2$ by assumption, we have $b\geq 4$. Let $A$ now denote $W_{\a}$. Then $[A,\yb]=1$. As $A\leq Q_{\a'-3}\leq G_{\a'-2}$, and as $Y_{\a'-2}=Y_{\a'-3}Y_{\a'-1}= Y_{\a'-3}\yb$, we have $A\leq Q_{\a'-2}$, and so $A\leq G_{\a'-1}$. We observe also that $\Phi(A)\leq \ya\leq Q_{\a'-1}$. If $A\leq Q_{\a'-1}$ then $[A,\yap]=[\ya,\yap]\leq \ya$. But $\eta(\ga,A)\geq 2$, as $\vb$ is not normal in $\ga$. Thus $A\nleq Q_{\a'-1}$. By analogy with (1) and (2) we may then choose $\l\in\Delta(\a'-1)$ so that: \vskip .1in \noindent (3) $|A:A\cap\gl|=p$, and $G_{\a'-1}=\lan a,G_{\a'-1}\cap\gl\ran$ for any $a\in A-\gl$. \vskip .1in Set $A_0=A\cap\gl$. Then $[A_0,\yl]\leq Y_{\a'-1}\leq\ya$. By 1.12, no element of $\qb-\qa$ induces a transvection on $A/\ya$, so we have $\yl\leq\qa$. We observe that, as a consequence of (3), $\yl$ is not $A$-invariant, and so $[A,\yl]\nleq\yb$. As $\yl\leq\qb$ we have $[\vb,\yl]\leq\yb$, so we may fix $\d\in\Delta(\a)-\{\b\}$ with $[\vd,\yl]\nleq\yb$. In particular, we have $[\vd,\yl]\neq 1$, while $\yl\leq\qa\leq\gd$. As $\yd\yb=\ya$ we have $[\yd,\yap]\neq 1$, and so $\yd\nleq V_{\a'-1}$. Then $\yd\nleq [\vd,\yl]$, so $[\vd,\yl\cap leq Q_{\d}]=1$. In particular, we have $\yl\nleq\qd$, and by analogy with (1) we may fix $\g\in\Delta(\d)$ so that $|\yl:\yl\cap\gg|=p$ and so that $\gd=\lan\gg\cap\gd,\yl\ran$. Then $[\yg,\yl]\neq 1$. But $[\yg,\yl\cap\gg]=1$ as $\yd\nleq V_{\a'-1}$. As $p1$. Notice that $C_{L_1}(\w Y_P)$ centralizes $\w V$, as $\w V=\lan (\w Y_P)^{L_0}\ran$. Thus $\bar L_1$ acts faithfully on $\w Y_P$, and since $|\w Y_P|=3$ we then have $s=2$, $Z(\bar K_1)=Z(\bar K_2)$, and $\w V=[\w V,D]$ is a natural $O_4^+(3)$-module for $\bar H$. Here $|\bar A|=3$ and $|\w V/C_{\w V}(\bar A)|=9$, so 2.3 shows that $b=3$, and if $(\a,\b,\g,\a')$ is a critical path in $\Gamma$ then $\vb$ is a quadratic $F2$-offender on $\wvap$. But $\vb\qap$ is normal in $(\gb\cap\gap)\qap$, and one may observe that $(\gb\cap\gap)\qap=G_{\g}\cap \gap=P\cap H$. But $\bar A$ is not normal in $P\cap H$ since $t$ interchanges $\bar K_1$ and $\bar K_2$. This contradiction proves that either the lemma holds holds or: \vskip .1in \noindent (1) We have $X\nleq D$. \vskip .1in We may now assume that (1) is the case. If $D\leq N_H(R)$ then $[D,R]\leq Q_H$, and then 1.10 yields $[\bar D,\bar O^p(H)]=1$. This implies that $O^p(H)\leq D$, contrary to (1), so we conclude that $D\nleq N_H(R)$. In particular, $D\nleq N_G(O^p(P))$. Minimality of $H$ implies that $\lan P,P_1\ran\in\Cal L$ for any $P_1\in\Cal P_{DS}(S)$, and then $P_1$-Uniqueness (1.2 and 1.3(4)) yields $q=2$ or $3$. Notice that, by 3.10, $\bar A$ is generated by elements $\bar a$ such that $|\w V/C_{\w V}(\bar a)|\leq p^2$. Then 3.12 yields $p=2$, whence also $q=2$ and $P\cap H=S$. Thus $H=\w P$, and $\bar D$ is a $3$-group, with $|[\bar D,\bar R]|=3$. Then $|\bar R|=2$, so $\bar R=\bar A$ and $|[\w V,R]|\leq 4$. We now appeal to 3.13, and find that $H/D$ is a symmetric group of degree $2^n+1$ for some $n$, and that $RD/D$ is generated by a transposition. Then $N_H(R)D$ is not maximal in $H$. This is contrary to $P_1$-Uniqueness, and the lemma is thereby proved. \qed \enddemo \proclaim {Lemma 4.4} Suppose that $F_p^*(H)\neq F_p(H)$. Then $H=(P\cap H)E_p(H)$, and $O^p(\w P)\leq E_p(H)$. Moreover, $H$ acts transitively on its set of $p$-components. \endproclaim \demo{Proof} Let $K$ be a $p$-component of $H$, and set $L=\lan K^H\ran$. Suppose that $O^p(\w P)\nleq L$. Then $S\cap L\nl \w P$, and since $P\cap H_0=O_p(P\cap H)$ we have also $S\cap L\nl P\cap H$. Thus $S\cap L\nl H$, and $\bar L$ is a $p'$-group. As $L\leq H_0$ we have $[L,H_0]\nleq Q_H$, and so $[L,O^p(\w P)]\nleq Q_H$. Set $R=O_p(O^p(P))$. As $R\nleq O_p(\w P)$, by 1.10, $[L,R]$ is not a $p$-group. Then 1.4 shows that $L=\lan N_L(R)\ran P_1$ where $P_1\in\Cal P_{LS}(S)$, and where $P_1$ involves $SL(2,q)$. As $\bar L$ is a $p'$-group we then have $q\leq 3$, $|P_1:N_{P_1}(R)|\leq 4$, and $\bar P_1\cap\bar L$ is an $r$-group, where $\{p,r\}=\{2,3\}$. Further, a Frattini Argument implies that $\bar S$ normalizes a Sylow subgroup of $\bar L$ for every prime divisor of $|\bar L|$, so 1.4 now yields $|L:N_L(R)|\leq 4$. As no non-abelian simple group has a subgroup of index less than $5$, we have a contradiction. Thus $O^p(\w P)\leq L$, and then $L=H_0$. This yields the lemma. \qed \enddemo In the following lemma there is some non-standard notation. For a group $X$ we write $Z_p(X)$ for the complete pre-image in $X$ of $Z(X/O_p(X))$. \proclaim {Lemma 4.5} Suppose that $H$ is non-solvable. Then the following hold. \roster \item"{(a)}" $O^p(\w P)$ is a product of $p$-components of $H$. \item"{(b)}" If $K$ is any $p$-component of $H$ then $K/Z_p(K)\in{\Cal Lie}(p)$ or $p=2$ and $K/Z_2(K)$ is an alternating group $Alt(2^n+1)$, $n\geq 3$. \endroster \endproclaim \demo{Proof} As $H$ is non-solvable, 4.3 implies that $F^*_p(H)\neq F_p(H)$, and the preceding lemma then yields $H=(P\cap H)E_p(H)$, with $O^p(\w P)\leq E_p(H)$. Set $T=C_S(Y_H)$. Then $T$ is a Sylow $p$-subgroup of $TH_0$, and $T\geq O_p(O^p(P))Q_H$. Recall the notation given in section 0, and set $r=r(T,\w V)$ and $\Cal A=\Cal A\Cal F^*(T,\w V)$. Further, set $T_0=\lan \Cal A\ran$. Notice that, by 2.3, we have $r\leq 2$. As $T_0\nleq Q_H$, we have $T_0\nleq O_p(\w P)$, and therefore we may fix $A\in\Cal A$ with $A\nleq O_p(\w P)$. As $F^*_p(H_0)=E_p(H_0)$ we may choose a $p$-component $K$ of $H$ such that $[\bar K,\bar A]\neq 1$. Then $H_0=\lan K^H\ran$, by 4.4. Set $L=\lan K,A\ran$, and suppose first that $K\cong SL(2,p^n)$ for some $n$. Then (ii) holds, and we may assume that (i) is false. Then $S\cap E_p(H)\nl \w P$. As $C_{P\cap H}(Y_H)=O_p(P\cap H)$, a Frattini Argument then shows that $S\cap E_p(H)\nl H$, which is absurd. Thus, the lemma holds in this case, and we may assume henceforth that $K$ is not isomorphic to $SL(2,p^n)$. Then 3.4 implies that $L=KA$, $K$ is invariant under $\lan A^H\ran$, and there exists a subgroup $B$ of $L$ with $|\bar B|>2$, such that $\bar B$ acts faithfully on $\bar K$ and quadratically on some non-trivial $L$-invariant section of $\w V$. Then 3.5 shows yields: \vskip .1in \noindent (1) We have $\bar K/Z(\bar K)\in{\Cal Lie}(p)$, or else $p=2$ and $\bar K/Z(\bar(K))$ is isomorphic to $Alt(m)$ for some $m$, or to $U_4(3)$, or a sporadic group. \vskip .1in Set $X=O^p(\w P)(S\cap E_p(H))$. As $O^p(\w P)\leq\lan A^H\ran$, $K$ is $X$- invariant, and indeed every $p$-component of $H$ is $X$-invariant. We may therefore assume that $K$ has been chosen so that $S\cap K\nleq O_p(X)$. Then $O^p(\w P)\leq [O^p(\w P),S\cap K]\leq K$. Set $X_1=(S\cap K)O^p(\w P)$ and set $N=N_{P\cap H}(K)$. As we have already seen, (ii) follows from (i) via 3.6, so we may assume that $K\neq X_1$. Denote by $\Cal M$ the set of maximal subgroups of $K$ containing $X_1$, and let $M\in\Cal M$. As $H_0Q_H=E_p(H)=\lan X_1,P\cap H\ran$ we have $\bar K=\lan (\bar M)^{\bar N}\ran$. Thus, $\bar K$ has an automorphism $\phi$ which fixes a Sylow $p$-subgroup of $K$, and which moves a maximal subgroup of $K$ containing that Sylow $p$-subgroup. In the case of the sporadic groups listed in (1), with $p=2$, there is no such automorphism, as one may check using [Aschbacher]. Suppose that $p=2$ and $\bar K/Z(\bar K)\cong U_4(3)$. Then $\bar M/Z(\bar K)$ is of the form $2^4:Alt(6)$, and $|N:N_N(\bar M)|=2$. The two maximal subgroups of $\bar M$ containing $Z(\bar K)(\bar S\cap\bar K)$ are invariant under all automorphisms of $\bar K$ which fix $\bar S\cap\bar K$, so we conclude in this case that $X_1$ is $N$-invariant. But $K=\lan (X_1)^N\ran$, and so we have a contradiction in this case. If $p=2$ and $\bar K$ is an alternating group $Alt(m)$ then $N\leq S$, and we again have $K=\lan (X_1)^N\ran=X_1$, for a contradiction. Thus, we are now reduced to the case where $\bar K$ is a group of Lie type in characteristic $p$. We shall say that a subgroup $\bar P_0$ of $\bar K$ is \lq\lq parabolic" if $P_0\geq Z(K\ mod\ C_H(\w V))$ and $\bar P_0/Z(\bar K)$ is a parabolic subgroup of $\bar K/Z(\bar K)$ in the ordinary sense. Thus, $\bar M$ is a maximal parabolic subgroup of $\bar K$, and as $\bar M$ is not $N$-invariant it follows that some element of $N$ induces a non-trivial automorphism on the Coxeter diagram associated with $\bar K$. Let $M_0$ be a parabolic subgroup of $M$ which is minimal subject to containing $\bar X_1$. Then $K=\lan (M_0)^N\ran$. As $\w P\in\Cal P$, it follows from inspection of the various Coxeter diagrams that $\bar M_0$ is in fact a minimal parabolic subgroup of $\bar K$, that $M_0=M$ is uniquely determined, and that $\bar K/Z(\bar K)$ is of Lie rank $2$. Then $S$ normalizes $M_0$, so $p$ is odd, and $\bar K/Z(\bar K)$ is isomorphic to $L_3(p^n)$ or $G_2(3^n)$ for some $n$. Moreover, we have $|N:N_N(M)|=2$. Set $D=N_H(S\cap E_p(H))$, and suppose next that $D\leq N_H(O^p(P))$. Let $N_1$ be a Hall $p'$-subgroup of $N$. Then $[D\cap K, N_1]\leq P\cap K\leq O_p(P\cap H)\cap K$, and thus $\bar N_1$ acts trivially on the group $\bar I=(\bar D\cap \bar K)/(\bar S\cap\bar K)$. But in fact, any automorphism of $\bar K$ which acts non-trivially on the diagram for $\bar K$ also acts non-trivially on $\bar I$, so we have a contradiction in this case. Thus $D\nleq N_H(O^p(P))$. Then 1.4 shows that there exists $P_1\in\Cal P_{DS}(S)$ such that $\lan P,P_1\ran\in\Cal L$, with $P_1\nleq N_H(O^p(P_1))$. Set $J=\lan P,P_1\ran$. By 1.3(4) we have $J/C_J(Y_J)\cong SL(3,q)$ or $Sp(4,q)$, and since $D$ is solvable, by 1.11(b), we conclude that $q=3$, $P_1/O_3(P_1)\cong SL(2,3)$, and $|P\cap H:S|=2$. As $O^3(\w P)\leq K$, $K$ is $S$-invariant, and we may now conclude that $N=P\cap H$ and that $Q_HK=E_p(H)$. Then $E(\bar H)\bar S/E(\bar H)$ may be identified with a subgroup of $Out(\bar K)$ covered by field automorphisms. It follows that $[\bar I,\bar S]\leq \bar S\cap \bar K$. As $[\bar P_1,\bar S]$ is not a $p$-group, we have a final contradiction at this point, proving 4.5. \qed \enddemo \proclaim {Lemma 4.6} There is then a unique maximal subgroup of $H$ containing $P\cap H$. \endproclaim \demo{Proof} Immediate from 4.5 and 4.3. \qed \enddemo For the remainder of this section, denote by $M$ the unique maximal subgroup of $H$ containing $P\cap H$, and set $N=N_H(Y_P)$. Also set $R=O_p(O^p(P))Q_H$, set $r=r(R,\w V)$, and fix $A\in\Cal A(R,\w V)$. \proclaim {Lemma 4.7} Let $K$ be a $p$-component of $H$. Then $\bar K/Z(\bar K)$ is of Lie type in characteristic $p$. Moreover, the Lie rank of $K$ is at most $2$, and if equal to $2$ then $p=2$ and $\bar K/Z(\bar K)\cong L_3(2^n)$ or $Sp(4,2^n)$ for some $n$. \endproclaim \demo{Proof} Evidently $N\leq M$. Suppose first that this inclusion is proper. Set $M_0=O^p(M\cap H_0)$. Then $M=(P\cap H)M_0$, and so $M_0\nleq N_G(Y_P)$. As $M_0S$ is a proper subgroup of $H$, we have $\lan P,P_1\ran\in\Cal L$ for every $P_1\in\Cal P_{M_0S}(S)$, and the $P_1!$-Theorem then says that there is a unique $P_1\in\Cal P_{M_0S}(S)$ such that $P_1\nleq N_G(Y_P)$, and moreover, we have $O^p(P_1/O_p(P_1))\cong SL(2,q)'$. Suppose now that a $p$-component $K$ of $H$ is of Lie type in characteristic $p$. Then $M_0/O_p(M_0)$ is abelian, and hence $q=2$, $P\cap H=S$, $H=\lan \w P,P\cap H\ran=\w P$, and the lemma holds. In view of ---- we may therefore assume that $p=2$ and that $\bar K\cong Alt(2^n+1)$ for any $2$-component $K$ of $H$. If $n\geq 4$ then $M_0S$ is generated by the elements of $\Cal P_{M0S}(S)-\{P_1\}$, by ----, and $M=N$ in that case. Thus, we have $n=3$, and $\bar M_0\cong Alt(8)\cong L_4(2)$. Again, $P_1!