ARBITRARY NILPOTENCY DEGREE OF

COHOMOLOGY RINGS IN CHARACTERISTIC ppp

Pham Anh Minh

29 September 2002

Abstract. Let p be an odd prime number and let n be an arbitrary positive integer. We prove that there exists a p-group G whose mod-p cohomology ring has a nilpotent element ξ ∈ H^{2}(G) satisfying ξ^{n} ≠ 0, ξ^{n+p−1} = 0. As a corollary, we get the existence of a p-group whose mod-p cohomology ring has an element of nilpotency degree n.

1. Statement of main results

For every p-group G, denote by H^{∗}(G) the mod-p cohomology algebra of G. We are interested in the nilpotency degrees of elements of H^{∗}(G). For the case p = 2, in [1, 4], it was shown that, given any positive integer n, there exists a 2-group whose cohomology ring has elements of nilpotency degree n + 1. It was also noted in [1] that, for p odd, we did not have any example of elements of H^{∗}(G) having nilpotency degrees greater than p.

Recently, such examples were given in [7, 8]. In [5], we also proved that, for any prime p and for any p-group G of order p^{m}, nilpotency degrees of nilpotent elements of H^{∗}(G) are bounded above by p^{m−1}. However, we could not find in the litterature any example of nilpotent mod-p cohomology classes of p-groups having arbitrarily large nilpotency degree.

The purpose of this paper is to prove the following

Theorem. Given a positive integer n, there exist a p-group G and a nilpotent mod-p cohomology class ξ ∈ H^{2}(G) such that ξ^{n} ≠ 0, ξ^{n+p−1} = 0.

Corollary. Given a positive integer n, there exists a p-group whose mod-p cohomology ring has an element of nilpotency degree n.

The case p = 2 of these results was proved by Avrunin-Carlson ([1]) and Inoue-Kono ([4]). From now on, we suppose that p is an arbitrary odd prime number, unless otherwise specified. We shall use the following notation. For homogeneous elements X, Y, … of a commutative graded ring R, (X, Y, …) denotes the ideal of R generated by X, Y, . . .. If S is a subset of a group G, then ⟨S⟩ denotes the subgroup of G generated by S. With some abuse of notation, for every ζ ∈ H^{∗}(G) and for every subgroup K of G, we consider ζ as an element of H^{∗}(K) via the restriction map; also, for every extension T of G, ζ is considered as an element of H^{∗}(T) via the inflation map.

Typeset by AMS-TEX

2000 Mathematics Subject Classification. 20J06; Secondary 20D15, 55R40.

2. The ring R_{m}

From now on, suppose that R_{m} = R_{m}/J_{m}, with R_{m} = ℤ/p[z_{0}, …, z_{m}] and J_{m} the ideal of R_{m} generated by the elements f_{1}, …, f_{m} given by

f_{i} = z_{i}^{p} − z_{i−1}^{p−1}z_{i}, 1 ≤ i ≤ m.

We first prove

Proposition 1. 0 ≠ z_{m}^{m(p−1)} in R_{m}/(z_{0}).

The case p = 2 of the above proposition was proved by Avrunin-Carlson ([1]) and Inoue-Kono ([4]), using either Groebner basis or the regularity of the sequence z_{0}, f_{0}, …, f_{m}. We shall give a short proof of it for the case p odd (this proof also holds for p = 2). The key lemma is Lemma 2. (I am indebted to N.V. Trung for providing a nice and surprising proof of this lemma.)

Lemma 1. For 1 ≤ ℓ ≤ m, as elements of R_{m},

z_{ℓ}^{j+i(p−1)} = \{ \begin{matrix}z_{ℓ−i+1}^{p−1} … z_{ℓ−1}^{p−1}z_{ℓ}^{p−1} & if j = 0 and 1 ≤ i ≤ ℓ + 1, \\ z_{ℓ−i}^{p−1} … z_{ℓ−1}^{p−1}z_{ℓ}^{j} & if 0 < j < p − 1 and 1 ≤ i ≤ ℓ. \end{matrix}

Proof. The lemma clearly holds for i = 1, as z_{ℓ}^{p} − z_{ℓ−1}^{p−1}z_{ℓ} = f_{ℓ} = 0 in R_{m}. Assume that it holds for i with 1 ≤ i < ℓ + 1 if j = 0, 1 ≤ i < ℓ if j > 0. It follows from the inductive hypothesis that z_{ℓ}^{i(p−1)} = z_{ℓ−1}^{(i−1)(p−1)}z_{ℓ}^{p−1}. Therefore

z_{ℓ}^{j+(i+1)(p−1)} = z_{ℓ}^{i(p−1)}z_{ℓ}^{j+p−1}
= z_{ℓ−1}^{(i−1)(p−1)}z_{ℓ}^{j+2(p−1)}
= z_{ℓ−1}^{(i−1)(p−1)}z_{ℓ}^{p}z_{ℓ}^{j+p−2}
= z_{ℓ−1}^{(i−1)(p−1)}z_{ℓ−1}^{p−1}z_{ℓ}z_{ℓ}^{j+p−2}
= z_{ℓ−1}^{i(p−1)}z_{ℓ}^{p−1}z_{ℓ}^{j}.

So z_{ℓ}^{j+(i+1)(p−1)} = z_{ℓ−i}^{p−1} … z_{ℓ−1}^{p−1}z_{ℓ}^{p−1} if j = 0. If 0 < j < p − 1, then

z_{ℓ}^{j+(i+1)(p−1)} = z_{ℓ−1}^{i(p−1)}z_{ℓ}^{p}z_{ℓ}^{j−1}
= z_{ℓ−1}^{i(p−1)}z_{ℓ−1}^{p−1}z_{ℓ}z_{ℓ}^{j−1}
= z_{ℓ−1}^{(i+1)(p−1)}z_{ℓ}^{j}
= z_{ℓ−i−1}^{p−1} … z_{ℓ−1}^{p−1}z_{ℓ}^{j}.  □

The proof of the following lemma is due to N.V. Trung.

