On the use of the Lazard correspondence in the classification of p-groups of maximal class ^{∗}

A. Jaikin-Zapirain, A. Vera-López

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

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

Abstract Let G be a p-group of maximal class of order p^{m}, p an odd prime and m ≥ 4. In this work we reduce the construction of this group to the consideration of certain elements of Hom_{S}(R/𝔞^{m−2} ∧ R/𝔞^{m−2}, R/𝔞^{m−4}), where R = ℤ[x]/(1+⋯+x^{p−1}) = ℤ[x̄], 𝔞 = (x̄−1) and S = ℤ[x]/(x^{p} −1). As an application of this result we prove that, if we consider in G the generators s, s_{1}, …, s_{m−1} introduced by Blackburn, then the structure of G is determinated by the (p − 3)/2 commutators [s_{i}, s_{i+1}] for 1 ≤ i ≤ p − 3)/2 and the three integers 0 ≤ u, v, w ≤ p − 1, where u and v are defined by the condition s^{p} = s_{m−1}^{u}, (ss_{1})^{p} = s_{m−1}^{v} and w = c_{m−1} is the exponent of s_{m−1} in the commutator [s_{(p−1)/2}, s_{(p+1)/2}] = s_{p}^{c_{p}} ⋯ s_{m−1}^{c_{m−1}}, 0 ≤ c_{i} ≤ p − 1.

1 Preliminaries

In this work, G will denote a finite p-group of maximal class of order p^{m}, p being an odd prime number and m ≥ 4. In addition, the following notation will be used throughout this paper without further comment. For i ≥ 2, Y_{i} will be the

i-th term of the lower central series: Y_{i} = [\overbrace{G, …, G}^{i}]. Thus Y_{m−1} ≠ 1 and Y_{i} = 1 for i ≥ m. If we set Y_{0} = G and define Y_{1} by the condition Y_{1} = C_{G}(Y_{2}/Y_{4}), it follows that |Y_{i} : Y_{i+1}| = p for i ∈ {0, …, m − 1}.

The degree of commutativity of G, c = c(G), is defined as

c(G) = max{k ≤ m − 2 | [Y_{i}, Y_{j}] ≤ Y_{i+j+k}, for all i, j ≥ 1}.

Following N. Blackburn [1], we take a couple of elements s ∈ G\(Y_{1}∪C_{G}(Y_{m−2})) and s_{1} ∈ Y_{1} \ Y_{2}, and define recursively s_{i} = [s_{i−1}, s] ∈ Y_{i} \ Y_{i+1}, for i = 2, …, m − 1.

We use the following notation: R and S are the rings R = ℤ[x]/(1 + ⋯ + x^{p−1}) = ℤ[x̄] and S = ℤ[x]/(x^{p}−1) = ℤ[x̃]; 𝔞 is the ideal of R generated by x̄−1. We have that 𝔞^{p−1} = pR and |R/𝔞^{j}| = p^{j}. The following ring isomorphisms hold:

R ≅ S/(1 + ⋯ + x̃^{p−1}), S ≅ ℤ[C_{p}], R/𝔞^{m−j} ≅ S/(S ∑_{i=0}^{p−1} x̃^{i} + S(x̃ − 1)^{m−j}).

In addition, we can define in R ∧ R an S-module structure by setting

x̃(a ∧ b) = (x̄a) ∧ (x̄b), where a ∧ b ∈ R ∧ R.

Since 𝔞 is actually an S-submodule of R, we can consider similarly R/𝔞^{k} ∧ R/𝔞^{k} as an S-module. In the exterior power R ∧ R we consider the S-submodules 𝔟_{k} = ⟨(x̄ − 1)^{i} ∧ (x̄ − 1)^{j} | i ≥ k or j ≥ k⟩. Thus, we have the natural S-module isomorphism

(R ∧ R)/𝔟_{k} ≅ R/𝔞^{k} ∧ R/𝔞^{k}.

The Lie product in a Lie ring will be denoted [ , ]_{L}, whereas [ , ] will be used to denote group commutators. Finally, Γ_{1}(x, y) (resp. γ_{1}(x, y)) will be the Lie ring (resp. the group) generated by x and y and, for i ≥ 2 we define inductively Γ_{i}(x, y) and γ_{i}(x, y) by

Γ_{i}(x, y) = [Γ_{1}(x, y), Γ_{i−1}(x, y)]_{L}, γ_{i}(x, y) = [γ_{1}(x, y), γ_{i−1}(x, y)].

In the next section we construct a correspondence between the p-groups of maximal class of order p^{m} and a certain subset of Hom_{S}(R/𝔞^{m−2}∧R/𝔞^{m−2}, R/𝔞^{m−4}). It will turn out that each of these groups is determined by the corresponding homomorphism and two additional integer parameters.

