Assume now that $v \not\in \langle x \rangle$ or $w \not\in \langle x \rangle$. If neither $v$ nor $w$ lies in $\langle x \rangle$, then $vw \in \langle x \rangle$. Consequently, $\lbrace v,w,vw \rbrace$ has an element $r$ of the form $x^{\ell 2^{k-1}}$ for some $1 \le \ell \le 4$ and an element $s$ of the form $x^i y$ for some $1 \le i \le 2^{k+1}$. If $\ell = 2$ or $4$, then (a) holds. Assume now that $\ell = 1$ or $3$. It is clear that $a^r = a$. Furthermore, we have

\begin{align*}
b^r &= (xy)^{x^{\ell 2^{k-1}}} \\
&= x y^{x^{\ell 2^{k-1}}} \\
&= x x^{-\ell 2^{k-1}} y x^{\ell 2^{k-1}} y^2 \\
&= x^{1-\ell 2^{k-1}} (x^y)^{\ell 2^{k-1}} y \\
&= x^{1-\ell 2^{k-1}}(x^{-1+2^k})^{\ell 2^{k-1}} y \\
&= x^{1-\ell 2^k+\ell 2^{2k-1}} y \\
&= x^{1-\ell 2^k} y \\
&\stackrel{\textnormal{$\ell$ odd}}{=} x^{1+2^k}y.
\end{align*}

On the other hand, we have

\begin{equation*}
b^3 = (xy)^3 = x^{2^k}xy = x^{1+2^k}y.
\end{equation*}

Consequently, $b^r = b^3$. Finally, we also have

\begin{equation*}
a^s = (x^2)^{x^i y} = (x^2)^y = (x^y)^2 = (x^{-1+2^k})^2 = x^{-2} = a^{-1}.
\end{equation*}

Thus (b) holds.

If $\alpha_j|_{Q,Q} = \mathrm{id}_Q$ for some $j \in \lbrace 1,2,3 \rbrace$, then $\alpha_j = \mathrm{id}_L$ by Lemma <ref>, which implies that

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

Suppose now that $\alpha_i$ acts nontrivially on $Q$ for all $i \in \lbrace 1,2,3 \rbrace$. Let $m \in \mathbb{N}$ with $|Q| = 2^m$. Using Lemma <ref> (together with Sylow’s theorem) and Lemma <ref>, we see that there exist $a,b \in Q$ and $i,j \in \lbrace 1,2,3 \rbrace$ such that the following hold:

(i) $\mathrm{ord}(a) = 2^{m-1}$, $\mathrm{ord}(b) = 4$, $a^b = a^{-1}$, $b^2 = a^{2^{m-2}}$;

(ii) $a^{\alpha_i} = a$, $b^{\alpha_i} = b^3$, $a^{\alpha_j} = a^{-1}$.

Clearly, $b^{\alpha_j} = a^{\ell}b$ for some $1 \le \ell \le 2^{m-1}$.

Assume that $q^{*} \equiv 1 \ \mathrm{mod} \ 4$. By Lemma <ref>, there is group isomorphism $\varphi: L \rightarrow SL_2(q^{*})$ with

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

for some generator $\lambda$ of the Sylow $2$-subgroup of $(\mathbb{F}_{q^{*}})^{*}$ and

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

Set $\beta_k := \varphi^{-1}\alpha_k \varphi$ for $k \in \lbrace 1,2,3 \rbrace$. Let $i$ and $j$ be as in (ii). Also, let

\begin{equation*}
m_i := \begin{pmatrix} 1 & \\ & -1 \end{pmatrix}.
\end{equation*}

Then $\beta_i$ and $c_{m_i}$ normalize $Q^{\varphi}$, and we have $\beta_i|_{Q^{\varphi},Q^{\varphi}} = c_{m_i}|_{Q^{\varphi},Q^{\varphi}}$. Applying Lemma <ref>, we conclude that $\beta_i = c_{m_i}$.

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

for some $2$-element $\mu$ of $(\mathbb{F}_{q^{*}})^{*}$. Set

\begin{equation*}
m_j := \begin{pmatrix} 0 & \mu \\ -1 & 0 \end{pmatrix}.
\end{equation*}

Then $\beta_j$ and $c_{m_j}$ normalize $Q^{\varphi}$, and we have $\beta_j|_{Q^{\varphi},Q^{\varphi}} = c_{m_j}|_{Q^{\varphi},Q^{\varphi}}$. Applying Lemma <ref>, we conclude that $\beta_j = c_{m_j}$.

It follows that $C_{SL_2(q^{*})}(\beta_i) \cap C_{SL_2(q^{*})}(\beta_j) = Z(SL_2(q^{*}))$. So we have $C_L(\alpha_i) \cap C_L(\alpha_j) = Z(L)$, and this implies that

\begin{equation*}
\bigcap_{k=1}^3 O(C_L(\alpha_k)) = 1
\end{equation*}

since $|Z(L)| = 2$.

If $q^{*} \equiv 3 \mod 4$, then a very similar argumentation shows that the same conclusion holds. Here, one has to use Lemma <ref> instead of Lemma <ref>, together with the fact that $SL_2(q^{*}) \cong SU_2(q^{*})$.

We bring this section to a close with a proof of the following lemma, which will play an important role in the proof of Theorem <ref>.

Let $q$ be a nontrivial odd prime power, $\varepsilon \in \lbrace +,- \rbrace$ and $n \ge 2$ a natural number. Set $T := \mathrm{Inn}(PSL_n^{\varepsilon}(q))$. Let $A$ be a subgroup of $\mathrm{Aut}(PSL_n^{\varepsilon}(q))$ such that $T \le A$ and such that the index of $T$ in $A$ is odd. Let $S$ be a Sylow $2$-subgroup of $T$. Then we have $\mathcal{F}_S(T) = \mathcal{F}_S(A)$.

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

Let $q$ be a nontrivial odd prime power, $\varepsilon \in \lbrace +,- \rbrace$, and let $r$ be positive integer. Also, let $W$ be a Sylow $2$-subgroup of $GL_{2^r}^{\varepsilon}(q)$. Then $\mathrm{Aut}(W)$ is a $2$-group.

We proceed by induction over $r$.

Suppose that $r = 1$. If $q \equiv -\varepsilon \mod 4$, then $W$ is semidihedral by Lemmas <ref> and <ref>, and so $\mathrm{Aut}(W)$ is a $2$-group by <cit.>. If $q \equiv \varepsilon \mod 4$, then $W \cong C_{2^k} \wr C_2$ for some positive integer $k$ by Lemmas <ref> and <ref>, and so $\mathrm{Aut}(W)$ is a $2$-group as a consequence of <cit.>.

Assume now that $r > 1$ and that the lemma is true with $r-1$ instead of $r$. Let $W_0$ be a Sylow $2$-subgroup of $GL_{2^{r-1}}^{\varepsilon}(q)$. Hence, $\mathrm{Aut}(W_0)$ is a $2$-group. By Lemma <ref>, we have $W \cong W_0 \wr C_2$. Applying <cit.>, we conclude that $\mathrm{Aut}(W)$ is a $2$-group.

Let $q$ be a nontrivial odd prime power, $\varepsilon \in \lbrace +,- \rbrace$, and let $n \ge 3$ be a natural number. Let $T := SL_n^{\varepsilon}(q)$, and let $S$ be a Sylow $2$-subgroup of $\mathrm{Inndiag}(T)$. Then $\mathrm{Aut}_{P\Gamma L_n^{\varepsilon}(q)}(S)$ is a $2$-group.

Let $\alpha \in N_{P\Gamma L_n^{\varepsilon}(q)}(S)$. It suffices to show that $c_{\alpha}|_{S,S}$ is a $2$-automorphism of $S$.

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 $1 \le i \le t$. By Lemma <ref>,

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

is a Sylow $2$-subgroup of $GL_n^{\varepsilon}(q)$.

Clearly, $\lbrace c_w|_{T,T} \ \vert \ w \in W \rbrace$ is a Sylow $2$-subgroup of $\mathrm{Inndiag}(T)$. Without loss of generality, we assume that $S = \lbrace c_w|_{T,T} \ \vert \ w \in W \rbrace$.

Let $p$ be the odd prime number and $f$ be the positive integer with $q = p^f$. Since $\alpha \in P \Gamma L_n^{\varepsilon}(q)$, there exist some $m \in GL_n^{\varepsilon}(q)$ and some natural number $\ell$, where $1 \le \ell \le f$ if $\varepsilon = +$ and $1 \le \ell \le 2f$ if $\varepsilon = -$, such that

\begin{equation*}
(a_{ij})^{\alpha} = ((a_{ij})^{p^{\ell}})^m
\end{equation*}

for all $(a_{ij}) \in T$.

\begin{equation*}
\widebar \alpha: GL_n^{\varepsilon}(q) \rightarrow GL_n^{\varepsilon}(q), (a_{ij}) \mapsto ((a_{ij})^{p^{\ell}})^m.
\end{equation*}

It is easy to note that $\widebar \alpha$ is an automorphism of $GL_n^{\varepsilon}(q)$. Using this, one can see that $\alpha^{-1}(c_w|_{T,T})\alpha = c_{w^{\widebar \alpha}}|_{T,T}$ for all $w \in W$.

Let $w \in W$. Since $\alpha$ normalizes $S$, there is some $\widetilde w \in W$ with $c_{w^{\widebar \alpha}}|_{T,T} = \alpha^{-1}(c_w|_{T,T})\alpha = c_{\widetilde w}|_{T,T}$. It follows that $w^{\widebar \alpha} \in \widetilde w Z(GL_n^{\varepsilon}(q)) \subseteq W Z(GL_n^{\varepsilon}(q))$. This implies $w^{\widebar \alpha} \in W$ since $W$ is the unique Sylow $2$-subgroup of $W Z(GL_n^{\varepsilon}(q))$. In particular, $\widebar \alpha$ induces an automorphism of $W$.

\begin{equation*}
d_i := \begin{pmatrix} I_{2^{r_1}} & & & & \\ & \ddots & & & \\ & & -I_{2^{r_i}} & & \\ & & & \ddots & \\ & & & & I_{2^{r_t}} \end{pmatrix}
\end{equation*}

for each $1 \le i \le t$. Then $d_i$ is a central involution of $W$ for each $1 \le i \le t$. So we have that $(d_i)^{\widebar \alpha} = (d_i)^m$ is a central involution of $W$ for each $1 \le i \le t$. As we see from Lemma <ref>, this already implies that $(d_i)^m = d_i$ for each $1 \le i \le t$. So there is some $m_i \in GL_{2^{r_i}}^{\varepsilon}(q)$ for each $1 \le i \le t$ such that

\begin{equation*}
m = \begin{pmatrix}
m_1 & & \\ & \ddots & \\ & & m_t
\end{pmatrix}.
\end{equation*}

\begin{equation*}
W_r \rightarrow W_r, (a_{ij}) \mapsto ((a_{ij})^{p^{\ell}})^{m_i}
\end{equation*}

is an automorphism of $W_r$ for each $1 \le r \le t$. Applying Lemma <ref>, we conclude that $\widebar{\alpha}|_{W,W}$ is a $2$-automorphism of $W$. Since $\alpha^{-1}(c_w|_{T,T})\alpha = c_{w^{\widebar \alpha}}|_{T,T}$ for all $w \in W$, it follows that $c_{\alpha}|_{S,S}$ is a $2$-automorphism of $S$, as required.

Let $q$ be a nontrivial odd prime power, $\varepsilon \in \lbrace +,- \rbrace$, and let $n \ge 3$ be a natural number. Let $T := PSL_n^{\varepsilon}(q)$, and let $S$ be a Sylow $2$-subgroup of $\mathrm{Inndiag}(T)$. Then $\mathrm{Aut}_{P \Gamma L_n^{\varepsilon}(q)}(S)$ is a $2$-group.

Let $q$ be a nontrivial odd prime power, $\varepsilon \in \lbrace +,- \rbrace$, and $n \ge 3$ be a natural number. Let $S$ be a Sylow $2$-subgroup of $SL_n^{\varepsilon}(q)Z(GL_n^{\varepsilon}(q))/Z(GL_n^{\varepsilon}(q))$, and let $S_1$ be a Sylow $2$-subgroup of $PGL_n^{\varepsilon}(q)$ containing $S$. Then $N_{PGL_n^{\varepsilon}(q)}(S) = N_{PGL_n^{\varepsilon}(q)}(S_1)$.

Let $T_1$ be a Sylow $2$-subgroup of $GL_n^{\varepsilon}(q))$ such that $S_1 = T_1 Z(GL_n^{\varepsilon}(q))/Z(GL_n^{\varepsilon}(q))$. Let $T := T_1 \cap SL_n^{\varepsilon}(q)$. Clearly, we have $S = T Z(GL_n^{\varepsilon}(q))/Z(GL_n^{\varepsilon}(q))$. It is rather easy to show $N_{PGL_n^{\varepsilon}(q)}(S) = N_{GL_n^{\varepsilon}(q)}(T)Z(GL_n^{\varepsilon}(q))/Z(GL_n^{\varepsilon}(q))$. By <cit.>, $N_{GL_n^{\varepsilon}(q)}(T) = T_1 C_{GL_n^{\varepsilon}(q)}(T_1) \le N_{GL_n^{\varepsilon}(q)}(T_1)$. It follows that $N_{PGL_n^{\varepsilon}(q)}(S) \le N_{PGL_n^{\varepsilon}(q)}(S_1)$. It is clear that we also have $N_{PGL_n^{\varepsilon}(q)}(S_1) \le N_{PGL_n^{\varepsilon}(q)}(S)$.

Let $q$ be a nontrivial odd prime power, $\varepsilon \in \lbrace +,- \rbrace$, and let $n \ge 3$ be a natural number. Let $T := PSL_n^{\varepsilon}(q)$, let $S$ be a Sylow $2$-subgroup of $\mathrm{Inn}(T)$, and let $S_1$ be a Sylow $2$-subgroup of $\mathrm{Inndiag}(T)$ containing $S$. Then $N_{\mathrm{Inndiag}(T)}(S) = N_{\mathrm{Inndiag}(T)}(S_1)$.

We are now ready to prove Lemma <ref>.

Assume that $n = 2$ and $q \equiv 3$ or $5 \mod 8$. Then $S \cong C_2 \times C_2$ by Lemma <ref>. There is only one non-nilpotent fusion system on $S$. Since $T$ and $A$ are not $2$-nilpotent, we have that $\mathcal{F}_S(T)$ and $\mathcal{F}_S(A)$ are not nilpotent (see <cit.>). It follows that $\mathcal{F}_S(T) = \mathcal{F}_S(A)$.

From now on, we assume that either $n \ge 3$, or $n = 2$ and $q \equiv 1$ or $7 \mod 8$. Let $P, Q \le S$ and $a \in A$ such that $P^a \le Q$. We are going to show that $c_a|_{P,Q}$ is a morphism in $\mathcal{F}_S(T)$. By the Frattini argument, we have $a = wu$ for some $w \in N_A(S)$ and some $u \in T$. We prove that $c_w|_{S,S} \in \mathrm{Inn}(S)$. This clearly implies that $c_a|_{P,Q}$ is a morphism in $\mathcal{F}_S(T)$.

Suppose that $n = 2$. Then $S$ is dihedral of order at least $8$ by Lemma <ref>, and so $\mathrm{Aut}(S)$ is a $2$-group by <cit.>. This implies that $\mathrm{Aut}_A(S) = \mathrm{Inn}(S)$, whence $c_w|_{S,S} \in \mathrm{Inn}(S)$.

Suppose now that $n \ge 3$. Let $S_1$ be a Sylow $2$-subgroup of $\mathrm{Inndiag}(PSL_n^{\varepsilon}(q))$ containing $S$. Since $T$ has odd index in $A$, we have that $A \le P \Gamma L_n^{\varepsilon}(q)$. By the Frattini argument, $w = w_1w_2$ for some $w_1 \in N_{P \Gamma L_n^{\varepsilon} (q)}(S_1)$ and some $w_2 \in \mathrm{Inndiag}(PSL_n^{\varepsilon}(q))$. Since $w_1$ normalizes both $S_1$ and $T$, we have that $w_1$ normalizes $S$. And since $w = w_1w_2$ normalizes $S$, we also have that $w_2$ normalizes $S$. So $w_2$ normalizes $S_1$ by Corollary <ref>. Consequently, $w = w_1 w_2 \in N_{P \Gamma L_n^{\varepsilon} (q)}(S_1)$. By Corollary <ref>, $c_w|_{S_1,S_1}$ is a $2$-automorphism of $S_1$. So $c_w|_{S,S}$ is a $2$-automorphism of $S$. Since $S \in \mathrm{Syl}_2(A)$ and $w \in A$, this implies that $c_w|_{S,S} \in \mathrm{Inn}(S)$, as required.

§ THE CASE $N \LE 5$

In this section, we verify Theorem <ref> for the case $n \le 5$.

Let $q$ be a nontrivial odd prime power, and let $G$ be a finite simple group. Then the following are equivalent:

(i) the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_2(q)$;

(ii) the Sylow $2$-subgroups of $G$ are isomorphic to those of $PSL_2(q)$;

(iii) $G \cong PSL_2^{\varepsilon}(q^{*})$ for some $\varepsilon \in \lbrace +,- \rbrace$ and some odd prime power $q^{*} \ge 5$ with $\varepsilon q^{*} \sim q$, or $\vert PSL_2(q) \vert_2 = 8$ and $G \cong A_7$.

In particular, Theorem <ref> holds for $n = 2$.

The implication (i) $\Rightarrow$ (ii) is clear.

(ii) $\Rightarrow$ (iii): Assume that the Sylow $2$-subgroups of $G$ are isomorphic to those of $PSL_2(q)$. Hence, $G$ has dihedral Sylow $2$-subgroups of order $\frac{1}{2}(q-1)_2(q+1)_2$. Applying a result of Gorenstein and Walter <cit.>, we may conclude that $G \cong PSL_2(q^{*})$ for some odd prime power $q^{*} \ge 5$, or $G \cong A_7$. Suppose that the former holds. Then $(q^{*}-1)_2(q^{*}+1)_2 = 2 \vert G \vert_2 = (q-1)_2(q+1)_2$, whence either $q^{*} \sim q$ or $-q^{*} \sim q$. Since $PSL_2(q^{*}) \cong PSU_2(q^{*})$, this implies that the first statement in (iii) is satisfied. If $G \cong A_7$, then $\vert PSL_2(q) \vert_2 = \vert G \vert_2 = 8$, so that the second statement in (iii) is satisfied.

(iii) $\Rightarrow$ (i): Assume that (iii) holds. Set $G_1 := G$ and $G_2 := PSL_2(q)$. For $i \in \lbrace 1,2 \rbrace$, let $S_i \in \mathrm{Syl}_2(G_i)$ and $\mathcal{F}_i := \mathcal{F}_{S_i}(G_i)$. Clearly, $S_1$ and $S_2$ are dihedral groups of the same order. Let $i \in \lbrace 1,2 \rbrace$. By <cit.>, any subgroup of $S_i$ is cyclic or dihedral. By <cit.>, a dihedral subgroup of $S_i$ with order greater than $4$ cannot be $\mathcal{F}_i$-essential. Since the automorphism group of a finite cyclic $2$-group is itself a $2$-group, a cyclic subgroup of $S_i$ cannot be $\mathcal{F}_i$-essential either. So we have that any $\mathcal{F}_i$-essential subgroup of $S_i$ is a Klein four group. Alperin’s fusion theorem <cit.> implies that

\begin{equation*}
\mathcal{F}_i = \langle \mathrm{Aut}_{\mathcal{F}_i}(P) \ \vert \ P \le S_i, P \cong C_2 \times C_2 \ \textnormal{or} \ P = S_i \rangle_{S_i}.
\end{equation*}

If $\vert S_i \vert = 4$, then $\mathrm{Aut}_{\mathcal{F}_i}(S_i)$ is the unique subgroup of $\mathrm{Aut}(S_i)$ with order $3$, because otherwise $\mathrm{Aut}_{\mathcal{F}_i}(S_i) = \mathrm{Inn}(S_i)$, so that <cit.> would imply that $G_i$ is $2$-nilpotent. If $\vert S_i \vert \ge 8$, then $\mathrm{Aut}_{\mathcal{F}_i}(S_i) = \mathrm{Inn}(S_i)$ since $\mathrm{Aut}(S_i)$ is a $2$-group by <cit.>, and for any Klein four subgroup $P$ of $S_i$, we have $\mathrm{Aut}_{\mathcal{F}_i}(P) = \mathrm{Aut}(P)$ by <cit.>. As $S_1 \cong S_2$ and as the preceding observations do not depend on whether $i$ is $1$ or $2$, we may conclude that $\mathcal{F}_1 \cong \mathcal{F}_2$, as required.

