On the structure of normal subgroups of potent p-groups

J. González-Sánchez ^{∗}

Departamento de Matemáticas, Facultad de Ciencias, Universidad del País Vasco e-mail: mtbgosaj@lg.ehu.es

A. Jaikin-Zapirain ^{†}

Departamento de Matemáticas, Facultad de Ciencias,

Universidad Autónoma de Madrid e-mail: ajaikin@uam.es

December 9, 2002

Abstract

Let G be a finite p-group satisfying [G, G] ≤ G^{4} for p = 2 and γ_{p−1}(G) ≤ G^{p} for p > 2. The main goal of this paper is to show that any normal subgroup of G lying in G^{2} is power abelian.

∗The work of the first author has been partially supported by the Government of the Basque Country.

†The work of the second author has been partially supported by the MCYT Grants BFM2001-0201, BFM2001-0180, FEDER and the Ramón y Cajal Program.

1 Introduction

After the work of A. Lubotzky and A. Mann ([6]) had appeared, the importance of powerful p-groups in the study of finite p-groups became clear. The structure of powerful p-groups is quite similar to that of abelian groups. For example, it is known that if G is a powerful p-group, then G^{p^{i}} is also powerful and G^{p^{i}} is precisely the set of p^{i}th powers of G. Recently, L. Wilson ([7]) proved that when p is odd, Ω_{i}(G) (the subgroup generated by the elements of order less than or equal to p^{i}) is precisely the set of elements of order less or equal than p^{i}; even more, he proved that the nilpotency class of Ω_{i}(G) in bounded by i + 1 (a short proof of these facts is also given by G. Fernández-Alcober in [3]). This, together with a result of L. Hethelyi and L. Levai ([4]), gives us that powerful p-groups are power abelian, that is

1. G^{p^{k}} = {g^{p^{k}} | g ∈ G},

2. Ω_{k}(G) = {g ∈ G | o(g) ≤ p^{k}},

3. |G^{p^{k}}| = |G : Ω_{k}(G)|.

In 1960s D. Arganbright ([1]) proved that if G is a p-group, p is odd and G satisfies that γ_{p−1}(G) ≤ G^{p}, then it follows that G is p-powered (G^{p} is precisely the set of p-powers of G). The aim of this paper is to extend these results to other families of finite p-groups.

We say that a finite p-group is potent if [G, G] ≤ G^{4} for p = 2 or γ_{p−1}(G) ≤ G^{p} for p > 2. Note that for p = 2 and p = 3 to be potent is the same as powerful. The main results of this paper are collected in the next theorem:

Theorem 1.1. Let G be a finite potent p-group.

1. If p = 2 then

(a) The exponent of Ω_{i}(G) is at most 2^{i+1}, even more, [Ω_{i}(G), G]^{2^{i}} = Ω_{i}(G^{2})^{2^{i}} = 1.

(b) The nilpotency class of Ω_{i}(G) is at most ⌊(i + 3)/2⌋.

(c) If N ◁ G and N ≤ G^{2} then N is power abelian.

(d) If N ◁ G and N ≤ G^{4} then N is powerful.

2. If p > 2 then

(a) The exponent of Ω_{i}(G) is at most p^{i}.

(b) The nilpotency class of Ω_{i}(G) is at most (p − 2)i + 1.

(c) If N ◁ G then N is power abelian.

(d) If N ◁ G and N ≤ G^{p} then N is powerful.

We want to note that a part of the previous theorem is also proved in the Ph. D. Thesis of L. Wilson ([8]) using different methods.

Acknowledgments. We would like to thank G. Fernández-Alcober, A. Mann and L. Wilson for some helpful comments on a preliminary version of this work.

2 Preliminaries

Before starting with the main part of this paper we will introduce some definitions and recall some properties about p-groups. To help us in the writing we will use the notation p = p if p is odd and p = 4 if p = 2 (note that G^{p/p} = G^{2}). We will write [N,_{k} M] for [N, M, ⋯, M], where M appears k times. The symbols ⌈r⌉ and ⌊r⌋ will denote the smallest integer greater than or equal to r and the greatest integer less than or equal to r respectively.

In the next theorem we collect the basic properties of potent p-groups (see [1, 6]).

Theorem 2.1. Let G be a potent p-group. Then the following properties hold

1. If p = 2, then γ_{k+1}(G) ≤ γ_{k}(G)^{4}, and if p > 2 then γ_{p−1+k}(G) ≤ (γ_{k+1}(G))^{p}.

2. γ_{i}(G) is potent.

3. ⟨x, [G, G]⟩ is potent, for all x ∈ G.

4. G is p-powered.

5. If N is a normal subgroup of G then G/N is potent.

The next two results are basic facts about finite p-groups.

Theorem 2.2. Let G be a finite p-group and N, M normal subgroups of G. If N ≤ M[N, G]N^{p}. Then N ≤ M.

Theorem 2.3 (Three Subgroups Lemma). Let G be a p-group, H, J and K subgroups of G and N a normal subgroup of G such that [H, J, K], [K, H, J] ≤ N. Then [J, K, H] ≤ N, in particular if N, L are normal subgroups of G, [N, N, L] ≤ [L, N, N].

Corollary 2.4. Let G be a p-group and N a normal subgroup of G. Then [γ_{k}(G), N] ≤ [N,_{k} G].

The following theorem given by P. Hall will be used several times in the paper. It is known as P. Hall’s collection formula.

Theorem 2.5 (P. Hall’s Formula). Let G be a group and x, y elements of G. Then

(xy)^{p^{n}} ≡ x^{p^{n}}y^{p^{n}} (mod γ_{2}(L)^{p^{n}}γ_{p}(L)^{p^{n−1}}γ_{p^{2}}(L)^{p^{n−2}}γ_{p^{3}}(L)^{p^{n−3}} ⋯ γ_{p^{n}}(L))

where L = ⟨x, y⟩, and

[x, y]^{p^{n}} ≡ [x^{p^{n}}, y] (mod γ_{2}(M)^{p^{n}}γ_{p}(M)^{p^{n−1}}γ_{p^{2}}(M)^{p^{n−2}} ⋯ γ_{p^{n}}(M))

where M = ⟨x, [x, y]⟩.