2 Construction of the correspondence

It is proved in [7] that the subgroup Y_{1} of G has nilpotent class at most (p + 1)/2 < p, so the Lazard correspondence (see [2]) allows us to define on Y_{1} a structure of Lie ring by means of two laws, + and [ , ]_{L}, which in addition satisfy

x + y = xyt_{1}(x, y), [x, y]_{L} = [x, y]t_{2}(x, y),  (1)

for suitable elements t_{1}(x, y) ∈ γ_{2}(x, y) and t_{2}(x, y) ∈ γ_{3}(x, y). Moreover, the group automorphism of Y_{1} induced by conjugation by the element s is also a Lie ring automorphism. As s^{p} ∈ Y_{m−1} ⊆ Z(Y_{1}), then (Y_{1}, +) can be considered as an S-module if we set

x̃t = t^{s} for all t ∈ Y_{1}.

It is clear that Y_{i} is an S-submodule of Y_{1} for any i ≥ 2. We note that the identity element in the additive group (Y_{1}, +) is the same as in (Y_{1}, ·). We shall denote it by 0 or 1 depending on the context.

Lemma 2.1. If t ∈ Y_{i} \ Y_{i+1}, 1 ≤ i ≤ m − 2, then (x̃ − 1)t ∈ Y_{i+1} \ Y_{i+2}.

Proof. If t ∈ Y_{i} \ Y_{i+1}, then by using (1) we have:

(x̃−1)t = t^{s}+(−t) = t^{s}+t^{−1} = t^{s}t^{−1}t_{1}(t^{s}, t^{−1}) = [s, t^{−1}]t_{1}(t^{s}, t^{−1}) ∈ Y_{i+1}\Y_{i+2}.

Lemma 2.2. For every i ≥ 1, Y_{i} is a cyclic S-module generated by any element t ∈ Y_{i} \ Y_{i+1}.

Proof. We prove by decreasing induction that if t ∈ Y_{i} \ Y_{i+1}, then Y_{j} ⊆ St for all j ≥ i, so that in particular Y_{i} = St is the S-submodule generated by t. For j = m we have Y_{m} = {0} ⊆ St. Now, we suppose that j ≥ i and Y_{j+1} ⊆ St.

We have to prove that any element u ∈ Y_{j} \ Y_{j+1} is in St. By Lemma 2.1, f = (x̃ − 1)^{j−i}t ∈ Y_{j} \ Y_{j+1}. Then, there exists n ∈ ℤ such that f^{n}u ∈ Y_{j+1}. By using (1), we obtain that

nf + u = f^{n} + u = f^{n}ut_{1}(f^{n}, u) ∈ Y_{j+1} ⊆ St.

Since f ∈ St, we have that u ∈ St.

Lemma 2.3. We have the following S-module isomorphisms:

Y_{1}/Y_{m−1} ≅ R/𝔞^{m−2}, Y_{j} ≅ R/𝔞^{m−j} for 2 ≤ j ≤ m − 1.

Proof. We note that R/𝔞^{m−j} ≅ S/(S ∑_{i=0}^{p−1} x̃^{i} + S(x̃ − 1)^{m−j}). We consider the element u = (∑_{i=0}^{p−1} x̃^{i})s_{1}. Obviously, x̃u = u, that is u^{s} = u, whence u ∈ Y_{m−1} because C_{Y_{1}}(s) = Y_{m−1} as G is a p-group of maximal class. Let ϕ: S → Y_{1}/Y_{m−1} be the S-epimorphism defined by the condition ϕ(1) = s̄_{1}. Clearly, we have

S ∑_{i=0}^{p−1} x̃^{i} + S(x̃ − 1)^{m−2} ⊆ Ker ϕ,

therefore, Y_{1}/Y_{m−1} = Ss̄_{1} is an epimorphic image of R/𝔞^{m−2}. Since both Y_{1}/Y_{m−1} and R/𝔞^{m−2} have the same order, they are actually isomorphic.

In order to see the second part of the lemma we take an element v ∈ Y_{j} \Y_{j+1}, 2 ≤ j ≤ m − 1, and consider the S-epimorphism ϕ_{v} : S → Y_{j} = Sv determined by the condition ϕ_{v}(1) = v. Note that

Y_{2} = S(x̃ − 1)s_{1} = (x̃ − 1)Y_{1}

so, (∑_{i=0}^{p−1} x̃^{i})Y_{2} = 0, whence (∑_{i=0}^{p−1} x̃^{i})Y_{j} = 0 for j ≥ 2. As (x̃−1)^{m−j}Y_{j} = 0 we conclude that S ∑_{i=0}^{p−1} x̃^{i} +S(x̃−1)^{m−j} is contained in Ker ϕ_{v} and ϕ_{v} induces an S-epimorphism from S/(S ∑_{i=0}^{p−1} x̃^{i} + S(x̃ − 1)^{m−j}) onto Y_{j}. Again, the equality of orders implies that this is indeed an isomorphism of modules.

