Proof. Assume that P is $\mathcal {F}$ -centric. The centralizer system $C_{\mathcal {F}}(P)$ is a fusion system over the abelian group $C_S(P) = Z(P)$ , and $Z(P)$ is normal in $C_{\mathcal {F}}(P)$ from the definitions. As each morphism between subgroups of $Z(P)$ in $C_{\mathcal {F}}(P)$ extends to act as the identity on P, each such morphism is an identity map.

Proof. Let $Q \in \mathcal {F}^{cr}$ . Then $\operatorname {Aut}_{PQ}(Q)$ is normal in $\operatorname {Aut}_{\mathcal {F}}(Q)$ , and so $\operatorname {Aut}_{PQ}(Q) \leqslant \operatorname {Inn}(Q)$ since Q is radical. Then $P \leqslant PQ \leqslant QC_S(Q) = Q$ with the equality because Q is centric.

Proof. See [Reference Aschbacher, Kessar and OliverAKO11, I.2.6(c)].

Proof. Consider the restriction map $\rho \colon \operatorname {Aut}_{\mathcal {F}}(P) \to N_{\operatorname {Aut}_{\mathcal {F}}(W)}(\operatorname {Aut}_P(W))$ , under which $\operatorname {Aut}_W(P)$ maps onto $\operatorname {Inn}(W)$ and under which $\operatorname {Inn}(P)$ maps onto $\operatorname {Aut}_P(W)$ . Since W is weakly closed, it is fully $\mathcal {F}$ -normalized by Lemma 2.6. A direct application of the Extension axiom [Reference Broto, Levi and OliverBLO03, Definition 1.2(II)] then gives that $\rho $ is surjective. Since W is $\mathcal {F}$ -centric, the centralizer in $\mathcal {F}$ of the centric subgroup W is the fusion system of $Z(W)$ by Lemma 2.4, so the kernel of $\rho $ is $\operatorname {Aut}_{Z(W)}(P)$ , which is contained in $\operatorname {Aut}_W(P) \subseteq \operatorname {Inn}(P)$ . The induced map

$$\begin{align*}\operatorname{Aut}_{\mathcal{F}}(P)/\operatorname{Aut}_W(P) \longrightarrow N_{\operatorname{Aut}_{\mathcal{F}}(W)}(\operatorname{Aut}_P(W))/\operatorname{Aut}_W(W) \cong N_{\operatorname{Out}_{\mathcal{F}}(W)}(\operatorname{Out}_P(W)) \end{align*}$$

is an isomorphism, and therefore upon factoring by $\operatorname {Aut}_P(P)/\operatorname {Aut}_W(P)$ , the induced map

(2.1) $$ \begin{align} \operatorname{Out}_{\mathcal{F}}(P) \longrightarrow N_{\operatorname{Aut}_{\mathcal{F}}(W)}(\operatorname{Aut}_P(W))/\operatorname{Aut}_P(W) \cong N_{\operatorname{Out}_{\mathcal{F}}(W)}(\operatorname{Out}_P(W))/\operatorname{Out}_P(W) \end{align} $$

is an isomorphism.

Observe that W is normal in S because it is weakly $\mathcal {F}$ -closed. So $\operatorname {Out}_P(W) \cong P/W$ since $C_S(W) \leqslant W$ . The map $P \mapsto \operatorname {Out}_P(W)$ is therefore a bijection between the subgroups containing W and the subgroups of $\operatorname {Out}_S(P)$ . By (2.1), $\operatorname {Out}_{\mathcal {F}}(P)$ corresponds to $N_{\operatorname {Out}_{\mathcal {F}}(W)}(\operatorname {Out}_P(W))/\operatorname {Out}_P(W)$ under the bijection, so P is $\mathcal {F}$ -radical if and only if $\operatorname {Out}_P(W)$ is p-radical in the group $\operatorname {Out}_{\mathcal {F}}(W)$ (Remark 2.2). The last statement now follows because the collection of $\mathcal {F}$ -centric subgroups is closed under passing to overgroups.

Proof. Part (a) is clear, and (b) follows since, for each i,

$$\begin{align*}(a^ib)^2=a^iba^ib=ba^{-i}a^ib=b^2 \end{align*}$$

has order 2. A general element $a^jb^m$ as in (a) conjugates $a^ib$ to

$$\begin{align*}b^{-m}a^{-j}a^iba^jb^{m}=(b^{-m}a^{i-2j}b^{m})b= \begin{cases} a^{i-2j}b, \mbox{ if } m=0\\ a^{2j-i}b, \mbox{ if } m=1 \end{cases} \end{align*}$$

from which the claim in (c) follows.

Let $\mathcal {Q}$ be the set of subgroups of R isomorphic to $Q_8$ as in (d), and fix $Q \in \mathcal {Q}$ . As $\langle a \rangle $ is cyclic of index $2$ in R, we have $Q\langle a \rangle = R$ , and so $Q \cap \langle a \rangle = \langle a^{2^l} \rangle $ by order considerations. This shows that Q is of the form $\langle a^{2^l},a^ib \rangle $ for some i. Conversely, for each i, the elements $a^{2^{l}}$ and $a^{i}b$ satisfy the relations (2.2), applied with $l=0$ , in place of a and b, respectively. Hence, $\langle a^{2^l},a^ib \rangle \cong Q_8$ , and so $\langle a^{2^l}, a^ib \rangle \in \mathcal {Q}$ . This completes the proof of (d).

Note that exactly four elements of the form $a^ib$ lie in a given member of $\mathcal {Q}$ . Since there are $2^{l+2}$ choices for i, $\mathcal {Q}$ has cardinality $2^{l+2}/4 = 2^l$ . Part (e) now follows from the conjugacy information in (c), while (f) follows from the observation that $b^a=a^{-2}b$ so that $b^{a^i}=a^{-2i}b$ .

Proof. See [Reference Levi and OliverLO02, Lemma A.4(b)].

Notation 2.11. We fix the following notation for certain subgroups of K.