Lemma 2. Let f be a non-zero element of R_{m−k}. Then, for 0 ≤ i_{1}, …, i_{k} ≤ p−1, f · z_{m−k+1}^{i_{1}} … z_{m}^{i_{k}} is a non-zero element of R_{m}.

Proof. Note that R_{m} = R_{m−k}[z_{m−k+1}, …, z_{m}]/(f_{m−k+1}, …, f_{m}). Suppose that g = f · z_{m−k+1}^{i_{1}} … z_{m}^{i_{k}} vanishes in R_{m} = R_{m−1}[z_{m}]/(f_{m}). Consider g and f_{m} as polynomials with variable z_{m} and with coefficients in R_{m−1}. Then deg(g) = i_{k} ≤ p − 1 < p = deg(f_{m}). So g = 0 in R_{m} implies f · z_{m−k+1}^{i_{1}} … z_{m−1}^{i_{k−1}} = 0 in R_{m−1}. Repeating this argument, it follows that f = 0 in R_{m−k}, a contradiction. □

The proof of the above lemma also yields Lemma 3. (i) Let f be a non-zero element of R_{m−1}/(z_{0}). Then, for 0 ≤ i ≤ p−1, f · z_{m}^{i} is a non-zero element of R_{m}/(z_{0}).

(ii) Let

f = ∑_{0≤i_{0},...,i_{m}≤p−1} μ_{i_{0},...,i_{m}}z_{0}^{i_{0}} … z_{m}^{i_{m}}

be an element of R_{m}, with μ_{i_{0},...,i_{m}} ∈ ℤ/p. If f vanishes in R_{m}, then all the μ_{i_{0},...,i_{m}}’s vanish. □

Proof of Proposition 1. We argue by induction on m. If m = 1, it is clear that z_{1}^{p−1} ≠ 0 in R_{1}/(z_{0}). Suppose that the lemma holds for m − 1. By Lemma 1 and by the inductive hypothesis, z_{m−1}^{(m−1)(p−1)} = z_{1}^{p−1} … z_{m−1}^{p−1} ≠ 0 in R_{m−1}/(z_{0}). By Lemmas 1 and 3 (i), z_{m}^{m(p−1)} = z_{1}^{p−1} … z_{m}^{p−1} ≠ 0 in R_{m}/(z_{0}). □

Let

𝔄 = R_{m−1} ⊗ ℤ/p[w_{0}, w_{1}]/(w_{1}^{p} − w_{0}^{p−1}w_{1}).

We have

Lermma 4. (i) Let A be the subalgebra of 𝔄 generated by w_{1}, z_{1}, …, z_{m}. Then A is free over ℤ/p[w_{1}].

(ii) 𝔄 is free, as a module over ℤ/p[w_{0}].

Proof. (i) is clear. For (ii), we need only prove that W = ℤ/p[w_{0}, w_{1}]/(w_{1}^{p}−w_{0}^{p−1}w_{1}) is free over ℤ/p[w_{0}]. As every element of W is a linear combination of the w_{0}^{i}w_{1}^{j}’s, 0 ≤ i, 0 ≤ j ≤ p − 1, W is free over ℤ/p[w_{0}]. □

Define

\begin{aligned}\overline{w}_{i} & = w_{i}^{p} − (z_{0} − w_{1})^{p−1}w_{i}, 0 ≤ i ≤ 1, \\ \overline{z}_{j} & = z_{j}^{p} − (z_{0} − w_{1})^{p−1}z_{j}, 0 ≤ j ≤ m − 1, \end{aligned}

and let h be an element of 𝔄 of form

h = _{∑_{0≤i_{2},...,i_{m}≤p−1} ; i≥0}μ_{i,i_{2},...,i_{m}}w_{0}^{i}(w_{1}^{p−1} − w_{0}^{p−1})z_{1}^{i_{2}} … z_{m−1}^{i_{m}}

with μ_{i,i_{2},...,i_{m}} ∈ ℤ/p. We have

Proposition 2. Suppose that

\overline{h} = _{∑_{0≤i_{2},...,i_{m}≤p−1} ; i≥0}μ_{i,i_{2},...,i_{m}} \overline{w}_{0}^{i}(\overline{w}_{1}^{p−1} − \overline{w}_{0}^{p−1})\overline{z}_{1}^{i_{2}} … \overline{z}_{m−1}^{i_{m}}

vanishes in 𝔄. Then all the μ_{i,i_{2},...,i_{m}}’s vanish.

Proof. Consider \overline{w}_{0}^{i}(\overline{w}_{1}^{p−1} − \overline{w}_{0}^{p−1}) as a polynomial with variable w_{0}. This polynomial is then of degree p(p − 1 + i) with leading term −w_{0}^{p(p−1+i)}. Also, for any i,

h_{i} = ∑_{0≤i_{2},...,i_{m}≤p−1} μ_{i,i_{2},...,i_{m}} \overline{z}_{1}^{i_{2}} … \overline{z}_{m−1}^{i_{m}},

as a polynomial with variable w_{1}, has leading term

− ∑_{0≤i_{2},...,i_{m}≤p−1} μ_{i,i_{2},...,i_{m}}z_{1}^{i_{2}} … z_{m−1}^{i_{m}}w_{1}^{(p−1)(i_{2}+···+i_{m})}.

By Lemma 4 (ii), in 𝔄, \overline{h} = 0 implies h_{i} = 0, for any i. Again, by Lemma 4 (i), h_{i} = 0 in 𝔄 implies

∑_{0≤i_{2},...,i_{m}≤p−1} μ_{i,i_{2},...,i_{m}}z_{1}^{i_{2}} … z_{m−1}^{i_{m}}.

vanishes in R_{m−1}. By Lemma 3 (ii), all the μ_{i,i_{2},...,i_{m}}’s vanish. □

Lemma 5. Let i, j be integers with 0 ≤ i, j ≤ p − 1, and let g = ∑_{i=0}^{p−1}z_{1}^{i}g_{i} be an

element of R_{1} with g_{i} ∈ ℤ/p[z_{0}]. Suppose that z_{1}^{j}g ∈ J_{1}. Then g ∈ (z_{1}^{p−1} − z_{0}^{p−1}).

