[2010]20D05, 20D06, 20D20

Let $q$ be a nontrivial odd prime power, and let $n \ge 2$ be a natural number with $(n,q) \ne (2,3)$. We characterize the groups $PSL_n(q)$ and $PSU_n(q)$ by their $2$-fusion systems. This contributes to a programme of Aschbacher aiming at a simplified proof of the classification of finite simple groups.

§ INTRODUCTION

The classification of finite simple groups (CFSG) is one of the greatest achievements in the history of mathematics. Its proof required around 15,000 pages and spreads out over many hundred articles in various journals. Many mathematicians from all over the world were involved in the proof, whose final steps were published in 2004 by Aschbacher and Smith, after it was prematurely announced as finished already in 1983. Because of its extreme length, a simplified and shortened proof of the CFSG would be very valuable. There are three programmes working towards this goal: the Gorenstein-Lyons-Solomon programme (see [27]), the Meierfrankenfeld-Stellmacher-Stroth programme (see [43]) and Aschbacher’s programme.

The goal of Aschbacher’s programme is to obtain a new proof of the CFSG by using fusion systems. The standard examples of fusion systems are the fusion categories of finite groups over $p$-subgroups ($p$ a prime). If $G$ is a finite group and $S$ is a $p$-subgroup of $G$ for some prime $p$, then the fusion category of $G$ over $S$ is defined to be the category $\mathcal{F}_S(G)$ given as follows: the objects of $\mathcal{F}_S(G)$ are precisely the subgroups of $S$, the morphisms in $\mathcal{F}_S(G)$ are precisely the group homomorphisms between subgroups of $S$ induced by conjugation in $G$, and the composition of morphisms in $\mathcal{F}_S(G)$ is the usual composition of group homomorphisms. Abstract fusion systems are a generalization of this concept. A fusion system over a finite $p$-group $S$, where $p$ is a prime, is a category whose objects are the subgroups of $S$ and whose morphisms behave as if they are induced by conjugation inside a finite group containing $S$ as a $p$-subgroup. For the precise definition, we refer to <cit.>. A fusion system is called saturated if it satisfies certain axioms motivated by properties of fusion categories of finite groups over Sylow subgroups (see <cit.>). If $G$ is a finite group and $S_1, S_2 \in \mathrm{Syl}_p(G)$ for some prime $p$, then $\mathcal{F}_{S_1}(G)$ and $\mathcal{F}_{S_2}(G)$ are easily seen to be isomorphic (in the sense of <cit.>). Given a finite group $G$, a prime $p$ and a Sylow $p$-subgroup $S$ of $G$, we refer to $\mathcal{F}_S(G)$ as the $p$-fusion system of $G$.

Originally considered by the representation theorist Puig, fusion systems have become an object of active research in finite group theory, representation theory and algebraic topology. It has always been a problem of great interest in the theory of fusion systems to translate group-theoretic concepts into suitable concepts for fusion systems. For example, there is a notion of normalizers and centralizers of $p$-subgroups in fusion systems, a notion of the center of a fusion system, a notion of factor systems, a notion of normal subsystems of saturated fusion systems and a notion of simple saturated fusion systems (see <cit.>). Roughly speaking, Aschbacher’s programme consists of the following two steps.

1. Classify the simple saturated fusion systems on finite $2$-groups. Use the original proof of the CFSG as a “template”.

2. Use the first step to give a new and simplified proof of the CFSG.

There is the hope that several steps of the original proof of the CFSG become easier when working with fusion systems. For example, in the original proof of the CFSG, the study of centralizers of involutions plays an important role. The $2'$-cores of the involution centralizers, i.e. their largest normal odd order subgroups, cause serious difficulties and are obstructions to many arguments. Such difficulties are not present in fusion systems since cores do not exist in fusion systems. This is suggested by the well-known fact that the $2$-fusion system of a finite group $G$ is isomorphic to the $2$-fusion system of $G/O(G)$, where $O(G)$ denotes the $2'$-core of $G$. For an outline of and recent progress on Aschbacher’s programme, we refer to [8].

So far, Aschbacher’s programme has focused mainly on Step 1, while not much has been done on Step 2. An important part of Step 2 is to identify finite simple groups from their $2$-fusion systems. The present paper contributes to Step 2 of Aschbacher’s programme by characterizing the finite simple groups $PSL_n(q)$ and $PSU_n(q)$ in terms of their $2$-fusion systems, where $n \ge 2$ and where $q$ is a nontrivial odd prime power with $(n,q) \ne (2,3)$.

In order to state our results, we introduce some notation and recall some definitions. Let $G$ be a finite group. A component of $G$ is a quasisimple subnormal subgroup of $G$, and a $2$-component of $G$ is a perfect subnormal subgroup $L$ of $G$ such that $L/O(L)$ is quasisimple. The natural homomorphism $G \rightarrow G/O(G)$ induces a one-to-one correspondence between the set of $2$-components of $G$ and the set of components of $G/O(G)$ (see <cit.>). We use $Z^{*}(G)$ to denote the full preimage of the center $Z(G/O(G))$ in $G$. In Step 2 of Aschbacher's programme, one may assume that a finite group $G$ is a minimal counterexample to the CFSG. Such a group $G$ has the following property.

\begin{align}
\label{CK}
\tag{$\mathcal{CK}$}
&\mbox{Whenever $x\in G$ is an involution and $J$ is a $2$-component of $C_G(x)$},\\
&\mbox{then $J/Z^*(J)$ is a known finite simple group.}\nonumber
\end{align}

By a known finite simple group, we mean a finite simple group appearing in the statement of the CFSG.

For each integer $n \ne 0$, we use $n_2$ to denote the $2$-part of $n$, i.e. the largest power of $2$ dividing $n$. Given odd integers $a, b$ with $|a|, |b| > 1$, we write $a \sim b$ provided that $(a-1)_2 = (b-1)_2$ and $(a+1)_2 = (b+1)_2$. If $q$ is a nontrivial prime power and if $n$ is a positive integer, then we write $PSL_n^{+}(q)$ for $PSL_n(q)$ and $PSL_n^{-}(q)$ for $PSU_n(q)$. With this notation, we can now state our main results.

Let $q$ be a nontrivial odd prime power, and let $n \ge 2$ be a natural number. Let $G$ be a finite simple group. Suppose that $G$ satisfies (<ref>) if $n \ge 6$. Then the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_n(q)$ if and only if one of the following holds:

(i) $G \cong PSL_n^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $\varepsilon q^{*} \sim q$;

(ii) $n = 2$, $\vert PSL_2(q) \vert_2 = 8$, and $G \cong A_7$;

(iii) $n = 3$, $(q+1)_2 = 4$, and $G \cong M_{11}$.

Our second main result is an extension of Theorem <ref>. In order to state it, we briefly mention some concepts from the local theory of fusion systems. Let $\mathcal{F}$ be a saturated fusion system on a finite $p$-group $S$ for some prime $p$, and let $\mathcal{E}$ be a normal subsystem of $\mathcal{F}$. In <cit.>, Aschbacher introduced a subgroup $C_S(\mathcal{E})$ of $S$, which plays the role of the centralizer of $\mathcal{E}$ in $S$. In <cit.>, he defined a normal subsystem $F^{*}(\mathcal{F})$ of $\mathcal{F}$, called the generalized Fitting subsystem of $\mathcal{F}$, and proved that $C_S(F^{*}(\mathcal{F})) = Z(F^{*}(\mathcal{F}))$, where the latter denotes the center of $F^{*}(\mathcal{F})$.

Let $q$ be a nontrivial odd prime power, and let $n \ge 2$ be a natural number. If $n = 2$, suppose that $q \equiv 1$ or $7 \mod 8$. Let $G$ be a finite simple group, and let $S$ be a Sylow $2$-subgroup of $G$. Suppose that $\mathcal{F}_S(G)$ has a normal subsystem $\mathcal{E}$ on a subgroup $T$ of $S$ such that $\mathcal{E}$ is isomorphic to the $2$-fusion system of $PSL_n(q)$ and such that $C_S(\mathcal{E}) = 1$. Then $\mathcal{F}_S(G)$ is isomorphic to the $2$-fusion system of $PSL_n(q)$. In particular, if $n \le 5$ or if $G$ satisfies (<ref>), then one of the properties (i)-(iii) from Theorem <ref> holds.

Let $q$ be a nontrivial odd prime power, and let $n \ge 2$ be a natural number. If $n = 2$, suppose that $q \equiv 1$ or $7 \mod 8$. Let $G$ be a finite simple group, and let $S$ be a Sylow $2$-subgroup of $G$. Suppose that $F^{*}(\mathcal{F}_S(G))$ is isomorphic to the $2$-fusion system of $PSL_n(q)$. Then $\mathcal{F}_S(G)$ is isomorphic to the $2$-fusion system of $PSL_n(q)$. In particular, if $n \le 5$ or if $G$ satisfies (<ref>), then one of the properties (i)-(iii) from Theorem <ref> holds.

The paper is organized as follows. In Sections <ref> and <ref>, we collect several results needed for the proofs of our main results. Preliminary results on abstract finite groups and abstract fusion systems are proved in Section <ref>. Section <ref> presents some results on linear and unitary groups over finite fields, mainly focussing on $2$-local properties and on the automorphisms of these groups.

In Section <ref>, we will verify Theorem <ref> for the case $n \le 5$. Our proofs strongly depend on work of Gorenstein and Walter [31] (for $n = 2$), on work of Alperin, Brauer and Gorenstein [2], [3] (for $n = 3$) and on work of Mason [40], [41], [42] (for $n = 4$ and $n = 5$).

For $n \ge 6$, we will prove Theorem <ref> by induction over $n$. In order to do so, we will consider a finite group $G$ realizing the $2$-fusion system of $PSL_n(q)$, where $q$ is a nontrivial odd prime power and where $n \ge 6$ is a natural number such that Theorem <ref> is true with $m$ instead of $n$ for any natural number $m$ with $6 \le m < n$. We will also assume that $O(G) = 1$ and that $G$ satisfies (<ref>). To prove that Theorem <ref> is satisfied for the natural number $n$, we will prove the existence of a normal subgroup $G_0$ of $G$ such that $G_0$ is isomorphic to a nontrivial quotient of $SL_n^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +, - \rbrace$ with $\varepsilon q^{*} \sim q$. This will happen in Sections <ref>-<ref>.

In Section <ref>, we will introduce some notation and prove some preliminary lemmas. Section <ref> describes the $2$-components of the centralizers of involutions of $G$. In Section <ref>, we will use signalizer functor methods to describe the components of the centralizers of certain involutions of $G$. This will be used in Section <ref> to construct the subgroup $G_0$ of $G$. One of the main tools here will be a version of the Curtis-Tits theorem <cit.> and a related theorem of Phan reproved by Bennett and Shpectorov in [14].

Finally, in Section <ref>, we will give a full proof of Theorem <ref> (basically summarizing Sections <ref>-<ref>), and we will prove Theorem <ref> and Corollary <ref>.

Notation and Terminology. Our notation and terminology are fairly standard. The reader is referred to [24], [28], [37] for unfamiliar definitions on groups and to [11], [19] for unfamiliar definitions on fusion systems.

However, we shall now explain some particularly important notation and definitions (before stating our main results, we already introduced some other important definitions).

Given a map $\alpha: A \rightarrow B$ and an element or a subset $X$ of $A$, we write $X^{\alpha}$ for the image of $X$ under $\alpha$. Also, if $C \subseteq A$ and $D \subseteq B$ such that $C^{\alpha} \subseteq D$, we use $\alpha|_{C,D}$ to denote the map $C \rightarrow D, c \mapsto c^{\alpha}$. Given two maps $\alpha: A \rightarrow B$ and $\beta: B \rightarrow C$, we write $\alpha\beta$ for the map $A \rightarrow C, a \mapsto (a^{\alpha})^{\beta}$.

Sometimes, we will interprete the symbols $+$ and $-$ as the integers $1$ and $-1$, respectively. For example, if $n$ is an integer and if $\varepsilon$ is assumed to be an element of $\lbrace +,- \rbrace$, then $n \equiv \varepsilon \mod 4$ shall express that $n \equiv 1 \mod 4$ if $\varepsilon = +$ and that $n \equiv -1 \mod 4$ if $\varepsilon = -$.

Let $G$ be a finite group. We write $G^{\#}$ for the set of non-identity elements of $G$. Given an element $g$ of $G$ and an element or a subset $X$ of $G$, we write $X^g$ for $g^{-1}Xg$. The inner automorphism $G \rightarrow G, x \mapsto x^g$ is denoted by $c_g$. For subgroups $Q$ and $H$ of $G$, we write $\mathrm{Aut}_H(Q)$ for the subgroup of $\mathrm{Aut}(Q)$ consisting of all automorphisms of $Q$ of the form $c_h|_{Q,Q}$, where $h \in N_H(Q)$.

We write $L(G)$ for the subgroup of $G$ generated by the components of $G$ and $L_{2'}(G)$ for the subgroup of $G$ generated by the $2$-components of $G$. We say that $G$ is core-free if $O(G) = 1$. If $G$ is core-free and if $L$ is a subnormal subgroup of $G$, then $L$ is said to be a solvable $2$-component of $G$ if $L \cong SL_2(3)$ or $PSL_2(3)$.

Let $n$ be a natural number. Then we use $E_{2^n}$ to denote an elementary abelian $2$-group of order $2^n$, and we say that $n$ is the rank of $E_{2^n}$. The maximal rank of an elementary abelian $2$-subgroup of a finite $2$-group $S$ is said to be the rank of $S$. It is denoted by $m(S)$.

Now let $p$ be a prime, and let $\mathcal{F}$ be a fusion system on a finite $p$-group $S$. Then $S$ is said to be the Sylow group of $\mathcal{F}$, and $\mathcal{F}$ is said to be nilpotent if $\mathcal{F} = \mathcal{F}_S(S)$. Given a fusion system $\mathcal{F}_1$ on a finite $p$-group $S_1$, we say that $\mathcal{F}$ and $\mathcal{F}_1$ are isomorphic if there is a group isomorphism $\varphi: S \rightarrow S_1$ such that

\begin{equation*}
\mathrm{Hom}_{\mathcal{F}_1}(Q^{\varphi},R^{\varphi}) = \lbrace (\varphi^{-1}|_{Q^{\varphi},Q})\psi(\varphi|_{R,R^{\varphi}}) \ \vert \ \psi \in \mathrm{Hom}_{\mathcal{F}}(Q,R) \rbrace
\end{equation*}

for all $Q,R \le S$. In this case, we say that $\varphi$ induces an isomorphism from $\mathcal{F}$ to $\mathcal{F}_1$. Let $Q$ be a normal subgroup of $S$. If $P$ and $R$ are subgroups of $S$ containing $Q$ and if $\alpha: P \rightarrow R$ is a morphism in $\mathcal{F}$ such that $Q^{\alpha} = Q$, we write $\alpha/Q$ for the group homomorphism $P/Q \rightarrow R/Q$ induced by $\alpha$. The fusion system $\mathcal{F}/Q$ on $S/Q$ with $\mathrm{Hom}_{\mathcal{F}/Q}(P/Q,R/Q) = \lbrace \alpha/Q \ \vert \ \alpha \in \mathrm{Hom}_{\mathcal{F}}(P,R), Q^{\alpha} = Q \rbrace$ for all $P, R \le S$ containing $Q$ is said to be the factor system of $\mathcal{F}$ modulo $Q$.

Suppose now that $\mathcal{F}$ is saturated. We write $\mathfrak{foc}(\mathcal{F})$ for the focal subgroup of $\mathcal{F}$ and $\mathfrak{hnp}(\mathcal{F})$ for the hyperfocal subgroup of $\mathcal{F}$. We say that $\mathcal{F}$ is quasisimple if $\mathcal{F}/Z(\mathcal{F})$ is simple and $\mathfrak{foc}(\mathcal{F}) = S$. A component of $\mathcal{F}$ is a subnormal quasisimple subsystem of $\mathcal{F}$. Given a normal subsystem $\mathcal{E}$ of $S$ and a subgroup $R$ of $S$, we write $\mathcal{E}R$ for the product of $\mathcal{E}$ and $R$, as defined in <cit.>.

§ PRELIMINARIES ON FINITE GROUPS AND FUSION SYSTEMS

In this section, we present some general results on finite groups and fusion systems.

§.§ Preliminaries on finite groups

Let $G$ be a finite group, and let $N$ be a normal $p'$-subgroup of $G$ for some prime $p$. Set $\widebar G := G/N$. If $R$ is a $p$-subgroup of $G$, then we have $N_{\overline{G}}(\overline{R}) = \overline{N_G(R)}$ and $C_{\overline{G}}(\overline{R}) = \overline{C_G(R)}$.

Let $G$ be a finite group, and let $N$ be a normal $p'$-subgroup of $G$ for some prime $p$. Set $\widebar G := G/N$. If $x \in G$ has order $p$, then we have $C_{\widebar G}(\widebar x) = \widebar{C_G(x)}$.

Let $G$ be a finite group, and let $Z$ be a cyclic central subgroup of $G$. Then each $E_8$-subgroup of $G/Z$ has an involution which is the image of an involution of $G$.

Let $Z \le E \le G$ such that $E/Z \cong E_8$. Let $R$ be a Sylow $2$-subgroup of $E$. Then $E = RZ$. It suffices to show that $R$ has an involution not lying in $R \cap Z$. Assume that any involution of $R$ is an element of $R \cap Z$. Then $R$ has a unique involution since $Z$ is cyclic. We have $R/(R \cap Z) \cong RZ/Z = E/Z \cong E_8$, and so $R$ is not cyclic. Applying <cit.>, we conclude that $R$ is generalized quaternion. In particular, $Z(R)$ has order $2$, and so we have $R \cap Z = Z(R)$. Since $R$ is a generalized quaternion group, $R/Z(R)$ is dihedral. In particular, $E/Z \cong R/(R \cap Z) = R/Z(R) \not\cong E_8$. This contradiction shows that $R$ has an involution not lying in $R \cap Z$, as required.

The following proposition is well-known. We include a proof since we could not find a reference in which it appears in the form given here.

Let $G$ be a finite group, and let $N$ be a normal subgroup of $G$ with odd order. If $L$ is a $2$-component of $G$, then $LN/N$ is a $2$-component of $G/N$. The map from the set of $2$-components of $G$ to the set of $2$-components of $G/N$ sending each $2$-component $L$ of $G$ to $LN/N$ is a bijection. Moreover, if $N \le K \le G$ and $K/N$ is a $2$-component of $G/N$, then $O^{2'}(K)$ is the associated $2$-component of $G$.

Let $L$ be a $2$-component of $G$. Hence, $L$ is a perfect subnormal subgroup of $G$ such that $L/O(L)$ is quasisimple. Clearly, $LN/N$ is perfect and subnormal in $G/N$. Also, we have $(LN/N)/O(LN/N) \cong L/O(L)$, and so $(LN/N)/O(LN/N)$ is quasisimple. It follows that $LN/N$ is a $2$-component of $G/N$.

Let $N \le K \le G$ such that $K/N$ is a $2$-component of $G/N$. In order to prove the second statement of the proposition, it is enough to show that there is precisely one $2$-component $L$ of $G$ such that $LN/N = K/N$.

Since $K/N$ is subnormal in $G/N$, we have that $K$ is subnormal in $G$. Therefore, $L := O^{2'}(K)$ is subnormal in $G$. Since $O^{2'}(K/N) = K/N$, we have that $K/N = LN/N$. Clearly, $O^{2'}(L) = L$. We have $L/O(L) \cong (LN/N)/O(LN/N) = (K/N)/O(K/N)$, and so $L/O(L)$ is quasisimple. Applying <cit.>, we conclude that $L$ is a $2$-component of $G$.

Now let $L_0$ be a $2$-component of $G$ such that $K/N = L_0 N /N$. Then $K = L_0 N$. In particular, $L_0$ is a subgroup of $K$ with odd index in $K$. Since $L_0$ is subnormal in $G$, we have that $L_0$ is subnormal in $K$. Applying <cit.>, we conclude that $L_0 = O^{2'}(L_0) = O^{2'}(K) = L$. The proof of the second statement of the proposition is now complete. The third statement also follows from the above arguments.

Let $G$ be a finite group, and let $n$ be a positive integer. Assume that $L_1$, …, $L_n$ are the distinct $2$-components of $G$, and assume that $L_i \trianglelefteq G$ for all $1 \le i \le n$. Let $x$ be a $2$-element of $G$, and let $L$ be a $2$-component of $C_G(x)$. Then $L$ is a $2$-component of $C_{L_i}(x)$ for some $1 \le i \le n$.

By <cit.>, we have $L_{2'}(C_G(x)) = L_{2'}(C_{L_{2'}(G)}(x))$, and by <cit.>, we have $L_{2'}(C_{L_{2'}(G)}(x)) = \prod_{i=1}^n L_{2'}(C_{L_i}(x))$. Using basic properties of $2$-components, as presented in <cit.>, it is not hard to deduce that $L$ is a $2$-component of $C_{L_i}(x)$ for some $1 \le i \le n$.

