Large Abelian Subgroups of Finite p-Groups

George Glauberman

August 19, 1997

1 Introduction

Let p be a prime and S be a finite p-group. Define d(S) to be the maximum of |A| as A ranges over the abelian subgroups of S, and A(S) to be the set of all abelian subgroups A for which |A| = d(S). Let J(S) be the subgroup of S generated by A(S).

In 1964, John Thompson introduced [T2] a subgroup similar to J(S) and used it to obtain an improved version of his first normal p-complement theorem [T1]. Since then, several variants of J(S) have been used to obtain related results. Most of their proofs have required a further result of Thompson, the Replacement Theorem ([Gor, p.273], [HB, III, p.21]):

Suppose A ∈ A(S), B is an abelian subgroup of S normalized by A, and B does not normalize A. Then there exists A^{∗} ∈ A(S) such that

A^{∗} ≤ AB, A ∩ B < A^{∗} ∩ B, and A^{∗} normalizes A.

It would be interesting to extend this result by allowing B to have nilpotence class 2 instead of necessarily being abelian. This cannot be done if p = 2 (Example 4.2), but perhaps it is possible for p odd. (It was done by the author ([Gor, p.274]; [HB, III, p.21]) for the special case in which p is odd and [B, B] ≤ A.) However, there is an application of Thompson’s Replacement Theorem that can be extended when p is odd and B has class 2, and this extension is our first main result. In order to state it, we require a definition.

Let S be a central series

1 = Z_{0} ≤ Z_{1} ≤ ⋯ ≤ Z_{n} = S

of a p−group S. For any abelian subgroups A, A^{∗} of S, define A <_{S} A^{∗} if

|A| = |A^{∗}|
|A ∩ Z_{i}| ≤ |A^{∗} ∩ Z_{i}|, for i = 1, 2, …, n,
and A ∩ Z_{i} < A^{∗} ∩ Z_{i}, for some 1 ≤ i ≤ n

Extending the notation of [Gor, pp.19,274] slightly, we define [X, Y, Z] to be [[X, Y], Z], for subsets or elements X, Y, Z of S, and define inductively

[T, u; 1] = [T, u]

and

[T, u; n + 1] = [ [T, u; n], u], for T ≤ S, u ∈ S, and n ≥ 1.

\underline{Theorem 1} Suppose p is an odd prime, S is a finite p-group, A ∈ A(S), and B is a subgroup of S that does not normalize A. Assume that

(i) B has nilpotence class at most 2 and is normalized by A, or

(ii) p ≥ 5, and A⊲ < A^{B} >, and [A, u; 3] = 1, for every u ∈ B. Then there exists A^{∗} ∈ A(S) such that

(a) A^{∗} ≤ < A^{B} > and A^{∗} normalizes A,

(b) for every central series S of S, A <_{S} A^{∗}, and

(c) if A normalizes B, then |A^{∗} ∩ B| ≥ |A ∩ B|.

\underline{Corollary} Suppose p is an odd prime, S is a finite p-group, and S is a central series of S. Let A be a maximal element of A(S) under <_{S}.

Then A is normalized by every subgroup B of S for which

(i) B has nilpotence class at most 2 and is normalized by A, or

(ii) p ≥ 5, A⊲ < A^{B} >, and [A, u; 3] = 1, for every u ∈ B.

Theorem 1 is proved by reducing to the case in which A itself is contained in a normal subgroup Â of class at most 3 (class at most 2, if p =3) in a subgroup of < A^{B} >. Then, by a result of M. Lazard, one may regard Â as a Lie ring. As in the proof of the Replacement Theorem, one explicitly constructs the subgroup A^{∗}. The ideas in the proofs of Lemma X.3.1 and Corollary X.3.2, pp. 19-20, of [HB, III] (applications of Thompson’s Replacement Theorem) are used to prove that A <_{S} A^{∗}.

We say that an abelian subgroup A of S is irreplaceable with respect to class 2 subgroups of S if A satisfies the conclusion of the corollary of Theorem 1. We write A_{∈}(S) for the set of all such subgroups in A(S). Thus, A_{∈}(S) is not empty.

Although there are examples in which no member of A(S) is normal in S (Example 4.1), our next main result and Corollary 1 show that, if p ≥ 5, every subgroup in A_{∈}(S) is close to being a normal subgroup in some sense:

\underline{Theorem 2} Suppose p is a prime and p ≥ 5, S is a finite p-group and A, A_{0} ∈ A_{∈}(S). Then

(a) A and A_{0} normalize each other,

and

(b) A is normalized by every subgroup B of S such that [A, u; 3] = 1 for every u ∈ B.

\underline{Corollary 1} Suppose p ≥ 5 and A ∈ A_{∈}(S) . Then

(a) A^{x} and A^{y} normalize each other, for every x, y ∈ S.

(b) ∏_{x∈S} A^{x} ⊲ S and A ⊲ ∏_{x∈S} A^{x}.

and

(c) [S, A, A, A] = 1.

These results allow us to define a characteristic subgroup of S, contained in J(S), in the next corollary. A similar characteristic subgroup is used in Theorems 3 and 4.

\underline{Corollary 2} Suppose p ≥ 5. Let H =< A | A ∈ A_{∈}(S) >. Then

(a) H is a characteristic subgroup of S.

(b) H = ∏_{A∈A_{∈}(S)} A.

( c ) A ⊲ H for every A ∈ A_{∈}(S).

We do not know whether analogous results are valid for p = 2 or p = 3, possibly for A maximal under <_{S} for some central series S, although we are dubious about p = 2. We have not attempted to generalize Theorem 1 to the case in which B has nilpotence class 3 or more, or [A, B; k] = 1 for some k ≥ 4. Some results of this sort for elementary abelian subgroups of maximal order in S are obtained by different methods in [AG].

Now let us consider the situation in which S is contained in a finite group G, not necessarily a p-group. Assume that

(1.1) S is a Sylow p-subgroup of G

and

(1.2) for T = O_{p}(G), C_{G}(T) ≤ T.

In 1968, using Thompson’s Replacement Theorem and other methods, we proved [G1, I] that

if J(S) ⋪G, then there exists u ∈ S − T such that

[X, u; 6] ≤ Y, for every chief factor X/Y of G such that X ≤ T.

Thompson asked whether, for some suitable characteristic subgroup in place of J(S), one could obtain the stronger condition

[X, u; 2] ≤ Y.

In 1970, we improved our 1968 result ([G2, p.35]; [S, p.231]; [HB, III, p.63]) to

[X, u; 4] ≤ Y if K_{∞}(S) ⋪G or K^{∞}(S) ⋪G,

for certain characteristic subgroups K_{∞}(S), K^{∞}(S) of S. Now, by using Theorems 1 and 2, we have found a characteristic subgroup, ZJ^{N}(S), for which we can obtain [X, u; 3] ≤ Y, which is closer to Thompson’s condition:

\underline{Theorem 3} Let p be a prime. Assume (1.1) and (1.2), and suppose that ZJ^{N}(S) ⋪G. Then

(a) G is not p-stable

and

(b) there exists u ∈ S − T such that [X, u; 3] ≤ Y, for every chief factor X/Y of G such that X ≤ T.

We note that if S > 1, then K_{∞}(S), K^{∞}(S), ZJ^{N}(S) ≥ Z(S) > 1. The definition of ZJ^{N}(S) is given in Definitions 6.1, 6.2 and 6.11.

Theorem 3 yields the following consequence:

\underline{Theorem 4} Suppose p ≥ 5, S is a Sylow p-subgroup of a group G, and Z = ZJ^{N}(S). Then

G/O^{p}(G) ≅ N_{G}(Z)/O^{p}(N_{G}(Z)).

This is proved from Theorem 3 by the same argument that yields the analogous results for K_{∞}(S) and K^{∞}(S) ([S, pp. 242-3]; [HB, III, p.65]).

Our 1970 result was applied by D. Holt [Hol] and M. Miyamoto [M] to prove results on cohomology. Because of the improvement from

[X, u; 4] ≤ Y in 1970

to

[X, u; 3] ≤ Y in Theorem 3,

their proofs remain valid almost word-for-word with the substitution of some smaller numbers. First, Holt’s results now assert:

Suppose p ≥ 7 and S is a Sylow p-subgroup of a group G. Then the p-primary parts of the Schur multipliers of G and N_{G}(ZJ^{N}(S)) are isomorphic. Moreover, for any G-module M on which G acts trivially, H^{2}(G, M) and H^{2}(N_{G}(ZJ^{N}(S)), M) have isomorphic p-primary parts.

(In pp.196 and 198 in [Hol], replace “[X, g^{(4)}] ≤ Y ” with “[X, g^{(3)}] ≤ Y ” and “[U, g^{(7)}] ≤ V ” with “[U, g^{(5)}] ≤ V.” )

Second, Miyamoto’s Theorem B immediately gives:

Suppose S is a Sylow p-subgroup of a group G, m is an integer, m ≥ 2, and p ≥ 3 + 8 · 6^{m−2}. Then the restriction map induces an isomorphism of cohomology

H^{m}(G, M) ≅ H^{m}(N_{G}(ZJ^{N}(S)), M)

for every trivial p-primary G-module M.

The main part of the above-mentioned reduction of Theorem 1 to a special case is given in Section 2. In Section 3, Lazard’s result and a result on Lie rings are used to finish the proof of Theorem 1. A short proof of Theorem 2, and some counterexamples, appear in Section 4. Theorem 3 is proved in Sections 5 and 6.

All of these results stem originally from a failed counterexample to Theorem 1. There is a natural choice for the subgroup A^{∗} in Theorem 1 (Remark 3.3) but the author constructed some examples in which this choice is not abelian. However, our attempt to show that no subgroup A^{∗} satisfied the theorem met with unexpected obstacles that eventually resolved themselves into Theorem 3.2 and Theorem 1.

Our notation is standard and generally taken from [Gor], except that we write A ≤ B to mean that A is a subgroup of B and A < B to mean that A ≤ B and A ≠ B. In particular, we usually denote the commutator subgroup [H, H] of a group H by H′.

All groups in this paper are finite, except possibly those appearing as additive groups of algebras. In addition,

throughout this paper p denotes a fixed but arbitrary prime, and S denotes a fixed but arbitrary finite p-group.

Often p will be assumed to be odd.

