On normalizers of nilpotent subgroups

Bernd Baumann

Mathematisches Institut, Justus-Liebig-Universität, Arndtstr. 2, 35392 Gießen, Deutschland Bernd.Baumann@math.uni-giessen.de

Ulrich Meierfrankenfeld

Department of Mathematics, Michigan State University East Lansing, MI 48840, USA meier@math.msu.edu

Introduction

Let G be a group and Γ a collection of nilpotent subgroups of G satisfying:

(C)  P^{g} ∈ Γ for P ∈ Γ and g ∈ G.
(I)  P ∩ Q ∈ Γ for P, Q ∈ Γ.
(P)  N_{P} (Q) · N_{Q}(P) ∈ Γ for P, Q ∈ Γ.

(MM) The minimum and the maximum condition hold for Γ (i.e. each non empty subset of Γ contains a minimal and a maximal element with respect to inclusion of sets).

Then we call Γ a nilpotent subgroup system of G (NSS for short) and the members of Γ we call Γ-subgroups of G (here P^{g} := {x^{g} | x ∈ P }, where x^{g} := g^{−1}xg, is a conjugate of P and N_{X}(Y) is the normalizer of Y in X).

The set of all nilpotent subgroups of a group is an example of a system satisfying (C), (I) and (P). Examples of NSS’s are the set of p-subgroups of a finite group (p a prime), the set of closed unipotent subgroups of an algebraic group, and the set of maximal cyclic subgroups plus the trivial group in a free group.

To state our main theorem we introduce a good portion of the notations used in this paper. Let Σ be a set of subgroups of G.

Σ^{∗} is the set of maximal elements of Σ (with respect to inclusion). The elements of Γ^{∗} are called maximal Γ-subgroups.

Σ_{∗} is the set of minimal non-trivial elements of Σ. The elements of Γ_{∗} are called minimal Γ-subgroups.

If U is a subgroup of G set ΣU := {A ∈ Σ | A ≤ U}. R(Γ) := ⋂_{P ∈Γ^{∗}} P is called the radical of Γ. If R(Γ) = 1 the NSS Γ is called reduced. Let P ∈ Γ. Then Γ_{P} := {T ∈ N_{Γ}(P) | T P ∈ Γ} is the residue of P in Γ.

It turns out that Γ_{P} is an NSS for N_{G}(P), see Proposition 2.8(1). Set P^{∘} := R(Γ_{P}) and call P closed if P = P^{∘}. Note that by (MM) any chain of Γ-subgroups is finite. Let rank(Γ) be

the supremum of the lengths of chains

P_{0} < P_{1} < … < P_{n}

of closed Γ-subgroups. (The length of such a chain is n).

Ω(P) := ⟨Γ_{∗}P ⟩ is the subgroup of P generated by the minimal Γ-subgroups of P.

P is called decomposable if P = Ω(P). μ(P) is the length of a maximal chain in ΓP. By Proposition 5.2 this is

well defined. μ(P) is called the measure of P. If Q ∈ ΓP, then μ(P/Q) = μ(P)−μ(Q). By Proposition 5.4(1), this is the length of any maximal Γ-chain from Q to P.

Let A ∈ Γ_{P}. If [[P, A]A] = 1, we say that A acts quadratically on P. If A and P both are decomposable abelian Γ-subgroups, [P, A] ≠ 1 and

μ(P/C_{P} (A)) ≤ μ(A/C_{A}(P))

then A is called a non-trivial Γ-offender on P. Note here that by Proposition 4.7 both C_{P} (A) and C_{A}(P) are Γ-subgroups.

Let V be a normal Γ-subgroup of G with V ≤ Ω(Z(R(Γ)) and put W = V/C_{V} (⟨Γ⟩). We say that W is a natural SL_{2}-module for Γ provided that

(i) W is the set of points and {wC_{W} (S) | S ∈ Γ^{∗}} is the set of lines of an affine Moufang plane;

(ii) For each S ∈ Γ^{∗}, C_{S}(W) = R(Γ) and S induces the group of shears on W with axis C_{W} (S); and

(iii) ⟨Γ⟩ induces on W the subgroup of a point stabilizer (of the point 1) generated by all shears.

We say that N ∈ Γ is large in Γ provided that N is closed and C_{P} (N) ≤ N for all P ∈ Γ_{N}.

A theorem of Glauberman’s [5, Theorem 2] characterizes finite two dimensional special linear groups as groups acting on p-groups with certain features. The object of the present paper is to prove the following generalization of Glauberman’s Theorem:

Theorem A Let G be a group with an NSS Γ. Assume:

(a) rank(Γ) = 1.

(b) V is a normal Γ-subgroup of G with V ≤ Ω(Z(R(Γ))).

(c) S ∈ Γ^{∗} and [C_{G}(V), S] ≤ R(Γ).

(d) S contains a non-trivial Γ-offender on V.

(e) R(Γ) is large in Γ.

Then V/C_{V} (⟨Γ⟩) is a natural SL_{2}-module for Γ.

It is well known that an affine Moufang plane is isomorphic to a plane whose point set consists of the ordered pairs of an alternative field or a skew field K and whose lines are the point sets L(a, b) := {(x, x · a + b) | x ∈ K} and L(c) := {(c, y) | y ∈ K}. Then for example shears with axis L(0) are the mappings (x, y) ↦ (x, x · d + y) (see [4] page 128 ff. and the literature quoted there).

For the proof of Theorem A see section 6 and 7 and for other main results of this paper see section 8.

We would like to thank G. Glauberman for pointing out an error in an earlier version of this paper.

1 Preliminaries

In this section we collect some elementary results about nilpotent groups. We start with some well known commutator properties (see for instance [6]).

Proposition 1.1 Let a, b, c be elements, A, B, C subgroups and N a normal subgroup of a group. Then

(1) [a, bc] = [a, c][a, b][[a, b], c] = [a, c][a, b]^{c} (2) [ab, c] = [a, c][[a, c], b][b, c] = [a, c]^{b}[b, c]

(3) [a, b] = [b, a]^{−1} = [b, a^{−1}][[b, a^{−1}], a]

(4) ab[b, a] = ba

(5) [[B, C], A] ⊆ N and [[C, A], B] ⊆ N imply [[A, B], C] ⊆ N.

(6) [A, B] is a normal subgroup of ⟨A, B⟩.

Proposition 1.2 Let G be a group, V an abelian normal subgroup of G, U a subgroup of V and g ∈ G with [[V, g], g] = 1. Then the following hold:

(1) {[u, g] | u ∈ U} is a subgroup of V

(2) UU^{g} = U[U, g]

(3) UU^{g} = U × [U, g] if and only if U ∩ U^{g} = C_{U}(g)

(4) C_{UU^{g}}(g) = C_{U∩U^{g}}(g)[U, g]

Proof. These properties are applications of Proposition 1.1. □

Proposition 1.3 Let A and B be subgroups and let N be a normal subgroup of the group G. Then

[N, ⟨A, B⟩] = ⟨[N, A], [N, B]⟩.

Proof. Obviously the right hand side is contained in the left hand side. Conversely, by Proposition 1.1(6) M := ⟨[N, A], [N, B]⟩ is a normal subgroup of ⟨A, B, N⟩ contained in N, as N is a normal subgroup. Now N/M is centralized by ⟨A, B⟩, whence [N, ⟨A, B⟩] ⊆ M. □

Let A be a group acting on a group D. We say that A acts nilpotently on D if [D, A, k] = 1 for some k (where [D, A, 0] := D and [D, A, i + 1] := [[D, A, i], A]). The minimal such k is called the nilpotence length of A on D. For a group G let L_{0}(G) = G and L_{i+1}(G) = [L_{i}(G), G].

Lemma 1.4 (1) Suppose A acts nilpotently on D. Then A/C_{A}(D) is nilpotent.

(2) Suppose G acts on D, A ≤ G and B ≤ N_{G}(A). If A and B act nilpotently on D, so does AB.

(3) Let N be normal in G. Then G is nilpotent if and only if G/N is nilpotent and G acts nilpotently on N.

(4) Let G = AB, where A and B are nilpotent subgroups of G, and A is normal in G. Assume N is a normal subgroup of G with N ≤ A ∩ B such that G/N is nilpotent. Then G is nilpotent.

(5) Let A, B be normal in G such that G/A and G/B are nilpotent. Then G/A ∩ B is nilpotent.

Proof. (1) See [7, Corollary to Theorem 3.8]

(2) By induction on the nilpotency length of A on D [[D, A], AB, i] = 1 for some i. Also if [D, B, j] = 1, then [D, AB, j] ⊆ [D, A] and so [D, AB, i+j] = 1.

(3) One direction is obvious. So suppose G/N is nilpotent and G acts nilpotently on N. Then L_{k}(G) ≤ N for some k and [N, G, i] = 1 for some i. Thus L_{k+i}(G) = 1.

(4) Since N ≤ A ∩ B, both A and B act nilpotently on N. By (2) G acts nilpotently on N and so (4) follows from (3).

(5) Let k be the maximum of the nilpotency classes of G/A and G/B. Then L_{k}(G) ≤ A ∩ B. □

Proposition 1.5 Let P and Q be nilpotent subgroups of the group G with Q ⊆ P C_{G}(P). Then P Q is a nilpotent subgroup of G.

Proof. Clearly P is normal in P Q and P Q acts nilpotently on P. Also P Q/P ≅ Q/Q∩P and so P Q/P is nilpotent. Hence the lemma follows from Lemma 1.4(3). □

Proposition 1.6 Let X be a proper subgroup of the nilpotent group G.

(1) X is contained in a proper normal subgroup of G.

(2) X is a proper subgroup of N_{G}(X).

(3) If N_{G}(X) = N_{G}(N_{G}(X)), then X is normal in G.

(4) ⟨X^{G}⟩ is a proper subgroup of G.

Proof. Well-known. □

Proposition 1.7 Let H be a nilpotent group of class k and x, y ∈ H, where x is an element of order p, p a prime. Then [x, y^{p^{k+1}}] = 1.

Proof. By induction on k , [x, y^{p^{k}}] ∈ Z(H). Then by Proposition 1.1

1 = [x^{p}, y^{p^{k}}] = [x, y^{p^{k}}]^{p} = [x, y^{p^{k+1}}].

□

Proposition 1.8 Let X be a subgroup of the group G and let U and A be subsets of G with U ⊆ X. Then (UA) ∩ X = U(A ∩ X).

Proof. Let u ∈ U and a ∈ A with ua ∈ X. Then a ∈ A ∩ X, hence (UA) ∩ X ⊆ U(A ∩ X). If d ∈ A ∩ X then ud ∈ (UA) ∩ X as U(A ∩ X) ⊆ X. Thus U(A ∩ X) ⊆ (UA) ∩ X and Proposition 1.8. □

2 Basic Properties of NSS’s

In this section G is a group with an NSS Γ with 1 ∈ Γ.

We remark that (MM) allows us to prove statements about Γ by induction. Namely suppose given a statement S about Γ-subgroups. Suppose also that if P ∈ Γ and S is true for all Q ∈ Γ with Q < P, then S is also true for P. Then S must be true for all P ∈ Γ. Indeed the set of Γ-subgroups for which S is false, does not have a minimal element and so is empty.

Note also that (I) and (MM) imply, that arbitrary intersections of Γsubgroups are Γ-subgroups.

Lemma 2.1 Let P, Q ∈ Γ. Then N_{P} (Q) ∈ Γ.

Proof. Note that N_{Q}(P) ∩ P ⊆ Q ∩ P ⊆ N_{P} (Q) and so by Proposition 1.8

(N_{P} (Q)N_{Q}(P)) ∩ P = N_{P} (Q)(N_{Q}(P) ∩ P) = N_{P} (Q).

By (P) and (I) the left hand side of this equation is in Γ. □

Proposition 2.2 Let P, Q ∈ Γ such that Q is a minimal element of {T ∈ Γ | P < T } or that P is a maximal element of {T ∈ Γ | T < Q}. Then P is normal in Q.

Proof. Note that the two conditions are actually equivalent. So suppose the first. By Lemma 2.1 P < N_{Q}(P) ∈ Γ and so Q = N_{Q}(P) by minimality of Q. □

Proposition 2.3 (1) If Δ is a nonempty subset of Γ, then ⋂_{X∈Δ} X ∈ Γ and ⋂_{X∈Δ} X = ⋂_{X∈Δ_{0}} X for some finite subset Δ_{0} of Δ.

(2) If Δ is a set of normal Γ- subgroups of G, then ⟨Δ⟩ ∈ Γ.

(3) If U is a subgroup of G, then ΓU is an NSS of U.

(4) P^{∘} ∈ Γ for all P ∈ Γ. In particular, R(Γ) is a normal Γ-subgroup of G.

(5) If S ∈ Γ and P ∈ ΓS\{S}, then P ⊂ ⟨Γ_{P} S⟩.

(6) If ⟨Δ⟩ is nilpotent for Δ ⊆ Γ, then ⟨Δ⟩ ∈ Γ.

(7) R(Γ) = ⟨⋂_{T ∈Γ^{∗}} N_{Γ}(T)⟩.

(8) Let S ∈ Γ^{∗} and A ∈ Γ(SC_{G}(S)). Then A ≤ S.

(9) Let S ⊆ G be nilpotent and put A = ⟨ΓS⟩. Then A ∈ Γ, Γ^{∗}S = {A} and A is normal in N_{G}(S).

Proof.

(1) By (I) intersections of the members of finite subsets of Δ are elements of Γ. Then (1) follows from the minimal condition for Γ applied to the set of intersections of the members of finite subsets of Γ.

(2) If N and M are normal Γ-subgroups then NM ∈ Γ by (P). Hence finite products of elements of Δ lie in Γ, and (2) follows from the maximal condition for Γ.