The concepts introduced by the following two definitions will play a crucial role in the proof of Theorem <ref> (see [32] for a detailed study of these concepts).

Let $G$ be a finite group, $k$ be a positive integer and $A$ be an elementary abelian $2$-subgroup of $G$.

(i) For each nontrivial elementary abelian $2$-subgroup $E$ of $G$, we define

\begin{equation*}
\Delta_G(E) := \bigcap_{a \in E^{\#}} O(C_G(a)).
\end{equation*}

(ii) We say that $G$ is $k$-balanced with respect to $A$ if whenever $E$ is a subgroup of $A$ of rank $k$ and $a$ is a non-trivial element of $A$, we have

\begin{equation*}
\Delta_G(E) \cap C_G(a) \le O(C_G(a)).
\end{equation*}

(iii) We say that $G$ is $k$-balanced if whenever $E$ is an elementary abelian $2$-subgroup of $G$ of rank $k$ and $a$ is an involution of $G$ centralizing $E$, we have

\begin{equation*}
\Delta_G(E) \cap C_G(a) \le O(C_G(a)).
\end{equation*}

(iv) By saying that $G$ is balanced (respectively, balanced with respect to $A$), we mean that $G$ is $1$-balanced (respectively, $1$-balanced with respect to $A$).

Let $G$ be a finite quasisimple group, and let $k$ be a positive integer. Then $G$ is said to be locally $k$-balanced if whenever $H$ is a subgroup of $\mathrm{Aut}(G)$ containing $\mathrm{Inn}(G)$, we have

\begin{equation*}
\Delta_H(E) = 1
\end{equation*}

for any elementary abelian $2$-subgroup $E$ of $H$ of rank $k$. We say that $G$ is locally balanced if $G$ is locally $1$-balanced.

We need the following proposition for the proof of Theorem <ref>. It includes <cit.> and some additional statements, which should be also known. We include a proof for the convenience of the reader.

Let $k$ be a positive integer, and let $G$ be a finite group. For each elementary abelian $2$-subgroup $A$ of $G$ of rank at least $k+1$, let

\begin{equation*}
W_A := \langle \Delta_G(E) \ \vert \ E \le A, m(E) = k \rangle.
\end{equation*}

Then, for any elementary abelian $2$-subgroup $A$ of $G$ of rank at least $k+1$, the following hold:

(i) $(W_A)^g = W_{A^g}$ for all $g \in G$.

(ii) Suppose that $A$ has rank at least $k + 2$ and that $G$ is $k$-balanced with respect to $A$. Then $W_A$ has odd order. Moreover, if $A_0$ is a subgroup of $A$ of rank at least $k+1$, then we have $W_A = W_{A_0}$ and $N_G(A_0) \le N_G(W_A)$.

In order to prove Proposition <ref>, we need the following theorem.

Let $k$ be a positive integer, $G$ be a finite group and $A$ be an elementary abelian $2$-subgroup of $G$ of rank at least $k+2$. Suppose that $G$ is $k$-balanced with respect to $A$. Then we obtain an $A$-signalizer functor on $G$ (in the sense of <cit.>) by defining

\begin{equation*}
\theta(C_G(a)) := \langle \Delta_G(E) \cap C_G(a) : \ E \le A, m(E) = k \rangle
\end{equation*}

for each $a \in A^{\#}$.

We also need the following lemma.

Let the notation be as in Theorem <ref>. Suppose that $A_0$ is subgroup of $A$ of rank $k+1$. Then we have

\begin{equation*}
\theta(G,A) := \langle \theta(C_G(a)) \ \vert \ a \in A^{\#} \rangle = \langle \Delta_G(E) \ \vert \ E \le A_0, m(E) = k \rangle =: W_{A_0}.
\end{equation*}

To prove this, we follow arguments found on pp. 40-41 of [40].

Since $\theta$ is an $A$-signalizer functor on $G$, $\theta(C_G(a))$ is $A$-invariant and in particular $A_0$-invariant for each $a \in A^{\#}$. Consequently, $\theta(G,A)$ is $A_0$-invariant. By the Solvable Signalizer Functor Theorem <cit.>, $\theta$ is complete (in the sense of <cit.>). In particular, $\theta(G,A)$ has odd order. Applying <cit.>, we conclude that

\begin{equation*}
\theta(G,A) = \langle C_{\theta(G,A)}(E) \ \vert \ E \le A_0, m(E) = k \rangle.
\end{equation*}

Since $\theta$ is complete, we have $C_{\theta(G,A)}(a) = \theta(C_G(a))$ for each $a \in A^{\#}$. By definition of $\theta$ and since $G$ is $k$-balanced with respect to $A$, we have $\theta(C_G(a)) \le O(C_G(a))$ for each $a \in A^{\#}$. So, if $E$ is a subgroup of $A_0$ of rank $k$, then

\begin{equation*}
C_{\theta(G,A)}(E) = \bigcap_{a \in E^{\#}}C_{\theta(G,A)}(a) = \bigcap_{a \in E^{\#}} \theta(C_G(a)) \le \bigcap_{a \in E^{\#}} O(C_G(a)) = \Delta_G(E).
\end{equation*}

It follows that $\theta(G,A) \le W_{A_0}$.

Let $E \le A_0$ with $m(E) = k$. Clearly, $\Delta_G(E)$ is $A$-invariant. As a consequence of <cit.>, we have

\begin{equation*}
\Delta_G(E) = \langle \Delta_G(E) \cap C_G(a) \ \vert \ a \in A^{\#} \rangle.
\end{equation*}

By definition of $\theta$, we have $\Delta_G(E) \cap C_G(a) \le \theta(C_G(a))$ for each $a \in A^{\#}$. It follows that $\Delta_G(E) \le \theta(G,A)$. Consequently, $W_{A_0} \le \theta(G,A)$.

It is straightforward to verify (i).

To verify (ii), let $A$ be an elementary abelian $2$-subgroup of $G$ of rank at least $k+2$ such that $G$ is $k$-balanced with respect to $A$. Let $\theta$ be the $A$-signalizer functor on $G$ given by Theorem <ref>, and let $\theta(G,A) := \langle \theta(C_G(a)) \ \vert \ a \in A^{\#} \rangle$. As a consequence of Lemma <ref>, we have $\theta(G,A) = W_A$. By the proof of Lemma <ref>, $W_A = \theta(G,A)$ has odd order.

Now let $A_0$ be a subgroup of $A$ of rank at least $k+1$. By Lemma <ref>, $W_A = \theta(G,A) \le W_{A_0} \le W_A$, and so $W_A = W_{A_0}$. Finally, if $g \in N_G(A_0)$, then $(W_A)^g = (W_{A_0})^g = W_{(A_0)^g} = W_{A_0} = W_A$, and hence $N_G(A_0) \le N_G(W_A)$.

§.§ Preliminaries on fusion systems

Let $p$ be a prime, $G$ be a finite group, $N$ be a normal subgroup of $G$ and $S \in \mathrm{Syl}_p(G)$. Then the canonical group isomorphism $S/(S \cap N) \rightarrow SN/N$ induces an isomorphism from $\mathcal{F}_S(G)/(S \cap N)$ to $\mathcal{F}_{SN/N}(G/N)$.

Let $\varphi$ denote the canonical group isomorphism $S/(S \cap N) \rightarrow SN/N$. Let $P$ and $Q$ be two subgroups of $S$ such that $S \cap N$ is contained in both $P$ and $Q$. Set $\widetilde P := P/(S \cap N)$, $\widetilde Q := Q/(S \cap N)$, $\widebar P := PN/N$ and $\widebar Q := QN/N$. Moreover, define $\widetilde{\mathcal{F}} := \mathcal{F}_S(G)/(S \cap N)$ and $\widebar{\mathcal{F}} := \mathcal{F}_{SN/N}(G/N)$. It is enough to show that

\begin{equation*}
\mathrm{Hom}_{\widebar{\mathcal{F}}}(\widebar P, \widebar Q) = \lbrace (\varphi^{-1}|_{\widebar P,\widetilde P})\alpha(\varphi|_{\widetilde Q, \widebar Q}) \mid \alpha \in \mathrm{Hom}_{\widetilde{\mathcal{F}}}(\widetilde P, \widetilde Q)\rbrace.
\end{equation*}

Let $\alpha \in \mathrm{Hom}_{\widetilde{\mathcal{F}}}(\widetilde P, \widetilde Q)$. Then there exists $g \in G$ with $P^g \le Q$ and $\alpha = (c_g|_{P,Q})/(S \cap N)$. By a direct calculation, $(\varphi^{-1}|_{\widebar P, \widetilde P})\alpha (\varphi|_{\widetilde Q, \widebar Q}) = c_{gN}|_{\widebar P, \widebar Q} \in \mathrm{Hom}_{\widebar{\mathcal{F}}}(\widebar P, \widebar Q)$.

Now let $\widebar \alpha \in \mathrm{Hom}_{\widebar{\mathcal{F}}}(\widebar P, \widebar Q)$. Then there exists $g \in G$ with $\widebar{P}^{gN} \le \widebar{Q}$ and $\widebar \alpha = c_{gN}|_{\widebar P, \widebar Q}$. Clearly, $P^g \le QN$. Since $S \cap N \le Q$, we have that $Q$ is a Sylow $p$-subgroup of $QN$. Since $P^g$ is a $p$-subgroup of $QN$, it follows that there exists an element $n \in N$ with $P^{gn} \le Q$. Set $\alpha := (c_{gn}|_{P,Q})/(S \cap N)$. Then a direct calculation shows that $\widebar{\alpha} = ({\varphi}^{-1}|_{\widebar P, \widetilde P})\alpha(\varphi|_{\widetilde Q,\widebar Q})$.

Let $p$ be a prime, $G$ be a finite group and $S \in \mathrm{Syl}_p(G)$. Then the canonical group isomorphism $S \rightarrow \widebar S := S O_{p'}(G)/O_{p'}(G)$ induces an isomorphism from $\mathcal{F}_S(G)$ to $\mathcal{F}_{\widebar S}(G/O_{p'}(G))$.

Let $K_1$ and $K_2$ be two quasisimple finite groups. If the $2$-fusion systems of $K_1$ and $K_2$ are isomorphic, then the $2$-fusion systems of $K_1/Z(K_1)$ and $K_2/Z(K_2)$ are isomorphic.

Suppose that the $2$-fusion systems of $K_1$ and $K_2$ are isomorphic. Let $S_i$ be a Sylow $2$-subgroup of $K_i$ and $\mathcal{F}_i := \mathcal{F}_{S_i}(K_i)$ for $i \in \lbrace 1,2 \rbrace$. As a consequence of <cit.>, we have $Z(\mathcal{F}_i) = S_i \cap Z^{*}(K_i)$ for $i \in \lbrace 1,2 \rbrace$. Since $K_1$ and $K_2$ are quasisimple, we have $Z^{*}(K_i) = Z(K_i)$ and hence $Z(\mathcal{F}_i) = S_i \cap Z(K_i)$ for $i \in \lbrace 1,2 \rbrace$. Since $\mathcal{F}_1 \cong \mathcal{F}_2$, it follows that

\begin{equation*}
\mathcal{F}_1/(S_1 \cap Z(K_1)) = \mathcal{F}_1/Z(\mathcal{F}_1) \cong \mathcal{F}_2/Z(\mathcal{F}_2) = \mathcal{F}_2/(S_2 \cap Z(K_2)).
\end{equation*}

Applying Lemma <ref>, we may conclude that the $2$-fusion system of $K_1/Z(K_1)$ is isomorphic to the $2$-fusion system of $K_2/Z(K_2)$.

Let $S$ be a finite $2$-group, and let $A$ and $B$ be normal subgroups of $S$ such that $S$ is the internal direct product of $A$ and $B$. Suppose that $A \cong Q_8$. Let $\mathcal{F}$ be a (not necessarily saturated) fusion system on $S$. Assume that $A$ and $B$ are strongly $\mathcal{F}$-closed and that there is an automorphism $\alpha \in \mathrm{Aut}_{\mathcal{F}}(S)$ such that $\alpha|_{A,A}$ has order $3$, while $\alpha|_{B,B} = \mathrm{id}_B$. Then each strongly $\mathcal{F}$-closed subgroup of $S$ contains or centralizes $A$.

Let $C$ be a strongly $\mathcal{F}$-closed subgroup of $S$ not containing $A$. Our task is to show that $C$ centralizes $A$.

Since $A$ and $C$ are strongly $\mathcal{F}$-closed, we have that $A \cap C$ is strongly $\mathcal{F}$-closed. In particular, $\alpha$ normalizes $A \cap C$. It is easy to see that an automorphism of $Q_8$ with order $3$ does not normalize any maximal subgroup of $Q_8$. So, as $\alpha|_{A,A}$ has order $3$ and normalizes $A \cap C$, we have that $A \cap C$ has order $1$ or $2$.

By <cit.>, we have

\begin{equation*}
[C,\langle \alpha \rangle] = [[C,\langle \alpha \rangle], \langle \alpha \rangle].
\end{equation*}

We claim that $[C,\langle \alpha \rangle] \le A \cap C$. Let $c \in C$ and $\beta \in \langle \alpha \rangle$. Let $a \in A$ and $b \in B$ such that $c = ab$. Since $A$ and $B$ commute and since $\beta$ normalizes $A$ and centralizes $B$, we have

\begin{equation*}
[c,\beta] = c^{-1}c^{\beta} = b^{-1}a^{-1}a^{\beta}b^{\beta} = a^{-1}a^{\beta} \in A \cap C.
\end{equation*}

Thus $[C,\langle \alpha \rangle] \le A \cap C$, as asserted.

Since $A \cap C$ has order $1$ or $2$, we have $[A \cap C, \langle \alpha \rangle] = 1$. So it follows that

\begin{equation*}
[C, \langle \alpha \rangle] = [[C,\langle \alpha \rangle],\langle \alpha \rangle] \le [A \cap C,\langle \alpha \rangle] =1.
\end{equation*}

Now we prove that $C$ centralizes $A$. Let $c \in C$ and $a \in A$, $b \in B$ with $c = ab$. We have $c^{-1}c^{\alpha} \in [C,\langle \alpha \rangle] = 1$, whence $c^{\alpha} = c$. Thus $ab = (ab)^{\alpha} = a^{\alpha}b$ and hence $a = a^{\alpha}$. As remarked above, $\alpha$ does not normalize any maximal subgroup of $A$. So $a$ cannot have order $4$. By the structure of $A \cong Q_8$, it follows that $a \in Z(A)$. This implies that $c = ab$ centralizes $A$.

We need the following definition in order to state the next proposition.

A nonabelian finite simple group $G$ is said to be a Goldschmidt group provided that one of the following holds:

(1) $G$ has an abelian Sylow $2$-subgroup.

(2) $G$ is isomorphic to a finite simple group of Lie type in characteristic $2$ of Lie rank $1$.

Let $G$ be a finite group, and let $S$ be a Sylow $2$-subgroup of $G$. Assume that for each $2$-component $L$ of $G$, the factor group $L/Z^{*}(L)$ is a known finite simple group. Let $\mathfrak{L}_{2'}$ denote the set of $2$-components $L$ of $G$ such that $L/Z^{*}(L)$ is not a Goldschmidt group. Then the following hold:

(i) Let $L$ be a $2$-component of $G$. Then $\mathcal{F}_{S \cap L}(L)$ is a component of $\mathcal{F}_S(G)$ if and only if $L \in \mathfrak{L}_{2'}$.

(ii) The map from $\mathfrak{L}_{2'}$ to the set of components of $\mathcal{F}_S(G)$ sending each element $L$ of $\mathfrak{L}_{2'}$ to $\mathcal{F}_{S \cap L}(L)$ is a bijection.

Let $L$ be a $2$-component of $G$. Set $\mathcal{G} := \mathcal{F}_{S \cap L}(L)$. Since $L$ is subnormal in $G$, we have that $\mathcal{G}$ is subnormal in $\mathcal{F}_S(G)$ (see <cit.>). Therefore, $\mathcal{G}$ is a component of $\mathcal{F}_S(G)$ if and only if $\mathcal{G}$ is quasisimple. We have $\mathfrak{foc}(\mathcal{G}) = S \cap L' = S \cap L$ by the focal subgroup theorem <cit.>, and so $\mathcal{G}$ is quasisimple if and only if $\mathcal{G}/Z(\mathcal{G})$ is simple. As a consequence of <cit.>, we have $Z(\mathcal{G}) = S \cap Z^{*}(L)$. Lemma <ref> implies that $\mathcal{G}/Z(\mathcal{G})$ is isomorphic to the $2$-fusion system of $L/Z^{*}(L)$. By <cit.>, the $2$-fusion system of $L/Z^{*}(L)$ is simple if and only if $L \in \mathfrak{L}_{2'}$. So $\mathcal{G}$ is a component of $\mathcal{F}_S(G)$ if and only if $L \in \mathfrak{L}_{2'}$, and (i) holds.

(ii) follows from <cit.>.

Let $G$ be a finite group with $O(G) = 1$, and let $S$ be a Sylow $2$-subgroup of $G$. Let $n \ge 1$ be a natural number, and let $L_1, \dots, L_n$ be pairwise distinct subgroups of $G$ such that $L_i$ is either a component or a solvable $2$-component of $G$ for each $1 \le i \le n$. Set $Q := (S \cap L_1)\cdots(S \cap L_n)$. Assume that $Q \trianglelefteq S$ and that $\mathcal{F}_S(G)/Q$ is nilpotent. Then, if $L_0$ is a component or a solvable $2$-component of $G$, we have $L_0 = L_i$ for some $1 \le i \le n$.

Let $L^s(G)$ denote the subgroup of $G$ generated by the components and the solvable $2$-components of $G$. By <cit.> and <cit.>, $L^s(G)$ is the central product of the subgroups of $G$ which are components or solvable $2$-components. Set $L := L_1 \cdots L_n \trianglelefteq L^s(G)$.

Let $\mathcal{G} := \mathcal{F}_{S \cap L^s(G)}(L^s(G))$. Clearly, $S \cap L = (S \cap L_1) \cdots (S \cap L_n) = Q$. Lemma <ref> implies that the $2$-fusion system of $L^s(G)/L$ is isomorphic to $\mathcal{G}/Q$. By hypothesis, $\mathcal{F}_S(G)/Q$ is nilpotent, and so $\mathcal{G}/Q$ is nilpotent. So the $2$-fusion system of $L^s(G)/L$ is nilpotent. Applying <cit.>, we conclude that $L^s(G)/L$ is $2$-nilpotent.

Now let $L_0$ be a component or a solvable $2$-component of $G$. If $L_0 \le L$, then we have $L_0 = L_i$ for some $1 \le i \le n$ since otherwise $L_0 \le Z(L)$, which is impossible. So it suffices to show that $L_0 \le L$.

If $L_0$ is a component of $G$, then $L_0/(L_0 \cap L)$ is both perfect and $2$-nilpotent, which implies that $L_0 \le L$, as needed.

Suppose now that $L_0$ is a solvable $2$-component of $G$. Assume that $L_0 \not\le L$. Then $L_0 \cap L \le Z(L_0)$. Since $L_0$ is a solvable $2$-component of $G$, it follows that $L_0/(L_0 \cap L)$ is isomorphic to $SL_2(3)$ or $PSL_2(3)$. On the other hand, $L_0/(L_0 \cap L)$ is $2$-nilpotent. This contradiction shows that $L_0 \le L$, as required.

Let $G$ be a finite group, and let $S$ be a Sylow $2$-subgroup of $G$. Let $n \ge 1$ be a natural number, and let $L_1, \dots, L_n$ be pairwise distinct $2$-components of $G$. Assume that $Q := (S \cap L_1)\cdots(S \cap L_n)$ is a normal subgroup of $S$ and that $\mathcal{F}_S(G)/Q$ is nilpotent. Then, if $L_0$ is a $2$-component of $G$, we have $L_0 = L_i$ for some $1 \le i \le n$.

Let $p$ be a prime, and let $\mathcal{E}$ be a simple saturated fusion system on a finite $p$-group $T$. Suppose that $\mathcal{E}$ is tamely realized (in the sense of <cit.>) by a nonabelian known finite simple group $K$ such that $\mathrm{Out}(K)$ is $p$-nilpotent. Assume moreover that $G$ is a nonabelian finite simple group containing a Sylow $p$-subgroup $S$ with $T \le S$ such that $\mathcal{E} \trianglelefteq \mathcal{F}_S(G)$ and $C_S(\mathcal{E}) = 1$. Then $\mathcal{F}_S(G)$ is tamely realized by a subgroup $L$ of $\mathrm{Aut}(K)$ containing $\mathrm{Inn}(K)$ such that the index of $\mathrm{Inn}(K)$ in $L$ is coprime to $p$.

