Proof. Let $Z \le E \le G$ such that $E/Z \cong E_8$ . Let R be a Sylow $2$ -subgroup of E. Then $E = RZ$ . It suffices to show that R has an involution not lying in $R \cap Z$ . Assume that any involution of R is an element of $R \cap Z$ . Then R has a unique involution since Z is cyclic. We have $R/(R \cap Z) \cong RZ/Z = E/Z \cong E_8$ , and so R is not cyclic. Applying [Reference Kurzweil and Stellmacher37, 5.3.7], we conclude that R is generalized quaternion. In particular, $Z(R)$ has order $2$ , and so we have $R \cap Z = Z(R)$ . Since R is a generalized quaternion group, $R/Z(R)$ is dihedral. In particular, $E/Z \cong R/(R \cap Z) = R/Z(R) \not \cong E_8$ . This contradiction shows that R has an involution not lying in $R \cap Z$ , as required.

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

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

Since $K/N$ is subnormal in $G/N$ , we have that K is subnormal in G. Therefore, $L := O^{2'}(K)$ is subnormal in G. Since $O^{2'}(K/N) = K/N$ , we have that $K/N = LN/N$ . Clearly, $O^{2'}(L) = L$ . We have $L/O(L) \cong (LN/N)/O(LN/N) = (K/N)/O(K/N)$ , and so $L/O(L)$ is quasisimple. Applying [Reference Gorenstein, Lyons and Solomon27, Lemma 4.8], we conclude that L is a $2$ -component of G.

Now let $L_0$ be a $2$ -component of G such that $K/N = L_0 N /N$ . Then $K = L_0 N$ . In particular, $L_0$ is a subgroup of K with odd index in K. Since $L_0$ is subnormal in G, we have that $L_0$ is subnormal in K. Applying [Reference Ballester-Bolinches, Esteban-Romero and Asaad12, Lemma 1.1.11], we conclude that $L_0 = O^{2'}(L_0) = O^{2'}(K) = L$ . The proof of the second statement of the proposition is now complete. The third statement also follows from the above arguments.

Proof. For each $1 \le i \le n$ , let $\mathfrak {L}_i$ denote the set of $2$ -components of $C_{L_i}(x)$ , and let $\mathfrak {L} := \mathfrak {L}_1 \cup \dots \mathfrak {L}_n$ . It suffices to show that $L \in \mathfrak {L}$ .

By [Reference Gorenstein and Walter31, Corollary 3.2], we have $L_{2'}(C_G(x)) = L_{2'}(C_{L_{2'}(G)}(x))$ , and by [Reference Gorenstein and Walter31, Lemma 2.18 (iii)], we have $L_{2'}(C_{L_{2'}(G)}(x)) = \prod _{i=1}^n L_{2'}(C_{L_i}(x))$ . Thus $L_{2'}(C_G(x)) = \langle \mathfrak {L} \rangle $ . Set .

Assume that $L \not \in \mathfrak {L}$ . Let J be an element of $\mathfrak L$ . Since $L \ne J$ and since L and J are $2$ -components of $C_G(x)$ , we have by Proposition 2.4. Also, since and are components of , we have by [Reference Kurzweil and Stellmacher37, 6.5.3]. Since , it follows that lies in the center of . This is impossible since is nontrivial and perfect. So we have $L \in \mathfrak {L}$ .

Proof. To prove this, we follow arguments found on pp. 40–41 of [Reference Mason40].

Since $\theta $ is an A-signalizer functor on G, $\theta (C_G(a))$ is A-invariant and in particular $A_0$ -invariant for each $a \in A^{\#}$ . Consequently, $\theta (G,A)$ is $A_0$ -invariant. By the solvable signalizer functor theorem [Reference Kurzweil and Stellmacher37, 11.3.2], $\theta $ is complete (in the sense of [Reference Gorenstein24, Definition 4.37]). In particular, $\theta (G,A)$ has odd order. Applying [Reference Gorenstein, Lyons and Solomon27, Proposition 11.23], we conclude that

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

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

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

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

Let $E \le A_0$ with $m(E) = k$ . Clearly, $\Delta _G(E)$ is A-invariant. As a consequence of [Reference Gorenstein, Lyons and Solomon27, Proposition 11.23], we have

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

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

Proof of Proposition 2.8

It is straightforward to verify (i).