During the preparation of this paper, the author enjoyed the support of a National Security Agency grant and the hospitality of the Institute for Advanced Study and the Drive Train Research Institute. He thanks each of these institutions. He also thanks the referee for a very careful reading of the original version and many corrections.

2 Reduction for Theorem 1

We now start toward the proof of Theorem 1. We will sometimes use the following lemma without explicitly mentioning it.

\underline{Lemma 2.1} [H, pp.254,257-8,292] Let G be a group generated by a subset X, and let

G = G_{1} ≥ G_{2} ≥ ⋯

be the lower central series of G.

(a) (E. Witt) For x, y, z ∈ G,

[x, y^{−1}, z]^{y} [y, x^{−1}, x]^{z} [z, x^{−1}, y]^{x} = 1

(b) (P. Hall: Three Subgroups Lemma) For H, K, L ≤ G and N ⊲ G,

if [H, K, L] ≤ N and [K, L, H] ≤ N, then [L, H, K] ≤ N.

( c ) For k = 1, 2, 3, …,

G_{k} = < [x_{1}, x_{2}, …, x_{k}]^{y} | x_{1}, …, x_{k} ∈ X and y ∈ G >
= < [x_{1}, x_{2}, …, x_{k}], G_{k+1} | x_{1}, …, x_{k} ∈ X >

(d) Suppose k, a_{1}, …, a_{k}, n_{1}, …, n_{k}, n are all integers and k ≥ 2, and n_{1}, …, n_{k} ≥ 1, and n = n_{1} + … + n_{k}. Let x_{i} ∈ G_{n_{i}} for i = 1, 2, … k. Then

[x_{1}^{a_{1}}, …, x_{k}^{a_{k}}] ≡ [x_{1}, …, x_{k}]^{a_{1}...a_{k}} (modulo G_{n+1}).

\underline{Lemma 2.2} Suppose A ≤ S and b ∈ S, and let

T =< A, b > and Â =< A^{T} > .

For each positive integer k, let R_{k} =< A^{b^{i}} | 0 ≤ i ≤ k >. Then (a) Â =< A^{b^{k}} | k an integer >,

(b) for each positive integer k,

R_{k} =< a, [a, b; i] | a ∈ A, 1 ≤ i ≤ k >,

(c) if k ≥ 2, and [A, b; k] = 1, then Â = R_{k−1} and, if in addition A is abelian and A ⊲ Â, then Â has nilpotence class at most k,

(d) if [A, b; 3] = 1, then [A, < b >; 3] = 1,

and

(e) if [A, b; 2] = 1, then [A, < b >; 2] = 1.

\underline{Proof:} (a) Let A^{∗} =< A^{b^{k}} | k ∈ ℤ >. Then A ≤ A^{∗} ≤ < A^{T} >= Â and N_{T} (A^{∗}) ≥< b, A >= T. So

Â =< A^{T} >≤< A^{∗T} >= A^{∗} ≤ Â, and A^{∗} = A.

(b) For every x ∈ T, x^{b} = x[x, b]. This proves (b) for k = 1. Since R_{k+1} =< R_{k}, R_{k}^{b} > for every k ≥ 1, the general case follows by induction.

( c ) Here, by (b), we have R_{k−1} = R_{i} for every i ≥ k − 1. So

R_{k−1}^{b} ≤ R_{k} = R_{k−1}; b ∈ N_{T} (R_{k−1}); and N_{T} (R_{k−1}) ≥< A, b >= T.

Arguing as in the proof of (a), we obtain that Â = R_{k−1}.

Suppose A is abelian and A ⊲ Â. Then A, A^{b}, ⋯, A^{b^{k−1}} are abelian normal subgroups of Â that generate Â. By Lemma 2.1, to complete the proof it will suffice to show that, for any x_{1}, …, x_{k+1} in the set-theoretic union of these subgroups,

[x_{1}, x_{2}, …, x_{k+1}] = 1.

However, this is obvious because at least two of the elements x_{i} must lie in the same abelian normal subgroup A^{b^{j}} of Â.

(d) Assume [A, b; 3] = 1. Let D_{1} = [A, < b >], D_{2} = [A, b], E = [A, b; 2], and R =< D_{2}, b >. Then E = [D_{2}, b], which is normalized by D_{2} (as in the proof of [Gor], Theorem 2.2.1(iii), p.18), and E is centralized by b. Hence R normalizes E. As E ≤ R, we have E ⊲ R.

We claim that D_{1} ≤ R. Since b has finite order,

D_{1} = < [a, b^{i}] | a ∈ A, i ≥ 1 > .

So we will show by induction on i that [a, b^{i}] ∈ R for all a and i. The proof is trivial for i = 1. Assume it is proved for some i ≥ 1. Then

[a, b^{i}] ∈ R = < D_{2}, b > and [a, b^{i+1}] = [a, b][a, b^{i}]^{b} ∈ R,

as desired. Thus

D_{1} ≤ R = < D_{2}, b > .

Now, E ⊲ R and, modulo E, the coset of b centralizes D_{2}E/E and R/E. Therefore,

[A, < b >; 2] = [D_{1}, < b >] ≤ [R, < b >] ≤ E = [A, b; 2] ≤ C_{R}(b),

and

[A, < b >; 3] ≤ [C_{R}(b), < b >] = 1,

as desired.

(e) The proof is similar to the proof of (d), but easier.

Q.E.D.

We now begin to reduce the proof of Theorem 1 to the case in which A is contained in a normal subgroup < A, B > of nilpotence class 2.

\underline{Theorem 2.3} Suppose p is odd, A ∈ A(S), and B is a subgroup of S of nilpotence class at most 2 normalized by A. Assume that B does not normalize A.

Then there exists b ∈ B − N(A) such that, for

T =< b, A > and Â =< A^{T} >=< A^{x}|x ∈ T >,

Â has nilpotence class 2.

\underline{Proof}

We may assume that B is minimal subject to the hypothesis of the theorem. Take b ∈ B − N(A) . Since S is nilpotent, B′ = [B, B] < B and [B, A] < B.

As A normalizes B′ and [B, A],

(2.1) B′[B, A] normalizes A

Let

T =< b, A > and Â =< A^{T} > .

Then B ∩ T is normal in T and hence is normalized by A. By the minimal choice of B,

B = B ∩ T ≤ T and T =< b, A > = BA.

Therefore, by (2.1),

T′ = [T, T] = B′[B, A] ≤ N_{B}(A). (2.2)

Let the lower central series of T be

T = T_{1} ≥ T_{2} ≥ T_{3} ≥ · · · .

\underline{Step 1}: We have T_{5} ≤ Z(T) and T_{6} = 1.

\underline{Proof:} From (2.1),

T_{3} = [T′, T] ≤ [N_{B}(A), BA] ≤ B′(A ∩ B),
T_{4} = [T_{3}, T] ≤ [B′(A ∩ B), BA] ≤ B′,
T_{5} ≤ [B′, BA] = [B′, A] ≤ B′ ∩ A ≤ Z(B) ∩ A ≤ Z(T) (by (2.1)),
T_{6} ≤ [Z(T), T] = 1.

Recall that, in a group of odd order, each element u has a unique square root u^{\frac{1}{2}} , and u^{\frac{1}{2}} = u^{i} for some integer i. For any u, v ∈ T, we write u^{−v} for (u^{−1})^{v}.

\underline{Step 2}: For all a, a′ ∈ A,

(a) [b, a, b, a′] ∈ Z(T),

and (b) [b, a, a′, b] = [b, a′, a, b] ∈ Z(T).

\underline{Proof}:

Take any a, a′ ∈ A. Then, by (2.1),

[b, a, b, a′] ∈ [B′, A] ≤ B′ ∩ A ≤ Z(T),

which proves (a). Moreover,

[b, a] = b^{−1}b^{a}, [b, a, a′] = (b^{−1}b^{a})^{−1}(b^{−1}b^{a})^{a′} = b^{−a}bb^{−a′}b^{aa′}.

Similarly,

[b, a′, a] = b^{−a′}bb^{−a}b^{aa′}.

Since B/B′ is abelian and aa′ = a′a,

[b, a, a′] ≡ bb^{−a}b^{−a′}b^{aa′} ≡ [b, a′, a] (modulo B′)

As B′ ≤ Z(B), [b, a, a′, b] = [b, a′, a, b]. (2.3)

By several applications of Lemma 2.1 we obtain (modulo T_{5})

1 ≡ [ [b, a], a′, b] [a′, b, [b, a] ] [b, [b, a], a′]
≡ [b, a, a′, a] [ [b, a′], [b, a] ]^{−1}[b, a, b, a′]^{−1},

and

[b, a, a′, b] ≡ [b, a, b, a′] [[b, a′], [b, a] ] (modulo T_{5}). (2.4)

Interchanging a with a′ yields

[b, a′, a, b] ≡ [b, a′, b, a] [ [b, a], [b, a′] ] (modulo T_{5}). (2.5)

However,

[ [b, a], [b, a′] ] = [ [b, a′], [b, a] ]^{−1} (modulo T_{5}).

Therefore, (2.3), (2.4), (2.5) yield

[b, a, a′, b]^{2} ≡ [b, a, b, a′][b, a′, b, a] (modulo T_{5}).

By (a) and Step 1, [b, a, a′, b]^{2} ∈ Z(T). Since T has odd order, this and (2.3) yield (b).

\underline{Step 3}: We have T_{4} ≤ Z(T) and T_{5} = 1.

\underline{Proof}:

Let X be the set-theoretic union of {b} with A. By Lemma 2.1, since T =< b, A >=< X >,

T_{4} =< [x_{1}, x_{2}, x_{3}, x_{4}], T_{5} | x_{i} ∈ X for i = 1, 2, 3, 4 > (2.6)

By Step 1, T_{5} ≤ Z(T) (2.7)

Suppose y is a commutator [x_{1}, x_{2}, x_{3}, x_{4}] of the form in (2.6). We must show that y ∈ Z(T). We may assume that y ≠ 1. Then [x_{1}, x_{2}] ≠ 1. Hence, one of x_{1}, x_{2} is b, and the other is in A. From Lemma 2.1,

[x_{2}, x_{1}, x_{3}, x_{4}] ≡ [[x_{1}, x_{2}]^{−1}, x_{3}, x_{4}] ≡ [x_{1}, x_{2}, x_{3}, x_{4}]^{−1}
≡ y^{−1} (modulo T_{5}).