3 Potently Embedded Subgroups

Let G be a p-group and N a normal subgroup, we say that N is potently embedded in G if [N, G] ≤ N^{4} for p = 2 and [N,_{p−2} G] ≤ N^{p} for p odd. In this section we will study potently embedded subgroups of potent p-groups. Since when p = 2, a potent 2-group is powerful, we will only consider potent p-groups for odd p. We will obtain similar results as in the case of powerful p-groups. The following theorem is a generalization of Interchanging Lemma (see [5]).

Theorem 3.1. Let N and M be potently embedded subgroups of G. Then [N^{p}, M] = [N, M]^{p}.

Proof. By P. Hall’s formula we have that

[N^{p}, M] ≤ [N, M]^{p}[M,_{p} N] ≤ [N, M]^{p}[[M^{p}, N], N].

Now we apply again P. Hall’s formula with [M^{p}, N] and we get that [N^{p}, M] ≤ [N, M]^{p}[N^{p}, M, M]. Therefore [N^{p}, M] ≤ [N, M]^{p}.

In order to prove the converse, we can assume that [N^{p}, M] = [[N, M]^{p}, G] = 1. Note that since M is potently embedded in G, [M,_{p} N] ≤ [M^{p}, N, N]. By the previous argument, [M^{p}, N] ≤ [M, N]^{p}, so

[M,_{p} N] ≤ [[M, N]^{p}, N] = 1.

Now, let us prove by the inverse induction on k that [N, M, γ_{k}(N)]^{p} = 1 for all k ≥ 1. It is clear when k is big enough. So, suppose that [N, M, γ_{k+1}(N)]^{p} = 1. Then we have

[N, M, γ_{k}(N)]^{p} ≤ [[N, M]^{p}, N][[N, M], γ_{k+1}(N)]^{p}[M,_{p} N] = 1.

Therefore we have that

[N, M]^{p} ≤ [N^{p}, M][N, M, N]^{p}[M,_{p} N] = 1.

Theorem 3.2. Let G be a potent p-group and N, M potently embedded subgroups of G. Then we have that

1. NM is potently embedded.

2. [N, G] is potently embedded.

3. N^{p} is potently embedded.

Proof. 1. It is obvious because [NM,_{p−2} G] ≤ [N,_{p−2} G][M,_{p−2} G].

2. We have that [[N, G],_{p−2} G] ≤ [N^{p}, G] and, by the previous theorem, [[N, G],_{p−2} G] ≤ [N, G]^{p}.

3. By the previous theorem, [N^{p},_{p−2} G] ≤ [N,_{p−2} G]^{p} ≤ (N^{p})^{p}, therefore N^{p} is potently embedded.

Now we introduce dimension subgroups of G by induction. We say that D_{1} = D_{1}(G) = G and for i ≥ 2 we put D_{i} = D_{i}(G) = D_{⌈i/p⌉}^{p}[D_{i−1}, G].

From Theorem 3.2 it follows that if G is a potent p-group, then D_{i} is potently embedded in G for all i. The theory of dimension subgroups is developed in [2]. If G is a potent p-group then they have the following structure:

Theorem 3.3. Suppose n ∈ ℕ and i = ⌊log_{p} n⌋ and let 1 ≤ k ≤ p be such that (k − 1)p^{i} < n ≤ kp^{i}. Then, if G is a potent p-group, we have

D_{n} = \{ \begin{matrix}D_{k}^{p^{i}} & if 1 ≤ k ≤ p − 2; \\ D_{1}^{p^{i+1}} & if p − 1 ≤ k ≤ p. \end{matrix}

4 Normal Subgroups of Potent p-Groups

The next result is fundamental in this paper. In some way, with this theorem we are going to be able to reduce problems about normal subgroups to a particular case.

Theorem 4.1. Let G be a potent p-group and N a normal subgroup of G. Then one of the following holds

1. For any i, s, t ≥ 0 such that n = i + s + t ≥ 1 if p is odd and n ≥ 2 if p = 2, [G, G]^{p^{n}} ≤ [N^{p^{i}}, G^{p^{s}}]^{p^{t}}.

2. There exists a proper potent subgroup T of G such that N is contained in T.

Proof. If [G, G]^{p} ≤ [N, G^{p}] then, using P. Hall’s formula, Theorem 2.1(1) and Theorem 3.1 we obtain that

[G, G]^{p} ≤ [N, G]^{p}[[G, G]^{p}, G].

Now, if p > 2, by Theorem 2.2, [G, G]^{p} ≤ [N, G]^{p}. If p = 2, then [[G, G]^{2}, G] ≤ [G, G]^{8} and so, using again Theorem 2.2, we get [G, G]^{4} ≤ [N, G]^{4}. Therefore, we have proved the base of induction for [G, G]^{p^{n}} ≤ [N, G]^{p^{n}} when p^{n} ≥ p.

Now, suppose that [G, G]^{p^{n}} ≤ [N, G]^{p^{n}}. We want to prove [G, G]^{p^{n+1}} ≤ [N, G]^{p^{n+1}}. We have

[G, G]^{p^{n+1}} ≤ ([N, G]^{p^{n}})^{p} ≤ [N, G]^{p^{n+1}}γ_{2}([N, G]^{p^{n}}) ≤ [N, G]^{p^{n+1}}[[G, G]^{p^{2n}}, G],

which implies [G, G]^{p^{n+1}} ≤ [N, G]^{p^{n+1}}.

