On almost regular automorphisms of finite p-groups. ^{∗}

A. Jaikin-Zapirain

Departamento de Matemáticas-Matematika Saila Facultad de Ciencias-Zientzi Fakultatea Universidad del País Vasco-Euskal Herriko Unibertsitatea Apdo. 644, 48080-Bilbao, Spain

August 23, 2000

Abstract

In this paper we prove that there are functions f(p, m, n) and h(m) such that any finite p-group with an automorphism of order p^{n}, whose centralizer has p^{m} points, has a subgroup of derived length ≤ h(m) and index ≤ f(p, m, n). This result gives a positive answer to a problem raised by E. I. Khukhro (see also Problem 14.96 from “Kourovka Notebook” [9]).

1 Introduction

By a classical result of J. Thompson and G. Higman a finite group with a regular (without non-trivial fixed points) automorphism of prime order p is nilpotent of nilpotency class bounded by a function depending only on p, h(p), known as Higman’s function. J. Thompson (see [21]) established the nilpotency and G. Higman (see [2]) showed the existence of such function h(p). Later in [11] V. A. Kreknin proved that a ℤ/nℤ-graded Lie ring L with L_{0} = 0 is soluble of derived length at most k(n), giving an explicit upper bound for k(n). Applying this theorem he showed that an arbitrary Lie ring admitting a regular automorphism of order n is soluble of derived length at most 2k(n). Also in [12] A. I. Kostrikin and V. A. Kreknin gave an upper bound for Higman’s function. The analog of Kreknin’s Theorem for finite groups remains unsettled: although a finite group with a regular automorphism of order n is known to be soluble, no bound for its derived length in terms of n only has been found at the moment for n composite, except for the particular case n = 4 (see [10]).

∗This work has been supported by a grant of the Government of the Basque Country and by the DGICYT Grant PB97-0604.

In a different direction, using Higman’s method, J. L. Alperin (see [1]) showed that if G is a finite p-group admitting an automorphism of order p with p^{m} fixed points, then the derived length of G is (p, m)-bounded (in this paper we say that a certain invariant is (a, b, …)-bounded if it is bounded above by some function of (a, b, …)). In [5, 14] this result was improved by E. I. Khukhro and N. Yu. Makarenko and it was shown that G has a subgroup H of (p, m)-bounded index and nilpotency class at most h(p).

In [18] A. Shalev considers a finite p-group G admitting an automorphism of order p^{n} with p^{m} fixed points, and proves that its derived length is (p, m, n)- bounded. E. I. Khukhro [6] precises this information and shows that G has a subgroup N of (p, m, n)-bounded index and derived length at most 2k(p^{n}). He also asked if the derived length of N can be bounded by some function depending only on m. Yu. Medvedev showed in [16] that this problem can be reduced to the analog question about finite Lie rings with an additive p-group and, in [17], he gave a positive answer to it for the particular case n = 1. In fact, he proved a bit more than this by showing that a subgroup of (p, m)-bounded index and m-bounded nilpotency class exists. Actually, R. Shepherd [20] and C.R. Leedham-Green and S. McKay [13] had proved that for n = 1 and m = 1, G has a subgroup of p-bounded index and nilpotency class at most 2. Also for m = 1, S. McKay [15] and I. Kiming [8] proved that G has a subgroup of (p, n)-bounded index and nilpotency class at most 2. In this paper we answer Khukhro’s question in the positive for arbitrary m and n. We refer the interested reader to [4] for more background on this subject.

Let ℤ_{p} be the ring of p-adic integers and ℚ_{p} its field of quotients. Set S = ℤ_{p}[x] and R = S/(1+⋯+x^{p^{n}−1}). Let M be a finitely generated R-module (or S-module). It is clear that if |M : (x − 1)M| is finite then it is a power of p. In this case we define the x-rank of M by rx M = log_{p}(|M : (x − 1)M|). Put m = rx M. Since (x−1)M is contained in the Jacobson radical of M, M can be generated by m elements. Hence, M is an epimorphic image of R^{m} and the rank of M as ℤ_{p}-module is (m, n)-bounded. In particular, if u is a (p, m, n)-bounded number then both the index of p^{u}M in M and the order of {t ∈ M | p^{u}t = 0} are (p, m, n)-bounded as well. In the next sections, we shall use several times these facts, sometimes without mentioning them explicitly. Note also that if M is finite then |C_{M}(x)| = p^{rx M}.

Throughout this paper we shall call Lie ℤ_{p}-(sub)algebras simply Lie (sub)rings for brevity. The main result of this paper is the following theorem.