So, by (2.7), we may assume that

x_{1} = b and x_{2} ∈ A.

Now, [x_{1}, x_{2}] ∈ [B, A] ≤ N_{B}(A), by (2.1). Hence,

y ∈ [N_{B}(A), A, A] ≤ [A, A] = 1, if x_{3}, x_{4} ∈ A;
y ∈ [N_{B}(A), B, B] ≤ [B′, B] = 1, if x_{3} = x_{4} = b.

Therefore, one of x_{3}, x_{4} is b, and the other is in A. By Step 2, y ∈ Z(T), as desired.

By (2.6) and (2.7), T_{4} ≤ Z(T). Thus, T_{5} ≤ [Z(T), T] = 1.

Recall that Â =< A^{T} >=< A^{x}|x ∈ T >.

\underline{Step 4}: We have

(a) T_{4} ≤ Z(T) ≤ Z(Â);

(b) Â =< A^{b^{k}}|k ∈ ℤ >;

(c) if a ∈ A and a^{b} ≡ a (modulo Z(Â)), then a ∈ Z(Â); (d) Â =< a, [a, b], [a, b, b]|a ∈ A > ;

(e) Â′ ≤ Z(Â).

\underline{Proof:} (a) By Step 3, T_{4} ≤ Z(T) ≤ C_{S}(A) = A ≤ Â, so Z(T) ≤ Z(Â).

(b) Apply Lemma 2.2(a).

( c ) Here, let C = CÂ(a). Then A ≤ C ≤ Â, and for every integer k,

C ≥ CÂ(aZ(Â)) = CÂ((aZ(Â))^{b^{k}}) = (CÂ(aZ(Â)))^{b^{k}} ≥ A^{b^{k}}.

By (b), C ≥ Â.

(d) Apply Lemma 2.2(c)

(e) Here, it suffices to show that the third term of the lower central series of Â reduces to the identity group. We take the set of generators of Â given in (d). So we merely need to show that

[x_{1}, x_{2}, x_{3}] = 1

if each x_{i} has the form a, [a, b], or [a, b, b] for some a ∈ A.

By Step 3, T_{5} = 1. Therefore, we may assume that each x_{i} has the form a or [a, b], and that only one has the form [a, b]. Since [A, A] = 1, there exist a_{1}, a_{2}, a_{3} ∈ A such that

[x_{1}, x_{2}, x_{3}] = [[a_{1}, b], a_{2}, a_{3}] ∈ [[B, A], A, A] ≤ [N_{T} (A), A, A] = 1

or

[x_{1}, x_{2}, x_{3}] = [a_{1}, [a_{2}, b], a_{3}] ∈ [A, [B, A], a] = [[B, A], A, A] = 1.

This completes the proof of (e), of Step 4, and of the theorem.

Q.E.D

3 Proof of Theorem 1

The goal of this seection is to prove Theorem 1. However, most of our proof consists of results on Lie rings. Recall that a derivation of a Lie ring δ is an additive endomorphism of L such that

δ[u, v] = [δu, v] + [u, δv], for all u, v ∈ L.

We say that an additive group A is divisible by 2 if, for each u ∈ A, there exists a unique element v of A such that 2v = u. In this case, for an automorphism α of A such that

0 = A(α − 1)^{3} = {uα^{3} − 3uα^{2} + 3uα − u | u ∈ A},

we define

log α = (α − 1) − \frac{1}{2}(α − 1)^{2}.

Similarly, for an endomorphism γ such that γ^{3} = 0, we define

exp(γ) = 1 + γ + \frac{1}{2}γ^{2}.

Note that A is divisible by 2 if A is finite of odd order, or if A is a vector space over a field of characteristic other than 2.

The following appears to be a special case of a result that is well known, but hard to find. Related results are proved in [AG, Propositions 2.1, 2.4 and 2.5].

\underline{Lemma 3.1}. Suppose L is a Lie ring divisible by 2, α ∈ Aut L, and L(α − 1)^{3} = 0. Let δ = log(α). Assume that

L = 3L or [ L(α − 1), L(α − 1)^{2} ] = 0. (3.1)

Then δ is a derivation of L, and α = exp δ.

\underline{Proof}:

We have

δ = (α − 1) − \frac{1}{2}(α − 1)^{2} = −\frac{3}{2} + 2α − \frac{α^{2}}{2} ,

and it is easy to show that δ^{3} = 0, α = exp δ, and α^{2} = exp 2δ.

Take any u, v ∈ L. Then

δ[u, v] = −\frac{3}{2}[u, v] + 2[u, v]^{α} − \frac{1}{2}[u, v]^{α^{2}}
= −\frac{3}{2}[u, v] + 2[u^{α}, v^{α}] − \frac{1}{2}[u^{α^{2}}, v^{α^{2}}]
= −\frac{3}{2}[u, v] + 2[u + uδ + \frac{1}{2}uδ^{2}, v + vδ + \frac{1}{2}vδ^{2}]

− \frac{1}{2}[u + 2uδ + 2uδ^{2}, v + 2vδ + 2vδ^{2}].

By performing the obvious calculations, one obtains,

δ[u, v] = [u, vδ] + [uδ, v] − [uδ, vδ^{2}] − [uδ^{2}, vδ] − \frac{3}{2}[uδ^{2}, vδ^{2}]. (3.2)

Since α^{2} = exp 2δ, we may substitute 2δ for δ in (3.2). Then, dividing by 2 yields δ[u, v] = [u, vδ] + [uδ, v] − 4[uδ, vδ^{2}] − 4[uδ^{2}, vδ] − 12[uδ^{2}, vδ^{2}]. (3.3)

Subtracting (3.3) from (3.2) yields

0 = 3( [uδ, vδ^{2}] + [uδ^{2}, vδ] ) + \frac{21}{2}[uδ^{2}, vδ^{2}]. (3.4)

Assume first that L ≠ 3L. By hypothesis,

[ L(α − 1), L(α − 1)^{2} ] = 0.

It follows that

[ L(α − 1)^{2}, L(α − 1)^{2} ] = 0.

Since α = exp δ, easy calculations give

[Lδ^{2}, Lδ^{2}] = 0 and 0 = [Lδ, Lδ^{2}] = [Lδ^{2}, Lδ]. By (3.2), δ is a derivation.

Now assume L = 3L. Take any u, v ∈ L. Then u = 3u′ for some u′ ∈ L. Substituting u′ for u in (3.4) yields (since 3[u′δ, vδ^{2}] = [uδ, vδ^{2}] etc.)

0 = [uδ, vδ^{2}] + [uδ^{2}, vδ] + \frac{7}{2}[uδ^{2}, vδ^{2}]. (3.5)

Substituting uδ for u in (3.5) yields 0 = [uδ^{2}, vδ^{2}]. Then (3.5) yields 0 = [uδ, vδ^{2}] + [uδ^{2}, vδ].

Now (3.2) reduces to the assertion that δ is a derivation.

Q.E.D.

We write Z(L) for the center of a Lie ring L. \underline{Theorem 3.2}. Let A be an abelian subring of a Lie ring L, and δ be a derivation of L. Assume that (L1) A and L are divisible by 2, (L2) δ^{3} = 0, (L3) A = 3A or [δ^{2}L, δL] = 0, (L4) L = A + δA + δ^{2}A.

and (L5) A ≥ Z(L).

Let (L6) A_{1} = {a − 2δb | a, b ∈ A and δa − δ^{2}b ∈ Z(L)} and A^{∗} = A_{1} + δ^{2}A.

Then

(a) A^{∗} is a δ-invariant abelian subring of L that contains Z(L);

(b) for each a ∈ A,

a ∈ Z(L) if and only if δa ∈ Z(L);

and (c) if a ∉ Z(L), then either

(c1) δ^{2}a ∈ A^{∗} − A,

or (c2) δ^{2}a ∈ Z(L) ≤ A and δa ∈ A^{∗} − A.

In addition, the following conditions are satisfied:

(d) Suppose A < L, and

0 = Z_{0} ≤ Z_{1} ≤ Z_{2} ≤ … ≤ Z_{n} = L

is a series of additive subgroups of L such that δ(Z_{i}) ≤ Z_{i−1}, for i = 1, 2, …, n.

Then A ∩ Z_{i} = Z(L) ∩ Z_{i} < A^{∗} ∩ Z_{i} for some i, 1 ≤ i ≤ n.

(e) Suppose L is finite of odd order. Then |A^{∗}| = |A| and |A^{∗} ∩ M| ≥ |A ∩ M| for every δ-invariant additive subgroup M of L.

\underline{Remark 3.3}. It would seem more natural to use a − δb to define A_{1} in (L6), but this sometimes yields a non-abelian group. Perhaps (L6) may seem more natural in view of [AG, Proposition 2.2].

The proof of (e) can be easily adapted to prove the following variation: Suppose in Theorem 3.2 that L is a finite-dimensional Lie algebra over

a field of characteristic other than 2, A is an F -subalgebra of L, and δ is a linear transformation over F. Then A^{∗} is an F -subalgebra of L,

dim_{F} (A^{∗}) = dim_{F} (A).

and

dim_{F} (A^{∗} ∩ M) ≥ dim_{F} (A ∩ M)

for every δ-invariant F -subspace M of L.

\underline{Proof of Theorem 3.2}:

Consider any a, b ∈ A. Since A is abelian, [a, b] = 0. Therefore,

0 = δ[a, b] = [δa, b] + [a, δb], for a, b ∈ A; (3.6)

0 = δ^{2}[a, b] = [δ^{2}a, b] + 2[δa, δb] + [a, δ^{2}b], for a, b ∈ A. (3.7)

Since δ^{3} = 0, similar calculations for any u, v ∈ L yield

0 = δ^{3}[u, v] = [δ^{3}u, v] + 3[δ^{2}u, δv] + 3[δu, δ^{2}v] + [u, δ^{3}v]
= 3{[δ^{2}u, v] + [δu, δ^{2}v]}.

