The Local Structure Theorem: The wreath product case

TL;DR

This paper investigates a specific class of finite groups with large p-subgroups, focusing on wreath product cases within the Local Structure Theorem, to enhance understanding of groups of Lie type and finite simple groups.

Contribution

It introduces a detailed analysis of wreath product cases in the Local Structure Theorem, providing new insights into the structure of groups with large p-subgroups.

Findings

Characterization of wreath product cases in the Local Structure Theorem

Enhanced understanding of groups of Lie type in characteristic p

Implications for the classification of finite simple groups

Abstract

Groups with a large $p$-subgroup, $p$ a prime, include almost all of the groups of Lie type in characteristic $p$ and so the study of such groups adds to our understanding of the finite simple groups. In this article we study a special class of such groups which appear as wreath product cases of the Local Structure Theorem.

Equations

$$C_G(O_p(L))\le O_p(L).$$

$${\mathcal{L}}_G(S)=\{L\le G\mid S\le L,O_p(L)\ne 1,C_G(O_p(L))\le O_p(L)\}.$$

$$L^{\circ }=⟨Q^L⟩,\overline{L}=L/C_L(Y_L)\text{ and }{V}_L=[Y_L,L^{\circ }]$$

$$V=\underset{K\in \mathcal{K}}{⨁}[V,K]\text{ and }⟨\mathcal{K}⟩={\text{bigtimes}}_{K\in \mathcal{K}}K,$$

$$\overline{L^{\circ }}\cong {\Omega }_4^{+}(2)\cong {\mathrm{SL}}_2(2)\times {\mathrm{SL}}_2(2)$$

$$W=⟨(V_L\cap Q{)}^D⟩,$$

$$U_L=⟨(W\cap Q_L{)}^L⟩$$

$$Z=C_{V_L}(Q).$$

$$Y_{L_0}=Y_L={\Omega }_1(Z(O_p(L_0)))$$

$$L^{\circ }=⟨Q^L⟩=⟨Q^{C_L(Y_L)L_0}⟩=⟨Q^{L_0}⟩=L_0^{\circ }.$$

$$Y_L={\Omega }_1(Z(Q_L)).$$

$$[O_p(X),O^p(R)]=[O_p(X),O^p(R),O^p(R)]\le [U,O^p(R)]\le [U,O_p(X);n]$$

$$[O_p(X),O^p(R),U]\le [[U,O_p(X);n],O_p(X)]=[U,O_p(X);n+1].$$

$$[U,O^p(R),O_p(X)]\le [[U,O_p(X);n],O_p(X)]=[U,O_p(X);n+1]$$

$$[U,O_p(X),O^p(R)]\le [U,O_p(X);n+1].$$

$$[U,O^p(R)]=[U,O^p(R),O^p(R)]\le [U,O_p(X),O^p(R)]\le [U,O_p(X);n+1].$$

$$[O_p(X),O^p(R)]=[O_p(X),O^p(R),O^p(R)]\le [U,O^p(R)]=1.$$

$$\theta :Y$$

$$y$$

$$(yz)\theta =[yz,x]=[y,x{]}^z[z,x]=[y,x][z,x]=(y)\theta (z)\theta ,$$

$$(y\theta {)}^{\ell }=[y,x{]}^{\ell }=[y^{\ell },x^{\ell }]=[y^{\ell },xc]=[y^{\ell },c][y^{\ell },x{]}^c=[y^{\ell },x]=(y^{\ell })\theta .$$

$$C_Y(x)\ge C_Y(R)\ge [Y,R]\ge [Y,x]$$

$$\theta :R$$

$$w$$

$$(uZ){\theta }_{w_1{w}_2}=[u,w_1{w}_2]=[u,w_2{]}^{w_1}[u,w_1]=[u,w_1][u,w_2]=(uZ){\theta }_{w_1}(uZ){\theta }_{w_2}$$

$$(uZ){\theta }_{w^{\ell }}=[u,w^{\ell }]=[u^{{\ell }^{-1}},w{]}^{\ell }=(u)({\theta }_w\cdot \ell ).$$

$$\mathrm{Hom}(U/Z,Z)\cong (U/Z{)}^{\ast }\otimes Z=\underset{i=1}{\overset{n}{⨁}}(U/Z{)}^{\ast }$$

$$N=N_1\times N_2\cdots \times N_s,$$

$$[Q_L,O^p(L)]\le [Q_L,⟨W,W^m⟩]=[Q_L,W][Q_L,W^m]\le (W\cap Q_L)(W^m\cap Q_L).$$

$$[Q_L,W,W]\le [W,W]\le V_L$$

$$[I,W]\le [W,W]\le V_L$$

Full text

Dedicated to the memory of Kay Magaard

1. Introduction

Throughout this article $p$ is a prime and $G$ is a finite group. We say that $L\leq G$ has characteristic $p$ if

[TABLE]

For $T$ a non-trivial $p$-subgroup of $G$, the subgroup $N_{G}(T)$ is called a * $p$-local* subgroup of $G$. By definition $G$ has local characteristic $p$ if all $p$-local subgroups of $G$ have characteristic $p$ and $G$ has parabolic characteristic $p$ if all $p$-local subgroups containing a Sylow $p$-subgroup of $G$ have characteristic $p$.

A group $K$ is called a ${\mathcal{K}}$-group if all its composition factors are from the known finite simple groups. So, if $K$ is a simple ${\mathcal{K}}$-group, then $K$ is a cyclic group of prime order, an alternating group, a simple group of Lie type or one of the 26 sporadic simple groups. A group $G$ is a ${\mathcal{K}}_{p}$-group, provided all subgroups of all $p$-local subgroups of $G$ are ${\mathcal{K}}$-groups. This paper is part of a programme to investigate the structure of certain $\mathcal{K}_{p}$-groups. See [MSS1, MSS2] for an overview of the project.

Of fundamental importance to the development of the programme are large subgroups of $G$: a $p$-subgroup $Q$ of $G$ is large if

• (i)

$C_{G}(Q)\leq Q$; and

• (ii)

$N_{G}(U)\leq N_{G}(Q)$ for all $1\neq U\leq C_{G}(Q)$.

For example, if $G$ is a simple group of Lie type defined in characteristic $p$, $S\in{\operatorname{Syl}}_{p}(G)$ and $Q=O_{p}(C_{G}(Z(S)))$, then $Q$ is a large subgroup of $G$ unless there is some degeneracy in the Chevalley commutator relations which define $G$. This means that $Q$ is a large subgroup of $G$ unless $G$ is one of ${\operatorname{Sp}}_{2n}(2^{k})$, $n\geq 2$, ${\operatorname{F}}_{4}(2^{k})$ or ${\operatorname{G}}_{2}(3^{k})$.

If $Q$ is a large subgroup of $G$, then it is easy to see that $O_{p}(N_{G}(Q))$ is also a large $p$-subgroup of $G$. Thus we also assume that

• (iii)

$Q=O_{p}(N_{G}(Q))$.

One of the consequences of $G$ having a large $p$-subgroup is that $G$ has parabolic characteristic $p$. In fact any $p$-local subgroup of $G$ containing $Q$ is of characteristic $p$ [MSS2, Lemma 1.5.5 (e)]. Further, if $Q\leq S\in\mathrm{Syl}_{p}(G)$, then $Q$ is weakly closed in $S$ with respect to $G$ ($Q$ is the unique $G$-conjugate of $Q$ in $S$) [MSS2, Lemma 1.5.2 (e)]. A significant part of the programme described in [MSS1] aims to determine the groups which possess a large $p$-subgroup. This endeavour extends and generalizes earlier work of Timmesfeld and others in the original proof of the classification theorem where groups with a so-called large extraspecial $2$-subgroup were investigated. The state of play at the moment is that the Local Structure Theorem has been completed and published [MSS2]. To describe this result we need some further notation.

For a finite group $L$, $Y_{L}$ denotes the unique maximal elementary abelian normal $p$-subgroup of $L$ with $O_{p}(L/C_{L}(Y_{L}))=1$. Such a subgroup exists [MSS1, Lemma 2.0.1(a)]. From now on assume that $G$ is a finite $\mathcal{K}_{p}$-group, $S$ a Sylow $p$-subgroup of $G$ and $Q$ a large $p$-subgroup of $G$ with $Q\leq S$ and $Q=O_{p}(N_{G}(Q))$. We define

[TABLE]

Under the assumption that $S$ is contained in at least two maximal $p$-local subgroups, for $L\in\mathcal{L}_{G}(S)$ with $L\not\leq N_{G}(Q)$, the Local Structure Theorem provides information about $L/C_{L}(Y_{L})$ and its action on $Y_{L}$. Given the Local Structure Theorem there are two cases to treat in order to fully understand groups with a large $p$-subgroup. Either there exists $L\in\mathcal{L}_{G}(S)$ with $Y_{L}\not\leq Q$ or, for all $L\in\mathcal{L}_{G}(S)$, $Y_{L}\leq Q$. Research in the first case has just started and, for this situation, this paper addresses the wreath product scenario in the Local Structure Theorem [MSS2, Theorem A (3)]. This case is separated from the rest because of the special structure of $L$ and $Y_{L}$. This structure allows us to use arguments measuring the size of certain subgroups to reduce to three exceptional configurations and has a distinct flavour from the remaining cases. For instance, the groups which are examples in the wreath product case typically have $Q$ of class 3 whereas in the more typical cases it has class at most 2. The configurations in the Local Structure Theorem which are not in the wreath product case and have $Y_{L}\not\leq Q$ will be examined in a separate publication as there are methods which apply uniformly to cover many possibilities at once. Contributions to the $Y_{L}\leq Q$ for all $L\in\mathcal{L}_{G}(S)$ are the subject of [PPS].

For $L\in\mathcal{L}_{G}(S)$ with $Q$ not normal in $L$ we set

[TABLE]

and use this notation throughout the paper. Set $q=p^{a}$. We recall from [MSS2, Remark A.25] the definition of a natural wreath ${\operatorname{SL}}_{2}(q)$-module for the group $X$ with respect to $\mathcal{K}$: suppose that $X$ is a group, $V$ is a faithful $X$-module and $\mathcal{K}$ is a non-empty $X$-invariant set of subgroups of $X$. Then $V$ is a natural ${\operatorname{SL}}_{2}(q)$-wreath product module for $X$ with respect to $\mathcal{K}$ if and only if

[TABLE]

and, for each $K\in\mathcal{K}$, $K\cong{\operatorname{SL}}_{2}(q)$ and $[V,K]$ is the natural ${\operatorname{SL}}_{2}(q)$-module for $K$.

We now describe the wreath product case in [MSS2, Theorem A (3)]. For $L\in\mathcal{L}_{G}(S)$ with $L\not\leq N_{G}(Q)$, $L$ is in the wreath product case provided

• •

there exists a unique $\overline{L}$-invariant set $\mathcal{K}$ of subgroups of $\overline{L}$ such that $V_{L}$ is a natural ${\operatorname{SL}}_{2}(q)$-wreath product module for $\overline{L}$ with respect to $\mathcal{K}$.

• •

$\overline{L^{\circ}}=O^{p}(\langle\mathcal{K}\rangle)\overline{Q}$ and $Q$ acts transitively on $\mathcal{K}$ by conjugation.

• •

$Y_{L}=V_{L}$ or $p=2$, $|Y_{L}:V_{L}|=2$, $\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(4)$ or $\Gamma{\operatorname{SL}}_{2}(4)$ and $V_{L}\not\leq Q$.

We say that $\overline{L}$ is properly wreathed if $|\mathcal{K}|>1$.

There are overlaps between the wreath product case and some other divisions in the Local Structure Theorem.

If $\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(q)$ with $V_{L}=Y_{L}$, then this situation can be inserted in the linear case of [MSS2, Theorem A (1)] by including $n=2$ is that case. Suppose that $|\mathcal{K}|=2$ and $K\cong{\operatorname{SL}}_{2}(2)$. If $\overline{Q}$ is a fours group, then, as $\overline{Q}$ conjugates $\overline{K_{1}}$ to $\overline{K_{2}}$,

[TABLE]

and $Y_{L}$ is the tensor product module. This is an example in the tensor product case of [MSS2, Theorem A (6)]. We declare $L$ to be in the unambiguous wreath product case if these two ambiguous configurations do not occur. The ambiguous cases will be handled in a more general setting in a forthcoming paper mentioned earlier.

Main Theorem**.**

Suppose that $p$ is a prime, $G$ is a finite group, $S$ a Sylow $p$-subgroup of $G$ and $Q\leq S$ is a large $p$-subgroup of $G$ with $Q=O_{p}(N_{G}(Q))$. If there exists $L\in\mathcal{L}_{G}(S)$ with $L$ in the unambiguous wreath product case and $V_{L}\not\leq Q$, then $G\cong{\operatorname{Mat}}(22)$, ${\operatorname{Aut}}({\operatorname{Mat}}(22))$, ${\operatorname{Sym}}(8)$, ${\operatorname{Sym}}(9)$ or ${\operatorname{Alt}}(10)$.

The proof of this theorem splits into four parts. First, in Section 3, we show that in the properly wreathed case we must have $q=|\mathcal{K}|=2$ and, as $L$ is unambiguous, $\overline{S}=\overline{Q}\cong{\operatorname{Dih}}(8)$ and $\overline{L^{\circ}}\cong\mathrm{O}_{4}^{+}(2)$. If $|\mathcal{K}|=1$, we show that $\overline{L^{\circ}}\cong\Gamma{\operatorname{SL}}_{2}(4)$ or ${\operatorname{SL}}_{2}(4)$ and $V_{L}$ is the natural module with $|Y_{L}:V_{L}|\leq 2$, where, if $\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(4)$, $|Y_{L}:V_{L}|=2$ holds. In the following three sections, we determine the groups corresponding to these three cases. Finally the Main Theorem follows by combining Propositions 3.5, 4.1, 5.1 and 6.2.