(3) is obvious by the definition of an NSS.

(4) is a consequence of (1).

(5) By Proposition 1.6(2) P < N_{S}(P) and by Lemma 2.1 N_{S}(P) ∈ Γ.

(6) Let S = ⟨Δ⟩ and without loss Δ = ΓS. Let P ∈ Δ^{∗}. If P is not normal in S, then Proposition 1.6(c) there exists x ∈ N_{S}(N_{S}(P)) with P ≠ P^{x}. By (C) and (P) we get P P^{x} ∈ ΓS, a contradiction to the maximality of P. So P is normal in S. Thus by (2) S = ⟨Δ⟩ = ⟨Δ^{∗}⟩ ∈ Γ.

(7) Let

Λ := ⋂_{T ∈Γ^{∗}} N_{Γ}(T) = {A ∈ Γ | A ≤ N_{G}(T) ∀T ∈ Γ^{∗}}.

We claim that |Λ^{∗}| = 1. Indeed, let X_{1}, X_{2} ∈ Λ^{∗} and pick T_{i} ∈ Γ^{∗} with X_{i} ≤ T_{i}. By (6), ⟨ΛT_{i}⟩ ∈ Γ and so the definition of Λ implies ⟨ΛT_{i}⟩ ∈ Λ. The maximality of X_{i} implies X_{i} = ⟨ΛT_{i}⟩. Hence X_{1} ≤ N_{G}(T_{2}) ≤ N_{G}(⟨ΛT_{2}⟩) ≤ N_{G}(X_{2}). So X_{1} normalizes X_{2} and X_{2} normalizes X_{1}. Thus by (P), X_{1}X_{2} ∈ Γ. Hence also X_{1}X_{2} ∈ Λ and X_{1} = X_{2}.

So indeed |Λ^{∗}| = 1. Let N be the unique element in Λ^{∗}. Then N is normal in G. Let T ∈ Γ^{∗}. The definition of Λ implies that N normalizes T. So by (P), NT ∈ Γ. Thus N ≤ T and N ≤ R(Γ). Clearly R(Γ) ≤ N and (7) holds.

(8) Obviously S is contained in the right hand side of this equation. Let P ∈ Γ(SC_{G}(S)). Then SP is nilpotent by Proposition 1.5 and therefore SP ∈ Γ by (6). Hence P ⊆ S because S is maximal.

(9) By (6) we get A ∈ Γ, which implies Γ^{∗}S = {A}, and by (C) A is normal in N_{G}(S).

□

Definition. A subset Δ of Γ is called a sub-NSS of Γ and we write Δ ≤ Γ provided that:

(Suba) If A ∈ Γ and B ∈ Δ with A ⊆ B then A ∈ Δ.

(Subb) If A, B ∈ Δ with ⟨A, B⟩ ∈ Γ then ⟨A, B⟩ ∈ Δ.

(Subc) If A, B ∈ Δ then A^{B} ⊆ Δ.

Lemma 2.4 Let Δ ≤ Γ, then Δ is an NSS for ⟨Δ⟩.

Proof. (C) follows from (Subc). Let P, Q ∈ Δ. Then since (I) holds for Γ, P ∩Q ∈ Γ. So by (Suba), P ∩Q ∈ Δ. So (I) holds. By Lemma 2.1, N_{P} (Q) and N_{Q}(P) are Γ-subgroups. So by (Suba), they are also Δ-subgroups. By (P) for Γ, N_{P} (Q)N_{Q}(P) ∈ Γ and so by (Subb), N_{P} (Q)N_{Q}(P) ∈ Δ. Thus (P) holds. (MM) follows from (MM) for Γ. □

Lemma 2.5 Let Δ ≤ Γ.

(1) R(Δ) ∈ Δ, and R(Δ) is normal in ⟨Δ⟩

(2) If A ∈ Δ then A R(Δ) ∈ Δ .

(3) Δ ≤ Γ_{R(Δ)}.

(4) Let Λ ≤ Δ. Then

(i) R(Δ) ∩ S = R(Δ) ∩ R(Λ) for all S ∈ Λ^{∗}.

(ii) R(Λ) ∩ R(Δ) is the unique maximal Λ-subgroup of R(Δ).

(iii) Λ-subgroups of R(Δ) are contained in R(Λ).

(5) Let Λ ≤ Δ with R(Δ) ∈ Λ. Then R(Δ) ≤ R(Λ).

(6) Suppose that Λ ≤ Δ ≤ Γ_{R(Λ)} and R(Δ) ∈ Λ. Then R(Λ) = R(Δ).

(7) R(Δ) is closed in Γ if and only if R(Δ)^{∘} ∈ Δ.

Proof.

(1) follows from Proposition 2.3(4) applied to the NSS Δ.

(2) By (MM) there exists S ∈ Δ^{∗} with A ⊆ S. By Proposition 2.3(6) A R(Δ) ∈ Γ and so by (Subb), A R(Δ) ∈ Δ.

(3) Follows from (1) and (2).

(4) Let S, T ∈ Λ^{∗}. By (2), T R(Δ) ∈ Δ and so by Proposition 2.3(6) also T (R(Δ) ∩ S) ∈ Δ. By (I) and (Suba), R(Δ) ∩ S ∈ Γ and so (Subb) implies T (R(Δ) ∩ S) ∈ Λ. Thus by maximality of T, R(Δ) ∩ S ⊆ T. So R(Δ) ∩ S ⊆ R(Λ). So (i) holds. (ii) and (iii) follow from (i).

(5) Follows from (4).

(6) By (5) R(Δ) ≤ R(Λ). Note that R(Λ) ∈ Λ ≤ Δ. Thus R(Λ) is a Δsubgroup of R(Γ_{R(Λ)}) and so by (4)(iii) applied to Δ ≤ Γ_{R(Λ)}, R(Λ) ≤ R(Δ).

(7) If R(Δ) = R(Δ)^{∘}, then R(Δ)^{∘} ∈ Δ by (1). So suppose R(Δ)^{∘} ∈ Δ. Then by (5) applied to Δ ≤ Γ_{R(Δ)}, R(Δ)^{∘} ≤ R(Δ). So R(Δ) is closed. □

Lemma 2.6 Let P ∈ Δ ≤ Γ such that P = R(Γ_{P} ∩ Δ). Then

(1) R(Δ) ⊆ P.

(2) If Γ_{P}^{∗} ∩ Δ ≠ ∅, then R(Δ) is closed.

Proof. Let T = R(Δ).

(1) Since P ∈ Δ, Lemma 2.5(2) implies P T ∈ Δ. Hence by Lemma 2.1, N_{T} (P) ∈ Δ. Let S ∈ (Γ_{P} ∩ Δ)^{∗}. Then again by Lemma 2.5(2), ST ∈ Δ. Hence by Proposition 2.3(6), N_{T} (P)S ∈ Γ_{P} ∩ Δ. By maximality of S, N_{T} (P) ⊆ S. Thus N_{T} (P) ≤ R(Γ_{P} ∩ Δ) = P. Since T P is nilpotent we conclude T ⊆ P.

(2) By Lemma 2.5(3), Δ ≤ Γ_{T}. Thus

(∗) Γ_{P} ∩ Δ ≤ Γ_{P} ∩ Γ_{T} ≤ Γ_{P} .

Let Q = R(Γ_{P} ∩ Γ_{T}). By assumption there exists S ∈ Γ_{P}^{∗} ∩ Δ. Then S ∈ (Γ_{P} ∩ Γ_{T})^{∗} and so Q ⊆ S. Hence by (Suba), Q ∈ Γ_{P} ∩ Δ. By (^{∗}) we can apply Lemma 2.5(6) (with Λ = Γ_{P} ∩ Δ and Δ = Γ_{P} ∩ Γ_{T}) Thus Q = R(Γ_{P} ∩ Δ) = P. So by (1) (applied to Γ_{T} in place of Δ), T^{∘} ⊆ P and thus T^{∘} ∈ Δ. By Lemma 2.5(7), T = R(Δ) is closed. □

Corollary 2.7 Suppose that N ∈ Γ is closed and Γ_{N} ≤ Δ ≤ Γ. Then R(Δ) ≤ N and R(Δ) is closed.

Proof. Since N is closed and Γ_{N} = Γ_{N} ∩Δ we have N = R(Γ_{N} ∩Δ). Also Γ_{N}^{∗} ⊆ Γ_{N} ⊆ Δ and so Γ_{N}^{∗} ∩ Δ ≠ ∅. Thus the Corollary follows from Lemma 2.6. □

Definition. If Q is a normal Γ-subgroup of G contained in R(Γ) we define

Γ/Q := {P Q/Q | P ∈ Γ}.

Note that Γ/Q = {P/Q | Q ≤ P ∈ Γ}. Proposition 2.8 Let L ∈ Γ. Then the following hold:

(1) Γ_{L} resp. Γ_{L}/L is an NSS of N_{G}(L) resp. N_{G}(L)/L.

(2) L ≤ L^{∘}.

(3) Γ = Γ_{R(Γ)}.

(4) R(Γ_{L}/L) = R(Γ_{L})/L.

(5) Γ/ R(Γ) is reduced.

(6) L is closed in Γ if and only if 1 is closed in Γ_{L}/L.

(7) If L is closed then L = ⋂{S ∈ Γ^{∗} | L ⊆ S}.

(8) Γ_{L} ⊆ Γ_{L^{∘}}.

(9) If M ∈ Γ with Γ_{L} ≤ Γ_{M}, then N_{M}(L) ≤ L^{∘}. If in addition L^{∘} ≤ M, then L^{∘} = N_{M}(L).

(10) There is some (not necessarily uniquely determined) closed Γ -subgroup M with L ⊆ M, Γ_{L} ⊆ Γ_{M} and L^{∘} = N_{M}(L).

(11) L^{∘} = N_{L^{∘∘}}(L).

(12) Let S ∈ Γ^{∗} and L be a normal Γ-subgroup of S. Then L^{∘} = L^{∘∘} is closed.

Proof.

(1) Let P, Q ∈ Γ_{L}. Then (P ∩ Q)L ⊆ P L ∩ QL ∈ Γ by (I). Hence (P ∩ Q)L ∈ Γ by Proposition 2.3(6) and P ∩ Q ∈ Γ_{L}. Similarly N_{P} (Q)N_{Q}(P)L ⊆ N_{P L}(QL)N_{QL}(P L) ∈ Γ by (P) and therefore N_{P} (Q)N_{Q}(P)L ∈ Γ implying N_{P} (Q)N_{Q}(P) ∈ Γ_{L}. Condition (MM) is satisfied for Γ_{L} as Γ_{L} ⊆ Γ, and (C) follows for Γ_{L} as (C) holds for Γ and thus P^{g}L ∈ Γ if P ∈ Γ_{L} and g ∈ N_{G}(L). Thus Γ_{L} and Γ_{L}/L are NSS’s.

(2) and (3) are obvious.

(4) follows from (Γ_{L}/L)^{∗} = Γ^{∗}/L := {S/L | S ∈ Γ_{L}^{∗}}.

(5) is a consequence of (4).

(6) is clear by (5) and (2).

(7) Put D := ⋂{S ∈ Γ^{∗} | L ⊆ S}. Let T ∈ Γ_{L}^{∗} and pick S ∈ Γ^{∗} with T ⊆ S. Then T ⊆ N_{S}(L) ∈ Γ_{L} by Lemma 2.1 and so T = N_{S}(L). Since D ⊆ S we conclude N_{D}(L) ⊆ T. As this is true for all T ∈ Γ_{L}^{∗}, N_{D}(L) ⊆ R(Γ_{L}) = L. Since L ⊆ D and D is nilpotent, L = D.

(8) If P ∈ Γ_{L} then there is Q ∈ Γ_{L}^{∗} with P ⊆ Q, hence P L^{∘} ∈ ΓQ ⊆ Γ by Proposition 2.3(6), and P ∈ Γ_{L^{∘}}.

(9) Note that N_{M}(L) ∈ Γ_{L} and N_{M}(L) ≤ M ≤ R(Γ_{M}). Thus by Lemma 2.5(4), N_{M}(L) ≤ R(Γ_{L}) = L^{∘}. If L^{∘} ≤ M, then L^{∘} ≤ N_{M}(L) ≤ L^{∘} and so L^{∘} = N_{M}(L).

(10) Let M in Γ be maximal with respect to L^{∘} ≤ M and Γ_{L} ⊆ Γ_{M}. Note that by (2) and (8) such an M exists. By (2) and (8) applied to M, L^{∘} ≤ M ≤ M^{∘} and Γ_{L} ⊆ Γ_{M} ⊆ Γ_{M^{∘}}. Thus the maximal choice of M implies M = M^{∘}. So M is closed. By (9), N_{M}(L) = L^{∘} and all parts of (10) are verified.

(11) Follows from (2),(8) and (9).

(12) As S ∈ Γ^{∗} we get S ∈ Γ_{L}^{∗}. It follows that L^{∘} is normal in S and thus L^{∘∘} ≤ S. Hence L is normal in L^{∘∘}. So by (12) L^{∘} = N_{L^{∘∘}}(L) = L^{∘∘}.

□

Lemma 2.9 Let N ∈ Γ and P, Q ∈ Γ_{N}. If [C_{P} (N), ⟨P, Q⟩] ⊆ N then N_{Q}(P)P ∈ Γ.