1. (a) $L_i \cong \operatorname {SL}_2(q)$ for $i = 1, 2, 3$ are the images in K of the subgroups $\widehat {L}_i$ of $\widehat {K}$ ;

2. (b) $L_0 := L_1L_2L_3$ ;

3. (c) $X \cong S_3$ is the image in K of the subgroup with the same name;

4. (d) $\tau \in X$ is the permutation $(1,2)$ on the indices of the $L_i$ ;

5. (e) S is a Sylow $2$ -subgroup of K containing $\tau $ ;

6. (f) $U = Z(L_0) = \langle [\pm 1, \pm 1, \pm 1] \rangle \cong C_2 \times C_2$ ; and

7. (g) $B := L_0S$ .

Notation 2.12. We fix the following additional notation for subgroups and elements of K.

1. (a) $R_i \cong Q_{2^{l+3}}$ is a Sylow $2$ -subgroup of $L_i$ for $i=1,2,3$ , chosen so that $X \cong S_3$ acts on the set $\{R_1,R_2,R_3\}$ ;

2. (b) $R_0:=R_1R_2R_3 \in \operatorname {Syl}_2(L_0)$ ;

3. (c) $\mathcal {Q}_i$ is the set of subgroups of $R_i$ isomorphic to $Q_8$ for $i=1,2,3$ ; thus, $\mathcal {Q}_i = \{R_i\}$ if $l = 0$ , while $\mathcal {Q}_i$ is a union of two $R_i$ -conjugacy classes of subgroups if $l> 0$ by Lemma 2.8(e);

4. (d) when $l> 0$ , $Q_i,Q_i' \in \mathcal {Q}_i$ are representatives for the two $R_i$ -conjugacy classes of subgroups chosen so that $X \cong S_3$ acts by permuting the sets $\{Q_1,Q_2,Q_3\}$ and $\{Q_1',Q_2',Q_3'\}$ ;

5. (e) $\mathbf {c} := [c,c,c]$ , where c is as in Section 2.2, so that $\mathbf {c}$ acts simultaneously on $L_i \cong SL_2(q)$ by conjugation in the way described there;

6. (f) $\mathbf {d} := [b,b,b]\mathbf {c} \in K$ , where b is as in Section 2.2, an involution commuting with $\tau $ ; and

7. (g) $\tau ' = \mathbf {d}\tau $ .

Proof. By [Reference Aschbacher and ChermakAC10, Lemma 4.9(c)], there is a unique homocyclic subgroup of S of rank $3$ and exponent $4$ , T is the centralizer in S of that subgroup, and $C_H(T)$ is a split maximal torus of H. Since T is abelian, this shows that T is the unique subgroup of S of its isomorphism type. Then [Reference Aschbacher and ChermakAC10, Lemmas 4.3 and 4.8] show that $S/T \cong C_2 \times D_8$ , and $\operatorname {Out}_{\mathcal {H}}(T) \cong C_2 \times S_4$ . The structure of the outer automorphism group $\operatorname {Out}_{\mathcal {F}}(T)$ follows from the construction of the Aschbacher-Chermak amalgam in [Reference Aschbacher and ChermakAC10, Lemma 5.2]. All other points follow.

Proof. Since $E = \Omega _1(T)$ , we have $C_S(E) \geqslant T$ . As T is $\mathcal {F}$ -centric, so is $C_S(E)$ . Let $\varphi \in \operatorname {Hom}_{\mathcal {F}}(C_S(E),S)$ . By Lemma 2.13, $T^\varphi = T$ , so also $E^\varphi = E$ . Hence, $C_S(E)^\varphi \leqslant C_S(E^\varphi ) = C_S(E)$ , and so $C_S(E)$ is weakly $\mathcal {F}$ -closed. From the description of $\operatorname {Out}_{\mathcal {F}}(T)$ in Lemma 2.13, the kernel of the action of $S/T$ on E is of order $2$ . Now $\mathbf {d} \in S$ inverts T, so centralizes $E = \Omega _1(T)$ . Hence, $\mathbf {d}$ represents the lone nontrivial coset of $C_H(T)$ in $C_H(E)$ , whose elements invert the maximal torus $C_H(T)$ of H containing T (see [Reference Aschbacher and ChermakAC10, Lemma 4.3(a,d)]). So $C_S(E) = T\langle \mathbf {d} \rangle $ , and $C_H(C_S(E)) \leqslant T$ from Lemma 2.13. Hence, the center $Z(C_S(E))$ is $C_H(C_S(E)) = C_{T}(\mathbf {d}) = E$ .

As $O_2(\operatorname {Out}_{\mathcal {F}}(T)) = \operatorname {Out}_{C_S(E)}(T)$ , the descriptions of the outer automorphism groups in $\mathcal {F}$ and $\mathcal {H}$ follow from Lemmas 2.7 and 2.13.

Proof. Since $C_S(E)$ contains its centralizer in H from Lemma 2.14, so does S. Then as the Sylow $2$ -subgroups of $S_4$ and $\operatorname {GL}_3(2)$ are self-normalizing, the lemma now follows from Lemmas 2.7 and 2.14.

Proof. From the structure of $\operatorname {Out}_{\mathcal {F}}(C_S(E))$ in Lemma 2.14, $\operatorname {Out}_{C_S(U)}(C_S(E)) = C_S(U)/C_S(E)$ is the unipotent radical of the stabilizer of U in the action of $\operatorname {Out}_{\mathcal {F}}(C_S(E))$ on E, so, in particular, $Z(C_S(U)) = C_{E}(C_S(U)) = U$ . The descriptions of the outer automorphism groups now follow from Lemmas 2.7 and 2.14 and the structure of $\operatorname {GL}_3(2)$ .

Proof. Observe that, $C_S(E) \leqslant C_S(E/Z)$ and that $C_S(E/Z)/C_S(E)$ is the group of transvections in $\operatorname {Out}_{\mathcal {F}}(C_S(E))$ on E with center Z. So $C_S(E/Z)/C_S(E)$ is the unipotent radical of the stabilizer of Z. Also, as $Z(C_S(E)) = Z$ from Lemma 2.14, we have $Z(C_S(E/Z)) = C_{E}(C_S(E/Z)) = Z$ . Since $C_S(E)$ is $\mathcal {F}$ -centric, weakly $\mathcal {F}$ -closed, and $\operatorname {Aut}_{\mathcal {H}}(C_S(E)) = C_{\operatorname {Aut}_{\mathcal {F}}(C_S(E))}(Z)$ , all points follow from Lemmas 2.7 and 2.14 as in the previous lemma.