$ implies that $q=2$ and then that $H=\w P$. Then $S$ acts transitively on the set of $2$-components of $H$, and since $P_1/O_2(P_1)\cong L_2(2)$ it follows that there is just one $2$-component in $H$. Suppose that $a^*<2$. Then ---- shows that $[\w V,K]$ is the irreducible permutation module for $\bar K$. Then $C_{[\w V,K]}(S\cap K)=C_{[\w V,K]}(M_0)$, and this group is of order $2$. But also $\w V=\lan (\w Y_P)^K\ran$, where $|\w Y_P|=2$, so it follows that $Y_P$ is $M_0$-invariant in this case. We conclude from ---- that $a^*=2$, $b=3$, and $A$ may be chosen so that $\bar A$ is normal in $\bar N$. Here $\bar N\cap \bar M_0$ is a maximal parabolic subgroup $\bar M_1$ of $\bar M_0$, and inspection of these then shows that $\bar A=O_2(\bar M_1)$ (of order $8$ or $16$). Now --- shows that there is no $\Bbb F_2$-module for $\bar K$ on which $\bar A$ is a quadratic $F2$-offender. By this contradiction, we may assume henceforth that $M=N$. Then ---- shows that $p=2$, $\bar K$ is isomorphic to $Alt(2^n+1)$ for some $n$, $n\geq 3$, and we have $M=N$. But one may observe that $O_2(M\cap H_0)=O_2(H)$, and hence $C_S(Z_0)=O_2(H)$. But this means that $S\cap H_0=O_2(H)$, which is evidently untrue. \qed \enddemo \proclaim {Lemma 4.8} Assume that $H$ is non-solvable. Then there are normal subgroups $L_i$ of $H_0$ and subgroups $V_i$ of $V$, $1\leq i\leq s$, such that \roster \item"{(1)}" $H_0=L_1\cdots L_s$, \item"{(2)}" $[L_i,L_j]\leq Q_H$ for all $i\neq j$, \item"{(3)}" $V_i=[V,L_i]$, and $[\w V_i,L_j]=1$ for all $i\neq j$, \item"{(4)}" Set $L=L_1$ and set $W=V_1/C_{V_1}(L)$. Assume that the indexing has been chosen so that $[\bar L,\bar A]\neq 1$, let $A_0$ be a complement in $A$ to $C_A(\bar L)$, and set $r_0=r(\bar A_0,W)$. Then one of the following holds. \itemitem{(a)} $\bar L$ is isomorphic to $SL(2,p^n)$, and $W$ is a natural module for $\bar L$. \itemitem{(b)} $r_0=2$, $\bar L\cong SU(3,p^n)$ or $Sz(p^n)$, and $W$ is a natural module for $\bar L$. \itemitem{(c)} $r_0=2$, $\bar L\cong SL(2,p^n)$, and $W$ involves two non-trivial constituents for the action of $\bar L$, each of which is a natural module. \itemitem{(d)} $r_0=p=2$, $\bar L\cong Sp(4,2^n)'$ and $W$ is the direct sum of a natural and a contragredient module for $\bar L$. \itemitem{(e)} $p=2$, $\bar L\cong SL(3,2^n)$ and $W$ is the direct sum of a natural and a dual module for $\bar L$. \itemitem{(f)} $\bar L\cong\Omega_4^{\e}(p^n)$ and $W$ is a natural orthogonal module for $\bar L$. \item"{(5)}" $\w V=\w V_1\cdots \w V_rC_{\w V}(H_0A)$. \endroster \endproclaim \demo{Proof} Choose $K$ so that $[\bar K, \bar A]\neq 1$, and with $K\leq \w P$. Set $L=[K,A]$ and $W=[V,L]$. Then 0.1 says that $C_A(\bar L)=C_A(\w W)$ and that $r(\bar A_0,\w W)\leq r(\bar A,\w V)$ for any complement $A_0$ to $C_{ A}(\bar K)$ in $A$. In particular, we have $r(\bar A_0,\w W)\leq 2$. Fix an irreducible $K$-submodule $\hat U$ of $\hat W$, and denote by $X$ the product of all of the $p$-components of $H$ which are not contained in $L$. Suppose first that $L\neq K$, and let $a\in A_0$ with $K\neq K^a$. Then $\hat U\cap(\hat U)^a=1$, and so, in particular, we have $|\w W/C_{\w W}(a)|>4$. Then $|\bar A_0|\neq 2$, and we may appeal to Theorem 3 of [$P_G(V)$-paper] for the structure of $\bar L\bar A_0$ and of $\w W$. Thus, $p=2$, $\bar L\cong \Omega_4^{\ +}(2^n)$ for some $n$, and $\w W$ is a direct sum of natural orthogonal modules for $\bar L$. As $r(\bar A_0,\w W)\leq 2$, one easily determines that $\eta(L,\w W)=1$, and thus $[\w W,X]=1$. Suppose next that $L=K$, and set $K^*=N_S(K)K$. Suppose that $\eta(K^*,\w W)>1$. Then 3.--- says that $\bar K\cong SL(2,p^n)$, that $\eta(\bar K,\w W)=2$, and that the $K$-irreducible constituents of $\hat W$ are natural $SL(2,p^n)$-modules for $\bar K$. It follows from ----- that there exists at most one $p$-component $K_1$ of $X$ such that $[\w W,K_1]\neq 1$, and that if such a $p$-component $K_1$ exists, then $\bar K\bar K_1\cong \Omega_4^{\ +}(p^n)$ and $\w W$ is a natural orthogonal module for $\bar K\bar K_1$. If $\eta(K^*,\w W)=1$ then also $\eta(K,\w W)=1$ and $[\w W,X]=1$. With 3.--- , the lemma now follows \qed \enddemo \proclaim {Lemma 4.9} Suppose that $H$ is non-solvable, and suppose that $M\neq N$. Set $U=\lan (Y_P)^M\ran$. Then $q=2$, $H=\w P$, $|Y_P|=4$, and $|U|=8$. \endproclaim \demo{Proof} As $M\neq N$, and since $M=(M\cap H_0)(P\cap H)$, it follows from 1.4 that there exists $P_1\in\Cal P_{(M cap H_0)S}(S)$ such that $\lan P,P_1\ran\in\Cal L$, and such that $O^p(P_1/O_p(P_1))$ is isomorphic to the commutator subgroup of $SL(2,q)$. On the other hand, $(M\cap H_0)/O_p(M\cap H_0)$ is abelian, by inspection of the set of outcomes in 4.8(4). Thus $q=2$, $P\cap H=S$, and $H=\w P$. Set $J=\lan P,P_1\ran$. Then 1.3(4) says that $J/C_J(Y_J)$ is isomorphic to $L_3(2)$ or $Sp(4,2)$, and that $Y_J$ is a natural module for $J/C_J(Y_J)$. Then $|U|=8$. \qed \enddemo \proclaim {Lemma 4.10} Assume that $H$ is non-solvable, let $K$ be a $p$-component of $H$, and set $\hat V=V/C_V(H_0)$. Then one of the following holds. \roster \item"{(i)}" $K$ is the unique $p$-component of $H$, $M=N$, $\bar K\cong SL(2,q)$ or $Sz(q)$, and $\hat V$ is a natural module for $\bar K$. Moreover, if $\bar K\cong Sz(q)$ then $b=3$. \item"{(ii)}" $q=2$ and $\bar K\cong L_3(2)$ or $Alt(6)$. \item"{(iii)}" $q=2$, $M\neq N$, $\bar H\cong Sym(5)$, and $\hat V$ is a natural $\Gamma L(2,4)$-module for $\bar H$. \item"{(iv)}" $q=2$, $M=N$, $b=3$, $\bar H\cong Alt(5)$ or $Sym(5)$, and $\hat V$ is an $\Omega_4^{\ -}(2)$-module for $\bar H$. \endroster \endproclaim \demo{Proof} Set $U=\lan (Y_P)^M\ran$ and set $M_i=O^p(M\cap L_i)$, $1\leq i\leq s$. If $M_1=1$ then it follows from 4.8(4) that $\bar L\cong L_3(2)$ or $Alt(6)$, so that outcome (ii) of the lemma holds in this case. We may therefore assume that $M_1\neq 1$. Suppose first that $[U,M_1]=1$. Then $[U,O^p(M\cap H_0)]=1$, and since $M=(M\cap H_0)(P\cap H)$ it follows that $Y_P$ is $M$-invariant. Further, as $C_{\w V}(M\cap H_0)\neq C_{\w V}(H_0)$, and since $M_1\neq 1$, it follows that 4.8(4)(f) holds with $p^n=2$ and with $\e=-1$. Here $R$ is $M$-invariant, by 1.7, so we have $A\leq H_0Q_H$, and then $r_0=2$. As $r_0\leq r$, by 0.1, 2.3 now implies that $b=3$. Thus (iv) holds, and we may assume henceforth that $[U,M_1]\neq 1$. Suppose next that we have $s>1$, and then suppose further that there exists an element $z$ in $([U,M_1]\cap Y_P)-Y_H$. Then $Q_P\cap Q_H=C_{Q_H}(z)$ is invariant under $\lan P\cap H, L_2\ran = H$, contrary to 1.9. Thus $[\w U,M_1]\cap \w Y_P=1$. This yields $U\neq Y_P$, and then 4.9 yields $|\w U|=4$. Since $[\w U,M_1]\neq 1$ we obtain $\w U\leq V_1$, contrary to $s>1$. We have therefore shown that $s=1$. As $P\cap H$ acts irreducibly on $\w Y_P$, and as $Y_P$ is not normal in $H$, we have $C_{Y_P}(O^p(H))=Y_H$. Thus $|\hat Y_H|=|\w Y_H|=q$. Further, we have $[Y_P,S\cap L]\leq Y_H$, since $S\cap L\leq O_p(P\cap H)$. Suppose that $\bar L\cong SL(2,p^n)$ and that $\hat V$ is a natural module for $\bar L$. Supposing that $M=N$, we obtain $q=p^n$, and so (i) holds. On the other hand, suppose that $M\neq N$. Then $q=2$, $H=\w P$, $|\w U|=4$ and $\bar L\cong SL(2,4)$, by 4.9. Further, there exists $P_1\in\Cal P_M(S)$, and therefore $\bar H\cong Sym(5)$. Thus (iii) holds in this case, and we may assume henceforth that the pair $(\bar L,\hat V)$ does not consist of $SL(2,p^n)$ and a natural module. In particular, $b$ is then odd, as follows from 2.1 and 3.9. Fix a critical path $(\a,\b,\cdots,\a')$, and take $H=\b$ and $P=\b+1$. As $b$ is odd, 1.15(b) yields $\yb\cap\yap=1$. Setting $V_0=V\cap\qap$, and taking $A=\vap$, we then have $[V_0,A\cap Q_H]=1$. Suppose next that $|\ya/(\ya\cap\qap)|2$, and then $M=N$ by 4.9. Observe that there exists a subgroup $X$ of $C_M(Y_P)$ with $|\bar X|=q-\e$. It follows that $[O^p(P),X]\leq Q_P$. Since $A=V^g$ for some $g\in O^p(P)$, $A$ is then $X$-invariant. It follows that $\e=+1$. As $\w P$ contains a $p$-component of $H$, by 4.5(a), there exists an element $f$ of $P\cap H$ such that $f$ interchanges the two $p$-components of $H$. If $p$ is odd then 1.16 shows that $f$ may be chosen in $\gb\cap\gap$, so that $f$ normalizes $A$. But for $p$ odd the only quadratic subgroups of $\bar L$ are contained in components of $\bar L$, so we now conclude that $p=2$. Then $f$ may be chosen to induce an $\Bbb F_q$-transvection on $\hat V$, centralizing $\hat Y_P$ in particular. Then $f\in O_2(P\cap H)=Q_PQ_H$, so we may take $f\in Q_P$. Then $f$ normalizes $A$, and indeed $[A,f]\leq [A,Q_P]\leq A\cap Q_H$, since ' $[W,Q_P]=[W,O_2(\gg\cap\gap)]$ is the unique $3$-dimensional $\Bbb F_q$-subspace of $W$ which is invariant under $Q_P$. Thus $[\bar A,\bar f]=1$. Let $D$ be a subgroup of $P\cap H$ of order $q-1$. Then $D$ acts regularly on $\hat Y_P$, and hence $\bar D$ acts as a group of inner automorphisms of $\bar L$. By 1.16 we may take $D$ to normalize $A$. We recall however that $\bar A$ is also invariant under a subgroup $\bar X$ of $\bar M$ of order $q-1$, such that $\bar X$ centralizes $\hat Y_P$. It follows that $\bar A$ is normal in $\bar M$. But the only proper $\bar M$-invariant subgroups of $\bar S\cap\bar L$ are contained in components of $\bar L$, so we have a contradiction at this point. We are reduced to the case where $\bar A\nleq \bar L$. Then $[\bar M,\bar A]$ is not a $p$-group. We recall however that $A\leq R$ and that $N$ normalizes $R$. Thus $M\neq N$, and 4.9 yields $q=2$ and $|U|=8$. As $Y_P$ is not normal in $M$ we then have $\hat U\neq \hat Y_P$, and so $|\hat U|=4$. This implies that $2^n=4$. Also, we note that $[Y_P,A]=1$, $\bar A$ acts as a group of $\Bbb F_4$-linear automorphisms of $\hat V$, and hence $|\bar A:\bar A\cap\bar H|=2$. Then also $|\bar A|\leq 8$. By symmetry we have also $|V/V_0|\leq 8$ Suppose that $|\bar A|>2$, and let $\bar a$ be a non-identity element of $\bar A\cap\bar L$. Then $|C_{\hat V}(\bar a)|=16$, and since $[V_0,A]\leq \yap$, of order $2$, it follows that $|V/V_0|\geq 8$. Again by symmetry, we conclude that $|\bar A|=8$ and that $|\bar A\cap\bar L|=4$ But $C_{\hat V}(\bar a)=C_{\hat V}(\bar b)$ for any non-identity element $\bar b$ of $\bar A\cap \bar L$, and this is inconsistent with $[\hat V_0,\bar A]$ being of order at most $2$, and with $|\hat V/\hat V_0|<16$. We therefore conclude that $|\bar A|=2$, and also, by symmetry, $|V/V_0|=2$. Then $\bar A$ induces an $\Bbb F_4$-transvection on $\hat V$, and similarly for $V$ on $W$. As $V_0$ and $A\cap Q_H$ are index-$2$ subgroups of $V$ and $A$, respectively, we obtain $[V_0,A]=\yap$ and $[V,A\cap Q_H]=Y_H$. Now 2.3 implies that $b=3$. As $\bar A$ is $\Bbb F_4$-linear on $\hat V$, and as $[V,A]\geq Y_H$, we have $|[V,A]|\geq 8$. Now $[\hat V,\bar Q_P]$ is the unique $Q_P$-invariant $3$-dimensional subspace of $\hat V$, and is therefore equal to $C_{\hat V}(\bar A)$. Thus $[V,Q_P,A]\leq C_V(O^p(H))$. But also $[V,Q_P,A]\leq [V_0,A]\leq \yap$, and since $Y_P=Y_H\yap$ we have $[\yap, O^p(H)]\neq 1$. Thus $[V,Q_P,A]=1$, and symmetry yields also $[A,Q_P,V]=1$. Then $[A,V]\leq \Omega_1(Z(Q_P))$, by the Three Subgroups Lemma, and so $[A,V]\leq Y_P$. As $|[A,V]|>4$ we have a contradiction, and have thereby succeeded in ruling out the case given by 4.8(4f). It remains now to consider the cases where $p=2$ and $\bar L\cong SL(3,2^n)$ or $Sp(4,2^n)$, with $n>1$. Here (1) implies that $|\w Y_P|>2$ and so $M=N$. Thus $\w Y_P$ is invariant under a Borel subgroup of $\bar L$, as well as being invariant under an element of $S$ which induces a diagram automorphism on $\bar L$. It follows that $\w Y_P$ covers $C_{\hat V}(\bar S\cap\bar L)$, and so $|\hat Y_P|=2^{2n}$. But $O^2(P\cap H )$ acts transitively on $\hat Y_P$ while preserving any non-trivial irreducible $\bar L$-invariant section of $\hat V$. This is contrary to $\eta(\bar L,V)=2$, so we have succeeded in eliminating cases (4d) and (4e) of 4.8. The proof of 4.10 is thereby complete. \qed \enddemo \vskip .3in \noindent {\bf Section 5: Two special cases} \vskip .1in Our goal in this section is to show that outcomes (ii) and (iv) in lemma 4.10 can not occur. \proclaim {Lemma 5.1} Suppose that $q=2$, and let $K$ be a $2$-component of $H$. Then $\bar K$ is not isomorphic to $L_3(2)$ or to $Alt(6)$. \endproclaim \demo {Proof} Suppose false. Choose a critical path $(\a,\b,\cdots,\a')$ in $\Gamma$, and take $H=\b$ and $P=\b+1$. Notice that $\w V$ is not an $F1$-module for $H$, as follows from 3.7, and then 2.1 implies that $b$ is odd. Note also that we have $\gb\cap G_{\b+1}=P\cap H=S$, and hence $H=\w P$. Denote by $\Cal K$ the set of $2$-components of $H$, and write $\Cal K=\{K_1,\cdots,K_s\}$. For any $i$, $1\leq i\leq s$, set $V_i=[V,K_i]$. Then 4.8 shows that we have $\w V=C_{\w V}(H_0)\times\w V_1\times\cdots\times\w V_r$, and $\w V_i$ is the direct sum of two non-isomorphic modules for $K_i$, each of dimension $3$ if $\bar K\cong L_3(2)$, or $4$ if $\bar K\cong Alt(6)$. We recall also that $N_S(K_i)$ contains an element which induces a diagram automorphism on $\bar K_i$ (viewed as a group of Lie type in characteristic $2$) and which interchanges the two irreducible submodules of $\w