Let $q$ be a nontrivial odd prime power, and let $G$ be a finite simple group. Then the following are equivalent:

(i) the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_3(q)$;

(ii) the Sylow $2$-subgroups of $G$ are isomorphic to those of $PSL_3(q)$;

(iii) $G \cong PSL_3^{\varepsilon}(q^{*})$ for some $\varepsilon \in \lbrace +,- \rbrace$ and some nontrivial odd prime power $q^{*}$ with $\varepsilon q^{*} \sim q$, or $(q+1)_2 = 4$ and $G \cong M_{11}$.

In particular, Theorem <ref> holds for $n = 3$.

The implication (i) $\Rightarrow$ (ii) is clear.

(ii) $\Rightarrow$ (iii): Assume that the Sylow $2$-subgroups of $G$ are isomorphic to those of $PSL_3(q)$. Hence, a Sylow $2$-subgroup of $G$ is wreathed (i.e. isomorphic to $C_{2^k} \wr C_2$ for some positive integer $k$) if $q \equiv 1 \mod 4$, and semidihedral if $q \equiv 3 \mod 4$. Applying work of Alperin, Brauer and Gorenstein, namely <cit.> and <cit.>, we may conclude that either $G \cong PSL_3^{\varepsilon}(q^{*})$ for some $\varepsilon \in \lbrace +,- \rbrace$ and some nontrivial odd prime power $q^{*}$ with $\varepsilon q^{*} \equiv q \mod 4$, or $q \equiv 3 \mod 4$ and $G \cong M_{11}$. If the former holds, then $((q^{*}-\varepsilon)_2)^2(q^{*}+\varepsilon)_2 = \vert G \vert_2 = ((q-1)_2)^2(q+1)_2$, and it easily follows that $\varepsilon q^{*} \sim q$. If $G \cong M_{11}$, then $16 = \vert G \vert_2 = ((q-1)_2)^2(q+1)_2$ and hence $(q+1)_2 = 4$.

(iii) $\Rightarrow$ (i): Assume that (iii) holds. If $q \equiv 1 \mod 4$, then Proposition <ref> implies that the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_3(q)$. Alternatively, this can be seen from <cit.>. Now suppose that $q \equiv 3 \mod 4$. If $(q+1)_2 \ne 4$, then we could apply Proposition <ref> again, but we are going to argue in a more elementary way. Let $G_1 := G$ and $G_2 := PSL_3(q)$. For $i \in \lbrace 1,2 \rbrace$, let $S_i \in \mathrm{Syl}_2(G_i)$ and $\mathcal{F}_i := \mathcal{F}_{S_i}(G_i)$. Clearly, $S_1$ and $S_2$ are semidihedral groups of the same order. Let $i \in \lbrace 1,2 \rbrace$. By <cit.>, any proper subgroup of $S_i$ is cyclic, dihedral or generalized quaternion. By <cit.>, dihedral subgroups of $S_i$ with order greater than $4$ and generalized quaternion subgroups of $S_i$ with order greater than $8$ cannot be $\mathcal{F}_i$-essential. Since the automorphism group of a finite cyclic $2$-group is itself a $2$-group, a cyclic subgroup of $S_i$ cannot be $\mathcal{F}_i$-essential either. Alperin’s fusion theorem <cit.> implies that

\begin{equation*}
\mathcal{F}_i = \langle \mathrm{Aut}_{\mathcal{F}_i}(P) \ \vert \ P \cong C_2 \times C_2, P \cong Q_8, \ \textnormal{or} \ P = S_i \rangle_{S_i}.
\end{equation*}

Since $\mathrm{Aut}(S_i)$ is a $2$-group by <cit.>, we have $\mathrm{Aut}_{\mathcal{F}_i}(S_i) = \mathrm{Inn}(S_i)$. From <cit.>, one can see that $\mathrm{Aut}_{\mathcal{F}_i}(P) = \mathrm{Aut}(P)$ for any subgroup $P$ of $S_i$ isomorphic to $C_2 \times C_2$ or $Q_8$. As $S_1 \cong S_2$ and as the preceding observations do not depend on whether $i$ is $1$ or $2$, we may conclude that $\mathcal{F}_1 \cong \mathcal{F}_2$, as required.

The next two lemmas are required to verify Theorem <ref> for the case $n = 4$.

Let $q$ be an odd prime power with $q \equiv 3 \mod 8$. Assume that $G=A_{10}$ or $A_{11}$. Then the $2$-fusion system of $G$ is not isomorphic to the $2$-fusion system of $PSL_4(q)$.

Set $x := (1 \ 2)(3 \ 4) \in G$ and $y := (1 \ 2)(3 \ 4)(5 \ 6)(7 \ 8) \in G$. Let $g \in G$ be an involution. Then the cycle type of $g$ is either that of $x$ or that of $y$. So, by <cit.>, $g$ is conjugate to $x$ or $y$ in the ambient symmetric group, which easily implies that $g$ is also $G$-conjugate to $x$ or $y$. The involutions $x$ and $y$ are not $G$-conjugate as they have different cycle types. It follows that $G$ has precisely two conjugacy classes of involutions with representatives $x$ and $y$.

By a direct calculation,

\begin{equation*}
S := \langle (1 \ 2 \ 3 \ 4)(9 \ 10),(1 \ 2)(3 \ 4),(5 \ 6 \ 7 \ 8)(9 \ 10),(5 \ 6)(7 \ 8),(1 \ 5)(2 \ 6)(3 \ 7)(4 \ 8) \rangle
\end{equation*}

is a Sylow 2-subgroup of $G$. Another calculation confirms that $S$ has precisely 14 involutions whose cycle type is that of $x$ and precisely 29 involutions whose cycle type is that of $y$. So there are precisely two $\mathcal{F}_S(G)$-conjugacy classes of involutions, one of which has 14 elements, while the other one has 29 elements. In order to prove that $\mathcal{F}_S(G)$ is not isomorphic to the 2-fusion system of $PSL_4(q)$, we show that the $2$-fusion system of $PSL_4(q)$ has a conjugacy class of involutions with precisely $17$ elements.

Let $W_1$ be a Sylow 2-subgroup 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>. Let $W := W_2 \cap SL_4(q) \in \mathrm{Syl}_2(SL_4(q))$, and let $R$ be the image of $W$ in $PSL_4(q)$. The involutions of $W_2$ are precisely the elements

\begin{equation*}
\begin{pmatrix}
a \ & \ \\
\ & \ b
\end{pmatrix} \ \ \textnormal{and} \ \
\begin{pmatrix}
\ & \ c \\
c^{-1} \ & \
\end{pmatrix},
\end{equation*}

where $a, b, c \in W_1$ and $\mathrm{max}\lbrace \mathrm{ord}(a), \mathrm{ord}(b) \rbrace = 2$. Bearing in mind that $W_1$ is semidihedral of order $16$, which holds because of $q \equiv 3 \mod 8$, we may see from Lemma <ref> that $W$ has precisely 35 involutions. As one of them is $-I_4$, and as the product of $-I_4$ with an involution of $W$ different from $-I_4$ is again an involution, we may conclude that $R$ has precisely 17 involutions that are images of involutions of $W$. Since any noncentral involution of $SL_4(q)$ is $SL_4(q)$-conjugate to a diagonal matrix having diagonal entries $1$,$1$,$-1$,$-1$, we have that all the noncentral involutions of $SL_4(q)$ are $SL_4(q)$-conjugate. Thus the 17 involutions of $R$ induced by involutions of $W$ are $PSL_4(q)$-conjugate. As an element of $PSL_4(q)$ induced by an involution cannot be conjugate to an element of $PSL_4(q)$ not induced by an involution, it follows that there is an $\mathcal{F}_R(PSL_4(q))$-conjugacy class of involutions with precisely 17 elements.

Let $q$ be an odd prime power with $q \equiv 5 \mod 8$. Assume that $G=M_{22}$, $M_{23}$ or $McL$. Then the $2$-fusion system of $G$ is not isomorphic to the $2$-fusion system of $PSL_4(q)$.

Let $S \in \mathrm{Syl}_2(G)$ and $\mathcal{F} := \mathcal{F}_S(G)$. Let $x$ be an element of $S$ with order $4$ such that $\langle x \rangle$ is fully $\mathcal{F}$-centralized. In other words, we have $C_S(x) \in \mathrm{Syl}_2(C_G(x))$. If $G=M_{22}$ or $M_{23}$, then by [1], $C_G(x)$ is a $2$-group, whence $C_{\mathcal{F}}(\langle x \rangle) = \mathcal{F}_{C_S(x)}(C_G(x)) = \mathcal{F}_{C_S(x)}(C_S(x))$. If $G=McL$, then by [1], $G$ has precisely one conjugacy class of elements of order $4$, so that all elements of $S$ with order $4$ are $\mathcal{F}$-conjugate.

Consequently, we either have that $C_{\mathcal{F}}(\langle x \rangle)$ is nilpotent for all elements $x \in S$ with order 4 such that $\langle x \rangle$ is fully $\mathcal{F}$-centralized, or all elements of $S$ with order 4 are $\mathcal{F}$-conjugate. We are going to show that the $2$-fusion system of $PSL_4(q)$ has neither of these properties.

Let $\lambda$ be an element of $\mathbb{F}_{q}^{*}$ of order $4$ and let $y$ be the image of $\mathrm{diag}(1,1,\lambda,\lambda^{-1})$ in $PSL_4(q)$. Clearly, $y$ has order $4$. Let $R$ be a Sylow 2-subgroup of $PSL_4(q)$ containing a Sylow $2$-subgroup of $C := C_{PSL_4(q)}(y)$. Clearly, $y \in R$. Let us denote $\mathcal{F}_R(PSL_4(q))$ by $\mathcal{G}$. Then $\langle y \rangle$ is fully $\mathcal{G}$-centralized. The centralizer $C$ is not $2$-nilpotent since it has a subgroup isomorphic to $SL_2(q)$. So, by <cit.>, $C_{\mathcal{G}}(\langle y \rangle) = \mathcal{F}_{C_R(y)}(C)$ is not nilpotent.

Let $m$ denote the matrix

\begin{equation*}
\begin{pmatrix}
0 & \lambda & \ & \ \\
1 & 0 & \ & \ \\
\ & \ & 0 & -\lambda \\
\ & \ & 1 & 0
\end{pmatrix} \in SL(4,q).
\end{equation*}

A direct calculation, using $q \equiv 5 \ \mathrm{mod} \ 8$, shows that $m$ has no eigenvalues, whence $m$ is in particular not diagonalizable. The image of $m$ in $PSL_4(q)$ has order 4, but it is not $PSL_4(q)$-conjugate to $y$. Therefore, $PSL_4(q)$ has more than one conjugacy class of elements with order $4$. Thus there is more than one $\mathcal{G}$-conjugacy class of elements with order $4$.

Let $q$ be a nontrivial odd prime power and let $G$ be a finite simple group. Then the following are equivalent:

(i) the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_4(q)$;

(ii) $G \cong PSL_4^{\varepsilon}(q^{*})$ for some $\varepsilon \in \lbrace +,- \rbrace$ and some nontrivial odd prime power $q^{*}$ with $\varepsilon q^{*} \sim q$.

In particular, Theorem <ref> holds for $n = 4$.

The implication (ii) $\Rightarrow$ (i) is given by Proposition <ref>.

(i) $\Rightarrow$ (ii): Assume that the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_4(q)$. Then the Sylow $2$-subgroups of $G$ are isomorphic to those of $PSL_4(q)$. Applying Mason’s results <cit.> and <cit.>, the latter together with <cit.>, we see that one of the following holds:

(1) $G \cong PSL_4^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $\varepsilon q^{*} \equiv q \mod 4$;

(2) $G \cong A_{10}$ or $A_{11}$, and $q \equiv 3 \ \mathrm{mod} \ 4$;

(3) $G \cong M_{22}$, $M_{23}$ or $McL$, and $q \equiv 5 \ \mathrm{mod} \ 8$.

Let $q_0$ be a nontrivial odd prime power, $\varepsilon_0 \in \lbrace +,- \rbrace$, and $k_0, s_0 \in \mathbb{N}$ such that $2^{k_0} = (q_0 - \varepsilon_0)_2$ and $2^{s_0} = (q_0 + \varepsilon_0)_2$. Then we have

\begin{equation*}
\vert PSL_4^{\varepsilon_0}(q_0) \vert_2 = \frac{\vert GL_4^{\varepsilon_0}(q_0) \vert_2}{2^{k_0}(4,2^{k_0})} = \frac{2 (\vert GL_2^{\varepsilon_0}(q_0) \vert_2)^2}{2^{k_0}(4,2^{k_0})} = \frac{2^{3k_0 + 2s_0 + 1}}{(4,2^{k_0})}.
\end{equation*}

Let $k, s \in \mathbb{N}$ such that $2^k = (q-1)_2$ and $2^s = (q+1)_2$.

Suppose that (1) holds, and let $k^{*}, s^{*} \in \mathbb{N}$ such that $2^{k^{*}} = (q^{*}-\varepsilon)_2$ and $2^{s^{*}} = (q^{*} + \varepsilon)_2$. Then we have

\begin{equation*}
\frac{2^{3k^{*} + 2s^{*} + 1}}{(4,2^{k^{*}})} = \vert G \vert_2 = \frac{2^{3k + 2s + 1}}{(4,2^{k})}.
\end{equation*}

Since $\varepsilon q^{*} \equiv q \mod 4$, this easily implies $\varepsilon q^{*} \sim q$.

Suppose that (2) holds. Then $2^7 = \vert G \vert_2 = 2^{3 + 2s}$, whence $s = 2$ and thus $q \equiv 3 \mod 8$. This is a contradiction to Lemma <ref>. So (2) does not hold.

Also, (3) cannot hold because of Lemma <ref>.

Let $q$ be a nontrivial odd prime power, and let $G$ be a finite simple group. Then the following are equivalent:

(i) the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_5(q)$;

(ii) the Sylow $2$-subgroups of $G$ are isomorphic to those of $PSL_5(q)$;

(iii) $G \cong PSL_5^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $\varepsilon q^{*} \sim q$.

In particular, Theorem <ref> holds for $n = 5$.

The implication (i) $\Rightarrow$ (ii) is clear, and the implication (iii) $\Rightarrow$ (i) is given by Proposition <ref>.

(ii) $\Rightarrow$ (iii): Assume that the Sylow $2$-subgroups of $G$ are isomorphic to those of $PSL_5(q)$. Applying work of Mason <cit.>, it follows that $G \cong PSL_5^{\varepsilon}(q^{*})$ for some $\varepsilon \in \lbrace +,- \rbrace$ and some nontrivial odd prime power $q^{*}$. In view of Lemma <ref>, it is easy to see that a Sylow $2$-subgroup of $G$ is isomorphic to a Sylow $2$-subgroup of $GL_4^{\varepsilon}(q^{*})$, while a Sylow $2$-subgroup of $PSL_5(q)$ is isomorphic to a Sylow $2$-subgroup of $GL_4(q)$. Now it is easy to deduce from Lemmas <ref>, <ref> and <ref> that a Sylow $2$-subgroup of $G$ has a center of order $(q^{*}-\varepsilon)_2$, while a Sylow $2$-subgroup of $PSL_5(q)$ has a center of order $(q-1)_2$. It follows that $(q^{*}-\varepsilon)_2 = (q-1)_2$. Let $k,s,k^{*},s^{*} \in \mathbb{N}$ with $2^k = (q-1)_2, 2^s = (q+1)_2, 2^{k^{*}} = (q^{*} - \varepsilon)_2$ and $2^{s^{*}} = (q^{*} + \varepsilon)_2$. Then

\begin{equation*}
2^{4k^{*} + 2s^{*} + 1} = \vert GL_4^{\varepsilon}(q^{*}) \vert_2 = \vert G \vert_2 = \vert GL_4(q) \vert_2 = 2^{4k + 2s + 1}.
\end{equation*}

Since $2^{k^{*}} = 2^k$, we thus have $k = k^{*}$ and $s = s^{*}$. This implies $\varepsilon q^{*} \sim q$.

§ THE CASE $N \GE 6$: PRELIMINARY DISCUSSION AND NOTATION

Given a natural number $k \ge 6$, we say that $P(k)$ is satisfied if whenever $q_0$ is a nontrivial odd prime power and $H$ is a finite simple group satisfying (<ref>) and realizing the $2$-fusion system of $PSL_k(q_0)$, we have $H \cong PSL_k^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $\varepsilon q^{*} \sim q_0$.

In order to establish Theorem <ref> for $n \ge 6$, we are going to prove by induction that $P(k)$ is satisfied for all $k \ge 6$. From now on until the end of Section <ref>, we will assume the following hypothesis.

Let $n \ge 6$ be a natural number such that $P(k)$ is satisfied for all natural numbers $k$ with $6 \le k < n$, and let $q$ be a nontrivial odd prime power. Moreover, let $G$ be a finite group satisfying the following properties:

(i) $G$ realizes the $2$-fusion system of $PSL_n(q)$;

(ii) $O(G) = 1$;

(iii) $G$ satisfies (<ref>).

We will prove the following theorem.

There is a normal subgroup $G_0$ of $G$ isomorphic to a nontrivial quotient of $SL_n^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +, - \rbrace$ with $\varepsilon q^{*} \sim q$. In particular, $P(n)$ is satisfied.

The proof of Theorem <ref> will occupy Sections <ref>-<ref>. In this section, we introduce some notation and prove some preliminary results needed for the proof.

For each $A \subseteq \lbrace 1, \dots, n \rbrace$ of even order, let $t_A$ be the image of the diagonal matrix $\mathrm{diag}(d_1,\dots,d_n)$ in $PSL_n(q)$, where

\begin{equation*}
d_i = \begin{cases}
-1 & \, \text{if $i \in A$} \\
1 & \, \text{if $i \not\in A$}
\end{cases}
\end{equation*}

for each $1 \le i \le n$. If $i$ is an even natural number with $2 \le i < n$ and $A = \lbrace n-i+1,\dots,n \rbrace$, then we write $t_i$ for $t_A$. We denote $t_2$ by $t$, and we write $u$ for $t_{\lbrace 1,2 \rbrace}$.

We assume $\rho$ to be an element of $\mathbb{F}_q^{*}$ of order $(n,q-1)$. If $\rho$ is a square in $\mathbb{F}_q$, then we assume $\mu$ to be a fixed element of $\mathbb{F}_q$ with $\rho = \mu^2$.

If $n$ is even, $\rho$ is a square in $\mathbb{F}_q$, and $i$ is an odd natural number with $1 \le i < n$, then

\begin{equation*}
\begin{pmatrix} \mu I_{n-i} & \\ & -\mu I_i \end{pmatrix}
\end{equation*}

is an element of $SL_n(q)$ by Proposition <ref>, and we will denote its image in $PSL_n(q)$ by $t_i$.

If $n$ is even and $\rho$ is a non-square element of $\mathbb{F}_q$, then we denote the matrix

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

by $\widetilde w$, and if $\widetilde w \in SL_n(q)$, then we use $w$ to denote its image in $PSL_n(q)$.

Note that, by Proposition <ref>, any involution of $PSL_n(q)$ is conjugate to $t_i$ for some $1 \le i < n$ such that $t_i$ is defined, or to $w$ (if defined).

Next, we construct a Sylow $2$-subgroup of $C_{PSL_n(q)}(t)$ containing some “nice” elements of $PSL_n(q)$. Take a Sylow $2$-subgroup $V$ of $GL_2(q)$ containing each diagonal matrix in $GL_2(q)$ with $2$-elements of $\mathbb{F}_q^{*}$ along the main diagonal. Similarly, we assume $V_2$ to be a Sylow $2$-subgroup of $GL_{n-4}(q)$ containing each diagonal matrix in $GL_{n-4}(q)$ with $2$-elements of $\mathbb{F}_q^{*}$ along the main diagonal. Now let $W$ be a Sylow $2$-subgroup of $GL_{n-2}(q)$ containing

\begin{equation*}
\left \lbrace \begin{pmatrix} A & \\ & B \end{pmatrix} \ : \ A \in V, B \in V_2 \right \rbrace.
\end{equation*}

If $n = 6$, then we assume that $V = V_2$ and that $W$ is the Sylow $2$-subgroup

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

of $GL_4(q)$.

Let $\widetilde t := \mathrm{diag}(1,\dots,1,-1,-1) \in SL_n(q)$. Then we have

\begin{equation*}
C_{SL_n(q)}(\widetilde t) = \left \lbrace \begin{pmatrix} A & \\ & B \end{pmatrix} \ : \ A \in GL_{n-2}(q), B \in GL_2(q), \mathrm{det}(A)\mathrm{det}(B) = 1  \right \rbrace.
\end{equation*}

It is easy to note that

\begin{equation*}
\widetilde T := \left \lbrace \begin{pmatrix} A & \\ & B \end{pmatrix} \ : \ A \in W, B \in V, \mathrm{det}(A)\mathrm{det}(B) = 1  \right \rbrace
\end{equation*}

is a Sylow $2$-subgroup of $C_{SL_n(q)}(\widetilde t)$. Let $T$ denote the image of $\widetilde T$ in $PSL_n(q)$. As the centralizer of $t$ in $PSL_n(q)$ is the image of $C_{SL_n(q)}(\widetilde t)$ in $PSL_n(q)$, we have that $T$ is a Sylow $2$-subgroup of $C_{PSL_n(q)}(t)$. We assume $S$ to be a Sylow $2$-subgroup of $PSL_n(q)$ containing $T$. Since $C_S(t) = T \in \mathrm{Syl_2}(C_{PSL_n(q)}(t))$, we have that $\langle t \rangle$ is fully $\mathcal{F}_S(PSL_n(q))$-centralized.

