Proof. (a) Assume $Q\le A$ and $R\le S$ are ${\mathcal {F}}$ -conjugate, and R is fully normalized in ${\mathcal {F}}$ . By the extension axiom, each $\psi \in {\operatorname {\mathrm {Iso}}_{{\mathcal {F}}}}(Q,R)$ extends to some $\overline {\psi }\in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(C_S(Q),S)$ . Then $C_S(Q)\ge A$ since A is abelian, $\overline {\psi }(A)=A$ since A is weakly closed in ${\mathcal {F}}$ , and so $R=\overline {\psi }(Q)\le A$ .

(b) Assume $P,Q\le A$ and $\varphi \in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(P,Q)$ , and choose $R\in Q^{\mathcal {F}}$ that is fully centralized in ${\mathcal {F}}$ . Thus $R\le A$ by (a), and there is $\psi \in {\operatorname {\mathrm {Iso}}_{{\mathcal {F}}}}(Q,R)$ . By the extension axiom again, $\psi $ extends to $\widehat {\psi }\in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(A,S)$ and $\psi \varphi $ extends to $\widehat {\varphi }\in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(A,S)$ , and $\widehat {\psi }(A)=A=\widehat {\varphi }(A)$ since A is weakly closed. Then $\widehat {\psi }^{-1}\widehat {\varphi }\in \operatorname {\mathrm {Aut}}_{{\mathcal {F}}}(A)$ , and $(\widehat {\psi }^{-1}\widehat {\varphi })|_P=\psi ^{-1}(\psi \varphi )=\varphi $ .

Proof. Choose $B_1\in A_1^{\mathcal {F}}$ that is fully centralized in ${\mathcal {F}}$ , fix $\chi \in {\operatorname {\mathrm {Iso}}_{{\mathcal {F}}}}(A_1,B_1)$ , and set $B_0=\chi (A_0)$ . By the extension axiom and since $A_0$ and $B_1$ are both fully centralized in ${\mathcal {F}}$ , there are $\varphi \in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(C_S(A_1),C_S(B_1))$ and $\psi \in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(C_S(B_0),C_S(A_0))$ such that $\varphi |_{A_1}=\chi $ and $\psi |_{B_0}=(\chi |_{A_0})^{-1}$ . Since $C_S(B_1)\le C_S(B_0)$ , the composite $\psi \varphi $ lies in $\operatorname {\mathrm {Hom}}_{C_{\mathcal {F}}(A_0)}(C_S(A_1),C_S(A_0))$ .

Since $A_1$ is fully centralized in $C_{\mathcal {F}}(A_0)$ ,

$$ \begin{align*} \psi\varphi(C_S(A_1))=C_{C_S(A_0)}(\psi(B_1)) =C_S(\psi(B_1))\ge\psi(C_S(B_1)), \end{align*} $$

and hence $\varphi (C_S(A_1))\ge C_S(B_1)$ . So $A_1$ is fully centralized in ${\mathcal {F}}$ since $B_1$ is.

Proof. By Lemma 1.5 and since , A is not strongly closed. So ${\mathscr {U}}\ne \emptyset $ and ${\mathscr {T}}\ne \emptyset $ if , and we show when proving (a) that this also holds if . The last statement, and the claim that ${\mathscr {W}}\ne \emptyset $ , follow from (a) when A is not weakly closed in ${\mathcal {F}}$ , and from (b) and (c) otherwise.

(a) If A is not weakly closed in ${\mathcal {F}}$ , then there is $\varphi \in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(A,S)$ such that $\varphi (A)\ne A$ . So by Theorem 1.3 (Alperin’s fusion theorem), there are $R\le S$ and $\alpha \in {\operatorname {\mathrm {Aut}}_{{\mathcal {F}}}}(R)$ such that $A\le R$ and $\alpha (A)\ne A$ . In the special case where , we take $R=N_S(A)$ , and set $\alpha =c_x^R$ for some $x\in N_S(R)\smallsetminus R$ . So in all cases, we can arrange that and hence $\alpha (A)\le N_S(A)$ .

Set $U=\alpha (A)\in {\mathscr {U}}$ and $A_*=U\cap A$ . Then $[A,U]\le A_*$ since A and U are both normal in R. So for each $t\in U\smallsetminus A\subseteq {\mathscr {T}}$ , we have $A_*\le C_A(U)\le C_A(t)$ and $|UA/A| = |U/A_*| = |A/A_*| = |C_{A/A_*}(t)|$ , proving that $(t,U,A_*)\in {\mathscr {W}}$ .

(b) Assume A is weakly closed in ${\mathcal {F}}$ (in particular, ). Fix $t\in {\mathscr {T}}$ , and let ${\mathscr {U}}_t$ be the set of all $U\in {\mathscr {U}}$ such that $t\in U$ . Choose $V\in {\mathscr {U}}_t$ such that $|V\cap A|$ is maximal among all $|U\cap A|$ for $U\in {\mathscr {U}}_t$ . Set $A_*=V\cap A$ and $U_2^*=N_A(A_*\langle {t}\rangle )$ . Then $A_*\langle {t}\rangle \cap A\le V\cap A=A_*$ , and so

(2-1) $$ \begin{align} U_2^*/A_* = \{x\in A\mid[x,t]\in A_*\} \big/ A_* = C_{A/A_*}(t) \ne 1, \end{align} $$

where $C_{A/A_*}(t)\ne 1$ since $A/A_*$ and t both have p-power order.

Choose $W\in (A_*\langle {t}\rangle )^{\mathcal {F}}$ such that W is fully normalized in ${\mathcal {F}}$ . Then $W\le A$ by Lemma 1.8(a) and since A is weakly closed. Let $\varphi \in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}}}(N_S(A_*\langle {t}\rangle ),S)$ be such that $\varphi (A_*\langle {t}\rangle )=W$ (see Proposition 1.4).

Set $U=\varphi (U_2^*)$ and $U_1^*=\varphi ^{-1}(U\cap A)$ . Then

$$ \begin{align*} \varphi(A_*) \le \varphi(U_2^*)\cap A = U\cap A = \varphi(U_1^*), \end{align*} $$

so $A_*\le U_1^*\le U_2^*\le A$ . Also, $U_1^*\langle {t}\rangle \in {\mathscr {U}}_t$ since $\varphi (U_1^*\langle {t}\rangle )=(U\cap A)\langle {\varphi (t)}\rangle \le A$ , and hence

$$ \begin{align*} |U_1^*| \le |U_1^*\langle{t}\rangle\cap A| \le |V\cap A| = |A_*| \end{align*} $$

by the maximality assumption on V. Thus $U_1^*=A_*<U_2^*$ where the strict inclusion holds by (2-1), and $A_*=U_1^*\in (U\cap A)^{\mathcal {F}}$ .