In [PPS] the authors proved that the unambiguous wreath product case does not lead to examples if for all $L\in\mathcal{L}_{G}(S)$ we have $Y_{L}\leq Q$, with the additional assumption that $G$ is of local characteristic $p$. In this paper we do not make the assumption that $G$ is of local characteristic $p$.

In the Local Structure Theorem there is also a possibility that $L\in\mathcal{L}_{G}(S)$ is of weak wreath type. Any such group is contained in one, which is of unambiguous wreath type. A corollary of our theorem is

Corollary**.**

Suppose that $p$ is a prime, $G$ is a finite group, $S$ a Sylow $p$-subgroup of $G$ and $Q\leq S$ is a large $p$-subgroup of $G$ with $Q=O_{p}(N_{G}(Q))$. If $L\in\mathcal{L}_{G}(S)$ is of weak wreath product type, then either $G$ is as in the Main Theorem or $V_{L}\leq Q$.

In addition to the notation already introduced, we will use the following

Notation**.**

For $p$ a prime, $G$ a group with a large $p$-subgroup $Q=O_{p}(N_{G}(Q))$ and $L\in\mathcal{L}_{G}(S)$, we set $Q_{L}=O_{p}(L)$ and assume that $V_{L}\not\leq Q$. Define $D=\langle V_{L}^{N_{G}(Q)}\rangle(L\cap N_{G}(Q))\in\mathcal{L}_{G}(S)$. Furthermore, set

[TABLE]

[TABLE]

and

[TABLE]

Notice that for $L_{0}=N_{L}(S\cap C_{L}(Y_{L}))$, we have $L=C_{L}(Y_{L})L_{0}$ and $C_{L}(Y_{L})\leq D$. Further

[TABLE]

by [MSS2, Lemma 1.2.4 (i)]. Since $C_{L}(Y_{L})$ normalizes $Q$,

[TABLE]

Therefore, if $L$ is in the unambiguous wreath product case, then so is $L_{0}$. Hence we also assume that $L=L_{0}$ and so

[TABLE]

2. Preliminaries

In this section we present some lemmas which will be used in the forthcoming sections.

Lemma 2.1**.**

Suppose that $X$ is a group, $E=O_{2}(X)$ is elementary abelian of order $16$ and $X/E\cong{\operatorname{Alt}}(6)$ induces the non-trivial irreducible part of the $6$-point permutation module on $E$. Then $X$ splits over $E$.

Proof.

Choose $R\leq X$ such that $R/E\cong{\operatorname{Sym}}(4)$ and $Z(R)=1$. Let $T\in\mathrm{Syl}_{3}(R)$. As $T$ acts fixed-point freely on $O_{2}(R)$, $N_{R}(T)\cong{\operatorname{Sym}}(3)$ and so there are involutions in $X/E$. Hence, as $X/E$ has one conjugacy class of involutions, there are involutions in $O_{2}(R)\setminus E$. Therefore $O_{2}(R)/Z(O_{2}(R))$ is elementary abelian of order $16$. Now we consider $O_{2}(R)$. The fixed-point free action of $T$ on $O_{2}(R)/Z(O_{2}(R))$ implies there is partition of this group into five $T$-invariant subgroups of order 4. As $T$ acts fixed-point freely on $O_{2}(R)$ the preimages of all these fours groups are abelian. As there are involutions in $O_{2}(R)\setminus E$, there is a $T$-invariant fours group $F^{*}\leq O_{2}(R)/Z(O_{2}(R))$ with $F^{*}\neq E/Z(O_{2}(R))$ and such that the preimage $F$ of $F^{*}$ is elementary abelian of order 16. Now the action of $X$ on $E$ shows that for any involution $i\in R\setminus E$ all involutions in the coset $Ei$ are conjugate to $i$ by an element of $E$. Hence all involutions in $O_{2}(R)\setminus E$ are in $F$. This shows that $F$ is invariant under $N_{R}(T)$.

Again there is a partition of $F$ into five groups of order four invariant under $T$. Let $t$ be an involution in $N_{R}(T)$. Then $|C_{F}(t)|=4$, where $|C_{E\cap F}(t)|=2$. Hence there is some fours group $F_{1}\leq F$, $F_{1}\not=E\cap F$ and $C_{F_{1}}(t)\not=1$. This shows that $F_{1}$ is normalized by $t$. Then $F_{1}\langle t\rangle\cong{\operatorname{Dih}}(8)$ is a complement to $E$. Using a result of Gaschütz [GLS2, Theorem 9.26] , $X$ splits over $E$. ∎

The next lemma is well-known.

Lemma 2.2**.**

Suppose that $X\cong{\operatorname{Sym}}(5)$, $F_{1}$ and $F_{2}$ are fours groups of $X$ with $F_{1}\leq{\operatorname{Alt}}(5)$ and $V$ is a non-trivial irreducible ${\operatorname{GF}}(2)X$-module. Then

1. (i)

$V$* is either the non-trivial irreducible part of the permutation module, which is the same as the natural $\mathrm{O}^{-}_{4}(2)$-module, or $V$ is the natural $\Gamma{\operatorname{L}}_{2}(4)$-module.* 2. (ii)

$F_{1}$* acts quadratically on $V$ if and only if $V$ is the natural $\Gamma{\operatorname{L}}_{2}(4)$-module.* 3. (iii)

$F_{2}$* acts quadratically on $V$ if and only if $V$ is the natural $\mathrm{O}_{4}^{-}(2)$-module.*

Lemma 2.3**.**

Suppose that $p$ is a prime, $X$ is a group of characteristic $p$ and $U$ is a normal $p$-subgroup of $X$. Let $R$ be a normal subgroup of $X$ with $R\leq C_{X}(U/[U,O_{p}(X)])$. If $[O_{p}(X),O^{p}(R)]\leq U$, then $R\leq O_{p}(X)$.

Proof.

It suffices to prove that $O^{p}(R)=1$. Suppose that $n\geq 1$ is such that $[U,O^{p}(R)]\leq[U,O_{p}(X);n]$. Then

[TABLE]

and so

[TABLE]

We also have

[TABLE]

and thus the Three Subgroups Lemma implies

[TABLE]

This yields

[TABLE]

Since $O_{p}(X)$ is nilpotent, we deduce $[U,O^{p}(R)]=1$. Hence

[TABLE]

As $X$ has characteristic $p$, $O^{p}(R)=1$ and so $R\leq O_{p}(X)$ as claimed. ∎

Lemma 2.4**.**

Assume that $X$ is a group, $Y$ is a normal subgroup of $X$ and $xC_{X}(Y)\in Z(X/C_{X}(Y))$. If $[Y,x]\leq Z(Y)$, then $Y/C_{Y}(x)\cong[Y,x]$ as $X$-groups.

Proof.

Define

[TABLE]

Then $\theta$ is independent of the choice of the coset representative in $xC_{X}(Y)$.

For $y,z\in Y$,

[TABLE]

and, for $y\in Y$ and $\ell\in X$, as $[x,\ell]\in C_{R}(Y)$, $x^{\ell}=xc$ for some $c\in C_{X}(Y)$, and so

[TABLE]

Thus $\theta$ is an $X$-invariant homomorphism from $Y$ to $[Y,x]$. As $\ker\theta=C_{Y}(x)$, we have $Y/C_{Y}(x)\cong[Y,x]$ as $X$-groups. ∎

Lemma 2.5**.**

Assume that $p$ is a prime, $X$ is a group, $Y$ is an abelian normal $p$-subgroup of $X$ and $R$ is a normal $p$-subgroup of $X$ which contains $Y$. Suppose that $Y=[Y,O^{p}(X)]$, $[R,O^{p}(X)]\leq C_{R}(Y)$ and $R$ acts quadratically or trivially on $Y$. Suppose that no non-central $X$-chief factor of $Y/C_{Y}(R)$ is isomorphic to an $X$-chief factor of $[Y,R]$. Then $Y\leq Z(R)$.

Proof.

Assume that $R>C_{R}(Y)$. Using $[R,O^{p}(X)]\leq C_{R}(Y)$, we may select $x\in R\setminus C_{R}(Y)$ such that $xC_{X}(Y)\in Z(X/C_{X}(Y))^{\#}$. As $Y$ is abelian, $[Y,x]\leq Z(Y)$ and so \freflem:abstracting 1.4 applies to give $Y/C_{Y}(x)\cong[Y,x]$ as $X$-groups. As $R$ acts quadratically on $Y$,

[TABLE]

and so the hypothesis on non-central $X$-chief factors now gives $Y/C_{Y}(x)$ and $[Y,x]$ only have central $X$-chief factors. In particular, $Y=[Y,O^{p}(X)]\leq C_{Y}(x)$ and this contradicts the initial choice of $x\in R\setminus C_{R}(Y)$. Hence $Y\leq Z(R)$. ∎

Lemma 2.6**.**

Suppose that $p$ is a prime, $X$ is a group, $V\leq U$ are normal $p$-subgroups of $X$, and $Q$ is a large $p$-subgroup of $X$ which is not normal in $X$. Assume that $V$ is a non-trivial irreducible ${\operatorname{GF}}(p)X$-module and $U/V$ is centralized by $O^{p}(X)$.Then

1. (i)

$U$* is elementary abelian; and* 2. (ii)

if $U\not\leq\Omega_{1}(Z(O_{p}(X)))$, then $O_{p}(X)/C_{O_{p}(X)}(U)$ contains a non-central chief factor isomorphic to $V$ as a ${\operatorname{GF}}(p)X$-module.

Proof.

Set $Z_{X}=\Omega_{1}(Z(O_{p}(X)))$. We have $[U,O^{p}(X)]\leq V\leq Z_{X}$ as $V$ is irreducible. As $O^{p}(X)$ does not centralize $U/\Phi(U)$ by Burnside’s Lemma [GLS2, Proposition 11.1] and $V$ is a non-trivial irreducible $X$-module, $V\not\leq\Phi(U)$ and $\Phi(U)$ is centralized by $O^{p}(X)$. Therefore $\Phi(U)\cap Z_{X}$ is centralized by $O^{p}(X)$ and is normalized by $Q$. Since $Q$ is large and $O^{p}(X)\not\leq N_{X}(Q)$, we deduce $\Phi(U)\cap Z_{X}=1$. Thus $\Phi(U)=1$ and so $U$ is elementary abelian. Hence (i) holds.

Set $Y=O_{p}(X)$ and assume that $U\not\leq Z_{X}$. Select $x\in U\setminus Z_{X}$ such that $[X,x]\leq U\cap Z_{X}\leq Z(Y)$. Then $xC_{X}(Y)\in Z(X/C_{X}(Y))$. Thus \freflem:abstracting 1.4 implies $Y/C_{Y}(x)\cong[Y,x]\leq U\cap Z_{X}$ and this isomorphism is as $X$-groups. Since $[Y,x]$ is normalized by $Q$, $[Y,x]\neq 1$ and $Q$ is large, $O^{p}(X)$ does not centralize $[Y,x]$. Thus $[Y,x]\geq V$ as $[U,O^{p}(X)]\leq V$. This proves (ii). ∎

Lemma 2.7**.**

Assume that $p$ is a prime, $X$ is a group, $U$ is an elementary abelian normal subgroup of $X$, $U=[U,O^{p}(X)]$ and $O_{p}(X)$ acts quadratically and non-trivially on $U$. Set $R=O_{p}(X)$, $W=R/C_{R}(U)$, and $Z=[U,R]$. Then $W$, $U/Z$ and $Z$ are $X/R$-modules and $W$ is isomorphic to an $X/R$-submodule of ${\operatorname{Hom}}(U/Z,Z)$. In particular, if $Z$ is centralized by $X$, then the set of $X$-chief factors of $W$ can be identified with a subset of the ${\operatorname{GF}}(p)$-duals of the $X$-chief factors of $U/Z$.

Proof.

Since $R$ acts quadratically on $U$, $W$ is elementary abelian. Furthermore, $R$ centralizes $W$, $U/Z$ and $Z$. Hence all of these groups can be regarded as ${\operatorname{GF}}(p)X/R$-modules. For $w\in R$, define

[TABLE]

The calculation in the proof of \freflem:abstracting 1.4 shows that the commutator $[u,w]$ defines a homomorphism from $U$ to $Z$ and, as $w$ centralizes $Z$, $\theta_{w}$ is a well-defined homomorphism from $U/Z$ to $Z$. Thus $\theta$ is a well-defined map. Consider $w_{1},w_{2}\in R$, $uZ\in U/Z$ and $\ell\in X$. Then

[TABLE]

which means $\theta_{w_{1}w_{2}}=\theta_{w_{1}}\theta_{w_{2}}$ and so $\theta$ is a group homomorphism. We show that $\theta$ is an $X$-module homomorphism. So let $\ell\in X$, $uZ\in U/Z$ and $w\in R$. Then $(w^{\ell})\theta=\theta_{w^{\ell}}$ and

[TABLE]

Since $\ker\theta=C_{R}(U)$, this completes the proof of the main claim.