Let $K_1$ be the image of

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

in $PSL_n(q)$, and let $K_2$ be the image of

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

in $PSL_n(q)$. Clearly, $K_1$ and $K_2$ are normal subgroups of $C_{PSL_n(q)}(t)$ isomorphic to $SL_{n-2}(q)$ and $SL_2(q)$, respectively. Define $X_1$ to be the image of

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

in $PSL_n(q)$, and define $X_2$ to be the image of

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

in $PSL_n(q)$.

Note that $X_1 = T \cap K_1 \in \mathrm{Syl}_2(K_1)$ and $X_2 = T \cap K_2 \in \mathrm{Syl}_2(K_2)$. Define

\begin{equation*}
\mathcal{C}_i := \mathcal{F}_{X_i}(K_i)
\end{equation*}

for $i \in \lbrace 1,2 \rbrace$. By <cit.>, $\mathcal{C}_1$ and $\mathcal{C}_2$ are normal subsystems of $\mathcal{F}_T(C_{PSL_n(q)}(t))$.

Let $\mathcal{F} := \mathcal{F}_S(PSL_n(q))$.
If $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$, then the components of $C_{\mathcal{F}}(\langle t \rangle)$ are precisely the subsystems $\mathcal{C}_1$ and $\mathcal{C}_2$. If $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$, then $\mathcal{C}_1$ is the only component of $C_{\mathcal{F}}(\langle t \rangle)$.

Set $C := C_{PSL_n(q)}(t)$. It is easy to note that the $2$-components of $C$ are precisely the quasisimple elements of $\lbrace K_1, K_2 \rbrace$. As $n \ge 6$ and as $K_1 \cong SL_{n-2}(q)$ and $K_2 \cong SL_2(q)$, it follows that the $2$-components of $C$ are $K_1$ and $K_2$ if $q \ne 3$, and that $K_1$ is the only $2$-component of $C$ if $q = 3$.

By Lemma <ref>, $K_1/Z(K_1)$ is not a Goldschmidt group. If $q \ne 3$, then the lemma just cited also shows that $K_2/Z(K_2)$ is a Goldschmidt group if and only if $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$.

Now we apply Proposition <ref> to conclude that $\mathcal{F}_{T \cap K_1}(K_1)$ and $\mathcal{F}_{T \cap K_2}(K_2)$ are precisely the components of $\mathcal{F}_T(C)$ if $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$, and that $\mathcal{F}_{T \cap K_1}(K_1)$ is the only component of $\mathcal{F}_T(C)$ if $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$. This completes the proof because $C_{\mathcal{F}}(\langle t \rangle) = \mathcal{F}_T(C)$, $\mathcal{C}_1 = \mathcal{F}_{T \cap K_1}(K_1)$ and $\mathcal{C}_2 = \mathcal{F}_{T \cap K_2}(K_2)$.

Let $\mathcal{F} := \mathcal{F}_S(PSL_n(q))$.
Then the factor system $C_{\mathcal{F}}(\langle t \rangle)/X_1X_2$ is nilpotent.

Set $C := C_{PSL_n(q)}(t)$. It is easy to note that $X_1 X_2 = K_1 K_2 \cap T$. By Lemma <ref>, $C_{\mathcal{F}}(\langle t \rangle)/X_1X_2$ is isomorphic to the $2$-fusion system of $C/K_1K_2$. The factor group $C/K_1K_2$ is abelian. This easily implies that $C/K_1K_2$ has a nilpotent $2$-fusion system. Hence $C_{\mathcal{F}}(\langle t \rangle)/X_1X_2$ is nilpotent.

Let $A \in W$ and $B \in V$ such that $\mathrm{det}(A)\mathrm{det}(B) = 1$. Let

\begin{equation*}
m := \begin{pmatrix}A & \\ & B \end{pmatrix}Z(SL_n(q)) \in T.
\end{equation*}

Then we have $m \in Z(\mathcal{C}_1 \langle m \rangle)$ if and only if $A \in Z(GL_{n-2}(q))$.

By <cit.>, we have $\mathcal{C}_1 \langle m \rangle = \mathcal{F}_{X_1 \langle m \rangle}(K_1 \langle m \rangle)$.

If $A \in Z(GL_{n-2}(q))$, then $m$ is central in $K_1\langle m \rangle$, which implies that $m$ lies in the center of $\mathcal{C}_1 \langle m \rangle$.

We show now that if $A \not\in Z(GL_{n-2}(q))$, then $m \not\in Z(\mathcal{C}_1 \langle m \rangle)$. Assume to the contrary that $A \not\in Z(GL_{n-2}(q))$, but $m \in Z(\mathcal{C}_1 \langle m \rangle)$. Clearly, $m \in Z(X_1 \langle m \rangle)$. So $m$ centralizes $X_1$. It easily follows that $A$ centralizes $W \cap SL_{n-2}(q)$. Using Sylow’s theorem, we may see from Lemma <ref> that any element $A_0$ of $W$ which centralizes $W \cap SL_{n-2}(q)$ without being central in $GL_{n-2}(q)$ is $SL_{n-2}(q)$-conjugate to an element of $W$ different from $A_0$. As $A$ centralizes $W \cap SL_{n-2}(q)$, but $A \not\in Z(GL_{n-2}(q))$, it follows that $A$ is $SL_{n-2}(q)$-conjugate to an element $A' \in W$ with $A \ne A'$. As $\mathrm{det}(A) = \mathrm{det}(A')$, we have $A' = A''A$ for some $A'' \in W \cap SL_{n-2}(q)$. Now, it follows that $m$ is $K_1$-conjugate to

\begin{equation*}
\begin{pmatrix} A' & \\ & B \end{pmatrix} Z(SL_n(q)) = \begin{pmatrix} A'' & \\ & I_2 \end{pmatrix} \begin{pmatrix}A & \\ & B \end{pmatrix} Z(SL_n(q)) \in X_1 \langle m \rangle.
\end{equation*}

So $m$ is $K_1$-conjugate to an element of $X_1 \langle m \rangle$ which is different from $m$. Therefore, $m \not\in Z(\mathcal{C}_1 \langle m \rangle)$, a contradiction.

Set $\mathcal{F} := \mathcal{F}_S(PSL_n(q))$ and $\mathcal{G} := C_{\mathcal{F}}(\langle t \rangle)$. Then $\mathfrak{hnp}(C_{\mathcal{G}}(X_1)) = X_2$.

Set $C := C_{PSL_n(q)}(t)$. Note that $C' = K_1 K_2$.

By <cit.>, we have $\mathfrak{foc}(C_{\mathcal{G}}(X_1)) = C_T(X_1) \cap C_C(X_1)' \le C_T(X_1) \cap C' = C_T(X_1) \cap X_1X_2 = Z(X_1)X_2$. As $\mathfrak{hnp}(C_{\mathcal{G}}(X_1)) \le \mathfrak{foc}(C_{\mathcal{G}}(X_1))$, it follows that $\mathfrak{hnp}(C_{\mathcal{G}}(X_1)) \le Z(X_1)X_2$.

Let $P$ be a subgroup of $C_T(X_1)$ and let $\varphi$ be a $2'$-element of $\mathrm{Aut}_{C_C(X_1)}(P)$. By <cit.>, we have

\begin{equation*}
[P,\langle \varphi \rangle] = [P, \langle \varphi \rangle, \langle \varphi \rangle] \le [\mathfrak{hnp}(C_{\mathcal{G}}(X_1)) \cap P,\langle \varphi \rangle] \le [Z(X_1)X_2 \cap P,\langle \varphi \rangle].
\end{equation*}

Since $\varphi \in \mathrm{Aut}_{C_C(X_1)}(P)$, $K_2 \trianglelefteq C$ and $X_2 = T \cap K_2$, it follows $[P,\langle \varphi \rangle] \le X_2$. Consequently, $\mathfrak{hnp}(C_{\mathcal{G}}(X_1)) \le X_2$.

On the other hand, since $K_2 \le O^2(C_C(X_1))$, we have $X_2 \le \mathfrak{hnp}(C_{\mathcal{G}}(X_1))$ by <cit.>.

Set $C := C_{PSL_n(q)}(t)$. Then $\mathrm{Aut}_{C}(X_1)$ is a $2$-group.

Let $m \in N_C(X_1)$. We have

\begin{equation*}
m = \begin{pmatrix} M_1 & \\ & M_2 \end{pmatrix} Z(SL_n(q))
\end{equation*}

for some $M_1 \in GL_{n-2}(q)$ and some $M_2 \in GL_2(q)$ with $\mathrm{det}(M_1)\mathrm{det}(M_2) = 1$. Let $A \in W \cap SL_{n-2}(q)$ and

\begin{equation*}
x := \begin{pmatrix} A & \\ & I_2 \end{pmatrix} Z(SL_n(q)) \in X_1.
\end{equation*}

As $m$ normalizes $X_1$, we have

\begin{equation*}
\begin{pmatrix} A^{M_1} & \\ & I_2 \end{pmatrix} Z(SL_n(q)) = x^m \in X_1.
\end{equation*}

This easily implies that $A^{M_1} \in W \cap SL_{n-2}(q)$. It follows that $M_1$ normalizes $W \cap SL_{n-2}(q)$. By <cit.>, we have $N_{GL_{n-2}(q)}(W \cap SL_{n-2}(q)) = W C_{GL_{n-2}(q)}(W)$. It follows that $c_m|_{X_1,X_1}$ is a $2$-automorphism.

Define $T_1$ to be the image of

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

in $PSL_n(q)$ and $T_2$ to be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} I_2 & & \\ & B & \\ & & I_2 \end{pmatrix} \ : \ B \in V_2 \cap SL_{n-4}(q) \right\rbrace
\end{equation*}

in $PSL_n(q)$. Clearly, $T_1$ and $T_2$ are subgroups of $X_1$. Recall that we use $u$ to denote $t_{\lbrace 1,2 \rbrace} \in X_1$. The following lemma sheds light on some properties of the centralizer fusion system $C_{\mathcal{C}_1}(\langle u \rangle)$.

The following hold.

(i) We have $C_{X_1}(u) \in \mathrm{Syl}_2(C_{K_1}(u))$. In particular, $\langle u \rangle$ is fully $\mathcal{C}_1$-centralized.

(ii) $\mathfrak{foc}(C_{\mathcal{C}_1}(\langle u \rangle)) = T_1 T_2$.

(iii) If $n = 6$ and $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$, then $T_1$ and $T_2$ are the only subgroups of $\mathfrak{foc}(C_{\mathcal{C}_1}(\langle u \rangle))$ which are isomorphic to $Q_8$ and strongly closed in $C_{\mathcal{C}_1}(\langle u \rangle)$.

(iv) If $n \ge 7$ and $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$, then $T_1$ is the only subgroup of the intersection $\mathfrak{foc}(C_{\mathcal{C}_1}(\langle u \rangle)) \cap C_{X_1}(T_2)$ which is isomorphic to $Q_8$ and strongly closed in $C_{\mathcal{C}_1}(\langle u \rangle)$.

(v) Let $C_1$ 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 $PSL_n(q)$ and $C_2$ be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} I_2 & &\\ & B & \\ & & I_2 \end{pmatrix} \ : \ A \in SL_{n-4}(q) \right\rbrace
\end{equation*}

in $PSL_n(q)$. Then any component of $C_{\mathcal{C}_1}(\langle u \rangle)$ lies in $\lbrace \mathcal{F}_{T_1}(C_1), \mathcal{F}_{T_2}(C_2) \rbrace$. Moreover, $\mathcal{F}_{T_1}(C_1)$ is a component if and only if $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$, and $\mathcal{F}_{T_2}(C_2)$ is a component if and only if $n \ge 7$ or $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$.

Clearly, $C_{K_1}(u)$ is the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} A & & \\ & B & \\ & & I_2 \end{pmatrix} \ : \ A \in GL_2(q), B \in GL_{n-4}(q), \mathrm{det}(A)\mathrm{det}(B) = 1 \right\rbrace
\end{equation*}

in $PSL_n(q)$. Let $\widetilde W$ be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} A & & \\ & B & \\ & & I_2 \end{pmatrix} \ : \ A \in V, B \in V_2, \mathrm{det}(A) \mathrm{det}(B) = 1 \right\rbrace
\end{equation*}

in $PSL_n(q)$. Clearly, we have $\widetilde W \le C_{X_1}(u)$. It is easy to note that $\widetilde W$ is a Sylow $2$-subgroup of $C_{K_1}(u)$. Thus $C_{X_1}(u) = \widetilde W \in \mathrm{Syl}_2(C_{K_1}(u))$. Hence (i) holds.

We have $C_{\mathcal{C}_1}(\langle u \rangle) = \mathcal{F}_{C_{X_1}(u)}(C_{K_1}(u)) = \mathcal{F}_{\widetilde W}(C_{K_1}(u))$. The focal subgroup theorem <cit.> implies that $\mathfrak{foc}(C_{\mathcal{C}_1}(\langle u \rangle)) = \widetilde W \cap (C_{K_1}(u))'$. It is easy to see that $(C_{K_1}(u))' = C_1C_2$, where $C_1$ and $C_2$ are as in (v). We thus have $\mathfrak{foc}(C_{\mathcal{C}_1}(\langle u \rangle)) = T_1T_2$. Hence (ii) holds.

Now we turn to the proofs of (iii) and (iv). Assume that $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$. Clearly, $C_1$ and $C_2$ are normal subgroups of $C_{K_1}(u)$ and we have $T_1 = C_1 \cap \widetilde W$, $T_2 = C_2 \cap \widetilde W$. This implies that $T_1$ and $T_2$ are strongly closed in $C_{\mathcal{C}_1}(\langle u \rangle)$. By Lemma <ref>, we have $T_1 \cong Q_8$ and, if $n = 6$, we also have $T_2 \cong Q_8$. Clearly, any strongly $C_{\mathcal{C}_1}(\langle u \rangle)$-closed subgroup of $\mathfrak{foc}(C_{\mathcal{C}_1}(\langle u \rangle)) = T_1T_2$ is strongly closed in $\mathcal{F}_{T_1T_2}(C_1C_2)$. Hence, in order to prove (iii), it suffices to show that if $n = 6$, then $T_1$ and $T_2$ are the only strongly $\mathcal{F}_{T_1T_2}(C_1C_2)$-closed subgroups of $T_1T_2$ which are isomorphic to $Q_8$. Similarly, in order to prove (iv), it suffices to show that if $n \ge 7$, then $T_1$ is the only subgroup of $T_1T_2$ which centralizes $T_2$, which is isomorphic to $Q_8$, and which is strongly closed in $\mathcal{F}_{T_1T_2}(C_1C_2)$.

Continue to assume that $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$. In order to prove the two statements just mentioned, we need some observations. As $C_1 \cong SL_2(q)$, we have that $C_1$ is not $2$-nilpotent. So $\mathcal{F}_{T_1}(C_1)$ is not nilpotent by <cit.>. Again by <cit.>, it follows that $\mathrm{Aut}_{C_1}(T_1)$ is not a $2$-group. So $\mathrm{Aut}_{C_1}(T_1)$ has an element of order $3$. Similarly, if $n = 6$, then $\mathrm{Aut}_{C_2}(T_2)$ has an element of order $3$. It follows that there is an element $\alpha \in \mathrm{Aut}_{C_1C_2}(T_1T_2)$ such that $\alpha|_{T_1,T_1}$ has order $3$, while $\alpha|_{T_2,T_2} = \mathrm{id}_{T_2}$. Moreover, if $n = 6$, then there is an element $\beta \in \mathrm{Aut}_{C_1C_2}(T_1T_2)$ such that $\beta|_{T_1,T_1} = \mathrm{id}_{T_1}$, while $\beta|_{T_2,T_2}$ has order $3$.

Continue to assume that $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$. If $n = 6$, then the observations in the preceding two paragraphs show together with Lemma <ref> that $T_1$ and $T_2$ are the only strongly $\mathcal{F}_{T_1T_2}(C_1C_2)$-closed subgroups of $T_1T_2$ which are isomorphic to $Q_8$. As observed above, this is enough to conclude that (iii) holds. If $n \ge 7$, then we may apply the observations in the preceding two paragraphs together with Lemma <ref> to conclude that if $T_0$ is a strongly $\mathcal{F}_{T_1T_2}(C_1C_2)$-closed subgroup of $T_1T_2$ such that $T_0 \cong Q_8$ and such that $T_0$ centralizes $T_2$, then $T_0 = T_1$. As observed above, this is enough to conclude that (iv) holds.

It remains to prove (v). It is easy to note that the $2$-components of $C_{K_1}(u)$ are precisely the quasisimple elements of $\lbrace C_1,C_2 \rbrace$. So (v) can be obtained from Proposition <ref> and Lemma <ref>.

Let $G$ be as in Hypothesis <ref>. The group $G$ realizes the $2$-fusion system of $PSL_n(q)$. So, if $R$ is a Sylow $2$-subgroup of $G$, then $\mathcal{F}_S(PSL_n(q)) \cong \mathcal{F}_R(G)$. For the sake of simplicity, we will identify $S$ with a Sylow $2$-subgroup $R$ of $G$ and $\mathcal{F}_S(PSL_n(q))$ with $\mathcal{F}_R(G)$. Hence we will work under the following hypothesis.

We will treat $G$ as a group with $S \in \mathrm{Syl}_2(G)$ and $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$.

The following lemma will play a key role in the proof of Theorem <ref>.

Let $x$ be an involution of $S$ such that $C_S(x) \in \mathrm{Syl}_2(C_G(x))$. Let $\mathcal{C}$ be a component of $\mathcal{F}_{C_S(x)}(C_G(x))$, and let $k$ be a natural number with $3 \le k < n$. Then the following hold.

(i) There is a unique $2$-component $Y$ of $C_G(x)$ such that $\mathcal{C} = \mathcal{F}_{C_S(x) \cap Y}(Y)$.

(ii) If $\mathcal{C}$ is isomorphic to the $2$-fusion system of $SL_k(q)$, then we either have that $Y/O(Y) \cong SL_k^{\varepsilon}(q^{*})/O(SL_k^{\varepsilon}(q^{*}))$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $q \sim \varepsilon q^{*}$; or $k = 3$, $(q+1)_2 = 4$, and $Y/Z^{*}(Y) \cong M_{11}$.

(iii) If $\mathcal{C}$ is isomorphic to the $2$-fusion system of a nontrivial quotient of $SL_k(q^2)$, then $Y/O(Y)$ is isomorphic to a nontrivial quotient of $SL_k^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,-\rbrace$ with $q^2 \sim \varepsilon q^{*}$.

In order to prove Lemma <ref>, we need the following observation.

Let $k \ge 6$ be a natural number satisfying $P(k)$. If $q_0$ is a nontrivial odd prime power and $H$ is a known finite simple group realizing the $2$-fusion system of $PSL_k(q_0)$, then $H \cong PSL_k^{\varepsilon}(q^{*})$ for some $\varepsilon \in \lbrace +,- \rbrace$ and some nontrivial odd prime power $q^{*}$ with $\varepsilon q^{*} \sim q_0$.

It suffices to show that any known finite simple group $H$ satisfies (<ref>). Without using the CFSG, this is a priori not clear. It can be deduced from <cit.> if $H$ is alternating, from <cit.> if $H$ is a finite simple group of Lie type in odd characteristic, and from <cit.> if $H$ is sporadic. If $H$ is a finite simple group of Lie type in characteristic $2$, then $H$ satisfies (<ref>) since, in this case, no involution centralizer in $H$ has a $2$-component (see <cit.>).

Since $G$ satisfies (<ref>), we have that $Y/Z^{*}(Y)$ is a known finite simple group for each $2$-component $Y$ of $C_G(x)$. Proposition <ref> implies that there is a unique $2$-component $Y$ of $C_G(x)$ with $\mathcal{C} = \mathcal{F}_{C_S(x) \cap Y}(Y)$. Thus (i) holds.

Suppose that $\mathcal{C}$ is isomorphic to the $2$-fusion system of $SL_k(q_0)/Z$, where either $q_0 = q$ and $Z = 1$, or $q_0 = q^2$ and $Z \le Z(SL_k(q^2))$. In order to prove (ii) and (iii), we need the following three claims.

(1) The $2$-fusion systems of $Y/Z^{*}(Y)$ and $PSL_k(q_0)$ are isomorphic.