Proof. There are four $2$ -radical subgroups in $\operatorname {GL}_3(2)$ inside a fixed Sylow $2$ -subgroup: the identity subgroup and the unipotent radicals of the three associated parabolics. So the lemma follows from the bijection of Lemma 2.7 together with Lemmas 2.14–2.17.

Proof. By Lemma 3.2(a) below, S contains an extraspecial subgroup with central quotient of rank $6$ , and hence $s(S) \geqslant 6$ . On the other hand, the sectional rank of a group is at most the sum of the sectional ranks of a normal subgroup and corresponding quotient, so Lemma 2.13 shows that $s(S) \leqslant s(T) + s(S/T) = 3 + 3 = 6$ .

Proof. As Q is a centric normal $2$ -subgroup of $\mathcal {K}$ , it is contained in every member of $\mathcal {K}^{cr}$ by Lemma 2.5. Now $S/Q$ is a four group (the four group $\langle \mathbf {d}, \tau \rangle $ is a complement to Q in S), so there are only five possible centric radical subgroups. Since $O^2(K) \cap S = Q$ , if two distinct subgroups of S containing Q were $\mathcal {K}$ -conjugate, then two distinct subgroups of the abelian group $S/Q$ would be $K/O^2(K) \cong S/Q$ -conjugate. Since this is not the case, no two distinct subgroups of S containing Q are $\mathcal {K}$ -conjugate. Next, from the definition of U, both Q and $\mathbf {d}$ centralize U while $\tau $ does not, so we must have $Q\langle \mathbf {d} \rangle =C_S(U)$ . This shows the equality in (e).

The structure of the outer automorphism groups are computable from knowledge of $\operatorname {Out}_{\mathcal {K}}(Q)$ : note that from the structure of K (cf. Lemma 2.9),

$$\begin{align*}\operatorname{Out}_{\mathcal{K}}(Q) \cong (C_3)^3 \rtimes (C_2 \times S_3) \end{align*}$$

is a split extension of the wreath product $C_3 \wr S_3$ by the group generated by the class $\left [c_{\mathbf {d}}\right ] \in \operatorname {Out}_{\mathcal {K}}(Q)$ of conjugation by $\mathbf {d}$ acting by inversion on the base. As Q is weakly $\mathcal {K}$ -closed and centric, $\operatorname {Out}_{\mathcal {K}}(P) \cong N_{\operatorname {Out}_{\mathcal {K}}(Q)}(\operatorname {Out}_P(Q))/\operatorname {Out}_P(Q)$ for each overgroup P of Q in S by Lemma 2.7. From a computation in the group $(C_3)^3 \overset {-1 \times \wr }{\rtimes } (C_2 \times S_3)$ , one sees, for example, that

$$\begin{align*}N_{\operatorname{Out}_{\mathcal{K}}(Q)}(\langle \left[ c_\tau \right] \rangle ) = C_{\operatorname{Out}_{\mathcal{K}}(Q)}(\langle \left[ c_\tau \right] \rangle ) \cong (C_3)^2 \overset{-1, 1}{\rtimes} (C_2 \times C_2), \end{align*}$$

where the acting group is given by $\langle \left [c_{\mathbf {d}} \right ]\rangle \times \langle \left [c_{\tau } \right ]\rangle $ . This shows that $\operatorname {Out}_{\mathcal {K}}(Q\langle \tau \rangle ) = N_{\operatorname {Out}_{\mathcal {K}}(Q)}(\langle \left [ c_\tau \right ] \rangle )/\langle \left [ c_\tau \right ] \rangle \cong (C_3 \times C_3) \overset {-1}{\rtimes } C_2$ as claimed. Cases (d) and (e) are handled similarly. Visibly, no resulting outer automorphism group has a nontrivial normal $2$ -subgroup, so all the candidate subgroups are $\mathcal {K}$ -centric radical.

Table 1

$\operatorname {Sol}(5)$ -conjugacy classes of $\operatorname {Sol}(5)$ -centric radical subgroups.

Proof. By Proposition 3.5, the column for $\mathcal {K}$ is correct. By [Reference Chermak, Oliver and ShpectorovCOS08, Lemma 2.1] and [Reference Levi and OliverLO02, Lemma A.11(e,f)], the column for $\mathcal {H}$ is correct. We work up to $\mathcal {F}$ -conjugacy in what follows. Let $P \in \mathcal {F}^{cr}$ . By Proposition 3.1, either P is listed in the last two rows of Table 1, or one of the following holds: (1) $P \in \mathcal {K}^{cr}$ and $\operatorname {Out}_{\mathcal {F}}(P) = \operatorname {Out}_{\mathcal {K}}(P)$ , or (2) $N_G(P) \nleq K$ , $P \in \mathcal {H}^{cr}$ , and $\operatorname {Out}_{\mathcal {F}}(P) = \operatorname {Out}_{\mathcal {H}}(P)$ . If (1) holds, then P is listed in the first five rows of the table by Proposition 3.5. If (1) does not hold, then (2) holds, P is listed in the next three rows of the table, where the entries follow from Proposition 3.2(a,b,d).

That no additional $\mathcal {F}$ -conjugacy can occur between these subgroups can be seen in several ways, one of which as follows. Only three subgroups have pairwise equal orders and isomorphic outer automorphism groups in $\mathcal {F}$ , namely, $Q\langle \tau ' \rangle $ , $C_S(U)$ , and $C_S(E/Z)$ .

By Lemma 2.14, $C_S(E/Z)$ has center Z. Likewise, since $Z(Q) = U$ , we have $Z(Q\langle \tau ' \rangle ) = C_{U}(\tau ')~=~Z$ . So as $U \leqslant Z(C_S(U))$ , it follows that $C_S(U)$ is not $\mathcal {F}$ -conjugate to either of the other two subgroups.

