Suppose that $q^{*} \ne 3$. Assume moreover that $q \equiv 1 \ \mathrm{mod} \ 4$ or $n \ge 7$. Then we have $\Delta_G(F) = 1$ for each Klein four subgroup $F$ of $S$.

We follow the pattern of the proof of <cit.>.

For each elementary abelian $2$-subgroup $A$ of $G$ of rank at least $3$, we define

\begin{equation*}
W_A := \langle \Delta_G(F) \ \vert \ F \le A, m(F) = 2 \rangle.
\end{equation*}

Let $P$ and $Q$ be elementary abelian subgroups of $S$ of rank at least $3$. We claim that $W_P = W_Q$. By Corollary <ref> (iii), $S$ is $3$-connected. So there exist a natural number $m \ge 1$ and a sequence

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

such that $P_i$, $1 \le i \le m$, is an elementary abelian subgroup of $S$ of rank at least $3$ and such that

\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>, $G$ is $2$-balanced. Proposition <ref> (ii) implies that $W_{P_i} = W_{P_{i+1}}$ for all $1 \le i < m$. Therefore, $W_P = W_Q$, as asserted.

We use $W_0$ to denote $W_P$, where $P$ is an elementary abelian subgroup of $S$ of rank at least $3$. Let $M := N_G(W_0)$. We accomplish the proof step by step.

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

Let $g \in N_G(S)$. Take an elementary abelian subgroup $P$ of $S$ with $m(P) \ge 3$. By Proposition <ref> (i), we have $(W_0)^g = (W_P)^g = W_{P^g} = W_0$. Thus $g \in M$.

(2) Let $x$ be an involution of $S$. Then $C_G(x) \le M$.

By Corollary <ref>, there is an elementary abelian subgroup $P$ of $S$ with $x \in P$ and $m(P) = 4$. Clearly, $P \le C_G(x)$. Let $R$ be a Sylow $2$-subgroup of $C_G(x)$ containing $P$. By Corollary <ref>, $C_G(x)$ is $3$-generated. Hence, $C_G(x)$ is generated by the normalizers $N_{C_G(x)}(U)$, where $U \le R$ and $m(U) \ge 3$. It suffices to show that each such normalizer lies in $M$.

So let $U$ be a subgroup of $R$ with $m(U) \ge 3$, and let $g \in N_{C_G(x)}(U)$. Let $Q$ be an elementary abelian subgroup of $U$ with $m(Q) = 3$, and let $h \in G$ with $R^h \le S$. Then $W_{Q^h} = W_{Q^{gh}} = W_{P^h} = W_0$. Proposition <ref> (i) implies that $W_Q = W_{Q^g} = W_P = W_0$. Applying Proposition <ref> (i) again, it follows that $(W_0)^g = (W_Q)^g = W_{Q^g} = W_0$. Hence $g \in M$ and thus $N_{C_G(x)}(U) \le M$.

(3) $M = G$.

Assume that $M \ne G$. By <cit.>, we may deduce from (1) and (2) 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 $M = G$.

(4) Conclusion.

Let $F$ be a Klein four subgroup of $S$. By Corollary <ref>, there is an elementary abelian subgroup $P$ of $S$ with $F \le P$ and $m(P) = 4$. Clearly, $\Delta_G(F) \le W_P$. Since $G$ is $2$-balanced, $W_P$ has odd order by Proposition <ref> (ii). Since $W_P = W_0$, we have $W_P \trianglelefteq G$ by (3). As $O(G) = 1$ by Hypothesis <ref>, it follows that $W_P = 1$. Hence $\Delta_G(F) = 1$.

Next, we deal with the case that $n = 6$, $q \equiv 3 \mod 4$ and $q^{*} \ne 3$. We show that, in this case, $\Delta_G(F) = 1$ for each Klein four subgroup $F$ of $S$ consisting of elements of the form $t_A$, where $A \subseteq \lbrace 1, \dots, n \rbrace$ has even order. We need the following lemma.

Suppose that $q^{*} \ne 3$. Set $\ell := n-4$. Let $E$ be the subgroup of $T$ consisting of all $t_A$, where $A \subseteq \lbrace 1, \dots, n \rbrace$ has even order. Let $E_1$ denote the subgroup of $X_1$ consisting of all $t_A$, where $A$ is a subset of $\lbrace 1,\dots,n-2 \rbrace$ of even order. Then we may choose elements $m_1, \dots, m_{\ell} \in N_K(E_1)$ and an $E_8$-subgroup $E_0$ of $E$ with

\begin{equation*}
K = \langle O(K), L_{2'}(C_K(E_0)), L_{2'}(C_K(E_0))^{m_1}, \dots, L_{2'}(C_K(E_0))^{m_{\ell}} \rangle.
\end{equation*}

Set $C := C_G(t)$ and $\widebar C := C/O(C)$. Let $H := SL_{n-2}^{\varepsilon}(q^{*})/O(SL_{n-2}^{\varepsilon}(q^{*}))$. Let $\widetilde{D}$ be the subgroup of $SL_{n-2}^{\varepsilon}(q^{*})$ consisting of all diagonal matrices in $SL_{n-2}^{\varepsilon}(q^{*})$ with diagonal entries in $\lbrace 1, -1 \rbrace$, and let $D$ denote the image of $\widetilde{D}$ in $H$. Denote by $H_1$ 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 $H$.

We claim that there is a group isomorphism $\psi: \widebar K \rightarrow H$ which maps $\widebar{E_1}$ to $D$ and $\widebar{A_1}$ to $H_1$. By Lemma <ref> (iii), there is a group isomorphism $\varphi: \widebar{K} \rightarrow H$ under which $\widebar{A_1}$ corresponds to $H_1$. Since $\widebar u$ is the only involution of $\widebar{A_1}$, we have that $\widebar u^{\varphi}$ is the image of $\mathrm{diag}(-1,-1,1,\dots,1) \in SL_{n-2}^{\varepsilon}(q^{*})$ in $H$. Clearly, $\widebar {E_1}$ is elementary abelian of order $2^{n-3}$. Using Lemma <ref>, we conclude that $\widebar{E_1}^{\varphi}$ is $H$-conjugate to $D$. So there is some $\alpha \in \mathrm{Inn}(H)$ mapping $\widebar{E_1}^{\varphi}$ to $D$. We may assume that $\alpha$ centralizes $\widebar{u}^{\varphi}$. Then ${H_1}^{\alpha} = H_1$, and the isomorphism $\psi := \varphi\alpha$ maps $\widebar{E_1}$ to $D$ and $\widebar{A_1}$ to $H_1$, as desired.

Using Lemma <ref>, we can find elements $x_1, \dots, x_{\ell} \in N_H(D)$ such that $H = \langle H_1$, ${H_1}^{x_1}$, …, ${H_1}^{x_{\ell}} \rangle$. Therefore, $K$ has elements $m_1$, $\dots$, $m_{\ell}$ such that

\begin{equation*}
\widebar K = \langle \widebar{A_1}, \widebar{A_1}^{\widebar{m_1}}, \dots, \widebar{A_1}^{\widebar{m_{\ell}}} \rangle
\end{equation*}

and $\widebar{m_1}, \dots, \widebar{m_{\ell}} \in N_{\widebar K}(\widebar{E_1})$. From Lemma <ref>, we see that $N_{\widebar{K}}(\widebar E_1) = \widebar{N_K(E_1)}$. So we may assume $m_i \in N_K(E_1)$ for $i \in \lbrace 1, \dots, \ell \rbrace$. Let $E_0 := \langle u, t_{\lbrace 3,4 \rbrace}, t_{\lbrace 4,5 \rbrace} \rangle$. By Lemma <ref>, we have $\widebar{A_1} \trianglelefteq C_{\widebar C}(\widebar u)$. In particular, $\widebar{E_0}$ normalizes $\widebar{A_1}$. Moreover, $\widebar{E_0}$ centralizes $\widebar{T_1}$. We have $\widebar{A_1} \cong SL_2(q^{*})$ and $\widebar{T_1} \in \mathrm{Syl}_2(\widebar{A_1})$ (see Lemma <ref>). Applying Lemma <ref>, we conclude that $\widebar{A_1} \le C_{\widebar K}(\widebar {E_0})$. As $\widebar{A_1} \trianglelefteq C_{\widebar K}(\widebar u)$ and $\widebar{A_1} \le C_{\widebar K}(\widebar{E_0}) \le C_{\widebar K}(\widebar u)$, we even have that $\widebar{A_1}$ is a component of $C_{\widebar K}(\widebar {E_0})$. It follows that

\begin{equation*}
\widebar K = \langle L_{2'}(C_{\widebar K}(\widebar{E_0})), L_{2'}(C_{\widebar K}(\widebar{E_0}))^{\widebar{m_1}}, \dots, L_{2'}(C_{\widebar K}(\widebar{E_0}))^{\widebar{m_{\ell}}} \rangle.
\end{equation*}