By using (1), we get

[Y_{1}, Y_{1}]_{L} ⊆ Y_{4} and [Y_{i}, Y_{j}]_{L} ⊆ Y_{i+j} for i + j > 3.  (2)

In particular, if t ∈ Y_{m−1}, then [t, Y_{1}]_{L} = [Y_{1}, t]_{L} = {0} and [ , ]_{L} can be viewed as a map from Y_{1}/Y_{m−1} ∧ Y_{1}/Y_{m−1} into Y_{4}. We denote by ϕ_{1} and ϕ_{2} the isomorphisms between Y_{1}/Y_{m−1} and R/𝔞^{m−2} and Y_{4} and R/𝔞^{m−4} constructed in the last lemma, namely,

ϕ_{1} : Y_{1}/Y_{m−1}  ⟶ R/𝔞^{m−2}
f(x̃)s̄_{1}  ↦ f(x̄) + 𝔞^{m−2},
ϕ_{2} : Y_{4}  ⟶ R/𝔞^{m−4}
f(x̃)(x̃ − 1)^{3}s_{1} ↦ f(x̄) + 𝔞^{m−4},

where f(x̃) ∈ S. We define now α: R/𝔞^{m−2} ∧ R/𝔞^{m−2} → R/𝔞^{m−4} by

α = ϕ_{2} ∘ [ , ]_{L} ∘ (ϕ_{1}^{−1} ∧ ϕ_{1}^{−1}).

Since multiplication by x̃ is a Lie ring automorphism, we have

α(x̃a ∧ x̃b) = x̃α(a ∧ b) for all a, b ∈ R/𝔞^{m−2},

which means that α ∈ Hom_{S}(R/𝔞^{m−2} ∧ R/𝔞^{m−2}, R/𝔞^{m−4}). Put h = (x̄ − 1)^{3} + 𝔞^{m−2}. The homomorphism α satisfies two additional properties: on the one hand, Jacobi’s identity on Y_{1} implies that

α((α(a ∧ b)h ∧ c) + α((α(c ∧ a)h ∧ b) + α((α(b ∧ c)h ∧ a) = 0,  (3)

for all a, b, c ∈ R/𝔞^{m−2} and, on the other hand, the inclusion [Y_{i}, Y_{j}]_{L} ⊆ Y_{i+j} means that

α(𝔞^{i}/𝔞^{m−2} ∧ 𝔞^{j}/𝔞^{m−2}) ⊆ 𝔞^{(i+j−2)}/𝔞^{m−4} for i + j > 1.  (4)

Of course the homomorphism α depends on s and s_{1}. When we want to emphasize this fact, we shall write α_{(G,s,s_{1})} instead of α.

Conversely, we suppose now that we have three integers u, v, m ≥ 4 and a homomorphism α ∈ Hom_{S}(R/𝔞^{m−2} ∧ R/𝔞^{m−2}, R/𝔞^{m−4}) satisfying (3) and (4). We want to construct a p-group of maximal class G of order p^{m} such that there exist s ∈ G \ (Y_{1} ∪ C_{G}(Y_{m−2})) and s_{1} ∈ Y_{1} \ Y_{2}, satisfying the rules:

1. s^{p} = s_{m−1}^{u} and (ss_{1})^{p} = s_{m−1}^{v},

2. α_{(G,s,s_{1})} = α.

Set M = R/𝔞^{m−1}. We can define on M a Lie ring structure by setting

[a, b]_{L} = α((a+𝔞^{m−2}/𝔞^{m−1})∧(b+𝔞^{m−2}/𝔞^{m−1}))((x̄−1)^{3}+𝔞^{m−1}) for all a, b ∈ M.

Jacobi’s identity follows from (3) and antisymmetry is consequence of the properties of the exterior power.

Lemma 2.4. M is a nilpotent Lie ring with nilpotency class less than or equal to (p + 1)/2.

Proof. For 1 ≤ i ≤ m, set M_{i} = 𝔞^{i−1}/𝔞^{m−1} and set M_{i} = 0 for i > m. We define

c = c(M) = max{k ≤ m − 2 | [M_{i}, M_{j}]_{L} ⊆ M_{i+j+k} for all i, j ≥ 1}.

We note that c ≥ 0 by (4). We set M̄ = ⊕_{i=1}^{m−1}M_{i}/M_{i+1} and turn M̄ into a Lie algebra as follows: if a ∈ M_{i} \ M_{i+1} and b ∈ M_{j} \ M_{j+1}, we define

[a + M_{i+1}, b + M_{j+1}]_{L} = [a, b]_{L} + M_{i+j+c+1} ∈ M_{i+j+c}/M_{i+j+c+1}.

Of course, the additive group M_{i}/M_{i+1} has order p and is generated by m_{i} = ((x̄ − 1)^{i−1} + 𝔞^{m−1}) + M_{i+1}. For i + j + c ≤ m − 1 we set

[m_{i}, m_{j}]_{L} = α_{i,j}m_{i+j+c},

where α_{i,j} ∈ 𝔽_{p}. If the equality in M

(x̃ − 1)[(x̄ − 1)^{i−1} + 𝔞^{m−1}, (x̄ − 1)^{j−1} + 𝔞^{m−1}]_{L} =
[(x̄ − 1)^{i−1} + 𝔞^{m−1}, (x̄ − 1)^{j} + 𝔞^{m−1}]_{L}+
[(x̄ − 1)^{i} + 𝔞^{m−1}, (x̄ − 1)^{j−1} + 𝔞^{m−1}]_{L}+
[(x̄ − 1)^{i} + 𝔞^{m−1}, (x̄ − 1)^{j} + 𝔞^{m−1}]_{L},

is read modulo M_{i+j+2+c}, it becomes α_{i,j} = α_{i,j+1} + α_{j,i+1}, for i + j + 1 + c ≤ m − 1.