Substituting δu for u gives 0 = 3[δ^{2}u, δ^{2}v]. By (L3) and the method for the last part of the proof of Lemma 3.1,

[δ^{2}u, δv] + [δu, δ^{2}v] = [δ^{2}u, δ^{2}v] = 0, for all u, v ∈ L. (3.8)

(a) Now we check the properties of A^{∗}. It is easy to see that A_{1} and A^{∗} are additive subgroups of L, and that δ(Z(L)) ≤ Z(L). For a ∈ Z(L) and for b = 0, we have (by (L5))

a, b ∈ A and δa − δ^{2}b = δa ∈ Z(L).

Hence, a = a − 2δb ∈ A_{1} ≤ A^{∗}, and Z(L) ≤ A^{∗}.

Take any a, b ∈ A such that δa − δ^{2}b ∈ Z(L), as in the definition of A_{1} in (L6). Then

δ(a − 2δb) ≡ δa − 2δ^{2}b ≡ δ^{2}b − 2δ^{2}b ≡ −δ^{2}b (modulo Z(L)),

and −δ^{2}b ∈ δ^{2}A. Thus, δA_{1} ≤ δ^{2}A + Z(L) ≤ A^{∗}, and

δA^{∗} = δA_{1} + δ(δ^{2}A) ≤ A^{∗} + 0 = A^{∗}.

So A^{∗} is invariant under δ.

To complete the proof of (a), we must show that A^{∗} is an abelian subring of L. By (3.8), [δ^{2}A, δ^{2}A] = 0. Next, suppose a, b, c ∈ A and δa−δ^{2}b ∈ Z(L). Then

[a − 2δb, δ^{2}c] = [a, δ^{2}c] − 2[δb, δ^{2}c] (by (3.7))
= −2[δa, δc] − [δ^{2}a, c] − 2[δb, δ^{2}c]
= −2[δ^{2}b, δc] − [δ^{3}b, δc] − 2[δb, δ^{2}c]
= −2([δ^{2}b, δc] + [δb, δ^{2}c]) = 0 (by (3.8)).

Thus [A_{1}, δ^{2}A] = 0.

Finally, assume a, b, a′, b′ ∈ A and δa − δ^{2}b, δa′ − δ^{2}b′ ∈ Z(L). Then

[a − 2δb, a′ − 2δb′] = −2[a, δb′] − 2[δb, a′] + 4[δb, δb′] by (3.6)
= 2[δa, b′] + 2[b, δa′] + 4[δb, δb′]
= 2[δ^{2}b, b′] + 2[b, δ^{2}b′] + 4[δb, δb′] = 0 (by (3.7)).

Thus [A_{1}, A_{1}] = 0. Together with the previous paragraph, this shows that A^{∗} is an abelian subring of L, and completes the proof of (a).

(b) Let a ∈ A. If a ∈ Z(L), then δa ∈ Z(L).

Conversely, assume δa ∈ Z(L). Then δ^{2}a ∈ Z(L). Take any b ∈ A. Then [a, b] = 0 and, by (3.6) and (3.7),

[a, δb] = −[δa, b] = 0 and [a, δ^{2}b] = −[δ^{2}a, b] − 2[δa, δb] = 0. Since L = A + δA + δ^{2}A by (L4), a ∈ Z(L).

( c ) By (L1) and a short argument, (3.9) whenever x ∈ L and 2x ∈ Z(L), then x ∈ Z(L).

Suppose a ∈ A − Z(L). Then δ(δ^{2}a) = δ^{3}a = 0 ∈ Z(L). So, by (b),

either δ^{2}a ∈ Z(L) or δ^{2}a ∉ A.

If δ^{2}a ∉ A, then δ^{2}a ∈ δ^{2}A ≤ A^{∗} and δ^{2}a ∈ A^{∗} − A, which gives (c1).

Now assume δ^{2}a ∈ Z(L). By (b), either δa ∈ Z(L) or δa ∉ A. But (b) also gives that δa ∉ Z(L) because a ∈ A − Z(L). Hence, δa ∉ A (3.10) By (L1), there exists a unique element b ∈ A such that 2b = a. Then 2δ^{2}b = δ^{2}(2b) = δ^{2}a ∈ Z(L). By (3.9), δ^{2}b ∈ Z(L).

Let a′ = 0. Then a′, b ∈ A and δa′ − δ^{2}b = −δ^{2}b ∈ Z(L).

So −δa = −2δb = a′ − 2δb ∈ A_{1} ≤ A^{∗}. By (3.10), δa ∈ A^{∗} − A, which gives (c2).

(d) Here, A < L. By (L4) and (L5), L = A + δA + δ^{2}A, and A ≥ Z(L). Therefore,

L > A ≥ Z(L) and A > Z(L).

Take i as large as possible such that A ∩ Z_{i} = Z(L) ∩ Z_{i}; i exists, since Z_{0} = 0. Since Z_{n} = L, i < n. Then

A ∩ Z_{i+1} > Z(L) ∩ Z_{i+1}.

Take a ∈ (A∩Z_{i+1})−Z(L). Then δa, δ^{2}a ∈ Z_{i}. By (c), either δa ∈ A^{∗}−A or δ^{2}a ∈ A^{∗} − A. Either way, Z_{i} contains an element of A^{∗} − A, and

A^{∗} ∩ Z_{i} > Z(L) ∩ Z_{i} = A ∩ Z_{i}.

(e) Here, L is finite and M is a δ-invariant additive subgroup of L. Let

B = A ∩ M, Z = Z(L) ∩ M, B_{0} = {a ∈ B | δa ∈ δ^{2}B + Z},

B_{1} = {a − 2δb | a, b ∈ B and δa − δ^{2}b ∈ Z}, and B^{∗} = B_{1} + δ^{2}B.

In addition, let M̄ = M/Z. For each element x and additive subgroup K of M, let

x̄ = x + Z and K̄ = (K + Z)/Z.

Since δ(Z) ≤ Z, δ induces an endormorphism on M̄, which we also denote by δ, for convenience. Clearly,

\{ \begin{matrix}Z ≤ B_{0}; Z ≤ B_{1} ≤ B^{∗} ≤ A^{∗} ∩ M; \\ and, if M = L, then B^{∗} = A^{∗}. \end{matrix} (3.11)

We may now divide the calculation of |B^{∗}| into several steps:

\underline{Step 1} B̄_{1} ∩ δ^{2}B̄ = δB̄_{0}, and |B̄_{1} ∩ δ^{2}B̄| = |B̄_{0}|.

\underline{Proof}

Take x ∈ B̄_{1} ∩ δ^{2}B̄. Then x ∈ δ^{2}B̄ and, for some a, b ∈ B, δā = δ^{2}b̄ and x = ā − 2δb̄.

Then

δx ∈ δ^{3}B̄ = 0 and 0 = δx = δā − 2δ^{2}b̄ = −δā,

i.e. δa ∈ Z = Z(L) ∩ M. By (b), a ∈ Z(L) ∩ B = Z(L) ∩ M = Z, and ā = 0. Therefore,

δ(−2b̄) = ā − 2δb̄ = x ∈ δ^{2}B̄, whence − 2b ∈ B_{0} and x ∈ δB̄_{0}.

Thus, B̄_{1} ∩ δ^{2}B̄ ≤ δB̄_{0}.

Conversely, let c ∈ B_{0}. Since |L| is odd, c = 2d for some multiple d of c. Then δd̄∈ δ^{2}B̄ (and δ^{2}d̄= 0) because δc̄ ∈ δ^{2}B̄. Moreover, δc̄ ∈ B̄_{1} because

0, −d ∈ B, δ¯0 = 0 = δ^{2}(−d̄), and δc̄ = ¯0 − 2δ(−d̄).

Therefore, B̄_{1} ∩ δ^{2}B̄ = δB̄_{0}.

By (b), δ maps B̄_{0} injectively onto δB̄_{0}. So |B̄_{1} ∩ δ^{2}B̄| = |B̄_{0}|.

\underline{Step 2} Define a surjection Ψ : B̄ → δ^{2}B̄ by Ψ(x) = δ^{2}x, for x ∈ B̄. Then

|B̄_{1}| = |B̄_{0}||ker Ψ|.

\underline{Proof}

An element of B̄_{1} has the form ā − 2δb̄, where ā, b̄ ∈ B̄ and δā = δ^{2}b̄. Given ā ∈ B̄, such an element b̄ exists precisely when ā ∈ B̄_{0}, by the definition of B_{0}. Thus, |B̄_{1}| is the number of distinct elements ā − 2δb̄ for which ā ∈ B̄_{0}, b̄ ∈ B̄, and δā = δ^{2}b̄. (3.12)

When can we have ā′ − 2δb̄′ = ā − 2δb̄? Then

ā − ā′ = 2δ(b̄ − b̄′)
2δ^{2}(b̄ − b̄′) = δ(ā − ā′) = δ^{2}(b̄ − b̄′), and
0 = δ^{2}(b̄ − b̄′) = δ(ā − ā′).

By (b), ā − ā′ = 0, i.e. ā′ = ā. But then 0 = ā − ā′ = δ(2(b̄ − b̄′)), 2(b̄ − b̄′) = 0 (by (b)), b̄ − b̄′ = 0 by (3.9), and b̄′ = b̄. Thus, |B̄_{1}| is the number of distinct pairs (ā, b̄) as in (3.12).

Take any ā ∈ B_{0}. As mentioned just before (3.12), there exists some element b̄_{0} satisfying (3.12) for b̄ = b̄_{0}. For any other element b̄ of B̄, the following are equivalent:

b̄ satisfies (3.12); δ^{2}b̄ = δā = δ^{2}b̄_{0}; δ^{2}(b̄ − b̄_{0}) = 0; b̄ ∈ b̄_{0} + ker Ψ.

Thus, there are altogether |ker Ψ| choices of b̄ for any given element ā of B_{0}, and

|B̄_{1}| = |B̄_{0}| |ker Ψ|.

\underline{Step 3} Conclusion.

\underline{Proof} By Steps 1 and 2,

|B̄^{∗}| = |B̄_{1} + δ^{2}B̄| = |B̄_{1}||δ^{2}B̄| / |B̄_{1} ∩ δ^{2}B̄|
= |B̄_{0}| |ker Ψ| |Image Ψ| / |B̄_{0}| = |B̄|.