Proof. If j = 0, as g ∈ (f_{1}), it follows that g ∈ (z_{1}^{p−1} − z_{0}^{p−1}). Suppose that j > 0 and set k = p − 1 − j. We have

z_{1}^{j}gj = ∑_{i=0}^{p−1} z_{1}^{i+j}g_{i}
= z_{1}^{j}g_{0} + ⋯ + z_{1}^{p−1+j}g_{p−1}
= (z_{1}g_{k+1} + ⋯ + z_{1}^{j−1}g_{p−2})z_{0}^{p−1} + z_{1}^{j}(g_{0} + g_{p−1}z_{0}^{p−1}) + z_{1}^{j+1}g_{1}
+ ⋯ + z_{1}^{p−2}g_{k−1} + z_{1}^{p−1}g_{k} + t

with t = (z_{1}^{p−1} − z_{0}^{p−1})(z_{1}g_{k+1} + ⋯ + z_{1}^{j}g_{p−1}). As t ∈ J_{1}, it follows that q = z_{1}^{j}g − t ∈ J_{1}. Since q, as a polynomial with variable z_{1}, is of degree ≤ p − 1, Lemma 2 shows that q = 0. Therefore g_{1} = ⋯ = g_{p−2} = g_{0} + z_{0}^{p−1}g_{p−1} = 0. So

\begin{aligned}g = g_{0} + z_{1}^{p−1}g_{p−1} & = g_{0} + z_{0}^{p−1}g_{p−1} + (z_{1}^{p−1} − z_{0}^{p−1})g_{p−1} \\ & = (z_{1}^{p−1} − z_{0}^{p−1})g_{p−1} \end{aligned}

which implies g ∈ (z_{1}^{p−1} − z_{0}^{p−1}). □

In the remaining of this section, g is supposed to be an element of R_{m} given by

g = _{∑_{0≤i_{1},...,i_{m}≤p−1} ; i≥0}λ_{i,i_{1},...,i_{m}}z_{0}^{i}z_{1}^{i_{1}} … z_{m}^{i_{m}}

with λ_{i,i_{1},...,i_{m}} ∈ ℤ/p.

Lemma 6. Suppose that m = 2 and g(z_{0}, z_{1}, λz_{1}) ∈ J_{1}, for every 0 ≠ λ ∈ ℤ/p. Then:

(i) for every i_{2}, g_{i_{2}}z_{1}^{i_{2}} ∈ J_{1}, with g_{i_{2}} = ∑_{i,i_{1}}λ_{i,i_{1},i_{2}}z_{0}^{i}z_{1}^{i_{1}};

(ii) g ∈ (z_{1}^{p−1} − z_{0}^{p−1}).

ARBITRARY NILPOTENCY DEGREE OF COHOMOLOGY RINGS 5

Proof. (i) For every 0 ≠ λ ∈ ℤ/p, it follows from the hypothesis that

g(z_{0}, z_{1}, λz_{1}) = ∑_{i_{2}} λ^{i_{2}}g_{i_{2}}z_{1}^{i_{2}} ∈ J_{1}.

As

\begin{aligned}∑_{0≠λ∈ℤ/p}(∑_{i_{2}} λ^{i_{2}}g_{i_{2}}z_{1}^{i_{2}}) & = ∑_{i_{2}}( ∑_{0≠λ∈ℤ/p} λ^{i_{2}})g_{i_{2}}z_{1}^{i_{2}} \\ & = −g_{p−1}z_{1}^{p−1} \end{aligned}

by Newton’s formula, g_{p−1}z_{1}^{p−1} also belongs to J_{1}.

Assume that g_{p−k}z_{1}^{p−k} ∈ J_{1}, for 1 ≤ k < p. For 0 ≠ λ ∈ ℤ/p, set

q_{λ} = ∑_{i_{2}=0}^{p−k−1} λ^{i_{2}}g_{i_{2}}z_{1}^{i_{2}}.

Then, by the inductive hypothesis, q_{λ} belongs to J_{1}, and hence so does

∑_{i_{2}=0}^{p−k−1} λ^{k+i_{2}}g_{i_{2}}z_{1}^{i_{2}} = λ^{k}q_{λ}.

Therefore

−g_{p−k−1}z_{1}^{p−k−1} = ∑_{λ∈ℤ/p}(λ^{k}g_{0} + λ^{k+1}g_{1}z_{1} + ⋯ + λ^{p−1}g_{p−k−1}z_{1}^{p−k−1})

also belongs to J_{1}. (i) is proved. (ii) It follows from (i) and Lemma 5 that, for any i_{2}, g_{i_{2}} ∈ (z_{1}^{p−1} − z_{0}^{p−1}). As

g = ∑_{i_{2}}g_{i_{2}}z_{2}^{i_{2}}, g also belongs to (z_{1}^{p−1} − z_{0}^{p−1}). □

Proposition 3. Suppose that g(z_{0}, z_{1}, λz_{1}, z_{2}, …, z_{m−1}) ∈ J_{m−1}, for any 0 ≠ λ ∈ ℤ/p. Then g ∈ (z_{1}^{p−1} − z_{0}^{p−1}).

Proof. For given i_{3}, …, i_{m}, set g_{i_{3},...,i_{m}} = ∑_{i,i_{1},i_{2}}λ_{i,i_{1},i_{2},...,i_{m}}z_{0}^{i}z_{1}^{i_{1}}z_{2}^{i_{2}}. So

g = ∑_{i_{3},...,i_{m}}g_{i_{3},...,i_{m}}z_{3}^{i_{3}} … z_{m}^{i_{m}}.