On the other hand, since −p = (x̄−1)^{p−1} (mod 𝔞^{p}), then for i+j+p−1+c ≤ m − 1,

[(x̄ − 1)^{i+p−2} + 𝔞^{m−1}, (x̄ − 1)^{j−1} + 𝔞^{m−1}]_{L} ≡
−p[(x̄ − 1)^{i−1} + 𝔞^{m−1}, (x̄ − 1)^{j−1} + 𝔞^{m−1}]_{L} ≡
(x̄ − 1)^{p−1}[(x̄ − 1)^{i−1} + 𝔞^{m−1}, (x̄ − 1)^{j−1} + 𝔞^{m−1}]_{L} (mod M_{i+j+p+c}),

which means that α_{i,j} = α_{i+p−1,j}.

Thus, the above algebra M̄ is of the same type as the one considered by Shepherd in [7] and therefore arguing as he does, it follows that the nilpotency class of M is less than or equal to (p + 1)/2.

From the above lemma, we conclude that the nilpotency class of M is less than p and so, we can apply the Lazard Correspondence, and define in M a group operation which, in addition, satisfies the condition

xy = x + y + ([x, y]_{L})/2 + t_{3}(x, y),  (5)

where t_{3}(x, y) ∈ Γ_{3}(x, y). We also have

[x, y] = [x, y]_{L} + t_{4}(x, y),  (6)

with t_{4}(x, y) ∈ Γ_{3}(x, y).

Since multiplication by x̃ is an automorphism of the Lie ring M, it is also a group automorphism of M. We denote it by s. Then, s^{p} = 1 because x̃^{p} = 1. We define H as the relative holomorph of M by ⟨s⟩. H is a p-group of maximal class because C_{H}(s) = ⟨s, M_{m−1}⟩ has order p^{2}. If we set s_{1} = 1 + 𝔞^{m−1}, then α_{(H,s,s_{1})} = α. Now, reasoning as in [5, p.178], beginning with the group H, we can construct another p-group of maximal class G of order p^{m} for which there exist s ∈ G \ (Y_{1} ∪ C_{G}(Y_{m−2})) and s_{1} ∈ Y_{1} \ Y_{2}, such that

1. s^{p} = s_{m−1}^{u} and (ss_{1})^{p} = s_{m−1}^{v},

2. α_{(G,s,s_{1})} = α. This proves the existence part in the following result. Theorem 2.5. Let u, v and m ≥ 4 be given integers and α ∈ Hom_{S}(R/𝔞^{m−2} ∧ R/𝔞^{m−2}, R/𝔞^{m−4}) satisfying (3) and (4). Then there exists one and only one pgroup of maximal class G of order p^{m} such that for some s ∈ G\(Y_{1}∪C_{G}(Y_{m−2})) and s_{1} ∈ Y_{1} \ Y_{2}, the following equalities hold:

1. s^{p} = s_{m−1}^{u} and (ss_{1})^{p} = s_{m−1}^{v},

2. α_{(G,s,s_{1})} = α. Besides, any p-group of maximal class with order p^{m}, m ≥ 4, arises in this way for some choice of u, v and α. Proof. Only uniqueness needs to be proved. Suppose that there exists another p-group of maximal class and order p^{m} H such that for some w ∈ H \ (Y_{1}(H) ∪ C_{H}(Y_{m−2}(H))) and w_{1} ∈ Y_{1}(H) \ Y_{2}(H), the following holds:

1. w^{p} = w_{m−1}^{u} and (ww_{1})^{p} = w_{m−1}^{v},

2. α_{(H,w,w_{1})} = α. Of course, we set here w_{i} = [w_{i−1}, w] for i > 1.

We know from Lemma 2.3 that the application

\begin{aligned}ϕ: & Y_{2}(G) & ⟶ & Y_{2}(H) \\ & f(x̃)(x̃ − 1)s_{1} & ↦ & f(x̃)(x̃ − 1)w_{1} \end{aligned}

is an S-isomorphism. We need the following lemma. Lemma 2.6. Let t(x, y) be a word in Γ_{2}(x, y). Then for any a, b ∈ S, ϕ(t(as_{1}, bs_{1})) = t(aw_{1}, bw_{1}). Proof. Let τ be the natural projection of S ∧ S onto R/𝔞^{m−2} ∧ R/𝔞^{m−2}. Since α_{(G,s,s_{1})} = α_{(H,w,w_{1})}, we obtain that

\begin{aligned}ϕ([as_{1}, bs_{1}]_{L}) & = & ϕ(α_{(G,s,s_{1})}(τ(a ∧ b))(x̃ − 1)^{3}s_{1}) \\ & = & α_{(H,w,w_{1})}(τ(a ∧ b))(x̃ − 1)^{3}w_{1} = [aw_{1}, bw_{1}]_{L}, \end{aligned}