If $Z$ is centralized by $X$, then

[TABLE]

where $n$ is such that $|Z|=p^{n}$. This completes the proof of the lemma. ∎

Lemma 2.8**.**

Suppose that $V$ is a $p$-group and $X$ is a group which acts faithfully on $V$ with $O_{p}(X)=1$. Assume $A\leq X$ is an elementary abelian $p$-subgroup of order at least $p^{2}$ which has the property $C_{V}(A)=C_{V}(a)$ for all $a\in A^{\#}$. If $L$ is a non-trivial subgroup of $X$ and $L=[L,A]$, then $A$ acts faithfully on $L$.

In particular, $A$ centralizes every $p^{\prime}$-subgroup which it normalizes, $[A,F(X)]=1$, $E(X)\neq 1$ and, if $L$ is a component of $X$ which is normalized but not centralized by $A$, then $A$ acts faithfully on $L$.

Proof.

Suppose that $L=[L,A]$ is a non-trivial subgroup of $X$. Assume that there is $b\in A^{\#}$ with $[L,b]=1$. Then $L$ normalizes $C_{V}(b)$ and so, as $C_{V}(b)=C_{V}(A)$, $L=[L,A]$ centralizes $C_{V}(b)$. Since $L=[L,A]$, $L=O^{p}(L)$ and the Thompson $A\times B$-Lemma implies $[L,V]=1$, a contradiction. Hence $A$ acts faithfully on $L$.

Let $F$ be a $p^{\prime}$-subgroup of $X$ which is normalized by $A$. Then $F=\langle C_{F}(a)\mid a\in A^{\#}\rangle$. If $A$ does not centralizes $F$, then there exists $a\in A^{\#}$ such that $1\neq[C_{F}(a),A]=[C_{F}(a),A,A]$. Hence, taking $L=[C_{F}(a),A]$, we have $L=[L,A]$ and $a\in C_{A}(L)$, a contradiction. Hence $[F,A]=1$. Now $A$ centralizes $F(X)$ and therefore $E(X)\neq 1$.

If $L$ is a component of $X$ which is normalized by $A$, then either $[L,A]=L$ or $[L,A]=1$. If $[L,A]\neq 1$, then we have $A$ acts faithfully on $L$. ∎

Lemma 2.9**.**

Let $X$ be a group, $N$ a normal subgroup of $G$ and $T\in\mathrm{Syl}_{p}(X)$. Assume that $X=NT$, $C_{T}(N)=1$, $q=p^{a}$ and

[TABLE]

where $N_{i}\cong{\operatorname{SL}}_{2}(q)$ for $1\leq i\leq s$. Then the $p$-rank of $G$ is $sa$.

Proof.

Assume first that $q=2$. Then $T$ acts faithfully on $O_{3}(N)$. As the $2$-rank of ${\operatorname{GL}}_{s}(3)$ is $s$, we are done. Similarly, if $q=3$, then $T$ acts faithfully on $O_{2}(N)/Z(N)$, which is elementary abelian of order $2^{2s}$ we are done as ${\operatorname{GL}}_{2s}(2)$ has $3$-rank $s$.

Thus we may assume that $q>3$. In particular, the subgroups $N_{i}$ are quasisimple and $T$ permutes the set $\{N_{i}\mid 1\leq i\leq s\}$.

Assume that $p$ is odd. Let $A$ be an elementary abelian subgroup in $T$ of maximal rank and assume that $A\not\leq N$. Then by Thompson replacement [GLS2, Theorem 25.2] we may assume that $A$ acts quadratically on $T\cap N$. This shows that $A$ has to normalize each $N_{i}$. As non-trivial field automorphisms are not quadratic on $T\cap N_{i}$, we get that $A$ centralizes $T\cap N$ and so $A\leq T\cap N$, the assertion.

Assume that $q=2^{a}$ with $a\geq 2$. Let $B=N_{N}(T\cap N)$. We have that $T$ normalizes $B$ and $T/(T\cap N)$ acts faithfully on $B/(T\cap N)$. Now the Thompson dihedral Lemma [GLS2, Lemma 24.1] says that for any elementary abelian subgroup $A$ of $T$ we have a $B$-conjugate $A^{g}$ such that $U=\langle A,A^{g}\rangle(T\cap N)/(T\cap N)$ is a direct product of $r$ dihedral groups where $2^{r}=|A/(A\cap N)|\leq 2^{s}$ and $A(T\cap N)/(T\cap N)$ is a Sylow 2-subgroup of $U$. Set $T_{1}=[O_{2^{\prime}}(U),T\cap N]$. As $U$ is generated by two conjugates of $A$ we see that $|T_{1}|=|C_{T_{1}}(A/A\cap N)|^{2}$. This now shows that $|A|\leq|T\cap N|$, the assertion again. This proves the lemma. ∎

In the next two lemmas we use the notation presented in the introduction though we do not assume that $L$ is unambiguous.

Lemma 2.10**.**

Suppose that $L\in\mathcal{L}_{G}(S)$, $L\not\leq N_{G}(Q)$ and $V_{L}=[Y_{L},L^{\circ}]$. Then

• (i)

$C_{Y_{L}}(L^{\circ})=1$.

• (ii)

$\Omega_{1}(Z(S))\leq V_{L}$.

• (iii)

If $V_{L}$ is an irreducible $L$-module, $V_{L}\not\leq Q$ and $\Omega_{1}(Z(Q_{L}))<Q_{L}$, then $V_{L}\leq Q_{L}^{\prime}\leq\Phi(Q_{L})$.

Proof.

As $C_{Y_{L}}(L^{\circ})\leq C_{G}(Q)$ is normalized by $L$, (i) is a consequence of $Q$ being large.

By [MSS2, Lemma 1.24 (g)], $\Omega_{1}(Z(S))\leq Y_{L}$ now Gaschütz Theorem [GLS2, Theorem 9.26] and (i) give (ii).

Assume that $N$ is a non-trivial normal $p$-subgroup of $L$. Then $\Omega_{1}(Z(S))\cap N\neq 1$. Since $V_{L}$ is irreducible as a $L$-module, (ii) gives $V_{L}\leq N$. In particular, as $V_{L}\not\leq Q$, $N\not\leq Q$.

Suppose that $Q_{L}$ is abelian. Then, as $Q=O_{p}(N_{G}(Q))$ and $[Q,Q_{L},Q_{L}]\leq Q_{L}^{\prime}=1$, $Q_{L}$ is quadratic on $Q$, and hence $Q_{L}Q/Q$ is elementary abelian and so $\Phi(Q_{L})\leq Q$. By the remark earlier taking $N=\Phi(Q_{L})$ we obtain $\Phi(Q_{L})=1$, contrary to $\Omega_{1}(Z(Q_{L}))<Q_{L}$. Hence $Q_{L}$ is non-abelian. Thus $Q_{L}^{\prime}\neq 1$ and so, as $V_{L}$ is irreducible, $V_{L}\leq Q_{L}^{\prime}\leq\Phi(Q_{L})$. This proves (iii). ∎

Lemma 2.11**.**

Suppose that $L\in\mathcal{L}_{G}(S)$, $L\not\leq N_{G}(Q)$ and $V_{L}=[Y_{L},L^{\circ}]$. Assume that $Y_{L}=\Omega_{1}(Z(Q_{L}))$, $m\in L$ and $O^{p}(L)Q_{L}\leq KQ_{L}$, where $K=\langle W,W^{m}\rangle$. Then $O^{p}(L)\leq K$ and the following hold

1. (i)

$[O^{p}(L),Q_{L}]\leq[W,Q_{L}][W^{m},Q_{L}]\leq(W\cap Q_{L})(W^{m}\cap Q_{L})=U_{L}.$** 2. (ii)

If $[W,W]\leq V_{L}$, then $W$ acts quadratically on the non-central chief factors of $Q_{L}/V_{L}$.

Assume, in addition, that $V_{L}$ is irreducible as a $K$-module, $[V_{L},W,W]\neq 1$, and $[W,W]\leq V_{L}$. Then

1. (iii)

$W\cap W^{m}\cap Q_{L}\leq Y_{L}$; 2. (iv)

$U_{L}/Y_{L}$* is elementary abelian or trivial; and* 3. (v)

either $Q_{L}=Y_{L}$ or $U_{L}^{\prime}\geq V_{L}$.

Proof.

Since $W$ and $W^{m}$ are normalized by $Q_{L}$, $K=\langle W,W^{m}\rangle$ is normalized by $Q_{L}K$ and so $O^{p}(L)\leq K$. Since $W$, $W^{m}$, $[Q_{L},W]$ and $[Q_{L},W^{m}]$ are normalized by $Q_{L}$, we have

[TABLE]

In particular, $A=(W\cap Q_{L})(W^{m}\cap Q_{L})$ is normalized by $O^{p}(L)$. Since $(W\cap Q_{L})^{L}=(W\cap Q_{L})^{SO^{p}(L)}=(W\cap Q_{L})^{O^{p}(L)}$, we have $A=U_{L}$. Thus (i) holds.

By the additional hypothesis,

[TABLE]

and so $W$ acts quadratically on all the non-central $L$-chief factors in $Q_{L}/V_{L}$, which is (ii).

Notice that part (ii), $V_{L}$ irreducible as a $K$-module and $[V_{L},W,W]\neq 1$ together imply that the non-central $K$-chief factors in $Q_{L}/V_{L}$ are not isomorphic to $V_{L}$.

Set $I=W\cap W^{m}\cap Q_{L}$. Then $I\leq W\cap W^{m}$ and so

[TABLE]

and

[TABLE]

Hence $IV_{L}/V_{L}$ is centralized by $\langle W,W^{m}\rangle=K$. As $W$ acts quadratically on all the non-central chief factors of $K$ in $Q_{L}/V_{L}$ by (ii) and by assumption, $W$ does not act quadratically on $V_{L}$, \freflem: U¿YL abelian implies that $I\leq\Omega_{1}(Z(Q_{L}))=Y_{L}$. This proves (iii).

Since $W$ is generated by elements of order $p$, $W/[W,W]$ is elementary abelian and therefore, as $[W,W]\leq V_{L}$, $WV_{L}/V_{L}$ is also elementary abelian. Since $W\cap Q_{L}$ and $Q_{L}\cap W^{m}$ normalize each other parts (i) and (iii) give (iv).

If $V_{L}\not\leq U_{L}^{\prime}$ and $Q_{L}\not=Y_{L}$, then, as $U_{L}/Y_{L}$ is elementary abelian by (iv), \freflem:VL sub QL’ (ii) implies $U_{L}$ is elementary abelian. Select $E$ with $Q_{L}\geq E>V_{L}$ of minimal order such that $E=[E,O^{p}(L)]$ and $E/V_{L}$ has a non-central $K$-chief factor. Then

[TABLE]

Furthermore, $V_{L}[E,Q_{L}]<E$ and so $[[E,Q_{L}],O^{p}(L)]\leq V_{L}$. Therefore \freflem: U¿YL abelian implies that $[E,Q_{L}]\leq Y_{L}$ and so $Q_{L}$ acts quadratically on $E$. Hence \freflem:isochieffactors implies that $E\leq Y_{L}$, a contradiction. Hence $U_{L}^{\prime}$ is non-trivial and it follows that $V_{L}\leq U_{L}^{\prime}$. ∎

3. The reduction

We use the notation presented in the introduction. For the rest of this article we have $L\in\mathcal{L}_{G}(S)$ with $Q$ not normal in $L$ and $L$ is in the unambiguous wreath product case. This means that $Y_{L}=V_{L}$ unless we are in the special case that $\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(4)$ or $\Gamma\mathrm{SL}_{2}(4)$, $|Y_{L}:V_{L}|=2$ and

[TABLE]

We start with a general result which just requires $V_{L}\not\leq Q$.

Lemma 3.1**.**

The following hold.

1. (i)

$\langle V_{L}^{D}\rangle$* is not a $p$-group;* 2. (ii)

$[Q,\langle V_{L}^{D}\rangle]\leq W$; and 3. (iii)

$W\not\leq C_{G}(V_{L})$.

Proof.

Let $\tilde{C}=N_{G}(Q)$ and $K=\langle V_{L}^{\tilde{C}}\rangle$. As $D=KN_{L}(Q)$ and $N_{L}(Q)$ acts on $V_{L}$ we have $\langle V_{L}^{D}\rangle=\langle V_{L}^{K}\rangle$ is subnormal in $H$. If $\langle V_{L}^{D}\rangle$ is a $p$-group, we obtain $V_{L}\leq O_{p}(N_{G}(Q))=Q$ which is a contradiction. This proves (i).

We have $[Q,V_{L}]\leq Q\cap V_{L}\leq W$. As $W$ and $Q$ are normalized by $D$, (ii) holds.

Assume $W\leq C_{G}(V_{L})$. Then $[W,V_{L}]=1$ and so $[W,\langle V_{L}^{D}\rangle]=1$. Hence $X=O^{p}(\langle V_{L}^{D}\rangle)$ centralizes $Q$ by (ii). Since $C_{G}(Q)\leq Q$, we have $X\leq Q$. Thus $X=1$ and $\langle V_{L}^{D}\rangle$ is a $p$-group, which contradicts (i). Hence $W\not\leq C_{G}(V_{L})$. ∎

We adopt the following notation. Let $B\geq C_{L}(V_{L})$ be such that $\overline{B}=\langle\mathcal{K}\rangle$ and let $S_{0}=S\cap B$. We write $B=K_{1}\dots K_{s}$ where $K_{i}\geq C_{L}(V_{L})$, $\overline{K_{i}}\in\mathcal{K}$, $\overline{K_{i}}\cong{\operatorname{SL}}_{2}(q)$ and, for $1\leq i\leq s$, put

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

We begin by showing that $\overline{W}$ is not contained in the base group $\overline{B}$.