As $\mathcal{C} = \mathcal{F}_{C_S(x) \cap Y}(Y)$, we have that the $2$-fusion system of $Y$ is isomorphic to the $2$-fusion system of $SL_k(q_0)/Z$. So, by Corollary <ref>, the $2$-fusion system of $Y/O(Y)$ is isomorphic to the $2$-fusion system of $SL_k(q_0)/Z$. Lemma <ref> implies that the $2$-fusion systems of $Y/Z^{*}(Y)$ and $PSL_k(q_0)$ are isomorphic.

(2) We have $Y/Z^{*}(Y) \cong PSL_k^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $q_0 \sim \varepsilon q^{*}$; or $k = 3$, $(q_0+1)_2 = 4$ and $Y/Z^{*}(Y) \cong M_{11}$.

If $k = 3$, then it follows from (1) and Proposition <ref>. If $k \in \lbrace 4, 5 \rbrace$, then it follows from (1) together with Propositions <ref> and <ref>. Assume now that $k \ge 6$. By Hypothesis <ref> and since $k < n$, we have that $k$ satisfies $P(k)$. Since $Y/Z^{*}(Y)$ is a known finite simple group, the claim follows from (1) and Lemma <ref>.

(3) Suppose that $Y/Z^{*}(Y) \cong PSL_k^{\varepsilon}(q^{*})$, where $q^{*}$ and $\varepsilon$ are as in (2). Then we have $Y/O(Y) \cong SL_k^{\varepsilon}(q^{*})/U$, where $U \le Z(SL_k^{\varepsilon}(q^{*}))$ and the index of $U$ in $Z(SL_k^{\varepsilon}(q^{*}))$ is equal to the $2$-part of $|Z(SL_k(q_0))/Z|$.

The group $Y/O(Y)$ is a perfect central extension of $PSL_k^{\varepsilon}(q^{*})$. Since $Y/O(Y)$ is core-free, the center of $Y/O(Y)$ is a $2$-group. So, by Lemmas <ref> and <ref>, there is a central subgroup $U$ of $SL_k^{\varepsilon}(q^{*})$ with $Y/O(Y) \cong SL_k^{\varepsilon}(q^{*})/U$. The claim now follows from

\begin{align*}
|PSL_k(q_0)|_2|Z(SL_k(q_0))/Z|_2 &= |SL_k(q_0)/Z|_2 \\
&= |Y|_2 \\
&= |Y/Z^{*}(Y)|_2 |Z(Y/O(Y))| \\
&= |PSL_k(q_0)|_2 |Z(SL_k^{\varepsilon}(q^{*}))/U|.
\end{align*}

Here, the second equality follows from the fact that $Y$ realizes $\mathcal{C}$, the third one holds since $|Z^{*}(Y)|_2 = |Z^{*}(Y)/O(Y)|_2 = |Z(Y/O(Y))|_2 = |Z(Y/O(Y))|$, and the fourth one follows from (1).

Assume that $q_0 = q$ and $Z = 1$. By (2) and (3), one of the following hold: either $k=3$, $(q+1)_2 = 4$ and $Y/Z^{*}(Y) \cong M_{11}$; or $Y/O(Y) \cong SL_k^{\varepsilon}(q^{*})/U$, where $q^{*}$ is a nontrivial odd prime power, $\varepsilon \in \lbrace +,- \rbrace$, $q \sim \varepsilon q^{*}$, $U \le Z(SL_k^{\varepsilon}(q^{*}))$ and the index of $U$ in $Z(SL_k^{\varepsilon}(q^{*}))$ is equal to the $2$-part of $|Z(SL_k(q))|$. Assume that the latter holds. As $q \sim \varepsilon q^{*}$, we have $(q-1)_2 = (\varepsilon q^{*} - 1)_2 = (q^{*}-\varepsilon)_2$. Since $|Z(SL_k(q))| = (k,q-1)$ and $|Z(SL_k^{\varepsilon}(q^{*}))| = (k,q^{*}-\varepsilon)$, it follows that the $2$-part of $|Z(SL_k(q))|$ is equal to the $2$-part of $|Z(SL_k^{\varepsilon}(q^{*}))|$. It follows that $U=O(Z(SL_k^{\varepsilon}(q^{*}))) = O(SL_k^{\varepsilon}(q^{*}))$. This completes the proof of (ii).

Assume now that $q_0 = q^2$. Then, since $q^2 \equiv 1 \ \mathrm{mod} \ 4$ , (2) und (3) imply that $Y/O(Y)$ is isomorphic to a nontrivial quotient of $SL_k^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $q^2 \sim \varepsilon q^{*}$. Thus (iii) holds.

§ $2$-COMPONENTS OF INVOLUTION CENTRALIZERS

In this section, we continue to assume Hypotheses <ref> and <ref>. We will use the notation introduced in the last section without further explanation.

The main goal of this section is to describe the $2$-components and the solvable $2$-components of the centralizers of involutions of $G$.

§.§ The subgroups $K$ and $L$ of $C_G(t)$

We start by considering $C_G(t)$. Let $\mathcal{F} := \mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$. Since $\langle t \rangle$ is fully $\mathcal{F}$-centralized, we have that $T = C_S(t) \in \mathrm{Syl}_2(C_G(t))$. Also, note that $\mathcal{F}_T(C_G(t)) = C_{\mathcal{F}}(\langle t \rangle) = \mathcal{F}_T(C_{PSL_n(q)}(t))$.

There is a unique $2$-component $K$ of $C_G(t)$ such that $\mathcal{C}_1 = \mathcal{F}_{T \cap K}(K)$. We have $K/O(K) \cong SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $q \sim \varepsilon q^{*}$. Moreover, $K$ is a normal subgroup of $C_G(t)$.

Set $\mathcal{F} := \mathcal{F}_S(G)$. By Lemma <ref>, $\mathcal{C}_1$ is a component of $C_{\mathcal{F}}(\langle t \rangle)$. Lemma <ref> (i) implies that there is a unique $2$-component $K$ of $C_G(t)$ such that $\mathcal{C}_1 = \mathcal{F}_{T \cap K}(K)$. By definition, the component $\mathcal{C}_1$ is isomorphic to the $2$-fusion system of $SL_{n-2}(q)$. Lemma <ref> (ii) implies that $K/O(K) \cong SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $q \sim \varepsilon q^{*}$.

It remains to show that $K$ is a normal subgroup of $C_G(t)$. Suppose that $\widetilde{K}$ is a $2$-component of $C_G(t)$ such that $K \cong \widetilde{K}$. Set $\widetilde{\mathcal{C}} := \mathcal{F}_{T \cap \widetilde{K}}(\widetilde{K})$. Since $\widetilde K$ is subnormal in $C_G(t)$, it easily follows from <cit.> that $\widetilde{\mathcal{C}}$ is subnormal in $C_{\mathcal{F}}(\langle t \rangle)$. Moreover, $\widetilde{\mathcal{C}} \cong \mathcal{C}_1$ as $\widetilde{K} \cong K$. Hence $\widetilde{\mathcal{C}}$ is a component of $C_{\mathcal{F}}(\langle t \rangle)$. But as a consequence of Lemma <ref>, there is no component of $C_{\mathcal{F}}(\langle t \rangle)$ which is isomorphic to $\mathcal{C}_1$ and different from $\mathcal{C}_1$. So we have $\mathcal{C}_1 = \widetilde{\mathcal{C}}$. The uniqueness in the first statement of the proposition implies that $K = \widetilde{K}$. Consequently, $C_G(t)$ has no $2$-component which is different from $K$ and isomorphic to $K$. So $K$ is characteristic and hence normal in $C_G(t)$.

From now on, $K$, $q^{*}$ and $\varepsilon$ will always have the meanings given to them by Proposition <ref>.

Our next goal is to prove the existence and uniqueness of a normal subgroup $\widebar L$ of $\widebar{C_G(t)} := C_G(t)/O(C_G(t))$ such that $\widebar L \cong SL_2(q^{*})$, and to show that the image $\widebar K$ of $K$ in $\widebar{C_G(t)}$ and $\widebar L$ are the only subgroups of $\widebar{C_G(t)}$ which are components or solvable $2$-components of $\widebar{C_G(t)}$. First, we need to prove some lemmas.

Let $A \in W$ and $B \in V$ such that $\mathrm{det}(A)\mathrm{det}(B) = 1$. Let

\begin{equation*}
m := \begin{pmatrix}A & \\ & B \end{pmatrix} Z(SL_n(q)) \in T.
\end{equation*}

Set $\widebar{C_G(t)} := C_G(t)/O(C_G(t))$. Then $\widebar m$ centralizes $\widebar K$ if and only if $A \in Z(GL_{n-2}(q))$.

By Lemma <ref>, we have $m \in Z(\mathcal{C}_1 \langle m \rangle)$ if and only if $A \in Z(GL_{n-2}(q))$. Let $\overline{\mathcal{C}_1}$ be the subsystem of $\mathcal{F}_{\widebar{T}}(\widebar{C_G(t)})$ corresponding to $\mathcal{C}_1$ under the isomorphism from $\mathcal{F}_{T}(C_G(t))$ to $\mathcal{F}_{\widebar{T}}(\widebar{C_G(t)})$ given by Corollary <ref>. Then we have $\widebar m \in Z(\overline{\mathcal{C}_1}\langle \widebar m \rangle)$ if and only if $A \in Z(GL_{n-2}(q))$. So it is enough to show that $\widebar m \in Z(\overline{\mathcal{C}_1}\langle \widebar m \rangle)$ if and only if $\widebar m$ centralizes $\widebar K$. It is easy to note that $\overline{\mathcal{C}_1}
= \mathcal{F}_{\widebar{X_1}}(\widebar K)$. As a consequence of Proposition <ref>, we have $\widebar K \trianglelefteq \widebar{C_G(t)}$. By <cit.>, we have

\begin{equation*}
\overline{\mathcal{C}_1}\langle \widebar m \rangle = \mathcal{F}_{\widebar{X_1}\langle \widebar m \rangle}(\widebar K \langle \widebar m \rangle).
\end{equation*}

Since $\widebar m$ is a $2$-element of $\widebar{C_G(t)}$, we have $O(\widebar K \langle \widebar m \rangle) = O(\widebar K) = 1$. Applying <cit.>, it follows that the center of the product $\overline{\mathcal{C}_1}\langle \widebar m \rangle$ is equal to the center of $\widebar K \langle \widebar m \rangle$. It follows that that $\widebar m \in Z(\overline{\mathcal{C}_1}\langle \widebar m \rangle)$ if and only if $\widebar m$ centralizes $\widebar K$, as required.

Suppose that $q^{*} = 3$. Let $C := C_G(t)$ and $\widebar C := C/O(C)$. Then:

(i) The factor group $\widebar C/\widebar{K}C_{\widebar C}(\widebar K)$ is a $2$-group.

(ii) The centralizer $C_{\widebar{C}}(\widebar u)$ is core-free.

(iii) The factor group $C_{\widebar C}(\widebar u)/C_{\widebar C}(\widebar K)$ is core-free.

Clearly, $\widebar{C}/\widebar{K}C_{\widebar C}(\widebar K)$ is isomorphic to a subgroup of $\mathrm{Out}(\widebar K)$. Since $q^{*} = 3$, we have that $\widebar{K} \cong SL_{n-2}^{\varepsilon}(3)$. By Propositions <ref> and <ref>, $\mathrm{Out}(\widebar K)$ is a $2$-group. So (i) holds.

Set $\widebar{C}_0 := \widebar K C_{\widebar C}(\widebar K)$. As a consequence of (i), $C_{\widebar C}(\widebar u)/C_{\widebar{C}_0}(\widebar u)$ is a $2$-group. Hence, in order to prove (ii), it suffices to show that $C_{\widebar{C}_0}(\widebar u)$ is core-free. As $\widebar u \in \widebar K$, we have $C_{\widebar{C}_0}(\widebar u) = C_{\widebar K}(\widebar u) C_{\widebar C}(\widebar K)$. It follows that $C_{\widebar{C}_0}(\widebar u)/C_{\widebar C}(\widebar K) \cong C_{\widebar K}(\widebar u)/(C_{\widebar K}(\widebar u) \cap C_{\widebar C}(\widebar K)) = C_{\widebar K}(\widebar u)/Z(\widebar K)$. By Corollary <ref>, $C_{\widebar K}(\widebar u)$ is core-free. This easily implies that $C_{\widebar K}(\widebar u)/Z(\widebar K)$ is core-free. It follows that $C_{\widebar{C}_0}(\widebar u)/C_{\widebar C}(\widebar K)$ is core-free. Consequently, $O(C_{\widebar{C}_0}(\widebar u)) = O(C_{\widebar C}(\widebar K)) = 1$. So (ii) follows.

Finally, (iii) is true since $C_{\widebar C}(\widebar u)/C_{\widebar{C}_0}(\widebar u)$ is a $2$-group and $C_{\widebar{C}_0}(\widebar u)/C_{\widebar C}(\widebar K)$ is core-free.

Let $\widebar{C_G(t)} := C_G(t)/O(C_G(t))$. Then there is a unique pair $({A_1}^{+},{A_2}^{+})$ of normal subgroups ${A_1}^{+}$, ${A_2}^{+}$ of $C_{\widebar K}(\widebar u)'$ such that $C_{\widebar K}(\widebar u)' = {A_1}^{+}  \times {A_2}^{+}$, ${A_1}^{+} \cong SL_2^{\varepsilon}(q^{*})$, ${A_2}^{+} \cong SL_{n-4}^{\varepsilon}(q^{*})$ and $\widebar u \in {A_1}^{+}$. Moreover, the following hold.

(i) ${A_1}^{+} \cap \widebar{X_1} = \widebar{T_1}$.

(ii) ${A_2}^{+} \cap \widebar{X_1} = \widebar{T_2}$.

(iii) There is a group isomorphism $\varphi: \widebar{K} \rightarrow SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$ under which ${A_1}^{+}$ corresponds to the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} A & \\ & I_{n-4} \end{pmatrix} \ : \ A \in SL_2^{\varepsilon}(q^{*}) \right\rbrace
\end{equation*}

in $SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$ and under which ${A_2}^{+}$ corresponds to the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} I_2 & \\ & B \end{pmatrix} \ : \ B \in SL_{n-4}^{\varepsilon}(q^{*}) \right\rbrace
\end{equation*}

in $SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$.

For each subsystem $\mathcal{G}$ of $\mathcal{F}_T(C_G(t))$, we use $\widebar{\mathcal{G}}$ to denote the subsystem of $\mathcal{F}_{\widebar T}(\widebar{C_G(t)})$ corresponding to $\mathcal{G}$ under the isomorphism from $\mathcal{F}_T(C_G(t))$ to $\mathcal{F}_{\widebar T}(\widebar{C_G(t)})$ given by Corollary <ref>. Note that $\overline{\mathcal{C}_1} = \mathcal{F}_{\widebar{X_1}}(\widebar K)$.

Set $H := SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$. For each even natural number $i$ with $2 \le i \le n-2$, let $h_i$ be the image of $\widetilde{h_i} := \mathrm{diag}(-1,\dots,-1,1,\dots,1) \in SL_{n-2}^{\varepsilon}(q^{*})$ in $H$, where $-1$ occurs precisely $i$ times as a diagonal entry.

We claim that there is a group isomorphism $\varphi: \widebar K \rightarrow H$ such that $\widebar{u}^{\varphi} = h_i$ for some even $2 \le i < n-2$. By Proposition <ref>, we have $\widebar{K} \cong K/O(K) \cong H$. As a consequence of Lemmas <ref> and <ref>, any involution of $SL_{n-2}^{\varepsilon}(q^{*})$ is conjugate to $\widetilde{h_i}$ for some even $2 \le i \le n-2$. Since any involution of $H$ is induced by an involution of $SL_{n-2}^{\varepsilon}(q^{*})$, it follows that any involution of $H$ is conjugate to $h_i$ for some even $2 \le i \le n-2$. As $\widebar u$ is an involution of $\widebar K$, it follows that there is an isomorphism $\varphi: \widebar K \rightarrow H$ mapping $\widebar u$ to $h_i$ for some even $2 \le i \le n-2$. Assume that $i = n-2$. Then $\widebar u$ is central in $\widebar K$. Thus $\widebar u \in Z(\overline{\mathcal{C}_1})$ and hence $u \in Z(\mathcal{C}_1)$. This is a contradiction to Lemma <ref> and the definition of $\mathcal{C}_1$. So we have $i < n-2$.

Set $h := \widebar u^{\varphi} = h_i$. Also, let $H_1$ be the image of

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

in $H$, and let $H_2$ be the image of

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

in $H$. For $j \in \lbrace 1,2 \rbrace$, let ${A_j}^{+}$ be the subgroup of $\widebar K$ corresponding to $H_j$ under $\varphi$.

We now proceed in a number of steps in order to complete the proof.

(1) We have $C_{\widebar K}(\widebar u)' = {A_1}^{+}{A_2}^{+}$, $[{A_1}^{+},{A_2}^{+}] = 1$, ${A_1}^{+}, {A_2}^{+} \trianglelefteq C_{\widebar K}(\widebar u)$, $\widebar u \in {A_1}^{+}$ and $\widebar u \not\in {A_2}^{+}$.

It is easy to note that $C_H(h)'$ is the central product of $H_1$ and $H_2$ and that $H_1$ and $H_2$ are normal in $C_H(h)$. Therefore, $C_{\widebar K}(\widebar u)'$ is the central product of ${A_1}^{+}$ and ${A_2}^{+}$, and ${A_1}^{+}, {A_2}^{+}$ are normal in $C_{\widebar K}(\widebar u)$. By definition of $H_1$ and $H_2$, we have $h \in H_1$ and $h \not\in H_2$. Thus $\widebar u \in {A_1}^{+}$ and $\widebar u \not\in {A_2}^{+}$.

(2) We have $C_{\widebar{X_1}}(\widebar u) \in \mathrm{Syl}_2(C_{\widebar K}(\widebar u))$, and $\lbrace \mathcal{F}_{\widebar{X_1} \cap {A_1}^{+}}({A_1}^{+}), \mathcal{F}_{\widebar{X_1} \cap {A_2}^{+}}({A_2}^{+}) \rbrace$ contains every component of $C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle)$.

By Lemma <ref> (i), we have that $\langle \widebar u \rangle$ is fully $\overline{\mathcal{C}_1}$-centralized. So we have $C_{\widebar{X_1}}(\widebar u) \in \mathrm{Syl}_2(C_{\widebar K}(\widebar u))$.

Set $P := C_{\widebar{X_1}}(\widebar u)^{\varphi} \in \mathrm{Syl}_2(C_H(h))$. It is easy to note that the $2$-components of $C_H(h)$ are precisely the quasisimple elements of $\lbrace H_1, H_2 \rbrace$. Proposition <ref> implies that the components of $\mathcal{F}_P(C_H(h))$ are precisely the quasisimple elements of $\lbrace \mathcal{F}_{P \cap H_1}(H_1), \mathcal{F}_{P \cap H_2}(H_2) \rbrace$.

Thus the components of $C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle) = \mathcal{F}_{C_{\widebar{X_1}}(\widebar u)}(C_{\widebar K}(\widebar u))$ are precisely the quasisimple elements of $\lbrace \mathcal{F}_{\widebar{X_1} \cap {A_1}^{+}}({A_1}^{+}), \mathcal{F}_{\widebar{X_1} \cap {A_2}^{+}}({A_2}^{+}) \rbrace$.

(3) $\widebar{X_1} \cap {A_1}^{+}$ and $\widebar{X_1} \cap {A_2}^{+}$ are subgroups of $\mathfrak{foc}(C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle))$ and are strongly closed in $C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle)$.

We have $\mathfrak{foc}(C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle)) = C_{\widebar{X_1}}(\widebar u) \cap C_{\widebar K}(\widebar u)'$ by the focal subgroup theorem <cit.>. So the claim follows from (1).

(4) Suppose that $n = 6$ and $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$. Then we have $i = 2$ and hence ${A_1}^{+} \cong SL_2^{\varepsilon}(q^{*}) \cong {A_2}^{+}$. Moreover, $\widebar{X_1} \cap {A_1}^{+} = \widebar{T_1}$ and $\widebar{X_1} \cap {A_2}^{+} = \widebar{T_2}$.