so the proof follows by an easy induction on the weight of t. Hence, in particular, ϕ is a Lie ring isomorphism, and also a group isomorphism. Now, we prove by induction that for each i ≥ 1 there exists f_{i} ∈ S such that s_{i} = f_{i}s_{1} and w_{i} = f_{i}w_{1}. If i = 1 we are done. Suppose that it holds for i = k. Then we have

\begin{aligned}s_{k+1} & = & [s_{k}, s] = (f_{k}s_{1})^{−1}(x̃f_{k}s_{1}) \\ & = & −f_{k}s_{1} + x̃f_{k}s_{1} + \frac{[−f_{k}s_{1},x̃f_{k}s_{1}]_{L}}{2} + t_{3}(−f_{k}s_{1}, x̃f_{k}s_{1}), \end{aligned}

and a similar equality can be written for w_{k+1}. Hence, by Lemma 2.6, ϕ(s_{k+1}) = w_{k+1}. Note that it is equivalent to the existence of an element f_{k+1} ∈ S such that s_{k+1} = f_{k+1}s_{1} and w_{k+1} = f_{k+1}w_{1}, which is what we had to prove in order to complete the induction. Now, using (6) and Lemma 2.6, we obtain that ϕ([as_{1}, bs_{1}]) = [aw_{1}, bw_{1}] and thus, ϕ([s_{i}, s_{j}]) = [w_{i}, w_{j}] for all i, j ≥ 1 (of course, this is clear for i, j > 1 since ϕ is a group isomorphism between Y_{2}(G) and Y_{2}(H)).

Let [s_{i}, s_{j}] = s_{i+j}^{c_{i+j,i,j}} ⋯ s_{m−1}^{c_{m−1,i,j}}. Since ϕ is a group isomorphism, then [w_{i}, w_{j}] = w_{i+j}^{c_{i+j,i,j}} ⋯ w_{m−1}^{c_{m−1,i,j}}. Note that the group G is determined up to isomorphism by the relations

s^{p} = s_{m−1}^{u}, (ss_{1})^{p} = s_{m−1}^{v}, [s_{i}, s_{j}] = s_{i+j}^{c_{i+j,i,j}} ⋯ s_{m−1}^{c_{m−1,i,j}}

because the powers s_{i}^{p} can be computed from them by using P. Hall’s compilation formula. Since the group H can be generated by elements satisfying exactly the same relations, we conclude that G and H are isomorphic groups.

Remark 2.7. Indeed, in the last theorem we have proved that there is an isomorphism τ: G → H such that τ(s) = w and τ(s_{1}) = w_{1}.

Remark 2.8. The construction given in [5] when Y_{1} is of class 2 can be considered as a particular case of the above theorem.

3 Relations defining a p-group of maximal class

As we have mentioned at the end of the last theorem, a p-group of maximal class G is determined once we know the p-powers s^{p} and (ss_{1})^{p} and the commutators [s_{i}, s_{j}]. In this section we shall prove that it is actually enough to specify the values of the commutators [s_{i}, s_{i+1}] for 1 ≤ i ≤ (p − 3)/2 and the parameters u, v and w defined by s^{p} = s_{m−1}^{u}, (ss_{1})^{p} = s_{m−1}^{v} and

[s_{(p−1)/2}, s_{(p+1)/2}] = s_{p}^{c_{p}} ⋯ s_{m−2}^{c_{m−2}}s_{m−1}^{w}, 0 ≤ c_{i} ≤ p − 1.

We recall that, in addition to the group operation, we have defined in the subgroup Y_{1} two laws + and [ , ]_{L}, and also an S-module structure. We maintain the notation of preceding section and, in particular, α will be the homomorphism α_{(G,s,s_{1})} ∈ Hom_{S}(R/𝔞^{m−2} ∧R/𝔞^{m−2}, R/𝔞^{m−4}). We define the elements w_{1} = s_{1} and w_{i} = (x̃ − 1)^{i−1}w_{1} for i > 1. It is easy to prove that w_{i} + Y_{i+1} = s_{i} + Y_{i+1}, and, in particular, w_{m−1} = s_{m−1}.

The structure of R ∧ R was studied in [5]. It was proved that R ∧ R is the direct sum of a free S-module F of rank (p − 3)/2, freely generated by {(x̄ − 1)^{i} ∧ (x̄ − 1)^{i−1} | 1 ≤ i ≤ (p − 3)/2}, and a cyclic module generated by an element z on which x̃ acts trivially. In [5, Proposition 5.6] an element z was given explicitly. We also consider the element z_{1} = (ϕ_{2}^{−1} ∘ α)(z + 𝔟_{m−2}) ∈ Y_{4}.

