On the abundance 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

May 4, 2000

Abstract In this paper we prove that there are only finitely many p-groups of any given abundance. This result gives a positive answer to a problem raised by Shalev in [6]. We also prove that the cobreadth of a p-group can be bounded by a function which depends only on the abundance of the group (and not on p).

1 Introduction

Let G be a finite p-group and let r(G) denote the number of conjugacy classes of G. By a result of P. Hall (for example, see [2, Chapter 5, Theorem 15.2]) if |G| = p^{m} and m = 2n + e, where e = 0, 1, then there exists a non-negative integer a = a(G), which we call the abundance of G, such that

r(G) = p^{e} + (p^{2} − 1)(n + a(p − 1)).

In [5] Poland proved that if a = 0 then G is a group of maximal class whose order is at most p^{p+2}. This bound is exact for p ≤ 7. For p ≥ 11 the exact bound was found in [1] and is p^{p+1}.

In this work we consider finite p-groups with arbitrary abundance and prove the following result.

Theorem 1.1. There are only finitely many p-groups of any given abundance.

Set rk G = max{d(H) | H ≤ G}, where d(H) is the minimum number of generators of the subgroup H. Then from [6, Proposition 4.8] we know that there is a number l = l(rk G) > 0 which depends only on rk G such that r(G) ≥ |G|^{l}. Therefore in order to prove Theorem 1.1 it suffices to show that rk G is (p, a)- bounded. In this paper we say that a certain invariant is (p, a)-bounded if it is bounded above by some function of (p, a). Let i ≥ 1 and |γ_{i}(G)/γ_{i+1}(G)| = p^{k}. By using Hall’s formula and bearing in mind that the abundance of G/γ_{i+1}(G) is less than or equal to that of G, we obtain

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

p^{k} − 1 ≤ r(G/γ_{i+1}(G)) − r(G/γ_{i}(G)) ≤ (p^{2} − 1)((k + 1)/2 + a(p − 1)).

Hence, it is clear now that the order of γ_{i}(G)/γ_{i+1}(G) is (p, a)-bounded for any i ≥ 1. In particular, d(G) is (p, a)-bounded.

Recall that a finite p-group H is powerful if p is odd and H/H^{p} is abelian, or p = 2 and H/H^{4} is abelian. Throughout this paper we shall use various facts about powerful p-groups, which can be found in [4]. For example, if H is a powerful p-group then rk H = d(H). By using this equality it is not difficult to see that in order to bound rk G it suffices to prove the existence of a powerful subgroup of (p, a)-bounded index. Therefore Theorem 1.1 follows from the next results.

Theorem 1.2. Let p be an odd prime and G a finite p-group. If a(G) < \frac{2p^{s}−p−2}{p^{2}−1} then γ_{p^{s}}(G) is powerful.

Theorem 1.3. Let G be a finite 2-group. If a(G) < 2^{s−1} − 1 then γ_{2^{s}}(G)^{2} is powerful.

Refining this argument we obtain the following quantitative version of Theorem 1.1.

Theorem 1.4. There is a function h = h(a), such that for any p-group G of order p^{m} and abundance a, the inequality m ≤ h · p^{6} holds.

Let x ∈ G and define cb(x) = log_{p} |C_{G}(x)|. Then the cobreadth of G is cb(G) = min_{x∈G} cb(x). If G is a p-group of maximal class then cb(G) = 2, and so a(G) = 0 implies cb(G) ≤ 2. Using Theorem 1.4 we shall prove that the cobreadth can be bounded by a function which depends only on the abundance (and not on p).

In the following, if T ⊆ G then r_{G}(T) is the number of conjugacy classes in G whose intersection with T is not empty.

In the next section we prove Theorem 1.2 and Theorem 1.3. Section 3 is devoted to the proof of Theorem 1.4.

2 Proof of the main result

For the rest of this paper, we assume that the group G satisfies the hypotheses of Theorems 1.2 or 1.3. First we prove that d(γ_{p^{s}}(G)) ≤ p^{s}. Note that by [4, Proposition 1.13] this inequality proves Theorem 1.2. Suppose that d(γ_{p^{s}}(G)) > p^{s}. Then there exists a normal subgroup N of G such that K = γ_{p^{s}}(G)/N is an elementary abelian group of order p^{p^{s}+1}. Put H = G/N. Since the group H acts on K, we have a homomorphism

ϕ: H → GL(p^{s} + 1, 𝔽_{p}).

Suppose that x ∈ H and o(ϕ(x)) = p^{k}. If we consider the Jordan form of ϕ(x) we obtain that |C_{K}(x)| ≥ p^{(p^{s}/p^{k})+1} for 0 ≤ k ≤ s and |C_{K}(x)| ≥ p for k ≥ s+1 ([3, Corollary 2.6]). Hence

|C_{H}(x)| ≥ |C_{K}(x)|p^{k} ≥ p^{s+2}.

