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The number of finite p-groups with bounded number of generators

Andrei Jaikin-Zapirain ^{∗}

Abstract. We establish new bounds for the number of finite p-groups generated by a fixed number of elements.

2000 Mathematics Subject Classification: Primary 20D15;Secondary 20E18.

1. Introduction

The investigation of the number of finite p-groups is a classical topic in group theory. Denoting by f(n) the number of groups of order n, we have the following result of G. Higman and C. C. Sims (see [2, 5]):

Theorem 1.1. (Higman, Sims)

f(p^{k}) = p^{\frac{2}{27} k^{3}+o(k^{3})},

where the term o depends only on k and not on p.

In this paper we consider the function f(n, d), which denotes the number of groups of order n which can be generated by d elements. By results of A. McIver, P.M. Neumann and A. Mann (see [3, 4]), we have

Theorem 1.2. (McIver, Neumann and Mann)

p^{c_{1}(d)k^{2}} ≤ f(p^{k}, d) ≤ p^{c_{2}(d)k^{2}},

for some positive constants c_{1}(d) and c_{2}(d).

Thus the following question seems reasonable:

Problem 1.3. (A. Mann) Is there a constant c(d) such that f(p^{k}, d) = p^{c(d)k^{2}+o(k^{2})}, and what is the value of this constant?

∗This work has been partially supported by the MCYT Grants BFM2001-0201, BFM2001-0180, FEDER and the Ramón y Cajal Program.

We have not been able to find an exact answer to this question, but we give rather reasonable estimates for c(d):

Theorem 1.4. We have that

p^{\frac{d−1}{4} k^{2}+o(k^{2})} ≤ f(p^{k}, d) ≤ p^{\frac{d−1}{2} k^{2}+o(k^{2})}.

Now, we explain the general strategy of the proof of these two bounds. Let F_{d} be a free pro-p group on d generators. Then every finite p-group G generated by d elements is a quotient of F_{d}: G ≅ F_{d}/N for some N ◁ F_{d}. We want to know how many such subgroups N there are. Of course, their number is less than or equal to the order of the group of automorphisms of G. On the other hand | Aut(G)| ≤ |G|^{d}. Thus we obtain:

Proposition 1.5. Let t_{p^{k}}(F_{d}) be the number of normal subgroups of index p^{t} in F_{d}. Then

f(p^{k}, d) ≤ t_{p^{k}}(F_{d}) ≤ p^{kd}f(p^{k}, d).

As we have seen in Theorem 1.2, f(p^{k}, d) is like p^{k^{2}}, and so it is much bigger than p^{kd} when k is big. Hence the main problem is to estimate t_{p^{k}}(F_{d}) (this function is called the normal subgroup growth of F_{d}).

Now, fix a normal subgroup N of index p^{t} in F_{d}. We want to know how many normal subgroups of F_{d} of index p in N there are. Define this set by M_{N}:

M_{N} = {K | K ⊴ F_{d}, K ≤ N, |N : K| = p}.

We have an easy characterization of the set M_{N}: if |N : K| = p then K ∈ M_{N} if and only if [N, F_{d}]N^{p} ≤ K. Thus, we have to estimate the rank of N/[N, F_{d}]N^{p}. We have the following well-known characterization of the rank of N/[N, F_{d}]N^{p}:

d(N/[N, F_{d}]N^{p}) = r_{p}(F_{d}/N) + d − d(F_{p}/N),  (1.1)

where d(G) and r_{p}(G) denotes the minimum number of generators and the minimum number of defining pro-p relations respectively of a finite p-group G. We set

g(p^{t}, d) = max{d(N/[N, F_{d}]N^{p}) | N ⊴ F_{d}, |F_{d}/N| = p^{t}}.

Then we have |M_{N}| ≤ \frac{p^{g(p^{t},d)}−1}{p−1} < p^{g(p^{t},d)}.

Now we come back to the function t_{p^{t}}(F_{d}). This number is clearly less than the number of chains F_{d} = N_{0} > N_{1} > N_{2} > ⋯ > N_{t}, where N_{i} is a normal subgroup of F_{d} and |N_{i} : N_{i+1}| = p, which is less than p^{∑_{s=0}^{t−1} g(p^{s},d)} by the previous observation. In this way we have an upper bound for f(p^{t}, d). We will see now that in the lower bound of f(p^{t}, d) the function g(p^{t}, d) also plays a crucial role.