Finally, note that $C_S(E/Z)$ contains the torus T. On the other hand, from the description of T in Section 2.5, we see that $Q \cap T = \langle [a,1,1],[1,a,1],[c^2,c^2,c^2] \rangle $ is of index $2$ in T. As each element in the coset $Q\tau '$ is nontrivial on $Q \cap T$ and T is abelian, it follows that $Q\langle \tau ' \rangle \cap T$ is still of index $2$ in T. So $C_S(E/Z)$ contains T, but $Q\langle \tau ' \rangle $ does not. Since T is weakly $\mathcal {F}$ -closed (Lemma 2.13), the subgroups $C_S(E/Z)$ and $Q\langle \tau ' \rangle $ are not $\mathcal {F}$ -conjugate.

Proof. The subgroup S is clearly weakly closed, and $C_S(E)$ , $C_S(U)$ , and $C_S(E/Z)$ were shown to be weakly $\mathcal {F}$ -closed in Lemmas 2.14, 2.16, and 2.17. Let P be one of the remaining subgroups, but not A. By Proposition 3.6, P is centric and radical in $\mathcal {H}$ , and either $P = Q$ or $Z(P) = Z$ . The quotient $P/Z$ is centric and radical in $\mathcal {H}/Z$ by [Reference Levi and OliverLO02, Lemma A.11(e)]. Hence, $P/Z$ is weakly $\mathcal {H}/Z$ -closed by [Reference Chermak, Oliver and ShpectorovCOS08, Lemma 2.1]. It follows that P is weakly $\mathcal {H}$ -closed. Since Q is normal in $\mathcal {K}$ , it is weakly $\mathcal {K}$ -closed. Hence, Q is weakly $\mathcal {F}$ -closed since $\mathcal {H}$ and $\mathcal {K}$ are fusion systems over S which generate $\mathcal {F}$ (end of Section 2.4).

We are reduced to the case in which $Z(P) = Z$ . Assume on the contrary that P is not weakly $\mathcal {F}$ -closed. By Alperin’s Fusion theorem [Reference Broto, Levi and OliverBLO03, Theorem A.10], there is an overgroup $Y \leqslant S$ of P and an automorphism $\alpha \in \operatorname {Aut}_{\mathcal {F}}(Y)$ , such that $P^\alpha \neq P$ . Then $Z(Y) \leqslant Z(P) = Z$ , as P is centric, so that $Z(Y) = Z$ is centralized by $\alpha $ . That is, $\alpha \in \mathcal {H}$ . But then $P^\alpha = P$ by the previous paragraph, a contradiction.

Proof. This is a statement depending on H only. Since $l=0$ , $q = 5$ . Write $\overline {H}$ for $H/Z$ . Fix a decomposition

$$\begin{align*}V = \ell_1 \perp \ell_2 \perp \ell_3 \perp \langle x_7 \rangle \end{align*}$$

with the following properties ([Reference Aschbacher and ChermakAC10, cf. Lemmas 4.4, 4.6]):

1. 1. each $\ell _i = \langle x_{2i-1}, x_{2i} \rangle $ is a hyperbolic line (i.e., $\mathfrak {q}(x_{2i-1}) = 0 = \mathfrak {q}(x_{2i})$ , $\mathfrak {b}(x_{2i-1}, x_{2i}) = 1$ ), and $\mathfrak {q}(x_7) = 1$ .

2. 2. $\ell _1 \perp \ell _2 = \langle x_1,x_4 \rangle \oplus \langle x_3,x_2 \rangle $ , with each summand on the right side a natural $\mathbb {F}_5L_1$ -module; in particular, $[a,1,1]$ and $[b,1,1]$ act via the matrices

   $$ \begin{align*} [a,1,1] &\mapsto \left[\begin{smallmatrix}2&0&0&0\\0&-2&0&0\\0&0&2&0\\0&0&0&-2\end{smallmatrix}\right] & [b,1,1] &\mapsto \left[\begin{smallmatrix}0&0&0&-1\\0&0&1&0\\0&-1&0&0\\1&0&0&0\end{smallmatrix}\right] \end{align*} $$
   with respect to the basis $\{x_1,x_2,x_3,x_4\}$ .

3. 3. $\ell _1 \perp \ell _2 = \langle x_1,x_3 \rangle \oplus \langle x_4,x_2 \rangle $ , with each summand on the right side a natural $\mathbb {F}_5L_2$ -module; in particular, $[1,a,1]$ and $[1,b,1]$ act via the matrices

   $$ \begin{align*} [1,a,1] &\mapsto \left[\begin{smallmatrix}2&0&0&0\\0&-2&0&0\\0&0&-2&0\\0&0&0&2\end{smallmatrix}\right] & [1,b,1] &\mapsto \left[\begin{smallmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\end{smallmatrix}\right] \end{align*} $$
   with respect to the basis $\{x_1,x_2,x_3,x_4\}$ .

4. 4. $\ell _3 \perp \langle x_7 \rangle $ is the three-dimensional orthogonal module for $L_3 \cong \operatorname {Spin}_3(5)$ . We may view it as the module in which $L_3$ acts by conjugation on $2\times 2$ trace zero matrices $M_2^0(\mathbb {F}_5)$ with quadratic form given by the determinant, via the isometry $M_2^0(\mathbb {F}_5) \longrightarrow \ell _3 \perp \{x_7\}$ defined by

   $$\begin{align*}\left\{\left[\begin{smallmatrix}0&0\\2&0\end{smallmatrix}\right], \left[\begin{smallmatrix}0&-1\\0&0\end{smallmatrix}\right], \left[\begin{smallmatrix}2&0\\0&-2\end{smallmatrix}\right]\right\} \longmapsto \{x_5,x_6,x_7\}. \end{align*}$$
   Under this identification, $[1,1,a]$ acts via the matrix $\operatorname {diag}(-1, -1, 1)$ with respect to the ordered basis $\{x_5,x_6,x_7\}$ , and $[1,1,b]$ acts via the matrix $\left [\begin {smallmatrix}0&2&0\\-2&0&0\\0&0&-1\end {smallmatrix}\right ]$ .