Set $\mathcal{F} := \mathcal{F}_S(G)$. By a result of Bob Oliver, namely by <cit.>, $\mathcal{F}$ is tamely realized by a subgroup $L$ of $\mathrm{Aut}(K)$ containing $\mathrm{Inn}(K)$. We are going to show that the index of $\mathrm{Inn}(K)$ in $L$ is coprime to $p$.

Let $S_0$ be a Sylow $p$-subgroup of $L$. Then $\mathcal{F} \cong \mathcal{F}_{S_0}(L)$. Clearly, $O^p(G) = G$, and so $\mathfrak{hnp}(\mathcal{F}) = S$ by the hyperfocal subgroup theorem <cit.>. It follows that $\mathfrak{hnp}(\mathcal{F}_{S_0}(L)) = S_0$.

By the hyperfocal subgroup theorem <cit.>, $S_0 = \mathfrak{hnp}(\mathcal{F}_{S_0}(L)) = O^p(L) \cap S_0$. Consequently, $O^p(L)$ has $p'$-index in $L$, whence $O^p(L) = L$. So we have $O^p(L/\mathrm{Inn}(K)) = L/\mathrm{Inn}(K)$. On the other hand, $L/\mathrm{Inn}(K)$ is $p$-nilpotent since $\mathrm{Out}(K)$ is $p$-nilpotent. It follows that $L/\mathrm{Inn}(K)$ is a $p'$-group, as claimed.

§ AUXILIARY RESULTS ON LINEAR AND UNITARY GROUPS

In this section, we collect several results on linear and unitary groups needed for the proofs of our main results. Some of the results stated here are known, while others seem to be new. For the convenience of the reader, we also include proofs of known results when we could not find a reference in which they appear in the form stated here.

§.§ Basic definitions

We begin with some basic definitions. Let $q$ be a nontrivial prime power, and let $n$ be a positive integer. The general linear group $GL_n(q)$ is the group of all invertible $n \times n$ matrices over $\mathbb{F}_q$ under matrix multiplication. The special linear group $SL_n(q)$ is the subgroup of $GL_n(q)$ consisting of all $n \times n$ matrices over $\mathbb{F}_q$ with determinant $1$. The center of $GL_n(q)$ consists of all scalar matrices $\lambda I_n$ with $\lambda \in (\mathbb{F}_q)^{*}$. We have $Z(SL_n(q)) = SL_n(q) \cap Z(GL_n(q))$. Set $PGL_n(q) := GL_n(q)/Z(GL_n(q))$ and $PSL_n(q) := SL_n(q)/Z(SL_n(q))$. By <cit.> and <cit.>, $SL_n(q)$ is quasisimple if $n \ge 2$ and $(n,q) \ne (2,2),(2,3)$.

As in <cit.>, we consider the general unitary group $GU_n(q)$ as the subgroup of $GL_n(q^2)$ consisting of all $(a_{ij}) \in GL_n(q^2)$ satisfying the condition $((a_{ij})^q)(a_{ij})^t = I_n$. The special unitary group $SU_n(q)$ is the subgroup of $GU_n(q)$ consisting of all elements of $GU_n(q)$ with determinant $1$. By <cit.>, we have $SL_2(q) \cong SU_2(q)$. The center of $GU_n(q)$ consists of all scalar matrices $\lambda I_n$, where $\lambda \in (\mathbb{F}_{q^2})^{*}$ and $\lambda^{q+1} = 1$. We have $Z(SU_n(q)) = SU_n(q) \cap Z(GU_n(q))$. Set $PGU_n(q) := GU_n(q)/Z(GU_n(q))$ and $PSU_n(q) := SU_n(q)/Z(SU_n(q))$. By <cit.>, $SU_n(q)$ is quasisimple if $n \ge 2$ and $(n,q) \ne (2,2),(2,3),(3,2)$.

We write $(P)GL_n^{+}(q)$ and $(P)SL_n^{+}(q)$ for $(P)GL_n(q)$ and $(P)SL_n(q)$, respectively. Also, we write $(P)GL_n^{-}(q)$ for $(P)GU_n(q)$ and $(P)SL_n^{-}(q)$ for $PSU_n(q)$.

§.§ Central extensions of $PSL_n(q)$ and $PSU_n(q)$

In the proofs of the following two lemmas, we use the terminology of <cit.>.

Let $n \ge 3$ be a natural number, and let $q$ be a nontrivial odd prime power. Let $H$ be a perfect central extension of $PSL_n(q)$. Then there is a subgroup $Z \le Z(SL_n(q))$ such that $H \cong SL_n(q)/Z$.

By <cit.>, the Schur multiplier of $PSL_n(q)$ is isomorphic to $C_{(n,q-1)} \cong Z(SL_n(q))$. From <cit.>, we see that this is just another way to say that $SL_n(q)$ is the universal covering group of $PSL_n(q)$. Applying <cit.> again, we conclude that $H \cong SL_n(q)/Z$ for some $Z \le Z(SL_n(q))$.

Let $n \ge 3$ be a natural number, and let $q$ be a nontrivial odd prime power. Let $H$ be a perfect central extension of $PSU_n(q)$. Assume that $(n,q) \ne (4,3)$ or that $Z(H)$ is a $2$-group. Then there is a subgroup $Z \le Z(SU_n(q))$ such that $H \cong SU_n(q)/Z$.

Suppose that $(n,q) \ne (4,3)$. By <cit.>, the Schur multiplier of $PSU_n(q)$ is isomorphic to $C_{(n,q+1)} \cong Z(SU_n(q))$. As in the proof of Lemma <ref>, we conclude that $H \cong SU_n(q)/Z$ for some $Z \le Z(SU_n(q))$.

Suppose now that $(n,q)=(4,3)$ and that $Z(H)$ is a $2$-group. Let $G := PSU_4(3)$, and let $\widetilde G$ be the universal covering group of $G$. Clearly, the Schur multiplier of $G$ is isomorphic to $Z(\widetilde G)$. By <cit.>, the Schur multiplier of $G$ is isomorphic to $C_4 \times C_3 \times C_3$. Thus $Z(\widetilde G) \cong C_4 \times C_3 \times C_3$. Clearly, if $Z \le Z(\widetilde G)$, then $Z(\widetilde G / Z) = Z(\widetilde G)/Z$. Let $Q$ be the unique Sylow $3$-subgroup of $Z(\widetilde G)$. By <cit.>, $\widetilde G$ is a central extension of $SU_4(3)$ and of $H$. Since $SU_4(3)$ has a center of order $4$, we have $SU_4(3) \cong \widetilde G / Q$. Let $Z \le Z(\widetilde G)$ with $H \cong \widetilde G / Z$. As $Z(H)$ is a $2$-group, we have $Q \le Z$, whence $H \cong \widetilde G / Z \cong (\widetilde G / Q) / (Z/Q)$ is isomorphic to a quotient of $SU_4(3)$ by a central subgroup.

§.§ Involutions

In this subsection, we collect several results on the involutions of the groups $(P)GL_n^{\varepsilon}(q)$ and $(P)SL_n^{\varepsilon}(q)$, where $q$ is a nontrivial odd prime power, $n \ge 2$ and $\varepsilon \in \lbrace +,- \rbrace$.

Let $q$ be a nontrivial odd prime power, and let $n \ge 2$. Let $T$ be an element of $GL_n(q)$ such that $T^2 = \lambda I_n$ for some $\lambda \in \mathbb{F}_q^{*}$. Then one of the following holds:

(i) There is some $\mu \in \mathbb{F}_q^{*}$ such that $\lambda = \mu^2$, and $T$ is $GL_n(q)$-conjugate to a diagonal matrix with diagonal entries in $\lbrace \mu, -\mu \rbrace$.

(ii) $n$ is even, $\lambda$ is a non-square element of $\mathbb{F}_q^{*}$, and $T$ is $GL_n(q)$-conjugate to the matrix

\begin{equation*}
\begin{pmatrix}  & I_{n/2} \ \\ \lambda I_{n/2} & \end{pmatrix}.
\end{equation*}

Moreover, we have $C_{GL_n(q)}(T) \cong GL_{\frac{n}{2}}(q^2)$.

We identify the field $\mathbb{F}_q$ with the subfield of $\mathbb{F}_{q^2}$ consisting of all $x \in \mathbb{F}_{q^2}$ satisfying $x^q = x$. It is easy to note that any element of $\mathbb{F}_q^{*}$ is the square of an element of $\mathbb{F}_{q^2}^{*}$. Let $\mu \in \mathbb{F}_{q^2}^{*}$ with $\lambda = \mu^2$.

If $\mu \in \mathbb{F}_q$, then basic linear algebra shows that $T$ is diagonalizable over $\mathbb{F}_q$, and it follows that (i) holds.

Assume now that $\mu \not\in \mathbb{F}_q$. Then $\lambda$ is a non-square element of $\mathbb{F}_q^{*}$.

Let $V$ be an $n$-dimensional vector space over $\mathbb{F}_q$, and let $B$ be an ordered basis of $V$. Let $\varphi$ be the element of $GL(V)$ such that $\varphi$ is represented by $T$ with respect to $B$. Clearly, $(1,\mu)$ is an $\mathbb{F}_q$-basis of $\mathbb{F}_{q^2}$. Using that $\varphi^2 = \lambda \mathrm{id}_V$, one can check that $V$ becomes a vector space over $\mathbb{F}_{q^2}$ by defining

\begin{equation*}
(x + y \mu) v := xv + y v^{\varphi}
\end{equation*}

for all $x,y \in \mathbb{F}_q$ and $v \in V$. Let $m$ be the dimension of $V$ over $\mathbb{F}_{q^2}$, and let $(v_1,\dots,v_m)$ be an $\mathbb{F}_{q^2}$-basis of $V$. Then $B_0 := (v_1,\dots,v_m,\mu v_1, \dots, \mu v_m)$ is an $\mathbb{F}_q$-basis of $V$. In particular, $n = 2m$ is even. For $1 \le i \le m$, we have $v_i^{\varphi} = \mu v_i$ and $(\mu v_i)^{\varphi} = (v_i)^{\varphi^2} = \lambda v_i$. So, with respect to $B_0$, $\varphi$ is represented by the matrix

\begin{equation*}
M := \begin{pmatrix}  & I_{n/2} \ \\ \lambda I_{n/2} & \end{pmatrix}.
\end{equation*}

It follows that $T$ and $M$ are $GL_n(q)$-conjugate.

Let $\psi$ be an automorphism of $V$ as an $\mathbb{F}_q$-vector space centralizing $\varphi$. For $x,y \in \mathbb{F}_q$ and $v \in V$, we have

\begin{equation*}
((x+y\mu)v)^{\psi} = (xv+yv^{\varphi})^{\psi} = xv^{\psi} + yv^{\psi\varphi} = (x+y\mu) v^{\psi},
\end{equation*}

whence $\psi$ is $\mathbb{F}_{q^2}$-linear. Conversely, if $\psi$ is $\mathbb{F}_{q^2}$-linear, then

\begin{equation*}
v_i^{\psi\varphi} = \mu v_i^{\psi} = (\mu v_i)^{\psi} = v_i^{\varphi\psi}
\end{equation*}

and hence $\psi\varphi = \varphi\psi$. It follows that the centralizer of $\varphi$ in the general linear group of $V$ as an $\mathbb{F}_q$-vector space is equal to the general linear group of $V$ as an $\mathbb{F}_{q^2}$-vector space. Thus $C_{GL_n(q)}(T) \cong GL_{\frac{n}{2}}(q^2)$. So (ii) holds.

Let $q$ be a nontrivial odd prime power, and let $n \ge 2$ be a natural number. Let $T \in GU_n(q)$.

(i) If $T^2 = \lambda I_n$ for some $\lambda \in \mathbb{F}_{q^2}^{*}$, then $\lambda$ is a square in $\mathbb{F}_{q^2}^{*}$.

(ii) If $T^2 = \rho^2 I_n$ for some $\rho \in \mathbb{F}_{q^2}^{*}$ with $\rho^{q+1} = 1$, then $T$ is $GU_n(q)$-conjugate to a diagonal matrix with diagonal entries in $\lbrace \rho,-\rho \rbrace$.

(iii) If $T^2 = \rho^2 I_n$ for some $\rho \in \mathbb{F}_{q^2}^{*}$ with $\rho^{q+1} \ne 1$, then $n$ is even, and we have $C_{GU_n(q)}(T) \cong GL_{\frac{n}{2}}(q^2)$.

Suppose that $T^2 = \lambda I_n$ for some $\lambda \in \mathbb{F}_{q^2}^{*}$. Since $T^2 \in GU_n(q)$, we have that $\lambda^{q+1} = 1$. It is easy to see that any element $x$ of $\mathbb{F}_{q^2}^{*}$ with $x^{q+1} = 1$ is a square in $\mathbb{F}_{q^2}^{*}$. So (i) holds.

A proof of (ii) and (iii) can be extracted from <cit.>.

Let $q$ be a nontrivial odd prime power, and let $n \ge 2$ be a natural number. Let $\rho$ be an element of $\mathbb{F}_q^{*}$ of order $(n,q-1)$. For each even natural number $i$ with $2 \le i < n$, let

\begin{equation*}
\widetilde{t_i} := \begin{pmatrix} I_{n-i} & \\  & -I_i \end{pmatrix} \in SL_n(q)
\end{equation*}

and let $t_i$ be the image of $\widetilde{t_i}$ in $PSL_n(q)$.

(i) Assume that $n$ is odd. Then each involution of $PSL_n(q)$ is $PSL_n(q)$-conjugate to $t_i$ for some even $2 \le i < n$.

(ii) Assume that $n$ is even and that there is some $\mu \in \mathbb{F}_q^{*}$ with $\rho = \mu^2$. For each odd natural number $i$ with $1 \le i < n$, the matrix

\begin{equation*}
\widetilde{t_i} := \begin{pmatrix} \mu I_{n-i} & \\  & -\mu I_i \end{pmatrix}
\end{equation*}

lies in $SL_n(q)$. Let $t_i$ denote the image of $\widetilde{t_i}$ in $PSL_n(q)$ for each odd $1 \le i < n$. Then each involution of $PSL_n(q)$ is $PSL_n(q)$-conjugate to $t_i$ for some (even or odd) $1 \le i \le \frac{n}{2}$.

(iii) Assume that $n$ is even and that $\rho$ is a non-square element of $\mathbb{F}_q$. Let

\begin{equation*}
\widetilde w := \begin{pmatrix}  & I_{n/2} \ \\ \rho I_{n/2} & \end{pmatrix}.
\end{equation*}

If $\widetilde w \in SL_n(q)$, then each involution of $PSL_n(q)$ is $PSL_n(q)$-conjugate to to $t_i$ for some even $2 \le i \le \frac{n}{2}$ or to $w := \widetilde{w} Z(SL_n(q)) \in PSL_n(q)$. If $\widetilde w \not\in SL_n(q)$, then each involution of $PSL_n(q)$ is $PSL_n(q)$-conjugate to $t_i$ for some even $2 \le i \le \frac{n}{2}$.

We follow arguments found in the proof of <cit.>.

Assume that $n$ is odd. Then $Z(SL_n(q))$ has odd order, and therefore, any involution of $PSL_n(q)$ is the image of an involution of $SL_n(q)$. As a consequence of Lemma <ref>, each involution of $SL_n(q)$ is $SL_n(q)$-conjugate to $\widetilde{t_i}$ for some even $2 \le i < n$. So (i) follows.

Assume now that $n$ is even and that $\rho = \mu^2$ for some $\mu \in \mathbb{F}_q^{*}$. Note that $Z(SL_n(q))$ equals $\langle \rho I_n \rangle$. We claim that $\mu^n = -1$. Since $\mu^{2n} = \rho^n = 1$, we have that $\mu^n = 1$ or $-1$. If $\mu^n = 1$, then $\mu \in \langle \rho \rangle$, and so $\rho$ is a square in $\langle \rho \rangle$, which is impossible. So we have $\mu^n = -1$. It follows that $\widetilde{t_i} \in SL_n(q)$ for each odd $1 \le i < n$. Now let $T \in SL_n(q)$ such that $TZ(SL_n(q)) \in PSL_n(q)$ is an involution. Then we have $T^2 = \rho^{\ell}I_n = \mu^{2\ell}I_n$ for some $1 \le \ell \le (n,q-1)$. Using Lemma <ref>, we conclude that $T$ is $SL_n(q)$-conjugate to a diagonal matrix $D \in SL_n(q)$ with diagonal entries in $\lbrace \mu^{\ell}, - \mu^{\ell} \rbrace$. Let $1 \le i < n$ such that $- \mu^{\ell}$ occurs precisely $i$ times as a diagonal entry of $D$. If $i$ is odd, we may assume that $D = \mu^{\ell-1} \widetilde{t_i}$, and if $i$ is even, we may assume that $D = \mu^{\ell} \widetilde{t_i}$. In either case, the image of $D$ in $PSL_n(q)$ is $t_i$. Hence, $TZ(SL_n(q))$ is $PSL_n(q)$-conjugate to $t_i$. Noticing that $t_i$ is $PSL_n(q)$-conjugate to $t_{n-i}$, we conclude that (ii) holds.

Now assume that $n$ is even and that $\rho$ is a non-square element of $\mathbb{F}_q$. Again let $T$ be an element of $SL_n(q)$ such that $TZ(SL_n(q)) \in PSL_n(q)$ is an involution. We have $T^2 = \rho^{\ell}I_n$ for some $1 \le \ell \le (n,q-1)$. Assume that $\ell$ is even. Then Lemma <ref> implies that $T$ or $-T$ is $SL_n(q)$-conjugate to $\rho^{\frac{\ell}{2}}\widetilde{t_i}$ for some even $2 \le i \le \frac{n}{2}$. It follows that $TZ(SL_n(q))$ is $PSL_n(q)$-conjugate to $t_i$ for some even $2 \le i \le \frac{n}{2}$. Assume now that $\ell$ is odd. As $\rho$ is not a square in $\mathbb{F}_q$, but $\rho^{\ell-1}$ is a square in $\mathbb{F}_q$, $\rho^{\ell}$ cannot be a square in $\mathbb{F}_q$. Using Lemma <ref>, we may conclude that $T$ is $GL_n(q)$-conjugate to the matrix

\begin{equation*}
M := \begin{pmatrix} 0 & \rho^{\ell} & & & \\
1 & 0 & & &\\
& & \ddots & & \\
& & & 0 & \rho^{\ell} \\
& & & 1 & 0
\end{pmatrix} \in SL_n(q).
\end{equation*}

It is rather easy to see that $T$ and $M$ are even conjugate in $SL_n(q)$. Let $k := \frac{\ell-1}{2}$. It is not hard to show that the matrices

\begin{equation*}
\begin{pmatrix} 0 & \rho^{\ell} \\ 1 & 0 \end{pmatrix} \ \textnormal{and} \ \begin{pmatrix} 0 & \rho^{k+1} \\ \rho^k & 0 \end{pmatrix}
\end{equation*}

are $SL_2(q)$-conjugate. So it follows that $M$ and hence $T$ is $SL_n(q)$-conjugate to $\rho^k M_2$, where

\begin{equation*}
M_2 := \begin{pmatrix} 0 & \rho & & & \\
1 & 0 & & &\\
& & \ddots & & \\
& & & 0 & \rho \\
& & & 1 & 0
\end{pmatrix} \in SL_n(q).
\end{equation*}

Consequently, the images of $T$ and $M_2$ in $PSL_n(q)$ are conjugate. Furthermore, as $\mathrm{det}(M_2) = \mathrm{det}(\widetilde w)$, we see that $\widetilde w \in SL_n(q)$. Also, $\widetilde w$ is $SL_n(q)$-conjugate to $M_2$, and so $TZ(SL_n(q))$ is $PSL_n(q)$-conjugate to $w$.

Let $q$ be a nontrivial odd prime power and let $n \ge 4$ be an even natural number. Let $\rho$ be an element of $\mathbb{F}_q^{*}$ of order $(n,q-1)$. Suppose that $\rho$ is a non-square element of $\mathbb{F}_q$ and that

\begin{equation*}
\widetilde w := \begin{pmatrix} & I_{n/2} \\ \rho I_{n/2} & \end{pmatrix}
\end{equation*}

lies in $SL_n(q)$. Denote the image of $\widetilde w$ in $PSL_n(q)$ by $w$. Set $C := C_{PSL_n(q)}(w)$. Let $P$ be a Sylow $2$-subgroup of $C$. Then the following hold:

(i) $C$ has a unique $2$-component $J$, and $J$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}(q^2)$.

(ii) We have $P \cap J \trianglelefteq P$, and the factor system $\mathcal{F}_P(C)/(P \cap J)$ is nilpotent.

(iii) If $n \ge 6$, then $P$ has rank at least $4$.

