NILPOTENCY DEGREE OF COHOMOLOGY

RINGS IN CHARACTERISTIC ppp

Pham Anh Minh

Abstract. Let p be an odd prime number. The purpose of this paper is to provide a p-group G whose mod-p cohomology ring has a nilpotent element ξ ∈ H^{∗}(G) satisfying ξ^{p} ≠ 0.

1. Statement of main result

For every p-group G, denote by H^{∗}(G) the mod-p cohomology algebra of G. We are now 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 an example was given in [8] for p = 3. In this case, G is chosen to be an extension of 𝔼 × 𝔄 by ℤ/p, where 𝔼 is the extraspecial 3-group of order 3^{3} and of exponent 3, and 𝔄 the elementary abelian 3-group of rank 4.

The purpose of this paper is to generalize the result in [8] to the case of any odd prime p. Let 𝔖_{p^{2},p} be the Sylow subgroup of the symmetric group on p^{2} letters (so 𝔖_{p^{2},p} = ℤ/p ≀ ℤ/p, the wreath product of ℤ/p and ℤ/p) and let Z be the center of 𝔖_{p^{2},p}. Define ℜ = 𝔖_{p^{2},p}/Z (so ℜ = 𝔼 for p = 3) and let 𝔄_{k} be the elementary abelian p-group of rank 2k. We shall prove

Theorem. For k ≥ 2p − 2, there exists a p-group G given by a central extension

0 → ℤ/p → G → ℜ × 𝔄_{k} → 1

whose mod-p cohomology ring has a nilpotent element ξ ∈ H^{2}(G) satisfying ξ^{p} ≠ 0.

We shall use the following notation. For homogeneous elements X, Y, … of a graded ring R, |X| denotes the degree of X and (X, Y, …) the ideal of R generated by X, Y, . . .. p is assumed from now on to be an arbitrary odd prime number. 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 maps.

Typeset by AMS-TEX

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

2. The groups ℜ and 𝔘

Let 𝔘 be the p-group of order p^{p+2} given by

\begin{aligned}𝔘 = ⟨a, a_{1}, …, a_{p+1}|a^{p} & = a_{3}^{p} = ⋯ = a_{p+1}^{p} = [a, a_{p+1}] = [a_{i}, a_{j}] = a_{1}^{p}a_{2}^{p(p−1)/2}a_{p} \\ & = a_{2}^{p}a_{p+1} = 1, [a, a_{ℓ}] = a_{ℓ+1}, 1 ≤ i, j ≤ p + 1, 1 ≤ ℓ ≤ p⟩. \end{aligned}

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

(𝔘)  0 → ℤ/p →𝔘 →^{p_{1}} 𝔗 → 1,
(𝔗)  0 → ℤ/p →^{j}𝔗 →^{p_{2}} ℜ → 1,
(ℜ)  0 → (ℤ/p)^{p−1} ≅ K →ℜ → ⟨a⟩ → 1.

Let x ∈ H^{1}(⟨a⟩) be the dual of a and consider x as an element of H^{1}(ℜ) via the inflation map. Define x_{1} ∈ H^{1}(ℜ) satisfying x_{1}(g_{1}) = 1, x_{1}(a) = 0 (x_{1} is welldefined, as ℜ is a 2-generator p-group and ℜ = ⟨a, g_{1}⟩). x, x_{1} is then a basis of H^{1}(ℜ). Set y = βx, y_{1} = βx_{1} with β the Bockstein homomorphism. Let w be the element of H^{2}(ℜ) classifying the extension (𝔗) and set U = w + y_{1}. We have

Lemma 1. (i) U|_{K} = 0; (ii) U^{p} + xy^{p−2} · βU − y^{p−1} · U = 0; (iii) x · U = xy_{1}; (iv) U^{p} = y^{p−1}y_{1}.

Proof. Set f = p_{1}(a_{1}). Since f^{p} = j(−1) and p_{2}^{−1}(K) is abelian, w|_{K} = −y_{1}. Hence U|_{K} = 0. (ii) follows from [6, Remark 1] (see also [7, Theorem 1.1]) and the fact that U|_{K} = 0.

Note that the term E_{2}(𝔗) of the Hochschild-Serre spectral sequence corresponding to the extension (𝔗) is of form

E_{2}(𝔗) = H^{∗}(ℜ) ⊗ H^{∗}(j(ℤ/p)).