Next, define $u_{j}$ and $v_{j}$ via

$$ \begin{align*} u_{2i-1} &= x_{2i-1}-2x_{2i}, & v_{2i-1} &= u_{2i-1}+u_{2i},\\ u_{2i} &= -2x_{2i-1}+x_{2i}, & v_{2i} &= u_{2i-1}-u_{2i},\\ u_7 &= x_7, & v_7 &= u_7. \end{align*} $$

Thus, $\{u_1,\dots ,u_7\}$ is an orthonormal basis for V, and $\{v_1,\dots ,v_7\}$ is an orthogonal basis, such that $\mathfrak {q}(v_i) = 2 \notin \mathbb {F}_5^{\times 2}$ for each $i = 1,\dots ,6$ . The decompositions

$$\begin{align*}\Lambda = \{\langle u_i \rangle \mid i \in \{1,\dots,7\}\}, \qquad \text{ and } \qquad \Lambda' = \{\langle v_i \rangle \mid i \in \{1,\dots,7\}\} \end{align*}$$

of V are therefore of the type appearing in Proposition 3.2(a) and (b), respectively. The centralizers of the decompositions are

$$ \begin{align*} C_H(\Lambda) &= \{e \in \operatorname{Spin}_7(5) \mid (u_i)\overline{e} = \pm u_i, i = 1,\dots,7\}, \text{ and}\\ C_{H}(\Lambda') &= \{f \in \operatorname{Spin}_7(5) \mid (v_i)\overline{f} = \pm v_i, i = 1,\dots,7\}. \end{align*} $$

Observe from the definition of the $v_i$ that $C_H(\Lambda ) \leqslant N_H(\Lambda ')$ . Similarly, it is a straightforward computation to see using (2)–(4) that Q acts on the sets $\Lambda $ and $\Lambda '$ , that is, $Q \leqslant N_H(\Lambda )$ and $Q \leqslant N_H(\Lambda ')$ . It follows that $QC_H(\Lambda )C_H(\Lambda )'$ is a $2$ -subgroup of H. Hence, we may choose $h \in H$ with $(QC_H(\Lambda )C_H(\Lambda '))^h \leqslant S$ . But $Q \leqslant S$ , and so $Q^h = Q$ by Lemma 3.7. Likewise, it follows from Lemma 3.7 that $C_H(\Lambda )^h = R_{1^7}$ and $C_H(\Lambda ')^h = R_{1^7}'$ . Replacing S with $S^{h^{-1}}$ if necessary, we may assume that $R_{1^7} = C_H(\Lambda )$ and $R_{1^7}' = C_H(\Lambda ')$ .

For a subset $I \subseteq \{1,\dots ,7\}$ , write $e_I$ for a fixed element of $C_H(\Lambda )$ which maps $u_i \mapsto -u_i$ if $i \in I$ , and which fixes $u_i$ otherwise. When $I \subseteq \{1,\dots ,6\}$ , denote by $f_I$ an analogous element of $C_H(\Lambda ')$ with respect to the $v_i$ ’s. A computation of the action of Q with respect to the bases $\{u_i \mid 1 \leqslant i \leqslant 7\}$ and $\{v_i \mid 1 \leqslant i \leqslant 7\}$ using (2)–(4) yields

$$ \begin{align*} Q \cap R_{1^7} &= \langle [-1,1,1], [b,ab,1], [ab,b,1], [1,1,a], [1,1,b] \rangle\\ &= \langle e_{1234}, e_{13}, e_{14}, e_{56}, e_{57} \rangle\\ &\cong C_2 \times (Q_8*Q_8), \end{align*} $$

and

$$ \begin{align*} Q \cap R_{1^7}' &= \langle [-1,1,1], [b,b,1], [ab,ab,1], [1,1,a] \rangle\\ &= \langle f_{1234}, f_{23}, f_{13}, f_{56} \rangle\\ &\cong C_2 \times (Q_8*C_4), \end{align*} $$

where here we have used Lemma 2.10 and the identity $[e,f] = (ef)^2$ to determine the isomorphism types. The order $|Q \cap R_{1^7}| = 2^6$ , and so $|QR_{1^7}| = \frac {|Q||R_{1^7}|}{|Q \cap R_{1^7}|}= 2^9$ . Similarly, $|Q \cap R_{1^7}'| = 2^5$ , so also $|QR_{1^7}'| = 2^9$ .

We have shown that $\{QR_{1^7}, QR_{1^7}'\} \subset \{Q\langle \mathbf {d} \rangle , Q\langle \tau \rangle , Q\langle \tau ' \rangle \}$ . The involution $e_{4567} \in R_{1^7}-Q$ (Lemma 2.10) acts as $-1$ on $\ell _3 \perp \langle x_7 \rangle $ , so centralizes $L_3$ . It also interchanges the one-dimensional subspaces $\langle x_3 \rangle $ and $\langle x_4 \rangle $ while centralizing the line $\ell _1$ , and hence from (2)–(3), it interchanges $L_1$ and $L_2$ by conjugation. It follows that $QR_{1^7} = Q\langle \tau \rangle $ , since neither $Q\langle \mathbf {d} \rangle $ nor $Q\langle \tau ' \rangle $ have such an element.

Finally, we show that $QR_{1^7}' = Q\langle \tau ' \rangle $ . First, since $f_{1234} \in U-Z$ does not commute with $f_{45}$ by Lemma 2.10, it follows that $QR_{1^7}'$ is not contained in $C_S(U) = Q\langle \mathbf {d} \rangle $ . Next, observe that in contrast to the previous case, $C_{R_{1^7}'}(L_3) = C_{R_{1^7}'}(\ell _3+\langle x_7 \rangle )$ (for example, note that “ $f_{4567}$ ” has nontrivial spinor norm). The group $C_{R_{1^7}'}(L_3) = \langle f_{12}, f_{13}, f_{14} \rangle $ induces the permutation group $\langle (1,2)(3,4) \rangle $ on $\{\langle x_1 \rangle ,\langle x_2 \rangle ,\langle x_3 \rangle ,\langle x_4 \rangle \}$ , and hence $C_{R_{1^7}'}(L_3)$ acts on $L_1$ and $L_2$ by (2)–(3). Therefore, $QR_{1^7}'$ has no element centralizing $L_3$ and interchanging $L_1$ and $L_2$ , and so $QR_{1^7}' = Q\langle \tau ' \rangle $ .