Now, B = A ∩ M ≥ Z(L) ∩ M = Z. By (3.11), Z ≤ B^{∗} ≤ A^{∗} ∩ M, and, if M = L, then B^{∗} = A^{∗} (and B = A ∩ M = A). So

|B^{∗}/Z| = |B̄^{∗}| = |B̄| = |(A ∩ M)/Z|

and

|A^{∗} ∩ M| ≥ |B^{∗}| ≥ |A ∩ M|

and the case of M = L yields |A^{∗}| = |A|. This completes the proof of (e) and of the theorem.

Q.E.D.

Now we may apply Theorem 3.2 to obtain part of Theorem 1. We first need a special case of an important theorem ([L] Theorem II.4.6, pp. 179-80; see pages 142-3 and results II.4.2 - II.4.3 and II.4.5) of M. Lazard that allows us to convert groups into Lie rings (and vice versa).

\underline{Theorem 3.4} (M. Lazard) Suppose S has nilpotence class at most p − 1. Then we may define operations + and [,] on S by canonical formulas using the group operation on S, under which S becomes a nilpotent Lie ring of the same class as S. A subset of S is a subgroup (respectively, normal subgroup) of S as a group if and only if it is a subring (respectively, ideal) of S as a Lie ring. Every automorphism of S as a group is an automorphism of S as a Lie ring, and vice versa.

Moreover, for x, y ∈ S and any integer n,

x^{n} in the group is equal to nx in the Lie ring.

and

xy = yx in the group if and only if [x, y] = 0 in the Lie ring.

For every abelian subgroup or quotient group in S, + coincides with the group operation.

If S has class 2 and p is odd, Lazard’s construction reduces to a construction of R. Baer ([HB, II, Lemma VIII.9.6, pages 347-9])

x + y = x^{\frac{1}{2}} yx^{\frac{1}{2}} and [x, y] = x^{−1}y^{−1}xy.

Now we finally obtain Theorem 1.

\underline{Theorem 1} Suppose p is odd, A ∈ A(S), and B is a subgroup of S that does not normalize A. Assume that

(3.13) B has nilpotence class at most 2 and is normalized by A, ,

or (3.14) p ≥ 5, A⊲ < A^{B} >, and [A, u; 3] = 1 for every u ∈ B.

Then there exists A^{∗} ∈ A(S) such that

(a) A^{∗} ≤< A^{B} > and A^{∗} normalizes A,

(b) for every central series S of S, A <_{S} A^{∗}, and

(c) if A normalizes B, then |A^{∗} ∩ B| ≥ |A ∩ B|.

\underline{Proof}:

We first choose an element b of B − N(A) as follows: if (3.14) holds, b can be any element of B − N(A); otherwise, take b as in

Theorem 2.3.

Then let T =< b, A > and Â =< A^{T} >. Then

\{ \begin{matrix}T =< A, B ∩ T >, Â =< A^{B∩T} >≤< A^{B} >, and \\ Z(Â) ≤ C_{S}(A) = A. \end{matrix} (3.15)

If (3.14) fails, then, by Theorem 2.3, Â has class 2 and Â′ ≤ Z(Â) ≤ A. So by (3.14),

(3.16) A ⊲ Â in all cases, and Â has class 2 if (3.14) fails.

Now we divide the proof into several steps.

\underline{Step 1}: The nilpotence class of Â is at most p − 1, Â =< A, A^{b}, A^{b^{2}} > and

[A, u; 3] = 1 for every u ∈ B.

\underline{Proof} If (3.14) fails, then B has class at most 2, and

[A, B; 3] ≤ [B, B, B] ≤ [B′, B] = 1.

Therefore, in all cases, [A, u; 3] = 1, for every u ∈ B. Now, by Lemma 2.2(c), Â =< A, A^{b}, A^{b^{2}} >. So, by (3.16), we are done if (3.14) fails. Assume (3.14) holds. Then p ≥ 5 and by Lemma 2.2(c), Â has class at most 3.

Q.E.D.

\underline{Step 2} Regard Â as a Lie ring as in Theorem 3.4. Let α be the mapping on Â given by

α(x) = x^{b} = b^{−1}xb (in the group T), for x ∈ Â.

Then:

(a) A and Â are divisible by 2 as Lie rings, and

(b) α is an automorphism of Â as a group and as a Lie ring.

Consider Â as a Lie ring. Then (c) Â(α − 1)^{3} = 0,

(d) Â = 3Â or [Â(α − 1), Â(α − 1)^{2}] = 0, and

(e) log α is well defined and is a derivation of Â, and α = exp(log α).

\underline{Proof}:

Inside the group T, let D = [A, < b >] and E = [A, < b >; 2]. By (3.15), D ≤ Â and D⊲ < A, b >= T. (3.17)

(a) Since |Â| is a power of an odd prime p, A and Â are divisible by 2.

(b) Clearly, α is an automorphism of Â as a group, hence as a Lie ring, by Theorem 3.4. Since D ⊲ T, we have α(D) = D.

(c) For each a ∈ A,

αa ≡ a^{b} ≡ a[a, b] ≡ a (modulo D);

similarly, a^{b^{2}} ≡ a (modulo D). By Step 1,

Â = < A, A^{b}, A^{b^{2}} > .

Therefore,

Â = < A, D > = AD and α acts trivially on Â/D.

For each x ∈ Â,

αx ≡ x and x(α − 1) ≡ xα − x ≡ x − x ≡ 0 (modulo D). Thus, Â(α − 1) ≤ D. A similar argument shows that

Â(α − 1)^{2} ≤ D(α − 1) ≤ E

and

Â(α − 1)^{3} ≤ [a, < b >; 3] ≤ [A, b; 3] = 1 by Lemma 2.2 and Step 1.

Hence, in the Lie ring notation, Â(α − 1)^{3} = 0.

(d) If p ≥ 5, then Â = 3Â because Â is a group under + and |Â| is a power of p.

Assume p = 3. Then (3.14) fails. Hence (3.13) holds, and B has nilpotence class at most 2 and is normalized by A. Then, in the group T,

D = [A, B ∩ T] ≤ B
E = [A, B ∩ T, B ∩ T] ≤ [B, B] = B′ ≤ Z(B),
and [D, E] = 1.

So E ≤ Z(D). From the proof of (c),

Â(α − 1)^{2} ≤ E ≤ Z(D), and Â(α − 1)^{2} ≤ Z(Â(α − 1)).

So, for x ∈ Â(α−1) and y ∈ Â(α−1)^{2}, xy = yx in the group Â; by Theorem 3.4, [x, y] = 0 in the Lie ring Â. This proves (d).

(e) As described in the beginning of this section, define

log(α) = (α − 1) − \frac{1}{2}(α − 1)^{2}.

By Lemma 3.1, log(α) is a derivation of Â, and α = exp(log α).

Q.E.D

\underline{Step 3}: Conclusion

\underline{Proof}:

We continue with the notation of Step 2. Let

δ = log α = (α − 1) − \frac{1}{2}(α − 1)^{2}. (3.18)

We first verify the hypothesis of Theorem 3.2 for L = Â.

By Step 2(e),

α = exp(δ) = 1 + δ + \frac{1}{2}δ^{2}.

This equation and (3.18) yield

Âδ = Â(α − 1), Âδ^{2} = Â(α − 1)^{2}, Âδ^{3} = Â(α − 1)^{3}. (3.19)

Now Step 2 gives (L1), (L2), and (L3).

By (3.16) and Theorem 3.4, Â contains A as a normal subgroup and as a Lie ideal. Since α is an automorphism ofÂ (both as a group and as a Lie ring), Â contains Aα(= A^{b}) and Aα^{2}(= A^{b^{2}}) similarly. Let

L_{0} = A + Aα + Aα^{2}.

Then L_{0} is a Lie ideal, and hence a normal subgroup of Â, and

L_{0} = A + A(α − 1) + A(α − 1)^{2} = A + Aδ + Aδ^{2} by (3.19).

By Step 2(c), (α − 1)^{3} = 0, and

L_{0}α ≤ L_{0} + L_{0}(α − 1) = A + A(α − 1) + A(α − 1)^{2} = L_{0}.

Thus, L_{0} = L_{0}α. In group-theoretic terms, L_{0} is a normal subgroup of Â and L_{0} = L_{0}^{b}. Since

T =< A, b > and N_{T} (L_{0}) ≥< Â, b >≥ T,

we have Â =< A^{T} >≤< L_{0}^{T} >= L_{0} ≤ Â. Hence L_{0} = Â, which gives (L4).

By Theorem 3.4, the center Z(Â) is the same whether Â is regarded as a group or as a Lie ring. Hence,

Z(Â) ≤ C_{S}(a) = A,

which gives (L5).

We have now verified the hypothess of Theorem 3.2 for L = Â, and we will apply the conclusion of the theorem. For A^{∗} as in (L6), Theorem 3.4 and part (a) of Theorem 3.2 assert that A^{∗} is an abelian subgroup of Â. By Theorem 3.2(e), |A^{∗}| = |A|, so A^{∗} ∈ A(S). By (3.15) and (3.16),

A^{∗} ≤ Â ≤ < A^{B} > ∩ N_{T} (A).

This proves (a) of Theorem 1.

Now take any central series S of S, say

1 = S_{0} ≤ S_{1} ≤ S_{2} ≤ … ≤ S_{n} = S.

Let Z_{i} = Â ∩ S_{i}, for i = 0, 1, 2, …, n. Then

[Z_{i}, b] ≤ Â ∩ [S_{i}, B] ≤ Â ∩ S_{i−1} = Z_{i−1}, 1 ≤ i ≤ n.

Going over to Lie ring notation for Â, we have, by Theorem 3.4 and the proof of Step 2,

Z_{i} is a Lie ideal of Â and Z_{i}δ = Z_{i}(α − 1) ≤ Z_{i−1}, 1 ≤ i ≤ n.

By Theorem 3.2(d),(e),

|A^{∗} ∩ Z_{i}| ≥ |A ∩ Z_{i}| for all i, and A^{∗} ∩ Z_{i} > A ∩ Z_{i} for some i. Therefore, A <_{S} A^{∗}, according to the definition in Section 1.

Suppose A normalizes B. Then B ⊲AB, and there exists a central series of AB in which B is one of the terms. The previous paragraph yields

|A^{∗} ∩ B| ≥ |A ∩ B|.

Now we have proved (b) and (c) of Theorem 1, which complete the proof of the theorem.

Q.E.D

4 Proof of Theorem 2

Consider the following conditions on a subgroup A of S.