Now $U\cap A=\varphi (U_1^*)<\varphi (U_2^*)=U$ , so $U\nleq A$ . Since $U=\varphi (U_2^*)$ , where $U_2^*\le A$ , this shows that $U\in {\mathscr {U}}$ . Also, $U\cap A=\varphi (A_*)$ , and so $UA/A\cong U/(U\cap A)\cong U_2^*/A_*=C_{A/A_*}(t)$ . Thus $(t,U,A_*)\in {\mathscr {W}}$ .

(c) Again assume A is weakly closed in ${\mathcal {F}}$ , and let Z be maximal among all subgroups of A fully centralized in ${\mathcal {F}}$ such that . Set ${\mathcal {F}}_0=C_{\mathcal {F}}(Z)$ and $S_0=C_S(Z)$ for short. Recall that ${\mathcal {F}}_0$ is saturated since Z is fully centralized in ${\mathcal {F}}$ (Theorem 1.7).

Fix $U\in {\mathscr {U}}_{{{\mathcal {F}}_0}}(A)$ , choose a morphism $\varphi \in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}_{0}}}(U,A)$ , and set $A_*=U\cap A$ . We must show that $A_*\le Z$ . Since $UZ\in {\mathscr {U}}_{{{\mathcal {F}}_0}}(A)$ , we can assume $U\ge Z$ .

Choose $B_*\in (A_*)^{{\mathcal {F}}_0}$ that is fully normalized in ${\mathcal {F}}_0$ . Then $B_*\le A$ by Lemma 1.8(a) and since A is weakly closed. By Proposition 1.4, there is $\chi \in \operatorname {\mathrm {Hom}}_{{\mathcal {F}}_0}(N_{S_0}(A_*),S_0)$ such that $\chi (A_*)=B_*$ . Then $\chi (A)=A$ since A is weakly closed, so $\chi \varphi (\chi |_U)^{-1}\in {\operatorname {\mathrm {Hom}}_{{\mathcal {F}}_{0}}}(\chi (U),A)$ where $Z\le \chi (U)\nleq A$ and $B_*=\chi (U\cap A)=\chi (U)\cap A$ , and where $B_*\le Z$ if and only if $A_*\le Z$ . Upon replacing U by $\chi (U)$ and $\varphi $ by $\chi \varphi (\chi |_U)^{-1}$ , we are now reduced to showing that $A_*\le Z$ when $A_*=U\cap A$ is fully centralized in ${\mathcal {F}}_0$ , and hence in ${\mathcal {F}}$ by Lemma 1.9.

By Lemma 1.8(b), there is an automorphism $\alpha \in {\operatorname {\mathrm {Aut}}_{{\mathcal {F}}_{0}}}(A)$ such that $\alpha |_{A_*}=\varphi |_{A_*}$ , hence such that $\alpha ^{-1}\varphi \in \operatorname {\mathrm {Hom}}_{C_{\mathcal {F}}(A_*)}(U,A)$ . Since $U\nleq A$ , this implies that , and so $A_*=Z$ by the maximality assumption on Z.

In particular, for each $(t,U,A_*)\in {\mathscr {W}}_{{{\mathcal {F}}_0}}(A)$ , since $U\cap A\le Z$ and $A_*\in (U\cap A)^{{\mathcal {F}}_0}$ , we have $U\cap A=A_*\le Z$ .

Proof. Let ${\mathcal {F}}$ be a saturated fusion system over a finite p-group S as above. Thus $A\le S$ is such that $C_S(A)=A$ and . Once we have proven (a), (b), and (c), it will then follow immediately that ${{\mathscr {R}}_{T}(A)} \ne \emptyset $ .

(a) Fix $(t,U,A_*)\in {\mathscr {W}}_{{\mathcal {F}}}(A)$ such that $A_*=U\cap A$ , and set $\tau =c_t^A\in T$ and $B=\operatorname {\mathrm {Aut}}_U(A)\le T$ . Then $A_*=U\cap A\le C_A(B)$ . Also, by definition of ${\mathscr {W}}_{{\mathcal {F}}}(A)$ , we have $A_*\le C_A(t)=C_A(\tau )$ and $|UA/A|=|C_{A/A_*}(t)|=|C_{A/A_*}(\tau )|$ .

By definition of ${\mathscr {T}}_{{\mathcal {F}}}(A)$ and ${\mathscr {U}}_{{\mathcal {F}}}(A)$ , the subgroups $\langle {\tau }\rangle $ and B are both isomorphic to subgroups of A. So to prove that $(\tau ,B,A_*)\in {\widehat {{\mathscr {R}}}_{T}(A)}$ , it remains only to show that $|UA/A|=|B|$ . But $C_S(A)=A$ by assumption, so $|B|=|\operatorname {\mathrm {Aut}}_U(A)|=|UA/A|$ .

(b) If A is not weakly closed in ${\mathcal {F}}$ , then by Proposition 2.2(a), there is $U\in A^{\mathcal {F}}\smallsetminus \{A\}$ such that $[U,A]\le U\cap A$ , and such that $(t,U,U\cap A) \in {\mathscr {W}}_{{\mathcal {F}}}(A)$ for each $t\in U\smallsetminus A$ . Thus $(c_t^A,\operatorname {\mathrm {Aut}}_U(A),U\cap A)\in {\widehat {{\mathscr {R}}}_{{\mathcal {F}}}(A)}$ for each $t\in U\smallsetminus A$ by (a).

Now set ${\mathscr {R}} = \{ (\tau ,\operatorname {\mathrm {Aut}}_U(A),U\cap A) \,|\, \tau \in B^\# \}\subseteq {\widehat {{\mathscr {R}}}_{{\mathcal {F}}}(A)}$ . Then ${\mathscr {R}}$ satisfies condition ( $*$ ) in Definition 2.3, so ${{\mathscr {R}}_{T}(A)} \supseteq {\mathscr {R}}\ne \emptyset $ .

(c) Assume A is weakly closed in ${\mathcal {F}}$ , and let $Z\le A$ be as in Proposition 2.2(c). Thus Z is fully centralized in ${\mathcal {F}}$ , , and $U\cap A\le Z$ for each $U\in {\mathscr {U}}_{C_{\mathcal {F}}(Z)}(A)$ .