To verify (ii), let A be an elementary abelian $2$ -subgroup of G of rank at least $k+2$ such that G is k-balanced with respect to A. Let $\theta $ be the A-signalizer functor on G given by Theorem 2.9, and let $\theta (G,A) := \langle \theta (C_G(a)) \ \vert \ a \in A^{\#} \rangle $ . As a consequence of Lemma 2.10, we have $\theta (G,A) = W_A$ . By the proof of Lemma 2.10, $W_A = \theta (G,A)$ has odd order.

Now let $A_0$ be a subgroup of A of rank at least $k+1$ . By Lemma 2.10, $W_A = \theta (G,A) \le W_{A_0} \le W_A$ , and so $W_A = W_{A_0}$ . Finally, if $g \in N_G(A_0)$ , then $(W_A)^g = (W_{A_0})^g = W_{(A_0)^g} = W_{A_0} = W_A$ , and hence, $N_G(A_0) \le N_G(W_A)$ .

Proof. Let $\varphi $ denote the canonical group isomorphism $S/(S \cap N) \rightarrow SN/N$ . Let P and Q be two subgroups of S such that $S \cap N$ is contained in both P and Q. Set $\widetilde P := P/(S \cap N)$ , $\widetilde Q := Q/(S \cap N)$ ,

and

. For any $g \in G$ , let

. Moreover, define $\widetilde {\mathcal {F}} := \mathcal {F}_S(G)/(S \cap N)$ and

. It is enough to show that

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

.

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

Proof. By Glauberman’s $Z^{*}$ -Theorem, more precisely by [Reference Glauberman22, Corollary 1], we have $Z(\mathcal {F}_S(G)) = S \cap Z^{*}(G)$ . Assume now that $Z^{*}(G)$ is $2$ -closed, and let $S_0 := S \cap Z^{*}(G)$ . Then $S_0 \trianglelefteq G$ and hence $[S_0,G] \le S_0 \cap [Z^{*}(G),G] \le S_0 \cap O(G) = 1$ . Thus, $Z(\mathcal {F}_S(G)) = S_0 = S \cap Z(G)$ .

Proof. Suppose that the $2$ -fusion systems of $K_1$ and $K_2$ are isomorphic. Let $S_i$ be a Sylow $2$ -subgroup of $K_i$ and $\mathcal {F}_i := \mathcal {F}_{S_i}(K_i)$ for $i \in \lbrace 1,2 \rbrace $ . Since $K_1$ and $K_2$ are quasisimple, we have $Z^{*}(K_i) = Z(K_i)$ for $i \in \lbrace 1,2 \rbrace $ . So, by Lemma 2.13, we have $Z(\mathcal {F}_i) = S_i \cap Z(K_i)$ for $i \in \lbrace 1,2 \rbrace $ . Since $\mathcal {F}_1 \cong \mathcal {F}_2$ , it follows that

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

Applying Lemma 2.11, we may conclude that the $2$ -fusion system of $K_1/Z(K_1)$ is isomorphic to the $2$ -fusion system of $K_2/Z(K_2)$ .

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

Since A and C are strongly $\mathcal {F}$ -closed, we have that $A \cap C$ is strongly $\mathcal {F}$ -closed. In particular, $\alpha $ normalizes $A \cap C$ . An automorphism of $A \cong Q_8$ of order $3$ is irreducible on $A/\Phi (A)$ . So, as $\alpha |_{A,A}$ has order $3$ and normalizes $A \cap C$ , we have that $A \cap C$ has order $1$ or $2$ .

By [Reference Kurzweil and Stellmacher37, 8.2.7], we have

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

By hypothesis $[S, \alpha ] = A$ , so $[C, \alpha ] \le [S, \alpha ] \cap C = A \cap C$ . As $|A \cap C| \le 2, [A \cap C, \alpha ] = 1$ , so $[C, \alpha ] = [C, \alpha , \alpha ] = [A \cap C, \alpha ] = 1$ . Hence, $C \le C_S (\alpha ) = Z(A)B = C_S(A)$ .