Since $n = 6$ and $2 \le i < n-2 = 4$, we have $i = 2$. Thus ${A_1}^{+} \cong H_1 \cong SL_2^{\varepsilon}(q^{*}) \cong H_2 \cong {A_2}^{+}$. By Proposition <ref>, we have $q \sim \varepsilon q^{*}$, whence $q^{*} \equiv 3$ or $5 \ \mathrm{mod} \ 8$. Clearly, $\widebar{X_1} \cap {A_1}^{+} \in \mathrm{Syl}_2({A_1}^{+})$ and $\widebar{X_1} \cap {A_2}^{+} \in \mathrm{Syl}_2({A_2}^{+})$. Lemma <ref> implies that $\widebar{X_1} \cap {A_1}^{+} \cong Q_8 \cong \widebar{X_1} \cap {A_2}^{+}$. By Lemma <ref> (iii), $\widebar{T_1}$ and $\widebar{T_2}$ are the only subgroups of $\mathfrak{foc}(C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle))$ which are isomorphic to $Q_8$ and strongly closed in $C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle)$. So, by (3), $\lbrace \widebar{X_1} \cap {A_1}^{+}, \widebar{X_1} \cap {A_2}^{+} \rbrace = \lbrace \widebar{T_1}, \widebar{T_2} \rbrace$. We have $\widebar u \in \widebar{T_1}$, and $\widebar u \not\in {A_2}^{+}$ by (1). It follows that $\widebar{X_1} \cap {A_1}^{+} = \widebar{T_1}$ and $\widebar{X_1} \cap {A_2}^{+} = \widebar{T_2}$.

(5) Suppose that $n = 6$ and $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$, or that $n \ge 7$. Then we have $i = 2$, and hence ${A_1}^{+} \cong SL_2^{\varepsilon}(q^{*})$ and ${A_2}^{+} \cong SL_{n-4}^{\varepsilon}(q^{*})$. Moreover, $\widebar{X_1} \cap {A_1}^{+} = \widebar{T_1}$ and $\widebar{X_1} \cap {A_2}^{+} = \widebar{T_2}$.

We begin by proving that $\widebar{X_1} \cap {A_2}^{+} = \widebar{T_2}$. As a consequence of Lemma <ref> (v), $C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle)$ has a component with Sylow group $\widebar{T_2}$. Applying (2), we may conclude that $\widebar{T_2} = \widebar{X_1} \cap {A_1}^{+}$ or $\widebar{X_1} \cap {A_2}^{+}$. Since $\widebar u \in A_1^{+}$ by (1), but $\widebar u \not\in \widebar{T_2}$, we have $\widebar{X_1} \cap {A_2}^{+} = \widebar{T_2}$.

We show next that $i = 2$. Using Proposition <ref>, or using the order formulas for $\vert SL_{n-4}(q^{*}) \vert$ and $\vert SU_{n-4}(q^{*}) \vert$ given by <cit.>, we see that

\begin{equation*}
|SL_{n-4}^{\varepsilon}(q^{*})|_2 = |SL_{n-4}(q)|_2 = |T_2| = |{A_2}^{+}|_2 = |H_2|_2 = |SL_{n-2-i}^{\varepsilon}(q^{*})|_2.
\end{equation*}

Using the order formulas cited above, we may conclude that $n-2-i = n-4$, whence $i = 2$. In particular, ${A_1}^{+} \cong SL_2^{\varepsilon}(q^{*})$ and ${A_2}^{+} \cong SL_{n-4}^{\varepsilon}(q^{*})$.

It remains to prove $\widebar{X_1} \cap {A_1}^{+} = \widebar{T_1}$. If $q \equiv 1$ or $7 \ \mathrm{mod} \ 8$, then Lemma <ref> (v) shows that $C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle)$ has a component with Sylow group $\widebar{T_1}$. Since $\widebar u \in \widebar{T_1}$, but $\widebar u \not\in {A_2}^{+}$, we have $\widebar{X_1} \cap {A_1}^{+} = \widebar{T_1}$ by (2).

Now suppose that $q \equiv 3$ or $5 \ \mathrm{mod} \ 8$. Then we have $q^{*} \equiv 3$ or $5 \ \mathrm{mod} \ 8$ since $q \sim \varepsilon q^{*}$. So, by Lemma <ref>, a Sylow $2$-subgroup of ${A_1}^{+}$ is isomorphic to $Q_8$. In particular, $\widebar{X_1} \cap {A_1}^{+} \cong Q_8$. By (3), $\widebar{X_1} \cap {A_1}^{+}$ is a subgroup of $\mathfrak{foc}(C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle))$ and is strongly closed in $C_{\overline{\mathcal{C}_1}}(\langle \widebar u \rangle)$. Moreover, by (1), $\widebar{X_1} \cap {A_1}^{+}$ centralizes $\widebar{X_1} \cap {A_2}^{+}  = \widebar{T_2}$. Lemma <ref> (iv) now implies that $\widebar{T_1} = \widebar{X_1} \cap {A_1}^{+}$.

(6) $C_{\widebar K}(\widebar u)' = {A_1}^{+} \times {A_2}^{+}$.

We have ${A_1}^{+} \cong SL_2^{\varepsilon}(q^{*})$ by (4) and (5), and $\widebar u \in Z({A_1}^{+})$ by (1). It follows that $Z({A_1}^{+}) = \langle \widebar u \rangle$. By (1), ${A_1}^{+} \cap {A_2}^{+} \le Z({A_1}^{+})$ and $\widebar u \not\in {A_1}^{+} \cap {A_2}^{+}$. It follows that ${A_1}^{+} \cap {A_2}^{+} = 1$. So (1) implies that $C_{\widebar K}(\widebar u)' = {A_1}^{+} \times {A_2}^{+}$.

(7) Assume that ${A_1}^{\circ}$ and ${A_2}^{\circ}$ are normal subgroups of $C_{\widebar K}(\widebar u)'$ such that $C_{\widebar K}(\widebar u)' = {A_1}^{\circ} \times {A_2}^{\circ}$, ${A_1}^{\circ} \cong SL_2^{\varepsilon}(q^{*})$, ${A_2}^{\circ} \cong SL_{n-4}^{\varepsilon}(q^{*})$ and $\widebar u \in {A_1}^{\circ}$. Then ${A_1}^{\circ} = {A_1}^{+}$ and ${A_2}^{\circ} = {A_2}^{+}$.

Let $j \in \lbrace 1,2 \rbrace$. As a consequence of (4) and (5), ${A_j}^{+}$ is either quasisimple or isomorphic to $SL_2(3)$. In either case, it is easy to see that ${A_j}^{+}$ is indecomposable, i.e. ${A_j}^{+}$ cannot be written as an internal direct product of two proper normal subgroups. Moreover, $\vert {A_1}^{+}/({A_1}^{+})' \vert$ and $\vert Z({A_{2}}^{+}) \vert$ as well as $\vert {A_2}^{+}/({A_2}^{+})' \vert$ and $\vert Z({A_{1}}^{+}) \vert$ are coprime. A consequence of the Krull-Remak-Schmidt theorem, namely <cit.>, implies that $\lbrace {A_1}^{+}, {A_2}^{+} \rbrace = \lbrace {A_1}^{\circ}, {A_2}^{\circ} \rbrace$. Since $\widebar u \in {A_1}^{+}$ and $\widebar u \not\in {A_2}^{\circ}$, we have ${A_1}^{+} = {A_1}^{\circ}$ and ${A_2}^{+} = {A_2}^{\circ}$.

(8) The isomorphism $\varphi: \widebar{K} \rightarrow H$ maps ${A_1}^{+}$ to the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} A & \\ & I_{n-4} \end{pmatrix} \ : \ A \in SL_2^{\varepsilon}(q^{*}) \right\rbrace
\end{equation*}

in $H$ and ${A_2}^{+}$ to the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} I_2 & \\ & B \end{pmatrix} \ : \ B \in SL_{n-4}^{\varepsilon}(q^{*}) \right\rbrace
\end{equation*}

in $H$.

By (4) and (5), we have $i = 2$. So the claim follows from the definitions of ${A_1}^{+}$ and ${A_2}^{+}$.

From now on, ${A_1}^{+}$ and ${A_2}^{+}$ will always have the meanings given to them by Lemma <ref>.

Let $C := C_G(t)$ and $\widebar C := C/O(C)$. Then ${A_1}^{+}$ and ${A_2}^{+}$ are normal subgroups of $C_{\widebar C}(\widebar u)$.

We have $C_{\widebar K}(\widebar u) \trianglelefteq C_{\widebar C}(\widebar u)$ as $\widebar K \trianglelefteq \widebar C$. Thus $C_{\widebar K}(\widebar u)' \trianglelefteq C_{\widebar C}(\widebar u)$. Having this observed, the lemma is immediate from Lemma <ref>.

Let $C := C_G(t)$ and $\widebar C := C/O(C)$. Next we introduce certain preimages of ${A_1}^{+}$ and ${A_2}^{+}$ in $C_{C}(u)$. By Corollary <ref>, we have $C_{\widebar{C}}(\widebar u) = \widebar{C_{C}(u)}$. We may see from Proposition <ref> that there is a bijection from the set of $2$-components of $C_C(u)$ to the set of $2$-components of $C_{\widebar C}(\widebar u)$ sending each $2$-component $A$ of $C_C(u)$ to $\widebar A$.

Suppose that $q^{*} \ne 3$. Then ${A_1}^{+}$ is a component and hence a $2$-component of $C_{\widebar C}(\widebar u)$. We use $A_1$ to denote the $2$-component of $C_{C}(u)$ corresponding to ${A_1}^{+}$ under the bijection described above.

Suppose that $q^{*} \ne 3$ or $n \ge 7$. Then ${A_2}^{+}$ is a component and hence a $2$-component of $C_{\widebar C}(\widebar u)$. We use $A_2$ to denote the $2$-component of $C_{C}(u)$ corresponding to ${A_2}^{+}$ under the bijection described above.

Suppose that $q^{*} = 3$. By Lemma <ref> (ii), $O(C_{\widebar{C}}(\widebar u)) = 1$. So the factor group $C_{C}(u)/(C_{C}(u) \cap O(C))$ is core-free, whence $O(C_{C}(u)) = C_{C}(u) \cap O(C)$. Let $O(C_{C}(u)) \le A_1 \le C_{C}(u)$ such that $A_1/O(C_{C}(u))$ corresponds to ${A_1}^{+}$ under the natural group isomorphism $C_{C}(u)/O(C_{C}(u)) \rightarrow C_{\widebar{C}}(\widebar u)$. Furthermore, if $n = 6$, let $O(C_{C}(u)) \le A_2 \le C_{C}(u)$ such that $A_2/O(C_{C}(u))$ corresponds to ${A_2}^{+}$ under the natural group isomorphism $C_{C}(u)/O(C_{C}(u)) \rightarrow C_{\widebar{C}}(\widebar u)$.

We have $T_1 \le A_1$ and $T_2 \le A_2$.

Let $i \in \lbrace 1,2 \rbrace$. Set $C := C_G(t)$ and $\widebar C := C/O(C)$.

Let $C_C(u) \cap O(C) \le \widetilde{A_i} \le C_C(u)$ such that $\widetilde{A_i}/(C_C(u) \cap O(C))$ corresponds to ${A_i}^{+}$ under the natural group isomorphism $C_C(u)/(C_C(u) \cap O(C)) \rightarrow C_{\widebar C}(\widebar u)$. We have $T_i \le C_C(u)$ and, by Lemma <ref>, $\widebar{T_i} \le {A_i}^{+}$. Thus $T_i \le \widetilde{A_i}$. If ${A_i}^{+} \cong SL_2(3)$, then we have $A_i = \widetilde{A_i}$ and thus $T_i \le A_i$. Assume now that ${A_i}^{+}$ is a component of $C_{\widebar C}(\widebar u)$. Then $A_i$ is the $2$-component of $C_C(u)$ associated to the $2$-component $\widetilde{A_i}/(C_C(u) \cap O(C))$ of $C_C(u)/(C_C(u) \cap O(C))$. So, by Proposition <ref>, $A_i = O^{2'}(\widetilde{A_i})$, and hence $T_i \le A_i$.

There is an element $g \in G$ such that ${T_1}^g = X_2$ and ${X_2}^g = T_1$. For each such $g \in G$, we have $u^g = t$ and $t^g = u$.

The first statement easily follows from $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$. By Lemma <ref>, the groups $T_1$ and $X_2$ are generalized quaternion. So $u$ is the only involution of $T_1$ and $t$ is the only involution of $X_2$. Thus $u^g = t$ and $t^g = u$ for any $g \in G$ with ${T_1}^g = X_2$ and ${X_2}^g = T_1$.

With the above lemmas at hand, we can now prove the following proposition.

Take an element $g \in G$ such that ${T_1}^g = X_2$ and ${X_2}^g = T_1$. Set $C := C_G(t)$ and $\widebar C := C/O(C)$. Let $L := {A_1}^g$. Then the following hold.

(i) $L \le C_C(u)$.

(ii) $\widebar L$ is subnormal in $\widebar C$ and $\widebar L \cong SL_2(q^{*})$.

(iii) The subgroups $\widebar K$ and $\widebar L$ are the only subgroups of $\widebar C$ which are components or solvable $2$-components of $\widebar C$. In particular, $\widebar K$ and $\widebar L$ are normal subgroups of $\widebar C$.

By Lemma <ref>, we have $t^g = u$ and $u^g = t$. Hence $C_C(u)^g = C_C(u)$. As $A_1$ is a subgroup of $C_C(u)$, we thus have $L = {A_1}^g \le C_C(u)$. So (i) holds.

Before proving (ii), we show that $C_{\widebar L}(\widebar K)$ is a normal subgroup of $\widebar L$ containing $\widebar{X_2}$. Since $C_{\widebar C}(\widebar K) \trianglelefteq \widebar{C}$, we have $C_{\widebar L}(\widebar K) = \widebar{L} \cap C_{\widebar C}(\widebar K) \trianglelefteq \widebar{L}$. Because of Lemma <ref>, we have $X_2 = {T_1}^g \le {A_1}^g = L$. Thus $\widebar{X_2} \le \widebar{L}$. By the definition of $X_2$ and by Lemma <ref>, we have $\widebar{X_2} \le C_{\widebar C}(\widebar K)$. Thus $\widebar{X_2} \le C_{\widebar L}(\widebar K)$.

Note that $\widebar{X_2}$ is generalized quaternion by Lemma <ref> and in particular nonabelian.

We now prove (ii) for the case $q^{*} \ne 3$. Then $A_1$ is a $2$-component of $C_C(u)$. As $g$ normalizes $C_C(u)$ and $L = {A_1}^g$, it follows that $L$ is a $2$-component of $C_C(u)$. So $\widebar{L}$ is a $2$-component of $C_{\widebar C}(\widebar u)$. Moreover, we have $A_1/O(A_1) \cong SL_2(q^{*})$ since $A_1/(A_1 \cap O(C)) \cong \widebar{A_1} = {A_1}^{+} \cong SL_2(q^{*})$. Hence $L/O(L)$ is isomorphic to $SL_2(q^{*})$. The group $C_{\widebar L}(\widebar K) O(\widebar L)/O(\widebar L)$ is normal in $\widebar{L}/O(\widebar L)$, and it is nonabelian since $\widebar{X_2} \le C_{\widebar L}(\widebar K)$. As $\widebar L/O(\widebar L)$ is quasisimple, it follows that $C_{\widebar L}(\widebar K)O(\widebar L) = \widebar L$. So $C_{\widebar L}(\widebar K)$ has odd index in $\widebar L$. Since $\widebar L$ is a $2$-component of $C_{\widebar C}(\widebar u)$, we have $O^{2'}(\widebar L) = \widebar L$. It follows that $\widebar{L} = C_{\widebar L}(\widebar K) \le C_{\widebar C}(\widebar K)$. Since $\widebar L$ is subnormal in $C_{\widebar C}(\widebar u)$ and $C_{\widebar C}(\widebar K) \le C_{\widebar C}(\widebar u)$, we have that $\widebar L$ is subnormal in $C_{\widebar C}(\widebar K)$. Hence $\widebar L$ is subnormal in $\widebar C$. As $\widebar C$ is core-free, we have $O(\widebar{L}) = 1$. It follows that $O(L) = L \cap O(C)$ and hence $\widebar L \cong L/O(L) \cong SL_2(q^{*})$. So we have proved (ii) for the case $q^{*} \ne 3$.

Assume now that $q^{*} = 3$. Then $O(C_C(u)) = C_C(u) \cap O(C)$, $O(C_C(u)) \le A_1 \le C_C(u)$, and $A_1/O(C_C(u))$ corresponds to ${A_1}^{+} \cong SL_2(3)$ under the natural isomorphism $C_C(u)/O(C_C(u)) \rightarrow C_{\widebar C}(\widebar u)$. By Lemma <ref>, ${A_1}^{+}$ is normal in $C_{\widebar C}(\widebar u)$. Hence, $A_1/O(C_C(u))$ is a normal subgroup of $C_C(u)/O(C_C(u))$ isomorphic to $SL_2(3)$. Since $g$ normalizes $C_C(u)$ and $L = {A_1}^g$, it follows that $O(C_C(u)) \le L$ and that $L/O(C_C(u))$ is a normal subgroup of $C_C(u)/O(C_C(u))$ isomorphic to $SL_2(3)$. Since $L/O(C_C(u))$ corresponds to $\widebar L$ under the natural isomorphism $C_C(u)/O(C_C(u)) \rightarrow C_{\widebar C}(\widebar u)$, it follows that $\widebar L$ is a normal subgroup of $C_{\widebar C}(\widebar u)$ isomorphic to $SL_2(3)$. Recall that $\widebar{X_2} \le C_{\widebar L}(\widebar K) \trianglelefteq \widebar{L}$. As $\widebar{L}$ has order $24$ and $\widebar{X_2}$ has order $8$, $C_{\widebar L}(\widebar K)$ either equals $\widebar L$ or has index $3$ in $\widebar L$. However, if the latter holds, then $\widebar L C_{\widebar C}(\widebar K)/C_{\widebar C}(\widebar K)$ is a normal subgroup of $C_{\widebar C}(\widebar u)/C_{\widebar C}(\widebar K)$ of order $3$, which is a contradiction to Lemma <ref> (iii). Thus $\widebar L = C_{\widebar L}(\widebar K) \le C_{\widebar C}(\widebar K)$. As $\widebar L \trianglelefteq C_{\widebar C}(\widebar u)$ and $C_{\widebar C}(\widebar K) \le C_{\widebar C}(\widebar u)$, it follows that $\widebar L$ is normal in $C_{\widebar C}(\widebar K)$ and hence subnormal in $\widebar C$. So we have proved (ii) for the case $q^{*} = 3$.

We now prove (iii). Clearly, $\widebar T \cap \widebar K = \widebar{X_1}$. Also $\widebar T \cap \widebar L = \widebar{X_2}$ since $\vert \widebar{X_2} \vert = \vert SL_2(q) \vert_2 = \vert SL_2(q^{*}) \vert_2 = \vert \widebar L \vert_2$ and $\widebar{X_2} \le \widebar{L}$. As a consequence of Lemma <ref>, the fusion system $\mathcal{F}_{\widebar{T}}(\widebar C)/(\widebar{X_1}\widebar{X_2})$ is nilpotent. Applying Lemma <ref>, we may conclude that $\widebar K$ and $\widebar L$ are the only subgroups of $\widebar C$ which are components or solvable $2$-components of $\widebar C$. As $\widebar K$ and $\widebar L$ are not isomorphic, both are characteristic and hence normal in $\widebar C$.

It is not difficult to observe that the definition of $L$ in Proposition <ref> is independent of the choice of $g$. From now on, $L$ will always have the meaning given to it by the above proposition.

§.§ $2$-components of centralizers of involutions conjugate to $t_i$, $i \ne 2$

Having described the components and the solvable $2$-components of the group $C_G(t)/O(C_G(t))$, we now turn our attention to centralizers of involutions of $G$ not conjugate to $t$.

First we recall some notation from Section <ref>. Let $1 \le i < n$. If $i$ is even, then $t_i$ denotes the image of

\begin{equation*}
\begin{pmatrix} I_{n-i} & \\ & -I_i \end{pmatrix}
\end{equation*}

in $PSL_n(q)$. We use $\rho$ to denote an element of $\mathbb{F}_q^{*}$ with order $(n,q-1)$, and if $\rho$ is a square in $\mathbb{F}_q$, then $\mu$ denotes an element of $\mathbb{F}_q^{*}$ with $\mu^2 = \rho$. If $n$ is even, $\rho$ is a square in $\mathbb{F}_q$ and $i$ is odd, then $t_i$ is defined to be the image of

\begin{equation*}
\begin{pmatrix} \mu I_{n-i} & \\ & -\mu I_i \end{pmatrix} \in SL_n(q)
\end{equation*}

in $PSL_n(q)$. It is easy to note that $t_i$ lies in $T$ and hence in $S$ whenever $t_i$ is defined.

Let $\mathcal{S}$ denote the set of all subgroups $E$ of $PSL_n(q)$ such that there is some elementary abelian $2$-subgroup $\widetilde E \le SL_n(q)$ with $E = \widetilde{E} Z(SL_n(q))/Z(SL_n(q))$. For each $3 \le i \le n$, we define $\mathcal{S}_i$ to be the set of all elements $E$ of $\mathcal{S}$ such that $E$ contains a $PSL_n(q)$-conjugate of $t_j$ for some even $2 \le j < i$.