Theorem 1.1. There is a function f = f(p, m, n) such that if M is a Lie ring and also a finitely generated R-module of x-rank m with x operating on it as a Lie automorphism, then M has a soluble subring N of index less than f and derived length at most 2^{m+1} − 2.

Suppose now that M is a Lie ring whose additive group is a finite p-group and M has an automorphism ϕ of order p^{n} with p^{m} fixed points. It is clear that p^{m}(1 + ⋯ + ϕ^{p^{n}−1})M = 0, and so p^{m}M can be viewed as an R-module of x-rank ≤ m. On the other hand, by [4, Corollary 2.7], the rank of M is (p, m, n)- bounded and so the index of p^{m}M is (p, m, n)-bounded. Hence, Theorem 1.1 yields the following corollary.

Theorem 1.2. Let M be a Lie ring whose additive group is a finite p-group and suppose that M has an automorphism ϕ of order p^{n} with p^{m} fixed points. Then there is a subring N of (p, m, n)-bounded index and derived length ≤ 2^{m+1} − 2.

Using the reduction theorem from [16] we obtain the following result.

Corollary 1.3. There are functions f(p, m, n) and g(m) such that any finite p-group with an automorphism of order p^{n} and p^{m} fixed points has a subgroup of derived length ≤ g(m) and index ≤ f(p, m, n).

I would like to thank Professor E. I. Khukhro for introducing me into this subject.

2 Preliminaries

In this paper all the exterior and tensor products are taken over ℤ_{p}. Let M be an R-module. Then we can define in M ∧ M an S-module structure by setting

x(a ∧ b) = (xa) ∧ (xb), where a ∧ b ∈ M ∧ M.

We define the category 𝕂 whose objects are the triples (M, +, ·) where

(i) (M, +) is an R-module.

(ii) · is an antisymmetric and bilinear form on M and if we put α(m_{1} ∧ m_{2}) = m_{1} · m_{2}, then α ∈ Hom_{S}(M ∧ M, M).

The morphisms of 𝕂 are defined in the natural way. Hence 𝕂 is the category of (nonassociative) rings with an additional compatible structure of R-modules. In the sequel we shall write (M, α) instead of M when we want to emphasize that the operation · in M ∈ 𝕂 is given by α ∈ Hom_{S}(M ∧ M, M). For any M_{1}, M_{2} ∈ 𝕂 we shall write M_{1} < M_{2} if M_{1} is a proper subring of M_{2} in the category 𝕂, i.e. M_{1} is a proper subset of M_{2} and the inclusion of M_{1} into M_{2} is a morphism in 𝕂. We use ≅ for isomorphism of R-modules and ≅_{𝕂} for isomorphism in the category 𝕂.

If M ∈ 𝕂 and A, B ⊆ M let A · B be the R-submodule generated by a · b, where a ∈ A and b ∈ B. Let Γ_{1}(M) be the R-submodule of M generated by the elements

a · (b · c) + c · (a · b) + b · (c · a), a, b, c ∈ M,

and, for i > 1, let Γ_{i}(M) = Γ_{i−1}(M) · M. Set Γ(M) = ∑_{i} Γ_{i}(M) (so Γ(M) is the ideal of M generated by Γ_{1}(M)). It is clear that M̄ = M/Γ(M) becomes a Lie ring if for ā, b̄ ∈ M̄ we define its Lie bracket by [ā, b̄] = (a · b) + Γ(M).

The polynomial 1 + ⋯ + x^{p^{n}−1} can be decomposed in ℤ_{p}[x] as the product of the cyclotomic polynomials f_{i} = ∏_{j}(x − w_{i,j}), i = 1, … n, where {w_{i,j}} is the set of p^{i}-th primitive roots of 1. Then R_{i} = ℤ_{p}[x]/(f_{i}) is a discrete valuation ring whose maximal ideal is generated by (x − 1). We call L ∈ 𝕂 a lattice if L ≅ ⊕_{i}R_{i}^{k_{i}} for some k_{1}, …, k_{n}. The key of the proof of Theorem 1.1 is to prove it for the Lie rings L/Γ(L) when L is a lattice. We shall deal with this case in Section 4. First, in Section 3, we show how the general case follows from this particular one.

3 The reduction theorem

The goal of this section is to show that in order to prove Theorem 1.1 there is no loss of generality if we take M to be a Lie ring of the type L̄ = L/Γ(L) where L is a lattice. This reduction is a consequence of the following theorem.

Theorem 3.1. There are functions g_{1} = g_{1}(p, m, n) and g_{2} = g_{2}(p, m, n) satisfying the following property: if M is a Lie ring and a finitely generated R-module of x-rank m with the element x acting as a Lie automorphism, then there exists a subring N_{1}, an ideal N_{2} of N_{1} and a lattice L of x-rank at most m such that

(i) |M : N_{1}| ≤ g_{1} and |N_{2}| ≤ g_{2}.