Proof. Let L be a $2$ -component of G. Set $\mathcal {G} := \mathcal {F}_{S \cap L}(L)$ . Since L is subnormal in G, we have that $\mathcal {G}$ is subnormal in $\mathcal {F}_S(G)$ (see [Reference Aschbacher, Kessar and Oliver10, Part I, Proposition 6.2]). Therefore, $\mathcal {G}$ is a component of $\mathcal {F}_S(G)$ if and only if $\mathcal {G}$ is quasisimple. We have $\mathfrak {foc}(\mathcal {G}) = S \cap L' = S \cap L$ by the focal subgroup theorem [Reference Gorenstein23, Chapter 7, Theorem 3.4], and so $\mathcal {G}$ is quasisimple if and only if $\mathcal {G}/Z(\mathcal {G})$ is simple. By Lemma 2.13, we have $Z(\mathcal {G}) = S \cap Z^{*}(L)$ . Lemma 2.11 implies that $\mathcal {G}/Z(\mathcal {G})$ is isomorphic to the $2$ -fusion system of $L/Z^{*}(L)$ . By [Reference Aschbacher9, Theorem 5.6.18], the $2$ -fusion system of $L/Z^{*}(L)$ is simple if and only if $L \in \mathfrak {L}_{2'}$ . So $\mathcal {G}$ is a component of $\mathcal {F}_S(G)$ if and only if $L \in \mathfrak {L}_{2'}$ , and (i) holds.

(ii) follows from [Reference Aschbacher8, (1.8)].

Proof. Let $L^s(G)$ denote the subgroup of G generated by the components and the solvable $2$ -components of G. By [Reference Kurzweil and Stellmacher37, 6.5.2] and [Reference Gorenstein, Lyons and Solomon27, Proposition 13.5], $L^s(G)$ is the central product of the subgroups of G which are components or solvable $2$ -components. Set $L := L_1 \cdots L_n \trianglelefteq L^s(G)$ .

Let $\mathcal {G} := \mathcal {F}_{S \cap L^s(G)}(L^s(G))$ . Clearly, $S \cap L = (S \cap L_1) \cdots (S \cap L_n) = Q$ . Lemma 2.11 implies that the $2$ -fusion system of $L^s(G)/L$ is isomorphic to $\mathcal {G}/Q$ . By hypothesis, $\mathcal {F}_S(G)/Q$ is nilpotent, and so $\mathcal {G}/Q$ is nilpotent. So the $2$ -fusion system of $L^s(G)/L$ is nilpotent. Applying [Reference Linckelmann39, Theorem 1.4], we conclude that $L^s(G)/L$ is $2$ -nilpotent.

Suppose $L_0 \ne L_i$ for any $1 \le i \le n$ . Then from paragraph one, $L_0$ centralizes L, so $L \cap L_0 \le Z(L_0)$ , and hence, $L_0 L/L \cong L_0 /(L \cap L_0)$ is quasisimple, $A_4$ , or $SL_2(3)$ . In particular, $L_0L/L$ is not $2$ -nilpotent, a contradiction.

Proof. Set $\mathcal {F} := \mathcal {F}_S(G)$ . By a result of Bob Oliver, namely by [Reference Oliver44, Corollary 2.4], $\mathcal {F}$ is tamely realized by a subgroup L of $\mathrm {Aut}(K)$ containing $\mathrm {Inn}(K)$ . We are going to show that the index of $\mathrm {Inn}(K)$ in L is coprime to p.

Let $S_0$ be a Sylow p-subgroup of L. Then $\mathcal {F} \cong \mathcal {F}_{S_0}(L)$ . We have $O^p(G) = G$ since G is nonabelian simple, and so $\mathfrak {hnp}(\mathcal {F}) = S$ by the hyperfocal subgroup theorem [Reference Craven18, Theorem 1.33]. It follows that $\mathfrak {hnp}(\mathcal {F}_{S_0}(L)) = S_0$ .

By the hyperfocal subgroup theorem [Reference Craven18, Theorem 1.33], $S_0 = \mathfrak {hnp}(\mathcal {F}_{S_0}(L)) = O^p(L) \cap S_0$ . Consequently, $O^p(L)$ has $p'$ -index in L, whence $O^p(L) = L$ . So we have $O^p(L/\mathrm {Inn}(K)) = L/\mathrm {Inn}(K)$ . On the other hand, $L/\mathrm {Inn}(K)$ is p-nilpotent since $\mathrm {Out}(K)$ is p-nilpotent. It follows that $L/\mathrm {Inn}(K)$ is a $p'$ -group, as claimed.

Proof. By [Reference Gorenstein, Lyons and Solomon28, pp. 312-313], the Schur multiplier of $PSL_n(q)$ is isomorphic to $C_{(n,q-1)} \cong Z(SL_n(q))$ . From [Reference Aschbacher5, 33.6], we see that this is just another way to say that $SL_n(q)$ is the universal covering group of $PSL_n(q)$ . Applying [Reference Aschbacher5, 33.6] again, we conclude that $H \cong SL_n(q)/Z$ for some $Z \le Z(SL_n(q))$ .

