0$ and some isometry $\rho$ of $\Sigma$. We are free to replace $x$ by any $Inndiag(K)$-conjugate of $x$, and then since $x$ is a field or graph-field automorphism of $K$ we may take $x=\tau_0|_K$, for some $\tau_0\in\Phi\Gamma$. Moreover, we have $[\sigma,\tau_0]=1$ by 3.5(b), and $(\tau_0)^p\in\lan\sigma\ran$ by 3.5(c). Suppose that $\tau_0\in\Gamma\lan\sigma\ran$. By assumption, $x$ is not a graph automorphism of $K$, so $K$ is not a Chevalley group. If $K$ is a Ree-Suzuki group then $\Gamma=1$, and since $\tau_0\notin\lan\sigma\ran$ we conclude that $K$ is a Steinberg variation. Then $\rho\neq 1$, and since $\tau_0$ and $\sigma$ commute it follows that $\tau_0\in\lan\rho,\sigma\ran$. Then $d=p$, contrary to assumption. Thus, we conclude that $\tau_0\notin\Gamma\lan\sigma\ran$. We have $\Phi\Gamma/\lan\sigma\ran\Gamma\cong \Bbb Z_n$, and since $x^p=1$ it now follows that $p$ divides $n$. Write $n=pm$ and set $\psi_1=\psi^m$. Suppose that $\rho=1$. We then have $(\psi_1)^p=\sigma$, and $\tau_0\in\lan\psi_1\ran\Gamma$. Write $\tau_0=(\psi_1)^k\gamma$, where $\gamma\in\Gamma$. As $[\Phi,\Gamma]=1$ we conclude that $|\gamma|=1$ or $p$, so there exists an integer $\ell$ with $\gamma^{k\ell}=\gamma$. We then take $\tau=\psi_1\gamma^{\ell}$, and obtain $\tau^p=\sigma$ and $\tau^k=\tau_0$. Thus, (1) holds in this case. On the other hand, suppose that $\rho\neq 1$. Then $p$ does not divide $|\rho|$, by assumption, and so there exists $\gamma\in\lan\gamma_{\rho}\ran$ with $\gamma^p=\rho$. Setting $\tau=\psi_1\gamma$, we then have $\tau^p=\sigma$. Any homomorphic image of $\Phi\times\lan\gamma_{\rho}\ran$ has at most one subgroup of order $p$, so $x\in\lan\tau|K\ran$, and thus (1) holds in any case. \qed \enddemo We next consider centralizers of semisimple elements. \proclaim {Lemma 3.7} Let $K$ be a simple group of Lie type in characteristic $r$, and let $(\bar K,\sigma)$ be a $\sigma$-setup of $K$. Identify $Inndiag(K)$ with $C_{\bar K}(\sigma)$, and let $x\in Inndiag(K)$ with $|x|$ prime to $r$. Then the following conditions are equivalent. \roster \item "{(1)}" $C_{\bar K}(x)$ contains a non-identity unipotent element. \item "{(2)}" $O^{r'}(C_K(x))\neq 1$. \item "{(3)}" $O^{r'}(C_K(x))$ is a product $L_1\cdots L_n$, ($n\geq 1$), where each $L_i$ is a group of Lie type in characteristic $r$, and where $[L_i,L_j]=1$ for all $i$ and $j$ with $i\neq j$. \endroster \endproclaim \demo {Proof} Set $\bar C=C_{\bar K}(x)$. As $|x|$ is relatively prime to $r$, x is a semisimple element of $\bar K$, and hence $\bar C$ is closed, connected and reductive. Set $\bar L=[\bar C,\bar C]$. Thus $\bar C=Z(\bar C)\bar L$, where $\bar L$ is closed, connected and semisimple, and where $Z(\bar C)$ is a torus. Then $\bar L$ contains all of the unipotent elements of $\bar C$. Denote by $\Cal M$ the set of normal, simple algebraic subgroups of $\bar M$. Then $\bar L$ is the commuting product of the members of $\Cal M$. If $\Cal M$ is non-empty, we write $\Cal M=\{\bar M_i\}_{1\leq i\leq t}$. Notice that if $L=1$ then $C_K(x)=C_{\bar C}(\sigma)$ consists of semisimple elements, so that $O^{r'}(C_K(x))=1$. Thus (2) implies (1). Suppose that $\Cal M$ is non-empty. That is, assume that (1) holds. As $\sigma$ commutes with $x$, $\bar C$ is $\sigma$-invariant and $\sigma$ then induces a permutation action on $\Cal M$. Let $\Cal M_1\cdots\Cal M_n$ be the set of orbits for $\sigma$ on $\Cal M$, and assume (without loss) that indices have been chosen so that $\Cal M_1=\{\bar M_i\}_{1\leq i\leq k}$. Since any positive power of a Steinberg endomorphism is again a Steinberg endomorphism, it follows that $\sigma^k$ induces a Steinberg endomorphism on each $\bar L_i$, $1\leq i\leq k$. For such $i$, set $M_i=O^{r'}(C_{\bar M_i}(\sigma^k))$. Then each $M_i$ is a group of Lie type in characteristic $r$, by definition. Now set $M=M_1\cdots M_k$. Then $M/Z(M)$ is the direct product of the images in $M/Z(M)$ of the groups $M_i$, $1\leq i\leq k$, and the action of $\sigma$ on $M/Z(M)$ is given by the transitive permuting of these factors. Set $L_1=O^{r'}(C_{\bar M}(\sigma))$. It now follows that $L_1=O^{r'}(C_M(\sigma))$ is isomorphic to a quotient of $M_1$ by a subgroup of $Z(M_1)$. We repeat this procedure for the remaining $\sigma$- orbits, obtaining the groups $L_1$ through $L_n$. Now set $L=O^{r'}(C_{\bar L}(\sigma))$. Since central quotients of groups in $Lie(r)$ are also in $Lie(r)$, we conclude that $L$ is the pairwise commuting product of the groups $L_j$, $1\leq j\leq n$, where each $L_j$ is a member of $Lie=(r)$. On the other hand, we have $C_{\bar K}(\sigma)=C_{\bar T}(\sigma)K$ for some $\sigma$-invariant maximal torus $\bar T$, and then $L=O^{r'}(C_{\bar K}(\sigma,x))=O^{r'}(C_{C_{\bar K}(\sigma)}(x))=O^{r'}(C_K(x))$. Thus (3) holds. Clearly, (3) implies (2), and thus the lemma is proved. \qed \enddemo \proclaim {Lemma 3.8} Let $r$ be a prime and let $\bar K$ be a simple linear algebraic group over an algebraic closure $\bar F$ of the field $F=\Bbb F_r$ of $r$ elements. Let $\sigma$ be a Steinberg endomorphism of $\bar K$, and set $K=O^{r'}(C_{\bar K}(\sigma))$ (so that $K$ is a group of Lie type in characteristic $r$). Let $g\in K$, and assume that either \roster \item "{(i)}" $|g|=2$ and $\bar K$ is not of type $A_1$, or \item "{(ii)}" $|g|=3$ and $\bar K$ is not of type $A_1$ or $A_2$. \endroster Then $C_{\bar K}(g)$ contains a non-identity unipotent element. \endproclaim \demo {Proof} We may assume that $|g|\neq r$ as otherwise the result holds trivially. Thus $g$ is a semisimple element of $\bar K$, and so there is a maximal torus $\bar T$ of $\bar K$ containing $g$. Let $\Sigma$ be the root system for $\bar K$ given by $\bar T$, $\Pi$ a fundamental system in $\Sigma$, $\bar B$ the corresponding Borel subgroup, and $\bar U$ the unipotent radical of $\bar B$. Recall that each $\alpha\in\Sigma$ is a homomorphism of $\bar T$ into $\bar F^{\times}$, and that there is then a $\bar T$-invariant subgroup $\bar U_{\alpha}$ of $\bar U$ and a parametrization $$ x_{\alpha}:\bar F^{\times}\longrightarrow \bar U_{\alpha} $$ such that $gx_{\alpha}(t)g^{-1}=x_{\alpha}(\alpha(g)t)$ for all $t\in \bar F^{\times}$. We aim to show that, under the conditions given in (a) and (b), there exists a root $\alpha$ such that $\bar U_{\alpha}\leq C_{\bar K}(g)$. Suppose false, and suppose first that $|g|=2$. Here $\alpha(g)^2=1$, so we have $\alpha(g)=-1$ for all $\alpha\in \Sigma$. Assuming that $\Sigma$ is not $A_1$, there exist two roots $\alpha$ and $\beta$ whose sum is a root, and we then have $(\alpha + \beta)(g)= \alpha(g)\beta(g)=1$, for a contradiction. Now suppose that $|g|=3$, and let $\omega$ be a primitive cube root of unity in $\bar F$. Then $\alpha(g)=\omega$ or $\omega^{-1}$ for all roots $\alpha$. It follows that $\alpha(g)=\beta(g)\neq 1$ whenever $\alpha$ and $\beta$ are roots whose sum is again a root. In particular, we may assume that $\alpha(g)=\omega$ for all $\alpha\in\Pi$. If $\Sigma$ has more than one root length then there are fundamental roots $\alpha$ and $\beta$ such that $\alpha + 2\beta$ is a root, and we obtain $(\alpha + 2\beta)(g)=1$ in that case. Also, if the rank of $\Sigma$ is at least $3$ then there exist fundamental roots $\alpha$, $\beta$, and $\gamma$ whose sum is a root, yielding $(\alpha +\beta +\gamma)(g)=1$. Thus, $\Sigma$ is of rank at most two, and $\Sigma$ has only one root length. That is, $\Sigma$ is $A_1$ or $A_2$. \qed \enddemo \proclaim {Lemma 3.9} Let $K$ be a simple group of Lie type in characteristic $r$, let $p$ be a prime different from $r$, and let $x$ be an element of order $p$ in $Inndiag(K)$. Suppose that $x$ is contained in a non-cyclic abelian $p$-subgroup of $Inndiag(K)$. Then there exists an element $y$ of order $p$ in $C_K(x)$ such that $O^{r'}(C_K(y))\neq 1$. \endproclaim \demo {Proof} Let $E$ be an elementary abelian subgroup of $Inndiag(K)$ of order $p^2$, containing $x$, and let $(\bar K,\sigma)$ be a $\sigma$-setup for $K$. Let $\bar T$ be a maximal torus of $\bar K$ containing $E$, let $\bar B$ be a Borel subgroup of $\bar K$ containing $\bar T$, let $\Sigma$ be the root system defined by $\bar T$ and $\bar B$, and let $\alpha\in\Sigma$. Then $\alpha$ is a homomorphism of $\bar T$ into the multiplicative group of an algrebaic closure of $\Bbb F_r$. The image of $\alpha$ is then cyclic, and so there exists a non-identity element $y\in E\cap Ker(\alpha)$. This means that $C_{\bar K}(y)$ contains the root subgroup of $\bar B$ corresponding to $\alpha$. The desired result then follows from 3.7. \qed \enddemo We next consider normalizers of $r$-groups in groups $K$, $K\in Lie(r)$. \proclaim {Lemma 3.10 (Borel-Tits)} Let $K\in Lie(r)$ and let $R$ be a non-identity $r$-subgroup of $K$. Then there is a parabolic subgroup $P$ of $K$ such that $R\leq O_r(P)$ and $N_K(R)\leq P$. \endproclaim \demo {Proof} This result, proved first in [BT], appears as [GLS3, Theorem 3.1.3(a)]. \enddemo Recall that a group $G$ is said to be {\bf $r$-constrained} if $C_G(O_r(G))\leq O_r(G)$. \proclaim {Lemma 3.11} Let $K\in Lie(r)$, let $X$ be a subgroup of $Aut(K)$ containing $Inn(K)$, and let $R$ be a non-identity $r$-subgroup of $K$. Then the following hold. \roster \item "{(a)}" Both $C_X(R)$ and $N_X(R)$ are $r$-constrained. \item "{(b)}" If $R=O_r(N_K(R))$ then the group $P=N_K(R)$ is a parablic subgroup of $K$, and $R=O_r(P)$. \endroster \endproclaim \demo {Proof} See [GLS3, Corollaries 3.1.4 and 3.1.5]. \qed \enddemo We now review the Schur multipliers of the groups of Lie type. \proclaim {Proposition 3.12} Let $K$ be a simple group of Lie type, in characteristic $r$, and let $\widehat K$ be the universal, perfect central extension of $K$. Set $Z=Z(K)$. Then $Z=Z_c\times Z_e$, where $Z_c$ (the \lq\lq canonical" part of $Z$ is isomorphic to the quotient group $Outdiag(K)=Inndiag(K)/Inn(K)$, and where $Z_e$ (the \lq\lq exceptional" part of $Z$) is equal to $O_r(Z)$. Moreover, we have $Z_e=1$ except in the following cases. \roster \item "{(a)}" $|Z_e|=2$, and $K$ is isomorphic to $L_2(4)$, $L_3(2)$, $Sp(4,2)'$, $L_4(2)$, $Sp(6,2)$, $U_4(2)$, $F_4(2)$, or $G_2(4)$. \item "{(b)}" $Z_e\cong\Bbb Z_2\times\Bbb Z_2$, and $K$ is isomorphic to $U_6(2)$, $D_4(2)$, $Sz(8)$, or $^2E_6(2)$. \item "{(c)}" $Z_e\cong\Bbb Z_4\times\Bbb Z_4$, and $K$ is isomorphic to $L_3(4)$. \item "{(d)}" $|Z_e|=3$, and $K$ is isomorphic to $L_2(9)$, $Sp(6,3)$, or $G_2(3)$. \item "{(e)}" $Z_e\cong\Bbb Z_3\times\Bbb Z_3$, and $K$ is isomorphic to $U_4(3)$. \endroster \endproclaim \demo{Proof} The relevant references are [St3], and [Gr]. See also chapter 6 of [GLS3]. \qed \enddemo \proclaim {Lemma 3.13} Let $K$ be one of the groups $Sz(8)$, $L_3(4)$, $D_4(2)$, or $U_6(2)$, and let $\a$ be an outer automorphism of $K$ of order $3$. Define $\wh K$ and $Z_e$ as in 3.12. Then $\a$ lifts to an automorphism of $\wh K$ which acts faithfully on $Z_e$. \endproclaim \demo {Proof} The result is contained in [GLS3, Theorem 6.3.1], but we present an altwernative proof here. Assume that $\alpha$ acts trivially on $Z(K^*)$, let $K_1$ be a perfect central extension of $K$ by $\Bbb Z_2$, and view $\alpha$ as an automorphism of $K_1$. Suppose first that $K\cong L_3(4)$. Let $P$ be an $\alpha$- invariant maximal parabolic subgroup of $K$. Then $P$ is a semidirect product of $SL(2,4)$ with the natural $SL(2,4)$-module, and $\alpha$ centralizes a complement to $O_2(P)$ in $P$. Let $P_1$ denote the pre-image of $P$ in $K_1$. Then $O_2(P_1)$ is elementary abelian, since $P$ acts transitively on the non-identity elements of $O_2(P)$. Further, we may choose $K_1$ so that $P_1$ has a subgroup isomorphic to $SL(2,4)$, since $SL(2,4)$ has no perfect central extension by $\Bbb Z_2 \times\Bbb Z_2$. But now $[O_2(P_1),\alpha](C_{P_1}(\alpha))'$ is isomorphic to $P$, and is a complement to $Z(K_1)$ in $P_1$. Gasch\"utz's Theorem [A2, result(10.4)] then implies that $K_1$ splits over $Z(K_1)$, and we have a contradiction. Suppose next that $K\cong U_6(2)$. Then $K\langle \alpha \rangle$ has a subgroup $P$ of the form $2^{1+8}_+:(U_4(2)\times 3)$. The Schur multiplier of $U_4(2)$ contains no fours group, so there is a perfect central extension of $K\langle \alpha \rangle$ by $\Bbb Z_2$ in which $P$ lifts to a group $P_1$ having a sugroup $U_4(2)\times 3$. By Lemma 3.2, above, $O_2(P_1)$ splits over $Z(K_1)$, and as in the case of $L_3(4)$ we find that $[O_2(P_1),\alpha](C_{P_1}(\alpha ))'$ is a complement to $Z(K_1)$ in $P_1$, andr a contradiction is reached as before. Suppose that $K\cong Sz(8)$. Let $S$ be an $\alpha$-invariant Sylow $2$-subgroup of $K$ and put $U=\Omega_1(S)$. Then $U=Z(S)$, and $N_K(S)$ acts transitively on the set of non-identity elements of $U$. Let $S_1$ and $U_1$ denote the pre-images of $S$ and $U$, respectively, in $K_1$. It follows at once that $U_1$ is elementary abelian. We have $C_S(\alpha)\cong \Bbb Z_4$, and since all involutions in $S_1$ lie in $U_1$ it then follows that $C_{S_1}(\alpha)\cong \Bbb Z_4 \times \Bbb Z_2$. Let $u\in C_{U_1}(\alpha)$ with $u\notin Z(K_1)$. Then $C_{S_1}(u)$ is $\alpha$-invariant, of index at most $2$ in $S_1$, and containing $C_{S_1}(\alpha)$. We conclude that in fact $u\in Z(S_1)$, and hence $U_1=Z(S_1)$. Now let $U^*$ and $S^*$ be the inverse images of $U$ and $S$ in the full covering group $K^*$ of $K$, and let $X$ be a subgroup of $N_{K^*}(S^*)$ of order $7$. It follows from the fore-going that $U^*=Z(S^*)$. Let $g\in S^*-U^*$. Then $g^2=yz$ where $y\in [U^*,X]$ and where $z\in Z(K^*)$. Without loss, we may assume that $K_1$ was chosen to begin with so that $z$ projects to the identity element of $K_1$. Taking $g_1$ for the image of $g$, and $X_1$ for the image of $X$ in $K_1$, we then have $(g_1)^2\in [U_1,X_1]$, and then $\Phi(S_1)=[U_1,X_1]$. Thus $S_1$ splits over $Z(K_1)$, with a contradiction as before. Suppose that $K\cong D_4(2)$. In order to analyze this group we will require the detailed structure of the group $P=V:L$, where $L\cong Alt(8)$ and where $V\cong 2^6$ is the unique non-trivial constituent in the permutation module for $L$ over $\Bbb F_2$. Here $V$ may be described as follows. Put $\Omega=\{1,2,\cdots,8\}$ and let $\Cal E$ be the $\Bbb F_2$-space of all even-cardinality subsets of $\Omega$, with addition given by symmetric difference. We may then identify $V$ with $\Cal E/\langle \Omega \rangle$, with natural action by $L$. We require the following facts. \vglue .2in \item {(1)} $P$ is isomorphic to a maximal parabolic subgroup of $D_4(2)$. \vglue .2in \item{(2)} $L\cong \Omega^+_6(2)$ and $V$ is isomorphic to the natural orthogonal module for $L$. Moreover, the singular vectors correspond to the four-element subsets of $\Omega$. \vglue .2in \item{(3)} We have $H^1(L,V)\cong \Bbb Z_2$. (Up to isomorphism, $\Cal E$ is the unique indecomposable $L$-module of order $2^7$ with quotient module $V$.) \vskip .1in The next two results are easily computed from the above information. \vglue .2in \item{(4)} Let $S$ be a Sylow $2$-subgroup of $P$. Then $S$ has exactly three elementary abelian subgroups of order $2^6$. They are $V$, $A_1$, and $A_2$, where $|A_i\cap V|=|A_i\cap L|=8$, and $N_L(A_i\cap L)\cong 2^3:L_3(2)$. \vglue .2in \item{(5)} In the semidirect product $\Cal E:L$, the pre-image of each $A_i$ is an extraspecial group. \vskip .1in From the $D_4$ diagram, and from (4), we obtain the following fact. \vglue .2in \item {(6)} Identify $S$ with an $\alpha$-invariant Sylow $2$-subgroup of $K$, and $P$ with a maximal parabolic subgroup of $K$. Then $\alpha$ permutes $\{V,A_1,A_2\}$ transitively. \vskip .1in With these facts in hand, one may prove that $\alpha$ acts non-trivially on the Schur multiplier of $K$. For, taking $K_1$ as in the previous cases, suppose first that $V$ lifts in $K_1$ to a group $V_1$ which is elementary abelian. Then (4) and (6) imply that the pre-image $L_1$ of $L$ is isomorphic to $L \times \Bbb Z_2$. The inverse image $P_1$ of $P$ is then isomorphic to $\Cal E:L$, as otherwise $Z(K_1)$ has a complement in $P$. Now (4) and (5) imply that the generalized Fitting subgroups of the pre-images in $K_1$ of the remaining two connected maximal parabolics over $S$ are extaspecial. Since $\alpha$ fuses these to $V_1$, we have a contradiction. We therefore conclude that $V_1$ is not abelian, and so $V_1$ is extraspecial. In the four-fold covering group $K^*$ the pre-image $V^*$ of $V$ is then of the form $2^{1+6}_+\times 2$, by Lemma 3.2. But then, taking $K^*/(V^*)'$ in place of $K_1$, we have a perfect double cover of $K$ in which the pre-image of $V$ is abelian, after all, and so we have a contradiction at this point. \qed \enddemo We end this section with a result which will be useful in determining the possible quadratic subgroups of sporadic groups. \proclaim {Lemma 3.14} Let $G$ be a finite group, put $X=O_2(G)$, and assume that $X=F^*(G)$ is an extraspecial $2$-group of width $n$ and sign $\epsilon$. Assume also that either: \roster \item $G/X\cong \Omega^{\epsilon}_{2n}(2)$, with $n\geq 3$ if $\epsilon=1$, or \item $G/X\cong SU(n,2)$, with $n\neq 2$, and with $\epsilon=(-1)^n$. \endroster Let $G^*$ be a group having a normal subgroup $\langle t \rangle$ of order $2$, with $G^*/\langle t \rangle \cong G$. Then $O_2(G^*)\cong X\times \langle t \rangle$. \endproclaim \demo{Proof} Put $M=X/Z(X)$. Then the squaring map from $M$ into $Z(X)$ defines a quadratic form $Q$ on $M$, with respect to which $M$ is a non-degenerate orthogonal space over $\Bbb F_2$, of Witt index $\epsilon$. If $G/X\cong \Omega^{\epsilon}_{2n}(2)$, it follows that $M$ may be identified with the natural $G/X$-module. If $G/X\cong SU(n,2)$, then $M$ may be identified with the natural $n$-dimensional hermitian module for $G/X$ over $\Bbb F_4$, whose hermitian form $h$ satisfies $h(v,v)=Q(v)$ for all $v\in M$. In both the cases (1) and (2), the singular vectors and the non-singular vectors in $M$ with respect to $Q$ each form a single orbit for the action of $G/X$. Denote by $Z$ the pre-image of $Z(X)$ in $G^*$. Also, denote by $\Cal D$ the set of subgroups $D$ of $G$ such that $[X,D]=[X,D,D]$ is a quaternion group. Thus, $\Cal D$ is a set of groups of order $3$, and since $n\geq 3$ if $G/X\cong \Omega^+(2n,2)$ it follows that $\Cal D$ is non-empty. Fix $D\in \Cal D$, and denote by $Y$ the inverse image of $\lan D^X\ran$ in $G^*$. Then $Y$ is isomorphic to $SL(2,3)\times \Bbb Z_2$. Denote by $z$ the involution in $Z(Y)$, set $G_0^*=C_{G^*}(z)$, and let $G_0$ be the image of $G_0^*$ in $G$. Then $|G:G_0|\leq 2$, and $z\neq t$. We now make the following claim. \vglue .1in \item {(*)} For any element $x^*$ of $O_2(G^*)$ whose image in $G$ has order $4$, we have $(x^*)^2=z$. \vglue .1in Suppose that (*) is not the case. As $G$ is transitive on the non-singular vectors in $M$, it follows that $G\neq G_0$, and that $G_0$ has two orbits on $\Cal D$. Let $\Cal D_0$ and $\Cal D_1$ be the two orbits for $G_0$ on $\Cal D$. Then, for any $D_0\in\Cal D_0$ and any $D_1\in\Cal D_1$, we have $|[X,D_0]\cap [X,D_1]|=2$, and hence $[X,D_0]$ commutes with $[X,D_1]$. Thus $[X,\lan\Cal D_0\ran]$ commutes with $[X,\lan\Cal D_1\ran]$, and so each $[X,\lan\Cal D_i\ran]$ is a proper subgroup of $X$. In particular, it follows that $\lan\Cal D_i\ran\neq O^2(G)$, and hence $G/X\cong \Omega_4^+(2)$ or $SU(2,2)$. These two cases are excluded by the conditions placed on $n$ in (1) and (2), so (*) holds. Now let $s\in X$ be an involution. By transitivity of $G/X$ on singular points, we may assume that $Y$ was chosen so that $\lan D^X,s \ran \cong SL(2,3)\circ\Bbb Z_4$ (the central product). Denote by $L$ the pre-image of $\lan D^X,s \ran$ in $G^*$. Then $O_2(L)$ is not isomorphic to $Q_8\times\Bbb Z_4$, by (*), and therefore $s$ lifts to an involution in $G^*$. This shows that $\{g^2 : g\in O_2(G^*)\}$ is of cardinality $2$, and hence $|\Phi(O_2(G^*))|=2$. This yields the lemma. \qed \enddemo \vskip .3in \noindent {\bf Section 4: Automorphisms, and alternating groups} \vskip .2in Our aim in this section is to prove the following result. \proclaim {Proposition 4.1} Let $G$ be a minimal counterexample to Theorem A. Then either $\bar H$ is a sporadic group, or $\bar H\in Lie(r)$, $r\neq p$, and $A$ induces a group of inner-diagonal automorphisms of $\bar H$. \endproclaim \proclaim {Lemma 4.2} Assume Hypothesis 1.1, set $\bar G=G/Z(G)$, and assume that $\bar H$ is a group of Lie type in characteristic $r$, possibly with $r=p$. Assume further that $G$ is a minimal counter-example to Theorem A, and let $a\in A$. Then $a$ induces an inner-diagonal automorphism of $\bar H$. \endproclaim \demo {Proof} Denote by $\alpha$ the automorphism of $\bar H$ induced by $a$. By 3.5 we have $\alpha=xfg$ where $x$ is an inner-diagonal automorphism, and where $f$ and $g$ are field and graph automorphisms, respectively. We assume that $\alpha\neq x$, and our aim will be to derive a contradiction from this assumption. We proceed by induction on $|G|$. Suppose first that $\bar a$ is not contained in any $r$-local subgroup of $\bar G$. Then $r\neq p$, and $\alpha$ is not conjugate to $f$ in $Aut(\bar H)$. Then 3.6 implies that $p=3$, and that $\bar H\cong D_4(q)$ or $^3D_4(q)$ for some power $q$ of $r$. As $a$ is contained in an $SL(2,3)$ subgroup of $G$, $\bar a$ is in a $2$-local subgroup of $\bar G$, and so $r\neq 2$. Let $b$ be an element of order $3$ in $C_H(a)$, and set $L=O^{r'}(C_H(b))$. Then 3.8 and 3.7 together imply that $L\neq 1$ and that $L=L_1\cdots L_k$ is a commuting product of groups $L_i\in Lie(r)$. Moreover, as $r>3$, each $L_i$ is quasisimple. If $k\geq 3$ and $a$ permutes the factors $L_1$, $L_2$, and $L_3$, then $L$ contains an abelian $3'$-subgroup on which $a$ acts non-trivially, and contrary to 2.9. Thus, $a$ fixes each of the factors $L_i$. If $[L_1,a]=1$ then $a$ is in an $r$-local subgroup, so in fact $[L_1,a]\neq 1$. We note that $L_1\neq H$ since $O_3(G)=1$. By the induction hypothesis, $a$ induces an inner-diagonal automorphism on $L_1$, and then since $r\neq 2$, induction in Theorem A implies that $r=5$ and $L_1\cong SL(2,5)$. Now 2.5 shows that either $L_1$ or $C_{L_1\lan a\ran}(L_1)$ contains a quadratic element of order $3$. As neither $D_4(q)$ nor $^3D_4(q)$ occur as outcomes in Theorem A, we conclude, by induction, that $C_{L_1\lan a\ran}(L_1)$ contains a quadratic element $a_1$ of order $3$. Now $a_1$ lies in an $r$-local subgroup of $G$, and we may replace $a$ by $a_1$. That is, we may assume from the beginning, and without loss of generality, that $a$ is in an $r$-local subgroup of $G$. As $r$-local subgroups of $\bar G$ are $r$-constrained, by 3.11, 2.9 implies that $r=p$, or $r=2$ and $p=3$. Write $\bar H={^d\Sigma(q)}$ as in section 3, and suppose first that $r=p$. As $O_p(G)=1$, 3.12 implies that $H\in Lie(r)$, and then any irreducible $\Bbb F_pH$-module is the restriction to $H$ of an irreducible module for $\Sigma(q)$, by [St1, Theorem 13.3]. Here $V$ is irreducible for $H$, by [Ch, Lemma 1.3], so we may now assume that $\bar H$ is a Chevalley group. If $g=1$ then $\bar H$ may be taken to be $L_2(q)$, $q=r^{pm}$, and we then violate 2.9 via the action of $a$ on a Cartan subgroup of $H$. On the other hand, suppose that $g\neq 1$, so that $\bar H\cong D_4(q)$. As $a$ normalizes a Sylow $3$-subgroup of $H$, there is then an $a$-invariant maximal subgroup $M$ of $H$ with $M/O_3(M)$ isomorphic to a commuting product of three copies of $SL(2,q)$, permuted transitively by $a$. There is a section $W$ of $V$ which centralizes $O_3(M)$ and on which $(M/O_3(M))\lan a\ran$ acts faithfully. By 2.9, applied to the action of $a$ on $Z(M/O_3(M))$, $M/O_3(M)$ is a central product (with center of order $2$). Then $a$ acts on a central product of three quaternion groups in $M/O_3(M)$, permuting the factors, and then once again there is an abelian $2$-group on which $a$ acts non-trivially. Thus 2.9 is violated in any case, and we conclude that $r\neq p$. We now have $r=2$ and $p=3$. Suppose next that $\alpha=f$. There is then an $a$-invariant subgroup $L$ of $H$, of Lie rank $1$, such that $a$ induces a field automorphism on $L$. By induction, it follows that $\bar H$ itself has Lie rank $1$. If $H\cong U_3(2^{3m})$ then again there is an $a$-invariant subgroup of $H$ isomorphic to $L_2(2^{3m})$, and on which $a$ induces a field automorphism, contrary to induction. If $H\cong L_2(2^{3m})$ or $Sz(2^{3m})$ then $a$ acts non-trivially on the center of a Sylow $2$-subgroup of $H$, contrary to 2.9. Thus $Z(H)\neq 1$, and indeed the preceding argument via 2.9 shows that $Z(H)$ contains a non-identity $2$-group. Then $\bar H\cong Sz(8)$, by 3.12, and then 3.13 implies that $a$ acts non-trivially on $Z(H)$. Again, this outcome is contrary to 2.9. We conclude that $\alpha\neq f$. Then 3.6 yields either $\bar H\cong D_4(q)$ and $g\neq 1$, or $\bar H\cong {^3D_4(q)}$. If $Z(H)\neq 1$ then 3.12 and 3.13 yield $\bar H\cong D_4(2)$, $Z(H)\cong\Bbb Z_2\times\Bbb Z_2$, and $a$ acts non-trivially on $Z(H)$, contrary to 2.9. Thus $Z(H)=1$. As $a$ lies in an $SL(2,3)$-subgroup of $G$, there exists a maximal $2$-local subgroup $M$ of $H\lan a\ran$ containing $a$. Set $L=O^{2'}(M)$. Then $M\cap H$ is a parabolic subgroup of $H$, and $[Z(O_2(M)),a]=1$. In particular, we have $[Z(S),a]=1$ for some Sylow $2$-subgroup $S$ of $L$, and $S$ is also a Sylow $2$-subgroup of $G$. We may then choose $M$ so that $M\cap H$ is the maximal parabolic subgroup $N_H(Z(S))$, and then $L/O_2(L)$ is isomorphic to a direct product of three copies of $L_2(q)$ permuted transitively by $a$ (in the $D_4(q)$ case), or to $L_3(q^3)$ (in the $^3D_4(q)$ case). A Sylow $3$-subgroup of $L\lan a\ran$ is then contained in a complement to $O_2(L)$, so in either case we find that $a$ acts non-trivially on an abelian $2$-subgroup of $L$. Again 2.9 yields a contradiction, and the lemma is thereby proved. \qed \enddemo \proclaim {4.3 Lemma} Assume Hypothesis 1.1, and suppose that $\bar G$ is an alternating group of degree $n$. Then $|Z(G)|=2$, and if $n$ is not equal to $6$ then $|A|=3$, and the non-identity elements of $A$ project to $3$-cycles in $\bar G$. \endproclaim \demo {Proof} Suppose first that $p>3$. There is then a quasisimple subgroup $K$ of $G$ with $K/Z(K)\cong Alt(p)$ and with $K= [K,a]$. Then two conjugates of $a$ suffice to generate $K\langle a\rangle$, and then 2.7(a) implies that $p=5$ and $K\cong SL(2,5)$. If $K$ is contained in an $a$-invariant subgroup $L$ of $G$ with $L/Z(L)\cong Alt(6)$ then two conjugates of $a$ will generate $L\lan a\ran$, which is contrary to 2.7(a). Thus we may assume that $n$ is divisible by $5$. If $n=5$ then there is nothing more to prove, so we may reduce to the case where $\bar G\cong Alt(10)$ and where $\bar a$ is a product of two disjoint $5$-cycles. Here $a$ lies in a subgroup $L$ of $G$ of the form $SL(2,5)\circ SL(2,5)$ (central product with amalgamated centers) and two conjugates of $a$ will then generate a subgroup of $L$ isomorphic to $Alt(5)$. This is again contrary to 2.7(a). Thus, we need now only consider the case where $p=3$. By Hypothesis 1.1, $3$ does not divide $|Z(G)|$, and so a classical result of Schur implies that $|Z(G)|\leq 2$. Let $a\in A$, $a\neq 1$. Let $k$ be the number of $3$-cycles in the standard notation for $\bar a$, and suppose first that $k>1$. As $a$ lies in no Frobenius subgroup of $G$ of order $21$, by 2.9, we then have $n=3k$. Suppose $k\geq 3$, and let $\bar L$ be the stabilizer in $\bar G$ of $n-9$ points which are permuted by $\bar a$ in three $3$-cycles. Then $\bar L$ has a subgroup $\bar K$ isomorphic to $SL(2,8)$, acting on the nine points of the projective line. Any element of $\bar L$ of order $3$ is fixed-point-free on these points, so we can choose $\bar L$ to be $\bar a$-invariant. Denote by $L$ the pre-image of $\bar L$ in $G$, and set $L_0=[L,L]$. Then $L_0\cong SL(2,8)$ and $L_0=[L_0,a]$. But $SL(2,8)$ has no subgroup isomorphic to $SL(2,3)$, so we violate 2.7(a). Assuming now that $n\ne 6$, we conclude that every non-identity element of $A$ projects to a $3$-cycle in $\bar G$. Then $|A|=3$, and since $A$ lies in no subgroup of $G$ which is isomorphic to $Alt(4)$, we obtain $|Z(G)|=2$. On the other hand, if $n=6$ and $Z(G)=1$ then every element of order $3$ in $G$ lies in an $Alt(4)$-subgroup of $G$. Thus we conclude that, in any case, we have $|Z(G)|=2$. \qed \enddemo Notice that 4.1 follows from lemmas 4.2 and 4.3, given our background hypothesis that $\bar H$ is a \lq\lq known" simple group. \vskip .3in \noindent {\bf Section 5: The case $p>3$} \vskip .2in Our aim in this section is to give a short proof of the following result (which is is proved also in [Sa]). \proclaim {Theorem 5.1} Assume Hypothesis 1.1 with $p>3$. Then $G/Z(G)$ is a group of Lie type in characteristic $p$. \endproclaim We fix notation as in section 1, so that $H=F^*(G)$ is a quasisimple group, and we have $\bar G=G/Z(H)$. Assume Hypothesis 1.1, and fix a non-identity element $a$ of $A$. Take $G$ to be a minimal counter-example to Theorem 5.1. The following result is then immediate. \proclaim {Lemma 5.2} Let $K$ be a proper quasisimple subgroup of $H$, with $K=[K,a]$. Then $K/Z(K)$ is a group of Lie type in characteristic $p$. \endproclaim \proclaim {Lemma 5.3} $\bar H$ is not a sporadic group. \endproclaim \demo {Proof} Suppose false. Then the outer automorphism group of $\bar H$ is of order at most $2$, and so $G=H$. Suppose first that $|C_{\bar G}(\bar a)|$ is even. Let $\bar t$ be an involution in $C_{\bar G}(\bar a)$, and set $\bar C=C_{\bar G}(\bar t)$. As $p>3$, 2.9 implies that $F^*(\bar C)\neq O_2(\bar C)$, and it follows from [GLS3, Table 5.3] that $\bar C$ has a component $\bar K$ with $\bar a\in\bar K$. By 5.2, $\bar K/Z(\bar K)$ is of Lie type in characteristic $p$, and then [GLS3, Table 5.3] yields $p=5$, $\bar K\cong Alt(5)$, and $\bar G\cong M_{12}$, $J_1$, or $J_2$. The inverse image $K$ of $\bar K$ in $G$ is then isomorphic to $SL(2,5)$, by 2.7, so $Z(G)\neq 1$, and so $G\cong 2^{\cdot}M_{12}$ or $2^{\cdot}J_2$. In fact, in both these cases the cited table in [GLS3] gives the extra information that $K\cong \Bbb Z_2\times Alt(5)$, and so we may obtain a contradiction in this way. Alternatively, one may note that $M_{12}$ has cyclic Sylow $5$-subgroups and contains $Alt(6)$, so that we contradict 2.7 in this case. In the case that $\bar G\cong J_2$, we have $\lan\bar a\ran$ contained in a subgroup isomorphic to $Alt(4)\times Alt(5)$ (the unique maximal subgroup of $\bar G$ containing $\bar C$), and thus $|C_{\bar G}(\bar a)|$ is divisible by $3$. Of the two classes of subgroups of order $5$ in $\bar G$, $\lan\bar a\ran$ is then identified as lying in a subgroup isomorphic to $3^{\centerdot}Alt(6)$, contrary to 2.7. We conclude that $C_{\bar G}(\bar a)$ is of odd order. Now 2.8 implies that $G$ has non-cyclic Sylow $p$-subgroups. Another trip through the cited table in [GLS3] shows, however, that for any element $g$ of prime order $p$ in a sporadic group $X$, if $X$ has non-cyclic Sylow $p$-subgroups then $C_X(g)$ is of even