Let i, s, t be non-negative integers such that i + s + t = n ≥ 1 if p is odd and n ≥ 2 if p = 2. Bearing again in mind P. Hall’s formula, Theorem 2.1(1) and Theorem 3.1 we conclude that

[G, G]^{p^{n}} ≤ [N, G]^{p^{n}} ≤ [N^{p^{i}}, G^{p^{s}}]^{p^{t}}[[G, G]^{p^{n+1}/p}, G],

which implies as before that [G, G]^{p^{n}} ≤ [N^{p^{i}}, G^{p^{s}}]^{p^{t}}.

If [G, G]^{p} ≰ [N, G^{p}] then there exists k ≥ 1 such that [G, D_{k}^{p}] ≰ [N, G^{p}] and [G, D_{k+1}^{p}] ≤ [N, G^{p}]. We take

T = {g ∈ G | [g, D_{k}^{p}] ≤ [N, G^{p}]}.

It is clear that, as N is normal in G, T is a proper normal subgroup of G and N is a normal subgroup of T. Let us see that T is potent. Take x ∈ [T, T] if p = 2 and x ∈ γ_{p−1}(T) if p is odd. Then we have that x ∈ G^{p} so, by Theorem 2.1(4), x = a^{p} for some a ∈ G. Now, by P. Hall’s formula and Theorem 3.1,

[a, D_{k}^{p}] ≤ [a, D_{k}]^{p}[G, D_{k}, D_{k}]^{p}[G,_{p} D_{k}]
≤ [a, D_{k}]^{p}[[G, D_{k}]^{p}, D_{k}][D_{k+1},_{p−2} D_{k}, D_{k}]
≤ [a, D_{k}]^{p}[D_{k+1}^{p}, D_{k}] ≤ [a, D_{k}]^{p}[N, G^{p}],

and

[a, D_{k}]^{p} ≤ [a^{p}, D_{k}][D_{k}, G, G]^{p}[D_{k},_{p} G]
≤ [a^{p}, D_{k}][[G, D_{k}]^{p}, G][D_{k+1},_{p−2} G, G]
≤ [a^{p}, D_{k}][N, G^{p}].

Now, suppose p = 2. Then by Theorem 2.3,

[a^{4}, D_{k}] ≤ [T, T, D_{k}] ≤ [D_{k}, T, T] ≤ [D_{k}^{4}, T] ≤ [N, G^{4}].

Therefore we deduce that [a, D_{k}^{4}] ≤ [N, G^{4}] and a ∈ T. Thus x = a^{4} ∈ T^{4}. So [T, T] ≤ T^{4} and T is potent.

If p > 2, then, by Corollary 2.4 we have

[a^{p}, D_{k}] ≤ [γ_{p−1}(T), D_{k}] ≤ [D_{k},_{p−1} T] ≤ [D_{k}^{p}, T] ≤ [N, G^{p}].

As before we obtain that a ∈ T and T is potent.

Now we are going to prove our first results for normal subgroups of powerful p-groups. Unfortunately, we cannot always prove for p = 2 the same properties as for odd p. Anyway, we will see that there exists a similarity between p = 2 and the odd case. As we will see in the last part of the paper, in the case p = 2 there are counterexamples for some results which hold if p is odd.

First we prove an auxiliary lemma.

Lemma 4.2. Let G be a potent p-group, N a normal subgroup of G and i ≥ 1. Then (N^{p^{i}})^{p} = N^{p^{i+1}} for every i ≥ 1.

Proof. The inclusion (N^{p^{i}})^{p} ≥ N^{p^{i+1}} is obvious. In order to prove the converse inclusion we use the previous theorem, so we can suppose that [G, G]^{p^{i+1}} ≤ [N, G]^{p^{i+1}}. Then we have

(N^{p^{i}})^{p} ≤ N^{p^{i+1}}[N^{p^{i}}, N^{p^{i}}] ≤ N^{p^{i+1}}[G, G]^{p^{2i}} ≤ N^{p^{i+1}}.

Theorem 4.3. Let G be a potent p-group and N a normal subgroup of G. Then

1. If p = 2, γ_{3}(N) ≤ [N, G]^{4}∩[N^{4}, G] and if p > 2, γ_{p}(N) ≤ [N, G]^{p}∩[N^{p}, G].

2. If p is odd, [N^{p}, N^{p}] ≤ [N, N]^{p^{2}} and if p = 2, [N^{2}, N^{2}] ≤ [N, N]^{4}[N, N, N]^{2} ≤ N^{8}. In particular, N^{p} is a powerful p-group.

3. The nilpotency class of N is at most e(N)(p − 2) + 1 if p is odd and ⌊(e(N) + 3)/2⌋ if p = 2 where e(N) = log_{p}(exp(N)).

Proof. In order to prove the theorem we argue by induction on the order of G. Thus, by Theorem 4.1, we may suppose that

[G, G]^{p^{n}} ≤ [N^{p^{i}}, G^{p^{s}}]^{p^{t}}

for n = i + s + t ≥ 1 if p is odd and n ≥ 2 if p = 2.

(1) Since G is potent, γ_{3}(G) ≤ [G, G]^{4} if p = 2 and γ_{p}(G) ≤ [G, G]^{p} if p > 2. Now, the statement follows directly from the assumption made in the previous paragraph.

(2) We claim that if p is odd then [N^{p}, N^{p}] ≤ [N, N]^{p^{2}}. Without loss of generality we can suppose that [N^{p}, N^{p}, G] = 1. First note that, by P. Hall’s formula,

[N^{p^{2}}, N] ≤ [N^{p}, N]^{p}[N,_{p} N^{p}] = [N^{p}, N]^{p},

and so

[N^{p}, N^{p}] ≤ [N^{p}, N]^{p}[N^{p},_{p} N] ≤ [N^{p}, N]^{p}[G^{p^{2}}, N, N]
≤ [N^{p}, N]^{p}[[G, N]^{p^{2}}, N] ≤ [N^{p}, N]^{p}.