Proof. Let $P \leqslant S$ be a centric radical subgroup of $\mathcal {K}$ , taken up to $\mathcal {K}$ -conjugacy. We proceed through the possibilities in the description of $\mathcal {K}^{cr}$ given by Proposition 3.4 and refer to the labelings of the three cases given there. If P occurs in (b)(i), then P is listed in the first two rows of the table. By Lemma 2.15, $\operatorname {Aut}_{\mathcal {K}}(S) = \operatorname {Inn}(S)$ , so that $\operatorname {Out}_{\mathcal {K}}(S) = 1$ . Also, $C_S(U) = R_0\langle \mathbf {d} \rangle $ , so that $\operatorname {Out}_{\mathcal {K}}(C_S(U)) \cong S_3$ is induced by X.

Table 2

$\mathcal {K}$ -conjugacy classes of $\mathcal {K}$ -centric radical subgroups, $l> 0$ .

Consider a subgroup P in (b)(ii). First, assume that $P_i \in \mathcal {Q}_i$ for all i. Upon conjugating in $L_0$ , we may assume that $P_i = Q_i$ or $Q_i'$ for each i. Conjugating by $\mathbf {d}$ , which interchanges $Q_i$ and $Q_i'$ for each i, we may assume that there is at most one $Q_i'$ among the $P_i$ ’s. Finally, we may conjugate by elements of X to see that P is one of the subgroups in rows 3 and 4 of the table.

To compute $\operatorname {Out}_{\mathcal {K}}(P)$ , observe that if $t \in L_0X$ , then $P^t$ and P have the same number of components $P_i^t$ which are $L_i$ -conjugate to $Q_i$ , while $P^{\mathbf {d}}$ has three minus the number for P. This shows that $N_{K}(P) = N_{L_{0}X}(P)$ . Thus, if $P = Q_1Q_2Q_3$ , then $N_K(P) = (N_{L_1}(Q_1)N_{L_2}(Q_2)N_{L_3}(Q_3))X$ , and we see that $\operatorname {Out}_{\mathcal {K}}(P) \cong S_3 \wr S_3$ by Lemma 2.9. Likewise, if $P = Q_1Q_2Q_3'$ , then $N_K(P) = N_{L_1}(Q_1)N_{L_2}(Q_2)N_{L_3}(Q_3')\langle \tau \rangle $ , so that $\operatorname {Out}_{\mathcal {K}}(P) \cong (S_3 \wr C_2) \times S_3$ .

Next, assume that $P_i \in \mathcal {Q}_i$ for exactly two indices i. Then as before, we may conjugate so that $P = Q_1Q_2R_3$ or $Q_1Q_2'R_3$ is on the table. Appealing to Lemma 2.9 again to see that $\operatorname {Out}_{L_3}(R_3) = 1$ , we have in the former case that $\operatorname {Out}_{\mathcal {K}}(P) \cong S_3 \wr C_2$ with the class of $\tau $ wreathing, while in the latter case, we have a similar situation with the class of $\tau '$ wreathing. This concludes the case (b)(ii).

Consider now a subgroup P in (b)(iii), and recall that $Z(L_0) = U$ . Thus, $P = P_0\langle s \rangle $ with $s \in P-C_P(U)$ normalizing $P_0$ . Set $N = N_K(P)$ and $M = N_K(P_0)$ . Denote quotients modulo $P_0$ with bars. We set $M^+ = \overline {M}/O_{2'}(\overline {M})$ and write quotients modulo $O_{2'}(\overline {M})$ with pluses. Thus, for any subgroup $Y \leqslant M$ , we write $Y^+$ for the image of Y modulo the preimage of $O_{3}(\overline {M})$ in M.

Since $L_0 \unlhd K$ , we see that $P_0 = P \cap L_0 \unlhd N$ , so that $N \leqslant M$ . In particular, $\overline {N}$ is defined. Also, since $\overline {s}$ is of order $2$ , N is the preimage in M of $C_{\overline {M}}(\overline {s})$ . As P is radical, we must have

(3.2) $$ \begin{align} \langle \overline{s} \rangle = O_2(C_{\overline{M}}(\overline{s})). \end{align} $$

We consider separately the cases where $P_0 \notin \mathcal {K}^{cr}$ and where $P_0 \in \mathcal {K}^{cr}$ . Assume first that $P_0 \notin \mathcal {K}^{cr}$ , the easier case. Upon comparing the conditions in (b)(ii) and (b)(iii), we have by our assumption that $P_3 \in \mathcal {Q}_3$ and $P_i = R_i$ for $i = 1, 2$ . Thus, $\overline {M} = \langle \overline {\tau } \rangle \times \overline {N_{L_3}(P_3)} \cong C_2 \times S_3$ , and so $P = P_0\langle \tau \rangle $ by (3.2). Hence, $\operatorname {Out}_{\mathcal {F}}(P) \cong S_3$ , and P appears in the last row of the table.

Assume next that $P_0 \in \mathcal {K}^{cr}$ , so that $P_0$ is conjugate to a subgroup considered in (b)(ii), rows 3–6 of the table. First, assume that $P_0$ itself appears in rows 3–6. Our description of the normalizer in K of $P_0$ in a previous paragraph together with order considerations imply that $N_{R_0}(P_0)\langle \tau \rangle /P_0$ is a Sylow $2$ -subgroup of $\operatorname {Out}_{\mathcal {K}}(P_0)$ if $P_0$ appears in rows 3–5, and that $N_{R_{0}}(P_0)\langle \tau ' \rangle /P_0$ is a Sylow $2$ -subgroup of $\operatorname {Out}_{\mathcal {K}}(P_0)$ if $P_0$ appears in row 6. This shows that $\operatorname {Aut}_{S}(P_0)$ is a Sylow $2$ -subgroup of $\operatorname {Aut}_{\mathcal {K}}(P_0)$ , that is, $P_{0}$ is fully $\mathcal {K}$ -automized [Reference Aschbacher, Kessar and OliverAKO11, I.2.2]. Since $P_0$ is $\mathcal {K}$ -centric, it is fully $\mathcal {K}$ -centralized ([Reference Aschbacher, Kessar and OliverAKO11, I.3.1]). Hence, $P_0$ is fully $\mathcal {K}$ -normalized by [Reference Aschbacher, Kessar and OliverAKO11, I.2.6(c)]. Thus, by Lemma 2.6 (and since $P_0 \unlhd P$ ), we may in any case replace P by a $\mathcal {K}$ -conjugate and assume that $P_0$ is in rows 3–6 of the table.