Lemma 3.2**.**

Suppose that $\overline{L}$ is either properly wreathed, or $q=p^{a}$ (where $p$ divides $a$) and some element of $\overline{L^{\circ}}$ induces a non-trivial field automorphism on $O^{p}(\overline{L^{\circ}})\cong{\operatorname{SL}}_{2}(q)$. Then $W$ is not contained in $S_{0}$. In particular, if $\overline{L}$ is properly wreathed with $q=s=2$, then $\overline{Q}$ is not cyclic of order $4$.

Proof.

Set $F=\bigcap_{g\in D}C_{Q}(V_{L})^{g}.$

Suppose that $W$ is contained in $S_{0}$. As $\overline{Q}$ normalizes $\overline{W}$ and acts transitively on $\mathcal{K}$ when $\overline{L}$ is properly wreathed and, as $V_{L}$ is the natural ${\operatorname{SL}}_{2}(q)$-module when $s=1$, and field automorphisms are present, the structure of $V_{L}$ yields

[TABLE]

Suppose that $g\in D$. Then using \freflem:w1(ii) and $(V_{L})^{g}=V_{L^{g}}$ yields

(3.2.1) **$[Z_{0},[V_{L^{g}},Q]]\leq[Z_{0},W]=1.$

We also remark that as $W\leq Q$, $Z_{0}\leq[V_{L},Q]\leq W=W^{g}\leq S_{0}^{g}$ and $Z_{0}\leq Z(W)$. In particular, as $S_{0}^{g}$ normalizes every element of $\mathcal{K}^{g}$, so does $Z_{0}$. Therefore, for $1\leq i\leq s$, $Z_{0}$ also normalizes each $K_{i}^{g}$ and so also $[Y_{L}^{g},K_{i}^{g}]=(V_{L}^{i})^{g}$.

If $s=1$ and we have field automorphisms in $\overline{L^{\circ}}$, then $[V_{L},Q]>Z_{0}$ and so \frefclm:notbase1 provides $Z_{0}\leq C_{Q}([V_{L^{g}},Q])=C_{Q}(V_{L^{g}})$. Thus

[TABLE]

in this case.

We will show that the same holds in the properly wreathed case. Because $Q$ acts transitively on $\mathcal{K}^{g}$,

[TABLE]

As $[Z_{0},[V_{L^{g}},Q]]=1$ by \frefclm:notbase1,

[TABLE]

Hence $Z_{0}\leq C_{Q}(V_{L^{g}})$ and this implies that

[TABLE]

in the properly wreathed case too. Therefore,

[TABLE]

Hence $V_{L}$ stabilizes the normal series $Q\geq W\geq W\cap F\geq 1$ in $D$ and so $V_{L}\leq O_{p}(D)$. But then $\langle V_{L}^{D}\rangle$ is a $p$-group contrary to \freflem:w1 (i). We conclude that $W\not\leq{S_{0}}$ as claimed.

If $q=s=2$ and $\overline{Q}$ is cyclic of order four, then, as $\overline{W}$ is generated by involutions, $\overline{W}=\overline{Q}\cap\overline{S}_{0}$, a contradiction. Thus $\overline{Q}$ is not cyclic of order $4$ in this case. ∎

We now reduce the properly wreathed case to one specific configuration which will be handled in detail in \frefsec:o4.

Proposition 3.3**.**

Assume that $\overline{L}$ is properly wreathed and unambiguous. Then $|\mathcal{K}|=2$, $q=2$, and $\overline{W}$ permutes $\mathcal{K}$ transitively by conjugation. Furthermore, $\overline{Q}=\overline{S}\cong{\operatorname{Dih}}(8)$, $\overline{L^{\circ}}\cong\mathrm{O}_{4}^{+}(2)$ and $Y_{L}=V_{L}$ is the natural $\mathrm{O}_{4}^{+}(2)$-module.

Proof.

Since $Q$ permutes $\mathcal{K}$ transitively by conjugation and $S_{0}$ normalizes $Q$, we have

(3.3.1) **

1. (i)

$\overline{Q\cap S_{0}}$ contains $[\overline{Q},\overline{S_{0}}]$; 2. (ii)

$|\overline{S_{0}}:\overline{Q\cap S_{0}}|\leq|\overline{S_{0}}:[\overline{Q},\overline{S}_{0}]|\leq q;$ and 3. (iii)

$\overline{[Q,S_{0}]}C_{\overline{L}}(\overline{K_{i}})/C_{\overline{L}}(\overline{K_{i}})\in\mathrm{Syl}_{p}(\overline{K_{i}}C_{\overline{L}}(\overline{K_{i}})/C_{\overline{L}}(\overline{K_{i}}))$.$\hfill\blacksquare$

As $W=\langle V_{L^{g}}\cap Q\mid g\in D\rangle$, \freflem:not base implies there exists $g\in D$ such that $V_{L^{g}}\cap Q\not\leq S_{0}$. We fix this $g$.

(3.3.2) ** We have $\overline{V_{L^{g}}\cap Q}\cap\overline{S_{0}}\neq 1$.

Suppose that $\overline{V_{L^{g}}\cap Q}\cap\overline{S_{0}}=1$. Then, as $\overline{Q\cap S_{0}}$ and $\overline{V_{L^{g}}\cap Q}$ normalize each other, $\overline{V_{L^{g}}\cap Q}$ centralizes $\overline{Q\cap S_{0}}$. If $\overline{V_{L^{g}}\cap Q}$ normalizes some $\overline{K_{i}}\in\mathcal{K}$, then, as $\overline{Q}$ acts transitively on $\mathcal{K}$ and normalizes $\overline{V_{L^{g}}\cap Q}$, $\overline{V_{L^{g}}\cap Q}$ normalizes every member of $\mathcal{K}$. As $\overline{V_{L^{g}}\cap Q}$ centralizes $\overline{[Q,S_{0}]}$, 3 (iii) implies that

[TABLE]

Since $Q$ acts transitively on $\mathcal{K}$, this is true for each $\overline{K_{i}}\in\mathcal{K}$. Thus

[TABLE]

which contradicts the choice of $g\in D$.

Hence $\overline{V_{L^{g}}\cap Q}$ does not normalize any member of $\mathcal{K}$. As $\overline{B}$ is a direct product we calculate that $C_{\overline{S}_{0}}(\overline{V_{L^{g}}\cap Q})$ has index at least $q^{p-1}$ in $\overline{S_{0}}$. However \frefclm:index [Q,S_0] (ii) states that $\overline{Q\cap S_{0}}$ has index at most $q$ in $\overline{S_{0}}$ and, as this subgroup is centralized by $\overline{V_{L^{g}}\cap Q}$, we deduce that

[TABLE]

Furthermore, as $\overline{V_{L^{g}}\cap Q}$ does not normalize any member of $\mathcal{K}$, if $s>2$, we have $C_{\overline{S}_{0}}(\overline{V_{L^{g}}\cap Q})$ has index at least $q^{2}$ in $\overline{S_{0}}$, and so we must have

[TABLE]

Since $\overline{V_{L^{g}}\cap Q}$ centralizes $[\overline{S_{0}},\overline{Q}]$ by \frefclm:index Q,S_0, no element in $\overline{V_{L^{g}}\cap Q}$ can act as a non-trivial field automorphism on $\overline{K_{1}}$ and so we infer from $\overline{V_{L^{g}}\cap Q}\cap\overline{S_{0}}=1$, that $|\overline{V_{L^{g}}\cap Q}|=2$. In particular, $|C_{V_{L}}(V_{L^{g}}\cap Q)|=q^{2}$ as $V_{L^{g}}\cap Q$ exchanges $V_{L}^{1}$ and $V_{L}^{2}$.

We know that $|V_{L^{g}}|=q^{4}$. As $|[V_{L^{g}},Q]|\geq q^{3}$, we have

[TABLE]

and we have just determined that

[TABLE]

Hence $V_{L^{g}}\cap Q\cap C_{G}(V_{L})$ has order at least $2^{3a-1}$, where $q=2^{a}$.

Assume that $a\neq 1$. Then, as $V_{L^{g}}^{1}$ has order $q^{2}$,

[TABLE]

It follows that $V_{L}\cap Q$ normalizes both $K_{1}^{g}$ and $K_{2}^{g}$. As $(V_{L}\cap Q)C_{L^{g}}(V_{L^{g}})/C_{L^{g}}(V_{L^{g}})$ is normalized by $Q$ and $Q$ permutes $\{K_{1}^{g},K_{2}^{g}\}$ transitively, $(V_{L}\cap Q)C_{L^{g}}(V_{L^{g}})/C_{L^{g}}(V_{L^{g}})$ does not centralize $K_{i}^{g}/C_{L^{g}}(V_{L^{g}})$ for $i=1,2$. Thus $|C_{V_{L^{g}}^{i}}(V_{L}\cap Q)|\leq q$ for $i=1,2$. But then

[TABLE]

which contradicts $a\neq 1$. We conclude that $q=s=2$ and $|\overline{V_{L^{g}}\cap Q}|=2$. Furthermore, $\overline{V_{L^{g}}\cap Q}$ is centralized by $\overline{Q}$ and so $\overline{Q}$ is elementary abelian of order $4$. It follows that $\overline{L^{\circ}}\cong\Omega_{4}^{+}(2)$ and $V_{L}$ is the natural module. Hence $L$ is ambiguous and we conclude that $\overline{V_{L^{g}}\cap Q}\cap\overline{S_{0}}\neq 1$.$\hfill\blacksquare$

(3.3.3) ** We have $|C_{V_{L}}(V_{L^{g}}\cap Q)|\leq q^{s/p}.$

We know $\overline{V_{L^{g}}\cap Q}\not\leq\overline{S_{0}}$ and $\overline{V_{L^{g}}\cap Q}\cap\overline{S}_{0}\neq 1$ by \frefclm:meet base. As $\overline{V_{L^{g}}\cap Q}$ is normalized by $\overline{Q}$, $\overline{V_{L^{g}}\cap Q}\cap\overline{S}_{0}\neq 1$ implies that

[TABLE]

If some element $d\in V_{L^{g}}\cap Q$ induces a non-trivial field automorphism on $\overline{K}_{i}$ for some $\overline{K_{i}}\in\mathcal{K}$, then $C_{V_{L}^{i}}({V_{L^{g}}\cap Q})\leq C_{Z_{i}}(d)$ has order $q^{1/p}$ and the result follows by transitivity of $\overline{Q}$ on $\mathcal{K}$. On the other hand, if $d\in V_{L^{g}}\cap Q$ has an orbit of length $p$ on $\mathcal{K}$, then $C_{\langle(V_{L}^{1})^{\langle d\rangle}\rangle}({V_{L^{g}}\cap Q})\leq C_{\langle Z_{1}^{\langle d\rangle}\rangle}(d)$ which has order $q$. Using the transitivity of $Q$ on $\mathcal{K}$, we deduce $|C_{V_{L}}(V_{L^{g}}\cap Q)|\leq q^{s/p}$. This proves the result. $\hfill\blacksquare$

As $Q$ acts transitively on the $\{V_{i}\mid 1\leq i\leq s\}$, we have $V_{L}=[V_{L},Q]V_{1}$. By \frefclm:meet base $\overline{Q}\cap\overline{S_{0}}\not=1$ and so $|[V_{1},Q]|\geq q$. In particular

[TABLE]

Since $V_{L}\cap Q\cap C_{L^{g}}(V_{L^{g}})\leq C_{V_{L}}(V_{L^{g}}\cap Q)$, \frefclm:cent bound and $|V_{L}|=q^{2s}$ together give

[TABLE]

On the other hand, by \freflem:prank the $p$-rank of $\overline{L}$ is $as$ where $q=p^{a}$. Hence

[TABLE]

and so

[TABLE]

In particular, \freflem:prank implies

(3.3.4) ** $|(V_{L}\cap Q)C_{L^{g}}(V_{L^{g}})/C_{L^{g}}(V_{L^{g}})|=q^{2}=2^{2a}$.

Assume that $q>2$. Since $S^{g}/S_{0}^{g}$ has $2$-rank $2$ and $V_{L}\cap Q$ is elementary abelian, $(V_{L}\cap Q\cap S_{0}^{g})C_{L^{g}}(V_{L^{g}})/C_{L^{g}}(V_{L^{g}})$ has rank at least $2a-2\neq 1$. Since $V_{L}\cap Q\cap S_{0}^{g}$ is normalized by $Q$ and $Q$ permutes $\{K_{1}^{g},K_{2}^{g}\}$ transitively, $V_{L}\cap Q\cap S_{0}^{g}$ contains an element which projects non-trivially on to both $S_{1}^{g}C_{L^{g}}(V_{L^{g}})/C_{L^{g}}(V_{L^{g}})$ and $S_{2}^{g}C_{L^{g}}(V_{L^{g}})/C_{L^{g}}(V_{L^{g}})$. Thus $V_{L}\geq[V_{L}\cap Q,[V_{L^{g}},Q]]\geq Z_{0}^{g}$. But then, using \frefclm:cent bound yields the contradiction

[TABLE]

Thus $q=s=2$. It follows from \freflem:not base that $W$ is transitive on $\mathcal{K}$ and $\overline{Q}\cong{\operatorname{Dih}}(8)$ or $\overline{Q}$ is elementary abelian of order $4$. The second possibility gives $\overline{L^{\circ}}\cong\Omega^{+}_{4}(2)$, which is ambiguous. This proves \frefprop:notwreath. ∎

Next we deal with the case $s=1$.