Proof. Suppose that $(n,q) \ne (4,3)$ . By [Reference Gorenstein, Lyons and Solomon28, pp. 312-313], the Schur multiplier of $PSU_n(q)$ is isomorphic to $C_{(n,q+1)} \cong Z(SU_n(q))$ . As in the proof of Lemma 3.1, we conclude that $H \cong SU_n(q)/Z$ for some $Z \le Z(SU_n(q))$ .

Suppose now that $(n,q)=(4,3)$ and that $Z(H)$ is a $2$ -group. Let $G := PSU_4(3)$ , and let $\widetilde G$ be the universal covering group of G. Then the Schur multiplier of G is isomorphic to $Z(\widetilde G)$ . By [Reference Gorenstein, Lyons and Solomon28, pp. 312-313], the Schur multiplier of G is isomorphic to $C_4 \times C_3 \times C_3$ . Thus, $Z(\widetilde G) \cong C_4 \times C_3 \times C_3$ . Since $\widetilde G$ is quasisimple, we have $Z(\widetilde G / Z) = Z(\widetilde G)/Z$ whenever $Z \le Z(\widetilde G)$ . Let Q be the unique Sylow $3$ -subgroup of $Z(\widetilde G)$ . By [Reference Aschbacher5, 33.6], $\widetilde G$ is a central extension of $SU_4(3)$ and of H. Since $SU_4(3)$ has a center of order $4$ , we have $SU_4(3) \cong \widetilde G / Q$ . Let $Z \le Z(\widetilde G)$ with $H \cong \widetilde G / Z$ . As $Z(H)$ is a $2$ -group, we have $Q \le Z$ , whence $H \cong \widetilde G / Z \cong (\widetilde G / Q) / (Z/Q)$ is isomorphic to a quotient of $SU_4(3)$ by a central subgroup.

Proof. We identify the field $\mathbb {F}_q$ with the subfield of $\mathbb {F}_{q^2}$ consisting of all $x \in \mathbb {F}_{q^2}$ satisfying $x^q = x$ . As $q + 1 = (q^2 - 1)/(q - 1)$ is even, any element of $\mathbb {F}_q^{*}$ is the square of an element of $\mathbb {F}_{q^2}^{*}$ . Let $\mu \in \mathbb {F}_{q^2}^{*}$ with $\lambda = \mu ^2$ .

If $\mu \in \mathbb {F}_q$ , then the minimal polynomial of T divides $(x - \mu )(x + \mu )$ , so T is diagonalizable over $\mathbb {F}_q$ , and it follows that (i) holds.

Assume now that $\mu \not \in \mathbb {F}_q$ . Then $\lambda $ is a nonsquare element of $\mathbb {F}_q^{*}$ . Let V be an n-dimensional vector space over $\mathbb {F}_q$ , and let B be an ordered basis of V. Let $\varphi $ be the element of $GL(V)$ such that $\varphi $ is represented by T with respect to B. Since $\mu \not \in \mathbb {F}_q$ , we have that $1$ and $\mu $ are linearly independent; so $(1,\mu )$ is an $\mathbb {F}_q$ -basis of $\mathbb {F}_{q^2}$ . Using that $\varphi ^2 = \lambda \mathrm {id}_V$ , one can check that V becomes a vector space over $\mathbb {F}_{q^2}$ by defining

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

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

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

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

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

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

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

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

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

Proof. Suppose that $T^2 = \lambda I_n$ for some $\lambda \in \mathbb {F}_{q^2}^{*}$ . Since $T^2 \in GU_n(q)$ , we have that $\lambda ^{q+1} = 1$ , so $\lambda $ is a square in $\mathbb {F}_{q^2}^{*}$ .

A proof of (ii) and (iii) can be extracted from [Reference Phan47, pp. 314-315].

Proof. We follow arguments found in the proof of [Reference Phan46, Lemma 1.1].

Assume that n is odd. Then $Z(SL_n(q))$ has odd order, and therefore, any involution of $PSL_n(q)$ is the image of an involution of $SL_n(q)$ . As a consequence of Lemma 3.3, each involution of $SL_n(q)$ is $SL_n(q)$ -conjugate to $\widetilde {t_i}$ for some even $2 \le i < n$ . So (i) follows.

