|E|^2|C_V(E)|$. This is contrary to the definition of $E$, so we conclude that $p\neq 2$. With $p$ odd, (1) shows that $G\cong \Gamma L(2,27)$ and that $|U|=3^8$. Here $|V/C_V(E)|\leq |E|^2=81$, while $|U/C_U(E)|=81$ since $G$ is generated by two conjugates of $E$. Thus $V=U$, by 1.2, and we have an $F_0$-linear action of $G$ on $U$. Then $G$ acts $F$-linearly on $\w U$. Identify $\w U$ with $N\otimes N^{\phi}\otimes N^{\phi^2}$, where $\phi$ is an automorphism of $F$ of order $3$, and identify $a$ with a standard field automorphism of $H$. Then $a$ acts on $\w U$ by permuting the three tensor factors. Explicitly, let $x=(1,0)$ and $y=(0,1)$ be the standard basis elements for $N$. One may then identify a standard basis for $\w U$ with the set of monomials of degree $3$ in the non-commuting variables $x$ and $y$, and then $C_{\w U}(a)$ is the $F$-linear span of $$ \{x^3, x^2y+xyx+yx^2, xy^2+yxy+y^2x, y^3\}. $$ Thus $dim_F(C_{\w U}(a)=4$. One observes that the given basis for $C_{\w U}(a)$ is not invariant under $E\cap T$, so $dim_F(C_{\w U}(E))\leq 3$. Then also $dim_{F_0}(C_U(E))\leq 3$, and so $|U/C_U(E)|\geq 3^5>|E|^2$. With this contradiction, we have shown that $m\neq 3$, and so $m\leq 2$. If $m=2$ then $U$ is an $\Omega_4^-(q_0)$-module, where $q_0=q^{1/2}$. On the other hand, suppose that $m=1$. Then (3) yields $d\leq 4$. If $d=4$ then $|U|=q^4$, so (1) and (2) imply that $E\nleq T$ and $q=4$. Then $U\cong N\otimes N^{\phi}$, where $N$ is the natural $SL(2,4)$-module for $H$ and where $\phi$ is the non-identity field automorphism. But then $U$ is reducible as an $\Bbb F_2H$-module, and contrary to the definition of $U$. Thus, we have $d\leq 3$. This shows the following. \vglue .2in \item {(4)} $U$ is a natural $SL(2,q)$-module, a natural $\Omega_3(q)$- module, or a natural $\Omega_4^-(q_0)$-module $(q_0=q^{1/2})$, for $H$. Moreover, if $E\neq T$ then $p=2$. \vglue .2in We observe that if $U$ is not $E$-invariant then $E\nleq T$ and $|C_U(E\cap T)|\geq p^2$. Then $|U/C_U(E\cap T)|\geq |E\cap T|$. For any minimal $HE$-invariant section $W$ of $V$ involving $U$ we then have $$ |E\cap T|^2|C_W(E\cap T)|>p^2|E\cap T|^2|C_W(E)|\geq |E|^2|C_W(E)|, $$ and $$ |W/C_W(E\cap T)|\geq |E\cap T|^2. $$ Then $|W|\geq |E|^2|C_W(E)|$, so $|W|=|V|=|E|^2|C_V(E)|$, and $$ |E\cap T|^2|C_V(E\cap T)|>|E|^2|C_V(E)|. $$ This is contrary to the definition of $E$, so we conclude that $U$ is $E$-invariant. Moreover, the preceding estimates show that $HE$ has at most two non-trivial irreducible constituents in $V$, and if there are two then for any choice of $U$ we have $|E|=|U/C_U(E)|$. If $H$ has only one non-trivial constituent in $V$ then 1.2 gives $V=U$. So assume that $H$ has two non-trivial constituents in $V$. If $E=T$ then $|U/C_U(E)|=|U|/|F_0|\leq q$, so $|U|\leq q\cdot q_0$, and then $q=q_0$ and $U$ is a natural $SL(2,q)$-module for $H$. On the other hand, suppose that $E\neq T$. Then (1) and (4) yield $p=2$ and $|U/C_U(E)|\leq |E|=2\cdot q^{1/2}$. As two conjugates of $E$ generate $G$ we then have $|U|\leq 4\cdot q$, and since $|U|\geq q^2$ we then have $q=4$ and $|U|=16$. If $U$ is a natural $\Gamma L(2,4)$-module for $G$ then $|U/C_U(E)|=8$, so in fact $U$ is a natural $O_4^-(2)$-module for $G$. Thus: \vglue .2in \item {(5)} Suppose that $U\neq V$. Then $H$ has exactly two non-trivial irreducible constituents in $V$. Moreover, these constituents are $E$- invariant, and for any such irreducible constituent $U$ we have $|E|=|U/C_U(E)|$. Either both irreducible constituents are natural $SL(2,q)$-modules for $H$, or else $q=4$ and both are natural $O_4^-(2)$- modules for $G$. \vglue .2in Suppose that $U\neq V$, and suppose that there is a trivial constituent for $H$ in $V$. As $C_V(H)=0$ and $V=[V,H]$, by hypothesis, there is then an indecomposable $H$-submodule $W$ of $V$ such that $[W,H]$ is irreducible for $H$ and such that $|W/[W,H]|=p$. Take $U=[W,H]$. Then (5) says that either $U$ is a natural $SL(2,q)$-module for $H$, or $q=4$ and $U$ is a natural $O_4^-(2)$-module for $G$. In either case, we have $W=U+C_W(E)$. Suppose that $U$ is a natural $SL(2,q)$- module. As $|E|=|U/C_U(E)|$ we then have $E=T$ and $W=U+C_W(T)$. As two conjugates of $T$ generate $H$ we then have $C_W(H)\neq 0$, and $W$ is decomposable. Thus $q=4$ and $U$ is an $O_4^-(2)$-module, but in that case we contradict 1.4. Thus, $H$ has no trivial constituents in $V$. Now let $A$ be any elementary abelian subgroup of $S$ such that $|A|^2\geq |V/C_V(A)|$. Suppose first that $|E|^2=|V/C_V(A)|$. Then we may take $A\leq E$ (by definition of $E$). In particular, if $V$ is a natural $\Omega_3(q)$-module for $H$ we have $A\leq E=T$, and then $A=E$. Thus, (iii) holds in this case. In the same way, if $H$ has two non-trivial irreducible constituents in $V$ then $A\leq E=T$. In the case that these constituents are natural $SL(2,q)$- modules we find that $A=E$, and (ii) holds. In the case that these constituents are natural $O_4^-(2)$-modules we observe that $E\cap H$ induces a $2$-transvection on each irreducible constituent, and so $|E\cap H|^2<|V/C_V(E\cap H|$. Thus $A\neq E\cap H$, and so (v) holds in this case. If $V$ is a natural $\Omega_4^-(q)$-module or a natural $SL(2,q)$-module for $H$ then we make no assertion about $A$, and we have (iv) or (i). In order to complete the proof of the lemma, it remains to show that $|A|\leq |V/C_V(A)|$. Suppose false. Then (i) or (iv) holds, by what has already been proved, and we then observe that $A\neq T$. By [REF] we may assume that $A$ is weakly closed in $C_G(A)$, and therefore $A\nleq T$. Setting $A_0=A\cap T$ we then have $T\neq A_0\neq 1$ and $|A_0|\geq |V/C_V(A_0)|$. This condition excludes (i), and so (iv) holds. But $\lan A_0^{N_H(T)}\ran=T$, and [REF] then implies that $|T|\geq |V/C_V(T)|$, which is not the case in (iv). Thus, we have a contradiction, and the lemma is proved. \qed \enddemo \proclaim {6.9 Lemma} Assume Hypothesis 2, with $H$ isomorphic to the commutator subgroup of a group of Lie type and of Lie rank equal to $1$, in characteristic $p$. Assume that $H/Z(H)\ncong PSL(2,q)$. If $G\cong{^2G_2(3)}$ set $T=S$, and otherwise set $T=S\cap H$. Then $|A|^2=|V/C_V(A)|$, $A\leq H$, and one of the following holds. \roster \item {(i)} $H\cong SU(3,q)$ and $V$ is a natural module for $H$. Moreover, either $A=Z(T)$ and $[V,A,A]=0$, or else $p$ is odd, $|A|=q^2$, and $[V,A,A]\neq 0$. \item {(ii)} $H\cong Sz(q)$ and $V$ is a natural module for $H$. Moreover, we have $A=Z(T)$ and $[V,A,A]=0$. \endroster \endproclaim \demo {Proof} As in the proof of the preceding lemma, we fix an elementary abelian subgroup $E$ of $S$ such that $|E|^2|C_V(E)|$ is as large as possible and, subject to this condition, so that $E$ maximal with respect to inclusion. Then 2.2 implies that $E$ is weakly closed in $C_G(E)$ with respect to $G$. Suppose first that $HT/Z(H)$ is isomorphic to $^2G_2(q)$, $q=3^{2n+1}$. Surveying the Schur multipliers of these groups, we find that $Z(H)$ is a $3$-group, and so $Z(H)=1$. We observe right away that $q^3+1$ divides $|G|$, and hence $|V|\geq q^{6}$ by 2.1. Let $D$ be a subgroup of $N_H(T)$ of order $q-1$, let $d$ be an involution in $D$, and set $L=O^2(C_H(t))$. Then $L\cong L_2(q)$, and $L\cap T$ is a complement in $\Omega_1(T)$ to $Z(T)$, by 6.1. Suppose that $E\nleq T$. Then 6.1(d) implies that $E=\Omega_1(C_S(a))$ for any $a\in E-T$. We may choose $a\in E-T$ so that $a$ centralizes an involution in $N_H(T)$, and so we may assume that $[a,d]=1$. Set $X=C_{E\cap T}(d)$. Then $|X|=q^{1/3}$. Here $q$ is an odd power of $3$, so $q\neq 9$, and a theorem of L.E. Dickson (2.8.4 in [Gor]) implies that two conjugates of $X$ suffice to generate $L$. As $E\cap T\nleq X$ we then conclude from 6.1(b) that two conjugates of $E$ suffice to generate $HE$. Then $|V|=|V/C_V(HE)|\leq 81\cdot q^{8/3}$. As $q\geq 3^{3}$ in this case, we then have $|V|\leq q^{4/3}q^{8/3}=q^4$, contrary to $|V|\geq q^6$. We therefore conclude that $E\leq T$. As $\Omega_1(T)$ is abelian, $E$ is normal in $N_H(T)$, and then $E=Z(T)$ or $E=\Omega_1(T)$. If $E=Z(T)$ then two conjugates of $E$ suffice to generate $H$, by 6.1(c), and so $|V|\leq q^4$. We therefore conclude that $E=\Omega_1(T)$. Now two conjugates of $E$ generate $HE$, and so $|V|\leq q^8$. Let $U$ be a non- trivial, irreducible $L$-invariant section of $V$. Then $|U/C_U(E\cap L)|\leq q^4$ and since $E\cap L$ is a Sylow subgroup of $L$ it follows from 6.4(b) that $|U|\leq q^5$. Set $F_0=End_L(U)$, $F=\Bbb F_{q}$, and denote by $\Gamma_0$ the Galois group of $F$ over $F_0$. Define the integers $t$ and $m$ by $|F:\Bbb F_p|=t$ and $|F:F_0|=m$, and set $r=t/m$. Set $\w U=U\otimes_{F_0}F$, and let modules $M_i$ and $N_j$, ($1\leq i,j\leq m$) be given as in 6.6. Write $d_i$ for the dimension of $M_i$ over $F$. As $p=3$ we have $d_i\leq 3$, and there are exactly three basic irreducible modules for $SL(2,q)$. Thus, any $M_i$ is either a trivial module, a natural $SL(2,q)$-module, or a natural $\Omega_3(q)$-module for the universal cover $SL(2,q)$ of $L$. As $|U|\leq q^5$, 4.6(c) yields $$ (d_1\cdots d_r)^m\leq 5m. $$ As $\w U$ affords a representation of $SL(2,q)$ with the scalar matrix $-I$ acting trivially, there are an even number of occurances of the natural $SL(2,q)$-module in the tensor decomposition of $\w U$ given by 6.6. As $|\Gamma_0|$ is odd, it follows that there are an even number (possibly $0$) of the non-trivial modules $N_i$ for which the associated basic irreducible modules $M_i$ are natural $SL(2,q)$-modules. Taking $d_1$ to be maximal among the numbers $d_i$, we then have either $d_1=3$ and $3^m\leq 5m$, or else $d_1=d_i=2$ for some $i\neq 1$ and $4^m\leq 5m$. As also $m$ is odd, we conclude that $d_1=3$, $m=1$, $F_0=F$, and $U$ is a natural $\Omega_3(q)$-module for $L$. As $|V|\leq q^8$ there are then at most two non-trivial irreducible constituents for $L$ in $V$. Suppose that there $L$ has two non-trivial constituents in $V$. As $|U/C_U(E\cap L)|=q^2$ we then have $|V/C_V(E\cap L)|=|V/C_V(E)|=q^4$. Then $C_V(L)\leq C_V(E)$, and since $\lan L,E\ran=H$, by 6.1(b), we conclude that $C_V(L)=0$. Then 1.5 implies that there are no trivial constituents for $L$ in $V$, and so $|V|=q^6$. As all involutions in $H$ are conjugate, all have non-trivial fixed points on $V$, and so the involution $d$ centralizes a subspace of $V$ of order $q^3$. But one observes that for any involution $x\in L$ we have $|C_U(x)|=q$, and so $|C_V(x)|=q^2$. With this contradiction we conclude that $L$ has a unique non-trivial constituent in $V$. Then $V=U\oplus C_V(L)$, by 1.3, and so $|E\cap L|^2=|V/C_V(L)|=q^2$. Then $|E|^2=|V/C_V(E)|=q^4$, and so $|C_V(E)|\geq q^2$. As $C_V(E)\leq C_V(E\cap L)=C_U(E\cap L)\oplus C_V(L)$, it follows that $C_V(L)\cap C_V(E)\neq 0$, and so $C_V(H)\neq 0$. We therefore conclude that $H$ is not isomorphic to $^2G_2(q)$. It remains to treat the cases where $H/Z(H)$ is a Suzuki group or a $3$-dimensional unitary group. Again, let $U$ be a non-trivial, irreducible $H$-invariant section of $V$ and set $F_0=End_H(U)$. If $H\cong Sz(q)$ set $F=\Bbb F_q$, while if $H/Z(H)\cong U_3(q)$ set $F=\Bbb F_{q^2}$. Set $\w U=U\otimes_{F_0}F$, and let $\Gamma$, $\Gamma_0$, $M_i$, $N_i$, $d_i$, $m$, $r$, and $t$ be given as in 6.6. Suppose that $p=2$ and that $H\cong Sz(q)$. Here we have $S=T$, by 6.2(c). As $\Omega_1(T)=Z(T)$, and since $N_H(T)$ acts irreducibly on $Z(T)$, we then have $E=Z(T)$. Then two conjugates of $E$ suffice to generate $H$, by 6.2(b), and since $C_V(H)=0$, it follows that $|V|\leq q^4$. Then 6.6 yields $(d_1\cdots d_r)^m\leq 4m$. But also, as $q^2+1$ divides $|H|$, it follows from 2.1 that $d_i\geq 4$ for any $i$ for which $d_i\neq 1$. As $m$ is odd, we then have $m=1$ and $d_i=1$ for all but one index $i$. That is, $U$ is an algebraic conjugate of a basic irreducible module for $H$, of dimension $4$ over $F$. The basic irreducible modules for $H$ are obtained by restriction from those for $Sp(4,q)$, by 4.5, so 4.7 implies that $U$ is a natural $Sp(4,q)$-module for $H$. One then observes that $|U/C_U(Z(T))|=|E|^2$, so $A=E$, and $V=U+C_V(E)$. As $H=O^2(H)$ we then have $U=[V,H]$, and then 1.2 implies that $U=V$. Thus, (ii) holds in this case. Suppose next that $H/Z(H)\cong U_3(q)$. Then $q^3+1$ divides $|H|$, and so $|V|\geq q^6$, by 3.1. Consider first the case where $E\nleq T$. Then 6.3(c) shows that $E=(E\cap T)\lan a\ran$ where $a$ induces a field automorphism on $H$. Suppose further that $p$ is odd, so that $C_T(a)$ is isomorphic to a Sylow $p$- subgroup of $U_3(q^{1/p})$. In particular, we note that $q\neq 9$. Let $X$ be the largest subgroup of $T$ for which $[X,a,a]=1$. Then $[X\cap Z(T),a]=C_{Z(T)}(a)=C_{Z(T)}(E)$, and since $E$ is weakly closed in $C_H(E)$ we then have $C_{Z(T)}(a)\leq E$. We claim that $E\cap T\nleq Z(T)$. Suppose false, and let $x\in X-Z(T)$. Then $[x,a]=[x,E]\leq C_T(E)$, and so $\lan E,E^x\ran$ is elementary abelian and properly contains $E$. This is contrary to $E$ being weakly closed in its centralizer, and proves the claim. We next claim that $HE$ is generated by two conjugates of $E$. Indeed, we have $E\cap Z(T)\leq L$ where $L$ is a subgroup of $H$ isomorphic to $SL(2,q)$. As $q$ is odd and $q\neq 9$, it follows from [Gor, 2.8.4] that $L$ is generated by two conjugates of $E\cap Z(T)$, and then 6.3(b) yields our claim. As $|E|\leq p\cdot q^{2/p}$ it follows that $|V|\leq p^4\cdot q^{8/p}$. As $q\geq p^3$ we then have $|V|\leq q^4$. But we have seen already that $|V|\geq q^6$, so we have a contradiction at this point. On the other hand, suppose that $E\nleq T$ and that $p=2$. Then $\Omega_1(T)=Z(T)$, and we conclude from 6.3(c) that $E\cap T=Z(T)$. Let $L$ be a subgroup of $H$, generated by $Z(T)$ and a conjugate of $Z(T)$, with $L\cong SL(2,q)$. If $L$ is $E$-invariant then $E$ induces inner automorphisms on $L$, and there is then an element $a_1$ of $E-Z(T)$ such that $[L,a_1]=1$, and we have $L_1=C_H(a_1)$. As $|E-Z(T)|=q$, and since there are at least $q^2$ conjugates of $L$ under $T$, we may therefore choose $L$ so that $L$ is not $E$-invariant. It then follows from 6.3(b) that two conjugates of $E$ suffice to generate $HE$. We then obtain a contradiction to $|V|\geq q^6$, just as in the case where $p$ is odd. We now have $E\leq T$. Suppose that $E\leq Z(T)$. As $N_H(T)$ acts irreducibly on $Z(T)$ we then have $E=Z(T)$. Now 6.3(b) implies that three conjugates of $E$ suffice to generate $H$, and so $|V|\leq q^6$. Here $F=\Bbb F_{q^2}$, so 6.6 yields $(d_1\cdots d_r)^m\leq 3m$. But also $d_i\geq 3$ if $d_i\neq 1$, so we obtain $m=1$, and $U$ is a natural $SU(3,q)$-module for $H$. Here $|E|^2=|V/C_V(E)|$, so $U=[V,H]$, and then $U=V$, by 1.2. Further, $E$ acts quadratically on $V$, and since $|E_0|^2<|V/C_V(E_0)|$ for every proper non- identity subgroup $E_0$ of $E$, we conclude that $A=Z(T)$ if $A\leq Z(T)$. On the other hand, suppose that $E\nleq Z(T)$. As already mentioned, $Z(T)=\Omega_1(T)$ if $p=2$, so we must have $p$ odd in this case. Now $q<|E|\leq q^2$, and two conjugates of $E$ suffice to generate $H$. Then $|V|\leq q^8$, and so $(d_1\cdots d_r)^m\leq 4m$. Again, as $d_i\geq 3$ for all $i$ for which $d_i\neq 1$, we obtain $m=1$ and $U=V$ is an irreducible $FH$-module with $dim_F(U)=3$ or $4$. By 4.5, $V$ is the restriction to $H$ of an irreducible module for the group $H^*=SL(3,F)$. Regard $H$ as a subgroup of $H^*$. Then $T$ is contained in two non- conjugate parabolic subgroups $P_1$ and $P_2$ of $H^*$, and we have $Z(T)\leq O_p(P_1)\cap O_p(P_2)$. Also, we have $P_i=\lan E^{P_i}\ran$ for both $i=1$ and $2$, and not both $P_1$ and $P_2$ centralize $C_V(Z(T))$. It follows that $C_V(Z(T))\neq C_V(E)$. But $dim_F(C_V(E)\geq 2$ since $|E|^2\geq |V/C_V(E)|$, so we conclude that $dim_F(C_V(Z(T))\geq 3$. Then $|V/C_V(Z(T))|=q^2$, and since three conjugates of $Z(T)$ generate $H$ we obtain $|V|\leq q^6$. This contradiction shows that $dim_F(V)=3$. That is, $V$ is the restriction to $H$ of a natural $SL(3,F)$-module, which is to say that $V$ is a natural $SU(3,q)$-module for $H$. Then $|E|=q^2$, in order to achieve $|E|^2\geq |V/C_V(E)|$. Since $E$ was chosen so that $|E|^2|C_V(E)|$ is as large as possible, we now conclude that $|A|^2|C_V(A)|=|V|$. The maximality of $E$ then yields $A\leq E$ for some choice of $E$. Then $A\leq H$, and since $|E_0|^2<|V/C_V(E_0)|$ for any non-identity subgroup $E_0$ of $E$ other than $E$ and $Z(T)$, we conclude that $A=Z(T)$ or $A=E$. Thus (i) holds, and the lemma is proved. \qed \enddemo We now consider the case in which $A$ acts quadratically on $V$. \proclaim {6.10 Lemma} Assume Hypothesis 4, and suppose that $H/Z(H)$ is a simple group of Lie type in characteristic $p$ and of Lie rank $1$. Let $A$ be a non-identity subgroup of $S$ such that $|A|^2\geq |V/C_V(A)|$ and such that $[V,A,A]=0$. Then one of the following holds. \roster \item "{(i)}" $H\cong SL(2,q)$, $V$ is a natural $SL(2,q)$-module for $H$, and $A\leq H$. If $|A|\geq |V/C_V(A)|$ then equality holds, and $A$ is a Sylow subgroup of $H$. \item "{(ii)}" $H\cong SL(2,q)$, $V$ is a direct sum of two natural $SL(2,q)$-modules for $H$, $A$ is a Sylow subgroup of $H$, and $|A|^2=|V/C_V(A)|$. \item "{(iii)}" $H\cong L_2(q^2)$ and $V$ is a natural $\Omega_4^-(q)$-module for $H$. If $A\leq H$ then $|A|^2=|V/C_V(A)|$ and $A$ is conjugate to a Sylow $p$-subgroup of a subgroup $K$ of $H$ with $K\cong L_2(q)$. If $|A|^2>|V/C_V(A)|$ then $p=2$, $|A|^2=2\cdot |V/C_V(A)|$ , and either: \itemitem {\rm{(a)}} $A=(A\cap H)\lan a\ran$, where $a$ induces a field automorphism on $H$ and where $A\cap H$ is a Sylow subgroup of $C_H(a)$, or \itemitem {\rm{(b)}} $q=4$, $|A|=2$, and $A$ induces a transvection on $V$. \item "{(iv)}" $G\cong Sym(5)$, $V$ is a direct sum of two natural $O_4^- (2)$-modules for $G$, $|A|^2=|V/C_V(A)|$, and $A\nleq H$. \item "{(v)}" $H\cong SU(3,q)$, $V$ is a natural module for $H$, $|A|^2=|V/C_V(A)|$, and $A$ is the center of a Sylow subgroup of $H$. \item "{(vi)}" $H\cong Sz(q)$, $V$ is a natural module for $H$, $|A|^2=|V/C_V(A)|$, and $A$ is the center of a Sylow subgroup of $H$. \endroster \endproclaim \demo {Proof} Suppose that $H\cong SL(2,q)$, that $H$ has more than one non- trivial irreducible constituent in $V$. Then 6.8 shows that there are just two such constituents, and no trivial constituents, for $H$ in $V$. Suppose that the constituents are natural $SL(2,q)$- modules for $H$. Then 6.8 says that $A$ is a Sylow subgroup of $H$, and then $V$ is a completely reducible $H$-module, by 1.6. Thus (ii) holds in this case. On the other hand, suppose that not both of the irreducible constituents for $H$ in $V$ are natural $SL(2,q)$-modules for $H$. Then 6.8 says that $q=4$, that $A\nleq H$, and that both irreducible