Whenever B is a subgroup of S and (4.1) A⊲ < A^{B} > and [A, u; 3] = 1 for every u ∈ B, then B normalizes A.

Whenever B is a subgroup of S and (4.2) [A, u; 3] = 1 for every u ∈ B, then B normalizes A. \underline{Theorem 4.1} Suppose A is an abelian subgroup of S that satisfies (4.1).

(a) If B is an abelian subgroup of S that satisfies (4.1), then A and B normalize each other.

Moreover,

(b) A satisfies (4.2).

\underline{Proof}:

(a) We use induction on |S|. We may assume that

A, B < S.

Let M be a maximal subgroup of S that contains A, and let Q =< A^{B} >. Since S is nilpotent,

M ⊲ S and A =< A^{B} >≤< M^{S} >= M.

For each x ∈ B, it is easy to see that A^{x} satisfies (4.1); since A, A^{x} ≤ M, induction yields that A and A^{x} normalize each other. Therefore,

A⊲ < A^{B} >= Q

and

[B, A; 3] = [ [B, A], A, A] ≤ [ [Q, A], A] ≤ [A, A] = 1.

By symmetry, [A, B; 3] = 1. Hence, by (4.1), B normalizes A. By symmetry, A normalizes B.

(b) Take B as in part (b). By part (a), for each x ∈ B, A and A^{x} normalize each other. Thus,

A⊲ < A^{B} > .

By (4.1), B normalizes A.

Q.E.D

\underline{Proof of Theorem 2}:

Now suppose A, A_{0} satisfy the hypothesis of Theorem 2. Then p ≥ 5 and A, A_{0} satisfy (4.1). By Theorem 4.1, we obtain (a) and (b) of Theorem 2.

Q.E.D

\underline{Example 4.1} (J. Alperin, [H, p.349]) Here, p may be any prime. Let T be the group generated by elements

x_{1}, y_{1}, x_{2}, y_{2}, …, x_{p}, y_{p},

(with subscripts taken modulo p) subject only to the relations

x_{i}^{p} = y_{i}^{p} = [x_{i}, y_{i}] = 1, for i = 1, 2, …, p

and

[x, y, z] = 1 for all x, y, z ∈ T.

(Thus, T has nilpotence class two, and, if p is odd, may be obtained from a Lie ring by using Theorem 3.4) .

Let α be the automorphism of T of order p given by

x_{i}^{α} = x_{i+1}, y_{i}^{α} = y_{i+1}, for all i.

Assume S is the semi-direct product of T by < α >. Then it is not difficult to see that

|T′| = p^{k} for k = ( \begin{matrix}2p \\ 2 \end{matrix} ) − p = 2p^{2} − 2p,

|T | = p^{2p^{2}}, |S| = p^{2p^{2}+1}, d(S) = p^{2p^{2}−2p+2},

and A(S) consists of the p different groups < x_{i}, y_{i}, T′ >. Therefore, no element of A(S) is normal in S.

\underline{Example 4.2} Here, we assume p = 2 and S and T are as in Example 4.1.

Since T/T′ is elementary abelian and p = 2,

S′ = [T, < α >]T′ =< uu^{α}, T′ | u ∈ T >
= < x_{1}x_{2}, y_{1}y_{2}, T′ >

Similarly, [T′, α] = {uu^{α} | u ∈ T′} ≤ C_{T′}(α). (4.3)

Let B =< α > S′. Clearly, B ⊲ S and B normalizes neither of the two elements of A(S). Therefore, S is a counterexample for p = 2 to Theorem 1 and its corollary and to Theorem 2(b), if B has nilpotence class 2. To show the latter, we will show that 1 < B′ ≤ C_{T′}(α). Let R = C_{T′}(α).

Now, for some v ∈ T′,

x_{1}x_{2}v = x_{2}x_{1} = x_{1}^{α}x_{2}^{α} = (x_{1}x_{2})^{α},

and

x_{1}x_{2} = (x_{1}x_{2}v)^{α} = (x_{1}x_{2})^{α}v^{α} = x_{1}x_{2}vv^{α}, v^{α} = v^{−1} = v.

So v ∈ C_{T′}(α) = R, and α centralizes x_{1}x_{2} (modulo R). Similarly, α centralizes y_{1}y_{2} (modulo R). Therefore, by (4.3),

(4.4) α centralizes S′/R.

By our choice of relations for defining T, we have 1 ≠ [x_{1}, x_{2}] = v ∈ B′.

Since T′ ≤ Z(T) and α centralizes x_{1}x_{2} and y_{1}y_{2} (modulo T′), α centralizes [x_{1}x_{2}, y_{1}y_{2}], which must be in C_{T′}(α)(= R). Therefore, (S′)′ ≤ R. As B =< α > S′, (4.4) yields

B′ ≤ R = C_{T′}(α) ≤ Z(B).

Hence, B has nilpotence class 2, as desired.

5 Preliminary Results for Characteristic Subgroups

In this section, we assume the following condition:

Hypothesis 5.1

(a) S is a Sylow p-subgroup of a group G.

(b) B, X and R are subgroups of S.

(c ) B and X normalize R, and X ⊲ G.

\underline{Lemma 5.2} Assume that R ⊲ G and R ≤ X and that X centralizes every abelian normal subgroup of G that is contained in R.

Then [R, X] ≤ Z(X).

\underline{Proof}:

Use induction on |R|. We may assume that R > 1. Let Q = [R, X]. Since S is nilpotent and R, X ⊲ G, we have

Q < R and Q ⊲ G.

By induction,

[Q, X] ≤ Z(X)

so [X, Q, X] = [Q, X, X] ≤ [Z(X), X] = 1. By the Three Subgroups Lemma

1 = [X, X, Q] ≥ [R, X, Q] = [Q, Q].

Thus Q is abelian. By hypothesis, X centralizes Q. So

[R, X] = Q ≤ Z(X).

Q.E.D

\underline{Lemma 5.3} Assume that R⊲G and that X centralizes every abelian normal subgroup of G contained in R ∩ X. Then

[R, X, X] ≤ Z(X).

\underline{Proof}:

By Lemma 5.2, [R ∩ X, X] ≤ Z(X). Since X, R ⊲ G,

[R, X, X] ≤ [R ∩ X, X] ≤ Z(X).

So [R, X; 3] = 1.

Q.E.D

Theorem 2 and the following results are the main tools for improving on our 1970 result (mentioned in the introduction) to obtain Theorem 3.

\underline{Proposition 5.4} Let Y be a normal subgroup of G contained in X. Assume that (a) X/Y is abelian and C_{X}(R) ≤ Y,

(b) R is abelian and [R, X, X] ≤ Z(X),

(c ) [R, B, B] = 1

and

(d) p ≥ 5.

Then

[X, B; 3] ≤ Y.

\underline{Proof}

Let X′ = [X, X]. By (b), [X, R, X] = [R, X, X] ≤ Z(X). Hence, by the Three Subgroups Lemma,

[R, X′] = [X, X, R] ≤ Z(X). (5.1)

Similarly,

[R, X, X′] = [ [R, X], X′] ≤ [ [R, X], X, X] ≤ [Z(X), X] = 1.

Then

[ [X′, X], R] ≤ [X, R, X′] [R, X′, X] ≤ [R, X, X′] [Z(X), X] = 1.

Thus, [X′, X] ≤ C_{X}(R). (5.2)

Let X̄ = X/C_{X}(R). Conjugation by X on R induces a faithful action of X̄ by automorphisms on R. Let H be the semi-direct product of R by X̄. We claim that (5.3) H has nilpotence class at most 3.

To prove (5.3), it suffices (by Lemma 2.1) to show that

[x_{1}, x_{2}, x_{3}, x_{4}] = 1 (5.4)

for all x_{1}, x_{2}, x_{3}, x_{4} in the set-theoretic union X̄ and R. If x_{1}, x_{2}, x_{3} ∈ X̄ then [x_{1}, x_{2}, x_{3}] = 1 by (5.2). So we may assume that x_{i} ∈ R for some i = 1, 2 or 3. As R is abelian and R ⊲ H, we obtain (5.4) except possibly if x_{j} ∈ X̄ for all j ≠ i. But now (5.4) follows from (b) if i = 1 or 2, and from (5.1) if i = 3. Hence (5.3) follows.

Now we invoke our assumption that p ≥ 5. By (5.3) we may use Lazard’s construction (Theorem 3.4), to obtain operations + and { , } under which H becomes a Lie ring. (We write { , } for the Lie bracket operation to avoid confusion with the group commutator [x, y] = x^{−1}y^{−1}xy). Moreover, since R ⊲ H, we obtain a bilinear mapping on X̄ and R (under +) into R (under +) given by

(x̄, y) ↦ {x̄, y}.

In the usual way, this mapping induces an additive group homomorphism

ϕ : X̄ → Hom (R, R).

We can make several observations at this point:

(5.5) Since R is abelian, the + operation on R coincides with the given group operation on R.

(5.6) Since X̄ acts faithfully on R, ϕ is an injection.

(5.7) Conjugation by B on R and on X induces automorphisms on the Lie rings R, X̄, and H, and ϕ is a B−module injection.

By hypothesis, [R, B, B] = 1. For y ∈ R and b ∈ B, (5.5) and (5.7) give

y^{b} − y = (b^{−1}yb)y^{−1} = y^{−1}b^{−1}yb.

Therefore, in the usual notation for modules, we obtain that the module commutator [y, b] coincides with the group commutator [y, b]. Hence, in module notation,

[ [R, B], B] = 0.

Now, a calculation ([S, pp.236-7] or [G2, Lemma A2.3, p.57]) shows

[ [ [ Hom (R, R), B], B], B] = 0. (5.8)

Consequently by (5.6),

[ [ [X̄, B], B], B] = 0.

We have assumed that X/Y is abelian and C_{X}(R) ≤ Y. Therefore, the arguments in the previous paragraph show that, in the usual notation for groups,

[X, B; 3] ≤ Y.

Q.E.D.

At this point, it is useful to quote the following result (with a minor change that requires no significant change in the proof).

\underline{Proposition 5.5} ([G2, Theorem A2.4, pp.57-9]) Suppose

R ≤ P ≤ S,  R, P ⊲ G,  C_{P} (R) ≤ R,

and X/Y is a chief factor of G such that X ≤ P.