Let ${\mathscr {T}}={\mathscr {T}}_{{C_{\mathcal {F}}(Z)}}(A)\ne \emptyset $ , ${\mathscr {U}}={\mathscr {U}}_{{C_{\mathcal {F}}(Z)}}(A)\ne \emptyset $ , and ${\mathscr {W}}={\mathscr {W}}_{{C_{\mathcal {F}}(Z)}}(A)\ne \emptyset $ be as in Proposition 2.2, and set

$$ \begin{align*} {\mathscr{R}} = \{ (c_t^A,\operatorname{\mathrm{Aut}}_U(A),U\cap A) | t\in{\mathscr{T}},~ U\in{\mathscr{U}},~ (t,U,A_*)\in{\mathscr{W}} \}, \end{align*} $$

where $A_*\in (U\cap A)^{C_{\mathcal {F}}(Z)}$ and hence $A_*=U\cap A$ since $U\cap A\le Z$ . By (a), ${\mathscr {R}}\subseteq {\widehat {{\mathscr {R}}}_{C_T(Z)}(A)}$ . By Proposition 2.2(b),(c), for each $t\in {\mathscr {T}}$ , there is $U\in {\mathscr {U}}$ such that $(t,U,U\cap A)\in {\mathscr {W}}$ . So ${\mathscr {R}}\ne \emptyset $ , and condition ( $*$ ) in Definition 2.3 holds for the pair ${\mathscr {R}}$ . Thus ${\mathscr {R}} \subseteq {{\mathscr {R}}_{C_T(Z)}(A)} \subseteq {{\mathscr {R}}_{T}(A)}$ .

Proof. Assume $A_1<A_2\le A$ are as above. If $(\tau ,B,A_*)\in {\widehat {{\mathscr {R}}}^{+}_{T}(A)}$ , then

$$ \begin{align*} |C_{A_2/(A_*A_1\cap A_2)}(\tau)|\le|C_{A_2/(A_*\cap A_2)}(\tau)| = |C_{A_2A_*/A_*}(\tau)|\le |C_{A/A_*}(\tau)| \le |B|, \end{align*} $$

the first inequality by Lemma A.4 and the second by inclusion. So we have ${(\tau ,B,(A_*A_1\cap A_2)/A_1)\in {\widehat {{\mathscr {R}}}^{+}_{T}(A_2/A_1)}}$ .

In particular, if ${\mathscr {R}}$ satisfies condition ( $*$ ) in Definition 2.3 for the pair $(T,A)$ , then ${\mathscr {R}}'$ satisfies ( $*$ ) for $(T,A_2/A_1)$ , where

$$ \begin{align*} {\mathscr{R}}' = \{(\tau,B,(A_*A_1\cap A_2)/A_1) \mid (\tau,B,A_*)\in{\mathscr{R}} \}.\\[-34pt] \end{align*} $$

Proof. For each $\tau \in T^\#$ , let $\varphi _{\tau }\in \operatorname {\mathrm {End}}(A)$ be the map $\varphi _{\tau }(a)=[\tau ,a]$ . For each $A_*\le C_A(\tau )$ , we have $C_A(\tau )=\operatorname {\mathrm {Ker}}(\varphi _{\tau })$ and $C_{A/A_*}(\tau )=\varphi _{\tau }^{-1}(A_*)/A_*$ , and hence

(2-5) $$ \begin{align} |C_{A/A_*}(\tau)| = \frac{|C_A(\tau)|\cdot|A_*\cap[\tau,A]|}{|A_*|} &= \frac{|C_A(\tau)|\cdot|[\tau,A]|}{|A_*[\tau,A]|} = \dfrac{|A|}{|A_*[\tau,A]|}\nonumber\\&=\dfrac{|C_A(\tau)[\tau,A]| \cdot|C_A(\tau)\cap[\tau,A]|} {|A_*[\tau,A]|}. \end{align} $$

Since $|B|\ge |C_{A/A_*}(\tau )|$ for each $(\tau ,B,A_*)\in {\widehat {{\mathscr {R}}}^{+}_{T}(A)}$ with equality if $(\tau ,B,A_*)\in {\widehat {{\mathscr {R}}}_{T}(A)}$ , points (2-2) and (2-3) follow immediately from (2-5) (and since $A_*\le C_A(\tau )$ ). Inequality (2-4) follows from (2-3), and the last statement holds since $[\tau ,A]\le C_A(\tau )$ if $p=2$ and A is elementary abelian.

Proof. Assume ${{\mathscr {R}}^{+}_{T}(A)}\ne \emptyset $ . Choose $(\tau _0,B_0,A_{*0})\in {{\mathscr {R}}^{+}_{T}(A)}$ such that $|C_A(\tau _0)\cap [\tau _0,A]|$ is the largest possible. By condition ( $*$ ) in Definition 2.3, for each $\tau \in B_0^\#$ , there is $(\tau ,B,A_*)\in {{\mathscr {R}}^{+}_{T}(A)}$ , and hence

$$ \begin{align*} |C_A(\tau)\cap[\tau,A]| \le |C_A(\tau_0)\cap[\tau_0,A]| \le |B_0|, \end{align*} $$

where the second inequality holds by (2-4).

Notation 2.8. Let A be an elementary abelian p-group, and let $\tau \in \operatorname {\mathrm {Aut}}(A)$ be an automorphism of p-power order. Set ${\mathscr {J}}_A(\tau )=\operatorname {\mathrm {rk}}(C_A(\tau )\cap [\tau ,A])$ , the number of nontrivial Jordan blocks for the action of $\tau $ on A.

Proof. Since ${\mathcal {F}}$ realizes $(\Gamma ,A)$ , we can arrange that $A\le S$ , , and ${\operatorname {\mathrm {Aut}}_{{\mathcal {F}}}} (A)=\Gamma $ . Set $T=\operatorname {\mathrm {Aut}}_S(A)\in \text {Syl}_{p}(\Gamma )$ . Then ${{\mathscr {R}}^{+}_{T}(A)}\ne \emptyset $ by Proposition 2.4. So by Corollary 2.7, there is an elementary abelian p-subgroup $B\le \Gamma $ such that $|B|\ge |C_{A}(\tau )\cap [\tau ,A]|$ for all $\tau \in B^\#$ . Thus $\operatorname {\mathrm {rk}}(B)\ge \operatorname {\mathrm {rk}}(C_{A}(\tau )\cap [A,\tau ]) = {\mathscr {J}}_{A}(\tau )$ for each $\tau \in B^\#$ .

Proof. The first equality is just a special case of Corollary 1.7.

To see the second equality, assume that $(\Gamma ,A)$ is realized by the fusion system ${\mathcal {F}}$ over $S\ge A$ . In particular, we can assume that $\operatorname {\mathrm {Aut}}_S(A)=T$ , and so $|N_S(A)/A|=|T|=p$ . Also, $|A/C_A(T)|=|[T,A]|>p$ by assumption, so A is the only abelian subgroup of index p in $N_S(A)$ . Hence, , since otherwise $A\ne {{}^{x}} A\le N_S(A)$ for $x\in N_S(N_S(A))\smallsetminus N_S(A)$ .