Now,

[N^{p}, N] ≤ [N, N]^{p}[N,_{p} N] ≤ [N, N]^{p}[[G, G]^{p}, N] ≤ [N, N]^{p}[[N, G]^{p}, N].

But

[[N, G]^{p}, N] ≤ [N, G, N]^{p}[N,_{p} [N, G]] ≤ [N, N]^{p}[[N, G]^{p}, N, N],

so we have that [[N, G]^{p}, N] ≤ [N, N]^{p}. Therefore, by Lemma 4.2,

[N^{p}, N^{p}] ≤ ([N, N]^{p})^{p} = [N, N]^{p^{2}}.

In particular, N^{p} is powerful.

When p = 2 we can suppose that [N, N]^{4}[N, N, N]^{2} = 1. From this we have that

γ_{4}(N) ≤ [[N, G]^{4}, N] ≤ [N, N]^{4}[N, N, N]^{2}γ_{5}(N) = γ_{5}(N),

and consequently γ_{4}(N) = 1. Now we have that

[N^{2}, N^{2}] ≤ [N, N^{2}]^{2}[N^{2}, N, N] = [N, N^{2}]^{2}[N, N, N]^{2} = [N, N^{2}]^{2}

and

[N, N^{2}]^{2} ≤ ([N, N]^{2}[N, N, N])^{2} = [N, N]^{4}[N, N, N]^{2} = 1.

Therefore [N^{2}, N^{2}] ≤ [N, N]^{4}[N, N, N]^{2}. And, as [N, N] ≤ N^{2}, by Lemma 4.2, we have that [N, N]^{4}[N, N, N]^{2} ≤ N^{8}. Hence [N^{2}, N^{2}] ≤ N^{8}[N, G]^{8} ≤ (N^{2})^{4} and so N^{2} is powerful.

(3) If p > 2 we have that γ_{e(N)(p−2)+2}(N) ≤ [G, G]^{p^{e(N)}} and by the assumption made at the beginning of the proof, this is equal to [G, N]^{p^{e(N)}}. In particular, if p is odd the class of N is bounded by e(N)(p − 2) + 1.

If p = 2, γ_{n+1}(N) ≤ [G, N]^{4^{n−1}}. Therefore, the class of N is bounded by ⌊(e(N) + 3)/2⌋.

Remark. If p > 2 and G is a finite p-group satisfying γ_{k}(G) ≤ G^{p} for some 2 ≤ k ≤ p − 1, then, using a similar argument, we can prove that the nilpotency class of a normal subgroup N is at most e(N)(k − 1) + 1.

Theorem 4.4. Let G be a potent p-group. Then

1. If p is odd the exponent of Ω_{i}(G) is at most p^{i}. In particular, the nilpotency class of Ω_{i}(G) is bounded by i(p − 2) + 2.

2. If p = 2 the exponent of Ω_{i}(G) is bounded by 2^{i+1} and γ_{2}(Ω_{i}(G))^{2^{i}} = 1. In particular the nilpotency class of Ω_{i}(G) is less than or equal to ⌊(i+4)/2⌋.

Proof. For ease of notation, let us write Ω_{i} instead of Ω_{i}(G). By Theorem 4.1, we can suppose that [G, G]^{p^{n}} ≤ [Ω_{i}^{p^{l}}, G^{p^{s}}]^{p^{t}} for any l, s, t ≥ 0 such that l + s + t = n ≥ 1 if p is odd and n ≥ 2 if p = 2.

First we suppose that p > 2. In this case it is enough to prove the theorem for i = 1. Since Ω_{1} is generated by elements of order p,

Ω_{1}^{p} ≤ [Ω_{1}, Ω_{1}]^{p}γ_{p}(Ω_{1}) ≤ [G, G]^{p} ≤ [Ω_{1}^{p}, G].

Hence Ω_{1}^{p} = 1.

Now, we suppose that p = 2. There exists j such that [Ω_{i}, G]^{2^{j+1}} = 1. We will see that if j ≥ i, then [Ω_{i}, G]^{2^{j}} = 1. Of course, it implies [Ω_{i}, G]^{2^{i}} = 1.

First we prove that the order of the elements [x, g], where x ∈ G has order 2^{i} and g ∈ G, is bounded by 2^{j}.

[x, g]^{2^{j}} ≡ [x^{2^{j}}, g] (mod γ_{3}(G)^{2^{j−1}}γ_{5}(G)^{2^{j−2}} ⋯).

But γ_{3}(G) ≤ [Ω_{i}, G]^{4} and in general γ_{2^{k}+1}(G) ≤ [Ω_{i}, G]^{4^{2^{k}−1}} for k > 1. From this we deduce that [x, g]^{2^{j}} = 1.

The set {[x, g]|x^{2^{i}} = 1, g ∈ G} is a generating set of [Ω_{i}, G]. Now we prove by induction on the length of h ∈ [Ω_{i}, G] respect to this generating set that h^{2^{j}} = 1. Suppose that x^{2^{j}} = y^{2^{j}} = 1 for some x, y ∈ [Ω_{i}, G]. We apply P. Hall’s formula and we have that

(xy)^{2^{j}} ≡ x^{2^{j}}y^{2^{j}} (mod γ_{2}([Ω_{i}, G])^{2^{j−1}}γ_{4}([Ω_{i}, G])^{2^{j−2}} ⋯).

Therefore, (xy)^{2^{j}} ∈ [Ω_{i}, G]^{2^{j+1}} = 1 and we conclude that [Ω_{i}, G]^{2^{j}} = 1.

