On the number of conjugacy classes of finite p-groups of class 2 ^{∗}

A. Jaikin-Zapirain

Departamento de Matemáticas, Facultad de Ciencias, Universidad Autónoma de Madrid

September 4, 2002

Abstract In this work we study the behaviour of the number of conjugacy

classes of finite p-groups of class 2.

1 Introduction

Let G be a finite p-group and let r(G) denote the number of conjugacy classes of G. If G is of class c then from [2] we know that

r(G) ≥ c|G|^{1/c} − c + 1.  (1)

This implies that

ϵ_{p}(c) = inf\{\frac{log_{p} r(G)}{log_{p} |G|}|G is a finite p-group of nilpotency class c\} ≥ \frac{1}{c}.

In [1] it is shown that actually we have the equality: Theorem 1.1. We have ϵ_{p}(c) = 1/c.

In this note we fix our attention to p-groups of class 2. Our main result is the following theorem. Theorem 1.2. Let α > (\sqrt{p}+1)\sqrt{p + 1}. Then there exist an infinite number of finite p-groups G with D_{3}(G) = 1 (the third dimension subgroup of G), such that r(G) < α|G|^{1/2}.

∗This work has been partially supported by the DGICYT Grant.

This result suggest us the following problem:

Question 1. Is there a constant α = α(p, c) such that there are infinitely many finite p-groups G of class c for which r(G) < α|G|^{1/c}?

2 Proof of the main result

First we present some examples which we will need later. These examples show also that ϵ_{p}(2) = \frac{1}{2}.

Lemma 2.1. Let G be a group generated by {x_{i}, y_{j}|1 ≤ n, 1 ≤ j ≤ n − 1} with the following relations:

x_{i}^{p} = y_{j}^{p} = [x_{i}, y_{j}] = 1, [x_{i}, x_{k}] = y_{i+k−2}, 1 ≤ i < k ≤ n, 1 ≤ j ≤ n − 1,

where we understand that y_{s} = 1 if s > n−1. Then r(G) ≤ p^{1/2} log_{p} |G||G|^{1/2}.

Proof. For each 1 ≤ k ≤ n + 1 define G_{k} as the normal subgroup of G generated by {x_{i}, y_{j}|k ≤ i ≤ n, 1 ≤ j ≤ n − 1}. Take an element g ∈ G_{k} \ G_{k+1}. Then C_{G}(g)| ≤ ⟨g, G_{n+2−k}⟩. Indeed let h ∈ C_{G}(g) and suppose that h ∉ ⟨g, G_{n+2−k}⟩. There exists s such that h ∈ G_{s} \ G_{s+1}. If s = k, then we replace h by hg^{m} such that hg^{m} ∈ G_{k+1}. Therefore we can suppose that s ≠ k and also s > n + 2 − k. Consider [h, g]. By the relations defining G, [h, g] ∈ ⟨y_{s+k−2}, …, y_{n−1}⟩ \ ⟨y_{s+k−3}, …, y_{n−1}⟩. Hence [g, h] ≠ 1 and our claim follows.

Since |C_{G}(g)| ≤ p^{n+k−1} for every g ∈ G_{k} \ G_{k+1}, we obtain that r_{G}(G_{k} \ G_{k+1}) ≤ (p−1)p^{n−1}. This implies that r(G) ≤ n(p−1)p^{n−1} +p^{n−1} ≤ np^{n} ≤ p^{1/2} log_{p} |G||G|^{1/2}.

Now we can prove Theorem 1.2.

Proof of Theorem 1.2 . We will prove the theorem by way of contradiction. Suppose that there exists n such that for any p-group G of order greater than p^{n} and D_{3}(G) = 3 the inequality r(G) ≥ α|G|^{1/2} holds. Put

ϵ_{1} = max\{\frac{log_{p} r(G)}{log_{p} |G|}|D_{3}(G) = 1, |G| ≤ p^{n}\}.

Note that \frac{log_{p} r(G)}{log_{p} |G|} is always greater than 1/2 if G has nilpotency class 2. Since there exists only a finite number of groups of order ≤ p^{n}, ϵ_{1} > 1/2. Let ϵ_{2} be the solution of the following equation in x:

\frac{(p − 1)\sqrt{p + 1}}{p^{1−x} − 1} = α.

Since α > (\sqrt{p} + 1)\sqrt{p + 1}, ϵ_{2} > 1/2.

Now let

ϵ_{3} = max\{\frac{log_{p} r(G)}{log_{p} |G|}|D_{3}(G) = 1, |G/Φ(G)| ≤ p^{2}\}

and

ϵ_{4} = max\{\frac{log_{p} r(G)}{log_{p} |G|}|D_{3}(G) = 1, |Φ(G)| ≤ p\}.

It is not difficult to see that ϵ_{3} and ϵ_{4} are greater than 1/2.