Set $C_0 := C_{SL_n(q)}(\widetilde w)/Z(SL_n(q)) \le C$. By a direct argument, $C_0$ has index $2$ in $C$. So the $2$-components of $C$ are precisely the $2$-components of $C_0$. One may deduce from Lemma <ref> that $C_{SL_n(q)}(\widetilde w)$ has a normal subgroup $\widetilde J$ isomorphic to $SL_{\frac{n}{2}}(q^2)$ such that the corresponding factor group is cyclic. Let $J$ be the image of $\widetilde J$ in $PSL_n(q)$. Then $J$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}(q^2)$. Moreover, $J \trianglelefteq C_0$ and $C_0/J$ is cyclic. Therefore, $J$ is the only $2$-component of $C_0$ and hence the only $2$-component of $C$. Thus (i) holds.

We have $P \cap J \trianglelefteq P$ because $J \trianglelefteq C$. By Lemma <ref>, the factor system $\mathcal{F}_P(C)/(P \cap J)$ is isomorphic to the $2$-fusion system of $C/J$. Since $C_0$ has index $2$ in $C$ and $C_0/J$ is abelian, we have that $C/J$ is $2$-nilpotent. So $C/J$ has a nilpotent $2$-fusion system, and (ii) follows.

We now prove (iii). Assume that $n \ge 6$. Let $u$ denote the image of

\begin{equation*}
\begin{pmatrix} 0 & \rho & & & \\
1 & 0 & & & \\
& & \ddots & & \\
& & & 0 & \rho \\
& & & 1 & 0  \end{pmatrix} \in SL_n(q)
\end{equation*}

in $PSL_n(q)$. It is easy to see that there exist $a, b \in \mathbb{F}_q$ with $a^2 \rho - b^2 \rho^2 = 1$. Let $s$ be the image of

\begin{equation*}
\begin{pmatrix}
-b \rho & a \rho & & & \\
-a & b\rho & & & \\
& & \ddots & & \\
& & & -b \rho & a \rho \\
& & & -a & b \rho
\end{pmatrix} \in SL_n(q)
\end{equation*}

in $PSL_n(q)$. By a direct calculation, $s \in C_{PSL_n(q)}(u)$. Another direct calculation shows that $s$ is an involution. Let $z_1$ denote the image of

\begin{equation*}
\begin{pmatrix} -I_2 & \\ & I_{n-2}\end{pmatrix} \in SL_n(q)
\end{equation*}

in $PSL_n(q)$, and let $z_2$ denote the image of

\begin{equation*}
\begin{pmatrix} I_2 & & \\ & - I_2 & \\ & & I_{n-4} \end{pmatrix} \in SL_n(q)
\end{equation*}

in $PSL_n(q)$. Then one can easily verify that $\langle s, u, z_1, z_2 \rangle \le C_{PSL_n(q)}(u)$ is isomorphic to $E_{16}$. So a Sylow $2$-subgroup of $C_{PSL_n(q)}(u)$ has rank at least $4$. This is also true for $P$ as $w$ and $u$ are conjugate (see Proposition <ref>).

Let $n \ge 2$ be a natural number and let $\varepsilon \in \lbrace +,- \rbrace$. Also, let $T \in GL_n^{\varepsilon}(3) \setminus Z(GL_n^{\varepsilon}(3))$ such that $T^2 \in Z(GL_n^{\varepsilon}(3))$. Then $C_{GL_n^{\varepsilon}(3)}(T)$ is core-free.

By Lemmas <ref> and <ref>, we either have $C_{GL_n^{\varepsilon}(3)}(T) \cong GL_i^{\varepsilon}(3) \times GL_{n-i}^{\varepsilon}(3)$ for some $1 \le i < n$, or $n$ is even and $C_{GL_n^{\varepsilon}(3)}(T) \cong GL_{n/2}(9)$. So we have that $C_{GL_n^{\varepsilon}(3)}(T)$ is core-free.

It is easy to deduce the following two corollaries from Lemma <ref>.

Let $n \ge 2$ be a natural number and let $\varepsilon \in \lbrace +,- \rbrace$. Then any involution centralizer in $SL_n^{\varepsilon}(3)$ is core-free.

Let $n \ge 2$ be a natural number and let $\varepsilon \in \lbrace +,- \rbrace$. Then any involution centralizer in $PGL_n^{\varepsilon}(3)$ is core-free.

§.§ Sylow $2$-subgroups and $2$-fusion systems

In this subsection, we consider several properties of Sylow $2$-subgroups and $2$-fusion systems of linear and unitary groups.

(<cit.>) Let $q$ be a nontrivial odd prime power. Let $k,s \in \mathbb{N}$ such that $2^k$ is the $2$-part of $q-1$ and that $2^s$ is the $2$-part of $q+1$. Then:

(i) Assume that $q \equiv 1 \ \mathrm{mod} \ 4$. Then

\begin{equation*}
\left \lbrace
\begin{pmatrix}
\lambda & \ \ \\
\ \ & \ \mu \\
\end{pmatrix}
\ : \ \lambda, \mu \ \textnormal{are $2$-elements of } \mathbb{F}_{q}^{*} \right \rbrace
\cdot
\left \langle
\begin{pmatrix}
0 \ & \ 1 \\
1 \ & \ 0 \\
\end{pmatrix} \right \rangle
\end{equation*}

is a Sylow $2$-subgroup of $GL_2(q)$. In particular, the Sylow $2$-subgroups of $GL_2(q)$ are isomorphic to the wreath product $C_{2^k} \wr C_2$.

(ii) If $q \equiv 3 \ \mathrm{mod} \ 4$, then the Sylow $2$-subgroups of $GL_2(q)$ are semidihedral of order $2^{s+2}$.

(<cit.>) Let $q$ be a nontrivial odd prime power. Let $k,s \in \mathbb{N}$ such that $2^k$ is the $2$-part of $q-1$ and that $2^s$ is the $2$-part of $q+1$. Then:

(i) If $q \equiv 1 \ \mathrm{mod} \ 4$, then the Sylow $2$-subgroups of $GU_2(q)$ are semidihedral of order $2^{k+2}$.

(ii) If $q \equiv 3 \ \mathrm{mod} \ 4$, then the Sylow $2$-subgroups of $GU_2(q)$ are isomorphic to the wreath product $C_{2^s} \wr C_2$. If $\varepsilon \in \mathbb{F}_{q^2}^{*}$ has order $2^s$, then a Sylow $2$-subgroup of $GU_2(q)$ is concretely given by

\begin{equation*}
W := \left \lbrace
\begin{pmatrix}
\lambda \ & \ \ \\
\ \ & \ \mu \\
\end{pmatrix}
\ : \ \lambda, \mu \in \langle \varepsilon \rangle \right \rbrace
\cdot
\left \langle
\begin{pmatrix}
0 \ & \ 1 \\
1 \ & \ 0 \\
\end{pmatrix} \right \rangle.
\end{equation*}

(<cit.>) If $q$ is a nontrivial odd prime power, then a Sylow $2$-subgroup of $SL_2(q)$ is generalized quaternion of order $(q^2-1)_2$.

(<cit.>) If $q$ is a nontrivial odd prime power, then $PSL_2(q)$ has dihedral Sylow $2$-subgroups of order $\frac{1}{2}(q^2-1)_2$.

(<cit.>) Let $q$ be a nontrivial odd prime power and let $\varepsilon \in \lbrace +,- \rbrace$. Let $r$ be a positive integer. Let $W_r$ be a Sylow $2$-subgroup of $GL_{2^r}^{\varepsilon}(q)$. Then $W_r \wr C_2$ is isomorphic to a Sylow $2$-subgroup of $GL_{2^{r+1}}^{\varepsilon}(q)$. A Sylow $2$-subgroup of $GL_{2^{r+1}}^{\varepsilon}(q)$ is concretely given by

\begin{equation*}
\left \lbrace
\begin{pmatrix}
A & \ \\
\ & B \\
\end{pmatrix}
\ : \ A, B \in W_r \right \rbrace
\cdot
\left \langle
\begin{pmatrix}
\ & I_{2^r} \\
I_{2^r} & \ \\
\end{pmatrix} \right \rangle.
\end{equation*}

(<cit.>) Let $q$ be a nontrivial odd prime power and let $n$ be a positive integer. Let $\varepsilon \in \lbrace +,- \rbrace$. Let $0 \le r_1 < \dots < r_t$ such that $n = 2^{r_1} + \dots + 2^{r_t}$. Let $W_i \in \mathrm{Syl}_2(GL_{2^{r_i}}^{\varepsilon}(q))$ for all $1 \le i \le t$. Then $W_1 \times \dots \times W_t$ is isomorphic to a Sylow $2$-subgroup of $GL_n^{\varepsilon}(q)$. A Sylow $2$-subgroup of $GL_n^{\varepsilon}(q)$ is concretely given by

\begin{equation*}
\left \lbrace
\begin{pmatrix}
A_1 & \ & \ \\
\ & \ddots & \ \\
\ & \ & A_t
\end{pmatrix}
\ : \ A_i \in W_i \right \rbrace.
\end{equation*}

Let $q$ be a prime power with $q \equiv 3 \mod 4$. Let $W$ be a Sylow $2$-subgroup of $GL_2(q)$, and let $m \in \mathbb{N}$ such that $|W| = 2^m$. Then:

(i) $W$ is semidihedral. In particular, there are elements $a, b \in W$ with $\mathrm{ord}(a)=2^{m-1}$ and $\mathrm{ord}(b)=2$ such that $a^b = a^{2^{m-2}-1}$.

(ii) We have $W \cap SL_2(q) = \langle a^2 \rangle \langle ab \rangle$.

(iii) Let $1 \le \ell \le 2^{m-1}$. If $\ell$ is odd, then $a^{\ell}$ has determinant $-1$, and $a^{\ell}b$ has determinant $1$. If $\ell$ is even, then $a^{\ell}$ has determinant $1$, and $a^{\ell}b$ has determinant $-1$.

(iv) The involutions of $W$ are precisely the elements $a^{2^{m-2}}$ and $a^{\ell}b$, where $2 \le \ell \le 2^{m-1}$ is even.

By Lemma <ref> (ii), we have (i).

Let $W_0 := W \cap SL_2(q)$. By Lemma <ref>, $W_0$ is generalized quaternion. Also, $W_0$ is a maximal subgroup of $W$ since $SL_2(q)$ has index $q-1$ in $GL_2(q)$ and $q \equiv 3 \mod 4$. By <cit.>, we have $\Phi(W) = \langle a^2 \rangle$. So the maximal subgroups of $W$ are precisely the groups $M_1 := \langle a \rangle$, $M_2 :=\langle a^2 \rangle \langle b \rangle$ and $M_3 := \langle a^2 \rangle \langle ab \rangle$. One can check that $M_1 \cong C_{2^{n-1}}$, $M_2 \cong D_{2^{n-1}}$ and $M_3 \cong Q_{2^{n-1}}$. Consequently, $W_0 = \langle a^2 \rangle \langle ab \rangle$, and (ii) holds.

(iii) follows from (ii) since any element of $W \setminus W_0$ has determinant $-1$.

The proof of (iv) is an easy exercise.

Let $q$ be a nontrivial odd prime power, $n$ a positive integer and $\varepsilon \in \lbrace +,- \rbrace$. Let $0 \le r_1 < \dots < r_t$ such that $n = 2^{r_1} + \dots + 2^{r_t}$. Then there is a Sylow $2$-subgroup $W$ of $G := GL_n^{\varepsilon}(q)$ containing all diagonal matrices in $G$ with $2$-power order such that $C_W(W \cap SL_n^{\varepsilon}(q))$ consists precisely of the matrices

\begin{equation*}
\begin{pmatrix}
\lambda_1 I_{2^{r_1}} \ & \ & \ \\
\ & \ddots \ & \ \\
\ & \ & \lambda_t I_{2^{r_t}}
\end{pmatrix},
\end{equation*}

where $\lambda_1, \dots, \lambda_t$ are $2$-elements of $\mathbb{F}_q^{*}$ if $G = GL_n(q)$ and $2$-elements of $\mathbb{F}_{q^2}^{*}$ with $\lambda_i^{q+1}=1$ (for each $1 \le i \le t$) if $G=GU_n(q)$.

Using Lemmas <ref> and <ref>, one can check that the centralizer of a Sylow 2-subgroup of $SL_2^{\varepsilon}(q)$ inside a Sylow 2-subgroup of $GL_2^{\varepsilon}(q)$ is the Sylow 2-subgroup of $Z(GL_2^{\varepsilon}(q))$. Applying Lemma <ref> and arguing by induction, one can see that a similar statement holds for the centralizer of a Sylow 2-subgroup of $SL_{2^r}^{\varepsilon}(q)$ inside a Sylow 2-subgroup of $GL_{2^r}^{\varepsilon}(q)$ for all $r \ge 0$. Now we may apply Lemma <ref> to obtain a Sylow 2-subgroup of $G$ with the desired properties.

Let $q$ be a nontrivial odd prime power, $n$ a positive integer and $\varepsilon \in \lbrace +,- \rbrace$. Let $G := SL_n^{\varepsilon}(q)$, and let $S$ be a Sylow $2$-subgroup of $G$. Then we have $Z(\mathcal{F}_S(G)) = S \cap Z(G)$.

Let $0 \le r_1 < \dots < r_t$ such that $n = 2^{r_1} + \dots + 2^{r_t}$. By Lemma <ref>, we may assume that $Z(S)$ consists precisely of the matrices

\begin{equation*}
\begin{pmatrix}
\lambda_1 I_{2^{r_1}} \ & \ & \ \\
\ & \ddots \ & \ \\
\ & \ & \lambda_t I_{2^{r_t}}
\end{pmatrix},
\end{equation*}

where $\lambda_1, \dots, \lambda_t$ are 2-elements of $\mathbb{F}_q^{*}$ with $\lambda_1^{2^{r_1}}\cdots \lambda_t^{2^{r_t}} = 1$ if $G = SL_n(q)$ and 2-elements of $\mathbb{F}_{q^2}^{*}$ with $\lambda_i^{q+1}=1$ (for each $1 \le i \le t$) and $\lambda_1^{2^{r_1}}\cdots \lambda_t^{2^{r_t}} = 1$ if $G=SU_n(q)$. Moreover, by Lemma <ref>, we may assume that $S$ contains each diagonal matrix in $G$ of $2$-power order.

Let $x$ be an element of $Z(S)$ with diagonal blocks $\lambda_1 I_{2^{r_1}}, \dots, \lambda_t I_{2^{r_t}}$. One can easily see that $x$ is $G$-conjugate to any diagonal matrix in $G$ that is obtained from $x$ by permuting its diagonal entries. It follows that, if $\lambda_i \ne \lambda_j$ for some $1 \le i \ne j \le t$, then $x \not\in Z(\mathcal{F}_S(G))$. This implies $Z(\mathcal{F}_S(G)) = S \cap Z(G)$.

Let $n$ be a positive integer. Let $q, q^{*}$ be nontrivial odd prime powers, and let $\varepsilon, \varepsilon^{*} \in \lbrace +,-\rbrace$. If $\varepsilon q \sim \varepsilon^{*}q^{*}$, then the $2$-fusion systems of $SL_n^{\varepsilon}(q)$ and $SL_n^{\varepsilon^{*}}(q^{*})$ are isomorphic.

Assume that $\varepsilon \ne \varepsilon^{*}$. From $\varepsilon q \sim \varepsilon^{*}q^{*}$, it is easy to deduce that $\varepsilon q \equiv \varepsilon^{*} q^{*} \mod 8$ and $(q^2-1)_2 = ((q^{*})^2-1)_2$. So, in view of the remarks at the bottom of p. 11 of [15], we may apply <cit.> to conclude that the 2-fusion system of $SL_n^{\varepsilon}(q)$ is isomorphic to the 2-fusion system of $SL_n^{\varepsilon^{*}}(q^{*})$.

Assume now that $\varepsilon = \varepsilon^{*}$. Using Dirichlet's theorem <cit.>, one can easily see that there is an odd prime $q_0$ with $\varepsilon q \sim \varepsilon q^{*} \sim -\varepsilon q_0$. By the preceding paragraph, both the 2-fusion system of $SL_n^{\varepsilon}(q)$ and the 2-fusion system of $SL_n^{\varepsilon}(q^{*})$ are isomorphic to the 2-fusion system of $SL_n^{-\varepsilon}(q_0)$. Consequently, the 2-fusion systems of $SL_n^{\varepsilon}(q)$ and $SL_n^{\varepsilon^{*}}(q^{*})$ are isomorphic.

Let $n$ be a positive integer. Let $q, q^{*}$ be nontrivial odd prime powers, and let $\varepsilon, \varepsilon^{*} \in \lbrace +,-\rbrace$. If $\varepsilon q \sim \varepsilon^{*} q^{*}$, then the $2$-fusion systems of $PSL_n^{\varepsilon}(q)$ and $PSL_n^{\varepsilon^{*}}(q^{*})$ are isomorphic.

Let $S$ and $S^{*}$ be Sylow $2$-subgroups of $G:= SL_n^{\varepsilon}(q)$ and $G^{*} := SL_n^{\varepsilon^{*}}(q^{*})$, respectively. By Proposition <ref>, $\mathcal{F} := \mathcal{F}_S(G)$ and $\mathcal{F}^{*} := \mathcal{F}_{S^{*}}(G^{*})$ are isomorphic. Therefore, $\mathcal{F}/Z(\mathcal{F})$ and $\mathcal{F}^{*}/Z(\mathcal{F}^{*})$ are isomorphic. Lemma <ref> implies that $\mathcal{F}/(S \cap Z(G))$ and $\mathcal{F}^{*}/(S^{*} \cap Z(G^{*}))$ are isomorphic. Now the proposition follows from Lemma <ref>.

The following lemma shows together with <cit.> that the $2$-fusion system of $PSL_n(q)$ is simple whenever $q$ is odd and $n \ge 3$.

Let $q$ be a nontrivial odd prime power and $n \ge 2$ a natural number such that $(n,q) \ne (2,3)$. Moreover, let $\varepsilon$ be an element of $\lbrace +,- \rbrace$. Then $PSL_n^{\varepsilon}(q)$ is a Goldschmidt group if and only if $n = 2$ and $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$.

Set $G := PSL_n^{\varepsilon}(q)$.

Assume that $n = 2$. Then $G \cong PSL_2(q)$. By Lemma <ref>, $G$ has dihedral Sylow $2$-subgroups of order $\frac{1}{2}(q-1)_2(q+1)_2$. So, if $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$, then $G$ has abelian Sylow $2$-subgroups and is thus a Goldschmidt group. If $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$, then the Sylow $2$-subgroups of $G$ are dihedral of order at least $8$ and hence nonabelian. Moreover, if $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$, then <cit.> shows that $G$ is not isomorphic to a finite simple group of Lie type in characteristic $2$ of Lie rank $1$. So $G$ is not a Goldschmidt group if $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$.

Assume now that $n \ge 3$. Again, we see from <cit.> that there is no finite simple group of Lie type in characteristic $2$ of Lie rank $1$ which is isomorphic to $G$. Also, $G$ has a subgroup isomorphic to $SL_2^{\varepsilon}(q) \cong SL_2(q)$, and therefore, the Sylow $2$-subgroups of $G$ are nonabelian. Consequently, $G$ is not a Goldschmidt group.

Let $n$ be a positive integer, $q$ a nontrivial odd prime power and $\varepsilon \in \lbrace +,- \rbrace$. Let $E$ be the subgroup of $SL_n^{\varepsilon}(q)$ consisting of the diagonal matrices in $SL_n^{\varepsilon}(q)$ with diagonal entries in $\lbrace 1,-1 \rbrace$. Then $\vert E \vert = 2^{n-1}$. Moreover, any elementary abelian $2$-subgroup of $SL_n^{\varepsilon}(q)$ is conjugate to a subgroup of $E$.

It is straightforward to check that $\vert E \vert = 2^{n-1}$.

Let $E_0$ be an elementary abelian $2$-subgroup of $SL_n^{\varepsilon}(q)$. We show that $E_0$ is conjugate to a subgroup of $E$. Using Dirichlet's theorem <cit.>, one can see that there is an odd prime number $q^{*}$ with $-q \sim q^{*}$, and Proposition <ref> shows that the $2$-fusion systems of $SU_n(q)$ and $SL_n(q^{*})$ are isomorphic. Therefore, it is enough to consider the case $\varepsilon = +$.

Since $E_0$ is an elementary abelian $2$-group, any two elements of $E_0$ commute, and any element of $E_0$ is diagonalizable (see Lemma <ref>). It follows that $E_0$ is simultaneously diagonalizable, and this implies that $E_0$ is conjugate to a subgroup of $E$.

Let $q$ be a nontrivial odd prime power, $n \ge 3$ a natural number and $S$ a Sylow $2$-subgroup of $PSL_n(q)$. Then $\mathrm{Aut}_{PSL_n(q)}(S) = \mathrm{Inn}(S)$.