Since (x̃−1)z = 0, it follows that α(z +𝔟_{m−2}) ∈ 𝔞^{m−5}/𝔞^{m−4} and z_{1} ∈ Y_{m−1}. Consequently, there exists 0 ≤ γ ≤ p − 1 such that

α(z + 𝔟_{m−2}) = γ(x̄ − 1)^{m−5} + 𝔞^{m−4}, z_{1} = γw_{m−1} = s_{m−1}^{γ}.  (7)

It is clear that if G ≅ H/Y_{m}(H) for some p-group of maximal class H of order at least p^{m+1} then γ = 0.

Finally, for each integer k, we write k = k_{1}(p − 1) + k_{2} with 0 ≤ k_{2} ≤ p − 2 and define the ideal I_{k} of S as follows:

I_{k} = \{ \begin{matrix}S & if k ≤ 0, \\ Sp^{k_{1}}(x̃ − 1)^{k_{2}} + Sp^{k_{1}+1} & if k > 0 and k_{2} ≠ 0, \\ Sp^{k_{1}} + Sp^{k_{1}−1}(x̃ − 1)^{p−1} & if k > 0 and k_{2} = 0. \end{matrix}

We need the following lemma.

Lemma 3.1. Let n > 0, t_{i} = (x̄ − 1)^{i−1} ∧ (x̄ − 1)^{i} for 1 ≤ i ≤ (p − 3)/2 and t_{(p−1)/2} = (x̄ − 1)^{(p−3)/2} ∧ (x̄ − 1)^{(p−1)/2} or t_{(p−1)/2} = z. Then there exist c_{i} ∈ I_{q+w+1−2i} such that

(x̄ − 1)^{q} ∧ (x̄ − 1)^{w} + 𝔟_{n} = ∑_{i=1}^{(p−1)/2} c_{i}t_{i} + 𝔟_{n}.  (8)

Proof. If t_{(p−1)/2} = (x̄ − 1)^{(p−3)/2} ∧ (x̄ − 1)^{(p−1)/2}, then Lemma follows from [6, Corollary 8.6].

By [5, Proposition 5.6], z ≡ (x̄ − 1)^{(p−1)/2} ∧ (x̄ − 1)^{(p−3)/2} mod C, where C is a S-submodule of R ∧ R generated by {(x̃ − 1)^{p−1−2i}((x̄ − 1)^{i−1} ∧ (x̄ − 1)^{i}) | i = 1, …, (p − 3)/2}. Hence Lemma is also true when t_{(p−1)/2} = z.

In the sequel, when we write the triple (G, s, s_{1}), we understand that G is a p-group of maximal class of order p^{m}, s ∈ G \ (Y_{1} ∪ C_{G}(Y_{m−2})) and s_{1} ∈ Y_{1} \ Y_{2}. Also, when we work with a group G′ we write all the notation related to this group with ′. In particular, α′ will be the homomorphism α_{(G′,s′,s′_{1})}.

First we prove the following theorem.

Theorem 3.2. Let G be a p-group of maximal class of order p^{m} and u, v, γ given integer numbers. Let v_{i}, 1 ≤ i ≤ (p − 3)/2, be words in s_{2i+1}, …, s_{m−1} and suppose the relations

s^{p}  = s_{m−1}^{u},
(ss_{1})^{p}  = s_{m−1}^{v},
[s_{i}, s_{i+1}] = v_{i},  1 ≤ i ≤ (p − 3)/2,
z_{1}  = s_{m−1}^{γ},

hold. Then, for any triple (G′, s′, s′_{1}) that satisfies the same conditions we have α_{(G′,s′,s′_{1})} = α_{(G,s,s_{1})} and, in particular, G′ is isomorphic to G.

Proof. Suppose now that we have a triple (G′, s′, s′_{1}) such that

(s′)^{p}  = (s′_{m−1})^{u},
(s′s′_{1})^{p}  = (s′_{m−1})^{v},
[s′_{i}, s′_{i+1}]  = v_{i}(s′_{2i+1}, …, s′_{m−1}), 1 ≤ i ≤ (p − 3)/2,
α′(z + 𝔟_{m−2}) = α(z + 𝔟_{m−2}).

We want to prove that α = α′. We proceed by induction on m. When m = 4, it is obvious. We suppose we have proved the claim for m − 1 and we shall prove it for m.

If we apply the inductive hypothesis to the groups G/Y_{m−1} and G′/Y′_{m−1} we obtain that α_{(G/Y_{m−1},s̄,s̄_{1})} = α_{(G′/Y′_{m−1},s̄′,s̄′_{1})} and so

α ≡ α′ (mod 𝔞^{m−5}/𝔞^{m−4}).  (9)

First we shall prove that if i + j ≥ p − 3 then

α((x̄ − 1)^{i} + 𝔞^{m−2} ∧ (x̄ − 1)^{j} + 𝔞^{m−2}) = α′((x̄ − 1)^{i} + 𝔞^{m−2} ∧ (x̄ − 1)^{j} + 𝔞^{m−2}). (10)

Using Lemma 3.1, we obtain that

(x̄ − 1)^{i} ∧ (x̄ − 1)^{j} + 𝔟_{m−2} = ∑_{k=1}^{(p−3)/2} c_{k}(x̄ − 1)^{k−1} ∧ (x̄ − 1)^{k} + c_{(p−1)/2}z + 𝔟_{m−2}, (11)

for some c_{k} ∈ I_{i+j+1−2k}. Since i+j ≥ p−3, then c_{k} ∈ I_{1} for 1 ≤ k ≤ (p−3)/2. It is clear that if a, b ∈ R, c ∈ I_{1} and a ≡ b (mod 𝔞^{m−5}) then ca ≡ cb (mod 𝔞^{m−4}). Therefore applying α and α′ to (11) and using (9) we obtain (10).