By Theorem 1.3 and since (recall that ${\mathcal {F}}$ realizes $(\Gamma ,A)$ ), there must be some ${\mathcal {F}}$ -essential subgroup $P\le S$ other than A, and by [Reference CravenCOS, Lemma 2.2(a)], $P\in {\mathcal {H}}\cup {\mathcal {B}}$ where the classes ${\mathcal {H}}$ and ${\mathcal {B}}$ of subgroups of S are defined in [Reference CravenCOS, Notation 2.1]. By [Reference CravenCOS, Lemma 2.6(a)] (and in terms of Notation 2.4 in [Reference CravenCOS]), we have $\mu ({\operatorname {\mathrm {Aut}}_{{\mathcal {F}}}}^{(P)}(S))=\Delta _t$ for $t=0$ or $-1$ , and from the definition of $\mu $ it then follows that $\operatorname {\mathrm {Aut}}_\Gamma (T)=\operatorname {\mathrm {Aut}}(T)$ and hence has order $p-1$ .

Proof. Let $n=12,24$ be such that $\Gamma \cong M_n$ , and let X be a $5$ -fold transitive $\Gamma $ -set of order n. In each case, $\Gamma $ has two classes of elements of order $2$ and two classes of elements of order $3$ , and they are distinguished by whether they act on X freely or with fixed points as described in Table 1. The outer automorphism of $M_{12}$ sends each of these classes to itself, and so the inclusion of $\operatorname {\mathrm {Aut}}(M_{12})$ into $M_{24}$ sends distinct classes to distinct classes. It thus suffices to prove the lemma when $\Gamma \cong M_{12}$ .

Table 1 The number of orbits in the action on X by each element of order $2$ or $3$ in $\Gamma $ . For example, a ${\mathbf {2B}}$ -element in $M_{12}$ acts with four orbits of length $2$ and four fixed points.

(a) Fix an element $g\in {\mathbf {2A}}$ . By [Reference GorensteinGL, page 41], $C_\Gamma (g)\cong C_2\times \Sigma _5$ , and the second factor must faithfully permute the six orbits under the action of g. Fix $N\le C_\Gamma (g)$ of order $5$ , and let $h\in C_\Gamma (g)\smallsetminus \langle {g}\rangle $ be such that $N\langle {h}\rangle \cong D_{10}$ . Then $N\langle {gh}\rangle \cong D_{10}$ , and we are done upon showing that h and $gh$ lie in different classes.

Set $X_0=C_X(N)$ , a subset of order $2$ whose elements are exchanged by g, and set $X_1=X\smallsetminus X_0$ . Of the two elements h and $gh$ , one fixes the two points in $X_0$ and the other exchanges them, and we can assume that h fixes them. Hence, $C_X(h)\ne \emptyset $ , so $h\in {\mathbf {2B}}$ . Also, $C_X(gh)\subseteq X_1$ , and since $gh$ freely permutes four of the five $\langle {g}\rangle $ -orbits in $X_1$ , we have $|C_X(gh)|\le 2$ . Since no involution in $M_{12}$ acts with exactly two fixed points, this shows that $gh\in {\mathbf {2A}}$ , finishing the proof of (a).

(b) Now fix an element $g\in {\mathbf {3A}}$ . Then $C_\Gamma (g)\cong C_3\times A_4$ by [Reference GorensteinGL, page 41]. Set $N=O_2(C_\Gamma (g))\cong E_4$ . The group $C_\Gamma (g)/\langle {g}\rangle \cong A_4$ acts faithfully on the set of four orbits of g, so the elements of order $2$ in N all act freely on X and hence lie in class 2A.

Fix $h\in C_\Gamma (g)$ such that $N\langle {h}\rangle \cong A_4$ . Then $N\langle {gh}\rangle $ and $N\langle {g^2h}\rangle $ are also isomorphic to $A_4$ . Also h freely permutes three of the four $\langle {g}\rangle $ -orbits in X, and the fourth orbit is fixed by exactly one of the elements h, $gh$ , or $g^2h$ . So one of these three elements lies in class ${\mathbf{3A}}$ , and the other two in class ${\mathbf{3B}}$ .

Proof. In the first part of the proof, we consider cases (a) and (b) together. Assume that the proposition is not true, and fix a triple $(\tau ,B,A_*)\in {{\mathscr {R}}^{+}_{T}(A)}$ . Thus $\tau \in T$ has order $2$ , $B\le T$ is an elementary abelian $2$ -subgroup, and $A_*\le C_A(\langle {B,\tau }\rangle )$ is such that $|B|\ge |C_{A/A_*}(\tau )|$ . By Proposition 2.6 and since $[\tau ,A]\le C_A(\tau )$ , we have

(3-1) $$ \begin{align} |B| \ge |[\tau,A]| \cdot |C_A(\tau) / A_*[\tau,A] | \ge |[\tau,A]|. \end{align} $$

Since $\Gamma \cong M_{22}$ , $M_{23}$ , or $3M_{22}$ has only one conjugacy class of involution, we have $|[\tau ,A]|=2^4$ : by Lemma B.3 in case (a), and by Lemma C.5(b) in case (b). Thus $|B|\ge 2^4$ , with equality since $\operatorname {\mathrm {rk}}_2(\Gamma )=4$ in all cases. So the inequalities in (3-1) are equalities, $C_A(\tau ) = A_*[\tau ,A]$ , and hence

(3-2) $$ \begin{align} \operatorname{\mathrm{rk}}(C_A(B))\ge \operatorname{\mathrm{rk}}(A_*)\ge\operatorname{\mathrm{rk}}(C_A(\tau)/[\tau,A]) = \operatorname{\mathrm{rk}}(A)-2\cdot\operatorname{\mathrm{rk}}([\tau,A]) = \operatorname{\mathrm{rk}}(A)-8. \end{align} $$

(a) Assume $T<\Gamma $ and A are as in Notations B.1 and B.2. Since $H_1$ and $H_2$ are the only subgroups of $T\in \text {Syl}_{2}(\Gamma _0)$ isomorphic to $E_{16}$ by Lemma B.3, B must be equal to one of them. Since $C_A(H_1)$ has rank $1$ by Lemma B.3 again, and $\operatorname {\mathrm {rk}}(C_A(B))\ge \operatorname {\mathrm {rk}}(A)-8\ge 2$ by (3-2), we have $B=H_2$ .

By condition ( $*$ ) in Definition 2.3, each element of $B^\#$ can appear as the first component in an element of ${{\mathscr {R}}^{+}_{T}(A)}$ . So we can assume that $(\tau ,B,A_*)$ was chosen such that $\tau =\mathfrak {tr}_{h_{1}} $ (and still $B=H_2$ ). Hence, by Tables 6 and 7,

$$ \begin{align*} \mathfrak{gr}_{h_{2}} +C_{56}\in C_A(\mathfrak{tr}_{h_{1}} ) = A_*[\mathfrak{tr}_{h_{1}} ,A] \le C_A(H_2)[\mathfrak{tr}_{h_{1}} ,A] = \langle{C_{12},C_{13},C_{14},C_{15},\mathfrak{gr}_{h_{1}} }\rangle, \end{align*} $$

a contradiction. We conclude that ${{\mathscr {R}}^{+}_{T}(A)}=\emptyset $ .

