Control of fixed points and existence and uniqueness of centric linking systems

Abstract

A. Chermak has recently proved that to each saturated fusion system over a finite p-group, there is a unique associated centric linking system. B. Oliver extended Chermak’s proof by showing that all the higher cohomological obstruction groups relevant to unique existence of centric linking systems vanish. Both proofs indirectly assume the classification of finite simple groups. We show how to remove this assumption, thereby giving a classification-free proof of the Martino–Priddy conjecture concerning the p-completed classifying spaces of finite groups. Our main tool is a 1971 result of the first author on control of fixed points by p-local subgroups. This result is directly applicable for odd primes, and we show how a slight variation of it allows applications for $p=2$ in the presence of offenders.

Appendices

Appendix 1: Modified norm argument

Here we give a proof of Theorem 4.1, which is the main technical result needed for the results of Sect. 4. We have postponed the proof until now so as to not interrupt the flow of that section, and because the proof is similar to that given in Theorem A1.4 of [9].

Given a group G and two nonempty subsets X and Y of G, define the product set

$$\begin{aligned} X \cdot Y = \{xy \mid x \in X, y \in Y\}. \end{aligned}$$

Definition 7.1

Let G be a finite group and V an abelian group on which G acts. Let X be a subset of G.

• (a) A subset Y of G is a transversal to X in G if for each $g \in G$, there are unique $x \in X$ and $y \in Y$ such that $g = xy$.

• (b) The norm from X to G relative to the transversal Y is the group homomorphism $\mathfrak {N}_{X;\,Y}^G:C_V(X) \rightarrow C_V(G)$ given by $v \mapsto \prod _{y \in Y}v^y$.

Given a subset X, a transversal Y to X in G, and an element $g \in G$, one sees that the map $y \mapsto y_g$ is a bijection $Y \rightarrow Y$, where $y_g \in Y$ is the unique element such that $yg = xy_g$ for some $x \in X$. Hence the image of $\mathfrak {N}_{X;\,Y}^G$ does indeed lie in $C_V(G)$.

Lemma 7.2

Let P be a finite p-group, let V an abelian group on which P acts, and let Q and R be subgroups of P. Then there exists a transversal to $Q \cdot R$ in P, and $\mathfrak {N}^P_{R} = \mathfrak {N}^P_{Q \cdot R;\, Y}\mathfrak {N}^Q_{Q \cap R}|_{\,C_V(R)}$ for any such transversal Y.

Proof

This is a combination of Lemmas A1.1 and A1.2 in [9], with the statement on norms following from Lemma A1.1(a) there and Definition 7.1(b) here. $\square$

Theorem 7.3

Suppose G is a finite group, S is a Sylow p-subgroup of G, and D is an abelian p-group on which G acts. Let $\mathcal {A}$ be a nonempty set of subgroups of S, and set $J = \langle \mathcal {A} \rangle$. Let H be a subgroup of G containing $N_G(J)$, and set $V = \Omega _1(D)$. Assume that J is weakly closed in S with respect to G, and that

$$\begin{aligned}&{\textit{whenever }} A \!\in \! \mathcal {A}, g \!\in \! G,\quad {\textit{ and }}\, A \nleq H^g,\; {\textit{then }}\, \mathfrak {N}_{A \cap H^g}^{A} = 1\; {\textit{ on }}\, V,\qquad \end{aligned}$$

(7.4)

or more generally,

$$\begin{aligned} {\textit{whenever }} g \in G\quad {\textit{ and }}\, J \nleq H^g,\; {\textit{ then }}\, \mathfrak {N}_{J \cap H^g}^{J} = 1\; {\textit{ on }}\, V. \end{aligned}$$

(7.5)

Then $C_D(H) = C_D(G)$.

Proof

We follow the argument from [9, Theorem A1.4]. Let H be a subgroup of G containing $N_G(J)$. Then $S \leqslant H$ since J is weakly closed in S with respect to G.

In the situation of (7.5), there is $A \in \mathcal {A}$ with $A \nleq J \cap H^g$, since $J = \langle \mathcal {A} \rangle$. Then $A \cap (J \cap H^g) = A \cap H^g$, and we see that (7.5) follows from (7.4) upon applying Lemma 7.2 with J, A, and $J \cap H^g$ in the roles of P, Q, and R, respectively.