(ii) The ring N_{1}/N_{2} is a epimorphic image of L̄ = L/Γ(L).

In order to prove this theorem we need some preliminary work. We introduce first some additional notation. Let f̄_{i} = ∏_{j≠i} f_{j}. The polynomials f_{i} and f̄_{i} are coprime in ℚ_{p}[x], so there exist r_{i}, q_{i} ∈ ℤ_{p}[x] and t_{i} ≥ 0 such that r_{i}f_{i} + q_{i}f̄_{i} = p^{t_{i}}. Define z_{1} = z_{1}(p, n) = max_{i} t_{i}.

Let L be an R-module and suppose that L = ⊕_{i=1}^{n}L_{i} with L_{i} ≅ R_{i}^{k_{i}}. The following lemma is an easy consequence of the definition of z_{1}.

Lemma 3.2. Let A be an R-submodule of L, and a = ∑ a_{i} ∈ A, a_{i} ∈ L_{i}. Then p^{z_{1}}a_{i} ∈ A.

Proof. We have q_{i}f̄_{i}a = q_{i}f̄_{i}a_{i} = p^{t_{i}}a_{i} ∈ A. Since z_{1} ≥ t_{i} then p^{z_{1}}a_{i} ∈ A.

Corollary 3.3. Let A be an R-submodule of L such that {l ∈ L | p^{k}l ∈ A for some k ∈ ℕ} = A. Then A ≅ ⊕_{i}R_{i}^{s_{i}} for some 0 ≤ s_{i} ≤ k_{i} and there exists B ≅ ⊕_{i}R_{i}^{k_{i}−s_{i}} such that L = A ⊕ B.

Consider now the ideal R̄ = ∑_{i} Rf̄_{i} of R. Note that this sum is direct and Rf_{i} ≅ R_{i}. It is easy to see that there exists z_{2} ≥ 0 such that p^{z_{2}}R ⊆ R̄. We can prove now the following lemma.

Lemma 3.4. Let M be a finitely generated R-module which is ℤ_{p}-torsion free. Then there exists a submodule N of finite index such that N ≅ ⊕_{i=1}^{n}R_{i}^{s_{i}} for some s_{i} ≥ 0 and p^{z_{2}}M ⊆ N.

Proof. Set N = R̄M. Then N = ∑_{i} N_{i}, where N_{i} = Rf̄_{i}M. Since N_{i} is ℤ_{p}-torsion free, then N_{i} ≅ R_{i}^{s_{i}} for some s_{i} ≥ 0. Suppose ∑_{i} a_{i} = 0, where a_{i} ∈ N_{i}. By using the same argument as in the proof of Lemma 3.2 we deduce that p^{z_{1}}a_{i} = 0 for every i, whence a_{i} = 0. Hence, N = ⊕_{i}N_{i} ≅ ⊕_{i}R_{i}^{s_{i}}.

Lemma 3.5. Let M be an S-module and finitely generated as ℤ_{p}-module. If rx(M/p^{m+1}M) ≤ m then rx M ≤ m.

Proof. Since rx(M/p^{m+1}M) ≤ m then p^{m}M ⊆ (x − 1)M + p^{m+1}M. Hence, by Nakayama’s Lemma, p^{m}M ⊆ (x − 1)M. Then we have

|M : (x − 1)M| = |M : (p^{m}M + (x − 1)M)| ≤ |M : (p^{m+1}M + (x − 1)M)| ≤ p^{m},

and so rx M ≤ m.

Lemma 3.6. Let M be an S-module of x-rank m, finitely generated as ℤ_{p}- module and N an S-submodule of M of finite index. Then rx N ≤ m.

Proof. Since the index of N in M is finite, there exists u such that p^{u}M ⊆ N. Set M̄ = M/p^{u+m+1}M and N̄ = N/p^{u+m+1}M. Then we have

rx(N/p^{m+1}N) ≤ rx N̄ = log_{p} |CN̄(x)| ≤ log_{p} |CM̄(x)| = rx M̄ ≤ m,

and so, by Lemma 3.5, rx N ≤ m.

Lemma 3.7. Let M be a finitely generated R-module of x-rank m. Then there exists a submodule N of (p, m, n)-bounded index which is an epimorphic image of ⊕_{i}R_{i}^{k_{i}} for some k_{i} ≥ 0 with ∑_{i} k_{i} ≤ m.