(b) Now assume $T<\Gamma $ and A are as in Notations C.2 and C.3. By Lemma C.5(a), $P_1$ and $P_2$ are the only subgroups of T isomorphic to $E_{16}$ . Since $\operatorname {\mathrm {rk}}(C_A(P_2))=2\cdot \dim _{\mathbb {F}_4}(C_A(P_2))=2$ by Lemma C.5(b), while $\operatorname {\mathrm {rk}}(C_A(B))\ge 4$ by (3-2), we have $B=P_1$ . By condition ( $*$ ) in Definition 2.3, we can assume that the triple $(\tau ,P_1,A_*)$ was chosen so that $\tau =\mu _{10}$ . But then

$$ \begin{align*} \langle{e_1,e_2,e_3,e_4}\rangle = C_A(\mu_{10}) = A_*[\mu_{10},A]\le C_A(P_1)[\mu_{10},A] = \langle{e_1,e_2,e_3}\rangle \end{align*} $$

by Lemma C.5(b), a contradiction.

Proof. Let $n\in \{11,12,22,23,24\}$ be such that $\Gamma _0/Z(\Gamma _0)\cong M_n$ . Fix $T\in \text {Syl}_{p}(\Gamma )=\text {Syl}_{p}(\Gamma _0)$ . We frequently refer to Tables 2 and 3 for our lower bounds on ${\mathscr {J}}_A(\tau )$ for $|\tau |=p$ , and they in turn are based on Lemmas 3.1 and A.1 and the character tables in the Atlas of Brauer characters [Reference Jansen, Lux, Parker and WilsonJLPW].

Table 2 In all cases, $A_0$ is an $\mathbb {F}_2\Gamma $ -module such that $C_{A_0}(\Gamma )=0$ and $[\Gamma ,A_0]=A_0$ , and the characters are taken with respect to $\mathbb {F}_2$ . The bounds for ${\mathscr {J}}_{A_0}(\tau )$ all follow from Lemmas 3.1(a) and A.1(a).

Table 3 In all cases, $A_0$ is an $\mathbb {F}_3\Gamma $ -module such that $C_{A_0}(\Gamma )=0$ and $[\Gamma ,A_0]=A_0$ , and the characters are taken with respect to $\mathbb {F}_3$ . Thus when $\Gamma \cong 2M_{22}$ , the character values for the simple 10-dimensional $\overline {\mathbb {F}}_3\Gamma $ -module are doubled here since it can only be realized over $\mathbb {F}_9$ . When $\Gamma \cong M_{11}$ , the bounds for ${\mathscr {J}}_{A_0}(\tau )$ apply only when $A_0$ is not the $10$ -dimensional permutation module. The bounds for ${\mathscr {J}}_{A_0}(\tau )$ all follow from Lemmas 3.1 and A.1(c), except when $\Gamma \cong M_{11}$ or $2M_{12}$ where Lemma A.1(d) is used.

Case 1. If $p>3$ , then $|T|=p$ in all cases. So by Lemma 2.10, we have $|N_\Gamma (T)/C_\Gamma (T)|=p-1$ and $|C_A(T)\cap [T,A]|=p$ . In the terminology of [Reference CravenCOS], this translates to saying that $\Gamma \in {\mathscr {G}}_p^\wedge $ and A is minimally active, and so the result follows from [Reference CravenCOS, Proposition 7.1].

Case 2. Assume $p=2$ . By Table 2, for $\tau \in \Gamma $ of order $2$ , we have ${\mathscr {J}}_{A_0}(\tau )>\operatorname {\mathrm {rk}}_2(\Gamma )$ (and hence ${{\mathscr {R}}^{+}_{T}(A_0)} =\emptyset $ ) for each nontrivial simple $\mathbb {F}_2\Gamma _0$ -module $A_0$ , except when $\Gamma _0\cong M_{22}$ , $M_{23}$ , or $M_{24}$ and $A_0$ is the Todd module or Golay module.

Thus if $Z(\Gamma _0)$ has odd order, then either $n\ge 22$ and $A_0$ is the Todd module or Golay module for $\Gamma $ , or $\Gamma _0\cong 3M_{22}$ and $A_0$ is the six-dimensional $\mathbb {F}_4\Gamma _0$ -module. In these cases, ${{\mathscr {R}}^{+}_{T}(A_0)} =\emptyset $ by Proposition 3.2, and so they are impossible by Propositions 2.4 and 2.5.

It remains to consider the cases where $Z(\Gamma )$ has even order. Assume first that $\Gamma _0\cong 2M_{12}$ . Then $\operatorname {\mathrm {rk}}_2(\Gamma )=4$ , and ${\mathscr {J}}_{A_0}(\tau )\ge 4$ for each $\mathbb {F}_2[\Gamma /Z(\Gamma )]$ -module $A_0$ with nontrivial action by Table 2. By the last statement in Lemma A.2 (applied with A in the role of V), for each elementary abelian 2-subgroup $B\le G$ of rank $4$ , since $Z(\Gamma )\le B$ , there is $\tau \in B$ of order $2$ such that ${\mathscr {J}}_{A}(\tau )\ge 5$ . So Corollary 2.9 again applies to show that $(\Gamma ,A)$ is not fusion realizable.

Now assume that $\Gamma _0/Z(\Gamma _0)\cong M_{22}$ , and let $Z\le Z(\Gamma )$ be the Sylow 2-subgroup. Thus $|Z|=2$ or $4$ , and $\operatorname {\mathrm {rk}}_2(\Gamma _0)\le 5$ . By Table 2 and since ${\mathscr {J}}_{A_0}(\tau )\le \operatorname {\mathrm {rk}}_2(\Gamma _0)$ , either $\Gamma _0/Z\cong M_{22}$ and $A_0$ is its Todd module or its dual, or $\Gamma _0/Z\cong 3M_{22}$ and $A_0$ is the six-dimensional $\mathbb {F}_4\Gamma /Z$ -module. By Lemma A.2(b) and since $\Gamma $ acts faithfully on A, there must be indecomposable extensions of $A_0$ by $\mathbb {F}_2$ and of $\mathbb {F}_2$ by $A_0$ . Thus $H^1(\Gamma /Z;A_0)\ne 0$ and $H^1(\Gamma /Z;A_0^*)\ne 0$ (where $A_0^*$ is the dual module), contradicting [Reference Meierfrankenfeld and StellmacherMS, Lemma 6.1]. We conclude that no such faithful $\mathbb {F}_2\Gamma $ -modules exist.