Let c be the nilpotency class of H. It is clear that c ≤ 2p^{s}. Choose T_{1} ≤ H such that |H: T_{1}| = p^{2} and γ_{2}(H) ⊆ T_{1} and put T_{i} = γ_{i}(H) for 2 ≤ i ≤ c + 1. For i ≥ 1 define

c_{i} = |{1 ≤ k ≤ c | |T_{k}: T_{k+1}| = p^{i}}|.

If |H| = p^{m} = p^{2n+e} where e = 0, 1, then we have

m = 2 + ∑_{i≥1} ic_{i}, p^{s} ≥ \frac{1}{2}c ≥ \frac{1}{2} ∑_{i≥1} c_{i}.

Note that if |T_{k}: T_{k+1}| = p^{i} then r_{H}(T_{k} \ T_{k+1}) ≥ p^{i} − 1. Also, since for any x ∈ H we have proved that |C_{H}(x)| ≥ p^{s+2} and H \ T_{1} is a normal set, we have

r_{H}(H \ T_{1}) = ∑_{x∈H\T_{1}} \frac{|C_{H}(x)|}{|H|} ≥ p^{s+2} \frac{|H \ T_{1}|}{|H|} = p^{s}(p^{2} − 1).

Therefore

r(H) = r_{H}(H \ T_{1}) + ∑_{k=1}^{c} r_{H}(T_{k} \ T_{k+1}) + 1
≥ p^{s}(p^{2} − 1) + ∑_{i≥1} c_{i}(p^{i} − 1) + 1

We distinguish now two cases depending on whether p is odd or even. If p > 2, then p^{i} − 1 − (i − 1)(p^{2} − 1)/2 ≥ i(p − 1) for any i ≥ 1 and so

r(H) ≥ (∑_{i≥1} c_{i})(p^{2} − 1)/2 + ∑_{i≥1} c_{i}(p^{i} − 1) + 1
= (m − 2)(p^{2} − 1)/2 + ∑_{i≥1} c_{i}(p^{i} − 1 − (i − 1)(p^{2} − 1)/2) + 1.
≥ (m − 2)(p^{2} − 1)/2 + ∑_{i≥1} ic_{i}(p − 1) + 1
= (2n + e − 2)(p^{2} − 1)/2 + (m − 2)(p − 1) + 1.

By using Hall’s formula for the group H, r(H) = p^{e} + (p^{2} − 1)(n + a(p − 1)), where a = a(H) is the abundance of H, the last inequality reads

(p − 1)(p^{2} − 1)a ≥ −(p^{2} − 1) + (m − 2)(p − 1) + e(p^{2} − 1)/2 − p^{e} + 1,

and so

(p^{2} − 1)a ≥ m − 2 − (p + 1).

Since the nilpotency class of H/K is equal to p^{s} − 1, then |H/K| ≥ p^{p^{s}} and so m ≥ 2p^{s} + 1. Therefore

a ≥ \frac{2p^{s} − p − 2}{p^{2} − 1} > a(G).

But H is a factor group of G, so a cannot be greater than the abundance of G. This contradiction proves Theorem 1.2.

If p = 2 we have

r(H) ≥ 3 · 2^{s} + ∑_{i≥1} c_{i}(2^{i} − 1) + 1
≥ 2^{s+1} + ∑_{i≥1} c_{i}(2^{i} − (1/2)) + 1
≥ 2^{s+1} + ∑_{i≥1}(3/2)ic_{i} + 1
= 2^{s+1} + (3/2)(m − 2) + 1
= 2^{s+1} − 2 + 3n + (3/2)e.

Writing r(H) = 2^{e} + 3(n + a), we obtain

3a ≥ 2^{s+1} − 2 + (3/2)e − 2^{e} ≥ 2^{s+1} − 3.

As in the previous case, a > a(G). This contradiction proves our claim that d(γ_{2^{s}}(G)) ≤ 2^{s}. Note that this is equivalent to |γ_{2^{s}}(G) : γ_{2^{s}}(G)^{2}| ≤ 2^{2^{s}}.

Now, in order to prove Theorem 1.3 it suffices to show that d(γ_{2^{s}}(G)^{2}) ≤ 2^{s} (see [4, Proposition 4.1.13]). Suppose that d(γ_{2^{s}}(G)^{2}) > 2^{s}. Then there exists a normal subgroup N of G such that K = γ_{2^{s}}(G)^{2}/N is an elementary abelian group of order 2^{2^{s}+1}. Put H = G/N. As in the previous case |C_{H}(x)| ≥ 2^{s+2}. The notation we used before related to H will be maintained now. Since |γ_{2^{s}}(H) : γ_{2^{s}}(H)^{2}| = |γ_{2^{s}}(G) : γ_{2^{s}}(G)^{2}| ≤ 2^{2^{s}}, it follows that c ≤ 3 · 2^{s} and so ∑_{i≥1} c_{i} ≤ 3 · 2^{s}. Therefore,

r(H) = r_{H}(H \ T_{1}) + ∑_{k=1}^{c} r_{H}(T_{k} \ T_{k+1}) + 1
≥ 3 · 2^{s} + ∑_{i≥1} c_{i}(2^{i} − 1) + 1
≥ 3 · 2^{s−1} + ∑_{i≥1}(2^{i} − (1/2))c_{i} + 1
≥ 3 · 2^{s−1} + ∑_{i≥1}(3/2)ic_{i} + 1
= 3 · 2^{s−1} + (3/2)m − 2
= 3 · 2^{s−1} − 2 + 3n + (3/2)e.