Thus, we assume (7.5) and prove $C_D(H) = C_D(G)$ by induction on the order of D. We may assume $D > 1$. The p-th power homomorphism on D has kernel V and image $\mho ^1(D)$, and so $D/V \cong \mho ^1(D)$. Since $\mho ^1(D) < D$ and $\Omega _1(\mho ^1(D)) \leqslant V$, the pair $(G, \mho ^1(D))$ satisfies the hypotheses of the theorem in place of (G, D). Thus

$$\begin{aligned} C_{D/V}(G) = C_{D/V}(H), \end{aligned}$$

(7.6)

by induction.

Let $z \in C_D(H)$ and suppose first that $\langle V,z \rangle < D$. The coset Vz is fixed by H, and so it is fixed by G by (7.6). Thus, $\langle V,z \rangle$ is G-invariant. Apply induction with $\langle V,z \rangle$ in the role of D to obtain that $z \in C_D(G)$ as required.

Next assume that $\langle V,z \rangle = D$ and $V < D$. Then $C_V(H) = C_V(G)$ by induction. Set $z' = \mathfrak {N}_H^G(z)$. Then $z' \in C_D(G) \leqslant C_D(H)$. Since Vz is G-invariant, $z' \equiv z^{|G:H|}$ modulo V. Then as |G : H| is prime to p, we see that $\langle V,z' \rangle = D$ and

$$\begin{aligned} z \in C_D(H) = C_V(H)\langle z' \rangle = C_V(G)\langle z' \rangle = C_D(G) \end{aligned}$$

as required.

Finally assume that $V = D$. Given a set $[H\backslash G/J]$ of H-J double coset representatives in G containing the identity, and a transversal $[J/J\cap H^g]$ to $J \cap H^g$ in J for each $g \in [H \backslash G/J]$, then the disjoint union of $g[J/J\cap H^g]$ as g ranges over $[H\backslash G/J]$ is a transversal to H in G. Further, $\mathfrak {N}^J_{J \cap H}(z) = \mathfrak {N}_J^J(z) = z$. Thus, the norm map decomposes as

$$\begin{aligned} \mathfrak {N}_H^G(z) = \prod _{g \in [H\backslash G/J]} \mathfrak {N}^J_{J \cap H^g}(z^g) = z \prod _{g \in [H\backslash G/J]-\{1\}} \mathfrak {N}_{J \cap H^g}^J(z^g). \end{aligned}$$

(7.7)

If $g \in [H\backslash G/J] - \{1\}$ and $J \leqslant H^g$, then we may choose $h \in H$ such that $J^{g^{-1}h} \leqslant S$. Then $g^{-1}h \in N_G(J) \leqslant H$, since J is weakly closed in S with respect to G, and so $HgJ = Hh^{-1}gJ = H$ yields $g = 1$ by our choice, a contradiction. Thus, $J \nleq H^g$ for each $g \in [H\backslash G/J]-\{1\}$. We conclude that $\mathfrak {N}_{J \cap H^g}^J(z^g) = 1$ for each such g from (7.5), and then $z = \mathfrak {N}_H^G(z) \in C_V(G)$ from (7.7). $\square$

Appendix 2: Conjugacy and conjugacy functors

We give here some elementary lemmas from finite group theory that are needed at various places in the paper. We also discuss the notion of a $\Gamma$-conjugacy functor W, and describe how it gives rise to the $\Gamma$-conjugation family of subgroups well-placed with respect to W, which is also a conjugation family for the fusion system of $\Gamma$.

Lemma 8.1

(Burnside) Let G be a finite group and S a Sylow p-subgroup of G. Assume that J is an abelian subgroup of S that is weakly closed in S with respect to G and that X and Y are subgroups of J. If X and Y are conjugate in G, then they are conjugate in $N_G(J)$.