V_i$. Suppose first that we have $b\geq 5$. By 2.3 we may then assume that $\vap$ is a quadratic $F2^*$-offender on $\wvb$, and further that $\yb\nleq \vap$, while $\yap\leq \vb$. In particular, 4.8 and 0.1 show that $\bar K_i\cong L_3(2)$ for all $i$. Fix a $2$-component $K$ of $H$ such that $[\bar K,\vap]\neq 1$, and set $V_K=[V,K]$. As $\vap$ acts quadratically on $\wvb$ we have $|\vap/C_{\vap}(\bar K)|\leq 4$. As $\yb\nleq \vap$ we have also $[V_K,C_{\vap}(\bar K)]=1$. Thus $|\wvap/C_{\wvap}(V_K)|\leq 4$. But also $|[\w V_K,\vap]\geq 4$ as $\eta(K,V_K)=2$, and so $V_K\nleq \qap$. It now follows that $|V_K/(V_K\cap\qap)|=2$ and that $|\wvap/C_{\wvap}(V_K)|=4$. Further, we have $\bar\vap\leq C_{\bar H}(\bar K)\bar K$. Let $\w U_1$ and $\w U_2$ be the two $K$-submodules of $\w V_K$ of order $8$, with the indexing chosen so that $\vap$ projects modulo $C_{\bar H}(\bar K)$ to the unipotent radical of the stabilizer of a point in $\w U_1$ and of a line in $\w U_2$. Set $U_0=V_K\cap\qap$. As we have seen, $\w U_0$ is a hyperplane of $\w V_K$, and $[\w U_0,\vap]=\w\yap$, of order $2$. Then $\w U_0=\w U_1\times [\w U_2,\vap]$, and $\w\yap\leq U_1$. Now $K$ is uniquely determined among the $2$-components of $H$ by the condition that $\yap$ lie inside $V_K$. Thus $[\bar H_0,\vap]=\bar K$, and $\bar K\geq\bar\vap$. In particular, we now have $|\vap/C_{\vap}(\vb)|=4$. Then also $|\vb/(\vb\cap\qap)|=2$, and $[O^2(\gap),\vb]$ contains a unique $2$-component $L$ of $\gap$. Set $X_{\b}=\lan (W_{\a})^{\gb}\ran$. As $b\geq 5$ we have $[W_{\a},X_{\b},W_{\a}]\leq [W_{\a},W_{\a}]=1$. Also $[\vb,X_{\b}]=1$ and so, recalling that $\yap\leq\vb$, we have $[\yap,X_{\b}]=1$. By the definition of $b$, we have $X_{\b}\leq G_{\a'-2}$, and $X_{\b}$ centralizes $Y_{\a'-2}\yap=Y_{\a'-1}$. Since $P\cap H=S$ is maximal in $H$, it now follows that $X_{\b}\leq Q_{\a'-1}$, and so $X_{\b}\leq \gap$. Then $L$ is $X_{\b}$-invariant. Set $\bar \gap=\gap/C_{\gap}(\wvap)$. As $[W_{\a},X_{\b},W_{\a}]=1$, where also $\Phi(W_{\a})=1$, it follows that either \vskip .1in \noindent (i) $|W_{\a}/C_{W_{\a}}(\bar L)|=2$, or \vskip .1in \noindent (ii) $|X_{\b}/C_{X_{\b}}(\bar L)|\leq 8$. \vskip .1in Set $V_L=[\vap,L]$, and suppose that (i) holds. Then $W_{\a}=C_{W_{\a}}(\bar L)\vb$, and hence $$ [V_L,W_{\a}]\leq \yap [V_L,\vb]\leq\vb\leq W_{\a}. $$ and hence $W_{\a}\nl \lan \ga,V_L\ran$. But $V_K=(V_K\cap\qap)\ya$, so $|[\ya,V_L]|\geq 4$, and thus $V_L\nleq \ga$. Then $\lan \ga,V_L\ran=\lan\ga,\gb\ran$, contrary to $O_2(\lan P,H\ran)= 1$. Thus (ii) holds. Then $|X_{\b}/C_{X_{\b}}(\bar L)\vb|\leq 4$. But $[C_{X_{\b}}(\bar L)\vb,V_L]\leq\vb$, so $X_{\b}/\vb$ is an $F1$-module for $\gb$ via the action of $V_L$. (Indeed, we have $|V_L/C_{V{L}}(\vb)|=4$.) As $\bar\gb$ has no $F1$-modules, we conclude that $[X_{\b},O^2(\gb)]\leq\vb$, so that $W_{\a}\nl\gb$, again contrary to $O_2(\lan P,H\ran)= 1$. This proves: \vskip .1in \noindent (1) $b=3$. \vskip .1in Set $\g=\b+1$. By 1.16 we have $\qb\vap\nl\gb\cap\gg$, and quadratic $F2$-action then implies that $\bar\vap=Z(\bar S\cap\bar H_0)$. As $b=3$, 1.17 implies that $[\vb,\vap]\leq D_{\g}$ and that $[D_{\g},\gg]\leq\yg$, where we recall from section $1$ that $D_{\g}$ denotes $\cap\{\vd\}_{\d\in\Delta(\g)}$. Thus $[\w V,Z(\bar S\cap\bar H_0),\bar S]$ is of order at most $2$. As $S$ acts transitively on the set $\Cal K$ of $2$-components of $H$, it follows that $|\Cal K|=1$. Write $V_0$ for $\vb\cap \qap$ and set $\hat V=V/C_V(H_0)$. Suppose that $\bar H_0\cong Alt(6)$. Then $|V/V_0|=4$ and $|[V_0,\vap]|\leq 2$. Further, we have $V_0\nl \qb(\gb\cap\ga')=S$, by 1.16, which implies that $\hat V_0$ intersects each of the two irreducible $H_0$-submodules of $\hat V$ in a hyperplane. But then $|[\hat V_0, \vap]|>2$, and we have a contradiction. As $\bar \vap \nl\bar S$ we conclude: \vskip .1in \noindent (2) $\bar H\cong Aut(L_3(2))$ and, setting $A=\vap$, we have $\bar A=Z(\bar S)$. \vskip .1in Observe that $[V,A]\geq \yb\yap=\yg=Y_P$. As $V=\lan (Y_P)^{H_0}\ran$ and as $\bar A\leq \bar H_0$, it follows that $\w V=[\w V,H_0]$. That is: \vskip .1in \noindent (3) $C_V(H_0)=Y_H$. \vskip .1in Choose $\l\in\Delta(\a)-\{\b\}$, and set $Q_{\a}^*=(\ql\cap\qa)(\qa\cap\qb)$. Then $\qa^*\nl\ga$, and $[O_2(\ga),O^2(\ga)]\leq \qa^*$. As usual, set $D_{\a}=\cap\{V_{\d}\}_{\d\in\Delta(\a)}$. We then claim: \vskip .1in \noindent (4) $[D_{\a},\ga]=\ya$. \vskip .1in \noindent Indeed, setting $X=\lan (\qb)^{\ga}\ran$, we have $[D_{\a},X]=\ya$ and $\qa^*\leq X$. There is a natural embedding of $\qa/C_{\qa}(D_{\a})$ into $Hom(D_{\a}/\ya,\ya)$, and so $\qa/C_{\qa}(D_{\a})$ is isomorphic to a direct sum of natural modules for $\ga/\qa$. Thus $\qa=C_{\qa}(D_{\a})[\qa,O^2(\ga)]=C_{\qa}(D_{\a})\qa^*$, and this proves (4). Notice that $\w Y_P=C_{\w V}(S)$. Notice further that $C_{\w V/\w Y_P}(S)=C_{\w V}(S\cap H_0)/\w Y_P$. As $D_{\a}/Y_P$ centralizes $S$, by (4), we then have: \vskip .1in \noindent (5) $|D_{\a}/\ya|\leq 2$. \vskip .1in Next, since $\ya$ does not induce a transvection on $\wvap$ we obtain $|\vap/C_{\vap}(\ya)|\geq 4$. We may then choose an element $t\in\vap\cap\qb -\qa$. Now $C_{\vl}(t)\ya\nl\lan Q_{\l},t\ran$, and $\lan Q_{\l},t\ran\geq O^2(\ga)$. Therefore $C_{\vl}(t)\ya\leq \vl\cap\vb$. Further, we have $\vl\cap\vb\nl O^2(\ga)\qb$, so that $\vl\cap\vb=D_{\a}$. With (5) we then have: \vskip .1in \noindent (6) $|C_{\vl}(t)\ya|\leq 8$, and $C_{\vl}(t)\ya\leq D_{\a}$. \vskip .1in \noindent On the other hand we have $|\vl/(\vl\cap\gap)|\leq 4$, so $\vl/(\vl\cap\qap)|\leq 16$, and so $|\vl/C_{\vl}(t)|\leq 32$. As $\ya\nleq C_G(t)$ we then have $|\vl/C_{\vl}(t)\ya|\leq 16$, and as $|\vl|=2^7$ we conclude from (6) that $|C_{\vl}(t)\ya|=8$, that $C_{\vl}(t)\ya=D_{\a}$, and that equality holds in each step in achieving this calculation. In particular, we have $[\vl\cap\qap,t]\neq 1$, and so \vskip .1in \noindent (7) $[\vl\cap\qap,t]=\yap$. \vskip .1in Similarly, we have $\vl\cap\gg\nleq Q_{\g}$. Choose $t'\in(\vl\cap\gg)-Q_{\g}$. By symmetry, we have $C_{\vap}(t')Y_{\g}=D_{\g}$, and $[\vap\cap Q_{\l},t']=Y_{\l}$. Now $$ \vap\cap Q_{\l}\leq \vap\cap G_{\l}\leq C_{\vap}(Y_{\l})\leq [\vap,\gg\cap\gap] $$ $$ =[\vap,Q_{\g}]\leq [W_{\g},Q_{\g}]. $$ As $t'\in\gg$ we then have $[t',\vap\cap Q_{\l}]\leq [W_{\g},Q_{\g}]$, and so $\yl\leq [W_{\g},Q_{\g}]$. For any $\d\in\Delta(\g)$ we have $\vd\qap/\qap\leq Z((\gd\cap\gap)/\qap)$, and so $W_{\g}\qap/\qap\leq Z((\gd\cap\gap)/\qap)$. Then $[W_{\g},Q_{\g}]\leq\qap$, and so $\ya\leq\qap$. As $(\a,\a')$ is a critical pair, we have a final contradiction, which then completes the proof of 5.1. \qed \enddemo \proclaim {Lemma 5.2} [Outcome (iv) of lemma 4.10 is out] \endproclaim \demo {Proof} To be provided. (It is in fact very easy, and very short. Maybe set it inside 4.10.) \qed \enddemo \vskip .2in \noindent {\bf Section 6: The Rudvalis Case} \vskip .1in We begin by warning the reader that we shall often write $Z$ for $Y_H$. \vskip .1in This section is concerned with outcome (iii) of lemma 4.10. We begin with two lemmas on modules for $L_3(2)$. \proclaim {Lemma 6.1} Let $V$ be a module for $L$ over $\Bbb F_2$, where $L\cong L_3(2)$. Assume that there exists a parabolic subgroup $P$ of $L$ and an element $x$ of $V$ such that $V=\lan x^L\ran$ and $dim(\lan x^P\ran)=2$. Then $\eta(L,V)\leq 2$. \endproclaim \demo{Proof} Set $S=C_P(x)$. Then $S$ is a Sylow $2$-subgroup of $L$, and so $S$ is contained in a (proper) parabolic subgroup $P_1$ of $L$, $P_1\neq P$. Set $U=\lan x^P\ran$ and set $U_1=\lan x^{P_1}\ran$. Let $W$ be a maximal $L$-submodule of $V$. Observe that $dim(U_1)\leq 3$. Suppose first that $U_1=\lan x\ran$. Then $|x^L|=7$ and so $dim(V)\leq 7$. As any non-trivial $L$-module has dimension at least $3$, we have $\eta(L,V)\leq 2$ in this case. Suppose next that $dim(U_1)>1$, and let $y$ be an element (possibly $0$) of $C_{U_1}(P_1)$. If $y\notin W$ for any choice of $W$ then $V=\lan y^L\ran$ and, again, we obtain $dim(V)\leq 7$ and $\eta(L,V)\leq 2$. As $dim(C_{U_1}(P_1))\leq 1$ we may therefore take $C_{U_1}(P_1)\leq W$. Setting $\bar V=V/W$, $bar V$ is then an irreducible $L$-module, with $dim(\lan\bar x^P\ran)= dim(\lan\bar x^{P_1}\ran)=2$. There are only four isomorphism classes of irreducible $\Bbb F_2L$-modules, and the given conditions identify $\bar V$ as the adjoint module (which is also the Steinberg module) for $L$. In particular $\bar V$ is projective, so $W=0$, and $\eta(L,V)=1$. \qed \enddemo \proclaim {Lemma 6.2} Take $L=L_3(2)$ and let $W=sl(3,2)$ be the $\Bbb F_2L$-module consisting of three-by-three matrices of trace zero. Then $L$ has exactly six orbits on the set of non-zero vectors of $W$. We may denote these orbits by $\Cal O_n$, $n=2,3,4,6,7,8$, where $n$ is the cardinality of $C_L(w_n)$, $w_n$ a representative of $\Cal O_n$. The elements $w_n$ may be taken as follows. $$ w_2=\pmatrix 0&0&1\\ 0&0&0\\ 1&0&0 \endpmatrix \qquad w_3=\pmatrix 0&1&0\\ 0&0&1\\ 1&0&0 \endpmatrix \qquad w_4=\pmatrix 0&1&0\\ 0&0&1\\ 0&0&0 \endpmatrix $$ $$ w_6=\pmatrix 0&0&0\\ 0&1&0\\ 0&0&1 \endpmatrix \qquad w_7=\pmatrix 0&1&0\\ 0&0&1\\ 1&1&0 \endpmatrix \qquad w_8=\pmatrix 0&0&1\\ 0&0&0\\ 0&0&0 \endpmatrix $$ \endproclaim \demo{Proof} To determine the cardinalities of the centralizers, notice that $w_3$ and $w_7$ are elements of $L_3(2)$ of order $3$ and $7$, respectively, and that $w_4+I$ and $w_8+I$ are elements of order $4$ and $2$, respectively. A slight knowledge of centralizers in $L_3(2)$ thus yields $|C_L(w_i)|$ for $i=3,4,7,8$. For $w_2$ and $w_6$, some straightforward matrix computation is called for. One observes that the summation of all the numbers $168/n$, $n=2,3,4,6,7,8$, is equal to $|W|-1$. \qed \enddemo \proclaim {Lemma 6.3} Suppose that $H$ is non-solvable and that $N_H(Y_P)$ is not a maximal subgroup of $H$. Then $q=2$, $|S|=2^{14}$, and the following hold. \roster \item"{(a)}" $H/O_2(H)\cong Sym(5)$, and $\w V$ is the $\Gamma L(2,4)$ module for $H/O_2(H)$. Further, $C_H(V)/V$ is a trivial module for $O^2(H)$ of order $4$, and $O_2(H)/C_H(V)$ is isomorphic to $V/Z$ as modules for $H$. \item"{(b)}" Let $P_1$ be the maximal subgroup of $H$ containing $S$, and set $L=\lan P,P_1\ran$. Then $L/O_2(L)\cong L_3(2)$, $\Phi(O_2(L))=Y_L$, $Y_L$ is a natural $L_3(2)$-module for $L/O_2(L)$, and $O_2(L)/Y_L$ is the adjoint module for $L/O_2(L)$. \endroster \endproclaim \demo{Proof} By 4.10, we have $q=2$, $H_0$ is a $2$-component of $H$, $\bar H$ is isomorphic to $Sym(5)$, and $V/C_V(H_0)$ is a natural $\Gamma L(2,4)$-module for $\bar H$. Thus $N_H(Y_P)$ has index $3$ in the maximal subgroup $M$ of $H$ containing $P\cap H$, and $|\lan (Y_P)^M\ran|=8$. Observe that $M\in\Cal P$. For this reason we shall henceforth write $P_1$ for $M$. Set $L=\lan P,P_1\ran$. Then 1.3(4) shows that $L/C_L(Y_L)$ is isomorphic to $L_3(2)$, or $Sp(4,2)$, with $Y_L$ a natural $L_3(2)$-module or a natural $Sp(4,2)$-module for $L$. Form the coset graph $\Gamma^*=\Gamma(L,H)$, and set $b^*=b(L,H)$. Take $\a=L$ and $\b=H$. For the remainder of this section, take $V=\vb=\lan (Y_P)^H\ran$ and set $W=\wb=\lan (Y_L)^H\ran$. Then $\vd$ and $W_{\d}$ are defined for any conjugate $\d$ of $\b$ in $\Gamma^*$. Also, set $U=U{\a,\b}=\lan (Y_P)^{P_1}\ran$. Then conjugation defines $U{\l,\m}$ for any edge $\{\l,\m\}$ of $\Gamma^*$. Notice that since $|V|>|Y_L|$, $Y_L\cap V$ is a proper $P_1$-invariant subspace of $V$. Then $Y_L\cap V=U$. Further, as $Y_P\nleq Z(O_2(H))$ we have also $U\nleq Z(O_2(H))$. Then $O_2(H)O_2(L)/O_2(L)$ is a non-trivial subgroup of $O_2(P_1)/O_2(L)$, and in the case that $L/C_L(Y_L)$ is isomorphic to $Sp(4,2)$ it follows that $O_2(H)O_2(L)/O_2(L)$ is not the transvection subgroup of order $2$ in $O_2(P_1)/O_2(L)$. Thus, in any case we have: \vskip .1in \noindent (1) $[O_2(H),O^2(P_1)]\nleq O_2(L)$. \vskip .1in We now assume that we have: \vskip .1in \noindent {\bf Case A. $L/C_L(Y_L)\cong Sp(4,2)$.