The number of finite p-groups with bounded number of generators 3

Let N be a normal subgroup of F_{d} of index p^{s} such that d(N/[N, F_{d}]N^{p}) = g(p^{s}, d). We have that every subgroup K, satisfying [N, F_{d}]N^{p} ≤ K ≤ N, is normal in F_{d}. Hence the number of normal subgroups of F_{d} of index p^{s+k} is at least the number of subspaces in N/[N, F_{d}]N^{p} of codimension k, which is greater than p^{k(g(p^{s},d)−k)}. Playing with k and s such that k + s = t, we obtain the lower bound for f(p^{t}, d). In Section 3 we will modify this method in order to obtain a better bound.

All this leeds us to study the function g(p^{t}, d). In view of (1.1), we should estimate the following function:

h(p^{t}, d) = max{r_{p}(G) | d(G) = d, |G| = p^{t}}.

Our result in this direction is the following:

Theorem 1.6. If d is fixed, then (d − 1)t(1 + o(1)) ≤ h(p^{t}, d) ≤ (d − 1)t − \frac{d(d−3)}{2} as t goes to infinity.

The proof of the upper bound is based on the following remark. Let G be a finite p-group. We can associate with G a Lie algebra L = L(G) = ⊕_{i}γ_{i}(G)/γ_{i+1}(G), where γ_{i}(G) is the ith term of the lower central series of G. Then it is well-known that the minimum number of generators of L (we denote it by d(L)) is equal to d(G) and that r_{p}(G) is less than or equal to the minimum number of relations needed to define L. Let F be a free Lie ℤ_{p}-algebra on d(G) generators and H an ideal of F such that L ≅ F/H. Then the minimum number of relations needed to define L is equal to d_{F} (H) = log_{p} |H/([H, F] + pH)|.

Hence the upper bound in the previous theorem follows from its analogue for Lie rings:

Theorem 1.7. Let F be a free Lie ℤ_{p}-algebra on d generators, H ≤ [F, F] + pF an ideal of F and L = F/H a nilpotent Lie ring of order p^{t}. Then d_{F} (H) ≤ (d − 1)t − \frac{d(d−3)}{2}.

Remark 1.8. If we remove the condition H ≤ [F, F] + pF in the theorem then we have that the minimum number of relations needed to define L is equal to d_{F} (H) − d + d(L) and so the theorem gives

d_{F} (H) ≤ (d(L) − 1)t − \frac{d(L)(d(L) − 3)}{2} + d − d(L).

We prove this theorem in Section 2. The lower bound in Theorem 1.6 is obtained by constructing an explicit family of p-groups {G_{t}} such that |G_{t}| = p^{t} and r_{p}(G_{t}) ≥ (d − 1)t + o(t). We present this family in Section 3. The notation is standard. If L is a Lie ℤ_{p}-algebra and M is a right L-module then d_{L}(M) = log_{p} |M/(ML+pM)|.

2. The upper bound in Theorem 1.4

In order to prove Theorem 1.7 we need to do some auxiliary work. First we consider the case when pF ⊆ H. We need the following lemma.

Lemma 2.1. Let F be a free Lie ℤ_{p}-algebra on d generators, H an ideal of F, pF ⊆ H and L = F/H a nilpotent Lie ring of order p^{t} > 1. Suppose Z = T/H ⊆ [L, L] is a minimal ideal of L. Then d_{F} (H/pF) ≤ d_{F} (T/pF) + (d − 1).

Proof. Note that

d_{F} (H/pF) = log_{p} |H/([H, F] + pF)| and d_{F} (T/pF) = log_{p} |T/([T, F] + pF)|.

We put F̄ = F/([H, F] + pF), T̄ = T/([H, F] + pF), H̄ = H/([H, F] + pF) and K̄ = ([T, F] + pF)/([H, F] + pF).

Choose z ∈ T̄ \ H̄. Then K̄ = [z, F̄]. Since [z, [F̄, F̄]] = 0, we obtain that |K̄| ≤ p^{d}. Hence |H̄| ≤ |T̄/K̄|p^{d−1}. ⊓ ⊔

Lemma 2.2. Let F be a free Lie ℤ_{p}-algebra on d generators, pF ≤ H ≤ [F, F] + pF an ideal of F and L = F/H a nilpotent Lie ring of order p^{t}. Then d_{F} (H/pF) ≤ t(d − 1) − \frac{d(d−1)}{2}.

Proof. When L is abelian, the result is trivial. The general case can be proved by induction on the order of L using the previous lemma. ⊓ ⊔

We will need the following auxiliary result. For the reader’s convenience we include the proof.