Proposition 3.4**.**

Suppose that $O^{p}(\overline{L^{\circ}})\cong{\operatorname{SL}}_{2}(q)$ where $q=p^{a}=r^{p}$, $V_{L}=Y_{L}$ is the natural $O^{p}(\overline{L^{\circ}})$-module and that some element of $\overline{L^{\circ}}$ induces a non-trivial field automorphism on $O^{p}(\overline{L^{\circ}})$. Then $p=2=r$.

Proof.

We may assume that $r^{p}>4$. By \freflem:not base we have that $W\not\leq S_{0}$ and, as $W$ is generated by elements of order $p$, we have that $|S_{0}W:S_{0}|=p$. As $Q$ is normal in $S$, $1\not=\overline{Q}\cap\overline{S}_{0}$, so $Z_{0}\leq Q\cap Y_{L}$. Furthermore, as $\overline{Q}$ contains elements which act as field automorphisms on $O^{p}(\overline{L^{\circ}})$,

[TABLE]

by assumption. Thus no element in $S\setminus Q_{L}$ centralizes a subgroup of index $p$ in $V_{L}\cap Q$.

Set $W_{1}=\langle Z_{0}^{D}\rangle$. As $Z_{0}$ centralizes $W\cap S_{0}$, every element of $Z_{0}$ centralizes a subgroup of index at most $p$ in $W$. As $W_{1}$ is generated by conjugates of $Z_{0}$, and these conjugates all contain elements which centralize a subgroup of index at most $p$ in $W$, $W_{1}$ is generated by elements which centralize a subgroup of index at most $p$ in $V_{L}\cap Q$. As no element in $S\setminus Q_{L}$ has this property, we conclude that $W_{1}\leq Q_{L}$. Hence $[V_{L},W_{1}]=1$. In particular $[V_{L}\cap Q,W_{1}]=1$ and so also $[W,Z_{0}]=[W,W_{1}]=1$. This shows $W\leq S_{0}$ and contradicts \freflem:not base. ∎

We collect the results of this section in the following proposition:

Proposition 3.5**.**

Suppose that $L\in\mathcal{L}_{G}(S)$, $L\not\leq N_{G}(Q)$, $V_{L}\not\leq Q$ and $L$ is in the unambiguous wreath product case. Then one of the following holds:

• (i)

$\overline{L^{\circ}}\cong\mathrm{O}_{4}^{+}(2)$* , $\overline{Q}=\overline{S}\cong{\operatorname{Dih}}(8)$ and $Y_{L}=V_{L}$ is the natural module.*

• (ii)

$\overline{L^{\circ}}\cong\Gamma{\operatorname{SL}}_{2}(4)$, $V_{L}$ is the natural ${\operatorname{SL}}_{2}(4)$-module and $|Y_{L}:V_{L}|\leq 2$.

• (iii)

$\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(4)$, $V_{L}$ is the natural module and $|Y_{L}:V_{L}|=2$.

Proof.

If $|\mathcal{K}|>1$, then (i) holds by \frefprop:notwreath, so we may assume that $|\mathcal{K}|=1$. As $L$ is unambiguous, either $Y_{L}\not=V_{L}$ or $\overline{L^{\circ}}\not\cong{\operatorname{SL}}_{2}(q)$. If $Y_{L}\not=V_{L}$, then by definition of the wreath product case, (ii) or (iii) holds. So we may assume $Y_{L}=V_{L}$ and $\overline{L^{\circ}}\not\cong{\operatorname{SL}}_{2}(q)$. Now (ii) holds by \frefprop:fieldauto. ∎

4. $\overline{L^{\circ}}\cong\mathrm{O}^{+}_{4}(2)$

In this section we analyse the configuration from \frefprop:unambiguous(i). We prove

Proposition 4.1**.**

Suppose that $L\in\mathcal{L}_{G}(S)$, $L\not\leq N_{G}(Q)$ and $L$ in the unambiguous wreath product case. If $Y_{L}\not\leq Q$ and $\overline{L^{\circ}}\cong\mathrm{O}^{+}_{4}(2)$, then $G\cong{\operatorname{Sym}}(8)$, ${\operatorname{Sym}}(9)$ or ${\operatorname{Alt}}(10)$.

Proof.

By \frefprop:unambiguous we have $\overline{Q}\cong{\operatorname{Dih}}(8)$. Since $Y_{L}$ is the natural $\mathrm{O}_{4}^{+}(2)$-module for $L/C_{L}(Y_{L})$ and $V_{L}$ is also the wreath product module for $L/C_{L}(Y_{L})$ with respect to $\{\overline{K_{1}},\overline{K_{2}}\}$, we have the following well known facts.

(4.1.1) **

1. (i)

$|[Y_{L},Q]|=2^{3}$, $|[Y_{L},Q,Q]|=2^{2}$ and $C_{Y_{L}}(Q)=[Y_{L},Q,Q,Q]$ has order $2$. 2. (ii)

$[Y_{L},S_{0}]=C_{Y_{L}}(S_{0})$ has order $2^{2}$; 3. (iii)

$|[Y_{L},Q^{\prime}]|=2^{2}$; 4. (iv)

$C_{L}([Y_{L},Q])\leq C_{L}(Y_{L})$.

Our first aim is to prove

(4.1.2) ** $\overline{W}$ is elementary abelian of order $2^{2}$, $[Y_{L},W]=[Y_{L},Q]=Y_{L}\cap Q$ and $[Y_{L},W,W]=C_{Y_{L}}(W)=C_{Y_{L}}(Q)=Z$.

Applying \freflem:w1, we consider $x\in D$ such that $Y_{L^{x}}\cap Q\not\leq C_{L}(Y_{L})$. Then $Y_{L^{x}}\cap Q$ is normalized by $Q$ and so

[TABLE]

In particular, \frefclm:YLprops(iii) gives

[TABLE]

As $Y_{L}$ is elementary abelian, $\overline{Y_{L^{x}}\cap Q}$ is elementary abelian.

Suppose that $[Y_{L},Y_{L^{x}}\cap Q,Y_{L^{x}}\cap Q]=1$. Then

[TABLE]

by \frefclm:YLprops (iv). Hence $[Y_{L},Y_{L^{x}}\cap Q,Y_{L^{x}}]=1$. Then as $|[Y_{L},Y_{L^{x}}\cap Q]|=2^{2}$ and $|Y_{L}\cap Q|=2^{3}$, we conclude that $(Y_{L}\cap Q)C_{{L^{x}}}(Y_{{L^{x}}})/C_{{L^{x}}}(Y_{{L^{x}}})$ has order $2$. Thus $[Y_{{L^{x}}},Y_{L}\cap Q,Y_{L}\cap Q]=1$. Now the argument just presented implies that $|\overline{Y_{L^{x}}\cap Q}|=2$ and so, as $Q$ normalizes $Y_{L^{x}}\cap Q$, $\overline{Y_{L^{x}}\cap Q}=Z(\overline{Q})$. In particular, as $[Y_{L},S_{0},S_{0}]=1$, we have proved that

if $\overline{Y_{L^{x}}\cap Q}\leq\overline{S_{0}}$, then $\overline{Y_{L^{x}}\cap Q}=Z(\overline{Q})$.

For a moment let $\overline{Q_{1}}$ be the fours subgroups of $\overline{Q}$ not equal to $\overline{S_{0}}$. Then as $\Phi(Y_{L^{x}}\cap Q)=1$ the displayed line implies that $\overline{W}\leq\overline{Q_{1}}$ and \freflem:not base and $\overline{Q}^{\prime}\leq\overline{Y_{L^{x}}\cap Q}$ imply $\overline{W}=\overline{Q_{1}}$. The remaining statements in \frefclm:W4 now follow from the action of $L$ on $Y_{L}$.$\hfill\blacksquare$

We have that $Z(Q)$ centralizes $[Y_{L},Q]$ and so $Z(Q)\leq S\cap C_{L}(Y_{L})=Q_{L}$. Hence using \frefclm:W4 we obtain

[TABLE]

(4.1.3) ** We have $Q_{L}=Y_{L}$.

Suppose that $Q_{L}>Y_{L}$. Let $m\in L$ be such that $\overline{K}\cong{\operatorname{SL}}_{2}(2)\times{\operatorname{SL}}_{2}(2)$, where $K=\langle W,W^{m}\rangle$. Recall that by the choice of $L$ in the Notation at the end of the introduction, we have $Y_{L}=\Omega_{1}(Z(Q_{L}))$ and by \frefprop:unambiguous and \frefclm:W4, $K$ acts irreducibly on $Y_{L}=V_{L}$. Hence we may apply \freflem:[WO_p][W^mO_p] (iii), (iv) and (v) which combined yield $U_{L}/Y_{L}$ is elementary abelian and

[TABLE]

Since $[Q_{L},W,W]\leq[W,W]=Z\leq Y_{L}$, we have $W$ acts quadratically on every chief factor of $L$ in $Q_{L}/Y_{L}$. In particular, no non-central $L$-chief factor of $Q_{L}/Y_{L}$ is isomorphic to $Y_{L}$.

Let $E$ be the preimage of $C_{U_{L}/Y_{L}}(K)$. Then $E$ is normal in $L$ and application of \freflem: U¿YL abelian implies that $E=Y_{L}$. Let $X\in{\operatorname{Syl}}_{3}(K)$. By \freflem:[WO_p]W^mO_p, $[K,C_{L}(Y_{L})]\leq U_{L}$, so $XU_{L}$ is normal in $L$. As $L$ is solvable, $C_{L}(Y_{L})=C_{X}(Y_{L})Q_{L}$ and either $C_{X}(Y_{L})=1$ or $X\cong 3^{1+2}_{+}$. The latter case is impossible as $W$ is quadratic on $U_{L}/Y_{L}$. Hence $U_{L}=[U_{L},O^{2}(L)]$ and $U_{L}/Y_{L}$ contains no central $L$-chief factors. We know that every $L$-chief factor in $U_{L}/Y_{L}$ is a wreath product module for ${\operatorname{SL}}_{2}(2)\wr 2$ with $\overline{W}$ acting quadratically. In particular, for every non-central chief factor $F$ of $L$ in $U_{L}/Y_{L}$ we have $[F,\overline{W}]=[F,Z(\overline{Q})]$. Set $W_{1}=[W,D]$. Then

[TABLE]

Hence $[F,W]=[F,W_{1}]$ for every non-central chief factor $F$ of $L$ in $U_{L}/Y_{L}$. Set $\widetilde{L}=L/Y_{L}$ and let $z\in Q$ with $Z(\overline{Q})=\langle\overline{z}\rangle$. As $C_{F}(Z(\overline{Q}))=[F,Z(\overline{Q})]$ for each $F$, we have $C_{\widetilde{U_{L}}}(z)=[\widetilde{U_{L}},z]$; then as $W$ acts quadratically on $\widetilde{U_{L}}$, we have $[W,\widetilde{U_{L}}]=C_{\widetilde{U_{L}}}(W)$. Thus $[U_{L},W]Y_{L}=[U_{L},W_{1}]Y_{L}$. In particular,

[TABLE]

and so $U_{L}$ acts quadratically on $W/W_{1}$. Therefore $U_{L}C_{D}(W/W_{1})/C_{D}(W/W_{1})$ is elementary abelian. Hence

[TABLE]

Set $R=\langle Y_{L}^{D}\rangle$. Then, as $Y_{L}\not\leq O_{2}(D)$ by \freflem:w1 (i), $Y_{L}\cap O_{2}(D)=Y_{L}\cap Q\leq W$ and so $R$ centralizes $O_{2}(D)/W$ and $W/W_{1}$. \freflem: cent W/[W,Q] yields $Y_{L}\leq O_{2}(D)$ and this contradicts \freflem:w1 (i). We have shown $Q_{L}=Y_{L}$. $\hfill\blacksquare$

(4.1.4) ** $|S|=2^{7}$ and $N_{G}(Q)/Q\cong{\operatorname{Sym}}(3).$

Since $Q_{L}=Y_{L}=V_{L}$ and $\overline{Q}\cong{\operatorname{Dih}}(8)$, $|S|=2^{7}$ and $|Q|=2^{6}$. Then $N_{G}(Q)=SX$, where $X$ is a Hall $2^{\prime}$-subgroup of $N_{G}(Q)$ and $QX$ is normal in $N_{G}(Q)$. Furthermore $W$ is extraspecial of order $2^{5}$. As $W/Z=J(Q/Z)$, we have $W$ is normal in $N_{G}(Q)$. Hence $X$ acts faithfully on $W$ and embeds in $\mathrm{O}^{+}_{4}(2)$. As $[\overline{W},\overline{Q}]=Z(\overline{Q})$, $S/W$ is faithful on $W/Z$, so $N_{G}(Q)/W$ embeds into $\mathrm{O}_{4}^{+}(2)$. Because $\mathrm{O}_{4}^{+}(2)\cong{\operatorname{Sym}}(3)\wr 2$, and $O_{2}(N_{G}(Q)/W)\not=1$, we get the claim. $\hfill\blacksquare$

Taking $T\in\mathrm{Syl}_{3}(L)$, we have $N_{L}(T)$ is a complement to $Q_{L}$ and so $L=Q_{L}N_{L}(T)$ is a split extension of $Q_{L}$ by $\mathrm{O}_{4}^{+}(2)$. In particular, the isomorphism type of $S$ is uniquely determined. As ${\operatorname{Sym}}(8)$ has a subgroup isomorphic to $L$ and ${\operatorname{Sym}}(8)$ has odd index in ${\operatorname{Alt}}(10)$, we have $S$ is isomorphic to a Sylow $2$-subgroup of ${\operatorname{Alt}}(10)$.