constituents are natural $O_4^-(2)$-modules for $G$. Let $a$ be an element of $A-H$. If $|A|=2$ then $A=\lan a\ran$ and $a$ induces a $2$-transvection on $V$. Suppose instead that $|A|=4$. Then $|C_V(A)|=16$ and for any irreducible constituent $U$ for $G$ in $V$ we observe that $|C_U(A)|=4$. Since $a$ commutes with an element $g$ of $H$ of order $3$, and $A$ is not $g$-invariant, it follows that $|C_V(a)|=64$, and that $a$ is a $2$-transvection in this case as well. Let $K$ be a subgroup of $G$ generated by three conjugates of $a$, with $K\cong Sym(4)$. Then, in any case, we have $|V/C_V(K)|=4$, and if $V_0$ is a fixed irreducible submodule of $V$ we may choose $v\in C_V(K)$ so that $v\notin V_0$. Now $|v^G|=5$, and so $\lan v^G\ran$ is a proper subspace of $V$. This shows that $V$ is completely reducible, and thus (iv) holds in this case. We are now reduced to the case where $V$ is irreducible. If $H\cong L_2(q)$ and $V$ is a natural $\Omega_3(q)$-module for $H$ then $A\leq H$, by 6.8. But here $p$ is odd and no non-identity element of $H$ acts quadratically on $V$, so Hypothesis (4) is violated in this case. If $H\cong SL(2,q)$ and $V$ is a natural $SL(2,q)$-module for $H$ then (i) follows. Assume that $H\cong L_2(q^2)$ and that $V$ is a natural $\Omega_4^- (q)$-module for $H$. Suppose further that $A\leq H$. No element of $H$ centralizes a $3$-dimensional $\Bbb F_q$-subspace of $V$, so we have $|A|\geq q$. Let $K$ be a subgroup of $H$ with $K\cong L_2(q)$ and with $K\cap S$ a Sylow $p$-subgroup of $K$. Set $B=K\cap S$. The conjugates of $B^{\#}$ under $N_H(T)$ then form a partition of $(H\cap S)^{\#}$. Here $B$ acts quadratically on $V$, and for any $b\in B^{\#}$ we have $C_V(b)=[V,b]=[V,B]=C_V(B)$ is a $2$-dimensional subspace of $V$. Let $C$ be a conjugate of $C$ under $N_H(H\cap S)$, $C\neq B$. Then $H\cap S=BC$, and so $C_V(BC)$ is of dimension $1$. Then $C_V(B)\neq C_V(C)$, and so $[V,b,c]\neq 0$ for any non-identity elements $b\in B$ and $c\in C$. We conclude that $A$ is conjugate to $B$ and that $|A|^2=|V/C_V(A)|$. On the other hand, suppose that $A\nleq H$ and that $|A|^2>|V/C_V(A)|$. Then the foregoing shows that either $A\cap H=1$ , or $|A\cap H|^2=|V/C_V(A\cap H)|=|V/C_V(A)|$ and that $A\cap H$ is conjugate to the group $B$ given as above. If $A\cap H\neq 1$ we then have (iii)(a), while if $A\cap H=1$ then $|V/C_V(A)|=2$ and we have (iii)(b). Thus (iii) holds if $V$ is a natural $\Omega_4^-(q)$-module for $H$. It remains to consider the cases in which $H/Z(H)$ is not a linear group. We then have (v) or (vi), as follows from 6.9 and from elementary properties of the natural modules for $SU(3,q)$ and for $Sz(q)$. \qed \enddemo \vskip .3in {\bf Section 7: Lie rank 2 in characteristic $p$} \vskip .2in \proclaim {7.1 Lemma} Let $G=L_3(2)$ and let $W$ be an indecomposable $G$-module of dimension $4$ over $\Bbb F_2$. Then there exists a fours group $A\leq G$ with $|C_W(A)|=2$. \endproclaim \demo{Proof} Since $G$ is generated by three involutions, and all involutions in $G$ are conjugate, no involution induces a transvection on $W$. Put $Z=C_W(G)$, and suppose first that $Z\neq 0$. Then $|Z|=2$ and $W/Z$ is a natural $G$-module. Let $P$ be a rank-$1$ parabolic subgroup of $G$ such that $C_{W/Z}(P)=0$, and put $A=O_2(P)$. Then $C_{W/Z}(A)$ is a fours group on which $P$ acts irreducibly, and since $A$ contains no transvections we conclude that $C_W(A)=Z$. On the other hand, suppose that $Z=0$. Then $W$ has a submodule $U$ of order $8$, on which every involution in $G$ act as a transvection. Now let $P$ be the rank-$1$ parabolic subgroup with $|C_U(P)|=2$, and put $A=O_2(P)$. Then $C_U(A)=C_U(P)$, and since $A$ contains no transvections on $W$ we have $C_W(A)\leq U$. Thus $|C_W(A)|=2$. \qed \enddemo \proclaim {7.2 Lemma} There are exactly four isomorphism classes of irreducible modules for $Alt(6)$ over $\Bbb F_2$. These classes are represented by the principal module $1$, a natural $Sp(4,2)$-module $U$, the module $U'$ contragredient to $V$, and a module $V$ of dimension $8$ which is a direct summand of the Steinberg module for $Sp(4,2)$ \endproclaim \demo {Proof} Set $H=Alt(6)$ and $G=Sp(4,2)$, and identify $H$ with $[G,G]$. There are four $2$-regular conjugacy classes in $H$, and so, by a well known result of Brauer, there are also four isomorphism classes of irreducible modules for $H$. The modules $1$, $U$, and $U'$ represent three of these classes (with $U$ and $U'$ distinguishable by their restrictions to subgroups of order $3$ in $H$). Let $W$ be the Steinberg module for $G$. Then $W$ is the reduction mod $2$ of an irreducible complex representation of $G$, written over $\Bbb Z$, and since $H$ has no irreducible characters of degree $16$ we conclude that $W$ is reducible for $H$. Let $S$ be a Sylow $2$-subgroup of $G$, and set $T=S\cap H$. Then $S$ acts freely on $W$, and so $dim(C_W(T))=2$. But $W$ is a direct sum of $G$-conjugate, irreducible $H$-submodules, so we conclude that $W$ has an $H$-submodule $V$ of dimension $8$. Now $\{1,U,U',V\}$ is a complete set of representatives of the isomorphism classes of $\Bbb F_2H$-modules. \qed \enddemo \proclaim {7.3 Lemma} Assume Hypothesis 2 with $p=2$, and assume that $H/Z(H)\cong L_3(2^n)$ or $Sp(4,2^n)'$. Suppose further that $Z(H)\neq 1$. Then $H\cong SL(3,2^n)$. \endproclaim \demo {Proof} Suppose false. As $O_2(H)=1$, a survey of the Schur multipliers of the groups in question yields $H/Z(H)\cong Alt(6)$ and $|Z(H)|=3$. Without loss, $V$ is irreducible, and then $V$ may be regarded as a vector space over the field $F=\Bbb F_4$. Evidently $dim_F(V)\geq 3$. Also, we may assume that $A$ has been taken to be as small as possible, subject to the condition that $|A|^2\geq |V/C_V(A)|$. Therefore: \vglue .1in \item {(1)} Let $B$ be a subgroup of index $2$ in $A$. Then $|C_V(B)/C_V(A)|\leq 2$, or $B=1$. \vglue .1in Let $X_1$ and $X_2$ be the two maximal subgroups of $H$ containing $S\cap H$, and set $T=N_S(X_1)$. Then also $T=N_S(X_2)$, and since $S$ is contained in a unique maximal subgroup of $G$ we have $|S:T|=2$. Set $Z=C_V(S\cap H)$, and set $U_i=\lan Z^{X_i}\ran$, ($i=1,2$). As $V$ is irreducible we have $C_V(H)=0$, and then since $X_1$ and $X_2$ are fused by $S$, we conclude that $dim_F(U_i/Z)\geq 1$, ($i=1,2$). As $[U_i,O_2(X_i)]=0$, we have $[U_1+U_2,Z(S\cap H)]=0$, and so $V\neq U_1+U_2$, and $dim_F(V)\geq 4$. Let $R$ be a Sylow $3$-subgroup of $H$. Then $N_H(R)$ contains a representative from each conjugacy of involutions in $HT$, and one observes that all involutions in $HT-H$ act non-trivially on $Z(H)$. That is, there does not exist a central extension of the form $3^{\cdot}Sym(6)$. Suppose that $A=\lan a\ran$ is of order $2$. Then $a$ is a $2$-transvection, and since $dim_F(V)\geq 3$ it follows from the preceding paragraph that $a$ commutes with $Z(H)$. Suppose $a\in T$. Then $a\in H$, and three conjugates of $a$ generate $H$, so $|V|=|V/C_V(H)|\leq 4^3$. This is contrary to $dim_F(V)\geq 4$, so we conclude that $a\notin T$. Then $a$ interchanges the two fours groups in $S\cap H$, and so $(S\cap H)A$ is dihedral of order $16$. Then $Z(H)(S\cap H)A$ is a maximal subgroup of $HA$, and so again, three conjugates of $A$ suffice to generate $HA$, with a contradiction as before. We therefore conclude that $|A|>2$. Suppose that $A\nleq T$, and let $a\in A-T$. Then $C_S(a)\leq Z(S\cap H)$, and so $[U_1+U_2,A\cap T]=0$. But $a$ interchanges $U_1$ and $U_2$, so $|(U_1+U_2)/C_{U_1+U_2}(a)|\geq 4$. This is contrary to (1), so we conclude that $A\leq T$. Now suppose that $A\nleq H$, and let $a\in A-H$. Then $a$ centralizes an $\Bbb F_2$-hyperplane of $C_V(A\cap H)$, by (1), and so $dim_F(C_V(A\cap H)=1$. Thus $|C_V(A)|=2$, and since $|A|\leq 8$ we obtain $|V|\leq 2^7$. but then $dim_F(V)=3$, and we have the same contradiction as before. Thus $A\leq H$, and $|A|=4$. Now $A\leq O_2(X_i)$ for some $i$, and we may take $i=1$. By conjugating $A$ within $X_1$ we may assume also that $A\nleq O_2(X_2)$. Set $B=A\cap O_2(X_2)$. Then $|B|=2$ and $C_V(B)\geq U_1+U_2$. As $O^{2'}(X_2)$ acts non-trivially on $U_2$, we have $[U_2,A]\neq 1$, and then $|U_2/C_{U_2}(A)|\geq 4$. This contradicts (1), and the lemma is thereby proved. \qed \enddemo \proclaim {7.4 Lemma} Let $L=L_3(q)$ or $Sp(4,q)$, $q=2^n$, and identify $L$ with the group of inner automorphisms of $L$. Let $t$ be a non-inner automorphism of $L$ of order $2$. Then all involutions in $Lt$ are conjugate via $L$. Moreover, the following hold. \roster \item "{(a)}" If $L=L_3(q)$ and $n$ is odd, and $t$ is conjugate to a standard graph automorphism of $L$, and $C_L(t)\cong L_2(q)$. \item "{(a)}" If $L=L_3(q)$ and $n$ is even, then $t$ is conjugate to a standard graph automorphism (with $C_L(t)\cong L_2(q)$), or a field automorphism (with $C_L(t)\cong L_3(q^{1/2})$), or a graph-field automorphism (with $C_L(t)\cong U_3(q^{1/2}$). \item "{(c)}" If $L=Sp(4,q)$ and $n$ is odd, then $t$ is conjugate to an automorphism of $L$ which induces a non-trivial polarity on the Dynkin diagram of $L$, and $C_L(t)\cong Sz(q)$. \item "{(d)}" If $L=Sp(4,q)$ and $n$ is even, then $t$ is conjugate to a field automorphism of $L$, and $C_L(t)\cong Sp(4,q^{1/2})$. \endroster \endproclaim \demo {Proof} Let $\bar F$ be an algebraic closure of $F=\Bbb F_q$, set $\bar L=L_3(\bar F)$ (resp. $Sp(4,\bar F)$), and view $L$ as the set of fixed-points for a Steinberg endomorphism $\sigma$ of $\bar L$. Fix a $\sigma$- invariant Borel subgroup $\bar B$ of $\bar L$, a $\sigma$-invariant maximal torus $\bar T$ of $\bar B$, let $\Sigma$ be the root system determined by $\bar T$, and let $\Pi$ be the fundamental system of roots determined by $\bar T$ and $\bar B$. Let $\bar S$ be the unipotent radical of $\bar B$, and set $S=C_{\bar S}(\sigma)$, so that $S$ is a Sylow $2$-subgroup of $L$. If $n=2m$ is even, set $r=2^m$, and let $\phi$ be the standard field automorphism of $L$ of order $2$, given by $(x_{\alpha}(v))^{\phi}= x_{\alpha}(v^r)$ for $\alpha\in \Sigma$ and $v\in F$. Let $\rho$ be the unique automorphism (resp. angle-preserving, length-changing bijection) of $\Sigma$ of order $2$ which preserves $\Pi$, and let $\gamma$ be the automorphism of $L$ given, in the case of $L=L_3(q)$, by $$ x_{\alpha}(v)^{\gamma}=x_{\alpha^{\rho}}(v), $$ and in the case of $L=Sp(4,q)$ by $$ x_{\alpha}(v)^{\gamma}=\left \{\aligned x_{\alpha^{\rho}}(v)\qquad \text{if $\alpha$ is long,} \\ x_{\alpha^{\rho}}(v^2)\qquad \text{if $\alpha$ is short.}\endaligned \right. $$ If $L=L_3(q)$ set $\tau=\gamma$, and if $L=Sp(4,q)$ set $\tau=\gamma^n$. Notice that if $L=Sp(4,q)$ we have $Aut(L)=Inn(L)\lan\gamma\ran$, $\tau=1$ if $n$ is even, and $\tau$ is an involution if $n$ is odd. Further, if $L=Sp(4,q)$ and $n$ is odd then $C_L(\tau)\cong Sz(q)$. (See [GLS3, 2.5.1].) It therefore only remains to show that all involutions in $St$ are conjugate via $L$. Identify $L$ with $Inn(L)$. Let $E$ and $F$ be the two maximal (under inclusion) elementary abelian subgroups of $S$, and set $Z=E\cap F$. Then $Z=Z(T)$. Suppose first that $F=E^t$. Then either $t\in S\tau$ or $L\cong L_3(q)$ and $t\in S\tau\phi$. Set $J=\{zxx^t\ |\ z\in C_Z(t),\ x\in E\}$. Then $J$ is the set of elements of $S$ inverted by $t$, and so $Jt$ is the set of involutions in $St$. Let $\mu:E\times C_Z(t)\longrightarrow J$ be the mapping given by $(x,z)\mapsto zxx^t$. Then for any $(x,z)\in E\times C_Z(t)$ we have $\mu^{-1}(zxx^t)=\{(xu,zuu^t)\ |\ u\in Z\}$. Thus $|J|=|E\times C_Z(t)|/|Z|$. If $t\in S\tau$ we then have $|J|=q^2$, while if $L=L_3(q)$ and $t\in S\tau\phi$ we have $|J|=q^{3/2}$. Thus, $|Jt|=|J|=|S:C_S(\tau)|$ if $t\in S\tau$, and again $|Jt|=|J|=|S:C_S(\tau\phi)|$ if $t\in S\tau\phi$. It follows that $t\in \tau^S$ or $(\tau\phi)^S$, and so we are done if $t$ interchanges $E$ and $F$. We are now reduced to the case where $t\in S\phi$. Let $P$ be a maximal parabolic subgroup of $L$ containing $S$, set $X=O^{2'}(P)$, and take $E=O_2(P)$. Then $P$ is $\phi$-invariant, and $|C_{S/E}(\phi)|=|S/E|^{1/2}$. It follows that the number of involutions in $(S/E)\phi$ is equal to the number of $S/E$-conjugates of $\phi$, and so we may take $t\in E\phi$. But also $|C_E(\phi)|=|E|^{1/2}$, and so the number of involutions in $E\phi$ is equal to $|\phi^E|$. Thus $t$ and $\phi$ are conjugate via $S$, and the lemma is proved. \qed \enddemo \proclaim {7.5 Proposition} Assume Hypothesis 3 with $p=2$, and assume that $H$ is isomorphic to the commutator subgroup of a group of Lie type in characteristic $p$, with Lie rank greater than $1$. Then $p=2$, and $HA\cong SL(3,q)$ or $Sp(4,q)$, with $q=2^n$. Moreover, there exists an element $s$ of $S$ which interchanges the two maximal parabolic subgroups of $HA$ containing $S\cap HA$, and one of the the following holds. \roster \item"{(i)}" $H\cong SL(3,q)$ and $V$ is the direct sum of $H$-submodules $U$ and $U^s$, where $U$ is a natural $SL(3,q)$-module for $H$. If $A\nleq H$ then $q=2$, $|A|=4$, $[V,A,A]\neq 0$, and $|A|^2=|V/C_V(A)|$. \item"{(ii)}" $HA\cong Sp(4,q)$ and $V$ is the direct sum of $HA$-submodules $U$ and $U^s$, where $U$ is a natural $Sp(4,2^n)$-module for $H$. Further, we have $|A|^{3/2}<|V/C_V(A)|$, and if $[V,A,A]=0$ then $|A|^2=|V/C_V(A)|$ and $A$ is conjugate in $G$ to $Z(S\cap HA)$. \endroster \endproclaim \demo {Proof} By 5.9 we have $p=2$ and $H/Z(H)\cong L_3(q)$ or $Sp(4,q)$ ($q=2^n$), or $Alt(6)$. If $H$ is not isomorphic to $Alt(6)$ set $T=S\cap H$, and if $H\cong Alt(6)$ set $T=(S\cap H)C_S(S\cap H)$. Thus, either $T=S\cap H$ or $HT\cong Sp(4,2)$. Let $P_1$ and $P_2$ be the two maximal subgroups of $HT$ containing $T$. We use the following notation: \vglue .2in \item\item{} $H_i=O^{2^{\prime}}(P_i)$, \ $Z_i=C_S(H_i)$, \ $Q_i=O_2(H_i)$, and $V_i=\langle C_V(T)^{H_i} \rangle$. \vglue .2in For the sake of easy reference, we record also the following basic facts. \vglue .2in \item {(1)} Either $Z_i\leq H_i$ or $HT\cong Sp(4,2)$. \vglue .2in \item {(2)} $H_i/Q_i\cong SL(2,q)$, and $Q_i/Z_i$ is a natural $SL(2,q)$-module for $H_i/Q_i$. \vglue .2in We list a few well known (and easily checked) facts about the Sylow $2$-subgroups of $SL(3,q)$, $Sp(4,q)$, and $Alt(6)$, as follows. \vglue .2in \item {(3)} We have $T=Q_1Q_2$, $Z(T)=Q_1\cap Q_2$, and every elementary abelian subgroup of $T$ is contained in $Q_1$ or in $Q_2$. \vglue .2in As $S$ is contained in a unique maximal subgroup of $G$, we have the following result. \vglue .2in \item {(4)} There exists $s\in S$ with $(H_1)^s=H_2$, and with $(H_2)^s=H_1$. In particular, the number $c$ of non-trivial irreducible constituents for $H_i$ in $V_i$ is independent of $i$. \vglue .2in Let $X$ be a non-central chief factor for some $H_i$ in $V$, and let $B$ be a non-identity subgroup of $T$. Then for any $i$, $i=1$ or $2$, $B$ is conjugate in $G$ to a subgroup $B_1$ of $T$ with $B_1 \nleq Q_1$. We have $|X|\geq q^2$ and $|X/C_X(B_1)|\geq q$, by 6.8. Thus: \vglue .2in \item {(5)} For any non-identity subgroup $B$ of $T$, we have $|V/C_V(B)|\geq q^c$. \vglue .2in Notice that $[V_1+V_2,\ Z(T)]=0$. Then $V\neq V_1+V_2$, and so $[V,O^2(H_i)]\nleq V_i$. Since $C_V(H)=0$, we also have $[V_i,O^2(H_i)]\neq 0$. Then (5) yields the following result. \vglue .2in \item{(6)} $c\geq 2$, and $|V/C_V(B)|\geq q^2$ for any non-identity subgroup $B$ of $T$. \vglue .2in We now choose an elementary abelian subgroup $E$ of $S$ in the following way. If there exists a non-identity elementary abelian subgroup $F$ of $T$, with $|F|^2\geq |V/C_V(F)|$, then we take $E\leq T$ so that $|E|^2|C_V(E)|$ is as large as possible and, subject to these conditions, so that $|E|$ is as large as possible. Otherwise, if no such subgroup $F$ of $T$ exists, we take $E$ so that $|E|^2|C_V(E)|$ and then $|E|$ are as large as possible. Then $E$ is weakly closed in $C_G(E)$, by 3.2. In particular, $E$ is invariant under $N_T(C_S(E))$. As $C_S(T)\leq T$ it follows that $E\cap T\neq 1$. Notice that our way of choosing $E$ then yields the following information. \vglue .2in \item {(7)} If $E\nleq T$ then $$ |V|>|F|^2|C_V(F)|<|E|^2|C_V(E)| $$ for any non-identity subgroup $F$ of $E\cap T$. \vglue .2in Let $W$ be a non-trivial, irreducible $G$-invariant section of $V$ and let $U$ be an irreducible $HT$-submodule of $W$. Set $F_0=End_{HT}(U)$, $F=\Bbb F_q$, and note that $F_0\leq F$, by 6.4(a). Denote by $\Gamma_0$ the Galois group of $F$ over $F_0$, and by $\Gamma$ the full automorphism group of $F$. Write $|F_0|=q_0=2^m$, and set $r=|\Gamma_0|$. Thus $n=mr$. Set $\w U=Y\otimes_{F_0}F$. Let the modules $M_i$ and $N_i$, $1\leq r$, be given as in 6.6, and set $d_i=dim_F(M_i)$ and $d=d_1\cdots d_r$. Further, let the indexing be given so that $d_1\geq \cdots \geq d_r$, and denote by $c_0$ the number of irreducible constituents for $HT$ in $W$. Suppose first that $HT=HE$ and that $HT$ is not isomorphic to $Alt(6)$. That is, suppose that $HE$ is a group of Lie type. With $E\leq T$, it follows from (3) that $E\leq Q_i$ for some $i$, (say $i=1$), and with $HT=HE$ we then have $\lan E^{H_1}\ran=Q_1$, and thus $E=Q_1$. Let $s\in S$ be as in (4). Then $EE^s=T$ and $E\cap E^s=Z(T)$. Then 2.2 implies that $|T|^2|C_V(T)|\geq |E|^2|C_V(E)|$. In particular, we have $|T|^2\geq |V/C_V(T)|$ and $|T|^2\geq |W/C_W(T)|$. As $|C_U(T)|=q_0$, by 6.4(b), we then have $|W|\leq (q_0)^{c_0}q^6$ (resp. $|W|\leq q_0^{c_0}q^8$) if $H/Z(H)\cong L_3(q)$ (resp. $HT\cong Sp(4,q)$). By 6.6(c) we have $d^m\leq 6m+1$ (resp. $8m+1$). Here $d_1=3$ or $d_1\geq 8$ (resp. $d_1=4$ or $d_1\geq 16$) by 6.7, so in fact $d_1=3$ (resp. $d_1=4$), and if $m>1$ then $m=2$ and $d_i=1$ for $11$ it follows that $r=2$. Thus, we have proved the following result. \vglue .2in \item {(8)} Suppose that $HT=HE\ncong Alt(6)$. Then $HT/Z(HT)\cong L_3(q)$ (resp $Sp(4,q)$) and one of the following holds. \itemitem{\rm{(i)}} $U$ is a natural module of dimension $3$ (resp. $4$) for $HA$. \itemitem{\rm{(ii)}} $q=q_0^2$, and $\w U\cong M\otimes_F M^{\phi}$ where $M$ is a natural module for $HT$, and where $\lan \phi\ran=\Gamma_0$. \vglue .2in With $U$ as in (8i) or (8ii) one observes that $C_U(Q_1)\ncong C_U(Q_2)$, so $U$ is not invariant under $s$. Then $W$ contains $U\oplus U^s$, and we obtain $$ 6m+2\geq 2\cdot 3^m\qquad \text {(resp. $8m+2\geq 2\cdot 4^m$)}. $$ It follows that $m=1$ and that (8i) holds. Moreover, we have shown that any non-trivial, irreducible, $HE\lan s\ran$-invariant section of $V$ is isomorphic to $U\oplus U^s$, and from this, and from the preceding estimates, it follows that there is a unique such section. Then 1.2 yields $V\cong U\oplus U^s$. Suppose now that $HE\cong Sp(4,q)$, and then suppose that $A\leq T$. Then $A\leq Q_i$ for some $i$, and we set $A*=\lan A^{H_i}\ran$. Thus $A*=Z(H_iT)$, or $O_2(H_iT)$, or $O_2(H_i)$, and it is now a straightforward matter to check that $|A*|^{3/2}|C_V(A*)|<|V|$. Then also $|A|^{3/2}|C_V(A)|<|V|$, by 2.2. Assume that $|A|^{3/2}|C_V(A)|\geq |V|$, and let $A$ be minimal for this property. Then $A\nleq T$ and we may take $A$ so that $|A_0|^{3/2}|C_V(A_0)|<|A|^{3/2}|C_V(A)|$ for any proper subgroup $A_0$ of $A$. Then $N_A(U)=N_A(U^s)=A$. Then any element $a$ of $A-T$ induces a non-identity field automorphism on $L$, by 7.4. Then $[C_V(T),a]\neq 0$, and so $A\cap T=A$, contrary to what has just been shown. Thus $|A|^{3/2}|C_V(A)|< |V|$. Continuing to assume that $HE\cong Sp(4,q)$, suppose now that $A\leq HE$ and that $[V,A,A]=0$. Inspection of the structure of $U$ and of $U^s$ allows us to fix the indices so that $C_U(H_1)\neq 0$ and $C_{U^s}(H_1)=0$. By (3), and possibly after conjugation by $s$, we may assume that $A\leq Q_1$. Let $R_i$, $1\leq i\leq 3$, be root subgroups of $Q_1$, with respect to a standard set-up of $HT$ as a group of Lie type, with $Q_1=R_1R_2R_3$, and taking $Z(H_1)=R_1$ and $Z(T)=R_1R_2$. Set $U_1=[U,H_1]$. Then $C_U(x)\leq U_1$ for any non-identity element $x$ of $Q_1$. Here both $U_1/C_U(H_1)$ and $Q_1/R_1$ are natural $SL(2,q)$-modules for $H_1/Q_1$, and the commutator map defines a pairing of $U_1/C_U(H_1)\times Q_1/R_1$ into $C_U(H_1)$, defined over $F$. Suppose that $AR_1\cap R_i\neq 1$ for both $i=2$ and $3$. Then $C_U(R_i)=C_U(ER_1\cap R_i)$ for any $i$, and thus $C_U(AR_1)=C_U(Q_1)=C_U(H_1)$ is of order $q$. But we also have $R_1=C_{Q_1}(U/C_U(H_1))$ so $[U,A]\nleq C_U(H_1)$, and thus $A$ fails to act quadratically on $U$. We therefore conclude that $AR_1\cap R_i=1$ for some $i$. In particular, we have $|AR_1|\leq q^2$. If $|C_U(A)|>q^2$ then $A\leq R_1$ and $C_V(A)= C_U(R_1)+C_{U^*}(R_1)$ is of index $q^3$ in $V$. In that case we have $|V/C_V(A)|>|A|^2$, so we conclude that $|C_U(A)|\leq q^2$. On the other hand, we have $|C_{U^*}(x)|=q^2$ for any non-identity element $x$ of $Q_1$, so we have $|V/C_V(A)|\geq q^4$, and hence $|A|=q^2$. It follows that $R_1\lq A$. Now view $Q_1/R_1$ as a $2$-dimensional vector space over $F$, for the action of $H_1/Q_1$. This action is doubly transitive on $1$-spaces, and the $1$-spaces partition the space. As $R_i\cap A=1$ for some $i$, and since the same is true for any $H_1$-conjugate of $A$, it follows that $A$ is conjugate in $H_1$ to $R_1R_2$. Thus, (ii) holds if $HE\cong Sp(4,q)$. Now assume that $HE\cong SL(3,q)$, and suppose that $A\nleq H$. Then 7.4 implies that $|A|\leq 2q$, and so $|V/C_V(A)|\leq 4q^2$. Suppose $q>2$. Then $|V/C_V(A)|\leq q^3$, while 7.4 shows that $|C_V(a)|=q^3$ for any $a\in A-H$. Then $|A|>4$, so $A\cap T\neq 1$, and $C_V(a)\leq C_V(A\cap T)$. But for $a\in A-H$, either $V=U+U^a$ or $a$ induces a field automorphism on $L$ (and on $V$), and one checks in either case that $[C_V(a),x]\neq 0$ for any $x\in T$. Thus, we have a contradiction in this case, and so $q=2$. Then $A=Z(T)\lan a\ran$, where $a$ induces a graph automorphism on $H$, and (i) holds. We are thus reduced to the following cases. \vglue .2 in \item {(9)} Either $E\nleq T$ or $HE\cong Alt(6)$. \vglue .2in Suppose that $HE\cong Alt(6)$, and set $T_0=S\cap H$. By (3) and 3.2 we have $E=T_0\cap Q_i$ for some $i$, and then $|V/C_V(E)|\leq 16$. Suppose that $U$ is a natural $Sp(4,2)$-module for $H$. Then $U\neq U^s$, and one checks that $|(U+U^s)/C_{U+U^s)}(E)|\geq 32$. Thus $|V/C_V(E)|>16$, and we conclude that $U$ is not a natural $Sp(4,2)$-module. Then 7.2 implies that $dim(U)=8$ and that $U$ is a direct summand of the Steinberg module for $Sp(4,2)$. It follows that $E$ acts freely on $U$, and thus $64=|U/C_U(A)|>|E|^2$. This contradiction shows that $HE\ncong Alt(6)$, and thus $E\nleq T$. Set $E_1=E\cap T$, and suppose next that $E$ normalizes $H_1$ and $H_2$. Then $|E/E_1|=2$, and by 7.4 there is an element $b\in E-E_1$ such that $b$ induces a non-identity field automorphism on $H$. In particular, we then have $q\geq 4$. By (3) we may choose indexing so that $E_1\leq Q_1$, and then $[Q_1,b]=C_{Q_1}(b)$. We then have $E=\lan b^{Q_1}\ran$, by 3.2. If $|V_1/C_{V_1}(b)|\geq 4$ then $|V/C_V(E_1)|\leq |E_1|^2$, in violation of (7), so in fact $|V_1/C_{V_1}(b)|=2$. Then 6.8 implies that $q=4$ and that $V_1/C_{V_1}(H_1)$ is a natural $O_4^-(2)$-module for $H_1\lan b\ran$. Now $|E|=8$ (resp. $16$) if $H/Z(H)\cong L_3(q)$ (resp. $(Sp(4,q))'$), and so $|V/C_V(E)|\leq 64$ (resp. $2^8$). Then also $|V/C_V(E_1)|\leq 2^5$ (resp $2^7$). Set $E_2=A\cap Z(T)$. Then $|E_1/E_2|=2$, and we observe that $[V_2,E_2]=[V_2,Z(T)]=0$, while $|V_2/C_{V_2}(E_1)|\geq 4$. Thus $|V/C_V(E_2)|\leq 8$ (resp $32$). If $H/Z(H)\cong L_3(4)$ we then contradict (6), so in fact we have $H\cong Sp(4,4)$ and $|E_2|=4$. Let $x\in C_{H_1}(b)-T$, and set $E_3=(E_2)^x$. Then $|E_3/(E_3\cap Z(T))|=2$, and it follows, by a repetition of the preceding argument, that $|V/C_V(E_3\cap Z(T)|\leq 8$. Thus, we violate (6) in this case as well. We have shown: \vglue .2in \item {(10)} There exists $a\in E$ such that $(H_1)^a=H_2$. \vglue .2in Fix $a$ as in (10). Set $E_0=N_E(H_1)$ and, as above, set $E_1=E\cap T$. Then $|E/E_0|=2$, $|E/E_1|\leq 4$, and $E_1=E\cap Z(T)$. Moreover, by 7.4 we have $|E/E_1|=2$ if $HT\cong Sp(4,q)$. Also, we have $E_1\leq C_T(a)=C_{Z(T)}(a)$, and so 3.2 implies that $E_1=[C_{Z(T)}(E_0),a]$. Notice that $|V_1/(V_1\cap V_2)|=|V_1/C_V(T)|\geq q$, by 6.8. If $|E/E_1|^2\leq q$ then $|V/C_V(E_1)|\leq |E_1|^2$, which is contrary to (7). Thus $|E/E_1|^2>q$. Then $q\leq 4$. Suppose $q=4$. Then $|E_1|=2$ and $|E|=8$. Moreover, we have $|V/C_V(E)|\leq 64$, while $|V/C_V(E_1)|\geq 16$ by (6). It follows that $|V_1/C_V(T)|=4$, and so $V_1/C_{V_1}(H_1)$ is a natural $SL(2,4)$-module for $H_1/Q_1$. Then an element of $E_0-E_1$ induces a field automorphism on $V_1/C_{V_1}(H_1)$, and so $|V_1/C_{V_1}(E)|\geq 8$. But then $|V/C_V(E_1)|\leq 8$, and so we have a contradiction in this case. We therefore conclude that $q=2$, and $H\cong Alt(6)$ or $L_3(2)$. Suppose that $H\cong Alt(6)$. Then $|A|=4$, and $C_H(a)$ is a dihedral group of order $10$. Let $X$ be the subgroup of $C_H(a)$ of order $5$, and let $c$ be the involution in $A\cap H$. Suppose that $X$ has more than one non- trivial constituent in $V$. Then $|[V,c]|\geq 16$, and since $|V/C_V(A)|\leq 16$ we obtain $|[V,c]|=16$ and $C_V(c)=C_V(A)$. Then $[V,A,A]=0$, and $[V,X,a]=0$. But also $C_V(X)\leq C_V(c)\leq C_V(a)$ in this case, and so $a$ centralizes $V$. So, we conclude that $X$ has a single non-trivial irreducible constituent in $V$. We again obtain $[V,X,a]=0$. As two conjugates of $X$ suffice to generate $H$, we have $|C_V(X)|\leq 16$, and so $|V|\leq 2^8$. Let $D$ be a dihedral group of order $8$ generated by two conjugates of $a$. Then $Z(D)=\lan c\ran$, and since $c$ is not a $2$-transvection, by (7), it follows that $a$ is not a transvection. Then $|C_V(X)|=16$ and $|V|=2^8$. In particular, we have $|[V,A]|\geq 16$. Moreover, we now have $|V/C_V(a)|=4$, and $|V/C_V(A)|=16$. Now also $C_V(A)=C_V(D)$ and, again since $D$ is generated by two conjugates of $a$, $[V,D]=[V,A]$ is of order $16$. Then $[V,A,A]=0$, and then also $[V,D,D]=0$, which implies that $D$ is elementary abelian. With this, we conclude that $H$ is not isomorphic to $Alt(6)$. Suppose finally that $H\cong L_3(2)$. Let $U$ be an irreducible $H$-submodule of $V$, and suppose first that $U$ is a natural $L_3(2)$-module. Then $U\ncong U^a$ as $H$-modules, and we find that $|(U+U^a)/C_{U+U^a}(A)|=16$. In view of 1.1, it follows that $V=U+U^a$, and so outcome (i) of the lemma holds in this case. On the other hand, suppose that $U$ is of dimension $8$. Then $U$ is the Steinberg module for $H$, so $|U:C_U(A\cap H)|=16$. As $|A|=4$ we then have $U=[V,H]$, $C_U(A)=C_U(A\cap H)$, and $A$ acts quadratically on $U$. Here $A\cap H=Z(T)$, so we have also $[C_U(Q_1)+C_U(Q_2),A]=0$. As $a$ interchanges $H_1$ and $H_2$, it follow that $[C_U(Q_1),Q_2]=0$, and then $[C_U(Q_1),H_1]=0$. Then $[C_U(T),\lan H_1,H_2\ran]=0$, contrary to $C_V(H)=0$. With this contradiction, the proof of the 7.5 is complete. \qed \enddemo \vskip .3in \noindent {\bf Section 8: Alternating groups, $p$ odd} \vskip .2in \proclaim {Theorem} Let $G=Alt(n)$, $n\geq 5$, and let $V$ be an irreducible $G$-module over $\Bbb F_3$. Suppose that there exists a $3$-cycle $a$ in $G$ such that $|V/C_V(a)|=9$ and such that $0=[V,a,a,a]\neq [V,a,a]$. Then $V$ is isomorphic to the unique non-trivial constituent in the natural permutation module for $G$ over $F$.\endproclaim \demo {Proof} For any $m$, denote by $P(m)$ the permutation module for $Alt(m)$ over $\Bbb F_3$, with basis vectors $x_1,\cdots x_m$ permuted in the natural way by $Alt(m)$. Denote by $P_0(m)$ the codimension-$1$ submodule of $P(m)$ consisting of the vectors $a_1x_1+\cdots+a_mx_m$ for which $a_1+\cdots+a_m=0$. Set $F(m)=C_{P(m)}(Alt(m))$, and set $\bar P(m)=P(m)/F(m)$. Take $a$ to be the $3$-cycle $(1\ 2\ 3)$, and let $H$ be the subgroup of $G$ fixing the point $1$ in the natural permutation representation for $G$. Similarly, let $K$ be the subgroup of $G$ fixing both $1$ and $2$, and let $L$ be the subgroup fixing each of $1$, $2$, and$3$. Thus $[a,L]=1$. \vglue .2in \item {(1)} We have $dim\ F(m)=1$. Further, $P(m)=P_0(m)\oplus F(m)$ if $n$ is not divisible by $3$, and otherwise $F(m)\leq P_0(m)$. \vglue .2in The irreducibility of $\bar P_0(m)$ will be assumed for $m\leq n$. We shall see that the irreducibility of $\bar P_0(n)$ will follow by induction. \vglue .2in \item {(2)} Let $\ell$ be the least integer such that $3+2\ell\geq n$. Then $\ell+1$ conjugates of $a$ generate $G$. \vglue .2in We will need the following result. \vglue .2in \item {(3)} Let $W$ be an $\Bbb F_3G$-submodule of $V$, such that $P_0(n)$ is a submodule of $W$, and such that with $dim(W/P_0(n))=1$. Suppose that there exists a $3$- cycle $a$ in $G$ such that $|V/C_V(a)|=9$ and such that $1=[V,a,a,a]\neq [V,a,a]$. Then $C_W(G)\neq 0$. \vglue .2in The proof of (3) is as follows. If $3$ divides $n$ then $P_0(n)\geq F(n)$ and there is nothing to prove. So assume that $3$ does not divide $n$. Suppose first that $3+2\ell=n$. Then $dim(W/C_W(G))\leq 2\ell+2=n-1$, by (2). As $dim(P_0(n))=n-1$ we then have $W=P_0(n)\oplus C_W(G)$, and $C_W(G)$ has dimension $1$. Suppose finally that $3+2\ell=n+1$. Then $3+2(\ell-1)=n-1$ and so $H$ is generated by $\ell$ $3$-cycles. Then $dim(W/C_W(H))\leq 2\ell=n-2$. As $dim(P_0(n)/C_{P_0(n)}(H))=n-2$ we conclude that $C_W(H)\nleq P_0(n)$. Let $x\in C_W(H)-P_0(n)$. Assuming that $[x,G]\neq 0$ we obtain $|x^G|=n$ and so $\lan x^G\ran$ is a homomorphic image of $P(n)$. Then $W\cong P(n)$, and thus $C_W(G)\neq 0$, proving (3). Suppose now that 8.1 is false, and let $n$ be minimal for this property. Suppose $C_V(H)\neq 0$ and let $0\neq x\in C_V(H)$. Then $|x^G|=n$ and $V=\lan x^G\ran$, and so $V$ is a homomorphic image of $P(n)$. As $V$ is irreducible we are done in this case. We may therefore assume henceforth that $C_V(H)=0$. Suppose that $3$ divides $n$. As $C_V(H)=0$ it follows from (3), and induction on $n$, that $V\cong P_0(n-1)$ as a module for $H$. Then also $V\cong P_0(K)\oplus C_V(K)$ as a module for $K$, where $dim\ C_V(K)=1$. Further, we then have $V\cong P(n-3)+C_V(L)$, where $C_V(L)$ is of dimension $2$. Here $[V,L]\cong P_0(n-3)$, and $[V,L,a]=0$. As $C_V(L)$ is $a$-invariant, we then have $C_V(L)=[V,a]$. \vglue .2in \item {(4)} If $3$ divides $n$ then $[V,L,a]=0$, $C_V(L)=[V,a]$, and $C_{[V,L]}(L)=[V,a,a]$. \vglue .2in We continue to suppose that $3$ divides $n$. Identify $G$ with the image of $G$ in $GL(V)$, and let $t$ be the element of order $2$ in $GL(V)$ given by $C_V(t)=[V,K]$ and $[V,t]=C_V(K)$. Then $[N_G(K),t]=1$, where $N_G(K)\cong Sym(n-2)$. Set $G^*=\lan t,G\ran$. We claim that $G^*\cong Sym(n)$. To prove this, it now suffices to show that $a^t=a^{-1}$, by the standard presentation of $Sym(n)$ as a Coxeter group. This calculation can be made as follows. Take $b=(3\ 4\ 5)$, set $J=\lan a,b\ran$, and set $U=[V,J]$. Then $J\cong Alt(5)$ and $U\cong P_0(5)$ as a module for $J$. Identify $J$ with $Alt(5)$, let $\{x_i\ |\ 1\leq i\leq 5\}$ be the standard basis of $P(5)$, and identify $U$ with $P_0(5)$. As $b\in K$, $t$ centralizes $[U,b]$. Also, we have $[V,a,a]=C_{[V,L]}(L)\leq [V,K]\leq C_V(t)$, and so $C_U(t)\geq [U,b]+[U,a,a]=\lan x_3-x_4,\ x_4-x_5,\ x_1+x_2+x_3\ran$. Now observe that $[U,a]=C_V(L)$, by (4). As $K=\lan b,L\ran$ we then have $C_{[U,a]}(b)=C_V(K)$, and thus $C_V(K)=\lan x_1-x_2\ran$. As $t$ acts as $-1$ on $C_V(K)$, the action of $t$ on $U$ has now been completely determined, and one may compute directly that $t$ inverts $a$. In detail: on the $3$-dimensional subspace of $U$ with ordered basis $(x_3+x_4+x_5, x_2-x_1, x_1+x_2+x_3)$ the matrices of $a$ and of $t$ are given respectively by $$ \pmatrix 1 & 1 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \endpmatrix \qquad \text{and} \qquad \pmatrix 1 & 0 & 0 \\ 0 & -1 & 0\\ 0 & 0 & 1 \endpmatrix, $$ and one then checks that the matrix for $t$ inverts the matrix for $a$. This shows: \vglue .2in \item {(5)} Suppose that $3$ divides $n$, let $t$ be the element of $GL(V)$ defined above, and identify $G$ with the image of $G$ in $GL(V)$. Set $G^*=\lan t,G\ran$. Then $G^*\cong Sym(n)$. \vglue .2in We continue to suppose that $3$ divides $n$, and we now form the semi-direct product $VG^*$. The usefulness of (5) lies in its providing us with a means to produce complements to $V$ in $VG^*$, by means of generators and relations. (This would be a more difficult task if we were forced to work with the smaller group $VG$.) The key to the proof of 8.1 is the following result. \vglue .2in \item {(6)} If $3$ divides $n$ then $H^1(G,V)\cong \Bbb F_3$. \vglue .2in The proof of (6) is as follows. Assume that $3$ divides $n$, and denote by $K^*$ the pointwise stabilizer in $G^*$ of $\{1,\ 2\}$. Then $K^*$ is generated by conjugates of $t$, i.e. by transvections, and it follows that $C_V(K)=C_V(K^*)$. Using multiplicative notation in $VG^*$, let $w$ be a non-identity element of $[V,t]=C_V(K^*)$ and set $t_i=tw^i$, $0\leq i\leq 3$. Then $[t_i,K^*]=1$. Identify $t$ with the transposition $(1\ 2)$ in $G^*$, and set $s=(2\ 3)$. Then each $t_i$ is conjugate to $t$ via an element of $C_V(s)$, and so $|t_is|=3$ for all $i$. Setting $G_i^*=\lan t_i, s, K^*\ran$, it now follows that $G_i^*\cong Sym(n)$, and that $G_i^*$ is a complement to $V$ in $VG^*$. Suppose next that there exists $x\in VG$ and indices $i$ and $j$ such that $(G_i^*)^x=G_j$. We may take $x\in V$. We have $G_i^*\cap (G_i^*)^x\geq H$, so that $H^{x^{-1}}\leq G_i^*\cap VH=H$. Thus $x\in N_V(H)=C_V(H)=1$ and so $G_i^*=G_j^*$. Thus there are at least three conjugacy classes of complements to $V$ in $VG^*$, and so $|H^1(G^*,V)|\geq 3$. Now let $\hat G$ be a complement to $V$ in $VG^*$. We have $H^1(H,V)=0$, as follows from (3), so we may assume (after conjugation) that $H\leq \hat G$, and then $\hat G=\lan tx,H\ran$ for some $x\in V$. Here $[tx,K^*]=1$, so $x\in C_V(K)=[V,t]$, and so $\hat G=G_i$ for some $i$. This shows that $|H^1(G^*,V)|=3$. But then also $|H^1(G,V)|=3$, as the reader may verify. This completes the proof of (6). We may now complete the proof of 8.1. Suppose first that $n\equiv 2\ (mod\ 3)$. As $C_V(H)=0$ it follows from (3), and from induction, that $V\cong P_0(n-1)$ as a module for $H$. In particular, we have $dim\ V=n-2$. As $5$ does not divide the order of $SL(3,3)$ we then have $n>5$. Now $V\cong P(n-2)$ as a module for $K$, and then $V\cong P_0(n-3)\oplus C_V(L)$ as a module for $L$. As $dim\ P_0(n-3)=n-4$ we then have $dim\ C_V(L)=2$. As $P_0(n-3)$ is irreducible for $L$ we then have $[V,L,a]=0$ and $a$ induces a transvection on $V$, contrary to hypothesis. Suppose next that $n\equiv 1\ (mod\ 3)$. Then $V$ has an $H$-submodule $V_0$ with $V_0\cong \bar P_0(n-1)$, and (6) implies that either $V=V_0$ or $V\cong \bar P(n- 1)$ as a module for $H$. In particular, we have $dim\ V=n-3$ or $n-2$. As an $L$- module we then have $V\cong P_0(n-3)\oplus C_V(L)$, where $dim\ C_V(L)\leq 2$. We therefore obtain a contradiction as in the preceding paragraph. Suppose that $n\equiv 0\ (mod\ 3)$. Then $V\cong P_0(n-1)$ as an $H$-module. By (6) $V$ is a submodule of an indecomposable $G$-module $W$ with $dim\ W/V=1$, and by (3) we have $C_W(H)\neq 0$. Then $W$ is spanned by $x^G$ for any non-zero $x\in C_W(H)$, and $W$ is then a homomorphic image of $P(n)$. Thus, $W\cong \bar P(n)$ and $V\cong \bar P_0(n)$, as required. \qed \enddemo \proclaim {Lemma 8.2} Let $H$ be the group $2^{\cdot}Alt(9)$ and let $V$ be a non-trivial, irreducible $H$-module over $\Bbb F_3$, of dimension at most $8$. Then either $dim(V)=8$ and $V$ is a spin module for $H$, or $dim(V)=7$ and $V$ is isomorphic to the non-trivial constituent in the natural permutation module for $H$. \endproclaim \demo {Proof} Let $x$ be an element of order $3$ in $H$ such that the image of $x$ in $H/Z(H)$ is a $3$-cycle. If $x$ acts quadratically on $V$ then $V$ is a spin module of dimension $8$, by [M]. So we may assume that $x$ does not act quadratically. Let $J$ be a matrix in Jordan canonical form which represents $x$ on $V$, let $m$ be the number of $3\times 3$ blocks of $J$, and let $n$ be the number of $2\times 2$ blocks. As $x$ is not quadratic, we have $m\geq 1$. Set $L=E(C_H(x))$. Then $L\cong SL(2,9)$ or $L_2(9)$, and so any non-trivial module for $L$ over $\Bbb F_3$ is of dimension at least $4$. Suppose that $m\geq 2$ or that $n\geq 1$. As $dim(V)\leq 8$, it follows that $L$ centralizes the chain $[V,x]>[V,x,x]>0$, and the chain $C_V(x)\geq C_V(x)\cap [V,x]\geq C_V(x)\cap [V,x,x]\geq 0$. But also, with $m\geq 2$ or $n\geq 1$ we have $dim(V/([V,x]+C_V(x))\leq 3$. As $[V,L]\neq 0$, we conclude that $m=1$ and $n=0$. Then $V$ is isomorphic to the non-trivial constituent in the natural permutation module, by 8.1. \qed \enddemo \proclaim {Theorem 8.3} Assume Hypothesis $4'$, with $p$ odd, and assume that $G/Z(G)\cong Alt(n)$, $n\geq 5$. If $p=5$ assume also that $n\neq 5$, and if $p=3$ assume $n\neq 6$. Then $p=3$, $|A|^2=|V/C_V(A)|$, and one of the following holds. \roster \item "{(i)}" $|A|=3$, $A$ acts non-quadratically on $V$, $G\cong Alt(n)$, and $V$ is isomorphic to the unique non-trivial irreducible constituent in the natural permutation module for $G$ over $\Bbb F_3$. \item "{(ii)}" $G\cong SL(2,5)$ and $V$ is isomorphic to the natural $SL(2,9)$-module for $H$. \item "{(iii)}" $G\cong 2^{\cdot}Alt(9)$, $|A|=27$, and $V$ is a spin module for $G$, of dimension $8$ over $\Bbb F_3$. \endroster \endproclaim \demo {Proof} If $A$ acts quadratically on $V$ then (ii) holds, by 5.6(iii). Thus, we may assume that $A$ is not quadratic. If $V$ is