Then either X/Y is G−isomorphic to a chief factor X_{1}/Y_{1} of G for which X_{1} ≤ Z(R), or there exist chief factors U/U_{1} and V/V_{1} of G such that

(i) U ≤ R,

(ii) V ≤ [U, P] ≤ [R, P], and

(iii) X/Y is G−isomorphic to a composition factor of Hom(U/U_{1}, V/V_{1}) under G.

\underline{Corollary 5.6} Suppose P = O_{p}(G), g ∈ G, R ⊲ G, C_{P} (R) ≤ R, and [R, g, g] = 1, and X/Y is a chief factor of G. Then

[X, g; 3] ≤ Y.

\underline{Proof}:

Since R and X are normal p−subgroups of G, they are contained in P. So we may apply Proposition 5.5. Since

[Z(R), g; 3] ≤ [R, g; 2] = 1,

we may assume that X/Y is G−isomorphic to a composition factor of Hom(U/U_{1}, V/V_{1}) under G, for U/U_{1}, and V/V_{1} as in the proposition. Then

[V, g; 2] ≤ [U, g; 2] ≤ [R, g; 2] = 1.

Now the proof of (5.8) (in the proof of Proposition 5.4) shows that

[ Hom (U/V_{1}, V/V_{1}), g; 3] = 0.

Therefore, [X, g; 3] = 1.

Q.E.D.

6 Characteristic Subgroups

Throughout this section we assume that p ≥ 5.

Take A_{∈}(S) as in Section 1, so that A_{∈}(S) is a non-empty subset of A(S). \underline{Definition 6.1} Let A be a non-empty subset of A(S). Define

J(A) = < A > and N_{S}^{∗}(A) = ∩_{A∈A} N_{S}(A).

Then A is an \underline{N-set} if A normalizes B for every A, B ∈ A.

\underline{Remark} Thus, A is an N-set if and only if J(A) ≤ N_{S}^{∗}(A), that is A⊲J(A) for every A ∈ A. \underline{Definition 6.2} Let A be a non-empty subset of A(S). Let J = J(A). Then A is an \underline{N^{∗}-set} (in S) if it satisfies the following conditions: (N1) A is an N-set and J ⊲ S; (N2) whenever x ∈ S and [J, x, x] = 1, then x ∈ N_{S}^{∗}(A); (N3) whenever B is an N-set in S, then J(B) ≤ N_{S}^{∗}(A), if the following

conditions are satisfied: (N3a) J normalizes J(B), and (N3b) N_{S}^{∗}(B) contains every abelian subgroup of J normalized by J(B).

\underline{Remark 6.3} Under this definition, the question of whether A is an N^{∗}-set in S depends on the whole group S and not merely on J(S). It would be desirable to have it depend on J(S) alone, but we do not see how to obtain our results in this case (in particular, how to prove (N2) in Proposition 6.9).

Let us recall the definition of a p-stable group.

\underline{Definition} ([Gor, pp.268-9]) Suppose that p is odd, G is a group, and that O_{p}(G) > 1. Then G is p-stable if it has the following property:

Whenever A and R are p-subgroups of G such that

O_{p′}(G)R ⊲ G, A ≤ N_{G}(R), and [R, A, A] = 1

then

AC_{G}(R)/C_{G}(R) ≤ O_{p}(N_{G}(R)/C_{G}(R)).

\underline{Definition 6.4} A \underline{special pair} for S is an ordered pair (G, T) satisfying the following conditions.

(P1) S is a Sylow p-subgroup of G;

(P2) T = O_{p}(G) and C_{G}(T) ≤ T;

(P3) G is p-stable or satisfies

(P3a) whenever u ∈ S and [X, u; 3] ≤ Y for every chief factor X/Y of G such that X ≤ T, then u ∈ T.

A special pair satisfying (P3a) will be called a \underline{3-special pair}.

\underline{Lemma 6.5} Suppose (G, T) is a special pair, x ∈ G, and U ≤ T. Assume that

U ⊲ G, C_{T} (U) ≤ U, and [U, x, x] = 1.

Then x ∈ T.

\underline{Proof}:

If (G, T) is a 3-special pair, apply Corollary 5.6 with T, U, and x in place of P, R and g. By (P3a), x ∈ T.

Henceforth, assume that G is p-stable. Let C = C_{G}(U). By Lemma 2.2, [U, < x >, < x >] = 1. Since G is p-stable,

< x > C/C ≤ O_{p}(G/C).

To complete the proof , it suffices to show that C ≤ T, for then

O_{p}(G/C) = T/C.

Take any p′-element y in C. Then y induces a p′-automorphism α on T by conjugation, and

[T, < y >] ≤ T ∩ C = C_{T} (U) ≤ U, [U, < y >] = 1.

Thus, α stabilizes the normal series T ≥ U ≥ 1 of T. Therefore, α has a p-power order [Gor, pp.178-9]. Since α is a p′-automorphism,

α = 1  and  y ∈ C_{G}(T) ≤ T.

As y is a p′-element, y = 1. This shows that 1 is the only p′-element of C.

Now C is a p-group. As C ⊲ G, we have C ≤ O_{p}(G) = T, as desired.

Q.E.D.

Proposition 6.6

(a) There exists an N^{∗}-set. In particular, A_{∈}(S) is an N^{∗}-set.

(b) If A_{∞} and A_{∈} are N^{∗}-sets, then so is A_{∞} ∪ A_{∈}.

\underline{Proof}:

(a) Let A = A_{∈}(S). Let J = J(A) =< A >. Then J ⊲ S, so J ⊲ J(S). By Theorem 2, A is an N-set. Thus A satisfies (N1).

Take x as in (N2). By Lemma 2.2, [J, < x >, < x >] = 1. Hence, for each A ∈ A,

[A, < x >; 3] = 1,

and x normalizes A, by the definition of A_{∈}(S). Thus, x ∈ N_{S}^{∗}(A), as desired.

Similarly, for B as in (N3), J(B) ≤ N_{S}^{∗}(A).

(b) Let A_{∞} and A_{∈} be N^{∗}-sets. Let

J_{i} = J(A_{⟩}) for ⟩ = ∞, ∈;  A = A_{∞} ∪ A_{∈};  and  J = J (A).

By (N1), J_{1}, J_{2} ⊲ S. Hence,

J = J_{1}J_{2} ⊲ S (6.1)

Suppose D is an abelian subgroup of J_{1}, normalized by J_{2}. Then

[J_{2}, D, D] ≤ [D, D] = 1.

By (N2), D ≤ N_{S}^{∗}(A_{∈}). Therefore, if we now replace A and B by A_{∞} and A_{∈}, (N3b) is satisfied; and (N3a) is also satisfied because J_{1} normalizes J_{2}. Since A_{∞} is an N^{∗}-set, (N3) gives

J_{2} = J(A_{∈}) ≤ N_{S}^{∗}(A_{∞}).

As A_{∞} is an N-set, J_{1} ≤ N_{S}^{∗}(A_{∞}). Hence, by (6.1),

J = J_{1} J_{2} ≤ N_{S}^{∗}(A_{∞}).

By symmetry, J ≤ N_{S}^{∗}(A_{∈}). Therefore, A is an N-set. By (6.1), A satisfies (N1).

Take x as in (N2). Then [J_{1}, x, x] ≤ [J, x, x] = 1. Since A_{∞} is an N^{∗}- set, x ∈ N_{S}^{∗}(A_{∞}). By symmetry, x ∈ N_{S}^{∗}(A_{∈}). Hence, x ∈ N_{S}^{∗}(A), and A satisfies (N2).

A similar argument shows that A satisfies (N3). Therefore, A is an N^{∗}-set.

Q.E.D.

Proposition 6.6 obviously yields:

\underline{Corollary 6.7} There exists a unique maximal N^{∗}-set A_{N} (S) in S.

Our strategy is to show that, for a special pair (G, T), A_{N} (S) ⊂ A_{N} (T ), and, if (G, T) is a 3-special pair, then A_{N} (S) = A_{N} (T ).

\underline{Proposition 6.8} Let (G, T) be a special pair for S. Then:

(a) d(T) = d(S), and

(b) every N^{∗}-set in S is an N^{∗}-set in T.

\underline{Proof}:

Take any A_{0} ∈ A_{∈}(S) and let U be a critical subgroup of T [Gor, p.186]. Then U has nilpotence class at most 2, C_{T} (U) ≤ U, and U ⊲ G. By the definition of A_{∈}(S) and Lemma 6.5,

[U, A_{0}, A_{0}] ≤ [A_{0}, A_{0}] = 1 and A_{0} ≤ T.

Therefore, d(T) = d(S) and A_{∈}(S) ≤ A(T ).

Clearly, A_{∈}(S) ≤ A_{∈}(T ). Take any N^{∗}-set A in S, and let

J = J(A),  B = A_{∈}(T ),  U = J (B).

By Theorem 2, B is an N-set. For every abelian subgroup D of J normalized by U,

1 = [U, D, D] = [J(A_{∈}(T )), D, D],

and D ≤ T by Lemma 6.5; then D ≤ N_{S}^{∗}(B) by the definition of A_{∈}(T ). Thus, A and B satisfy condition (N3b) of the definition of a N^{∗}-set. Since J(B) ⊲ G, they satisfy (N3a) as well. Therefore, by (N3),

U = J(B) ≤ N_{S}^{∗}(A).

For every A ∈ A, U normalizes A; hence,

[U, A, A] = 1,

and A ≤ T by Lemma 6.5. This shows that A ⊆ A(T ). Since A satisfies conditions (N1), (N2), and (N3) with respect to S, it is easy to see that A satisfies them with respect to T. Thus, A is an N^{∗}-set in T.

Q.E.D.

\underline{Proposition 6.9} Suppose (G, T) is a 3-special pair, A is an N^{∗}-set in T that contains an element of A_{∈}(S), and J(A) ⊲ G.

Then A is an N^{∗}-set in S.

\underline{Proof}:

Take A_{0} ∈ A_{∈}(S) ∩ A, and let J = J(A). Then

d(T) = d(S), A ⊆ A(T ) ⊆ A(S)

and

C_{S}(J) ≤ C_{S}(A_{0}) ≤ A_{0} ≤ J.

Since A is an N^{∗}-set in T, A is an N-set By hypothesis, J ⊲ G. Hence, A satisfies condition (N1) of the definition of an N^{∗}-set in S (rather than T).