Proof. Since (x − 1)M is contained in the Jacobson radical of M, M can be generated by m elements. Hence, M is an epimorphic image of R^{m} and the index of K = R̄M in M is (p, m, n)-bounded. Using Lemma 3.6 we obtain that rx K ≤ m.

From the definition of K there is an R-submodule T of R̄^{m} = ⊕_{i}R_{i}^{m} such that K ≅ R̄^{m}/T. Let T_{i} be the image of the projection of T to (R_{i})^{m}. Put T̄ = ⊕_{i}T_{i}. Then T ⊆ T̄. Set D = R̄^{m}/T̄ = ⊕_{i}D_{i}, where D_{i} ≅ R_{i}^{m}/T_{i} and denote the x-rank of D_{i} by k_{i}. It is clear that D_{i} is an epimorphic image of R_{i}^{k_{i}}. Since D is a quotient of K we have

∑_{i} k_{i} = ∑_{i} rx D_{i} = rx D ≤ m.

Set N = p^{z_{1}}K. By Lemma 3.2, p^{z_{1}} T̄ ⊆ T and so N is an epimorphic image of D, whence it is also a quotient of ⊕_{i}R_{i}^{k_{i}}. Finally, we have that p^{z_{1}+z_{2}}M ⊆ N and so the index of N in M is (p, m, n)-bounded.

Corollary 3.8. There is a function f = f(p, m, n) such that for every Lie ring M which is a finitely generated R-module of x-rank m, there exists a subring N of index ≤ f which is, as R-module, an epimorphic image of ⊕_{i}R_{i}^{k_{i}} for some k_{i} ≥ 0 with ∑_{i} k_{i} ≤ m.

Proof. From Lemma 3.7 there exists a submodule K of (p, m, n)-bounded index which is a quotient of ⊕_{i}R_{i}^{k_{i}} for some k_{i} ≥ 0 with ∑_{i} k_{i} ≤ m. Let p^{u} be the index of K in M. Then the Lie subring N = p^{u}K satisfies the desired conditions.

Proof of Theorem 3.1. By Corollary 3.8 we can suppose that there exist L = ⊕_{i}L_{i}, L_{i} ≅ R_{i}^{k_{i}}, ∑_{i} k_{i} ≤ m, and T ⊆ L, such that M ≅ L/T. Let β be the natural projection from L onto M and define ᾱ: L ∧ L → M by ᾱ(a ∧ b) = [β(a), β(b)]. Consider H = (L ∧ L)/(1 + ⋯ + x^{p^{n}−1})(L ∧ L). Since (1 + ⋯ + x^{p^{n}−1})M = 0, ᾱ((1 + ⋯ + x^{p^{n}−1})(L ∧ L)) = 0 and so we can view ᾱ as an R-homomorphism from H to M. Let U be the R-submodule of H of ℤ_{p}-torsion elements. Since H is ℤ_{p}-finitely generated, U is finite. Its order depends on the numbers k_{i} and p, so it is (p, m, n)-bounded. Hence, without loss of generality, we can suppose that ᾱ(U) = 0 and ᾱ can be regarded as an R-homomorphism from H̄ = H/U to M.

Since H̄ is ℤ_{p}-torsion free, by Lemma 3.4, there exists a submodule D of H̄, such that D = ⊕_{i}D_{i}, and D_{i} ≅ R_{i}^{s_{i}}. Let d ∈ D_{k} for some k and a = ᾱ(d). There exist l_{i} ∈ L_{i} such that a = β(∑ l_{i}). By the definition of z_{1}, there exist q, r ∈ ℤ_{p}[x] such that p^{z_{1}} = qf_{k} + rf̄_{k}, whence

p^{z_{1}} ᾱ(d) = rf̄_{k}ᾱ(d) = rf̄_{k}β(∑ l_{i}) = rf̄_{k}β(l_{k}) = p^{z_{1}}β(l_{k}).

We fix now a set of free R_{k}-generators {d_{k,i} ∈ D_{k}} of D_{k}. We have that p^{z_{1}} ᾱ(d_{k,i}) = p^{z_{1}}β(l_{k,i}), where l_{k,i} ∈ L_{k}. Therefore if we define the R-module homomorphism γ: p^{z_{1}}D → p^{z_{1}}L by setting γ(p^{z_{1}}d_{k,i}) = p^{z_{1}}l_{k,i}, we obtain that γ is well-defined and β ∘ γ = ᾱ on p^{z_{1}}D. Let u = z_{1} + z_{2}, so that p^{u}H̄ ⊆ p^{z_{1}}D. Set N = p^{u}M and let τ be the natural projection from p^{u}L ∧ p^{u}L onto p^{2u}H̄ and α = γ ∘ τ. Then β: p^{u}L → N, α: p^{u}L ∧ p^{u}L → p^{u}L, ᾱ ∘ τ: p^{u}L ∧ p^{u}L → N and β ∘ α = ᾱ ∘ τ. Hence β(Γ(p^{u}L, α)) = 0 and so N is a ( Lie ring) quotient of p^{u}L/Γ(p^{u}L, α).