Assume now that n is even and that $\rho = \mu ^2$ for some $\mu \in \mathbb {F}_q^{*}$ . Note that $Z(SL_n(q))$ equals $\langle \rho I_n \rangle $ . We claim that $\mu ^n = -1$ . Since $\mu ^{2n} = \rho ^n = 1$ , we have that $\mu ^n = 1$ or $-1$ . If $\mu ^n = 1$ , then $\mu \in \langle \rho \rangle $ , and so $\rho $ is a square in $\langle \rho \rangle $ , which is impossible. So we have $\mu ^n = -1$ . It follows that $\widetilde {t_i} \in SL_n(q)$ for each odd $1 \le i < n$ . Now let $T \in SL_n(q)$ such that $TZ(SL_n(q)) \in PSL_n(q)$ is an involution. Then we have $T^2 = \rho ^{\ell }I_n = \mu ^{2\ell }I_n$ for some $1 \le \ell \le (n,q-1)$ . Using Lemma 3.3, we conclude that T is $SL_n(q)$ -conjugate to a diagonal matrix $D \in SL_n(q)$ with diagonal entries in $\lbrace \mu ^{\ell }, - \mu ^{\ell } \rbrace $ . Let $1 \le i < n$ such that $- \mu ^{\ell }$ occurs precisely i times as a diagonal entry of D. If i is odd, we may assume that $D = \mu ^{\ell -1} \widetilde {t_i}$ , and if i is even, we may assume that $D = \mu ^{\ell } \widetilde {t_i}$ . In either case, the image of D in $PSL_n(q)$ is $t_i$ . Hence, $TZ(SL_n(q))$ is $PSL_n(q)$ -conjugate to $t_i$ . Noticing that $t_i$ is $PSL_n(q)$ -conjugate to $t_{n-i}$ , we conclude that (ii) holds.

Now assume that n is even and that $\rho $ is a nonsquare element of $\mathbb {F}_q$ . Again, let T be an element of $SL_n(q)$ such that $TZ(SL_n(q)) \in PSL_n(q)$ is an involution. We have $T^2 = \rho ^{\ell }I_n$ for some $1 \le \ell \le (n,q-1)$ . Assume that $\ell $ is even. Then Lemma 3.3 implies that T or $-T$ is $SL_n(q)$ -conjugate to $\rho ^{\frac {\ell }{2}}\widetilde {t_i}$ for some even $2 \le i \le \frac {n}{2}$ . It follows that $TZ(SL_n(q))$ is $PSL_n(q)$ -conjugate to $t_i$ for some even $2 \le i \le \frac {n}{2}$ . Assume now that $\ell $ is odd. As $\rho $ is not a square in $\mathbb {F}_q$ , but $\rho ^{\ell -1}$ is a square in $\mathbb {F}_q$ , $\rho ^{\ell }$ cannot be a square in $\mathbb {F}_q$ . Using Lemma 3.3, we may conclude that T is $GL_n(q)$ -conjugate to the matrix

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

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

$$ \begin{align*} \begin{pmatrix} 0 & \rho^{\ell} \\ 1 & 0 \end{pmatrix} \ \text{and} \ \begin{pmatrix} 0 & \rho^{k+1} \\ \rho^k & 0 \end{pmatrix} \end{align*} $$

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

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

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

Proof. Set $C_0 := C_{SL_n(q)}(\widetilde w)/Z(SL_n(q)) \le C$ .

Let $y \in C \setminus C_0$ , and let $\widetilde y$ be a preimage of y in $SL_n(q)$ . Then $\widetilde w^{\widetilde y} = \lambda \widetilde w$ for some $1 \ne \lambda \in \langle \rho \rangle $ . The characteristic polynomial of $\widetilde w$ is $(x^2-\rho )^{\frac {n}{2}}$ , and $\lambda \widetilde w$ has the characteristic polynomial $(x^2 - \lambda ^2 \rho )^{\frac {n}{2}}$ . Since $\widetilde w^{\widetilde y} = \lambda \widetilde w$ , both polynomials are equal, and so we have $\lambda ^2 = 1$ . Thus, $\lambda = -1$ and hence $\widetilde w^{\widetilde y} = - \widetilde w$ . If z is another element of $C \setminus C_0$ and if $\widetilde {z}$ is a preimage of z in $SL_n(q)$ , then we have $\widetilde w^{\widetilde y} = - \widetilde w = \widetilde w^{\widetilde z}$ , and so $\widetilde y \widetilde {z}^{-1}$ centralizes $\widetilde w$ . This implies that $y{z}^{-1} \in C_0$ . It follows that $|C : C_0| \le 2$ (and one can show that in fact $|C : C_0| = 2$ ).