Let w′ be the element of H^{2}(𝔗) classifying the extension (𝔘) and let u be the element of H^{1}(j(ℤ/p)) satisfying d_{2}(u) = w. Since ⟨a_{p}, a_{p+1}⟩ = p_{1}^{−1}(Im j) is of exponent p, w′ restricts trivially on j(ℤ/p). So w′ represents a non-zero element of E_{∞}^{2,0}(𝔗) ⊕ E_{∞}^{1,1}(𝔗). Note that X|_{⟨a,j(ℤ/p)⟩} (resp. X|_{⟨a_{1},j(ℤ/p)⟩}) is a scalar multiple of y (resp. y_{1}), for any X ∈ E_{∞}^{2,0}(𝔗) ⊂ Im Inf_{𝔗}^{ℜ}. As [a, a_{p}] = a_{p+1} and [a_{1}, a_{p}] = 1, it follows that w′ represents a non-zero element θ of E_{2}^{1,1}(𝔗) and w′|_{⟨a,j(ℤ/p)⟩} ≠ 0, w′|_{⟨f,j(ℤ/p)⟩} = 0. So θ is a non-zero scalar multiple of x ⊗ u. This means that xd_{2}(u) = xw = 0. Hence x · U = x(w + y_{1}) = xy_{1}.

Finally, as

y · U − x · βU = β(x · U)
= β(xy_{1})  by (iii)
= yy_{1},

we have, by (ii),  0 = U^{p} + xy^{p−2} · βU − y^{p−1} · U
= U^{p} + y^{p−2}(x · βU − y · U)
= U^{p} − y^{p−2} · yy_{1}
= U^{p} − y^{p−1}y_{1}.

The lemma is proved. □

Remarks. 1. It is known that Inf_{𝔖_{p^{2},p}}^{ℜ} : H^{2}(ℜ) → H^{2}(𝔖_{p^{2},p}) is surjective (this comes from the fact that 𝔖_{p^{2},p} is a terminal p-group, see [3]). Furthermore, as H^{∗}(𝔖_{p^{2},p}) is detected by elementary abelian subgroups ([9]) and U|_{⟨a⟩} = 0, U|_{K} = 0, it follows that Inf_{𝔖_{p^{2},p}}^{ℜ}(U) = 0. Hence U is nothing other than the cohomology class classifying the central extension 1 → Z → 𝔖_{p^{2},p} → ℜ → 1.

2. The group 𝔘 can be obtained from algebraic number theory, as follows. Let ϑ be a primitive pth root of unity and let ℤ_{p} be the ring of p-adic integers. Then

ℤ_{p}[ϑ] = ⊕^{p−2}_{j=0}ϑ^{j}ℤ_{p}.

Let ⟨a⟩ be the cyclic group of order p. Multiplication by ϑ induces an action of ⟨a⟩ on ℤ_{p}[ϑ], hence on ℨ = ℤ_{p}[ϑ]/((ϑ − 1)^{p+1}). We can check that the map

{a, a_{1}, …, a_{p+1}} → ⟨a⟩ ⋉ ℨ

which maps a to a, a_{i} to (ϑ − 1)^{i−1}, 1 ≤ i ≤ p + 1, can be extended to a homomorphism of groups from 𝔘 to ⟨a⟩ ⋉ ℨ which is bijective. So 𝔘 ≅ ⟨a⟩ ⋉ ℨ.

3. Proof of the theorem

Pick a basis c_{1}, …, c_{2k} of the elementary abelian p-group 𝔄_{k} of rank 2k. Let u_{1}, …, u_{2k} be the basis of H^{1}(𝔄_{k}), dual to that of 𝔄_{k}. Define v_{i} = βu_{i}, 1 ≤ i ≤ 2k. Set G = ℜ × 𝔄_{k}. By Künneth formula,

\begin{aligned}H^{∗}(G) & = H^{∗}(ℜ) ⊗ H^{∗}(𝔄_{k}) \\ & = H^{∗}(ℜ) ⊗ Λ[u_{1}, …, u_{2k}] ⊗ 𝔽_{p}[v_{1}, …, v_{2k}]. \end{aligned}

Let

\begin{aligned}z & = y + u_{1}u_{2} + ⋯ + u_{2k−1}u_{2k}, \\ α & = u_{1}u_{2} + ⋯ + u_{2k−1}u_{2k}, \\ γ & = u_{3}u_{4} + ⋯ + u_{2k−1}u_{2k} \end{aligned}

be elements of H^{∗}(G). Set Λ = Λ[u_{1}, …, u_{2k}] and Λ′ = Λ[u_{3}, …, u_{2k}]. For i ≥ 0, let Λ_{i} be the degree i-part of Λ. We have

Lemma 2. Let 0 ≠ X be a homogeneous element of H^{∗}(G) and let n, k be positive integers.

(i) If n ≤ min(p − 1, k) and X · α^{n} = 0, then |X| ≥ min(2p − 2n, k − n + 1). (ii) If X · z = 0:

(iia) |X| ≥ min(2p − 2, k); (iib) |X| > 2p − 1, provided that k ≥ 2p − 2 and X ∈ (v_{1}, …, v_{2k}).