4 The proof of the main result

If L is a lattice, the tensor product 𝕃 = ℚ_{p} ⊗ L belongs to the category 𝕂. We call 𝕃 simple if 𝕃^{2} ≠ 0 and there is no proper ℚ_{p}[x]-submodule A of 𝕃 such that 𝕃 · A ⊆ A. Also L is called maximal if there are no lattice N and injective morphism ϕ: L → N in 𝕂 such that ϕ(L) < N and L ≅ N. The first step in the proof of our main result is to show that if L is a lattice and 𝕃 is simple then there exists a maximal lattice M such that L ≤ M < 𝕃.

First we need some preliminary lemmas.

Lemma 4.1. Let M, N be R-submodules of 𝕃 such that ℚ_{p}M = 𝕃, M ⊆ N and N is finitely generated. Then |N : M| is finite.

Proof. Since ℚ_{p}M = 𝕃, N/M is a ℤ_{p}-torsion module. By using that N is finitely generated, we obtain that |N : M| is finite.

Lemma 4.2. Suppose M ≤ N are lattices. Then |N : M| is finite if and only if M ≅ N.

Proof. We can decompose M and N as follows: M = ⊕_{i}M_{i} and N = ⊕_{i}N_{i}, where M_{i} ≅ R_{i}^{s_{i}} and N_{i} ≅ R_{i}^{t_{i}} for some non-negative integers s_{i} and t_{i}. It is clear from the hypothesis M ≤ N that M_{i} ⊆ N_{i}, whence s_{i} ≤ t_{i}. Moreover s_{i} = t_{i} if and only if N_{i}/M_{i} is finite. Now the result is clear bearing in mind that the condition M ≅ N amounts to the equalities s_{i} = t_{i} for all i.

Now we give a criterion for a ℤ_{p}-submodule of ℚ_{p}^{k} to be finitely generated.

Lemma 4.3. Let N be a ℤ_{p}-submodule of ℚ_{p}^{k} and suppose that {n ∈ N | p^{−k}n ∈ N for every k ∈ ℕ} = 0. Then N is finitely generated.

Proof. We prove it by induction on k. For k = 1 it is trivial. Suppose that we have proved the result for k − 1.

Of course we can suppose N ≠ 0. We choose an element x ∈ N with p^{−1}x ∉ N and define

N̄ = N/ℤ_{p}x = N/(ℚ_{p}x ∩ N) ≅ (N + ℚ_{p}x)/ℚ_{p}x.

Suppose there exists n ∈ N \ ℤ_{p}x such that p^{−k}(n + ℚ_{p}x) ⊆ N + ℚ_{p}x for every k ∈ ℕ. Then for every k ≥ 0 there exists a_{k} ∈ ℚ_{p} such that p^{−k}(n + a_{k}x) ∈ N. Since p^{−1}x ∉ N, a_{k} ∈ ℤ_{p} and a_{k+1} − a_{k} ∈ p^{k}ℤ_{p}. Let a = lim_{k↦∞} a_{k}. Then p^{−k}(n+ax) ∈ N for every k ∈ ℕ, which is impossible because n+ax ≠ 0. Thus, no such an element n exists and this allows us to apply the inductive hypothesis on N̄. Then N̄ is finitely generated and so is N.

Lemma 4.4. Suppose 𝕃 is simple, M is a lattice and L < M < 𝕃. Define t(L) = min{t | p^{t}L ⊆ L^{2}} and k = max{l | L ⊆ p^{l}M}. Then t(L) is finite and k ≤ t(L).

Proof. Since 𝕃 is simple, 𝕃^{2} = 𝕃 and ℚ_{p}L^{2} = 𝕃^{2} = 𝕃. By Lemma 4.1 |L : L^{2}| is finite and so t(L) is also finite. By the definitions of t(L) and k, we obtain

p^{t(L)}L ⊆ L^{2} ⊆ p^{2k}M^{2} ⊆ p^{2k}M.

By Lemma 4.1, k is finite, so by the maximality of k it follows that 2k−t(L) ≤ k, whence k ≤ t(L).

Proposition 4.5. Suppose 𝕃 is simple. Then there is no proper ascending series of lattices L < L_{1} < L_{2} … < 𝕃.