Lemma 2.3. Let L be a Lie ℤ_{p}-Lie algebra and M a right L-module. Then, if N is a L-submodule of M,

d_{L}(M) ≤ d_{L}(M/N) + d_{L}(N).

Proof. Since NL + pN ≤ (ML + pM) ∩ N, then

|M : (ML + pM)| = |M : (ML + pM + N)||(ML + pM + N) : (ML + pM)|
= |M : (ML + pM + N)||N : (N ∩ (ML + pM)|
≤ |M : (ML + pM + N)||N : (NL + pN)|.
⊓ ⊔

Proof of Theorem 1.7. We prove Theorem 1.7 by induction on the order of L. When L is abelian, the result is trivial.

By Lemma 2.3, we have that

d_{F} (H) ≤ d_{F} (H ∩pF)+d_{F} (H/(H ∩pF)) ≤ d_{F} (H ∩pF)+d_{F} ((H +pF)/pF). (2.1)

Consider first d_{F} ((H + pF)/pF). Applying Lemma 2.2, we obtain that

d_{F} ((H + pF)/pF) ≤ (d − 1) log_{p}(|F/(H + pF)|) − \frac{d(d − 1)}{2}.

Now, consider d_{F} (H ∩ pF). We can divide by p in pF. Hence

d_{F} (H ∩ pF) = d_{F} (p^{−1}(H ∩ pF)).

Note that |F : p^{−1}(H ∩ pF)| = |pF/(H ∩ pF)| = |(pF + H)/H|. Put

M = F/p^{−1}(H ∩ pF).

Then, by induction, using Remark 1.8, we have that

\begin{aligned}d_{F} (H ∩ pF) & ≤ & (d(M) − 1) log_{p}(|(pF + H)/H|) − \frac{d(M)(d(M)−3)}{2} + d − d(M) \\ & ≤ & (d − 1) log_{p}(|(pF + H)/H|) + d. \end{aligned}

(The equality occurs only when M = 0.) Therefore from (2.1), d_{F} (H) ≤ (d − 1) log_{p} |F/H| − \frac{d(d−3)}{2}. ⊓ ⊔

Proof of the upper bound in Theorem 1.4. As we have seen before, Theorem 1.7 implies the upper bound in Theorem 1.6. Hence we have

g(p^{t}, d) ≤ (d − 1)t + d.  (2.2)

Now, as was showed in Introduction f(p^{t}, d) ≤ p^{∑_{i=o}^{t−1} g(p^{i},d)}. From 2.2 we get

∑_{i=1}^{t} g(p^{i}, d) ≤ \frac{d − 1}{2}(t − 1)t + td.

⊓ ⊔

3. The lower bound in Theorem 1.4

Now we prove the second part of Theorem 1.4. First we need the following technical lemma:

Lemma 3.1. Let M_{d}(n) = \frac{1}{n} ∑_{k|n} μ(k)d^{n/k}, where μ(m) is the Mobius function which is defined for positive integers by the rules μ(1) = 1, and for n = p_{1}^{e_{1}} ⋯ p_{s}^{e_{s}}; p_{1}, ⋯, p_{s} being distinct primes, μ(n) = 0 if e_{i} > 1, and μ(p_{1} ⋯ p_{s}) = (−1)^{s}. Then when d is fixed and n goes to infinity the following holds

1. M_{d}(n) = \frac{d^{n}}{n} (1 + o(1));

2. s_{d}(n) = ∑_{i=1}^{n} M_{d}(i) = \frac{d^{n+1}}{n(d−1)}(1 + o(1)).

3. ∑_{i=1}^{n−1} s_{d}(i) = \frac{d^{n+1}}{n(d−1)^{2}} (1 + o(1)).

Proof. The first statement is well-known approximation of M_{d}(n) (see [1, Chapter VIII]). The second and third follow from the next result. Suppose that lim_{n→∞} a_{n} = lim_{n→∞} b_{n} = ∞ and there exists lim_{n→∞} \frac{a_{n+1}−a_{n}}{b_{n+1}−b_{n}} = k. Then lim_{n→∞} \frac{a_{n}}{b_{n}} = k. ⊓ ⊔

Let F be a free pro-p group on d generators x_{1}, ⋯, x_{d}. The lower p-series P_{i} = P_{i}(F) of F is defined inductively by

P_{1} = F and P_{i+1} = [P_{i}, F]P_{i}^{p}.