Proof. By Lemma 2.1 we may assume that Q = N_{Q}(P). So Q normalizes P. Since P N and QN are in Γ they are both nilpotent. So P and Q act nilpotently on N. By Lemma 1.4(2) P Q acts nilpotently on N. Thus by Lemma 1.4(1), P Q/C_{P Q}(N) is nilpotent. Also P Q/P ≅ Q/Q∩P is nilpotent and so by Lemma 1.4(5) P Q/C_{P} (N) is nilpotent. Since [C_{P} (N), P Q] ⊆ N we get that P Q acts nilpotently on C_{P} (N). Thus the assertion follows from Lemma 1.4(3). □

Proposition 2.10 Let G = ⟨A, B⟩, where A and B are nilpotent subgroups of G. Assume A ∈ Γ, N is a normal subgroup of G, N ⊆ A ∩ B and G/N is nilpotent. Then G is nilpotent.

Proof. By (MM) A can be chosen maximal fulfilling the assumptions of the Proposition. Then by nilpotency of G/N and (P), A is normal in G and Proposition 2.10 follows from Lemma 1.4(4). □

Proposition 2.11 Let S ∈ Γ^{∗} be fixed.

(1) Let T ∈ Γ^{∗}\{S} such that S ∩ T is maximal. Then S ∩ T is closed.

(2) Let T ∈ Γ^{∗}\{S}, then there exists a closed Γ-subgroup P with S ∩ T ≤ P < S.

Proof. (1) Set P := S ∩ T Then P^{∘}N_{S}(P) ∈ Γ by definition of Γ_{P} and 1.10(6). Therefore there is X ∈ Γ^{∗} with P^{∘}N_{S}(P) ⊆ X. By maximality of S, S ≰ T and so P < S. Hence by Proposition 1.6(3), P < N_{S}(P) ≤ X ∩ S. By maximality of P, X = S. Thus P^{∘} ≤ S. Note also that N_{T} (P)P^{∘} ≤ Y for some Y ∈ Γ^{∗}. Since P < N_{T} (P), N_{T} (P) ≰ S and so Y ≠ S. Thus by maximality of P, Y ∩ S = P. Since P^{∘} ≤ Y ∩ S we get P^{∘} = P and P is closed.

(2) Let T^{∗} ∈ Γ^{∗}\{S} with S ∩ T ≤ S ∩ T^{∗} and S ∩ T^{∗} maximal. Then S ∩ T^{∗} is closed by (1). □

The following statement is a variant of Baer’s famous theorem [1].

Theorem 2.12 Let X ∈ Γ such that ⟨X, X^{g}⟩ ∈ Γ for all g ∈ G, then ⟨X^{G}⟩ ∈ Γ.

Proof. Set Δ := X^{G} and assume ⟨Δ⟩ ∉ Γ. Then there are Q = ⟨ΔQ⟩ ∈ Γ and R = ⟨ΔR⟩ ∈ Γ with ⟨Q, R⟩ ∉ Γ. Choose Q and R such that D := ⟨Δ(Q ∩ R)⟩ is maximal. Suppose that ΔN_{Q}(D) = Δ(Q ∩ R). Then

N_{Q}(N_{Q}(D)) ≤ N_{Q}(⟨ΔN_{Q}(D)⟩) = N_{Q}(D)

and so by Proposition 1.6(3), N_{Q}(D) = Q. But ΔQ ≠ Δ(Q ∩ R), a contradiction. Thus there exists A ∈ Δ with A ≰ N_{Q}(D) and A ≤ D. Similarly there exists B ∈ Δ with B ≤ N_{R}(D) and B ≰ D. By assumption ⟨A, B⟩ ∈ Γ. By Proposition 2.10, applied with AD,BD and D in place of A, B and N, P := ⟨A, B, D⟩ is nilpotent. Since D < AD ≤ Q ∩ P, the maximality of D implies ⟨Q, P ⟩ ∈ Γ. Similarly ⟨R, P ⟩ ≤ Γ. But ⟨R, P, Q⟩ ∉ Γ and D < P ≤ ⟨R, P ⟩ ∩ ⟨Q, P ⟩. This contradiction to the maximality of D completes the proof of Theorem 2.12. □

3 NSS’ of rank 1 and 2

As in the previous section let G be a group with an NSS Γ.

Theorem 3.1 Suppose |Γ^{∗}| > 1. Then following properties are equivalent:

(a) rank(Γ) = 1.

(b) S ∩ T = R(Γ) for S, T ∈ Γ^{∗} with S ≠ T.

(c) S ∩ S^{g} = R(Γ) for S ∈ Γ^{∗} and g ∈ G\N_{G}(S).

Proof. Suppose (a) holds. Let S, T ∈ Γ^{∗} with P = S ∩ T maximal. Then P is closed by Proposition 2.11. and so R(Γ) ≤ P < S is a chain of closed Γ-subgroups. Since Γ has rank 1, we get P = R(Γ). Thus S ∩ T = R(Γ) for all S ≠ T ∈ Γ^{∗} and so (b) holds.

From (C) we get S^{g} ∈ Γ^{∗} for S ∈ Γ^{∗} and g ∈ G. Thus (b) implies (c). Suppose that (c) holds. Let P be a closed Γ subgroup. We will show that

P = R(Γ) or P ∈ Γ^{∗} and note that this implies (a).

Assume that |Γ_{P}^{∗} | = 1. Since P is closed we get P ∈ Γ_{P}^{∗}. Let P ≤ S ∈ Γ^{∗}. Then P ≤ N_{S}(P) ∈ Γ_{P} and so P = N_{S}(P) and P = S.

Suppose next that |Γ_{P}^{∗} | > 1 and let Q ≠ T ∈ Γ_{P}^{∗}. By (P) applied to the NSS Γ_{P}, we may assume that T does not normalize Q. Let Q ≤ S ∈ Γ^{∗}. Then Q ≤ N_{S}(P) ∈ Γ_{P} and so by maximality of Q, Q = N_{S}(P). Thus N_{G}(P) ∩ N_{G}(S) ≤ N_{G}(Q). Since T normalizes P but not Q we get T ≰ N_{G}(S). Pick g ∈ T with S ≠ S^{g}. Then P ≤ S ∩ S^{g} = R(Γ) and so P ≤ R(Γ). By Corollary 2.7 R(Γ) ≤ P and so P = R(Γ). □

Lemma 3.2 Suppose that N is large in Γ and P, Q ∈ Γ_{N}

(1) N_{Q}(P)P ∈ Γ_{N}.

(2) If P ∈ Γ_{N}^{∗}, then N_{Q}(P) ≤ P and Γ_{N} ∩ ΓN_{G}(P) = ΓP.

Proof. (1) By definition of large, C_{P} (N) ≤ N. Hence [C_{P} (N), ⟨P, Q⟩] ≤ N and (1) follows from Lemma 2.9.

(2) By (1) and maximality of P^{∗}, N_{Q}(P) ≤ P. The second statement in (2) just rephrases the first. □

Lemma 3.3 Suppose that N ≤ P ∈ Γ, N is large and P is closed. Then P is large.

Proof. Let P ≤ T ∈ Γ_{P}. Then C_{T} (P) ≤ N_{T} (N) ∈ Γ_{N} and since N is large, C_{T} (P) ≤ N_{T} (N) ∩ C_{G}(N) ≤ N ≤ P. Thus P is large. □

Lemma 3.4 Let Γ be an NSS of rank 1 and P ∈ Γ with P ≰ R(Γ).

(1) P is contained in a unique maximal Γ-subgroup P^{∗}.

(2) Suppose R(Γ) is large and x ∈ Q ∈ Γ. If ⟨P, P^{x}⟩ ∈ Γ then x ∈ P^{∗}.

Proof. (1) By Theorem 3.1, S ∩ T = R(Γ) for all S ≠ T ∈ R(Γ).

(2) By (1) P^{∗} = ⟨P, P^{x}⟩^{∗} = P^{x∗} = P^{∗x}. Thus x ∈ N_{Q}(P^{∗}). By Lemma 3.2(2) N_{Q}(P^{∗}) ≤ P^{∗} and (2) holds. □

Lemma 3.5 Let Γ be an NSS of rank 1 and S ∈ Γ^{∗}. Define Π = ⋃_{g∈G} ΓS^{g}. Then Π ≤ Γ, Π has rank at most one and Π^{∗} = S^{G} ⊆ Γ^{∗}. If in addition R(Γ) is large then Π has rank 1 and R(Π) = R(Γ).

Proof. Clearly Π fulfils (Suba) and (Subc). Now let A, B ∈ Π with ⟨A, B⟩ ∈ Γ. If A ≤ R(Γ), then ⟨A, B⟩ ≤ B R(Γ) ∈ Π and so also ⟨A, B⟩ ∈ Π. So suppose A ≰ R(Γ) and B ≰ R(Γ). Then by Lemma 3.4(1)

A^{∗} = ⟨A, B⟩^{∗} = B^{∗}.

Thus ⟨A, B⟩ ≤ A^{∗} and ⟨A, B⟩ ∈ Π. Thus Π ≤ Γ. Clearly Π^{∗} = S^{G} ⊆ Γ^{∗}.

Suppose first that |Π^{∗}| > 1. By Theorem 3.1, A ∩ B = R(Γ) for all A, B ∈ Π^{∗} and R(Π) = R(Γ). Hence by Theorem 3.1 , Π has rank 1. So the lemma holds in this case.

Suppose next that |Π^{∗}| = 1. Then Π^{∗} = {S}, S is normal in G and Π has rank 0. So we may now assume that that R(Γ) is large. Since S is normal in G, Lemma 3.2 implies P S ∈ Γ for all P ∈ Γ. But then P ≤ S by maximality of S and Γ^{∗} = S, a contradiction to rank(Γ) = 1. □

Lemma 3.6 Suppose that Γ has rank 1. Let K ≤ G with ⟨ΓK⟩ ∉ Γ and P ∈ ΓK with P ≰ R(Γ). Then ⟨P, P^{x}⟩ ∉ Γ for some x ∈ K.

Proof. Since ⟨ΓK⟩ ≰ Γ, Proposition 2.3(6) implies Q ≰ P^{∗} for some Q ∈ ΓK. Let x ∈ Q\P^{∗}. Then by Lemma 3.4(2), ⟨P, P^{x}⟩ ∉ Γ. □

Proposition 3.7 Let N ∈ Γ be closed of co-rank 1, (here the co-rank of N is the supremum of the lengths of chains of closed Γ-subgroup starting with N).

(1) Let N ≤ S_{1} ∩ S_{2} with S_{1} ≠ S_{2} ∈ Γ^{∗}. Then N = S_{1} ∩ S_{2}.

(2) Γ_{N} has rank 1.

(3) Let N < P ∈ Γ. Then P lies in a unique maximal Γ-subgroup P^{∗}. Moreover, N_{G}(P) ≤ N_{G}(P^{∗}),

(4) Let P, S ∈ Γ with S ∈ Γ^{∗} and N < S ∩ P. Then P ⊆ S.

Proof. (1) By Proposition 2.11(2) N ≤ S_{1} ∩ S_{2} ≤ T < S for some closed T ∈ Γ and some S ∈ Γ^{∗}. Since N has co-rank 1 we conclude that N = T and so N = S_{1} ∩ S_{2}.

(2) Let Q_{1} ≠ Q_{2} ∈ Γ^{∗}P and Q_{i} ≤ S_{i} ∈ Γ^{∗}. Since ⟨Q_{1}, Q_{2}⟩ ∉ Γ, S_{1} ≠ S_{2}. So by (1) and Theorem 3.1, Γ_{N} has rank at most 1. Suppose that Γ_{N} has rank 0. Then since N is closed {N} = Γ_{N}. Let N ≤ S ∈ Γ^{∗}. Then N ≤ N_{S}(N) ∈ Γ_{N} and so N = N_{S}(N). Hence S = N, a contradiction to N ∉ Γ^{∗}.

(3) and (4) are easy consequences of (1) and (2). □

Theorem 3.8 If Γ is reduced of rank 2 then one of the following holds:

1. There are S ∈ Γ^{∗} and closed P, Q ∈ ΓS\{S, 1} such that Γ⟨Γ_{P} , Γ_{Q}⟩ is reduced.

2. There is an reduced NSS Δ of G with rank(Δ) = 1 and Δ ≤ Γ.

Proof. Suppose first that there are S ∈ Γ^{∗} and closed P, Q ∈ ΓS\{1, S} with P ≠ Q. Let N := R(Γ⟨Γ_{P} , Γ_{Q}⟩). By Corollary 2.7, N ⊆ P ∩ Q and N is closed. Since rank(Γ) = 2 we get N = 1 and 1. holds.

Suppose next that for all S ∈ Γ^{∗} there is at most one closed P ∈ ΓS with 1 ≠ P ≠ S. If such a P exists we denote it by P (S). Otherwise let P (S) = 1. We will show that

(^{∗}) P (S) = P (T) ≠ 1 for all S, T ∈ Γ^{∗} with S ∩ T ≠ 1.

If S ∩ T is closed, P (S) = S ∩ T = P (T). So we may assume that S ∩ T is not closed. Then by Proposition 2.8(10) there exists a closed M ∈ Γ with S ∩ T ⊂ M and Γ_{S∩T} ⊆ Γ_{M}. By Lemma 2.5(2) N_{S}(S ∩ T)M ∈ Γ. So there exists S̃ ∈ Γ^{∗} with N_{S}(S ∩ T)M ⊆ S̃ and similarly choose T̃. Then S∩T ⊂ N_{S}(S∩T) ⊆ S∩S̃, S∩T ⊂ M ⊆ S̃∩T̃ and S∩T ⊂ N_{T} (S∩T) ⊆ T ∩T̃. So by downwards induction on S ∩ T, P (S) = P (S̃) = P (T̃) = P (T) ≠ 1. Thus (^{∗}) holds.

Put Δ = ⋃{ΓP (S) | S ∈ Γ^{∗}}. We claim that Δ ≤ Γ. (Suba) and Sub(c) are obvious from the definition of Δ. Let A, B ∈ Δ and S, T ∈ Γ^{∗} with A ⊆ P (S) and B ⊆ P (T).