Proof. Suppose there exists an infinite series of lattices L < L_{1} < L_{2}, … < 𝕃 and K = ∪_{i≥1}L_{i}. Then it is clear that K ∈ 𝕂. Define A = {a ∈ K | p^{−k}a ∈ K for every k ∈ ℕ}. Then A is a ℚ_{p}[x]-submodule of 𝕃. For any l ∈ K, a ∈ A and k ∈ ℕ we have

p^{−k}(l · a) = l · (p^{−k}a) ∈ K.

Hence K · A ⊆ A. Since 𝕃 = ℚ_{p}K, 𝕃 · A ⊆ A and either A = 0 or 𝕃.

In the former case, by Lemma 4.3, K is a finitely generated R-module and so, by Lemma 4.1, |K : L| is finite, which is a contradiction.

If A = 𝕃 fix a_{1}, …, a_{m} an R-system of generators of L. Since p^{−t(L)−1}a_{i} ∈ K, there exists k ≥ 1 such that p^{−t(L)−1}a_{i} ∈ L_{k} for all i. Hence L ⊆ p^{(t(L)+1)}L_{k}, which contradicts Lemma 4.4.

Corollary 4.6. Suppose 𝕃 is simple. Then there exists a maximal lattice M such that L ≤ M < 𝕃.

Proof. By Proposition 4.5 there exists a lattice M such that L ≤ M < 𝕃 and M is maximal with this property. We shall prove that M is a maximal lattice. Suppose by way of contradiction that there exist a lattice N and an injective morphism ϕ such that ϕ(M) < N and M ≅ N. By Lemma 4.2, N/ϕ(M) is finite and so for every n ∈ N there exists k ∈ ℕ such that p^{k}n = ϕ(m) for some m ∈ M. We define ψ: N → 𝕃 by means of ψ(n) = p^{−k}m ∈ 𝕃. This map is well defined and it is a morphism in the category 𝕂. Hence L ≤ M < ψ(N) < 𝕃, against the choice of M.

For k = (k_{1}, …, k_{n}), with the number k_{i} non-negative integer and ∑_{i} k_{i} = m, we set T^{(k)} = ⊕_{i}R_{i}^{k_{i}}, 𝔼^{(k)} = {(T^{(k)}, α) ∈ 𝕂 | α ∈ Hom_{S}(T^{(k)} ∧ T^{(k)}, T^{(k)})} and 𝔼_{n}^{(k)} = {(T^{(k)}, α) ∈ 𝔼^{(k)} | α(T^{(k)} ∧ T^{(k)}) ⊆ p^{n−1}T^{(k)}}. We can define a structure of compact metric space on 𝔼^{(k)} by putting ρ((T^{(k)}, α_{1}), (T^{(k)}, α_{2})) = 1/n if (T^{(k)}, α_{1} − α_{2}) is in 𝔼_{n}^{(k)} but not in 𝔼_{n+1}^{(k)}.

Let L ∈ 𝕂 and define γ(L) = max{n | Γ_{1}(L) ⊆ p^{n}L}. Put L^{(0)} = L and, for i > 0, L^{(i)} = L^{(i−1)} · L^{(i−1)}.

Proposition 4.7. There exists a = a(p, m, n) ≥ 0 such that for every maximal lattice L of x-rank m and for every integer s ≥ 0, p^{3s+a}L/Γ(p^{s}L) has x-rank less than m.

Proof. Suppose for every j there exists a maximal lattice L_{j} of x-rank m such that γ(L_{j}) > j. Since the number of possible choices for the numbers k_{i} is finite, we can suppose that for all j, L_{j} ≅ T^{(k)} for some k = (k_{1}, …, k_{n}), whence L_{j} ≅_{𝕂} (T^{(k)}, α_{j}) for some α_{j} ∈ Hom_{S}(T^{(k)} ∧ T^{(k)}, T^{(k)}). Since 𝔼^{(k)} is compact, for some {j_{i}} there exists lim_{i↦∞}(T^{(k)}, α_{j_{i}}) = (T^{(k)}, α) = L. It is clear that γ(L) = 0 and so L is a Lie ring. By Kreknin’s Theorem L is soluble. Hence, there exists an abelian ideal A ≠ 0 (for example, A = L^{(d−1)}, where d is the derived length of L) which is also an R-submodule. Let Ā = {l ∈ L | p^{k}l ∈ A for some k ∈ ℕ}. It is clear that Ā is an abelian ideal of L and satisfies the conditions of Corollary 3.3. Thus, Ā ≅ ⊕_{i}R_{i}^{s_{i}} and there exists B ≅ ⊕_{i}R_{i}^{k_{i}−s_{i}} such that T^{(k)} = Ā ⊕ B. Since (T^{(k)}, α − α_{j}) ∈ 𝔼_{3}^{(k)} for some j, we have α_{j}(T^{(k)} ∧ Ā) ⊆ Ā + p^{2}T^{(k)} and α_{j}(Ā ∧ Ā) ⊆ p^{2}T^{(k)}. Let M = (1/p)Ā⊕B ≅ T^{(k)} and denote by β the natural extension of α_{j} to M ∧M. Then (T^{(k)}, α_{j}) < (M, β), against the maximality of (T^{(k)}, α_{j}). We conclude that there exists a_{1} = a_{1}(p, m, n) such that γ(L) ≤ a_{1} for every maximal lattice L of x-rank m.