We know that log_{p} |P_{i}/P_{i+1}| = s_{d}(i). Put B_{n} = F/P_{n}. Now, we calculate the number of defining relations of B_{n}.

Lemma 3.2. We have that r_{p}(B_{n}) = (d − 1) log_{p} |B_{n}|(1 + o(1)) as n goes to infinity.

Proof. Note that [P_{n}, F]P_{n}^{p} = P_{n+1}. Hence the result follows from (1.1) and Lemma 3.1. ⊓ ⊔

Now, we want to construct a series {P_{n,k}} of normal subgroups of F, such that

P_{1} = P_{1,d} > P_{1,d−1} > ⋯ > P_{n−1,1} > P_{n} = P_{n,s_{d}(n)} > P_{n,s_{d}(n)−1} > ⋯,

the index of P_{n,k} in P_{n,k+1} is p and

r_{p}(F/P_{n,k}) = (d − 1) log_{p} |F/P_{n,k}|(1 + o(1))

as n goes to infinity. Clearly it will give the lower bound in Theorem 1.6. Define inductively a basis W_{i} of P_{i}/P_{i+1} in the following way:

W_{1} = {w_{1,1}P_{2}, ⋯, w_{1,d}P_{2}}, w_{1,i} = x_{i}.

Now suppose we have a basis W_{n} = {w_{n,i}P_{n+1}|1 ≤ i ≤ s_{d}(n)} of P_{n}/P_{n+1}. Put w_{n+1,i} = w_{n,i}^{p}, when 1 ≤ i ≤ s_{d}(n). We know that

w_{n+1,1}P_{n+2}, ⋯, w_{n+1,s_{d}(n)}P_{n+2}

are independent in P_{n+1}/P_{n+2} and we can complete this set to a basis W_{n+1} of P_{n+1}/P_{n+2} by adding some vectors

w_{n+1,s_{d}(n)+1}P_{n+2}, ⋯, w_{n+1,s_{d}(n+1)}P_{n+2},

where w_{n+1,k} (when k ≥ s_{d}(n) + 1) is of type [x_{i_{1}}, ⋯, x_{i_{n+1}}]. Let P_{n,k} be the normal subgroup of F generated by P_{n+1} and w_{n,i} with i ≤ k.

Lemma 3.3. Suppose that n ≥ 3 and k ≥ \frac{d(d+1)}{2}. Then

w_{n,k+1}^{p} ∈ [P_{n,k}, F]P_{n,k}^{p}.

Proof. Since k ≥ \frac{d(d+1)}{2}, we have

w_{n,k+1} = [[x_{i_{1}}, ⋯, x_{i_{s−1}}], x_{i_{s}}]^{p^{n−s}}

for some 1 ≤ i_{1}, ⋯, i_{s} ≤ d and s ≥ 3 which can be calculated explicitly from k and n. Hence w_{n,k+1}^{p}P_{n+2} = [[x_{i_{1}}, ⋯, x_{i_{s−1}}]^{p^{n−s+1}}, x_{i_{s}}]P_{n+2}. Now, note that [x_{i_{1}}, ⋯, x_{i_{s−1}}]^{p^{n−s+1}} ∈ P_{n,k} and P_{n+2} ≤ [P_{n,k}, F]P_{n,k}^{p}. ⊓ ⊔

Put B_{n,k} = F/P_{n,k}. In the next lemma we relate the number of defining relations of B_{n,k+1} and B_{n,k}:

Lemma 3.4. Suppose n ≥ 3. We have that r_{p}(B_{n,k}) ≤ r_{p}(B_{n,k+1}) + d − 1 if k ≥ \frac{d(d+1)}{2} and r_{p}(B_{n,k}) ≤ r_{p}(B_{n,k+1}) + d in general.

Proof. Use a similar argument as in Lemma 2.1 bearing in mind Lemma 3.3. ⊓ ⊔

Corollary 3.5. We have that

r_{p}(B_{n+1}) ≤ r_{p}(B_{n,k}) + (d − 1)(log_{p} |B_{n+1}| − log_{p} |B_{n,k}|) + \frac{d(d + 1)}{2}.

Theorem 3.6. We have that r_{p}(B_{n,k}) ≥ (d − 1) log_{p} |B_{n,k}|(1 + o(1)) as n goes to infinity.

Proof. Suppose that there are ϵ and an infinite number of pairs (n, k) with 1 ≤ k ≤ s_{d}(n) such that

r_{p}(B_{n,k}) ≤ (d − 1 − ϵ) log_{p} |B_{n,k}|.