It is clear now that, when p = 2, [Ω_{i}, Ω_{i}]^{2^{i}} = 1. We want to see that Ω_{i}(G)^{2^{i+1}} = 1. By P. Hall’s formula

Ω_{i}^{2^{i+1}} ≤ γ_{2}(Ω_{i})^{2^{i}}γ_{4}(Ω_{i})^{2^{i−1}} · · · .

But, as we have that γ_{2^{k}}(Ω_{i}) ≤ [Ω_{i}, G]^{4^{2^{k}−2}}, it follows that the exponent of Ω_{i} is less than or equal to 2^{i+1}.

Corollary 4.5. Let G be a potent p-group. Then Ω_{1}(G^{p}) is abelian and Ω_{i}(G^{p}) has exponent less than or equal to p^{i}.

Proof. It is clear that it is enough to prove that Ω_{1}(G^{p}) is abelian, because that implies that the exponent of Ω_{1}(G^{p}) is p and, by induction on i, that Ω_{i}(G^{p})^{p^{i}} = 1.

Note that Ω_{1}(G^{p}) ≤ Ω_{2}(G)^{p}, and by Theorem 4.3 and Theorem 4.4,

[Ω_{2}(G)^{p}, Ω_{2}(G)^{p}] ≤ Ω_{2}(G)^{pp} = {1}.

Hence, Ω_{1}(G^{p}) is abelian.

Corollary 4.6. Let G be a potent p-group and N, L ≤ G^{2} normal subgroups of G. Then (NL)^{p} = N^{p}L^{p}.

Proof. It is obvious that N^{p}L^{p} ≤ (NL)^{p}. Now, we will prove that N^{p}L^{p} ≥ (NL)^{p}. Note that G/N^{p}L^{p} is a potent p-group and NL/N^{p}L^{p} ≤ Ω_{1}(G^{2}/N^{p}L^{p}). But the exponent of Ω_{1}(G^{2}/N^{p}L^{p}) is p, whence (NL)^{p} ≤ N^{p}L^{p} and therefore (NL)^{p} = N^{p}L^{p}.

Corollary 4.7. Let G be a potent p-group and N a normal subgroup of G such that N ≤ G^{p}. Then N is powerful.

Proof. We take

T = ⟨x ∈ G^{2} | x^{p} ∈ N⟩.

It is clear that T is a normal subgroup of G. Then, as the exponent of T/N = Ω_{1}(G^{2}/N) is p, we have that T^{p} ≤ N. From Theorem 4.3 we know that T^{p} is powerful. Therefore, in order to prove that N is powerful, it is enough to see that N ≤ T^{p}. For any n ∈ N there is x ∈ G^{2} such that x^{p} = n, and by the definition of T, x ∈ T and we have that N ≤ T^{p}.

5 Normal subgroups of powerful p-groups

It is clear that every powerful p-group is also potent and all results of the previous section are valid for these groups. But for powerful p-groups we can obtain some additional results. The proofs are based on an analog of Theorem 4.1 for powerful groups:

Theorem 5.1. Let G be a powerful p-group and N a normal subgroup of G. Then one of the following two properties holds:

1. for any i, s, t ≥ 0 such that n = i+s+t ≥ 1 if p is odd and n ≥ 2 if p = 2, [G, G]^{p^{n}} ≤ [N^{p^{i}}, G^{p^{s}}]^{p^{t}}.

2. There is a proper powerful subgroup T of G such that N ≤ T.

Proof. Arguing in the same way as in the proof of Theorem 4.1

Corollary 5.2. Let G be a powerful p-group. Then

1. if p is odd, [Ω_{i}(G), G] is powerfully embedded in Ω_{i}(G) and [Ω_{i}(G), G]^{p} ≤ [Ω_{i−1}(G), G],

2. if p = 2, [G, Ω_{i}(G), Ω_{i}(G)] ≤ [Ω_{i−1}(G), G] and [Ω_{i}(G), G]^{2} ≤ [Ω_{i−1}(G), G]. In particular the exponent of [Ω_{i}(G), G] is bounded by 2^{i}.

Proof. If p is odd we have that [Ω_{i}, G] ≤ Ω_{i} ∩ G^{p} = Ω_{i+1}^{p}. We apply Theorem 5.1 to N = Ω_{i+1} and we obtain that there exists a potent subgroup T of G such that for any i, s, t ≥ 0 with n = i + s + t ≥ 1, [T, T]^{p^{n}} ≤ [Ω_{i+1}^{p^{s}}, T^{p^{i}}]^{p^{t}}. Hence we have that

[Ω_{i+1}^{p}, Ω_{i}] ≤ [Ω_{i+1}, Ω_{i}]^{p}[Ω_{i}, Ω_{i+1}, Ω_{i+1}]^{p}γ_{4}(Ω_{i+1}).

But γ_{4}(Ω_{i+1}) ≤ [T, T]^{p^{2}} = [Ω_{i+1}^{p}, T]^{p} and we obtain that

[Ω_{i+1}^{p}, Ω_{i}] ≤ [Ω_{i}, T]^{p}.

We conclude that

[G, Ω_{i}, Ω_{i}] ≤ [Ω_{i+1}^{p}, Ω_{i}] ≤ [Ω_{i}, T]^{p},

and, in particular, [Ω_{i}, G] is powerfully embedded in Ω_{i}, and also

[G, Ω_{i}]^{p} ≤ [G, Ω_{i}^{p}][G, Ω_{i}, Ω_{i}]^{p}[G,_{p} Ω_{i}] ≤ [G, Ω_{i−1}][Ω_{i}^{p^{2}}, T] ≤ [G, Ω_{i−1}].