Let l ∈ Γ_{1}(L) \ p^{a_{1}}L. Write L as ⊕L_{i}, where L_{i} ≅ R_{i}^{k_{i}} and let l = ∑_{i} l_{i}, where l_{i} ∈ L_{i}. Since l ∉ p^{a_{1}}L, there is an index i such that l_{i} ∉ p^{a_{1}}L. By Lemma 3.2, t = p^{z_{1}}l_{i} ∈ Γ_{1}(L) \ p^{a_{1}+z_{1}}L. Put a = a_{1} + z_{1}. Then p^{3s}t ∈ L_{i} ∩ Γ_{1}(p^{s}L) and p^{3s}t ∉ p^{a+3s}L_{i}. Hence, the x-rank of p^{a+3s}L/Γ(p^{s}L) is less than m.

Theorem 4.8. Theorem 1.1 is true for m = 1.

Proof. By Theorem 3.1 we can suppose that M ≅_{𝕂} L/Γ(L), where L ≅ R_{k} for some 1 ≤ k ≤ n. If L^{2} = 0 then we are done. If L^{2} ≠ 0 then 𝕃 is simple, and by Corollary 4.6, there exists a maximal lattice N such that L ≤ N < 𝕃. Since N ≅ R_{k}, we have L = (x − 1)^{t}N for some t ≥ 0. Let s = min{i | p^{i}N ⊆ L}. (Actually, s = [(t − 1)/(p^{k−1}(p − 1))] + 1.) Since |L : p^{s}N| ≤ p^{p^{n}}, we can suppose L = p^{s}N. Hence, from Proposition 4.7 there is f = f(p, n) such that p^{f}γ_{3}(M) = 0.

Now we prove Theorem 1.1 by induction on m. The case m = 1 has proved in the previous theorem. Suppose now that m > 1 and also that Theorem 1.1 is true for m − 1.

Proposition 4.9. Suppose 𝕃 is simple, M ∈ 𝕂 is a maximal lattice and L ≤ M < 𝕃. We define s = min{k | p^{k}M ⊆ L}. Then there are functions f_{1} = f_{1}(p, n, m) and f_{2} = f_{2}(p, m, n) such that p^{f_{1}}L^{(2^{m}−2)} ⊆ p^{s}M + Γ(L) and (p^{3s+f_{2}}M)^{(2^{m}−2)} ⊆ Γ(L).

Proof. We can decompose L and M as follows: L = ⊕_{i=1}^{n}L_{i} and M = ⊕_{i=1}^{n}M_{i}, where L_{i} ≅ M_{i} ≅ R_{i}^{k_{i}}. It is clear that L_{i} ⊆ M_{i} and s = max_{i} min{k | p^{k}M_{i} ⊆ L_{i}}. From this observation it follows that if we define L̄ = L/(((p^{s−1}M) ∩ L) + Γ(L)) then the x-rank of L̄ is less than m. By the inductive hypothesis, there exists a function f_{1} = f_{1}(p, n, m) such that p^{f_{1}}L^{(2^{m}−2)} ⊆ p^{s}M + Γ(L).

Now, by Proposition 4.7, there exists a = a(p, m, n) such that the Lie ring p^{3s+a}M/Γ(p^{s}M) has x-rank less than m. Since p^{s}M ⊆ L, we have that rx(p^{3s+a}M + Γ(L))/Γ(L) < m and the proof is finished by applying again the inductive hypothesis.

Proof of Theorem 1.1. By Theorem 3.1 we can suppose that M ≅_{𝕂} L/Γ(L), where L is a lattice of x-rank m. If 𝕃 is simple then, by Proposition 4.9, there exists f = f(p, m, n) such that (γ_{3}((p^{f}M)^{(2^{m}−2)}))^{(2^{m}−2)} = 0.