By Lemma 2, it follows that, for any i_{3}, …, i_{m} and for 0 ≠ λ ∈ ℤ/p, g_{i_{3},...,i_{m}}(z_{0}, z_{1}, λz_{1}) ∈ J_{1};

therefore g_{i_{3},...,i_{m}} ∈ (z_{1}^{p−1} − z_{0}^{p−1}), by Lemma 6. So g ∈ (z_{1}^{p−1} − z_{0}^{p−1}). □

3. The group 𝔐_{n}

Let 𝔗 be the p-group of order p^{p+1} given by

\begin{matrix}𝔗 = ⟨a_{0}, a_{1}, …, a_{p}|a_{0}^{p} = a_{2}^{p} = ⋯ = & a_{p}^{p} = [a_{0}, a_{p}] = [a_{i}, a_{j}] = a_{1}^{p}a_{p} = 1, \\ [a_{0}, a_{ℓ}] = a_{ℓ+1}, 1 ≤ i, j ≤ p, 1 ≤ ℓ ≤ p − 1⟩. \end{matrix}

Then 𝔗 is a 2-generator p-group of maximal class (see e.g. [2, Section 4]). Set ℜ = 𝔗/⟨a_{p}⟩ and g_{i} = a_{i}⟨a_{p}⟩ ∈ ℜ, 0 ≤ i ≤ p − 1. We then have the extension of groups

(𝔗)  0 → ℤ/p → 𝔗 →^{p_{2}} ℜ → 1.

Set S = ℜ/Φ(ℜ). S is then elementary abelian of rank 2 and is generated by r_{0}, r_{1} with r_{ℓ} = g_{ℓ}Φ(ℜ), 0 ≤ ℓ ≤ 1. Let s_{0}, s_{1} be elements of H^{∗}(S) defined by s_{k}(r_{ℓ}) = δ_{kℓ} with δ_{kℓ} the Kronecker symbol, 0 ≤ k, ℓ ≤ 1. Set t_{0} = βs_{0}, t_{1} = βs_{1} with β the Bockstein homomorphism. We then have

H^{∗}(S) = Λ[s_{0}, s_{1}] ⊗ ℤ/p[t_{0}, t_{1}].

For 0 ≤ i ≤ 1, s_{i} and t_{i} are considered as elements of H^{∗}(ℜ), hence of H^{∗}(𝔗), via the inflation maps. Let w be the element of H^{2}(ℜ) classifying the extension (𝔗). By [8, Lemma 1], we have

Lemma 7. As elements of H^{∗}(ℜ), w^{p} + t_{1}^{p} = t_{0}^{p−1}t_{1}. □

We then have

Lemma 8. As elements of H^{∗}(𝔗), t_{1}^{p} = t_{0}^{p−1}t_{1}.

Proof. Straightforward from Lemma 7, by noting that Inf_{𝔗}^{ℜ}(w) = 0. □

According to [2, Section 4], we also have

Lemma 9. For λ, μ ∈ ℤ/p with μ ≠ 0, (a_{0}^{μ}a_{1}^{λ})^{p} = 1 in 𝔗. □

Let T_{1}, …, T_{m} be m groups isomorphic to 𝔗 via the isomorphisms θ_{i} : 𝔗 → T_{i}, 1 ≤ i ≤ m, and set 𝔗_{m} = T_{1} × ⋯ × T_{m}. For each i, 1 ≤ i ≤ m, and for 0 ≤ j ≤ 1, set

\begin{aligned}a_{j}^{(i)} & = θ_{i}(a_{j}), \\ u_{j}^{(i)} & = (θ_{i}^{∗})^{−1}(s_{j}). \end{aligned}

Embed each T_{i} into 𝔗_{m} via the monomorphism

\begin{matrix}φ_{i} : T_{i} & → 𝔗_{m} \\ a & ↦ (\underbrace{1, …, 1}_{i−1 times}, a, 1 …, 1). \end{matrix}

By Künneth formula, the u_{j}^{(i)}’s are also considered as elements of H^{∗}(𝔗_{m}). We then have u_{k}^{(i)}(a_{ℓ}^{(j)}) = δ_{ij}δ_{kℓ}, 1 ≤ i, j ≤ m, 0 ≤ k, ℓ ≤ 1.

Let 𝔐_{m}, m ≥ 1, be the subgroup of 𝔗_{m} given by

𝔐_{m} = \{ \begin{matrix}𝔗_{1} & for m = 1, \\ ∩^{m−1}_{i=1}ker (u_{1}^{(i)} − u_{0}^{(i+1)}) & for m > 1, \end{matrix}

and, for 0 ≤ j ≤ m, define

b_{j} = \{ \begin{matrix}a_{0}^{(1)} & if j = 0, \\ a_{1}^{(j)}a_{0}^{(j+1)} & if 1 ≤ j ≤ m − 1, \\ a_{1}^{(m)} & if j = m, \end{matrix}

x_{j} = \{ \begin{matrix}Res_{𝔐_{m}}^{𝔗_{m}}(u_{0}^{(1)}) & if j = 0, \\ Res_{𝔐_{m}}^{𝔗_{m}}(u_{1}^{(j)}) & if 1 ≤ j ≤ m, \end{matrix}

and set y_{j} = βx_{j}. We have

Lemma 10. (i) Let λ_{1}, …, λ_{k} be non-zero elements of ℤ/p, 1 ≤ k ≤ m. Then b_{0}b_{1}^{λ_{1}} … b_{k−1}^{λ_{k−1}}b_{k}^{λ_{k}} is an element of order p of 𝔐_{m}.