Proof

Assume X and Y are conjugate in G, and fix $g \in G$ with $X^g = Y$. Since $\langle J, J^g \rangle \leqslant C_G(Y)$, we may choose $h \in C_G(Y)$ such that $\langle J^h, J^g \rangle$ is a p-group. Choose $g_1 \in G$ with $\langle J^{hg_1}, J^{gg_1} \rangle \leqslant S$. Then $J^{hg_1} = J = J^{gg_1}$ since J is weakly closed in S with respect to G. Thus, $gh^{-1} \in N_G(J)$, and $X^{gh^{-1}} = Y^{h^{-1}} = Y$ since $h^{-1}$ centralizes Y. $\square$

The following lemma gives an alternative, more elementary, argument for Lemma 5.2(a) using the norm map. One should apply it there by taking $(N_\Gamma (Q), N_S(Q), Q, \tau )$ in the role of $(\Gamma , S, Y, \tau )$ below.

Lemma 8.2

Let $(\Gamma , S, Y)$ be a general setup for the prime p and $\tau$ an automorphism of $\Gamma$ that centralizes S. Then $\tau$ is conjugation by an element of Z(S).

Proof

Set $D = Z(Y)$ for short. Denote by $\hat{\Gamma } = \Gamma \langle \tau \rangle$ the semidirect product of $\Gamma$ by $\langle \tau \rangle$, and set $\hat{S} = S\langle \tau \rangle = S \times \langle \tau \rangle$ and $\hat{D} = D\langle \tau \rangle = D \times \langle \tau \rangle$. Then Y and D are normal in $\hat{\Gamma }$, and $\hat{D} = C_{\hat{\Gamma }}(Y)$, so that $\hat{D}$ is normal abelian in $\hat{\Gamma }$.

Let $k = |D|$. Now take any element g of $\Gamma$, and let $z = g^{-1}g^\tau = (\tau ^{-1})^g \tau = g^{-1} \tau ^{-1} g \tau$. Since $\hat{\Gamma }/\Gamma$ is abelian and $\hat{D}$ is normal in $\hat{\Gamma }$, we have $z \in \Gamma \cap \hat{D} = D \leqslant S$, so $z^\tau = z$. By induction, $g^{\tau ^i} = gz^i$ for all $i \geqslant 1$. Hence $g^{\tau ^k} = gz^k = g$. Since g is arbitrary and $\tau$ is an automorphism, $\tau ^k =1$. Thus, the order of $\tau$ is a power of p.

Consider the norm $\mathfrak {N}:= \mathfrak {N}_{\hat{S}}^{\hat{\Gamma }}:C_{\hat{D}}(\hat{S}) \rightarrow C_{\hat{D}}(\hat{\Gamma })$, and set $n = |\Gamma :S|$. Since $\hat{\Gamma }$ centralizes $\hat{D}/D$ and since n is prime to p, the restriction $\mathfrak {N}|_{\langle \tau \rangle }$ is injective. Choose an integer m such that $mn=1$ (mod k). Then $\mathfrak {N}(\tau ) \equiv \tau ^n$ (mod D) and $\mathfrak {N}(\tau )^m \equiv \tau$ (mod D). Let $\sigma = \mathfrak {N}(\tau )^m$, and choose $d \in D$ with $\sigma = \tau d$. Then $\tau \equiv d^{-1}$ modulo $Z(\hat{\Gamma })$ since $\hat{\Gamma }$ centralizes $\sigma$. Further, since $\hat{S}$ centralizes $\tau$, we see that $d \in C_{\hat{\Gamma }}(\hat{S}) \cap D = Z(S)$, as desired. $\square$

Lemma 8.3

Let $n \geqslant 2$ and let G be the symmetric group $S_{2n+1}$. Set

$$\begin{aligned} R_1&= \langle (1,2),(3,4),\ldots ,(2n-1,2n) \rangle ,\quad \text { and}\\ R_2&= \langle (1,2),(3,4),\ldots ,(2n-3,2n-2) \rangle . \end{aligned}$$

Then G is generated by $N_G(R_1)$ and $N_G(R_2)$.