order. This yields the desired contradiction. \qed \enddemo \proclaim {Lemma 5.4} $G$ is not isomorphic to $SL(2,r^n)$ for any $n$. \endproclaim \demo {Proof} Suppose false, and put $q=r^n$. From L.E. Dickson's determination of the subgroups of $L_2(q)$ (for which one may see [Suz]), we know that, for $p\neq r$, the only possible $SL(2,p)$ subgroups of $G$ are given by $p=5$. Here $5$ must divide $q+1$, as otherwise $a$ acts non-trivially on a Sylow $r$-subgroup of $G$, in violation of 2.9. Put $N=N_G(\langle a \rangle)$, and let $\Cal L$ be the set of all subgroups $L$ of $G$ containing $a$ and with $L\cong SL(2,5)$. Each $L\in \Cal L$ has precisely $50$ elements $g$ such that $L=\langle a,a^g \rangle$. From this we obtain: $$ |\Cal L|=(|G|-|N|)/50 =(q^3-3q-2)/50 $$ Since each $L\in \Cal L$ has exactly five Sylow $5$-subgroups other than $\langle a\rangle$ it then follows that $G$ has exactly $1+(q^3-3q-2)/10$ Sylow $5$-subgroups. But this number is then equal to $|G:N|$, so we obtain: $$ 1+\frac{1}{10}(q^3-3q-2) = \frac{1}{2}(q^2-q) $$ which yields $q^3-5q^2+2q+8=0$. That is: $(q-4)(q-2)(q+1)=0$, so that $q=4$. As $SL(2,4)$ does not contain $SL(2,5)$ we have a contradiction at this point. \qed \enddemo \proclaim {Lemma 5.5} $\bar H$ is not a group of Lie type in characteristic $r$ different from $p$. \endproclaim \demo {Proof} Suppose false. By 4.1, $A$ induces a subgroup of $Inndiag(\bar H)$. Let $1\neq a\in A$, and suppose first that $|C_{\bar H}(\bar a)|$ is of even order. Let $\bar t$ be an involution in $C_{\bar H}(\bar a)$. If $r=2$ then 3.11 implies that $C_{\bar G}(\bar t)$ is $2$-constrained, and we contradict 2.9. Thus $r\neq 2$. If $\bar H\cong PSL(2,r^n)$ then $H\cong SL(2,r^n)$ since $H$ involves $SL(2,p)$, and we then contradict 5.4. Thus $\bar H$ is not isomorphic to $PSL(2,r^n)$, and then by 3.8 and 3.7 there is a subnormal subgroup $\bar K$ of $C_{\bar H}(\bar t)$, with $\bar K$ of Lie type in characteristic $r$. As $p>3$ and $r\neq 2$ there are no isomorphisms between any members of $Lie(r)$ and $Lie(p)$, and so $\bar K\notin Lie(p)$. If $[\bar K,\bar a]=1$ then $\bar a$ is in an $r$-local subgroup of $\bar G$, and we again contradict 2.9 via the Borel-Tits theorem. Thus, $[\bar K,\bar a]\neq 1$. Then 5.2 implies that $\lan(\bar K)^{\lan \bar a\ran}\ran$ is a product of $p$ components of $C_{\bar H}(\bar t)$, or a commuting product of $p$ copies of $SL(2,3)$ or of $L_2(3)$. Again, the result is that $\bar a$ lies in an $r$-local subgroup of $\bar G$, and a contradiction ensues. We therefore conclude that $C_{\bar G}(\bar a)$ is of odd order. Now 2.8 shows that $G$ has non-cyclic Sylow $p$-subgroups. As $p$ is odd, there is then an elementary abelian subgroup $\bar B$ of $\bar G$ of order $p^2$, with $\bar a\in\bar B$. Then 3.9 implies that $O^{r'}(C_{\bar H}(\bar b))\neq 1$ for some $\bar b\in\bar B$. By 3.7 we then have $O^{r'}(C_{\bar H}(\bar b)=\bar L_1\cdots\bar L_m$, where each $\bar L_i$ is a group of Lie type in characteristic $r$, and where $[\bar L_i,\bar L_j]=1$ for all $i\neq j$. Here $O^{r'}(C_{\bar H}(\bar b))$ is $\bar a$-invariant, and since we have already seen that $\bar a$ lies in no $r$-local subgroup of $\bar G$, we conclude that each $\bar L_i$ is $\bar a$-invariant and that $[\bar L_i,\bar a]\neq 1$. Now 5.2 implies that each $\bar L_i$ is solvable, and since $p>3$ it then follows that $\bar L_i$ has no automorphisms of order $p$. Then $[\bar L,\bar a]=1$, and we have a contradiction. \qed \enddemo Notice that lemma 4.3, and lemmas 5.2 through 5.5, yield Theorem 5.1. \vskip .3in \noindent {\bf Section 6: Cross-Characteristic Lie Type Groups, $p=3$} \vskip .2in In this section we assume Hypothesis 1.1 with $p=3$. As always, we set $\bar G=G/Z(G)$ and $H=E(G)$ We shall assume further that $\bar H$ is a group of Lie type in characteristic different from $3$. Indeed, we even wish to assume that there exists no exceptional isomorphism of $\bar H$ with a group of Lie type in characteristic $3$. Thus, $\bar H$ is not isomorphic to $Sp(4,2)'\ (\cong L_2(9))$, $G_2(2)'\ (\cong U_3(3))$, or $U_4(2)\ (\cong PSp(4,3))$. By a \lq\lq parabolic subgroup" of $H$, we mean the complete inverse image in $H$ of a bona fide parabolic subgroup of $H/Z(H)$. Similarly, we have the notions of \lq\lq Borel subgroup", \lq\lq Cartan subgroup", and of \lq\lq root group" in $H$. \vskip .1in Our goal, in this section, is the following result. \proclaim {6.1 Theorem} Assume Hypothesis 1.1, with $H/Z(H)$ a group of Lie type, and not isomorphic to a group of Lie type in characteristic $3$. Then either $G$ is isomorphic to one of the groups $PGU(n,2)$, $n\geq 5$, or else $|Z(G)|=2$, and $\bar G$ is isomorphic to one of the groups $L_2(4)$, $L_4(2)$, $Sp(6,2)$, $D_4(2)$, or $G_2(4)$. Moreover, we have $|A|=3$ in every case. \endproclaim For the remainder of this section, let $G$ be a minimal counter-example to Theorem 6.1. Throughout, let $r$ denote the defining characteristic of $\bar H$, $r\neq 3$, and fix a non-identity element $a\in A$. \proclaim {Lemma 6.2} Suppose that the Lie rank of $H/Z(H)$ is equal to $1$. Then $G\cong 2^{\centerdot}L_2(4)$. \endproclaim \demo {Proof} By 4.2, $A$ induces inner-diagonal automorphisms on $\bar H$. Thus $|Inndiag(\bar H)|$ is divisible by $3$, and so $\bar H$ is not isomorphic to $Sz(2^n)$. Also, as $r\neq 3$, by assumption, $\bar H$ is not a Ree group in characteristic $3$. Thus $\bar H\cong PSL(2,q)$ or $PSU(3,q)$ for some $q$, $q=r^n$. Suppose first that $\bar H\cong PSL(2,q)$. Then $|Inndiag(H):\bar H|\leq 2$, so $A\leq H$, and so $H=G$. As $G$ involves $SL(2,3)$, we conclude that $|Z(G)|=2$. Assuming that $\bar G$ is not isomorphic to $SL(2,4)$, it follows from 3.12 that $r$ is odd. Thus $r\geq 5$, and since $2^{\cdot}SL(2,4)\cong SL(2,5)$ we have $q>5$. Put $d=q-1$, and let $\lambda$ be a primitive $d^{th}$ root of unity in $\Bbb F_q$. Then take: $$ a=\pmatrix \format\r&\quad\r\\ -1 & -1 \\ 1 & 0 \endpmatrix \qquad b=\pmatrix 0 & \lambda^{-1}\\ -\lambda & -1 \endpmatrix $$ and obtain: $$ ab=\pmatrix \lambda & 1-\lambda^{-1}\\ 0&\lambda^{-1} \endpmatrix $$ Here $b$ is of order $3$, so $b$ is conjugate to $a$ in $G$. By 2.6, we have $\lan a,b\ran$ isomorphic to $SL(2,3)$ or $SL(2,5)$, and so $|ab|\leq 5$. But $ab$ has order $q-1$, so we conclude that $q\leq 6$, and then $q=5$, contrary to our choice of $q$. Suppose next that $\bar H\cong PSU(3,q)$. Let $L_0$ be a subgroup of $H$ with $L_0=\lan a^{L_0}\ran\cong SL(2,3)$. Let $t$ be the involution in $L_0$, and set $L=O^{r'}(C_H(t))$. Then $L\cong SL(2,q)$, and $[L,a]\neq 1$. By what has already been shown in the preceding paragraph, we then have $q=5$. Both $SU(3,5)$ and $PGU(3,5)$ have extraspecial Sylow $3$-subgroups of exponent $3$. In particular, all subgroups of order $3$ in $PGU(3,5)$ which are contained in $PSU(3,5)$ are conjugate. By the Frattini argument, the normalizer in $PGU(3,5)$ of a Sylow $5$-subgroup contains such an \lq\lq outer" subgroup of order $3$, so we must conclude from 2.9 that $a\in H$. One may deduce from the action of $SU(3,5)$ on its natural module that all subgroups of order $3$ in $PSU(3,5)$ are conjugate. As $PSU(3,5)$ contains a Frobenius group of order $21$, we again contradict 2.9. \qed \enddemo \proclaim {Lemma 6.3} $\bar H$ is not isomorphic to $PSL_3(q)$ for any power $q$ of $r$. \endproclaim \demo{Proof} Suppose $\bar H\cong PSL_3(q)$, $q=r^n$. As always, we have a subgroup $X$ of $G$ containing $a$, with $X\cong SL(2,3)$. Consider first the case where $Z(H)=1$. If $r$ is even, then the centralizer in $G$ of any non-identity $2$-subgroup of $G$ is contained in a parabolic subgroup of $G$, while if $r$ is odd then $G$ has a unique conjugacy class of involutions. In either case we find that $C_G(Z(X))$ is contained in a proper parabolic subgroup $P$ of $G$. Here $O_r(P)$ is abelian, and then 2.9 yields $a\in C_G(O_r(P))$, whereas $O_r(P)\geq C_G(O_r(P))$. We conclude that $Z(H)\neq 1$. As $G$ acts irreducibly on $V$, $|Z(H)|$ is prime to $3$, so 3.12 implies that $\bar H\cong L_3(4)$ and $Z(H)$ is a $2$-group. Since $a$ centralizes $Z(H)$, we conclude from 3.13 that $a\in H$. As all elements of order $3$ in $L_3(4)$ are conjugate, and as $L_3(4)$ contains a Frobenius group of order $21$, we contradict 2.9. This proves the lemma. \qed \enddemo \proclaim {Lemma 6.4} If the Lie rank of $\bar H$ is at least $2$ then $r=2$, and $a$ normalizes a maximal parabolic subgroup of $\bar H$. \endproclaim \demo {Proof} Assume that the Lie rank of $\bar H$ is at least $2$. Then $\bar a$ lies in an $r$-local subgroup $\bar N$ of $\bar G$, by 6.3 and 3.8. By 3.11(a), $\bar N$ is $r$-constrained, so 2.9 implies that $r=2$. By 3.11(b) we may choose $\bar N$ so that $\bar N\cap \bar H$ is a parabolic subgroup of $\bar H$. Let $\bar P$ be an $a$-invariant, proper parabolic subgroup of $\bar H$. As $a$ induces an inner-diagonal automorphism of $\bar H$ we may write $a=xd$ where $x\in P$ and where $d\in N_G(\bar S)$ where $\bar S$ is a Sylow $r$-subgroup of $\bar P$. Then any maximal parabolic subgroup of $\bar H$ containing $\bar P$ is $a$-invariant. \qed \enddemo For the remainder of this section we assume that the Lie rank of $G$ is at least $2$. Thus $r=2$, by 6.4. Fix a Borel subgroup $B$ of $H$, and let $\Sigma$ (resp. $\Sigma^+$) be the root system (resp. the positive subsystem) associated with $H$ and with $B$, so that $O_2(B)$ is generated by the root groups $X_{\a}$, $\a\in\Sigma^+$. If $P$ is a parabolic subgroup of $H$ containing $B$ then the set of simple roots $\a\in\Sigma^+$ such that $X_{-\a}\leq P$ will be denoted $\Cal D(P)$. We take $\Cal D(P)$ to have also the structure of a graph, with incidence induced from the Coxeter diagram of $\Sigma$, and we say that $P$ is {\bf connected} if $\Cal D(P)$ is connected. More generally, let $\Cal D_1,\cdots,\Cal D_r$ be the connected components of $\Cal D(P)$, and for each $i$, $1\leq i\leq r$, put $$ L_i=\lan X_{\a_i},X_{-\a_i}\ :\ \a \in\Cal D_i\ran \quad\text{and}\quad\Lambda=\Lambda(P)=\{L_1,\cdots,L_r\}. $$ We will refer to the members of $\Lambda(P)$ as the {\bf Levi complements} of $P$, relative to $\Sigma$. \proclaim {Lemma 6.5} Assume that the Lie rank of $G$ is at least $2$. Then there is a Sylow $2$-subgroup $S$ of $H$, and a proper parabolic subgroup $P$ of $G$ containing $\lan N_H(S),a\ran$, for which the following condition holds. $$ \text {For every $L\in\Lambda(P)$ we have $1\neq [a,L]\leq L$.