Hence, by Corollary 3.5,

r_{p}(B_{n+1}) ≤ (d − 1) log_{p} |B_{n+1}| − ϵ log |B_{n,k}| + \frac{d(d + 1)}{2}.

Bearing in mind that log_{p} |B_{n,k}| ≥ (log_{p} |B_{n+1}|)/d, we obtain that

r_{p}(B_{n+1}) ≤ (d − 1 − ϵ/d) log_{p} |B_{n+1}| + \frac{d(d + 1)}{2},

which contradicts to Lemma 3.2. ⊓ ⊔

Corollary 3.7. If d is fixed, then h(p^{t}, d) ≥ (d−1)t(1+o(1)) as t goes to infinity.

Proof of the lower bound in Theorem 1.4. In order to prove the lower bound in Theorem 1.4 we have to show that for any α < 1 there exists T, such that f(p^{t}, d) ≥ p^{\frac{α(d−1)}{4} t^{2}} for any t ≥ T.

Choose a natural number a such that \frac{p^{a}(d−1)−1}{p^{a}(d−1)} = β > α, and put e = p^{a}(d − 1) + 1. Let F be a free pro-p group on d generators x_{1}, ⋯, x_{d} and H a normal subgroup of F generated by x_{1}^{p^{a}}, x_{2}, ⋯, x_{d}. Then F/H is a cyclic group of order p^{a}. The group H is a free pro-p group on e generators

y_{1} = x_{1}^{p^{a}}, y_{(i−2)p^{a}+k+2} = x_{i}^{x_{1}^{k}}, 2 ≤ i ≤ d, 0 ≤ k ≤ p^{a} − 1.

Put P_{i} = P_{i}(H). Note that x_{1} acts on P_{1}/P_{2} as an automorphism of order p^{a} permuting the elements {y_{i}P_{2}}, whence x_{1} also acts on P_{i}/P_{i+1} as an automorphism of order p^{a}. As before we can construct a series {P_{n,k}} of normal subgroups of H, such that

P_{1} = P_{1,e} > P_{1,e−1} > ⋯ > P_{n−1,1} > P_{n} = P_{n,s_{e}(n)} > P_{n,s_{e}(n)−1} > ⋯,

the index of P_{n,k} in P_{n,k+1} is p and

r_{p}(H/P_{n,k}) = (e − 1) log_{p} |H/P_{n,k}|(1 + o(1))

as n goes to infinity. Moreover, without loss of generality we can assume that P_{n,k} are x_{1}-invariant. Our purpose is to calculate the number of x_{1}-invariant subgroups of M_{n,k} = P_{n,k}/[P_{n,k}, H]P_{n,k}^{p}. It will give an estimate of the normal subgroup growth of F. For this we need to understand the structure of M_{n,k} as 𝔽_{p}[x_{1}]-module. Note that x_{1}^{p^{a}} acts trivially on M_{n,k}, whence we can consider M_{n,k} as 𝔽_{p}[C_{p^{a}}]-module.

Claim 1 The 𝔽_{p}[C_{p^{a}}]-module M_{n,k} is isomorphic to a direct sum A_{n,k} ⊕ D_{n,k}, where A_{n,k} is a maximal free 𝔽_{p}[C_{p^{a}}]-submodule and D_{n,k} is some complement. Moreover, the 𝔽_{p}[C_{p^{a}}]-rank of A_{n,k} is (d − 1) log_{p} |F/P_{n,k}|(1 + o(1)) as n goes to infinity.

Proof. Let A_{n,k} be a maximal free 𝔽_{p}[C_{p^{a}}]-submodule of M_{n,k}. It is clear that there exists a 𝔽_{p}[C_{p^{a}}]-submodule D_{n,k} such that M_{n,k} = A_{n,k} ⊕ D_{n,k}. Now, the 𝔽_{p}[C_{p^{a}}]-rank of M_{n,k} is equal to d(P_{n,k}/[P_{n,k}, F]P_{n,k}^{p}) and so, by Theorem 1.6, it is less than (d − 1) log_{p} |F/P_{n,k}| + 1. Hence if r_{1} is the 𝔽_{p}[C_{p^{a}}]-rank of A_{n,k} and r_{2} is the 𝔽_{p}[C_{p^{a}}]-rank of D_{n,k}, then

r_{1} + r_{2} ≤ (d − 1) log_{p} |F/P_{n,k}| + 1.  (3.1)