(ii) For m ≥ 2, set U_{λ} = ker (x_{2} − λx_{1}) with 0 ≠ λ ∈ ℤ/p. Then the map

\begin{matrix}𝔐_{m−1} & →^{ρ_{λ}} U_{λ} \\ b_{0} & ↦ b_{0} \\ b_{1} & ↦ b_{1}b_{2}^{λ} \\ b_{i} & ↦ b_{i+1}, 2 ≤ i ≤ m − 1, \end{matrix}

is an isomorphism. (iii) As elements of H^{∗}(𝔐_{m}),

(iiia) y_{k} is not nilpotent, 0 ≤ k ≤ m; (iiib) y_{j}^{p} = y_{j−1}^{p−1}y_{j}, 1 ≤ j ≤ m; (iiic) for 1 ≤ k ≤ m, 0 ≤ j < p − 1,

y_{k}^{j+i(p−1)} = \{ \begin{matrix}y_{k−i+1}^{p−1} … z_{k−1}^{p−1}z_{k}^{p−1} & if j = 0 and 1 ≤ i ≤ k + 1, \\ z_{k−i}^{p−1} … z_{k−1}^{p−1}z_{k}^{j} & if 0 < j < p − 1 and 1 ≤ i ≤ k. \end{matrix}

Proof. (i) Set a = b_{0}b_{1}^{λ_{1}} … b_{k−1}^{λ_{k−1}}b_{k}^{λ_{k}}. It follows that 1 ≠ a. Set a_{0}^{(m+1)} = 1. We have

a^{p} = (a_{0}^{(1)}(a_{1}^{(1)}a_{0}^{(2)})^{λ_{1}} … (a_{1}^{(k)}a_{0}^{(k+1)})^{λ_{k}})^{p}

= (a_{0}^{(1)}(a_{1}^{(1)})^{λ_{1}})^{p}((a_{0}^{(2)})^{λ_{1}}(a_{1}^{(2)})^{λ_{2}})^{p} … ((a_{0}^{(k)})^{λ_{k−1}}(a_{1}^{(k)})^{λ_{k}})^{p}(a_{0}^{(k+1)})^{λ_{k+1}})^{p} = 1

by Lemma 9. Since u_{0}^{(i+1)}(a) = u_{1}^{(i)}(a), 1 ≤ i ≤ m − 1, a belongs to 𝔐_{m}. (i) is proved.

(ii) Note that, by (i), a_{0}^{(2)}(a_{1}^{(2)})^{λ} is of order p. If m = 2, as a_{0}^{(2)}(a_{1}^{(2)})^{λ} ∈ C_{𝔗_{2}}(T_{1}), it follows that ρ_{λ} : 𝔐_{1} → U_{λ} = ⟨a_{0}^{(1)}, a_{1}^{(1)}a_{0}^{(2)}(a_{1}^{(2)})^{λ}⟩ is an isomorphism. Suppose

8 PHAM ANH MINH

that m > 2. Set c = (a_{0}^{(3)})^{λ}, c_{1} = a_{1}^{(3)}. By (i), it follows that ⟨c, c_{1}⟩ ≅ 𝔗. Set c_{0} = a_{0}^{(2)}(a_{1}^{(2)})^{λ}(a_{0}^{(3)})^{λ}. As a_{0}^{(2)}(a_{1}^{(2)})^{λ} is of order p and belongs to C_{𝔗_{m}}(⟨c, c_{1}⟩), the map 𝔗 →^{ϕ_{λ}} ⟨c_{0}, c_{1}⟩, with ϕ_{λ}(a_{i}) = c_{i}, 0 ≤ i ≤ 1, is an isomorphism. Since U_{λ} = ⟨a_{0}^{(1)}, a_{1}^{(1)}c_{0}, b_{3}, …, b_{m}⟩, it follows that ρ_{λ} is an isomorphism. (iii) As Res_{⟨a⟩}^{𝔐_{m}}(x_{k}) ≠ 0 and ord(a) = p, by (i), it follows that Res_{⟨a⟩}^{𝔐_{m}}(y_{k}) ≠ 0. So y_{k} is not nilpotent. (iiia) is proved. (iiib) follows from Lemma 8, by noting that Res_{𝔐_{m}}^{𝔗_{m}}(u_{1}^{(i)} − u_{0}^{(i+1)}) = 0. (iiic) is straightforward from Lemma 1 and (iiib). □

Set S_{m} = 𝔐_{m}/Φ(𝔐_{m}). Pick elements ϕ_{i}, ψ_{i}, 0 ≤ i ≤ m, of H^{∗}(S_{m}) satisfying Inf_{𝔐_{m}}^{S_{m}}(ϕ_{i}) = x_{i}, ψ_{i} = βϕ_{i}. The polynomial algebra P_{m} = ℤ/p[ψ_{0}, …, ψ_{m}] is then a subalgebra of H^{∗}(S_{m}) and Inf_{𝔐_{m}}^{S_{m}}(ψ_{i}) = y_{i}, 0 ≤ i ≤ m. Let I_{m} be the ideal of P_{m} generated by the elements ψ_{i+1}^{p} − ψ_{i}^{p−1}ψ_{i+1}, 0 ≤ i ≤ m − 1. By Lemma 10 (iiib), I_{m} ⊂ ℑ_{m} = P_{m} ∩ ker Inf_{𝔐_{m}}^{S_{m}}. Lemma 11. I_{1} = ℑ_{1}. Proof. For λ ∈ ℤ/p, set U_{λ} = ⟨b_{0}b_{1}^{−λ}⟩. By Lemma 10 (i), we have ℤ/p ≅ U_{λ} ⊂ ker (x_{1} + λx_{0}). Let f be an element of ℑ_{1}. From the commutativity of the diagram

U_{λ} ⟶ 𝔐_{1}
∥ ↓
U_{λ} ⟶ S_{1}

in which horizontal (resp. vertical) arrows are inclusion (resp. projection) maps, it follows that

Res_{U_{λ}}^{S_{1}}(f) = Res_{U_{λ}}^{𝔐_{1}} ∘ Inf_{𝔐_{1}}^{S_{1}}(f)
= 0.

Hence f ∈ P_{1} ∩ ker Res_{U_{λ}}^{S_{1}} = (ψ_{1} + λψ_{0}). So f ∈ ∩_{λ∈ℤ/p}(ψ_{1} + λψ_{0}). This implies that f contains every ψ_{1} + λψ_{0} as a factor. Therefore f contains ∏_{λ∈ℤ/p}(ψ_{1} + λψ_{0}) = ψ_{1}^{p} − ψ_{0}^{p−1}ψ_{1} as a factor. So f ∈ I_{1}. Hence I_{1} = ℑ_{1}. □

We shall prove Proposition 4. I_{m} = ℑ_{m}.