Let $R \in \mathrm{Syl}_2(SL_n(q))$ such that $S$ is the image of $R$ in $PSL_n(q)$. Let $T$ be a Sylow $2$-subgroup of $GL_n(q)$ with $R \le T$. By <cit.>, we have $N_{GL_n(q)}(R) = T C_{GL_n(q)}(T)$. So we have that $\mathrm{Aut}_{SL_n(q)}(R)$ is a $2$-group. Since the image of $N_{SL_n(q)}(R)$ in $PSL_n(q)$ equals $N_{PSL_n(q)}(S)$ (see <cit.>), it follows that $\mathrm{Aut}_{PSL_n(q)}(S)$ is a $2$-group, and this implies $\mathrm{Aut}_{PSL_n(q)}(S) = \mathrm{Inn}(S)$.

§.§ $k$-connectivity

In this subsection, we prove some connectivity properties of the Sylow $2$-subgroups of $SL_n(q)$ and $PSL_n(q)$, where $q$ is a nontrivial odd prime power and $n \ge 6$. We will work with the following definition (see <cit.>):

Let $S$ be a finite $2$-group, and let $k$ be a positive integer. If $A$ and $B$ are elementary abelian subgroups of $S$ of rank at least $k$, then $A$ and $B$ are said to be $k$-connected if there is a sequence

\begin{equation*}
A = A_1, A_2, \dots, A_n = B \ \ \ \ (n \ge 1)
\end{equation*}

of elementary abelian subgroups $A_i$, $1 \le i \le n$, of $S$ with rank at least $k$ such that

\begin{equation*}
A_i \subseteq A_{i+1} \ \textnormal{or} \ A_{i+1} \subseteq A_i
\end{equation*}

for all $1 \le i \le n-1$. The group $S$ is said to be $k$-connected if any two elementary abelian subgroups of $S$ of rank at least $k$ are $k$-connected.

Let $S$ be a finite $2$-group, and let $k$ be a positive integer. If $S$ has a normal elementary abelian subgroup of rank at least $2^{k-1}+1$, then $S$ is $k$-connected.

Let $q$ be a nontrivial odd prime power with $q \equiv 1 \ \mathrm{mod} \ 4$, and let $n \ge 6$ be a natural number. Then the Sylow $2$-subgroups of $PSL_n(q)$ and those of $SL_n(q)$ are $3$-connected.

Let $W_0$ be the unique Sylow $2$-subgroup of $GL_1(q)$, and let $W_1$ be the Sylow $2$-subgroup of $GL_2(q)$ given in Lemma <ref> (i). For each $r \ge 2$, let $W_r$ be the Sylow $2$-subgroup of $GL_{2^r}(q)$ obtained from $W_{r-1}$ by the construction given in the last statement of Lemma <ref>. Let $0 \le r_1 < \dots < r_t$ such that $n = 2^{r_1} + \dots + 2^{r_t}$, and let $W$ be the Sylow $2$-subgroup of $GL_n(q)$ obtained from $W_{r_1}, \dots, W_{r_t}$ by using the last statement of Lemma <ref>.

Let $R$ denote the subgroup of $GL_n(q)$ consisting of all diagonal matrices $D \in GL_n(q)$, where $D^2 \in Z(GL_n(q))$ and any diagonal element of $D$ is a $2$-element of $\mathbb{F}_q^{*}$. It is easy to note that $R \trianglelefteq W$.

Set $R_0 := R \cap SL_n(q)$. Then $\Omega_1(R_0)$, the subgroup of $R_0$ generated by all involutions of $R_0$, is elementary abelian of order $2^{n-1} \ge 2^5$, and $\Omega_1(R_0) \trianglelefteq W \cap SL_n(q)$. Also, $R_0 Z(SL_n(q))/Z(SL_n(q))$ is a normal elementary abelian subgroup of $(W \cap SL_n(q))Z(SL_n(q))/Z(SL_n(q))$, and one can easily check that the order of $R_0 Z(SL_n(q))/Z(SL_n(q))$ is at least $2^5$. Lemma <ref> implies that $W \cap SL_n(q)$ and its image in $PSL_n(q)$ are $3$-connected.

Lemma <ref> and the proof of Lemma <ref> show that we also have the following:

Let $q$ be a nontrivial odd prime power with $q \equiv 1 \ \mathrm{mod} \ 4$, and let $n \ge 6$ be a natural number. Then the Sylow $2$-subgroups of $PSL_n(q)$ and those of $SL_n(q)$ are $2$-connected.

We now study the case $q \equiv 3 \mod 4$.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \ \mathrm{mod} \ 4$, and let $n \ge 6$ be a natural number. Then the Sylow $2$-subgroups of $PSL_n(q)$ and those of $SL_n(q)$ are $2$-connected. If $n \ge 10$, then we even have that the Sylow $2$-subgroups of $PSL_n(q)$ and those of $SL_n(q)$ are $3$-connected.

Let $W_0$ denote the unique Sylow $2$-subgroup of $GL_1(q)$, and let $W_1$ be a Sylow $2$-subgroup of $GL_2(q)$. By Lemma <ref> (ii), $W_1$ is semidihedral. Let $m \in \mathbb{N}$ with $|W_1| = 2^m$. Also, let $h,a \in W_1$ such that $\mathrm{ord}(h) = 2^{m-1}$, $\mathrm{ord}(a) = 2$ and $h^a = h^{2^{m-2}-1}$. Set $z := -I_2 = h^{2^{m-2}}$. For each $r \ge 2$, let $W_r$ be the Sylow $2$-subgroup of $GL_{2^r}(q)$ obtained from $W_{r-1}$ by the construction given in the last statement of Lemma <ref>. Let $0 \le r_1 < \dots < r_t$ such that $n = 2^{r_1} + \dots + 2^{r_t}$, and let $W$ be the Sylow $2$-subgroup of $GL_n(q)$ obtained from $W_{r_1}, \dots, W_{r_t}$ by using the last statement of Lemma <ref>.

Given a natural number $\ell \ge 1$ and elements $x_1,\dots,x_{\ell} \in GL_2(q)$, we write $\mathrm{diag}(x_1,\dots,x_{\ell})$ for the block diagonal matrix

\begin{equation*}
\begin{pmatrix} x_1 & & \\ & \ddots & \\ & & x_{\ell} \end{pmatrix}.
\end{equation*}

For each natural number $r \ge 1$, let $A_r$ denote the subgroup of $GL_{2^r}(q)$ consisting of the matrices $\mathrm{diag}(x_1, \dots, x_{2^{r-1}})$, where either $x_i \in \langle z \rangle$ for all $1 \le i \le 2^{r-1}$ or $x_i$ is an element of $\langle h \rangle$ with order $4$ for all $1 \le i \le 2^{r-1}$. By induction over $r$, one can see that $A_r \trianglelefteq W_r$ for all $r \ge 1$. Also, let $\widetilde{A_r} := \Omega_1(A_r)$ for all $r \ge 1$. Clearly, $\widetilde{A_r} \trianglelefteq W_r$ for all $r \ge 1$.

We now consider two cases.

Case 1: $n$ is even.

Let $E$ be the subgroup of $GL_n(q)$ consisting of the matrices $\mathrm{diag}(x_1,\dots,x_{\frac{n}{2}})$, where either $x_i \in \langle z \rangle$ for all $1 \le i \le \frac{n}{2}$ or $x_i$ is an element of $\langle h \rangle$ with order $4$ for all $1 \le i \le \frac{n}{2}$. Let $\widetilde{E} := \Omega_1(E)$. Since $A_{r_i} \trianglelefteq W_{r_i}$ for all $1 \le i \le t$, we have that $E$ and $\widetilde E$ are normal subgroups of $W$. Lemma <ref> (iii) shows that $E \le W \cap SL_n(q)$.

As $\widetilde E$ is elementary abelian of order $2^{\frac{n}{2}}$, Lemma <ref> implies that $W \cap SL_n(q)$ is $2$-connected, and even $3$-connected if $n \ge 10$. Since $EZ(SL_n(q))/Z(SL_n(q))$ is a normal elementary abelian subgroup of $(W \cap SL_n(q))Z(SL_n(q))/Z(SL_n(q))$ with order $2^{\frac{n}{2}}$, Lemma <ref>  also shows that a Sylow $2$-subgroup is $2$-connected, and even $3$-connected if $n \ge 10$.

Case 2: $n$ is odd.

Now let $E$ denote the subgroup of $GL_n(q)$ consisting of the matrices

\begin{equation*}
\left( \begin{array}{c|cc} 1 & \begin{matrix} & &  \end{matrix} \\ \hline \begin{matrix} & \\ & \\ & \end{matrix} & \begin{matrix} x_1 & & \\ & \ddots & \\ & & x_{\frac{n-1}{2}} \end{matrix} \end{array} \right),
\end{equation*}

where $x_i \in \langle z \rangle$ for all $1 \le i \le \frac{n-1}{2}$. Since $\widetilde{A_{r_i}} \trianglelefteq W_{r_i}$ for all $2 \le i \le t$, we have that $E$ is a normal subgroup of $W \cap SL_n(q)$. Moreover, $E$ is elementary abelian of order $2^{\frac{n-1}{2}}$. Lemma <ref> implies that $W \cap SL_n(q)$ is $2$-connected, and even $3$-connected if $n \ge 11$. There is nothing else to show since the Sylow $2$-subgroups of $PSL_n(q)$ are isomorphic to those of $SL_n(q)$ (as $n$ is odd).

We show next that the groups $SL_n(q)$, where $6 \le n \le 9$ and $q \equiv 3 \mod 4$, and the groups $PSL_n(q)$, where $7 \le n \le 9$ and $q \equiv 3 \mod 4$, also have $3$-connected Sylow $2$-subgroups.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \mod 4$. Then the Sylow $2$-subgroups of $SL_6(q)$ and those of $SL_7(q)$ are $3$-connected.

Let $W_1$ be a Sylow $2$-subgroup of $GL_2(q)$, let $W_2$ be the Sylow $2$-subgroup of $GL_4(q)$ obtained from $W_1$ by the construction given in the last statement of Lemma <ref>, and let $W$ be the Sylow $2$-subgroup of $GL_6(q)$ obtained from $W_1$ and $W_2$ by using the last statement of Lemma <ref>.

From Lemma <ref>, we see that the Sylow $2$-subgroups of $SL_7(q)$ are isomorphic to those of $GL_6(q)$. So it is enough to show that $W$ and $W \cap SL_6(q)$ are $3$-connected. Given elements $x_1,x_2,x_3 \in GL_2(q)$, we write $\mathrm{diag}(x_1,x_2,x_3)$ for the block diagonal matrix

\begin{equation*}
\begin{pmatrix} x_1 & & \\ & x_2 & \\ & & x_3 \end{pmatrix}.
\end{equation*}

Let $A$ be the subgroup of $W \cap SL_6(q)$ consisting of the matrices $\mathrm{diag}(x_1,x_2,x_3)$, where $x_i \in \langle -I_2 \rangle$ for $1 \le i \le 3$. Clearly, $A \cong E_8$. We prove the following:

(1) If $E$ is an elementary abelian subgroup of $W$ of rank at least $3$, then $E$ is $3$-connected to an elementary abelian subgroup of $W \cap SL_6(q)$ of rank at least $3$.

(2) If $E$ is an elementary abelian subgroup of $W \cap SL_6(q)$ of rank at least $3$, then $E$ is $3$-connected to $A$ in $W \cap SL_6(q)$.

By (1) and (2), any elementary abelian subgroup of $W$ of rank at least $3$ is $3$-connected to $A$, and so $W$ is $3$-connected. Similarly, (2) implies that $W \cap SL_6(q)$ is $3$-connected.

Let $Z := \langle \mathrm{diag}(-I_2, I_2, I_2), \mathrm{diag}(I_2,-I_2,-I_2) \rangle$. Since $Z \le Z(W)$, we have that any elementary abelian subgroup of $W$ of rank at least $3$ is $3$-connected to an $E_8$-subgroup of $W$ containing $Z$. Also, any elementary abelian subgroup of $W \cap SL_6(q)$ of rank at least $3$ is $3$-connected (in $W \cap SL_6(q)$) to an $E_8$-subgroup of $W \cap SL_6(q)$ containing $Z$. Therefore, we only need to consider $E_8$-subgroups containing $Z$ in order to prove (1) and (2).

So let $E$ be an $E_8$-subgroup of $W$ with $Z \le E$, and let $s \in E \setminus Z$. Suppose that $s = \mathrm{diag}(s_1,s_2,s_3)$, where $s_1,s_2,s_3 \in W_1$. Then $[E,A] = 1$, and it is easy to deduce that $E$ is $3$-connected to $A$, so that $E$ satisfies (1). Also, if $E \le W \cap SL_6(q)$, it is easy to deduce that $E$ satisfies (2).

Suppose now that

\begin{equation*}
s = \begin{pmatrix}
s_1 & & \\
& & s_2 \\
& s_3 &
\end{pmatrix}
\end{equation*}

for some $s_1, s_2, s_3 \in W_1$. Since $s^2 = I_6$, we have $s_2 = s_3^{-1}$. Let $a$ be an involution of $W_1$ with $a \ne -I_2$. Set $s^{*} := \mathrm{diag}(I_2,a,a^{s_2})$ and $E^{*} := \langle Z, s^{*} \rangle \cong E_8$. Clearly, $E^{*} \le W \cap SL_6(q)$. It is easy to check that $[E,E^{*}] = 1$, which implies that $E$ is $3$-connected to $E^{*}$. So $E$ satisfies (1). If $E \le W \cap SL_6(q)$, then $E$ is $3$-connected to $E^{*}$ in $W \cap SL_6(q)$, and $E^{*}$ is $3$-connected to $A$ in $W \cap SL_6(q)$ since $[E^{*},A] = 1$. Therefore, $E$ satisfies (2) when $E \le W \cap SL_6(q)$.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \mod 4$. A Sylow $2$-subgroup of $PSL_7(q)$ is isomorphic to a Sylow $2$-subgroup of $SL_7(q)$. So, by Lemma <ref>, the Sylow $2$-subgroups of $PSL_7(q)$ are $3$-connected.

We need the following lemma in order to prove that the Sylow $2$-subgroups of $SL_n(q)$ and $PSL_n(q)$ are $3$-connected when $n \in \lbrace 8,9 \rbrace$.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \ \mathrm{mod} \ 4$, and let $V$ be a Sylow $2$-subgroup of $GL_4(q)$. Let $u \in V$ with $u^2 = I_4$ or $u^2 = -I_4$. Then there is an involution $v \in V \setminus \langle u, -I_4 \rangle$ which commutes with $u$.

Fix a Sylow $2$-subgroup $W_1$ of $GL_2(q)$, and let $W_2$ be the Sylow $2$-subgroup of $GL_4(q)$ obtained from $W_1$ by the construction given in the last statement of Lemma <ref>. By Sylow’s Theorem, we may assume that $V = W_2$. Let $a$ be an involution of $W_1$ with $a \ne -I_2$.

First, we consider the case that

\begin{equation*}
u = \begin{pmatrix} x & \\ & y \end{pmatrix}
\end{equation*}

with elements $x,y \in W_1$. If $x \not\in \langle -I_2 \rangle$ or $y \not\in \langle -I_2 \rangle$, then

\begin{equation*}
\begin{pmatrix} -I_2 & \\ & I_2 \end{pmatrix} \in W_2
\end{equation*}

is an involution commuting with $u$ and not lying in $\langle u, -I_4 \rangle$. If $x,y \in \langle -I_2 \rangle$, then we may choose

\begin{equation*}
v := \begin{pmatrix} a & \\ & a \end{pmatrix}.
\end{equation*}

Assume now that

\begin{equation*}
u = \begin{pmatrix} & x \\ y & \end{pmatrix}
\end{equation*}

with elements $x,y \in W_1$. Let

\begin{equation*}
v := \begin{pmatrix}a & \\ & a^x \end{pmatrix}.
\end{equation*}

As $a$ is an involution of $W_1$, we have that $v$ is an involution of $W_2$. By a direct calculation (using that $xy \in \langle -I_2 \rangle$), $v$ has the desired properties.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \ \mathrm{mod} \ 4$. Then the Sylow $2$-subgroups of $SL_8(q)$ and those of $SL_9(q)$ are $3$-connected.

Fix a Sylow $2$-subgroup $W_1$ of $GL_2(q)$, let $W_2$ be the Sylow $2$-subgroup of $GL_4(q)$ obtained from $W_1$ by the construction given in the last statement of Lemma <ref>, and let $W$ be the Sylow $2$-subgroup of $GL_8(q)$ obtained from $W_2$ by the construction given in the last statement of Lemma <ref>. Set $S := W \cap SL_8(q)$.

From Lemma <ref>, we see that the Sylow $2$-subgroups of $SL_9(q)$ are isomorphic to those of $GL_8(q)$. So it is enough to show that $W$ and $S$ are $3$-connected.

Given a natural number $\ell \ge 1$ and $x_1, \dots, x_{\ell}$ of $GL_2(q) \cup GL_4(q)$, we write $\mathrm{diag}(x_1,\dots,x_{\ell})$ for the block diagonal matrix

\begin{equation*}
\begin{pmatrix}
x_1 & & \\
& \ddots & \\
& & x_{\ell}
\end{pmatrix}.
\end{equation*}

\begin{equation*}
A := \left \lbrace \mathrm{diag}(x_1,x_2,x_3,x_4) \ \vert \ x_i \in \langle -I_2 \rangle \ \forall \ 1 \le i \le 4 \right \rbrace \le S
\end{equation*}

\begin{equation*}
Z := \langle - I_8 \rangle \le S.
\end{equation*}

Clearly, $A \cong E_{16}$. Since $Z \le Z(W)$, we have that any elementary abelian subgroup of $W$ of rank at least $3$ is $3$-connected to an $E_8$-subgroup of $W$ containing $Z$. Similarly, any elementary abelian subgroup of $S$ of rank at least $3$ is $3$-connected to an $E_8$-subgroup of $S$ containing $Z$. So it suffices to prove that any $E_8$-subgroup $E$ of $W$ with $Z \le E$ is $3$-connected to $A$, where $E$ is even $3$-connected in $S$ to $A$ if $E \le S$. Thus let $E$ be an $E_8$-subgroup of $W$ containing $Z$, and let $x,y \in E$ with $E = \langle Z, x, y \rangle$.

We consider a number of cases. Below, $a$ will always denote an involution of $W_1$ with $a \ne -I_2$.

Case 1: $x = \mathrm{diag}(-I_4,I_4)$ and $y = \mathrm{diag}(b_1,b_2)$ for some $b_1,b_2 \in W_2$.

We determine an involution $y_1 \in C_W(E) \setminus \langle Z, x \rangle$ such that $\langle Z, x, y_1 \rangle \cong E_8$ is $3$-connected to $A$. In the case that $E \le S$, we determine $y_1$ such that $y_1 \in S$ and such that $\langle Z, x, y_1 \rangle$ is $3$-connected to $A$ in $S$. The existence of such an involution $y_1$ easily implies that $E$ is $3$-connected to $A$, and even $3$-connected to $A$ in $S$ if $E \le S$. The involution $y_1$ is given by the following table in dependence of $y$. In each row, $r_1, r_2, r_3, r_4$ are assumed to be elements of $W_1$ such that $y$ is equal to the matrix given in the column “$y$” and such that the conditions in the column “Conditions” (if any) are satisfied. The column “$y_1$” gives the involution $y_1$ with the desired properties. For each row, one can verify the stated properties of $y_1$ by a direct calculation or by using the previous rows.

Case     $y$     Conditions     $y_1$

1.1     $\begin{pmatrix} r_1 & & & \\ & r_2 & & \\ & & r_3 & \\ & & & r_4 \end{pmatrix}$          $y$

1.2     $\begin{pmatrix} r_1 & & & \\ & r_2 & & \\ & & & r_3 \\ & & r_4 & \end{pmatrix}$     $\langle r_1, r_2 \rangle \not\le \langle -I_2 \rangle$     $\begin{pmatrix} r_1 & &  \\ & r_2 & \\ & & I_4 \end{pmatrix}$

1.3     $\begin{pmatrix} r_1 & & & \\ & r_2 & & \\ & & & r_3\\ & & r_4 &  \end{pmatrix}$     $r_1, r_2 \le \langle -I_2 \rangle$     $\begin{pmatrix} a & & \\ & a & \\ & & I_4 \end{pmatrix}$

1.4     $\begin{pmatrix} & r_1 & & \\ r_2 & & & \\ & & r_3 &\\ & & & r_4 \end{pmatrix}$     $\langle r_3, r_4 \rangle \not\le \langle -I_2 \rangle$     $\begin{pmatrix} I_4 & & \\ & r_3 & \\ & & r_4 \end{pmatrix}$

1.5     $\begin{pmatrix} & r_1 & & \\ r_2 & & & \\ & & r_3 &\\ & & & r_4 \end{pmatrix}$     $r_3, r_4 \le \langle -I_2 \rangle$     $\begin{pmatrix} I_4 & & \\ & a & \\ & & a \end{pmatrix}$

1.6     $\begin{pmatrix} & r_1 & & \\r_2 & & & \\ & & & r_3\\ & & r_4 & \end{pmatrix}$          $\begin{pmatrix} & r_1 & \\ r_2 & & \\ & & I_4 \end{pmatrix}$

Case 2: $x = \mathrm{diag}(a_1,a_2)$ and $y = \mathrm{diag}(b_1,b_2)$ for some $a_1, a_2, b_1, b_2 \in W_2$.