Case 3. Assume $p=3$ . We claim that ${\mathscr {J}}_{A_0}(\tau )>\operatorname {\mathrm {rk}}_3(\Gamma _0)$ (and hence $(\Gamma ,A)$ is not fusion realizable) in all cases except when $\Gamma _0\cong M_{11}$ or $2M_{12}$ and $A_0$ is the Todd module for $\Gamma _0$ or its dual. This follows from Table 3 except when $\Gamma _0\cong M_{11}$ , $\dim (A_0)=10$ , and $A_0\oplus \mathbb {F}_3$ is the $11$ -dimensional permutation module. But in that case, ${\mathscr {J}}_{A_0}(\tau )=3$ whenever $|\tau |=3$ since $\tau $ acts on an $11$ -set with three free orbits.

Finally, if $\Gamma _0\cong M_{11}$ or $2M_{12}$ and A is the Todd module or its dual, then A is absolutely irreducible by [Reference OliverO2, Lemmas 4.2 and 5.2], and hence $\Gamma \cong M_{11}$ , $M_{11}\times C_2$ , or $2M_{12}$ .

Proof. Assume otherwise: assume ${\mathcal {F}}$ is a saturated fusion system over $\widehat {S}$ for which . Thus some element $t\in \widehat {S}\smallsetminus \widehat {A}$ is ${\mathcal {F}}$ -conjugate to an element of $\widehat {A}$ , and upon replacing t by $t^2$ if necessary, we can arrange that $t\in \sigma ^2\widehat {A}$ . Since $|C_{\widehat {A}}(\sigma )|=2$ and $|C_{\widehat {A}}(\sigma ^2)|=4$ , each abelian subgroup of $\widehat {S}$ not contained in $\widehat {A}$ has order at most $8$ , and hence $\widehat {A}$ is weakly closed in ${\mathcal {F}}$ .

Table 4 Let $S=A\langle {s,t}\rangle $ , where $A=\langle {v_1,v_2,v_3}\rangle \cong C_{2^n}\times C_{2^n}\times C_{2^n}$ , the elements s and t act on A as described in the table, and also $t^2=1$ and $s^4\in \langle {v_1v_3}\rangle $ . Set $T=\operatorname {\mathrm {Aut}}_S(A)=\langle {c_s,c_t}\rangle \cong D_8$ .

By Proposition 2.2(b),(c) and since $\widehat {A}$ is weakly closed in ${\mathcal {F}}$ , there is $U\le \widehat {S}$ that is ${\mathcal {F}}$ -conjugate to a subgroup of $\widehat {A}$ and such that $(t,U,U\cap \widehat {A})\in {\mathscr {W}}_{{\mathcal {F}}}(\widehat {A})$ . In particular, $|U\widehat {A}/\widehat {A}|=|C_{\widehat {A}/(U\cap \widehat {A})}(t)|$ .

Since conjugation by t sends each element of $\widehat {A}$ to its inverse, $U\cap \widehat {A}\le C_{\widehat {A}}(t)=\Omega _1(\widehat {A})$ , and hence $C_{\widehat {A}/(U\cap \widehat {A})}(t)=\Omega _1(\widehat {A}/(U\cap \widehat {A}))$ has order $4$ . Thus $|U\widehat {A}/\widehat {A}|=4$ , and so there is $u\in U$ such that $u\in \sigma \widehat {A}$ .

We claim that for each $U^*\in U^{\mathcal {F}}$ , either $U^*\widehat {A}=\widehat {S}$ or $U^*\le \widehat {A}$ . Assume otherwise; then $U^*\widehat {A}=\widehat {A}\langle {\sigma ^2}\rangle $ . So $U^*\cap \widehat {A}\le C_{\widehat {A}}(\sigma ^2)=\Omega _1(\widehat {A})$ , and $U^*$ is elementary abelian since each element of $\sigma ^2 \widehat {A}$ has order $2$ . Since $U\cong U^*$ is not elementary abelian (recall that $|u|=4$ ), this is impossible.

By Theorem 1.3 (Alperin’s fusion theorem), there is a subgroup $R\le \widehat {S}$ , together with an automorphism $\alpha \in {\operatorname {\mathrm {Aut}}_{{\mathcal {F}}}}(R)$ and subgroups $A_1$ and $U_1=\alpha (A_1)$ , such that $A_1,U_1\in U^{\mathcal {F}}$ , $A_1\le \widehat {A}$ , and $U_1\nleq \widehat {A}$ . We just saw that this implies $U_1\widehat {A}=\widehat {S}$ . So $\widehat {A}\cap R$ contains a cyclic subgroup of order $4$ and is normalized by $\sigma $ . Hence, $R\ge \langle {v^{2^{n-1}},(vw)^{2^{n-2}}}\rangle $ , and so $[R,R]\ge \Omega _1(\widehat {A})$ . Since $\alpha $ sends some element of $\Omega _1(\widehat {A})$ to an element in the coset $\sigma ^2\widehat {A}\nsubseteq [R,R]$ , this is impossible.

Proof. This follows immediately from the centralizers listed in Table 5, since if $A_1<S$ were abelian of index $8$ and $A_1\ne A$ , then for $x\in A_1\smallsetminus A$ the subgroup $C_A(x)\ge A\cap A_1$ would have index at most $4$ in A.

Proof. Assume otherwise: assume ${\mathcal {F}}$ is such that . By Proposition 2.4(c) and since A is weakly closed in ${\mathcal {F}}$ by Lemma 4.2, there is a subgroup $Z\le A$ fully centralized in ${\mathcal {F}}$ such that , and such that for each $u\in {\mathscr {T}}_{C_{\mathcal {F}}(Z)}(A)$ there is $U\in {\mathscr {U}}_{C_{\mathcal {F}}(Z)}(A)$ such that $U\cap A\le Z$ and $(c_u^A,\operatorname {\mathrm {Aut}}_U(A),U\cap A)\in {{\mathscr {R}}_{T}(A)}$ . Set $\tau =c_u^A$ ; we can assume that $|\tau |=2$ . Set $B=\operatorname {\mathrm {Aut}}_U(A)$ and $A_*=U\cap A$ .

Table 5 Centralizers and commutators involving some of the abelian subgroups $H\le \langle {s,t}\rangle $ . Here, $\varepsilon =2^{n-1}$ and $\delta =2^{n-2}$ .

By Table 5, we have $|C_A(\tau )\cap [\tau ,A]|=4$ . So $|B|\ge 4$ by inequality (2-4) in Proposition 2.6, with equality since $T\cong D_8$ has no abelian subgroups of order $8$ . Hence,

(4-1) $$ \begin{align} C_A(B)[\tau,A] \ge A_*[\tau,A] = C_A(\tau)[\tau,A], \end{align} $$

where the equality follows from (2-2) in Proposition 2.6.