Let $1 \le i < n$ such that $t_i$ is defined. Assume that $i \ne 2$, and that $i \le \frac{n}{2}$ if $n$ is even. Let $P$ be a Sylow $2$-subgroup of $C_{PSL_n(q)}(t_i)$ and $\mathcal{F} := \mathcal{F}_P(C_{PSL_n(q)}(t_i))$. Then the following hold.

(i) Assume that $i \not\in \lbrace 1,n-1 \rbrace$. Then $\mathcal{F}$ has precisely two components. Denoting them in a suitable way by $\mathcal{E}_1$ and $\mathcal{E}_2$, the following hold.

(a) $\mathcal{E}_1$ is isomorphic to the $2$-fusion system of $SL_{n-i}(q)$.

(b) $\mathcal{E}_2$ is isomorphic to the $2$-fusion system of $SL_i(q)$.

(c) Let $Y_1$ be the Sylow group of $\mathcal{E}_1$ and let $Y_2$ be the Sylow group of $\mathcal{E}_2$. Then $Y_1Y_2$ is normal in $P$ and $\mathcal{F}/Y_1Y_2$ is nilpotent. The group $Y_i$, where $i \in \lbrace 1,2 \rbrace$, contains a $PSL_n(q)$-conjugate of $t$. Moreover, any elementary abelian subgroup of $Y_1$ of rank at least $2$ is contained in $\mathcal{S}_{n-i}$, and any elementary abelian subgroup of $Y_2$ of rank at least $2$ is contained in $\mathcal{S}_i$.

(ii) Assume that $i = 1$ or $i = n-1$. Then $\mathcal{F}$ has a unique component. This component is isomorphic to the $2$-fusion system of $SL_{n-1}(q)$. If $Y$ is its Sylow group, then $Y \trianglelefteq P$ and $\mathcal{F}/Y$ is nilpotent. Moreover, any elementary abelian subgroup of $Y$ of rank at least $2$ is contained in $\mathcal{S}_{n-1}$.

Assume that $i \not\in \lbrace 1,n-1 \rbrace$. By hypothesis, we have $i \ne 2$, and $i \le \frac{n}{2}$ if $n$ is even. It follows that $i \ge 3$ and $n-i \ge 3$. Let $J_1$ be the image of

\begin{equation*}
\left\lbrace \begin{pmatrix} A & \\ & I_i \end{pmatrix} \ : \ A \in SL_{n-i}(q) \right\rbrace
\end{equation*}

in $PSL_n(q)$, and let $J_2$ be the image of

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

in $PSL_n(q)$. It is easy to note that $J_1$ and $J_2$ are the only $2$-components of $C_{PSL_n(q)}(t_i)$. Applying Proposition <ref> and Lemma <ref>, we may conclude that $\mathcal{E}_1 := \mathcal{F}_{P \cap J_1}(J_1)$ and $\mathcal{E}_2 := \mathcal{F}_{P \cap J_2}(J_2)$ are the only components of $\mathcal{F} = \mathcal{F}_P(C_{PSL_n(q)}(t_i))$. Clearly, $\mathcal{E}_1$ is isomorphic to the $2$-fusion system of $SL_{n-i}(q)$, while $\mathcal{E}_2$ is isomorphic to the $2$-fusion system of $SL_i(q)$. Set $Y_1 := P \cap J_1$ and $Y_2 := P \cap J_2$. It is easy to note that $Y_1Y_2 = P \cap J_1J_2$. As $J_1J_2 \trianglelefteq C_{PSL_n(q)}(t_i)$, it follows that $Y_1Y_2 \trianglelefteq P$. By Lemma <ref>, $\mathcal{F}/Y_1Y_2$ is isomorphic to the $2$-fusion system of $C_{PSL_n(q)}(t_i)/J_1J_2$, and it is easy to note that $C_{PSL_n(q)}(t_i)/J_1J_2$ is $2$-nilpotent. So $\mathcal{F}/Y_1Y_2$ is nilpotent by <cit.>.  It is clear from the definitions of $J_1$ and $J_2$ that both $J_1$ and $J_2$ contain a $PSL_n(q)$-conjugate of $t$. Hence $Y_k$ has an element which is $PSL_n(q)$-conjugate to $t$ for $k \in \lbrace 1,2 \rbrace$. Clearly, any elementary abelian $2$-subgroup of $J_k$, $k \in \lbrace 1,2 \rbrace$, lies in $\mathcal{S}$. Moreover, any noncentral involution of $J_1$ is $PSL_n(q)$-conjugate to $t_j$ for some even $2 \le j < n-i$, and any noncentral involution of $J_2$ is $PSL_n(q)$-conjugate to $t_j$ for some even $2 \le j < i$. This implies that any elementary abelian subgroup of $Y_1$ of rank at least $2$ is contained in $\mathcal{S}_{n-i}$, and that any elementary abelian subgroup of $Y_2$ of rank at least $2$ is contained in $\mathcal{S}_i$. This completes the proof of (i).

We omit the proof of (ii) since it is very similar to the one of (i).

Let $1 \le i < n$ such that $t_i$ is defined. Assume that $i \not\in \lbrace 1, 2, n-1 \rbrace$, and that $i \le \frac{n}{2}$ if $n$ is even. Let $x$ be an involution of $S$ which is $G$-conjugate to $t_i$. Then $C_G(x)$ has precisely two $2$-components. Denoting them in a suitable way by $J_1$ and $J_2$, the following hold.

(i) $J_1/O(J_1)$ is isomorphic to $SL_{n-i}^{\varepsilon}(q^{*})/O(SL_{n-i}^{\varepsilon}(q^{*}))$, where $\varepsilon$ and $q^{*}$ are as in Proposition <ref>.

(ii) $J_2/O(J_2) \cong SL_i^{\varepsilon}(q^{*})/O(SL_i^{\varepsilon}(q^{*}))$, where $\varepsilon$ and $q^{*}$ are as in Proposition <ref>.

(iii) Any elementary abelian $2$-subgroup of $J_1$ of rank at least $2$ is $G$-conjugate to a subgroup of $S$ lying in $\mathcal{S}_{n-i}$, and any elementary abelian $2$-subgroup of $J_2$ of rank at least $2$ is $G$-conjugate to a subgroup of $S$ lying in $\mathcal{S}_i$.

Let $\mathcal{F} := \mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$. It suffices to prove the proposition under the assumption that $\langle x \rangle$ is fully $\mathcal{F}$-centralized, and we will assume that this is the case. So we have $C_S(x) \in \mathrm{Syl}_2(C_G(x))$ and $C_S(x) \in \mathrm{Syl}_2(C_{PSL_n(q)}(x))$. Also, $\mathcal{F}_{C_S(x)}(C_G(x)) = C_{\mathcal{F}}(\langle x \rangle) = \mathcal{F}_{C_S(x)}(C_{PSL_n(q)}(x))$.

Clearly, $x$ is $PSL_n(q)$-conjugate to $t_i$. So Lemma <ref> (i) shows together with Lemma <ref> (i) that there exist two distinct $2$-components $J_1$ and $J_2$ of $C_G(x)$ satisfying the following conditions, where $Y_1 := C_S(x) \cap J_1$ and $Y_2 := C_S(x) \cap J_2$.

(1) $\mathcal{F}_{Y_1}(J_1)$ is isomorphic to the $2$-fusion system of $SL_{n-i}(q)$.

(2) $\mathcal{F}_{Y_2}(J_2)$ is isomorphic to the $2$-fusion system of $SL_i(q)$.

(3) $Y_1Y_2$ is normal in $C_S(x)$, and $C_{\mathcal{F}}(\langle x \rangle)/Y_1Y_2$ is nilpotent.

(4) For $k \in \lbrace 1,2 \rbrace$, $Y_k$ contains a $G$-conjugate of $t$.

(5) Any elementary abelian abelian subgroup of $Y_1$ of rank at least $2$ lies in $\mathcal{S}_{n-i}$, and any elementary abelian subgroup of $Y_2$ of rank at least $2$ lies in $\mathcal{S}_i$.

By (3) and Corollary <ref>, $J_1$ and $J_2$ are the only $2$-components of $C_G(x)$. It remains to show that $J_1$ and $J_2$ satisfy (i)-(iii). As $Y_k \in \mathrm{Syl}_2(J_k)$ for $k \in \lbrace 1,2 \rbrace$, (5) implies (iii).

We now prove (ii). The proof of (i) will be omitted since it is very similar to the proof of (ii).

Let $s$ be an element of $J_1$ which is $G$-conjugate to $t$. Set $C := C_G(s)$, $\widehat C := C/O(C)$ and $\widebar{C_G(x)} := C_G(x)/O(C_G(x))$.

Since $\widebar{J_1}$ and $\widebar{J_2}$ are distinct components of $\widebar{C_G(x)}$, we have $[\widebar{J_1},\widebar{J_2}] = 1$ by <cit.>. As $\widebar s \in \widebar{J_1}$, it follows that $\widebar{J_2}$ is a component of $C_{\widebar{C_G(x)}}(\widebar s)$. As a consequence of Corollary <ref> and Proposition <ref>, $C_G(x) \cap C$ has a $2$-component $H$ with $\widebar{H} = \widebar{J_2}$.

By assumption, $s$ is $G$-conjugate to $t$. So, by Proposition <ref>, $\widehat{C}$ has a unique normal subgroup $K^{+}$ isomorphic to $SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$ and a unique normal subgroup $L^{+}$ isomorphic to $SL_2(q^{*})$. Moreover, $K^{+}$ and $L^{+}$ are the only subgroups of $\widehat C$ which are components or solvable $2$-components of $\widehat C$.

Clearly, $\widehat{H}$ is a $2$-component of $C_{\widehat{C}}(\widehat x)$. Lemma <ref> implies that $\widehat H$ is a $2$-component of $C_{K^{+}}(\widehat x)$ or of $C_{L^{+}}(\widehat x)$. By Corollary <ref> (i), we even have that $\widehat H$ is a component of $C_{K^{+}}(\widehat x)$ or $C_{L^{+}}(\widehat x)$. It is easy to note that $\widehat{H}/Z(\widehat{H}) \cong H/Z^{*}(H) \cong \widebar{J_2}/Z(\widebar{J_2})$. By Corollary <ref> (ii), we have $\widehat{H}/Z(\widehat H) \not\cong M_{11}$, and so $\widebar{J_2}/Z(\widebar{J_2}) \not\cong M_{11}$. Now (2) and Lemma <ref> (ii) imply that $\widebar{J_2} \cong SL_i^{\varepsilon_0}(q_0)/O(SL_i^{\varepsilon_0}(q_0))$ for some nontrivial odd prime power $q_0$ and some $\varepsilon_0 \in \lbrace +,- \rbrace$ with $q \sim \varepsilon_0 q_0$. Hence $\widehat{H}/Z(\widehat H) \cong \widebar{J_2}/Z(\widebar{J_2}) \cong PSL_{i}^{\varepsilon_0}(q_0)$. Note that $\varepsilon q^{*} \sim q \sim \varepsilon_0 q_0$ and in particular $({q^{*}}^2-1)_2 = ({q_0}^2-1)_2$. Applying Corollary <ref> (iii), we may conclude that $q_0 = q^{*}$ and $\varepsilon_0 = \varepsilon$. Consequently, we have $J_2/O(J_2) \cong SL_i^{\varepsilon}(q^{*})/O(SL_i^{\varepsilon}(q^{*}))$. So we have proved (ii).

The proof of the following proposition runs along the same lines as that of the previous result.

Suppose that $n$ is odd and $i = n-1$, or that $n$ is even, $i = 1$ and $t_1$ is defined. Let $x$ be an involution of $S$ which is $G$-conjugate to $t_i$. Then $C_G(x)$ has precisely one $2$-component $J$. We have $J/O(J) \cong SL_{n-1}^{\varepsilon}(q^{*})/O(SL_{n-1}^{\varepsilon}(q^{*}))$, where $\varepsilon$ and $q^{*}$ are as in Proposition <ref>. Moreover, any elementary abelian $2$-subgroup of $J$ of rank at least $2$ is $G$-conjugate to a subgroup of $S$ lying in $\mathcal{S}_{n-1}$.

Let $\mathcal{F} := \mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$. It suffices to prove the proposition under the assumption that $\langle x \rangle$ is fully $\mathcal{F}$-centralized, and we will assume that this is the case. So we have $C_S(x) \in \mathrm{Syl}_2(C_G(x))$ and $C_S(x) \in \mathrm{Syl}_2(C_{PSL_n(q)}(x))$. Also, $\mathcal{F}_{C_S(x)}(C_G(x)) = C_{\mathcal{F}}(\langle x \rangle) = \mathcal{F}_{C_S(x)}(C_{PSL_n(q)}(x))$.

Clearly, $x$ is $PSL_n(q)$-conjugate to $t_i$. Lemma <ref> (ii) implies that $C_{\mathcal{F}}(\langle x \rangle)$ has a unique component $\mathcal{E}$, and that $\mathcal{E}$ is isomorphic to the $2$-fusion system of $SL_{n-1}(q)$. Applying Lemma <ref> (i), we may conclude that $C_G(x)$ has a unique $2$-component $J$ with $\mathcal{E} = \mathcal{F}_{C_S(x) \cap J}(J)$. By Lemma <ref> (ii), $J/O(J) \cong SL_{n-1}^{\varepsilon_0}(q_0)/O(SL_{n-1}^{\varepsilon_0}(q_0))$ for some nontrivial odd prime power $q_0$ and some $\varepsilon_0 \in \lbrace +,- \rbrace$ with $\varepsilon_0 q_0 \sim q$.

Set $Y := C_S(x) \cap J$. By Lemma <ref> (ii), $Y \trianglelefteq C_S(x)$ and $C_{\mathcal{F}}(\langle x \rangle)/Y$ is nilpotent. Applying Corollary <ref>, we may conclude that $J$ is the only $2$-component of $C_G(x)$. Using Lemma <ref> (ii), we see that any elementary abelian subgroup of $Y$ of rank at least $2$ lies in $\mathcal{S}_{n-1}$. As $Y \in \mathrm{Syl}_2(J)$, it follows that any elementary abelian $2$-subgroup of $J$ of rank at least $2$ is $G$-conjugate to a subgroup of $S$ lying in $\mathcal{S}_{n-1}$.

It remains to show that $\varepsilon_0 = \varepsilon$ and $q_0 = q^{*}$. Define $s := t_i$ if $i = 1$ and $s := t_A$, where $A := \lbrace 1, \dots, n-1 \rbrace$, if $i = n-1$. Then we have $s \in C_G(t)$, and $s$ is $G$-conjugate to $x$. Set $\widebar{C_G(t)} := C_G(t)/O(C_G(t))$. Lemma <ref> shows that $\widebar s$ centralizes $\widebar K$. Hence, $\widebar K$ is a component of $C_{\widebar{C_G(t)}}(\widebar s)$. As a consequence of Corollary <ref> and Proposition <ref>, $C_G(t) \cap C_G(s)$ has a $2$-component $H$ with $\widebar{H} = \widebar K$. Set $C := C_G(s)$ and $\widehat C := C/O(C)$. Then $\widehat H$ is a $2$-component of $C_{\widehat C}(\widehat t)$. Since $s$ is $G$-conjugate to $x$, $\widehat C$ has precisely one component $J^{+}$, and $J^{+}$ is isomorphic to $SL_{n-1}^{\varepsilon_0}(q_0)/O(SL_{n-1}^{\varepsilon_0}(q_0))$. By Lemma <ref>, $\widehat H$ is a $2$-component of $C_{J^{+}}(\widehat t)$. We see from Corollary <ref> (i) that $\widehat H$ is in fact a component of $C_{J^{+}}(\widehat t)$. It is easy to see that $\widehat H/Z(\widehat H) \cong H/Z^{*}(H) \cong \widebar{K}/Z(\widebar K) \cong PSL_{n-2}^{\varepsilon}(q^{*})$. Note that $\varepsilon_0 q_0 \sim q \sim \varepsilon q^{*}$ and in particular $({q_0}^2-1)_2 = ({q^{*}}^2-1)_2$. Using this, we may deduce from Corollary <ref> (iii) that $q_0 = q^{*}$ and $\varepsilon_0 = \varepsilon$.

§.§ $2$-components of centralizers of involutions conjugate to $w$

Recall that we assume $\rho$ to be an element of $\mathbb{F}_q^{*}$ with order $(n,q-1)$. Recall moreover that if $n$ is even and $\rho$ is a non-square element of $\mathbb{F}_q$, then $\widetilde w$ denotes the matrix

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

and, if $\widetilde w \in SL_n(q)$, then $w$ denotes its image in $PSL_n(q)$.

Suppose that $w$ is defined. Let $P$ be a Sylow $2$-subgroup of $C_{PSL_n(q)}(w)$, and let $\mathcal{F}$ denote the fusion system $\mathcal{F}_P(C_{PSL_n(q)})(w))$. Then $\mathcal{F}$ has precisely one component. This component is isomorphic to the $2$-fusion system of a nontrivial quotient of $SL_{\frac{n}{2}}(q^2)$. If $Y$ is its Sylow subgroup, then we have $Y \trianglelefteq P$, and $\mathcal{F}/Y$ is nilpotent.

By Lemma <ref> (i), $C_{PSL_n(q)}(w)$ has precisely one $2$-component $J$, and $J$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}(q^2)$. Applying Proposition <ref> and Lemma <ref>, we may conclude that $\mathcal{F}_{P \cap J}(J)$ is the only component of $\mathcal{F}$. The last statement of the lemma is given by Lemma <ref> (ii).

Suppose that $w$ is defined. Let $x$ be an involution of $S$ which is $PSL_n(q)$-conjugate to $w$. Then $C_G(x)$ has precisely one $2$-component, say $J$. The group $J/O(J)$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}^{\varepsilon_0}(q_0)$ for some nontrivial odd prime power $q_0$ and some $\varepsilon_0 \in \lbrace +,-\rbrace$ with $q^2 \sim \varepsilon_0 q_0$.

Let $\mathcal{F} := \mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$. It suffices to prove the proposition under the assumption that $\langle x \rangle$ is fully $\mathcal{F}$-centralized, and we will assume that this is the case. So we have $C_S(x) \in \mathrm{Syl}_2(C_G(x))$ and $C_S(x) \in \mathrm{Syl}_2(C_{PSL_n(q)}(x))$. Also, $\mathcal{F}_{C_S(x)}(C_G(x)) = C_{\mathcal{F}}(\langle x \rangle) = \mathcal{F}_{C_S(x)}(C_{PSL_n(q)}(x))$.

As $x$ is $PSL_n(q)$-conjugate to $w$, Lemma <ref> implies that $C_{\mathcal{F}}(\langle x \rangle)$ has precisely one component, say $\mathcal{E}$, and that $\mathcal{E}$ is isomorphic to the $2$-fusion system of a nontrivial quotient of $SL_{\frac{n}{2}}(q^2)$. By Lemma <ref> (i), $C_G(x)$ has a unique $2$-component $J$ such that $\mathcal{E} = \mathcal{F}_{C_S(x) \cap J}(J)$. Set $Y := C_S(x) \cap J$. As a consequence of Lemma <ref>, we have $Y \trianglelefteq C_S(x)$, and the factor system $C_{\mathcal{F}}(\langle x \rangle)/Y$ is nilpotent. So, by Corollary <ref>, $J$ is the only $2$-component of $C_G(x)$. Lemma <ref> (iii) shows that $J/O(J)$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}^{\varepsilon_0}(q_0)$ for some nontrivial odd prime power $q_0$ and some $\varepsilon_0 \in \lbrace +,- \rbrace$ with $q^2 \sim \varepsilon_0 q_0$.

§ THE COMPONENTS OF $C_G(T)$

The goal of this section is to determine the isomorphism types of $K$ and $L$. In order to do so, we will apply the signalizer functor techniques introduced by Gorenstein and Walter in [32]. In particular, we will see that $L$ is isomorphic to $SL_2(q^{*})$. This will enable us in Section <ref> to prove that a certain collection of conjugates of $L$ generates a subgroup $G_0$ of $G$ which is isomorphic to a nontrivial quotient of $SL_n^{\varepsilon}(q^{*})$ and normal in $G$. This will complete the proof of Theorem <ref>.

§.§ $3$-generation of involution centralizers

For each $3 \le i \le n$, we define $\mathcal{U}_i$ to be the set of all subgroups $U$ of $PSL_n(q)$ such that $U$ has a subgroup $E$ with $E \in \mathcal{S}_i$ and $m(E) \ge 3$. The following lemma will be important later in this section.