Now, let p = 2. As exp(Ω_{i}) ≤ 2^{i+1}, we have that [G, Ω_{i}] ≤ Ω_{i} ∩ G^{4} ≤ Ω_{i+3}^{4}. Then if we apply Theorem 5.1 to N = Ω_{i+3}, we obtain that there exists a potent subgroup T of G such that for any i, s, t ≥ 0 with n = i + s + t ≥ 2, [T, T]^{2^{n}} ≤ [Ω_{i+3}^{2^{s}}, T^{2^{i}}]^{2^{t}}. We have

[Ω_{i+3}^{4}, Ω_{i}] ≤ [Ω_{i+3}, Ω_{i}]^{4}[Ω_{i}, Ω_{i+3}, Ω_{i+3}]^{2}γ_{5}(Ω_{i+3}).

But γ_{5}(Ω_{i+3}) ≤ [Ω_{i+3}^{2^{6}}, T] ≤ [Ω_{i−1}, T] and, since Ω_{i}^{2} ≤ Ω_{i−1}[Ω_{i}, Ω_{i}], we obtain

[Ω_{i}, Ω_{i+3}, Ω_{i+3}]^{2} ≤ [Ω_{i−1}, T][Ω_{i}, Ω_{i}, Ω_{i+3}, Ω_{i+3}]
≤ [Ω_{i−1}, T][Ω_{i}, Ω_{i+3}, Ω_{i}, Ω_{i+3}] ≤ [Ω_{i−1}, T][Ω_{i+3}^{4}, Ω_{i}, Ω_{i+3}].

From this we have

[Ω_{i+3}^{4}, Ω_{i}] ≤ [Ω_{i+3}, Ω_{i}]^{4}[Ω_{i+3}^{4}, Ω_{i}, Ω_{i+3}][Ω_{i−1}, T].

Therefore [Ω_{i+3}^{4}, Ω_{i}] ≤ [Ω_{i+3}, Ω_{i}]^{4}[Ω_{i−1}, T]. Now, by P. Hall’s Formula,

[Ω_{i+3}, Ω_{i}]^{4} ≤ [Ω_{i+3}, Ω_{i}^{4}][Ω_{i+3}, Ω_{i}, Ω_{i}]^{2}γ_{5}(Ω_{i+3}) ≤ [Ω_{i+3}, Ω_{i}, Ω_{i}]^{2}[Ω_{i−1}, T].

Hence we have [G, Ω_{i}, Ω_{i}] ≤ [Ω_{i+3}^{4}, Ω_{i}] ≤ [G, Ω_{i}, Ω_{i}]^{2}[Ω_{i−1}, G] and we finally obtain that [G, Ω_{i}, Ω_{i}] ≤ [Ω_{i−1}, G].

Now, since Ω_{i}^{2} ≤ Ω_{i−1}[Ω_{i}, Ω_{i}], it follows that

[Ω_{i}, G]^{2} ≤ [Ω_{i−1}, G][G, Ω_{i}, Ω_{i}] ≤ [Ω_{i−1}, G].

Now, we are ready to obtain the final bound on the nilpotency class of Ω_{i}(G) where G is a powerful 2-group.

Corollary 5.3. Let G be a powerful 2-group. Then the nilpotency class of Ω_{i}(G) is at most ⌊(i + 3)/2⌋.

Proof. Applying Theorem 5.1 for N = Ω_{i}(G), we obtain that γ_{n+2}(Ω_{i}(G)) ≤ [Ω_{i}(G), G]^{4^{n}}. By the previous theorem, the nilpotency class of Ω_{i}(G) is at most ⌊(i + 3)/2⌋.

6 Normal subgroups of potent p-groups are power abelian

In this section we will finish our study of the structure of normal subgroups of potent p-groups. Due to a theorem given by L. Hethelyi and L. Levai together with L. Wilson’s result we know that powerful p-groups are power abelian when p is odd. Now we will extend this result to normal subgroups of potent p-groups. First we show that normal subgroups of potent p-groups are p-powered.

Theorem 6.1. Let G be a potent p-group and N ≤ G^{2} a normal subgroup of G. Then N is a p-powered group.

Proof. We prove the theorem by induction on |N| and |G|. Let x, y be two elements of N. Then, if p is odd,

x^{p}y^{p} ≡ (xy)^{p} (mod γ_{2}(N)^{p}γ_{p}(N)).

But, by Theorem 4.3, γ_{p}(N) ≤ [N, G]^{p}, and therefore

x^{p}y^{p} ≡ (xy)^{p} (mod [N, G]^{p}).

If p = 2, let T = {g ∈ G | g^{2} ∈ N}. It follows from Theorem 4.4 that T^{4} ≤ N. Since N ≤ T^{2} and [T^{2}, T, T] ≤ [T, T, T]^{2}γ_{4}(T),

[N, N] ≤ [N, T^{2}] ≤ [N, T]^{2}[N, T, T] = [N, T]^{2}[T^{2}, T, T] ≤ [N, T]^{2}γ_{3}(T)^{2}γ_{4}(T).

But γ_{3}(T) ≤ [T^{4}, G] ≤ [N, G] and γ_{4}(T) ≤ [T^{4}, G]^{4} ≤ [N, G]^{4}, whence [N, N] ≤ [N, G]^{2} and therefore

x^{2}y^{2} ≡ (xy)^{2} (mod [N, G]^{2}).

Since [N, G] < N, by induction we have that [N, G] is p-powered. Thus x^{p}y^{p} = (xy)^{p}c^{p} with c ∈ [N, G]. Take H = ⟨xy, [G, G]⟩, which can be a cyclic group or a potent p-group of smaller order than G (see Theorem 2.1). If it is cyclic we are done. Otherwise, as c ∈ H and xy ∈ H, by induction there exists z ∈ H ∩ N such that (xy)^{p}c^{p} = z^{p}. Thus N^{p} = {n^{p} | n ∈ N}. Now, as N^{p} is powerful, by Theorem 4.3, it follows that N^{p^{i}} = {n^{p^{i}} | n ∈ N} and therefore N is p-powered.