Let $z\in C_{Y_{L}}(Q)^{\#}$, then as $Y_{L}$ is a $+$-type space for $L$, there is a fours group $A$ of $Y_{L}$ which has all non-trivial elements $L$-conjugate to $z$. Since $C_{G}(z)$ has characteristic $2$, $C_{O(G)}(z)=1$ and so by coprime action

[TABLE]

Assume that $G$ has no subgroup of index two. Then $S$ is isomorphic to a Sylow $2$-subgroup of ${\operatorname{Alt}}(10)$. Therefore [Mas, Theorem 3.15] implies that $F^{\ast}(G)\cong{\operatorname{Alt}}(10)$, ${\operatorname{Alt}}(11)$, ${\operatorname{PSL}}_{4}(r)$, $r\equiv 3\pmod{4}$, or ${\operatorname{PSU}}_{4}(r)$, $r\equiv 1\pmod{4}$. Notice that $Z(Q)=C_{Y_{L}}(Q)=\langle z\rangle$ and so $C_{G}(z)=N_{G}(Q)$ has characteristic $2$. In ${\operatorname{Alt}}(11)$, $z$ corresponds to $(12)(34)(56)(78)$ and so $C_{G}(z)\leq({\operatorname{Alt}}(8)\times Z_{3}):2$, which implies that $C_{G}(z)$ is not of characteristic 2. In the linear and unitary groups $C_{G}(z)$ has a normal subgroup isomorphic to ${\operatorname{SL}}_{2}(r)\circ{\operatorname{SL}}_{2}(r)$, and this contradicts \frefclm:NGQ. Hence $G\cong{\operatorname{Alt}}(10)$.

Assume now that $G$ has a subgroup of index two. As $V_{L}\leq G^{\prime}$ we also have $W\leq G^{\prime}$. Therefore $(G^{\prime}\cap L)/Y_{L}\cong\Omega^{+}_{4}(2)$ and so $G^{\prime}$ has Sylow 2-subgroups isomorphic to those of ${\operatorname{Alt}}(8)$. Applying [GH, Corollary A*] we have $F^{\ast}(G)\cong{\operatorname{Alt}}(8)$, ${\operatorname{Alt}}(9)$ or ${\operatorname{PSp}}_{4}(3)$. Again in $G^{\prime}\cong{\operatorname{PSp}}_{4}(3)$, we have that $G^{\prime}$ contains a subgroup of shape ${\operatorname{SL}}_{2}(3)\circ{\operatorname{SL}}_{2}(3)$. This contradicts \frefclm:NGQ and proves the proposition. ∎

5. $\overline{L^{\circ}}\cong\Gamma\mathrm{SL}_{2}(4)$

In this section we attend to the case from \frefprop:unambiguous(ii). Hence we have $p=2$, $\overline{L^{\circ}}\cong\Gamma{\operatorname{SL}}_{2}(4)$, $V_{L}$ is the natural ${\operatorname{SL}}_{2}(4)$-module and either $Y_{L}=V_{L}$ or $|Y_{L}/V_{L}|=2$. Notice that as $L\not\leq N_{G}(Q)$ and $L$ centralizes $Y_{L}/V_{L}$, if $Y_{L}>V_{L}$, $Y_{L}$ does not split over $V_{L}$ and $C_{Y_{L}}(Q)=C_{V_{L}}(Q)$ has order $2$. Furthermore, $C_{S}([Y_{L},Q])=Q_{L}$.

Our aim is to prove

Proposition 5.1**.**

Suppose $L\in\mathcal{L}_{G}(S)$ and $L\not\leq N_{G}(Q)$ with $\overline{L}$ in the unambiguous wreath product case. If $Y_{L}\not\leq Q$ and $\overline{L^{\circ}}\cong\Gamma\mathrm{SL}_{2}(4)$, then $G\cong{\operatorname{Mat}}(22)$ or ${\operatorname{Aut}}({\operatorname{Mat}}(22))$.

Notice that as $Q_{L}\in\mathrm{Syl}_{2}(C_{L}(Y_{L}))$, $C_{L}(Y_{L})/Q_{L}$ is centralized by $L^{\circ}$, and so $C_{L^{\circ}}(Y_{L})=Q_{L}\cap L^{\circ}$ as the Schur multiplier of ${\operatorname{SL}}_{2}(4)$ has order $2$. We also have $|\overline{Q}|\geq 4$ and $|Z(Q)\cap V_{L}|=2$.

Lemma 5.2**.**

For $N=N_{G}(Q_{L})$ we have $(Z(Q)\cap V_{L})^{N}\cap Y_{L}\subseteq V_{L}$. In particular, $N$ normalizes $V_{L}$.

Proof.

If $V_{L}=Y_{L}$, there is nothing to prove. Assume that $|Y_{L}:V_{L}|=2$. Choose $g\in N$, put $U=(Z(Q)\cap V_{L})^{g}$ and assume that $U\not\leq V_{L}$. Recall that $Y_{L}=\Omega_{1}(Z(Q_{L}))$ and so $U\leq Y_{L}$ and $Y_{L}$ is normalized by $N$. Then $C_{L}(U)C_{N}(Y_{L})/C_{N}(Y_{L})\cong 5:4$ or $2\times{\operatorname{Sym}}(3)$. As $C_{N}(U^{g^{-1}})$ normalizes $Q\cap Y_{L}$, $C_{N}(U^{g^{-1}})$ is not irreducible on $Y_{L}/U^{g^{-1}}$. This excludes the possibility $C_{L}(U)C_{N}(Y_{L})/C_{N}(Y_{L})\cong 5:4$ which is irreducible on $Y_{L}/U$. Hence we see that $Z(Q)\cap V_{L}$ has exactly $15+10=25$ conjugates under $N$, but 25 does not divide the order of ${\operatorname{SL}}_{5}(2)={\operatorname{Aut}}(Y_{L})$. This contradiction proves the lemma. ∎

Lemma 5.3**.**

We have $Q_{L}=Y_{L}$ and either

1. (i)

$|S|=2^{7}$, $L/Q_{L}\cong\Gamma{\operatorname{SL}}_{2}(4)$, $N_{G}(Q)/Q\cong{\operatorname{SL}}_{2}(2)$, there exists a subgroup $E\leq S$ of order $2^{4}$ which is normalized by $N_{G}(Q)$ such that $N_{G}(E)/E\cong{\operatorname{Alt}}(6)$ and $N_{L}(E)$ has index $5$ in $L$. Furthermore all the involutions in $\langle N_{G}(E),L\rangle$ are conjugate. 2. (ii)

$G$* has a subgroup of index $2$ which satisfies the conditions in *(i).

Proof.

We have $\overline{S}\cong{\operatorname{Dih}}(8)$ and $\overline{Q}\not\leq\overline{S}_{0}$ as $\overline{L^{\circ}}\cong\Gamma\mathrm{SL}_{2}(4)$. \freflem:not base implies that $\overline{W}\not\leq\overline{S_{0}}$. By assumption, we either have $Y_{L}=V_{L}$ or $|Y_{L}:V_{L}|=2$. In particular, $2^{4}\leq|Y_{L}|\leq 2^{5}$. Since $\overline{Q}$ is normal in $\overline{S}$ and contains $\overline{W}$ we know

(5.3.1) ** Either $\overline{Q}$ is elementary abelian of order $4$ or $\overline{Q}=\overline{S}$

As $V_{L}$ is a natural ${\operatorname{SL}}_{2}(4)$-module and $L\not\leq N_{G}(Q)$, we have $C_{Y_{L}}(Q)=C_{Y_{L}}(S)$ has order $2$ and $[Y_{L},Q]=[V_{L},Q]$ has order $8$. Furthermore, as $W$ is normal in $S$ and is not contained in $S_{0}$, we have $[Y_{L},Q,W]=Z$ where $Z=C_{V_{L}}(S)$ has order $2$. Thus, arguing exactly as before \frefclm:qlyl and in the proof of \frefclm:W4 we obtain

(5.3.2) ** $|\overline{W}|=4$, $[W,W]=Z$ and $[Q_{L},W,W]\leq Y_{L}$.

(5.3.3) ** Assume that $Q_{L}>Y_{L}$. Then $[Q_{L},O^{2}(L)]\not\leq Y_{L}$.

Suppose that $[Q_{L},O^{2}(L)]\leq Y_{L}$. Then $V_{L}\not\leq\Phi(Q_{L})$ by Burnside’s Lemma [GLS2, Proposition 11.1], which contradicts \freflem:VL sub QL’(iii). This proves the claim$\hfill\blacksquare$

(5.3.4) ** If $V_{L}<Y_{L}$, then $\overline{Q}=\overline{S}$.

If $\overline{Q}$ has order $4$, then $\overline{Q}=\overline{W}$ by \frefclm:S5-W quad, so $\overline{Q}$ normalizes a Sylow $3$-subgroup $\overline{T}$ of $\overline{L}$ and so $Q$ normalizes $C_{Y_{L}}(T)$ which has order $2$ and complements $V_{L}$. Hence $C_{Y_{L}}(T)\leq Z(Q)$, so $T\leq N_{G}(Q)$ and therefore $L=\langle T,S\rangle\leq N_{G}(Q)$, a contradiction. Thus $\overline{Q}=\overline{S}$ has order $8$.$\hfill\blacksquare$

(5.3.5) ** We have $Q_{L}=Y_{L}$.

Suppose false. By \frefclm:S5-W quad $W$ acts quadratically on $Q_{L}/Y_{L}$ and $|\overline{W}|=4$. Also $\overline{W}\not\leq\overline{S_{0}}$, so \freflem:Sym5-modules implies that the non-central $L$-chief factors in $Q_{L}/Y_{L}$ are orthogonal modules for $\overline{L}\cong\mathrm{O}_{4}^{-}(2)$. In particular, as $L$-modules, the non-central $L$-chief factors of $Q_{L}/Y_{L}$ are not isomorphic to $V_{L}$.

Choose $E\leq Q_{L}$ normal in $L$ and minimal so that $E/Y_{L}$ contains a non-central $L$-chief factor and let $F$ be the preimage of $C_{E/Y_{L}}(O^{2}(L))$. Then $[F,O^{2}(L)]\leq Y_{L}$ and \freflem: U¿YL abelian applies to yield $F\leq Y_{L}$. In particular, $[E,E]\leq Y_{L}$.

We claim $E^{\prime}\leq V_{L}$. This is obviously the case if $V_{L}=Y_{L}$. So suppose that $|Y_{L}:V_{L}|=2$. If $E^{\prime}\not\leq V_{L}$. Then the minimal choice of $E$ and $E^{\prime}V_{L}=Y_{L}$ implies that $E/V_{L}$ is extraspecial of order $2^{5}$. Notice that $[E,W]\leq W$ and $W/Z$ is elementary abelian as $[W,W]=Z$ by \frefclm:S5-W quad. Hence, as $[E,W]Y_{L}/V_{L}$ has order $2^{3}$, we infer that $E/V_{L}$ has $+$-type contrary to $\overline{L}\cong\Gamma{\operatorname{SL}}_{2}(4)$. Hence $E/V_{L}$ is elementary abelian. If $[Q_{L},E]=V_{L}$, then $E/V_{L}$ has order $2^{4}$ by \freflem:Sym5-modules and so $Q_{L}/C_{Q_{L}}(E)$ embeds into

[TABLE]

by \freflem:dualchieffactors. Since $Q_{L}/C_{Q_{L}}(E)$ involves only trivial and orthogonal modules this contradicts [Pr, Lemma 2.2].

Thus $[E,Q_{L}]=Y_{L}>V_{L}$.

By \frefclm:S5-V_LnotY_L

$\overline{Q}=\overline{S}$ has order $8$.

In summary we now know $|\overline{W}|=4$ and $\overline{[W,Q]}=\overline{[W,S]}=Z(\overline{S})$.

We calculate using $Z$ is normal in $D$ by \frefclm:S5-W quad that

[TABLE]

Therefore

[TABLE]

As $|[Z(\overline{S}),E/Y_{L}]|=4$ and $\overline{Q}=\overline{S}$, this implies that $|C_{E/Y_{L}}(\overline{S})|=4$. As $E/Y_{L}$ is the orthogonal $\mathrm{O}_{4}^{-}(2)$-module for $L$, this is impossible. We have proved the claim. $\hfill\blacksquare$

(5.3.6) ** Suppose that $Y_{L}=V_{L}$. Then $L$ is a maximal 2-local subgroup of $G$, $N_{G}(Q)/Q\cong{\operatorname{SL}}_{2}(2)$, there exists a subgroup $E\leq S$ of order $2^{4}$ which is normalized by $N_{G}(Q)$ such that $N_{G}(E)/E\cong{\operatorname{Alt}}(6)$ and $N_{L}(E)$ has index $5$ in $L$.