We prove now by simultaneous decreasing induction on k ≥ 1 the following two claims:

P_{k}: For every i ≥ k + 2 there exists f_{i} ∈ (x̃ − 1)^{k+1}S such that s_{i} = f_{i}w_{1} and s′_{i} = f_{i}w′_{1}.

Q_{k}: If i + j ≥ k then α((x̄ − 1)^{i} + 𝔞^{m−2} ∧ (x̄ − 1)^{j} + 𝔞^{m−2}) = α′((x̄ − 1)^{i} + 𝔞^{m−2} ∧ (x̄ − 1)^{j} + 𝔞^{m−2}).

The basis of our induction is (10), which simply says that Q_{k} is true for k ≥ p−3. Also we have that P_{k} is true for k = m − 3. Note that Q_{k} is equivalent to

R_{k}: If i + j ≥ k + 2 then there exists h_{i,j} ∈ S such that [w_{i}, w_{j}]_{L} = h_{i,j}w_{4} and [w′_{i}, w′_{j}]_{L} = h_{i,j}w′_{4}.

We need the following lemma.

Lemma 3.3. Suppose Q_{k} holds and let t(x, y) be a word in Γ_{2}(x, y). Let h_{1} ∈ 𝔞^{i}, h_{2} ∈ 𝔞^{j}, i + j ≥ k, and suppose also that r_{1} ≡ h_{1}w_{1} (mod Y_{m−1}), r_{2} ≡ h_{2}w_{1} (mod Y_{m−1}), r′_{1} ≡ h_{1}w′_{1} (mod Y′_{m−1}) and r′_{2} ≡ h_{2}w′_{1} (mod Y′_{m−1}). Then there exists h ∈ (x̃ − 1)^{k+1}S such that t(r_{1}, r_{2}) = hw_{1} and t(r′_{1}, r′_{2}) = hw′_{1}.

Proof. Note that [r_{1}, r_{2}]_{L} = α(h_{1} + 𝔞^{m−2} ∧ h_{2} + 𝔞^{m−2})w_{4}, so the proof follows by an easy induction on the weight of t.

Now we prove that if k ≥ 2, then {P_{k}, Q_{k}} ⇒ P_{k−1}. Suppose that i ≥ k + 1. By (5), we have that there exists t ∈ Γ_{2}(s_{i−1}^{−1}, x̃s_{i−1}) = Γ_{2}(s_{i−1}, (x̃ − 1)s_{i−1}) such that

s_{i} = s_{i−1}^{−1}s_{i−1}^{s} = s_{i−1}^{−1} + s_{i−1}^{s} + ([s_{i−1}^{−1}, s_{i−1}^{s}]_{L})/2 + t_{3}(s_{i−1}^{−1}, s_{i−1}^{s}) = (x̃ − 1)s_{i−1} + t.

We can write the same equality for s′_{i}: s′_{i} = (x̃−1)s′_{i−1}+t′. Since α_{(G/Y_{m−1},s̄,s̄_{1})} = α_{(Ḡ′/Y′_{m−1},s̄′,s̄′_{1})}, there exists g_{i−1} ∈ 𝔞^{i−2} such that s_{i−1} ≡ g_{i−1}w_{1} (mod Y_{m−1}) and s′_{i−1} ≡ g_{i−1}w′_{1} (mod Y′_{m−1}). Hence (x̃ − 1)s_{i−1} = g_{i−1}(x̃ − 1)w_{1} and (x̃ − 1)s′_{i−1} = g_{i−1}(x̃ − 1)w′_{1}. We can apply Lemma 3.3 because i − 2 + i − 1 ≥ 2k − 1 ≥ k so that t = hw_{1} and t′ = hw′_{1} for some h ∈ (x̃ − 1)^{k+1}S. This proves P_{k−1}.

Finally, we prove that if 1 ≤ k ≤ p − 4, then {Q_{k+1}, P_{k}} ⇒ Q_{k}. Let k + 2 ≤ 2l + 1 ≤ p − 2. We have

[s_{l}, s_{l+1}] = v_{l}(s_{2l+1}, …, s_{m−1}), [s′_{l}, s′_{l+1}] = v_{l}(s′_{2l+1}, …, s′_{m−1}).

Applying P_{k}, (5) and Lemma 3.3 to the right side of these equalities, we obtain that there exists h_{l} ∈ (x̃ − 1)^{k+1}S such that [s_{l}, s_{l+1}] = h_{l}w_{1} and [s′_{l}, s′_{l+1}] = h_{l}w′_{1}.