By the construction of P_{n,k}, the 𝔽_{p}-rank of M_{n,k} is equal to (e−1) log_{p} |F/P_{n,k}|(1+ o(1)) when n goes to infinity. On the other hand, it is less than or equal to p^{a}r_{1} + (p^{a} − 1)r_{2}. Hence we obtain that

p^{a}r_{1} + (p^{a} − 1)r_{2} ≥ p^{a}(d − 1) log_{p} |F/P_{n,k}|(1 + o(1))  (3.2)

Comparing (3.1) and (3.2) we obtain r_{1} = (d − 1) log_{p} |F/P_{n,k}|(1 + o(1)). ⊓ ⊔

Claim 2 Let M be a free 𝔽_{p}[C_{p^{a}}]-module of rank m. Then there are at least p^{p^{a}s(m−s)} submodules N of M such that M/N is a free 𝔽_{p}[C_{p^{a}}]-module of rank s.

Proof. The proof goes in the same lines as the count of the number of subspaces of codimension r in an m-dimensional 𝔽_{p}-space. We have that there are

p^{(p^{a}−1)m(m−s)}(p^{m} − 1)(p^{m} − p) ⋯ (p^{m} − p^{m−s+1})

(m − s)-tuples, generating free 𝔽_{p}[C_{p^{a}}]-submodules of rank (m − s). On the other hand, every free 𝔽_{p}[C_{p^{a}}]-module of rank (m − s) has

p^{(p^{a}−1)(m−s)^{2}}(p^{m−s} − 1)(p^{m} − p) ⋯ (p^{m−s} − p^{m−s−1})

generating (m−s)-tuples. Hence the number of 𝔽_{p}[C_{p^{a}}]-submodules of rank (m−s) is equal to

p^{(p^{a}−1)(m−s)s} \frac{p^{m} − 1}{p^{m−s} − 1} \frac{p^{m−1} − 1}{p^{m−s−1} − 1} ⋯ \frac{p^{s+1} − 1}{p − 1}.

This proves the claim. ⊓ ⊔

Hence, using Claim 1 and Claim 2, we obtain that the group M_{n,k} has at least

p^{p^{a}s((d−1) log_{p} |F/P_{n,k}|(1+o(1))−s)}

x_{1}-invariant subgroups of index p^{p^{a}s}.

Let s be the integer part of t/2p^{a}. Consider T = P_{n,k} such that the order of F/T is p^{t−sp^{a}}. By the previous observation, the number of x_{1}-invariant subgroups of T/[T, H]T^{p} of index p^{sp^{a}} is at least

p^{sp^{a}((d−1)(t−sp^{a})−s)+o(t^{2})} = p^{\frac{β(d−1)}{4} t^{2}+o(t^{2})}.

Since every x_{1}-invariant normal subgroup of H is a normal subgroup of F, the normal subgroup growth of F is at least p^{\frac{β(d−1)}{4} t^{2}+o(t^{2})}. It implies that there exists T, such that f(p^{t}, d) ≥ p^{\frac{α(d−1)}{4} t^{2}} for any t ≥ T. Since α is an arbitrary number less than 1, we obtain that f(p^{t}, d) ≥ p^{\frac{(d−1)}{4} t^{2}(1+o(1))}. ⊓ ⊔

Acknowledgments. I thank A. Mann for helpful conversations and D. Segal for reading a previous version of this work. A part of this work was done while I was visiting the Hebrew University of Jerusalem. I would like to take this opportunity to thank the Institute of Mathematics and personally A. Lubotzky for his hospitality.

References

[1] B Huppert and N. Blackburn, Finite groups II, Springer-Verlag, Berlin – Heidelberg, 1982.

[2] G. Higman, Enumerating p-groups I. Inequalities., Proc. London Math. Soc. 10 (1960), 24–30.

[3] A. McIver and P. M. Neumann, Enumerating finite groups, Quart. J. Math. Oxford Ser.(2) 38 (1987), 473–488.

[4] A Mann, Enumerating finite p-groups and their defining relations, J. Group Theory 1 (1998), 59–64.

[5] C. C. Sims, Enumerating p-groups, Proc. London Math. Soc. 15 (1965), 151–166.

Departamento de Matemáticas, Facultad de Ciencias, Universidad Autónoma de Madrid, Cantoblanco, Ciudad Universitaria,28049 Madrid, Spain.

Email: ajaikin@uam.es