Let $k \in K$ such that $\widebar{k} \in C_{\widebar K}(\widebar{E_0})$. As $K \trianglelefteq C$, we have $[k,E_0] \le O(C) \cap K = O(K)$. Thus $k O(K) \in C_{C/O(K)}(E_0O(K)/O(K))$. By Lemma <ref>, there is an element $z \in C_C(E_0)$ such that $kO(K) = zO(K)$. Observing that $z \in C_K(E_0)$ and that $\widebar{k} = \widebar{z}$, we may conclude that $C_{\widebar K}(\widebar{E_0}) = \widebar{C_K(E_0)}$. If $1 \le i \le \ell$, then $L_{2'}(C_{\widebar K}(\widebar{E_0}))^{\widebar{m_i}} = L_{2'}(\widebar{C_K(E_0)})^{\widebar{m_i}} = \widebar{L_{2'}(C_K(E_0))}^{\widebar{m_i}} = \widebar{L_{2'}(C_K(E_0))^{m_i}}$, where the second equality follows from Proposition <ref>. It follows that

\begin{equation*}
K = \langle O(K), L_{2'}(C_K(E_0)), L_{2'}(C_K(E_0))^{m_1}, \dots, L_{2'}(C_K(E_0))^{m_{\ell}} \rangle.
\end{equation*}

This completes the proof.

Suppose that $n = 6$, $q \equiv 3 \ \mathrm{mod} \ 4$ and $q^{*} \ne 3$. Let $E$ denote the subgroup of $S$ consisting of all $t_A$, where $A$ is a subset of $\lbrace 1, \dots, n \rbrace$ of even order. Then $\Delta_G(F) = 1$ for any Klein four subgroup $F$ of $E$.

We follow the pattern of the proof of <cit.>.

Set $W_0 := \langle \Delta_G(F) \ \vert \ F \le E, m(F) = 2 \rangle$ and $M := N_G(W_0)$. Since $T$ is the image of

\begin{equation*}
\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*}

in $PSL_n(q)$, we have $T \in \mathrm{Syl}_2(PSL_n(q))$ by Lemma <ref>. Hence $S = T$ and thus $t \in Z(S)$. By choice of $W$ (see Section <ref>), we have

\begin{equation*}
W = \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*}

We accomplish the proof step by step.

(1) For each subgroup $E_0$ of $E$ with order at least $8$, we have $N_G(E_0) \le M$.

Clearly, $E \cong E_{16}$. Therefore, the statement follows from the $2$-balance of $G$ (see Lemma <ref>) and Proposition <ref> (ii).

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

First we prove $S \le M$. By (1), we have $E \le M$. As $q \equiv 3 \ \mathrm{mod} \ 4$ and $S = T$, any element of $S$ can be written as a product of an element of $E$ and an element of $S$ induced by a matrix of the form

\begin{equation*}
\begin{pmatrix}
A & \\ & B
\end{pmatrix}
\end{equation*}

with $A \in W \cap SL_4(q)$ and $B \in V \cap SL_2(q)$. So, in order to prove that $S \le M$, it suffices to show that each element of $S$ induced by a matrix of this form lies in $M$. If $B \in V \cap SL_2(q)$, then the image of

\begin{equation*}
\begin{pmatrix}
I_4 & \\ & B
\end{pmatrix}
\end{equation*}

in $S$ centralizes the group $\langle t_{\lbrace 1,2 \rbrace}, t_{\lbrace 2,3 \rbrace}, t_{\lbrace 3,4 \rbrace} \rangle \cong E_8$. So it is contained in $M$ by (1). Hence, in order to prove that $S \le M$, it suffices to show that if $A \in W \cap SL_4(q)$, then the image of

\begin{equation*}
\begin{pmatrix}
A & \\ & I_2
\end{pmatrix}
\end{equation*}

in $S$ lies in $M$. So assume that $A \in W \cap SL_4(q)$. By the structure of $W$, there are elements $M_1$, $M_2$ of $V$ such that $\mathrm{det}(M_1) = \mathrm{det}(M_2)$ and

\begin{equation*}
A =
\begin{pmatrix}
M_1 & \\ & M_2
\end{pmatrix} \
\textnormal{or} \
A = \begin{pmatrix}
M_1 & \\ & M_2
\end{pmatrix}
\begin{pmatrix}
& I_2 \\ I_2 &
\end{pmatrix}.
\end{equation*}

The image of

\begin{equation*}
\begin{pmatrix}
M_1 & & \\ & M_2 & \\ & & I_2
\end{pmatrix}
\end{equation*}

in $S$ can be written as a product of an element of $E$ and an element of $S$ induced by a matrix of the form

\begin{equation*}
\begin{pmatrix}
\widetilde{M_1} & & \\ & \widetilde{M_2} & \\ & & I_2
\end{pmatrix}
\end{equation*}

with $\widetilde{M_1},\widetilde{M_2} \in V \cap SL_2(q)$. The images of

\begin{equation*}
\begin{pmatrix}
\widetilde{M_1} & \\ & I_4
\end{pmatrix} \ \textnormal{and} \ \begin{pmatrix}
I_2 & & \\ & \widetilde{M_2} & \\ & & I_2
\end{pmatrix}
\end{equation*}

in $S$ centralize the groups $\langle t_{\lbrace 3,4 \rbrace}, t_{\lbrace 4,5 \rbrace}, t_{\lbrace 5,6 \rbrace} \rangle$ and $\langle t_{\lbrace 1,2 \rbrace}, t_{\lbrace 2,5 \rbrace}, t_{\lbrace 5,6 \rbrace} \rangle$, respectively. So they are elements of $M$. It follows that the image of

\begin{equation*}
\begin{pmatrix}
M_1 & & \\ & M_2 & \\ & & I_2
\end{pmatrix}
\end{equation*}

in $S$ lies in $M$. The image of the block matrix

\begin{equation*}
\begin{pmatrix}
& I_2 & \\
I_2 & & \\
& & I_2
\end{pmatrix}
\end{equation*}

in $S$ normalizes $E$ and is thus contained in $M$. It follows that the image of

\begin{equation*}
\begin{pmatrix}
A & \\ & I_2
\end{pmatrix}
\end{equation*}

in $S$ lies in $M$. Consequently, $S \le M$.

By Lemma <ref>, $\mathrm{Aut}_{PSL_n(q)}(S) = \mathrm{Inn}(S)$. As $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$, it follows that $\mathrm{Aut}_G(S) = \mathrm{Inn}(S)$, and so $N_G(S) = SC_G(S)$. We have seen above that $S \le M$, and we have $C_G(S) \le M$ by (1). Hence $N_G(S) \le M$.

(3) $C_G(t) \le M$.