By \frefclm:Q_L=Y_L we have $|S|=2^{7}$, and $|\overline{W}|=2^{2}$. Also $|[W,Y_{L}]|=8$ and $Y_{L}\not\leq Q$, so $Q\cap Y_{L}=[W,Y_{L}]=W\cap Y_{L}$, Therefore $|W|=2^{5}$. Set $C=C_{Q}(W)$. Then $C$ centralizes $[Y_{L},Q]$ which has order $2^{3}$ and so $C\leq C_{L}([Y_{L},Q])=Y_{L}$. Thus $C\leq C_{Y_{L}}(W)$ which has order $2$. Then, by \frefclm:S5-W quad, $W^{\prime}=Z=C$ and, as $W$ is generated by involutions, we have $W$ is extraspecial. Since $[Y_{L},Q]\leq W$, $W$ has $+$-type.

Observe $W/Z=J(Q/Z)$, so $W$ is normal in $N_{G}(Q)$ and $N_{G}(Q)/Z$ embeds into ${\operatorname{Aut}}(W)\cong 2^{4}{:}\mathrm{O}_{4}^{+}(2)$.

Assume that $Y_{L}Q/Q$ normalizes a subgroup $T$ of $O_{3}(N_{G}(Q))/Q$ which has fixed points on $W/Z$. Then $W=[W,T]C_{W}(T)$ and $[W,T]\cong C_{W}(T)\cong\mbox{\rm Q}_{8}$ and these subgroups are normalized by $Y_{L}$. But then

[TABLE]

Since $[W,Y_{L}]$ is elementary abelian and $\Omega_{1}(P)=Z(P)$ for $P\cong\mbox{\rm Q}_{8}$, we conclude that

[TABLE]

and then $[W,Y_{L}]$ has order $2$ which is nonsense as $Y_{L}$ is the natural module. Therefore $Y_{L}$ normalizes no such subgroup.

Let $F=O_{2,3}(N_{G}(Q))$. Assume that $|F/Q|=9$. Then the previous argument implies that $C_{F/Q}(Y_{L})\not=1$. Let $T_{1}$ be the preimage of this subgroup. Then $[Y_{L},Q]$ is normalized by $T_{1}$. Hence $Y_{L}=C_{Y_{L}Q}([Y_{L},Q])$ is normalized by $T_{1}$. Using the fact that $Q$ is weakly closed in any $2$-group which contains it, for $w\in Y_{L}^{\#}$, we let $Q_{w}$ be the unique conjugate of $Q$ in $O_{2}(C_{G}(w))$. Then $T_{1}$ permutes the elements of $Y_{L}$ and so $T_{1}$ normalizes $L^{\circ}=\langle Q_{w}\mid w\in Y_{L}^{\#}\rangle$. Since $L=L^{\circ}Y_{L}$, we have that $T_{1}$ normalizes $L$. On the other hand, $WY_{L}$ is normalized by $T_{1}$ and, as $T_{1}$ acts fixed-point freely on $W/Z$, $T_{1}$ acts transitively on $WY_{L}/Y_{L}\cong W/[Y_{L},Q]\cong 2^{2}$ and this is impossible as $W\cap O^{2}(L)$ is a maximal subgroup of $W$ and is normalized by $T_{1}$.

Hence $|F/Q|=3$, $N_{G}(Q)=FS$ and $N_{G}(Q)/Q\cong{\operatorname{SL}}_{2}(2)$. In particular, $|Q|=2^{6}$, $S=Y_{L}Q$, and $FY_{L}/W\cong 2\times{\operatorname{SL}}_{2}(2)$. It follows that

[TABLE]

et $E=C_{S}([W,Q])$. As $W$ is normal in $N_{G}(Q)$, so is $E$. As $|S|=2^{7}$ and $|{\operatorname{GL}}_{3}(2)|_{2}=2^{3}$, we have $|E|\geq 2^{4}$. Since $F$ acts fixed-point freely on $W/Z$ (being normalized by $Y_{L}$), we have $E\leq Q$ and then $E$ is normal in $N_{G}(Q)$. Since $E\cap W=[W,Q]$, we find $|E|=2^{4}$. Let $S\leq L_{1}\leq L$ be such that $L_{1}/Q_{L}\cong{\operatorname{Sym}}(4)$ has index $5$ in $L$. Notice that $O_{2}(L_{1})=S_{0}$. Then $E\leq C_{L}([Y_{L},Q,Q])=Y_{L}S_{0}$. Also $Y_{L}\leq S_{0}$, so $S_{0}=Y_{L}S_{0}$. Therefore $E\leq S_{0}$. Now $EY_{L}/Y_{L}$ acts as a Sylow $2$-subgroup of ${\operatorname{SL}}_{2}(4)$ on the natural module. In particular for any involution $e\in E\setminus Y_{L}$ we have that $C_{Y_{L}}(e)=E\cap Y_{L}$. This implies that all involutions in $EY_{L}$ are contained in $Y_{L}\cup E$ and therefore $E$ and $Y_{L}$ are the only elementary abelian subgroups of $S_{0}$ of order $2^{4}$. In particular, $L_{1}$ normalizes $E$. Now $N_{G}(E)\geq\langle L_{1},N_{G}(Q)\rangle\in\mathcal{L}_{G}(S)$. Notice that $L_{1}$ has orbits of lengths $3$, and $12$ on $E$ and that $N_{G}(Q)$ does not preserve these orbits. Hence $N_{G}(E)$ acts transitively on $E^{\#}$. As $N_{G}(Q)=C_{G}(Z)$, we now have that $|N_{G}(E)|=15|N_{G}(Q)|=2^{7}\cdot 3^{2}\cdot 5$. We have that $X=N_{G}(E)/E$ is isomorphic to a subgroup of ${\operatorname{GL}}_{4}(2)\cong{\operatorname{Alt}}(8)$ of order $2^{3}\cdot 3^{2}\cdot 5$. We consider the action of $X$ on a set of size $8$. As ${\operatorname{Alt}}(8)$ has no subgroups of order 45, $X$ is not transitive. Hence $X$ is isomorphic to a subgroup of ${\operatorname{Alt}}(7)$, ${\operatorname{Sym}}(6)$ or $X\cong({\operatorname{Alt}}(5)\times 3){:}2$. Suppose that $X\cong({\operatorname{Alt}}(5)\times 3){:}2$. As $N_{G}(Q)/Q\cong{\operatorname{Sym}}(4)$, we see that $EQ/E\leq{\operatorname{Alt}}(5)$. Since $E$ is the natural ${\operatorname{SL}}_{2}(4)$-module, we get that $|Z(Q)|=4$. But, by \frefclm:S5-W quad, $|Z(Q)|=2$. Hence we have one of the first two possibilities and then obviously $X=N_{G}(E)/E\cong{\operatorname{Alt}}(6)$.

We just have to show that $L$ is a maximal 2-local subgroup. Let $M$ be a 2-local subgroup with $L\leq M$. As $Q\leq M$, we have that $M$ is of characteristic 2. Then $Y_{L}=Y_{M}$ and $C_{G}(Y_{L})=Y_{L}$. As $|N_{G}(Q):S|=3$ and $Y_{L}$ is not normal in $N_{G}(Q)$, we have $N_{M}(Q)=S=N_{L}(Q)$. As $L$ acts transitively on $Y_{L}^{\#}$, we conclude $M=N_{M}(Q)L=N_{L}(Q)L=L$. $\hfill\blacksquare$

(5.3.7) ** If $Y_{L}=V_{L}$, then $G$ has just one conjugacy class of involutions.

By \frefclm:structure $N_{G}(E)/E\cong{\operatorname{Alt}}(6)$. As $Y_{L}\not\leq E$, there is an involution $y\in Y_{L}\setminus E$. Now $y$ inverts an element of order 5 in $N_{G}(E)$ and so $|[E,y]|=|C_{E}(y)|=4$. This shows that all involutions in $Ey$ are conjugate. As all involutions in $S/E$ are conjugate in ${\operatorname{Alt}}(6)$ and all the involutions in $Y_{L}$ are $L$-conjugate, this proves the claim.$\hfill\blacksquare$

We have now proved that (i) holds when $Y_{L}=V_{L}$.

(5.3.8) ** Suppose that $Y_{L}>V_{L}$. Then $G$ has a subgroup of index $2$.

We have that $|S|=2^{8}$. By \frefclm:S5-V_LnotY_L and \frefclm:Q_L=Y_L, $S=QY_{L}$. We are going to show that $J(S)=Y_{L}$. For this let $A\leq S$ be elementary abelian of maximal order and assume that $A\neq Y_{L}$. Then $|AY_{L}/Y_{L}|\leq 4$. As there are no transvections on $V_{L}$, we get $|AY_{L}/Y_{L}|=4$ and we may assume that $A$ acts quadratically on $Y_{L}$ by [GLS2, Theorem 25.2]. As $W\not\leq S_{0}$ by \freflem:not base and $|\overline{W}|=4$ by \frefclm:S5-W quad, $W$ does not act quadratically on $Y_{L}$, $AY_{L}/Y_{L}\leq S_{0}/Y_{L}$ and $S_{0}=AY_{L}$. Now $A\cap Y_{L}$ has order 8 and so $|C_{Y_{L}}(S_{0})|=8$. But $(L^{\circ})^{\prime}$ is generated by two conjugates of $S_{0}$, which gives $C_{Y_{L}}(L^{\circ})\not=1$ a contradiction to \freflem:VL sub QL’(i). Thus $Y_{L}=J(S)$ is the Thompson subgroup of $S$. In particular, $N_{G}(Y_{L})$ controls $G$-fusion of elements in $Y_{L}$. As $S\in\mathrm{Syl}_{2}(G)$ and $C_{S}(Y_{L})=Q_{L}$, $Q_{L}\in\mathrm{Syl}_{2}(C_{G}(Y_{L}))$ and we have $N_{G}(Y_{L})=C_{G}(Y_{L})N_{N_{G}(Y_{L})}(Q_{L})$. By \freflem:fusioncontrol

[TABLE]

Suppose that $O^{2}(L)\geq Y_{L}$. Then $O^{2}(L)/V_{L}\cong{\operatorname{SL}}_{2}(5)$ has quaternion Sylow $2$-subgroups and $|L:O^{2}(L)|=2$. On the other hand, there exists $g\in N_{G}(Q)\setminus N_{G}(Y_{L})$ with $WY_{L}\geq(Y_{L}^{g}\cap Q)Y_{L}\neq Y_{L}$ and $(Y_{L}^{g}\cap Q)V_{L}/V_{L}$ is elementary abelian, which is a contradiction. Therefore $O^{2}(L)/V_{L}\cong{\operatorname{SL}}_{2}(4)$ and, as $W$ does not act quadratically on $Y_{L}$, we see that $|W:W\cap O^{2}(L)|=2$ and thus $O^{2}(L)W/V_{L}\cong\Gamma{\operatorname{SL}}_{2}(4)$. Hence $L$ has a subgroup $L_{0}=O^{2}(L)W$ of index $2$ with $Y_{L}\cap L_{0}=V_{L}$.

Let $T\in\mathrm{Syl}_{2}(L_{0})$ and $w\in Y_{L}\setminus T$. Suppose that for some $x\in G$, $w^{x}\in T$ and $|C_{S}(w^{x})|\geq|C_{S}(w)|$. As $L^{\circ}$ has orbits of length 6 and 10 on $Y_{L}\setminus V_{L}$, we may assume $|C_{S}(w^{x})|\geq|S|/2$. But then as $V_{L}$ is the natural module, it does not admit transvections and so $w^{x}\in V_{L}$. As $N_{G}(Y_{L})=N_{G}(V_{L})$ and $N_{G}(Y_{L})$ controls fusion in $Y_{L}$, this is not possible. Hence the supposed condition cannot hold. Application of [GLS2, Proposition 15.15], shows that $G$ has a subgroup of index $2$. This proves \frefclm:56.$\hfill\blacksquare$

Let $G_{0}$ be a subgroup of $G$ of index $2$, and set $Q_{0}=Q\cap G_{0}$. We have $V_{L}\leq L^{\circ}\leq G_{0}$. Hence $W=\langle[V_{L},Q]^{D}\rangle\leq G_{0}$. In particular, $W\leq Q_{0}$ and so $Z(Q_{0})=Z$ and $Q_{0}$ is large in $G_{0}$. Set $L_{0}=O^{2}(L)Q_{0}=O^{2}(L)W$. Then $L_{0}^{\circ}/V_{L}\cong\Gamma{\operatorname{SL}}_{2}(4)$ and $Y_{L_{0}}=V_{L_{0}}=V_{L}\not\leq Q_{0}$. Thus $(G_{0},L_{0})$ satisfies the hypotheses of (i). This proves (ii) holds if $V_{L}\not=Y_{L}$. ∎

Proof of \frefprop:S5-1:.

By \freflem:S5-2 we just have to examine the structure in \freflem:S5-2(i), so we may assume that \freflem:S5-2(i) holds.