reducible, then 2.2 shows that $G$ has more than one non-trivial constituent in $V$, and then, by induction, we may assume that a proper submodule $U$ satisfies the conclusion of the lemma. But then $V=C_V(A)+U$ and $[V,G]=U$, so we may in fact assume that $V$ is irreducible. Suppose first that $|A|=p$, and then suppose further that $p\geq 5$. Then $A$ is not quadratic on $V$, as follows from [Sa] or from [C3]. Let $a\in A^{\#}$, let $a_1$ be a $p$-cycle of $a$, and put $b=a(a_1)^{-1}$. Then two conjugates of $a$ generate a group $L=K\times\lan b\ran$, where $K/Z(K)\cong Alt(p)$. As $|V/C_V(a)|\leq p^2$ there is then a homomorphism of $L$ into $SL(4,p)$ which is non-trivial on $K$. If $p\geq 13$ then $K$ contains an elementary abelian subgroup of order $81$, whereas $SL(4,p)$ contains no such subgroup. Thus $5\leq p\leq 11$, and one observes in these three cases that $K$ has a subgroup $K_0$ with $K_0/Z(K_0)\cong L_2(p)$. As $A$ acts non-quadratically, and as $[V,A,A,A]=0$, it follows from 6.8 that $K_0$ has a unique non-trivial constituent $U$ in $V$, and that $U$ is a natural $O_3(p)$-module for $K_0$. In particular, we have $Z(K_0)=1$, and so $K\cong Alt(p)$ or $3^{\cdot}Alt(7)$. As $Alt(6)$ has no faithful complex representation of degree $4$, we conclude that $p=5$ or $K\cong 3^{\cdot}Alt(7)$. It is easy to check for any element $d$ of order $3$ in $X=SL(4,7)$, that any component of $C_X(d)$ is isomorphic $SL(2,7)$ or $SL(3,7)$, and then since $|SL(3,7)|$ is not divisible by $5$, we conclude that $p=5$ and $K\cong Alt(5)$. Here $K=K_0$, and with $U$ as above we have $H^1(K,U)=0$ by 1.4. Thus $V=U\oplus C_V(K)$. But $|U/C_U(A)|=25$, so $[C_V(K),A]=0$, and so $A\leq K$. That is, $a$ is a $5$-cycle. Since $n\neq 5$, by hypothesis, it follows that there is a subgroup $J$ of $H$ with $J\cong Alt(6)$ or $3^{\cdot}Alt(6)$, and with $J$ generated by two conjugates of $A$. Set $W=[V,J]$. As $|SL(3,5)|$ is not divisible by $9$ we have $W\neq U$, and so $|W|=5^4$. As $|SL(4,5)|$ is not divisible by $27$ we have $J\cong Alt(6)$. There is a subgroup $K_1$ of $J$ with $N_J(A)\leq K_1\cong Alt(5)$ and with $\lan K,K_1\ran=J$. Just as with $K$, we find that $[V,K_1]$ is an $O_3(5)$-module for $K_1$, and $W=[V,K_1]\oplus C_W(K_1)$. As $C_W(K)\neq C_W(K_1)$ it follows that $C_W(N_J(A))$ is of dimension $2$. Then $C_U(N_J(A))\neq 0$. But this is contrary to the case, as one may check that an involution in $K$ inverting $A$ acts non-trivially on $C_U(A)$. We conclude that $|A|=3$. Fix an identification of $G/Z(G)$ with $Alt(n)$, and suppose that $aZ(G)$ corresponds to a $3$-cycle. Then two conjugates of $a$ generate a subgroup $X$ of $G$ containing $Z(G)$ and with $X/Z(G)\cong Alt(5)$. Then $dim([V,X])=4$, and it follows from 3.1 that $X\cong Alt(5)$ and $Z(G)=1$. Then 8.1 yields (i). So assume that $a$ does not correspond to a $3$-cycle. Then $n\geq 7$, and one may observe that two conjugates of $a$ suffice to generate a subgroup of $G$ whose order is divisible by $7$. (Indeed, if $a$ corresponds to a product of two disjoint $3$-cycles then $a$ is contained in a Frobenius subgroup of $G$ of order $21$, and if $a$ corresponds to a product of three or more pairwise disjoint $3$-cycles then one has only to observe that $(1\ 2\ 3)(4\ 5\ 6)(7\ 8\ 9)(1\ 3\ 5)(2\ 6\ 9)(4\ 8\ 7)$ is a seven-cycle.) As $7$ does not divide the order of $SL(4,3)$, we have a contradiction at this point. Thus $|A|>p$. We continue to fix an identification of $G/Z(G)$ with $Alt(n)$. Fix $a\in A^{\#}$ so as to minimize the number $k$ of $p$-cycles in writing the permutation $aZ(G)$ as a product of pairwise disjoint $p$-cycles. Set $C_a=O^2(C_G(a))$. Then $C_a=EK_1\times K_2$, where $E$ is a normal elementary abelian $p$-subgroup of $C_a$ of order $p^k$, $K_1\cong Alt(k)$, and $K_2/Z(K_2)\cong Alt(n-pk)$. The minimality of $k$ implies that $E\cap A=\lan a\ran$, and so $p$ divides $|C_a/E|$. Suppose that $A$ acts non-trivially on some $K\in\{K_1E/E,K_2\}$, and if possible choose $K$ so that $K=K_1E/E$. (Notice that $A$ acts non-trivially on some such $K$ if $p\geq 5$.) As $[V,A,A,A]=0$ we have $[V,A,A]\leq C_V(a)$, and so $A$ acts quadratically on $V/C_V(a)$. Also, $A$ acts quadratically on $C_V(a)$, by Hypothesis $4'$. There is then a non-trivial irreducible constituent $U$ for $K$ in $V$, admitting $A$, and on which $A$ acts quadratically. As $K/Z(K)$ is an alternating group it follows from [C3] that $p=3$, or that $p=5$ and $K\cong SL(2,5)$. Suppose $p=5$. Then $K$ may be identified with a component of $C_a$, and we have $1\neq Z(K)\leq Z(H)$ Then every non-trivial constituent for $K_0$ in $V$ is a natural $SL(2,5)$-module. A Sylow $5$-subgroup $B$ of $K$ acts quadratically on every irreducible constituent for $K$ in $V$, and it then follows from 1.6 that $B$ acts quadratically on $V$. This is contrary to what has aleady been shown, so we conclude that $p=3$. Then [M] yields $K\cong 2^{\cdot}Alt(m)$ for some $m$, $m\geq 4$, so $K=K_2$, $Z(K)= Z(G)\neq 1$, and either $|A/C_A(K)|=3$ or $m=6$. As $V$ is irreducible for $G$, $Z(K)$ acts as $-1$ on $V$, and so every irreducible constituent for $K$ in $V$ is non-trivial. Set $W=C_V(C_A(K))$. Suppose that $|A/C_A(K)|^2\leq |W/C_W(A)|$. Then 3.3 implies that $C_A(K)$ is an $F2$-offender on $V$. Induction on $|A|$ then yields $|C_A(K)|=3$ and that (i) holds with $C_A(K)$ in place of $A$. This is contrary to $1\neq Z(H)$, so we conclude that, in fact, we have $|A/C_A(K)|^2> |W/C_W(A)|$. Then $W$ is irreducible, $m=4$ or $6$, and $W$ is a natural $SL(2,3)$-module (if $m=4$) or a natural $SL(2,9)$-module (if $m=6$). Moreover, if $m=6$ then $|A/C_A(K)|=9$. In either of our two remaining cases, the choice of $K$ implies that $k\leq 3$, and so $|C_A(K)|\leq 9$. Suppose $m=6$. Then $|A|\leq 81$ and $C_V(A)=C_W(A)$ is of order $9$, so $|V|\leq 3^10$. Since all irreducible constituents for $K$ in $V$ are natural $SL(2,9)$-modules we then have $|V|=3^8$ and there are just two such constituents. Then $C_a\cong \lan a\ran\times K$, so $k=1$ and $n=9$, and $|C_V(a)|=3^4$. Here $[V,a]=C_V(a)$, so $a$ acts quadratically on $V$. Now [M] implies (iii). Suppose $m=4$. If $k=3$ then $n=13$, and the minimality of $k$ implies that $A\cap K=1$, and so $|A|=9$. If $k<3$ then $A\cap EK_1=\lan a\ran$, and so $|A|=9$ in any case. As $C_V(A)=C_W(A)$ is of order $3$, we then have $|V|\leq 3^5$. Then $|V|=3^4$ (and $V/W$ is a natural $SL(2,3)$-module for $K$). As $7$ doesn't divide the ordr of $SL(4,3)$ we then have $n\leq 6$, and with $|A|=9$ we get $n=6$, contrary to hypothesis. We have now reduced to the case where no $K$ as above exists. Thus $p=3$, and $A$ acts trivially on $C_a/E$, whence also $[K_2,A]=1$. Suppose $k>3$. Then $C_{C_a}(EK_1/E)=EK_2$, so $A\leq EK_2$, and then $n-3k=3$. Then $|A|=9$ and, by the minimality of $k$, any $b\in A-\lan a\ran$ is a product of at least $k$ disjoint $3$-cycles, $k-1$ of which represent orbits of $\lan a\ran$. But with $k>3$ we can then choose $b$ so that two at least of the $3$-cycles in $b$ are $3$- cycles of $a^{-1}$, and then $ab$ has fewer than $k$ $3$-cycles. With this contradiction we conclude that $k\leq 3$. Suppose $k=1$. Then $C_a/Z(H)\cong 3\times Alt(n-3)$, and since $|A|>3$ and $[A,K_2]=1$ we have $n=6$, contrary to hypothesis. Suppose that $k=2$. Then $C_a\cong 3^2\times Alt(n-6)$, and $EK_1=E$. As $A\cap E=\lan a\ran$ and $[A,K_2]=1$, we then have $n=9$ and $|A|=9$. Here we may identify $Z(H)A/Z(H)$ with $$ \lan (1\ 2\ 3)(4\ 5\ 6),\ (4\ 5 \ 6)(7\ 8\ 9)\ran. $$ There is then a unique element $c$ of $A$ such that $c$ can be represented as aproduct of three disjoint $3$-cycles. By [M], $c$ is not quadratic on $V$, and so $c$ acts non-trivially on $V/C_V(c)$. Here $\lan c\ran=Z(S)$ for some Sylow $3$-subgroup $S$ of $H$, and so $S$ acts faithfully on $V/C_V(c)$. Then $|V/C_V(c)|\geq 3^4$, and we conclude that $C_V(c)=C_V(A)$ has index $3^4$ in $V$. Now observe that since $a$ is represented as a product of two disjoint $3$-cycles, there exists a subgroup $J$ of $H$ with $a\in J\cong Alt(4)$, and such that every involution $t$ in $J$ is represented as a product of four disjoint transpositions. Suppose $C_V(a)\neq C_V(A)$. Then $|V/C_V(a)|\leq 27$, from which it follows that $J$ has a unique non-trivial constituent in $V$. That is, we have $|[V,O_2(J)]|=27$, and $|V/C_V(t)|=9$. On the other hand, $t$ inverts an element of order $7$ in $H$, whereas $7$ does not divide $|GL(4,3)|$. This shows that, in fact, we have $C_V(a)=C_V(A)$, and thus $C_V(A)=C_V(d)$ for all $d\in A^{\#}$. Set $X=N_H(EA)$. Here $EA=J(S)$ for a Sylow $3$-subgroup $S$ of $H$ (so $EA$ is elementary abelian of order $27$), and we have $X/EA\cong Sym(4)$, with $X$ acting irreducibly on $EA$. Now $$ X\leq\lan g\in H\ :\ A\cap A^g\neq 1\ran $$ and so $C_V(A)$ is invariant under $X$. Then $C_V(A)=C_V(EA)$. Let $x$ be an element of $EA$ such that $x$ is represented by a $3$-cycle, and let $L$ be the component in $C_H(x)$ ($L/Z(H)\cong Alt(6)$). No $3$-element of $L$ induces a transvection on $C_V(x)$, by [M] so we have $C_V(x)=C_V(EA)$, and indeed we have shown that $C_V(EA)=C_V(y)$ for all $y\in (EA)^{\#}$. Then $C_V(EA)$ is invariant under the group $$ X^*=\lan g\in H\ :\ EA\cap (EA)^g\neq 1\ran. $$ We now ask the reader to perform the fairly straightforward exercise of showing that $X^*=H$. With this we have $H$ acting on the $4$-space $V/C_V(EA)$, and an evident contradiction. Finally, suppose that $k=3$. Then $EK_1\cong \Bbb Z_3\wr\Bbb Z_3$, and the minimality of $k$ implies that $|A\cap EK_1|\leq 9$. Suppose that $|A|>9$. Then $A\nleq EK_1$, and since $[A,K_2]=1$ we conclude that $|K_2|=3$ and $n=12$. Let $S$ be a Sylow $3$-subgroup of $H$ containing $A$. Then $S\cong (3\wr 3)\times 3$, and the minimality of $k=3$ implies that $A\nleq J(S)$. Then $A=C_S(A)\geq Z(S)$, so $A$ contains a $3$- cycle, and so $k=1$. We conclude from this contradiction that $|A|=9$. For any $b\in A$ such that $b$ is represented as a product of three disjoint $3$-cycles, we have $|V/C_V(b)|\geq 3^4$, by the same reasoning as in the treatment of the case $k=2$. Then, for such an element $b$, we have $C_V(b)=C_V(A)$, of index $3^4$ in $V$. Suppose next that $A\leq EK_1$. In this case we may identify $Z(H)A/Z(H)$ with the group $$ \lan (1\ 2\ 3)(4\ 5\ 6)(7\ 8\ 9),\ (1\ 4\ 7)(2\ 5\ 8)(3\ 6\ 9)\ran.\tag1 $$ Here every non-identity element of $A$ has a representation as a product of three disjoint $3$-cycles. We may assume that $n=9$, by restricting to an appropriate subgroup of $H$. Set $X=\lan C_H(b)\ :\ b\in A^{\#}\ran$. Then $C_V(A)$ is $X$-invariant. On the other hand, $Z(H)X/Z(H)$ contains all the $3$-cycles in the two generators given in (1), and one may check that these six $3$-cycles generate $Alt(9)$. Then $C_V(A)=C_V(H)$, and indeed $[V,H]=0$. We conclude that $A\nleq EK_1$. Now $n=12$ (as $[A,K_2]=1$) and there are, up to conjugation, two possibilities for $Z(H)A/Z(H)$. These are as follows: $$ \lan (1\ 2\ 3)(4\ 5\ 6)(7\ 8\ 9),\ (4\ 6\ 5)(7\ 8\ 9)(10\ 11\ 12)\ran, \qquad \text{or}\tag2 $$ $$ \lan (1\ 2\ 3)(4\ 5\ 6)(7\ 8\ 9),\ (1\ 4\ 7)(2\ 5\ 8)(3\ 6\ 9)(10\ 11\ 12)\ran.\tag3 $$ Suppose that $A$ is as in (2). Then every element of $A^{\#}$ is represented as a product of three disjoint $3$-cycles. Now set $X=\lan C_H(b)\ :\ b\in A^{\#}\ran$, as before. Then $C_V(A)$ is $X$-invarian, and $X$ contains a subgroup $X_0\cong 3^4:Alt(4)$, and $O_3(X_0)=\lan A^{X_0}\ran$. Then $C_V(A)=C_V(O_3(X_0))$, and so $O_3(X_0)$ is an $F1$- offender on $V$. By Thompson Replacement there then exists a quadratic $F1$- offender $B$ on $V$. This is contrary to [M], and we thereby conclude that $A$ is as in (3). Take $Z(H)a=(1\ 2\ 3)(4\ 5\ 6)(7\ 8\ 9)$ and let $b\in A$ with $Z(H)b=(1\ 4\ 7)(2\ 5\ 8)(3\ 6\ 9)(10\ 11\ 12)$. Then $\lan b\ran$ is contained in a subgroup $J$ of $H$ with $J/Z(J)\cong Alt(4)$. If $C_V(b)\neq C_V(A)$ then, as in the case $k=2$, we find that an involution $t$ in $J$ satisfies $|[V,t]=9$ and that $t$ inverts an element of $H$ of order $7$, contrary to $7$ not being a divisor of $|GL(4,3)|$. Thus $C_V(b)=C_V(A)$. Set $Y=O^2(C_H(b))$. Then $Y\cong 3^4:Alt(4)$, and $O_3(Y)Z(H)/Z(H)$ is generated by the $3$-cycles of $Z(H)b$. Then $a\notin O_3(Y)$, and since we now have $C_V(b)=C_V(a)$ we conclude that $[C_V(b),Y]=0$. Then $O_3(Y)$ is an $F1$-offender on $V$, and again, Thompson Replacement yields a quadratic $F1$-offender and a contradiction to [M]. This completes the proof of the theorem. \qed \enddemo \vskip .3in \noindent {\bf Section 9: Alternating groups, $p=2$} \vskip .2in This section contains a single result, which is as follows. \proclaim {9.1 Proposition} Assume Hypothesis $4'$, with $p=2$. Suppose that $F^*(G)=H\cong Alt(n)$, $n$ odd, $n\geq 7$. Then there exists $a\in A^{\#}$ and an involution $t\in G-\lan a\ran$ such that $[V,a,t]=0$. \endproclaim \demo {Proof} Fix $A$ as in Hypothesis $4'$, and assume that 9.1 is false. The following is then obvious. \vglue .2in \item {(1)} For any involution $a\in A$ and for any subgroup $D$ of $C_G(a)$ of even order with $a\notin D$, we have $[V,a,D]\neq 0$. \vglue .2in Suppose first that $V$ is reducible for $G$. Then 1.2 implies that $G$ has at least two non-trivial constituents in $V$. As $|A|^{3/2}\geq |V/C_V(A)|$, by Hypothesis $4'$, there then exists a non-trivial irreducible constituent $W$ for $G$ in $V$ such that $|A|^{3/4}\geq|W/C_W(A)|$. By Timmesfeld Replacement, there is then a quadratic subgroup $B$ of $A$ with $|B|^{3/4}\geq |W/C_W(B)|$, and then 9.4 yields $n$ even, contrary to hypothesis. Thus: \vglue .2in \item {(2)} $V$ is irreducible for $G$. \vglue .2in Let $a\in A^{\#}$. As $n\geq 7$ it follows from (1) and from 5.5 that $|V/C_V(a)|\geq 8$. Suppose that we have $|V/C_V(a)|=8$. Then $|[V,a]|=8$, and since $L_3(2)$ has $2$-rank equal to $2$ it follows from (1) that the $2$-rank of $C_G(a)$ is at most $3$. Then $n=7$, and we may record this result as follows. \vglue .2in \item {(3)} We have $|V/C_V(a)|\geq 8$ for any involution $a$ in $A$, with equality only if $n=7$. \vglue .2in In particular, (3) implies that $|A|>2$. As $A$ contains no quadratic fours group, we have $C_V(a)\neq C_V(A)$ for $a\in A^{\#}$. If $|A|=4$ then $|V/C_V(A)|\leq 8$ and $|V/C_V(a)|\leq 4$, contrary to (3). Suppose that $|A|=8$. Then $|V/C_V(A)|\leq 16$ and $|V/C_V(a)|\leq 8$. Then (3) yields $n=7$, and since the $2$-rank of $Alt(7)$ is just $2$ we obtain $G\cong Sym(7)$. Then $a$ may be chosen to be a transposition, whence $C_G(a)$ contains a subgroup $X$ isomorphic to $Alt(5)$, and we have $[V,a,X]=0$. This is contrary to (1), so: \vglue .2in \item {(4)} We have $|A|\geq 16$. \vglue .2in Let $\Omega=\{1,2,\cdots,n\}$ be the set supporting the natural permutation representation of $G$. We now fix $a\in A^{\#}$ so that the number of orbits of length $2$ for $\lan a\ran$ on $\Omega$ is as small as possible. Denote this number by $k$, and identify $a$ with the permutation $(1\ 2)\cdots(2k-1\ 2k)$. Set $m=n-2k$. Set $C_a=O^2(C_G(a))$. We then have $$ C_a\cong 2^{k-1}:Alt(k)\times Alt(m). $$ Denote by $K_1^*$ the pointwise stabilizer in $C_G(a)$ of $\{2k+1,\cdots,n\}$, and by $K_2^*$ the pointwise stabilizer in $C_G(a)$ of $\{1,2,\cdots,2k\}$. Denote by $E^*$ the subgroup of $C_G(a)$ generated by the set of pairwise disjoint $2$-cycles whose product is $a$. Set $K_i=K_i^*\cap C_a$, and $E=E^*\cap C_a$. Then $E$ is elementary abelian of order $2^{k-1}$, $K_1/E\cong Alt(k)$, $E/\lan a\ran$ is a natural $Sym(k)$-module for $K_1/E$, and $K_2\cong Alt(m)$. The minimality of $k$ yields: \vglue .2in \item {(5)} $A\cap E^*=\lan a\ran$. \vglue .2in Set $L=O^2(\lan A^{C_G(a)}\ran$. Suppose that $L$ is solvable, and then suppose that $K_1$ is non-solvable. Then $[K_1,A]\leq E$, and so $A\leq E^*K_2^*$. Then (4) and (5) imply that the $2$-rank of $K_2^*$ is at least $3$, so $K_2$ is non-solvable, $K_2\leq L$, and $L$ is non-solvable, contrary to assumption. On the other hand, suppose that $K_2$ is non-solvable. Then $A\leq K_1^*$, and then (4) and (5) imply that $K_1$ is non-solvable and $K_1\leq L$. Thus, under the assumption that $L$ is solvable we conclude that $C_a$ is solvable. That is, we have $k\leq 4$ and, as $n$ is odd, $m\leq 3$. Suppose that $k=3$ or $4$. No involution in $K_2^*$ is the product of $3$ or $4$ pairwise disjoint transpositions, so the minimality of $k$ yields $A\cap K_2^*=1$. Let $b$ be an involution in $E^*K_2^*$. As $m\leq 3$, either $b$ or $ab$ is a product of fewer than $k$ disjoint transpositions, and so $A\cap E^*K_2^*=\lan a\ran$. Then $C_A(K_1/E)=\lan a\ran$, and $|A|\leq 8$, contrary to (4). We conclude that $k\leq 2$. Then $n\leq 7$, again contrary to (4). This proves: \vglue .2in \item {(6)} $L$ is non-solvable. \vglue .2in Set $L_i=L\cap K_i$, $i=1,2$. Then $L=L_1\times L_2$. Set $X=C_V(a)$ and $Y=[V,a]$. By Hypothesis $4'$, $A$ acts quadratically on $X$, and hence also on $Y$. Set $A_0=C_A(L/O_2(L))$. Then $\lan a\ran\leq A_0=A\cap O_2(LA)$. Suppose that $A_0\neq \lan a\ran$. Then $O_2(LA)\neq E^*$, and it follows that either $L_1$ is solvable (which includes the possibility that $L_1=1$) or that $L_2=1$. Suppose that $L_2=1$. Then $A\leq K_1^*$, and $L_1$ is non-solvable, by (6), and then $A_0=A\cap E^*=\lan a\ran$. So in fact $L_2\neq 1$, and $L_1$ is solvable. Then $A_0\leq K_1^*$. If $k=1$ we have $K_1^*=\lan a\ran$, while if $k=3$ we have $C_{K_1^*}(K_1)=\lan a\ran$. So $k=2$ or $4$. Suppose $k=2$. Then $L_1=1$, $K_1^*$ is dihedral of order $8$, and $|A_0|\leq 4$. On the other hand, suppose that $k=4$. If $L_1\neq 1$ then $A\nleq O_2(K_1^*)K_2^*$, and $|C_{O_2(K_1^*)}(A)E^*/E^*|=2$. Therefore $|A_0|\leq 4$ in this case. If instead we have $L_1=1$ then $|A_0|\leq 8$ since the $2$-rank of $O_2(K_1^*)/E^*$ is $2$. Thus: \vglue .2in \item {(7)} Suppose $A_0\neq\lan a\ran$. Then $A_0\leq O_2(K_1^*)$, $k=2$ or $4$, and $|A_0|\leq 8$. Moreover, we have $|A_0|=4$ unless $k=4$ and $L_1=1$. \vglue .2in Fix a complement $B$ to $A_0$ in $A$. Set $B_1=C_B(K_2)$ and let $B_2$ be a complement to $B_1$ in $B$. Set $Y_i=[Y,L_i]$, and set $U_i=Y_i/C_{Y_i}(L_i)$, $(i=1,2)$. We will obtain the following result. \vglue .2in \item {(8)} Suppose that $C_B(L_2)\neq 1$. Then $[Y_1,L_2]= [Y_2,L_1]=0$, and $$ |Y/C_Y(B)|\geq |U_1/C_{U_1}(B_1)||U_2/C_{U_2}(B_2)|. $$ \vglue .2in Indeed, set $D=C_B(L_2)$. If $L_2=1$ then $B_1=B$ and (8) follows trivially. So we may assume that $L_2\neq 1$. As $L_2=[A,L_2]=[B,L_2]$, we have $$ 0=[Y,B,D]=[Y,\lan B^{L_2}\ran,D]\geq [Y,L_2,D]\geq [Y,L_2,\lan D^{L_1}\ran]. $$ As $D\neq 1$, by assumption, we have $L_1\leq \lan D^{L_1}\ran$, and thus $[Y,L_2,L_1]=0$. The Three Subgroups Lemma yields also $[Y,L_1,L_2]=0$, so it remains only to prove the second assertion in (8). Set $U=U_1\oplus U_2$. Then $U$ is a homomorphic image of the submodule $Y_1+Y_2$ of $Y$, so $$ |Y/C_Y(B)|\geq |U/C_U(B)|\geq |U_1/C_{U_1}(B_1)||U_2/C_{U_2}(B_2)|, $$ and we have (8). The next step is to show: \vglue .2in \item {(9)} $|B|\leq |Y/C_Y(B)|$. \vglue .2in Suppose that (9) is false. If $C_B(L_2)=1$ then, since $m$ is odd, 5.8 yields $|Y_2/C_{Y_2}(B)|\geq |B|$, and then (9) follows. Thus, we may assume that $C_B(L_2)\neq 1$, and then $|Y/C_Y(B)|\geq |U_1/C_{U_1}(B_1)||U_2/C_{U_2}(B_2)|$, by (8). Again, as $m$ is odd, 5.8 implies that $|U_2/C_{U_2}(B_2)|\geq |B_2|$, and hence $|U_1/C_{U_1}(B_1)| <|B_1|$. In particular, we have $L_1\neq 1$, and $L_1=K_1$. Notice that there exists $t\in C_G(a)$ such that $K_1\lan t\ran/E\cong Sym(k)$. It then follows from 5.8 that $k$ is even and that if $k>4$ then $U_1$ is a natural $Sym(k)$-module for $K_1/E$ and $B_1$ is generated by elements which correspond to transpositions in $Sym(k)$. That is, either $k=4$ or $B$ contains elements which are the product of two disjoint transpositions. The minimality of $k$ then yields $k=4$. Then $|B_1|=2$, and so $|U_1/C_{U_1}(B_1)|\geq |B_1|$. This completes the proof of (9). \vskip .1in We have $|A|=|A_0||B|$, so $$ |A_0|^{3/2}|B|^{3/2}\geq |V/C_V(A)|.