Let $E_1$ be the subgroup of $X_1$ consisting of all $t_A$, where $A$ is a subset of $\lbrace 1, \dots, n-2 \rbrace$ of even order. As a consequence of Lemma <ref>, there is an $E_8$-subgroup $E_0$ of $E$ such that $K = \langle O(K), C_K(E_0), N_K(E_1) \rangle$. By (1), $C_K(E_0)$ and $N_K(E_1)$ are subgroups of $M$. By <cit.>, we have

\begin{equation*}
O(K) = \langle C_{O(K)}(B) \ \vert \ B \le E, m(B) = 3 \rangle.
\end{equation*}

Therefore, $O(K) \le M$ by (1). Consequently, $K \le M$. By the Frattini argument,

\begin{equation*}
C_G(t) = K N_{C_G(t)}(X_1).
\end{equation*}

So it suffices to show that $N_{C_G(t)}(X_1) \le M$. Since $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$, we may conclude from Lemma <ref> that $\mathrm{Aut}_{C_G(t)}(X_1)$ is a $2$-group. Hence, $N_{C_G(t)}(X_1)/C_{C_G(t)}(X_1)$ is a $2$-group. As $X_1 \trianglelefteq T = S \in \mathrm{Syl}_2(C_G(t))$, it follows that $N_{C_G(t)}(X_1) = S C_{C_G(t)}(X_1)$. We have $S \le M$ by (2), and $C_{C_G(t)}(X_1) \le C_G(E_1) \le M$ by (1). Consequently, $N_{C_G(t)}(X_1) \le M$, as required.

(4) Let $x$ be an involution of $S$ which is $G$-conjugate to $t$. Then $x$ is $M$-conjugate to $t$.

It is easy to see that if an element of $T$ is $PSL_n(q)$-conjugate to $t$, then it is $C_{PSL_n(q)}(t)$-conjugate to an element of $E$. As $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$ and $S = T$, it follows that $x$ is $C_G(t)$-conjugate and hence $M$-conjugate to an element $y$ of $E$. It is rather easy to show that if an element of $E$ is $PSL_n(q)$-conjugate to $t$, then it is $N_{PSL_n(q)}(E)$-conjugate to $t$. So, as $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$, we have that $y$ is $N_G(E)$-conjugate to $t$. By (1), $N_G(E) \le M$, and so $x$ is $M$-conjugate to $t$.

(5) Let $x$ be an involution of $S$. Then $C_G(x) \le M$.

Let $R$ be a Sylow $2$-subgroup of $C_G(x)$ with $C_S(x) \le R$. We have $t \in Z(S) \le C_S(x)$ and $t \in M$. Thus $t \in R \cap M$. Let $r \in N_R(R \cap M)$. Then $y := t^r \in R \cap M$. As a consequence of (4), $y$ is $M$-conjugate to $t$. So there is an element $m$ of $M$ such that $t^r = y = t^m$. We have $rm^{-1} \in C_G(t) \le M$ by (3), and so $r \in R \cap M$. Hence, $N_R(R \cap M) = R \cap M$, and thus $R = R \cap M$.

By Corollary <ref>, $C_G(x)$ is $3$-generated. Therefore, $C_G(x)$ is generated by the normalizers $N_{C_G(x)}(U)$, where $U \le R$ and $m(U) \ge 3$. It suffices to show that each such normalizer lies in $M$.

So let $U \le R$ with $m(U) \ge 3$, and let $g \in N_{C_G(x)}(U)$. Take an elementary abelian subgroup $Q$ of $U$ of rank $3$. Lemma <ref> shows that any $E_8$-subgroup of $S$ has an involution which is the image of an involution of $SL_n(q)$. This implies that $Q$ has an element $s$ which is $G$-conjugate to $t$. Since $s, s^g \in U \le R \le M$, we see from (4) that $s$ and $s^g$ are $M$-conjugate to $t$. So there are elements $m,m' \in M$ such that $s = t^m$ and $s^g = t^{m'}$. We have $t^{m'} = s^g = (t^m)^g = t^{mg}$. Thus $mgm'^{-1} \in C_G(t) \le M$, and hence $g \in M$. It follows that $N_{C_G(x)}(U) \le M$.

(6) $M = G$.

Assume that $M \ne G$. By <cit.>, we may deduce from (2) and (5) 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 precisely two conjugacy classes of involutions. This contradiction shows that $M = G$.

(7) Conclusion.

Let $F$ be a Klein four subgroup of $E$. Clearly, $\Delta_G(F) \le W_0$. By (6), we have $W_0 \trianglelefteq G$. Since $G$ is $2$-balanced, $W_0$ has odd order by Proposition <ref> (ii). As $O(G) = 1$ by Hypothesis <ref>, it follows that $W_0 = 1$. Hence $\Delta_G(F) = 1$.

§.§ Quasisimplicity of the $2$-components of $C_G(t)$

In this subsection, we determine the isomorphism types of $K$ and $L$.

Let $x$ and $y$ be two commuting involutions of $G$. Set $C := C_G(x)$ and $\widebar C := C/O(C)$. Then any $2$-component of $C_{\widebar C}(\widebar y)$ is a component of $C_{\widebar C}(\widebar y)$.

By <cit.>, $L_{2'}(C_{\widebar C}(\widebar y)) = L_{2'}(C_{L(\widebar C)}(\widebar y))$. We know from Section <ref> that $L(\widebar C)$ is a $K$-group, i.e. the composition factors of $L(\widebar C)$ are known finite simple groups. Applying <cit.>, we conclude that $L_{2'}(C_{L(\widebar C)}(\widebar y)) = L(C_{L(\widebar C)}(\widebar y))$. Therefore, any $2$-component of $C_{L(\widebar C)}(\widebar y)$ is a component of $C_{L(\widebar C)}(\widebar y)$. So any $2$-component of $C_{\widebar C}(\widebar y)$ is a component of $C_{\widebar C}(\widebar y)$.

Instead of using <cit.>, the lemma could be proved directly by using Corollary <ref> (i) and the results of Section <ref>.

$K$ is isomorphic to a quotient of $SL_{n-2}^{\varepsilon}(q^{*})$ by a central subgroup of odd order.

The proof is inspired from the proof of <cit.>.

For $q^{*} = 3$, the proposition follows from Proposition <ref>. From now on, we assume that $q^{*} \ne 3$.

Set $C := C_G(t)$. Let $E$ denote the subgroup of $T$ consisting of all $t_A$, where $A \subseteq \lbrace 1, \dots, n \rbrace$ has even order. We assume $m_1, \dots, m_{\ell}$, where $\ell := n-4$, to be elements of $K$ and $E_0$ to be an $E_8$-subgroup of $E$ with

\begin{equation*}
K = \langle O(K), L_{2'}(C_K(E_0)), L_{2'}(C_K(E_0))^{m_1}, \dots, L_{2'}(C_K(E_0))^{m_{\ell}} \rangle.
\end{equation*}

Such elements $m_1, \dots, m_{\ell}$ and such a subgroup $E_0$ exist by Lemma <ref>.

The proof will be accomplished step by step.

(1) Let $f$ be an involution of $E_0$. Then $L_{2'}(C_K(E_0)) \le L_{2'}(C_C(f))$.

As $K \trianglelefteq C$, we have $C_K(E_0) \trianglelefteq C_C(E_0)$. This implies $L_{2'}(C_K(E_0)) \le L_{2'}(C_C(E_0))$. By <cit.>, we have $L_{2'}(C_{C_C(f)}(E_0)) \le L_{2'}(C_C(f))$. Clearly, $C_{C_C(f)}(E_0) = C_C(E_0)$. It follows that $L_{2'}(C_K(E_0)) \le L_{2'}(C_C(E_0)) \le L_{2'}(C_C(f))$.

(2) Let $F$ be a Klein four subgroup of $E_0$. Set $D := [C_{O(K)}(F),L_{2'}(C_K(E_0))]$. Then $D = 1$.