By Lemma 11, the above proposition holds for m = 1. Suppose that it holds for m − 1 ≥ 1. Lemma 12. Let f be an element of ℑ_{m} given by

f = f(ψ_{0}, …, ψ_{m}) = _{∑_{0≤i_{1},...,i_{m}≤p−1} ; i≥0}λ_{i,i_{1},...,i_{m}}ψ_{0}^{i}ψ_{1}^{i_{1}} … ψ_{m}^{i_{m}}

with λ_{i,i_{1},...,i_{m}} ∈ ℤ/p. Then f ∈ (ψ_{1}^{p−1} − ψ_{0}^{p−1}). Proof. By Lemma 11, we may suppose that the lemma holds for m−1. Let U_{λ}, 0 ≠ λ ∈ ℤ/p, be subgroups of 𝔐_{m} given as in Lemma 10 and let \overline{ρ_{λ}} : S_{m−1} → U_{λ}/Φ(U_{λ}) be the map induced by ρ_{λ}. Note that the diagram

S_{m−1} ⟶^{\overline{ρ_{λ}}} U_{λ}/Φ(U_{λ}) ⟶^{\overline{i_{λ}}} S_{m}
↑  ↑  ↑
𝔐_{m−1} ⟶_{ρ_{λ}}  U_{λ}  ⟶_{i_{λ}} 𝔐_{m}

commutes, in which vertical arrows are projection maps, i_{λ} is the inclusion map, and \overline{i_{λ}} is the map induced by i_{λ} on quotient. So

Inf_{𝔐_{m−1}}^{S_{m−1}} ∘ \overline{ρ_{λ}}^{∗} ∘ Res_{U_{λ}/Φ(U_{λ})}^{S_{m}}(f) = ρ_{λ}^{∗} ∘ Res_{U_{λ}}^{𝔐_{m}} ∘ Inf_{𝔐_{m}}^{S_{m}}(f)
= 0.

Since

(\overline{ρ_{λ}}^{∗} ∘ Res_{U_{λ}/Φ(U_{λ})}^{S_{m}}(f))(ψ_{0}, …, ψ_{m−1}) = f(ψ_{0}, ψ_{1}, λψ_{1}, ψ_{2}, …, ψ_{m−1}),

f(ψ_{0}, ψ_{1}, λψ_{1}, ψ_{2}, …, ψ_{m−1}) belongs to ℑ_{m−1} = I_{m−1}, as Proposition 4 holds for m − 1. Hence, by Lemma 6 and Proposition 3, f ∈ (ψ_{1}^{p−1} − ψ_{0}^{p−1}). □

Set G = 𝔗 × 𝔐_{m−1}. Consider every element of H^{∗}(𝔗) (resp. H^{∗}(𝔐_{m−1})) as an element of H^{∗}(G), via the Künneth formula. Since Proposition 4 holds for m − 1,

B = ℤ/p[t_{0}, t_{1}]/(t_{1}^{p} − t_{0}^{p−1}t_{1}) ⊗ ℤ/p[y_{0}, …, y_{m−1}]/(y_{i}^{p} − y_{i−1}^{p−1}y_{i}|1 ≤ i ≤ m − 1)

is a subalgebra of H^{∗}(G). Set

\begin{aligned}\overline{t}_{i} & = t_{i}^{p} − (y_{0} − t_{1})^{p−1}t_{i}, 0 ≤ i ≤ 1, \\ \overline{y}_{j} & = y_{j}^{p} − (y_{0} − t_{1})^{p−1}y_{j}, 0 ≤ j ≤ m − 1. \end{aligned}

Define K = ker (x_{0} − s_{1}). It is clear that K ≅ 𝔐_{m}. We have

Lemma 13. If f(t_{0}, t_{1}, y_{0}, …, y_{m−1}) is an element of B which restricts trivially to H^{∗}(K), then f(\overline{t}_{0}, \overline{t}_{1}, \overline{y}_{0}, …, \overline{y}_{m−1}) = 0.

Proof. Identify K with 𝔐_{m}. Let N_{𝔐_{m}→G}, N_{S_{m}→G/Φ(G)} be the Evens norm maps. Note that G/Φ(G) is elementary abelian and contains S_{m} = K/Φ(K) as a maximal subgroup. Furthermore, Res_{S_{m}}^{G/Φ(G)} maps t_{0} (resp. t_{1}, ψ_{0}, ψ_{1}, …, ψ_{m−1}) to ψ_{0} (resp. ψ_{1}, ψ_{1}, ψ_{2}, …, ψ_{m}). Set g = g(t_{0}, t_{1}, ψ_{0}, …, ψ_{m−1}) ∈ H^{∗}(G/Φ(G)). By [3, Theorem 6.1.1] and [6, Corollaries 1.3, 1.5], we have

N_{S_{m}→G/Φ(G)}Res_{S_{m}}^{G/Φ(G)}(g) = g(\overline{t}_{0}, \overline{t}_{1}, \overline{ψ}_{0}, …, \overline{ψ}_{m−1})

with

\begin{aligned}\overline{t}_{i} & = N_{S_{m}→G/Φ(G)}(ψ_{i}) = t_{i}^{p} − (ψ_{0} − t_{1})^{p−1}t_{i}, 0 ≤ i ≤ 1, \\ \overline{ψ}_{j} & = N_{S_{m}→G/Φ(G)}(ψ_{j+1}) = ψ_{j}^{p} − (ψ_{0} − t_{1})^{p−1}ψ_{j}, 0 ≤ j ≤ m − 1 \end{aligned}

(note that \overline{ψ}_{0} = \overline{t}_{1}). Since Inf_{G}^{G/Φ(G)}(g) = f, it follows from the commutative diagram

H^{∗}(G/Φ(G)) ⟶^{Inf_{G}^{G/Φ(G)}} H^{∗}(G)
N_{Sm→G/Φ(G)}↑  ↑^{N_{𝔐m→G}}
H^{∗}(S_{m})  ⟶_{Inf_{𝔐m}^{Sm}} H^{∗}(𝔐_{m})

(see [3, Theorem 6.1.1]) that

Inf_{G}^{G/Φ(G)}N_{S_{m}→G/Φ(G)}Res_{S_{m}}^{G/Φ(G)}(g) = N_{𝔐_{m}→G}Inf_{𝔐_{m}}^{S_{m}}Res_{S_{m}}^{G/Φ(G)}(g)
= N_{𝔐_{m}→G}Res_{𝔐_{m}}^{G}Inf_{G}^{G/Φ(G)}(g)
= N_{𝔐_{m}→G}Res_{𝔐_{m}}^{G}(f)
= 0.