} \vskip .1in Notice that, in this case, (1) implies that $[Y_L,O_2(H)]=[Y_LV,O_2(H)]=U$. As $U$ is not normal in $H$ it then follows that $Y_LV$ is not normal in $H$, and so $Y_LV\neq W$. Thus: \vskip .1in \noindent (2) $\eta(H,W)\geq 2$. \vskip .1in We also note the following. \vskip .1in \noindent (3) There is a unique element element $\bar t$ of $O_2(P_1)/O_2(L)$ such that $\bar t$ induces a transvection on $Y_L$. Moreover, we have $[Y_L,\bar t]=Y_H$. \vskip .1in Suppose next that $Y_lV\nl L$. As $|Y_LV/Y_L|= |V/(Y_L\cap V)|=4$, we then have $[V,O^2(L)]\leq Y_L$. But evidently $[V,O^2(P_1)]=V\nleq Y_L$. Thus: \vskip .1in \noindent (4) $Y_LV$ is not normal in $L$, and if $V\leq O_2(L)$ then $\eta(L,\lan V^L\ran)\geq 2$. \vskip .1in Fix a critical path $(\a,\b,\cdots,\a')$ in $\Gamma^*$, and suppose first that $b^*$ is odd. Then $b^*\geq 3$, by 1.3(6). [OR MAYBE BY AN ARGUMENT TO BE SUPPLIED LATER]. Suppose that $\vap\leq\qb$. Then $[\vb,\vap]\leq\yb$, where $\yb$ is of order $2$. As $V/Z$ is a $\Gamma L(2,4)$-module for $H$, no element of $H$ acts as a transvction on $V/Z$, and so $\vb\leq\qap$. Then $[\vb,\vap]\leq\yb\cap\yap$, and so $[\vb,\vap]=1$, by 1.15(b). Then $\vap$ centralizes the hyperplane $\ya\cap\vb$ of $\ya$, and so $|\vap/C_{\vap}(\ya)|\leq 2$. This is contrary to $\ya\nleq\qap$, so we conclude that $\vap\nleq\qb$. In particular, $(\a',\cdots,\b)$ is a pre-critical path, and so also $\vb\nleq\qap$, by symmetry. Notice that $\vb\leq Q_{\a'-1}\leq O_2(G_{\a'-1}\cap\gap)$. For any element $x$ of $O_2(P_1)-O_2(H)$ we have $[V,x]Z=U$, so now $[\vb,\vap]\yap=U_{\a'-1,\a'}$. For the same reason, we have also $[\wb,\vap]\yap=U_{\a'-1,\a'}$. In view of (2), it follows that $\vap$ acts as a transvection group on $\wb/\vb$. But $vap\leq O_2(\gb\cap G_{\b+1})\leq O^2(H)O_2(H)$, and since no element of $SL(2,4)$ induces a transvection on any $SL(2,4)$-module over $\Bbb F_2$, we have a contradiction at this point. We conclude that $b^*$ is even, $b^*\geq 2$. Let $P_0=(P_1)^g$ be an $L$-conjugate of $P_1$ such that $(P_0\cap P_1)/O_2(L)\cong SL(2,2)$. (The existence of such a conjugate $P_0$ of $P_1$ can be read off easily from a picture of the $B_2$ root system.) One then has $\lan t,P_1\ran=L$ for any $t\in O_2(P_0)\O_2(L)$. By edge-transitivity on $\Gamma^*$ we may then fix a vertex $\l\in\Delta(\a')$ such that: \vskip .1in \noindent (5) $\lan \gap\cap\gl,t\ran=\gap$, for any $t\in O_2(G_{\a'-1}\cap\gap)-\qap$. \vskip .1in We next show: \vskip .1in \noindent (6) $\yb=Y_{\a'-1}$. \vskip .1in In order to prove (6), suppose first that $[\vb,\yap]\neq 1$. As $yap\leq\qb$ we then have $[\vb,\yap]=\yb$, and so $\vb$ induces a transvection group on $\yap$. Then (2) yields $[\vb,\yap]=Y_{\a'-1}$, and so (6) holds in this case. So assume that $[\vb,\yap]=1$. Then $C_{\ya}(\yap)=\ya\cap\vb$ is a hyperplane of $\ya$, whence also $C_{\yap}(\ya)$ is a hyperplane of $\yap$. Then (3) yields $[\vb,\yap]=\yb=Y_{\a'-1}$, and (6) holds in any case. The next step will be to show: \vskip .1in \noindent (7) We have $b^*=2$. \vskip .1in Assume that (7) is false, so that $b^*\geq 4$. Suppose that $[Y_{\b+1},\wl]\neq 1$. There is then a critical pair $(\b+1,\m)$ with $\m\in\Delta(\l)$. Applying (6) to this critical pair, we obtain $Y_{\b+2}=\yl$, and then $C_G(\yl)\geq \lan \ya,\gl\ran$. But (5) then yields $C_G(\yl)\geq \lan \gap,\gl\ran$, contrary to $\lan L,H\ran\notin\Cal L$. We therefore conclude that $[Y_{\b+1},\wl]=1$. Now $\wl\leq Q_{\b+1}\leq\gb$. Since $\yap\leq \qb$ we have $|[\vb,\yap]|\leq 2$, and then $|\vb/(\vb\cap\qap)|\leq 2$. Suppose that $\vl\nleq\qb$. Then $[\vb,\vl]\yb=U_{\b,\b+1}\geq [\vb,\wl]$. Since $O_2(P_1)$ contains no transvections on $W/V$, it follows that $\vb\cap\gl\leq \ql$. As $[\ya,\yl]\neq 1$, as we have seen, we have $\yl\nleq\vb$, and then $[\vb\cap\ql,\vl]=1$. Thus $\vl$ centralizes a hyperplane of $\vb$, and so $\vl\leq\qb$. We now have $[\vb,\vl]\leq\yb=Y_{\a'-1}\leq\yap$. If $\vb\nleq\gl$ it then follows from (5) that $\yap\vl$ is normal in $\gap$, contrary to (4). Thus $\vb\leq\gl$, and since $|\vb,\vl]|\leq 2$ we obtain $\vb\leq\ql$ and $[\vb,\vl]\leq\vb\cap\yl$. We have already seen that $[\ya,\yl]\neq 1$, so in fact $[\vb,\vl]=1$. Then $\vl$ induces a transvection on $\ya$, and $[ya,\vl]=\yb\leq\yap$. This again yields $\yap\yl\nl\gap$, contrary to (4), and completing the proof of (7). Since $[Y_L,O_2(P_1)]\leq V$, we now have $[\ya,\yap]\leq \ya\cap\vb\cap\yap$. But $(\ya\cap\vb)/\yb$ and $(\yap\cap\vb)/\yb$ are two different $1$-dimensional $\Bbb F_4$-subspaces of $V/Z$ (different since $\ga\cap\gb\neq \gb\cap\gap$, by maximality of $P_1$ in $H$). Therefore $[\ya,\yap]=\yb$. This yields: \vskip .1in \noindent (8) $[W_{\b},W_{b}]=\yb$. \vskip .1in \noindent Further, $\ya\cap\vb$ is the hyperplane of $\ya$ which centralizes $\wb$. The normal closure of $\ya\cap\vb$ in $\gb$ is $\vb$, so we have: \vskip .1in \noindent (9) $[\vb,\wb]=1$. \vskip .1in Applying (9) at $\l$, we have $[\yap,\vl]=1$, so $\vl\leq\qap\leq\gb$. Suppose $\vl\leq\qb$. Then $[\vb,\vl]\leq\yb$, and since $\vb\leq\qap\leq\gl$ we get $\vb\leq\ql$ and $[\vb,vl]\leq\yb\cap\yl=1$. As in the proof of (7), we then have $[\ya,\vl]=\yb$ and $\yap\vl\nl \gap$, contrary to (4). We therefore conclude that $\vl\nleq\qb$. As $b^*=2$ we have $\wl\nleq\qap$. Let $t\in\wl-\qap$. Then $[\vb,(\vb)^t]\leq \vb\cap (\vb)^t\cap \yap=(\vb\cap \yap)\cap (\vb\cap\yap)^t$. By our choice of $\l$ in (5), we have $O_2(\gap\cap\gl) \cap\gb=\qap$, so $t\notin\gb$. As $t$ induces a transvection on $\vap$, by (8), we then have $\yap=C_{\yap}(t)\yb$ and $(\vb\cap \yap)\cap (\vb\cap\yap)^t=C_{\vb\cap\yap}(t)$ is of order $4$. Then $C_{\vb\cap\yap}(t)=[\vb,\vl]\leq\vl$, and so we have shown that $[\vb,(\vb)^t]\leq\vl$. But $[\wl,\vb]=[(\wl\cap\gb)\lan t\ran,\vb]\leq \vb (\vb)^t$, and so $[\wl,\vb,\vb]\leq [\vb,(\vb)^t]\leq\vl$. As $|[\vb,\vl]|>2$ we have $\vb\nleq\ql$, and we have shown that $\vb$ acts quadratically on $\wl/\vl$. Set $W_0=\wl\cap\qap$. Then $|\wl/W_0|=2$ and we have $[\vb,W_0]\leq U_{\b,\a'}=[\vb,\vl]\yb$. Setting $\hat\wl=\wl/\vl$, we then see that $\vb$ acts on $\hat \wl$, with $|[\hat W_0,\vb]|\leq 2$. It follows that $\hat \wl$ involves no natural $\Gamma L(2,4)$-module for $\gap$, and then quadratic action implies that $|\vb/C_{\vb}(\hat \wl)|=2$. Thus $\vb\cap\ql$ is a hyperplane of $\vb$. We have seen that $[\vb,\vl]=\vb\cap\yap\cap\vl$ is of order $4$, so $\yb\nleq [\vb,\vl]$ and $\yl\nleq [\vb,\vl]$. Then $[\vb\cap\ql,\vl]=1$, which yields $\vl\leq\qb$ and a final contradiction in Case A. Thus, we are reduced to \vskip .1in \noindent {\bf Case B: $L/C_L(Y_L)\cong L_3(2)$.} \vskip .1in We may begin by observing that in this case we have $U=Y_L\leq V=W$. In particular, $W$ is abelian, and so $b^*\geq 3$. Suppose that $b^*$ is even. As $\vb\leq Q_{\a'-1}$ and $V_{\a'-1}\leq\qb$, we have $[\vb, V_{\a'-1}]=[\ya,\yap]=\yb=Y_{\a'-1}$. As $C_L(Z)=P_1\leq H$ we have $\yb\neq Y_{\b+2}$, and so $b*\geq 6$. Set $X=\lan (\vb)^{\ga}\ran$. As $V$ is not normal in $L$ we have $X\neq \vb$ and so $\eta(\ga,X)>1$. Now $X\leq Q_{\a'-3}\leq G_{\a'-2}$, and $[X,Y_{\a'-1}]=[X,\yb]=1$. It follows that $X\leq G_{\a'-1}$. If $X\leq Q_{\a'-1}$ then $[X,V_{\a'-1}]\leq \yb\leq\ya$, contrary to $\eta(\ga,X)>1$. Thus $X\nleq Q_{\a'-1}$. But $X\leq O_2(G_{\a'-1}\cap G_{\a'-1})\leq O^2(G_{\a'-1})O_2(G_{\a'-1})$, and since Sylow $2$-subgroups of $O^2(\bar H)$ are $TI$-sets it follows that $X\nleq\gap$. Next, observe that $\vl\leq Q_{\b+2}\leq G_{\b+1}$, and $[\yb,\vl]=[Y_{\a'-1},\vl]=1$, so $\vl\leq\gb$. If $\vl\leq\qb$ then $[\vl,\vb]\leq\yb\leq \yap\leq\vl$, and then $\vl\nl\lan \vb,\gap\cap\gl\ran=\gap$, whereas $V$ is not normal in $L$. Thus $\vl\nleq\qb$, and it follows that $C_{\vb}(\vl)=Y_{\b+1}$, of index $4$ in $\vb$. As $L/O_2(L)\cong L_3(2)$ we have $O_2(P_1)\cap (P_1)^g\nleq O_2(L)$ for any $g\in L$. As $\vb\leq O_2(G_{\a'-1}\cap\gap$, and as $vb\nleq\gl$ by definition of $\l$, we then have $|\vb/(\vb\cap\gl)|=2$. Now $[\vb\cap\ql,\vl]\leq \vb\cap\yl$. But $yl$ is not centralized by $\gap$, so $[\vb,\yl]\neq 1$, and so $[\vb\cap\ql,\vl]=1$. Thus $\vb\cap\ql=C_{\vb}(\vl)$ which, we have seen, has index $4$ in $\vb$. We may then conclude that $\vb\cap\gl\nleq \ql$. This yields $[\vb\cap\gl,\vl]\yl=\yap$. As $Y_{\a'-2}\cap\yap=Y_{\a'-1}$, of order $2$, we have $[\vb\cap\gl,\vl]\nleq Y_{\a'-2}$. Now $X$ centralizes $Y_{\a'-2}$ and, as $b^*\geq 6$, $X$ centralizes $\vb$. As no element of $H$ induces a transvection on $V/Z$ we have the desired contradiction at this point. That is: \vskip .1in \noindent (10) $b^*$ is odd. \vskip .1in [AS BEFORE, ASSUME FROM GLOBAL HYPOTHESIS or 1.3(6) THAT $b^*\neq 1$.] As $\vb$ does not induce a transvection on $\vap/\yap$ we have $\vap\nleq\qb$, and since $\vap\leq O^2(\gb)O_2(\gb)$ we get also $\vap\nleq\ga$. Suppose that $\yap\leq\vb$. As $Y_{\a'-1}=[\vb,\vap]\yap$, we then have $Y_{\a'-1}\leq\vb$. Set $X=X_{\a}=\lan (\vb)^{\ga}\ran$, and assume now that $b^*\geq 5$. Then $X\leq Q_{\a'-3}\leq G_{\a'-2}$, and $X$ centralizes $Y_{\a'-3}Y_{\a'-1}=V_{\a'-2}$. We then have $X\leq Q_{\a'-1}\leq\gap$, and so $[X,\vap]\leq Y_{\a'-1}\vb\leq X$. Thus $X$ is normal in $\lan \ga,\vap\ran$, and since $\vap\nleq \ga$ we get $X$ normal in $\gb$, contrary to $\lan L,H\ran\notin\Cal L$. We may now conclude that $yap\nleq\vb$. As also $(\a',\cdots,\b)$ is a pre-critical path, we obtain also $\yb\nleq\vap$, by symmetry. Then $|\vap/(\vap\cap\qb)|=|\vb/(\vb\cap\qap)|=4$. Set $D=\lan (X)^{\gb}\ran$. Then $D\leq \qb$ and indeed $[D,\vb]=1$. Set $F=[\vb,\vap]$. Then $F$ is a complement to $\yap$ in $Y_{\a'-1}$, contained in $C_{V_{\a'-2}}(D)$. As $D$ does not induce a transvection on $V_{\a'-2}/Y_{\a'-2}$ we get $D\leq Q_{\a'-2}\leq G_{\a'-1}$. As $D\leq O_2(G_{\a'-2}\cap G_{\a'-1})$, we have $|D/(D\cap\gap)|\leq 2$. As $X$ is not normal in $\gb$ we have $\eta(\gb,X)\geq 2$, and then $|D/C_D(\vap)|\geq 16$, and $|[D,\vl]|\geq 16$. Then $D\nleq \gap$. Suppose that $D$ is abelian. Then $\yap\nleq D$, so that $D\cap\qap\leq C_D(\vap)$, and then since a Sylow $2$-subgroup of $\gap/\qap$ is non-abelian of order $8$ we get $|D/C_D(\vap)|<16$. This contradiction shows that $D$ is non-abelian, and so $b^*=5$. Then $F\leq Y_{\b+1}\cap Y_{\b+3}=Y_{\b+2}$, which is contrary to $|F|=4$. Thus, we have shown: \vskip .1in \noindent (11) $b^*=3$. \vskip .1in Set $\g=\b+1$, so that also $\g=\a'-1$. Here $\vap\nleq\qb$ since $\vb$ is not a transvection group on $\vap/\yap$. We have $$ \vb\cap\vap\geq\yg[\vb,\vap]\nl\lan\qb,\qap\ran\qg=\gg. $$ Suppose that $|\vb|>32$. That is, suppose that there are non-trivial fixed points for $H$ on $V/Z$. Since $V=\lan (Y_L)^H\ran$ where $Y_L=[Y_L,P_1]$, it then follows that $|C_{V/Z}(H_0)|=2$ or $4$, and that $V=[V,H]$. Thus, $V/Z$ is an indecomposable module for $H_0$, and we have $|V/[V,O_2(P_1)]|=|V/[V,x]Y_L|=4$ for any element $x$ of $O_2(P_1)-O_2(H)$. Thus $|\vb/[\vb,\vap]\yg|=4$, and since $[\vb,\vap]\yg\leq\vap$ it follows that $\yg$ is a proper subgroup of $\vb\cap\vg$. [SHORTER ARGUMENT FOR THIS ?] Let $y\in \vb\cap\vap$ with $[y,O^2(\gb)]=1$ and with $\yg\lan y\ran$ normal in $\gb\cap\gg$. Then $\yg\lan y\ran$ is normal in $\lan\qb,\qap\lan\qg=\gg$, and $y$ centralizes $[\qg,O^2(\gb\cap\gg)]$. But the commutator map defines a pairing of $\lan y\ran\times O_2(H)/C_{O_2(H)}(y)$ onto $Y_L$, as $y\notin Y_L=\Omega_1(Z(O_2(L)))$, and so $y$ does not centralize any conjugate of $O^2(P_1)$ in $L$. This contradiction shows that, in fact, we have $|V|=32$, and $V/Z$ is irreducible for $H$. Observe that $|\qb/(\qb\cap\qg|\leq 4$, and that $[\qb\cap\qg,\vap]\leq [\qg,\vap]\leq\yg\leq\vb$. Since no element of $Alt(5)$ induces a transvection on any $\Bbb F_2$-module for $Alt(5)$, it follows that $\eta(H,O_2(H)/V)\leq 1$. Set $Q=O_2(H)$ and set $R=C_Q(V)$. If $[Q,O^2(H)]\leq R$ then $V\leq Z(Q)$ by the Thompson $P\times Q$ Lemma, whereas $[V,O_2(H)]\geq [Y_L,O_2(H)]=Z$, by (1). Thus $\eta(H,O_2(H)/V)=1$, and $[R,O^2(H)]=V$. The commutator map defines a pairing of $Q/R\times V/Z$ onto $Z$, so $Q/R$ is isomorphic to the dual of $V/Z$ as a module for $H$. We record the results so far as follows. \vskip .1in \noindent (12) Both $V/Z$ and $O_2(H)/C_H(V)$ are natural $\Gamma L(2,4)$-modules for $H/O_2(H)$, and $[C_H(V),O^2(H)]=V$. \vskip .1in For any vertex $\d$ of $\Gamma$ and for any positive integer $n$, define $\gd^{(n)}$ to be the point-wise stabilizer in $\gd$ of all the vertices at distance at most $n$ from $\d$. Observe that $\ya\cap\ga^{(4)}=1$ as $b^*=3$. As $\ya=Y_L=\Omega_1(Z(O_2(L)))$, we then have $\ga^{(4)}=1$. In a similar vein, observe that since $[\qa,\vb,\qa]\leq [\ya,\qa]=1$, we have $[\qa,\qa]\leq C_{\qa}(X_{\a})$ by the Three Subgroups Lemma, and so $[\qa,\qa]\leq\ga^{(3)}$. We note also that $C_{\gb}(\vb)=\gb^{(2)}$. Let us write $Q_L$ for $O_2(L)$, and $Q_H$ for $O_2(H)$. Write $X=X_{\a}$. Then $Q_L=(Q_L\cap Q_H)X$. Now $(O_2(L)\cap O_2(H))/C_H(V)$ is a $1$-dimensional $\Bbb F_4$-subspace of $Q_H/C_H(V)$ as seen in (12). As $Q_HX=O_2(P_1)$ it then follows that $Q_L\cap Q_H=(X\cap Q_H)C_H(V)$. Thus $$ \qd=(\qa\cap\qd)X=\gd^{(2)}X $$ for all $\d\in\Delta(\a)$, and we have the following result. \vskip .1in \noindent (13) $\qa=\cap_{\d\in\Delta(\a)}C_{Q_{\a}}(\vd)X$ \vskip .1in\noindent We now have $$ [\ga^{(3)},\qa]=\bigcap_{\d\in\Delta(\a)}[\ga^{(3)},C_{\qd}(\vd)X]= \bigcap_{\d\in\Delta(\a)}[\ga^{(3}),\gd^{(2})] $$ $$ \leq\bigcap_{\d\in\Delta(\a)}[\gd^{(2)},\gd^{(2)}]\leq \bigcap_{\d\in\Delta(\a)}\gd^{(3)}\leq\ga^{(4)}=1. $$ This shows: \vskip .1in \noindent (14) We have $\ga^{(3)}=\ya$. \vskip .1in Suppose next that there exists an element $h\in C_H(O^2(H))-Z$. We may then choose $h$ so that $[H,h]=Z$. By (12) we have $h\in C_H(V)$, and then $[h,X]\leq [O_2(L),X]\leq Y_L$. As $C_{Y_L}(O^2(H))=Z$ we then have $[h,X]\leq Z$. For any $\m\in\Delta(\a)$ we have $h\in\gm$, and since no element of $\gm$ induces a transvection on $V_{\m}/Y_{\m}$ we further have $h\in Q_{\m}$. But then $[V_{\m},h]\leq Y_{\m} \neq \ya=Z$, and so $[V_{\m},h]=1$. Thus $h\in\ga^{(3)}$, and so $h\in \ya$ by (14). As $C_{Y_L}(O^2(H))=Z$ we have a contradiction. This proves: \vskip .1in \noindent (15) $C_H(O^2(H))=Z$. \vskip .1in Here is a further consequence of (13). Namely, we have $$ [\qa,\qb]\leq [\gb^{(2)}X,\qb]\leq [\gb^{(2)},\qb]X =[C_H(V),Q_H]X. $$ But (15) implies that $C_H(V)/Z$ is abelian, so $C_H(V)/V$ is abelian, and we then have $[C_H(V),Q_H]\leq V$, by (12). We then conclude that $[\qa,\qb]\leq X$, and so $X\geq [Q_L,O^2(L)]$. As $X=\lan V^L\ran$, where $V=[V,O^2(P_1)]$, we then have $X= [Q_L,O^2(L)]$. On the other hand, 6.1 shows that $\eta(L,X/Y_L)\leq 2$. If $X$ involves no Steinberg module for $L/O_2(L)$ it follows that $\eta(P_1,X/Y_L)=2$ and that $\eta(P_1,O_2(P_1))=4$. But it is evident from the structure of $H$ that $\eta(P_1,O_2(P_1))=5$. Thus $X$ has a constituent $W$ which is a Steinberg module for $L/O_2(L)$. Then $\eta(P_1,W)=3$ and it follows that $W$ is the unique non-trivial constituent for $L$ in $X/Y_L$. Projectivity of $W$ now yields $W=X/Y_L$. As $X=[Q_L,O^2(L)]$ it follows that also $Q_L=C_{Q_L}(X)X$. But $C_{Q_L}(X)=\ga^{(3)}\leq X$ by (14). Thus: \vskip .1in \noindent (16) $Q_L/Y_L$ is a Steinberg module for $L/O_2(L)$, and $|S|=2^{14}$. \vskip .1in Finally, we observe from $|S|$ that $|C_H(V)/V|=4$. This completes the proof of lemma 6.3. \qed \enddemo \proclaim {Lemma 6.4} Set $W=Q_L/Y_L$, identify $W$ with $sl(3,2)$, and identify $L/Q_L$ with $L_3(2)$, with action on $W$ given by matrix conjugation. Let the orbits for $L$ on $W-\{0\}$ be given as in 6.2. For any $x\in Q_L$, write $\hat x$ for the image of $x$ in $W$. Then $|x|\leq 2$ if $x\in Y_L$ or if $\hat x\in\Cal O_7\cup\Cal O_8$, and otherwise $|x|=4$. \endproclaim \demo{Proof} We have maps $\Cal S:W\longrightarrow Y_L$ and $\Cal B:W\times W\longrightarrow Y_L$, given by squaring and commutation, respectively. Then, for any $x$ and $y$ in $Q_L$, we have $$ \Cal S(\hat x+\hat y)=\Cal S(\hat x)+\Cal S(\hat y)+\Cal B(\hat x,\hat y).\tag1 $$ We shall also identify $Y_L$ with a natural $SL(3,2)$-module, written additively. Let $t\in V-Y_L$ with $[t,S]\leq Y_L$. Then $\lan\hat t\ran=C_W(S)$. We take $S$ to induce conjugation by upper triangular matrices on $W$. Adopting the notation of 6.2, we then have $\hat t=w_8$. As $\Phi(V)=1$, and as $\Cal S$ is constant on $L$-orbits, we see that $\Cal S$ is trivial on $\Cal O_8$. As $\Cal S(w_7)$ is invariant under $C_L(w_7)$ (which is of order $7$) it follows that $\Cal S(w_7)= (0,0,0)$, and thus $\Cal S$ is trivial on $\Cal O_7$. Denote by $\hat t'$ the transpose of $\hat t$. The set of all matrices in $W$ whose lower-left entry is $0$ is the unique maximal $S$-invariant subspace of $W$, and it therefore contains $C_{Q_L}(t)/Y_L$. It follows that $\Cal B(\hat t,\hat t')\neq (0,0,0)$. Then $\Cal S(w_2)= \Cal S(\hat t+\hat t')\neq (0,0,0)$, by (1). Thus, $\Cal S$ is non-trivial on $\Cal O_2$. Next, observe that $w_3+\hat t\in\Cal O_7$. (To see this, either check that, as an invertible matrix, $w_3+\hat t$ is of order $7$, or observe that conjugation of $w_3+\hat t$ by $w_3$ yields $w_7$.) Then $(0,0,0)=\Cal S(w_3+\hat t)=\Cal S(w_3)+\Cal B(w_3,\hat t)$. But evidently $w_3$ does not lie in the unique maximal $S$-invariant subspace of $W$, so $\Cal B(w_3,\hat t)\neq (0,0,0)$, and so $\Cal S$ is non-trivial on $\Cal O_3$. Further, as $\Cal S(w_3)$ is invariant under $C_L(w_3)$ we obtain $\Cal S(w_3)=(1,1,1)$. For the same reason, we have $\Cal S(w_3')=(1,1,1)$, where $w_3'$ is the transpose of $w_3$. We next check that $w_3'+w_6\in\Cal O_7$. (Indeed, one observes that $w_3'+w_6$ is the square of the matrix $w_7$.) Suppose that $\Cal S(w_6)=(0,0,0)$. Then $(0,0,0)=\Cal S(w_3'+w_6)=\Cal S(w_3')+\Cal B(w_3',w_6)$, and thus $\Cal B(w_3',w_6)=(1,1,1)$. Thus $\Cal B(w_3',w_6)$ is invariant under any element $g$ of $L$ whose matrix for the $L$-action on $W$ is given by $w_3'$. The matrix of $g^2$ is then $w_3$, and setting $u=(w_6)^g$ and $v=(w_6)^{g^2}$, we have $w_6=u+v$. Then $$ (1,1,1)=\Cal B(w_3',w_6)=\Cal B(w_3',u+v)=\Cal B(w_3',u)+\Cal B(w_3',v) $$ $$ =\Cal B(w_3',w_6)^g+\Cal B(w_3',w_6)^{g^2}= (1,1,1)+(1,1,1)=(0,0,0). $$ This contradiction shows that, in fact, $\Cal S$ is non-trivial on $\Cal O_6$. It only remains to treat $\Cal O_4$. For this, define matrices $y$ and $g$ as follows. $$ y=\pmatrix 0&0&0\\ 0&0&0\\ 1&1&0 \endpmatrix \qquad g=\pmatrix 0&1&0\\ 1&1&0\\ 0&0&1 \endpmatrix $$ Then $y^g=w_2'$, and we have $w_4=y+w_7$. Further, we have $$ (w_7)^g=\pmatrix 1&1&1\\ 1&1&0\\ 1&0&0 \endpmatrix $$ and hence $\Cal B(y,w_7)\neq (0,0,0)$. Now $\Cal S(w_4)=\Cal S(y+w_7)=\Cal B(y,w_7)$, and so $\Cal S$ is non-trivial on $\Cal O_4$. \qed \enddemo \proclaim {Corollary 6.5} Set $R=C_H(V)$, and let $x$ be an element of order $3$ in $P_1$. Then $R=C_R(x)V$, and $C_R(x)$ is a quaternion group. \endproclaim \demo{Proof} As $V$ is a natural $\Gamma L(2,4)$-module for $H/Q_H$, we have $V=[V,x]Z$. By 6.3(a), $R/V$ centralizes $O^2(H)$, so $R=C_R(x)V$ with $C_R(x)\cap V=Z$, and with $|C_R(x)|=8$. We have $R\leq Q_L$ since $P_1/Q_L\cong Sym(4)$. As $C_R(x)-Y_L= C_R(x)-Z$, it now follows from 6.4 that $C_R(x)-Y_H$ consists entirely of elements of order $4$, and hence that $C_R(x)$ is a quaternion group. \qed \enddemo \proclaim {Proposition 6.6} Both $L$ and $H$ are maximal $2$-local subgroups of $G$. \endproclaim \demo{Proof} Let $L^*$ be a $2$-local subgroup of $G$ containing $L$. Suppose first that $Q_L$ properly contains $O_2(L^*)$. Then $O_2(L^*)=Y_L$ and thus $C_L(O_2(L^*))\nleq O_2(L^*)$, contrary to the Main Hypothesis concerning $\Cal L(S)$. Thus $O_2(L^*)=Q_L$. But then $L^*=C_{L^*}(Y_L)L$, and $C_{L^*}(Y_L)/Q_L$ is of odd order, centralized by $\lan Q^L\ran$ (where we recall that $Q$ denotes $O_2(\w C)$). Thus, we have $L^*/Q_L=L/Q_L)\times O_{2,2'}(L)/Q_L$. As $L$ acts irreducibly on $Q_L/Y_L$, it then follows that $L^*=L$. Now let $H^*$ be a $2$-local subgroup of $G$ containing $H$. Suppose first that $O_2(H^*)$ does not contain $[Q_H,O^2(H)]$. Then $O_2(H^*)\leq C_H(V)$, and since $H^*$ is $2$-constrained we have $V$ properly contained in $O_2(H^*)$. The preceeding lemma then shows that $V=\Omega_1(Z(O_2(H^*)))$ and that $Z=\Phi(O_2(H^*)$. Then $H^*=C_{H^*}(V)H$, as $Sym(5)$ is maximal among subgroups of $L_4(2)$ having a Sylow $2$-subgroup of order $8$. Now $C_{H^*}(V)\leq C_G(Y_L)\leq L$, by what has already been proved, and then $C_{H^*}(V)\leq P_1\leq H$. Thus, we may assume that $[Q_H,O^2(H)]\leq O_2(H^*)$. But this yields $V=Z(O_2(H^*))$, and $Z=[V,O_2(H^*)]$, with the result, as just argued, that $H=H^*$. \qed \enddemo [NOW COMES THE HARD PART: IDENTIFYING Ru. I have some notes which yield $|Ru|$. Better yet, I have Michael's notes which do the same thing, with a clever idea for avoiding some of the computation required for a strict applicaion of the Thompson order formula. He also tells how to get an amalgam for $^2F_4(2)$ from $(H,L)$, and with $|G|$ (and much of the local structure for odd primes) in hand, one can obtain $^2F_4(2)$ using a result of Bennett and Shpectorov. At that point, he produces $Ru$ as a rank-3 permutation group.] \vskip .2in {\bf Section 7: The Solvable Case, $p=2$, Part I} \vskip .1in In this section we consider the case where $p=2$ and where both $P$ and $H$ are solvable. In particular, we then have $q=2$, so that $H=\w P$. Our treatment may be viewed as a re-working of parts of Stellmacher's N-Group Paper [St2] (specifically, sections $9$ and $10$). In the case $b=3$, we will stick closely to the original (section $10$). This section and the next will be devoted to the proof of the following result. \proclaim {Theorem 7.0} Assume that $b\geq 3$, $q=2$, and $H=\w P$ is solvable. Then $(P,\w P)$ is a weak $BN$-pair. Moreover, we have $b=3$, and $\w P/O_2(\w P)$ is isomorphic to either $SL(2,2)$ or $Sz(2)$. \endproclaim \vskip .1in \noindent {\bf Remark:} From the data given by the Theorem 7.1, we will recover the amalgam for $M_{12}$ or $Aut(M_{12})$ (when $\w P/O_2(\w P)$ is isomorphic to $SL(2,2)$), or the amalgam for the Tits group or for $^2F_4(2)$ (when $\w P/O_2(\w P)$ is isomorphic to $Sz(2)$). The actual construction of these groups from the local data promises to be an interesting exercise. Or not: it being likely that one has only to copy the relevant portions of the Aschbacher-Smith Opus. \redefine\za{Z_{\alpha}} \redefine\zap{Z_{\alpha'}} \redefine\zb{Z_{\beta}} \redefine\zm{Z_{\m}} \redefine\zr{Z_{\rho}} \redefine\ywm{Y_{\widetilde{\mu}}} \redefine\zt{Z_{\tau}} \redefine\ywt{Y_{\widetilde\tau}} \redefine\zd{Z_{\delta}} \redefine\zg{Z_{\g}} \redefine\zl{Z_{\l}} \redefine\zs{Z_{\s}} \redefine\ga{G_{\alpha}} \redefine\gb{G_{\beta}} \redefine\gd{G_{\delta}} \redefine\gr{G_{\rho}} \redefine\gm{G_{\mu}} \redefine\gap{G_{\alpha'}} \redefine\gwm{G_{\w{\m}}} \redefine\gg{G_{\gamma}} \redefine\ggp{G_{\gamma'}} \redefine\gl{G_{\lambda}} \redefine\gs{G_{\s}} \redefine\cga{C_{G_{\alpha}}} \redefine\cgap{C_{G_{\alpha'}}} \redefine\cgb{C_{G_{\beta}}} \redefine\czm{C_{\ym}} \redefine\d{\delta} \redefine\zp{Z_{\phi}} \redefine\gdp{G_{\delta'}} \redefine\qdp{Q_{\delta'}} \redefine\zdp{Z_{\delta'}} \redefine\t{\tau} \redefine\gt{G_{\tau}} \redefine\vl{V_{\l}} \redefine\vb{V_{\b}} \redefine\vap{V_{\a'}} \redefine\vg{V_{\g}} \redefine\vd{V_{\d}} \redefine\vr{V_{\r}} \redefine\vs{V_{\s}} \redefine\vt{V_{\t}} \redefine\wm{W_{\m}} \redefine\ws{W_{\s}} \redefine\wa{W_{\a}} \proclaim {Lemma 7.1} Denote by $\Cal T$ the set of elements $t$ of $S$ such that $|\w V/C_{\w V}(t)|=2$. Let $\Cal T_0$ be a subset of $\Cal T$, and put $T_0=\lan \Cal T_0\ran$. Suppose that $Y_P\leq [V,T_0]Y_H$. Then $C_V(O^2(H))=Y_H$ and $T_0Q_H=\lan \Cal T\ran Q_H$. \endproclaim \demo{Proof} By assumption we have $[V,T_0]\nleq Y_H$, so $T_0\nleq Q_H$. Set $T=\lan \Cal T\ran$. Then 3.11 implies that $O^2(\bar H)\bar T$ is a direct product of groups $\bar K_i$, $1\leq i\leq r$, where $\bar K_i\cong SL(2,2)$ and where $|[\w V,\bar K_i]|=4$ for all $i$. It follows that $[\w V, T]\leq [\w V,O^2(H)]$, and thus $Y_P\leq [V,O^2(H)]Y_H$, by assumption. This shows that $C_V(O^2(H))=Y_H$ and that $\w V$ is the direct product of the fours groups $[\w V, \bar K_i]$, $1\leq i\leq r$. Suppose that $\bar{\Cal T_0}$ is a proper subset of $\bar{\Cal T}$. Then $[\w V,\bar T_0]$ centralizes at least one of the groups $\bar K_i$. As $\bar H=\lan \bar K_i,\bar S\ran$ for any $i$, we then have $Y_P\nl H$, which one recognizes as a contradiction, and which proves the lemma. \qed \enddemo \proclaim {Lemma 7.2} Suppose that there exists an element $t$ of $[Q_P,O^2(P)]O_2(H)$ such that $|\w V/C_{\w V}(t)|=2$. Then $C_H(\w V)=O_2(H)$. \endproclaim \demo{Proof} Set $N=C_H(\w V)$. As $H\in \Cal P$ (or, more generally, by 1.11(b)) $Q_H$ is a Sylow $2$-subgroup of $N$. As $N$ normalizes $Y_P$, 1.7 implies that $N\leq N_G(O^2(P))$. Then $[N,t]$ is a $2$-group, and thus $[N,t]\leq Q_H$. Set $\hat H=H/Q_H$. Define subgroups $\bar K_i$ of $\bar H$, $1\leq i\leq s$, as in the proof of Lemma 7.1. Then $\bar K_i=\lan \bar t_i,\bar a_i\ran$, where $|\bar a_i|=3$ and where $\bar t_i\in\bar S$. As $H\in\Cal P$, $\bar S$ acts transitively on $\{\bar