By \freflem:splitA6

[TABLE]

As $N_{G}(Q)\leq N_{G}(E)$, for a 2-central involution $z$ we have that $C_{G}(z)$ is a split extension of $E$ by ${\operatorname{Sym}}(4)$. As $O(C_{G}(z))=1$ coprime action yields $O(G)=\langle C_{O(G)}(e)\mid e\in E^{\#}\rangle=1$. In particular $F(G)=1$ and $E(G)\neq 1$. Suppose that $J^{*}$ is a non-trivial subnormal subgroup of $G$ normalized by $\langle L,N_{G}(E)\rangle$. Then $S\cap J^{*}\neq 1$. Since $1\neq J^{*}\cap N_{G}(E)$ is normal in $N_{G}(E)$ and $1\neq J^{*}\cap L$ is normal in $L$, it follows that $J^{*}\cap N_{G}(E)\geq J^{*}\cap S\geq EY_{L}$. Hence $J^{*}\geq\langle Y_{L}^{N_{G}(E)}\rangle=N_{G}(E)\geq S$ and $J^{*}\geq\langle S^{L}\rangle=L$. Therefore there is a unique non-trivial subnormal subgroup of $G$ of minimal order normalized by $\langle L,N_{G}(E)\rangle$. It follows that $\langle L,N_{G}(E)\rangle$ is contained in a component $J$ of $G$. Since $O(G)=1$ and $S\leq J$, $J=E(G)$. As $J$ has just one conjugacy class of involutions by \freflem:S5-2(i) and, for $z\in E^{\#}$, $C_{G}(z)\leq N_{G}(E)$, it follows that $G=J$ is simple. Using $G$ has just one conjugacy class of involutions and applying [J, Theorem] yields $G\cong{\operatorname{Mat}}(22)$. This proves the proposition when \freflem:S5-2(i) holds. If \freflem:S5-2(ii) holds, then $G\cong{\operatorname{Aut}}({\operatorname{Mat}}(22))$. ∎

6. $\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(4)$

In this section we investigate the configuration in \frefprop:unambiguous(iii). Thus $\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(4)$, $|Y_{L}:V_{L}|=2$ and $V_{L}$ is the natural ${\operatorname{SL}}_{2}(4)$-module.

As $Q\leq L^{\circ}$, $C_{V_{L}}(S_{0})=C_{V_{L}}(Q)\leq Z(Q)$, so $Q$ is normal in $N_{L^{\circ}}(C_{V_{L}}(S_{0}))$ and hence $\overline{Q}=\overline{S_{0}}$ is a Sylow 2-subgroup of $\overline{L^{\circ}}$. In particular $Z(Q)\cap Y_{L}=Z(Q)\cap V_{L}$ is of order 4.

Lemma 6.1**.**

The subgroup $Q$ is elementary abelian. In particular, $Q\cap Y_{L}=Q\cap V_{L}=C_{Y_{L}}(Q)=Z$, $|Y_{L}Q/Q|=2^{3}$ and $|V_{L}Q/Q|=2^{2}$.

Proof.

We know that $[Q,V_{L}]=C_{V_{L}}(Q)=Q\cap V_{L}$ and, as $\overline{Q}$ is elementary abelian, $\Phi(Q)\leq Q_{L}$. If $\Phi(Q)\not=1$, then, since $Z(S)\cap\Phi(Q)\not=1$, we deduce $\Phi(Q)\cap V_{L}\not=1$. As $N_{L}(QQ_{L})$ normalizes $Q$ and is irreducible on $[V_{L},Q]$, $[V_{L},Q]\leq\Phi(Q)$. But then $V_{L}$ centralizes $Q/\Phi(Q)$, so $V_{L}\leq O_{p}(N_{G}(Q))=Q$, a contradiction. This shows $Q$ is elementary abelian and then also $Y_{L}\cap Q=V_{L}\cap Q=C_{Y_{L}}(Q)$. ∎

Proposition 6.2**.**

Suppose $L\in\mathcal{L}_{G}(S)$ and $L\not\leq N_{G}(Q)$ with $\overline{L}$ in the unambiguous wreath product case. If $Y_{L}\not\leq Q$, $\overline{L^{\circ}}\cong{\operatorname{SL}}_{2}(4)$ and $|Y_{L}:V_{L}|=2$, then $G$ is ${\operatorname{Aut}}({\operatorname{Mat}}(22))$.

Proof.

We start by observing that the action of $L$ on $Y_{L}$ gives

(6.2.1) **

1. (i)

$|V_{L}Q/Q|=|Q:C_{Q}(V_{L})|=2^{2}$; 2. (ii)

for all $v\in V_{L}\setminus Q$, $C_{Q}(v)=C_{Q}(V_{L})$; and 3. (iii)

for all $w\in Q\setminus Q_{L}$, $[w,V_{L}]=[Q,V_{L}]$.

Let $B=N_{L}(QQ_{L})$. Then $B$ contains an element $\beta$ of order $3$ which acts fixed-point freely on $V_{L}$ and irreducibly on $[V_{L},Q]=C_{Y_{L}}(Q)$.

Using \frefclm:1 (ii) and \freflem:soldual yields $[V_{L},F(N_{G}(Q)/Q)]=1$. Let $K\geq Q$ be the preimage of

[TABLE]

Then $K$ is non-trivial, normalized by $B$ and \freflem:soldual implies $V_{L}Q/Q$ acts faithfully on $K/Q$.

The three involutions of $QQ_{L}/Q_{L}$ each centralize a subgroup of $Y_{L}$ of order $2^{3}$ and by \freflem:VL sub QL’(i), there are three elements of $Y_{L}Q/Q$ which act on $Q$ as ${\operatorname{GF}}(2)$-transvections, they generate $Y_{L}Q/Q$ and are permuted transitively by $B/Q$. As $B$ normalizes $K$ and as $V_{L}Q/Q$ acts faithfully on $K/Q$, at least one and hence all of the transvections in $Y_{L}Q/Q$ act faithfully on $K/Q$.

If $C_{Q}(K)\neq 1$, then $C_{C_{Q}(K)}(S)\neq 1$. As $\Omega_{1}(Z(S))=C_{V_{L}}(S)$ by \freflem:VL sub QL’ (ii), and $C_{Q}(K)$ is normalized by $B$, we have $[Q,V_{L}]\leq C_{Q}(K)$. But then $K=\langle V_{L}^{K}\rangle Q$ centralizes $Q/C_{Q}(K)$ contrary to $C_{K}(Q)=Q$. Hence $C_{Q}(K)=1$.

Let $V$ be a non-trivial minimal $KY_{L}$-invariant subgroup of $Q$. Then $KY_{L}$ acts irreducibly on $V$. Moreover, as $Y_{L}$ does not centralize $V$, $V\not\leq Q_{L}$ and, as $V_{L}$ is the natural $\overline{L^{\circ}}$-module we have $[Y_{L},V]=[Y_{L},Q]=Y_{L}\cap Q\leq V$. It follows that $K$ centralizes $Q/V$ and so $K/Q$ acts faithfully on $V=[Q,K]$ which is normalized by $B$. Hence $C_{Y_{L}}(V)=Y_{L}\cap V=Y_{L}\cap Q$ and $Y_{L}Q/Q$ acts faithfully on $V$. Recall that $Y_{L}Q/Q$ is generated by elements which operate as transvections on $Q$ and hence on $V$. Therefore [McL, Theorem] applies to give $KY_{L}/Q\cong{\operatorname{SL}}_{m}(2)$ with $m\geq 3$, ${\operatorname{Sp}}_{2m}(2)$ with $m\geq 2$, $\mathrm{O}_{2m}^{\pm}(2)$ with $m\geq 2$, or ${\operatorname{Sym}}(m)$ with $m\geq 7$. Furthermore, $V=[Q,K]$ is the natural module in each case.

Since $C_{Y_{L}Q/Q}(S/Q)$ contains a transvection and has order $2^{2}$, $KY_{L}/Q\not\cong{\operatorname{SL}}_{m}(2)$ with $m\geq 3$ or $\mathrm{O}_{2m}^{\pm}(2)$ with $m\geq 2$. Suppose that $KY_{L}/Q\cong{\operatorname{Sym}}(m)$ with $m\geq 7$. Then, as $Y_{L}Q/Q$ is generated by three transvections, we see that $Y_{L}Q/Q$ is generated by three commuting transpositions in $KY_{L}/Q$. Let $t$ be the product of these transpositions. Then, as $m\geq 7$, $|[V,t]|=2^{3}$. However, $|[V,Y_{L}]|=2^{2}$, and so we have a contradiction. We have demonstrated

(6.2.2) ** $KY_{L}/Q\cong{\operatorname{Sp}}_{2m}(2)$, $m\geq 2$ and $[Q,K]=[Q,KY_{L}]$ is the natural module.

Since $[Q,K]$ is the natural $KY_{L}/Q$-module and $[V_{L},Q]\leq[Q,K]$ has order $2^{2}$, we have $[[V_{L},Q],S]\neq 1$. In particular, $QQ_{L}/Q_{L}<S/Q_{L}\cong{\operatorname{Dih}}(8)$ and $SQ/Q\cap K/Q$ acts non-trivially on $[Q,V_{L}]$.

Consider $Q^{*}=O_{2}(KS)$. Since $Q^{*}$ centralizes $[Q,K]$, $Q^{*}$ centralizes $[V_{L},Q]$ and so $Q^{*}Q_{L}=QQ_{L}$. Hence $\Phi(Q^{*})\leq Q_{L}$. If $\Phi(Q^{*})\neq 1$, then

[TABLE]

and so also $[Q^{*},K]=[Q^{*},K,K]\leq[Q,K]\leq\Phi(Q^{*})$ which is impossible. Hence $Q^{*}$ is elementary abelian and it follows that $Q\leq Q^{*}=C_{Q^{*}}(Q)\leq Q$. Since $KS$ acts on $[Q,K]$ and $KY_{L}/Q\cong{\operatorname{Sp}}_{2m}(2)$, we now deduce $S\leq KY_{L}$ from the structure of $\mbox{\rm Out}(K/Q)$. Hence $B=\langle S^{B}\rangle\leq KY_{L}$ as $B$ normalizes $KV_{L}$. It follows that $B/Q$ is the minimal parabolic subgroup $P$ of $K/Q$ irreducible on $[Y_{L},V]$ and with $O^{2}(P)$ centralizing $[Y_{L},V]^{\perp}/[Y_{L},V]=C_{Y_{L}}(V)/[Y_{L},V]$. Therefore there is $\beta\in K$ of order three such that $\langle\beta\rangle$ is transitive on the transvections in $Y_{L}Q/Q$ and normalizes $Q_{L}Q/Q$ which has index $2$ in $S/Q$. In particular, from the structure of the natural ${\operatorname{Sp}}_{2m}(2)$-module $\beta$ centralizes

[TABLE]

As $V$ is abelian, $V$ acts quadratically on $Q_{L}/V_{L}$. By \freflem:Sym5-modules, $Q_{L}/V_{L}$ involves only natural ${\operatorname{SL}}_{2}(4)$-modules and trivial modules as $L$-chief factors. We know $\beta$ acts fixed-point freely on the natural module and so, as $\beta$ centralizes $[Q_{L},V]Y_{L}/Y_{L}$, all the $L$-chief factors of $Q_{L}/V_{L}$ are centralized by $L$. In particular, $V_{L}$ is the unique non-central $L$-chief factor in $Q$ and so $Y_{L}\cap\Phi(Q_{L})=1$. As $\Omega_{1}(Z(S))\leq V_{L}$ by \freflem:VL sub QL’ (ii), $\Phi(Q_{L})=1$, so $Q_{L}=\Omega_{1}(Z(Q_{L}))=Y_{L}$, which together with $S/Q_{L}\cong{\operatorname{Dih}}(8)$ implies

(6.2.3) ** $Y_{L}=Q_{L}$ has order $2^{5}$ and $|S|=2^{8}$.

Together 6 and 6 give

(6.2.4) **$|Q|=2^{4}$ and $N_{G}(Q)/Q\cong{\operatorname{Sym}}(6)$.

We next show that $G$ has a subgroup of index two. In $N_{G}(Q)$ we have a subgroup $U$ of index 2 of shape $2^{4}.{\operatorname{Alt}}(6)$. Furthermore $Y_{L}\not\leq U$ and $V_{L}\leq U$. Since $[v,Q]=C_{Q}(v)$ for $v\in V_{L}\setminus Q$ and $U/Q$ has one conjugacy class of involutions, all the involutions in $U\setminus Q$ are $U$-conjugate. Since $L$ acts transitively on $V_{L}$ and $U$ is transitive on $Q^{\#}$, we have that all the involutions in $U$ are $G$-conjugate. As $Q$ is large, we have $C_{G}(z)\leq N_{G}(Q)$ for $z\in Q^{\#}$. Hence all the involutions in $U$ have centralizer which is a $\{2,3\}$-group. There is an involution $t$ in $Y_{L}\setminus V_{L}$, which is not in $U$ and centralized by an element of order 5 in $L$. Hence $t$ is not conjugate to any involution of $U$. Application of [GLS2, Proposition 15.15] gives a subgroup $G_{1}$ of index two in $G$. We have $N_{G_{1}}(Q)/Q\cong{\operatorname{Alt}}(6)$. By \freflem:splitA6 this extension splits and we have that the centralizer of a $2$-central involution $z\in G_{1}$ is a split extension of an elementary abelian group of order 16 by ${\operatorname{Sym}}(4)$. In particular $O(C_{G}(z))=1$ and so by coprime action $O(G)=\langle C_{O(G)}(e)\mid e\in Q^{\#}\rangle=1$. As $Y_{L}\not\leq Q$, there is an involution $y\in N_{G_{1}}(Q)\setminus Q$. Since all involutions in $Qy$ and in $N_{G_{1}}(Q)/Q$ are conjugate, $G_{1}$ has just one conjugacy class of involutions. In particular $F^{\ast}(G_{1})$ is simple. Application of [J, Theorem] gives that $F^{\ast}(G_{1})\cong{\operatorname{Mat}}(22)$ and so $G\cong{\operatorname{Aut}}({\operatorname{Mat}}(22))$. ∎

Acknowledgment

We thank the referee for numerous comments which have improved the readability and clarity of our work. The second author was partially supported by the DFG.