Since $|B|=4$ , we have $c_{s^2}\in B$ . So we can choose $u\in s^2A$ with $u\in {\mathscr {T}}_{C_{\mathcal {F}}(Z)}(A)$ (thus $C_{\mathcal {F}}(Z)$ -conjugate to an element of A), and hence $\tau =c_u^A=c_{s^2}$ . By Table 5,

$$ \begin{align*} [\tau,A]=\langle{v_1v_3^{-1},v_1^2v_2^{-2}}\rangle \quad\text{and}\quad C_A(\tau)[\tau,A]=\langle{v_1v_3,v_1^2,v_2^2}\rangle. \end{align*} $$

So by Table 5, inequality (4-1) fails when $B=\langle {s^2,t}\rangle $ or $\langle {s^2,st}\rangle $ , and holds only when $B=\langle {s}\rangle $ and $A_*=C_A(s)=\langle {v_1v_3}\rangle $ . Since $A_*\le Z\le C_A(B)$ by assumption, we have $Z=\langle {v_1v_3}\rangle $ .

Set $\widehat {{\mathcal {F}}}=C_{\mathcal {F}}(Z)/Z$ , $\widehat {A}=A/Z$ , and $\widehat {S}=C_S(Z)/Z$ (see Definition 1.10). Then by assumption, hence is not strongly closed by Lemma 1.5, and so $A/Z$ is not strongly closed in $C_{\mathcal {F}}(Z)/Z$ . Thus . Let $v,w,\sigma \in \widehat {S}$ be the classes (modulo Z) of $v_1,v_2,s\in S$ . Then are as in Lemma 4.1, so by that lemma, giving a contradiction.

Proof. (b) Since $\langle {a,x}\rangle $ is nonabelian of order $pq$ , where $p\mid (q-1)$ and $|a|=q$ , we have

$$ \begin{align*} \dim(V/C_V(a)) = \chi_V(1) - \frac1q\sum_{i=0}^{q-1}\chi_V(a^i) = \frac1q\sum_{i=1}^{q-1}(\chi_V(1)-\chi_V(a^i)). \end{align*} $$

The action of x on $\overline {\mathbb {F}}_p\otimes _{\mathbb {F}_p}(V/C_V(a))$ freely permutes the eigenspaces for a, corresponding to the primitive q th roots of unity in $\overline {\mathbb {F}}_p$ . So all Jordan blocks for this action have length p, and the same holds for Jordan blocks for the action of x on $V/C_V(a)$ . So ${\mathscr {J}}_V(x)\ge {\mathscr {J}}_{V/C_V(a)}(x)=\frac 1p\dim (V/C_V(a))$ .

(a) Since $|a|=q$ and $\operatorname {\mathrm {ord}}_q(2)=q-1$ , we have $\chi _V(a^i)=\chi _V(a)$ for all i prime to q. So this is a special case of (b).

(c) Let $b\in \langle {a,x}\rangle \cong A_4$ be such that $\langle {a,b}\rangle \cong E_4$ . Since a, b, and $ab$ are permuted cyclically by x, they all have the same character. Hence, each of the three nontrivial irreducible characters for $\langle {a,b}\rangle \cong E_4$ appears with multiplicity

$$ \begin{align*} n = \tfrac13\dim(V/C_V(\langle{a,b}\rangle)) = \tfrac13( \chi_V(1) - \tfrac14(\chi_V(1)+3\chi_V(a)) ) = \tfrac14(\chi_V(1)-\chi_V(a)). \end{align*} $$

Since x permutes those three characters cyclically, we have ${\mathscr {J}}_V(x)\ge n$ .

(d) Set $H=\langle {a,x}\rangle \cong 2A_4$ where $|a|=4$ , and set $z=a^2\in Z(H)$ . Then $V=V_+\oplus V_-$ as $\mathbb {F}_3H$ -modules, where $V_\pm $ are the eigenspaces for the action of z, and it suffices to prove the claim when $V=V_+$ or $V=V_-$ . The case $V=V_+$ was shown in (c).

Now assume $V=V_-$ , and set $m=\dim (V)=\chi _V(1)$ and $H_0=O_2(H)\cong Q_8$ . Let W be the (unique) irreducible two-dimensional $\mathbb {F}_3H_0$ -module. Then $V|_{H_0}\cong W^{m/2}$ , and $\operatorname {\mathrm {Hom}}_{\mathbb {F}_3H_0}(W,V)\cong \mathbb {F}_3^{m/2}$ since $\operatorname {\mathrm {End}}_{\mathbb {F}_3H_0}(W)\cong \mathbb {F}_3$ . So there are $\frac 12(3^{m/2}-1)$ submodules of $V|_{H_0}$ isomorphic to W, they are permuted by $\langle {x}\rangle \cong C_3$ , and hence there is at least one two-dimensional $\mathbb {F}_3H$ -submodule $W_1\le V$ . By applying the same argument to $V/W_1$ and then iterating, we get a sequence of $\mathbb {F}_3H$ -submodules ${0=W_0<W_1<\cdots <W_k=V}$ such that $\dim (W_i/W_{i-1})=2$ for each $1\le i\le k$ . Then $\dim (C_{W_i/W_{i-1}}(x))=1$ for each i, so $\dim (C_{V}(x))\le m/2$ , and $\dim ([x,V])\ge m/2$ . Each nontrivial Jordan block in V has dimension $2$ or $3$ , and intersects with $[x,V]$ with dimension $1$ or $2$ , respectively. Thus

$$ \begin{align*} {\mathscr{J}}_{V}(x) \ge \tfrac12\dim([x,V]) \ge \tfrac14m = \tfrac14\chi_V(1) = \tfrac14(\chi_V(1)-\chi_V(a)), \end{align*} $$

the last equality since $\chi _V(a)=0$ (recall that $a^2=z$ acts on V via $-\operatorname {\mathrm {Id}}$ ).

Proof. Assume (a) does not hold. Thus all but one of the composition factors in V have trivial G-action, and there are $\mathbb {F}_pG$ -submodules $V_0<V_1\le V$ such that $V_1/V_0$ is simple (hence Z acts trivially) and all composition factors of $V_0$ and of $V/V_1$ are trivial. Since $G=O^p(G)$ is generated by $p'$ -elements, it acts trivially on $V_0$ and on $V/V_1$ .

Let $W\le V_1$ be the submodule generated by the $[g,V_1]$ for all $p'$ -elements $g\in G$ . For each such g, $[g,V_1]\cap V_0\le [g,V_1]\cap C_{V_1}(g)=0$ since g acts trivially on $V_0$ , so projection onto $V_1/V_0$ sends $[g,V_1]$ injectively, and Z acts trivially on $[g,V_1]$ since it acts trivially on $V_1/V_0$ . Thus $[Z,W]=0$ , and $V_1=W+V_0$ since $V_1/V_0$ is simple and $W\nleq V_0$ . So Z acts trivially on $V_1$ .

By a similar argument, Z acts trivially on the dual $(V/V_0)^*$ , and hence acts trivially on $V/V_0$ . Since Z acts nontrivially on V, we have $V_1<V$ and $V_0\ne 0$ .