}\tag* $$ Moreover, we can choose $P$ so that the Lie rank of each $L$ in $\Lambda(P)$ is equal to $1$. \endproclaim \demo {Proof} By 6.4, $a$ normalizes a maximal parabolic subgroup $N$ of $H$. If $[N,a]\leq O_2(N)$ then $a$ normalizes a Sylow $2$-subgroup of $N$ (hence of $H$), and then $a$ normalizes every parabolic subgroup of $H$ containing $S$. In particular, there is then a rank-1 parabolic subgroup $P$ of $H$, invariant under $a$, and with $[O^{2'}(P),a]\nleq O_2(P)$. Thus, the desired conclusion holds in this case, and we may therefore assume that $[N,a]\nleq O_2(N)$. Among all $a$-invariant parabolic subgroups $N$ with $[N,a]\nleq O_2(N)$, choose $N$ so that the Lie rank of $N$ is as small as possible. We then construct the set $\Lambda(N)=\{L_1,\cdots,L_r\}$ of Levi complements in $N$, relative to a fixed Borel subgroup of $N$. Then $N=O_2(N) L_1\cdots L_rK$, for some Cartan subgroup $K$ of $B$, and we may assume (possibly after replacing $a$ by a conjugate) that $a$ normalizes $L_1\cdots L_rK$. As $a$ is inner-diagonal, $a$ normalizes each $L_i$, and if $[a,L_i]=1$ for some $i$ we contradict the minimality of $N$. This proves the first part of the lemma. But further, if the Lie rank of some $L_i$ is bigger than $1$, then we may apply induction on the Lie rank, with $L_i\lan a\ran$ in place of $G$, to conclude that $L_i\lan a\ran$ has a proper parabolic subgroup containing $a$. From this we again contradict the minimality of $N$, and thus each $L_i$ has Lie rank equal to $1$. \qed \enddemo \proclaim {Lemma 6.6} Assume that the Lie rank of $\bar H$ is at least $2$, and assume that the field of definition for $\bar H$ $($in the sense of a $\sigma$-setup, as in section 3$)$ is larger than $\Bbb F_2$. Then $G\cong 2^{\cdot}G_2(4)$, and $|A|=3$. Moreover, we have $A=Z(R)$ for some Sylow $3$-subgroup $R$ of $G$, and $C_G(A)\cong SL(3,4)$. \endproclaim \demo {Proof} By 6.5 there exists a proper parabolic subgroup $P$ of $G$, and a Levi complement $L$ in $P$, such that $L\geq [L,a]\neq 1$. Since the field of definition of $\bar G$ is larger than $\Bbb F_2$, we may apply 6.2 to $L\lan a\ran$ and obtain $L\cong 2^{\cdot}L_2(4)$. Here $L\notin Lie(2)$, so $Z(L)\leq Z(H)$. Then 3.12 and 6.3 yield $G\cong 2^{\cdot}G_2(4)$. Let $R$ be a Sylow $3$-subgroup of $G$ containing $A$. Then $|R|=27$, and $R$ is contained in an $SL(3,4)$ subgroup $X$ of $G_2(4)$. Every element of $R-Z(R)$ is contained in a Frobenius subgroup of $X$ of order $21$, so 2.9 implies that $A=Z(R)$ is of order $3$. We observe that $A$ is contained in a Cartan subgroup $D$ of $X$, which is a Cartan subgroup of $G$. The Chevalley relations imply that $X$ is generated by the set of root subgroups centralized by $A$, relative to the root system determined by $D$. Then $X=O^{2'}(C_G(A))$, and $C_G(A)=XD=X$. \qed \enddemo For the remainder of this section we assume that the Lie rank of $\bar H$ is at least $2$, and that $\Bbb F_2$ is the field of definition for $\bar H$. Further, we assume that there exists no exceptional isomorphism between $\bar H$ and a group in $Lie(3)$. By 6.5, we may fix a parabolic subgroup $P$ of $G$ containing $a$, such that condition (*) in 6.5 holds, and such that every member of $\Lambda(P)$ is of Lie rank $1$. Let $\Cal M$ be the set of of all maximal parabolic subgroups of $H$ containing $P$, and having the property that every connected component of the diagram $\Cal D(M)$ contains at least one vertex of $\Cal D(P)$. One readily verifies that $\Cal M$ is non-empty, and we fix $M\in\Cal M$. \proclaim {Lemma 6.7} The following hold. \roster \item "{(a)}" We have $[L,a]\neq 1$ for any $L\in\Lambda(M)$. \item "{(b)}" We have $\lan a^M\ran\geq O^{2'}(M)$. \item "{(c)}" Let $S$ be a Sylow $2$-subgroup of $M$, and suppose that $Z(S)\nleq Z(H)$. Then $M=N_H(Z(S))$, and $\Cal M=\{M\}$. \endroster \endproclaim \demo {Proof} Part (a) is immediate from the definition of $\Cal M$. Then $L\leq \lan a^L\ran$ for any $L\in\Lambda(M)$, and (b) follows. Suppose that $Z(S)\nleq Z(H)$. We have $[Z(O_2(M)),a]=1$ by 2.9, and it follows from part (b) that $[Z(O_2(M)),O^{2'}(M)]=1$. Then $Z(S)\nl M$, and then since $M$ is a maximal parabolic we have $M=N_H(Z(S))$. This yields (c). \qed \enddemo \proclaim {Lemma 6.8} Suppose that $\bar H$ is isomorphic to $PSU(n,2)$ $n\geq 5$. Let $\phi$ be the canonical homomorphism from $GU(n,2)$ onto $PGU(n,2)$, and let $U$ be the natural module for $GU(n,2)$ over $\Bbb F_4$. Then $G\cong PGU(n,2)$, $|A|=3$, and $a=\phi(a*)$ for some element $a^*\in GU(n,2)$ such that $C_U(a*)$ has codimension $1$ in $U$. \endproclaim \demo {Proof} Let $S$ be a Sylow $2$-subgroup of $M$. It follows from 3.14 that $Z(S)\nleq Z(G)$, so 6.7(b) yields $M=N_H(Z(S))$. Then $O_2(\bar M)$ is extraspecial of width $n-2$, and $O^{2'}(\bar M/O_2(\bar M))$ is isomorphic to $SU(n-2,2)$. Further, as $O_3(G)=1$ and $Z(G)$ is cyclic, it follows from 3.12 that $|Z(G)|\leq 2$, and that $Z(G)=1$ if $n\neq 6$. Then 3.14 implies that $O_2(M)=X\times Z(G)$, where $X$ is a central product of $n-2$ quaternion groups. Set $Y=[O_2(M),a]$. We have $\Phi(Y)\leq Z(X)$, so $Y$ is contained in an extraspecial subgroup of $O_2(M)$, and then 2.9 implies that $Y$ is a quaternion group. From this we may conclude that $M/O_2(M)Z(G)$ is isomorphic to $GU(n-2,2)$, and then $\bar G\cong PGU(n,2)$. In particular, if $3$ divides $n$ then $a\notin H$, and so 3.13 yields $Z(G)=1$. Thus $O_2(M)=X$, and $C_X(a)$ is a central product of $n-3$ quaternion groups. It follows that $O^{2'}(C_G(a))\cong SU(n-1,2)$. Set $G^*=GU(n,2)$, let $U$ be the natural module for $G^*$ over $\Bbb F_4$, and let $a^*$ be a pre-image of $a$ in $G^*$. Then $O^{2'}(C_{G^*}(a^*))\cong SU(n-1,2)$, and we may choose $a^*$ so that $C_U(a^*)$ has codimension $1$ in $U$. Suppose that $|A|>3$. Then $C_G(a)=\lan A^{C_G(a)}\ran$, and so $$ 0=[V,a,C_G(a)]=[V,C_G(a),C_G(a)]. $$ Then $[C_G(a),C_G(a)]$ centralizes $V$, by the Three Subgroups Lemma. But $C_G(a)$ is non-abelian, as $n>3$. Thus $|A|=3$, and all parts of the lemma have been established. \qed \enddemo \proclaim {Lemma 6.9} Assume that $\bar H$ is defined over $\Bbb F_2$, that $\bar H$ is not a unitary group $PSU(n,2)$ with $n\geq 5$, and that $\bar H$ cannot be viewed (via an exceptional isomorphism) as a group of Lie type in characteristic $3$. Then $|Z(G)|=2$, and $\bar G$ is isomorphic to $\Omega_4^-(2)$, $L_4(2)$, $Sp(6,2)$, or $D_4(2)$. \endproclaim \demo {Proof} As $A$ induces inner-diagonal automorphisms on $\bar H$, it follows that $H=G$ or that $\bar H\cong {^2E_6(2)}$. Suppose first that $\bar G$ is isomorphic to $L_n(2)$ or $Sp(2n,2)$. If $Z(G)\neq 1$ then 3.12 and 6.3 yield $G\cong 2^{\cdot}L_4(2)$ or $2^{\cdot}Sp(6,2)$, and thus the lemma holds in this case. On the other hand, if $Z(G)=1$ then $C_G(Z(S))$ is not a maximal parabolic subgroup of $G$, and we contradict 6.7(b). Thus, we may assume that $\bar G$ is not isomorphic to $L_n(2)$ or $Sp(2n,2)$. Suppose that $\bar G$ is an orthogonal group $\Omega_{2n}^{\epsilon}(2)$, and let $U$ be a natural module for $\bar G$ over $\Bbb F_2$, of dimension $2n$. As $G$ is non-solvable we have $n\geq 2$, and $n\geq 3$ if $\epsilon=+1$. In view of 6.2, and the isomorphism of $\Omega_4^-(2)$ with $SL(2,4)$, we need only consider the cases where $n\geq 3$. As $\Omega_6^+(2)\cong L_4(2)$, and $\Omega_6^-(2)\cong PSp(4,3)$, we may in fact take $n\geq 4$. If $Z(H)\neq 1$, then 3.12 yields $\bar G\cong D_4(2)$ (which is isomorphic to $\Omega_8^+(2)$), and then since $Z(G)$ is cyclic, 3.12 yields $|Z(G)|=2$. Thus, the lemma holds in this case, and so we may assume that $Z(G)=1$. Let $U_0$ be a totally singular subspace of $U$, of dimension $2$, and denote by $L$ the stabilizer in $G$ of $U_0$. Without loss, we may assume that a Sylow $2$-subgroup $S$ of $L$ is contained in $M$. With the aid of Witt's Theorem on extensions of isometries, we find that $L=X(K_1\times K_2)$, where $X=O_2(L)$, $K_1\cong\Omega_{2n-4}^{\epsilon}(2)$ and $K_2\cong L_2(2)$. Further, $X$ is extraspecial, of width $2n-4$, and $X/Z(X)$ is isomorphic, as a module for $K_1K_2$, to a tensor product $N_1\otimes N_2$, where $N_i$ is a natural module for $K_i$ over $\Bbb F_2$. In particular, $L$ is a maximal parabolic subgroup of $G$, and $Z(S)=Z(L)$, so $L=M$ by 6.7(b), and $a\in L$. Now let $N_0$ be an irreducible $K_1$-submodule of $X/Z(X)$. Then $X/Z(X)=N_0\oplus (N_0)^g$ for any $g\in K_1K_2-K_1$. For any element $d$ of $K_1$ of order $3$ we then have $|[X/Z(X),d]|\geq 16$, and so $[X,d]$ is not a quaternion group. Thus $a\notin K_1$, by 2.9. But, for any element $d$ of $K_1K_2-K_1$ of order $3$, we have $|[X/Z(X),d]|\geq |N_0|$, where $|N_0|\geq 16$ as $n\geq 4$. As $a$ is conjugate to an element of $K_1K_2$, we have a contradiction at this point. Thus, we may assume that $\bar H$ is not an orthogonal group. As $\bar H$ is not a unitary group (the case of $U_4(2)\cong \Omega_6^-(2)$ having been treated above), we now conclude that $\bar G$ is not a classical group. If $\bar G\cong E_n(2)$, ($n=6,7,8$), then $|Z(G)|$ is odd, and so $|\Cal M|=1$, by 6.7. Recall, however, that $\Cal M$ is the set of maximal parabolic subgroups $M$ of $H$ containing $P$, where $P$ is a totally disconnected parabolic subgroup of $H$, and where each connected component of $M$ contains at least one vertex of $\Cal D(P)$. One has only to glance at the diagrams for the groups $E_n(2)$, however, to see that in fact $|\Cal M|>1$ for any choice of $P$. Thus, $\bar G\ncong E_n(2)$. Suppose that $\bar G\cong$ $^2F_4(2)'$. Then again $Z(G)=1$ and $\Cal M=\{C_G(Z(S))\}$. Then $M/O_2(M)\cong Sz(2)$, and so $|M|$ is prime to $3$, contrary to $a\in M$. By a similar argument, if $\bar H\cong {^3D_4(2)}$ then $|Z(G)|=1$ and $a\in M=C_G(Z(S))$, so that $M$ has an $a$-invariant Levi complement isomorphic to $SL(2,8)$. But this result is excluded by 6.2. As $G_2(2)'$ may be viewed as a group in characteristic $3$, we come finally to $\bar G\cong F_4(2)$ or $\bar H\cong {^2E_6(2)}$. Then $\Cal D(G)$ is the $F_4$ diagram, and since $\Cal D(P)$ is totally disconnected we can choose $M\in \Cal M$ so that $\Cal D(M)$ contains a subdiagram of type $A_2$. There then exists $L\in \Lambda(M)$ with $L/Z(L)\cong L_3(2)$ or $L_3(4)$. Replacing $G$ by $L\langle a rangle$, we obtain a contradiction from 6.3. \qed \enddemo \proclaim {Lemma 6.10} If $\bar G\cong L_4(2)$, $Sp(6,2)$, $G_2(4)$, or $D_4(2)$ then $|A|=3$. \endproclaim \demo {Proof} Suppose, by way of contradiction, that $|A|=9$. If $\bar G\cong L_4(2)$ it follows that both classes of elements of order $3$ in $G$ are represented in $A$, and since $G$ contains a Frobenius group of order $21$, we contradict 2.9. Suppose next that $\bar G\cong Sp(6,2)$, and let $U$ be the natural module for $\bar G$ over $\Bbb F_2$. If there exists $a$ in $A$ with $|[U,\bar a]|\neq 16$ then $\bar a$ is contained in an $L_3(2)$-subgroup of $\bar G$, contrary to 6.3. On the other hand, we have $U=\lan C_U(\bar a)\ :\ 1\neq \bar a\in \bar A \ran$, so there exist $a,b\in A$ such that $|[U,\bar a]|=|[U,\bar b]|=4$, and with $\lan a,b \ran=A$. Then $|[U,\bar{ab}]|=16$, and so we have a contradiction in this case. Finally, suppose that $\bar G\cong D_4(2)$, and let $U$ be a natural $O_8^+(2)$-module for $\bar G$. We claim that there exists $a\in A$ with $16\leq |[U,a]|\leq 64$. Suppose false. Then $|[U,a]|=4$ or $2^8$ for every non-identity element $a\in A$. Let $a$ and $b$ generate $A$. If $|[U,a]|=|[U,b]|=4$ then $|[U,ab]|=16$, while if $|[U,a]|=4$ and $|[U,b]|=2^8$ then either $[U,ab]$ or $[U,ab^2]$ is of order $64$. The only other case is that in which $C_U(a)=0$ for every non-identity $a\in A$, which is absurd. The claim is therefore established. Now fix $a\in A$ with $|[U,a]|=16$ or $64$. There is then a non-degenerate $a$-invariant subspace $W$ of $V$, of type $O_6^+(2)$, with $|[W,a]|=16$. Let $H$ be the point-wise stabilizer in $G$ of $W^{\perp}$. Then $H\cong \Omega_6^+(2) \cong Alt(8)$, and we have $H=[H,a]$. Identifying $H$ with $Alt(8)$, and identifying $W$ with the non-trivial irreducible constituent in the natural permutation module for $Alt(8)$, it follows that $a$ induces on $W$ the action of a product of two disjoint $3$-cycles. There is then a $7$-cycle $x$ in $H$ with $x^a=x^2$. This contradicts 2.9, so the lemma is proved. \qed \enddemo Theorem 6.1 follows from lemmas 6.2 through 6.10. \vskip .3in \noindent {\bf Section 7: Sporadic Groups, $p=3$} \vskip .2in We continue to assume Hypothesis 1.1, with $p=3$. Further, we assume that $\bar H$ is among the $26$ sporadic simple groups. The index of $\bar H$ in its automorphism group is then at most $2$, and then since $G=\lan A^G \ran$ we have $G=H$. Also, since $O_3(G)=1$, the only cases in which $Z(G)\neq 1$ occur when $Z(G)$ is of order $2$ or (in the unique case of $M_{22}$) of order $4$). We will obtain the following result. \proclaim {7.1 Theorem} Assume Hypothesis 1.1, with $\bar G$ a sporadic simple group. Then $G\cong 2^{\cdot}J_2$, $2^{\cdot}Suz$, or $2^{\cdot}Co_1$, and we have $|A|=3$. \endproclaim We will make free use of the tables in section 5.3 of [GLS3], in which, for each sporadic group $X$, and each subgroup $Y$ of $X$ of prime order, the normalizer $N=N_X(Y)$ is determined, in the sense that a chief series for $N$ is given, along with the action of $N$ on the various chief factors. Also, we will draw on the character tables in the ATLAS of Finite Groups [CCNPW], in order to establish that $2^{\cdot}Co_1$ contains a perfect central extension $6^{\cdot}Suz$, and that $2^{\cdot}Suz$ contains a perfect central extension $6^{\cdot}U_4(3)$. \vskip .1in Six cases may be eliminated right away. Namely, by 2.9, if $G$ has a unique conjugacy class of subgroups of order $3$, then $G$ does not contain a Frobenius subgroup of order $21$. In this way, we obtain the following result. \proclaim {7.2} $\bar G$ is not isomorphic to $M_{22}$, $M_{23}$, $J_1$, $HS$, $Ru$, or $O'N$. \endproclaim We next observe, that the centralizer of any element of order $3$ in any sporadic group is of even order. In particular, $|C_{\bar G}(a)|$ is even. For the remainder of this section, we fix an element $t$ of $C_{G}(a)$ with $\bar t$ of order $2$. Set $\bar C=C_{\bar G}(\bar t)$, and denote by $C$ the inverse image of $\bar C$ in $G$. Also, set $K=\lan a^C\ran$, and set $R=F^*(K)$. We will proceed by induction on $|G|$. \proclaim {Definition 7.3} Let $X$ be a group, and set $R_0=F^*(X)$. We say that $X$ is of {\bf extraspecial type} if the following three conditions hold. \roster \item "{(i)}" $R_0=Z(R_0)E$ where $R_0$ is an extraspecial group of width $n\geq 2$. \item "{(ii)}" $X/R_0$ is isomorphic to one of the groups $Alt(2n+1)$, $Alt(2n+2)$, $GU(n,2)$, $\Omega_{2n}^{\ \epsilon}(2)$ (for some sign $\epsilon$), or $Sp(2n,2)$, and \item "{(iii)}" $R_0/Z(R_0)$ is a natural (irreducible) $\Bbb F_2$-module for $K/R_0$. \endroster \endproclaim \proclaim {Lemma 7.4} The following hold. \roster \item"{(a)}" If $K$ is quasisimple then $K/O_3(K)$ is in the list of quasisimple groups which are outcomes in Theorem 1.2. \item"{(b)}" If $R=Z(R)E$ is a $2$-group where $E$ is extraspecial of width $n\geq 2$, then $K$ is of extraspecial type, in the sense of 7.3. \endroster \endproclaim \demo {Proof} Part (a) is by induction on $|G|$. Part (b) is immediate from [Ch, Theorem A]. \qed \enddemo Before going to work with 7.4, it will be convenient to eliminate eight more groups by considering $5$-local subgroups. \proclaim {Lemma 7.5} $\bar G$ is not isomorphic to $Mc$, $Co_3$, $Co_2$, $Ly$, $F_5$, $F_3$, $F_2$, or $F_1$. \endproclaim \demo {Proof} We first show that in each of the above possibilities for $\bar G$ we have $|C_G(a)|$ divisible by $5$. Indeed, in the cases other than $\bar G\cong Mc$, $F_3$, or $Co_3$, one checks that the centralizer of every element of order $3$ has a subgroup of order $5$. Suppose that $\bar G\cong Mc$. Then $Z(G)=1$, $G$ has one class of involutions, and then $C_G(t)\cong 2^{\centerdot}A_8$. Then $C_G(\lan a,t \ran)\cong \Bbb Z_3\times SL(2,5)$, by 4.3, and thus $|C_G(a)|$ is divisible by $5$ in this case. Suppose next that $\bar G\cong F_3$. Then $Z(G)=1$, $G$ has just one conjugacy class of involutions, and we have $C_G(t)$ of the form $2^{1+8}_+.Alt(9)$. Now [Ch] shows that $a$ is incident with a $3$-cycle in $C_G(t)/O_2(C_G(t))$, and so we again get $5$ dividing the order of $C_G(a)$. Suppose that $\bar G\cong Co_3$ and that $5$ does not divide the order of $C_G(a)$. Again, we have $Z(G)=1$, and we find that $C_G(a)\cong \Bbb Z_3\times L_2(8):3$. In particular, $a$ is not contained in the commutator subgroup of $C_G(a)$. Now consider $C=C_G(t)$. By 7.4(a), $C$ is not isomorphic to $\Bbb Z_2\times M_{12}$. This leaves only the case $C\cong 2^{\cdot}Sp(6,2)$. Let $U$ denote the natural $Sp(6,2)$-module for $C$. As $5$ does not divide $|C_G(a)|$ we have $|[U,a]|>4$, and since $a$ is not in the commutator subgroup of $C_G(a)$ we have $|[U,a]|\neq 64$. This leaves $|[U,a]|=16$. But then $a$ lies in a Frobenius $21$-subgroup of $G$, and we have a contradiction via 2.9. Thus, we have found that $|C_G(a)|$ is divisible by $5$ in all cases under consideration. Let $F$ be a subgroup of $C_G(a)$ of order $5$. Then $C_{\bar G}(\bar F)$ is not $5$-constrained, by 2.9. We consult [GLS3, Table 5.3] for the structure of centralizers of elements of order $5$. setting $D= O^{3'}(C_G(F))$, we have $D\ne 1$. Further, $D$ is not isomorphic to $Alt(5)$ (as follows from 2.4) or to $U_3(5)$ (by theorem 6.1), or to $HS$ or $F_5$ (by induction in 7.1). But in fact, as one checks, this exhausts the list of possibilities for the structure of $C_G(F)$, and so 7.5 is proved. \qed \enddemo \proclaim {7.6} $\bar G$ is isomorphic to $J_2$, $Suz$, or $Co_1$. \endproclaim \demo {Proof} We shall go through the list of groups, and check the conditions in 7.3 against the structure of the centralizers of involutions in the sporadic groups that remain to be considered. In view of 7.2 and 7.5, these are (aside from the three groups mentioned in the statement of the lemma) the nine groups $Fi_{24}'$, $Fi_{23}$, $Fi_{22}$, $He$, $J_4$, $J_3$, $M_{24}$, $M_{12}$, and $M_{11}$. We note that, among these nine groups, only $Fi_{22}$, and $M_{12}$ have non-trivial Schur multipliers, and in these two cases the multiplier is of order $2$. We begin with $G\cong Fi_{24}'$. Here there are two classes of involutions, and we find that either $K$ is double cover of $Fi_{22}$ or $R$ is an extraspecial $2$-group of width $6$ with $K/R$ isomorphic to $3U_4(3)$. In both these cases, we violate 7.4. Suppose next that $G\cong Fi_{23}$. In view of 7.4(a), $K$ is not a Schur extension of $Fi_{22}$ or $U_6(2)$. This leaves only the possibility that $C$ is of the form $$ (2^2\times Q_8^{(4)})((GU_4(2))2).\tag1 $$ Then $Z(C/O_2(C))$ is of order $3$, acting non-trivially on $Z(O_2(C))$, as follows from the structure of the corresponding involution-centralizer in $Fi_{24}'$. Thus $K/R\cong U_4(2)$. Here $Z(C/O_2(C))$ acts non-trivially on $R/Z(R)$, so that $R/Z(R)$ is the natural unitary module for $K/R$. This violates 7.4(b). Suppose that $\bar G\cong Fi_{22}$. By 7.4(a), $K$ is not a Schur extension of $U_6(2)$ and examination of the remaining classes of involution centralizers then yields $F^*(C)=O_2(C)$. Further, for any involution $\bar s$ of $\bar G$ such that $C_{\bar G}(\bar s)$ is $2$- constrained, either $C_{\bar G}(\bar s)$ is of the form $$ (2\times Q_8^{*4})U_4(2)\tag2 $$ or $C_{\bar G}(\bar s)$ does not contain a Sylow $2$-subgroup of $G$. It follows from 2.9 that $[Z(S),a]=1$ for some Sylow $2$-subgroup $S$ of $G$. Since the group in (2) lifts to a subgroup of the group in (1) in $Fi_{23}$, one observes that there exists a $2$-central involution $s$ of $G$ with $s\notin Z(G)$. We may then take $s=t$, whence $\bar C$ is as in (2). This violates 7.4. In the group $He$ there are are two classes of involutions, and we find that either $K$ is a Schur extension of $L_3(4)$ or $K$ is of the form $D_8^{*3}:L_3(2)$. Both these possibilities are excluded by 7.4 (or by noticing that in both these groups, each element of order $3$ lies in a Frobenius group of order $21$). In $J_4$ there are two classes of involutions, and we find that $K$ is of the form $Q_8^{*6}(3M_{22})$ or $2^{11}M_{22}$, in each case violating of 7.4. If $G$ is isomorphic to $J_1$ or $J_3$, we obtain $C/O_2(C)\cong Alt(5)$, and $C$ contains a subgroup isomorphic to $Alt(5)$. This violates 2.7. In $M_{24}$ there are two classes of involutions, with centralizers of the form $(D_8^{*3})L_3(2)$ and $(2^6)Sym(5)$. Thus, we violate 7.4 if $G\cong M_{24}$. If $G$ is isomorphic to $M_{11}$ then $G$ has a single conjugacy class of elements of order $3$, and since $M_{11}\geq M_{10}\geq Alt(6)\geq Alt(4)$, we contradict 2.9. Finally, suppose that $\bar G$ is isomorphic to $M_{12}$, and let $S$ be a Sylow $3$-subgroup of $G$ containing $a$. Then $S$ is extraspecial of order $27$. If $a\in Z(S)$ then every elementary abelian subgroup of order $9$ in $G$ contains a conjugate of $a$, and hence $a$ lies in an $M_{11}$-subgroup of $G$, contrary to the preceeding paragraph. Thus $a\notin Z(S)$, and one then has $C_{\bar G}(\bar a)\cong \Bbb Z_3\times A_4$. Let $\bar s$ be an involution in $C_{\bar G}(\bar a)$. If $\bar s$ is $2$-central then $O^2(C_{\bar G}(\bar s))\cong 2^{1+4}_+:\Bbb Z_3$, whereas $C_{\bar G}(\lan \bar a,\bar s) \ran)\cong \Bbb Z_6\times \Bbb Z_2$. Thus $\bar s$ is not $2$-central, and so $C_{\bar G}(\bar s)\cong \Bbb Z_2\times\Sigma_5$. Then also $\bar a$ normalizes a fours group $\bar F\subseteq E(C_{\bar G}(\bar s))$, where every involution in $\bar F$ is $2$-central. Now 2.9 implies that $G\cong 2^{\cdot}M_{12}$. But also, we have $N_{\bar G}(\bar F)\cong 4^2:D_{12}$. Let $\bar X$ be the normal subgroup of $N_{\bar G}(\bar F)$ with $\bar X \cong \Bbb Z_4\times \Bbb Z_4$, let $X$ be the pre-image of $\bar X$ in $G$, and let $F$ be the pre-image of $\bar F$ in $G$. Then $X=\lan x,y \ran$, where $F=\lan x^2,y^2 \ran$ and $F\lan a \ran\cong SL(2,3)$. Thus $[x^2,y^2]\ne 1$. But $[x,y]\in Z(X)$, so $[x^2,y^2]=[x,y]^4=[x^4,y]=1$, for a final contradiction. \qed \enddemo \proclaim {7.7} If $\bar G\cong J_2$, $Suz$, or $Co_1$ then $|Z(G)|=2$ and $|A|=3$, and we have $C_G(A)\cong 2^{\cdot}Alt(6)$, $6^{\cdot}U_4(3)$, or $6^{\cdot}Suz$, respectively. \endproclaim \demo {Proof} Suppose first that $\bar G\cong J_2$. There are two classes of involutions in $\bar G$, with centralizers isomorphic to either $2^{1+4}_-:Alt(5)$ or $2^2\times Alt(5)$. Thus, $\bar C$ has a subgroup $L$ containing $a$ and isomorphic to $Alt(5)$, and so 2.6 implies that $Z(G)\neq 1$. There are two conjugacy classes of subgroups of order $3$ in $\bar G$, with centralizers $3^{\cdot}Alt(6)$ and $3\times SL(2,3)$. Suppose that $C_{\bar G}(\bar a)\cong 3\times SL(2,3)$. In the notation of [GLS3, Table 5.3g] (which is the same as ATLAS notation, cf. [CCNPW]) we then have $\bar a\in 3B$, and the table gives $[3B,2C]=1$ where $2C$ is an outer involution of $\bar G$ satisfying $C_{\bar G}(2C)\cong L_3(2)$. Thus $\bar a$ lies in a Frobenius group of order $21$, and we violate 2.9. This shows that $\bar a$ is in the class $3A$. Now suppose that $|A|>3$. A Sylow $3$-subgroup $S$ of $G$ is extraspecial of order $27$, and $N_{\bar G}(\bar S)$ contains a dihedral subgroup $D$ of order $8$ which acts faithfully on $\bar S/\Phi(\bar S)$. It follows that $D$ acts transitively on the set of maximal elementary abelian subgroups of $S$, and so $A$ contains representatives from each conjugacy class of subgroups of order $3$ in $G$. But we have seen that $A$ contains representatives of only one class, so in fact $|A|=3$. Suppose next that $\bar G\cong Suz$. Then $\bar G$ has two classes of involutions. One of these has a corresponding centralizer $\bar C_0$ with $O^{3'}(\bar C_0)\cong L_3(4)$. It follows from theorem 6.1 that $\bar t$ represents the other class, with $\bar C$ an extension of an extraspecial group $2^{1+6}_-$ by $\Omega_6^-(2)$. By 3.14, $Z(G)$ is a direct factor of $O_2(C)$, and then 2.9 implies that $C_{O_2(\bar C)}(\bar a)\cong 2^{1+4}_+$. Thus $C_{\bar G}(\lan \bar a,\bar t\ran)$ is an extension of $\lan \bar a\ran\times C_{O_2(\bar C)}(\bar a)$ by $\Omega_4^+(2)$, and so $2^7$ divides $|C_{\bar G}(\bar a)|$. This information suffices to single out the conjugacy class of $\lan \bar a\ran$, and to yield $C_{\bar G}(\bar a)\cong 3^{\cdot}U_4(3)$. Let $\bar f$ be an element of order $5$ in $C_{\bar G}(\bar a)$ and set $Y=O^{3'}(C_G(f))$. Then $a\in \bar Y\cong Alt(6)$ or $Alt(5)$, and then 2.6 implies that $|Z(Y)|=2$. But $Z(Y)\leq Z(G)$, and so $|Z(G)|=2$. Suppose that $|A|>3$. One checks from the character table for $2^{\cdot}Suz$ in [CCNPW] that $C_{\bar G}(\bar a)$ lifts to a completely nonsplit extension $6^{\cdot}U_4(3)$ in $G$, so $C_G(a)=\lan A^{C_G(a)}\ran$. Then $[V,a,C_G(a)]=0$, whereas $Z(G)$ is fixed-point-free on $V$. This contradiction shows that $|A|=3$. Suppose finally that $\overline G\cong Co_1$. Then $\bar G$ has three classes of involutions, with corresponding centralizers $\bar C_i$, ($1\leq i\leq 3$), where the groups $\bar C_i$ have the structure given as follows. $$ O_2(\bar C_1)\cong 2^{11}\quad{\text and}\quad \bar C_1/O_2(\bar C_1)\cong M_{11}, $$ $$ \bar C_2\cong 2^2\times G_2(4), $$ $$ O_2(\bar C_3)\cong 2^{1+8}_+\quad{\text and}\quad \bar C_3/O_2(\bar C_3)\cong D_4(2). $$ Let $C_i$ denote the inverse image of $\bar C_i$ in $G$. If $C=C_1$ we obtain a faithful quadratic module either for $M_{11}$ or for $C_1$, and we contradict 7.4. Suppose that $C=C_2$. Then 5.1 implies that $Z(G)\neq 1$, and it only remains to show that $|A|=3$. Further, it follows from 6.6 that $C_{\bar G}(\lan \bar a,\bar t\ran)$ contains a subgroup isomorphic to $2^2\times SL(3,4)$, and this serves to identify $\lan\bar a\ran$ among the three conjugacy classes of subgroups of order $3$ in $G$, and to yield $C_{\bar G}(\bar a)\cong 3^{\centerdot}Suz$. On the other hand, suppose that $C=C_3$. We then appeal 3.14 to conclude that $O_2(C_{\bar G}(\lan\bar a,\bar t\ran))\cong 2^{1+6}_-$, and then also $C_{\bar G}(\lan\bar a,\bar t\ran)/O_2(C_{\bar G}(\lan\bar a,\bar t\ran))\cong \Omega_6^-(2))$. This information again serves to identify $\lan\bar a\ran$, among the three conjugacy classes of subgroups of order $3$ in $\bar G$, and we again obtain $O^{3'}(C_{\bar G}(\bar a))\cong 3^{\cdot}Suz$. Let $\bar g$ be an element of order $7$ in $C_{\bar G}(\bar a)$. Then $O^{3'}(C_{\bar G}(\bar g))\cong L_3(2)$ or $Alt(7)$, and then 6.3 and 4.2 yield $O^{3'}(C_G(g))\cong 2^{\centerdot}Alt(7)$, and $Z(G)\neq 1$. Thus, we have shown that, in any case, we have $Z(G)\neq 1$, and $O^{3'}(C_{\bar G}(\bar a))\cong 3^{\cdot}Suz$. The character table for $2^{\cdot}Co_1$ in [CCNPW] then yields $C_G(a)\cong 6^{\cdot}Suz$ (with no non-trivial direct factors). As in the case of $Suz$, we obtain $C_G(a)\leq \lan A^{C_G(a)}$ if $|A|>3$, and in that case we contradict the fact that $C_V(Z(G))=1$. Thus $|A|=3$ and the lemma is proved. \qed \enddemo Notice that results 7.2 through 7.6 yield theorem 7.1. Theorem A is then given by the union of the results 4.1, 5.1, 6.1, and 7.1. \vskip .2in \noindent {\bf Section 8: Theorem B and Corollary C} \vskip .2in \proclaim {Hypothesis 8.1} Assume Hypothesis 1.1 and assume also that $G$ is not a group of Lie type in characteristic $p$. \endproclaim By Theorem A, Hypothesis 8.1 implies that $|A|=p=3$, and $G$ is one of the exceptional groups listed in Theorem A. We aim first of all to determine which subgroups of order $3$ in $G$ can possibly be quadratic subgroups, with respect to some irreducible $G$-module $V$. Some of these identifications have already been made, in 4.3, 6.6, 6.8, and 7.7. Whenever Hypothesis 8.1 is in effect, let $a$ be a generator of $A$, set $\bar G=G/Z(G)$, and set $\bar C=C_{\bar G}(\bar A)$. We note that, by 2.4, $\bar A$ is contained in a $2$-local subgroup of $\bar G$, and we may fix a subgroup $M$ of $G$, containing $Z(G)A$, such that $\bar M$ is a maximal $2$-local subgroup of $G$. \proclaim {Lemma 8.2} Assume Hypothesis 8.1, and suppose that $\bar G$ is isomorphic to $D_4(2)$. Then $\bar C\cong GU(4,2)$, and $A$ is contained in a subgroup $L$ of $G$ of the form $(2^{1+6}_+)L_4(2)$. These conditions determine $A$ up to conjugacy in $Aut(G)$. \endproclaim \demo {Proof} Identify $\bar G$ with $\Omega_8^+(2)$ and let $U$ be the natural module for $G$ over $\Bbb F_2$. Then $|[U,\bar A]|=2^{2k}$ for some $k$, $1\leq k\leq 4$. The integer $k$ determines the structure of $\bar C$, and we have: \vglue .1in \item {(1)} If $k=1$ then $\bar C\cong 3\times \Omega_6^-(2)$. \vglue .1in \item {(2)} If $k=2$ then $\bar C\cong GU(2,2)\times \Omega_4^+(2)$. \vglue .1in \item {(3)} If $k=3$ then $\bar C\cong GU(3,2)\times 3$, and $\bar A\leq [\bar C,\bar C]$. \vglue .1in \item {(4)} If $k=4$ then $\bar C\cong GU(4,2)$ \vglue .1in The maximal $2$-local subgroup $\bar M$ of $\bar G$ is a maximal parabolic subgroup. Suppose first that $\bar M$ is of the form $2^6:\Omega_6^+(2)$. Then 3.14 implies that $M$ is of the form $(2^{1+6}_+)\Omega_6^+(2)$, and that $[O_2(M),A]$ is a quaternion group. Then $C_{O_2(M)}(A)$ is of order $32$, and so $|\bar C|$ is divisible by $16$. In this case we have $k=1$ or $4$. Let $S$ be a Sylow $2$-subgroup of $M$. There are then three maximal parabolic subgroups of $\bar G$ containing $\bar S$ and of the form $2^6:\Omega_6^+(2)$. In the full covering group ${2^2}^{\cdot}D_4(2)$ these parabolics lift to subgroups of the form $(2\times 2^{1+6}_+)\Omega_6^+(2)$, as follows from 3.14. Since $Out(D_4(2))$ acts faithfully on the Schur multiplier of $D_4(2)$, by 3.13, it follows that, in $G$, two of these maximal parabolics lift to groups which are isomorphic to $M$, and that one lifts to a group $N$ such that $O_2(N)$ is elementary abelian. Let $M$ and $M_1$ be the two which are isomorphic to $M$. Then $M$ and $M_1$ are fused in $Aut(G)$, and thus $A$ is determined up to conjugacy in $Aut(G)$ in this case. On the other hand, suppose that $\bar A$ is not contained in a maximal parabolic subgroup of $\bar G$ of the form $2^6:\Omega_6^+(2)$. Then $k=2$ or $3$, and $\bar M$ is of the form $(2^{1+8}_+):(Sym(3)\times Sym(3)\times Sym(3))$. Set $R=[O_2(M),A]$, and let $V_1$ be an irreducible $RA$-submodule of $V$. Then $RA/C_R(V_1)\cong SL(2,3)$, as follows from Theorem A of [Ch]. Set $R_1=[C_R(V_1),A]$. If $R_1=1$ then $|\bar C|$ is divisible by $2^7$, which is contrary to having $k=2$ or $3$. Thus $R_1\neq 1$. Let $V_2$ be a non-trivial irreducible section for $R_1A$ in $V$. Then $R_1A/C_{R_1}(V_2)\cong SL(2,3)$. We have $Z(G)\cap R_1=1$, so $R_1$ is isomorphic to a subgroup of $\bar R$. As $R_1/C_{R_1}(V_2)$ is a quaternion group, it follows that $C_{R_1}(V_2)$ is elementary abelian, and then $[C_{R_1}(V_2),A]=1$, by 4.3. Thus, $|R/C_R(A)|=16$, and so $|\bar C|$ is divisible by $32$. This is again contrary to $k=2$ or $3$, and the lemma is thereby proved. \qed \enddemo \proclaim {Lemma 8.3} Assume Hypothesis 8.1, and suppose that $\bar G\cong Sp(6,2)$. Then $\bar C\cong 3\times Sp(4,2)$, and this condition determines $A$ up to conjugacy in $G$. \endproclaim \demo {Proof} There are three conjugacy classes of subgroups of order $3$ in $\bar G$, two of which are represented in a subgroup $\bar L$ of $\bar G$ of the form $L_2(8):3$. Theorem 1.1 implies that $\bar A\nleq \bar L$, so the conjugacy class of $A$ in $G$ is uniquely determined. Let $U$ be the natural module for $\bar G$ over $\Bbb F_2$, and let $b$ be an element of order $3$ in $L-E(L)$. Then $b$ lies in a Frobenius subgroup of $L$ of order $21$, and hence $|[U,\bar b]|=16$. Let $c$ be an element of order $3$ in $E(L)$. Then $c$ is contained in a cyclic group of order $9$, and so $[U,\bar c]=U$. Thus, $[U,\bar a]$ is of order $4$, and the lemma follows. \qed \enddemo Theorem B now follows from the results 4.3 (concerning the alternating groups), 6.6 (concerning $2^{\cdot}G_2(4)$), 6.8 (concerning the groups $PGU((n,2)$), 7.7 (concerning $2^{\cdot}J_2$, $2^{\cdot}Suz$, and $2^{\cdot}Co_1$), 8.2, and 8.3. \vskip .1in We end this section with the proof of Corollary C. Thus, assume Hypothesis 8.1, and assume that $|A|^2\geq |V/C_V(A)|$. That is, assume that $|V/C_V(A)|\leq 9$. Denote by $\Cal L$ the set of all pairs $(L,B)$ where $L$ is a quasisimple subgroup $L$ of $C_G(A)$ and $B$ is a $G$-conjugate of $A$ contained in $L$, with $B\nleq Z(L)$. Then $[V,A,B]=0$, and so $AB$ acts quadratically on $V$. This is contrary to Theorem B, so $\Cal L$ is empty. If $G\cong PGU(n,2)$ with $n\geq 5$ then the conditions given by 6.8 guarantee that $\Cal L$ is non-empty. This will also be the case if $\bar G\cong Alt(n)$ with $n\geq 8$, by 4.3. If $\bar G\cong D_4(2)$ or $Sp(6,2)$, we again get $\Cal L$ non-empty, by 8.2 and 8.3. Suppose that $\bar G\cong Co_1$. Then $\bar C\cong 3^{\cdot}Suz$, and $\bar A$ is not contained in the center of a Sylow $3$-subgroup of $G$. Then $\bar A$ is not weakly closed in $\bar C$ with respect to $\bar G$, and so $\Cal L$ is non-empty in this case as well. Thus, none of these cases occur. We have $2^{\cdot}Suz\geq 2^{\cdot}G_2(4)\geq 2^{\cdot}J_2$, and this descending series of groups corresponds to a descending chain of values for $\bar C$: $3^{\cdot}U_4(3)\geq SL(3,4)\geq 3^{\cdot}Alt(6)$. These conditions guarantee that the class of quadratic elements in $2^{\cdot}Suz$ restricts to the class of quadratic elements in the groups farther down the chain. Thus, to eliminate these groups it will suffice to eliminate the case $\bar G\cong J_2$. In that case $\bar A$ is contained in a subgroup $\bar M$ of $\bar G$ of the form $(2^{1+4}_-)Alt(5)$, where the extension is split. Thus, $A$ is contained in a subgroup $K$ of $G$ with $K\cong SL(2,5)$ and with $Z(K)\leq Z(G)$. Then $C_V(Z(K))=0$, and $V$ is a direct sum of $2$-dimensional subspaces $V_i$, $1\leq i\leq m$, where each $V_i$ is an irreducible module for a fixed quaternion subgroup $K_1$ of $K$. We may choose $K_1$ to be $A$-invariant, so $m\leq 2$. But evidently $G\nleq SL(4,3)$, so we have a contradiction at this point. It remains to consider the cases $\bar G\cong Alt(n)$, $n=5$ or $7$. In these cases, there is an $A$-invariant quaternion subgroup $K_1$ of $G$ with $Z(K_1)=Z(G)$, so we obtain an embedding of $G$ in $SL(4,3)$. As $7$ does not divide the order of $SL(4,3)$ we conclude that $n=5$, and then 2.1 implies that $V$ is a natural $SL(2,9)$-module for $G$. This completes the proof of Corollary C. \vskip .2in \noindent {\bf Section 9: Examples} \vskip .1in As mentioned in the introduction, the classification of the irreducible quadratic modules for $2^{\cdot}Alt(n)$ is given in [M], where it is shown that all such modules are \lq\lq spin modules" and that all spin modules are quadratic. In this section we will show, by example, that all of the groups mentioned in parts (a) and (c) of Theorem B have quadratic modules. In order to do this, it will be convenient to have available the information given by the following lemma. \proclaim {Lemma 9.1} Let $X$ be an extraspecial $2$-group, expressed as the central product of subgroups $X_i$, $1\leq i\leq n$, where each $X_i$ is a quaternion group or a dihedral group of order $8$. Let $F$ be a field of characteristic different than $2$, and let $U$ be a faithful irreducible module for $X$ over $F$. Then the following hold. \roster \item "{(a)}" The module $U$ is uniquely determined up to isomorphism. It has dimension $2^n$, and it is the tensor product module $U=U_1\otimes\cdots\otimes U_n$, where $U_i$ is the (unique) faithful $2$-dimensional module for $X_i$ over $F$. \item "{(b)}" We have $N_{GL(U)}(X)/C_{GL(U)}(X)\cong Aut(X)$. \item "{(c)}" If the characteristic of $F$ is $3$, and $a$ is an automorphism of $X$ such that $[X,a]$ is a quaternion group, then $a$ induces an $F$-linear automorphism of $U$ with $[U,a,a]=0$, and with $dim([U,a])=2^{n-1}$. \endroster \endproclaim \demo {Proof} Each $X_i$ has four linear characters and one irreducible character of degree $2$. Since $F$ is a splitting field for $X_i$, there is then a unique faithful irreducible representation of $X_i$ over $F$, and it has degree $2$. Any irreducible representation of $X$ over $F$ factors through a representation of the direct product $X_1\times\cdots\times X_n$, and is therefore a tensor product of irreducible representations of the groups $X_i$. If the representation is also faithful then each of its tensor factors is faithful, and so (a) holds. Part (b) is immediate from the uniqueness of $U$. Let $a$ be an automorphism of $X$ such that $[X,a]$ is a quaternion group. Then $|a|=3$, and (b) implies that $a$ induces a non-trivial automorphism of $U$ over $F$. Here $[X,a]$ commutes with $C_X(a)$, by the Three Subgroups Lemma, and so we may take $[X,a]=X_1$. The tensor decomposition in (a) then implies that $U$ is a direct sum of isomorphic two-dimensional modules for the group $L=\lan a^X\ran=[X,a]\lan a\ran$. Here $L\cong SL(2,3)$, and if $F$ has characteristic $3$ then $a$ acts quadratically on each irreducible $L$ invariant summand of $U$. This yields (c). \qed \enddemo Now for the examples. \vskip .2in \noindent $G=PGU(n,2):$ \noindent Let $X$ be the central product of $n$ quaternion groups. Then the semidirect product $K=X:GU(n,2)$ is contained in a maximal parabolic subgroup of $SU(n+2,2)$. It follows from the preceding lemma that there is a quadratic module $U$ for $K$, of dimension $2^n$ over $\Bbb F_3$. \vskip .2in \noindent $\bar G=D_4(2)$ or $Sp(6,2):$ \noindent Take $G=2^{\cdot}D_4(2)$. Then $G$ is the commutator subgroup of the Weyl group of the $E_8$-root lattice $\Lambda$. Set $V=\Lambda/3\Lambda$. Then $G$ acts faithfully on $V$. Choose a maximal subgroup $M$ of $G$, of the form $(2^{1+6}_+)\Omega_6^+(2)$, and let $A$ be a subgroup of order $3$ in $M$, such that $[O_2(M),A]$ is a quaternion group. By 9.1, we may identify $V$ with the unique faithful irreducible module for $O_2(M)$, and $A$ acts quadratically on $V$. We may identify $\bar G$ with $\Omega_8^+(2)$, in such a way that $\bar A$ centralizes a $6$-dimensional non-degenerate subspace of the natural $\Bbb F_2$-module $U$ for $\bar G$. Let $\bar G_0$ be the stabilizer in $\bar G$ of a non-singular point in $U$. Then $\bar G_0\cong 2\times Sp(6,2)$. Let $G_1$ be the inverse image in $G$ of the commutator subroup of $\bar G_0$. Then $A\leq G_1$, and since $A$ acts quadratically on $V$, Theorem 1.2 implies that $G_1\cong 2^{\cdot}Sp(6,2)$. \vskip .2in \noindent $\bar G=Co_1$, $Suz$, $G_2(4)$, or $J2$: \noindent Next consider the case where $\Lambda$ is the Leech lattice and where $G=2^{\cdot}Co_1$ -- the automorphism group of $\Lambda$. Again, take $V=\Lambda/3\Lambda$. Then $G$ acts faithfully on $\Lambda$. Let $M$ be a maximal subgroup of $G$, such that $\bar M$ is of the form $(2^{1+8}_+)D_4(2)$. Then 4.1 implies that $Z(G)$ is a direct factor of $O_2(M)$. There then exists a subgroup $A$ of $M$, of order $3$, such that $[O_2(M),A]$ is a quaternion group. Let $R$ be a complement to $Z(G)$ in $O_2(M)$, chosen so that $R$ is invariant under an elementary abelian subgroup $E$ of $M$ of order $81$. Then $R$ is generated by four conjugates of $A$, and $R=[R,E]$. Set $W=[V,Z(R)]$. Then 9.1 implies that $dim(W)\geq 16$ and that $A$ acts quadratically on $W$, with $dim([W,A])=1/2 dim(W)$. We now have $dim(C_V(Z(R))\leq 8$, and evidently neither $M/Z(R)$ nor $M/Z(R)Z(G)$ has a faithful representation of degree $8$ over $\Bbb F_3$. Thus $C_V(Z(R))=C_V(R)$, and then also $[V,Z(R)]=[V,R]$. This implies that both $C_V(R)$ and $[V,R]$ are $M$-invariant, and so $R$ is normal in $M$. As $Z(G)$ is not a direct factor of $G$, it now follows from Gasch\"utz's Theorem that $M/R$ is a non-split central extension of $D_4(2)$, and thus $M$ has a subgroup $K$ with $K\cong 2^{cdot}D_4(2)$, and with $A\leq K$. As $[V,R]$ is a quadratic module for $K$, Theorem 1.3 implies that $A$ is contained in a subgroup $N$ of $K$ of the form $(2^{1+6}_+)L_4(2)$, where $[O_2(N),A]$ is a quaternion group. Then 9.1 implies that $dim(C_V(R))=8$, that $A$ acts quadratically on $C_V(R)$, and $dim([C_V(R),A])=4$. Then also $dim([V,R])=16$, $dim([V,A])=12$, and $A$ acts quadratically on $V$. We now have $C_G(A)\cong 6^{\cdot}Suz$, by Theorem 1.2. Then $A$ is not contained in the center of a Sylow $3$-subgroup of $G$, and so there exists a subgroup $G_1$ of $G$ with $G_1\cong 6^{\cdot}Suz$, such that $A\leq G_1$ and $A\nleq Z(G_1)$. Set $V_1=[V,Z(G_1)]$. As $Z(G_1)$ is conjugate to $A$, we have $dim(V_1)=12$, and $V_1$ is then a quadratic module for $G_1/O_3(G_1)$. There are subgroups $G_2$ and $G_3$ of $G_1$, with $G_2\cong 2^{\cdot}G_2(4)$ and with $G_2\geq G_3\cong 2^{\cdot}J_2$. By considering the structure of centralizers of elements of order $3$ in the groups $G_i$, one finds that $G_2$ and $G_3$ contain conjugates of $A$, and thus $V_1$ is a quadratic module for $G_i$, $1\leq i\leq 3$. As $O_2(G_i)=Z(G)$, for all $i$, all irreducible constituents for $G_i$ in $V_1$ are non-trivial. As $25$ divides the order of $G_3$ and does not divide the order of $SL(6,3)$, we conclude that $V_1$ is irreducible for each $G_i$. 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