The next lemma is an analogue of Lemma 3 from [4] for potent p-groups.

Lemma 6.2. Let G be a potent p-group and x ∈ D_{i} such that x^{p} ∈ D_{ip+1}. Then there exists w ∈ D_{i} such that w^{p} = 1 and xD_{i+1} = wD_{i+1}.

Proof. The case p = 2 is proved in [4]. We suppose that p > 2. If x ∈ D_{i+1} then w = 1 and we are done. Otherwise, by Theorem 3.3, there is j ≥ i + 1 such that x^{p} ∈ D_{j}^{p}. We are going to argue by inverse induction on j, if j is big enough x^{p} = 1 and we are done. Since D_{j} is p-powered, we obtain that x^{p} ∈ D_{j}^{p} implies the existence of s ∈ D_{j} such that s^{p} = x^{p}. Hence we have that

(xs^{−1})^{p} ≡ 1 (mod [D_{j}, D_{i}]^{p}[D_{j},_{p−1} G]).

Since D_{j} is potently embedded subgroup of G, [D_{j},_{p−1} G] ≤ [D_{j}, G]^{p}. From this we obtain that (xs^{−1})^{p} ∈ D_{j+1}^{p}. On the other hand s ∈ D_{i+1}, whence xD_{i+1} = xs^{−1}D_{i+1} and we can apply the induction hypothesis because (xs^{−1})^{p} ∈ D_{j+1}^{p}.

Proposition 6.3. Let G be a potent p-group and N ≤ G^{2} a normal subgroup of G such that Ω_{1}(G^{2}) ≤ N. Then N^{p} ∩ D_{pi} = (N ∩ D_{i})^{p} for every i ≥ 1.

Proof. It is obvious that (N ∩ D_{i})^{p} ≤ N^{p} ∩ D_{pi}. Let us see that

N^{p} ∩ D_{pi} ≤ (N ∩ D_{i})^{p}

by inverse induction on i. Then we assume that N^{p} ∩ D_{p(i+1)} ≤ (N ∩ D_{i+1})^{p}. Let x ∈ N^{p} ∩ D_{pi}. By Theorem 6.1, we have that N is p-powered, whence there exists y ∈ N satisfying x = y^{p}. Let j ≤ i − 1 be such that y ∈ D_{i−j}. If j ≤ 0 we are done.

If j ≥ 1, we work by decreasing induction on j. Let us suppose that if y ∈ N ∩ D_{i−j+1} there exists z ∈ N ∩ D_{i} such that z^{p} = y^{p}.

Since y ∈ D_{i−j} for some j ≥ 1 and y^{p} ∈ D_{pi}, we can apply Lemma 6.2. So, there exists w ∈ D_{i−j} of order p such that yD_{i−j+1} = wD_{i−j+1}. Let g ∈ D_{i−j+1} be such that y = wg. Since y, w ∈ N, we also have g ∈ N.

Now, suppose first that p > 2. Let L be a normal subgroup of G generated by w and y. Since y^{p} and w^{p} lie in D_{pi}, L^{p} ≤ D_{pi}. Then we have

y^{p} = (wg)^{p} ≡ g^{p} (mod [L, L]^{p}γ_{p}(L)).

From Theorem 4.3 we obtain that y^{p} ≡ g^{p} (mod [L^{p}, G]). Hence y^{p} ≡ g^{p} (mod N^{p} ∩ D_{i+1}^{p}).

If p = 2 we have that y^{2} = (wg)^{2} = g^{2}[g, w]^{g}. But [g, w] ∈ N^{2} ∩ D_{i+1}^{2} and we also have y^{2} = g^{2} (mod N^{2} ∩ D_{i+1}^{2}).

Then, by the first induction hypothesis, y^{p} = g^{p}c^{p} for some c ∈ N ∩ D_{i+1}. From this we have that y^{p} = g^{p}c^{p} = y_{1}^{p} for some y_{1} ∈ N ∩ D_{i−j+1}. Now we can apply the second induction hypothesis and we get that there is z ∈ N ∩ D_{i} such that x = z^{p}. Therefore we have that N^{p} ∩ D_{pi} = (N ∩ D_{i})^{p}.

Corollary 6.4. Let G be a potent p-group and N ≤ G^{2} a normal subgroup of G such that Ω_{1}(G^{2}) ≤ N. Then |Ω_{1}(N)| = |N/N^{p}|.

Proof. Let us consider the following abelian group:

L(G) = (D_{1}/D_{2}) ⊕ (D_{2}/D_{3}) ⊕ (D_{3}/D_{4}) ⊕ ….

We define a structure of 𝔽_{p}[t]-module on L(G) by putting t(xD_{i+1}) = x^{p}D_{pi+1} if x ∈ D_{i} and extending this action by linearity.

If M is finitely generated 𝔽_{p}[t]-module, we define d_{𝔽_{p}[t]}(M) to be the minimal number of 𝔽_{p}[t]-generators of M. Note that if t^{k}M = 0 for some k, then

d_{𝔽_{p}[t]}(M) = log_{p} |M/tM| = log_{p} | Ann_{M}(t)| = d_{𝔽_{p}[t]}(Ann_{M}(t)),

where Ann_{M}(t) = {x ∈ M|tx = 0}.

For any normal subgroup K of G we consider the following submodule

L(K) = (KD_{2})/D_{2} ⊕ ((KD_{3}) ∩ D_{2})/D_{3} ⊕ ((KD_{4}) ∩ D_{3})/D_{4} ⊕ ….

Now, we have L(Ω_{1}(G^{2})) ≤ Ann_{L(G^{2})}(t), because Ω_{1}(G^{2}) has exponent p, and on the other hand, Lemma 6.2 implies that L(Ω_{1}(G^{2})) ≥ Ann_{L(G^{2})}(t). Hence we obtain that L(Ω_{1}(G^{2})) = Ann_{L(G^{2})}(t).