Assume $V_1$ is not indecomposable. Thus $V_1=W_0\oplus W_1$ , where $W_0$ and $W_1$ are nontrivial $\mathbb {F}_p\overline {G}$ -submodules of $V_1$ and $W_0\le V_0$ . The action of G on $V/W_1$ is trivial (an extension of $W_0$ by $V/V_1$ ), so $[G,V]\le W_1$ , and $W_0$ splits off as a direct summand of V, contradicting the assumption that V be indecomposable. Thus $V_1$ is indecomposable as an $\mathbb {F}_p\overline {G}$ -module, and a similar argument involving the dual module $V^*$ shows that $V/V_0$ is also indecomposable, finishing the proof of (b).

Now fix $g\in G\smallsetminus Z$ , and assume that $h_1,h_2\in gZ$ are distinct elements such that $\operatorname {\mathrm {rk}}([h_i,V])=\operatorname {\mathrm {rk}}([h_i,V_1/V_0])$ for $i=1,2$ . Set $z=h_1^{-1}h_2\in Z^\#$ . Since G acts faithfully on V by assumption, there is some $a_0\in V$ such that $[z,a_0]\ne 0$ . By (b), we have $a_0\notin V_1$ and $[z,a_0]\in V_0$ .

Set $h=h_1$ for short, so that $h_2=zh$ . Then $[h,V_1/V_0]=[hz,V_1/V_0]$ , so $\operatorname {\mathrm {rk}}([h,V])=\operatorname {\mathrm {rk}}([h,V_1/V_0])=\operatorname {\mathrm {rk}}([hz,V])$ , and hence $[h,V]=[h,V_1]=[hz,V]$ and $[h,V_1]\cap V_0=0$ . In particular, $[h,a_0]$ and $[hz,a_0]$ are both in $[h,V_1]$ . Also,

$$ \begin{align*} [hz,a_0] = z(h(a_0)-a_0)+(z(a_0)-a_0) = z([h,a_0])+[z,a_0], \end{align*} $$

so $0\ne [z,a_0]\in [h,V_1]\cap V_0$ , a contradiction.

The last statement now follows since if $p=2$ and $|h|=2$ , then ${\mathscr {J}}_V(h)=\operatorname {\mathrm {rk}}([h,V])$ and ${\mathscr {J}}_{V_1/V_0}(h)=\operatorname {\mathrm {rk}}([h,V_1/V_0])$ .

Proof. Set

$$ \begin{align*} \sigma_G = \sum_{gH\in G/H}gH \in C_{\widehat{V}}(G)=V_0 \quad\text{and}\quad \sigma_Z= \sum_{z\in Z}zH \in C_{\widehat{V}}(Z)=V_2. \end{align*} $$

Note that $Z\cap H=1$ since it is normal in G and contained in H.

Since no nontrivial normal subgroup of G is contained in H, the group G acts faithfully on $\widehat {V}$ and $G/Z$ acts faithfully on $V_2$ . So G acts faithfully on V if Z does.

Fix an element $1\ne z\in Z$ ; we show that $[z,V]\ne 0$ . Let $Z_0<Z$ and $x\in Z\smallsetminus Z_0$ be such that $Z=Z_0\times \langle {x}\rangle $ and $z\notin Z_0$ , and set $p^\ell =|x|$ (thus $\ell \le k$ ). Choose $\lambda \in \mathbb {Z}/p^k$ of order $p^\ell $ , let $g_1,\ldots ,g_m\in G$ be representatives for the left cosets of $HZ$ in G, and set

$$ \begin{align*} v = \sum_{i=1}^m\,\sum_{t\in Z_0}\,\sum_{s=0}^{p^\ell-1} s\lambda \cdot(tx^sg_iH) \in \widehat{V}. \end{align*} $$

Let $z_0\in Z_0$ and $0<r<p^\ell $ be such that $z=z_0x^r$ . Then

$$ \begin{align*} zv = \sum_{i=1}^m\,\sum_{t\in Z_0}\,\sum_{s=0}^{p^\ell-1} s\lambda \cdot(tz_0x^{s+r}g_iH) = v - r\lambda\cdot \sigma_G, \end{align*} $$

and $[z,v]\ne 0$ since $r\lambda \ne 0$ .

For each $g\in G$ , let $z_1,\ldots ,z_m\in Z_0$ and $r_1,\ldots ,r_m\in \mathbb {Z}$ be such that for each i, $gg_iH=z_jx^{r_j}g_jH$ for some j. Then

$$ \begin{align*} gv = \sum_{i=1}^m\,\sum_{t\in Z_0}\,\sum_{s=0}^{p^\ell-1} s\lambda \cdot(tx^sgg_iH) = \sum_{j=1}^m\,\sum_{t\in Z_0}\,\sum_{s=0}^{p^\ell-1} s\lambda \cdot(tz_jx^{s+r_j}g_jH) = v - \sum_{j=1}^m r_j\lambda\cdot g_j\sigma_Z, \end{align*} $$

and so $[g,v]\in C_{\widehat {V}}(Z)=V_2$ . Thus $v\in V$ , finishing the proof that Z acts faithfully on V.

Since $[Z,[G,V]]=1$ by definition and $[Z,G]=1$ , we have $[G,[Z,V]]=1$ by the three-subgroup lemma (see [Reference Gorenstein and LyonsGo, Theorem 2.2.3]). Hence, $[Z,V]\le V_0$ , so Z acts trivially on $V/V_0$ .

If $V_1<V_2$ , then G acts trivially on $V_2/V_1$ and on $V/V_2$ , and hence acts trivially on $V/V_1$ (recall that G is generated by $p'$ -elements). So $V/V_1=(V_2/V_1)\times (V'/V_1)$ for some $\mathbb {Z}/p^kG$ -submodule $V'<V$ containing $V_1$ with $V'/V_1\cong V/V_2$ . Also, Z acts faithfully on $V'$ since it acts faithfully on $V=V'+V_2$ and trivially on $V_2$ , so G acts faithfully on $V'$ since $G/Z$ acts faithfully on $[G,V_2]\le V_1=V'\cap V_2$ .

Proof. Set $G=\langle {\alpha }\rangle \le \operatorname {\mathrm {Aut}}(A)$ . The short exact sequence $0\to A_0\longrightarrow A\longrightarrow A/A_0~\to ~0$ induces an exact sequence in cohomology

and hence

$$ \begin{align*} |C_A(G)| \ge |C_{A/A_0}(G)|\cdot |C_{A_0}(G)| \big/ |H^1(G;A_0)|. \end{align*} $$

Since $G=\langle {\alpha }\rangle $ and $A_0$ is finite, we have $|H^1(G;A_0)|=|H^2(G;A_0)|$ where $H^2(G;A_0)$ is a quotient group of $C_{A_0}(G)$ (see [Reference WeibelW, Theorem 6.2.2]). So $|C_A(G)|\ge |C_{A/A_0}(G)|$ .