Clearly, $L_{2'}(C_K(E_0))$ normalizes $C_{O(K)}(F)$. Also, $O^{2'}(L_{2'}(C_K(E_0))) = L_{2'}(C_K(E_0))$, and $C_{O(K)}(F)$ is a $2'$-group. Applying <cit.>, we conclude that $D = [D,L_{2'}(C_K(E_0))]$.

Now let $f$ be an involution of $F$. We are going to show that $D \le O(C_G(f))$. Set $M := L_{2'}(C_C(f))$. By (1), $L_{2'}(C_K(E_0)) \le M$. Also, $D \le C_C(F) \le C_C(f)$ and $M \trianglelefteq C_C(f)$. It follows that $D = [D,L_{2'}(C_K(E_0))] \le [C_C(f), M] \le M$.

Let $\widebar{C_G(f)} := C_G(f)/O(C_G(f))$. By Corollary <ref>, $C_{\widebar{C_G(f)}}(\widebar t) = \widebar{C_C(f)}$. As a consequence of Proposition <ref>, $L_{2'}(C_{\widebar{C_G(f)}}(\widebar t)) = \widebar{M}$. Lemma <ref> implies that $\widebar M = L(C_{\widebar{C_G(f)}}(\widebar t))$. It easily follows that $O(\widebar M)$ is central in $\widebar{M}$.

From the definition of $D$, it is clear that $D \le O(K)$. So we have $D \le M \cap O(K) \le O(M)$. It follows that $\widebar D \le \widebar{O(M)} \le O(\widebar M) \le Z(\widebar M)$. In particular, $\widebar{L_{2'}(C_K(E_0))}$ centralizes $\widebar D$. Thus $D = [D,L_{2'}(C_K(E_0))] \le O(C_G(f))$.

Since $f$ was arbitrarily chosen, it follows that $D \le \Delta_G(F)$. By Lemmas <ref> and <ref>, we have $\Delta_G(F) = 1$. Consequently, $D = 1$, as wanted.

(3) $O(K) \le Z(K)$.

By <cit.>, we have

\begin{equation*}
O(K) = \langle C_{O(K)}(F) : F \le E_0, m(F) = 2 \rangle.
\end{equation*}

Because of (2), it follows that $O(K)$ centralizes $L_{2'}(C_K(E_0))$. By choice of $E_0$, we have

\begin{equation*}
K = \langle O(K), L_{2'}(C_K(E_0)), L_{2'}(C_K(E_0))^{m_1}, \dots, L_{2'}(C_K(E_0))^{m_{\ell}} \rangle
\end{equation*}

for some $m_1, \dots, m_{\ell} \in K$. It follows that $K = O(K)C_K(O(K))$. Therefore, $C_K(O(K))$ has odd index in $K$. We have $O^{2'}(K) = K$ since $K$ is a $2$-component of $C$. It follows that $K = C_K(O(K))$. Consequently, $O(K) \le Z(K)$.

(4) Conclusion.

Applying <cit.>, we deduce from (3) that $K$ is a component of $C$. Therefore, $K$ is quasisimple. We have

\begin{equation*}
K/Z(K) \cong (K/O(K))/Z(K/O(K)) \cong PSL_{n-2}^{\varepsilon}(q^{*}).
\end{equation*}

Applying Lemmas <ref> and <ref>, we conclude that $K \cong SL_{n-2}^{\varepsilon}(q^{*})/Z$ for some central subgroup $Z$ of $SL_{n-2}^{\varepsilon}(q^{*})$. Using Proposition <ref>, or using the order formulas for $\vert SL_{n-2}^{\varepsilon}(q^{*}) \vert$ and $\vert SL_{n-2}(q) \vert$ given by <cit.>, we see that

\begin{equation*}
\vert SL_{n-2}^{\varepsilon}(q^{*}) \vert_2 = \vert SL_{n-2}(q) \vert_2 = \vert X_1 \vert = \vert K \vert_2 = \vert SL_{n-2}^{\varepsilon}(q^{*})/Z \vert_2.
\end{equation*}

Thus $Z$ has odd order.

We have $L \cong SL_2(q^{*})$ and $L \trianglelefteq C_G(t)$. Moreover, $L$ is the only normal subgroup of $C_G(t)$ which is isomorphic to $SL_2(q^{*})$.

For $q^{*} = 3$, this follows from Propositions <ref> and <ref>.

Assume now that $q^{*} \ne 3$. Let $\widetilde K := KO(C_G(t))$. By the last statement in Proposition <ref>, $K = O^{2'}(\widetilde K)$. Let $i \in \lbrace 1,2 \rbrace$. Since $A_i$ is a $2$-component of $C_{C_G(t)}(u)$, we have $A_i = O^{2'}(A_i)$. Also, $A_i \le \widetilde K$, and so $A_i \le O^{2'}(\widetilde K) = K$. It follows that $A_i$ is a $2$-component of $C_K(u)$.

By Proposition <ref>, we have $K \cong SL_{n-2}^{\varepsilon}(q^{*})/Z$ for some central subgroup $Z$ of $SL_{n-2}^{\varepsilon}(q^{*})$ with odd order. It is easy to see that if $m$ is a non-central involution of $SL_{n-2}^{\varepsilon}(q^{*})/Z$ and $J$ is a $2$-component of its centralizer in $SL_{n-2}^{\varepsilon}(q^{*})/Z$, then $J \cong SL_k^{\varepsilon}(q^{*})$ for some $k \ge 2$. Since $u$ is a non-central involution of $K$ and $A_1/O(A_1) \cong SL_2(q^{*})$, it follows that $A_1 \cong SL_2(q^{*})$. By definition of $L$ (see Proposition <ref>), $L$ is isomorphic to $A_1$. So we have $L \cong SL_2(q^{*})$.

Let $L_0$ be the $2$-component of $C_G(t)$ associated to $LO(C_G(t))/O(C_G(t))$. By <cit.>, we have $[L_0,K] = 1$. Hence $L_0 \le C_{C_G(t)}(u)$. So $L_0$ is a $2$-component of $C_{C_G(t)}(u)$. Clearly $A_1 \ne L_0 \ne A_2$. Lemma <ref> implies that $L_0 = L$. From Proposition <ref> (iii), we see that $L = L_0 \trianglelefteq C_G(t)$.

Proposition <ref> (iii) also shows that $K$ and $L$ are the only $2$-components of $C_G(t)$. So $L$ is the only normal subgroup of $C_G(t)$ isomorphic to $SL_2(q^{*})$.

§ THE SUBGROUP $G_0$

Let $A$ be a subset of $\lbrace 1, \dots, n \rbrace$ with order $2$. Then $t_A$ is $G$-conjugate to $t$. Proposition <ref> implies that $C_G(t_A)$ has a unique normal subgroup isomorphic to $SL_2(q^{*})$. We denote this subgroup by $L_{A}$, and we define $G_0$ to be the subgroup of $G$ generated by the groups $L_A$, where $A = \lbrace i,i+1 \rbrace$ for some $1 \le i < n$. We are going to prove that $G_0 \trianglelefteq G$ and that $G_0$ is isomorphic to a nontrivial quotient of $SL_n^{\varepsilon}(q^{*})$. This will complete the proof of Theorem <ref>.

By Proposition <ref>, $K$ is isomorphic to a quotient of $SL_{n-2}^{\varepsilon}(q^{*})$ by a central subgroup of odd order. By the proof of Proposition <ref>, $A_1$ and $A_2$ are $2$-components of $C_K(u)$ if $q^{*} \ne 3$.

Let $Z \le Z(SL_{n-2}^{\varepsilon}(q^{*}))$ with $K \cong H := SL_{n-2}^{\varepsilon}(q^{*})/Z$. Let $H_1$ be 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 $H_2$ the image of

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

in $H$. Then there is a group isomorphism $\varphi: K \rightarrow H$ which maps $A_1$ to $H_1$ and $A_2$ to $H_2$.

For $q^{*} = 3$, this follows from Proposition <ref> and Lemma <ref> (iii).