Let $1 \le i < n$ such that $t_i$ is defined. Suppose that $i \le \frac{n}{2}$ if $n$ is even. Let $x$ be an involution of $S$ such that $x$ is $G$-conjugate to $t_i$ and such that $\langle x \rangle$ is fully $\mathcal{F}_S(G)$-centralized. Then $C_G(x)$ is $3$-generated in the sense of Definition <ref>. Moreover, if $i \ge 4$, then we have

\begin{equation*}
C_G(x) = \langle N_{C_G(x)}(U) \ \vert \ U \le C_S(x), U \in \mathcal{U}_i \rangle.
\end{equation*}

If $i = 2$, then we have

\begin{equation*}
C_G(x) = \langle N_{C_G(x)}(U) \ \vert \ U \le C_S(x), U \in \mathcal{U}_{n-2} \rangle.
\end{equation*}

Set $C := C_G(x)$ and $\widebar C := C/O(C)$. Recall that $L_{2'}(C)$ denotes the subgroup of $C$ generated by the $2$-components of $C$ and that $L(\widebar C)$ denotes the product of all components of $\widebar C$. Clearly, $\widebar{L_{2'}(C)} = L(\widebar C)$.

First we consider the case $(n,i) \ne (6,3)$. Then, by Propositions <ref>, <ref> and <ref>, $C$ has a $2$-component $J$ such that $\widebar J \cong SL_k^{\varepsilon}(q^{*})/O(SL_k^{\varepsilon}(q^{*}))$ for some $k \ge 4$ and such that any elementary abelian subgroup of $Y := C_S(x) \cap J$ of rank at least $2$ lies in $\mathcal{S}_k$. If $i \ge 4$, then we may assume that $k = i$, and if $i = 2$, then $k = n-2$.

Clearly, $Y \in \mathrm{Syl}_2(J)$. By Lemma <ref>, we have that $\widebar J$ is $3$-generated. So we have

\begin{equation*}
\widebar J = \langle N_{\widebar J}(\widebar U) \ \vert \ U \le Y, m(U) \ge 3 \rangle.
\end{equation*}

Set $X := C_S(x) \cap L_{2'}(C)$. By the Frattini argument, $L(\widebar C) = \widebar J N_{L(\widebar C)}(\widebar Y)$ and $\widebar C = L(\widebar C) N_{\widebar C}(\widebar X)$. It follows that

\begin{equation*}
\widebar C = \langle N_{\widebar C}(\widebar U) \ \vert \ \textnormal{$U = X$, or $U \le Y$ and $m(U) \ge 3$} \rangle.
\end{equation*}

Lemma <ref> implies that $C$ is generated by $O(C)$ together with the normalizers $N_C(U)$, where $U = X$, or $U \le Y$ and $m(U) \ge 3$.

Let $E$ denote the subgroup of $S$ generated by $t$, $t_{\lbrace n-2,n-1 \rbrace}$, $t_{\lbrace n-3,n-2 \rbrace}$ and $t_{\lbrace n-4,n-3 \rbrace}$. Clearly, $E \cong E_{16}$. Since $x$ is $G$-conjugate to $t_i$ and $E \le C_G(t_i)$, there is a subgroup $E_x$ of $C_S(x)$ which is $G$-conjugate to $E$. By <cit.>, we have

\begin{equation*}
O(C) = \langle C_{O(C)}(D) \ \vert \ D \le E_x, D \cong E_8 \rangle.
\end{equation*}

As remarked above, any elementary abelian subgroup of $Y$ of rank at least $2$ lies in $\mathcal{S}_k$. So, if $U \le Y$ and $m(U) \ge 3$, then $U \in \mathcal{U}_k$. Also $X \in \mathcal{U}_k$. Clearly, any $E_8$-subgroup of $E_x$ lies in $\mathcal{S}_k$ and hence in $\mathcal{U}_k$. We therefore have

\begin{equation*}
C = \langle N_C(U) \ \vert \ U \le C_S(x), U \in \mathcal{U}_k \rangle.
\end{equation*}

Consequently, $C$ is $3$-generated, and the last two statements of the lemma are satisfied.

Suppose now that $(n,i) = (6,3)$. By Proposition <ref>, $C$ has precisely two $2$-components $J_1$ and $J_2$, and we have $\widebar{J_1} \cong PSL_3^{\varepsilon}(q^{*}) \cong \widebar{J_2}$. Set $Y_1 := C_S(x) \cap J_1$ and $Y_2 := C_S(x) \cap J_2$. Since $\widebar{J_1}$ is $2$-generated by Lemma <ref>, we have

\begin{equation*}
\widebar{J_1} = \langle N_{\widebar{J_1}}(\widebar U) \ \vert \ U \le Y_1, m(U) \ge 2 \rangle.
\end{equation*}

Let $y$ be an involution of $Y_2$. We have $[\widebar{J_1}, \widebar{J_2}] = 1$ by <cit.>, and so $\widebar y$ centralizes $\widebar{J_1}$. As $Z(\widebar{J_1}) = 1$, we have $\widebar y \not\in \widebar{J_1}$. Now let $U \le Y_1$ with $m(U) \ge 2$. Then $\langle \widebar U, \widebar y \rangle$ has rank at least $3$. Moreover, it is clear that $N_{\widebar{J_1}}(\widebar U)$ normalizes $\langle \widebar U, \widebar y \rangle$. Thus

\begin{equation*}
\widebar{J_1} = \langle N_{\widebar{J_1}}(\widebar U) \ \vert \ U \le Y_1Y_2, m(U) \ge 3 \rangle.
\end{equation*}

Interchanging the roles of $J_1$ and $J_2$, we also see that

\begin{equation*}
\widebar{J_2} = \langle N_{\widebar{J_2}}(\widebar U) \ \vert \ U \le Y_1Y_2, m(U) \ge 3 \rangle.
\end{equation*}

By the Frattini argument, $\widebar{C} = \widebar{J_1}\widebar{J_2}N_{\widebar C}(\widebar{Y_1} \widebar{Y_2})$. Lemma <ref> implies that $C$ is generated by $O(C)$ together with the normalizers $N_C(U)$, where $U \le Y_1Y_2$ and $m(U) \ge 3$. For any $E_{16}$-subgroup $A$ of $C_S(x)$, we have

\begin{equation*}
O(C) = \langle C_{O(C)}(B) \ \vert \ B \le A, B \cong E_8 \rangle.
\end{equation*}

by <cit.>. It follows that $C$ is $3$-generated. The proof is now complete.

Suppose that $w$ is defined. Let $x$ be an involution of $S$ which is $PSL_n(q)$-conjugate to $w$. Then $C_G(x)$ is $3$-generated.

Set $C := C_G(x)$ and $\widebar C := C/O(C)$. By Proposition <ref>, $C$ has a unique $2$-component $J$, and $\widebar J$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}^{\varepsilon_0}(q_0)$ for some nontrivial odd prime power $q_0$ and some $\varepsilon_0 \in \lbrace +,-\rbrace$ with $q^2 \sim \varepsilon_0 q_0$. Note that $q_0 \equiv \varepsilon_0 \mod 8$.

First we prove that $\widebar C$ is $3$-generated. Let $R$ be a Sylow $2$-subgroup of $C$ and $Y := R \cap J$. We consider two cases.

Case 1: $n \ge 8$.

By Lemma <ref>, $\widebar J$ is $3$-generated. Hence

\begin{equation*}
\widebar J = \langle N_{\widebar J}(\widebar U) \ \vert \ U \le Y, m(U) \ge 3 \rangle.
\end{equation*}

By the Frattini argument, $\widebar C = \widebar J N_{\widebar C}(\widebar Y)$. So $\widebar C$ is $3$-generated.

Case 2: $n = 6$.

We have $\widebar J \cong PSL_3^{\varepsilon_0}(q_0)$. By Lemma <ref>, $\widebar J$ is $2$-generated. Applying the Frattini argument, we may conclude that

\begin{equation*}
\widebar C = \langle N_{\widebar C}(\widebar U) \ \vert \ U \le Y, m(U) \ge 2 \rangle.
\end{equation*}

Now let $U \le Y$ with $m(U) \ge 2$. Since $\widebar x$ is a central involution of $\widebar C$ and $Z(\widebar J)$ is trivial, we have $\widebar x \not\in \widebar J$ and hence $\widebar x \not\in \widebar U$. It follows $\langle \widebar U, \widebar x \rangle$ has rank at least $3$. Moreover, as $\widebar x$ is central in $\widebar C$, we have $N_{\widebar C}(\widebar U) \le N_{\widebar C}(\langle \widebar U, \widebar x \rangle)$. Clearly, $\langle \widebar U, \widebar x \rangle \le \widebar R$. It follows that

\begin{equation*}
\widebar C = \langle N_{\widebar C}(\widebar U) \ \vert \ U \le R, m(U) \ge 3 \rangle.
\end{equation*}

Hence $\widebar C$ is $3$-generated.

Applying Lemma <ref>, we may conclude that $C$ is generated by $O(C)$ together with the normalizers $N_C(U)$, where $U \le R$ and $m(U) \ge 3$. By Lemma <ref> (iii), $R$ has an elementary abelian $2$-subgroup of rank $4$, say $A$. By <cit.>, we have

\begin{equation*}
O(C) = \langle C_{O(C)}(B) \ \vert \ B \le A, B \cong E_8 \rangle.
\end{equation*}

So $C$ is $3$-generated.

Let $x$ be an involution of $S$. Then $C_G(x)$ is $3$-generated.

As a consequence of Proposition <ref>, $x$ is $G$-conjugate to $t_i$ for some $1 \le i < n$ such that $t_i$ is defined or $PSL_n(q)$-conjugate to $w$ (if defined). So the statement follows from Lemmas <ref> and <ref>.

§.§ The case $q^{*} = 3$

Recall that our goal is to determine the isomorphism types of $K$ and $L$. First we will deal with the case $q^{*} = 3$. We will prove that, in this case, $O(C_G(t)) = 1$.

Let $x$ be an involution of $S$, and let $J$ be a $2$-component of $C_G(x)$. Let $1 \le i < n$ such that $t_i$ is defined. Suppose that $q^{*} = 3$ and that $x$ is $G$-conjugate to $t_i$. Then $J/O(J)$ is locally balanced.

By Propositions <ref> (iii), <ref> and <ref>, we have $J/O(J) \cong SL_k^{\varepsilon}(3)$ for some $3 \le k < n$. So $J/O(J)$ is locally balanced by Lemma <ref>.

Let $P$ and $Q$ be subgroups of $S$.

(i) If $P \in \mathcal{S}$ and $m(P) \le 2$, then there is a subgroup $\widebar P$ of $S$ such that $P < \widebar P$, $\widebar P \in \mathcal{S}$ and $m(\widebar P) = 3$.

(ii) If $P$ and $Q$ are elements of $\mathcal{S}$ of rank at least $3$, then there exist some $m \ge 1$ and a sequence

\begin{equation*}
P = P_1, \dots, P_m = Q,
\end{equation*}

where $P_i$, $1 \le i \le m$, is a subgroup of $S$ of rank at least $2$ lying in $\mathcal{S}$ and where

\begin{equation*}
P_i \subseteq P_{i+1} \ \textnormal{or} \ P_{i+1} \subseteq P_i
\end{equation*}

for all $1 \le i < m$.

Suppose that $P \in \mathcal{S}$ and $m(P) \le 2$. Let $\widetilde S$ be a Sylow $2$-subgroup of $SL_n(q)$ such that $S$ is the image of $\widetilde S$ in $PSL_n(q)$. Note that $\widetilde S$ is unique. Since $P$ is an element of $\mathcal{S}$, there exists some elementary abelian $2$-subgroup $\widetilde P$ of $SL_n(q)$ such that $P$ is the image of $\widetilde P$ in $PSL_n(q)$. Clearly, $\widetilde P \le \widetilde S$. We have $m(\widetilde P) \le 3$ as $m(P) \le 2$. By Corollary <ref>, $\widetilde P$ is contained in an $E_{16}$-subgroup of $\widetilde S$. This implies (i).

We now prove (ii). Suppose that $P$ and $Q$ are elements of $\mathcal{S}$ of rank at least $3$. There are elementary abelian subgroups $\widetilde P$ and $\widetilde Q$ of $SL_n(q)$ such that $P$ is the image of $\widetilde P$ in $PSL_n(q)$ and such that $Q$ is the image of $\widetilde Q$ in $PSL_n(q)$. Clearly, $\widetilde P, \widetilde Q \le \widetilde S$. Also $m(\widetilde P), m(\widetilde Q) \ge 3$. Since $\widetilde S$ is $3$-connected by Corollary <ref>, there exist some $m \ge 1$ and a sequence

\begin{equation*}
\widetilde P = \widetilde{P}_1, \dots, \widetilde{P}_n = \widetilde Q,
\end{equation*}

where $\widetilde{P}_i$ ($1 \le i \le m$) is an elementary abelian subgroup of $\widetilde S$ of rank at least $3$ and where

\begin{equation*}
\widetilde{P}_i \subseteq \widetilde{P}_{i+1} \ \textnormal{or} \ \widetilde{P}_{i+1} \subseteq \widetilde{P}_i
\end{equation*}

for all $1 \le i < m$. Let $P_i$, $1 \le i \le m$, denote the image of $\widetilde{P}_i$ in $S$. Then the sequence

\begin{equation*}
P = P_1, \dots, P_m = Q
\end{equation*}

has the desired properties.

Suppose that $q^{*} = 3$. For each elementary abelian subgroup $E$ of $S$ of rank at least $2$, let

\begin{equation*}
W_E := \langle O(C_G(x)) \ \vert \ x \in E^{\#} \rangle.
\end{equation*}

Let $P$ and $Q$ be subgroups of $S$ with $P, Q \in \mathcal{S}$ and $m(P), m(Q) \ge 3$. Then $W_P = W_Q$.

By Lemma <ref> (ii), there exist some $m \ge 1$ and a sequence

\begin{equation*}
P = P_1, \dots, P_m = Q,
\end{equation*}

where $P_i$, $1 \le i \le m$, is a subgroup of $S$ of rank at least $2$ lying in $\mathcal{S}$ and where

\begin{equation*}
P_i \subseteq P_{i+1} \ \textnormal{or} \ P_{i+1} \subseteq P_i
\end{equation*}

for all $1 \le i < m$. By Lemma <ref> (i), there is a subgroup $\widebar{P_i}$ of $S$ such that $\widebar{P_i} \in \mathcal{S}$, $m(\widebar{P_i}) \ge 3$ and $P_i \le \widebar{P_i}$ for each $1 \le i \le m$.

Let $1 \le i \le m$ and let $x$ be an involution of $\widebar{P_i}$. Also let $J$ be a $2$-component of $C_G(x)$. As $\widebar{P_i} \in \mathcal{S}$, we have that $x$ is $G$-conjugate to $t_j$ for some even $2 \le j < n$. Therefore, by Lemma <ref>, $J/O(J)$ is locally balanced. Applying <cit.>, we may conclude that $G$ is balanced with respect to $\widebar{P_i}$.

Let $1 \le i < m$. We have $m(P_i \cap P_{i+1}) \ge 2$ since $P_i \subseteq P_{i+1}$ or $P_{i+1} \subseteq P_i$ and $m(P_i), m(P_{i+1}) \ge 2$. Hence $m(\widebar{P_i} \cap \widebar{P_{i+1}}) \ge 2$. Proposition <ref> (ii) implies

\begin{equation*}
W_{P_i} = W_{\widebar{P_i}} = W_{\widebar{P_i} \cap \widebar{P_{i+1}}} = W_{\widebar{P_{i+1}}} = W_{P_{i+1}}.
\end{equation*}

Consequently, $W_P = W_Q$, as wanted.

Suppose that $q^{*} = 3$. Let $x$ be an involution of $S$ which is $G$-conjugate to $t_i$ for some even $2 \le i < n$. Then we have $O(C_G(x)) = 1$. In particular, $O(C_G(t)) = 1$.

We follow the pattern of the proof of <cit.>. Let $E$ be the subgroup of $S$ consisting of all $t_A$, where $A \subseteq \lbrace 1, \dots, n \rbrace$ has even order. For each elementary abelian $2$-subgroup $A$ of $G$ of rank at least $2$, let

\begin{equation*}
W_A := \langle O(C_G(y)) \ \vert \ y \in A^{\#} \rangle.
\end{equation*}

Set $W_0 := W_E$ and $M := N_G(W_0)$. We accomplish the proof step by step.

(1) $N_G(S) \le M$.

Let $g \in N_G(S)$. Clearly, $E \in \mathcal{S}$, and it is easy to note $E^g$ still lies in $\mathcal{S}$. Lemma <ref> implies that $W_0 = W_{E^g}$. On the other hand, we have $(W_0)^g = W_{E^g}$ by Proposition <ref> (i). So we have $(W_0)^g = W_0$ and hence $g \in M$.

(2) Let $y$ be an involution of $S$ such that $y$ is $G$-conjugate to $t_j$ for some even $2 \le j < n$. Then $y$ is $M$-conjugate to $t_j$.

We have $\langle y \rangle \in \mathcal{S}$. By Lemma <ref> (i), there is a subgroup $A$ of $S$ with $\langle y \rangle \le A$, $A \in \mathcal{S}$ and $m(A) = 3$. As a consequence of Lemma <ref>, there is an element $g$ of $G$ with $A^g \le E$. By Lemma <ref> and Proposition <ref> (i), we have $(W_0)^g = (W_A)^g = W_{A^g} = W_0$. Thus $g \in M$.

We have $y^g \in E$, and $y^g$ is $G$-conjugate and hence $PSL_n(q)$-conjugate to $t_j$. It is rather easy to show that an element of $E$ is $N_{PSL_n(q)}(E)$-conjugate to $t_j$ if it is $PSL_n(q)$-conjugate to $t_j$. So $y^g$ is $N_{PSL_n(q)}(E)$-conjugate and hence $N_G(E)$-conjugate to $t_j$. As $N_G(E) \le M$, it follows that $y^g$ is $M$-conjugate to $t_j$. Hence $y$ is $M$-conjugate to $t_j$.

(3) Let $y$ be an involution of $S$ such that $y$ is $G$-conjugate to $t_j$ for some even $2 \le j < n$. Then $C_G(y) \le M$.

Because of (2), we may assume that $\langle y \rangle$ is fully $\mathcal{F}_S(G)$-centralized. Then, by Lemma <ref>, $C_G(y)$ is generated by the normalizers $N_{C_G(y)}(U)$, where $U$ is a subgroup of $C_S(y)$ such that there exists $B \le U$ with $B \in \mathcal{S}$ and $m(B) \ge 3$. It suffices to show that each such normalizer lies in $M$.

Let $U$ and $B$ be as above and let $g \in N_{C_G(y)}(U)$. By Lemma <ref> and Proposition <ref> (i), we have $(W_0)^g = (W_B)^g = W_{B^g} = W_0$. Thus $g \in M$ and hence $N_{C_G(y)}(U) \le M$, as required.

(4) Let $y$ be an involution of $S$. Then $C_G(y) \le M$.

We can see from Lemmas <ref> and <ref> that $Z(S)$ has an involution $s$ which is $G$-conjugate to $t_j$ for some even $2 \le j < n$. Let $P$ be a Sylow $2$-subgroup of $C_G(y)$ with $s \in P$. By (1), $s \in M$ and hence $s \in P \cap M$. Now let $r \in N_P(P \cap M)$. Then $s^r \in P \cap M$. As a consequence of (1) and (2), $s^r$ and $s$ are $M$-conjugate to $t_j$. Therefore, there is some $m \in M$ with $s^r = s^m$. We have $rm^{-1} \in C_G(s)$, and so $rm^{-1} \in M$ by (3). Hence $r \in M$. Consequently, $N_P(P \cap M) = P \cap M$. It follows that $P = P \cap M$.

Let $U \le P$ with $m(U) \ge 3$ and let $g \in N_{C_G(y)}(U)$. By Lemma <ref>, any $E_8$-subgroup of $S$ has an involution which is the image of an involution of $SL_n(q)$. Since $m(U) \ge 3$, it follows that $U$ has an element $u$ which is $G$-conjugate to $t_k$ for some even $2 \le k < n$. By the preceding paragraph, $u, u^g \in U \le P \le M$. As a consequence of (1) and (2), $u$ and $u^g$ are $M$-conjugate to $t_k$. So there is some $m \in M$ with $u^g = u^m$. Hence $gm^{-1} \in C_G(u)$. From (3), we see that $C_G(u) \le M$, and so $gm^{-1} \in M$. Thus $g \in M$ and hence $N_{C_G(y)}(U) \le M$. Since $C_G(y)$ is $3$-generated by Corollary <ref>, it follows that $C_G(y) \le M$.

(5) $M = G.$

Assume that $M \ne G$. By <cit.>, we may deduce from (1) and (4) that $M$ is strongly embedded in $G$, i.e. $M \cap M^g$ has odd order for any $g \in G \setminus M$. Applying <cit.>, it follows that $G$ has only one conjugacy class of involutions. On the other hand, we see from Proposition <ref> that $G$ has at least two conjugacy classes of involutions. This contradiction shows that $M = G$.

(6) Conclusion.