To show (Subc) we assume A ≠ 1 ≠ B and ⟨A, B⟩ ≤ Γ. Pick Q ∈ Γ^{∗} with ⟨A, B⟩ ≤ Q. Then A ≤ S ∩ Q and B ≤ Q ∩ T and (^{∗}) implies P (S) = P (Q) = P (T). Thus ⟨A, B⟩ ≤ P (Q) and ⟨A, B⟩ ∈ Δ. Thus (Subb) holds. Thus Δ ≤ Γ and by Lemma 2.4, Δ is an NSS.

Suppose that |Δ^{∗}| > 1. Let A, B ∈ Γ^{∗} with A ∩ B ≠ 1 and let S, T be as above. Then by (^{∗}), A = P (S) = P (T) = B and by Theorem 3.1, Δ is reduced of rank 1. Thus 2. holds in this case.

Suppose that |Δ^{∗}| = 1 and let A be the unique member of Δ^{∗}. Assume that A = 1. Then P (S) = 1 for all S ∈ Γ^{∗} and so Γ has rank 1, a contradiction. Thus A ≠ 1. Let Λ = Γ \ Γ_{A} ∪ {1}. We claim that Λ ≤ Γ. Let P ≤ Q ≤ S with 1 ≠ P ∈ Γ, Q ∈ Λ and S ∈ Γ^{∗}. Since Γ_{A} ≤ Γ and Q ∉ Γ_{A}, S ∉ Γ_{A}. Thus S ∈ Λ. Suppose that P ∈ Γ_{A}. Then P A ∈ Γ. Put P A ∩ S ≠ 1 and (^{∗}) implies A ≤ S. Thus S ∈ Γ_{A}, a contradiction. So P ∈ Λ and we conclude that (Suba) holds for Λ. Clearly (Subb) and (Subc) hold.

We proved Λ ≤ Γ. Since A ≰ R(Γ), Λ ≠ {1}. Suppose that Λ has a unique maximal element B. Then B ∈ Γ^{∗} and by (^{∗}), B ∩ A = 1. Since both A and B are normal in G, [A, B] = 1. Thus AB is nilpotent and AB ∈ Γ, a contradiction to B ∉ Γ_{A}. Thus |Λ^{∗}| > 1. By (^{∗}) X ∩ Y = 1 for any two maximal members of Λ and so Theorem 3.1 implies that Λ is a reduced NSS of rank 1. Thus 2. holds for Λ in place of Δ. □

4 Minimal Γ-subgroups

In this section we continue to assume that a G is group with an NSS Γ and 1 ∈ Γ. We consider elements X ∈ Γ_{∗}. Recall that this just means that X is a minimal non-trivial element of Γ. In particular for two different elements X, Y ∈ Γ_{∗} we have X ∩ Y = 1.

Proposition 4.1 Assume P ∈ Γ and X, Y ∈ ΓP_{∗} with X ≠ Y. If N_{X}(Y) ≠ 1 or [x, y] = 1 for some x ∈ X^{#} and y ∈ Y^{#}, then ⟨X, Y ⟩ = X × Y.

Proof. If [x, y] = 1, then y ∈ Y ∩ Y^{x} and so Y = Y^{x}. So we may assume N_{X}(Y) ≠ 1. Using Lemma 2.1 we get X = N_{X}(Y). Since XY ⊆ P, XY is nilpotent. As Y is normal in XY, C_{Y} (X) ≠ 1. Hence N_{Y} (X) ≠ 1 and Y = N_{Y} (X). So [X, Y] ≤ X ∩ Y = 1. □

Proposition 4.2 Let P ∈ Γ with P = ⟨X, Y ⟩ where X, Y ∈ ΓP_{∗}\{P }. If X′ ≠ 1 then X is a normal subgroup of P (here X′ := [X, X] is the commutator subgroup of X).

Proof. Consider a counterexample with P minimal. Then there is y ∈ Y with X^{y} ≠ X. Set E := ⟨X, X^{y}⟩. So by Proposition 2.3(6) and Proposition 1.6, E ∈ Γ and E < P. Of course X, X^{y} ≠ E and by minimality of P, X and X^{y} are normal in E. Therefore by Proposition 4.1 E = X × X^{y}. Let Q := ⟨Y^{P} ⟩. Then Q is a proper Γ-subgroup of P by Proposition 2.3(6) and Proposition 1.6. Since P = ⟨X, Y ⟩, X ≰ Q and Q ∩ X = 1 as X ∈ Γ_{∗} and Q ∩ X ∈ Γ by (I). Now [X, y] ≤ E ∩ Q and E ∩ Q is normal in E. Since X × X^{y} = X[X, y] we have 1 ≠ [X^{y}, X^{y}] = [X^{y}, [X, y]] ≤ X^{y} ∩ Q. Hence also X ∩ Q ≠ 1, a contradiction. □

Corollary 4.3 Let P ∈ Γ and Δ := {X ∈ ΓP_{∗} | X′ ≠ 1}. Then Δ is finite and ⟨Δ⟩ = X_{1} × … × X_{n}, where Δ = {X_{1}, …, X_{n}}.

Proof. Let X ≠ Y ∈ Δ. Then by Proposition 4.2 and Proposition 4.1, [X, Y] = 1. Let Z = ⟨Δ \ {X}⟩. Then Z ∈ Γ and [X, Z] = 1. Thus X ∩ Z is a proper Γ-subgroup of X and so X ∩ Z. Thus the Corollary holds by the definition of the direct product. (Note also that Δ is finite by (MM).) □

Define an elementary abelian p-group to be an an abelian group so that all non-trivial elements have order p. Note that this makes sense for p a prime or p = ∞. Indeed, an elementary abelian ∞-group is just a torsion free abelian group.

Proposition 4.4 Let X, Y ∈ Γ_{∗}, X ≠ Y, H := ⟨X, Y ⟩ ∈ Γ and [X, Y] ≠ 1. Then X and Y are both elementary abelian p-groups, p = ∞ or a prime.

Proof. Suppose first Y is not elementary abelian. Let M ∈ Γ maximal with respect to X ≤ M < H. Then by Proposition 2.2, M is normal in H. Also Y ≰ M. Since Y ∩ M ∈ Γ and Y ∈ Γ_{∗}, Y ∩ M = 1. Let 1 ≠ x ∈ X. By Proposition 4.1, N_{X}(Y) = 1 and so Y ≠ Y^{x}. Hence by Proposition 4.2, Y is abelian. Since ⟨Y, Y^{x}⟩ ≠ H we get by induction that [Y, Y^{x}] = 1. Let D = Y Y^{x} ∩ M. Since [Y, x] ⊆ D, Y Y^{x} = Y D = Y^{x}D. Let E ∈ Γ_{∗}D. Then 1 ≠ Y ∩ (EY^{x}) ∈ Γ and so Y ⊆ EY^{x}. Thus E = D. Note that D is isomorphic to Y and ⟨D, X⟩ ≤ M. In particular, Y is not elementary abelian and so by induction [D, X] = 1. Since [Y, x] ≤ D we get [Y, x] ≤ Z(⟨Y, x⟩). Let y ∈ Y has order p, p a prime. Then by Proposition 1.1, [y, x^{p}] = [y^{p}, x] = 1 and so by Proposition 4.1 x^{p} = 1. Hence for all z ∈ Y, [z^{p}, x] = [z, x^{p}] = 1 and so by Proposition 4.1 z^{p} = 1.

Hence Y is an elementary abelian p-group and by symmetry X is an elementary abelian q-group. To show p = q we may assume p ≠ ∞. Then by Proposition 1.7 [y, x^{p^{k}}] = 1 for some positive integer k. So by Proposition 4.1, x^{p^{k}} = 1 and q = p. □

Proposition 4.5 Let A_{1} be a Γ-subgroup of the decomposable abelian Γsubgroup A. Then there is a decomposable Γ-subgroup A_{2} of A with A = A_{1} × A_{2}.

Proof. Let K be a decomposable Γ-subgroup maximal with A_{1}K = A_{1} × K. If A = A_{1}K we are done. So suppose A_{1}K < A. Since A is decomposable, there exists X ∈ Γ_{∗}A with X ≰ A_{1}K. Then A_{1}K ∩ X = 1 and A_{1}KX = (A_{1} × K) × X = A_{1} × (K × X). But K < KX and we obtain a contradiction to the maximal choice of K. □

Proposition 4.6 Γ-subgroups of decomposable abelian Γ-subgroups are decomposable.

Proof. Let A be a decomposable abelian Γ-subgroup and B a Γ-subgroup of A. By Proposition 4.5 there exists D ∈ ΓA with A = Ω(B) × D. By Proposition 1.8 B = Ω(B) × (B ∩ D). Also Ω(B ∩ D) ≤ Ω(B) ∩ D = 1 and since B ∩ D ∈ Γ, B ∩ D = 1 and B = Ω(B). □

Proposition 4.7 Let A, B ∈ Γ such that A is decomposable abelian and B is generated by abelian Γ-subgroups. If ⟨A, B⟩ ∈ Γ, then C_{A}(B) is a decomposable abelian Γ-subgroup.

Proof. Since C_{A}(B) = ⋂{C_{A}(E) | E ∈ ΓB, E abelian } we may by (I) assume that B is abelian. By Proposition 4.6 we only need to show C_{A}(B) ∈ Γ. By Proposition 2.3(1) we get C_{A}(B) ≤ D := ⋂_{b∈B} A^{b} ∈ Γ. Note that C_{A}(B) = C_{D}(B) and that B normalizes D. By Proposition 4.6 D is decomposable.

If D = C_{D}(B) = C_{A}(B) we are done. So suppose [D, B] ≠ 1. Since DB is nilpotent, there exists d ∈ D with 1 ≠ [d, B] ≤ C_{D}(B). Then B^{d} ≤ C_{D}(B)B ≤ C_{G}(B). Thus BB^{d} is abelian and BB^{d} ∈ Γ. Thus

1 ≠ [d, B] ≤ BB^{d} ∩ D ≤ C_{D}(B).

Put E := BB^{d} ∩ D. Then E is a non-trivial Γ subgroup of C_{D}(B). By Proposition 4.5, D = E × F for some decomposable Γ subgroup F of D. Then C_{D}(B) = E × C_{F} (B). Since F < A, induction on A shows C_{F} (B) ∈ Γ. Hence also C_{D}(B) ∈ Γ and the Proposition is proved. □

Proposition 4.8 Let A, B ∈ Γ such that A is decomposable, B an abelian Γ-subgroup and A ∈ Γ_{B}. Then [B, A] is a Γ-subgroup of G.

Proof. Since [B, A] = ⟨[B, E] | E ∈ Γ_{∗}A⟩ we may by Proposition 2.3(6) assume that A ∈ Γ_{∗}. If A ≤ B, then since B is abelian [A, B] = 1 ∈ Γ. We therefore may assume A ⊈ B and so A ∩ B = 1 by minimality of A. Note that ⟨A^{B}⟩ = A[B, A] and so

⟨A^{B}⟩ ∩ B = [B, A](A ∩ B) = [B, A].

By Proposition 2.3(6) ⟨A^{B}⟩ ∈ Γ and so by (I), [B, A] ∈ Γ. □

5 Measure and the Thompson subgroup

G continues to be a group with an NSS Γ with 1 ∈ Γ. We define a measure function and use it to state and prove a variant of the Thompson Replacement Theorem.

Proposition 5.1 Let X, Y ∈ Γ with XY ∈ Γ. Let

X = X_{0} < X_{1} < … < X_{r} = XY

be any maximal chain of Γ-subgroups from X to XY. Then

X ∩ Y = X_{0} ∩ Y < X_{1} ∩ Y < … X_{r} ∩ Y = Y

is a maximal chain of Γ-subgroups from X ∩ Y to Y.

Proof. Let A be a Γ subgroup with X_{i} ∩ Y ≤ A ≤ X_{i+1} ∩ Y. Since X ≤ X_{i} ≤ XY, Proposition 1.8 implies X_{i} = X(X_{i} ∩ Y). Thus

X_{i} ≤ AX ≤ X_{i+1}.

X_{i} is a maximal Γ-subgroup of X_{i+1} and so by Proposition 2.2 X_{i} is normal in X_{i+1}. Thus AX = AX_{i} is a subgroup of X_{i+1}. Since X_{i+1} is nilpotent, Proposition 2.3(6) implies AX ∈ Γ. By the maximality of the X_{i}-chain, AX = X_{k} for some k ∈ {i, i + 1}. Thus X_{k} ∩ Y = AX ∩ Y = A(X ∩ Y) = A. □

Proposition 5.2 Let X ∈ Γ. Then there exists a maximal chain of Γsubgroups from 1 to X and any two such chains have the same length. We denote this common length by μ(X).

Proof. The existence of a maximal Γ-chain from 1 to X follows from (MM). Let A and B be maximal Γ-subgroups of X. By induction any maximal Γ-chain from 1 to X through A has unique length μ(A) + 1. It remains to show that μ(A) = μ(B). Without loss A ≠ B. By maximality of A and B, A is normal in X, AB ∈ Γ and AB = X. Note that A ≤ X is a maximal chain from A to X and so by Proposition 5.1, A ∩ B < B is a maximal chain from A ∩ B to B. Thus μ(B) = μ(A ∩ B) + 1 = μ(A). □

Abusing the term we call μ of Proposition 5.2 a measure function on Γ and μ(A) is called the measure of A.

Proposition 5.3 μ(P) = μ(P^{g}) for all P ∈ Γ and g ∈ G.

Proof. This follows from (C) and Proposition 5.2. □

Proposition 5.4 Assume X, Y ∈ Γ .