Put ϵ = min{ϵ_{1}, ϵ_{2}, ϵ_{3}, ϵ_{4}}. Now we will prove by induction on order of G that if G is a finite p-group and D_{3}(G) = 1, then r(G) ≥ |G|^{ϵ}. This will give us the final contradiction because it contradicts the previous lemma.

From choice of ϵ if |G| ≤ p^{n} or |G/Φ(G)| ≤ p^{2} or |Φ(G)| ≤ p, then r(G) ≥ |G|^{ϵ}. This is the base for the induction. Now suppose that |G| ≥ p^{n+1}, |G/Φ(G)| > p^{2} and |Φ(G)| > p. In particular, r(G) ≥ α|G|^{1/2}.

Let U = G/Φ(G) and V = Φ(G). Then U and V are elementary abelian p-groups and V ≤ Z(G). First suppose that

(p^{1−ϵ} − 1)|G|^{ϵ} < (p − 1)|U| and  (2)
(p^{1−ϵ} − 1)|G|^{ϵ} < (p^{2} − 1)|V |.  (3)

Multiplying this two inequalities, we have

(p^{1−ϵ} − 1)^{2}|G|^{2ϵ} < (p − 1)(p^{2} − 1)|G|,

and so

|G|^{ϵ} < α|G|^{1/2} ≤ r(G).

Now, suppose that the reversing inequality holds in (2):

(p^{1−ϵ} − 1)|G|^{ϵ} ≥ (p − 1)|U|  (4)

Of course we have r(G) = | Irr(G|V)| + |U|. Let T be the set of subgroups of V of order p. If λ ∈ Irr(G|V), then there are \frac{(|V |/p)−1}{p−1} subgroups of V of order p contained in Ker λ. Hence

|Irr(G|V)| = \frac{p − 1}{(|V |/p) − 1} ∑_{W ∈T} |Irr(G/W |V/W)| = \frac{p − 1}{(|V |/p) − 1}( ∑_{W ∈T} r(G/W)−|U|).

Applying the inductive hypothesis, we obtain

|Irr(G|V)| ≥ \frac{p − 1}{(|V |/p) − 1} \frac{|V | − 1}{p − 1}(|G|^{ϵ}p^{−ϵ} − |U|) ≥ |G|^{ϵ}p^{1−ϵ} − p|U|,

and so

r(G) ≥ |G|^{ϵ} + (p^{1−ϵ} − 1)|G|^{ϵ} − (p − 1)|U|.

Therefore using the inequality (4), we obtain r(G) ≥ |G|^{ϵ}.

Now, we deal with the final case

(p^{1−ϵ} − 1)|G|^{ϵ} ≥ (p^{2} − 1)|V |.  (5)

Let M be a subgroup of G containing V and A_{M} be the subset of M × M consisting of the pairs (x, y), such that x and y commute and the group ⟨x, y, V ⟩/V has order p^{2}. Then we have

r(M) = \frac{1}{|M|}|\{(x, y) ∈ M×M|[x, y] = 1\}| = \frac{1}{|M|}(|A_{M}|+(p+1)|M||V |−p|V |^{2}).

Let S be the set of maximal subgroups of G. If (x, y) ∈ A_{G}, then there are \frac{(|U|/p^{2})−1}{p−1} maximal subgroups of G such that (x, y) ∈ A_{M}. Hence we have

|A_{G}| = \frac{p − 1}{(|U|/p^{2}) − 1} ∑_{M∈S} |A_{M}| =

\frac{p − 1}{(|U|/p^{2}) − 1} ∑_{M∈S}(|M|r(M) − (p + 1)|M||V | + p|V |^{2}).

Applying the inductive hypothesis we obtain

|A_{G}| ≥ \frac{p − 1}{(|U|/p^{2}) − 1} \frac{|U| − 1}{p − 1}(p^{−1−ϵ}|G|^{1+ϵ} − (p + 1)|G||V |/p + p|V |^{2}) ≥

p^{1−ϵ}|G|^{1+ϵ} − (p + 1)p|G||V | + p^{3}|V |^{2},

and so

r(G) = \frac{1}{|G|}(|A_{G}| + (p + 1)|G||V | − p|V |^{2}) ≥

\frac{1}{|G|}(p^{1−ϵ}|G|^{1+ϵ} − (p + 1)p|G||V | + p^{3}|V |^{2} + (p + 1)|G||V | − p|V |^{2}) ≥

|G|^{ϵ} + (p^{1−ϵ} − 1)|G|^{ϵ} − (p^{2} − 1)|V |.

Therefore using the inequality (5), we obtain r(G) ≥ |G|^{ϵ}.

Problem 2. Give an explicit construction of the groups in the previous theorem.

References

[1] A. Jaikin-Zapirain, On the number of conjugacy classes of finite pgroups, preprint 2002.

[2] G. Sherman, A lower bound for the number of conjugacy classes in a finite nilpotent group, Pacific J. Math., 80 (1979), 253–254.