Hence

f(\overline{t}_{0}, \overline{t}_{1}, \overline{y}_{0}, …, \overline{y}_{m−1}) = Inf_{G}^{G/Φ(G)}(g(\overline{t}_{0}, \overline{t}_{1}, \overline{ψ}_{0}, …, \overline{ψ}_{m−1}))
= Inf_{G}^{G/Φ(G)}N_{S_{m}→G/Φ(G)}Res_{S_{m}}^{G/Φ(G)}(g)
= 0.  □

Proof of Proposition 4. Let g be an element of ℑ_{m}. We need only prove that g ∈ I_{m}.

Without loss of generality, we may suppose that, for any 1 ≤ j ≤ m, by considering g as a polynomial with variable ψ_{j}, g is a polynomial of degree ≤ p − 1. By Lemma 12, g contains ψ_{1}^{p−1} − ψ_{0}^{p−1} as a factor. Since ψ_{1}(ψ_{1}^{p−1} − ψ_{0}^{p−1}) ∈ I_{m}, we may assume that g is of form

g = _{∑_{0≤i_{2},...,i_{m}≤p−1} ; i≥0}μ_{i,i_{2},...,i_{m}}ψ_{0}^{i}(ψ_{1}^{p−1} − ψ_{0}^{p−1})ψ_{2}^{i_{2}} … ψ_{m}^{i_{m}}

with μ_{i,i_{2},...,i_{m}} ∈ ℤ/p.

Let h be the element of B given by

h = _{∑_{0≤i_{2},...,i_{m}≤p−1} ; i≥0}μ_{i,i_{2},...,i_{m}}t_{0}^{i}(t_{1}^{p−1} − t_{0}^{p−1})y_{1}^{i_{2}} … y_{m−1}^{i_{m}}.

Identify K with 𝔐_{m}. As Res_{𝔐_{m}}^{G} maps t_{0} (resp. t_{1}, y_{1}, …, y_{m−1}) to y_{0} (resp. y_{1}, y_{2}, …, y_{m}), it follows that Res_{𝔐_{m}}^{G}(h) = Inf_{𝔐_{m}}^{S_{m}}(g) = 0. So h ∈ ker Res_{𝔐_{m}}^{G}. By Lemma 13, this implies h(\overline{t}_{0}, \overline{t}_{1}, \overline{y}_{0}, …, \overline{y}_{m−1}) = 0. Therefore, by Proposition 2, all the μ_{i,i_{2},...,i_{m}}’s vanish. So g = 0. □

Set Y_{m} = ℤ/p[y_{0}, …, y_{m}]/(y_{i+1}^{p} − y_{i}^{p−1}y_{i+1}|0 ≤ i ≤ m − 1). We end this section by the following

Proposition 5. Y_{m} is a subalgebra of H^{∗}(𝔐_{m}).

Proof. Via the inflation map, H^{∗}(S_{m})/ker Inf_{𝔐_{m}}^{S_{m}} is a subalgebra of H^{∗}(𝔐_{m}). As P_{m}/ℑ_{m} is a subalgebra of H^{∗}(S_{m})/ker Inf_{𝔐_{m}}^{S_{m}}, it is also a subalgebra of H^{∗}(𝔐_{m}). The proposition follows from Proposition 4. □

4. Proofs of main results

Lemma 14. In H^{∗}(𝔐_{m}), if y_{m}^{i} ∈ (y_{0}) with i ≥ 1, then

y_{m}^{pi} = ∑_{j=1}^{i}(−1)^{j−1}\binom{i}{j}y_{m}^{pi−j(p−1)}y_{0}^{(p−1)j}.

Proof. y_{m}^{i} ∈ (y_{0}) implies y_{m}^{i}|_{ker x_{0}} = 0. From [5, Remark 1] (see also [6, Theorem 1.1]) and [6, Corollary 1.5], it follows that

0 = (y_{m}^{p} − y_{0}^{p−1}y_{m})^{i}
= y_{m}^{pi} + ∑_{j=1}^{i}(−1)^{j}\binom{i}{j}y_{m}^{pi−j(p−1)}y_{0}^{(p−1)j}.  □

Proposition 6. Let s, m be integers such that ps < m(p − 1). Then y_{m}^{s}, as an element of H^{∗}(𝔐_{m}), does not belong to (y_{0}).

Proof. By Proposition 1, as elements of Y_{m}, y_{m}^{m(p−1)} ∉ (y_{0}), hence y_{m}^{ps} ∉ (y_{0}). By Proposition 5 and Lemma 14, , as an element of H^{∗}(𝔐_{m}), y_{m}^{s} ∉ (y_{0}). □

Let G be given by the central extension

(G)  0 → ℤ/p →^{ι} G → 𝔐_{m} → 1

classified by the cohomology class y_{0} ∈ H^{2}(𝔐_{m}). As usual, each element of H^{∗}(𝔐_{m}) can be considered as an element of H^{∗}(G) via the inflation map.

Denote by {E_{r}, d_{r}} the Hochschild-Serre spectral sequence corresponding to the extension (G). Pick an element u of H^{1}(ι(ℤ/p)) satisfying d_{2}(u) = y_{0}, and set v = βu. It follows that

E_{2} = H^{∗}(𝔐_{m}) ⊗ Λ[u] ⊗ ℤ/p[v].