Set $x_1 := \mathrm{diag}(-I_4,I_4)$. Since $E = \langle Z, x, y \rangle \cong E_8$, the elements $x$ and $y$ cannot be both contained in $\langle Z, x_1 \rangle$. Without loss of generality, we may assume that $y \not\in \langle Z, x_1 \rangle$. Then $E_1 := \langle Z, x_1, y \rangle \cong E_8$. The group $E_1$ is $3$-connected to $A$ by Case 1, and it is $3$-connected to $E$ since $E$ and $E_1$ commute. Hence, $E$ is $3$-connected to $A$. Clearly, if $E \le S$, then $E$ is even $3$-connected in $S$ to $A$.

Case 3: There are $a_1, a_2, b_1, b_2 \in W_2$ with

\begin{equation*}
\lbrace x, y \rbrace = \left \lbrace \begin{pmatrix} a_1 & \\ & a_2 \end{pmatrix}, \begin{pmatrix} & b_1 \\ b_2 & \end{pmatrix} \right \rbrace.
\end{equation*}

Without loss of generality, we assume that

\begin{equation*}
x = \begin{pmatrix} a_1 & \\ & a_2 \end{pmatrix} \ \textnormal{and} \ y = \begin{pmatrix} & b_1 \\ b_2 & \end{pmatrix}.
\end{equation*}

Since $x$ and $y$ are commuting involutions, we have $b_1 = b_2^{-1}$ and $a_2 = {a_1}^{b_1}$. By Lemma <ref>, there is an involution $\widetilde{a_1} \in W_2 \setminus \langle a_1, -I_4 \rangle$ which commutes with $a_1$. Set

\begin{equation*}
y_1 := \begin{pmatrix} \widetilde{a_1} & \\ & {\widetilde{a_1}}^{b_1} \end{pmatrix}.
\end{equation*}

It is easy to see that $y_1 \in S$, and $y_1$ is an involution since $\widetilde{a_1}$ is an involution of $W_2$. We have $[x,y_1] = 1$ since $\widetilde{a_1}$ commutes with $a_1$ and $\widetilde{a_1}^{b_1}$ commutes with ${a_1}^{b_1} = a_2$. A direct calculation using that $b_1 = b_2^{-1}$ shows that we also have $[y,y_1] = 1$. Thus $E = \langle Z,x,y \rangle$ commutes with $E_1 := \langle Z, x, y_1 \rangle$. Since $\widetilde{a_1} \not\in \langle a_1, -I_4 \rangle$, we have $y_1 \not\in \langle Z,x \rangle$ and hence $E_1 \cong E_8$. Applying Case 2, it follows that $E$ is $3$-connected to $A$ (and even $3$-connected in $S$ to $A$ when $E \le S$).

Case 4: There are $a_1, a_2, b_1, b_2 \in W_2$ with

\begin{equation*}
x = \begin{pmatrix} & a_1 \\ a_2 & \end{pmatrix} \ \textnormal{\textit{and}} \ y = \begin{pmatrix} & b_1 \\ b_2 & \end{pmatrix}.
\end{equation*}

This case can be reduced to Case 3 since $E = \langle Z, x, y \rangle = \langle Z, x, xy \rangle$.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \mod 4$. A Sylow $2$-subgroup of $PSL_9(q)$ is isomorphic to a Sylow $2$-subgroup of $SL_9(q)$. So, by Lemma <ref>, the Sylow $2$-subgroups of $PSL_9(q)$ are $3$-connected.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \ \mathrm{mod} \ 4$. Then the Sylow $2$-subgroups of $PSL_8(q)$ are $3$-connected.

Let $W_1$ be a Sylow $2$-subgroup of $GL_2(q)$. Let $W_2$ be the Sylow $2$-subgroup of $GL_4(q)$ obtained from $W_1$ by the construction given in the last statement of Lemma <ref>, and let $W_3$ be the Sylow $2$-subgroup of $GL_8(q)$ obtained from $W_2$ by the construction given in the last statement of Lemma <ref>. Set $S := W_3 \cap SL_8(q)$. For each subgroup or element $X$ of $SL_8(q)$, let $\widebar X$ denote the image of $X$ in $PSL_8(q)$. We prove that $\widebar S$ is $3$-connected.

Given a natural number $\ell \ge 1$ and $x_1, \dots, x_{\ell}$ of $GL_2(q) \cup GL_4(q)$, we write $\mathrm{diag}(x_1,\dots,x_{\ell})$ for the block diagonal matrix

\begin{equation*}
\begin{pmatrix}
x_1 & & \\
& \ddots & \\
& & x_{\ell}
\end{pmatrix}.
\end{equation*}

\begin{equation*}
A := \left\lbrace \mathrm{diag}(x_1,x_2,x_3,x_4) \ \vert \ x_i \in \langle -I_2 \rangle \ \forall \ 1 \le i \le 4 \right\rbrace \le S.
\end{equation*}

We have $\widebar A \cong E_8$.

\begin{equation*}
Z := \left \langle \mathrm{diag}(-I_4,I_4) \right \rangle.
\end{equation*}

We have $\widebar{Z} \le Z(\widebar S)$. Using this, it is easy to note that any elementary abelian subgroup of $\widebar S$ of rank at least $3$ is $3$-connected to an $E_8$-subgroup of $\widebar S$ containing $\widebar Z$. Hence, it suffices to prove that any $E_8$-subgroup of $\widebar S$ containing $\widebar Z$ is $3$-connected to $\widebar A$.

Let $x, y \in S$ and $B := \langle \widebar Z, \widebar x, \widebar y \rangle$. Suppose that $B \cong E_8$. Considering a number of cases, we will prove that $B$ is $3$-connected to $\widebar A$. Below, $a$ will always denote an involution of $W_1$ with $a \ne -I_2$.

Case 1: $x = \mathrm{diag}(r_1,r_2,r_3,r_4) \ \textnormal{\textit{and}} \ y = \mathrm{diag}(m_1,m_2)$ for some $r_1,r_2,r_3,r_4 \in W_1$ and $m_1,m_2 \in W_2$.

We consider a number of subcases. These subcases are given by the rows of the table below. In each row, we assume that $s_1, s_2, s_3, s_4$ are elements of $W_1$ such that $y$ is equal to the matrix given in the column “$y$”. We also assume that the conditions in the column “Conditions” (if any) are satisfied. The column “$y_1$” gives an element of $S$ such that $\widebar{y_1}$ is an involution in $C_{\widebar S}(\widebar E) \setminus \langle \widebar Z, \widebar x \rangle$ and such that $\langle \widebar Z, \widebar x, \widebar{y_1} \rangle$ is $3$-connected to $\widebar A$. The existence of such an element $y_1$ easily implies that $B$ is $3$-connected to $\widebar A$.

Case     $y$     Conditions     $y_1$

1.1     $\begin{pmatrix} s_1 & & & \\ & s_2 & & \\ & & s_3 & \\ & & & s_4 \end{pmatrix}$          $y$

1.2     $\begin{pmatrix} & s_1& & \\ s_2& & & \\ & & s_3 & \\ & & & s_4\end{pmatrix}$     $x \not\in A$     $\begin{pmatrix} I_4 & &  \\ & -I_2 & \\ & & I_2 \end{pmatrix}$

1.3     $\begin{pmatrix} & s_1& & \\ s_2& & & \\ & & s_3 & \\ & & & s_4\end{pmatrix}$     $x \in A$     $\begin{pmatrix} a & &  \\ & a^{s_2^{-1}} & \\ & & I_4 \end{pmatrix}$

1.4     $\begin{pmatrix} & s_1 & & \\s_2 & & & \\ & & & s_3\\ & & s_4 & \end{pmatrix}$     $x \not\in A$     $\begin{pmatrix} I_2 & & & \\ & -I_2 & & \\ & & I_2 & \\ & & & -I_2\end{pmatrix}$

1.5     $\begin{pmatrix} & s_1 & & \\s_2 & & & \\ & & & s_3\\ & & s_4 & \end{pmatrix}$     $x \in A$     $\begin{pmatrix} a & & & \\ & a^{s_2^{-1}} & & \\ & & a & \\ & & & a^{s_4^{-1}} \end{pmatrix}$

The subcase that $y$ has the form

\begin{equation*}
\begin{pmatrix} s_1 &  & & \\ & s_2 & & \\ & & & s_3\\ & & s_4 & \end{pmatrix}
\end{equation*}

can be easily reduced to Cases 1.2 and 1.3.

Case 2: There are $r_1,r_2,r_3,r_4 \in W_1$ and $m_1, m_2 \in W_2$ with

\begin{equation*}
x = \begin{pmatrix} & r_1 & & \\
r_2 & & & \\
& & r_3 & \\
& & & r_4
\end{pmatrix}  \ \textnormal{\textit{and}} \ y = \begin{pmatrix} m_1 & \\ & m_2 \end{pmatrix}.
\end{equation*}

Case 2.1: There are $s_1,s_2,s_3,s_4 \in W_1$ with

\begin{equation*}
y = \begin{pmatrix} s_1 & & & \\
& s_2 & & \\
& & s_3 & \\
& & & s_4
\end{pmatrix} \ \textnormal{\textit{or}} \
y = \begin{pmatrix} & s_1 & & \\
s_2 & & & \\
& & s_3 & \\
& & & s_4
\end{pmatrix}.
\end{equation*}

Noticing that $\langle \widebar Z, \widebar x, \widebar y \rangle = \langle \widebar Z, \widebar x, \widebar x \widebar y \rangle$, this case can be reduced to Case 1.

Case 2.2: There are $s_1,s_2,s_3,s_4 \in W_1$ with

\begin{equation*}
y = \begin{pmatrix} s_1 & & & \\
& s_2 & & \\
& &  & s_3 \\
& & s_4 &
\end{pmatrix}.
\end{equation*}

Since $B \cong E_8$, we have $\varepsilon x^y = x$, where $\varepsilon \in \lbrace +,- \rbrace$. By a direct calculation, we have

\begin{equation*}
x^y = \begin{pmatrix} & s_1^{-1}r_1 s_2 & & \\
s_2^{-1} r_2 s_1 & & & \\
& & r_4^{s_4} & \\
& & & r_3^{s_3} \end{pmatrix}.
\end{equation*}

As $x = \varepsilon x^y$, we have $r_1 = \varepsilon s_1^{-1} r_1 s_2$, $r_2 = \varepsilon s_2^{-1}r_2s_1$, $r_3 = \varepsilon r_4^{s_4}$ and $r_4 = \varepsilon r_3^{s_3}$. Note that $\varepsilon s_1^{r_1} = s_2$ and $\varepsilon s_2^{r_2}=s_1$.

We now consider a number of subsubcases. These subsubcases are given by the rows of the table below. The columns “Condition 1” and “Condition 2” describe the subsubcase under consideration. The column “$y_1$” gives an element $y_1 \in S$ such that $\widebar{y_1}$ is an involution in $C_{\widebar S}(\widebar E) \setminus \langle \widebar Z, \widebar x \rangle$ and such that $\langle \widebar Z, \widebar x, \widebar{y_1} \rangle$ is $3$-connected to $\widebar A$. In each subsubcase, one can see from the above calculations and from the previous cases that $y_1$ indeed has the stated properties. The existence of such an element $y_1$ easily implies that $B$ is $3$-connected to $\widebar A$ in all subsubcases.

Case     Condition 1     Condition 2     $y_1$

2.2.1     $x^2 = I_8 = y^2$     $\langle r_3, r_4 \rangle \not\le \langle -I_2 \rangle$     $\begin{pmatrix} \varepsilon s_1 & & & \\ & s_2 & & \\ & & \varepsilon r_3 & \\ & & & r_4 \end{pmatrix}$

2.2.2     $x^2 = I_8 = y^2$     $\langle r_3, r_4 \rangle \le \langle -I_2 \rangle$     $\begin{pmatrix} & r_1 & & \\ r_2 & & & \\ & & \varepsilon a & \\ & & & a^{s_3} \end{pmatrix}$

2.2.3     $x^2 = -I_8 = y^2$         $\begin{pmatrix} \varepsilon s_1 & & & \\ & s_2 & & \\ & & \varepsilon r_3 & \\ & & & r_4 \end{pmatrix}$

2.2.4     $x^2 = I_8, y^2 = -I_8$     $\langle r_3, r_4 \rangle \not\le \langle -I_2 \rangle$     $\begin{pmatrix} I_4 & & \\ & \varepsilon r_3 & \\ & & r_4\end{pmatrix}$

2.2.5     $x^2 = I_8, y^2 = -I_8$     $\langle r_3, r_4 \rangle \le \langle -I_2 \rangle$     $\begin{pmatrix} I_4 & & \\ & \varepsilon a & \\ & & \varepsilon a^{s_3}\end{pmatrix}$

The case that $x^2 = -I_8$ and $y^2 = I_8$ can be easily reduced to Cases 2.2.4 and 2.2.5.

Case 2.3: There are $s_1,s_2,s_3,s_4 \in W_1$ with

\begin{equation*}
y = \begin{pmatrix}  & s_1 & & \\
s_2 & & & \\
& &  & s_3 \\
& & s_4 &
\end{pmatrix}.
\end{equation*}

Since $\langle \widebar Z, \widebar x, \widebar y \rangle = \langle \widebar Z, \widebar x, \widebar x \widebar y \rangle$, this case can be reduced to Case 2.2.

Case 3: There are $r_1,r_2,r_3,r_4 \in W_1$ and $m_1,m_2 \in W_2$ with

\begin{equation*}
x = \begin{pmatrix} r_1 & & & \\
& r_2 & & \\
& &  & r_3 \\
& & r_4 &
\end{pmatrix} \ \textnormal{\textit{and}} \
y = \begin{pmatrix} m_1 & \\ & m_2 \end{pmatrix}.
\end{equation*}

This case can be reduced to Case 2.

Case 4: There are $r_1,r_2,r_3,r_4 \in W_1$ and $m_1,m_2 \in W_2$ with

\begin{equation*}
x = \begin{pmatrix} & r_1 & & \\
r_2 & & & \\
& &  & r_3 \\
& & r_4 &
\end{pmatrix} \ \textnormal{\textit{and}} \
y = \begin{pmatrix} m_1 & \\ & m_2 \end{pmatrix}.
\end{equation*}

In view of Cases 1-3, we may assume that

\begin{equation*}
y = \begin{pmatrix} & s_1 & & \\
s_2 & & & \\
& &  & s_3 \\
& & s_4 &  \end{pmatrix}
\end{equation*}

for some $s_1,s_2,s_3,s_4 \in W_1$. Since $\langle \widebar Z, \widebar x, \widebar y \rangle = \langle \widebar Z, \widebar x, \widebar x \widebar y \rangle$, we can now reduce the given case to Case 1.

Case 5: There are $a_1,a_2,b_1,b_2 \in W_2$ with

\begin{equation*}
\lbrace x,y \rbrace =
\left \lbrace
\begin{pmatrix} a_1 & \\ & a_2 \end{pmatrix},
\begin{pmatrix} & b_1 \\ b_2 & \end{pmatrix}
\right\rbrace.
\end{equation*}

Without loss of generality, we assume that

\begin{equation*}
x = \begin{pmatrix} a_1 & \\ & a_2 \end{pmatrix} \ \textnormal{and} \ y = \begin{pmatrix} & b_1 \\ b_2 & \end{pmatrix}.
\end{equation*}

We have $x^2 \in \langle -I_8 \rangle$ since $B = \langle \widebar Z, \widebar x, \widebar y \rangle \cong E_8$, and hence ${a_1}^2 \in \langle -I_4 \rangle$. So, by Lemma <ref>, there is an involution $\widetilde{a_1} \in W_2 \setminus \langle a_1, -I_4 \rangle$ which commutes with $a_1$. Set

\begin{equation*}
y_1 := \begin{pmatrix} \widetilde{a_1} & \\ & \widetilde{a_1}^{b_1} \end{pmatrix}.
\end{equation*}

Clearly, $\widebar{y_1}$ is an involution of $\widebar S$. As $[x,y] \in \langle -I_8 \rangle$, we have ${a_1}^{b_1} \in \lbrace a_2, -a_2 \rbrace$. Since $a_1$ and $\widetilde{a_1}$ commute, it follows that $\widetilde{a_1}^{b_1}$ and $a_2$ commute. So we have $[x,y_1] = 1$ and hence $[\widebar{x},\widebar{y_1}] = 1$. Using that $y^2 \in \langle -I_8 \rangle$, one can easily verify that $[y,y_1] = 1$ and hence $[\widebar y, \widebar{y_1}] = 1$. As $\widetilde{a_1} \not\in \langle a_1, -I_4 \rangle$, we have $\widebar{y_1} \not\in \langle \widebar Z, \widebar x \rangle$.

Now $\langle \widebar Z, \widebar x, \widebar{y_1} \rangle$ is an $E_8$-subgroup of $\widebar S$ which commutes with $B$ and which is $3$-connected to $\widebar A$ by Cases 1-4. Thus $B$ is $3$-connected to $\widebar A$.

Case 6: There are $a_1,a_2,b_1,b_2 \in W_2$ with

\begin{equation*}
x = \begin{pmatrix} & a_1 \\ a_2 & \end{pmatrix} \ \textnormal{\textit{and}} \
y = \begin{pmatrix} & b_1 \\ b_2 & \end{pmatrix}.
\end{equation*}

Noticing that $\langle \widebar Z, \widebar x, \widebar y \rangle = \langle \widebar Z, \widebar x, \widebar x \widebar y \rangle$, we can reduce this case to Case 5.

We summarize the above lemmas in the following corollary.

Let $q$ be a nontrivial odd prime power and $n \ge 6$. Then the following hold:

(i) The Sylow $2$-subgroups of $SL_n(q)$ and those of $PSL_n(q)$ are $2$-connected.

(ii) The Sylow $2$-subgroups of $SL_n(q)$ are $3$-connected.

(iii) If $q \equiv 1 \ \mathrm{mod} \ 4$ or $n \ge 7$, then the Sylow $2$-subgroups of $PSL_n(q)$ are $3$-connected.

Unfortunately, the Sylow $2$-subgroups of $PSL_6(q)$ are not $3$-connected when $q \equiv 3 \mod 4$ (this is not terribly difficult to observe).

Let $q$ be a nontrivial odd prime power and $n \ge 6$. Let $G = SL_n(q)$, or $G = PSL_n(q)$ and $n \ge 7$ if $q \equiv 3 \mod 4$. For any Sylow $2$-subgroup $S$ of $G$ and any elementary abelian subgroup $A$ of $S$ with $m(A) \le 3$, there is some elementary abelian subgroup $B$ of $S$ with $A < B$ and $m(B) = 4$.

By Corollary <ref>, $S$ is $2$-connected and $3$-connected. Applying <cit.>, the claim follows.

§.§ Generation

Next we discuss some generational properties of $(P)SL_n(q)$ and $(P)SU_n(q)$, where $n \ge 3$ and $q$ is a nontrivial odd prime power. We need the following definition (see <cit.>).

Let $G$ be a finite group, let $S$ be a Sylow $2$-subgroup of $G$, and let $k$ be a positive integer. We say that $G$ is $k$-generated if

\begin{equation*}
G = \Gamma_{S,k}(G) := \langle N_G(T) \ | \ T \le S, m(T) \ge k \rangle.
\end{equation*}

The following two lemmas will later prove to be useful.

(see [6]) Let $q$ be a nontrivial odd prime power. Then the groups $SL_3(q)$, $PSL_3(q)$, $SU_3(q)$ and $PSU_3(q)$ are $2$-generated.

Let $q$ be a nontrivial odd prime power, and let $n \ge 4$ be a natural number. Moreover, let $\varepsilon \in \lbrace +,- \rbrace$ and $Z \le Z(SL_n^{\varepsilon}(q))$. Assume that one of the following holds:

(i) $n \ge 5$,

(ii) $q \equiv \varepsilon \ \mathrm{mod} \ 8$,

(iii) $Z = 1$.

Then $SL_n^{\varepsilon}(q)/Z$ is $3$-generated.

We need the following lemma in order to prove Lemma <ref>.