K_i\}_{1\leq i\leq s}$. Then each $\bar t_i$ is the image in $\bar H$ of a conjugate $t_i$ of $t$. Set $X_i=[O^2(H),t_i]$. As $[N,t_i]\leq Q_H$, it follows that $|\hat X_i|=3$. In particular, we have $[\hat X_i,\hat X_j]=1$ for all $i$ and $j$. Setting $T=\lan t_1,\cdots,t_s\ran$, we now have $[O^2(\hat H),\hat T]$ elementary abelian of order $3^s$, and where $1\neq \hat T\nl\hat S$. But $H\in\Cal P$, so $[O^2(\hat H),\hat T]=O^2(\hat H)$. This completes the proof that $N=Q_H$. \qed \enddemo \proclaim {Lemma 7.3} Let $(\r,\s,\t)$ be a path in $\Gamma$ with $\s$ of type $P$. Set $D=\vr\cap\vt$, $T=\qr\cap\qs$, and assume: \roster \item"{(i)}" $|\vt/D|=2$, and \item"{(ii)}" there exists $t\in T-\qt$ with $[D,t]=1$. \endroster Then $H/O_2(H)\cong SL(2,2)$, and $|V|=8$. \endproclaim \demo{Proof} We take $S=\gs\cap\gt$, and set $\bar G_{\t}=\gt/C_{\gt}(\vt/\yt)$. Thus, we have $\bar S\cong S/\qt$. As $S=\qs\qt$ and $T\nl\qs$, we then have $\bar T\nl\bar Q_{\s}=\bar S$. We have $[D,t]=1$, by assumption, and thus $t$ induces a transvection on $\vt/\yt$. Proposition 3.11 then yields the structure of $\vt/\yt$ as a module for the group $X=\langle t^{\gt}\rangle$. Thus, $\bar X$ is a direct product of $s$ copies of $Sym(3)$ and, setting $\w V=\vt/\yt$, we have $\w V= C_{\w V}(O^2(X))\times V_1\times\cdots\times V_s$, where each $V_i$ is an $X$-invariant fours group. As $t\in T$, $\bar T$ contains all the transvections in $\bar S$. As $[D,T]\leq Y_{\r}$, of order $2$, it follows that $s\leq 2$. But also $\w D$ is $\bar S$-invariant, as $D\nl Q_{\s}$. If $s=2$ (so that $\bar S$ is dihedral of order $8$) this condition uniquely determines the hyperplane $\w D$ of $\w V$, and one sees that in this case $|[D,T]|>2$. Thus $s=1$, and $8\leq |\vs|\le 16$. Set $W=C_{\vt}(O^2(\gt))$, and suppose that $W\nleq D$. Then $\vt=WD$, and so $[\w V,T]=\w Y_{\r}$. In this case, Lemma 7.1 yields $|V|=8$. On the other hand, suppose that $W\leq D$, and suppose further that $W\neq \ys$. Set $W_1=C_{V_{\r}}(O^2(G_{\r}))$. We then have $D=WW_1$ and $W\cap W_1\neq 1$. But $O_2(O^2(\gt))\nleq Q_{\s}$, as $Q_{\t}\cap\qs$ is not normal in $\gr$. Then $[W\cap W_1,O^2(G_{\s})]=1$, which is contrary to $Z(P)=1$. This contradiction now shows that $|V|=8$ in any case. Now $P/Q_P\cong SL(2,2)$, by Lemma 7.2. \qed \enddemo \proclaim {Lemma 7.4} Suppose that $b$ is even. Then $|V|=8$. \endproclaim \demo{Proof} Fix a critical pair $(\a,\a')$ and a path $\pi=(\a,\b,\b+1,\b+2,\cdots,\a'-1,\a')$ of length $b$. Fix $\l\in\Delta(\a')-\{\a'-1\}$. As $b$ is even, we have $\yb=[\ya,\yap]=[\ya,\yl]=Y_{\a'-1}$. As $\yb\neq Y_{\b+2}$, we then have $b\geq 6$. In particular, it then follows that $[V_{\b+2},\ya]=1$. It will be convenient to establish, and to list, (and to prove) the following facts. \vskip .1in \noindent (1) $[V_{\b+2},\vl]=1$. \vskip .1in \noindent (2) $\vl\leq\gb$ and $\vl\nleq \qb$. \vskip .1in \noindent In order to prove (1) and (2), notice that we have $[\ya,\yl]\neq 1$. As $V_{\b+2}\leq\ql$, we have $[V_{\b+2},\vl]\leq\yl$, while also $[V_{\b+2},\ya]=1$. This yields (1). Then $[\yb,\vl]=1$, and so $\vl\leq\gb$. Supposing $\vl\leq\qb$, we obtain $[\ya,\vl]=[\ya,\yl]=Y_{\a'-1}\leq\vl$, and so $\vl\nl\langle\ya,\gl\rangle= \langle \gap,\gl\rangle$, which contradicts our basic hypothesis that $\lan P,H\ran\notin \Cal L$. Thus (2) holds. By Lemma 1.2 in [Bernd's N-Groups] there exists $g\in \vb-\gl$ with $|\vl/C_{\vl}(g)|\leq 4$. Set $D=C_{\vl}(g)\yl$. Then $D\nl\langle g,\ql\rangle$. Thus $D$ is invariant under $O^2(\gap)$, and so $D\leq V_{\a'-1}\cap\vl$. Thus $|\vl/V_{\a'-1}\cap\vl|= 2$, and so also $|\vb/\vb\cap V_{\b+2}|=2$. Now (1) and (2) and Lemma 7.3 together yield $|V|=8$. \qed \enddemo We now take up the case where $b$ is odd. \proclaim{Reminder} For any vertex $\d$ of $\Gamma$ and for any non-negative integer $n$, we set $V_{\d}^{(n)}=\lan \yg\ :\ dist(\g,\d)\leq n\ran$. \endproclaim \proclaim {Lemma 7.5} Assume $b$ odd, $b\geq 5$. Let $(\r,\s,\t,\l)$ be a path of length $3$ in $\Gamma$, with $\r$ of type $P$ (and $\s$ of type $H$). Set $B=V_{\r}^{(b-1)}$, and assume that there exists $t\in B-\gs(\t)$ with $|W_{\t}/C_{W_{\t}}(t)|\leq 4$. Then $H/O_2(H)\cong SL(2,2)$, and $(P,H)$ is a weak $BN$-pair. \endproclaim \demo{Proof} Set $U=C_{\vl}(t)\yl$, $U^*=U(\yl)^t$, $R=N_{\qs}(U)$, and $R^*=N_{\qs}(U^*)$. As $t\notin \gs(\t)$, by assumption, we have $\yl\nleq U$, and so $|\vl/U|\leq 2$. Suppose first that $\qs\cap\qt\nleq R$. Then $\lan U^{\qs\cap\qt}\ran=\vl$. Since $(\yl)^t\leq\vs$ we then have $$ \lan (U^*)^{\qs}\ran\vs=\lan U^{\qs}\ran\vs=\lan U^{(\qs\cap\qt)\qs}\ran\vs=\lan \vl^{\qs}\ran\vs =W_{\t}. $$ But this implies that $W_{\t}$ is $t$-invariant, which is contrary to $t\notin\gs(\t)$. We therefore conclude that $\qs\cap\qt\leq R$. Suppose next that $R\nleq\qt$. Then $O^2(\gt)\leq \lan R,\ql\ran$, and hence $U\leq D_{\t}$. Thus $|\vs/D_{\t}|\leq 2$, and then 7.3 yields the conclusion of the lemma. Thus, we may assume that $R\leq\qt$, and hence that $R=\qs\cap\qt$. Then $R$ is not $t$-invariant, by 1.9, and so $R\neq R^*$. As $|\qs/(\qs\cap\qt)|=2$, we have thus shown: \vskip .1in \noindent (1) $R^*=\qs$, $R=\qs\cap\qt$, and $U\neq U^*$. \vskip .1in Let $g\in R^*-R$. As $|U^*/U|=2$ we then have $|U/(U\cap U^g)|=2$, and since $|\vl/U|/\leq 2$ we obtain $|\vl/(\vl\cap (\vl)^g)|\leq 4$. As $g\notin\gl$, by (1), 1.17 yields \vskip .1in \noindent (2) $|\vs/D_{\t}|=|\vs/D_{\r}|\leq 4$. \vskip .1in Recall that $B$ denotes $V_{\r}^{(b-1)}$. For any vertex $\g$ with $dist(\g,\r)\leq b-1$ we have $[\yg,\vd]=1$ for some $\d\in\Delta(\r)$, and thus $[B,D_{\r}]=1$. Take $P=\r$ and $H=\s$, so that $S=\gr\cap\gs$. We shall also write $D$ for $D_{\r}$. We have $t\notin N_G(W_{\t})$, so $W_{\t}\neq V_{\s}C_{W_{\t}}(t)$. As $|W_{\t}/C_{W_{\t}}(t)|\leq 4$, by assumption, it then follows that $|\vs/C_{\vs}(t)|=2$. In particular, $t$ induces a transvection on $\w\vs$. As $B\nl S$, it now follows from 3.11 and from (2) that $H/Q_H\cong SL(2,2)$ or $O_4^+(2)$, with $\w V/C_{\w V}(O^2(H))$ a natural module for $H/Q_H$. In the $SL(2,2)$-case we are done, so we assume henceforth that $H/Q_H$ is isomorphic to $O_4^+(2)$. We next show: \vskip .1in \noindent (3) $\w V$ is a natural $O_4^+(2)$-module for $H/Q_H$. \vskip .1in Suppose (3) is false. Then $C_V(O^2(H))$ properly contains $Y_H$, and so there exists $x\in V$ with $[x,H]=Y_H$. As $[B,D]=1$, it follows from (2) that $x\in D$, and then $[x,P]\leq Y_P$ by 1.17(b). Here $x\notin Y_P$ as $Y_P$ is not normal in $H$. As $Y_P= \Omega_1(Z(Q_P))$ and as $V$ is elementary abelian, we then have $[x,Q_P]=Y_P$ and $|S:C_S(x)|>2$, for a contradiction. This proves (3). Notice that $D$ is now identified as $[V,S]Y_H$, of order $8$. Since $|V/C_V(t)|=2$, we then have $C_S(D/Y_H)=O_2(H)B$, and $Q_HB/Q_H=Q_H)C_S(D)/Q_H$ is the subgroup of $S/Q_H$ induced by the elements of $S/Q_H$ which induce transvections on $\w V$. Set $J=\lan (Q_P\cap Q_H))^P\ran$. Then $J\geq [Q_P,O^2(P)]$. As $D\nleq Y_P$ we have $Q_P/C_P(D)$ isomorphic to $Y_P$ as modules for $P$, and so $Q_P=C_{Q_P}(D)J$. As $B\leq J$, this says that $J$ induces the whole of the action of $Q_P$ on $V/Y_H$. Then, since $S=Q_PQ_H)$, we obtain $S=Q_HJ$. Translating this information to the edge $(\t,\l)$, and setting $\qt^*=(\qs\cap\qt)(\qt\cap\ql)$ we observe that $\qt^*$ is conjugate to $J$, by 1.17(c). We have thus shown: \vskip .1in \noindent (4) $\gt\cap\gl=\qt^*\ql=(\qs\cap\qt)\ql$. \vskip .1in As $U$ is normal in $\qs\cap\qt$, by (1), it now follows from (3) and (4) that \vskip .1in \noindent (5) $U=[\vl,\gt\cap\gl]=[\vl,\qt\ql]$. \vskip .1in Note that (5) also holds if $\l$ is replaced by the vertex $\l'\in\Delta(\t)-\{\s,\l\}$, and $U$ is replaced by $C_{V_{\l'}}(t)Y_{\l'}$. Set $X=[W_{\t},\qt]\vs$. Then $X=\lan (U)^{\gt}\ran\vs$, by (5). It follows that $X$ is $t$-invariant, and so $X\nl \lan t,\gs\cap\gt\ran=\gs$. Thus $X\leq W_{\d}$ for any $\d\in\Delta(\s)$, and thus $[W_{\t},\qt]\leq W_{\g}$ for every vertex $\g$ at distance $2$ from $\t$. For every such $\g$ we have $[V_{\g}^{(b-3)},W_{\g}]=1$, and we obtain $[W_{\t},\qt,V_{\t}^{(b-1)}]=1$. Translating back to the edge $(\r,\s)$, we then have $[V,Q_H),B]=1$, and so $B$ induces a transvection on $V/Z$. This is contrary to $B\nl S$, and the lemma is thereby proved. \qed \enddemo \proclaim {Lemma 7.6} Suppose that $\bar H\cong SL(2,2)$. Then $b\neq 5$. \endproclaim \demo{Proof} We'll just copy this from [9.11 in Weak $(B,N)$-pairs of Rank $2$]. \qed \enddemo For the remainder of this section we will assume that $b$ is odd, $b\geq 5$. (The case $b=3$ will be taken up in the next section.) Under this assumption, it follows from 2.3 that there is a subgroup $A$ of $[Q_P,O^2(P)]O_2(H)$ such that $A$ is a quadratic $F2^*$-offender on $\w V$. Then 3.11 implies that $\bar H_0$ is a direct product $$ \bar H_0=\bar L_1\times\cdots\bar L_r $$ where $|\bar L_i|=3$ and where $|[\w V,\bar L_i]|=4$ for all $i$, $1\leq i\leq r$. For each such $i$, set $\w V_i=\prod\{[\w V,L_j]\}_{j\neq i}$, and set $K_i=C_H(\w V_i)$. (Thus, we have $\bar K_i=\bar L_i$ unless there exist elements of $H$ which induce transvections on $\w V$.) We set $$ \Cal K=\{K_1,\cdots,K_r\} $$ and for any $H$-vertex $\d=H^g$ of $\Gamma$ we define $\Cal K_{\d}$ to be $\Cal K^g$. The parameter $r$ will be fixed for the remainder of this section. \vskip .1in A {\bf path} in $\Gamma$ will always be understood to be non-empty, without stammering. For any path $\pi=(\d,\cdots,\d')$ in $\Gamma$, we write $|\pi|$ for the {\bf length} of $\pi$, and we write $\overline\pi$ for the {\bf opposite path} $(\d',\cdots,\d)$. If $\g$ and $\g'$ are vertices of $\pi$, with $\g$ preceding $\g'$, then $\pi(\g,\g')$ denotes the {\bf sub-path} $(\g,\cdots,\g')$ of $\pi$. If $\pi'$ is a path whose initial vertex is incident with the terminal vertex of $\pi$, then we write $\pi\circ\pi'$ to denote the path obtained by {\bf concatenation} of $\pi$ and $\pi'$, in the given order. Let $H(\pi)$ denote the set of $H$-vertices of $\pi$ which are not terminal vertices of $\pi$. Then, denote by $\Cal S(\pi)$ the set of vertices $\d$ in $H(\pi)$ which satisfy the following two conditions, for both $\e=1$ and $\e=-1$. \vskip .1in \noindent (1) There exists $K\in\Cal K_{\d}$, and an element $g$ of $H$ such that either $g$ is in $K-C_K(\w V)$ or $g$ induces a transvection on $\w V$ which inverts $\bar K$, and such that $\d+\e$ is fused to $\d-\e$ by $g^{\e}$. \vskip .1in \noindent (2) We have $N_{\gd\cap G_{\d+\e}}(Y_{\d-\e})\leq G_{\d-\e}$. \vskip .1in \noindent We will refer to the group $K$ in (1) as the {\bf linkage group} for $\pi$ at $\d$. \proclaim {Lemma 7.7} Assume that $b$ is odd, $b\geq 5$. \roster \item"{(a)}" Let $K\in\Cal K$ and let $g\in K-C_K(\w V)$. Set $T=N_S((Y_P)^g)$. Then there exists $g'\in N_K(T)$ such that $(Y_P)^g=(Y_P)^{g'}$. \item"{(b)}" Let $(\r,\s)$ be an edge of $\Gamma$ with $\r$ of type $P$, $\s$ of type $H$. Then $\vs=\lan \yt\ :\ \s\in\Cal S(\r,\s,\t)\ran\yr$. \endroster \endproclaim \demo{Proof} We have $[\w V,H_0]$ the direct product of $\{[\w V, K_i]\ :\ 1\leq i\leq r\}$, and the action of $H$ on $\Cal K$ induces an equivalent action on this set of subgroups of $\w V$. Let $y\in Y_P- Y_H$. Then $T$ centralizes both $\w y$ and $\w y^g$, so $T$ centralizes $[\w y,g]$, and therefore $\bar K$ is $T$-invariant. Let $\phi$ be the endomorphism of $[\w V,H_0]$ which restricts to the identity on $[\w V,K]$ and which is the zero mapping on any $[\w V,K_i]$ for $K_i\neq K$. Then $N_S(\bar K)$ centralizes also $\phi(\w y)$, and so $T$ centralizes $[\w V,K]$. Setting $N=C_H(\w V)$, we now have $[T,K]\leq N$. As $T\geq O_2(H)$, $T$ is a Sylow $2$-subgroup of $NT$, and the Frattini argument now shows that there exists $g'\in N_{NT\lan g\ran}(T)-NT$. This proves (a). In order to prove (b), take $\r=P$ and $\s=H$. Set $V_i=\lan (Y_P)^{K_i}\ran$. Then $V=V_1\cdots V_r$, and (b) then follows from (a). \qed \enddemo \proclaim {Lemma 7.8} Assume that $b$ is odd, $b\geq 5$. Let $\pi=(\r,\s,\t)$ be a path in $\Gamma$, with $\s\in\Cal S(\pi)$, and let $A$ be an elementary abelian subgroup of $Q_{\r}$. Assume that there exists $a\in A-G_{\t}$ with $[\vs,a,A]=1$. Then $|A/A\cap G_{\t}|= 2$. \endproclaim \demo{Proof} Let $K$ be the linkage group for $\pi$ at $\s$, and let $g\in K$ with $\t=\r^g$. Identify $H$ with $\s$, and set $N=C_H(\w V)$. Then $[C_A(\bar K),g]\leq N$, and so $C_A(\bar K)$ normalizes $\yt$. As $\s\in\Cal S(\pi)$ we then have $C_A(\bar K)\leq \gt$, and it therefore suffices to show that $|A/C_A(\bar K)|\leq 2$. This follows from quadratic action, and from the structure of $\bar H_0$ and of $\w V$ given after 7.6. \qed \enddemo \proclaim {Lemma 7.9} Assume that $b$ is odd, $b\geq 5$. Let $\pi=(\d,\cdots,\g)$ be a path with $\d$ of type $P$ and with $\g$ of type $H$. There then exists $\m\in\Delta(\g)$ such that $\g\in\Cal S(\pi\circ(\m))$. Moreover, if $(\d,\g)$ is a critical pair and $|\pi|=b$, then $\m$ may be chosen so that $\yd\nleq \gm$. \endproclaim \demo{Proof} Take $H=\g$ and take $P=\g-1$, where $\g-1$ denotes the vertex of $\pi$ which is adjacent to $\g$. Thus, we have $S=G_{\g-1}\cap\gg$. Choose $K\in\Cal K$, with $[\bar K,\bar\yd]\neq 1$ if $\yd\nleq \qg$. Set $N=C_H(\w V)$, choose $g\in K- N$, and set $m=(\g-1)^g$. Set $T=N_S(\ym)$. By Lemma 7(a) we may assume that $g$ normalizes $T$, and hence that $T\leq\gm\cap (G_{\g-1})^{g^{-1}}$. Thus $\g\in\Cal S(\pi\circ(\m))$. Now suppose that $\yd\nleq \qd$ so that, by choice, $[\yd,K]\nleq N$. We then have $\yd\nleq T$. Thus, $\yd\nleq\gm$, and this completes the proof of the lemma. \qed \enddemo \proclaim {Lemma 7.10} Assume that $b$ is odd, $b\geq 5$. Let $\pi=(\d,\cdots,\g)$ be a path of length $b-1$ with both $\d$ and $\g$ of type $H$. Suppose that $V_{\d}\nleq Q_{\g}$. There then exists $\l\in\Delta(\d)$ such that $(\l,\g)$ is a critical pair, and such that $\d\in\Cal S((\l)\circ\pi)$. \endproclaim \demo{Proof} Immediate from Lemma 7.7(b). \qed \enddemo \proclaim {Lemma 7.11} Assume that $b$ is odd, $b\geq 5$, and let $\pi=(\r,\s,\t,\l)$ be a path in $\Gamma$, with $\s\in\Cal S(\pi)$. Then the following hold. \roster \item"{(a)}" $\yr\cap\vl=\ys$. \item"{(b)}" $W_{\r}=\langle x\in W_{\r}$ $:$ $|\vl/C_{\vl}(x)|\leq 2 \rangle$. \item"{(c)}" $[W_{\r},W_{\t}]\cap\yt\leq \ys$. \endroster \endproclaim \demo{Proof} Suppose that (a) is false. As $\ys\leq\vl$ we then have $\yr\leq \vl$, and thus $[\yr,\ql]\leq \yl$. As $\gt$ is $2$-transitive on $\Delta(\t)$, there exists a vertex $\a$ at distance $b-2$ from $\l$ such that $(\a,\s)$ is a critical pair. As $\s\in\Cal S(\pi)$, we have $\r=\t^x$ for some $x\in\gs-\gt$, such that $x$ lies in the linkage group $K$ for $\pi$ at $\s$. Take $H=\s$ and take $P=\t$, so that $S=\gs\cap\gt=\qs\qt$. Set $R=\langle (\ya)^{\qt}$. Then $[\yr,R]\leq[\vl,R]=1$, and since $\s\in\Cal S(\pi)$ we then have $R\leq\gr$. Notice that $1\neq\bar R\nl\bar S$. As $\bar S$ is transitive on $\Cal K$, we then have $[\bar R,\bar K]\neq 1$. Then $H=\lan S^x,R\ran$, and we may conclude that $\qr\cap\qs=C_{\qs}(\yr)$ is normal in $H$. This contradicts ---, and so (a) is proved. Suppose next that (b) is false. Then $W_{\r}\nleq \ql$, and so $b=5$. Choose $\g\in\Delta(\r)$ with $\vg\neq\langle x\in\gg$ $:$ $|\vl/C_{\vl}(x)|\leq 2\rangle$. It then folows from Lemma 5(b) that there exists $\d\in\Delta(\g)$ such that $\g\in\Cal S(\d,\g,\r)$ and such that, for some $x\in\gd$, we have $|\vl/C_{\vl}(x)|>2$. Here $\vl\leq \gg$, $\vl$ acts quadratically on $\vg$, and $\vl\nleq\gd$. In particular, $\vl$ does not centralize the linkage group $K$ for $(\d,\g,\r)$ modulo $C_{\gg}(\w\vg)$. 7.6 then yields $|\vl/\vl\cap\gd|=2$, and then also $[\yd,\vl\cap\gd]\neq 1$. Thus, we have $\yg = [\yd,\vl\cap\gd]\leq\vl$, which contradicts (a). Finally, suppose that (c) is false. Again, we have $b=5$, as otherwise $b\geq 7$ and $[W_{\r},W_{\t}]=1$. Write $\Delta(\r)=\{\s,\g,\g'\}$ and $\Delta(\t)=\{\s,\l,\l'\}$. If $W_{\r}\leq \ql$ and $[W_{\r},\vl]\neq 1$ then $[V_{\g},\vl][V_{\g'},\vl]=\yl$. and so $\yl$ is contained either in $V_{\g}$ or in $V_{\g'}$, contrary to (a). Thus, if $W_{\r}\nleq\ql$ then $[W_{\r},\vl]=1$. As $[W_{\r},W_{\t}]\neq 1$, we may then fix notation (replacing $\l$ by $\l'$, or $\g$ by $\g'$, if necessary) so that $V_{\g}\nleq\ql$. By Lemma 7.9 we may choose $\m\in\Delta(\l)$ so that $\l\in\Cal S(\t,\l,\m)$, and so that $\vg\nleq \gm$. Suppose that $\ym\leq\qg$. Then $[\vg,\ym]=\yg$, and thus $\yg\leq\vl$, which is contrary to (a). We conclude that $\ym\nleq\qg$, and then 7.9 implies that there exists $\d\in\Delta(\g)$ such that $\g\in\Cal S(\d,\g,\r)$ and with $\ym\nleq \gd$. Suppose that $[W_{\d},\yl]=1$. As $\s\in\Cal S(\pi)$ we then have $W_{\d}\leq\qt\leq \gl$. As $[W_{\d},V_{\g}]=1$, we have $[\vl,\vg,W_{\d}]=1$. Then $|W_{\d}/W_{\d}\cap\gm|=2$, by Lemma 8, and then also $|W_{\d}/C_{W_{\d}}(\ym)|\leq 4$. Now 7.5 yields $r=1$ and $|V|=8$. Since also $b=5$, we have a contradiction to 7.6. We therefore conclude that $[W_{\d},\yl]\neq 1$. On the other hand, we have $\yl\leq \yt\leq\ys[W_{\r},W_{\t}]$, by assumption in (c). Thus $\yl\leq\ys[\vg,W_{\t}][V_{\g'},W_{\t}]$, and so $1\neq [\yl,W_{\d}]\leq [V_{\g'},W_{\t},W_{\d}]$. By (b), above, both $W_{\t}$ and $W_{\d}$ are generated by elements which centralize hyperplanes of $V_{\g'}$. Since $W_{\t} W_{\d}$ is contained in the $2$-group $\gr\cap G_{\g'}$ it follows that $[V_{\g'},W_{\t},W_{\d}]\leq Y_{\g'}$. Thus, $[\yl,W_{\d}]=Y_{\g'}$. Write $\Delta(\d)=\{\g,\b,\b'\}$. As $\yl\leq\qd$ we conclude that, up to a possible permutation of $\{\b,\b'\}$, we have $Y_{\g'}\leq V_{\b}$. But this contradicts (a), with $(\r,\g,\d,\b)$ in place of $\pi$. The proof of (c) is thereby complete. \qed \enddemo \proclaim {Lemma 7.12} Assume that $b$ is odd, $b\geq 5$. There then exists a critical pair $(\a,\a')$, a path $\pi=(\a,\b,\cdots,\a')$ of length $b$, and vertices $\m\in\Delta(\a')-\{\a'-1\}$ and $\l\in\Delta(\m)-\{\a'\}$ such that the following hold. \roster \item"{(i)}" Both $\b$ and $\a'$ are in $\Cal S(\pi\circ(\m))$. \item"{(ii)}" $\ya$ is not contained in $\gm$. \item"{(iii)}" If $W_{\m}\nleq G_{\b+1}$, then $\b+2\in\Cal S(\pi)$ and $Y_{\b+2}\nleq [V_{\b+2},W_{\m}]$. \endroster \endproclaim \demo{Proof} Suppose false, and choose a critical path $\pi=(\a,\b,\cdots,\a')$ so that $|\Cal S(\pi)|$ is as large as possible. Then $\b\in\Cal S(\pi)$ by 7.10 (applied to $\pi(\b,\a')$). By 7.9, we may choose $\m\in\Delta(\a')-\{\a'-1\}$ so that $\a'\in\Cal S(\pi\circ(\m))$ and so that $\ya\nleq\gm$. As the lemma is assumed false, we have $W_{\m}\nleq G_{b+1}$. Choose $\l\in\Delta(\m) -\{\a'\}$ so that $\vl\nleq G_{\b+1}$. Suppose next that $V_{\b+2}\leq \ql$. Then $[V_{\b+2},\vl]=\yl$, and since $\vl$ acts non-trivially on $V_{\b+2}/Y_{\b+2}$ we conclude that $Y_{\b+2}\nleq [V_{\b+2},\vl]$. As the lemma is false, we then have $\b+2\notin\Cal S(\pi)$. Now set $\pi_0=(\l,\m)\circ \overline{\pi}(\a',\b+2)$ and apply 7.10 to $\pi_0$. Thus, there exists $\g_0\in\Delta(\l)-\{\m\}$ such that $(\g_0,\b+2)$ is a critical pair and such that, for $\pi'=(\g_0)\circ \pi_0$, we have $\l\in\Cal S(\pi')$. Notice however that $|\Cal S(\pi')|> |\Cal S(\pi)|$, which is contrary to our choice of $\pi$. We conclude that $V_{\b+2}\nleq \ql$. We now apply 7.10 to the path $\pi_1=\pi(\b+2,\a')\circ(\m,\l)$. Thus, there exists a vertex $\g_1\in\Delta(\b+2)$ such that $(\g_1,\l)$ is a critical pair, and such that, upon setting $\pi^*=(\g_1)\circ \pi_1$, we have $\b+2\in\Cal S(\pi^*)$. Observe that $|\Cal S(\pi^*)|\geq |\Cal S(\pi)|$. Thus, we may replace $\pi$ by $\pi^*$, and since $\a'\in\Cal S(\pi^*)$ we then obtain $\a'-2\in\Cal S(\pi)$. But then maximality of $|\Cal S(\pi)|$ shows that, to begin with, and before making the above replacement, we had $\a'-2\in\Cal S(\pi)$. By iteration it now follows that $\Cal S(\pi)=\{\b,\b+2,\cdots,\a'-2\}$. Moreover, since the lemma fails to hold for $\pi^*$, we have (prior to replacement) $Y_{\b+2}\leq [V_{\b+2},W_{\m}]$, and thus (after replacement) $\yb\leq [\vb,W_{\a'-1}]\leq W_{\a'-1}$. Lemma 7.11(c) then yields $b>5$, so that $[W_{\a'-1},W_{\m}]=1$, whence $[\yb,W_{\m}]=1$. But with $\b+2\in\Cal S(\pi)$ we then have $W_{\m}\leq Q_{\b+1}$. Thus $W_{\m}\leq G_{\b+1}$ after all, and the lemma is proved. \qed \enddemo \proclaim {Proposition 7.13} Assume that $b$ is odd, $b\geq 5$. Then $|V|=8$. \endproclaim \demo{Proof} Assume that $|V|>8$. Choose $\pi$ and $\m$ as in 7.12, and fix $t\in \ya-\yb$. Suppose first that $\yb\leq[\vb,\vap]$. Then $\yb\leq \vap\leq W_{\m}$, and it then follows from 7.12 that $W_{\m}\leq\gb$. Suppose that $W_{\m}$ acts quadratically on $\vb$, or that $\vap\nleq\ga$. As $\b\in\Cal S(\pi)$, 7.8 then implies that $|W_{\m}/W_{\m}\cap \ga|\leq 2$, and so $|W_{\m}/C_{W_{\m}}(t)|\leq 4$. As 7.5 then yields $r=1$, we conclude that $W_{\m}$ is not quadratic on $\vb$ and that $\vap\le\ga$. In particular, it follows that $[W_{\m},(W_{\m})^g]\neq 1$ for some $g\in\vb$, and so $b=5$. Further, with $\vap\le\ga$ we have $|\vap/C_{\vap}(t)|=2$. Now $\vb=(\vb\cap\gm)\ya$, and $W_{m}$ acts quadratically on the hyperplane $\vb\cap\gm$ of $\vb$. It follows that the $W_{\m}$-orbits on $\Cal K_{\b}$ are of length at most $2$, and that $|W_{\m}/W_{\m}\cap\ga|\leq 4$. Thus, we have shown that $[t,W_{\m},W_{\m}]\neq 1$, and $|W_{\m}/C_{W_{\m}}(t)|\leq 8$ in the case that $\yb\leq \vap\leq W_{\m}$. Suppose, on the other hand, that $\yb\nleq [\vb,\vap]$. Then $[t,\vap\cap\ga]=1$, and so $|\vap/C_{\vap}(t)|=2$, as in the preceding case. Further, we have $\vap\nleq\qb$. Notice that $|W_{\m}/W_{\m}\cap\gb|\leq 2$, by 7.12. As $[\vb,\vap,W_{\m}]=1$, 7.8 then yields $|W_{\m}/W_{\m}\cap\ga|\leq 4$, and so once again $|W_{\m}/C_{W_{\m}}(t)|\leq 8$. Moreover, we either have $|W_{\m}/C_{W_{\m}}(t)|\leq 4$ or $\yb\leq [t,W_{\m}\cap \ga]$. If now $[t,W_{\m},W_{\m}]=1$ we thus obtain $[\yb,W_{\m}]=1$ and $|W_{\m}/C_{W_{\m}}(t)|\leq |W_{\m}/(W_{\m}\cap\qa)|\leq 4$, again contrary to 7.5. We have thus shown that, in any case, we have the following information. \vskip .1in \noindent (1) $b=5$, $[t,W_{\m},W_{\m}]\neq 1$, $|W_{\m}/C_{W_{\m}}(t)|= 8$, and $|\vap/C_{\vap}(t)|=2$. \vskip .1in Set $\s=\m^t$, and observe that since $t$ induces a transvection on $\w\vap$, we have $\a'\in\Cal S(\s,\a',\m)$. Further, we have the following consequence of 7.2. \vskip .1in \noindent (2) $\a'\in\Cal S(\s,\a',\m)$. \vskip .1in Suppose next that $W_{\m}\leq \qr$ for all $\r\in\Delta(\s)$. Then $[W_{\m},W_{\s}]\leq\ys$, and then 7.11(c) yields $[W_{\m},W_{\s}]\leq\yap$. For any $\r\in\Delta(\s)-\{\a'\}$ we then have $[\vr,W_{\m}]\leq]yr\cap\yap=1$, and so $[W_{\m},W_{\s}]=1$. This implies that $[t,W_{\m},W_{\m}]=1$, contrary to (1). We conclude that there exists $\r\in\Delta(\s)$ with $W_{\m}\nleq \qr$. Fix such a vertex $\r$ and set $l=\r^t$. Then $W_{\s}\nleq \ql$. Notice that $W_{\m}\cap W_{\s}\geq C_{\wm}(t)\vap$. Then (1) yields: \vskip .1in \noindent (3) $\wm\cap\ws$ is of index at most $4$ in $\wm$ and in $\ws$. \vskip .1in We note also that 7.11(b) yields the following. \vskip .1in \noindent (4) $\ws=\lan x\in\ws\ :\ |\vl/C_{\vl}(x)|\leq 2\ran$. \vskip .1in Suppose that $|\vl/C_{\vl}(W_{\s})|=2$. Noting that $\yl\nleq [W_{\s},\vl]$ by 7.11(c), we then have $|[W_{\s},\vl]|=2$. Set $U=[\ws,\vl]$. Since $\qap\cap\qm$ is not normal in $\gap$, by 1.9, where $\gap=\lan \gs\cap\gap,\gap\cap\gm\l\ran$, it follows that $\qap\cap\qm\neq\qap\cap\qs$, and hence $(\qap\cap\qm)(\qap\cap\qs)=\qap$. In particular, $\qap\cap\qm$ is transitive on $\Delta(\s)- \{\a'\}$. As $\ws\nleq\ql$ we then have $\vr\nleq\ql$, and so $U=[\vr,\vl]$ is $t$- invariant. As $U$ lies in the center of $\qap\cap\qm$ we then have $[U,\qap]=1$, and then 1.1(a) shows that $U=\yap$. Lemma 7.1 then implies that $|V|=8$, which is contrary to the hypothesis. Thus $|\vl/C_{\vl}(W_{\s})|>2$. Now (3) implies that $W_{\m}=C_{W_{\m}}(W_{\s})\vl$. Then $[W_{\s},\vl]=[W_{\s},W_{\m}]\nl\qap$, and so $\yap\leq [W_{\s},\vl]$, by 1.8. Then (4) and Lemma 1 imply that $\ws\ql/\ql$ is of order $2^r$, and (3) then yields $r\leq 2$. As $|V|>8$, Lemmas 1 and 2 now give: \vskip .1in \noindent (5) $|V|=32$ and $H/Q_H\cong\Omega_4^+(2)$. \vskip .1in For any edge $(\g,\d)$ of $\Gamma$, with $\g$ of type $H$, set $T(\g,\d)=\langle g\in\gg\cap\gd\ :\ |[\vg/\yg,g]|\leq 2\rangle$. For any vertices $\phi$ and $\psi$ in $\Delta(\a')$, set $F(\phi,\psi)= [W_{\phi},W_{\psi}]$. Set $F=F(\s,\m)$. We have seen that $|\ws/C_{\ws}(\vl)|=4$, that $\ws$ is generated by $C_{\ws}(\vl)$ together with two elements that induce transvections on $\vl$, and that $[\ws,\wm]= [\ws,\vl]\geq\yap$. Thus $F$ is a fours group containing $\yap$, and we have $F\yl/\yl=[\vl/\yl,T(\l,\m)]$. Setting $U=F\yl$, it follows that $U\nl \gm\cap\gl$. But also $F$, and hence also $U$, is normal in $\qap$, and so $U\nl\gm$. Then $U\leq \vap\cap\vl$, and $U$ has index $4$ in $\vl$. Since $|\vl/C_{\vl}(W_{\s}|=4$, it follows that $U=\vap\cap\vl$, and so $U=D_{\a'}$, by 1.17. Similarly, we have $F\yr=D_{\s}$, and 7.11(a) yields $F=D_{\m}\cap D_{\s}$. Write $\Cal K_{\a'}=\{K_1,K_2\}$ and define $X_i=X_{\a'}^{(i)}$ to be the inverse image in $\vap$ of $[\vap/\yap,K_i]$, $(i=1,2)$. As $U\nl\gm$ we have $U=[\vap,T(\a',\m)]$, and since $\yl\nleq F$ it follows that $F\leq X_i$ for some $i$. For definiteness, take $F\leq X_1$. Then $[F,N_S(X_1)]\leq \yap$. On the other hand we have $U=F\yl=F\ym$, and so $[U,\qm]=[F,\qm]=\ym$. As $D_{\m}\neq\ym$, $\qm/C_{\qm}(F)$ is a fours group admitting non-trivial action by $O^2(\gm)$. Setting $R=O_2(O^2(\gm))$, we then have $[F,R]=\ym$, and thus $R\nleq N_S(X_1)$. As $R\nl S$, $R/R\cap\qap$ then contains a fours group. Let us record these results. \vskip .1in \noindent (6) We have $F(\s,\m):=[W_{\s},W_{\m}]=\vr\cap\vap\cap\vl=[X_1,T(\a',\m)]$. \vskip .1in \noindent (7) $R/R\cap\qap$ contains a fours group, where $R:=O_2(O^2(\gm))$. \vskip .1in We have $|\Delta(\a')|=9$, and we may identify $\Delta(\a')$ with the set of singular points in the orthogonal space $\vap/\yap$, for the action of $\gap$. Define a map $$ B:\Delta(\a')\times\Delta(\a')\longrightarrow\Bbb F_2 $$ by the formula $$ B(\phi,\psi)=\cases 0,&\text{if $[W_{\phi},W_{\psi}]=1$}\\ 1,&\text{if $[W_{\phi},W_{\psi}]\neq1$}.