Case 1. Assume $P_0 = Q_1Q_2Q_3$ , and recall Lemma 2.8(e).

Here, $\overline {M} = \overline {N_{L_0}(P_0)}\overline {X} \cong S_3 \wr S_3$ , $\overline {N_{R_0}(P_0)}^+ = O_{2}(M^+)$ , and $\overline {N_{S}(P_0)} = \overline {N_{R_0}(P_0)}\langle \overline {\tau } \rangle $ . By assumption, s is not in $C_P(U)$ , so it is not in $R_0$ . Hence, $\overline {s}$ is not in $\overline {N_{R_0}(P_0)}$ . Thus,

(3.3) $$ \begin{align} \overline{s} \in \overline{N_{R_0}(P_0)}\overline{\tau}. \end{align} $$

Write $\overline {s} = \overline {[t_1,t_2,t_3]}\overline {\tau }$ , where each $t_i \in R_i$ , and where we take $t_i = 1$ if $t_i \in P_i$ and $t_i = a^{2^{l-1}}$ if $t_i \notin P_i$ . As $\tau $ acts by swapping $R_1$ and $R_2$ and centralizing $R_3$ , it follows from (3.3) that $C_{\overline {N_{R_0}(P_0)}}(\overline {s})$ is of order 4 generated by $\overline {[a^{2^{l-1}}, a^{2^{l-1}}, 1]}$ and $\overline {[1,1,a^{2^{l-1}}]}$ .

Note that $t_1 = t_2$ since $\overline {s}$ is of order $2$ . We claim that (3.2) and (3.3) imply $t_3 = 1$ . Assume on the contrary that $t_3 = a^{2^{l-1}}$ . Then $O_{2'}(C_{\overline {M}}(\overline {s})) = C_{O_{2'}(\overline {M})}(\overline {s}) \leqslant \overline {N_{L_1}(Q_1)}\overline {N_{L_2}(Q_2)}$ . By (3.3), $C_{O_{2',2,2'}(\overline {M})}(\overline {s}) = C_{O_{2',2}(\overline {M})}(\overline {s})$ . So since $\langle \overline {[1,1,a^{2^{l-1}}]} \rangle $ is a normal $2$ -subgroup of $\overline {N_S(P_0)}$ , it follows that $\langle \overline {[1,1,a^{2^{l-1}}]} \rangle $ is a normal $2$ -subgroup of $C_{\overline {M}}(\overline {s})$ . By (3.2), $\overline {s} = \overline {[1,1,a^{2^{l-1}}]}$ . This contradicts (3.3).

We conclude that $\overline {s} = \overline {\tau }$ or $\overline {s} = \overline {[a^{2^{l-1}}, a^{2^{l-1}}, 1]}\overline {\tau }$ , and hence

$$\begin{align*}C_{\overline{M}}(\overline{s}) = \langle \overline{s} \rangle \times C_{\overline{N_{L_1}(Q_1)N_{L_2}(Q_2)}}(\overline{s}) \times C_{\overline{N_{L_3}(Q_3)}}(\overline{s}) \cong C_2 \times S_3 \times S_3. \end{align*}$$

However, since the two possibilities for $\overline {s}$ are conjugate under $\overline {N_{R_1}(Q_1)}$ , we may take $\overline {s} = \overline {\tau }$ , as desired.

Case 2. Assume $P_0 = Q_1Q_2Q_3'$ . In this case, $\overline {M} \cong S_3 \wr \langle \overline {\tau } \rangle \times S_3$ . Replacing all occurrences of $Q_3$ by $Q_3'$ , we argue verbatim as in Case 1, except that in the present situation, we have $O_{2',2,2'}(\overline {M}) = O_{2',2}(\overline {M}) = \overline {M}$ , obviating the need to observe that $C_{O_{2',2,2'}(\overline {M})}(\overline {s}) = C_{O_{2',2}(\overline {M})}(\overline {s})$ . Again, we may take $\overline {s}=\overline {\tau }$ .

Case 3. Assume $P_0 = Q_1Q_2R_3$ . We have $\overline {M} = \overline {N_{L_0}(P_0)}\overline {X} \cong S_3 \wr \langle \overline {\tau } \rangle $ , $\overline {N_{R_0}(P_0)}^+ = O_{2}(M^+)$ , and $\overline {N_{S}(P_0)} = \overline {N_{R_0}(P_0)}\langle \overline {\tau } \rangle $ . By assumption, s is not in $C_P(U)$ , so it is not in $R_0$ . Hence, $\overline {s}$ is not in $\overline {N_{R_0}(P_0)}$ . Thus, $\overline {s} \in \overline {N_{R_0}(P_0)}\overline {\tau }.$ As in the previous cases, (3.2) forces $\overline {s} = \overline {\tau }$ or $\overline {[a^{2^{l-1}}, a^{2^{l-1}},1]}\overline {\tau },$ so that

$$ \begin{align*} C_{\overline{M}}(\overline{s}) = \langle \overline{s}\rangle \times C_{\overline{N_{L_1}(Q_1)N_{L_2}(Q_2)}}(\overline{s}) = C_2 \times S_3.\end{align*} $$

Again, these possibilities for $\overline {s}$ are conjugate under $\overline {N_{R_1}(Q_1)}$ , and we see that we may take $\overline {s} = \overline {\tau }$ , as needed.

Case 4. Assume $P_0 = Q_1Q_2'R_3$ . This time, replace $\tau $ by $\tau '$ , and repeat the argument from the previous case.