Suppose now that 𝕃 is not simple. Then there exists a ℚ_{p}[x]-submodule 0 ≠ A ≠ 𝕃 such that A · 𝕃 ⊆ A. Let B = A ∩ L. From Corollary 3.3 there exists C such that L = B ⊕ C. Note that the x-rank of B and C is less than m and B̄ = (B + Γ(L))/Γ(L) is an ideal of L/Γ(L), so the theorem follows from the inductive hypothesis.

5 Final remarks

In fact, the proof of Theorem 1.1 gives us the following result.

Theorem 5.1. Under the hypothesis of Theorem 1.1, there exists a subring N of (p, m, n)-bounded index such that γ_{3}(γ_{3}(… γ_{3}(N) …) = 0, where γ_{3} appears 2^{m} − 1 times.

We expect that this result can be improved and pose the next conjecture.

Conjecture. The number of times that γ_{3} appears in Theorem 5.1 can be reduced to m.

We think that the method of this paper will be useful in another situations. For example, in [3] a similar argument is used to prove that a finite p-group of rank r admitting an automorphism with p^{m} fixed points, has a subgroup H of (p^{m}, r)-bounded index and r-bounded derived length. For background on this problem see [19, 7] and also [9, Problem 13.56].

References

[1] J. L. Alperin, Automorphisms of solvable groups, Proc. Amer. Math. Soc. 13 (1962), 175–180.

[2] G. Higman, Groups and rings which have automorphisms without nontrivial fixed elements, J. London Math. Soc. (2) 32 (1957), 321–334.

[3] A. Jaikin Zapirain, Finite groups of bounded rank with an almost regular automorphism, in preparation.

[4] E. I. Khukhro, p-Automorphisms of Finite p-groups, Cambridge University Press, Cambridge, 1998.

[5] E. I. Khukhro, Finite p-groups admitting an automorphism of order p with a small number of fixed points, Mat. Zametki 38 (1985), 625-657; English. transl., Math. Notes 38 (1986), 867–870.

[6] E. I. Khukhro, Finite p-groups admitting p-automorphisms with few fixed points, Mat. Sbornik 184 (1993), 53–64; English. transl., Russian Acad Sci. Sbornik Math. 80 (1995), 435–444.

[7] E. I. Khukhro, Almost regular automorphisms of finite groups of bounded rank, Sibirsk. Math. Zh. 37 (1996), 1407–1412; English. transl., Siberian Math. J. 37 (1996), 1237–1241.

[8] I. Kiming, Structure and derived length of finite p-groups possessing an automorphism of p-power order having exactly p fixed points, Math. Scand. 62 (1988), 153–172.

[9] The Kourovka notebook: unsolved problems in group theory, E. I. Khukhro (ed.), V. D. Mazurov (ed.), Novosibirsk, 14th ed., 1999.

[10] L. G. Kovacs, Groups with regular automorphism of order four, Math. Z. 75 (1961), 277–294.

[11] V. A. Kreknin, The solubility of the Lie algebras with a regular automorphism, Sibirsk. Math. Zh. 8 (1967), 715–716; English. transl., Siberian Math. J. 8 (1968), 536–537.

[12] V. A. Kreknin and A. I. Kostrikin, Lie algebras with regular automorphism, Dokl. Akad. Nauk SSSR 149 (1967), 249–251; English. transl., Math. USSR Doklady 4 (1968), 355–358.

[13] C. R. Leedham-Green and S. McKay, On p-groups of maximal class, I, Quart. J. Math. Oxford Ser. (2) 27 (1976), 297–311.

[14] N. Yu. Makarenko, Nilpotent groups with an almost regular automorphism of prime order, Sibirsk. Math. Zh. 35 (1994), 630–632; English. transl., Siberian Math. J. 35 (1994), 564–565.

[15] S. McKay, On the structure of a special class of p-groups, Quart. J. Math. Oxford Ser. (2) 38 (1987), 489–502.

[16] Yu. Medvedev, p-Groups, Lie p-rings and p-automorphisms, J. London Math. Soc. (2) 58 (1998), 27-37.

[17] Yu. Medvedev, p-Divided Lie rings and p-groups, J. London Math. Soc. (2) 59 (1999), 787-798.

[18] A. Shalev, On almost fixed point free automorphisms, J. Algebra 157 (1993), 271–282.

[19] A. Shalev, Automorphisms of finite groups of bounded rank, Israel J. Math. 82 (1993), 395–404.

[20] R. Shepherd, p-groups of maximal class, Ph.D. Thesis, University of Chicago, 1970.

[21] J. G. Thompson, Finite groups with fixed-point-free automorphisms of prime order, Proc. Nat. Acad. Sci. U.S.A. 45 (1959), 578–581.