Proof

Let $H = \langle N_G(R_1), N_G(R_2) \rangle$ and $\Omega = \{1,2,\ldots ,2n+1\}$. Now $N_G(R_1)$ is transitive on $\Omega -\{2n+1\}$. Similarly, $N_G(R_2)$ is transitive on $\Omega -\{2n-1,2n,2n+1\}$ and contains a subgroup inducing the symmetric group on $\{2n-1,2n,2n+1\}$. Therefore, H is transitive on $\Omega$ and the stabilizer of $2n+1$ in H is transitive on $\Omega -\{2n+1\}$. Since H contains the transposition $(2n,2n+1)$ and is 2-transitive on $\Omega$, it contains all transpositions. Hence, $H = G$. $\square$

We next give the background on conjugacy functors and well-placed subgroups, which are used in Sects. 5 and 6.

Definition 8.4

Let $\Gamma$ be a finite group with Sylow p-subgroup S. A $\Gamma$-conjugacy functor on $\mathscr {S}(S)$ is a mapping $W:\mathscr {S}(S) \rightarrow \mathscr {S}(S)$ such that for all $P \leqslant S$,

• (a) $W(P) \leqslant P$;

• (b) $W(P) \ne 1$ whenever $P \ne 1$; and

• (c) $W(P)^g = W(P^g)$ whenever $g \in \Gamma$ with $P^g \leqslant S$.

Lemma 8.5

Let $\Gamma$ be a finite group, S a Sylow p-subgroup of $\Gamma$, and W a $\Gamma$-conjugacy functor on $\mathscr {S}(S)$. Then for all $P \leqslant S$,

• (a) $N_S(P) \leqslant N_S(W(P))$;

• (b) $W(P) = W(N_S(W(P)))$ if and only if $W(P) = W(S)$; and

• (c) $P = N_S(W(P))$ if and only if $P = S$.

Proof

Let $P \leqslant S$ and $T = N_S(W(P))$. Part (a) holds by Definition 8.4(c). If $W(T) = W(P)$, then by (a), $N_S(T) \leqslant N_S(W(T)) = N_S(W(P)) = T$, so that $T = S$ and $W(P) = W(T) = W(S)$. Now (b) holds since the converse is clear.

If $P = T$, then again $N_S(P) \leqslant T = P$, so that $P = S$. Now (c) holds since the converse is clear. $\square$

A $\Gamma$-conjugacy functor W on $\mathscr {S}(S)$ can be uniquely extended to a $\Gamma$-conjugacy functor $\widehat{W}$ in the sense of [9, §5]: given a p-subgroup P of $\Gamma$, choose $g \in \Gamma$ with $P^g \leqslant S$ and define $\widehat{W}(P) = W(P^g)^{g^{-1}}$. Then $\widehat{W}$ is a mapping on all p-subgroups of $\Gamma$ that is uniquely determined, by Lemma 8.4(c).

Each $\Gamma$-conjugacy functor W gives rise to a conjugation family via its well-placed subgroups. A conjugation family for the fusion system $\mathcal {F}$ over S is a collection $\mathcal {C}$ of subgroups of S such that every morphism in $\mathcal {F}$ is a composition of restrictions of $\mathcal {F}$-automorphisms of the members of $\mathcal {C}$. A conjugation family $\mathcal {C}$ for S in $\Gamma$ in the sense of [9, §3] is itself a conjugation family for $\mathcal {F}_S(\Gamma )$ in the above sense.

For a $\Gamma$-conjugacy functor W on $\mathscr {S}(S)$ and a subgroup $P \leqslant S$, define $W_1(P) = P$ and, for all $i \geqslant 2$, define inductively $W_{i}(P) = W(N_S(W_{i-1}(P)))$. Then P is said to be well-placed (with respect to W) if $W_i(P)$ is fully $\mathcal {F}_S(\Gamma )$-normalized for all $i \geqslant 1$.

Theorem 8.6

Let $\Gamma$ be a finite group, S a Sylow p-subgroup of $\Gamma$, and W a $\Gamma$-conjugacy functor on $\mathscr {S}(S)$. Then every subgroup of S is $\Gamma$-conjugate to a well-placed subgroup of S. The set of well-placed subgroups of S forms a conjugation family for $\mathcal {F}_S(\Gamma )$.

Proof

This is a combination of Lemma 5.2 and Theorem 5.3 of [9], given the above remarks. $\square$