Proof. (i) By induction on k. It obviously holds for k = 1. Assume that it holds for k − 1. Without loss of generality, we may assume that X ∈ Λ. If n = k ≤ p − 1, then 0 = X · α^{n} = k!X · u_{1} … u_{2k} implies |X| ≥ 1. We may then suppose that n ≤ k − 1.

Write X = X_{1} + X_{2} · u_{1}u_{2} + X_{3} · u_{1} + X_{4} · u_{2} with X_{i} homogeneous in Λ′. So 0 = X · α^{n} = X · (γ + u_{1}u_{2})^{n} implies

X_{1} · γ^{n} = 0,  (1)
(nX_{1} + X_{2} · γ) · γ^{n−1} = 0,  (2)
X_{3} · γ^{n} = X_{4} · γ^{n} = 0.  (3)

If X_{3} ≠ 0 or X_{4} ≠ 0, it follows from (3) and the inductive hypothesis that min(|X_{3}|, |X_{4}|) ≥ min(2p − 2n, k − n), hence |X| = min(|X_{3}|, |X_{4}|) + 1 ≥ min(2p − 2n + 1, k − n + 1). So we may suppose that X_{3} = X_{4} = 0. By (2) and by the inductive hypothesis, |nX_{1} + X_{2} · γ| ≥ min(2p − 2n + 2, k − n + 1). Hence |X| = |nX_{1}+X_{2}·γ| ≥ min(2p−2n, k−n+1) if nX_{1}+X_{2}·γ ≠ 0. If nX_{1}+X_{2}·γ = 0, then nX_{1} = −X_{2} · γ. By (1), X_{2} · γ^{n+1} = 0. By the inductive hypothesis, |X_{2}| ≥ min(2p − 2n − 2, k − n − 1). Hence |X| = |X_{2}| + 2 ≥ min(2p − 2n, k − n + 1).

(i) is then proved.

(ii) Write X = ∑_{i=0}^{m}X_{i} · Y_{i} with Y_{i} ∈ Λ_{i}, X_{i} ∈ H^{∗}(ℜ) ⊗ 𝔽_{p}[v_{1}, …, v_{2k}] and X_{m} ·

Y_{m} ≠ 0. Then X_{m} · Y_{m} · α = 0. By (i), it follows that |X| = |X_{m} · Y_{m}| ≥ min(2p − 2, k). (iia) is proved.

Suppose that k ≥ 2p − 2, |X| ≤ 2p − 1 and X ∈ (v_{1}, …, v_{2k}). Fix i with 1 ≤ i ≤ 2k and write X = v_{i}X′ + X′′ with X′′ free of v_{i}. Then X′ · z = 0. As |X′| ≤ 2p − 3, it follows from (iia) that X′ = 0. X is then free of v_{i}. Hence by repeating the argument, we see that X is free of v_{1}, …, v_{2k}, a contradiction.

The lemma is proved. □

Suppose from now on that k ≥ 2p − 2. We have

Lemma 3. If X is homogeneous in H^{∗}(G) of degree ≤ 2p − 5 and X · βz ∈ (z), then X ∈ (z, βz).

Proof. Write X · βz = A · z with A homogeneous in H^{∗}(G). Then A · βz · z = 0. As |A · βz| = |X| + 4 ≤ 2p − 1 and A · βz ∈ (v_{1}, …, v_{2k}), it follows from Lemma 2 (ii) that A · βz = 0. By [5, Lemma 2.1], A ∈ (βz, u_{1} … u_{2k}). So A ∈ (βz), as |A| ≤ 2p − 4 < 2k. Write A = B · βz. We then have (X − B · z)βz = 0. Again, by [5, Lemma 2.1], X − B · z ∈ (βz). Therefore X ∈ (z, βz). The lemma is proved. □

For every 1 ≤ j ≤ p − 1, let B_{j} be a set of elements of Λ_{2j} such that the disjoint union B_{j} ⊔ {α^{j}} forms a basis of Λ_{2j}. We then have a decomposition

H^{∗}(ℜ) ⊗ Λ_{2j} = H^{∗}(ℜ) ⊗ (⟨B_{j}⟩ ⊕ ⟨α^{j}⟩), 1 ≤ j ≤ p − 1.  (4)

For 1 ≤ i < p − 1 and for 0 ≠ b ∈ ⟨B_{i}⟩, μ ∈ ℤ/p, as |b − μα^{i}| = 2i < 2p − 2, it follows from Lemma 2 that (b − μα^{i}) · α ≠ 0. The B_{j}’s can be then chosen such that, for every 1 ≤ i < p − 1,

{b · α|0 ≠ b ∈ ⟨B_{i}⟩} ⊂ ⟨B_{i+1}⟩.  (5)

Lemma 4. y^{p−1}y_{1} ∉ (z, βz).