\tag* $$ Notice that $V/X\cong Y$ as modules for $C_G(a)$. Then $$ |V/C_V(A)|=|V/X||X/C_X(B)|\geq |Y||Y/C_Y(B)|=|Y/C_Y(B)|^2|C_Y(B)|. \tag** $$ As $|Y/C_Y(B)|\leq |B|$, by (9), we then have \vglue .2in \item {(10)} $|A_0|^3\geq |B||C_Y(B)|^2$, and if $|Y/C_Y(B)|>|B|$ then $|A_0|^3\geq 16|B||C_Y(B)|^2$. \vglue .2in Suppose that $A_0=\lan a\ran$. Then (10) yields $|B|=2$, $|C_Y(B)|=2$, and $|Y/C_Y(B)|=2$. But $L/\lan a\ran$ has a non- solvable direct factor which, by (1), acts faithfully on $Y$, and so we have a contradiction in this case. Now (6) and (7) yield: \vglue .2in \item {(11)} $|A_0|>2$, $k=2$ or $4$, and $m\geq 5$. \vglue .2in Suppose next that $k=4$ and that $|Y|\leq 32$. Then (1) implies that $L_5(2)$ has a subgroup of the form $2^4:3\times Alt(m)$. As $m\geq 5$ we may choose $x\in L_2$ of order $5$, and find that $[Y,x]$ is a hyperplane of $Y$, and that $[Y,x,O_2(K_1)]=0$. Here $O_2(K_1)/\lan a\ran$ is elementary abelian of order $16$, and contains its centralizer in $L_5(2)$, so we have a contradiction. Thus: \vglue .2in \item {(12)} If $k=4$ then $|Y|\geq 64$. \vglue .2in Suppose that $|Y/C_Y(B)|>|B|$. If also $|A_0|=4$ then (10) yields $|B|=1$, which is contrary to (4). Thus $|A_0|=8$, and (7) implies that $k=4$, $L_1=1$, and $A_0E^*=O_2(K_1^*)$. Now (10) yields $|B|\leq 8$ so $|A|\leq 64$ and $|V/C_V(A)|\leq 2^9$. Note also that if $|B|=4$ then $|V/C_V(A)|\leq 2^7$, and if $|B|=2$ then $|V/C_V(A)|\leq 2^6$. But $|V/C_V(A)|\geq |Y||Y/C_Y(B)|$, by (**), and so $|Y|\leq 32$, contrary to (12). Thus: \vglue .2in \item {(13)} We have $|Y/C_Y(B)|=|B|$. \vglue .2in Suppose that $L_1=1$. Then $L=L_2$, and $B$ acts faithfully on $L_2$. Set $Y_0=U_2/C_{U_2}(L_2)$. Notice that there exists an involution $t$ in $C_H(a)$ such that $L_2\lan t\ran\cong Sym(m)$. As $m$ is odd it follows from (13) and 5.7 that either $Y_0$ is a natural module for $LB$, or $m=5$ and $Y_0$ is a spin module for $LB$, of dimension $4$. Suppose that $m=5$ and $Y_0$ is a spin module (i.e. a natural $SL(2,4)$-module). Then $|B|=4$, and since $|A_0|\leq 8$, by (7), we get $|A|\leq 2^5$. Moreover, if $|A|=2^5$ then $|A\cap L_2|=4$ and so $k=2$. But with $k=2$ we have $|A_0|\leq 4$, so we conclude that, in fact, $|A|\leq 16$, and thus $|V/C_V(A)|\leq 64$. As $C_V(A)\neq C_V(a)$, by (1), it follows that $|Y|\leq 32$. By (11), we may choose $a_0\in A_0-\lan a\ran$, and then (1) implies that $\lan a_0\ran\times L_2$ acts faithfully on $Y$. But $2\times L_2(4)$ has no faithful action on a $5$-dimensional space over $\Bbb F_2$, so we conclude that $Y_0$ is not a spin module. Suppose that $k=2$. Then $L_1=1$, and the preceding paragraph applies. As $m$ is odd, (13) and 9.4 imply that $B$ contains an element $b$ which acts as a transvection on $U_2$. As $k\neq 1$, $b\notin K_2^*$. Notice that either $K_1^*=A_0$ or $K_1^*$ is dihedral of order $8$. In either case we have $C_G(A_0)=A_0K_2^*$. Thus, there exists $c\in A_0$ so that $bc\in K_2^*$. But $[U_2,c]=0$, so $bc$ is a transvection on $U_2$, and then $bc$ is a transposition in $K_2^*$. This is contrary to the minimality of $k$, so we now conclude that $k=4$. If also $|A_0|=4$ then (10) yields $|B|\leq 16$, $|A|\leq 64$, and $|V/C_V(A)|\leq 2^9$. Then also $|Y|\leq 32$, and we contradict (12). Thus $|A_0|=8$. As $A_0\leq O_2(K_1^*)$, by (7), it follows that $E^*A_0=O_2(K_1^*)$, and since $[B,A_0]=1$ we get $[L_1,B]\leq E$. Then $L_1=1$. As in the case $k=2$, there then exists $b\in B$ such that $b$ induces a transvection on $U_2$, and we have $b=xt$ where $x\in O_2(K_1^*)$ and $t$ is a transposition in $K_2^*$. Then $a_0x\in E^*$ for some $a_0\in A_0$. Then either $a_0x$ or $aa_0x$ is the product of fewer than three pairwise disjoint transpositions. Replacing $a_0$ by $aa_0$ if necessary, it follows that $a_0b=a_0xt$ is the product of fewer than four pairwise disjoint transpositions, and we again contradict the minimality of $k$. This completes the proof of 9.1. \qed \enddemo \vskip .3in \noindent {\bf Section 10: Lie type groups in cross characteristic} \vskip .2in \proclaim {10.1 Hypothesis} Hypothesis $4'$ holds, and $S$ is contained in a unique maximal subgroup of $G$. Further, $H/Z(H)$ is a quasisimple group of Lie type, whose defining characteristic $r$ is different from $p$. Indeed, it is assumed that there exists no isomorphism between $H/Z(H)$ and a group of Lie type in characteristic $p$ \endproclaim Our aim in this section is to prove the following result. \proclaim {10.2 Theorem} Assume Hypothesis 10.1. Then $|A|=3$, and one of the following holds. \roster \item "{(a)}" $G/Z(G)\cong L_2(5)$. \item "{(b)}" $G\cong Sp(6,2)$, $C_G(A)\cong 3\times Sp(4,2)$, and $|V|=3^7$. \endroster \endproclaim We remark that there is a natural embedding of $Sp(6,2)$ in $\Omega_7(3)$, which one obtains by identifying $2\times Sp(6,2)$ with the Weyl group of $E_7$. Further, it is the case that in $Sp(6,2)$ a Sylow $3$-subgroup is contained in a unique maximal subgroup (isomorphic to $Aut(U_4(2)$). Whenever Hypothesis 10.1 is in effect we set $\bar G=G/Z(H)$. We begin the proof of 10.2 by considering the case where $\bar H\cong Sp(6,2)$. \proclaim {10.3 Lemma} Assume Hypothesis 10.1, and assume that $\bar H\cong Sp(6,2)$. Then $p=3$ and one of the following holds. \roster \item "{(i)}" $|A|=3$, $G\cong Sp(6,2)$, $C_G(A)\cong 3\times Sp(4,2)$, and $|V|=3^7$. \item "{(ii)}" $|A|=27$, $|Z(G)|=2$, and $|V|=3^8$. \endroster \endproclaim \demo {Proof} We have $G=H$ and $|Z(G)|\leq 2$. Let $U$ be the natural $\Bbb F_2$-module for $\bar G$. There are three conjugacy classes of subgroups of order $3$ in $G$, which are distinguished by the dimensions of the commutators $[U,X]$, $X$ a representive of the class. We will say that a subgroup (or element) $X$ of order $3$ is in the class $3A$ (resp. $3B$, resp. $3C$) if $dim([U,X])=1$ (resp. $2$, resp. $3$). There is a subgroup $M$ of $G$ with $\bar M\cong L_2(8):3\cong {^2G_2(3)}$, and $M$ contains representatives of the classes $3C$ (in $[M,M]$) and $3B$ (in $M-[M,M]$). Suppose first that $|A|=3$, so that $|V/C_V(A)|\leq 9$. By 5.6, $A$ is not quadratic on $V$, and so $|V/C_V(A)|=9$. Then $A\nleq M$, as follows from 6.9. Thus, $A$ is of type $3A$. Then there is a maximal subgroup $K^*$ of $G$ containing $A$, with $\bar K^*\cong Sym(8)$, and such that $\bar A$ is generated by a $3$-cycle in $\bar K^*$. Let $K$ be a subgroup of $K^*$ containing $A$, with $\bar K\cong Alt(7)$. Then 8.3 implies that $K\cong Alt(7)$ and that $[V,K]$ is a natural module for $K$, of dimension $6$ over $\Bbb F_3$. In particular, we now have $Z(G)=1$. Let $K_0$ be a subgroup of $K$ containing $A$, with $K_0\cong Alt(6)$, and let $b$ be an element of of order $3$ in $C_G(K_0)$. By [C2, Theorem 4.2], $[K^*,K^*]$ is the unique proper subgroup of $G$ which properly contains $K$ and which is generated by conjugates of $A$. Since $\lan b\ran$ is conjugate to $A$ we then have $\lan K,b\ran=G$. We may then assume that $K^b\nleq K^*$, and so also $G=\lan K,K^b\ran$. Here $b$ centralizes $[V,K_0]$, which has codimension $1$ in $[V,K]$, so we get $V=[V,K]+[V,K^b]$ of dimension at most $7$. But $Sp(6,2)$ has no faithful representation on degree $6$ over $\Bbb F_3$, by [SZ], so $dim(V)=7$. Thus, (i) holds in this case. We next consider the various conjugacy classes of subgroups of $\bar G$ of order $9$. Any such class is characterized by the conjugacy classes of its four cyclic subgroups, and in this way we find that there are exactly four such classes, which we list as follows, and where we list also a subgroup of $\bar G$ containing a representative of the given class. \vglue .1in $Y_1=(3A,3A,3B,3B)\leq Alt(7)$, \vglue .1in $Y_2=(3C,3B,3B,3B)\leq L_2(8):3$, \vglue .1in $Y_3=(3A,3B,3C,3C)\leq Sym(3)\times Sym(6)$, \vglue .1in $Y_4=(3C,3C,3C,3C)\leq 3^{1+2}:GL(2,3)$. \vglue .1in Suppose that $|A|=9$, so that $|V/C_V(A)|\leq 3^4$. Then $A$ is not of type $Y_1$, by 8.3, and $A$ is not of type $Y_2$ by 6.9. Thus $A$ is of type $Y_3$ or $Y_4$, and so $A$ contains a subgroup $\lan a\ran$ of type $3C$. Set $K=C_G(a)$, and observe that $\bar K\cong 3^{1+2}:SL(2,3)$. Theorem 1.3 of [C3] implies that any quadratic element of order $3$ in $G$ is of type $3A$, so $a$ is not quadratic. Set $W=[V,a]$. Then $dim(W)\leq 4$, and $[W,a]\neq 0$, so $K$ acts faithfully on $W$. It follows that $dim(W)=4$, and then $C_V(a)=C_V(A)$. As $\lan A^K\ran\geq O_3(K)$ we then have $C_W(a)=C_W(O_3(K))$. Also, by Hypothesis $4'$ we have $[W,A,A]=0$, so $[W,A,O_3(K)]=0$, and then $O_3(K)$ acts quadratically on $W$, contrary to $\Phi(O_3(K))\neq 1$. We therefore conclude that $|A|\neq 9$. Suppose finally that $|A|=27$. Let $Y$ be a subgroup of $A$ of type $Y_2$. Then two conjugates of $Y$ suffice to generate a maximal subgroup $M$ of $G$ with $M\cong L_2(8):3$, and since $A\nleq M$ it follows that two conjugates of $A$ suffice to generate $G$. As $dim(V/C_V(A))\leq 6$ we then have $dim(V)\leq 12$. Let $A\leq L\leq G$ with $\bar L\cong Sym(3)\times Sym(6)$, set $L_0=E(L)$, set $A_0=C_A(L_0)$, and set $V_0=[V,A_0]+C_V(A_0)$. Then $A\cap L_0$ acts quadratically on $V_0$, by Hypothesis $4'$, and so 1.7 implies that $[V_0,L_0]$ is a direct sum of natural $SL(2,9)$-modules for $L_0$. As $V/C_V(A_0)\cong [V,A_0]$ as $L_0$-modules, we have $[V_0,L_0]\neq 0$, so we conclude that $Z(L_0)=Z(G)\neq 1$. We may assume that $V$ is irreducible for $G$, so $V=[V,Z(G)]$, and since $dim(V)\leq 12$, $V_0$ is a direct sum of one or two natural modules for $L_0$. Then also $V/V_0$ is a natural module for $L_0$. If $A_0$ is not quadratic on $V$ we then have $dim(V)=12$ and $|C_V(A)|=|C_{V_0}(A)|= 3^4$, so that $|V/C_V(A)|>|A|^2$. Thus, $A_0$ is quadratic on $V$, and $dim(V)=8$. Thus, (ii) holds in this case. \qed \enddemo The following lemma is given by Table I in [SZ]. \proclaim {10.4 Lemma} Let $X$ be a simple group of Lie type, in characteristic different from $p$, and let $V$ be a non-trivial projective $\Bbb F_pX$-module. Assume that $X$ is not isomorphic to any of the groups in the following list $\Cal L_0$. $$\align \Cal L_0&=\{L_2(4),\ L_2(9),\ L_3(2),\ L_3(4),\ L_4(2)\,\ L_4(3),\ PSp(4,2), \ U_4(3), \\ & P\Omega_8^+(2),\ P\Omega_7(3),\ F_4(2),\ G_2(3),\ G_2(4),\ Sz(8)\} \endalign $$ Then the dimension of $V$ is at least $\ell$, where $\ell$ is given as follows. \vglue .1in \item {(1)} $L_2(q)$: \ $\ell=(1/d)(q-1)$, $d=(2,q-1)$. \vglue .1in \item {(2)} $L_n(q)$, $n\geq 3$: \ $\ell=(q^n-1)/(q-1)-n$. \vglue .1in \item {(3)} $PSp(2n,q)$, $n\geq 2$: $$ \ell = \left\{ \aligned & (q^n-1)/2\qquad \text {if $q$ is odd, and}\\ & q(q^n-1)(q^{n-1}-1)/2(q+1) \qquad \text {if $q$ is even.} \endaligned \right. $$ \vglue .1in \item {(4)} $U_n(q)$, $n\geq 3$: $$ \ell=\left\{ \aligned & q(q^{n-1}-1)/(q+1)\qquad \text {if $n$ is odd, and}\\ & (q^n-1)/(q+1) \qquad \text {if $n$ is even.} \endaligned \right. $$ \vglue .1in \item {(5)} $P\Omega_{2n+1}(q)$, $q$ odd, $n\geq 3$: $$ \ell=\left\{ \aligned & (q^{2n}-1)/(q^2-1)-n \qquad \text {if $q\neq 3$, and}\\ & (3^{2n}-1)/8-(3^{n}-1)/2 \qquad \text {if $q=3$.} \endaligned \right. $$ \vglue .1in \item {(6)} $P\Omega_{2n}^+(q)$, $n\geq 4$: $$ \ell= \left\{ \aligned & q(q^{2n-2}-1)/(q^2-1)+q^{n-1}-n \qquad \text {if $q\neq 2,3$, and}\\ & q(q^{2n-2}-1)/(q^2-1)-(q^{n-1}-1)/(q-1)-7\delta_{2,p} \qquad \text {if $q$ is even.} \endaligned \right. $$ \vglue .1in \item {(7)} $P\Omega_{2n}^-(q)$, $n\geq 4$: \ $\ell=q(q^{2n-2}-1)/(q^2-1)-q^{n-1}-n+2$. \vglue .1in \item {(8)} $E_6(q)$: \ $\ell=q^9(q^2-1)$. \vglue .1in \item {(9)} $E_7(q)$: \ $\ell=q^{15}(q^2-1)$. \vglue .1in \item {(10)} $E_8(q)$: \ $\ell=q^{27}(q^2-1)$. \vglue .1in \item {(11)} $F_4(q)$: \ $\ell=q^6(q^2-1)$. \vglue .1in \item {(12)} $^2E_6(q)$: \ $\ell=q^9(q^2-1)$. \vglue .1in \item {(13)} $G_2(q)$: \ $\ell=q(q^2-1)$. \vglue .1in \item {(14)} $^3D_4(q)$: \ $\ell=q^3(q^2-1)$. \vglue .1in \item {(15)} $^2F_4(q)$: \ $\ell=(\sqrt{q/2})q^4(q-1)$. \vglue .1in \item {(16)} $Sz(q)$: \ $\ell=(\sqrt{q/2})(q-1)$. \vglue .1in \item {(17)} $^2G_2(q)$: \ $\ell=q(q-1)$. \qed \endproclaim We insert the following lemma, as support for the case where $H\cong L_2(q)$. \proclaim {10.5 Lemma} Set $L=L_2(q)$, $q$ odd, and identify $L$ with the group $Inn(L)$ of inner automorphisms of $L$. Let $A$ be an elementary abelian $2$-subgroup of $Aut(L)$. Then one of the following holds. \roster \item "{(i)}" $q$ is a perfect square, $A\nleq L$, and there exists an element $t$ of $A-L$ such that $t$ is conjugate via $Aut(L)$ to a field automorphism of $L$. Further, we have $A=(A\cap L)\lan t\ran$ and $|A|\leq 8$. \item "{(ii)}" $A\nleq L$, and $LA=L\lan d\ran$, where $d$ is a diagonal automorphism of $L$. Here $|A/(A\cap L)|=2$, $|A|\leq 4$, and all involutions in $LA-L$ are conjugate. If $|A|=4$ then $LA=\lan A,A^g\ran$ for some $g\in L$. \item "{(iii)}" $A\leq L$, $|A|\leq 4$, and if $|A|=4$ then $LA=\lan A,A^g\ran$ for some $g\in L$. \endroster \endproclaim \demo {Proof} Set $G=LA$, and let $T$ be a Sylow $2$-subgroup of $G$ containing $A$. Also, set $G_1=PGL(2,q)$, and set $G^*=G_1T$. Further let $T^*$ be a Sylow $2$-subgroup of $G^*$ containing $T$ and set $T_1=T^*\cap G_1$. We first note the following. \vglue .2in \item {(1)} $T_1$ is dihedral, and $G_1$ has exactly two classes of involutions. \vglue .2in Suppose that $q=r^2$ is a perfect square, and let $\alpha$ be the standard field automorphism of $L$ of order $2$, induced by the automorphism of $GL(2,q)$ which raises matrix entries to the power $r$ (and which we denote also by $\alpha$). Our first step will be to provide a proof of the following (well known) fact. \vglue .2in \item {(2)} We have $C_{G_1}(\alpha)=C_L(\alpha)\cong PGL(2,r)$. \vglue .2in The proof is as follows. Set $\w G_1=GL(2,q)$, let $Z$ be the group of scalar matrices in $\w G_1$, and set $Z_0=\{\lambda^{q-1}I\ \|\ \lambda \in\Bbb F_q\}$. Denote by $\w C_1$ the set of matrices $\w x\in \w G_1$ such that $[\w x,\alpha]\in Z$. As $Z\leq Z(\w C_1)$, there is a homomorphism $\phi: \w C_1\longrightarrow Z$ given by $\phi(\w x)=[\w x,\alpha]$. Then the image of $phi$ is $Z_0$, and the kernel of $\phi$ is $C_{\w G_1}(\alpha)=GL(2,r)$. One observes that $\phi|_Z$ maps $Z$ onto $Z_0$, so $\w C_1=Ker(\phi)Z$. We then have $$ C_{G_1}(\alpha)=Ker(\phi)Z/Z\cong Ker(\phi)/(Ker(\phi)\cap Z)\cong PGL(2,r). $$ On the other hand, let $\omega\in \Bbb F_q-\Bbb F_r$ so that $\lan \omega\ran=\Bbb F_r^{\times}$, and let $d$ be the linear fractional transformation $x\mapsto \omega^2x$. Then $\omega^r=-\omega$, and so $[d,\alpha]=1$. Also, $d\in L$, and $d$ induces an outer automorphism on the subgroup $PSL(2,r)$ of $L$. Thus $C_L(\alpha)$ contains a subgroup isomorphic to $PGL(2,r)$, and this yields (2). Next, suppose that there exists an element $t\in A-G_1$. As $|t|=2$ we have $g^{\alpha} =g^{-1}$, and then $t$ is conjugate either to $\alpha$ or to $z\alpha$, where $z$ is an involution in $C_{G_1}(\alpha)$. Suppose that $t$ is not conjugate to $\alpha$. Then $z\in L$, by (2). We now apply (1) to $C_L(t)$ in place of $G_1$, and find that, up to conjugation in $L$, there are two choices for $z$. One (in $L_2(r)$) is given by the linear fractional transformation $z_1:x\mapsto -1/x^{-1}$, and the other (in $C_L(\alpha)-L_2(r)$) is given by $z_2:x\mapsto -\omega^2x$, where $\omega$ is as above. To show that $\alpha$ and $z_i\alpha$ are conjugate, we suffices to display elements $g_i$ of $G_1$ such that $g_i^{\alpha}=g_iz_i$. Let $\rho\in \Bbb F_q$ with $\rho^{1+r}=-1$. The transformations $g_1:x\mapsto (x+\rho^r)/(\rho x+\rho^{2r})$ and $g_2: x\mapsto (x-\omega)/(\omega x+\omega^2)$ perform this trick, as the careful reader may check. Thus (i) holds if $A\nleq G_1$. Suppose next that $A\leq G_1$, suppose that $A\nleq L$, and let $t\in A-L$. As $T_1$ is dihedral, we have $|C_T(t)|=2$, and thus $|A|\leq 4$. Let $s$ be the involution in $C_T(t)$, and set $D=C_L(s)$. Then $D$ is dihedral of order $q-\epsilon$, $\epsilon=1$ or $-1$, and where $q\cong \epsilon (mod\ 4)$. Further, $D\lan t\ran$ is dihedral, and so $C_D(t)=\lan s\ran$. It follows that $C_L(t)$ is dihedral, of twice-odd order. Now let $D_1$ be a dihedral subgroup of $L$, containing $s$, of order $q+\epsilon$. Suppose that $D_1^x$ is $t$-invariant, for every $x\in C_L(s)$. Then $[D,t]\leq N_L(D_1)$. As $D_1$ is maximal in $L$, and $|D\cap D_1|=2$, we conclude that $[D,t]=\lan s\ran$, $|D|=4$, and $q=5$. In this case we have $G_1\cong Sym(5)$, and one checks that two conjugates of $\lan (1\ 2), (3\ 4)\ran$ suffice to generate $Sym(5)$. On the other hand, suppose that there exists $x\in C_L(s)$ such that $D_1^x$ is not $t$-invariant. We may then take $x=1$, and we conclude that $G=\lan s,t,s^g,t^g\ran$, where $g$ is a generator for the maximal cyclic subgroup of $D_1$. Thus (ii) holds in this case. Finally, suppose that $A\leq L$, and that $|A|>2$. Then $A$ is a fours group. Let $s$ be an involution in $A$, and let $D$ be a dihedral subgroup of $L$ containing $s$, of twice-odd order $q-1$ or $q+1$. Then $D$ is maximal, so $D$ is not $A$-invariant, and so $G=L=\lan A,A^g\ran$ for any generator $g$ of $[D,D]$. Thus, (iii) holds in this case, and the lemma is proved. \qed \enddemo \proclaim {10.6 Lemma} Assume Hypothesis 10.1, and suppose that $H/Z(H)\cong L_2(q)$, $q$ a power of $r$. Then $q=5$, $r=3$, and $|V|=81$. \endproclaim \demo {Proof} First of all, if $|Z(H)|>2$ then $|Z(H)|$ is divisible by $3$, and $H/Z(H)\cong L_2(9)$. Then $p=2$ or $5$, and since $L_2(9)\cong (Sp(4,2))'$ we have $p=5$. Then $|A|=5$ and $H$ is generated by two conjugates of $A$, so that $|V|\leq 5^4$. As $3$ divides $|Z(H)|$, $V$ may then be regarded as a $2$-dimensional space over $\Bbb F_{25}$. But $L_2(9)$ is not a subgroup of $L_2(25)$, so we conclude that $|Z(H)|\leq 2$. Suppose that $G$ contains a quadratic subgroup of order at least $3$. If $p$ is odd then 3.1 yields $H\cong SL(,2,5)$, $p=3$, and $V$ is a natural $SL(2,9)$-module for $H$. Thus, the lemma holds in this case. If $p=2$ then the main result of [MS1] implies that $H/Z(H)\cong U_4(3)$, contrary to $H/Z(H)\cong L_2(q)$. Thus, we may assume henceforth that any quadratic subgroup of $G$ has order at most $2$. Suppose first that $p=2$. Then $q$ is odd, and since $H/Z(H)$ is not isomorphic to a group of Lie type in characteristic $2$ we have $q\geq 11$. By 5.4 we have $|V/C_V(a)|\geq 8$ for any $a\in A^{\#}$. Also, $|A|^{3/2}\geq |V/C_V(A)|$ by Hypothesis $4'$, and so $|A|\neq 2$. If $|A|=4$ we have $|V/C_V(A)|\leq 8$, and then $C_V(a)=C_V(A)$ for all $a\in A^{\#}$, and $A$ is a quadratic fours group. Thus $|A|\neq 4$. Then 10.5 implies that $|A|=8$, $|A\cap H|=4$, and there exists $a\in A-H$ such that $a$ induces a field automorphism on $H/Z(H)$. Further, if we apply 10.5 to $A\cap H$ then we find that $H$ is generated by two conjugates of $A\cap H$. Here $|V/C_V(A)|\leq 16$, so $dim(V)\leq 8$. But $dim(V)\geq (q-1)/2$, by 10.4(1), and so $q\leq 17$. As $q$ is odd and a perfect square we then have $q=9$, and thus $H/Z(H)\cong (Sp(4,2))'$. We therefore conclude that $p$ is odd. Suppose that $|A|=p$, so that $|V/C_V(A)|\leq p^2$. Suppose also that $A\leq H$. The group $L_2(r)$ is a homomorphic image of $L_2(\Bbb Z)$, and $L_2(\Bbb Z)$ contains the free product $\Bbb Z_3*\Bbb Z_3$ as a subgroup of index $2$, so $L_2(r)$ is generated by two elements of order $3$. Thus, if $p=3$ then two conjugates of $A$ generate a subgroup $Y$ of $H$ with $Y/Z(Y)\cong L_2(r)$, and 2.2 then yields $r=5$. Then in any case, we may choose a conjugate $B$ of $A$ so that $\lan A,B\ran$ is nonsolvable, and then $\lan A,B\ran$ is quasisimple. By 2.2 we then have $p\geq 5$ (resp. $p=3$), $\lan A,B\ran\cong L_2(p)$ (resp. $L_2(5)$), and $|[V,\lan A,B\ran]|=p^3$ (resp. $81$). Any quasisimple subgroup of $H/Z(H)$ is isomorphic to $L_2(5)$ or $L_2(q_1)$ for some power $q_1$ of $r$, so we have $p=3$ or $5$ at this point. If $p=3$ and $q=5$ we have the conclusion of the lemma. Assuming that the lemma is false, choose a conjugate $C$ of $A$, not contained in $\lan A,B\ran$, so that $\lan A,C\ran$ is quasisimple, and set $X=\lan A,B,C\ran$. Then $|[V,X]|\leq p^4$ (resp. $3^6$), and $X/Z(X)\cong L_2(q_1)$ for some power $q_1$ of $r$ (with $q_1\geq 25$ if $p=3$). Then $q_1>5$. If $q_1=7$ then $p=3$ and $X$ contains a Frobenius group of order $21$, which is contrary to 4.2. If $q_1=9$ then $p=5$ and we may choose $B$ so that $\lan A,B\ran=X$, which is contrary to 2.2. Thus $q\geq 11$. If $p=5$ then $dim([V,X])\leq 4$, and 10.4(1) yields $q\leq 9$. So in fact $p=3$, $dim([V,X])\leq 6$, and 10.4(1) yields $q\leq 13$. But we have seen that $q_1\geq 25$ if $p=3$, so we have a contradiction at this point. Thus, $A\nleq H$. Let $a\in A-H$. By Theorem 4.9.1 of [GLS3], $a$ induces a field automorphism on $H$, and thus $A$ normalizes a Sylow $r$-subgroup $R$ of $H$. Then also $[R,A]$ is non-cyclic. Suppose that $|A|=p$. Then 2.1 implies that $p=3$ and that $|[R,A]|=4$. It follows that $q=8$ and that $HA\cong {^2G_2(3)}$, contrary to $H\notin Lie(p)$. We therefore conclude that $|A|=p^2$. Set $X=C_H(a)$. Then $X/Z(H)\cong L_2(q_0)$ where $(q_0)^p =q$. Hypothesis $4'$ implies that $A\cap X$ acts quadratically on the subspace $W=[V,a]+C_V(a)$ of $V$, and since $V/C_V(a)$ is $X$-isomorphic to $[V,a]$ it follows that $X$ acts non-trivially on $W$. Then 5.6 yields $p=3$ and $X\cong SL(2,5)$, so that $q=5^3$. Evidently, three conjugates of $A$ suffice to generate $HA$, so $dim(V)\leq 12$. But 10.4(1) yields $dim(V)\geq 62$, so we have a final contradiction, and the lemma is proved. \qed \enddemo In the following lemma we shall eliminate from consideration a further number of \lq\lq small" groups, including groups having exceptional Schur multipliers, and the groups in the list $\Cal L_0$ from 10.4. Conspicuously missing from this expanded list are any of the groups $PSp(4,q)$. These will be addressed later, in 10.10. \proclaim {10.7 Lemma} Assume Hypothesis 10.1, and let $\Cal L$ be the set of groups whose members are as follows: \vglue .1in $L_3(q)$, $3\leq q\leq 5$, \vglue .1in $L_n(q)$, with $n=4$ or $5$, and with $n+q\leq 7$, \vglue .1in $U_3(q)$, $q\leq 7$, \vglue .1in $U_4(q)$, $q\leq 3$, \vglue .1in $U_n(2)$, $n\leq 7$, \vglue .1in $PSp(2n,q)$, with $n=3$ or $4$, and with $q\leq 3$, \vglue .1in $\Omega_7(3)$, $\Omega_8^{+}(2)$, $\Omega_8^-(2)$, and $^3D_4(2)$ \vglue .1in $^2E_6(2)$, $F_4(2)$, $^2F_4(2)$, $G_2(3)$, $G_2(4)$, and $Sz(8)$. \vglue .1in Assume that $\bar H$ is isomorphic to a group in the list $\Cal L$. Then $\bar H\cong Sp(6,2)$. \endproclaim \demo {Proof} Assume that $\bar H$ is not isomorphic to $Sp(6,2)$. We shall consider the various groups individually, and we shall make free use of the ATLAS, in deciding whether a Sylow $p$-subgroup of a particular group $G$ is contained in a unique maximal subgroup of $G$. This task will be simplified by the following reduction. \vglue .1in \item {(a)} Set $H^*=HS$. Then there exists at most one maximal subgroup of $\bar H^*$ containing $\bar S$, and at most one $\bar H$-invariant maximal subgroup of $\bar S$. \vglue .1in Indeed, if (a) is false then the unique maximal subgroup $M$ of $G$ containing $S$ contains $H$. But $M$ contains also $N_G(S\cap H)$, and the Frattini argument then yields $M=G$. We may also employ the following consequence of 2.2. \vglue .1in \item {(b)} A Sylow $p$-subgroup of $G$ has order greater than $p$. \vglue .1in Most groups in $\Cal L$, and most primes, may be eliminated at once by recourse to (a) and (b). The only exceptions will be given by the groups $L_3(4)$ and $^3D_4(2)$ with $p=3$, and the groups $L_3(3)$ and $L_4(3)$ with $p=2$. Apart from the treatment of these four exceptions, the argument will consist simply of a list, in which we give the isomorphism type of $\bar H$, the order of $\bar H$ and of $Out(\bar H)$, and then, for each prime $p$ different from $r$, and for which $p^2$ divides the order of $Aut(\bar H)$, a pair of distinct maximal subgroups of $\bar H^*=\bar H\bar S$ containing $\bar S$. \vglue .2in \item {(1)} $\bar H\cong L_3(3)$, $|\bar H|=2^4\cdot 3^3\cdot 13$, $|Out(\bar H)|=2$, ($p=2$). \noindent There is a unique maximal subgroup containing $S$ if $S\nleq H$. This case will be treated below. \vglue .2in \item {(2)} $\bar H\cong L_3(4)$, $|\bar H|=2^6\cdot 3^2\cdot 5\cdot 7$, $|Out(\bar H)|=12$, ($p=3$). \noindent There is a unique maximal subgroup of $\bar H^*$ containing $\bar S$ if $S\nleq H$, so this case will be considered separately below. \noindent Maximal subgroups when $S\leq H$: $$ M_{10},\qquad U_3(2). $$ \vglue .2in \item {(3)} $\bar H\cong L_3(5)$, $|\bar H|=2^5\cdot 3\cdot5^3 \cdot 31$, $|Out(H)|=2$, ($p=2$). \noindent Maximal subgroups when $S\leq H$: $$ 5^2:GL(2,5),\qquad 4^2:Sym(3). $$ Maximal subgroups when $S\nleq H$: $$ GL(2,5).2,\qquad 4^2.(Sym(3)\times 2). $$ \vglue .2in \item {(4)} $\bar H\cong L_4(2)$, $|\bar H|=2^6\cdot 3^2\cdot 5\cdot 7$, $|Out(H)|=2$, ($p=3$). \noindent Maximal subgroups: $$ Alt(7), \qquad 2^4:(Sym(3)\times Sym(3)). $$ \vglue .2in \item {(5)} $\bar H\cong L_4(3)$, $|\bar H|=2^7\cdot 3^6\cdot 5\cdot 13$, $|Out(\bar H)|=4$, ($p=2$). \noindent There may be a unique maximal subgroup of $\bar H^*$ containing $\bar S$, if $S\nleq H$. This will be treated below. \vglue .2in \item {(6)} $\bar H\cong L_5(2)$, $|\bar H|=2^{10}\cdot 3^2\cdot 5\cdot 7 \cdot 31$, $|Out(\bar H)|=2$, ($p=3$). \noindent Maximal subgroups: $$ 2^4:L_4(2), \qquad 2^6:(L_3(2)\times L_2(2)). $$ \vglue .2in \item {(7)} $\bar H\cong U_3(3)\cong (G_2(2))'$, $|\bar H|=2^5\cdot 3^3\cdot 7$, $|Out(\bar H)|=2$, cyclic Sylow subgroups in cross characteristic. \vglue .2in \item {(8)} $\bar H\cong U_3(4)$, $|\bar H|=2^6\cdot 3\cdot 5^2\cdot 13$, $|Out(\bar H)|=2$, ($p=5$). \noindent Two maximal subgroups $GU(2,4)$. \vskip .2in \item {(9)} $\bar H\cong U_3(5)$, $|\bar H|=2^4\cdot 3^2\cdot 5^4\cdot 7$, $|Out(\bar H)|=6$, ($p=2$ or $3$). \noindent Maximal subgroups for $p=3$, with $S\leq H$: $$ Alt(7),\qquad M_{10}. $$ Maximal subgroups of $\bar H^*$, for $p=3$, with $S\nleq H$: $$ 6^2:Sym(3),\qquad 3^2:SL(2,3). $$ $\bar S$-invariant maximal subgroups of $\bar H$, for $p=2$: $$ M_{10},\qquad 2.Sym(5). $$ \vglue .2in \item {(10)} $\bar H\cong U_4(2)\cong PSp(4,3)$, $|\bar H|=2^6\cdot 3^4\cdot 5$, $|Out(\bar H)|=2$, cyclic Sylow subgroups in cross characteristic. \vglue .2in \item {(11)} $\bar H\cong U_4(3)$, $|\bar H|=2^7\cdot 3^6\cdot 5\cdot 7$, $|Out(\bar H)|=8$, ($p=2$). \noindent In this case, the existence of a pair of proper subgoups of $G$, containing $S$, and generating $G$, is evident from the $\tilde B_2$ geometry of $2$-local subgroups of $\bar G$. See [K]. \vglue .2in \item {(12)} $\bar H\cong U_5(2)$, $|\bar H|=2^{10}\cdot 3^5\cdot 5\cdot 11$, $|Out(\bar H)|=2$, ($p=3$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$: $$ Sym(3)\times GU(3,2),\qquad GU(4,2). $$ \vglue .2in \item {(13)} $\bar H\cong U_6(2)$, $|\bar H|=2^{15}\cdot 3^6\cdot 5\cdot 7\cdot 11$, $|Out(\bar H)|=6$, ($p=3$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$, if $\bar S\leq \bar H$: $$ 3^{1+4}.[2^7.3], \qquad U_4(3):2. $$ Maximal subgroups of $\bar H^*$ containing $\bar S$, if $\bar S\nleq \bar H$. $$ 3^{1+4}.[2^7.3], \qquad 3^5:Sym(6). $$ \vglue .2in \item {(14)} $\bar H\cong U_7(2)$, $|\bar H|=2^{21}\cdot 3^8\cdot 5\cdot 7\cdot 11\cdot 43$, $|Out(\bar H)|=2$, ($p=3$). \noindent Subgroups of $\bar H$ containing $\bar S$ and which generate $\bar H$: $$ GU(6,2),\qquad U_4(2)\times GU(3,2). $$ \vglue .2in \item {(16)} $\bar H\cong PSp(6,3)$, $|\bar H|=2^9\cdot 3^9\cdot 5\cdot 7 \cdot 13$, $|Out(\bar H)|=2$, ($p=2$). \noindent Maximal subgroups of $\bar H$ invariant under $\bar S$: $$ 2(Alt(4)\times U_4(2)), \qquad 2^{2+6}:3^3:Sym(3). $$ \vglue .2in \item {(19)} $\bar H\cong Sp(8,2)$, $|\bar H|=2^{16}\cdot 3^5\cdot 5^2 \cdot 7\cdot 17$, $|Out(\bar H)|=1$, ($p=3$ or $5$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$ if $p=5$: $$ D_4(2), \qquad Sp(4,4):2. $$ Maximal subgroups of $\bar H$ containing $\bar S$ if $p=3$: $$ D_4(2), \qquad Sym(3)\times Sp(6,2). $$ \vglue .2in \item {(20)} $\bar H\cong PSp(8,3)$, $|\bar H|=2^{14}\cdot 3^{16}\cdot 5^2 \cdot 7\cdot 13\cdot 41$, $|Out(\bar H)|=2$, ($p=2$ or $5$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$ if $p=5$: $$ (Sp(4,3)\circ Sp(4,3))2, \qquad PSp(4,9):2. $$ Maximal subgroups of $\bar H$ invariant under $\bar S$ if $p=2$: $$ (Sp(4,3)\circ Sp(4,3))2, \qquad \bar M, $$ where $\bar M$ is the image in $\bar H$ of a subgroup of $Sp(8,3)$ of the form $SL(2,3)\wr Sym(4)$. \vglue .2in \item {(21)} $\bar H\cong \Omega_7(3)$, $|\bar H|=2^9\cdot 3^9\cdot 5\cdot 7 \cdot 13$, $|Out(\bar H)|=2$, ($p=2$). \noindent Maximal subgroups of $\bar H$ invariant under $\bar S$: $$ 2U_4(3):2, \qquad 2^6:Alt(7). $$ \item {(22)} $\bar H\cong \Omega_8^+(2)$, $|\bar H|=2^{12}\cdot 3^5\cdot 5^2\cdot 7$, $|Out(\bar H)|=6$, ($p=3$ or $5$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$ if $p=5$: $$ (Alt(5)\times Alt(5)):2^2 \quad \text {in three different ways}. $$ Maximal subgroups of $\bar H$ containing $\bar S$ if $p=3$ and $S\leq H$: $$ (3\times U_4(2)):2, \qquad 3^4:2^3.Sym(4). $$ Maximal subgroups of $\bar H^*$ containing $\bar S$ if $p=3$ and $S\nleq H$: $$ 3^{1+4}_+:GL(2,3), \qquad (3^4:2^3.Sym(4)).3. $$ \item {(23)} $\bar H\cong \Omega_8^-(2)$, $|\bar H|=2^{12}\cdot 3^4\cdot 5\cdot 7\cdot 17$, $|Out(\bar H)|=2$, ($p=3$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$: $$ 2^6:U_4(2), \qquad Sp(6,2). $$ \vglue .2in {(24)} $\bar H\cong {^3D_4(2)}$, $|\bar H|=2^{12}\cdot 3^4\cdot 7^2 \cdot 13$, $|Out(\bar H)|=3$, ($p=3$ or $7$). Maximal subgroups of $\bar H$ containing $\bar S$ if $p=7$: $$ (7\times L_3(2)):2, \qquad 7^2:SL(2,3). $$ There is a unique maximal subgroup of $\bar H^*$ containing $\bar S$ if $p=3$. This case will be treated below. \vglue .2in \item {(25)} $\bar H\cong G_2(3)$, $|\bar H|=2^6\cdot 3^6\cdot 7\cdot 13$, $|Out(\bar H)|=2$, ($p=2$). \noindent Maximal subgroups of $\bar H$ invariant under $\bar S$: $$ (2^3)L_3(2), \qquad 2^{1+4}_+:3^2.2. $$ \vglue .2in \item {(26)} $\bar H\cong G_2(4)$, $|\bar H|=2^{12}\cdot 3^3\cdot 5^2\cdot 7\cdot 13$, $|Out(\bar H)|=2$, ($p=3$ or $5$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$ if $p=5$: $$ U_3(4):2, \qquad Alt(5)\times Alt(5). $$ Maximal subgroups of $\bar H$ containing $\bar S$ if $p=3$: $$ SL(3,4):2, \qquad J_2. $$ \vglue .2in \item {(27)} $\bar H\cong F_4(2)$, $|\bar H|=2^{24}\cdot 3^6\cdot 5^2\cdot 7^2\cdot 13\cdot 17$, $|Out(\bar H)|=2$, ($p=3$, $5$, or $7$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$ if $p=7$: $$ ^3D_4(2), \quad \text {in two different ways}. $$ Maximal subgroups of $\bar H$ containing $\bar S$ if $p=5$: $$ ^2F_4(2), \qquad Sp(8,2). $$ Maximal subgroups of $\bar H$ containing $\bar S$ if $p=3$: $$ Aut(D_4(2)), \qquad L_4(3):2. $$ \vglue .2in \item {(28)} $\bar H\cong {(^2F_4(2))'}$, $|\bar H|=2^{11}\cdot 3^3\cdot 5^2\cdot 13$, $|Out(\bar H)|=2$, ($p=3$ or $5$). \noindent Maximal subgroups containing $\bar S$ if $p=5$: $$ L_2(25), \qquad 5^2:4Alt(4). $$ Maximal subgroups containing $\bar S$ if $p=3$: $$ L_3(3):2 \quad \text{in two different ways}. $$ \vglue .2in \item {(29)} $\bar H\cong Sz(8)$, $|\bar H|=2^6\cdot 5\cdot 7\cdot 13$, $|Out(\bar H)|=3$. This case contradicts (2). \vglue .2in \item {(30)} $\bar H\cong {^2E_6(2)}$, $|\bar H|=2^{36}\cdot 3^9\cdot 5^2 \cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$, $|Out(\bar H)|=6$, ($p=3$, $5$, or $7$). \noindent Maximal subgroups of $\bar H$ containing $\bar S$ if $p=7$: $$ F_4(2), \quad \text {in three different ways}. $$ Maximal subgroups of $\bar H$ containing $\bar S$ if $p=5$: $$ F_4(2), \quad \text {in three different ways}. $$ Maximal subgroups of $\bar H$ containing $\bar S$ if $p=3$ and $S\leq H$: $$ Fi_{22} \quad \text {in three different ways}. $$ Maximal subgroups of $\bar H^*$ containing $\bar S$ if $p=3$ and $S\nleq H$. $$ 3^6:U_4(2):2, \qquad 3^{1+6}.2^{3+6}.(Sym(3)\times 3).3. $$ It now remains to take up the four exceptional cases: $\bar H\cong L_3(4),\ {^3D_4(2)},\ L_3(3),$ and $ L_3(4)$. Suppose first that $\bar H\cong L_3(4)$, of order $2^6\cdot 3^2\cdot 5\cdot 7$. Then $p=3$, by (2). Then $Z(H)$ is a $2$-group, the $3$-rank of $G$ is $2$, and so $|A|\leq 9$. We claim that four conjugates of $A$ suffice to generate $HA$ if $|A|=3$, and that three conjugates suffice if $|A|=9$. Indeed, suppose that $A\cap H\neq 1$, and let $1\neq a\in A\cap H$. Then two conjugates of $a$ generate a subgroup $K$ of $H$ with $Z(H)K/Z(H)\cong L_3(2)$, and $A\cap K$ is a Sylow $3$-subgroup of $N_G(K)$. Here $Z(H)K$ is maximal in $H$, so the claim holds in this case. On the other hand, suppose that $A\cap H=1$, and set $X=C_H(A)$. Referring to the ATLAS, we find that $Z(H)X/Z(H)\cong SL(2,4)$ or $Fr(21)$. Let $B$ be a Sylow $3$-subgroup of $C_G(A)$. Then $A$ is conjugate to every cyclic subgroup of $B$ not contained in $X$, and so there are conjugates $A_1$ and $A_2$ of $A$ such that $Z(H)\lan A_1, A_2\ran=Z(H)XA$. It is left to the reader to demonstrate that two conjugates of $XA$ suffice to generate $HA$ (or to find a different argument) and thus to prove the claim. We conclude that $|V|\leq 3^8$. If $S\leq H$ then there are two different maximal subgroups $M_1$ and $M_2$ of $H$ containing $S$ (with $\bar M_1\cong U_3(2)$ and with $\bar M_2 \cong M_{10}$), contrary to (1). Thus $S\nleq H$. There then exists a subgroup $Y$ of $G$ with $Y\cong Fr(21)\times 3$. Let $b$ be a non-identity element of $O_3(Y)$ and $x$ a non-identity element of $O_7(Y)$. Then $dim([V,x])=6$, and $[V,x,b]=0$, so $|V/C_V(b)|\leq 9$. But, as we have seen, $b$ is conjugate to an element of $Y$ not contained in $\lan b\ran$, and in this way we violate 2.2. Thus, $H/Z(H)\ncong L_3(4)$. Suppose next that $\bar H\cong {^3D_4(2)}$ and that $p=3$. It so happens that $Z(H)=1$, so $\bar G=G$. There are exactly two classes of elements of order $3$ in $H$. To be consistent with ATLAS notation, we label a pair of fixed representatives of these two classes by $3A$ and $3B$, where $N_H(3A)\cong Sym(3)\times L_2(8)$, and where $N_H(3B)$ is of the form $3^{1+2}_+)GL(2,3)$. Set $L= E(C_H(3A))$. We may assume that $[3A,3B]=1$, and then $3B\in E(C_H(3A))$ since $3A$ is not contained in a cyclic group of order $9$. Let $t$ be an involution in $N_H(3A)$ with $[t,L]=1$. Then $C_H(t)$ is a maximal parabolic subgroup of $H$, of the form $2^{1+8}_+:L_2(8)$. As two conjugates of $3B$ suffice to generate $L$, it follows that three conjugates of $3B$ suffice to generate $C_H(t)$, and then four conjugates suffice to generate $H$. On the other hand, we observe that $3A$ is not central in a Sylow $3$-subgroup of $H$, and hence $3A$ is not weakly closed in a Sylow $3$- subgroup of $N_H(3A)$. Thus, there is a conjugate of $3A$ which is \lq\lq diagonal" in $\lan 3A\ran\times L$, and so $C_H(3A)$ is generated by two conjugates of $3A$. Then four conjugates of $3A$ suffice to generate a subgroup of $H$ containing $C_H(t)$. As $C_H(t)$ contains no conjugate of $3A$, it follows that four conjugates of $3A$ suffice to generate $H$. Thus, we have shown that for any element $x\in H$ of order $3$, four conjugates of $x$ suffice to generate $H$. No involution-centralizer in $H$ contains a copy of $\Bbb Z_3\times \Bbb Z_3$, and then it follows from the given structure of $N_H(3B)$ that the $3$-rank of $H$ is two, and the $3$-rank of $Aut(H)$ is then at most three. In particular, we have $|A|\leq 27$, and so $|V/C_V(A)|\leq 3^6$. We now appeal to 10.4, where we find that $dim(V)\geq 24$. It follows that four conjugates of $A$ do not suffice to generate $HA$, and therefore $A\cap H=1$. Thus $|A|=3$, $|V/C_V(A)|\leq 9$, and twelve conjugates of $A$ do not suffice to generate $HA$. But $A$ is not weakly closed in $S$, and therefore eight conjugates of $A$ suffice to generate $HA$. Thus, we have a contradiction, and $H\ncong {^3D_4(2)}$. Suppose next that $\bar H\cong L_3(3)$. Then $Z(G)=1$, $|G/H|\leq 2$, and by (2) we have $p=2$. A Sylow $2$-subgroup of $Aut(H)$ is a direct product of a semidihedral with $\Bbb Z_2$, so the $2$-rank of $A$ is at most $3$. By [MS1], $G$ contains no quadratic fours groups, and then no element of $A$ is a $2$-transvection on $V$, by 5.6. Also, as $A$ contains no quadratic