We have

Lemma 15. E_{∞} = E_{3} = H^{∗}(𝔐_{m})/(y_{0}) ⊗ ℤ/p[v] ⊕Ann_{H^{∗}(𝔐_{m})}(y_{0}) ⊗ ℤ/p[v]u.

Proof. As d_{2}(u) = y_{0}, u is transgressive. v = βu is also transgressive, d_{2}(v) = 0 and d_{3}(v) = d_{3}(βu) = β(d_{2}(u)) = β(y_{0}) = 0. Calculating ker d_{2}/Im d_{2} yields E_{3}. As d_{3}(v) = 0, it follows from the structure of E_{3} that d_{r} = 0, r ≥ 3. Therefore E_{∞} = E_{3}. □

Proof of the theorem. Suppose that n < p. Set G = (ℤ/p)^{2n}. Let u_{1}, …, u_{2n} be a basis of Hom(G, ℤ/p) and set v_{i} = βu_{i}, 1 ≤ i ≤ 2n. Then

H^{∗}(G) = Λ[u_{1}, …, u_{2n}] ⊗ ℤ/p[v_{1}, …, v_{2n}].

Set ξ = u_{1}u_{2} + ⋯ + u_{2n−1}u_{2n}. Then ξ^{n} = n!u_{1} … u_{2n} ≠ 0, since n! ≠ 0(mod p). Furthermore, as ξ^{n+1} = 0, ξ^{n+p−1} also vanishes..

Suppose that n ≥ p. Write n = (p − 1)k + j with 0 ≤ j < p − 1. Define

t(n) = \{ \begin{matrix}n & if j = 0, \\ (p − 1)(k + 1) & if j ≠ 0. \end{matrix}

Let m be the smallest integer such that, as an element of H^{∗}(𝔐_{m}), y_{m}^{t(n)} ∉ (y_{0}). Note that the existence of m follows from Proposition 6. Since t(n) ≥ n ≥ p, it follows that y_{m−1}^{p−1}y_{m} = y_{m}^{p} ∉ (y_{0}). Hence m > 1.

Set V = 𝔐_{m}/⟨b_{m}, [b_{m−1}, b_{m}]⟩. It is clear that V ≅ 𝔐_{m−1}. Via this isomorphism, 𝔐_{m−1} is now considered as a quotient of 𝔐_{m}.

Suppose that y_{m}^{n+p−1} ∉ (y_{0}), as an element of H^{∗}(𝔐_{m}). By Lemma 10 (iiic), y_{m}^{n+p−1} = y_{m−1}^{k(p−1)}y_{m}^{p−1+j}. Since Inf_{𝔐_{m}}^{𝔐_{m−1}} maps y_{i} ∈ H^{∗}(𝔐_{m−1}) to y_{i} ∈ H^{∗}(𝔐_{m}), 0 ≤ i ≤ m − 1, this implies that, as an element of H^{∗}(𝔐_{m−1}), y_{m−1}^{k(p−1)} ∉ (y_{0}). If j = 0, this implies y_{m−1}^{t(n)} = y_{m−1}^{n} = y_{m−1}^{k(p−1)} ∉ (y_{0}). If j > 0, by Lemma 10 (iiic), y_{m−1}^{(k+1)(p−1)}y_{m}^{j} = y_{m}^{n+p−1} ∉ (y_{0}) which implies y_{m−1}^{t(n)} ∉ (y_{0}). So, in any case, y_{m−1}^{t(n)} ∉ (y_{0}) in H^{∗}(𝔐_{m−1}). This contradicts the fact that m is the smallest integer such that, as an element of H^{∗}(𝔐_{m}), y_{m}^{t(n)} ∉ (y_{0}).

Therefore y_{m}^{n+p−1} ∈ (y_{0}), as an element of H^{∗}(𝔐_{m}). Since n ≤ t(n), in H^{∗}(𝔐_{m}), y_{m}^{t(n)} ∉ (y_{0}) also implies y_{m}^{n} ∉ (y_{0}).

Let G be given as above and set ξ = y_{m}. It follows from the structure of E_{∞} given in Lemma 15 that ker Inf_{G}^{𝔐_{m}} = (y_{0}). Hence, in H^{∗}(G), ξ^{n} ≠ 0 and ξ^{n+p−1} = 0. The theorem is proved. □

Proof of the corollary. Set ℓ = (p − 1)n. By the theorem, there exist a p-group G and ξ ∈ H^{2}(G) such that ξ^{ℓ} ≠ 0, ξ^{ℓ+p−1} = 0. Set χ = ξ^{p−1}. It follows that χ^{n} = ξ^{ℓ} ≠ 0 and χ^{n+1} = ξ^{ℓ+p−1} = 0. The corollary follows. □

References

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2. N. Blackburn, On a special class of p-groups, Acta Math. 100 (1958), 45–92.

3. L. Evens, The cohomology of groups, Oxford Mathematical Monographs, Clarendon Press, 1991.

4. K. Inoue, A. Kono, Nilpotency of a kernel of the Quillen map, J. Math. Kyoto Univ. 33 (1993), 1047-1055.

5. P. A. Minh, Essential mod-p cohomology classes of p-groups: an upper bound for the nilpotency degree, Bull. London Math. Soc. 32 (2000), 285-291.

6. , Evens norm and restriction maps in mod-p cohomology of p-groups, Math. Proc. Camb. Phil. Soc. 129 (2000), 253-262.

7. , Nilpotency degree of cohomology rings in characteristic 3, Proc. Amer. Math. Soc. 130 (2002), 307-310.

8. , Nilpotency degree of cohomology rings in characteristic p, to appear in Proc. Amer. Math. Soc.

Department of Mathematics, College of Sciences, University of Hue, Dai hoc Khoa hoc, Hue, Vietnam

   E-mail address: paminh@dng.vnn.vn