Since Ω_{1}(G^{2}) ≤ N ≤ G^{2} we have that L(Ω_{1}(G^{2})) ≤ L(N) ≤ L(G^{2}). Hence

d_{F_{p}[t]}(L(Ω_{1}(G^{2}))) ≤ d_{F_{p}[t]}(L(N)) ≤ d_{F_{p}[t]}(L(G^{2}))
= d_{F_{p}[t]}(Ann_{L(G^{2})}(t)) = d_{F_{p}[t]}(L(Ω_{1}(G^{2}))).

From this we can see that d_{F_{p}[t]}(L(N)) = d_{F_{p}[t]}(L(G^{2})). On the other hand

tL(N) = L(N^{p}),

because L(N^{p}) ≤ tL(N) follows from Proposition 6.3 and the inclusion tL(N) ≤ L(N^{p}) is obvious. Thus we obtain that

|Ω_{1}(N)| = |Ω_{1}(G^{2})| = |L(Ω_{1}(G^{2}))| = | Ann_{L(G^{2})}(t)| = p^{d_{Fp[t]}(L(G^{2}))}
= p^{d_{Fp[t]}(L(N))} = |L(N)/tL(N)| = |L(N) : L(N^{p})| = |N : N^{p}|.

Theorem 6.5. Let G be a potent p-group and N ≤ G^{2} a normal subgroup of G. Then |Ω_{1}(N)| = |N/N^{p}|.

Proof. By Corollary 4.6 and Corollary 4.5 we have that

(Ω_{1}(G^{2})N)^{p} = Ω_{1}(G^{2})^{p}N^{p} = N^{p}.

Then we have that, by the previous corollary,

|Ω_{1}(G^{2})| = |Ω_{1}(Ω_{1}(G^{2})N)| = \frac{|Ω_{1}(G^{2})N|}{|(Ω_{1}(G^{2})N)^{p}|} = \frac{|Ω_{1}(G^{2})N|}{|N|} \frac{|N|}{|N^{p}|}.

From here it follows that

\frac{|N|}{|N^{p}|} = \frac{|Ω_{1}(G^{2})||N|}{|Ω_{1}(G^{2})N|} = |Ω_{1}(G^{2}) ∩ N| = |Ω_{1}(N)|.

Theorem 6.6. Let G be a potent p-group and N ≤ G^{2} a normal subgroup of G. Then N is power abelian.

Proof. We have already proven that N^{p^{i}} = {n^{p^{i}} | n ∈ N} (Theorem 6.1), Ω_{i}(N) = {n ∈ N | n^{p^{i}} = 1} (Corollary 4.5) and |N^{p}| = |N : Ω_{1}(N)| (Theorem 6.5). So we just need to see that for any i ≥ 1 |N^{p^{i}}| = |N : Ω_{i}(N)|. We work by induction on i. Then

|N^{p^{i+1}}| = |(N^{p})^{p^{i}}| = |N^{p} : Ω_{i}(N^{p})|.

But Ω_{i}(N^{p}) = Ω_{i}(N) ∩ N^{p}, and we also have that

N^{p}/(Ω_{i}(N) ∩ N^{p}) ≅ N^{p}Ω_{i}(N)/Ω_{i}(N) = (N/Ω_{i}(N))^{p}.

Hence |N^{p^{i+1}}| = |(N/Ω_{i}(N))^{p}|. By applying the case i = 1 to the potent p-group G/Ω_{i}(N) and its normal subgroup N/Ω_{i}(N), we get that

|(N/Ω_{i}(N))^{p}| = |N/Ω_{i}(N) : Ω_{1}(N/Ω_{i}(N))| =
= |N/Ω_{i}(N) : Ω_{i+1}(N)/Ω_{i}(N)| = |N : Ω_{i+1}(N)|,

which concludes the proof.

7 Counterexamples

We will finish this paper by giving some counterexamples that show us why some properties are different for p = 2. The first example is a powerful 2-group that satisfies that exp(Ω_{i}(G)) = 2^{i+1} and has a normal subgroup that is not 2-powered. The second one is a powerful 2-group that has a normal subgroup such that [N^{2}, N^{2}] ≰ [N, N]^{4}.

Example. Let us consider the semidirect product

G = ⟨a⟩[⟨b⟩ × ⟨c⟩] ≅ C_{2^{i}}[C_{2^{i}} × C_{2^{i+2}}],

where a acts trivially on c and b^{a} = bc^{4}. It is quite easy to see that G is powerful. We also have that o(a) = o(b) = 2^{i} but, since [b, a] commutes with both a and b,

(ab)^{2^{i}} = a^{2^{i}}b^{2^{i}}[b, a]^{\binom{2^{i}}{2}} = c^{2^{2}\binom{2^{i}}{2}} = c^{2^{2}2^{i−1}(2^{i}−1)} = c^{2^{i+1}} ≠ 1.

So we have two elements of order 2^{i} whose product is of order 2^{i+1}, in particular exp(Ω_{i}(G)) = 2^{i+1}.

Now, if we define N = ⟨a, b, c^{4}⟩ then N is not a 2-powered.

Example. Let us consider the semidirect product

G = ⟨a⟩[⟨b⟩ × ⟨c⟩ × ⟨d⟩] ≅ C_{2^{i+2}}[C_{2^{i}} × C_{2^{i+2}} × C_{2^{i+4}}],

where b^{a} = bc^{4}, c^{a} = cd^{4} and a acts trivially on d. It is quite easy to see that G is powerful and if we take N = ⟨a, b, c^{4}, d^{4}⟩ then this normal subgroup satisfies that [N^{2}, N^{2}] ≰ [N, N]^{4}.

References

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[3] G. Fernández-Alcober, Omega subgroups of powerful p-groups, private comunication.

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