Proof. Let $P \in \mathcal {H}^{cr}$ . If $N_H(P) \nleq K$ , then using Proposition 3.2(a)–(d), we obtain the groups in the last four rows of Table 3. Hence, for the remainder of the proof, we may assume that

(3.4) $$ \begin{align} N_H(P) \leqslant K. \end{align} $$

By [Reference Levi and OliverLO02, Lemma 3.3(a)], P is $\mathcal {F}$ -centric, so that P is also $\mathcal {K}$ -centric.

Suppose first that P is $\mathcal {K}$ radical, so that $P \in \mathcal {K}^{cr}$ . In this case, we appeal to Proposition 3.9 to obtain the first 13 entries in Table 3, as follows. A case-by-case check shows that for each $\mathcal {K}$ -conjugacy class $\mathcal {C} = Y^{\mathcal {K}}$ of a subgroup Y listed in Table 2, one of the following holds: either no member of $\mathcal {C}$ is $\mathcal {H}$ -radical ( $Y = C_S(U)$ ), $\mathcal {C}$ meets exactly one $\mathcal {H}$ -radical conjugacy class ( $Y = S, Q_1Q_2Q_3, Q_1Q_2Q_3\langle \tau \rangle , Q_1Q_2Q_3'\langle \tau \rangle , Q_1Q_2R_3\langle \tau \rangle , Q_1Q_2'R_3\langle \tau ' \rangle $ , $R_1R_2Q_3\langle \tau \rangle $ ), or $\mathcal {C}$ is the class of one of the entries in rows $4$ through $6$ of Table 2 ( $Y = Q_1Q_2Q_3', Q_1Q_2R_3$ , or $Q_1Q_2'R_3$ ). In this last case, $\mathcal {C}$ meets one of two $\mathcal {H}$ -classes of $\mathcal {H}$ -radical subgroups, and corresponding representatives of these $\mathcal {H}$ -classes appear in rows $3$ through $8$ of Table 3. In each of the three cases, $\operatorname {Out}_{\mathcal {H}}(P)$ is computed using (3.4) and the descriptions $K = C_H(U)X$ and $H \cap K = C_K(Z) = N_H(U) = C_H(U)\langle \tau \rangle = C_H(U)\langle \tau ' \rangle $ .

We illustrate the argument of the previous paragraph with four examples. First, consider $Y = C_S(U)$ . As $C_H(U)$ is normal in K with Sylow $C_S(U)$ , we have $C_S(U)$ is strongly $\mathcal {K}$ -closed. In particular, $C_S(U)^{\mathcal {K}} = \{C_S(U)\}$ . Appealing to Lemma 2.16, we see that $\operatorname {Out}_{\mathcal {H}}(C_S(U)) \cong C_2$ , so $C_S(U)$ is not $\mathcal {H}$ -radical.

Next, consider $Y = Q_1Q_2Q_3$ . Since X normalizes Y, we have that $Y^{\mathcal {K}} = Y^{\mathcal {H}}$ , and $N_H(Y)$ is of index $3$ in $N_K(Y)$ . Since $N_H(U) = C_H(U)\langle \tau \rangle = C_H(U)\langle \tau ' \rangle $ , it follows that $\operatorname {Out}_{\mathcal {H}}(Y) \cong S_3 \wr C_2 \times S_3$ is of index $3$ in $\operatorname {Out}_{\mathcal {K}}(Y) \cong S_3 \wr S_3$ .

Third, consider $Y = Q_1Q_2'R_3\langle \tau ' \rangle $ . This time, no element of $K-C_K(Z)$ normalizes Y, so $N_H(Y) = N_K(Y)$ by (3.4). However, we claim that $Y^{\mathcal {K}} = Y^{\mathcal {H}}$ . For the proof, assume on the contrary that there is $k \in K$ with $Y^k \leqslant S$ and $Y^k$ not H-conjugate to Y. Then $k \notin K-N_H(U)$ , and k normalizes $R_0 = L_0 \cap S$ , so $\langle k \rangle $ permutes the set $\{R_1,R_2,R_3\}$ transitively. It follows that the element $\tau ^{\prime k} \in S$ interchanges $R_3$ and some other $R_i$ by conjugation. This is a contradiction, as $R_3$ is a fixed point in the action of S on $\{R_1,R_2,R_3\}$ (see Section 2.4). Hence, $Y^{\mathcal {K}}$ meets exactly one $\mathcal {H}$ -conjugacy class as claimed, and $\operatorname {Out}_{\mathcal {H}}(Y) = \operatorname {Out}_{\mathcal {K}}(Y)$ , so Y is $\mathcal {H}$ -radical.

Finally, consider $Y = Q_1Q_2R_3$ . Then $N_X(Y) = \langle \tau \rangle $ is of order $2$ . As $L_3$ is $C_H(U)$ -invariant, $R_3$ is strongly closed in $C_S(U)$ with respect to $C_H(U)$ , so no element of $C_H(U)O_3(X)-C_H(U)$ normalizes Y. It follows that $N_H(Y) = N_K(Y)$ from (3.4). This also shows that if we fix a nontrivial element $x \in O_3(X)$ , then representatives for the $\mathcal {H}$ -conjugacy classes in $Y^{\mathcal {K}}$ may be taken as a subset of $\{Y, Y^x, Y^{x^2}\} = \{Q_1Q_2R_3, Q_1R_2Q_3, R_1Q_2Q_3\}$ . As $\tau \in S$ interchanges $R_1Q_2Q_3$ and $Q_1R_2Q_3$ , it follows that $Y^{\mathcal {K}}$ meets two $\mathcal {H}$ -conjugacy classes (at most), with representatives Y and $Q_1R_2Q_3$ . From $N_H(Y) = N_K(Y)$ , we have $\operatorname {Out}_{\mathcal {H}}(Y) = \operatorname {Out}_{\mathcal {K}}(Y) \cong S_3 \wr C_2$ , while $\operatorname {Out}_{\mathcal {H}}(Q_1R_2Q_3) \cong S_3 \times S_3$ is induced by $N_{L_1}(Q_1)N_{L_3}(Q_3)$ . So indeed $Q_1R_2Q_3$ is $\mathcal {H}$ -radical and not $\mathcal {H}$ -conjugate to Y.