(see [45], [14]) Let $q > 2$ be a prime power, and let $n \ge 3$ be a natural number. Let $\varepsilon \in \lbrace +,- \rbrace$. Define

\begin{equation*}
U_1 := \left\lbrace \begin{pmatrix} A & \\ & I_{n-2} \end{pmatrix} \ : \ A \in SL_2^{\varepsilon}(q) \right\rbrace
\end{equation*}

\begin{equation*}
U_{n-1} := \left\lbrace \begin{pmatrix} I_{n-2} & \\ & A \end{pmatrix} \ : \ A \in SL_2^{\varepsilon}(q) \right\rbrace.
\end{equation*}

Moreover, for each $2 \le i \le n-2$, let

\begin{equation*}
U_i := \left\lbrace \begin{pmatrix} I_{i-1} & & \\ & A & \\ & & I_{n-i-1} \end{pmatrix} \ : \ A \in SL_2^{\varepsilon}(q) \right\rbrace.
\end{equation*}

Then the following hold:

(i) We have $SL_n^{\varepsilon}(q) = \langle U_i \ : \ 1 \le i \le n-1 \rangle$.

(ii) For each $1 \le i \le n-2$, there is a monomial matrix $m_i$ in $SL_n^{\varepsilon}(q)$ with $U_i^{m_i} = U_{i+1}$.

Let $q$ be a nontrivial odd prime power, and let $n \ge 4$ be a natural number. Moreover, let $\varepsilon \in \lbrace +,- \rbrace$ and $Z \le Z(SL_n^{\varepsilon}(q))$. Suppose that one of the conditions $n \ge 5$, $q \equiv \ \varepsilon \ \mathrm{mod} \ 8$ or $Z = 1$ is satisfied. We have to show that $SL_n^{\varepsilon}(q)/Z$ is $3$-generated.

Let $U_1, \dots, U_{n-1}$ denote the $SL_2^{\varepsilon}(q)$-subgroups of $SL_n^{\varepsilon}(q)$ corresponding to the $2 \times 2$ blocks along the main diagonal (as in Lemma <ref>). Let $E$ be the subgroup of $SL_n^{\varepsilon}(q)$ consisting of the diagonal matrices in $SL_n^{\varepsilon}(q)$ with diagonal entries in $\lbrace -1,1 \rbrace$.

Assume that $n \ge 5$. Then one can easily see that, for each $i \in \lbrace 1,\dots,n-1\rbrace$, there is an $E_8$-subgroup $E_i$ of $E$ with $E_i \cap Z(SL_n^{\varepsilon}(q)) = 1$ and $[E_i,U_i] = 1$. Hence, $U_iZ/Z$ centralizes $E_iZ/Z \cong E_8$ for each $i \in \lbrace 1,\dots,n-1 \rbrace$. Now, if $S$ is a Sylow $2$-subgroup of $SL_n^{\varepsilon}(q)/Z$ containing $EZ/Z$, we have $U_iZ/Z \le \Gamma_{S,3}(SL_n^{\varepsilon}(q)/Z)$ for each $i \in \lbrace 1,\dots, n-1 \rbrace$, and Lemma <ref> (i) implies that $SL_n^{\varepsilon}(q)/Z$ is $3$-generated.

We now consider the case $n = 4$. By hypothesis, $Z = 1$ or $q \equiv \varepsilon \ \mathrm{mod} \ 8$. Let

\begin{equation*}
U := \left \lbrace \left( \begin{array}{c|c} A & \begin{matrix} 0 \\ 0 \end{matrix} \\ \hline \begin{matrix} 0 & 0 \end{matrix} & 1 \end{array} \right) \ : \ A \in SL_3^{\varepsilon}(q) \right\rbrace.
\end{equation*}

If $Z = 1$, set $y := -I_4$. If $q \equiv \varepsilon \ \mathrm{mod} \ 8$, let $\lambda$ be an element of $\mathbb{F}_{q^2}^{*}$ of order $8$ such that $\lambda^{q-\varepsilon} = 1$. Note that $\lambda \in \mathbb{F}_q^{*}$ if $\varepsilon = +$. Also, if $q \equiv \varepsilon \ \mathrm{mod} \ 8$ and $\vert Z \vert = 2$, let $y := \lambda^2 I_4 \in SL_4^{\varepsilon}(q)$, and if $q \equiv \varepsilon \ \mathrm{mod} \ 8$ and $\vert Z \vert = 4$, let $y := \mathrm{diag}(\lambda, \lambda, \lambda, - \lambda) \in SL_4^{\varepsilon}(q)$.

Let $S_0$ be a Sylow $2$-subgroup of $U$ containing $E \cap U$. Let $\widetilde S$ be a Sylow $2$-subgroup of $SL_4^{\varepsilon}(q)$ containing $S_0$ and $y$. Denote the image of $\widetilde S$ in $SL_4^{\varepsilon}(q)/Z$ by $S$. We have $S \cap UZ/Z = S_0Z/Z \in \mathrm{Syl}_2(UZ/Z)$. By Lemma <ref>, $UZ/Z \cong U \cong SL_3^{\varepsilon}(q)$ is $2$-generated. So we have

\begin{equation*}
UZ/Z = \Gamma_{S_0 Z/Z,2}(UZ/Z) = \langle N_{UZ/Z}(T) \ \vert \ T \le S_0 Z/Z, m(T) \ge 2 \rangle.
\end{equation*}

Let $T \le S_0 Z/Z$ with $m(T) \ge 2$ and $\widehat T := \langle T, yZ \rangle$. Clearly, $yZ$ is an involution of $S$ not contained in $UZ/Z$ and centralizing $UZ/Z$. Therefore, we have that $m(\widehat T) \ge 3$ and $N_{UZ/Z}(T) \le N_{SL_n^{\varepsilon}(q)/Z}(\widehat T)$. It follows that $UZ/Z \le \Gamma_{S,3}(SL_n^{\varepsilon}(q)/Z)$. In particular, $U_iZ/Z \le \Gamma_{S,3}(SL_n^{\varepsilon}(q)/Z)$ for $i \in \lbrace 1,2 \rbrace$.

From Lemma <ref> (ii), we see that there is some $m \in SL_4^{\varepsilon}(q)$ such that ${U_2}^m = U_3$ and such that $m$ normalizes $\langle E, y \rangle$. So $mZ$ normalizes $\langle EZ/Z, yZ \rangle$. It is easy to note that $\langle EZ/Z, yZ \rangle \cong E_8$, and so we have $mZ \in \Gamma_{S,3}(SL_n^{\varepsilon}(q)/Z)$. It follows that $U_3 Z/Z = (U_2 Z/Z)^{mZ} \le \Gamma_{S,3}(SL_n^{\varepsilon}(q)/Z)$.

So we have $U_iZ/Z \le \Gamma_{S,3}(SL_n^{\varepsilon}(q)/Z)$ for $i \in \lbrace 1,2,3 \rbrace$, and Lemma <ref> (i) implies that $SL_n^{\varepsilon}(q)/Z$ is $3$-generated.

§.§ Automorphisms of $(P)SL_n(q)$

Fix a prime number $p$, a positive integer $f$ and a natural number $n \ge 2$. Set $q := p^f$ and $T := SL_n(q)$. We now briefly describe the structure of $\mathrm{Aut}(T/Z)$, where $Z \le Z(T)$, referring to [20] and <cit.> for further details.

Let $\mathrm{Inndiag}(T) := \mathrm{Aut}_{GL_n(q)}(T)$. Note that

\begin{equation*}
\mathrm{Inndiag}(T)/\mathrm{Inn}(T) \cong C_{(n,q-1)}.
\end{equation*}

The map

\begin{equation*}
\phi: T \rightarrow T, (a_{ij}) \mapsto ({a_{ij}}^p)
\end{equation*}

is an automorphism of $T$ with order $f$. One can check that $\phi$ normalizes $\mathrm{Inndiag}(T)$. Set

\begin{equation*}
P\Gamma L_n(q) := \mathrm{Inndiag}(T) \langle \phi \rangle.
\end{equation*}

It is easy to note that $\langle \phi \rangle \cap \mathrm{Inndiag}(T) = 1$, so that $P \Gamma L_n(q)$ is the inner semidirect product of $\mathrm{Inndiag}(T)$ and $\langle \phi \rangle$.

The map

\begin{equation*}
\iota: T \rightarrow T, a \mapsto (a^t)^{-1}
\end{equation*}

is an automorphism of $T$ with order $2$. If $n = 2$, then $\iota$ turns out to be an inner automorphism of $T$, while we have $\iota \not\in P\Gamma L_n(q)$ when $n \ge 3$. By a direct calculation, $\iota$ normalizes $\mathrm{Inndiag}(T)$ and commutes with $\phi$. In particular, $A := P \Gamma L_n(q) \langle \iota \rangle$ is a subgroup of $\mathrm{Aut}(T)$, and we have

\begin{equation*}
A/\mathrm{Inndiag}(T) \cong C_f \times C_a,
\end{equation*}

where $a = 1$ if $n = 2$ and $a = 2$ if $n \ge 3$.

Now let $Z$ be a central subgroup of $T$. It can be easily checked that the natural homomorphism $\mathrm{Aut}(T) \rightarrow \mathrm{Aut}(T/Z)$ is injective. The image of $\mathrm{Inndiag}(T)$ under this homomorphism will be denoted by $\mathrm{Inndiag}(T/Z)$. By abuse of notation, we denote the image of $P\Gamma L_n(q)$ in $\mathrm{Aut}(T/Z)$ again by $P \Gamma L_n(q)$ and the images of $\iota$ and $\phi$ again by $\iota$ and $\phi$, respectively.

With this notation, we have

\begin{equation*}
\mathrm{Aut}(T/Z) = P \Gamma L_n(q) \langle \iota \rangle.
\end{equation*}

Note that the natural homomorphism $\mathrm{Aut}(T) \rightarrow \mathrm{Aut}(T/Z)$ is an isomorphism and that it induces an isomorphism $\mathrm{Out}(T) \rightarrow \mathrm{Out}(T/Z)$.

The elements of $\mathrm{Inndiag}(T/Z) \setminus \mathrm{Inn}(T/Z)$ are said to be the (non-trivial) diagonal automorphisms of $T/Z$. An automorphism of $T/Z$ is called a field automorphism if it is conjugate to $\phi^i$ for some $1 \le i < f$. The automorphisms of the form $\alpha \iota$, where $\alpha \in \mathrm{Inndiag}(T/Z)$, are said to be the graph automorphisms of $T/Z$. An automorphism of $T/Z$ is said to be a graph-field automorphism if it is conjugate to an automorphism of the form $\phi^i \iota$ for some $1 \le i < f$. We remark that these definitions are particular cases of more general definitions, see <cit.>.

Let $q$ be a nontrivial prime power, and let $n \ge 2$. Then $\mathrm{Out}(PSL_n(q))$ is $2$-nilpotent.

From the above remarks, it is easy to see that $\mathrm{Out}(PSL_n(q))$ is supersolvable. By <cit.>, any supersolvable finite group is $2$-nilpotent, and so the proposition follows.

The following proposition also follows from the above remarks.

Let $n \ge 2$ be a natural number. Then $\mathrm{Out}(SL_n(3))$ is a $2$-group.

§.§ Automorphisms of $(P)SU_n(q)$

Let $p$ be a prime number, $f$ be a positive integer and $n \ge 3$ be a natural number. Set $q := p^f$ and $T := SU_n(q)$. We now briefly describe the structure of $\mathrm{Aut}(T/Z)$, where $Z \le Z(T)$, referring to [20] and <cit.> for further details.

Let $\mathrm{Inndiag}(T) := \mathrm{Aut}_{GU_n(q)}(SU_n(q))$. It is rather easy to note that

\begin{equation*}
\mathrm{Inndiag}(T)/\mathrm{Inn}(T) \cong C_{(n,q+1)}.
\end{equation*}

The map

\begin{equation*}
\phi: T \rightarrow T, (a_{ij}) \mapsto ({a_{ij}}^p)
\end{equation*}

is an automorphism of $T$ with order $2f$. One can check that $\phi$ normalizes $\mathrm{Inndiag}(T)$. Set

\begin{equation*}
P \Gamma U_n(q) := \mathrm{Inndiag}(T) \langle \phi \rangle.
\end{equation*}

It is rather easy to note that $\langle \phi \rangle \cap \mathrm{Inndiag}(T) = 1$, so that $P\Gamma U_n(q)$ is the inner semidirect product of $\mathrm{Inndiag}(T)$ and $\langle \phi \rangle$. Note that

\begin{equation*}
P \Gamma U_n(q) / \mathrm{Inndiag}(T) \cong C_{2f}.
\end{equation*}

Now let $Z$ be a central subgroup of $T$. It can be easily checked that the natural homomorphism $\mathrm{Aut}(T) \rightarrow \mathrm{Aut}(T/Z)$ is injective. The image of $\mathrm{Inndiag}(T)$ under this homomorphism will be denoted by $\mathrm{Inndiag}(T/Z)$. By abuse of notation, we denote the image of $P \Gamma U_n(q)$ in $\mathrm{Aut}(T/Z)$ again by $P \Gamma U_n(q)$ and the image of $\phi$ again by $\phi$.

With this notation, we have

\begin{equation*}
\mathrm{Aut}(T/Z) = P \Gamma U_n(q).
\end{equation*}

Note that the natural homomorphism $\mathrm{Aut}(T) \rightarrow \mathrm{Aut}(T/Z)$ is an isomorphism and that it induces an isomorphism $\mathrm{Out}(T) \rightarrow \mathrm{Out}(T/Z)$.

The elements of $\mathrm{Inndiag}(T/Z) \setminus \mathrm{Inn}(T/Z)$ are said to be the (non-trivial) diagonal automorphisms of $T/Z$. An automorphism of $T/Z$ is called a field automorphism if it is conjugate to $\phi^i$ for some $1 \le i < 2f$ such that $\phi^i$ has odd order. The automorphisms of the form $\alpha \phi^i$, where $\phi^i$ has even order and $\alpha \in \mathrm{Inndiag}(T/Z)$, are said to be the graph automorphisms of $T/Z$. There are no graph-field automorphisms of $T/Z$.

Let $q$ be a nontrivial prime power, and let $n \ge 3$. Then $\mathrm{Out}(PSU_n(q))$ is $2$-nilpotent.

We see from the above remarks that $\mathrm{Out}(PSU_n(q))$ is supersolvable. So $\mathrm{Out}(PSU_n(q))$ is $2$-nilpotent by <cit.>.

The following proposition also follows from the above remarks.

Let $n \ge 3$ be a natural number. Then $\mathrm{Out}(SU_n(3))$ is a $2$-group.

§.§ Some lemmas

We now prove several results on the automorphism groups of $(P)SL_n(q)$ and $(P)SU_n(q)$, where $n \ge 2$ and $q$ is a nontrivial odd prime power.

Let $q$ be a nontrivial odd prime power. Also, let $T := SL_2(q)$ and $S \in \mathrm{Syl}_2(T)$. Suppose that $\alpha$ and $\beta$ are $2$-elements of $\mathrm{Aut}(T)$ such that $S^{\alpha} = S = S^{\beta}$ and $\alpha|_{S,S} = \beta|_{S,S}$. Then $\alpha = \beta$.

Let $\gamma := \alpha \beta^{-1} \in C_{\mathrm{Aut}(T)}(S)$. We have $C_{\mathrm{Inndiag}(T)}(S) = 1$ by <cit.>. Therefore, it suffices to show that $\gamma \in \mathrm{Inndiag}(T)$. Clearly, the images of $\alpha$ and $\beta^{-1}$ in $\mathrm{Aut}(T)/\mathrm{Inndiag}(T)$ are $2$-elements of $\mathrm{Aut}(T)/\mathrm{Inndiag}(T)$. Since $\mathrm{Aut}(T)/\mathrm{Inndiag}(T)$ is abelian,

\begin{equation*}
\gamma \cdot \mathrm{Inndiag}(T) = (\alpha \cdot \mathrm{Inndiag}(T)) \cdot (\beta^{-1} \cdot \mathrm{Inndiag}(T))
\end{equation*}

is still a $2$-element of $\mathrm{Aut}(T)/\mathrm{Inndiag}(T)$. By <cit.>, $C_{\mathrm{Aut}(T)}(S)$ is a $2'$-group, and so $\gamma$ has odd order. Therefore, $\gamma \cdot \mathrm{Inndiag}(T)$ has odd order. It follows that $\gamma \in \mathrm{Inndiag}(T)$, as required.

Let $q = p^f$, where $p$ is an odd prime and $f$ is a positive integer. Let $T := PSL_2(q)$, and let $\alpha$ be an involution of $\mathrm{Aut}(T)$. Suppose that $C_T(\alpha)$ has a $2$-component $K$. Then we have $2 \mid f$, $(f,p) \ne (2,3)$ and $K \cong PSL_2(p^{\frac{f}{2}})$. In particular, $K$ is a component of $C_T(\alpha)$.

Note that $C_T(\alpha) \cong C_{\mathrm{Inn}(T)}(\alpha)$.

Assume that $\alpha \in \mathrm{Inndiag}(T)$. Noticing that $\mathrm{Inndiag}(T) \cong PGL_2(q)$, we see from Lemma <ref> that $C_{\mathrm{Inndiag}(T)}(\alpha)$ is solvable. Thus $C_T(\alpha) \cong C_{\mathrm{Inn}(T)}(\alpha)$ is solvable, and $C_T(\alpha)$ has no $2$-components, a contradiction to the choice of $\alpha$.

So we have $\alpha \not\in \mathrm{Inndiag}(T)$. By the structure of $\mathrm{Aut}(PSL_2(q))$ and since $\alpha$ has order $2$, we can write $\alpha$ as a product of an inner-diagonal automorphism and a field automorphism of order $2$. In particular, $f$ must be even. Consulting <cit.>, we see that $\alpha$ itself is a field automorphism. So we can apply <cit.> to conclude that $C_{\mathrm{Inndiag}(T)}(\alpha) \cong \mathrm{Inndiag}(PSL_2(p^{\frac{f}{2}})) \cong PGL_2(p^{\frac{f}{2}})$. Consequently, $K$ is isomorphic to a $2$-component of $PGL_2(p^{\frac{f}{2}})$. It follows that $(f,p) \ne (2,3)$ and $K \cong PSL_2(p^{\frac{f}{2}})$.

Before we state the next lemma, we introduce some notational conventions for adjoint Chevalley groups. Given a nontrivial prime power $q$, we denote $A_1(q)$ also by $B_1(q)$ and by $C_1(q)$. Moreover, $B_2(q)$ will be also denoted by $C_2(q)$, and $A_3(q)$ will be also denoted by $D_3(q)$. We also set $D_2(q) := A_1(q) \times A_1(q)$ and $^2D_2(q) := A_1(q^2)$.

Let $q = p^f$, where $p$ is an odd prime and $f$ is a positive integer. Let $n \ge 3$ be a natural number and $\varepsilon \in \lbrace +,- \rbrace$. Let $T := PSL_n^{\varepsilon}(q)$, and let $\alpha$ be an involution of $\mathrm{Aut}(T)$. Suppose that $C_T(\alpha)$ has a $2$-component $K$. Then $K$ is in fact a component, and one of the following holds:

(i) $K \cong SL_i^{\varepsilon}(q)$ for some $2 \le i < n$, where $i > 2$ if $q = 3$;

(ii) $n$ is even, and $K$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}(q^2)$;

(iii) $\varepsilon = +$, $f$ is even, $K \cong PSL_n(p^{\frac{f}{2}})$ or $K \cong PSU_n(p^{\frac{f}{2}})$;

(iv) $q \ne 3$, $n = 3$ or $4$, and $K \cong PSL_2(q)$;

(v) $n$ is odd, $n \ge 5$ and $K \cong B_{\frac{n-1}{2}}(q)$;

(vi) $n$ is even and $K \cong C_{\frac{n}{2}}(q)$;

(vii) $n$ is even, $n \ge 6$ and $K \cong D_{\frac{n}{2}}(q)$;

(viii) $n$ is even, $n \ge 6$ and $K \cong$ $^2D_{\frac{n}{2}}(q)$.

Here, the (twisted) Chevalley groups appearing in (v)-(viii) are adjoint.

It can be shown that any involution of $\mathrm{Aut}(T)$ is an inner-diagonal automorphism, a field automorphism, a graph automorphism, or a graph-field automorphism (see <cit.> or <cit.>).

Case 1: $\alpha \in \mathrm{Inndiag}(T)$, or $\alpha$ is a graph automorphism.

Set $C^{*} := C_{\mathrm{Inndiag}(T)}(\alpha)$ and $L^{*} := O^{p'}(C^{*})$. One can see from <cit.> that $C^{*}/L^{*}$ is solvable and that one of the following holds:

(1) $L^{*}$ is the central product of two subgroups isomorphic to $SL_i^{\varepsilon}(q)$ and $SL_{n-i}^{\varepsilon}(q)$ for some natural number $i$ with $1 \le i \le \frac{n}{2}$,

(2) $n$ is even and $L^{*}$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}(q^2)$,

(3) $n$ is odd and $L^{*} \cong B_{\frac{n-1}{2}}(q)$,

(4) $n$ is even and $L^{*} \cong C_{\frac{n}{2}}(q)$,

(5) $n$ is even and $L^{*} \cong D_{\frac{n}{2}}(q)$,

(6) $n$ is even and $L^{*} \cong$ $^2D_{\frac{n}{2}}(q)$,

where the (twisted) Chevalley groups appearing in the last four cases are adjoint. Since $C_T(\alpha)$ is isomorphic to $C_{\mathrm{Inn}(T)}(\alpha) \trianglelefteq C^{*}$, we have that $K$ is isomorphic to a $2$-component of $C^{*}$ and thus isomorphic to a $2$-component of $L^{*}$. Therefore, one of the conditions (i)-(viii) is satisfied.