Let $y \in E^{\#}$ and let $J$ be a $2$-component of $C_G(y)$. By Lemma <ref>, $J/O(J)$ is locally balanced. So, by <cit.>, $G$ is balanced with respect to $E$. Proposition <ref> (ii) implies that $W_0$ has odd order. By (5), we have $M = G$ and hence $W_0 \trianglelefteq G$. As $O(G) = 1$ by Hypothesis <ref>, it follows that $W_0 = 1$. So we have $O(C_G(y)) = 1$ for all $y \in E^{\#}$, and the statement of the proposition follows.

Proposition <ref> implies that if $q^{*} = 3$, then $K \cong SL_{n-2}^{\varepsilon}(3)$ and $L \cong SL_2(3)$. Our next goal is to find the isomorphism types of $K$ and $L$ for the case $q^{*} \ne 3$.

In general, $O(C_{PSL_n(q)}(t))$ is not trivial. So, if $q^{*}$ is not assumed to be $3$, we have no chance to prove that $O(C_G(t)) = 1$. However, we will be able to show that

\begin{equation*}
\Delta_G(F) = \bigcap_{a \in F^{\#}} O(C_G(a)) = 1
\end{equation*}

for any Klein four subgroup $F$ of $G$ consisting of elements of the form $t_A$, where $A \subseteq \lbrace 1, \dots, n \rbrace$ has even order. This will later enable us to determine the isomorphism types of $K$ and $L$ for the case $q^{*} \ne 3$.

§.§ $2$-balance of $G$

In this subsection, we prove that $G$ is $2$-balanced when $q^{*} \ne 3$.

Set $C := C_G(t)$ and $\widebar C := C/O(C)$. Let $F$ be a Klein four subgroup of $C$. Then $[\Delta_{\overline C}(\overline F), \widebar K] = 1$.

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

First we consider the case that $F$ has a nontrivial element $y$ such that $\widebar y$ centralizes $\widebar K$. Then $\widebar K$ normalizes $O(C_{\widebar C}(\widebar y))$ and, as $\widebar K \trianglelefteq \widebar C$, $O(C_{\widebar C}(\widebar y))$ also normalizes $\widebar K$. It follows that

\begin{equation*}
[\widebar K,O(C_{\widebar C}(\widebar y))] \le \widebar K \cap O(C_{\widebar C}(\widebar y)).
\end{equation*}

Hence, $[\widebar K,O(C_{\widebar C}(\widebar y))]$ is a subgroup of $\widebar K$ with odd order. By <cit.>, $\widebar{K}$ normalizes $[\widebar K,O(C_{\widebar C}(\widebar y))]$. It follows that

\begin{equation*}
[\widebar K,O(C_{\widebar C}(\widebar y))] \le O(\widebar K).
\end{equation*}

As $O(\widebar K) = 1$, this implies that $O(C_{\widebar C}(\widebar y))$ centralizes $\widebar K$. By definition of $\Delta_{\widebar C}(\widebar F)$, we have $\Delta_{\widebar C}(\widebar F) \le O(C_{\widebar C}(\widebar y))$. Consequently, $\Delta_{\widebar C}(\widebar F)$ centralizes $\overline K$.

Now we treat the case that $C_{\widebar F}(\widebar K) = 1$. For each subgroup or element $X$ of $C$, let $\widehat X$ denote the image of $\widebar X$ in $\widebar{C}/C_{\widebar C}(\widebar K)$. Since $C_{\widebar F}(\widebar K) = 1$, we have $\widehat{F} \cong \widebar{F}$, and so $\widehat F$ is a Klein four subgroup of $\widehat C$. As $\widebar{K} \cong SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$, we have that $\widebar{K}$ is locally $2$-balanced (see Lemma <ref>). Using this together with the fact that the group $\widehat C = \widebar{C}/C_{\widebar{C}}(\widebar K)$ is isomorphic to a subgroup of $\mathrm{Aut}(\widebar K)$ containing $\mathrm{Inn}(\widebar K)$, we may conclude that $\Delta_{\widehat C}(\widehat F) = 1$. By <cit.>, if $X$ is a finite group, $B$ a $2$-subgroup of $X$ and $N \trianglelefteq X$, then the image of $O(C_X(B))$ in $X/N$ lies in $O(C_{X/N}(BN/N))$. Thus, if $y$ is an involution of $F$, then the image of $O(C_{\widebar C}(\widebar y))$ in $\widehat C$ lies in $O(C_{\widehat C}(\widehat y))$. It follows that the image of $\Delta_{\widebar C}(\widebar F)$ in $\widehat C$ is contained in $\Delta_{\widehat C}(\widehat F) = 1$. Hence $\Delta_{\widebar C}(\widebar F) \le C_{\widebar{C}}(\widebar K)$.

Let $C := C_G(t)$ and $\widebar C := C/O(C)$. Then $C_{\widebar C}(\widebar K) \cap C_{\widebar C}(\widebar L)$ is a $2$-group.

For convenience, we denote $C_{\widebar C}(\widebar K) \cap C_{\widebar C}(\widebar L)$ by $C_{\widebar C}(\widebar K, \widebar L)$. Since $\widebar C$ is core-free, we have that $C_{\widebar C}(\widebar K, \widebar L)$ is core-free. So it is enough to prove that $C_{\widebar C}(\widebar K, \widebar L)$ is $2$-nilpotent. By <cit.>, it suffices to show that $C_{\widebar C}(\widebar K, \widebar L)$ has a nilpotent $2$-fusion system.

Let $X$ denote the subgroup of $T$ consisting of all elements of $T$ of the form

\begin{equation*}
\begin{pmatrix}A & \\ & B \end{pmatrix} Z(SL_n(q))
\end{equation*}

with $A \in W \cap Z(GL_{n-2}(q))$, $B \in V \cap Z(GL_2(q))$ and $\mathrm{det}(A) \mathrm{det}(B) = 1$.

Let $A \in W$ and $B \in V$ with $\mathrm{det}(A)\mathrm{det}(B) = 1$ and

\begin{equation*}
m := \begin{pmatrix}A & \\ & B \end{pmatrix} Z(SL_n(q)) \in T.
\end{equation*}

Assume that $\widebar m$ centralizes $\widebar K$ and $\widebar L$. Then we have $A \in Z(GL_{n-2}(q))$ by Lemma <ref>. Since $\widebar m$ centralizes $\widebar L$, $\widebar m$ also centralizes $\widebar{X_2}$. Thus $m$ centralizes $X_2$, and so $B$ centralizes $V \cap SL_2(q)$. Lemma <ref> implies that $B \in Z(GL_2(q))$. So we have $m \in X$. Conversely, if $A \in Z(GL_{n-2}(q))$ and $B \in Z(GL_2(q))$, then $\widebar{m} \in C_{\widebar C}(\widebar K, \widebar L)$ as a consequence of Lemmas <ref> and <ref>. It follows that $\widebar T \cap C_{\widebar C}(\widebar K, \widebar L) = \widebar X$.

Let $\mathcal{F} := \mathcal{F}_S(PSL_n(q)) = \mathcal{F}_S(G)$. Since $X$ is central in $C_{PSL_n(q)}(t)$, the only subsystem of $C_{\mathcal{F}}(\langle t \rangle)$ on $X$ is the nilpotent fusion system on $X$. It follows that $\mathcal{F}_{\widebar X}(C_{\widebar C}(\widebar K, \widebar L))$ is nilpotent. So $C_{\widebar C}(\widebar K, \widebar L)$ has a nilpotent $2$-fusion system, as required.

In the following lemma, $A_1$ and $A_2$ have the meanings given to them after Lemma <ref>.

Set $C := C_G(t)$. Suppose that $q^{*} \ne 3$. Then $A_1$, $A_2$ and $L$ are the only $2$-components of $C_C(u)$. Moreover, the following hold:

(i) $A_1$ is the only $2$-component of $C_C(u)$ containing $u$.

(ii) $A_2$ is the only $2$-component of $C_C(u)$ containing neither $u$ nor $t$.

(iii) $L$ is the only $2$-component of $C_C(u)$ containing $t$.

By definition, $A_1$ and $A_2$ are $2$-components of $C_C(u)$. Also, it is clear from the definition of $L$ (see Proposition <ref>) that $L$ is a $2$-component of $C_C(u)$.

Set $\widebar C := C/O(C)$. As a consequence of Lemma <ref>, $\widebar{A_1}$ and $\widebar{A_2}$ are the only $2$-components of $C_{\widebar K}(\widebar u)$. Moreover, $\widebar L$ is a component of $C_{\widebar C}(\widebar u)$. So Lemma <ref> shows that $\widebar{A_1}$, $\widebar{A_2}$ and $\widebar L$ are the only $2$-components of $C_{\widebar C}(\widebar u)$. As we have observed after Lemma <ref>, there is a bijection from the set of $2$-components of $C_C(u)$ to the set of $2$-components of $C_{\widebar C}(\widebar u)$ sending each $2$-component $A$ of $C_C(u)$ to $\widebar A$. Therefore, $A_1$, $A_2$ and $L$ are the only $2$-components of $C_C(u)$.

It remains to prove (i), (ii) and (iii). We have $T_1 \le A_1$ by Lemma <ref> and thus $u \in A_1$. From the definition of $L$, it is clear that $t \in L$. Moreover, $u \not\in L$ since $\widebar t$ is the only involution of $\widebar L$. Similarly, $t \not\in A_1$. Also, it is easy to see from Lemma <ref> that $u$ and $t$ cannot be elements of $A_2$.

Suppose that $q^{*} \ne 3$. Let $F$ be a Klein four subgroup of $T$. Then we have $\Delta_G(F) \cap C_G(t) \le O(C_G(t))$.

Set $C := C_G(t)$, $D := \Delta_G(F) \cap C$ and $\widebar C := C/O(C)$. We are going to show that $\widebar D$ is trivial.

A direct calculation shows that $D \le \Delta_C(F)$. For each $a \in F^{\#}$, we have $\widebar{O(C_C(a))} \le O(C_{\widebar C}(\widebar a))$ as a consequence of Corollary <ref>. Therefore, we have $\widebar{\Delta_C(F)} \le \Delta_{\widebar C}(\widebar F)$, and hence $\widebar D \le \Delta_{\widebar C}(\widebar F)$. Lemma <ref> implies that $[\widebar D, \widebar K] = 1$. In particular, $\widebar D \le C_{\widebar C}(\widebar u) = \widebar{C_C(u)}$. Fix a subgroup $D_0$ of $C_C(u)$ with $\widebar{D_0} = \widebar{D}$. Also, let $g \in G$ with $u^g = t$ and $t^g = u$ (such an element exists by Lemma <ref>). Note that $(D_0)^g \le (C_C(u))^g = C_C(u)$.

We accomplish the proof step by step.

(1) $A_1$, $A_2$ and $L$ are normal subgroups of $C_C(u)$.

This is immediate from Lemma <ref>.

(2) There is a group isomorphism $\mathrm{Aut}(\widebar{A_1}) \rightarrow \mathrm{Aut}(\widebar L)$ which maps $\mathrm{Inn}(\widebar{A_1})$ to $\mathrm{Inn}(\widebar L)$ and $\mathrm{Aut}_{\widebar{(D_0)^g}}(\widebar{A_1})$ to $\mathrm{Aut}_{\widebar D}(\widebar L)$.

Let $\mathrm{Aut}_{D_0}(L/O(L))$ denote the image of $\mathrm{Aut}_{D_0}(L)$ under the natural group homomorphism $\mathrm{Aut}(L) \rightarrow \mathrm{Aut}(L/O(L))$. Also, let $\mathrm{Aut}_{(D_0)^g}(A_1/O(A_1))$ denote the image of $\mathrm{Aut}_{(D_0)^g}(A_1)$ under the natural group homomorphism $\mathrm{Aut}(A_1) \rightarrow \mathrm{Aut}(A_1/O(A_1))$.

From Lemma <ref>, it is clear that $(A_1)^{g^{-1}} = L$. The group isomorphism $c_{g^{-1}}|_{A_1,L}$ induces a group isomorphism $A_1/O(A_1) \rightarrow L/O(L)$, and this group isomorphism induces a group isomorphism $\mathrm{Aut}(A_1/O(A_1)) \rightarrow \mathrm{Aut}(L/O(L))$. By a direct calculation, the group isomorphism just mentioned maps $\mathrm{Aut}_{(D_0)^g}(A_1/O(A_1))$ to $\mathrm{Aut}_{D_0}(L/O(L))$ and $\mathrm{Inn}(A_1/O(A_1))$ to $\mathrm{Inn}(L/O(L))$.

We have $A_1/(A_1 \cap O(C)) \cong \widebar{A_1} \cong SL_2(q^{*})$. As $SL_2(q^{*})$ is core-free, it follows that $A_1 \cap O(C) = O(A_1)$. So the natural group homomorphism $A_1 \rightarrow \widebar{A_1}$ induces a group isomorphism $A_1/O(A_1) \rightarrow \widebar{A_1}$. This group isomorphism induces a group isomorphism $\mathrm{Aut}(A_1/O(A_1)) \rightarrow \mathrm{Aut}(\widebar{A_1})$. By a direct calculation, the group isomorphism just mentioned maps $\mathrm{Aut}_{(D_0)^g}(A_1/O(A_1))$ to $\mathrm{Aut}_{\widebar{(D_0)^g}}(\widebar{A_1})$ and $\mathrm{Inn}(A_1/O(A_1))$ to $\mathrm{Inn}(\widebar{A_1})$. In a very similar way, we obtain an isomorphism $\mathrm{Aut}(L/O(L)) \rightarrow \mathrm{Aut}(\widebar{L})$ which maps $\mathrm{Aut}_{D_0}(L/O(L))$ to $\mathrm{Aut}_{\widebar{D_0}}(\widebar L) = \mathrm{Aut}_{\widebar D}(\widebar{L})$ and $\mathrm{Inn}(L/O(L))$ to $\mathrm{Inn}(\widebar{L})$.

As a consequence of the preceding observations, there is a group isomorphism $\mathrm{Aut}(\widebar{A_1}) \rightarrow \mathrm{Aut}(\widebar L)$ which maps $\mathrm{Inn}(\widebar{A_1})$ to $\mathrm{Inn}(\widebar L)$ and $\mathrm{Aut}_{\widebar{(D_0)^g}}(\widebar{A_1})$ to $\mathrm{Aut}_{\widebar D}(\widebar L)$, as asserted.

(3) $\mathrm{Aut}_{\widebar{(D_0)^g}}(\widebar{A_1}) \le \mathrm{Inn}(\widebar{A_1})$.

As observed above, $\widebar{D_0} = \widebar D$ centralizes $\widebar K$. In particular, $\widebar D$ centralizes $\widebar{A_2}$. This implies that $[D_0,A_2] \le O(C)$. As $D_0$ normalizes $A_2$ by (1), we also have that $[D_0,A_2] \le A_2$. Consequently, $[D_0,A_2] \le O(A_2)$. Because of Lemma <ref>, we have $(A_2)^g = A_2$. It follows that $[(D_0)^g,A_2] \le O(A_2)$. This easily implies $[\widebar{(D_0)^g},\widebar{A_2}] \le O(\widebar{A_2})$. As $\widebar{A_2} \cong SL_{n-4}^{\varepsilon}(q^{*})$ by Lemma <ref>, we have $O(\widebar{A_2}) \le Z(\widebar{A_2})$. It follows that $[\widebar{A_2}, \widebar{(D_0)^g},\widebar{A_2}] = [\widebar{(D_0)^g},\widebar{A_2},\widebar{A_2}] \le [Z(\widebar{A_2}),\widebar{A_2}] = 1$. The Three Subgroups Lemma <cit.> implies $[\widebar{A_2},\widebar{(D_0)^g}] = [\widebar{A_2},\widebar{A_2},\widebar{(D_0)^g}] = 1$. Hence, $\widebar{(D_0)^g}$ centralizes $\widebar{A_2}$. By (1), $\widebar{(D_0)^g}$ normalizes $\widebar{A_1}$. Clearly, $\mathrm{Aut}_{\widebar{(D_0)^g}}(\widebar K)$ has odd order. The assertion now follows from Lemmas <ref> (iii), <ref> and <ref>.

(4) $\widebar D \le \bigcap_{y \in F^{\#}} O(C_{\widebar L}(\widebar y))$.

As a consequence of (2) and (3), we have $\mathrm{Aut}_{\widebar D}(\widebar L) \le \mathrm{Inn}(\widebar L)$. This implies $\widebar D \le \widebar L C_{\widebar C}(\widebar L)$. By <cit.>, $\widebar L \le C_{\widebar C}(\widebar K)$. As observed above, $[\widebar D, \widebar K] = 1$ and hence $\widebar D \le C_{\widebar C}(\widebar K)$. It follows that $\widebar D$ is a subgroup of $\widebar L (C_{\widebar C}(\widebar L) \cap C_{\widebar C}(\widebar K))$. By Lemma <ref>, $C_{\widebar C}(\widebar L) \cap C_{\widebar C}(\widebar K)$ is a $2$-group. As $\widebar D$ has odd order and $\widebar L \trianglelefteq \widebar C$, this implies that $\widebar D \le \widebar L$. Now we see that

\begin{align*}
\widebar D &\le \widebar L \cap \Delta_{\widebar C}(\widebar F) \\
&= \bigcap_{y \in F^{\#}} (\widebar L \cap O(C_{\widebar C}(\widebar y))) \\
&= \bigcap_{y \in F^{\#}} (C_{\widebar L}(\widebar y) \cap O(C_{\widebar C}(\widebar y))) \\
&= \bigcap_{y \in F^{\#}} O(C_{\widebar L}(\widebar y)).
\end{align*}

(5) Conclusion.

As $F$ is a Klein four subgroup of $T$, we have $F = \langle y_1, y_2 \rangle$ for two commuting involutions $y_1$ and $y_2$ of $T$. For $i \in \lbrace 1,2\rbrace$, we have

\begin{equation*}
y_i = \begin{pmatrix} A_i & \\ & B_i \end{pmatrix} Z(SL_n(q))
\end{equation*}

for some $A_i \in W$ and $B_i \in V$ with $\mathrm{det}(A_i) \mathrm{det}(B_i) = 1$. Let $y_3 := y_1y_2, A_3 := A_1A_2$ and $B_3 := B_1B_2$. As $y_1, y_2, y_3$ are involutions, we have $(B_i)^2 \in Z(GL_2(q))$ for each $i \in \lbrace 1,2,3 \rbrace$.

It is easy to note that $\widebar{X_2} \in \mathrm{Syl}_2(\widebar L)$. If $B \in V \cap SL_2(q)$ and

\begin{equation*}
y := \begin{pmatrix} I_{n-2} & \\ & B \end{pmatrix} Z(SL_n(q)) \in X_2,
\end{equation*}

\begin{equation*}
y^{y_i} = \begin{pmatrix} I_{n-2} & \\ & B^{B_i} \end{pmatrix} Z(SL_n(q))
\end{equation*}

for each $i \in \lbrace 1,2,3 \rbrace$. Applying Lemma <ref>, we deduce that

\begin{equation*}
\bigcap_{y \in F^{\#}} O(C_{\widebar L}(\widebar y)) = 1.
\end{equation*}

So we have $\widebar D = 1$ by (4). This completes the proof.

Suppose that $q^{*} \ne 3$. Then $G$ is $2$-balanced.

Let $F$ be a Klein four subgroup of $G$ and let $a$ be an involution of $G$ centralizing $F$. We have to show that $\Delta_G(F) \cap C_G(a) \le O(C_G(a))$.

Assume that $a$ is $G$-conjugate to $t$. Then there is some $g \in G$ with $a^g = t$ and $F^g \le T$. By Lemma <ref>, we have $\Delta_G(F^g) \cap C_G(t) \le O(C_G(t))$. Clearly $\Delta_G(F)^g = \Delta_G(F^g)$. It follows that $\Delta_G(F) \cap C_G(a) \le O(C_G(a))$.

Assume now that $a$ is not $G$-conjugate to $t$. Let $J$ be a $2$-component of $C_G(a)$. By Propositions <ref>, <ref> and <ref>, either $J/O(J) \cong SL_k^{\varepsilon}(q^{*})/O(SL_k^{\varepsilon}(q^{*}))$ for some $k \ge 3$, or $J/O(J)$ is isomorphic to a nontrivial quotient of $SL_{\frac{n}{2}}^{\varepsilon_0}(q_0)$ for some nontrivial odd prime power $q_0$ and some $\varepsilon_0 \in \lbrace +,- \rbrace$. So $J/O(J)$ is locally $2$-balanced by Lemma <ref>. Applying <cit.>, we may conclude that $\Delta_{C_G(a)}(F) \le O(C_G(a))$. A direct calculation shows that $\Delta_G(F) \cap C_G(a) \le \Delta_{C_G(a)}(F)$. Hence $\Delta_G(F) \cap C_G(a) \le O(C_G(a))$.

§.§ The case $q^{*} \ne 3$: Triviality of $\Delta_G(F)$