By the preceding paragraph, $C/C_0$ is abelian, and so the $2$ -components of C are precisely the $2$ -components of $C_0$ . One may deduce from Lemma 3.3 that $C_{SL_n(q)}(\widetilde w)$ has a normal subgroup $\widetilde J$ isomorphic to $SL_{\frac {n}{2}}(q^2)$ such that the corresponding factor group is cyclic. Let J be the image of $\widetilde J$ in $PSL_n(q)$ . Then J is isomorphic to a nontrivial quotient of $SL_{\frac {n}{2}}(q^2)$ . Moreover, $J \trianglelefteq C_0$ and $C_0/J$ is cyclic. Therefore, J is the only $2$ -component of $C_0$ and hence the only $2$ -component of C. Thus, (i) holds.

Since $J \trianglelefteq C$ , we have that $P \cap J$ is strongly closed in P with respect to $\mathcal {F}_P(C)$ . By Lemma 2.11, the factor system $\mathcal {F}_P(C)/(P \cap J)$ is isomorphic to the $2$ -fusion system of $C/J$ . Since $C_0$ has index $\le 2$ in C and $C_0/J$ is abelian, we have that $C/J$ is $2$ -nilpotent. So $C/J$ has a nilpotent $2$ -fusion system, and (ii) follows.

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

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

in $PSL_n(q)$ .

We claim that there exist $a, b \in \mathbb {F}_q$ with $a^2 \rho - b^2 \rho ^2 = 1$ . The field $\mathbb {F}_q$ has precisely $\frac {q+1}{2}$ square elements. Therefore, each of the sets $M_1 := \lbrace a^2 \rho \mid a \in \mathbb {F}_q \rbrace $ and $M_2 := \lbrace 1 + b^2 \rho ^2 \mid b \in \mathbb {F}_q \rbrace $ has cardinality $\frac {q+1}{2}$ . It follows that $M_1 \cap M_2 \ne \emptyset $ . So there exist $a, b \in \mathbb {F}_q$ with $a^2 \rho = 1 + b^2 \rho ^2$ , or in other words $a^2 \rho - b^2 \rho ^2 = 1$ .

Let s be the image of

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

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

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

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

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

in $PSL_n(q)$ . Then one can easily verify that $\langle s, u, z_1, z_2 \rangle \le C_{PSL_n(q)}(u)$ is isomorphic to $E_{16}$ . So a Sylow $2$ -subgroup of $C_{PSL_n(q)}(u)$ has rank at least $4$ . This is also true for P as w and u are conjugate (see Proposition 3.5).

Proof. By Lemmas 3.3 and 3.4, we either have $C_{GL_n^{\varepsilon }(3)}(T) \cong GL_i^{\varepsilon }(3) \times GL_{n-i}^{\varepsilon }(3)$ for some $1 \le i < n$ , or n is even and $C_{GL_n^{\varepsilon }(3)}(T) \cong GL_{n/2}(9)$ . So we have that $C_{GL_n^{\varepsilon }(3)}(T)$ is core-free.

Proof. By Lemma 3.10 (ii), we have (i).

Let $W_0 := W \cap SL_2(q)$ . By Lemma 3.12, $W_0$ is generalized quaternion. Also, $W_0$ is a maximal subgroup of W since $SL_2(q)$ has index $q-1$ in $GL_2(q)$ and $q \equiv 3 \mod 4$ . By [Reference Gorenstein23, Chapter 5, Theorem 4.3 (ii) (b)], we have $\Phi (W) = \langle a^2 \rangle $ . So the maximal subgroups of W are precisely the groups $M_1 := \langle a \rangle $ , $M_2 :=\langle a^2 \rangle \langle b \rangle $ and $M_3 := \langle a^2 \rangle \langle ab \rangle $ . One can check that $M_1 \cong C_{2^{n-1}}$ , $M_2 \cong D_{2^{n-1}}$ and $M_3 \cong Q_{2^{n-1}}$ . Consequently, $W_0 = \langle a^2 \rangle \langle ab \rangle $ , and (ii) holds.

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

The proof of (iv) is an easy exercise.