fours group we have $C_V(a)\neq C_V(A)$ for any $a\in A^{\#}$. As $|A|^{3/2}\geq |V/C_V(A)|$, by Hypothesis $4'$, we conclude that $|A|=8$, that $|V/C_V(A)|=16$, and that $|V/C_V(a)|=8$ for all $a\in A^{\#}$. Let $a\in A-H$. Then $C_H(a)\cong GL(2,3)$, and since $|[V,a]|=8$ we have $[V,a,Z(C_H(a))]=0$. Thus $A$ contains a quadratic fours group, and we have a contradiction in this case. Finally, suppose that $\bar H\cong L_4(3)$. Then $p=2$, as seen in (5). As $Z(H)$ is a $2$-group, we then have $Z(H)=1$ and $H\cong L_4(4)$. A Sylow $2$ subgroup of $Aut(H)$ is contained in a unique maximal subgroup of $Aut(G)$, of the form $2\times(GL(2,3)\circ GL(2,3))2$, so we have $|A|\leq 32$. As in the preceding paragraph, $G$ contains no quadratic fours groups, so $|V/C_V(a)|\geq 8$ and $C_V(a)\neq C_V(A)$ for any $a\in A^{\#}$, and we have $|A|\geq 8$. In particular, we then have $A\cap H\neq 1$. There are exactly two conjugacy classes of involutions in $H$, with representatives $2A$ and $2B$, and where $C_H(2A)$ and $C_H(2B)$ have the form $(4\times L_2(9)):2$ and $(SL(2,3)\circ SL(2,3)):2^2$, respectively. Moreover, an element of $C_H(2B)$ interchanges the two subnormal $SL(2,3)$- subgroups of $C_H(2B)$. (The involution $2B$ lifts to an involution in $SL(4,3)$, while $2A$ lifts to an element of order $4$.) Let $1\neq a\in A\cap H$, and set $K=C_H(a)$. As $H$ contains no quadratic fours groups, it follows that $a$ is the unique involution in $C_K([V,a])$, and hence $dim([V,a])\geq 4$. As $|A|^{3/2}\geq |V/C_V(A)|>|V/C_V(a)|$, we then have $|A|=16$ or $32$. Suppose $|A|=16$. Then $|V/C_V(A)|\leq 2^6$, and so $|V/C_V(a)|\leq 2^5$. We have $dim(V)\geq 26$, by 10.4(2), so it follows that $H$ cannot be generated by five conjugates of $a$. Suppose that $a$ is of type $2B$, and let $L_1$ and $L_2$ be two copies of $L_3(3)$ in $H$ such that $\lan L_1,L_2\ran=H$, with $L_1\cap L_2\cong GL(2,3)$. Then $L_1\cap L_2$ is generated by three conjugates of $a$, and since $L_1\cap L_2$ is maximal in each $L_i$ it follows that five conujugates of $a$ suffice to generate $H$. On the other hand, suppose that $a$ is of type $2A$. Then $K$ lifts in $SL(4,4)$ to a group of the form $(8\circ SL(2,9)):2$, so $O^2(K)$ contains a conjugate of $a$, and then three conjugates of $a$ suffice to generate $O^2(K)$. We have $O^2(K)\leq L$, where $L\cong PSp(4,3)$, and $N_L(O^2(K))$ is the unique maximal subgroup of $L$ containing $O^2(K)$. It follows that four conjugates of $a$ suffice to generate $L$. As $N_H(L)$ is the unique maximal subgroup of $H$ containing $L$, it follows that five conjugates of $a$ generate $H$. We conclude that $|A|=32$. Here $|V/C_V(A)|\leq 2^7$, so $|V/C_V(a)|\leq 2^6$. Then $HA$ is not generated by $A$ together with three conjugates of $a$. Further, we now have $HA=G\cong Aut(H)$, so we may choose $a\in G-H$ so that $H\lan a\ran\cong PGL(4,3)$, and so that $a$ inverts an element $x$ of order $13$ in $H$. Then three conjugates of $a$ generate a subgroup $N$ of $G$ of the form $3^3:13:2$, while a conjugate of $A$ contains an element $b$ which induces a transpose-inverse automorphism of $H$, and such that $\lan O_3(N),O_3(N)^b\ran=H$. Thus, we have a contradiction in this case, and the lemma is proved. \qed \enddemo Having disposed of so many individual groups in 10.7, we may now start in on the general case. To this end, the following elementary lemma will be helpful. \proclaim {10.8 Lemma} Let $1\leq n_1\leq n_2<\cdots\leq n_k$ be a non-decreasing sequence of natural numbers, and let $q>1$. Set $N=\sum_{i=1}^k n_i$. Then $\prod_{i=1}^k(q^{n_i}+(-1)^i)q(q-1)^2/2. $$ This inequality evidently fails if $q\geq 8$, so we have $q=2$ or $4$. The case $q=2$ has already been eliminated in 10.5, so $\bar H\cong Sp(4,4)$. Then $|\bar H|=2^8\cdot 3^2\cdot 5^2\cdot 17$, and $Out(\bar H)$ is a $2$-group. The criteria (a) and (b) in the proof of 10.7 then reduce the problem to finding a pair of distinct maximal subgroups of $\bar H$ containing $\bar S$, in the case that $p=3$ or that $p=5$. According to the ATLAS, there are in fact two different maximal subgroups of $\bar H$ of the form $(SL(2,4)\times SL(2,4)):2$, and so we conclude that $q$ is odd. Now 10.4 yields $$ 22 log_p(q)>(q-1)^2/2.\tag1 $$ Suppose that $p=2$. Then Hypothesis $4'$ yields $|A|^{3/2}\geq |V/C_V(A)|$, and so (1) improves to $$ 33 log_2(q)>(q-1)^2.\tag2 $$ This yields $q\leq 11$. Here $Z(H)$ is a $2$-group, so $Z(H)=1$ and $H=\bar H$. There is an $S$-invariant maximal subgroup $L$ of $H$ of the form $(SL(2,q)\circ SL(2,q)):2$, where the components of $L$ are interchanged by an involution in $L$. If $q$ is congruent to $3$ or $5$ mod $8$ then there is also a maximal $S$-invariant subgroup $M$ of $H$ of the form $2^4:Alt(5)$ (and which may be described as the set-wise stabilizer of an orthonormal basis in the $\Omega_5(q)$-module for $H$). Thus, we are reduced to the cases where $q=7$ or $9$. Denote by $L^*$ the normalizer in $Aut(H)$ of $L$. If $q=7$ then $L^*=L\lan d\ran$ where $d$ is an involution which induces a diagonal automorphism on each of the components of $L$. If $q=9$ then $L^*=L\lan d,\alpha\ran$, where $d$ is as just described and where $\alpha$ induces a field automorphism on each component of $L$. It follows that the $2$-rank of $S$ is $4$ if $q=7$, and at most $5$ if $q=9$. In particular, we have $|A|\leq 32$, and $|A|\leq 16$ if $q=7$. By 5.6, $G$ contains no quadratic fours groups, and so 5.5 implies that $|A|\geq 8$ and that $C_V(A)\neq C_V(a)$ for any $a\in A^{\#}$. As $|A|^{3/2}\geq |V/C_V(A)|$ we have $|V/C_V(A)|\leq 2^7$, and $|V/C_V(A)|\leq 2^6$ if $q=5$. Then $|[V,a]|\leq 2^6$ (and $|[V,a]|\leq 2^5$ if $q=5$) for any $a\in A^{\#}$. Suppose that $A\cap H$ contains an involution $a$ with $C_H(a)=L$. As $A$ contains no quadratic fours groups, $L/\lan a\ran$ acts faithfully on $[V,a]$, whereas $[V,a]$ is too small to support such an action. Thus $A\cap Z(L)=1$. There are only two conjugacy classes of subgroups of order $2$ in $H$, and one of these lifts to a class of cyclic subgroups of order $4$ in $SP(4,q)$. It follows from this that $|A\cap H|\leq 4$. Thus $|A|\leq 16$, and $|A|=8$ if $q=7$. Fix an involution $a\in A\cap H$. Then $|V/C_V(a)|\leq 32$, and $|V/C_V(a)|\leq 8$ if $q=7$. On the other hand, 8.4 yields $dim(V)\geq 24$ if $q=7$, and $dim(V)\geq 40$ if $q=9$. As $C_V(H)=0$ it follows that $H$ cannot be generated by eight conjugates of $a$. We leave to the reader the task of verifying the absurdity of this outcome. \qed \enddemo We may now complete the proof of Theorem 10.2. Thus, assume Hypothesis 10.1. In view of 10.3, 10.5, 10.6, and 10.9, we may assume that $\bar H$ is not isomorphic to $L_2(q)$, or to any of the groups in the list $\Cal L$ given by 10.6. Among the groups in $\Cal L$ one finds all the groups of Lie type (other than those which are isomorphic to $L_2(q)$) which have an exceptional Schur multiplier. Having excluded these groups, it follows that $H$ is a homomorphic image of the universal version, in the Lie-theoretic sense, of $\bar H$. Set $G_0=HA$. By Hypothesis 10.1 we have $|A|^2\geq |V/C_V(A)|$, and then Theorem 2.3 of [CD1] implies that $|G_0|^2\geq |V|$. An upper bound for $G_0$ is given by 10.8, while a lower bound for $dim(V)$ is given by 10.4. The proof of 10.2 now reduces to a systematic comparison of $log_p(|G_0|^2)$ with $dim(V)$. Suppose first that $\bar H\cong L_n(q)$, $n\geq 3$. Then $|G_0|\leq 2pq^{n^2-1}$, and the factor $p$ is required only if $q$ is a $p^{th}$ power. Thus $|G_0|\leq q^{n^2}$. By 10.4 we then have $$ 2n^2log_p(q)>(q^n-1)/(q-1)-n.\tag1 $$ Suppose that $q=2$. Then $p\geq 3$, and so $2 log_p(q)<1.5$. Then (1) yields $1.5 n^2>2^n-n-1$, and then $n\leq 5$. Thus $\bar H\in\Cal L$ in this case, and thus $q\geq 3$. From (1) we have $$ 2n^2q+n>q^{n-1}, $$ and then $$ 2n^2+n>q^{n-2}.\tag2 $$ We observe that (2) is false if $q\geq 3$ and $n\geq 6$. As $q>2$, it follows that (2) holds only if $n\leq 5$. For $n=3$ one checks that (1) is false for $q\geq 7$, and for $n=4$ we find that (1) is false if $q\geq 5$. Also, (1) fails to hold if $n=5$ and $q\geq 3$. Thus, $\bar H\in\Cal L$. Suppose next that $\bar H\cong U_n(q)$, $n\geq 3$. As in the preceding case, we find $|G_0|\leq q^{n^2}$. Referring to 10.8, we then have $$ 2n^2log_p(q)> \left\{ \aligned & (q^n-q)/(q+1)\qquad \text {if $n$ is odd, and}\\ & (q^n-1)/(q+1) \qquad \text {if $n$ is even.} \endaligned \right.\tag3 $$ Suppose $q=2$. Then we replace (3) by the estimate $$ 2n^2log_3(2)> (2^n-2)/3.\tag4 $$ If $n=8$ then the left side of (4) is less than $81$ and the right side is bigger than $84$, so (4) fails in this case. Taking the derivative of each side of (4) with respect to $n$, we observe that $$ 4n log_3(2)<(ln(2))2^n $$ for all $n\geq 8$, and so (4) holds only if $n\leq 7$. Thus $\bar H\in\Cal L$ if $q=2$, and so $q\geq 3$. Suppose $n=3$. We may take $p=2$ in (3), and obtain $18 log_2(q)> (q^3-q)/4$, which is false if $q=7$. Taking derivatives with respect to $q$ we find that $18/(ln(2)q)<(3q^2-1)/4$ for $n\geq 7$, so $q\leq 5$ if $n=3$. Now suppose that $n=4$. Then (3) yields $32 log_2(q)> (q^4-1)/4$, which is false if $q=5$. We leave it to the reader to perform the derivative test which establishes that (3) is false if $n=4$ and $q\geq 5$. We also leave to the reader the task of verifying that if $n=5$ then $q=2$. Thus, in all these cases we have $\bar H\in \Cal L$. We now have $n\geq 6$, and we obtain $$ 2n^2log_p(q)>q(q^{n-1}-1)/(q+1) $$ from (3). Then also $$ 2n^2>(q^{n-1}-1)/(q+1)>(q^{n-1}-1)/2(q-1)>q^{n-2}/2, $$ and so $4n^2>q^{n-2}$. This result fails to hold if $q=4$ and $n=6$, and it is a trivial matter to check that $q^{n-2}$ grows more rapidly than $4n^2$ as a function of $q$ and $n$, independently, for $q\geq 4$ and for $n\geq 6$. One also checks directly that (3) fails to hold if $n=6$ and $q=3$. This completes the proof that $\bar H\in\Cal L$ if $\bar H\cong U_n(q)$. Suppose next that $\bar H\cong PSp(2n,q)$, $n\geq 3$ (the case $n=2$ having been dealt with in 10.9). Then 10.8 yields $|G_0|\left\{ \aligned & (q^n-1)/2\qquad \text {if $q$ is odd, and}\\ & q(q^n-1)(q^{n-1}-1)/2(q+1) \qquad \text {if $q$ is even.} \endaligned \right.\tag5 $$ If $p=2$ then there is more than one maximal subgroup of $G$ containing $S$, by Theorem A in [A1]. Thus $p\geq 3$. Suppose $n=3$, and then suppose that $q$ is even. Then (5) yields $44 > (q^3-1)(q-1)/2$, which is false if $q\geq 4$. On the other hand, suppose that $q$ is odd. Then (5) yields $44 log_p(q)> (q^3-1)/2$, which is false if $q\geq 7$. If $q=5$ then $q$ is not a $p^{th}$ power, and then $|G_0|<5^{10}$. One then checks that $40 log_3(5)<60<(5^3-1)/2$. We conclude that $q=2$ if $n=3$. Suppose $n=4$. Then $q\geq 3$, by 10.6, and (5) then reduces us to the case $q=3$. The only prime greater than $3$ whose square divides $|PSp(8,3)|$ is $5$, so 2.2 yields $p=5$. Then $S$ is contained in a pair of subgroups $L_1$ and $L_2$ of $H$, with $L_1\cong PSp(4,3)\circ PSp(4,3)$ and with $L_2\cong PSp(4,9)$, such that $H=\lan L_1, L_2\ran$. Thus $n\geq 5$. Suppose that $n=5$. If $q=2$ then (5) yields $228 log_p(2)>155$, which is false for $p=3$ (as one checks on one's pocket calculator) and which is therefore false for all $p$ since $p\neq 2$. Then also (5) fails to hold for all even $q$, when $n=5$. If instead $q$ is odd then (5) yields $228 log_p(q)> q^5-2$, and one quickly verifies that this is false for $q=3$ (where $p\geq 5$) and then for all odd $q$ and $p$. Suppose $q=2$. Then (5) yields $$ 2(2n^2+n+1)log_3(2)>(2^n-1)(2^{n-1}-1)/3, $$ which fails to hold for $n\geq 6$. Thus, $q>2$. In general, (5) yields $$ 2(2n^2+n+1)>(q^n-1)/2log_p(q)>(q^n-1)/q> q^{n-1}-1. $$ This fails for $n=6$ and $q=3$, and since $q^{n-1}-1$ grows more rapidly than $2(2n^2+n+1)$ as a function of $n$, we have thereby eliminated all of the groups $PSp(2n,q)$ as possibilities for $\bar H$, with the exception of $Sp(6,2)$. Suppose next that $\bar H\cong P\Omega_{2n+1}(q)$, $q$ odd, $n\geq 3$. Then 10.8 and 10.4 yield $$ 2(2n^2+n+1)log_p(q) > \left\{ \aligned & (q^{2n}-1)/(q^2-1)-n\qquad \text {if $q\neq 3$, and}\\ & (3^{2n}-1)/8-(3^n-1)/2 \qquad \text {if $q=3$.} \endaligned \right.\tag6 $$ Supppose $q\neq 3$. If $n=3$ then (6) yields $44 log_2(q)> q^4+q^2-2$, which fails to hold for any odd $q$. Also, in the case $n=4$ we have $78 log_2(q)>q^6+q^4+q^2-3$, which also fails for odd $q$. In general, (6) yields $$ 2(2n^2+n+1)>(q^{2n-2}-1)/(q^2-1)log_p(q)-n/log_p(q)>q^{2n-4}/log_p(q)-n, $$ and so $4n^2+3n+2>q^{2n-5}$. This fails for $n\geq 5$ and all odd $q$, so we have a contradiction in the case $q\neq 3$. On the other hand, if $q=3$ then $log_p(q)<2$, and so (6) implies that $4(2n^2+n+1)>3^{2n-2}-3^{n-2}>3^{2n-3}$. This fails for $n\geq 4$, and then since $\Omega_7(3)\in\Cal L$ we have a contradiction. Suppose next that $\bar H\cong P\Omega_{2n}^+(q)$, $n\geq 4$. Then 10.8 and 10.4 yield $$ 2(2n^2-n+2) log_p(q)>\left\{ \aligned & q(q^{2n-2}-1)/(q^2-1)+q^{n-1}-n \qquad \text {$q\neq 2,3$, and}\\ & q(q^{2n-2}-1)/(q^2-1)-(q^{n-1}-1)/(q-1)-7\delta_{2,q} \qquad \text {$q$ even.} \endaligned \right.\tag7 $$ Suppose that $q\neq 2$ or $3$. As $log_p(q)q^{2n-4}+q^{n-2}, $$ This fails to hold if $n=4$ and $q=4$, and then also if $n\geq 4$ and $q\geq 4$. Thus, $q=2$ or $3$. As $log_p(q)q^{2n-4}-q^{n-3}-7\delta{2,q}/q>q^{2n-5}-4. $$ These inequalities fail if $n=6$ and $q=2$, and then also if $n\geq 6$ and $q\geq 2$. They also fail if $n=5$ and $q=3$. As $\Omega_8^+(2)\in\Cal L$, we are then reduced to the case $n=4$ and $q=3$, and to the case $n=5$ and $q=2$. One may then check directly from (7) that neither of these cases may occur. Suppose next that $\bar H\cong P\Omega_{2n}^-(q)$, $n\geq 4$. Then 10.8 and 10.4 yield $$ 2(2n^2-n+2) log_p(q) > q(q^{2n-2}-1)/(q^2-1)-q^{n-1}+2.\tag8 $$ Then $2(2n^2-n+2) > q^{2n-4}-q^{n-2}$, and this fails if $n\geq 6$, or if $q\geq 3$. As $P\Omega_8^-(2)\in\Cal L$, we are reduced to the case where $n=5$ and $q=2$, and in this case one checks directly that (8) fails to hold. It remains to consider the exceptional groups and the groups $^3D_4(q)$. The estimates that we obtain from 10.4 and 10.8 are as follows. \vglue .1in $^3D_4(q)$: $58 log_p(q) > q^3(q^2-1)$. \vglue .1in $E_6(q)$ or $^2E_6(q)$: $158 log_p(q) > q^9(q^2-1)$. \vglue .1in $E_7(q)$: $268 log_p(q) > q^{15}(q^2-1)$. \vglue .1in $E_8(q)$: $498 log_p(q) > q^{27}(q^2-1)$. \vglue .1in $F_4(q)$: $106 log_p(q) > q^6(q^2-1)$ if $q$ is odd, or $q^7(q^3-1)(q-1)/2$ if $q$ is even. \vglue .1in $^2F_4(q)$: $54 log_p(q) > \sqrt{q/2}(q^4)(q-1)$. \vglue .1in $G_2(q)$ ($q\geq 3$): $30 log_p(q) > q(q^2-1)$. \vglue .1in $^2G_2(q)$ ($q\geq 27$ an odd power of $3$): $14 log_p(q) > q(q-1)$. \vglue .1in $Sz(q)$ ($q\geq 8$ an odd power of $2$): $12 log_p(q) > \sqrt{q/2}(q-1)$. \vskip .1in In the case of $Sz(q)$ we may take $q\geq 32$, as $Sz(8)\in\Cal L$. None of the above inequalities holds for $q=5$, and then none holds for $q\geq 5$. For $p=3$, only the inequality for $G_2(3)$ is valid, and for $q=2$, only the inequalities for $^3D_4(2)$ and $^2F_4(2)$ are valid. All three of these groups are in $\Cal L$, so we have a final contradiction at this point, proving Theorem 10.2. \vskip .3in \noindent {\bf Section 11: Theorems 1 through 6} \vskip .2in We ask the reader to recall the definitions of $\Cal Q(Y,V)$, $q(Y,V)$ and $\Cal Q^*(Y,V)$, from the Introduction. Recall also that, for any group $G$, a {\it component} of $G$ is a subnormal quasisimple subgroup of $G$, and that $E(G)$ is the product of all the componentsof $G$. \proclaim {11.1 Lemma} Let $p$ a prime, $G$ a group whose order is divisible by $p$, and let $S$ be a Sylow $p$-subgroup of $G$. Assume that $O_p(G)=1$ and that $S$ is contained in a unique maximal subgroup of $G$. Set $H=O^p(G)$ and set $\bar G=G/\Phi(G)$. Then one of the following holds. \roster \item "{(i)}" $H$ is an $r$-group for some prime $r\neq p$, and $S$ acts irreducibly on $H/\Phi(H)$. \item "{(ii)}" $\bar H$ is a direct product of non-abelian simple groups, permuted transitively by $\bar S$. Moreover, for any component $\bar L$ of $\bar G$, $N_{\bar S}(\bar L)$ is contained in a unique maximal subgroup of $N_{\bar S}(\bar L)\bar L$. \endroster Moreover, if there exists a faithful $\Bbb F_pG$-module $V$ with $q(S,V)\leq 2$ then, in case (ii), we have $H=E(G)$. \endproclaim \demo {Proof} Denote by $M$ be the unique maximal subgroup of $G$ containing $S$, and set $F=\cap\{M^g\}_{g\in G}$. Then $G=N_G(S\cap F)F$. As $O_p(G)=1$ and $G\neq M$ it follows that $S\cap F=1$, and so $F$ is a $p'$-group. Let $M^*$ be any maximal subgroup of $G$. If $F\nleq M^*$ then $G=FM^*$, so $S^g\leq M^*$ for some $g\in G$, and this implies that $M^*$ is conjugate to $M$. Thus $F\leq M^*$, and $F=\Phi(G)$. Now let $N$ be the inverse image in $G$ of a minimal normal subgroup of $\bar G$. Then $NS=G$, and so $N=H$. Suppose that $N$ is a $p'$-group. For any prime divisor $r$ of $|N|$ there then exists an $S$-invariant Sylow $r$-subgroup of $N$, and it follows that $N$ is an $r$-group. Uniqueness of $M$ then implies that $S$ acts irreducibly on $N/\Phi(N)$, and so (i) holds in this case. On the other hand, suppose that $N$ is not a $p'$-group. Here $\bar N$ is not a $p$-group, as $N\neq F$, so $\bar N$ is a direct product of simple groups. Suppose that (ii) is false. We may then assume without loss of generality that $F=1$. The uniqueness of $M$ implies that $S$ acts transitively on the set of components of $N$. Fix a component $L$ of $N$, and suppose that there exist distinct maximal subgroups $X_1$ and $X_2$ of $N_S(L)L$ containing $N_S(L)$. Set $X_i^* =\lan S^{X_i}\ran$. Then $X_i^*$ is a proper subgroup of $G$, and $\lan X_1^*, X_2^*\ran=G$. As this contradicts the uniqueness of $M$, (ii) is proved. Suppose now that $G$ is non-solvable, and that $[\Phi(G),H]\neq 1$. Setting $Q=\Phi(G)$, we then have $Q=F(G)=F^*(G)$. Suppose also that there is a faithful $\Bbb F_pG$-module $V$ with $q(G,V)\leq 2$, and set $q=q(S,V)$. Without loss of generality, we may assume that $V$ is irreducible for $G$. Denote by $\Cal A$ the set of elements $A$ of $\Cal Q^*(S,V)$ of minimal order, and set $G_0=\lan\Cal A\ran$. Then $H\leq G_0$. Let $\Cal K$ the set of subgroups of $Q$ given by 4.6, and set $D=[Q,G_0]$. Every member of $\Cal K$ has a solvable automorphism group, so if $H$ fixes every member of $\Cal K$ we obtain $[D,H]=1$. But in that case we have $[Q,H,H]=1$, and then $[Q,O^{p'}(H)]=1$, and so $H=1$. We conclude that $H$ acts non-trivially on $\Cal K$, and then 4.6(d) implies that $|A|=p$ for any $A\in\Cal A$. We have $p\leq 3$, as follows from 4.1. Suppose first that $p=2$, and let $K\in\Cal K$. By the definition of $\Cal K$ (at the start of the proof