Assume now that $q^{*} \ne 3$. Let $\varphi : K \rightarrow H$ be a group isomorphism. For each even natural number $k$ with $2 \le k < n-2$, let $h_k$ be the image of

\begin{equation*}
\begin{pmatrix}
-I_k & \\ & I_{n-2-k}
\end{pmatrix}
\end{equation*}

in $H$. It is easy to note that each non-central involution of $H$ is conjugate to $h_k$ for some even $2 \le k < n-2$. As $u$ is a non-central involution of $K$, we may assume that $u^{\varphi} = t_k$ for some even $2 \le k < n-2$.

Let $\widetilde{H_1}$ be the image of

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

in $H$ and $\widetilde{H_2}$ be the image of

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

in $H$. It is easy to note that the $2$-components of $C_H(t_k)$ are precisely the quasisimple elements of $\lbrace \widetilde{H_1}, \widetilde{H_2} \rbrace$. Also, $t_k \in \widetilde{H_1}$, but $t_k \not\in \widetilde{H_2}$. On the other hand, $A_1$ and $A_2$ are the $2$-components of $C_K(u)$, and we have $u \in A_1$. This implies $(A_1)^{\varphi} = \widetilde{H_1}$ and $(A_2)^{\varphi} = \widetilde{H_2}$. Since $A_1 \cong L \cong SL_2(q^{*})$, we have $k = 2$, and hence $\widetilde{H_1} = H_1$ and $\widetilde{H_2} = H_2$.

Let $1 \le i < j < n$. Set $A := \lbrace i,i+1 \rbrace$ and $B:= \lbrace j,j+1 \rbrace$. Then:

(i) If $i+1 < j$, then $[L_A,L_B] = 1$.

(ii) Suppose that $j = i+1$. Then there is a group isomorphism from $\langle L_A, L_B \rangle$ to $SL_3^{\varepsilon}(q^{*})$ under which $L_A$ corresponds to the subgroup

\begin{equation*}
\left \lbrace \left( \begin{array}{c|c} M & \begin{matrix} 0 \\ 0 \end{matrix} \\ \hline \begin{matrix} 0 & 0 \end{matrix} & 1 \end{array} \right) \ : \ M \in SL_2^{\varepsilon}(q^{*}) \right\rbrace
\end{equation*}

of $SL_3^{\varepsilon}(q^{*})$ and under which $L_B$ corresponds to the subgroup

\begin{equation*}
\left \lbrace \left( \begin{array}{c|cc} 1 & \begin{matrix} 0 & 0 \end{matrix} \\ \hline \begin{matrix} 0 \\ 0 \end{matrix} & M \end{array} \right) \ : \ M \in SL_2^{\varepsilon}(q^{*}) \right\rbrace
\end{equation*}

of $SL_3^{\varepsilon}(q^{*})$.

(iii) Suppose that $1 \le i \le n-3$ and that $j = i+1$. Set $k := i+2$ and $C := \lbrace k,k+1 \rbrace$. Then $\langle L_A, L_B, L_C \rangle$ is isomorphic to $SL_4^{\varepsilon}(q^{*})$.

Let $H$, $H_1$, $H_2$ and $\varphi$ be as in Lemma <ref>. For each $D \subseteq \lbrace 1, \dots, n-2\rbrace$ of even order, let $h_D$ be the image of the matrix $\mathrm{diag}(d_1,\dots, d_{n-2}) \in SL_{n-2}^{\varepsilon}(q^{*})$ in $H$, where $d_{\ell} = -1$ if $\ell \in D$ and $d_{\ell} = 1$ if $\ell \in \lbrace 1, \dots, n-2 \rbrace \setminus D$. Note that $u^{\varphi} = h_{\lbrace 1,2 \rbrace}$. Let $J$ be the subgroup of $H$ consisting of all $h_D$, where $D \subseteq \lbrace 1, \dots, n-2 \rbrace$ has even order, and let $E_1$ denote the subgroup of $X_1$ consisting of all $t_D$, where $D \subseteq \lbrace 1, \dots, n-2 \rbrace$ has even order. From Lemma <ref>, we see that $(E_1)^{\varphi}$ is $C_H(u^{\varphi})$-conjugate to $J$. Upon replacing $\varphi$ by a composite of $\varphi$ and an inner automorphism of $H$, we may (and will) assume that $(E_1)^{\varphi} = J$.

From the definition of $L$ (Proposition <ref>), it is easy to see that $L_{\lbrace 1,2 \rbrace} = A_1$.

We now prove (i). Assume that $i+1 < j$. Since $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$, there is some $g \in G$ with $(t_A)^g = t_{\lbrace 1,2 \rbrace} = u$ and $(t_B)^g = t_{\lbrace 3,4 \rbrace}$. So it suffices to show that $[L_{\lbrace 1,2 \rbrace},L_{\lbrace 3,4 \rbrace}] = 1$. Let $h$ denote the image of $t_{\lbrace 3,4 \rbrace}$ under $\varphi$. Then $h \in H_2$ since $t_{\lbrace 3,4 \rbrace} \in T_2 \le A_2$. Therefore, and since $h$ is conjugate to $u^{\varphi} = h_{\lbrace 1,2 \rbrace}$, we may choose $\varphi$ such that $h = h_{\lbrace 3,4 \rbrace}$ (and for the rest of the proof of (i), we will assume that $\varphi$ has been chosen in this way). We see from Lemma <ref> (ii) that there is an $a \in H$ with $(h_{\lbrace 1,2 \rbrace})^a = h_{\lbrace 3,4 \rbrace}$ and $(H_1)^a \le H_2$. In particular, $[H_1,(H_1)^a] = 1$. If $k$ is the preimage of $a$ under $\varphi$, then $u^k = t_{\lbrace 3,4 \rbrace}$ and $[A_1,(A_1)^k] = 1$. We also have $(A_1)^k = L_{\lbrace 3,4 \rbrace}$ and thus $[L_{\lbrace 1,2 \rbrace},L_{\lbrace 3,4 \rbrace}] = 1$.

We now prove (ii). Assume that $j = i+1$. Since $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$, there is some $g \in G$ with $(t_A)^g = t_{\lbrace 1,2 \rbrace}$ and $(t_B)^g = t_{\lbrace 2,3 \rbrace}$. Therefore, it is enough to prove (ii) under the assumption that $i = 1$, and we will assume that this is the case. We see from Lemmas <ref> (ii) and <ref> that $X_1 \cap A_2 = T_2$. Thus $t_{\lbrace 2,3 \rbrace} \not\in A_2$. Let $h$ denote the image of $t_{\lbrace 2,3 \rbrace}$ under $\varphi$. Then $h \not\in H_2$. Therefore, and since $h$ is conjugate to $u^{\varphi} = h_{\lbrace 1,2 \rbrace}$, we may choose $\varphi$ such that $h = h_{\lbrace 2,3 \rbrace}$ (and for the rest of the proof of (ii), we will assume that $\varphi$ has been chosen in this way). Let $\widetilde{H_1}$ be the image of

\begin{equation*}
\left \lbrace \begin{pmatrix} 1 & & \\ & M & \\ & & I_{n-5} \end{pmatrix} \ : \ M \in SL_2^{\varepsilon}(q^{*}) \right \rbrace
\end{equation*}

in $H$. By Lemma <ref> (ii), there is some $a \in H$ with $(h_{\lbrace 1,2 \rbrace})^a = h_{\lbrace 2,3 \rbrace}$ and $(H_1)^a = \widetilde{H_1}$. Let $k$ be the preimage of $a$ under $\varphi$. Then $u^k = t_{\lbrace 2,3 \rbrace}$ and hence $ L_{\lbrace 2,3 \rbrace} = (L_{\lbrace 1,2 \rbrace})^k = (A_1)^k$. We see now that $\varphi$ induces an isomorphism from $\langle L_{\lbrace 1,2 \rbrace}, L_{\lbrace 2,3 \rbrace} \rangle$ to $\langle H_1, \widetilde{H_1} \rangle$ mapping $L_{\lbrace 1,2 \rbrace}$ to $H_1$ and $L_{\lbrace 2,3 \rbrace}$ to $\widetilde{H_1}$. With this observation, it is easy to complete the proof of (ii).