Proof. Using Lemmas 3.10 and 3.11, one can check that the centralizer of a Sylow 2-subgroup of $SL_2^{\varepsilon }(q)$ inside a Sylow 2-subgroup of $GL_2^{\varepsilon }(q)$ is the Sylow 2-subgroup of $Z(GL_2^{\varepsilon }(q))$ . Applying Lemma 3.14 and arguing by induction, one can see that a similar statement holds for the centralizer of a Sylow 2-subgroup of $SL_{2^r}^{\varepsilon }(q)$ inside a Sylow 2-subgroup of $GL_{2^r}^{\varepsilon }(q)$ for all $r \ge 0$ . Now we may apply Lemma 3.15 to obtain a Sylow 2-subgroup of G with the desired properties.

Proof. This follows from Lemma 2.13.

Proof. Assume that $\varepsilon \ne \varepsilon ^{*}$ . From $\varepsilon q \sim \varepsilon ^{*}q^{*}$ , it is easy to deduce that $\varepsilon q \equiv \varepsilon ^{*} q^{*} \mod 8$ and $(q^2-1)_2 = ((q^{*})^2-1)_2$ . So, in view of the remarks at the bottom of p. 11 of [Reference Broto, Møller and Oliver14], we may apply [Reference Broto, Møller and Oliver14, Proposition 3.3 (a)] to conclude that the 2-fusion system of $SL_n^{\varepsilon }(q)$ is isomorphic to the 2-fusion system of $SL_n^{\varepsilon ^{*}}(q^{*})$ .

Assume now that $\varepsilon = \varepsilon ^{*}$ . Using Dirichlet’s theorem [Reference Fine and Rosenberger20, Theorem 3.3.1], one can easily see that there is an odd prime $q_0$ with $\varepsilon q \sim \varepsilon q^{*} \sim -\varepsilon q_0$ . By the preceding paragraph, both the 2-fusion system of $SL_n^{\varepsilon }(q)$ and the 2-fusion system of $SL_n^{\varepsilon }(q^{*})$ are isomorphic to the 2-fusion system of $SL_n^{-\varepsilon }(q_0)$ . Consequently, the 2-fusion systems of $SL_n^{\varepsilon }(q)$ and $SL_n^{\varepsilon ^{*}}(q^{*})$ are isomorphic.

Proof. Let S and $S^{*}$ be Sylow $2$ -subgroups of $G:= SL_n^{\varepsilon }(q)$ and $G^{*} := SL_n^{\varepsilon ^{*}}(q^{*})$ , respectively. By Proposition 3.19, $\mathcal {F} := \mathcal {F}_S(G)$ and $\mathcal {F}^{*} := \mathcal {F}_{S^{*}}(G^{*})$ are isomorphic. Therefore, $\mathcal {F}/Z(\mathcal {F})$ and $\mathcal {F}^{*}/Z(\mathcal {F}^{*})$ are isomorphic. Lemma 3.18 implies that $\mathcal {F}/(S \cap Z(G))$ and $\mathcal {F}^{*}/(S^{*} \cap Z(G^{*}))$ are isomorphic. Now the proposition follows from Lemma 2.11.

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

Assume that $n = 2$ . Then $G \cong PSL_2(q)$ . By Lemma 3.13, G has dihedral Sylow $2$ -subgroups of order $\frac {1}{2}(q-1)_2(q+1)_2$ . So, if $q \equiv 3$ or $5 \ \mathrm {mod} \ 8$ , then G has abelian Sylow $2$ -subgroups and is thus a Goldschmidt group. If $q \equiv 1$ or $7 \ \mathrm {mod} \ 8$ , then the Sylow $2$ -subgroups of G are dihedral of order at least $8$ and hence nonabelian. Moreover, if $q \equiv 1$ or $7 \ \mathrm {mod} \ 8$ , then [Reference Steinberg49, Theorem 37] shows that G is not isomorphic to a finite simple group of Lie type in characteristic $2$ of Lie rank $1$ . So G is not a Goldschmidt group if $q \equiv 1$ or $7 \ \mathrm {mod} \ 8$ .

Assume now that $n \ge 3$ . Again, we see from [Reference Steinberg49, Theorem 37] that there is no finite simple group of Lie type in characteristic $2$ of Lie rank $1$ which is isomorphic to G. Also, G has a subgroup isomorphic to $SL_2^{\varepsilon }(q) \cong SL_2(q)$ , and therefore, the Sylow $2$ -subgroups of G are nonabelian. Consequently, G is not a Goldschmidt group.