Since r(H) = 2^{e} + 3(n + a), we conclude that

3a ≥ 3 · 2^{s−1} − 2 + (3/2)e − 2^{e} ≥ 3 · 2^{s−1} − 3

and, again, a > a(G), the final contradiction which proves Theorem 1.3.

3 The relation between the abundance and the cobreadth

Proof of Theorem 1.4. Without loss of generality we may suppose that p is odd. Let s be the smalest integer satisfying a = a(G) < \frac{2p^{s}−p−2}{p^{2}−1}. Put H = γ_{p^{s}}(G) and Ḡ = G/H. From the proof of Theorem 1.2 we know that d(H) ≤ p^{s} and H is powerful. Let |Ḡ| = p^{m_{1}} and |H| = p^{m_{2}}. First we shall analyze m_{1}. If we put d_{i} = |{1 ≤ k ≤ p^{s} − 1 | |γ_{k}(Ḡ)/γ_{k+1}(Ḡ)| = p^{i}}|, we obtain

m_{1} = ∑_{i} id_{i}, ∑_{i} d_{i} ≤ p^{s} − 1.

Hence we have

r(Ḡ) ≥ ∑_{i} d_{i}(p^{i} − 1) + 1 ≥ ∑_{i} d_{i}(p^{i} − 1 + p^{2} − 1) + 1 − (p^{s} − 1)(p^{2} − 1)
≥ ∑_{i} id_{i}(p^{2} − 1) + 1 − (p^{s} − 1)(p^{2} − 1)
= (m_{1} + 1)(p^{2} − 1) − p^{s}(p^{2} − 1) + 1.

On the other hand,

r(Ḡ) ≤ (m_{1} + 1)(p^{2} − 1)/2 + (p − 1)(p^{2} − 1)a
≤ (m_{1} + 1)(p^{2} − 1)/2 + (p − 1)(2p^{s} − p − 2)

and we obtain the following inequality for m_{1},

(m_{1} + 1)(p^{2} − 1)/2 ≤ p^{s}(p^{2} − 1) + (p − 1)(2p^{s} − p − 2) − 1.

Now, it is not dificult to see that m_{1} ≤ 3p^{s}.

Since H is powerful, we know from [6, Lemma 4.7] that r(H) ≥ \frac{p−1}{p}|H|^{1/d(H)}. Hence

r_{G}(H) ≥ \frac{1}{|G : H|}r(H) ≥ \frac{p − 1}{p^{3p^{s}+1}} p^{m_{2}/p^{s}} = (p − 1)p^{m_{2}/p^{s}−3p^{s}−1}.

By Hall’s formula

r_{G}(H) ≤ r(G) − r(Ḡ) + 1 ≤ (m_{2} + 1)(p^{2} − 1)/2 + (p − 1)(p^{2} − 1)a
≤ (m_{2} + 1)(p^{2} − 1)/2 + (p − 1)(2p^{s} − p − 2).

If we compare these two inequalities for r_{G}(H) we obtain

p^{m_{2}/p^{s}−3p^{s}−1} ≤ (m_{2} + 1)(p + 1)/2 + 2p^{s} − p − 2.

Now, writing m_{2} = cp^{2s} and analyzing c, we see that m_{2} ≤ 5p^{2s}.

We have chosen the integer s such that a ≥ \frac{2p^{s−1}−p−2}{p^{2}−1}. Therefore p^{s} ≤ (a + 1)p^{3}. Hence m = m_{1} + m_{2} ≤ 3p^{s} + 5p^{2s} ≤ 3(a + 1)p^{3} + 5(a + 1)^{2}p^{6}, and the proof is finished.

Corollary 3.1. The cobreadth of a p-group G is bounded above in terms of its abundance only.

Proof. From the previous theorem and Hall’s formula we obtain that there exists f = f(a) such that r(G) ≤ f(a)p^{8}. On the other hand

r(G) = ∑_{x∈G} \frac{|C_{G}(x)|}{|G|} ≥ p^{cb G}.

Hence cb G ≤ 8 + log_{p} f(a). In particular, cb G ≤ 8 + log_{2} f(a).

References

[1] G. Fernández-Alcober and R. Shepherd, On the order of p-groups of abundance zero, J. Algebra, 201 (1998), 392-400.

[2] B. Huppert, Endliche Gruppen I, Springer, Berlin-Heidelberg-New York, 1982.

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

[4] A. Lubotzky and A. Mann, Powerful p-groups. I. Finite groups, J. Algebra, 105 (1987), 484-505.

[5] J. Poland, Two problems with finite groups with k conjugate classes, J. Austr. Math. Soc.(Ser. A), 8 (1968), 49-55.

[6] A. Shalev, Finite p-groups, Collection: Finite and locally finite groups (Istanbul, 1994), NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 471 (1995), 401–450.