Proof. Note that, as ℜ is of exponent p, y^{p−1}y_{1}|_{⟨ag_{1}⟩} ≠ 0. So y^{p−1}y_{1} ≠ 0.

Suppose that y^{p−1}y_{1} = A · z + B · βz with A, B homogeneous in H^{∗}(G). Write A = A′+A′′ with A′ ∈ H^{∗}(ℜ)⊗Λ, A′′ ∈ (v_{1}, …, v_{2k}). It follows that y^{p−1}y_{1} = A′·z and A′′ · z + B · βz = 0. Hence, without loss of generality, we may assume that y^{p−1}y_{1} = A · z with A ∈ H^{∗}(ℜ) ⊗ Λ. By the decomposition (4), we have

A = A_{2p−2} + ∑_{i=1}^{p−1}(A_{2p−2i−2} · α^{i} + ∑_{b∈B_{i}} A_{2p−2i−2,b} · b)

with A_{j}, A_{j,b} ∈ H^{∗}(ℜ). By (5), it follows that

A_{2p−2}y = y^{p−1}y_{1},
(A_{2p−2} + A_{2p−4}y) · α = 0,
…
(A_{2} + A_{0}y) · α^{p−1} = 0.

Therefore, for every i ≥ 1, A_{2i} = −A_{2i−2}y. So y^{p−1}y_{1} = A_{0}y^{p}. As y^{p−1}y_{1}|_{⟨a⟩} = 0, this implies A_{0} = 0. Hence y^{p−1}y_{1} = 0, a contradiction. Thus y^{p−1}y_{1} ∉ (z, βz). The lemma is proved. □

Let {E_{r}, d_{r}} be the Hochschild-Serre spectral sequence corresponding to the central extension

0 → ℤ/p →^{i} G → G → 1

classified by z ∈ H^{2}(G). Pick s (resp. t) in H^{1}(i(ℤ/p)) (resp. H^{2}(i(ℤ/p)) satisfying d_{2}(s) = z (resp. d_{3}(t) = βz). It is known that

E_{2} = H^{∗}(G) ⊗ H^{∗}(i(ℤ/p)),
E_{3} = H^{∗}(G)/(z) ⊗ 𝔽_{p}[t]
⊕ Ann_{H^{∗}(G)}(z) ⊗ 𝔽_{p}[t]s.

We have

Lemma 5. There exists no element η ∈ E_{n}^{2p−n,n−1}, n ≥ 3, satisfying d_{n}(η) = y^{p−1}y_{1}.

Proof. Suppose that η is an element of E_{2r}^{2p−2r,2r−1}, r ≥ 2, represented by X ⊗ t^{r−1}s ∈ E_{2}^{2p−2r,2r−1}. Then 0 = d_{2}(X ⊗ t^{r−1}s) = X · z ⊗ t^{r−1}. So X · z = 0. Since |X| = 2p − 2r ≤ 2p − 4, X = 0 by Lemma 2. Hence 0 = d_{2r}(η) ≠ y^{p−1}y_{1}.

Suppose now that η ∈ E_{2r+1}^{2p−2r−1,2r} is represented by X⊗t^{r} ∈ E_{3}^{2p−2r−1,2r}, p−1 ≥ r ≥ 1. If r = 1, then y_{1}^{p−1}y_{2} ≠ X · βz = −d_{3}(X ⊗ t) by Lemma 4. If r ≥ 2, then −rX·βz⊗t^{r−1} = d_{3}(X⊗t^{r}) = 0 in E_{3}. So X·βz ∈ (z). As |X| = 2p−2r−1 ≤ 2p−5, X ∈ (z, βz) by Lemma 3. This implies X ⊗ t^{r} = 0 in E_{4} unless r = p − 1 and X ∈ (βz). But r = p − 1 also implies |X| = 2p − 2r − 1 = 1 and X ∈ (βz), hence X = 0. So 0 = d_{2r+1}(η) ≠ y^{p−1}y_{1}. The lemma is proved. □

Proof of the theorem. Set ξ = Inf_{G}^{G}(U). By Lemma 1, ξ^{p} = y^{p−1}y_{1}. Since z vanishes in H^{∗}(G), y^{p} = −α^{p} = 0 in H^{∗}(G). So ξ^{2p} = y^{2p−2}y_{1}^{2} = 0. Hence ξ is nilpotent. Besides, Lemmas 4 and 5 show that y^{p−1}y_{1} ∉ Ker Inf_{G}^{G}, which means that ξ^{p} = y^{p−1}y_{1} ≠ 0. The theorem is proved. □

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

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