We now prove (iii). Assume that $1 \le i \le n-3$ and that $j = i+1$. Let $k$ and $C$ be as in the statement of (iii). Since $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$, there is some $g \in G$ with $(t_A)^g = t_{\lbrace 1,2 \rbrace} = u$, $(t_B)^g = t_{\lbrace 2,3 \rbrace}$ and $(t_C)^g = t_{\lbrace 3,4 \rbrace}$. Therefore, it is enough to show that $\langle L_{\lbrace 1,2 \rbrace}, L_{\lbrace 2,3 \rbrace}, L_{\lbrace 3,4 \rbrace} \rangle$ is isomorphic to $SL_4^{\varepsilon}(q^{*})$. Let $h := (t_{\lbrace 2,3 \rbrace})^{\varphi}$ and $\widetilde h := (t_{\lbrace 3,4 \rbrace})^{\varphi}$. As in the proof of (ii), we can choose $\varphi$ such that $h = h_{\lbrace 2,3 \rbrace}$. Also, $\widetilde h = h_D$ for some $D \subseteq \lbrace 1,\dots,n-2 \rbrace$ of order $2$. We have $t_{\lbrace 3,4 \rbrace} \in T_2 \le A_2$ and hence $h_D = \widetilde h \in H_2$. Therefore, $D \cap \lbrace 1,2 \rbrace = \emptyset$. We claim that $D \cap \lbrace 2,3 \rbrace = \lbrace 3 \rbrace$. Assume not. Then $D \cap \lbrace 2,3 \rbrace = \emptyset$, and it is easy to find an element $a \in N_H(J)$ with $h^a = h_{\lbrace 1,2 \rbrace} = u^{\varphi}$ and $(\widetilde h)^a = h_{\lbrace 3,4 \rbrace} \in H_2$. So there is some $k \in N_K(E_1)$ with $(t_{\lbrace 2,3 \rbrace})^k = u$ and $(t_{\lbrace 3,4 \rbrace})^k \in T_2$. On the other hand, it is easy to see from $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$ that there is no $g \in K$ with $(t_{\lbrace 2,3 \rbrace})^g = u$ and $(t_{\lbrace 3,4 \rbrace})^g \in T_2$. This contradiction shows that $D \cap \lbrace 2,3 \rbrace = \lbrace 3 \rbrace$. So we can choose $\varphi$ such that $h = h_{\lbrace 2,3 \rbrace}$ and $\widetilde h = h_{\lbrace 3,4 \rbrace}$. Now the proof of (iii) can be completed by using similar arguments as in the proof of (ii).

$G_0$ is isomorphic to a nontrivial quotient of $SL_n^{\varepsilon}(q^{*})$.

Assume that $\varepsilon = +$. By Lemma <ref>, the groups $L_{\lbrace 1,2 \rbrace}, \dots, L_{\lbrace n-1,n \rbrace}$ form a weak Curtis-Tits system in $G$ of type $SL_n(q^{*})$ (in the sense of <cit.>). Applying a version of the Curtis-Tits theorem, namely <cit.>, we conclude that $G_0$ is isomorphic to a quotient of $SL_n(q^{*})$.

Assume now that $\varepsilon = -$. Then Lemma <ref> shows that $G_0$ has a weak Phan system of rank $n-1$ over $\mathbb{F}_{{q^{*}}^2}$ (in the sense of <cit.>). If $q^{*} \ne 3$, then <cit.> implies that $G_0$ is isomorphic to a quotient of $SU_n(q^{*})$. If $q^{*} = 3$, the same follows from <cit.> and Lemma <ref> (iii).

Let $R$ be a Sylow $2$-subgroup of $G_0$. Then $R \in \mathrm{Syl}_2(G)$ and $\mathcal{F}_R(G_0) = \mathcal{F}_R(G)$.

Since $q \sim \varepsilon q^{*}$, we have that the $2$-fusion system of $PSL_n^{\varepsilon}(q^{*})$ is isomorphic to the $2$-fusion system of $PSL_n(q)$ (see Proposition <ref>). Clearly, $G_0/Z(G_0) \cong PSL_n^{\varepsilon}(q^{*})$. So the $2$-fusion system of $G_0/Z(G_0)$ is isomorphic to the $2$-fusion system of $G$. It easily follows that $\vert G_0 \vert_2 = \vert G_0/Z(G_0) \vert_2 = \vert G \vert_2$, and Lemma <ref> shows that the $2$-fusion system of $G_0$ is isomorphic to that of $G_0/Z(G_0)$ and hence to that of $G$. This completes the proof.

The following hold.

(i) If $q^{*} \ne 3$, then $O^{2'}(O^{2}(C_G(t))) = KL$.

(ii) If $q^{*} = 3$, then $O^2(C_G(t)) = KL$.

Set $C := C_G(t)$.

Assume that $q^{*} \ne 3$. Then $KL$ is perfect. This implies that $KL = O^{2'}(O^2(KL)) \le O^{2'}(O^{2}(C))$. Since $T \cap KL = (T \cap K)(T \cap L) = X_1 X_2$, Lemmas <ref> and <ref> show that $C/KL$ has a nilpotent $2$-fusion system. So $C/KL$ is $2$-nilpotent by <cit.>. This implies $O^{2'}(O^{2}(C)) \le KL$.

We assume now that $q^{*} = 3$. Then $KL = O^2(KL)$ since $K$ is perfect and $L \cong SL_2(3)$. Thus $KL \le O^2(C)$. In order to prove equality, it suffices to show that $C/KL$ is a $2$-group. By Proposition <ref> and Lemma <ref> (i), $C/KC_C(K)$ is a $2$-group. By <cit.>, we have $L \le C_C(K)$. It is enough to show that $C_C(K)/L$ is a $2$-group.

We have $O^2(C_{C}(K)) \cap T \le O^2(C_{C}(X_1)) \cap T = X_2$ by Lemma <ref> and the hyperfocal subgroup theorem <cit.>. On the other hand, $X_2 \le L = O^2(L) \le O^2(C_{C}(K))$. Consequently, $X_2 = O^2(C_{C}(K)) \cap T \in \mathrm{Syl}_2(O^2(C_C(K)))$. Set $U := C_{O^2(C_C(K))}(X_2)$. We have $X_2 \trianglelefteq C$ since $X_2$ is the unique Sylow $2$-subgroup of $L \cong SL_2(3)$. So we have $U \trianglelefteq C$. Hence $Z(X_2) = X_2 \cap U \in \mathrm{Syl}_2(U)$. Applying <cit.>, we conclude that $U$ is $2$-nilpotent. We have $O(U) = 1$ since $U \trianglelefteq C$ and $O(C) = 1$ by Proposition <ref>. It follows that $U = Z(X_2)$.

Clearly, $O^{2}(C_C(K))/U$ is isomorphic to a subgroup of $\mathrm{Aut}(X_2)$. We have $|O^{2}(C_C(K))/U|_2 = 4$ since $Q_8 \cong X_2 \in \mathrm{Syl}_2(O^2(C_C(K)))$ and $U = Z(X_2)$. Also, $|O^{2}(C_C(K))/U| \ge 12$ since $L \le O^2(C_C(K))$. As $\mathrm{Aut}(X_2) \cong \mathrm{Aut}(Q_8) \cong S_4$ by <cit.>, it follows that $|O^{2}(C_C(K))/U| = 12$. This implies $O^2(C_C(K)) = L$. So $C_C(K)/L$ is a $2$-group, as required.

We have $KL \le G_0$.