\endcases $$ We require the following elementary lemma (for which we need provide no proof). \proclaim {Lemma 7.14} Let $\w V$ be an $O_4^+(2)$-space with associated bilinear form $\w B$, let $t$ be an orthogonal transvection on $\w V$, and let $x$ and $v$ be singular points with $x=x^t$ and $v\neq v^t$. Then the following hold. \roster \item"{(a)}" $\w B(v,v^t)\neq 0$. \item"{(b)}" There exists a singular point $y$, with $y\neq y^t$, and such that $\w B(x,y)=0$. \endroster \endproclaim Notice that $S$ has two orbits on $\Delta(\a')-\{\m\}$, each orbit being of length $4$. Let $\t\in\Delta(\a')$ with $\t\notin \s^S$. Thus, we have $\t=\m^g$ where $g$ is an element of order $3$ in $\gap$ which is fixed-point-free on $\vap/\yap$. For any $\d\in\Delta(\t)-\{\a'\}$ one may then observe that $\vd\cap\vap\cap\vl=\yap$, and then for any $\l'\in\Delta(\m)$ we have $[\vd,V_{\l'}]\leq \yap$. Lemma 7.1 then implies that $[\vd,V_{\l'}]=1$, and thus $B(\t,\m)=0$. Since $B(\s,\m)=1$, where $\s=\m^t$, it follows from 7.13(a) that $B$ is indeed the bilinear form associated with the orthogonal space $\vap/\yap$. Then 7.13(b) says that we may choose $\t$ as above, with $B(\a'-1,\t)=B(\m,\t)=0$, and with $t\notin\gt$. In particular, we have $[W_{\t},V_{\b+2}]=1$, and so $W_{\t}$ stabilizes the path $\pi(\b,\a')$. As $t\notin\gt$, 7.5 shows that $|W_{\t}/C_{W_{\t}}(t)|\geq 8$. Then $|W_{\t}/W_{\t}\cap\ga|=4$ (so that $W_{\t}$ induces a non-quadratic fours group on $\vb/\yb$) and $[\ya,W_{\t}\cap\ga]=\yb$. As $t\in X_{\b}^{(i)}$ for some $i$, ($i=1$ or $2$), one may then compute that $[t,W_{\t}]=[\vb,\gb\cap G_{\b+1}]$, and that $[t,W_{\t},W_{\t}]=Y_{\b+1}$. On the other hand, 7.13(a) shows that $[(W_{\t})^t,W_{\t}]\neq 1$, and so $[t,W_{\t},W_{\t}]=[(W_{\t})^t,W_{\t}]=F(\t^t,\t)\leq \vap$, of order $4$. Thus $F(\t^t,\t)=Y_{\b+1}\leq\vap$. This contradicts 7.11(a), thus completing the proof of 7.12. \qed \enddemo [THERE MUST BE AN EASIER ARGUMENT FOR 7.13] \vskip .2in \noindent {\bf Section 8: The solvable case, $b=3$} \vskip .1in \proclaim {Lemma 8.1} Suppose that $P$ and $H$ are solvable, with $p=2$. Then $b=3$, and one of the following holds. \roster \item"{(i)}" $H/O_2(H)\cong SL(2,2)$ and $|V|=8$. \item"{(ii)}" $H/O_2(H)\cong Sz(2)$ and $|V|=32$. \item"{(iii)}" $\bar H$ is isomorphic to a subgroup of $\cong O_4^+(2)$ of index at most $2$, and $|V|=32$. \item"{(iv)}" $O_3(\bar H)$ is extraspecial of order $27$ and of exponent $3$, and \itemitem (a) There exists $P_1\in\Cal P_H(S)$ such that $O_3(\bar P_1)=Z(O_3(\bar H))$ and such that $\lan (Y_P)^{P_1}\ran$ is invariant under $P$. \itemitem (b) $\w V$ is a natural $SU(3,2)$-module for $O^2(\bar H)O_2(\bar P_1)$. \itemitem (c) $O^2(\bar H)O_2(\bar P_1)$ is isomorphic to a subgroup of $SU(3,2)$ of index at most $2$. \endroster \endproclaim \demo{Proof} Suppose $b>3$. Then 7.12 yields $|V|=8$ and $H/O_2(H)\cong SL(2,2)$. Now [AN ARGUMENT EXTRACTED FROM THE WEAK BN-PAIRS PAPER] yields a contradiction. Thus, we have $b=3$. Set $\g=\b+1=\a'-1$, and take $H=\b$ and $P=\g$. Denote by $\d$ the unique element of $\Delta(\g)-\{\b,\a'\}$. As $b>2$ we have $V\leq Q_P$, and so 1.10 implies that $[Q_H,O^2(H)]\nleq V$. Thus, there is at least one non-central chief factor $X$ for $H$ in $Q_H/V$. Suppose first that $\qb\cap\qg\leq\vb\qap$. Then $[\qb\cap\qg,\vap]\leq\vb\yap=\vb$. As $|\qb/(\qb\cap\qg)|=2$, $\vap$ induces a transvection on $X$. But $\vap\qb\nl \qb\qg=Q_HQ_P=S$, and it follows that $H/C_H(X)\cong SL(2,2)$. Then $H/Q_H$ is dihedral of order $2\cdot 3^m$ for some $m$, $m\geq 1$. As $\w V$ is a quadratic $F2$-module for $H$, it follows from 3.11 that $\bar H\cong SL(2,2)$. As $\w V$ is the closure under $H$ of $\w Y_P$, we then have $|\w V|\leq 8$, and then 7.2 implies that $H/Q_H\cong SL(2,2)$. Suppose that $|\w V|=8$. As $P$ is doubly transitive on $\Delta(\g)$ we have $[\w V,\vd]=[\w V,\vap]$, and 7.1 implies that neither $\yb$ nor $yd$ is contained in $[vb,\vap]$. But $[\vb,\vap]\nl\qg$, so $\yg\cap[\vb,\vap]\neq 1$. Then $\yg\cap[\vb,\vap]=\yb$. This contradicts 7.2, applied to the action of $\vb$ on $\vap$, so we conclude that $|V|=8$ and that outcome (i) of the lemma holds. We may therefore assume that $\qb\cap\qg\nleq \vb\qap$. Then also: \vskip .1in \noindent (1) $\qg\cap\qap\nleq \vap\qb$. \vskip .1in Set $A=\vap$. As $[A,\qap]=\yap\leq\qb$, (1) yields: \vskip .1in \noindent (2) $C_{\bar S}(\bar A)\nleq \bar A$. \vskip .1in Suppose that $A$ is an $F1$-offender on $\w V$. Then 3.11 shows that $\bar A$ is generated by transvections, and since $\bar A\nl\bar S$ it follows that $\bar A$ contains all the transvections in $\bar S$. This contradicts (2), and so, in fact, $\bar A$ is not an $F1$-offender on $\w V$, and similarly $V\qap/\qap$ is not an $F1$-offender on $\vap/\yap$. Then by 1.18 and symmetry, we may assume that $[V\cap\qap,A]=\yap$. We then have: \vskip .1in \noindent (3) There exists $v\in\w V$ such that $[v,A]=\w Y_P$. \vskip .1in \noindent We note also that $A$ is a quadratic $F2$-offender on $\w V$, by 2.3, and then 3.10 shows that $\bar A$ is generated by $2$-transvections. By 3.11 there is an $\bar S$-transitive collection $\Cal D=\{\bar D_i\}_{1\leq i\leq r}$ of subgroups $\bar D$ of $\bar H$, such that $O^2(\bar H)=\bar D_1\times\cdots\times\bar D_r$, with $[\w V,O^2(\bar H)]=[\w V,\bar D_1]\times\cdots\times [\w V,\bar D_r]$, and such that $\bar D_i$ is isomorphic to $\Bbb Z_3$, $\Bbb Z_5$, or $O_3(SU(3,2))$. Set $U_i= [\w V,\bar D_i]$. The transitivity of $\bar S$ on $\Cal D$ then implies that there are non-identity elements $u_i$ of $U_i$ such that $\w Y_P=\lan u_1\cdots u_r\ran$. On the other hand, (3) shows that $u_1\cdots u_r=[v,\bar a]$ for some $v\in \w V$ and some $2$-transvection $\bar a$ in $\bar A$. It follows that either $r=1$ or that $|U_i|=4$ and $r= 2$. In any case we have $|\bar A|=2$ and $\bar A$ induces a $2$-transvection on $[\w V,O^2(H)]$, so (3) implies that $[\w V,O^2(H)]=\w V$. If $r=2$ we have the desired outcome (iii) of the lemma, so we assume now that $r=1$. Then $|O^2(\bar H)|>3$, by (2). Suppose that $|O^2(\bar H)|=5$. Then (2) implies that $\bar H\cong Sz(2)$, while $|V|=32$. Then also $H/Q_H\cong Sz(2)$, by 1.11(b), and we have outcome (ii). Finally, suppose that $O^2(\bar H)$ is an extraspecial group of order $27$ and exponent $3$, with $|\w V|=64$. Denote by $H_1$ the inverse image in $H$ of $Z(O^2(\bar H))$. Then $\lan (Y_P)^{H_1}\ran$ is of order $8$, and 1.7 then shows that $O^2(P)$ is not invariant under $H_1$. By 1.4 there then exists $P_1\in\Cal P_{H_1S}(S)$ such that $O^2(P)$ is not $P_1$-invariant, and such that $\lan P,P_1\ran\in\Cal L$. Then $O^2(\bar P_1)=Z(O^2(\bar H))$. Notice that $O_2(\bar P_1)$ is a subgroup of index at most $2$ in a quaternion group. Thus $|\Omega_1(\bar{\qg})|=2$. For any two distinct vertices $\d$ and $\d'$ in $\Delta(\g)$ we then have $[\vd,V_{\d'}]=[\vd,\Omega_1(\bar {\qg})]$, and thus $[\vd,V_{\d'}]$ is $\gg$-invariant. Thus $[\vb,\vap]$ is invariant under both $P$ and $P_1$, and (iv) holds. \qed \enddemo \proclaim {Lemma 8.2} Outcome (iv) of Lemma 8.1 does not hold. \endproclaim \demo{Proof} Suppose false, and let $P_1$ be as in 8.1(iv). Set $L=\lan P,P_1\ran$. As we have seen, we have $L\in\Cal L$, and $Y_P$ is not normal in $L$. Form the amalgam $\Gamma^*=\Gamma(H,L)$ and set $b^*=b(H,L)$. It follows from 1.3(4) and from 8.1(iv) that $|Y_L|=8$ and that $L/O_2(L)\cong SL(3,2)$. In particular, we have $Y_L\leq \lan (Y_P)^H\ran$, and so we have $V=\lan (Y_L)^H\ran$. Then $b^*\geq 3$, and since $b^*\leq b$ we conclude that $b^*=3$. We may then label a critical path $(\a,\b,\g,\a')$, and we shall take $H=\b$ and $L=\g$, so that $P_1=\gb\cap\gg$. For $\d$ a vertex with stabilizer $H^g$ we write $\vd$ for $V^g$. We note that, since $L$ is doubly transitive on $Y_L^{\sharp}$, $\gg$ is doubly transitive on $\Delta(\g)$, and hence $\gb$ is transitive on $\Delta^{(2)}(\b)$. In particular, any path of length $2$ from $\b$ is a critical path. Wishing to avoid needless repetition, we ask the reader to check that the argument at the relevant point in the proof of 8.1, above, shows: \vskip .1in \noindent (1) $|V/C_V(\vap)|=4$, $C_V(\vap)=[V,O_2(P_1)]$, and $[V,\vap]=Y_L$. \vskip .1in Set $X=X_{\b}=\lan C_{\vd}(V)\ :\ \d\in\Delta^{(2)}(\b)\ran$. Then $X_{\a'}=\Omega_1(X_{\a'})$ and $X_{\a'}\leq\qg\leq O_2(\gg\cap\gb)=O_2(P_1)$. Here $\w V$ is a natural $SU(3,2)$-module for $O^2(H)O_2(P_1)$, and $\Omega_1(O_2(\bar P_1))$ is of order $2$, so $[\w V,X_{\a'}]\leq [\w V,\vap]$, and then $[V,X_{\a'}]Y_L$. Thus $[V,X_{\a'}]\leq\vap$, and so also $[X,\vap]\leq V$, by symmetry. It follows that $[X,O^2(H)]\leq V$. Further, we have $[C_{Q_H}(V),\vap]\leq[\qg,\vap]\leq C_{\vap}(V)$, by (1), and so $[C_{Q_H}(V),O^2(H)]\leq X$. This now shows: \vskip .1in \noindent (2) $[C_{Q_H}(V),O^2(H)]=V$. \vskip .1in We have $[V,Q_H]=Y_H$, so: \vskip .1in \noindent (3) $Q_H/C_{Q_H}(V)$ is isomorphic to the dual of $\w V$ as a module for $H$. \vskip .1in \noindent Set $D=\lan [\vd,\qg]\ :\ \d\in\Delta(\g)\ran$. Then (1) shows that $D\leq X\cap X_{\a'}$, so $[D,\vap]=1$, and so $V\nleq D$. As $\eta(P_1,V)=3$, it follows from (2) that $\eta(P_1,D/Y_L)=1$. As $(V\cap D)/Y_L$ is a $P_1$-invariant subgroup of $D/Y_L$ of order $4$, it follows that $D/Y_L$ is an $L$-module of order $8$, dual to $Y_L$. Thus $V\cap D$ is a hyperplane of $D$. Set $W=W_{\g}$. Then $D\leq \Omega_1(Z(W))$. We have $C_S(W)\leq C_{Q_H}(V)$, and since $V\nleq C_S(W)$ we then have $\eta(P_1,C_S(W))=2$. Thus $[C_S(W),O^2(L)]=D$. Suppose that there exists an element $x$ in $Q_H$ such that $[x,H]=Y_H$. From (3) we have $x\in C_H(V)$, so $x\in\gg^{(2)}\leq\gap$. As $|[x,\vap]|\leq 2$ we have $x\in\qap$, and since $[x,\vap]\neq\yap$ we conclude that $[x,\vap]=1$. In this way we have $x\in\gg^{(4)}$ and, in particular, $x\in C_S(W)$. Set $R=\lan x^L\ran$. Then $Y_L\leq R\leq\lan x\ran D$. As $[x,P_1]\leq Y_H\leq Y_L$, and as $C_{D/Y_L}(P_1)=1$, lemma --- [FROM J-MODULES PRELIM LEMMAS] implies that $R/Y_L$ is of order $2$. But then $[x,Q_L]=Y_L$, contrary to $|S/C_S(x)|=2$. We conclude that no such element $x$ exists, and this yields the following result. \vskip .1in \noindent (4) $C_H(O^2(H))=Y_H$. \vskip .1in Let $d\in D-V$. In particular, we have $d\in C_{Q_H}(V)$. If $[d,Q_H]=Y_H$ then (2) implies that there exists $d'\in Vd$ such that $[d',O^2(H)]=1$, and we contradict (4). Thus $[d,Q_H]=V$, and it follows from (3) that $|[Q_L\cap Q_H,d]|\geq 16$. As $[d,Q_L]\leq Y_L$, of order $8$, we have a contradiction, and the lemma is proved. \qed \enddemo [THE REMAINDER OF THIS SECTION (AND THE NEXT SECTION) WILL BE A CLOSE COPY OF BERND'S PAPER ON N-GROUPS, SECTION 10, TO PRODUCE THE AMALGAM FOR $M_{12}$, $Aut(M_{12})$, the Tits group, OR $^2F_4(2)$. AFTER THAT, IMITATE MICHAEL IN ORDER TO PRODUCE THE GROUPS.] \enddocument