The proof that A satisfies (N2) in S follows easily from Lemma 6.5, since C_{S}(J) ≤ J and A satisfies (N2) in T. Therefore, we need only prove that A satisfies (N3) in S.

Take B as in (N3). Since A is an N-set in T, it suffices to show that J(B) ≤ T . Now,

[J, B; 3] ≤ [J(B), B, B] = ∞, forB ∈ B. (6.2)

and [D, B, B] = 1 (6.3)

for every B ∈ B and every abelian subgroup D of J normalized by J(B).

We must show that J(B) ≤ T . As (G, T) is a 3-special pair,

(6.4) it suffices to show that [X, B; 3] ≤ Y

for every chief factor X/Y of G such that X ≤ T, and every B ∈ B.

Now consider an arbitrary chief factor X/Y of G for which X ≤ T. We may assume that X is minimal in the sense that (6.5) If X_{1}/Y_{1} is a chief factor of G isomorphic to X/Y as a G-module, and

if X_{1} ≤ X, then X_{1} = X.

We must show that [X, B; 3] ≤ Y for every B ∈ B.

\underline{Step 1}. Whenever N ⊲ G and N < X, then N ≤ Y.

\underline{Proof}: Suppose not. Then Y < NY ≤ X and NY ⊲ G. By the definition of a chief factor, NY = X. Then, as G-modules,

N/(N ∩ Y) ≅ NY/Y = X/Y,

contrary to (6.5).

\underline{Step 2}. We may assume that

X ∩ J ≤ Y (6.6)

and that

(6.7) X centralizes every normal abelian subgroup of G contained in J.

\underline{Proof}:

Take any B ∈ B. If X ≤ J, then [X, B; 3] = 1 ≤ Y, by (6.2). If X is not contained in J, then X ∩ J ≤ Y by Step 1. Thus, we may assume (6.6).

Suppose R is a normal abelian subgroup of G contained in J, and X does not centralize R. We may assume that R is minimal subject to these conditions. Since

[R, X] ⊲ G  and  [R, X] < R,

X centralizes [R, X]. So

[R, X, X] = 1 ≤ Z(X).

As C_{X}(R) ⊲ G and C_{X}(R) < X, Step 1 yields that C_{X}(R) ≤ Y.

By (6.3), [R, B, B] = 1, and by the hypothesis of this entire section, p ≥ 5. Therefore, Hypothesis 5.1 is satisfied and Proposition 5.4 yields

[X, B; 3] ≤ Y.

This completes the proof of Step 2.

Henceforth, we will assume conditions (6.6) and (6.7) in Step 2.

\underline{Step 3}. Let B ∈ B. Then

(a) [J, X, X] ≤ Z(X) and [J, X; 3] = [J, B; 3] = 1,

(b) A_{0} ≤ J, and A_{0} is normalized by X and B, and

(c ) [A_{0}, B, B] = 1.

\underline{Proof}:

Observe that Hypothesis 5.1 is satisfied for R = J.

Therefore, by (6.7) and Lemma 5.3,

[J, X, X] ≤ Z(X) and (hence) [J, X; 3] = 1.

By (6.2), we obtain (a).

Since A_{0} ∈ A, A_{′} ≤ J (A) = J. By (a),

[A_{0}, X; 3] = [A_{0}, B; 3] = 1.

Now the definition of A_{∈}(S) gives (b). Thus, J(B) normalizes A_{0}, and (6.3) gives (c).

\underline{Step 4}. Let B ∈ B. Then

[X, B; 3] ≤ Y.

\underline{Proof}:

Since A_{0} ∈ A(S), C_{X} (A_{′}) = X ∩ A_{′} ≤ X ∩ J ≤ Y, by (6.6). Now observe that Hypothesis 5.1 is satisfied for R = A_{0}. Finally, Step 3 and Proposition 5.4 yield that [X, B; 3] ≤ Y, as desired.

By Step 4 and (6.4), we have verified that A satisfies (N3) in S, and not merely in T. Therefore, A is an N^{∗}-set in S, not merely in T.

Q.E.D

Now we obtain our main result about 3-special pairs from Corollary 6.7 and Propositions 6.8 and 6.9.

\underline{Theorem 6.10} Let (G, T) be a 3-special pair for S, and let A_{N} be the unique maximal N^{∗}-set in S.

Then A_{N} is the unique maximal N^{∗}-set in T, and J(A_{N}) ⊲ G.

\underline{Definition 6.11} Let A_{N} (S) denote the unique maximal N^{∗}-set in S, and let J_{N}(S) = J(A_{N} (S)) =< A_{N} (S) >.

Let A^{N} (S) be the set of all A ∈ A(S) such that J_{N}(S) normalizes A, and let

J^{N}(S) = < A^{N} (S) > and ZJ^{N} (S) = Z(J^{N} (S)).

It is easy to see that

A_{N} (S) ≤ A^{N} (S), J_{N} (S) ≤ J^{N} (S)

and

Z(S) ≤ ZJ(S) ≤ ZJ^{N}(S) ≤ Z(J_{N}(S)).

Suppose (G, T) is a 3-special pair for S. Theorem 6.10 asserts that A_{N} (S) = A_{N} (T ). Part (b) of our next result asserts that A^{N} (S) = A^{N} (T ) as well. We do not know whether these results extend to all special pairs for S. Part (a) of our next result gives the weaker result that ZJ^{N}(S) ⊲ G for all special pairs (G, T).

\underline{Theorem 6.12} Let (G, T) be a special pair for S. Then

(a) ZJ^{N}(S) ⊲ G, and

(b) if (G, T) is a 3-special pair, then

A^{N} (S) = A^{N} (T ) and J^{N} (S) = J^{N} (T ) ⊲ G.

\underline{Proof}:

Let J = J_{N}(S).

First, assume (G, T) is a 3-special pair. By Theorem 6.10, J = J_{N}(T). Take any A ∈ A^{N} (S). Then

[J, A, A] ≤ [A, A] = 1,

and A ≤ T by Lemma 6.5. It is easy to see now that

A^{N} (S) = A^{N} (T ), J^{N} (S) = J^{N} (T ) ⊲ G,

and

ZJ^{N}(S) = ZJ^{N}(T) ⊲ G.

This proves (b), and proves (a) for 3-special pairs.

From the definition of a special pair, we may now assume that G is p-stable. Let

Z = Z(J_{N}(T))  and  C = C_{G}(Z)

and define D ≤ G by

D ≥ C  and  D/C = O_{p}(G/C).

By Proposition 6.8, J = J_{N}(S) ≤ J_{N}(T). Since A_{N} (T ) is a N-set,

A_{N} (T ) ≤ A^{N} (S) and (6.8)

ZJ^{N}(S) ≤ C_{S}(J_{N}(T)) = Z ≤ Z(J) ≤ J,

and ZJ^{N}(S) = C_{Z}(J^{N}(S)). (6.9)

Take any A ∈ A^{N} (S). Then J normalizes A. By (6.9),

[Z, A, A] ≤ [J, A, A] = 1.

By the definition of p-stability, A ≤ D. Thus,

J^{N}(S) ≤ D ∩ S. (6.10)

Let A = A_{N} (D ∩ S). Since A_{N} (S) is an N^{∗}-set contained inside T, hence inside D ∩ S,

A_{N} (S) ⊆ A.

We claim that A_{N} (S) = A. It will suffice to show that A is an N^{∗}-set with respect to S. Condition (N1) is obvious. For x as in (N2), [Z, x, x] ≤ [J(A), §, §] = ∞; so x ∈ D ∩ S, and (N2) for D ∩ S yields (N2) for S. For B as in (N3) and B ∈ B,

[Z, B, B] = 1, and hence B ≤ D ∩ S, for every B ∈ B.

So we get (N3) for S, and A is an N^{∗}-set in S. Therefore

A_{N} (S) = A = A_{N} (D ∩ S).

It is easy to see from (6.10) that

J^{N}(S) = J^{N}(D ∩ S) ⊲ N_{G}(D ∩ S). (6.11)

As D/C is a normal p-subgroup of G/C,

D ∩ S is a Sylow p-subgroup of D, and D = (D ∩ S)C = C(D ∩ S).

By the Frattini argument, G = DN_{G}(D ∩ S) = C(D ∩ S)N_{G}(D ∩ S) = CN_{G}(D ∩ S). Recall equation (6.9),

ZJ^{N}(S) = C_{Z}(J^{N}(S)).

As C = C_{G}(Z), the subgroup ZJ^{N}(S) is centralized by C. By (6.11), it is normalized by N_{G}(D ∩ S) Therefore, it is normalized by G. This proves (a).

Q.E.D

References

[AG] J. Alperin and G. Glauberman, “Limits of abelian subgroups in finite p-groups,” in preparation.

[G1] G. Glauberman, “Prime-power factor groups of finite groups, I, II,” Math. Zeit. 107 (1968), pp. 159-72; 117 (1970), pp. 45-56.

[G2] G. Glauberman, “Global and local properties of finite groups,” Chap. I of Finite Simple Groups, M.B. Powell and G. Higman, ed., Academic Press, (1971) New York.

[Gor] D. Gorenstein, Finite Groups, Second Edition, Chelsea, (1980), New York.

[Hol] D. Holt, “More on the Local Control of Schur Multipliers,” Quart. J. Math. Oxford, (2), 31, (1980), pp. 191-208.

[H] B. Huppert, Endliche Gruppen I, Springer-Verlag, (1967), Berlin.

[HB] B. Huppert and N. Blackburn, Finite Groups II, III, Springer-Verlag, (1982), Berlin.

[L] M. Lazard, “Sur les groupes nilpotents et les anneaux de Lie,” Ann. Sci. Ecole Norm. Sup., (3), 71, (1954), 101-90.

[M] M. Miyamoto, “An affirmative answer to Glauberman’s conjecture,” Pacific J. Math., 102, (1982), pp. 89-105.

[S] M. Suzuki, Group Theory II, Springer-Verlag, (1986), Berlin.

[T1] J.G. Thompson, “Normal p-complements for finite groups,” Math. Zeit, 72, (1968), pp. 332-54.

[T2] J.G. Thompson, “Normal p-complements for finite groups,” J. Alg., 1, (1964), pp. 43-46.