R_Y\leq \op Z(R)$, a
contradiction. So $R_Y$ is not abelian. Each $R_Y$ is normal in $R$
and so $RA$ acts on $\ca Y$ and on $\{ R_Y\mid Y\in \ca Y\}$ by
conjugation. Since $V$ is a simple $\E R$-module \ref{tensor decomposition for homogenous} yields a $RA$
invariant tensor decomposition $\bigotimes_{Y\in \ca Y}^\E V_Y$, where each
$V_Y$ is a faithful simple $\E R_Y$-module. Since $R_Y$ is non-abelian
$\dim_\E V_Y\geq 2$. Thus \ref{nearly quadratic and tensor
decompositions} shows that $A$ is abelian. Thus $X=A$, $\o R$ is a
simple $X$-module and $|\ca Y|=1$, a contradiction.
We proved that $|\ca Y|=1$. Hence $\ca Y=\{Y\}$ for some $Y\in \ca
Y$. Then $[\o R,Y]=1$. Since $[\op Z(R),A]=1$ we have that $C_A(\o R)\leq
C_A(R)\subseteq A\cap \E=1$. Thus $Y=1$ and so $X$ is cyclic. As $X$
is cubic on $V$, the Hall-Higman Theorem B \cite[11.1.1]{Gor} shows
that $|X|\leq 4$ and if $|X|=4$ then $R$ is a $3$-group.
Suppose $|X|=2$, then $X=A$ and $A$ is quadratic, a contradiction.
Suppose $|X|=3$, then $X=A$. Since $\o R$ is a simple $A$-module we get
$|R/\op Z(R)|=4$ and $[R,A]\cong Q_8$. Since $Q_8$ has a unique faithful simple
module over $\E$ and this module has dimension 2, there
exists a $2$-dimensional $\E[R,A]$-submodule $V_0$ in $V$. On the other
hand, $R = \op Z(R)[R,A]$ and $\op Z(R)\subseteq \E$. Thus $R$ normalizes $V_0$ and
$V_0 = V$. But then $A$ acts quadratically on $V$, a contradiction.
Thus $|X|=4$ and so $R$ is a $3$-group. Suppose $A=X$. The Jordan Canonical Form for $A$ on $V$
over $\K$ shows that $V\cong V_0\otimes \F_2\K$ for some $F_2
A$-module $V_0$. By \rf{15} $\F_2\neq \K$ and we obtain a contradiction
to \ref{binherit3}.
Thus $A\neq X$ and so $A\cong D_8$ or $Q_8$. In the first case $A$ has
an maximal abelian subgroup isomorphic to $C_2\times C_2$, a
contradiction. Thus $A\cong Q_8$ and since $A$ act simply on $\o R$,
$|\o R|=3^2$ and $[R,A]$ is extra special of order $3^3$ and exponent
3. Since $V$ is a faithful simple $\E R$-module and
$R=\op Z(R)[R,A]=(R\cap \E)[R,A]$ we have $\dim_\E V=3$. Thus $FA\leq
N_{\GL_\E(V)}(R)\sim \E^\sharp\circ 3^{1+2}.\SL_2(3)$. Hence $F\cap
\SL_\E(V)=R$ and so $R\unlhd H$. Since $H=O^{p^\prime}(H)$ we conclude
that $H=RA\cong SU_3(2)^\prime$. Since $V$ is a simple $\F_p
H$-module, $\E=\F_4$ and Case 13 of the Theorem holds.
\qed
\be{theorem}{components} Let $G$ be a finite group $K$ a
component of $G$ and $V$ be
a faithful $\F_p G$-module. Suppose that there exists
a $p$-subgroup $A\leq G$ with $|A/C_A(K)|>2$ acting nearly
quadratically on $V$. Then $|A/N_A(K)| \leq 2$ and either $A\leq N_G(K)$, or
$p = 2$ and $K/O_p(K) \cong \SL_n(2)$ or $\SL_2{2^m}$.
\en{theorem}
\proof We may assume without loss that $A\nleq N_G(K)$. Let $W$ be
composition factor for $H:=\$ on $V$ with $[W,K]\neq 0$. By
\rf[trivial property] c, $W$ is a nearly quadratic $A$-module. Note
also that $H=\$.
If $A$ is not quadratic on $W$ we conclude
from \ref{nearly quadratic module theorem} that $K/C_K(W)\cong \SL_n(2)$.
Suppose that $A$ is quadratic on $W$. Let $L=\$. Then
$H=LA$. Let $U$ be a Wedderburn component for $L$ on $W$. Then
$W=\**__$. From $|A/C_A(K)|> 2$ we have $|A/C_A(W)|>2$ and
\ref{quadratic imprimitivity} implies that $U=V$. Hence
$W=V$. So $W$ is a homogenous $\F_pL$-module. For example by
\rf[simple tensor] c , the number of simple $\F_p L$-modules in $W$ is
not divisible by $p$ and since $H=LA$ acts simply on $W$, $W$ is a
simple $\F_p A$-module. Let $K^A=\{K_1,K_2,\ldots K_r\}$ and
$I=\{1,\ldots r\}$. Then $|I|>2$ and $A$ acts transitively on $I$ via
$K_{ia}=K_i^a$. By \ref{tensor decomposition for homogenous} there
exists a $H$-invariant tensor decomposition $V\cong \bigotimes_{i\in
I}^\K V_i$, where $V_i$ is a simple $\E K_i$-module. Thus by
\ref{quadratic and tensor2}, $p=2$, $|I|=2$, $\dim_\K V_i=2$ and
$[V_i,C_A(I)]\neq 1$. \ref{char slm} now shows that $K/C_K(W)\cong
SL_2(2^m)$.
Thus $K/C_K(W)\cong \SL_n(2)$ or $\SL_2(2^m)$. Since this holds for all
non-trivial composition factors of $K$ on $V$, $K/O_p(K)\cong \SL_n(2)$
or $\SL_2(2^m)$.\qed
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\bibitem[Hu]{Huppert} B. Huppert, {\it Endliche Gruppen I} Springer
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\end{document}__