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

Let $E_0$ be an elementary abelian $2$ -subgroup of $SL_n^{\varepsilon }(q)$ . We show that $E_0$ is conjugate to a subgroup of E. Using Dirichlet’s theorem [Reference Fine and Rosenberger20, Theorem 3.3.1], one can see that there is an odd prime number $q^{*}$ with $-q \sim q^{*}$ , and Proposition 3.19 shows that the $2$ -fusion systems of $SU_n(q)$ and $SL_n(q^{*})$ are isomorphic. Therefore, it is enough to consider the case $\varepsilon = +$ .

Since $E_0$ is an elementary abelian $2$ -group, any two elements of $E_0$ commute, and any element of $E_0$ is diagonalizable (see Lemma 3.3). It follows that $E_0$ is simultaneously diagonalizable, and this implies that $E_0$ is conjugate to a subgroup of E.

Proof. Let V be the defining module for Y. Let $B = (v_1,\dots ,v_n)$ be a basis for V with B orthonormal if V is unitary. For any $A \subseteq \lbrace 1,\dots ,n \rbrace $ , let $e_A$ be the linear map $V \rightarrow V$ represented by $t_A$ with respect to B. Then $e_A \in GL^{\varepsilon }(V)$ .

Let $\pi \in S_n$ . To prove (i), it suffices to find some $\alpha _{\pi } \in SL^{\varepsilon }(V)$ such that $(e_A)^{\alpha _{\pi }} = e_{A^{\pi }}$ for any $A \subseteq \lbrace 1, \dots , n \rbrace $ of even order. Let $\widetilde {\alpha }_{\pi }$ be the linear map $V \rightarrow V$ sending $v_i$ to $v_{i^{\pi }}$ for each $1 \le i \le n$ . Then $\mathrm {det}(\widetilde {\alpha }_{\pi }) = \mathrm {sgn}(\pi ) \in \lbrace -1, 1 \rbrace $ . Set $\alpha _{\pi } := \widetilde {\alpha }_{\pi }$ if $\mathrm {det}(\widetilde {\alpha }_{\pi }) = 1$ and $\alpha _{\pi } := e_{\lbrace 1 \rbrace } \widetilde {\alpha }_{\pi }$ if $\mathrm {det}(\widetilde {\alpha }_{\pi }) = -1$ . Then $\alpha _{\pi } \in SL^{\varepsilon }(V)$ . Also, if $A \subseteq \lbrace 1, \dots , n \rbrace $ and $1 \le i \le n$ , then

$$ \begin{align*} (v_i)^{(\alpha_{\pi})^{-1}e_A \alpha_{\pi}} = v_i^{(\widetilde{\alpha}_{\pi})^{-1}e_A\widetilde{\alpha}_{\pi}} = \begin{cases} -v_i & \, \text{if } i \in A^{\pi} \\ v_i & \, \text{if } i \not\in A^{\pi} \end{cases} \end{align*} $$

and hence $(e_A)^{\alpha _{\pi }} = e_{A^{\pi }}$ . The proof of (i) is now complete.

We now prove (ii). If $n \in \lbrace 1,2 \rbrace $ , then $\mathrm {Aut}_Y(E) = \lbrace \mathrm {id}_E \rbrace = \lbrace \varphi _{\pi } \mid \pi \in S_n \rbrace $ . Assume now that $n \ge 3$ . Let $\varphi \in \mathrm {Aut}_Y(E)$ , and let $y \in Y$ with $\varphi = c_y|_{E,E}$ . We are going to show that y is a generalized permutation matrix, which implies the desired conclusion that $\varphi = \varphi _{\pi }$ for some $\pi \in S_n$ . Let $y_1, \dots , y_n$ denote the columns of y, and let $1 \le j \le n$ . To prove that y is a generalized permutation matrix, it suffices to show that $y_j$ has precisely one nonzero entry. Let $1 \le k \ne \ell \le n$ with $k \ne j \ne \ell $ . Let $A := \lbrace j,k \rbrace $ and $C := \lbrace j, \ell \rbrace $ . As y normalizes E, there exist distinct subsets $A_0, C_0 \subseteq \lbrace 1, \dots , n \rbrace $ with $|A_0| = 2 = |C_0|$ and $(t_{A_0})^y = t_A$ , $(t_{C_0})^y = t_C$ . Hence, $t_{A_0} \cdot y = y \cdot t_A$ and $t_{C_0} \cdot y = y \cdot t_C$ , and so $y_j$ is an eigenvector of $t_{A_0}$ and of $t_{C_0}$ with eigenvalue $-1$ . Together with the fact that $|A_0| = 2 = |C_0|$ and $A_0 \ne C_0$ , it follows that $y_j$ has only one nonzero entry, as required.