(1) Suppose X ≤ Y, then any maximal Γ-chain from X to Y has length μ(Y/X) := μ(Y) − μ(X).

(2) Suppose XY ∈ Γ. Then μ(XY) = μ(X) + μ(Y) − μ(X ∩ Y).

Proof. (1) follows from Proposition 5.2.

(2) By (1) μ(XY/X) = μ(XY) − μ(X). By (1) and Proposition 5.1, μ(XY/X) = μ(Y/X ∩ Y). Again by (1) μ(Y/X ∩ Y) = μ(Y) − μ(X ∩ Y). Thus (2) holds.

□

Definition. For P ∈ Γ let A(P) be the set of all decomposable abelian Γsubgroups of P with maximal measure. Let J(P) := ⟨A(P)⟩, the Thompsonsubgroup of P (compare with the introduction of [5]).

Then J(P) is a Γ-subgroup of P by Proposition 2.3(6).

Proposition 5.5 Let V be a decomposable abelian Γ-subgroup of G and A ∈ Γ_{V} with A ∈ A(AV). Then C_{V} (A) = V ∩A and μ(A/C_{A}(V)) ≥ μ(V/C_{V} (A))

Proof. By Proposition 4.7 and (P), C_{V} (A)A is an decomposable abelian Γ-subgroup of P. The maximality of μ(A) implies C_{V} (A) ≤ A and thus C_{V} (A) = V ∩ A. Thus V ∩ C_{A}(V) = V ∩ A = C_{V} (A) and by maximality of A, Proposition 4.7 and Proposition 5.4:

μ(A) ≥ μ(V C_{A}(V)) = μ(V C_{A}(V)/C_{A}(V)) + μ(C_{A}(V))
= μ(V/C_{V} (A)) + μ(C_{A}(V)).
□

The next lemma is our version of the Thompson Replacement Theorem.

Lemma 5.6 Let A, V be decomposable abelian Γ-groups with A ∈ Γ_{V} ∩ A(AV). Let x ∈ N_{V} (N_{V} (A)A) and define

D = ((AA^{x}) ∩ V)(A ∩ A^{x}).

Then

(1) D ∈ A(AV) and ⟨x⟩N_{V} (A)A ⊆ N_{G}(D).

(2) If [V, A] ≠ 1, then [V, D] ≠ 1.

Proof. (1) Let P = N_{V} (A)A. Since x normalizes P, both A and A^{x} are normal in P. Thus AA^{x} = ⟨A, A^{x}⟩. Since A is abelian, A ∩ A^{x} ⊆ Z(AA^{x}). By Proposition 4.6 both AA^{x} ∩ V and A ∩ A^{x} are decomposable Γ-groups and so D is an abelian decomposable Γ-group. Also [x, A] ⊆ V ∩ (AA^{x}) ⊆ D and so x ∈ N_{G}(D). Note that

(^{∗}) μ(AA^{x}) = μ(A) + μ(A^{x}) − μ(A ∩ A^{x}) = 2μ(A) − μ(A ∩ A^{x}).

Also AA^{x} ⊆ V A and so AA^{x} = AA^{x} ∩ V A = A(V ∩ AA^{x}) = AD. Moreover, D ∩ A = (V ∩ A)(A ∩ A^{x}) and V ∩ A ⊆ C_{A}(x) ⊆ A ∩ A^{x}. Thus D ∩ A = A ∩ A^{x}. Hence μ(AA^{x}) = μ(DA) = μ(D) + μ(A) − μ(A ∩ A^{x}). Comparing with (*) we obtain μ(A) = μ(D) and so D ∈ A(AV).

(2) Suppose that [V, D] = 1. Then A∩A^{x} ≤ C_{A}(V) and so by Proposition 5.5, A ∩ A^{x} = A ∩ V. Hence D ≤ V. Since D ∈ A(AV) we get V = D ⊆ P ⊆ N_{G}(A). Thus A = A^{x} = A ∩ A^{x} ≤ D and [V, A] = 1. Thus (2) holds.

□

Proposition 5.7 Let V be a decomposable abelian Γ-subgroup of G and P ∈ Γ_{V} with V ⊆ P and J(P) ≰ C_{G}(V). Then there exists A ∈ A(P) such that [[V, A], A] = 1 ≠ [V, A] ≤ A.

Proof. Since J(P) ≰ C_{G}(V) there exists A ∈ A(P) with 1 ≠ [V, A]. Choose such an A with N_{V} (A) maximal.

Suppose that V does not normalize A. Then V ≰ N_{V} (A)A and so by Proposition 1.6(2) there exists x ∈ N_{V} (N_{V} (A)A) with x ∉ N_{V} (A). Let D be defined as in Lemma 5.6. Then D ∈ A(AV), [V, D] ≠ 1 and ⟨x⟩N_{V} (A) ≤ N_{V} (D), contradiction to the maximal choice of N_{V} (A).

Thus V normalizes A, [V, A] ≤ A and [[V, A], A] = 1. □

Lemma 5.8 Let A, B be abelian Γ-subgroups with [A, B] ≤ A ∩ B. Let a ∈ A. Suppose that B is decomposable and C_{B}(a) ∈ Γ. Then [a, B] ∈ Γ and μ([a, B]) = μ(B/C_{B}(a)).

Proof. By Proposition 4.5 there exists a Γ-subgroup D of B with B = C_{B}(a) × D. Then [D, a] ≤ A ∩ B ≤ C_{A}(D) and so by Proposition 1.2(2) DD^{a} = D[D, a] ∈ Γ. Moreover, DD^{a} ∩ A = (D ∩ A)[D, a] and D ∩ A ≤ C_{D}(a) = 1. [D, a] = DD^{a} ∩ A ∈ Γ. In particular, D ∩ [D, a] = 1 and so by Proposition 1.2(3), D ∩ D^{a} = C_{D}(a) = 1. Thus 2μ(D) = μ(DD^{a}) = μ(D) + μ([D, a]) and

μ([a, B]) = μ([a, D]) = μ(D) = μ(B/C_{B}(a)).
□

6 Glauberman’s Theorem, Part I

In this section we begin the proof of Theorem A stated in the introduction. Assume G, V, S, A, Γ have the meaning and the properties mentioned there.

Proposition 6.1 Set Π := ⋃_{g∈G} ΓS^{g}. Let T ∈ Π^{∗}.

(1) Π is an NSS of rank 1.

(2) Π^{∗} = S^{G} ⊆ Γ^{∗}.

(3) R(Π) = R(Γ).

(4) [C_{G}(V), ⟨Π⟩] ≤ R(Γ).

(5) Let P ∈ Γ(T C_{G}(V)), then P ≤ T.

(6) R(Γ) = C_{T} (V).

Proof. By (a) (that is assumption (a) of Theorem A), Γ has rank 1. By (e) R(Γ) is large. So (1),(2) and (3) follow from Lemma 3.5.

(4) By (c) [C_{G}(V), S] ≤ R(Γ). Thus (4) follows by conjugation.

(5) By (4) [P, T] ≤ [T C_{G}(V), T] ≤ T R(Γ) ≤ T. Thus P ≤ N_{G}(T). By (e), R(Γ) is large and so by Lemma 3.2 P T ∈ Γ. Since T ∈ Π^{∗} = S^{G} ⊆ Γ^{∗}, P ≤ T.

(6) Let R ∈ Π^{∗}. By (4) [T ∩ C_{G}(V), R] ≤ R(Γ) ≤ R. Thus C_{T} (V) ≤ E := ⋂_{R∈Π^{∗}} N_{T} (R). By (I) and (MM) E ∈ Γ and by Proposition 2.3(8), E ≤ R(Π) = R(Γ). □

Lemma 6.2 There exists a non-trivial quadratic Γ-offender E in S on V with C_{V} (E) = V ∩ E.

Proof. By (d) there exists a non-trivial Γ-offender A in S on V. Since V ≤ R(Γ), A ∈ Γ_{V}. Let B = C_{A}(V)V and D = C_{V} (A)A. By Proposition 4.7, B ∈ Γ. Since A is an offender on V, μ(V/C_{V} (A)) ≤ μ(A/C_{A}(V)). But this is equivalent to μ(B) ≤ μ(D). We will show that

(^{∗}) J(AV) ≰ C_{AV} (V).

Since AV = AB, C_{AV} (V) = C_{A}(V)B = B. From μ(D) ≥ μ(B), we get D ∈ A(AB) and D ≰ B, since 1 ≠ [A, V] ≤ [B, D].

Thus (*) holds. The existence of E now follows from Proposition 5.7 and Proposition 5.5. □

Notation.

Δ := E^{G}, where E is as in Lemma 6.2 E, F ∈ Δ such that μ([V, E][V, F]) is minimal with respect to ⟨E, F ⟩ ∉ Γ.

W := [V, E][V, F]. H := ⟨E, F ⟩ R(Γ). Z := [V, E] ∩ [V, F]. Λ := ΔH. q := μ(A/C_{A}(V)) for A ∈ Δ. m := μ([V, A]).

Notice that C_{A}(V), [V, A], W and Z are decomposable abelian subgroups by Proposition 4.5 - Proposition 4.8, (P) and (I). Hence the measure of these groups is defined. Note that by Proposition 5.4 and the choice of E, F, μ(Z) is maximal with respect to ⟨E, F ⟩ ∉ Γ. The existence of F ∈ Δ with ⟨E, F ⟩ ∉ Γ is guaranteed by Lemma 3.6. In view of Lemma 3.4(1) we denote by D^{∗} the unique member of Γ^{∗} which contains D provided D ∈ Γ with D ≰ R(Γ). Observe that by Proposition 6.1(6), A ∉ R(Γ) for all A ∈ Δ.

Proposition 6.3 Let A, B ∈ Δ

(1) ⟨A, B⟩ ∉ Γ if and only if A^{∗} ≠ B^{∗}.

(2) If A^{∗} ≠ B^{∗}, then ⟨A, A^{b}⟩ ∉ Γ for all b ∈ B\R(Γ).

(3) If A^{∗} ≠ B^{∗}, then [V, A] ≠ [V, B].

Proof. (1) If ⟨A, B⟩ ∈ Γ then A^{∗} = ⟨A, B⟩^{∗} = B^{∗}. If A^{∗} = B^{∗} then ⟨A, B⟩ ∈ Γ by Proposition 2.3(6). Hence (1).

(2) Let b ∈ B with ⟨A, A^{b}⟩ ∈ Γ. Then by Lemma 3.4(2), b ∈ A^{∗} ∩ B^{∗}. So by Theorem 3.1(2), b ∈ R(Γ).

(3) Assume [V, A] = [V, B]. By (c), A and B are quadratic and so [[V, A], B] = [[B, V], A] = 1 . Thus [[A, B], V] = 1 by Proposition 1.1 (5) and [A, B] ⊆ C_{G}(V). Let b ∈ B\ R(Γ). Then A^{b} ∈ AC_{G}(V) ≤ A^{∗}C_{G}(V) and so by Proposition 6.1(5), A^{b} ≤ A^{∗}, a contradiction to (2). □

Proposition 6.4 Let A ∈ Λ and a ∈ A\ R(Γ). Then

(1) [V, A] = Z × [W, a] = C_{W} (a) = C_{W} (A).

(2) ⟨B, B^{a}⟩ ∉ Γ and [V, B] ∩ C_{V} (a) = [V, A] ∩ [V, B] = Z for all B ∈ Λ with ⟨A, B⟩ ∉ Γ.

Proof. By Lemma 3.6 there is B ∈ Λ with ⟨A, B⟩ ∉ Γ. So by Proposition 6.3 ⟨B, B^{a}⟩ ∉ Γ. Hence W = [V, A][V, B] = [V, B][V, B^{a}] by minimality of μ(W). Put D := C_{[V,B]}(a). Then by Proposition 5.4, Proposition 1.2, and quadratic action,

C_{W} (a) = D × [[V, B], a] = [V, A] = C_{W} (A).

Also [W, a] = [[V, B], a] and

D = [V, B] ∩ C_{W} (a) = [V, B] ∩ [V, A] = C_{W} (B) ∩ C_{W} (A) = C_{W} (⟨A, B⟩).

Since H centralizes Z, Z ≤ D. The maximality of μ(Z) now implies Z = D. □

Proposition 6.5 Let A, B ∈ Λ with ⟨A, B⟩ ∉ Γ, w ∈ [V, B]\Z, and a ∈ A\ R(Γ). Then

(1) W = [V, B] × [w, A].

(2) V = W C_{V} (A).

(3) q = μ(V/C_{V} (A)) = μ(W/C_{W} (A)) = μ(A/C_{A}(V)) = μ([w, A]) = μ([V, a]).

Proof. (1) By Proposition 4.8, W ∈ Γ. Since V is decomposable, also W is decomposable and by Proposition 4.7 C_{W} (A) ∈ Γ. Hence also C_{W} (A)A ∈ Γ and we can assume [V, A] = C_{W} (A) ⊆ A. Then A ∩ W = [V, A]. By Proposition 6.4(2), [w, a] ≠ 1 for all a ∈ A\R(Γ) and so C_{A}(w) = A∩R(Γ) ∈ Γ. From Lemma 5.8 we conclude [w, A] ∈ Γ and (^{∗}) μ([w, A]) = μ(A/C_{A}(V)) = q. Note that μ([V, B]) = m = μ([V, A]) = μ(C_{W} (A)) and so (^{∗∗}) μ(V/C_{V} (A)) ≥ μ(W/C_{W} (A)) = μ(W/[V, B]). By (^{∗}),(^{∗∗}) and since A is an offender, μ([w, A]) ≥ μ(W/[V, B]). By

Proposition 6.4 [w, A] ∩ [V, B] ≤ [w, A] ∩ Z = 1 and we conclude that μ([w, A]) = μ(W/[V, B]) and W = [V, B] × [w, A]. So (1) holds.