On the other hand,

[s_{l}, s_{l+1}] = [s_{l}, s_{l+1}]_{L} + t_{4}(s_{l}, s_{l+1}) = [w_{l}, w_{l+1}]_{L} + t, [s′_{l}, s′_{l+1}] = [w′_{l}, w′_{l+1}]_{L} + t′,

where t = [s_{l}−w_{l}, s_{l+1}−w_{l+1}]_{L}+[w_{l}, s_{l+1}−w_{l+1}]_{L}+[s_{l}−w_{l}, w_{l+1}]_{L}+t_{4}(s_{l}, s_{l+1}). We can apply Lemma 3.3 to each summand of t to conclude that there exists g_{l} ∈ (x̃ − 1)^{k+1}S such that

[w_{l}, w_{l+1}]_{L} = g_{l}w_{1}, [w′_{l}, w′_{l+1}] = g_{l}w′_{1},  (12)

which yields the equality α((x̄ − 1)^{l−1} + 𝔞^{m−2} ∧ (x̄ − 1)^{l} + 𝔞^{m−2}) = α′((x̄ − 1)^{l−1} + 𝔞^{m−2} ∧ (x̄ − 1)^{l} + 𝔞^{m−2}). Hence Q_{k} follows from (11). This proves the theorem because Q_{1} means that α = α′.

Now we can prove the result announced, in the beginning of this section.

Theorem 3.4. Let G be a group of maximal class of order p^{m} and u, v, w given integer numbers. Let v_{i}, 1 ≤ i ≤ (p − 3)/2, be words in s_{2i+1}, …, s_{m−1}. Suppose the relations

s^{p}  = s_{m−1}^{u},
(ss_{1})^{p}  = s_{m−1}^{v},
s_{i}  = [s_{i−1}, s], i > 1,
[s_{i}, s_{i+1}] = v_{i},  1 ≤ i ≤ (p − 3)/2,

hold and also that [s_{(p−1)/2}, s_{(p+1)/2}] = s_{p}^{c_{p}} ⋯ s_{m−2}^{c_{m−2}}s_{m−1}^{w} for some 0 ≤ c_{i} ≤ p − 1. Then, for any triple (G′, s′, s′_{1}) that satisfies the same conditions we have α_{(G′,s′,s′_{1})} = α_{(G,s,s_{1})} and, in particular, G′ is isomorphic to G.

Proof. Suppose now that there exists a triple (G′, s′, s′_{1}) such that

(s′)^{p}  = (s′_{m−1})^{u},
(s′s′_{1})^{p}  = (s′_{m−1})^{v},
[s′_{i}, s′_{i+1}] = v_{i}(s′, s′_{1}), 1 ≤ i ≤ (p − 3)/2,
[s′_{(p−1)/2}, s′_{(p+1)/2}] = (s′_{p})^{b_{p}} ⋯ (s′_{m−2})^{b_{m−2}}(s′_{m−1})^{w},

where 0 ≤ b_{i} ≤ p − 1.

By the previous theorem we have that α_{(G/Y_{m−1},s̄,s̄_{1})} = α_{(G′/Y′_{m−1},s̄′,s̄′_{1})}. Hence, by Remark 2.7, we obtain that b_{i} = c_{i} for p ≤ i ≤ m − 2. As in the previous theorem we can prove the statement Q_{p−1} and P_{p−2}. Then there exists h ∈ S such that [s_{(p−1)/2}, s_{(p+1)/2}] = hw_{4} and [s′_{(p−1)/2}, s′_{(p+1)/2}] = hw′_{4}. Now using the same argument as in the proof of (12) we obtain that α((x̄−1)^{(p−3)/2}+ 𝔞^{m−2} ∧(x̄−1)^{(p−1)/2} +𝔞^{m−2}) = α′((x̄−1)^{(p−3)/2} +𝔞^{m−2} ∧(x̄−1)^{(p−1)/2} +𝔞^{m−2}).

By [5, Proposition 5.6], there exist d_{k} ∈ I_{p−1−2k} such that

z = ∑_{k=1}^{(p−1)/2} d_{k}(x̄ − 1)^{k−1} ∧ (x̄ − 1)^{k}.  (13)

Therefore applying α and α′ to (13) we obtain that α(z) = α′(z) and, by the previous theorem, α = α′.

It should be remarked that this theorem had been proved by Shepherd in [8] using different methods under the hypotheses m ≤ 2c + p or Y_{1} of class 2.

Corollary 3.5. Let T_{m} be the number of p-groups of maximal class of order p^{m}. Then

T_{m} ≤ \{ \begin{matrix}p^{(m/2)+3} & if p = 5 \\ p^{(m+p−6)(p−3)/4+3} & if p > 5 \end{matrix}

Proof. We know if p = 5, then m ≤ 2c + 6 and if p > 5 then m ≤ 2c + 2p − 5 ([3]). By above theorem we obtain the result desired.

Acknowledgements. We wish to thank Prof. J. Sangroniz Gomez for his helpful comments and suggestions in the writing of this paper.

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[8] R. Shepherd, On p-groups of maximal class, unpublished.