We have $t \in X_2 \le L = L_{\lbrace n-1, n \rbrace} \le G_0$. Let $R \in \mathrm{Syl}_2(G_0)$ with $t \in R$ such that $\langle t \rangle$ is fully centralized in $\mathcal{G} := \mathcal{F}_R(G_0)$. By Lemma <ref>, $R \in \mathrm{Syl}_2(G)$ and $\mathcal{G} = \mathcal{F}_R(G)$. Therefore, $C_R(t) \in \mathrm{Syl}_2(C_G(t))$ and $C_{\mathcal{G}}(\langle t \rangle) = \mathcal{F}_{C_R(t)}(C_G(t))$. Also, $T = C_S(t) \in \mathrm{Syl}_2(C_G(t))$ and $C_{\mathcal{F}_S(G)}(\langle t \rangle) = \mathcal{F}_T(C_G(t))$. So, by Lemma <ref>, $C_{\mathcal{G}}(\langle t \rangle)$ has a component isomorphic to the $2$-fusion system of $SL_{n-2}(q)$.

Let $Z \le Z(SL_n^{\varepsilon}(q^{*}))$ with $G_0 \cong SL_n^{\varepsilon}(q^{*})/Z$. By the proof of Lemma <ref>, $Z(G_0)$ has odd order.

Let $\widetilde x$ be an element of $SL_n^{\varepsilon}(q^{*})$ such that $x := \widetilde x Z$ is an involution of $SL_n^{\varepsilon}(q^{*})/Z$. Set $C := C_{SL_n^{\varepsilon}(q^{*})/Z}(x)$. It is easy to note that the $2$-components of $C$ are precisely the images of the $2$-components of $C_{SL_n^{\varepsilon}(q^{*})}(\widetilde x)$ in $SL_n^{\varepsilon}(q^{*})/Z$. Using this, it is not hard to see from Lemmas <ref> and <ref> that one of the following holds:

(1) $q^{*} \ne 3$, $O^{2'}(O^2(C)) = K_0L_0$, where $K_0$ and $L_0$ are subnormal subgroups of $C$ such that $K_0 \cong SL_{n-i}^{\varepsilon}(q^{*})$ and $L_0 \cong SL_i^{\varepsilon}(q^{*})$ for some $1 \le i < n$. Moreover, the $2$-components of $C$ are precisely the quasisimple elements of $\lbrace K_0, L_0 \rbrace$.

(2) $q^{*} = 3$, $O^2(C) = K_0L_0$, where $K_0$ and $L_0$ are subnormal subgroups of $C$ such that $K_0 \cong SL_{n-i}^{\varepsilon}(q^{*})$ and $L_0 \cong SL_i^{\varepsilon}(q^{*})$ for some $1 \le i < n$. Moreover, the $2$-components of $C$ are precisely the quasisimple elements of $\lbrace K_0, L_0 \rbrace$.

(3) $C$ has precisely one $2$-component, and this $2$-component is isomorphic to a nontrivial quotient of $SL_{n/2}((q^{*})^2)$.

As seen above, $C_{\mathcal{G}}(\langle t \rangle) = \mathcal{F}_{C_R(t)}(C_{G_0}(t))$ has a component isomorphic to the $2$-fusion system of $SL_{n-2}(q)$. By Proposition <ref>, this component is induced by a $2$-component of $C_{G_0}(t)$. In view of the preceding observations, we can conclude that $C_{G_0}(t)$ has subgroups $K_0$ and $L_0$ with $K_0 \cong SL_{n-2}^{\varepsilon}(q^{*})$ and $L_0 \cong SL_2(q^{*})$ such that $O^{2'}(O^2(C_{G_0}(t))) = K_0L_0$ if $q^{*} \ne 3$ and $O^2(C_{G_0}(t)) = K_0 L_0$ if $q^{*} = 3$.

Clearly, $O^{2'}(O^2(C_{G_0}(t))) \le O^{2'}(O^2(C_G(t)))$ and $O^2(C_{G_0}(t)) \le O^2(C_G(t))$. Lemma <ref> implies that $K_0 L_0 \le KL$. If $n$ is odd, then it is easy to see that $\vert K_0 L_0 \vert = \vert K_0 \vert \vert L_0 \vert \ge \vert K \vert \vert L \vert = \vert KL \vert$. If $n$ is even, then one can easily see that $\vert K_0 L_0 \vert = \frac{1}{2} |K_0||L_0| \ge \frac{1}{2} |K||L| = |KL|$. Consequently, $K_0L_0 \le KL$ and $|K_0L_0| \ge |KL|$. It follows that $KL = K_0L_0 \le G_0$.

Let $x$ be an involution of $G_0$ which is $G$-conjugate to $t$. Let $L_0$ be the unique normal $SL_2(q^{*})$-subgroup of $C_G(x)$, and let $K_0$ be the component of $C_G(x)$ different from $L_0$. Then we have $K_0 L_0 \le G_0$.

Since $t \in G_0$, we see from Lemma <ref> that there is some $g \in G_0$ with $x = t^g$. Clearly, $(K_0L_0) = (KL)^g$, and so $K_0L_0 \le G_0$ by Lemma <ref>.

We have $N_G(S) \le N_G(G_0)$.

Set $M := N_G(G_0)$. Let $s \in N_S(S \cap M)$, and let $1 \le i \le n-1$. We have $t_{\lbrace i,i+1 \rbrace} \in S \cap L_{\lbrace i,i+1 \rbrace} \le S \cap G_0 \le S \cap M$, and hence $(t_{\lbrace i,i+1 \rbrace})^s \in S \cap M \le M$. Since $G_0$ has odd index in $M$ by Lemma <ref>, we even have $(t_{\lbrace i,i+1 \rbrace})^s \in G_0$. Corollary <ref> implies that $(L_{\lbrace i,i+1 \rbrace})^s \le G_0$. So we have $s \in M$ by the definition of $G_0$. Thus $N_S(S \cap M) = S \cap M$ and hence $S \le M$. It is clear that $C_G(S) \le M$. Using Lemma <ref>, we conclude that $N_G(S) = SC_G(S) \le M$.

If $x$ is an involution of $S$, then $C_G(x) \le N_G(G_0)$.

Set $M := N_G(G_0)$.

We begin by proving that $C_G(t) \le M$. We have $K \le G_0 \le M$ by Lemma <ref> and $C_G(t) = K N_{C_G(t)}(X_1)$ by the Frattini argument. Also, $N_{C_G(t)}(X_1) = T C_{C_G(t)}(X_1)$ as a consequence of Lemma <ref>, and $T \le M$ by Lemma <ref>. So it suffices to show that $C_{C_G(t)}(X_1) \le M$.

Let $z \in C_{C_G(t)}(X_1)$. In order to prove $z \in M$, it is enough to show that $(L_{\lbrace i, i+1 \rbrace})^z \le G_0$ for all $1 \le i < n$. If $1 \le i < n$ and $i \ne n-2$, we have $z \in C_G(t_{\lbrace i,i+1 \rbrace})$ and hence $(L_{\lbrace i,i+1 \rbrace})^z = L_{\lbrace i,i+1 \rbrace} \le G_0$. It remains to show that $(L_{\lbrace n-2,n-1 \rbrace})^z \le G_0$. Since $\mathcal{F}_S(G) = \mathcal{F}_S(PSL_n(q))$, there is some $g \in G$ with $t^g = u$, $u^g = t$ and $(t_{\lbrace 2,3 \rbrace})^g = t_{\lbrace n-2,n-1 \rbrace}$. From the definition of $L$ (Proposition <ref>), it is easy to see that $L_{\lbrace 1,2 \rbrace} = A_1 \le K$. Since $u = t_{\lbrace 1,2 \rbrace}$ and $t_{\lbrace 2,3 \rbrace}$ are $K$-conjugate, we thus have $L_{\lbrace 2,3 \rbrace} \le K \le L_{2'}(C_G(t))$. Hence $L_{\lbrace n-2,n-1 \rbrace} = (L_{\lbrace 2,3 \rbrace})^g \le L_{2'}(C_G(t))^g = L_{2'}(C_G(u))$. Since $z$ centralizes $u$, it follows that $(L_{\lbrace n-2,n-1 \rbrace})^z \le L_{2'}(C_G(u))$. From Corollary <ref>, we see that $L_{2'}(C_G(u)) \le G_0$. So we have $(L_{\lbrace n-2,n-1 \rbrace})^z \le G_0$, and it follows that $C_{C_G(t)}(X_1) \le M$. Consequently, $C_G(t) \le M$.