(2) We also conclude that the inequalilty in (**) actually is an equality. So μ(V/C_{V} (A)) = μ(W/C_{W} (A)). Hence (2) holds.

(3) By Proposition 6.4, C_{W} (a) = C_{W} (A) ∈ Γ. So by Lemma 5.8, [W, a] ∈ Γ and μ([W, a]) = μ(W/C_{W} (A)) = q. By (2), [V, a] = [W, a] and all parts of (3) are proved. □

Proposition 6.6 Let A, B ∈ Λ with ⟨A, B⟩ ∉ Γ and Σ := [V, A]^{H}. Then:

(1) If B ∈ Λ with ⟨A, B⟩ ∉ Γ then Σ = {[V, A]} ∪ [V, B]^{A}.

(2) If M, N ∈ Σ with M ≠ N then M ∩ N = Z.

(3) W = ⋃_{M∈Σ} M.

(4) For D ∈ Δ put D̂ = D R(Γ). Let D ∈ Λ with D^{∗} = A^{∗}, then D̂ = Â.

(5) Let Λ̂ = {B̂ | B ∈ Λ}. Then Λ̂ = {Â} ∪ {B̂^{A}}.

(6) H = ⟨C, D⟩ R(Γ) for all C, D ∈ ΔH with Ĉ ≠ D̂.

(7) V = C_{V} (H)W and V = ⋃_{D∈Λ} C_{V} (D).

Proof. Let A, B ∈ Λ with ⟨A, B⟩ ∉ Γ. Then W = [V, A][V, B] = [V, B] × [w, A] for w ∈ [V, B]\Z by Proposition 6.5 and [V, A] ∩ [V, B] = Z by Proposition 6.4. Therefore w^{A}Z = w[V, A], which shows

(^{∗}) W = [V, A] ∪ ⋃_{a∈A}[V, B]^{a}.

By Lemma 3.6 we can apply (^{∗}) to an element of A^{H} in the role of B and so (3) holds.

Also [V, B]^{a_{1}} ∩ [V, B]^{a_{2}} = Z for a_{1}, a_{2} ∈ A with a_{1}a_{2}^{−1} ∉ R(Γ) by Proposition 6.4. Let C ∈ Λ. Then there is D ∈ {A} ∪ B^{A} with [V, C] ∩ [V, D] ⊃ Z.

Hence by maximality of μ(Z), ⟨C, D⟩ ∈ Γ, C^{∗} = D^{∗} and ⟨C, K⟩ ∉ Γ for K ∈ ({A} ∪ B^{A})\{C}. But then by (^{∗})

[V, C]\Z ⊆ W \ ⋃{[V, K] | D ≠ K ∈ {A} ∪ B^{A}} ⊆ [V, D]

and [V, C] = [V, D]. Thus (1) and (2) hold.

(4) Let d ∈ D. By (1) [V, B]^{da} = [V, B] for some a ∈ A. Thus da ∈ N_{G}(B^{∗}) and so da ∈ R(Γ). Hence d ∈ A R(Γ) = Â. Thus (4) holds.

(5) Let C ∈ Λ with Ĉ ≠ Â. By (4), C^{∗} ≠ A^{∗} and by Proposition 6.3(3), [V, C] ≠ [V, A]. So by (1), [V, C] = [V, B]^{a} for some a ∈ A. By Proposition 6.3(3), C^{∗} = B^{a∗} and so by (4) Ĉ = B̂^{a}.

(6) By (5), H is doubly transitive on Λ̂. Since H = ⟨Ê, F̂⟩, (6) holds.

(7) Since H = ⟨A, B⟩ R(Γ), we have C_{V} (H) = C_{V} (A) ∩ C_{V} (B). Since μ(V/C_{V} (A)) = q we get μ(V/C_{V} (H)) ≤ 2q. Since μ(W/C_{W} (H)) = 2q, the first part of (7) holds.

Let v ∈ V. Then v = cw with c ∈ C_{V} (H) and w ∈ W. By (1), w ∈ [V, C] for some C ∈ Λ. So v ∈ C_{V} (H)[V, C] ≤ C_{V} (C). □

Lemma 6.7 Let t ∈ G and B ∈ Δ. Suppose that one of the following holds:

1. t ∈ A ∈ Δ and [V, t] ∩ C_{V} (B) ≠ 1.

2. μ(C_{V} (B)/(C_{V} (B) ∩ C_{V} (B)^{t})) < q.

Then ⟨B, B^{t}⟩ ∈ Γ.

Proof. Suppose that 1. holds. Then by Proposition 1.2(2) C_{V} (B^{t}) ⊆ [V, t]C_{V} (B). By Proposition 6.5(3), μ([V, t]) = q and so 1. implies 2.

So we may assume that 2. holds. Then

μ([V, B]/([V, B] ∩ C_{V} (T))) < q.

Since [V, B]∩C_{V} (t) ≤ [V, B]∩[V, B]^{t} and μ([V, B]/Z) = q, the maximality of μ(Z) implies ⟨B, B^{t}⟩ ∈ Γ.

Lemma 6.8 Let A ∈ Δ. Then A ⊆ B R(Γ) ⊆ H for some B ∈ Λ.

Proof. Let a ∈ A\C_{A}(V). By Proposition 6.6(7) there exists B ∈ Λ with [V, a] ∩ C_{V} (B) ≠ 1. By Lemma 6.7, ⟨B, B^{a}⟩ ∈ Γ. Thus by Lemma 3.4(2), a ∈ B^{∗}. Hence A ⊆ B^{∗} and A^{∗} = B^{∗}. Since R(Γ) ≤ C_{G}(V), Proposition 6.6(5) implies [V, a] ⊆ C_{V} (B). By Lemma 6.7 a ∈ B^{∗} and so A^{∗} = B^{∗}. So by Proposition 6.6(5), B R(Γ) is independent from the choice of a. Hence [V, A] ⊆ C_{V} (B). Let A = B^{g} for g ∈ G. Then A ∈ Λ^{g} and so by symmetry [V, B] ≤ C_{V} (A). Thus [V, A][V, B] ⊆ C_{V} (AB).

Let D ∈ Λ\ΛB^{∗}.

Put T = ⟨A, B, D⟩, U = [V, T] = [V, A][V, B][V, D] and Y = [V, A][V, B]∩ C_{V} (D). Then Y is centralized by A, B and D and so Y ⊆ C_{U}(T). Since μ(V/C_{V} (D)) = q = μ([V, B]C_{V} (D)/C_{V} (D)), [V, A][V, B] = [V, B]Y. Let a ∈ A and w ∈ [V, D]\Z. Note that Z ⊆ [V, D]∩C_{V} (B) ≤ Y. By Proposition 6.5(1), [w, B]Z = [V, B] and so [V, A][V, B] = [w, B]Y. Hence [w, a]Y = [w, b]Y for some b ∈ B. Let t = b^{−1}a. Then w^{t}Y = wY. Since wY ⊆ [V, D]Y ⊆ C_{V} (D), wY = w^{t}Y ⊆ C_{V} (D^{t}). Hence Z < ⟨w⟩Z ⊆ [V, D] ∩ C_{V} (D^{t}) and so μ([V, D]/([V, D] ∩ C_{V} (D^{t}))) < q.

Thus by Lemma 6.7, ⟨D, D^{t}⟩ ∈ Γ and by Lemma 3.2(2), t ∈ D^{∗}. Hence t ∈ D^{∗} ∩ B^{∗} = R(Γ). So a = bt ∈ B R(Γ) ⊆ H. □

Theorem 6.9 ⟨Γ⟩ = H and Γ^{∗} = {A R(Γ) | A ∈ Δ}.

Proof. Let P ∈ Γ^{∗}. By Lemma 6.8, H = ⟨Δ⟩ R(Γ) and so H is normal in G. So P normalizes W = [V, H] and Z = C_{W} (H). As P V is nilpotent, P centralizes some 1 ≠ wY in W/Z. By Proposition 6.6(3), w ∈ [V, A] for some A ∈ Λ. Thus P ⊆ N_{G}([V, A]). By Proposition 6.3(3) P ⊆ N_{G}(A^{∗}) and so by Lemma 3.2, P = A^{∗}. By Proposition 6.6(5), A acts transitively on Λ̂ \ Â, whence P = AN_{P} (B̂) for B ∈ Λ with ⟨A, B⟩ ∉ Γ. But N_{P} (B̂) ≤ N_{P} (B̂^{∗}) = P ∩ B^{∗} = R(Γ) and so P = A R(Γ). □

7 Glauberman’s Theorem, Part II

In this section we complete the proof of Theorem A. We continue to use the notations from the previous section. In addition we define:

V_{0} = W/Z, written additively.

V_{1} = [V, E]/Z and V_{2} = [V, F]/Z.

We view V_{0} as a left module over the endomorphism ring End(V_{0}). In particular if α, δ ∈ End(V_{0}) and v ∈ V_{0}, then (αδ)(v) = α(δ(v)). For h ∈ H define σ_{h} ∈ End(V_{0}) by σ_{h}(wZ) = w^{h}Z for w ∈ W. Note that σ_{hh′} = σ_{h′}σ_{h}. From Proposition 6.6 we obtain:

(i) V_{1} = C_{V_{0}}(E) = [V_{0}, E] = [V_{0}, a] for all a ∈ E\C_{E}(V_{0}). (ii) V_{2} = C_{V_{0}}(F) = [V_{0}, F] = [V_{0}, b] for all b ∈ F \C_{F} (V_{0}). (iii) V_{0} = V_{1} ⊕ V_{2}. (iv) For g ∈ H with σ_{g}(V_{1}) ≠ V_{1} there is a ∈ E with σ_{ga}(V_{1}) = V_{2}.

Take b ∈ F fixed such that σ_{b}(V_{1}) ≠ V_{1} and set β := σ_{b} − 1 ∈ End(V_{0}). Similarly set χ_{a} = σ_{a} − 1. Moreover for i = 1, 2 let π_{i} be the projection from V_{0} on V_{i} according to the direct sum decomposition V_{0} = V_{1} ⊕ V_{2}. Proposition 7.1 The following equations hold, where a, c ∈ E:

(1) σ_{b} = π_{1} + π_{2} + β.

(2) σ_{a} = π_{1} + π_{2} + χ_{a}.

(3) χ_{a}π_{1} = π_{2}χ_{a} = βπ_{2} = π_{1}β = 0.

(4) βπ_{1} = π_{2}β = β and π_{1}χ_{a} = χ_{a}π_{2} = χ_{a}.

(5) β^{2} = χ_{a}χ_{c} = 0.

(6) π_{1}σ_{a} = π_{1} + χ_{a}.

(7) χ_{ac} = χ_{a} + χ_{c} and χ_{a^{−1}} = −χ_{a}. Proof. Straightforward.. □

Proposition 7.2 There exists a_{1} ∈ E such that (χ_{a_{1}}β) |_{V_{1}}= id_{V_{1}}. Proof. By (iv) there exists a ∈ E such that σ_{ba}(V_{1}) = V_{2}. Now Proposition 7.1 affords

π_{1}σ_{ba} = (π_{1}σ_{a})σ_{b} = (π_{1} + χ_{a})(π_{1} + π_{2} + β) = π_{1} + χ_{a} + χ_{a}β and 0 = π_{1}σ_{ba} |_{V_{1}}= id_{V_{1}} + (χ_{a}β) |_{V_{1}}. Let a_{1} = a^{−1}. □

Proposition 7.3 For every a ∈ E\C_{E}(V_{0}) there exists â ∈ A such that (χâβ) |_{V_{1}}= ((χ_{a}β) |_{V_{1}})^{−1}.

Proof. Let g = b^{−1}ab. A straightforward calculation shows

(^{∗}) σ_{g} = (π_{1} + χ_{a} − χ_{a}β) + (π_{2} + β χ_{a} − βχ_{a}β).

By Proposition 6.6 σ_{g}(V_{1}) ≠ V_{1}. Hence by (iv) there is c ∈ E such that σ_{gc}(V_{1}) = V_{2}. Then π_{1}σ_{gc} = (π_{1}σ_{c})σ_{g} = (π_{1} + χ_{c})σ_{g}. Using (^{∗}) we compute

0 = π_{1}σ_{gc} |_{V_{1}}= id_{V_{1}} − (χ_{a}β) |_{V_{1}} −(χ_{c}βχ_{a}β) |_{V_{1}} .

Multiplying this equation with ((χ_{a}β) |_{V_{1}})^{−1} from the right we obtain

(χ_{c}β) |_{V_{1}}= ((χ_{a}β) |_{V_{1}})^{−1} − id_{V_{1}}.

By Proposition 7.2 there exists a_{1} ∈ E such that (χ_{a_{1}}β) |_{V_{1}}= id. Let â = ca_{1}. Then χâ = χ_{c} + χ_{a_{1}} we compute (χâβ) |_{V_{1}}= ((χ_{a}β) |_{V_{1}})^{−1}. □

In Proposition 7.4 and Proposition 7.5 we pick a fixed v_{1} ∈ V_{1} with v_{1} ≠ 0.

Proposition 7.4 Let a, a′ ∈ E. Define

\overline{χ}_{a} = (σ_{a} − 1)β ∈ End(V_{0}) and x_{a} := \overline{χ}_{a}(v_{1}).

There is a unique coset a′′C_{E}(V_{0}) with \overline{χ}_{a′}(x_{a}) = x_{a′′}. Define

x_{a} + x_{a′} := x_{aa′} and x_{a} · x_{a′} := x_{aa′′}.

Set D := {x_{a} | a ∈ E}. Then (D, +, ·) is a Cayley-Dickson-Division-Algebra or a skew field with (D, +) ≃ E/C_{E}(V_{0}).