of 4.6) there then exists $A\in\Cal A$ with $K=[K,A]$. Then 4.6(f) says that every element of $\Cal K$ is $A$-invariant. But $H\leq\lan A^G\ran$, so it follows that, in fact, $H$ acts trivially on $\Cal K$. Thus, we have a contradiction if $p=2$. Suppose that $p=3$. Fix $A\in\Cal A$, and set $L=\lan A^H\ran$. Then $L=\lan A^L\ran$ and $F^*(L)=F(L)$. There then exists an irreducible $L$-submodule $U$ of $V$ such that $[F(L),L]$ acts non-trivially on $U$. Setting $\bar L= L/C_L(U)$, we then have $F^*(\bar L)=F(\bar L)$, and we have $\bar A\cong A$ as $|A|=3$. Set $\bar Y=[F(\bar L),\bar A]$. Then [C2, Theorem A] implies that $F(\bar L)$ is a $2$-group of symplectic type, that $\bar Y$ is a quaternion group, and that $U$ is an irreducible module for $F(\bar L)$. Further, if the width of a largest extraspecial subgroup $\bar X$ of $F(\bar L)$ is $n$, then $U$ is a direct sum of $2^{n-1}$ natural $SL(2,3)$-modules for $\bar Y\bar A$ if $\bar X= F(\bar L)$, and a direct sum of $2^n$ such modules if $\bar X\neq F(\bar L)$. As $q(G,V)\leq 2$, $\bar A$ is generated by a $2$-transvection, so we conclude that $n\leq 2$, and that $\bar X=F(\bar L)$ if $n=2$. As $\bar L$ is non-solvable, it follows that $n=2$, $F(\bar L)\cong Q_8\circ D_8$, and $\bar L/F(\bar L)\cong Alt(5)$. Here $\bar L/F(\bar L)$ is incident with a component of $G/Q$, so outcome (ii), above, implies that $\bar A$ is contained in a unique maximal subgroup of $\bar L/F(\bar L)$. But such is not the case, and we have a contradiction at this point. Thus, the final statement in the lemma holds for all $p$. \qed \enddemo \proclaim {11.2 Proposition} Let $G$ be a group with $O_p(G)=1$ and let $V$ be a faithful $\Bbb F_pG$-module. Let $A\in \Cal Q(S,V)$, let $K$ be a component of $G$, put $E=\langle K^A \rangle$, and assume that $K\neq E$. Then $p=2$, and one of the following holds: \roster \item "{(i)}" $|A/C_A(E)|=2$, or \item "{(ii)}" $E\cong \Omega_4^+(2^n)$ for some $n$, $n\geq 2$, and $[V,E]$ is a direct sum of natural $O_4^+(2^n)$-modules for $EA$. \endroster \endproclaim \demo{Proof} Without loss, $G=\langle K,A \rangle$ and $C_A(E)=1$. Let $\Omega=\{K_1,\cdots ,K_t\}$ be the set of components of $G$, with $K=K_1$. Suppose first that $|A|=p$, and let $1\neq a\in A$. Then $t=p$, and for each prime factor $q$ of $|K|$ there is an $A$-invariant Sylow $q$-subgroup $Q$ of $E$ with $[Q,A]\neq 1$. If $p$ is odd then Theorem 3.8.1 in [G] says that $QA$ involves $SL(2,p)$, and so $p=3$ and $q=2$. But then $K$ is a $\{2,3\}$-group, and hence $K$ is solvable by a theorem of Burnside. Thus $p=2$, and (i) holds. On the other hand, if $|A|>p$, then Theorem 2 of [C1] yields (ii). \qed \enddemo \proclaim {11.3 Lemma} Assume Hypothesis 4, and let $A\in \Cal Q_*(S,V)$. Suppose further that there exists a component $X$ of $G$ not normalized by $A$. Set $K= \lan X^A\ran$, and let $K^S=\{K_1,\cdots,K_r\}$. Set $V_i=[V,K_i]$, $1\leq i\leq r$. Then the following hold: \roster \item "{(i)}" $p=2$, $H=K_1\times \cdots \times K_r$, and $V=C_V(H)\oplus V_1\oplus \cdots \oplus V_r$, where $K_i\cong \Omega_4^+(2^n)$, and $V_i$ is a natural orthogonal module for $K_i$. \item "{(ii)}" Each $K_i$ is invariant under $\langle \Cal Q(S,V) \rangle$, and $K_iA/C_{K_iA}(K_i)\cong O_4^+(2^n)$. \endroster \endproclaim \demo {Proof} Set $U=[V,K]$, and let $A_0$ be a complement to $C_A(K)$ in $A$. As $q(A,V)\leq 2$, 3.4 implies that $q(A_0,U)\leq 2$. Suppose that $|A_0|>2$. Then 11.2 yields $p=2$, $K\cong \Omega_4^+(2^n)$, and $V=C_V(K)\oplus U$, where $U$ is a direct sum of natural orthogonal modules for $KA_0$. Set $q=2^n$, and let $U_0$ be an irreducible $KA_0$-submodule of $U$. Then $|U_0|=q^4$, and $|A_0|\leq 2q$. Further, no element of $A_0\cap K$ induces a transvection on $U_0$ over $\Bbb F_q$, and so $|U_0/C_{U_0}(A_0)|\geq q^2$. Thus $|A_0|^2\leq 4|U_0/C_{U_0}(A_0)|$, and if $U\neq U_0$ it follows that $|A_0|^2\leq 4|U/C_U(A_0)|^{1/2}$. But if $U\neq U_0$ we have $4|U/C_U(A_0)|^{1/2}<|U/C_U(A_0)|$, and so $q(A,V)>2$. We therefore conclude that $U=U_0$. We have $H=\langle X^S \rangle$, by 11.2, so $H=\langle K^S \rangle$. Let $\{K_1,\cdots ,K_r\}$ be the set of conjugates of $K$ in $G$, and put $V_i=[V,K_i]$. Then $[U_i,K_j]=0$ for $i\neq j$, and so (i) holds. Part (ii) follows from 11.2. Suppose next that $|A_0|=2$, and let $a$ be the involution in $A_0$. Then $|U/C_U(a)|\leq 4$. As $O_2(G)=1$, there is a Sylow $2$-subgroup $T$ of $K$ of the form $R\times R^a$, where $R$ is a Sylow $2$-subgroup of $X$, and $R$ has $2$-rank at least $2$. Then $C_R(a)$ has $2$-rank at least $2$, and so $C_R(a)$ contains an involution $b$ such that $[U,a,b]=0$. Setting $A_1=\lan a,b\ran$, we then have $[U,A_1,A_1]=0$, and so 11.2 implies that $K\cong \Omega_4^+(2^n)$ and that $U$ is a direct sum of natural orthogonal modules for $KA_0$. As in the preceding paragraph, let $U_0$ be an irreducible $KA_0$-submodule of $U$. As $2^n\geq 4$, we then have $|A_0|^2\leq |U_0/C_{U_0}(A_0)|$, and hence $U=U_0$. Now (i) and (ii) follow, as above. \qed \enddemo We may now assemble the proofs of Theorems 1 through 6. \proclaim {Theorem 1} Assume Hypothesis 4. Then there exists a subgroup $K$ of $H$, unique up to conjugation, such that, upon setting $U=[V,K]$, the following conditions hold. \roster \item"{(a)}" $H = K_1 \times \cdots \times K_r$ where $\{K_1, \cdots, K_r\} = K^S$. \item"{(b)}" We have $[V,K_i,K_j]=0$ whenever $i\neq j$. \item"{(c)}" One of the following holds: \itemitem {\rm{(i)}} $K\cong O^p(SL(2,p^n))$, $n \geq 1$, and $U/C_U (K)$ is a natural $SL(2,p^n)$-module for $K$, or a direct sum of two natural modules for $K$. \itemitem {\rm{(ii)}} $K\cong O^p(O_4^\epsilon (p^n))$, $n \geq 1$, $\epsilon = \pm 1$, and $U$ is a natural orthogonal module for $K$. \itemitem {\rm{(iii)}} $K\cong O^p(SU(3,p^n))$ and $U$ is a natural module for $K$. \itemitem {\rm{(iv)}} $p=2$, $K\cong O^2(Sz(2^n))$, and $U$ is a natural module for $K$. \itemitem {\rm{(v)}} $p=2$, $A\leq C_G(K)K$, $K\cong SL(3,2^n)$ (resp. $O^2(Sp(4,2^n)$) and $U$ is the direct sum of a natural and a dual module (resp. a natural and a contragredient module) for $K$. Moreover, there exists $g\in N_S(K)$ such that $g$ interchanges, by conjugation, the two maximal subgroups of $K$ containing $S\cap K$, and if $K\cong O^2(Sp(4,2^n))$ then $A$ is conjugate to $Z(S)$. \itemitem {\rm{(vi)}} $p = 2$, $K\cong Alt(2^n + 1)$, $n \geq 3$, and $U$ is a natural module for $K$, or a direct sum of two natural modules for $K$. \itemitem {\rm{(vii)}} $p=2$, $K\cong Alt(9)$, and $U$ is a spin module for $K$, of dimenson $8$ over $\Bbb F_2$. \itemitem {\rm{viii}} $p=3$, $K\cong SL(2,5)$, and $U$ is a natural $SL(2,9)$-module for $K$. \endroster Moreover, if $K$ is not invariant under $\Cal Q^*(S,V)$ then $p=2$, $K\cong \Bbb Z_3$, $|U|=4$, and $q(S,V)=2$. \endproclaim \demo {Proof} If $G$ is solvable then the theorem follows from 11.1 and 4.6. So assume that $G$ is non-solvable. Then 11.1 yields $H=E(G)$. Let $A\in\Cal Q(S,V)$. If there is a component $X$ of $G$ such that $X$ is not $A$-invariant, then the theorem follows from 11.3. So assume that $X$ is $A$-invariant. If $X\cong SL(2,p^n)$, $[V,X]$ is a direct sum of two natural modules for $X$, and there exists a unique conjugate $Y$ of $X$ such that $[X,Y]=1$ and $[V,X,Y]\neq 0$, set $K=XY$, and otherwise set $K=X$. Let $\{K_1,\cdots,K_r\}$ be the set of conjugates of $K$ under $S$. By 11.1 we then have $H= K_1\cdots K_r$. Set $V_i=[V,K_i]$, and set $U=[V,K]$. As $O_p(G)=1$ we may assume that $A$ and $X$ have been chosen so that $[X,A]\neq 1$. Choose a complement $A_0$ to $C_A(K)$ in $A$. Then $q(A_0,U)\leq 2$, by 3.4. If $K=X$ set $A_1=A_0$, and otherwise take $A_1$ to be a complement in $A_0$ to $C_{A_0}(X)$. Then also $q(A_1,[V,X])\leq 2$. If $p=3$, $X\cong SL(2,5)$, and $[V,X]$ is a natural $SL(2,9)$-module for $X$ then $K=X$ and the theorem holds (with (c)(viii)), so we may assume that this special case does not obtain. Then 5.5 implies that $X/Z(X)$ is a group of Lie type in characteristic $p$, or that $p=2$ and $X$ is an alternating group $Alt(2^n+1)$. If $X/Z(X)$ is of Lie type in characteristic $p$ then 5.9 says that the pair $(X,[V,X])$ is described by 6.10 or by 7.5, while if $X\cong Alt(2^n+1)$ then 5.7 provides a description of $(X,[V,X])$. In particular, if $[V,X]$ is reducible, and is not a direct sum of two non-isomorphic irreducible submodules, then $X\cong SL(2,p^n)$ or $Alt(2^n+1)$ and $X$ is a direct sum of two natural modules for $X$. It follows that if $End_X([V,X])$ contains a subgroup isomorphic to $X$, then $[V,X]$ is a direct sum of two isomorphic irreducible modules, and if $W$ is an irreducible submodule of $[V,X]$ then the dimension of $W$ over $End_X(W)$ is equal to $2$. This is the case only if $X\cong SL(2,p^n)$ and $[V,X]$ is a direct sum of two natural modules. Suppose that $[V,X,Y]\neq 0$ for some conjugate $Y$ of $X$ with $Y\neq X$. Then the preceding discussion yields $XY\cong\Omega_4^+(p^n)$, $[V,X]$ is a natural orthogonal module for $XY$, and $X$ and $Y$ are the only conjugates of $X$ which act non-trivially on $[V,X]$. By definition, we then have $K=XY$ and $U=[V,X]$. The theorem follows in this case (with (c)(ii)), so we may assume henceforth that $[V,X,Y]=0$ for any conjugate of $X$ distinct from $X$. Then also $K=X$. If $K/Z(K)$ is a group of Lie type in characteristic $p$, the theorem now follows from 6.10 and 7.5. Thus, we are reduced to the case where $p=2$ and $K\cong Alt(2^n+1)$. Since $Alt(5)$ is a group of Lie type in characteristic $2$, we may assume that $n\geq 3$. The theorem then follows from 5.7 (with either part (vi) or part (vii) of (c) obtaining). \qed \enddemo In order to prove Theorems 2 and 3, let the hypotheses be as in Theorem 1, and fix $A\in\Cal Q^*(S,V)$. Suppose that $q^*(A,V)<2$, and let $K$ and $U$ be as in Theorem 1, with $[K,A]\neq 1$. Let $A_0$ be a complement in $A$ to $C_A(K)$. Then $q(A_0,U)<2$ by 3.4, and so we do not have the case given by (c)(viii) in Theorem 1. Thus, $K$ is of Lie type in characteristic $p$, or $p=2$ and $K\cong Alt(2^n+1)$ with $n\geq 3$. Suppose that $K$ is quasisimple. If the Lie rank of $K/Z(K)$ is $1$ then 6.10 implies that $H\cong SL(2,p^n)$, or that $p=2$ and $H\cong\Omega_4^-(2^n)$ or $Alt(5)$, with $U$ a natural module for $K$ in each of these cases. Moreover, if $H\cong Omega_4^-(2^n)$ then 6.10 says that $KA/C_{KA}(K)\cong O_4^-(2^n)$. Thus, Theorem 2 holds in these cases. If the Lie rank of $K/Z(K)$ is greater than $1$ then 5.9 and 7.5 yield $p=2$ and $K\cong SL(3,2^n)$, and Theorem 2 holds in this case. Now suppose that $K\cong Alt(2^n+1)$, $n\geq 3$. Then 5.7 applies. In particular, the condition $q(A_0,U)<2$ excludes the case where $n=3$ and $U$ is a spin module for $K$, and excludes also the case where $U$ is a direct sum of two natural modules for $K$. Thus, $U$ is a natural module for $K$, and we have Theorem 2 in this case. Now assume that $K$ is not quasisimple, and that $K$ is non-solvable. Then Theorem 1 yields $K\cong \Omega_4^+(p^n)$. Suppose that $A$ fixes each of the components of $K$. As $q(A_0,U)<2$ it is then easy to see that $|A|>p$, and so $A_0\cap C_G(K)K\neq 1$. Quadratic action then forces $A_0\leq C_G(K)K$, and we again appeal to the quadratic action of $A_0$, to conclude that $|A_0|\leq p^n$. But $dim_{\Bbb F_{p^n}}(C_U(a))\leq 2$ for any $a\in A_0^{\#}$, so $q(A_0,U)\geq 2$ in this case. We therefore conclude that $p=2$ and that $A_0$ interchanges the two components of $K$. Then 11.2 yields $KA/C_{KA}(K)\cong O_4^+(2^n)$. This is outcome (ii) of Theorem 2. Suppose finally that $K$ is solvable. If $K$ is not $A$-invariant then Theorem 1 yields $K\cong \Bbb Z_3$ and $|U|=4$, and this result is contained in outcome (i) of Theorem 2. So assume that $K$ is $A$-invariant. As $q(A,V)<2$, 4.5(a) implies that $|U|=p^2$, and that $K\cong O^p(SL(2,p))$ ($p=2$ or $3$). We have thus eliminated all the cases in Theorem 1 which do not remain in the statement of Theorem 2, and thus Theorem 2 is proved. Now suppose that $q(S,V)\leq 1$. If $K$ is solvable then Theorem 3 follows from 4.5(b), so assume that $K$ is non-solvable. We have $q(A_0,U)\leq 1$, so 7.5 implies that $K$ is not isomorphic to $SL(3,2^n)$ with $p=2$. If $p=2$ and $K\cong \Omega_4^+(2^n)$, $n\geq 2$, then Theorem 2 yields $|A_0|=2^{n+1}$ and $|U/C_U(A_0)|=2^{2n}$, and so $q(A_0,U)>1$ in this case. If $p=2$ and $K\cong Alt(2^n+1)$, $n\geq 3$, then 5.7 implies that $A_0$ is generated by a set of commuting transvections on the natural module $U$, so that $KA/C_{KA}(K)\cong Sym(2^n+1)$ and $q(A_0,U)=1$. Then $1=q(A,V)=q(S,V)$, and $\lan A^H\ran$ is isomorphic to a direct product of copies of $Sym(2^n+1)$. Then also $\lan A^G\ran$ is the point-wise stabilizer in $G$ of $\{K_1,\cdots,K_r\}$, and is a direct product of copies of $Sym(2^n+1)$. Thus, Theorem 3 holds in this case. Finally, the last case given by Theorem 2 is that in which $K\cong SL(2,p^n)$ and $U$ is a natural module for $K$. Again, we have $q(A_0,U)\leq 1$, and it follows that $A_0$ is incident with a Sylow $p$-subgroup of $KA/C_{KA}(K)$ and that $q(A_0,U)=1$. We then obtain $q(S,V)=1$, and $\lan A^G\ran=H$. This completes the proof of Theorem 3. \proclaim {Theorem 4} Assume Hypothesis 3, and assume that $H$ is a quasisimple group of Lie type in characteristic $p$. Let $A$ be an $F2$-offender on $V$. Then one of the following holds. \roster \item "{(a)}" $H\cong SL(2,p^n)$ and one of the following holds. \itemitem {\rm{(i)}} $V$ is a natural $SL(2,p^n)$-module for $H$. \itemitem {\rm{(ii)}} $V$ has an $H$- submodule $U$ such that both $U$ and $V/U$ are natural $SL(2,p^n)$-modules for $H$. \itemitem {\rm{(iii)}} $n$ is even and $V$ is a natural $\Omega_4^- (p^{n/2})$-module for $H$. \item "{(b)}" $H\cong SU(3,p^n)$ and $V$ is a natural module for $H$. \item "{(c)}" $p=2$, $H\cong Sz(2^n)$, and $V$ is a natural module for $H$. \item "{(d)}" $p=2$, $H \cong SL(3,2^n)$ (resp. $O^2(Sp(4,2^n)$) and $V$ is the direct sum of a natural and a dual module (resp. a natural and a contragredient module) for $H$. Moreover, there exists $g\in N_S(K)$ such that $g$ interchanges, by conjugation, the two maximal subgroups of $H$ containing $S\cap H$. If $A$ is a quadratic $F2$-offender then $A\leq H$, and if also $H\cong Sp(4,2^n)$ then $A$ is conjugate to $Z(S)$. \endroster \endproclaim \demo {Proof} Immediate from 5.9, 6.8, 6.9, and 7.5. \qed \enddemo \proclaim {Theorem 5} Assume Hypothesis $4'$, and assume that $H/Z(H)$ is isomorphic to $Alt(n)$, $n\geq 5$. If $p=2$ assume that $n$ is odd. Then one of the following holds. \roster \item "{(i)}" $p=2$, $H\cong Alt(n)$, $n\geq 5$, and $V$ is a natural module for $H$. \item "{(ii)}" $p=2$, $H\cong Alt(n)$, $n=5,7,$ or $9$, and $V$ is a spin module for $H$ (of dimension $4$, $4$, or $8$, respectively). Moreover, if $n=9$ then $A$ is the direct product of two quadratic fours groups in $H$. \item "{(iii)}" $|A|=p=3$, $G\cong Alt(n)$, $n\neq 6$, and $V$ is a natural module for $G$. Moreover, $A$ is generated by a $3$-cycle. \item "{(iv)}" $|A|=p=3$, $H\cong SL(2,5)$, and $V$ is isomorphic to the natural $SL(2,9)$-module for $H$. \item "{(v)}" $p=3$, $H\cong Alt(9)$, $|A|=27$, and $V$ is a spin module for $H$ $($of dimension $8$ over $\Bbb F_3)$. Moreover, we have $|A|^2=|V/C_V(A)|$. \endroster \endproclaim \demo {Proof} If $p$ is odd then 8.3 yields one of the outcomes (iii) through (v) above. So assume that $p=2$ (and hence also that $n$ is odd). Suppose that there is more than one non-trivial irreducible constituent for $HA$ in $V$. As $q(A,V)\leq 3/2$, there there then exists a non-trivial irreducible constituent $U$ for $HA$ in $V$ such that $q(A,U)\leq 3/4$. By the Timmesfeld Replacement Theorem there then exists $B\in\Cal Q(S,U)$ with $q(B,U)\leq 3/4$, and this is contrary to 5.7. Thus, there is a unique non-trivial constituent in $V$ for $HA$, and then $V$ is irreducible for $HA$, and hence also for $G$, by 1.2. If $n=5$ then $H\cong L_2(4)$, and we obtain (i) or (ii) from Theorem 4. So we may assume that $n\geq 7$. By 9.1 there then exists a fours group $F$ in $G$ with $A\cap F\neq 1$ and with $[V,F,F]=0$. Assuming that $V$ is not a natural module for $G$, it then follows from [MS2, Theorem 4] that $V$ is a spin module for $HF$. Let $a$ be a non-identity element of $A\cap F$, and identify $H$ with $Alt(n)$. After conjugation, we have $F=\lan (1\ 2)(3\ 4),\ (1\ 3)(2\ 4)\ran$, and $F$ is the unique quadratic fours group in $G$ containing $a$. Denote by $k$ the greatest integer in $n/3$. Then $H$ contains a subgroup $E$ generated by $k$ pairwise disjoint $3$-cycles. For any $3$-cycle $x$ in $H$ we have $C_V(x)=0$, so the dimension of $V$ over $\Bbb F_2$ is a multiple of $2^k$. In particular, if $dim(V)<8$ then $n=7$, and we have (ii). So we may assume that $dim(V)\geq 8$ (and that $n\geq 9$). As $a$ inverts a $3$-cycle, we have also $dim(V)=2dim(C_V(a))$. Then $|V/C_V(a)|\geq 16$, and then since $|A|^{3/2}\geq |V/C_V(A)|\geq |V/C_V(a)|$ we have $|A|\geq 8$. If $C_V(a)=C_V(A)$, and then every fours group in $A$ which contains $a$ is quadratic, contrary to the uniqueness of $F$. We therefore conclude that $C_V(a)\neq C_V(A)$, and this yields $|A|\geq 16$. Let $K$ be the component in $C_H(a)$, and let $A_0$ be a complement to $C_A(K)$ in $A$. Then $|A_0|\geq 4$. For any $3$-cycle $y$ in $K$ we have $C_{[V,a]}(y)=0$, so $[V,a]=[V,a,K]$, and Theorem 4 in [MS2] implies that every irreducible constituent for $K$ in $[V,a]$ is a spin module. As $A_0$ acts quadratically on $[V,a]$, by hypothesis, we conclude that $|A_0|=4$. Then $F\leq A$, we may take $A_0$ to be a quadratic fours group contained in $K$, and we have $A=F\times A_0$. Now $|A|^{3/2}=64$, and since $C_V(a)\neq C_V(A)$ we have $dim(C_V(a))\leq 5$. Then $dim(C_V(a))=4$, and the faithful action of $K$ on $C_V(a)$ yields $n=9$ or $11$. Now $dim(V)=8$, and since $11$ is not a divisor of $|L_8(2)|$ we have $n=9$, and (ii) holds. \qed \enddemo \proclaim {Theorem 6} Assume Hypothesis $4'$, and assume that $S$ is contained in a unique maximal subgroup of $G$. Assume also that $H/Z(H)$ is a quasisimple group of Lie type in characteristic different from $p$, or a sporadic group, and that there exists no isomorphism of $H/Z(H)$ with a group of Lie type in characteristic $p$. Then $|A|=p=3$, $G\cong Sp(6,2)$, $V$ has dimension $7$ over $\Bbb F_3$, and $|A|^2=|V/C_V(A)|$. \endproclaim \demo {Proof} Immediate from 2.2 and 10.2. \qed \enddemo \Refs \widestnumber\key{CCNPW} \ref \key A1 \by M. Aschbacher \paper On finite groups of Lie type and odd characteristic \endref \ref \key A2 \bysame \paper $GF(2)$-representations of finite groups \jour Amer. 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