Since $G_0$ has odd index in $M$ by Lemma <ref>, we see from Lemma <ref> that $S \le G_0$. Also, $\mathcal{F}_S(G_0) = \mathcal{F}_S(G)$ by Lemma <ref>. As $C_G(t) \le M$, it follows that $C_G(x) \le M$ for any involution $x$ of $S$ which is $G$-conjugate to $t$.

Assume now that $x$ is an involution of $S$ which is $G$-conjugate to $t_i$ for some even natural number $i$ with $4 \le i < n$ such that $i \le \frac{n}{2}$ if $n$ is even. We are going to show that $C_G(x) \le M$. Arguing by induction over $i$ and using the preceding observations, we may assume that for each even $2 \le j < i$ and each involution $y$ of $S$ which is $G$-conjugate to $t_j$, we have $C_G(y) \le M$. Furthermore, we may assume that $\langle x \rangle$ is fully $\mathcal{F}_S(G)$-centralized since $\mathcal{F}_S(G) = \mathcal{F}_S(G_0)$.

As a consequence of Lemma <ref>, $C_G(x)$ is generated by the normalizers $N_{C_G(x)}(U)$, where $U$ is a subgroup of $C_S(x)$ containing a $G$-conjugate of $t_j$ for some even $2 \le j < i$. We show that each such normalizer is contained in $M$. Thus let $U$ be a subgroup of $C_S(x)$ and let $y$ be an element of $U$ which is $G$-conjugate to $t_j$ for some even $2 \le j < i$. Also, let $g \in N_{C_G(x)}(U)$. Then $y^g \in U \le C_S(x) \le S$. Since $\mathcal{F}_S(G_0) = \mathcal{F}_S(G)$, we have that $y$ and $y^g$ are $G_0$-conjugate. Hence, there is some $m \in G_0$ with $y^g = y^m$. We have $mg^{-1} \in C_G(y) \le M$. This implies $g \in M$ since $m \in G_0 \le M$. So we have $N_{C_G(x)}(U) \le M$ and hence $C_G(x) \le M$.

Assume now that $x$ is an arbitrary involution of $S$. We are going to prove that $C_G(x) \le M$. Since $\mathcal{F}_S(G) = \mathcal{F}_S(G_0)$, we may assume that $\langle x \rangle$ is fully $\mathcal{F}_S(G)$-centralized. By Corollary <ref>, $C_G(x)$ is $3$-generated. Therefore, $C_G(x)$ is generated by the normalizers $N_{C_G(x)}(U)$, where $U \le C_S(x)$ and $m(U) \ge 3$. Take some $U \le C_S(x)$ with $m(U) \ge 3$. By Lemma <ref>, any $E_8$-subgroup of $S$ has an involution which is the image of an involution of $SL_n(q)$. It follows that $U$ has an element $y$ which is $G$-conjugate to $t_k$ for some even $2 \le k < n$. By the preceding observations, $C_G(y) \le M$. Arguing as above, we can conclude that $N_{C_G(x)}(U) \le M$. It follows that $C_G(x) \le M$.

We have $G_0 \trianglelefteq G$.

Suppose that $M := N_G(G_0)$ is a proper subgroup of $G$. By <cit.>, we may deduce from Lemmas <ref> and <ref> that $M$ is strongly embedded in $G$. Therefore, by <cit.>, $G$ 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 $M = G$. Hence $G_0 \trianglelefteq G$.

With Propositions <ref> and <ref>, we have completed the proof of Theorem <ref>.

§ PROOFS OF THE MAIN RESULTS

By Section <ref>, Theorem <ref> is true for $n \le 5$.

Suppose now that $n \ge 6$. Let $q$ be a nontrivial odd prime power, and let $G$ be a finite simple group satisfying (<ref>).

Recall that a natural number $k \ge 6$ is said to satisfy $P(k)$ 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 $\varepsilon q^{*} \sim q_0$. Theorem <ref> shows that $P(k)$ is satisfied for all natural numbers $k \ge 6$.

Therefore, if the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_n(q)$, then condition (i) of Theorem <ref> is satisfied.

Conversely, if one of the conditions (i), (ii), (iii) of Theorem <ref> is satisfied, then this can only be condition (i), and Proposition <ref> implies that the $2$-fusion system of $G$ is isomorphic to the $2$-fusion system of $PSL_n(q)$.

Let $q$ be a nontrivial odd prime power and let $n \ge 2$ be a natural number, where $q \equiv 1$ or $7 \mod 8$ if $n = 2$. Let $G$ be a finite simple group and $S \in \mathrm{Syl}_2(G)$. Suppose that $\mathcal{F}_S(G)$ has a normal subsystem $\mathcal{E}$ on a subgroup $T$ of $S$ such that $\mathcal{E}$ is isomorphic to the $2$-fusion system of $PSL_n(q)$ and such that $C_S(\mathcal{E}) = 1$. We have to show that $\mathcal{F}_S(G)$ is isomorphic to the $2$-fusion system of $PSL_n(q)$.

By Lemma <ref>, $PSL_n(q)$ is not a Goldschmidt group. Applying <cit.>, we conclude that $\mathcal{E}$ is simple. We see from <cit.> that $\mathcal{E}$ is tamely realized by some finite simple group of Lie type $K$.

By Theorem <ref>, we have $K \cong PSL_n^{\varepsilon}(q^{*})$ for some nontrivial odd prime power $q^{*}$ and some $\varepsilon \in \lbrace +,- \rbrace$ with $\varepsilon q^{*} \sim q$.

By Propositions <ref> and <ref>, we have that $\mathrm{Out}(K)$ is $2$-nilpotent. Now Proposition <ref> implies that $\mathcal{F}_S(G)$ is tamely realized by a subgroup $L$ of $\mathrm{Aut}(K)$ containing $\mathrm{Inn}(K)$ such that the index of $\mathrm{Inn}(K)$ in $L$ is odd. By Lemma <ref>, the $2$-fusion system of $L$ is isomorphic to the $2$-fusion system of $\mathrm{Inn}(K) \cong K$ and hence isomorphic to the $2$-fusion system of $PSL_n(q)$. So $\mathcal{F}_S(G)$ is isomorphic to the $2$-fusion system of $PSL_n(q)$.

Let $q$ be a nontrivial odd prime power and let $n \ge 2$ be a natural number, where $q \equiv 1$ or $7 \mod 8$ if $n = 2$. Let $G$ be a finite simple group and let $S$ be a Sylow $2$-subgroup of $G$. Suppose that $F^{*}(\mathcal{F}_S(G))$ is isomorphic to the $2$-fusion system of $PSL_n(q)$.

We have $F^{*}(\mathcal{F}_S(G)) \trianglelefteq \mathcal{F}_S(G)$ and $C_S(F^{*}(\mathcal{F}_S(G))) = Z(F^{*}(\mathcal{F}_S(G))) = 1$. So Theorem <ref> implies that $\mathcal{F}_S(G)$ is isomorphic to the $2$-fusion system of $PSL_n(q)$.

§ ACKNOWLEDGEMENTS

This paper is based on the author’s PhD thesis, written at the University of Aberdeen under the supervision of Professor Ellen Henke and Professor Benjamin Martin. The author is deeply grateful to them for their guidance and support. Moreover, the author would like to thank Professor Ron Solomon for helpful discussions.

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