Proof. For each v ∈ V_{1}^{#} we have \overline{χ}_{E}(v) := \{\overline{χ}_{a}(v) | a ∈ E\} = V_{1} by Proposition 6.5. As elements of \overline{χ}_{A} are not singular we get a^{−1}a′ ∈ C_{E}(V_{0}) if χ_{a}(v) = \overline{χ}_{a′}(v). Hence for v, v′ ∈ V_{1}\\{0\} there is a unique coset a′′C_{E}(V_{0}) with \overline{χ}_{a′′}(v) = v′. Thus the product x_{a} · x_{a′} for a, a′ ∈ E is well defined. Now the proof of Glauberman [5, (IX) on page 7 f] shows that (D, +, ·) is an alternative division ring or a skew field. Thus Proposition 7.4 follows from [2]. □

Proposition 7.5 Let D be as in Proposition 7.4. Then {V_{1}} ∪ V_{2}^{E} is a congruence partition of an affine plane over D.

Proof. By Proposition 6.6 {V_{1}} ∪ V_{2}^{E} is a congruence partition. Let a_{0} ∈ E. Then (\overline{χ}_{a_{0}} \overline{χ}_{a})(v_{1}) = −(β\overline{χ}_{a})(v_{1}) + (σ_{a_{0}}β\overline{χ}_{a})(v_{1}) for a ∈ E. Hence σ_{a_{0}}(V_{2}) = \{(\overline{χ}_{a_{0}} \overline{χ}_{a})(v_{1}) + (β\overline{χ}_{a})(v_{1}) | a ∈ E\}. Now \overline{χ}_{a}(v_{1}) ↔ aC_{E}(V_{0}) ↔ (β\overline{χ}_{a})(v_{1}) define bijective maps between D ≃ E/C_{E}(V_{0}), V_{1} and V_{2} which induce a bijective map between V_{0} and D×D. Then σ_{a_{0}}(V_{2}) is mapped on \{(\overline{χ}_{a_{0}} \overline{χ}_{a}, \overline{χ}_{a}) | a ∈ E} and we get Proposition 7.5 (see [4, page 131 f]). □

Proposition 7.6 By Proposition 7.5 we may view V_{0} as an affine plane over D. Then E induces the group of shears with axis V_{1} on V_{0} and H = L induces the subgroup of a point-stabilizer of V_{0} generated by all shears.

Proof. Since E is transitive on all lines through 0 different from V_{1} by Proposition 6.6, E contains all shears by [4, page 122]. As H is transitive on the lines through 0 we get Proposition 7.6. □

Theorem A now follows from Proposition 7.6 and Theorem 6.9.

8 Strong NSS’s

We say that an NSS Γ is strong provided that

(Z) Ω(Z(N)) ≠ 1 for all 1 ≠ N ∈ Γ.

Throughout this section we assume that G is a group with a reduced strong NSS Γ with 1 ∈ Γ. In addition to our previous notations we let

Θ := {N ∈ Γ\Γ^{∗} | N is large in Γ}.

Lemma 8.1 Let rank Γ = 2, N ∈ Θ, V = Ω(Z(N)), P ∈ Γ_{N}^{∗} and Z ∈ Γ with Z normal in P. Then:

(1) Let 1 ≠ D ∈ Γ be normal in N_{G}(N). Then D^{∘} = N and N_{G}(D) = N_{G}(N)

(2) N_{G}(V) = N_{G}(N).

(3) If R(Γ_{N} ∩ Γ_{Z}) ≠ N then (Γ_{N} ∩ Γ_{Z})^{∗} = {P }.

(4) If Z ⊆ V and R(Γ_{N} ∩ Γ_{Z}) ≠ N, then [C_{G}(V), P] ⊆ C_{P} (V) = N.

Proof. (1) Note that Γ_{N} ≤ Γ_{D}. So by Corollary 2.7, D^{∘} is closed and contained in N. Since 1 ≠ D ≤ D^{∘} ≤ N < S ∈ Γ^{∗} and Γ is reduced of rank 2, D^{∘} = N. So N_{G}(D) ⊆ N_{G}(N). By assumption N_{G}(N) ≤ N_{G}(D) and (1) holds.

(2) follows from (1) applied to D = V.

(3) Put T = R(Γ_{N} ∩ Γ_{Z}) and suppose T ≰ N. Since N ≤ T we get N < T. Since Γ_{N} has rank 1, Proposition 3.7(3) implies (Γ_{N} ∩ Γ_{Z})^{∗} = {P }.

(4) Since Z ⊆ V, C_{G}(V) ⊆ N_{G}(N)∩N_{G}(Z) and so C_{G}(V) ⊆ N_{G}(((Γ_{N})_{Z})^{∗}) = N_{G}(P). Thus

[C_{G}(V), P] ⊆ C_{P} (V).

Suppose that N_{G}(N) ≤ N_{G}(P) and let Q ∈ Γ_{N}. By definition of Θ, C_{P} (N) ≤ N and thus [C_{P} (N), ⟨P, Q⟩] ≤ N. So by Lemma 2.9, QP ∈ Γ. Thus Q ∈ Γ_{P} and Γ_{N} ≤ Γ_{P}. Corollary 2.7 implies P ≤ P^{∘} ≤ N, a contradiction. Thus N_{G}(N) ≰ N_{G}(P).

Let g ∈ N_{G}(N)\N_{G}(P). Then C_{P} (V) ⊆ N_{P} (N) ∩ C_{G}(Z^{g}) ⊆ N_{P} (P^{g}), whence C_{P} (V)C_{P} (V)^{g} ⊆ N_{P} (P^{g})N_{P^{g}}(P) ∈ Γ. Pick Q ∈ Γ_{N}^{∗} with

N_{P} (P^{g})N_{P^{g}}(P) ⊆ Q.

If Q ≠ P, then C_{P} (V) ⊆ P ∩ Q ⊆ N, by Proposition 3.7(1).

If Q = P, then C_{P} (V)^{g} ⊆ P^{g} ∩P = N, again by Proposition 3.7(1). Since N = N^{g} we get C_{P} (V) ≤ N. □

Theorem 8.2 Let G be a group with a reduced strong NSS Γ of rank 2. Let N ∈ Θ, S ∈ Γ_{N}^{∗}, V := Ω(Z(N)) and Z := C_{V} (J(S)). Then 1 ≠ Z ∈ Γ. Moreover,

(1) If J(S) ≤ N, then J(S)^{∘} = N and N_{G}(J(S)) = N_{G}(N).

(2) If J(S) ≰ N and N = Z^{∘}, then N = Ω(Z(P))^{∘} for any P ∈ Γ^{∗} with S ≤ P.

(3) If N ≠ Z^{∘}, then V/C_{V} (⟨Γ_{N}⟩) is a natural SL_{2}-module for Γ_{N} and S = J(S)N.

Proof. Since V ≠ 1 and V J(S) is nilpotent, Z ≠ 1. By Proposition 4.7 Z ∈ Γ.

(1) Follows from Lemma 8.1(1).

(2) Suppose that (ΓC_{G}(Z))^{∗} = {T } for some T. Then T is normal in N_{G}(Z). Since N = Z^{∘} is large, we get from Lemma 3.2 that T ≤ Q for all Q ∈ Γ_{Z}^{∗}. Thus T ≤ Z^{∘} = N, a contradiction since J(S) ≤ T and J(S) ≠ N.

Thus there exist L, Q ∈ ΓC_{G}(Z)^{∗} with L ≠ Q. Then ⟨L, Q⟩ ∉ Γ. Put M = Ω(Z(P))^{∘}. Note that N ⊆ L ∩ Q, and both LN_{M}(N) and QN_{M}(N) are in Γ. Thus NN_{M}(N) ⊆ LN_{M}(N) ∩ QN_{M}(N) ⊆ N, by Proposition 3.7(1). Thus 1 ≠ M ⊆ N. Since M ⊴ P, Proposition 2.8(12) implies M is closed and as rank G = 2, M = N.

(3) From rank(Γ) = 2, Proposition 3.7(1) and Theorem 3.1 we get P ∩Q = N and rank(Γ_{N}) = 1 for P, Q ∈ Γ_{N}^{∗} with P ≠ Q. Suppose N = R(Γ_{N} ∩ Γ_{Z}). Then by Lemma 2.6 (applied with Λ = Γ_{N} ∩ Γ_{Z}, Δ = Γ_{Z} and P = N), Z^{∘} = R(Γ_{Z}) ⊆ N and Z^{∘} is closed. Thus

1 ≠ Z ⊆ Z^{∘} ≤ N ∉ Γ^{∗}.

Since rank(Γ) = 2 we get N = Z^{∘}, a contradiction. Therefore N ≠ R((Γ_{N} ∩ Γ_{Z})). In particular, Γ_{N} ∩ Γ_{Z} ≠ Γ_{N} and so Γ_{N} ⊈ Γ_{Z}.

Moreover, by Lemma 8.1(4) [C_{G}(V), S] ⊆ N = C_{S}(V).

Assume J(S) ⊆ N. Then Z = V and Γ_{N} ⊆ Γ_{Z}, a contradiction. Thus J(S) ⊈ N = C_{S}(V). Pick A ∈ A(S) with A ≰ C_{S}(V). Then by Proposition 5.5 A is a non-trivial Γ-offender on V. By Proposition 3.7(2) Γ_{N} has rank 1. By definition of Θ, N is large in Γ_{N}.

We verified that all the the assumptions of Theorem A are satisfied for N_{G}(N), Γ_{N}, S,A and V. Hence V/C_{V} (⟨Γ_{N}⟩) is a natural SL_{2}-module for Γ_{N}. By Theorem 6.9, S = A R(Γ_{N}) = AN and so S = J(S)N. □

Theorem 8.3 Suppose rank(Γ) = 2, N ∈ Θ and S ∈ Γ_{N}^{∗} with N_{G}(S) ⊈ N_{G}(N). Put V = Ω(Z(N)). Then V/C_{V} (⟨Γ_{N}⟩) is a natural SL_{2}-module for Γ_{N}.

Proof. Suppose that J(S) ≤ N. Then using Theorem 8.2(1)

N_{G}(S) ≤ N_{G}(J(S)) ≤ N_{G}(J(S)^{∘}) = N_{G}(N),

a contradiction to the assumptions.

Hence J(S) ⊈ N. Set Z := C_{V} (J(S)). Suppose that Z^{∘} = N. By Proposition 3.7(3) S lies in a unique maximal Γ-subgroup P. Then by Theorem 8.22, N_{G}(S) ⊆ N_{G}(P) ≤ N_{G}(Ω(Z(P))^{∘} = N_{G}(N), a contradiction.

Hence Z^{∘} ≠ N and Theorem 8.3 follows from Theorem 8.2(3) □

Theorem 8.4 Suppose rank(Γ) = 2, S ∈ Γ^{∗} and |ΘS| > 2. Then there is N ∈ ΘS such that V/C_{V} (⟨Γ_{N}⟩) is a natural SL_{2}-module for Γ_{N}, where V = Ω(Z(N))

Proof. Let N ∈ ΘS. By Proposition 3.7(3), S is the unique maximal Γsubgroup containing N_{S}(N). Hence N_{S}(N) ∈ Γ_{N}^{∗}. If N_{G}(N_{S}(N)) ≰ N_{G}(N) we are done by Theorem 8.3.

So we may assume that N_{G}(N_{S}(N)) ≤ N_{G}(N) for all N ∈ ΘS. In particular N_{S}(N_{S}(N)) ≤ N_{S}(N) and so N_{S}(N) = S. Thus S ∈ Γ_{N} and N_{G}(S) ≤ N_{G}(N).

Since |ΘS| ≥ 3 there exists N ∈ ΘS with N ≠ J(S)^{∘} and N ≠ ΩZ(S)^{∘}. Thus by Theorem 8.2 J(S) ≰ N and N ≠ Z^{∘}. So Theorem 8.4 follows from Theorem 8.2(c) □

The following theorem deals with a situation which had been considered more detailed for finite groups in [3].

Theorem 8.5 Let rank(Γ) = 2, S ∈ Γ^{∗} and M, N ∈ ΘS with M ≠ N. Assume there is P ∈ Γ_{M}^{∗} ∩ Γ_{N}^{∗} with

(^{∗}) Z ∩ Z^{g} = 1 for all g ∈ G\N_{G}(P)

where Z := Ω(Z(J(P))). Then N is a natural SL_{2}-module for Γ_{N}. Moreover P = MN and P is of nilpotency class 2.

Proof. For L ∈ {M, N} set V_{L} := Ω(Z(L)). As ⟨M, N⟩ ⊆ P ∩ S and rank(Γ) = 2 we get P ⊆ S by Proposition 3.7(4).

Since rank(Γ) = 2, ⟨M, N⟩ ∉ {M, N}. Thus by Lemma 3.2(2),

Γ_{L} ∩ ΓN_{G}(P) = ΓP.

Suppose that J(P) ⊆ L. Then V_{L} ⊆ Z and so by (^{∗}) N_{G}(L) ⊆ N_{G}(P). Thus Γ_{L} ⊆ Γ_{L} ∩ ΓN_{G}(P) = ΓP and L = P, a contradiction.

Thus J(P) ⊈ L. Let X = Ω(Z(P)). Then

1 ≠ X ≤ Z ∩ V_{L} ⊆ V_{L}.

By (^{∗}) N_{G}(X) ⊆ N_{G}(P), and so Γ_{L}∩Γ_{X} ⊆ Γ_{L}∩ΓN_{G}(P) and R(Γ_{L}∩Γ_{X}) = P. Thus by Lemma 8.1(4), C_{S}(V_{L}) ⊆ L. So we can apply Theorem 8.2(c) and V_{L}/C_{V_{L}}(⟨Γ_{L}⟩) is a natural SL_{2}-module for Γ_{L}.