Case 2: $\alpha$ is a field automorphism or a graph-field automorphism.

Again, let $C^{*} := C_{\mathrm{Inndiag}(T)}(\alpha)$. Since the field automorphisms of $PSU_n(q)$ have odd order and $PSU_n(q)$ has no graph-field automorphisms, we have $\varepsilon = +$. Also, $f$ is even since $\alpha$ is a field automorphism or a graph-field automorphism of order $2$. From <cit.>, we see that $C^{*} \cong PGL_n(p^{\frac{f}{2}})$ if $\alpha$ is a field automorphism and that $C^{*} \cong PGU_n(p^{\frac{f}{2}})$ if $\alpha$ is a graph-field automorphism. Since $K$ is isomorphic to a $2$-component of $C^{*}$, it follows that (iii) is satisfied.

Let $q = p^f$, where $p$ is an odd prime and $f$ is a positive integer. Let $n \ge 2$ be a natural number and $\varepsilon \in \lbrace +,- \rbrace$. Let $Z$ be a central subgroup of $SL_n^{\varepsilon}(q)$ and let $T := SL_n^{\varepsilon}(q)/Z$. Let $\alpha$ be an involution of $\mathrm{Aut}(T)$, and let $K$ be a $2$-component of $C_T(\alpha)$. Then the following hold:

(i) $K$ is a component of $C_T(\alpha)$, and $K/Z(K)$ is a known finite simple group.

(ii) $K/Z(K) \not\cong M_{11}$.

(iii) Assume that $K/Z(K) \cong PSL_k^{\varepsilon^{*}}(q^{*})$ for some positive integer $2 \le k \le n$, some nontrivial odd prime power $q^{*}$ and some $\varepsilon^{*} \in \lbrace +,- \rbrace$. Then one of the following holds:

(a) $q^{*} = q$;

(b) $q^{*} = q^2$, $n \ge 4$ is even, $k = \frac{n}{2}$, and $\varepsilon^{*} = +$ if $n \ge 6$;

(c) $f$ is even, $k = n$, $q^{*} = p^{\frac{f}{2}}$.

Set $\widebar T := T/Z(T) \cong PSL_n^{\varepsilon}(q)$. Let $\overline \alpha$ be the automorphism of $\overline T$ induced by $\alpha$.

Clearly, $\widebar K$ is a $2$-component of $\widebar{C_T(\alpha)}$. It is easy to note that $\widebar{C_T(\alpha)}$ is a normal subgroup of $C_{\widebar T}(\widebar{\alpha})$. So $\widebar K$ is a $2$-component of $C_{\widebar T}(\widebar{\alpha})$. Lemmas <ref> and <ref> imply that $\widebar K$ is a component of $C_{\widebar T}(\widebar \alpha)$ and that $\widebar{K}/Z(\widebar K)$ is a known finite simple group. Applying <cit.>, we conclude that $K'$ is a component of $C_T(\alpha)$. We have $K = K'$ since $K$ is a $2$-component of $C_T(\alpha)$, and so it follows that $K$ is a component of $C_T(\alpha)$. Also, $K/Z(K) \cong \widebar{K}/Z(\widebar K)$, so that $K/Z(K)$ is a known finite simple group. Hence (i) holds.

If $K/Z(K) \cong M_{11}$, then $\widebar K/Z(\widebar K) \cong M_{11}$, which is not possible by Lemmas <ref> and <ref>. So (ii) holds.

Suppose that $K/Z(K) \cong PSL_k^{\varepsilon^{*}}(q^{*})$ for some positive integer $2 \le k \le n$, some nontrivial odd prime power $q^{*}$ and some $\varepsilon^{*} \in \lbrace +,- \rbrace$. By Lemmas <ref> and <ref>, one of the following holds:

(1) $\widebar K/Z(\widebar K) \cong PSL_i^{\varepsilon}(q)$ for some $2 \le i < n$;

(2) $n$ is even and $\widebar K/Z(\widebar K)$ is isomorphic to $PSL_{\frac{n}{2}}(q^2)$;

(3) $f$ is even, $\widebar K \cong PSL_n(p^{\frac{f}{2}})$ or $PSU_n(p^{\frac{f}{2}})$;

(4) $q \ne 3$, $n = 3$ or $4$, $\widebar K \cong PSL_2(q)$;

(5) $n$ is odd, $n \ge 5$, $\widebar K \cong B_{\frac{n-1}{2}}(q)$;

(6) $n$ is even, $n \ge 4$, $\widebar K \cong C_{\frac{n}{2}}(q)$;

(7) $n$ is even, $n \ge 6$, $\widebar K \cong  D_{\frac{n}{2}}(q)$;

(8) $n$ is even, $n \ge 6$, $\widebar K \cong$ $^2D_{\frac{n}{2}}(q)$.

Here, the (twisted) Chevalley groups appearing in (5)-(8) are adjoint. On the other hand, we have $\widebar K/Z(\widebar K) \cong PSL_k^{\varepsilon^{*}}(q^{*})$. Now, if (1) holds, then $PSL_k^{\varepsilon^{*}}(q^{*}) \cong PSL_i^{\varepsilon}(q)$ for some $2 \le i < n$, and <cit.> shows that this is only possible when $q^{*} = q$, so that (a) holds. Similarly, if (2) holds, then we have (b). Moreover, (3) implies (c) and (4) implies (a). As Theorem <cit.> shows, the cases (5) and (6) cannot occur, while (7) and (8) can only occur when $n = 6$. As above, one can see that if $n = 6$ and (7) or (8) holds, then we have (a).

Let $n \ge 3$ and $\varepsilon \in \lbrace +,- \rbrace$. Then $SL_n^{\varepsilon}(3)$ is locally balanced (in the sense of Definition <ref>).

Set $T := SL_n^{\varepsilon}(3)$. Let $H$ be a subgroup of $\mathrm{Aut}(T)$ containing $\mathrm{Inn}(T)$, and let $x$ be an involution of $H$. It is enough to show that $O(C_H(x)) = 1$.

Assume that $O(C_H(x)) \ne 1$. Then $x \in \mathrm{Inndiag}(T)$ by <cit.>. By Propositions <ref> and <ref>, $\mathrm{Out}(T)$ is a $2$-group. This implies $O(C_H(x)) = O(C_{\mathrm{Inn}(T)}(x)) = O(C_{\mathrm{Inndiag}(T)}(x))$. Since $x$ is an involution of $\mathrm{Inndiag}(T) \cong PGL_n^{\varepsilon}(3)$, we have $O(C_{\mathrm{Inndiag}(T)}(x)) = 1$ by Corollary <ref>. Thus $O(C_H(x)) = 1$. This contradiction completes the proof.

Let $n \ge 3$ be a natural number, let $q$ be a nontrivial odd power, and let $\varepsilon \in \lbrace +,- \rbrace$. Then any non-trivial quotient of $SL_n^{\varepsilon}(q)$ is locally $2$-balanced (in the sense of Definition <ref>).

By <cit.> or <cit.>, $PSL_n^{\varepsilon}(q)$ is locally $2$-balanced. Let $K$ be a non-trivial quotient of $SL_n^{\varepsilon}(q)$. As we have seen, there is an isomorphism $\mathrm{Aut}(K) \rightarrow \mathrm{Aut}(PSL_n^{\varepsilon}(q))$ mapping $\mathrm{Inn}(K)$ to $\mathrm{Inn}(PSL_n^{\varepsilon}(q))$. So the local $2$-balance of $K$ follows from the local $2$-balance of $PSL_n^{\varepsilon}(q)$.

Let $q$ be a nontrivial odd prime power and $n \ge 4$ be a natural number. Let $Z \le Z(SL_n(q))$ and $T := SL_n(q)/Z$. Let $K_1$ be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} A & \\ & I_{n-2} \end{pmatrix} \ : \ A \in SL_2(q) \right\rbrace
\end{equation*}

in $T$, and let $K_2$ be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} I_2 & \\ & B \end{pmatrix} \ : \ B \in SL_{n-2}(q) \right\rbrace
\end{equation*}

in $T$. Let $\alpha$ be an automorphism of $T$ with odd order such that $\alpha$ normalizes $K_1$ and centralizes $K_2$. Then $\alpha|_{K_1,K_1}$ is an inner automorphism.

By hypothesis, $q = p^f$ for some odd prime number $p$ and some positive integer $f$. We have $\alpha \in P \Gamma L_n(q)$ since $\alpha$ has odd order and $|\mathrm{Aut}(T)/P \Gamma L_n(q)| = 2$. So there are some $m \in GL_{n}(q)$ and some $1 \le r \le f$ such that, for each element $(a_{ij})$ of $SL_n(q)$, $\alpha$ maps $(a_{ij})Z$ to $((a_{ij})^{p^r})^mZ$.

Let $x$ be the image of $\mathrm{diag}(-1,-1,1,\dots,1) \in SL_n(q)$ in $T$. Then $x$ is the unique involution of $K_1$, and so we have $x^{\alpha} = x$. This easily implies that

\begin{equation*}
m = \begin{pmatrix} m_1 & \\ & m_2 \end{pmatrix}
\end{equation*}

for some $m_1 \in GL_2(q)$ and some $m_2 \in GL_{n-2}(q)$.

Since $\alpha$ centralizes $K_2$, we have $((a_{ij})^{p^r})^{m_2} = (a_{ij})$ for all $(a_{ij}) \in SL_{n-2}(q)$. Therefore, the automorphism $SL_{n-2}(q) \rightarrow SL_{n-2}(q), (a_{ij}) \mapsto (a_{ij})^{p^r}$ is an element of $\mathrm{Inndiag}(SL_{n-2}(q))$. This implies $r = f$.

Thus, under the isomorphism $\mathrm{Aut}(SL_2(q)) \rightarrow \mathrm{Aut}(K_1)$ induced by the canonical isomorphism $SL_2(q) \rightarrow K_1$, the automorphism $\alpha|_{K_1,K_1}$ of $K_1$ corresponds to the inner-diagonal automorphism $\widetilde \alpha: SL_2(q) \rightarrow SL_2(q), a \mapsto a^{m_1}$, and this automorphism has odd order since $\alpha$ has odd order. The index of $\mathrm{Inn}(SL_2(q))$ in $\mathrm{Inndiag}(SL_2(q))$ is $2$, and so it follows that $\widetilde \alpha \in \mathrm{Inn}(SL_2(q))$. Consequently, $\alpha|_{K_1,K_1} \in \mathrm{Inn}(K_1)$.

By using similar arguments as in the proof of Lemma <ref>, one can prove the following lemma.

Let $q$ be a nontrivial odd prime power and $n \ge 4$ be a natural number. Let $Z \le Z(SU_n(q))$ and $T := SU_n(q)/Z$. Let $K_1$ be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} A & \\ & I_{n-2} \end{pmatrix} \ : \ A \in SU_2(q) \right\rbrace
\end{equation*}

in $T$, and let $K_2$ be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} I_2 & \\ & B \end{pmatrix} \ : \ B \in SU_{n-2}(q) \right\rbrace
\end{equation*}

in $T$. Let $\alpha$ be an automorphism of $T$ with odd order such that $\alpha$ normalizes $K_1$ and centralizes $K_2$. Then $\alpha|_{K_1,K_1}$ is an inner automorphism.

Our next goal is to prove the following lemma.

Let $q$ and $q^{*}$ be nontrivial odd prime powers. Let $L$ be a group isomorphic to $SL_2(q^{*})$. Let $Q$ be a Sylow $2$-subgroup of $L$. Moreover, let $V$ be a Sylow $2$-subgroup of $GL_2(q)$ and $V_0 := V \cap SL_2(q)$. Suppose that there is a group isomorphism $\psi: V_0 \rightarrow Q$. Let $v_1, v_2, v_3$ be elements of $V$ such that $v_3 = v_1v_2$ and such that the square of any element of $\lbrace v_1,v_2,v_3 \rbrace$ lies in $Z(GL_2(q))$. For each $i \in \lbrace 1,2,3 \rbrace$, let $\alpha_i$ be a $2$-element of $\mathrm{Aut}(L)$ normalizing $Q$ such that

\begin{equation*}
\alpha_i|_{Q,Q} = \psi^{-1} (c_{v_i}|_{V_0,V_0}) \psi.
\end{equation*}

Then we have

\begin{equation*}
\bigcap_{i=1}^3 O(C_L(\alpha_i)) = 1.
\end{equation*}

To prove Lemma <ref>, we need to prove some other lemmas.

Let $q$ be a nontrivial odd prime power with $q \equiv 1 \ \mathrm{mod} \ 4$, and let $k \in \mathbb{N}$ with $(q-1)_2 = 2^k$. Let $G$ be a group isomorphic to $SL_2(q)$ and $Q \in \mathrm{Syl}_2(G)$. Then the following hold:

(i) There are elements $a, b \in Q$ such that $\mathrm{ord}(a) = 2^k$, $\mathrm{ord}(b) = 4$, $a^b = a^{-1}$ and $b^2 = a^{2^{k-1}}$.

(ii) Let $a$ and $b$ be as in (i). Then there is a group isomorphism $\varphi: G \rightarrow SL_2(q)$ such that

\begin{equation*}
a^{\varphi} = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}
\end{equation*}

for some $\lambda \in \mathbb{F}_q^{*}$ with order $2^k$ and

\begin{equation*}
b^{\varphi} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
\end{equation*}

(i) follows from Lemma <ref>.

We now prove (ii). Assume that $k \ge 3$. By Lemma <ref> (i),

\begin{equation*}
\left\lbrace \begin{pmatrix} \mu & 0 \\ 0 & \mu^{-1} \end{pmatrix} \ : \ \textnormal{$\mu$ is a $2$-element of $\mathbb{F}_q^{*}$} \right\rbrace \left\langle \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right\rangle
\end{equation*}

is a Sylow $2$-subgroup of $SL_2(q)$. Choose a group isomorphism $\psi: G \rightarrow SL_2(q)$ such that $Q^{\psi} = R$. Clearly, since $k \ge 3$, $Q$ has only one cyclic subgroup of order $2^k$. This implies that

\begin{equation*}
a^{\psi} = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}
\end{equation*}

for some $\lambda \in \mathbb{F}_q^{*}$ with order $2^k$. Since $b \not\in \langle a \rangle$, we have

\begin{equation*}
b^{\psi} = \begin{pmatrix} 0 & \mu \\ -\mu^{-1} & 0 \end{pmatrix}
\end{equation*}

for some $2$-element $\mu$ of $\mathbb{F}_q^{*}$. Composing $\psi$ with the automorphism

\begin{equation*}
SL_2(q) \rightarrow SL_2(q), \ A \mapsto \begin{pmatrix} \mu^{-1} & 0 \\ 0 & 1 \end{pmatrix}A\begin{pmatrix} \mu & 0 \\ 0 & 1 \end{pmatrix}
\end{equation*}

we get a group isomorphism $\varphi: G \rightarrow SL_2(q)$ with the desired properties. This completes the proof of (ii) for the case $k \ge 3$.

Assume now that $k = 2$. Let $\psi: G \rightarrow SL_2(q)$ be a group isomorphism. We have $(a^\psi)^2 = -I_2$ since $-I_2$ is the only involution of $SL_2(q)$ and $\mathrm{ord}(a^2) = 2$. So, by Lemma <ref>, we may assume that

\begin{equation*}
a^{\psi} = \begin{pmatrix}\lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}
\end{equation*}

for some $\lambda \in \mathbb{F}_q^{*}$ with order $4$. Since $a^b = a^{-1}$, we have

\begin{equation*}
\begin{pmatrix}\lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}^{b^{\psi}} = \begin{pmatrix}\lambda^{-1} & 0 \\ 0 & \lambda \end{pmatrix}.
\end{equation*}

This implies that

\begin{equation*}
b^{\psi} = \begin{pmatrix} 0 & \mu \\ -\mu^{-1} & 0 \end{pmatrix}
\end{equation*}

for some $\mu \in \mathbb{F}_q^{*}$. Again we may compose $\psi$ with a suitable diagonal automorphism of $SL_2(q)$ to obtain a group isomorphism $\varphi: G \rightarrow SL_2(q)$ with the desired properties.

By using similar arguments as in the proof of Lemma <ref>, one can prove the following lemma.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \mod 4$, and let $s \in \mathbb N$ with $(q+1)_2 = 2^s$. Let $G$ be a group isomorphic to $SU_2(q)$ and $Q \in \mathrm{Syl}_2(G)$. Then the following hold:

(i) There are elements $a, b \in Q$ such that $\mathrm{ord}(a) = 2^s$, $\mathrm{ord}(b) = 4$, $a^b = a^{-1}$ and $b^2 = a^{2^{s-1}}$.

(ii) Let $a$ and $b$ be as in (i). Then there is a group isomorphism $\varphi: G \rightarrow SU_2(q)$ such that

\begin{equation*}
a^{\varphi} = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix}
\end{equation*}

for some $\lambda \in \mathbb{F}_{q^2}^{*}$ with order $2^s$ and

\begin{equation*}
b^{\varphi} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
\end{equation*}

Let $q$ be a nontrivial odd prime power with $q \equiv 1 \ \mathrm{mod} \ 4$. Let $\rho$ be a generating element of the Sylow $2$-subgroup of $\mathbb{F}_q^{*}$, and let

\begin{equation*}
a := \begin{pmatrix} \rho & \\ & \rho^{-1} \end{pmatrix}, \ \ \ \ b := \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.
\end{equation*}

Let $V$ be the Sylow $2$-subgroup of $GL_2(q)$ given by Lemma <ref> (i), and let $v,w \in V$ such that $v^2,w^2,(vw)^2 \in Z(GL_2(q))$. Then one of the following holds:

(i) $\lbrace v,w,vw \rbrace \cap Z(GL_2(q)) \ne \emptyset$.

(ii) There exist $r,s \in \lbrace v,w,vw \rbrace$ with $a^{r} = a$, $b^{r} = b^3$ and $a^{s} = a^{-1}$.

It is easy to note that (i) holds if $v$ and $w$ are diagonal matrices.

Suppose now that $v$ or $w$ is not a diagonal matrix. If neither $v$ nor $w$ is a diagonal matrix, then $vw$ is a diagonal matrix. So there exist $r, s \in \lbrace v,w,vw \rbrace$ such that

\begin{equation*}
r = \begin{pmatrix} \lambda_1 & \\ & \lambda_2 \end{pmatrix}, \ \ \ \ s = \begin{pmatrix} & \mu_1 \\ \mu_2 \end{pmatrix},
\end{equation*}

where $\lambda_1$, $\lambda_2$, $\mu_1$ and $\mu_2$ are $2$-elements of $\mathbb{F}_q^{*}$.

If $\lambda_1 = \lambda_2$, then (i) holds. Assume now that $\lambda_1 \ne \lambda_2$. Then $\lambda_2 = - \lambda_1$ since $r^2 \in Z(GL_2(q))$, and a direct calculation shows that $a^{r} = a$, $b^{r} = b^3$ and $a^{s} = a^{-1}$.

Let $q$ be a nontrivial odd prime power with $q \equiv 3 \ \mathrm{mod} \ 4$, and let $k \in \mathbb{N}$ with $(q+1)_2 = 2^k$. Let $V$ be a Sylow $2$-subgroup of $GL_2(q)$.

(i) There exist $x, y \in V$ with $\mathrm{ord}(x) = 2^{k+1}$, $\mathrm{ord}(y) = 2$ and $x^y = x^{-1+2^k}$. We have $V \cap SL_2(q) = \langle x^2 \rangle \langle xy \rangle$.

(ii) Let $x$ and $y$ be as above, and let $a := x^2$ and $b := xy$. Let $v, w \in V$ with $v^2,w^2,(vw)^2 \in Z(GL_2(q))$. Then one of the following holds:

(a) $\lbrace v,w,vw \rbrace \cap Z(GL_2(q)) \ne \emptyset$.

(b) There exist $r,s \in \lbrace v,w,vw \rbrace$ such that $a^r = a$, $b^r = b^3$ and $a^s = a^{-1}$.

(i) follows from Lemma <ref> (i), (ii).

We now prove (ii). We have $Z(V) = \langle x^{2^k} \rangle$ by Lemma <cit.>. Thus $Z(GL_2(q)) \cap V = \langle x^{2^k} \rangle$. Clearly, $\lbrace v,w,vw \rbrace \cap \langle x \rangle \subseteq \langle x^{2^{k-1}} \rangle$.

If $v,w \in \langle x \rangle$, then $v,w \in \langle x^{2^{k-1}} \rangle$, and it easily follows that (a) holds.