Let {K, L} = {M, N}. By Theorem 6.9 KL = P = AL and L = C_{P} (V_{L}) for all A ∈ A(P) with A ≰ L. Moreover X = C_{V_{L}}(A) = V_{L} ∩ V_{K} = C_{V_{L}}(⟨Γ_{L}⟩)[V_{L}, A]. Since X ∩ X^{g} = 1 for g ∈ G\N_{G}(P) we conclude C_{V_{L}}(⟨Γ_{L}⟩) = 1. Thus by Proposition 6.5(3)

q := μ(X) = \frac{1}{2}μ(V_{L}) = μ(X) = μ(A/A ∩ L).

In particular, μ(V_{L}(A ∩ L) = μ(A) and so V_{L}(A ∩ L) ∈ A(L) ∩ A(P). Using this and symmetry in K and L, A(K)∪A(L) ⊆ A(P). Suppose that A(K) = A(L), then Γ_{L} ∪ Γ_{K} ⊆ Γ_{J(K)}. Thus by Corollary 2.7 R(Γ_{J(K)}) is closed and contained in L ∩ K, a contradiction to rank(Γ) = 2. So A(K) ≠ A(L) and interchanging K and L if necessary we assume A(K) ⊈ A(L).

So we can choose A ∈ A(K).

Suppose for a contradiction that [V_{K}, V_{L}] = 1. Then V_{K}V_{L} ≤ K ∩ L. As AL = P ⊈ C_{G}(V_{K}) we get [V_{K}, L] = X. Let W ∈ V_{K}^{⟨Γ_{L}⟩}\{V_{K}}. Then [V_{K}, W] ⊆ (V_{K} ∩ V_{L}) ∩ (W ∩ V_{L}) = 1. Since A normalizes [A ∩ W, L] and [A∩W, L] ≤ W ∩V_{L} we get [A∩W, L] = 1 and so A∩W ≤ A∩W ∩V_{L} = 1. Now μ(W) = q = 2μ(A/A∩L) implies μ(W (A∩L)) > μ(A). Thus [A∩L, W] ≠ 1 and so by Proposition 6.5(3) applied to (A ∩ L)V_{L}, [A ∩ L, W] = V_{L} ∩ W and μ(A)/μ(C_{A}(W)) = μ(W) = 2μ(X). Thus C_{A}(W)W ∈ A(L).

Let a ∈ A\L. Since W centralizes C_{A}(W), also W^{a} centralizes C_{A}(W). Since [W, W^{a}] ⊆ V_{L} ∩ W ∩ W^{a} = 1 we conclude that C_{A}(W)W W^{a} is a decomposable abelian Γ-subgroup. Since C_{A}(W)W ∈ A(P), C_{A}(W)W = C_{A}(W)W^{a}. Thus

V_{L} ∩ W = [A ∩ L, W] = [A ∩ L, W^{a}] = V_{L} ∩ W^{a},

a contradiction to V_{L} ∩ W ∩ W^{a} = 1.

Therefore [V_{K}, V_{L}] ≠ 1 and so V_{N} ≰ M and V_{M} ≰ N.

Let h ∈ ⟨Γ_{M}⟩\N_{G}(P). Note that M = V_{M}(N∩N^{h}). Hence ΩZ(N∩N^{h}) ≤ ΩZ(M) = V_{M}. But V_{N} centralizes N and so

ΩZ(N ∩ N^{h}) ≤ C_{V_{M}} (V_{N}) ∩ C_{V_{M}} (N^{h}) = V_{M} ∩ V_{N} ∩ V_{N}^{h} = 1.

By the assumptions of this section, Γ is strong and so N ∩ N^{h} = 1. Thus M = V_{M}, N = V_{N} and P = V_{M}V_{N} = MN. Now P′ = X = M ∩ N and P has class 2. □

Theorem 8.6 Suppose that Π is a G-invariant subset of Θ such that

(i) ⋂_{g∈G} A^{g} = 1 for all A ∈ Π.

(ii) If S ∈ Γ^{∗} with |ΘS ∩ Π| ≥ 2, then |ΘS| = 2.

(iii) Whenever X, Y ∈ Π with X ∈ Y^{G} and X ≠ Y then R(Γ⟨X, Y ⟩) ∈ Π.

Let Π_{p} be an arbitrary orbit for G on Π and define Π̌_{p} = {R(Γ⟨A, B⟩) | A, B ∈ Π_{p} with A ≠ B}. Then

(1) ˇ Π̌_{p} = Π_{p}.

(2) Π_{p} is the set of points, Π̌_{p} is the set of lines of a projective Moufang plane π and ⟨Π_{p}⟩ = ⟨Π̌_{p}⟩ induces the group generated by all the elations on π.

(3) C_{G}(π) ≤ C_{G}(⟨Π_{p}⟩).

We remark that using knowledge of the automorphism group of a Moufang plane it should not be to difficult to show that G only has two orbits on Π.

Proof. From (i) we get

(1.) N ⋬ G for all N ∈ Π.

We say X, Y ∈ Π are incident if X ≠ Y and ⟨X, Y ⟩ ∈ Γ. We show next

(2.) If X, Y are incident then X ∈ Γ_{Y} and Y ∈ Γ_{X}.

Indeed by (ii) Θ⟨X, Y ⟩ = {X, Y } and so X and Y are normal in ⟨X, Y ⟩.

For X, Y ∈ Π with X ≠ Y write X̂Y := R(Γ⟨X, Y ⟩).

(3.) F ≮ E for all E, F ∈ Π.

Otherwise let g ∈ G \ N_{G}(E). Then |ΘE{ \atop EE^{g}}| = 2 and so by (iii) F = { \atop EE^{g}} = F^{g}. Since \{F \} = ΘE \ \{E\}, F^{g} = F for all g ∈ G. Thus F ⊴ G, a contradiction to (1.), proving (3.)

Let A ∈ Π. By (1.) there exists B ∈ A^{G} with A ≠ B. By (iii) ÂB ∈ Θ. Let D = ÂB ∈ Γ.

Suppose that A and B are incident. Then ⟨A, B⟩ ∈ Γ and D = ⟨A, B⟩. By (ii), |ΘD| = 2. Since A ≠ B we may assume A = D. Hence B ≤ A. Since A and B are conjugate μ(A) = μ(B) and we conclude that A = B, a contradiction.

We proved

(4.) No two distinct conjugate elements of Π are incident.

Suppose C ∈ Π is incident with A and B, Then ⟨A, B⟩ ≤ Γ_{C} and so C ≤ D. Since ΘAD = {A, D} and A ≰ D, C = D. Thus

(5.) ÂB is the unique element of Π incident with A and B.

Let Σ(A) = Π ∩ Γ_{A} \ {A}, the set of elements of Π incident with A. Let Ξ_{A} := ⋃{ΓAE | E ∈ Σ(A)}.

(6.) Let A < X ∈ Ξ_{A}. Then there exists a unique X^{∗} ∈ Γ^{∗} with X ≤ X^{∗} and a unique E ∈ Σ(A) with Θ(X^{∗}) = {E, A} and X ≤ AE.

Pick E ∈ Σ(A) with X ≤ EA and P ∈ Γ^{∗} with EA ≤ P. Suppose there exists Q ∈ Γ^{∗} with X ≤ Q but Q ≠ P. Choose such a Q with P ∩Q maximal. Then by Proposition 2.11(1), P ∩ Q is closed. Since A ≤ P ∩ Q, Lemma 3.3 implies that P ∩ Q is large. So P ∩ Q ∈ ΘP. But ΘP = {A, E} and we conclude that E = P ∩ Q, but then A < E, a contradiction to (3.).

(7.) Ξ_{A} ≤ Γ_{A}, Ξ_{A} is an NSS of rank 1 for N_{G}(A), Ξ_{A}^{∗} = {AE | E ∈ Σ(A)} and R(Ξ_{A}) = A.

Clearly Ξ_{A}^{∗} = {AE | E ∈ Σ(A)}, and (Suba) and (Subc) are fulfilled. Let X, Y ∈ Ξ_{A} with ⟨X, Y ⟩ ∈ Γ. We need to show that ⟨X, Y ⟩ ∈ Ξ_{A}. If X ≤ A or Y ≤ A this is obvious. We may assume A < X and A < Y. Pick Q ∈ Γ^{∗} with ⟨X, Y ⟩ ≤ Q. Let E, F ∈ Σ(A) with X ≤ EA and Y ≤ F A. By (6.), Θ(Q) = {A, E} = {A, F } and so E = F. Thus ⟨X, Y ⟩ ≤ EA and ⟨X, Y ⟩ ∈ Ξ_{A}.

By Theorem 3.1 it remains to show that |Ξ_{A}^{∗}| > 1. Otherwise we conclude that Σ_{A} = {K} for some K, and K = ÂA^{g} for all g with A ≠ A^{g} and then K ⊴ G, a contradiction to (1.). This completes the proof of (7.).

(8.) Let 1 ≠ X ∈ ΓA. Then N_{Γ}(X) ⊆ N_{Γ}(A).

Suppose not and pick Q ∈ N_{Γ}(X) with Q ≰ N_{G}(A). Pick g ∈ Q with A ≠ A^{g}. Let E ∈ Σ(A).

Suppose that X ≰ L for some L ∈ Σ(A).

If A^{g} is incident with L, then both A and A^{g} are incident with L and so L = ÂA^{g}. By (6) applied to L in place of A, X ≤ A ∩ A^{g} ≤ L.

Thus neither A^{g} nor A^{g^{−1}} are incident with L. In particular L ≠ L^{g}. Note that X normalizes L and L^{g} and so also F := L̂L^{g}. By Lemma 3.2(1) applied to L in the place of N, we get XLF ∈ Γ. By X ≰ L and (6.), XL lies in a unique maximal Γ-subgroup of G. Hence ⟨AL, XLF ⟩ ∈ Γ and (ii) implies A = F. Thus L^{g} is incident with A and so A^{g^{−1}} is incident with L, a contradiction.

Thus X ≤ L for all L ∈ Σ(A). Let Y := ⋂ Σ(A). Then X ≤ Y and so Y ≠ 1. Since N_{G}(A) ≤ N_{G}(Y) we have N_{G}(Y) ≰ N_{G}(L).The claim we just proved applied to (Y, L) in place of (X, A) yields Y ≤ K for all K ∈ Σ(L) and all L ∈ Σ(A). Thus Y ≤ A^{g} for all g ∈ G and (i) implies Y = 1, a contradiction.

(9.) Let E ∈ Σ(A) and V_{A} = Ω(Z(A)). Then [C_{G}(V_{A}), EA] ≤ A.

Let F ∈ Σ(A) and put X = Ω(Z(AF)). Since A and F are large, X ≤ A ∩ F. Since Γ is strong, X ≠ 1. By (8.) applied to F in place of A, N_{G}(X) ≤ N_{G}(F). Since X ≤ V_{A} we conclude that C_{G}(V_{A}) ≤ N_{G}(F) for all F ∈ Σ(A). So by (6.) C_{G}(V_{A}) ≤ N_{G}(P) for all P ∈ Ξ_{A}^{∗}. Define U := ⋂{N_{EA}(P) | P ∈ Ξ_{A}^{∗}}. Then U ∈ Ξ_{A} and by Proposition 2.3(8), U = R(Ξ_{A}) = A. But [C_{G}(V_{A}), EA] ≤ C_{G}(V_{A}) ∩ EA ≤ U and so (9.) holds.

(10.) Let E ∈ Σ(A). Then J(EA) ≰ E ∩ A.

If J(EA) ≤ E ∩ A then J(E) = J(A). Then N_{G}(E) ≤ N_{G}(J(A)) and so N_{G}(J(A)) ≰ N_{G}(A), a contradiction to (8.)

By (10.) and interchanging A and E if necessary

(11.) we can choose A ∈ Π and E ∈ Σ(A) with J(EA) ≰ A.

By (8.) and Proposition 5.5 there exists a non-trivial offender in EA on V_{A}. Let H_{A} =: ⟨Ξ_{A}⟩. Note that C_{V_{A}}(H_{A}) ≤ C_{V_{A}}(E) ≤ E and so by (8.) C_{V_{A}}(H_{A}) = 1. We conclude that the Hypothesis of Theorem A holds for N_{G}(A), Ξ_{A}, V_{A} and EA. So V_{A} is a natural SL_{2}-module for Ξ_{A}. In particular,

(12.) E acts transitively on Σ(A) \ {E}, N_{G}(A) acts transitively on Σ(A) and C_{EA}(V_{A}) = A.

By (8.), V_{A} ≰ E. Let X ∈ A(AE). By Proposition 6.5(3), (X ∩ A)V_{A} ∈ A(AE) and we reestablish symmetry in A and E. Let R := ⟨V_{E}, V_{E}^{h}⟩ for some h ∈ H_{A} with V_{E} ≠ V_{E}^{h}. Then A ∩ E ∩ E^{h} ≤ C_{A}(R) and so by (8.), A ∩ E ∩ E^{h} = 1. It follows that N = V_{N} and so N is a natural SL_{2}-module for Ξ_{N} for all N ∈ Π.

It follows from (12.) that ÂA^{g} ∈ E^{G} and { \atop EE^{g}} ∈ A^{G} for all g ∈ G. Thus A^{G} and E^{G} form a projective plane.

By (12.) we have (N, N)-transitivity. Then [4, page 130] shows that we have got a projective Moufang plane.

Let C_{A} be the kernel of the action of G on A^{G}. Then clearly C_{A} also acts trivially on E^{G}. Moreover [C_{A}, A] ≤ A ∩ C_{A} ≤ C_{A}(E) ≤ A ∩ E and (8.) implies [C_{A}, A] = 1. □

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